2017-2018上海复旦附中高一上期 中考试

2017-2018年上海市复旦附中⾼⼀上学期数学期中试卷带答案2017-2018学年上海市复旦附中⾼⼀（上）期中数学试卷⼀.填空题1．（3分）已知全集U=Z，A={﹣1，0，1，2}，B={x|x2=x}0则A∩（C U B）=．2．（3分）命题“如果a+b＞0，那么a＞0且b ＞0”的否命题是命题（填“真”或“假”）3．（3分）已知集合A={y|y=x2﹣2x﹣3}，集合B={y|y=﹣x2+2x+13}，则A∩B=．4．（3分）已知α：，β：1﹣2a＜x＜3a+2，若α是β的充分不必要条件，则实数a的取值范围是．5．（3分）设M={a，b}，则满⾜M∪N?{a，b，c}的集合N的个数为．6．（3分）函数f（x）=的定义域为[﹣2，1]，则a的值为．7．（3分）已知函数f（x）=（m﹣1）x+2m﹣3，⽆论m取什么实数，函数f（x）的图象始终过⼀个定点，该定点的坐标为．8．（3分）已知关于x的⽅程x2+kx+k2+k﹣4=0有两个实数根，且⼀根⼤于1，⼀根⼩于1，则实数k的取值范围为．9．（3分）给出下列四个命题：（1）若a＞b，c＞d，则a﹣d＞b﹣c；（2）若a2x＞a2y，则x＞y；（3）a＞b，则；（4）若，则ab＜b2．其中正确命题是．（填所有正确命题的序号）10．（3分）若x∈（﹣∞，2），则的最⼩值为．11．（3分）设函数f（x）=x﹣2，若不等式|f（x+3）|＞|f（x）|+m对任意实数x恒成⽴，则m的取值范围是．12．（3分）对于实数A和正数B，称满⾜不等式|x﹣A|＜B（A∈R，B＞0）的实数x的集合叫做A的B领域，已知t为给定的正数，a、b为正数，若a+b﹣t的a+b领域是⼀个关于原点对称的区间，则a2+b2的最⼩值为．⼆.选择题13．（3分）设实数a1，a2，b1，b2均不为0，则“成⽴”是“关于x的不等式a1x+b1＞0与a2x+b2＞0的解集相同”的（）A．充分⾮必要条件 B．必要⾮充分条件C．充要条件D．⾮充分⾮必要条件14．（3分）解析式为y=2x2+1，值域为{5，19}的函数有（）个．A．4 B．6 C．8 D．915．（3分）设f（x）是定义在正整数集上的函数，且f（x）满⾜：“当f（x）＞x2成⽴时，总可以推出f（x+1）＞（x+1）2成⽴”．先给出以下四个命题：（1）若f（3）≥9，则f（4）≥16；（2）若f（3）=10，则f（5）＞25；（3）若f（5）=25，则f（4）≤16；（4）若f（x）≥（x+1）2，则f（x+1）≥x2．其中真命题的个数为（）A．1个 B．2个 C．3个 D．4个16．（3分）设a，b，c为实数，f（x）=（x+a）（x2+bx+c），g（x）=（ax+1）（cx2+bx+1）记集合S={x|f（x）=0，x∈R}，T={x|g（x）=0，x∈R}．若|S|，|T|分别为集合S，T的元素个数，则下列结论不可能的是（）A．|S|=1且|T|=0 B．|S|=1且|T|=1 C．|S|=2且|T|=2 D．|S|=2且|T|=3三.解答题17．已知集合A={x|（m﹣1）x2+3x﹣2=0}，是否存在这样的实数m，使得集合A 有且仅有两个⼦集？若存在，求出所有的m 的值组成的集合M；若不存在，请说明理由．18．我校第⼆教学楼在建造过程中，需建⼀座长⽅体形的净⽔处理池，该长⽅体的底⾯积为200平⽅⽶，池的深度为5⽶，如图，该处理池由左右两部分组成，。

2017学年复旦附中高一上期中一.填空题1.已知全集U =R ，{1,0,1,2}A =-，2{|}B x x x ==，则U A C B = __________2.命题“如果0a b +>，那么0a >且0b >”的否命题是__________命题（填“真”或“假”）3.已知集合2{|23}A y y x x ==--，2{|213}B y y x x ==-++，则A B = __________4.已知“12a x a ≤≤+”是“1232a x a -<<+”的充分不必要条件，则实数a 的取值范围是__________5.设M={a,b}，则满足M ∪N ⊆{a,b,c}的非空集合N 的个数为______________．6.函数()f x =的定义域为[2,1]-，则a 的值为__________7.已知函数()(1)23f x m x m =-+-，无论m 取什么实数，函数()f x 的图像始终过一个定点，该定点的坐标为__________8.已知关于x 的方程2240x kx k k +++-=有两个实数根，且一根大于1，一根小于1，则实数k 的取值范围为__________9.给出下列四个命题：（1）若a b >，c d >，则a d b c ->-；（2）若22a x a y >，则x y >；（3）若a b >，则11a b a >-；（4）110a b <<，则2ab b <.其中正确命题是________.（填所有正确命题的序号）10.若(,2)x ∈-∞，则2542-+-x x x 的最小值为__________11.设函数()2f x x =-，若不等式|(3)|()f x f x m +>+对任意实数x 恒成立，则m 的取值范围是__________12.对于实数A 和正数B ，称满足不等式||x A B -<(,0)A B ∈>R 的实数x 的集合叫做A 的B 邻域，已知t 为给定的正数，a 、b 为正数，若a b t +-的a b +领域是一个关于原点对称的区间，则22a b +的最小值为__________二.选择题13.已知1a ，1b ，2a ，2b R ∈且都不为零，则“1122a b a b =”是“110a x b +>与220a x b +>解集相同”的A.充分非必要条件B.必要非充分条件C.充要条件D.既非充分又非必要条件14.解析式为221y x =+，值域为{}5,19的函数有A .4 B.6 C.8 D.915.设()f x 是定义在正整数集上的函数，且()f x 满足：“当2()f x x >成立时，总可以推出2(1)(1)f x x +>+成立”，给出以下四个命题：①若(3)9f ≥，则(4)16f ≥；②若(3)10f =，则(5)25f >；③若(5)25f =，则(4)16f ≤；④若2()(1)f x x ≥+，则2(1)f x x +≥.其中真命题的个数为（）个A.1 B.2 C.3 D.416.设a ，b ，c 为实数，22()()(),()(1)(1),f x x a x bx cg x ax cx bx =+++=+++记集合{}{}|()0,,|()0,,S x f x x R T x g x x R ==∈==∈若{S }，{T }分别为集合S ，T 的元素个数，则下列结论不可能的是（）A.{S }=1且{T }=0 B.{S }=1且{T }=1 C.{S }=2且{T }=2 D.{S }=2且{T }=3三.解答题17.已知集合2{|(1)320}=-+-=A x m x x ，是否存在这样的实数m ，使得集合A 有且仅有两个子集？若存在，求出所有的m 的值组成的集合M ；若不存在，请说明理由.18.我校第二教学楼在建造过程中，需建一座长方体形的净水处理池，该长方体的底面积为200平方米，池的深度为5米，如图，该处理池由左右两部分组成，中间是一条间隔的墙壁，池的外围周壁建造单价为400元/平方米，中间的墙壁（不需考虑该墙壁的左右两面）建造单价为100元/平方米，池底建造单价为60元/平方米，池壁厚度忽略不计，问净水池的长AB 为多少时，可使总造价最低？最低价为多少？19.已知a ∈R ，集合26{|0}1x x A x x --=≤+，集合{||2|1}B x x a a =+≤+.（1）求集合A 与集合B ；（2）若A B B = ，求实数a 的取值范围.20.已知函数2|1|()4x m f x x +-=-，0m >，满足(2)2f =-.（1）求实数m 的值；（2）在平面直角坐标系中，作出函数()f x 的图像，并且根据图像判断：若关于x 的方程()f x k =有两个不同实数解，求实数k 的取值范围（直接写结论）21.已知M 是满足下列性质的所有函数()f x 组成的集合：对任何12,f x x D ∈（其中f D 为函数()f x 的定义域），均有1212()()||f x f x x x -≤-成立.（1）已知函数2()1f x x =+，11[,]22x ∈-，判断()f x 与集合M 的关系，并说明理由；（2）是否存在实数a ，使得()2a p x x =+，[1,)x ∈-+∞属于集合M ？若存在，求a 的取值范围，若不存在，请说明理由；（3）对于实数a 、b ()a b <，用[,]a b M 表示集合M 中定义域为区间[,]a b 的函数的集合.定义：已知()h x 是定义在[,]p q 上的函数，如果存在常数0T >，对区间[,]p q 的任意划分：011n n p x x x x q -=<<⋅⋅⋅<<=，和式11|()()|ni i i h x h x T -=-≤∑恒成立，则称()h x 为[,]p q 上的“绝对差有界函数”，其中常数T 称为()h x 的“绝对差上界”，T 的最小值称为()h x 的“绝对差上确界”，符号121n i n i tt t t ==++⋅⋅⋅+∑；求证：集合[1009,1008]M -中的函数()h x 是“绝对差有界函数”，并求()h x 的“绝对差上确界”.2017学年复旦附中高一上期中一.填空题1.已知全集U =R ，{1,0,1,2}A =-，2{|}B x x x ==，则U A C B = __________【答案】{1,2}-【分析】先求出集合B ，再求出U C B ，最后求出U A C B ⋂．【详解】由题意得{}{}2|0,1B x x x ===，∴()()(),00,11,U C B ∞=-⋃⋃+∞，∴{}1,2U A C B ⋂=-．故答案为{}1,2-．【点睛】本题考查集合的运算，解题时根据集合运算的顺序进行求解即可，属于基础题．2.命题“如果0a b +>，那么0a >且0b >”的否命题是__________命题（填“真”或“假”）【答案】真【分析】根据原命题的逆命题和其否命题为等价命题判断命题的真假．【详解】由题意得命题“如果0a b +>，那么0a >且0b >”的逆命题为“如果0a >且0b >，那么0a b +>”，其真命题，所以否命题为真命题．故答案为“真”．【点睛】判断命题的真假时，可通过命题直接进行判断也可通过其等价命题的真假来判断，解题时要根据条件选择合理的方法进行求解．3.已知集合2{|23}A y y x x ==--，2{|213}B y y x x ==-++，则A B = __________【答案】[4,14]-【分析】分别求出集合,A B ，然后再求出A B ⋂即可．【详解】由题意得{}(){}{}22|23|14|4A y y x x y y x y y ==--==--=≥-，{}{}{}22|213|(1)14|14B y y x x y y x y y ==-++==--+=≤，∴[]4,14A B ⋂=-．故答案为[]4,14-．【点睛】本题考查集合的交集运算，解题的关键是正确求出集合,A B ，属于简单题．4.已知“12a x a ≤≤+”是“1232a x a -<<+”的充分不必要条件，则实数a 的取值范围是__________【答案】13a >【分析】将充分不必要条件转化为集合间的包含关系求解可得结论．【详解】设{}1|,|12322A x a x a B x a x a ⎧⎫=≤≤+=-<<+⎨⎬⎩⎭，∵“12a x a ≤≤+”是“1232a x a -<<+”的充分不必要条件，∴A B ，∴1232121322a a a a a a ⎧⎪-<+⎪-<⎨⎪⎪+<+⎩，解得13a >，∴实数a 的取值范围是1,3⎛⎫+∞ ⎪⎝⎭．故答案为1,3⎛⎫+∞ ⎪⎝⎭．【点睛】根据充要条件求解参数范围的方法步骤(1)把充分条件、必要条件或充要条件转化为集合间的关系；(2)根据集合关系画数轴或Venn 图，由图写出关于参数的不等式(组)，求解．注意：求解参数的取值范围时，一定要注意区间端点值的检验，尤其是利用两个集合之间的关系求解参数的取值范围时，不等式是否能够取等号决定端点值的取舍，处理不当容易出现漏解或增解的现象．5.设M={a,b}，则满足M ∪N ⊆{a,b,c}的非空集合N 的个数为______________．【答案】4【详解】根据M ∪N ⊆{a ，b ，c}而M 中没有c 元素，所以N 集合中一定要有c 元素，可能有a,b 元素且N 为非空集合，所以N 可以为{c}，{a ，c}，{b ，c}，{a ，b ，c}共4个．故答案为46.函数()f x =的定义域为[2,1]-，则a 的值为__________【答案】2【分析】由题意得不等式()()2213160ax a x -+-+≥的解集为[]2,1-，然后根据“三个二次”间的关系求解即可得到结论．【详解】∵函数()f x =的定义域为[]2,1-，∴不等式()()2213160a x a x -+-+≥的解集为[]2,1-，∴2,1x x =-=是方程()()2213160a x a x -+-+=的两个根，∴()()()()2241616013160a a a a ⎧---+=⎪⎨-+-+=⎪⎩，整理得2223203100a a a a ⎧--=⎨+-=⎩，解得2a =．故答案为2．【点睛】本题以函数的定义域为载体，考查一元二次方程、二次函数、二次不等式间的关系，解题的关键是根据题意得到方程的两根，然后再根据方程的有关概念求出a 的值，考查转化能力和运算能力，属于基础题．7.已知函数()(1)23f x m x m =-+-，无论m 取什么实数，函数()f x 的图像始终过一个定点，该定点的坐标为__________【答案】()2,1--【分析】将函数解析式变形为()230x m x y +---=，然后令20x +=且30x y ---=，求得方程组的解后即可定点的坐标．【详解】由()123y m x m =-+-变形得()230x m x y +---=，解方程组2030x x y +=⎧⎨---=⎩得21x y =-⎧⎨=-⎩，所以函数()f x 的图象过的定点的坐标为()2,1--．故答案为()2,1--．【点睛】本题考查一次函数的图象过定点的问题，解题时可把函数解析式化为(,)(,)0kf x y g x y +=（k 为参数）的形式，则以方程组(,)0(,)0f x y g x y =⎧⎨=⎩的解为坐标的点即为定点．8.已知关于x 的方程2240x kx k k +++-=有两个实数根，且一根大于1，一根小于1，则实数k 的取值范围为__________【答案】(3,1)-【分析】根据一元二次方程根的分布求解，令()224f x x kx k k =+++-，则有()10f <，解不等式可得所求范围．【详解】令()224f x x kx k k =+++-，∵方程的一个实数根大于1，另一个实数根小于1，∴()21140f k k k =+++-<，即2230k k +-<，解得31k -<<，∴实数k 的取值范围为()3,1-．故答案为()3,1-．【点睛】本题考查根据方程根的情况求参数的取值范围，解题时根据方程根的分布将问题转化为不等式求解，体现了转化和数形结合的思想方法在解题中的应用．9.给出下列四个命题：（1）若a b >，c d >，则a d b c ->-；（2）若22a x a y >，则x y >；（3）若a b >，则11a b a>-；（4）110a b <<，则2ab b <.其中正确命题是________.（填所有正确命题的序号）【答案】（1）（2）（4）【分析】根据不等式的性质，以及特殊值验证，逐项判断，即可得出结果.【详解】（1）若a b >，c d >，则a c b d +>+，因此a d b c ->-，即（1）正确；（2）若22a x a y >，根据不等式性质，可得x y >；即（2）正确；（3）若1a =，1b =-，满足a b >，但不满足11a b a>-；（3）错误；（4）若110a b <<，则0b a <<，因此()20ab b b a b -=-<，即2ab b <；故（4）正确；故答案为：（1）（2）（4）【点睛】本题主要考查判定命题的真假，考查由不等式性质判定所给结论是否正确，属于基础题型.10.若(,2)x ∈-∞，则2542-+-x x x的最小值为__________【答案】2【分析】将原式变形后根据基本不等式求解．【详解】∵2x <，∴20x ->．