复旦附中2018学年第一学期高一上期中考卷

上海市杨浦区复旦大学附属中学2018-2019学年高一上学期期中考试数学试题一．填空题（共12小题，满分54分）1.若实数a满足：a2∈{1，4，a}，则实数a的取值集合为_____．【答案】{﹣1，﹣2，2，0}【解析】∵实数满足：∈{1，4，}，∴=1或=4，或=a，解得=﹣2或=2或=﹣1或=1或=0，当=1时，集合为{1，4，1}，不合题意；当=﹣1，或=±2，或=0时，满足题意．∴实数的取值集合为{﹣1，﹣2，2，0}．故答案为：{﹣1，﹣2，2，0}．2.函数的定义域为_____．【答案】[﹣2，3）【解析】由题意得，解得．∴函数的定义域为：[﹣2，3）．故答案为：[﹣2，3）．3.命题“若ab＝0，则b＝0”的逆否命题是______．【答案】“若b≠0，则ab≠0”【解析】因为一个命题的逆否命题，是将原命题逆命题的条件与结论同时否定得到，所以命题“若ab＝0，则b＝0”的逆否命题是“若b≠0，则ab≠0”.故答案为：“若b≠0，则ab≠0”.4.函数y=+2的单调区间是_____．【答案】（﹣∞，0）和（0，+∞）【解析】由题意得函数的定义域为，又函数在和上单调递减，所以函数的单调减区间是和．故答案为：（∞，0）和（0，+∞）．5.已知为定义在上的奇函数，当时，，则当时，__________．【答案】【解析】设，则，由已知当时，，当时，可得，．6.已知符号函数sgn（x），则函数f（x）=sgn（x）﹣2x的所有零点构成的集合为_____．【答案】【解析】①当x＞0时，函数f（x）=sgn（x）﹣2x =1﹣2x，令1﹣2x=0，得x=，即当x＞0时，函数f（x）的零点是；②当x=0时，函数f（x）=0，故函数f（x）的零点是0；③当x＜0时，函数f（x）=﹣1﹣2x，令﹣1﹣2x=0，得x=，即当x＜0时，函数f（x）的零点是．综上可得函数f（x）=sgn（x）﹣x的零点的集合为：．7.函数的值域为_______.【答案】【解析】由指数函数的性质可知：，据此可知：，函数的值域为.8.已知a＞0，b＞0，则的最小值为_____．【答案】4【解析】由题意得，∵，∴，∴，当且仅当，即时等号成立．∴的最小值为4．故答案为：4．9.设集合A={1，2，6}，B={2，4}，C={x∈R|﹣1≤x≤5}，则（A∪B）∩C=_____【答案】{1，2，4}【解析】∵A={1，2，6}，B={2，4}，∴A∪B={1，2，4，6}，又C={x|﹣1≤x≤5，x∈R}，∴（A∪B）∩C={1，2，4}．故答案为：{1，2，4}．10.若y=f（x）是定义在（﹣∞，+∞）上的单调减函数，且f（x）＜f（2x﹣2），则x的取值范围_____．【答案】（﹣∞，2）【解析】∵f（x）＜f（2x﹣2），且y=f（x）是定义在（﹣∞，+∞）上的单调减函数，∴x＞2x﹣2，解得x＜2．∴x的取值范围为（﹣∞，2）．故答案为：（﹣∞，2）．11.若函数，则_____．【答案】1【解析】由题意得．故答案为：1．12.定义：若平面点集A中的任一个点（x0，y0），总存在正实数r，使得集合，则称A为一个开集．给出下列集合：①{（x，y）|x2+y2=1}；②{（x，y）|x+y+2＞0}；③{（x，y）||x+y|≤6}；④．其中不是开集的是_____．（请写出所有符合条件的序号）【答案】①③【解析】对于①，集合A={（x，y）|x2+y2=1}表示以原点为圆心，1为半径的圆，则在该圆上任意取点（x0，y0），以任意正实数r为半径的圆面，均不满足，故①不是开集．对于②，集合A={（x，y）|x+y+2＞0}，对于A中的任一点（x0，y0），设该点到直线x+y+2=0的距离为d，取r=d，则满足，故②是开集．对于③，集合A={（x，y）||x+y|≤6}，在曲线|x+y|=6任意取点（x0，y0），以任意正实数r为半径的圆面，均不满足，故该集合不是开集．对于④，集合A=表示以点为圆心，以1为半径除去圆心和圆周的圆面，在该平面点集A中的任一点（x0，y0），则该点到圆周上的点的最短距离为d，取r=d，则满足，故该集合是开集．综上可得①③中的集合不是开集．故答案为：①③．二．选择题（共4小题，满分20分，每小题5分）13.设x∈R，则“|x﹣2|＜1”是“x2﹣x﹣6＜0”的（）A. 充分而不必要条件B. 必要而不充分条件C. 充分必要条件D. 既不充分也不必要条件【答案】A【解析】由|x﹣2|＜1得﹣1＜x﹣2＜1，解得1＜x＜3，由x2﹣x﹣6＜0，得﹣2＜x＜3．因为，，所以“1＜x＜3”是“﹣2＜x＜3”的充分不必要条件，即“|x﹣2|＜1”是“x2﹣x﹣6＜0”的充分不必要条件．故选A．14.已知函数f（x）=3x+x，g（x）=log3x+x，h（x）=sin x+x的零点依次为x1，x2，x3，则以下排列正确的是（）A. x1＜x2＜x3B. x1＜x3＜x2C. x3＜x1＜x2D. x2＜x3＜x1【答案】B【解析】函数f（x）=3x+x，g（x）=log3x+x，h（x）=sin x+x的零点依次为x1，x2，x3，在坐标系中画出y=3x，y=log3x，y=sin x与y=﹣x的图象，如下图所示：由图形可知x1＜0，x2＞0，x3=0，所以x1＜x3＜x2．故选B．15.已知非空集合M满足：若x∈M，则∈M，则当4∈M时，集合M的所有元素之积等于（）A. 0B. 1C. －1D. 不确定【答案】C【解析】依题意，得当4∈M时，有，从而，，于是集合M的元素只有4，，所有元素之积等于4×（）×=-1．16.已知函数f（x）是定义在R上的奇函数，对任意的x∈R，均有f（x+2）=f（x），当x∈[0，1）时，f（x）=2x﹣1，则下列结论正确的是（）A. f（x）的图象关于x=1对称B. f（x）的最大值与最小值之和为2C. 方程f（x）﹣lg|x|=0有10个实数根D. 当x∈[2，3]时，f（x）=2x+2﹣1【答案】C【解析】由函数f（x）是定义在R上的奇函数，可得．又当x∈[0，1）时，f（x）=2x﹣1，所以，当x∈[﹣1，0）时，﹣x∈[0，1），则f（﹣x）=2﹣x﹣1=﹣f（x），∴．又f（x+2）=f（x），∴函数f（x）是周期为2的周期函数．画出函数y=f（x）与y=lg|x|的图象，如图所示，对于A，结合图象可得函数f（x）的图象无对称轴，所以A不正确．对于B，由图象可得，函数f（x）没有最大值和最小值，所以B不正确．对于C，结合图象可得当x＞0时，函数y=f（x）与y=lg|x|的图象有4个交点，当x＜0时，函数y=f（x）与y=lg|x|的图象有6个交点，故方程f（x）﹣lg|x|=0有10个实数根．