由题意得2254(2)11==(2)+2222x x x x x x x -+-+-≥=---，当且仅当122x x-=-，即1x =时等号成立．∴2542x x x-+-的最小值为2.故答案为2.【点睛】应用基本不等式求最值时一定要注意“一正二定三相等”这三个条件缺一不可，当不满足不等式使用的条件时，可通过适当的变形使得出现定值的形式，这是解题中常遇到的情形．11.设函数()2f x x =-，若不等式|(3)|()f x f x m +>+对任意实数x 恒成立，则m 的取值范围是__________【答案】3m <-【分析】12x x +--表示数轴上的x 对应点到-1对应点的距离减去它到2对应点的距离，其最小值为-3，故有m<-3，由此求得m 的取值范围．【详解】∵()2f x x =-，不等式()()3f x f x m +>+对任意实数x 恒成立，∴12m x x <+--对任意实数x 恒成立，又12x x +--表示数轴上的x 对应点到-1对应点的距离减去它到2对应点的距离，∴123x x +--≥-，∴3m <-，∴实数m 的取值范围是(),3-∞-．故答案为(),3-∞-．【点睛】本题考查恒成立问题，解题的关键是根据绝对值的几何意义求出12x x +--的最小值，考查转化和数形结合思想的运用能力．12.对于实数A 和正数B ，称满足不等式||x A B -<(,0)A B ∈>R 的实数x 的集合叫做A 的B 邻域，已知t 为给定的正数，a 、b 为正数，若a b t +-的a b +领域是一个关于原点对称的区间，则22a b +的最小值为__________【答案】22t 【分析】先根据条件求出()2t x a b t -<<+-；再结合邻域是一个关于原点对称的区间得到a b t +=，最后结合不等式的知识可求出22a b +的最小值．【详解】∵A 的B 邻域在数轴上表示以A 为中心，B 为半径的区域，∴()x a b t a b -+-<+，∴()a b x a b t a b --<-+-<+，解得()2t x a b t -<<+-．∵邻域是一个关于原点对称的区间，∴()220a b t +-=，∴a b t +=．∵222a b ab +≥，∴()()22222222a ba b ab a b t +≥++=+=，∴2222t a b +≥，当且仅当a b =时等号成立，∴22a b +的最小值为22t ．故答案为22t ．【点睛】本题以新概念为载体考查重要不等式的应用，考查变换能力和阅读理解能力．解题的关键是根据题意得到a b t +=这一结论，然后再通过变形得到所求的最小值．二.选择题13.已知1a ，1b ，2a ，2b R ∈且都不为零，则“1122a b a b =”是“110a x b +>与220a x b +>解集相同”的A.充分非必要条件B.必要非充分条件C.充要条件D.既非充分又非必要条件【答案】B【分析】根据充分条件和必要条件的定义，结合不等式的性质进行判断即可.【详解】若1122a b a b =，取111a b ==，221a b ==-，则10x +>与10x -->的解集不同，所以“1122a b a b =”不是“110a x b +>与220a x b +>解集相同”的充分条件；若1a ，1b ，2a ，2b R ∈且都不为零，且110a x b +>与220a x b +>的解集相同，此时必有1212b b a a -=-，所以1122a b a b =成立，所以“1122a b a b =”是“110a x b +>与220a x b +>解集相同”的必要条件.综上，“1122a b a b =”是“110a x b +>与220a x b +>解集相同”的必要不充分条件.故选：B.【点睛】本题主要考查充分条件和必要条件的判断，属于常考题.14.解析式为221y x =+，值域为{}5,19的函数有A.4B.6C.8D.9【答案】D【分析】根据y 的值求出相应的x 的值，再根据函数的有关概念得到定义域的不同形式，进而可得结论．【详解】由2215x +=，解得x =；由22119x +=，解得3x =±．所以函数的定义域可为}}{}{}{}{},3,,3,,3,----{}}{}3,3,3,3,3,3---，共9种情况．故选D ．【点睛】本题考查函数的概念，考查分析理解问题的能力，解题的关键是深刻理解函数的概念，根据对应关系求出x 的取值，然后再根据定义域中元素的个数确定出函数定义域的不同情形．15.设()f x 是定义在正整数集上的函数，且()f x 满足：“当2()f x x >成立时，总可以推出2(1)(1)f x x +>+成立”，给出以下四个命题：①若(3)9f ≥，则(4)16f ≥；②若(3)10f =，则(5)25f >；③若(5)25f =，则(4)16f ≤；④若2()(1)f x x ≥+，则2(1)f x x +≥.其中真命题的个数为（）个A.1B.2C.3D.4【答案】C【分析】根据题意对给出的四个命题分别进行分析、排除后可得正确的结论．【详解】对于①，由于f(3)=9时，可以使得f(4)<16，这并不与题设矛盾，所以当f(3)≥9时，由题设不一定得到f(4)≥16成立，所以①为假命题．对于②，∵f(3)=10>9，∴f(4)>4²，∴f(5)>5²=25，所以②为真命题；对于③，若f(4)>16，则f(5)>25，这与f(5)=25矛盾，所以f(4)≤16，所以③为真命题；对于④，∵f(x)≥(x+1)²>x ²，∴f(x+1)>(x+1)²>x ²，即有f(x+1)≥x²，所以④为真命题．综上可得②③④为真命题．故选C ．【点睛】本题考查推理论证能力，解题的关键是根据条件“当()2f x x >成立时，总可以推出()()211f x x +>+成立”进行判断，注意解题方法的选择，如直接推理、利用反证法判断等．16.设a ，b ，c 为实数，22()()(),()(1)(1),f x x a x bx cg x ax cx bx =+++=+++记集合{}{}|()0,,|()0,,S x f x x R T x g x x R ==∈==∈若{S }，{T }分别为集合S ，T 的元素个数，则下列结论不可能的是（）A.{S }=1且{T }=0B.{S }=1且{T }=1C.{S }=2且{T }=2D.{S }=2且{T }=3【答案】D【详解】∵2()()(),f x x a x bx c =+++当()0f x =时至少有一个根x a =-，当240b c -=时，()0f x =还有一根2b x =-，只要b ≠﹣2a ，()0f x =就有2个根；当b =﹣2a ，()0f x =是一个根当240b c -<时，()0f x =只有一个根；当240b c ->时，()0f x =只有二个根或三个根；当a =b =c =0时{S }=1，{T }=0当a ＞0，b =0，c ＞0时，{S }=1且{T }=1当a =c =1，b =﹣2时，有{S }=2且{T }=2故选：D 三.解答题17.已知集合2{|(1)320}=-+-=A x m x x ，是否存在这样的实数m ，使得集合A 有且仅有两个子集？若存在，求出所有的m 的值组成的集合M ；若不存在，请说明理由.【答案】11,8M ⎧⎫=-⎨⎬⎩⎭【分析】若集合A 有且仅有两个子集，则A 有且仅有一个元素，即方程()21320m x x -+-=只有一个根，进而可得答案【详解】存在11,8M ⎧⎫=⎨⎬⎩⎭满足条件．理由如下：若集合A 有且仅有两个子集，则A 有且仅有一个元素，即方程()21320m x x -+-=只有一个根，①当10m -=，即=1m 时，由320x -=，解得23x =，满足题意．②当10m -≠，由A 有且仅有一个元素得()10Δ=9+81=0m m -≠-⎧⎨⎩，解得18m =-．综上可得=1m 或18m =-，∴所有的m 的值组成的集合11,8M ⎧⎫=-⎨⎬⎩⎭．【点睛】本题考查集合元素个数的问题，考查分析问题的能力，解题的关键是由题意得到方程根的个数，然后通过对方程类型的分类讨论得到所求的参数．18.我校第二教学楼在建造过程中，需建一座长方体形的净水处理池，该长方体的底面积为200平方米，池的深度为5米，如图，该处理池由左右两部分组成，中间是一条间隔的墙壁，池的外围周壁建造单价为400元/平方米，中间的墙壁（不需考虑该墙壁的左右两面）建造单价为100元/平方米，池底建造单价为60元/平方米，池壁厚度忽略不计，问净水池的长AB 为多少时，可使总造价最低？最低价为多少？【答案】15AB =时，总造价最低为132000元.【分析】设AB 的长为x 米，进而得到宽BC 为200x 米，根据题意得到总造价的表达式，然后根据基本不等式求出造价的最小值即可．【详解】设AB 的长为x 米，则宽BC 为200x 米，由题意得总造价为200200400(22)5100560200y x x x =+⨯⨯+⨯⨯+⨯450(2)12000x x=++12000≥+132000=，当且仅当4502x x=，即15x =时等号成立．所以当净水池的长15AB =米时，可使总造价最低，最低价为132000元．【点睛】基本不等式为求最值提供了工具，在利用基本不等式求最值时，一定要注意使用基本不等式的条件，即“一正二定三相等”，且三个条件缺一不可，当题目中不满足使用不等式的条件时，则需经过变形得到所需要的形式及条件．19.已知a ∈R ，集合26{|0}1x x A x x --=≤+，集合{||2|1}B x x a a =+≤+.（1）求集合A 与集合B ；（2）若A B B = ，求实数a 的取值范围.【答案】（1）(,2](1,3]A =-∞-⋃-，当1a >-，[31,1]B a a =---+，当1a =-，{2}B =，当1a <-，B =∅；（2）(,0)[3,)-∞⋃+∞.【分析】（1）解不等式得出集合A 、B ；（2）根据A∩B=B 得出B ⊆A ，讨论B=∅和B≠∅时，求出满足条件的实数a 的取值范围．【详解】（1）由题意得()()(](]2236|0|0,21,311x x x x A x x x x ⎧⎫+-⎧⎫--=≤=≤=-∞-⋃-⎨⎬⎨⎬++⎩⎭⎩⎭．当10a +<，即1a <-时，B =∅；当10a +=，即1a =-时，{}2B =；当10a +>，即1a >-时，{}[]|12131,1B x a x a a a a =--≤+≤+=---+．（2）∵A B B ⋂=，∴B ⊆A ．①当1a <-时，B =∅，满足B ⊆A ；②当1a =-时，{}2B =，满足B ⊆A ；③当1a >-时，[]31,1B a a =---+，由B ⊆A 得31113a a -->-⎧⎨-+≤⎩或12a -+≤-，解得20a -≤<或3a ≥，又1a >-，∴10a -<<或3a ≥．综上可得0a <或3a ≥，∴实数a 的取值范围为()[),03,-∞⋃+∞．【点睛】根据集合间的包含关系求参数的取值范围时，一般要借助于数轴进行求解，根据集合端点值的大小关系转化为不等式（组）求解，解题时要注意不等式中的等号是否成立，这是解题中容易出现错误的地方．20.已知函数2|1|()4x m f x x +-=-，0m >，满足(2)2f =-.（1）求实数m 的值；（2）在平面直角坐标系中，作出函数()f x 的图像，并且根据图像判断：若关于x 的方程()f x k =有两个不同实数解，求实数k 的取值范围（直接写结论）【答案】（1）1m =；（2）图象见解析，()2,0-.【分析】（1）直接由f （2）=-2求得m 的值；（2）把m 值代入函数解析式，写出分段函数，根据函数的单调性作出图象，然后利用数形结合即可求得使关于x 的方程f （x ）=k 有两个不同实数解的实数k 的取值范围．【详解】（1）∵()214x m f x x +-=-，0m >，且()22f =-，∴221224m +-=--，即12m +=，解得1m =或3m =-，又0m >，∴1m =．（2）由（1）得()2,042424,04x x x x x f x x x x x ⎧≥≠⎪⎪-==⎨-⎪-<⎪-⎩且，当04x x ≥≠且时，()22(4)882444x x f x x x x -+===+---，∴函数()f x 在[0,4)和(4,)+∞上为减函数；当0x <时，()22(4)882444x x f x x x x -+=-=-=-----，∴函数()f x 在(,0)-∞上为增函数，且()()200f x f -<<=．画出函数图象如下图：由图可知，要使关于x 的方程()f x k =有两个不同实数解，则20k -<<，∴实数k 的取值范围是()2,0-．【点睛】（1）描点法画函数图象的步骤：①确定函数的定义域；②化简函数的解析式；③讨论函数的性质，即奇偶性、周期性、单调性、最值(甚至变化趋势)等；④描点连线，画出函数的图象．（2）利用函数图象确定方程或不等式的解，形象直观，体现了数形结合思想，解题的关键是正确的作出函数的图象．21.已知M 是满足下列性质的所有函数()f x 组成的集合：对任何12,f x x D ∈（其中f D 为函数()f x 的定义域），均有1212()()||f x f x x x -≤-成立.（1）已知函数2()1f x x =+，11[,]22x ∈-，判断()f x 与集合M 的关系，并说明理由；（2）是否存在实数a ，使得()2a p x x =+，[1,)x ∈-+∞属于集合M ？若存在，求a 的取值范围，若不存在，请说明理由；（3）对于实数a 、b ()a b <，用[,]a b M 表示集合M 中定义域为区间[,]a b 的函数的集合.定义：已知()h x 是定义在[,]p q 上的函数，如果存在常数0T >，对区间[,]p q 的任意划分：011n n p x x x x q -=<<⋅⋅⋅<<=，和式11|()()|ni i i h x h x T -=-≤∑恒成立，则称()h x 为[,]p q 上的“绝对差有界函数”，其中常数T 称为()h x 的“绝对差上界”，T 的最小值称为()h x 的“绝对差上确界”，符号121n i n i tt t t ==++⋅⋅⋅+∑；求证：集合[1009,1008]M -中的函数()h x 是“绝对差有界函数”，并求()h x 的“绝对差上确界”.【答案】（1）()f x 属于集合M ；（2）[1,1]-；（3）略.【分析】（1）利用已知条件，通过任取1211,,22x x ⎡⎤∈-⎢⎥⎣⎦，证明()()1212f x f x x x -≤-成立，说明f （x ）属于集合M ．（2）若p （x ）∈M ，则有121222a a x x x x -≤-++，然后可求出当[]1,1a ∈-时，p （x ）∈M ．（3）直接利用新定义加以证明，并求出h （x ）的“绝对差上确界”T 的值．【详解】（1）设1211,,22x x ⎡⎤∈-⎢⎥⎣⎦，则()()2212121212f x f x x x x x x x -=-=-+，∵121111,2222x x -≤≤-≤≤，∴1211x x -≤+≤，∴1201x x ≤+≤∴()()221212121212f x f x x x x x x x x x -=-=-+≤-，∴函数()f x 属于集合M ．（2）若函数()2a p x x =+，[)1,x ∈-+∞属于集合M ，则当[)12,1,x x ∈-+∞时，()()1212p x p x x x -≤-恒成立，即121222a a x x x x -≤-++对[)12,1,x x ∈-+∞恒成立，∴12(2)(2)a x x ≤++对[)12,1,x x ∈-+∞恒成立．∵[)12,1,x x ∈-+∞，∴12(2)(2)1x x ++≥，∴||1a ≤，解得11a -≤≤，∴存在实数a ，使得()2a p x x =+，[)1,x ∈-+∞属于集合M ，且实数a 的取值范围为[1,1]-．（3）取1009,1008p q =-=，则对区间[]1009,1008-的任意划分：01110091008n n x x x x --=<<⋅⋅⋅<<=，和式()()()()()()()()1110211i i i n n n h x h x h x h x h x h x h x h x =--∑-=-+-++-10211n n x x x x x x -≤-+-++- 10211=()()()n n x x x x x x --+-++- 0n x x =-1008(1009)=--2017=，∴集合[]1009,1008M -中的函数()h x 是“绝对差有界函数”，且()h x 的“绝对差上确界”2017T =．【点睛】本题考查新信息问题，考查阅读理解和应用能力，具有一定的综合性，解题的关键是弄懂给出的定义，解题时始终要围绕着给出的定义进行验证、求解等．。

复旦大学附属中学2017学年度第一学期高一年级物理期中考试试卷（时间：60分钟，满分100分,g取10m/s2）一、单项选择题（每题4分，共40分）1.下列说法正确的是（C）A.若物体的加速度均匀增加，则物体做匀加速直线运动B.若物体的加速度增加，则物体不可能做减速运动C.只要物体的加速度不为零，它的速度总是在发生变化的D.只要物体做直线运动，位移的大小和路程就一定相等2.三个质点A、B、C的运动轨迹如图2所示,同时从N点出发,同时到达M点,下列说法中正确的是(C)A.三个质点任意时刻的速度方向都相同B.三个质点从N点出发到M的任意时刻速度大小都相同C.三个质点从N点到M点的平均速度大小和方向均相同D.三个质点从N点到M点的平均速率相同【解析】平均速度等于位移比时间，三个质点的位移相同，故平均速度相同。

而平均速率等于路程比时间，三个质点的路程不相等。

3.如图所示，各接触面均光滑，A、B间无弹力作用的是（A）A．B．C．D．4.如图所示,某人用手握住一个保温杯,则下列说法中正确的是(B)A.若保持保温杯始终静止，则手握得越紧，保温杯受到摩擦力越大B.握着保温杯匀速向上运动，保温杯所受的摩擦力向上C.握着保温杯匀速向下运动，保温杯所受的摩擦力向下D.握着保温杯水平匀速运动，保温杯不受摩擦力5.某质点以20m/s 的初速度竖直向上运动,其加速度保持不变,经2s 到达最高点,上升高度为20m,又经过2s 回到出发点时,速度大小仍为20m/s,关于这一运动过程的下列说法中不正确的是(C )A ．质点运动的加速度大小为10m/s 2,方向竖直向下B ．最高点时速度为零,加速度不为零C ．质点在上升过程中速度的变化量大小相等20m/s ，方向竖直向上D ．质点在上升过程中和下降过程中速度的变化量大小相等方向相同6.用轻绳系住一小球静止在光滑斜面上,如图所示.若要按力的实际作用效果来分解小球的重力,则重力的两个分力的方向分别是图中的(C )A .1和5B .2和5C .3和5D.3和2【解析】小球重力产生两个效果。

2017学年复旦附中高一上期中填空题1.已知全集U=R, A={T,0,l,2}, B = {x|x2 = x},则A A =【答案】{-1,2}【解析】【分析】先求出集合B,再求出CuB,最后求出AHCuB.【详解】由题意得B={X|X2= X}=(0.1},■,- CuB = (.")u((M)u(]. + s),.•.AnCyBT-1,2}.故答案为{-1.2).【点晴】本题考査集合的运算，解题时根据集合运算的顺序进行求解即可，属于基础題.2.命题“如果a + b>0,那么a>0且b>0”的否命题是命题(填“真”或“假”)【答案】真【解析】【分析】根据原命题的逆命题和其否命题为等价命题判断命题的真假.【详解】由题意得命题"如果a+b>0,那么a>0且b>0”的逆命题为“如果a>0且b>0,那么a + b>0” 其真命题，所以否命题为真命题. 故答案为“真”.【点晴】判断命题的真假时，可通过命题直接进行判断也可通过其等价命题的真假来判断，解題时要根据条件选择合理的方法进行求解.3.已知集合A = {y|y = x2-2x-3}, B = (y|y = -x2 + 2x+13},则A AB-【答案】[-4,14]【解析】【分析】分别求出集合A.B,然后再求出AQB即可.【详解】由题意得A-{y|y = x2・2x・3}-{y|y=(x-l)2.4} = {y|yN -4},B = {y I y = . x2 + 2x + 13} = {y I y = - (x -1)2 + 14j = {y I y < 14},•・ AC1B = [.4,14].故答案为[.4,14].【点睛】本题考査集合的交集运算，解题的关键是正确求出集合AB,属于简单题.4.已知“aga +孑是“ l・2a<x<3a + 2”的充分不必要条件，则实数a的取值范围是【答案】aR3【解析】【分析】将充分不必要条件转化为集合间的包含关系求解可得结论.【详解】设A= x a<x<a + - B = {x| 1 - 2a <x <3a + 2},*a<x<a + ^是'l.2a<x〈3a + 2”的充分不必要条件,l-2a〈3a + 2l-2a <a 1•• 1 ，解得a>;,a+ —〈3a + 2 32实数a的取值范围是（? + 8）.故答案为@ + oo）.【点睛】根据充要条件求解参数范围的方法步骤⑴把充分条件、必要条件或充要条件转化为集合间的关系；（2）根据集合关系画数轴或Venn图，由图写出关于参数的不等式（组），求解.注意：求解参数的取值范围时，一定要注意区间端点值的检验，尤其是利用两个集合之间的关系求解参数的取值范围时，不等式是否能够取等号决定端点值的取舍，处理不当容易出现漏解或增解的现象.5.设M = {&b},则满足MUNG^b.c}的集合N的个数为【答案】8【解析J盼析】分别写出満足条件的集合N后可得所求集合的个数.【详解】由题意得，满足题意得集合N为0, {a}, {b},{c}, (a,c}, {b,c}, (a,b} , {a,b,c},共8个.故答案为8.【点睛】解题时要根据集合N中元素的个数为标准进行求解，考査理解能力和判断能力，属于基础題.6.函数f(x) = ((12)x2+3(1*+ 6的定义域为[-2,1 J,则a的值为【答案】2【解析】【分析】由题意得不等式(l-a2)x2+ 3(l.a)x + 6>0的解集为[-2,1],然后根据“三个二次”间的关系求解即可得到结论.【详解】.•函数f(x) = J(1 - a* + 3(1 - a)x + 6的定义域为[-2,1],••不等式(1“浓2 + 3(1.必+ 6?0的解集为[-2,1],.. x = - 2,x = I是方程(]. a2)x2 + 3(1 - a)x + 6 = 0的两个根,.4(l-a2)-6(l-a) + 6 = 0(l-a2) + 3(l-a) + 6 = 0，整理得！矿了#「解得a = 2 .Ia~ + 3a-10 = 0故答案为2.【点睛】本题以函数的定义域为载体，考査一元二次方程、二次函数、二次不等式间的关系，解题的关蚀是根据题意得到方程的两根，然后再根据方程的有关概念求出a的俏，考査转化能力和运算能力，属于基础题.7.已知函数f(x)=(m-l)x + 2m-3,无论m取什么实数，函数f(x)的图像始终过一个定点，该定点的坐标为【答案】(-2,-1)【解析】【分析】将函数解析式变形为(x + 2)m・x.y.3 = O,然后令x + 2 =0旦・x . y . 3 = 0,求得方程组的解后即可定点的坐标.【详解】由y・(m・l)x+2m-3变形得(x+2)m - x-y- 3-0,解方程组&篇％得疝彳，所以函数f(x)的图象过的定点的坐标为(-2,-1).故答案为(-2,-1).【点睛】本题考査一次函数的图象过定点的问题，解题时可把函数解析式化为kf(x.y) + g(%y) = 0 (k为参数)的形式，则以方程组｛；修与号的解为坐标的点即为定点.8.已知关于x的方程x2 + kx + k2 + k-4 = 0有两个实数根，且一根大于1, 一根小于1,则实数k的取值范围为【答案】(-3.1)【解析】【分析】根据一元二次方程根的分布求解，令f(x)=x2 + kx + k2+k-4,则有解不等式可得所求范围【详解】令f(x)=x2+kx + k2 + k-4,方程的一个实数根大于1,另一个实数根小于1,1 + k+k2 + k-4<0.即k2+2k-3<0>解得-3<k<l,实数g取值范围为(-3,1).故答案为(-3,1).【点晴】本题考査根据方程根的情况求参数的取值范围，解题时根据方程根的分布将问题转化为不等式求解，体现了转化和数形结合的思想方法在解题中的应用.9-给出下列四个命题：(1)若a > b,c a d,则a-d > b-c;(2)若a2x>a2y .则x>y;(3)a>b,则二a-b a(4)若以<0,则abvb，.a b其中正确命题的是.(填所冇正确命题的序号)【答案J (1) (2) (4)【解析】试题分析：(3)中a = 0时不等式不成立，故正确的只有(1) (2) (4).考点：不等式的基本性质10.若xe(-oo,2),则5—4X + X 的最小值为2-x【答案】2【解析】【分析】将原式变形后根据基本不等式求解.【详解】..•x<2, •••2-x>0.当且仅当2-x = d-，即x = l 时等号成立. 2-x••5~4X + X2的最小值为2. 2-x故答案为2.【点睛】应用基本不等式求最值时一定要注意“一正二定三相等.这三个条件缺一不可，当不满足不等式使用的条件时，可通过适当的变形使得出现定值的形式，这是解题中常遇到的情形. 11.设函数f(x)=x-2,若不等式|Rx+3)1 > |f(x)|十m 对任意实数x 恒成立，则m 的取值范围是【答案】mv-3 【解析】【分析】 |x+l|-|x-2|表示数轴上的x 对应点到-1对应点的距离减去它到2对应点的距离，其最小值为-3,故有m<-3, 由此求得m 的取值范围.【详解】I f(x) = x - 2,不等式|f(x + 3)|> |f(x)| + m 对任意实数x 恒成立，二n 】Y|xi 1| -|x-2对任意实数X 恒成立，乂 |x+l|・X ・2|表示数轴上的x 对应点到・1对应点的距离减去它到2对应点的距离，.・.|x+l|・|x ・2|N ・3,二 m v • 3二实数m 的取值范围是(・皿・3).故答案为(・s ，・3).【点睛】本题考査恒成立问题，解题的关键是根据绝对值的儿何意义求出|x +l|-|x-2|的最小值，考査转化和数形结合思想的运用能力. 12.对于实数A 和正数B,称满足不等式|x-A|<B (AGR,B>0)9!I 实数x 的集合叫做A 的B 邻域，已知t 为给定的 正数，a 、b 为正数，若a +由题意得2—x 5—4x + x~ (2-x)~+ 1 —=^=(2-x)+ 2,b-t的a + b领域是一个关于原点对称的区间，则a2 + b2的最小值为【答案】L2【解析】【分析】先根据条件求出-t<x<2(a+b)-t;再结合邻域是一个关于原点对称的区间得到 a + b = t ,最后结合不等式的知识可求出a2+ b2的最小值.【详解】.. A的B邻域在数轴上表示以A为中心，B为半径的区域，|x - (a + b -1)| < a + b,二- a- bvx-(a + b-t)va + b,解得-t<x<2(a + b)-t.邻域是一个关于原点对称的区间，二2(a + b) - 2t = 0,二a + b = t., a2 + b* > 2ab,.•・ 2(a2 + b2) > a2 + b2 + 2ab = (a+ b)2 = t2,•,•a2+ b2>-,当且仅当a = b时等号成立，2二a2 + b2M最小值为2故答案为2【点睛】本题以新概念为载体考査重要不等式的应用，考査变换能力和阅读理解能力.解题的美綻足根据题意得到a + b-t这-结论，然后再通过变形得到所求的最小值.二.选择題侣.设实数勺、全、b卜3不为0,则了禹成如是“关于满不等式"心。

2017学年复旦附中高一上期中2017.11一. 填空题1. 已知全集U =R ，{1,0,1,2}A =-，2{|}B x x x ==，则U A C B =2. 命题 “如果0a b +>，那么0a >且0b >”的否命题是 命题（填“真”或“假”）3. 已知集合2{|23}A y y x x ==--，2{|213}B y y x x ==-++，则AB = 4. 已知“12a x a ≤≤+”是“1232a x a -<<+”的充分不必要条件，则实数a 的取值范 围是5. 设{,}M a b =，则满足{,,}M N a b c ⊆的集合N 的个数为6.函数()f x 的定义域为[2,1]-，则a 的值为7. 已知函数()(1)23f x m x m =-+-，无论m 取什么实数，函数()f x 的图像始终过一个定点，该定点的坐标为8. 已知关于x 的方程2240x kx k k +++-=有两个实数根，且一根大于1，一根小于1，则实数k 的取值范围为9. 给出下列四个命题：① 若a b >，c d >，则a d b c ->-；② 若22a x a y >，则x y >；③ 若a b >，则11a b a >-；④ 若110a b<<，则2ab b <. 其中正确命题是 （填所有正确命题的序号）10. 若(,2)x ∈-∞，则2542x x x-+-的最小值为 11. 设函数()2f x x =-，若不等式|(3)||()|f x f x m +>+对任意实数x 恒成立，则m 的取值范围是12. 对于实数A 和正数B ，称满足不等式||x A B -<(,0)A B ∈>R 的实数x 的集合叫做A 的B 领域，已知t 为给定的正数，a 、b 为正数，若a b t +-的a b +领域是一个关于原点对称的区间，则22a b +的最小值为二. 选择题13. 设实数1a 、2a 、1b 、2b 均不为0，则“1122a b a b =成立”是“关于x 的不等式110a x b +> 与220a x b +>的解集相同”的（ ）条件A. 充分不必要B. 必要不充分C. 充要D. 既不充分也不必要14. 解析式为221y x =+，值域为{5,19}的函数有（ ）个A. 4B. 6C. 8D. 915. 设()f x 是定义在正整数集上的函数，且()f x 满足：“当2()f x x >成立时，总可以推出2(1)(1)f x x +>+成立”，给出以下四个命题：① 若(3)9f ≥，则(4)16f ≥； ② 若(3)10f =，则(5)25f >；③ 若(5)25f =，则(4)16f ≤； ④ 若2()(1)f x x ≥+，则2(1)f x x +≥.其中真命题的个数为（ ）个A. 1B. 2C. 3D. 416. 设a 、b 、c 为实数，2()()()f x x a x bx c =+++，2()(1)(1)g x ax cx bx =+++，记集合{|()0,}S x f x x ==∈R ，{|()0,}T x g x x ==∈R ，若||S 、||T 分别为集合S 、T 的元素个数，则下列结论不可能是（ ）A. ||1S =且||0T =B. ||1S =且||1T =C. ||2S =且||2T =D. ||2S =且||3T =三. 解答题17. 已知集合2{|(1)320}A x m x x =-+-=，是否存在这样的实数m ，使得集合A 有且仅有两个子集？若存在，求出所有的m 的值组成的集合M ；若不存在，请说明理由.18. 我校第二教学楼在建造过程中，需建一座长方体形的净水处理池，该长方体的底面积为200平方米，池的深度为5米，如图，该处理池由左右两部分组成，中间是一条间隔的墙壁，池的外围周壁建造单价为400元/平方米，中间的墙壁（不需考虑该墙壁的左右两面）建造单价为100元/平方米，池底建造单价为60元/平方米，池壁厚度忽略不计，问净水池的长AB 为多少时，可使总造价最低？最低价为多少？19. 已知a ∈R ，集合26{|0}1x x A x x --=≤+，集合{||2|1}B x x a a =+≤+. （1）求集合A 与集合B ；（2）若AB B =，求实数a 的取值范围.20. 已知函数2|1|()4x m f x x +-=-，0m >，满足(2)2f =-. （1）求实数m 的值；（2）在平面直角坐标系中，作出函数()f x 的图像，并且根据图像判断：若关于x 的方程()f x k =有两个不同实数解，求实数k 的取值范围（直接写结论）21. 已知M 是满足下列性质的所有函数()f x 组成的集合：对任何12,f x x D ∈（其中f D 为函数()f x 的定义域），均有1212|()()|||f x f x x x -≤-成立.（1）已知函数2()1f x x =+，11[,]22x ∈-，判断()f x 与集合M 的关系，并说明理由；（2）是否存在实数a ，使得()2a p x x =+，[1,)x ∈-+∞属于集合M ？若存在，求a 的取值 范围，若不存在，请说明理由；（3）对于实数a 、b ()a b <，用[,]a b M 表示集合M 中定义域为区间[,]a b 的函数的集合.定义：已知()h x 是定义在[,]p q 上的函数，如果存在常数0T >，对区间[,]p q 的任意划分：011n n p x x x x q -=<<⋅⋅⋅<<=，和式11|()()|ni i i h x h x T -=-≤∑恒成立，则称()h x 为[,]p q 上的“绝对差有界函数”，其中常数T 称为()h x 的“绝对差上界”，T 的最小值称为()h x 的“绝对差上确界”，符号121n i n i tt t t ==++⋅⋅⋅+∑；求证：集合[1009,1008]M -中的函数()h x 是“绝对差有界函数”，并求()h x 的“绝对差上确界”.参考答案一. 填空题1. {1,2}-2. 真3. [4,14]-4. 13a > 5. 8 6. 2 7. (2,1)- 8. (3,1)- 9. ①②④ 10. 2 11. 3m <- 12. 22t二. 选择题13. B 14. D 15. C 16. D三. 解答题 17. 1{1,}8M =.18. 15AB =时，总造价最低为132000元. 19.（1）(,2](1,3]A =-∞--，当1a >-，[31,1]B a a =---+，当1a =-，{2}B =，当1a <-，B =∅；（2）(,0)[3,)-∞+∞.20.（1）1m =；（2）(2,0)-.21.（1）()f x 属于集合M ；（2）[1,1]-；（3）略.。