所以C 正确．对于D，当x∈[2，3)时，x﹣2∈[0，1)，所以．故D不正确．故选C．三．解答题（共5小题，满分76分）17.设p：实数x满足x2－4ax＋3a2＜0，其中a＞0；q：实数x满足x2－x－6≤0.(1)若a＝1，p且q为真，求实数x的取值范围；(2)若¬q是¬p的充分不必要条件，求实数a的取值范围．解：(1)由x2－4ax＋3a2＜0得(x－3a)(x－a)＜0，又a＞0，所以a＜x＜3a，当a＝1时，1＜x＜3，即p为真时，实数x的范围是1＜x＜3；由q为真时，实数x的范围是－2≤x≤3，若p且q为真，则p真且q真，所以实数x的取值范围是(1,3)．(2)¬p：x≤a或x≥3a，¬q：x＜－2或x＞3，由¬q是¬p的充分不必要条件，有得0＜a≤1，显然此时¬p¬q，即a的取值范围为(0,1]．18.已知函数y=f（x）为定义在（﹣∞，0）∪（0，+∞）上的奇函数，且当x＞0时，（1）试求f（﹣2）的值；（2）指出f（x）的单调递增区间（直接写出结论即可）；（3）求出f（x）的零点．解：（1）∵函数为奇函数，∴．（2）当x＞0时，函数在（3，+∞）上单调递增，又函数y=f（x）为定义在（﹣∞，0）∪（0，+∞）上的奇函数，∴函数y=f（x）在（﹣∞，﹣3）上也单调递增，∴函数的单调递增区间为（﹣∞，﹣3）和（3，+∞）.（3）当时，由，得，解得，∴是函数的零点．又函数为奇函数，∴也为函数的零点．综上可得函数的零点为和．19.已知函数.（1）求不等式的解集；（2）若对恒成立，求的取值范围.解：（1）因为，，所以当时，由得；当时，由得；当时，由得.综上，的解集为.（2）法一：由得，因为，当且仅当取等号，所以当时，取得最小值.所以当时，取得最小值，故，即的取值范围为.法二：设，则，当时，取得最小值，所以当时，取得最小值，故时，即的取值范围为.20.函数f(x)的定义域为D＝{x|x≠0}，且满足对任意x1，x2∈D，有f(x1·x2)＝f(x1)＋f(x2)．(1)求f(1)的值；(2)判断f(x)的奇偶性并证明你的结论；(3)如果f(4)＝1，f(x－1)<2，且f(x)在(0，＋∞)上是增函数，求x的取值范围．解：(1)∵对于任意x1，x2∈D，有f(x1·x2)＝f(x1)＋f(x2)，∴令x1＝x2＝1，得f(1)＝2f(1)，∴f(1)＝0.(2)令x1＝x2＝－1，有f(1)＝f(－1)＋f(－1)，∴f(－1)＝f(1)＝0.令x1＝－1，x2＝x有f(－x)＝f(－1)＋f(x)，∴f(－x)＝f(x)，∴f(x)为偶函数．(3)依题设有f(4×4)＝f(4)＋f(4)＝2，由(2)知，f(x)是偶函数，∴f(x－1)<2⇔f(|x－1|)<f(16)．又f(x)在(0，＋∞)上是增函数．∴0<|x－1|<16，解之得－15<x<17且x≠1.∴x的取值范围是{x|－15<x<17且x≠1}．21.已知函数，．(1)若函数是奇函数，求实数的值；(2)在(1)的条件下，判断函数与函数的图象公共点个数，并说明理由；(3)当时，函数的图象始终在函数的图象上方，求实数的取值范围．解：（1）因为为奇函数，所以对于定义域内任意，都有，即，，显然，由于奇函数定义域关于原点对称，所以必有.上面等式左右两边同时乘以得，化简得，.上式对定义域内任意恒成立，所以必有，解得.（2）由（1）知，所以，即，由得或，所以函数定义域.由题意，要求方程解的个数，即求方程在定义域上的解的个数. 令，显然在区间和均单调递增，又，且，.所以函数在区间和上各有一个零点，即方程在定义域上有2个解，所以函数与函数的图象有2个公共点.（附注：函数与在定义域上的大致图象如图所示）（3）要使时，函数的图象始终在函数的图象的上方，必须使在上恒成立，令，则，上式整理得在恒成立.方法一：令，.①当，即时，在上单调递增，所以，恒成立；②当，即时，在上单调递减，只需，解得与矛盾.③当，即时，在上单调递减，在上单调递增，所以由，解得，又，所以综合①②③得的取值范围是.方法二：因为在恒成立. 即，又，所以得在恒成立令，则，且，所以，由基本不等式可知（当且仅当时，等号成立.）即，所以,所以的取值范围是.。

2017学年复旦附中高一上期中一. 填空题1.已知全集，，，则__________【答案】【解析】【分析】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．2.命题“如果，那么且”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，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．3.已知集合，，则__________【答案】【解析】【分析】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．4.已知“”是“”的充分不必要条件，则实数的取值范围是__________【答案】【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，∵“”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．5.设，则满足的集合的个数为__________【答案】8【解析】【分析】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．6.函数的定义域为，则的值为__________【答案】2【解析】【分析】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．7.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】【分析】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．8.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】【分析】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．9. 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.10.若，则的最小值为__________【答案】2【解析】【分析】将原式变形后根据基本不等式求解．【详解】∵，∴．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．