【试卷分析】1、考点分布：复旦大学附属中学2017 学年第一学期高一年级化学期中考试试卷原子结构：1、2、3、8、24、25物质的量及物质的量浓度：17、21、31、33气体摩尔体积：4、5、6、15、18、26、27、28、34氯气及氯化氢（含粗盐提纯）：9、12、13、14、16、20、22、30、32离子反应方程式：7氧还：10、11、19、23、29集中考察氯气和氯化氢，包括氯气基础性质、氯碱工业、氯化氢的制备和喷泉实验，其中氯气和碱的反应考察较多（选择题22、实验题32），选择的考察较基础，填空题个别小题结合氧还对学生要求较高，实验题32 题将氯气的实验室制备和氯气与碱的反应结合考察，较综合；气体摩尔体积部分比重仅次于氯气，考察学生计算能力以及十字交叉法等计算方法；氧还部分考察四类物质和简单计算，整体不难。

2、本试卷的难点及易错点：第18 题气体摩尔体积结合物质的量计算，较综合第21 题涉及字母的物质的量浓度计算，学生易错第22 题考察守恒思想第32 题实验探究题，考察学生信息读取和运用能力，对学生要求较高1 复旦大学附属中学 2017 学年第一学期高一年级化学期中考试试卷（总分 100 分，时间 60 分钟）一、选择题（每题只有一个正确答案，共 36 分）1. 1993 年 8 月，我国科学家利用超高真空扫描隧道显微镜，在一块晶体硅的表面通过探针的作用搬走原子，写下了“中国”两个字。

下列说法中，错误的是A. 上述操作中发生了化学变化B. 这两个汉字是目前世界上最小的汉字C. 上述操作中只发生了物理变化D. 这标志着我国科学家已进入操作原子的阶段 【答案】 A2. 下列说法中正确的是A. 人们已知道了 107 种元素，即已经知道了 107 种原子B. 互为同位素的原子，它们原子核内的中子数一定不相等C. 40K 和 40Ca 原子中的中子数相等D. 质子数相同的微粒，核外电子数一定相同，但化学性质不一定相同 【答案】B3. 化学科学需要借助化学专用语言描述，下列有关化学用语正确的是A. 硫离子的电子式：S 2-B. Cl 原子结构示意图：C. 质量数为 3 的超重氢原子： 3 HD. 次氯酸钙的化学式：CaClO 【答案】C4. 下列物质中质量最大的是A. 1mol 硫酸B. 22.4L （S.T.P ）水C. 44.8L(S.T.P)O 2D. 12.04×10 23 个 NH 3 分子【答案】 B5.下列示意图中，白球代表氢原子，黑球代表氦原子，方框代表容器，容器中间有一个可以上下滑动的隔板（其质量忽略不记）。

2017学年复旦附中第一学期期中考试试卷初一年级数学（满分100 时间：90分钟）一、选择题：（本大题共6题，每题2分，满分12分）1.下列代数式a xy x x x ,4,1,2,3732--中单项式的个数是 （ ） A.1个 B.2个 C.3个 D.4个2.下列运算正确的是( )A. 32)(aB.4223432y y =⎪⎭⎫ ⎝⎛- C.236a a a =÷ D.532a a a =⋅ 3.下列各式从左到右各式是因式分解的是（ ）A. 1)1)(1(2-=-+x x xB.6)5(652+-=+-x x x xC.22)1(12-=+-x x xD.)()(y x b y x a by bx ay ax +++=+++4.下列分式化简正确的是（ ） A. b a b a b a -=--3)(32 B.32322a a a a =- C.b a b ab a 31236142+=-- D.ba b a b a +=++122 5.数学老师上课出了一道因式分解的思考题，题意是1622++mx x 能在有理数的范围内因式分解，则整数m 的值有几个，小军和小华为此争论不休，请你判断整数m 的值有几个？（ ） A.4个 B.5个 C. 6个 D.8个6.从边长为a 的大正方形纸板中挖去一个边长为b 的小正方形后，将其裁成四个相同的等腰梯形（图甲），然后拼成一个平行四边形（图乙），那么通过计算两个图形的阴影部分的面积，可以验证成立的公式是（ ）A. 222)(b a b a -=-B.))((22b a b a b a +-=-C.2222)(b ab a b a +-=-D.b ab a b a ++=+2)(2二、填空题：（本大题共12题，每题3分，满分36分）7.“12减去y 的41的差”用代数式表示是： 8.将多项式153322---y y x xy 按字母y 升幂排列是： 9.使分式25-+x x 有意义的条件是： 10.计算：=÷-22332)453(y xy y 11.计算：=+--+-21212a a a a 12.若,3293211⨯=⨯⨯m 则=m13.如果,4,5==+ab b a 则=-b a14.要使)12)(3(22---x x x ax 的展开式中不含2x ，则=a 15.c b a ,,为三角形三边长，022=--+bc b ac a ，则该三角形的形状为16.现在规定一种运算ab b b a --=2*，期中ab 为有理数，=-m n m *)(17.李丽从家到学校的路程为S 米，无风时她以平均a 米/秒的速度骑行，便能按时到达；当风速为b 米/秒时，她若顶风按时到校，请用代数式表示她必须提前出发的时间是18.在图中取阴影等边三角形各边的中点,连成一个等边三角形,将其挖去,得到图;对图中的每个阴影等边三角形仿照先前的做法,得到图,如此继续.如果图的等边三角形面积为,则第个图形中所有阴影三角形面积的和为_________.三、简答题：（本大题共7小题，每小题4分，满分28分）19.计算：ax by axy b by ax 41)2(2)221(22÷⎥⎦⎤⎢⎣⎡-+- 20.计算：)52)(52(++-+-+z y x z y x21.解方程：2229)31)(13()12()3(4x x x x x +-+=+--22.因式分解：222222)53()35(n m n m +-+23.因式分解：72)(18)(222+---x x x x24.因式分解：ab b a 4)1)(1(22--- 25.先化简，再求值：11)1112(22+÷+-+-x x x x x ，期中.2-=x 四、解答题（本大题共5小题，第26-29题每题5分，第30题4分，满分24分）26.已知311=-y x ，求yxy x y xy x ---+2535的值。

2017-2018学年上海复旦大学附属中学高一上学期期中考试数学试题一、单选题1．设实数、、、均不为0，则“成立”是“关于的不等式与的解集相同”的（）条件A．充分不必要B．必要不充分C．充要D．既不充分也不必要【答案】B【解析】根据充分条件和必要条件的定义结合不等式的性质进行判断即可．【详解】（1）若，则，所以不等式即为，若，则可化为，所以两个不等式的解集相同，若，则可化为，此时两个不等式的解集不相同，所以充分性不成立．（2）若关于x的不等式与的解集相同，则，由于、、、均不为0，①若，则不等式的解为，由两不等式的解集相同可得，可得，即必要性成立．②若，同理可得，即必要性成立．综上可得“成立”是“关于的不等式与的解集相同”的的必要不充分条件．故选B．【点睛】解答本题的关键有两个：一个是准确把握充分必要条件的判断方法，解题时要结合定义求解；二是注意分类讨论思想方法在解题中的应用．本题具有综合性，考查分析问题和解决问题的能力．2．解析式为，值域为的函数有（）A．4 B．6 C．8 D．9【答案】D【解析】根据的值求出相应的的值，再根据函数的有关概念得到定义域的不同形式，进而可得结论．【详解】由，解得；由，解得．所以函数的定义域可为，共9种情况．故选D．【点睛】本题考查函数的概念，考查分析理解问题的能力，解题的关键是深刻理解函数的概念，根据对应关系求出x的取值，然后再根据定义域中元素的个数确定出函数定义域的不同情形．3．设是定义在正整数集上的函数，且满足：“当成立时，总可以推出成立”，给出以下四个命题：① 若，则；② 若，则；③ 若，则；④ 若，则.其中真命题的个数为（）个A．1 B．2 C．3 D．4【答案】C【解析】根据题意对给出的四个命题分别进行分析、排除后可得正确的结论．【详解】对于①，由于f(3)=9时，可以使得f(4)<16，这并不与题设矛盾，所以当f(3)≥9时，由题设不一定得到f(4)≥16成立，所以①为假命题．对于②，∵f(3)=10>9，∴f(4)>4²，∴f(5)>5²=25，所以②为真命题；对于③，若f(4)>16，则f(5)>25，这与f(5)=25矛盾，所以f(4)≤16，所以③为真命题；对于④，∵f(x)≥(x+1)²>x²，∴f(x+1)>(x+1)²>x²，即有f(x+1)≥x²，所以④为真命题．综上可得②③④为真命题．故选C．【点睛】本题考查推理论证能力，解题的关键是根据条件“当成立时，总可以推出成立”进行判断，注意解题方法的选择，如直接推理、利用反证法判断等．4．设、、为实数，，，记集合，，若、分别为集合、的元素个数，则下列结论不可能是（）A．且B．且C．且D．且【答案】D【解析】分和两种情况对方程根的个数进行进行分析后可得正确的结论，进而得到不可能的结论．【详解】①若，,，当时，；当时，；当时，．②若，,，则当时，；当时，；当时，．所以只有D不可能．故选D．【点睛】解答本题的关键是由方程根的情况得到、取值的所有可能，然后再根据选项进行判断，考查分析问题和分类讨论在解题中的应用，具有一定的综合性和难度．二、填空题5．已知全集，，，则__________【答案】【解析】先求出集合，再求出，最后求出．【详解】由题意得，∴，∴．故答案为．【点睛】本题考查集合的运算，解题时根据集合运算的顺序进行求解即可，属于基础题．6．命题“如果，那么且”的否命题是__________命题（填“真”或“假”）【答案】真【解析】根据原命题的逆命题和其否命题为等价命题判断命题的真假．【详解】由题意得命题“如果，那么且”的逆命题为“如果且，那么”，其真命题，所以否命题为真命题．故答案为“真”．【点睛】判断命题的真假时，可通过命题直接进行判断也可通过其等价命题的真假来判断，解题时要根据条件选择合理的方法进行求解．7．已知集合，，则__________【答案】【解析】分别求出集合，然后再求出即可．【详解】由题意得，，∴．故答案为．【点睛】本题考查集合的交集运算，解题的关键是正确求出集合，属于简单题．8．已知“”是“”的充分不必要条件，则实数的取值范围是__________【答案】【解析】将充分不必要条件转化为集合间的包含关系求解可得结论．【详解】设，∵“”是“”的充分不必要条件，∴ ，∴，解得，∴实数的取值范围是．故答案为．【点睛】根据充要条件求解参数范围的方法步骤(1)把充分条件、必要条件或充要条件转化为集合间的关系；(2)根据集合关系画数轴或Venn图，由图写出关于参数的不等式(组)，求解．注意：求解参数的取值范围时，一定要注意区间端点值的检验，尤其是利用两个集合之间的关系求解参数的取值范围时，不等式是否能够取等号决定端点值的取舍，处理不当容易出现漏解或增解的现象．9．设，则满足的集合的个数为__________【答案】8【解析】分别写出满足条件的集合后可得所求集合的个数．【详解】由题意得，满足题意得集合为，，，，，，共8个．故答案为8．【点睛】解题时要根据集合中元素的个数为标准进行求解，考查理解能力和判断能力，属于基础题．10．函数的定义域为，则的值为__________【答案】2【解析】由题意得不等式的解集为，然后根据“三个二次”间的关系求解即可得到结论．【详解】∵函数的定义域为，∴不等式的解集为，∴是方程的两个根，∴，整理得，解得．故答案为2．【点睛】本题以函数的定义域为载体，考查一元二次方程、二次函数、二次不等式间的关系，解题的关键是根据题意得到方程的两根，然后再根据方程的有关概念求出的值，考查转化能力和运算能力，属于基础题．11．已知函数，无论取什么实数，函数的图像始终过一个定点，该定点的坐标为__________【答案】【解析】将函数解析式变形为，然后令且，求得方程组的解后即可定点的坐标．【详解】由变形得，解方程组得，所以函数的图象过的定点的坐标为．故答案为．【点睛】本题考查一次函数的图象过定点的问题，解题时可把函数解析式化为（为参数）的形式，则以方程组的解为坐标的点即为定点．12．已知关于的方程有两个实数根，且一根大于1，一根小于1，则实数的取值范围为__________【答案】【解析】根据一元二次方程根的分布求解，令，则有，解不等式可得所求范围．【详解】令，∵方程的一个实数根大于1，另一个实数根小于1，∴，即，解得，∴实数的取值范围为．故答案为．【点睛】本题考查根据方程根的情况求参数的取值范围，解题时根据方程根的分布将问题转化为不等式求解，体现了转化和数形结合的思想方法在解题中的应用．13．给出下列四个命题：（1）若，则；（2）若，则；（3），则；（4）若，则．其中正确命题的是．（填所有正确命题的序号）【答案】（1）（2）（4）【解析】试题分析：（3）中时不等式不成立，故正确的只有（1）（2）（4）.【考点】不等式的基本性质.14．若，则的最小值为__________【答案】2【解析】将原式变形后根据基本不等式求解．∵，∴．由题意得，当且仅当，即时等号成立．∴的最小值为2.故答案为2.【点睛】应用基本不等式求最值时一定要注意“一正二定三相等”这三个条件缺一不可，当不满足不等式使用的条件时，可通过适当的变形使得出现定值的形式，这是解题中常遇到的情形．15．设函数，若不等式对任意实数恒成立，则的取值范围是__________【答案】【解析】表示数轴上的对应点到1对应点的距离减去它到2对应点的距离，其最小值为3，故有m<-3，由此求得m的取值范围．【详解】∵，不等式对任意实数恒成立，∴对任意实数恒成立，又表示数轴上的对应点到1对应点的距离减去它到2对应点的距离，∴，∴，∴实数的取值范围是．故答案为．本题考查恒成立问题，解题的关键是根据绝对值的几何意义求出的最小值，考查转化和数形结合思想的运用能力．16．对于实数和正数，称满足不等式的实数的集合叫做的邻域，已知为给定的正数，、为正数，若的领域是一个关于原点对称的区间，则的最小值为__________【答案】【解析】先根据条件求出；再结合邻域是一个关于原点对称的区间得到，最后结合不等式的知识可求出的最小值．【详解】∵A的B邻域在数轴上表示以A为中心，B为半径的区域，∴，∴，解得．∵邻域是一个关于原点对称的区间，∴，∴．∵，∴，∴，当且仅当时等号成立，∴的最小值为．故答案为．【点睛】本题以新概念为载体考查重要不等式的应用，考查变换能力和阅读理解能力．解题的关键是根据题意得到这一结论，然后再通过变形得到所求的最小值．三、解答题17．已知集合，是否存在这样的实数，使得集合有且仅有两个子集？若存在，求出所有的的值组成的集合；若不存在，请说明理由.【答案】【解析】若集合A有且仅有两个子集，则A有且仅有一个元素，即方程只有一个根，进而可得答案【详解】存在满足条件．理由如下：若集合A有且仅有两个子集，则A有且仅有一个元素，即方程只有一个根，①当，即时，由，解得，满足题意．②当，由A有且仅有一个元素得，解得．综上可得或，∴所有的的值组成的集合．【点睛】本题考查集合元素个数的问题，考查分析问题的能力，解题的关键是由题意得到方程根的个数，然后通过对方程类型的分类讨论得到所求的参数．18．我校第二教学楼在建造过程中，需建一座长方体形的净水处理池，该长方体的底面积为200平方米，池的深度为5米，如图，该处理池由左右两部分组成，中间是一条间隔的墙壁，池的外围周壁建造单价为400元/平方米，中间的墙壁（不需考虑该墙壁的左右两面）建造单价为100元/平方米，池底建造单价为60元/平方米，池壁厚度忽略不计，问净水池的长为多少时，可使总造价最低？最低价为多少？【答案】时，总造价最低为132000元.【解析】设的长为米，进而得到宽为米，根据题意得到总造价的表达式，然后根据基本不等式求出造价的最小值即可．【详解】设的长为米，则宽为米，由题意得总造价为，当且仅当，即时等号成立．所以当净水池的长米时，可使总造价最低，最低价为132000元．【点睛】基本不等式为求最值提供了工具，在利用基本不等式求最值时，一定要注意使用基本不等式的条件，即“一正二定三相等”，且三个条件缺一不可，当题目中不满足使用不等式的条件时，则需经过变形得到所需要的形式及条件．19．已知，集合，集合.（1）求集合与集合；（2）若，求实数的取值范围.【答案】（1），当，，当，，当，；（2）.【解析】（1）解不等式得出集合A、B；（2）根据A∩B=B得出B⊆A，讨论B=和B≠时，求出满足条件的实数的取值范围．【详解】（1）由题意得．当，即时，；当，即时，；当，即时，．（2）∵，∴B⊆A．①当时，，满足B⊆A；②当时，，满足B⊆A；③当时，，由B⊆A得或，解得或，又，∴或．综上可得或，∴实数的取值范围为．【点睛】根据集合间的包含关系求参数的取值范围时，一般要借助于数轴进行求解，根据集合端点值的大小关系转化为不等式（组）求解，解题时要注意不等式中的等号是否成立，这是解题中容易出现错误的地方．20．已知函数，，满足.（1）求实数的值；（2）在平面直角坐标系中，作出函数的图像，并且根据图像判断：若关于的方程有两个不同实数解，求实数的取值范围（直接写结论）【答案】（1）；（2）图象见解析，.【解析】（1）直接由f（2）=-2求得m的值；（2）把m值代入函数解析式，写出分段函数，根据函数的单调性作出图象，然后利用数形结合即可求得使关于x的方程f（x）=k有两个不同实数解的实数k的取值范围．【详解】（1）∵，，且，∴，即，解得或，又，∴．（2）由（1）得，当时，，∴函数在和上为减函数；当时，，∴函数在上为增函数，且．画出函数图象如下图：由图可知，要使关于x的方程有两个不同实数解，则，∴实数k的取值范围是．【点睛】（1）描点法画函数图象的步骤：①确定函数的定义域；②化简函数的解析式；③讨论函数的性质，即奇偶性、周期性、单调性、最值(甚至变化趋势)等；④描点连线，画出函数的图象．（2）利用函数图象确定方程或不等式的解，形象直观，体现了数形结合思想，解题的关键是正确的作出函数的图象．21．已知是满足下列性质的所有函数组成的集合：对任何（其中为函数的定义域），均有成立.（1）已知函数，，判断与集合的关系，并说明理由；（2）是否存在实数，使得，属于集合？若存在，求的取值范围，若不存在，请说明理由；（3）对于实数、，用表示集合中定义域为区间的函数的集合.定义：已知是定义在上的函数，如果存在常数，对区间的任意划分：，和式恒成立，则称为上的“绝对差有界函数”，其中常数称为的“绝对差上界”，的最小值称为的“绝对差上确界”，符号；求证：集合中的函数是“绝对差有界函数”，并求的“绝对差上确界”.【答案】（1）属于集合；（2）；（3）略.【解析】（1）利用已知条件，通过任取，证明成立，说明f（x）属于集合M．（2）若p（x）∈M，则有，然后可求出当时，p（x）∈M．（3）直接利用新定义加以证明，并求出h（x）的“绝对差上确界”T的值．【详解】（1）设，则，∵，∴，∴∴，∴函数属于集合．（2）若函数，属于集合，则当时，恒成立，即对恒成立，∴对恒成立．∵，∴，∴，解得，∴存在实数，使得，属于集合，且实数的取值范围为．（3）取，则对区间的任意划分：，和式，∴集合中的函数是“绝对差有界函数”，且的“绝对差上确界”．【点睛】本题考查新信息问题，考查阅读理解和应用能力，具有一定的综合性，解题的关键是弄懂给出的定义，解题时始终要围绕着给出的定义进行验证、求解等．。

上海中学高一上期中2017.11一. 填空题1. 若集合2{|20}M x x x =+=，{0,1,2}N =，则M N =2. 不等式4021x x -<-的解是 3. 函数241y x x =-++（[1,1]x ∈-）的最大值等于 4. 命题“若1x =且2y =，则3x y +=”的逆否命题是5. 若集合7{||||}5x x Z x m ∈-<中只有一个元素，则实数m 的取值范围是 6. 已知“125m x m -<<+”是“23x <<”成立的必要非充分条件，请你写出符合条件的实数m 的一个值7. 已知正实数x 、y 满足22x y xy +=，则x y +的最小值为8. 若集合2{|320}A x x x =-+≤，{|}B x x a =<，若A B ⊆，则最小的整数a 为 9. 若关于x 的不等式|1|||2x x a a -+-<上的解集为∅，则实数a 的取值范围是 10. 关于x 的不等式210x kx k -+-<，当(1,2)x ∈时恒成立，则实数k 的取值范围是 11. 若三个二次函数2443y x ax a =+-+，22(1)y x a x a =+-+，222y x ax a =--+表示 的图像中至少有一条与x 轴有交点，则实数a 的取值集合是12. 上海中学在每学年的上学期会举行体育嘉年华活动，假设在今年的活动中共设了8个体育项目，高一某班的班主任参加了其中的若干个项目，甲、乙、丙三位同学猜测该老师参加的项目见下表：（“×”表示未参加，“√”表示参加）√ 老师告诉甲、乙、丙：“你们分别猜对5次、5次、6次”，由此请你猜测该老师参加的体 育项目编号依次为 二. 选择题13. 设,x y R ∈，“||||1x y +>”的一个充分条件是（ ）A. ||1x ≥B. ||1x y +≥C. 2y ≤-D. 1||2x ≥或1||2y ≥ 14. 不等式||x x x >的解集是（ ）A. {|01}x x <<B. {|1x x >或1}x <-C. {|1x x >或10}x -<<D.{|1x x <-或01}x <<15. 对三个正实数a 、b 、c ，下列说法正确的是（ ） A. 存在（a 、b 、c ）的一组值，使得1a b +、1b c +、1c a +均小于2 B. 存在（a 、b 、c ）的一组值，使得1a b +、1b c +、1c a+中恰有两个小于2C. 对（a 、b 、c ）任意值，1a b +、1b c +、1c a +都不小于2D. 对（a 、b 、c ）任意值，1a b +、1b c +、1c a+中至多有两个不小于216. 已知非空集合M 满足：对任意x M ∈，总有2x M ∉M ，若{0,12,3,4,5}M ⊆，则满足条件的M 的个数是（ ）A. 11B. 12C. 15D. 16三. 解答题17. 设二次函数2()f x x ax b =++的图像过原点，且集合{|()}A x f x x ==为单元集，求a 、b 的值.18. 解下列不等式：（1）21331x x -<-；（2）2|22||21|x x x -+>-.19. 已知集合2{|20}A x x px q =+-=，22{|420}B x x qx q p =+-+=（其中p 、q 为实数），判断“1p q ==”是“1A B ∈”的什么条件，并说明理由.20. 已知函数2()(1)||f x x x x a =+--. （1）若1a =-，解不等式()1f x >；（2）是否存在实数a ，使不等式()23f x x ≥-对一切实数x R ∈恒成立？若存在，求出a 的取值范围，若不存在，请说明理由.21. 已知二次函数2()f x ax bx c =++（0a >）.（1）若1a =，图像()f x 与x 轴的两个不同交点的横坐标都在(2,1)--内， 求证：(2)(0,1)f -∈；（2）若存在0x Z ∈，满足01|()|4f x ≤，则称0x 为函数()f x 的一个“近似整零点”， 若()f x 有四个不同的“近似整零点”，求a 的最大值.参考答案一. 填空题1. {2,0,1,2}-2. 1(,4)23. 44. “若3x y +≠，则1x ≠或2y ≠”5. 23(,]556. 0m =7.32+ 8. 3 9. 13a ≤ 10. 3k ≥11. 3(,][1,)2-∞--+∞ 12. 1、3二. 选择题13. C 14. C 15. B 16. A三. 解答题 17. 1a =，0b =. 18.（1）(,2)(1,5)-∞-；（2）(,1)(3,)-∞+∞.19. 充分非必要条件. 20.（1）(1,)+∞；（2）[3,1]-. 21.（1）略；（2）14.。

复旦大学附属中学2017-2018学年第一学期高三年级英语期中考试试卷第I卷（共103分）I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversation and the question will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. This afternoon. B. Tomorrow. C. Next week. D. Next month.2. A. She doesn‟t play tennis well. B. She likes other sports as well.C. She is an enthusiastic tennis player.D. She is a professional athlete.3. A. At a paint store. B. At an oil market.C. At a science museum.D. At a gallery.4. A. Work in the yard. B. Buy some wood.C. Go to the bookstore.D. Take a walk.5. A. A taxi driver. B. A passenger. C. A car cleaner. D. A mechanic.6. A. Call a repairman. B. Get out the paper stuck.C. Turn to her colleague for help.D. Restart the machine.7. A. There are not enough gardens. B. Parking areas are full before 10:00.C. Parking areas are closed after 10:00.D. All classes begin at 10:00.8. A. The presentation will begin at noon.B. She‟ll present her work to the man.C. She‟d like to invite the man for lunch.D. She suggests working on the presentation at 12:00.9. A. The dormitory hours. B. The problem with the rules.C. The door number of the dormitory.D. The time to open the dormitory.10. A. The chairs didn‟t need to be painted.B. He doesn‟t like the color of the chairs.C. The park could have avoided the problem.D. The woman should have been more careful.Section BDirections: In section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Worried B. Surprised. C. Satisfied. D. Uninterested.12. A. It spoiled Juana‟s reputation. B. It copied her ideas without permission.C. It bought Juana‟s dishwashers.D. It wanted to share the dishwasher market.13. A. A successful business case. B. Juana‟s waterless laundry.C. A case against a global company.D. The worldwide dishwasher market.Questions 14 through 16 are based on the following passage.14. A. Footprints. B. Food. C. Living insects. D. Orange seeds.15. A. Don‟t touch animals under any circumstances.B. Don‟t take away any natural objects from the park.C. Don‟t leave litter in the park or throw any off the boat.D. Don‟t transport animals from one island to another.16. A. To protect the guide‟s interest. B. To improve the unique environment.C. To ensure a trouble-free visit.D. To get rid of illegal behaviors.Section CDirections: In section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. (本题做在答题纸上)Write onlyII. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)Libraries are my world. I've been a patron all my life, and for the past nine years I (25) ______ (work) at multiple libraries and archives in and around Detroit. The library (26) ______ an institution has many roles, but as our country struggles through an economic crisis, I have watched the library where I work evolve into a career and business center, a community gathering place and a bastion for hope.In the spring of 2007 I got a library internship at the Southfield Public Library, just north of Detroit. Summers at SPL (27) ______ (be) usually slow, but that year, we experienced a library that hustled and bustled like science-fair project week, midterms or tax season. Yet patrons weren't looking for Mosby's Nursing Drug Reference or 1040 forms. They were coming for information on entrepreneurship and growing their small business.I interpreted people's interest in our business collection as the first step to (28) ______ (pursue) their dreams, but these patrons were not motivated by dreams. They were responding to reality, and they were looking for Plan B.Things worsened in 2008, and in 2009 the economic crisis continues to suffocate Michigan. Last year, we put up a display with a variety of job resources that we restocked every hour. Each night the library closed, the display was bare. (29) ______ we normally keep displays up for a week, we kept the job resources display up for months.Then there's the tightening credit market. People see the writing on the wall and they want to get educated. They can't afford a financial adviser, but checking books out is free. Some of (30) ______(popular) titles now are "Rich Dad, Poor Dad," "Think and Grow Rich," and "Suze Orman's 2009 Action Plan."The economic downturn affects us all. I have had to work long hours and don't get to see (31) ______ of my boyfriend or experience any kind of social life lately, but I am thankful to be in a position where I can help people overcome this struggle. In Michigan, we haven't lost hope. (32) ______ ______ ______ there are libraries here, there will always be hope.(B)It‟s estimated that 300 million people in China are studying, or (33) ______ (study), English. That‟s an impressive number and I can‟t think of any other country in the world where one quarter of the population is so dedicated to (34) ______ (learn) a second language. But some people are questioning whether this “craze” for studying English is worthwhile.Professor Zhang Shuhua of the Chinese Academy of Social Sciences says that too much emphasis is placed on learning English and that it is a waste of education resources as well as a threat to the study of Chinese. He says that having English as a compulsory course in university “has distracted much of students‟attention (35) ______ specialized subjects,”and that some students have been denied access to postgraduate education because they failed English. Others have admitted that studying so much English has made them (36) ______ (poor) Chinese speakers.Both of these criticisms are legitimate, but they beg the question of why so many Chinese still want to learn English. English, (37) ______ recognize, is the lingua franca of the modern world. It is the language of business and has become the language of international relations and culture. When people from different countries get together, they frequently speak in English rather than try to translate their native languages. It seems that everyone everywhere can speak at least some English.For China to be part of that international conversation, it is necessary that some level ofEnglish proficiency (38) ______ be achieved. But what, you may ask, about those who will never speak a word of English once they leave school? Well, for good or ill, they will still be surrounded by English. It is there in signs, in music, in movies and in the casual conversations they overhear of the increasing number of foreigners on the city streets. To know English is to be included in the rest of the world, (39) ______ ______ your world is limited to China.I agree with Professor Zhang on one point, (40) ______. English should not be a compulsory subject in university. For most, passing the CET is just the endless drudgery of memorizing word lists. There is little emphasis placed on communication. And if you can‟t communicate in English after years of study in primary school, middle school and high school, a few more years in university probably won‟t help.Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.For writers in Western cultures, autumn is a difficult season to describe. On the one hand, it is the end of the summer, and therefore a little sad. The nights draw in, and when you wake in the morning, there‟s mist and it‟s co oler: Winter is around the ___41___. American writer Ernest Hemingway wrote in his book A Moveable Feast: “You expected to be sad in the fall. Part of you died each year when the leaves fell from the trees and their branches were ___42___against the wind a nd the cold, wintry light.”On the other hand, autumn has its good side. There are so many changes in nature at this time of year, such as the reds and browns that the leaves change to, and the ___43___ they fall from the trees. French writer Albert Camus even though autumn was a second spring: “Autumn is a second spring when every leaf is a flower.” It‟s a view you can also find in the most famous autumn poem in English literature, To Autumn by John Keats. In that poem Keats says that the autumn has its own songs, just like spring.Another autumn theme is wisdom. The arrival of the season is thought to be similar to a person becoming ___44___. Their summer peak may have been and gone, but old age has not yet come. At this time it‟s thought that people have ___45___ a thing or two about life. The great Irish poet W.B. Yeats takes up this theme in his poem The Wild Swans at Coole. Yeats puts together a picture for the reader out of the ___46___ of the changing seasons in Coole Park in the west of Ireland, a place he knew well. Seeing and counting 59 swans, he remembers first making the count 19 years ago. He ___47___ whether he can still love like the lover swans do.Of course, many other themes and subject matters can play a part in the literature of autumn. For example, it‟s the beginning of a new term of the school year. As you would expect, autumncan___48___ in writing for children and young people. But autumn writing usually ___49___ on the changes in nature that we see, which writers often use as a ___50___ for changes in human life.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.After the college-board examinations in June, Basil Duke Lee and five other boys from St. Regis School ___51___ the train for the West. Two got out at Pittsburgh, one slanted south toward St. Louis and two stayed in Chicago; from then on Basil was alone. It was the first time in his life that he had ever felt the need of tranquility, but now he took long breaths of it; for, though things had gone better toward the end, he had had a / an ___52___ year at school.He wore one of those extremely flat derbies (常礼帽) in vogue during the twelfth year of the century, and a blue business suit became a little too short for his constantly ___53___ body. Within he was by turns a disembodied (空洞的) spirit, almost ___54___ of his person and moving in a mist of impressions and emotions, and a fiercely competitive individual trying ___55___ to control the rush of events that were the steps in his own ___56___ from child to man. He believed that everything was a matter of ___57___ — the current principle of American education — and his fantastic ___58___ was continually leading him to expect too much. He wanted to be a great athlete, popular, brilliant and always happy. During this year at school, where he had been punished for his “freshness,” for fifteen years of thorough spoiling at hom e, he had grown uselessly introspective, and this ___59___ with that observation of others which is the beginning of wisdom. It was apparent that before he obtained much success in dealing with the world he would know that he‟d been in a fight.Fifteen is of all ages the most difficult to ___60___———to put one‟s fingers on and say, “That‟s the way I was.” And all one can know is that somewhere between thirteen, boyhood‟s___61___, and seventeen, when one is a sort of counterfeit young man, there is a time when youth ___62___ hourly between one world and another —— pushed ceaselessly forward into unprecedented experiences and ___63___ trying to struggle back to the days when nothing had to be ___64___ for. Fortunately none of our contemporaries remember much more than we do of how we behaved in those days; nevertheless the ___65___ is about to be drawn aside for an inspection of Basil‟s madness that summer.51. A. boarded B. missed C. jumped D.followed52. A. happy B. unhappy C.memorableD.favourable53. A. swelling B. bending C. lengthening D.strengthening54. A. aware B.fond C. critical D.unconscious55. A. randomly B. desperately C. particularly D.indifferently56. A. evolution B. revolution C. solutionD.introduction57. A. fact B. opinion C. course D. effort58. A. fashion B. ambition C. character D.treasure59. A. contacted B. associated C. interfered D.smashed60. A. digest B.describe C. deal D. locate61. A. majority B. minority C. senior D. junior62. A. floats B. varies C. fluctuates D. ranges63. A. successfully B. vainly C. wildly D. gently64. A. hunted B. provided C. compensated D. paid65. A.curtain B. adolescence C. portrait D. ceilingSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)There are people in Italy who can‟t stand soccer. Not all Canadians love hockey. A similar situation exists in America, where there are those individuals you may be one of them who yawn or even frown when somebody mentions baseball. Baseball to them means boring hours watching grown men in funny tight outfits standing around in a field staring away while very little of an ything happens. They tell you it‟s a game better suited to the 19th century, slow, quiet, gentlemanly. These are the same people you may be one of them who love football because there‟s the sport that glorifies “the hit”.By contrast, baseball seems abstract, cool, silent, still.On TV the game is fractured into a dozen perspectives, replays, close ups. The geometry of the game, however, is essential to understanding it. You will contemplate the game from one point as a painter does his subject; you may, of course, project yourself into the game. It is in this projection that the game affords so much space and time for involvement. The TV won‟t do it for you.Take, for example, the third baseman. You sit behind the third base dugout and you watch him watching home plate. His legs are apart, knees flexed. His arms hang loose. He does a lot of this. The skeptic still cannot think of any other sports so still, so passive. But watch what happens every time the pitcher throws: the third baseman goes up on his toes, flexes his arms or brings the glove to a point in front of him, takes a step right or left, backward or forward, perhaps he glances across the field to check his first baseman‟s position. Suppose the pitch is a ball. “Nothing happened,” you say. “I could have had my eyes closed.”The skeptic and the innocent must play the game. And this involvement in the stands is no more intellectual than listening to music is. Watch the third baseman. Smooth the dirt in front of you with one foot; smooth the pocket in your glove; watch the eyes of the batter, the speed of the bat, the sound of horsehide on wood. If football is a symphony of movement and theatre, baseball is chamber music, a spacious interlocking of notes, chores and responses.66. Those who don‟t like baseball may complain that ______.A. it is only to the taste of the oldB. it involves fewer players than footballC. it is not exciting enoughD. it is pretentious and looks funny67. The author admits that ______.A. baseball is too peaceful for the youngB. baseball may seem boring when watched on TVC. football is more attracting than baseballD. baseball is more interesting than football68. By stating …I could have had my eyes closed.‟ the author means (4th paragraph last sentence):A. The third baseman would rather sleep than play the game.B. Even if the third baseman closed his eyes a moment ago, it could make no difference to theresult.C. The third baseman is so good at baseball that he could finish the game with eyes closed allthe time and do his work well.D. The consequence was too bad he could not bear to see it.69. We can safely conclude that the author ______.A. likes footballB. hates footballC. hates baseballD. likes baseball(B)Some of the world‟s most signifi cant problems never hit headlines. One example comes from agriculture. Food riots and hunger make news. But the trend lying behind these matters is rarely talked about. This is the decline in the growth in yields of some of the world‟s major crops. A new study by the University of Minnesota and McGill University in Montreal looks at where, and how far, this decline is occurring.The authors take a vast number of data points for the four most important crops: rice, wheat corn and soybeans. They find that on between 24% and 39% of all harvested areas, the improvement in yields that took place before the 1980s slowed down in the 1990s and 2000s.There are two worrying features of the slowdown. One is that it has been particularly sharp in the world‟s most p opulous countries, India and China. Their ability to feed themselves has been an important source of relative stability both within the countries and on world food markets. That self-sufficiency cannot be taken for granted if yields continue to slow down or reverse.Second, yield growth has been lower in wheat and rice than in corn and soya beans. This is problematic because wheat and rice are more important as foods, accounting for around half of all calories consumed. Corn and soybeans are more important as feed grains. The authors note that “we have preferentially focused our crop improvement efforts on feeding animals and cars rather than on crops that feed people and are the basis of food security in much of the world.”The report qualifies the more optimistic findings of another new paper which suggests that the world will not have to dig up a lot more land for farming in order to feed 9 billion people in 2050, as the Food and Agriculture Organization has argued.Instead, it says, thanks to slowing population growth, land currently ploughed up for crops might be able to revert to forest or wilderness. This could happen. The trouble is that the forecast assumes continued improvements in yields, which may not actually happen.70. What does the author try to draw attention to?A. Food riots and hunger in the world.B. News headlines in the leading media.C. The decline of the grain yield growth.D. The food supply in populous countries.71. Why does the author mention India and China in particular?A. Their self-sufficiency is vital to the stability of world food markets.B. Their food yields have begun to decrease sharply in recent years.C. Their big populations are causing worldwide concerns.D. Their food self-sufficiency has been taken for granted.72. What does the new study by the two universities say about recent crop improvement efforts?A. They fail to produce the same remarkable results as before the 1980s.B. They contribute a lot to the improvement of human food production.C. They play a major role in guaranteeing the food security of the world.D. They focus more on the increase of animal feed than human food grains.73. What does the Food and Agriculture Organization say about world food production in the coming decades?A. The growing population will greatly increase the pressure on world food supplies.B. The optimistic prediction about food production should be viewed with caution.C. The slowdown of the growth in yields of major food crops will be reversed.D. The world will be able to feed its population without increasing farmland.(C)Among the more colorful characters of Leadville‟s golden age were H.A.W. Tabor and his second wife, Elizabeth McCourt, better known as “Baby Doe”. Their history is fast becoming one of the legends of the Old West. Horace Austin Warner Tabor was a school teacher in Vermont. With his first wife and two children he left Vermont by covered wagon in 1855 to homestead in Kansas. Perhaps he did not find farming to his liking, or perhaps he was tempted by rumors of fortunes to be made in Colorado mines. At any rate, a few years later he moved west to the small Colorado mining camp known as California Gulch, which he later renamed Leadville when he became its leading citizen. “Great deposits of lead are sure to be found here.” he said.As it turned out, it was silver, not lead, that was to make Leadville‟s fortune and wealth. Tabor knew little about mining himself, so he opened a general store, which sold everything from boots to salt, flour, and tobacco. It was his custom to “grubstake” prospective miners, in other words, to supply them with food and supplies, or “grub”, while they looked for ore, in return for which he would get a share in the mine if one was discovered. He did this for a number of years, but no one that he aided ever found anything of value.Finally one day in the year 1878, so the story goes, two miners came in and asked for “grub”. Tabor had decided to quit supplying it because he had lost too much money that way. These werepersistent, howe ver, and Tabor was too busy to argue with them. “Oh help yourself. One more time won‟t make any difference,” He said and went on selling shoes and hats to other customers.The two miners took $17 worth of supplies, in return for which they gave Tabor a one-third interest in their findings. They picked a barren place on the mountain side and began to dig. After nine days they struck a rich vein of silver. Tabor bought the shares of the other two men, and sothe mine belonged to him alone. This mine, known as the “Pittsburgh Mine,” made 1 300 000 for Tabor in return for his $17 investment.Later Tabor bought the Matchless Mine on another barren hillside just outside the town for $117 000. This turned out to be even more fabulous than the Pittsburgh, yielding $35 000 worth of silver per day at one time. Leadville grew. Tabor became its first mayor, and later became lieutenant governor of the state.74. Leadville got its name for the following reasons EXCEPT ______.A. because Tabor became its leading citizenB. because great deposits of lead is expected to be found thereC. because it could bring good fortune to TaborD. because Tabor renamed it so75. The word “grubstake” in paragraph 2 means ______.A. to supply miners with food and suppliesB. to open a general storeC. to do one‟s contribution to the development of the mineD. to supply miners with food and supplies and in return get a share in the mine, if one wasdiscovered76. Tabor made his first fortune ______.A. by supplying two prospective miners and getting in return a one-third interest in the findingsB. because he was persuaded by the two miners to quit supplyingC. by buying the shares of the otherD. as a land speculator77. The underlying reason for Tabor‟s life career is ______.A. purely accidentalB.based on the analysis of miner‟s being very poor and their possibility of discoveringprofitable mining siteC. through the help from his second wifeD. he planned well and accomplished targets step by stepSection CDirections: Read the passage carefully. Then answer the questions or complete the statements inthe fewest possible words.When the Internet powerhouse Yahoo wanted to teach ethics to its employees, it faced a challenge familiar to multinational companies.Yahoo employs nearly 14,000 people at 25 sites worldwide. They would feel bored at sitting down in front of a dated video in which actors with 1980s haircuts tell them what to do. So it hired a company called The Network to design a game. In the game, the truck where Yahoo wasfounded traveled the world, turning into a boat and a helicopter along the way as it visited some of Yahoo's foreign offices. Participants play in game show-like scenarios that quiz them about conflicts of interest and doing business fairly. And employees note: Yahoo is tracking how well they do.Such activities draw more enthusiastic participation and teach more effectively than traditional methods. They are described as alternative-reality games (ARGs), involving both interactive and real-world elements. Besides teaching employees, ARGs have also been used in many areas for a number of different purposes.From a marketing perspective, a number of very successful ARGs have been written as a way to build product awareness. A very popular ARG called I Love Bees was produced to market the 2004 video game Halo 2. At its height, I Love Bees received between two to three million unique visitors over the course of three months.ARGs are more than just a fun way to learn. They have also been used to solve real world problems. An ARG called World Without Oil was created to obtain collective input from players about dealing with the world's dependency on oil. World without Oil simulates the first 32 days of a global oil crisis and anybody could play by creating a personal story that recorded the imagined reality of their life in the crisis. World Without Oil's success on a small budget has opened the door for similar games to engage mainstream Internet users with climate change, education reform, governmental policy and other timely, vital issues.(Note: Answer the questions or complete the statements in NO MORE THAN TEN WORDS.)78.What challenge did yahoo face in teaching ethics to its employees?79.In the game designed for yahoo, participants had to answer questions about ________.80.What are the three major functions of ARG mentioned in the passage?81.The success of World Without Oil suggests that ARGs can ________.第II卷（共47分）I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1.没有人不希望和平。