11.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】【分析】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．【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】【分析】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．二. 选择题13.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】【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】（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．14.解析式为，值域为的函数有（）A. 4B. 6C. 8D. 9【答案】D【解析】【分析】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．15.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. 1B. 2C. 3D. 4【答案】C【解析】【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>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．16.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．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】【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】【分析】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】【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】【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）∵，，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】【分析】（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．。

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．。

2017-2018 上海市复旦附中高一上学期期中考试Grammar and Vocabulary: (18%)Section A (8%)Directions: Beneath each of the following sentences there are four choices marked A, B, C and D. Choose the one answer that best completes the sentence.21. The discovery of gold in Australia led thousands to believe that a fortune_________.A, had been making B. was making C. was to be made D. would make 22. The president hopes that people will be better off when he quits than when he _________.A. has startedB. startsC. will startD. started23. Travellers to that area can carry disease to their own countries that have never experienced _________.A. themB. itC. themselvesD. itself24. No one can tell the exact number of U.S. pilots and fighter planes ________were lost in the Pearl Harbor air attack.A. whoB. whichC. thatD. what25. ____________, the climbers, who had already conquered many high mountains before, determined to reach the top of the mountain.A. However it was highB. How high it wasC. However high it wasD. However high was it26. They will not allow others to decide the future of their country, _________.A. which is to be knownB. as is to be knownC. as is known to allD. as what is known to all27. We have come to realize that the brain must “ forget” some pieces of information ________it can remember others.A. whenB. sinceC. so thatD. if28. They are our school’s volunteers _______to help elderly people cross the streets every day.A. whose task it isB. it is whose taskC. to whom is the taskD. whose task wasSection B (10%)Directions: Complete the following passage by using the words in the box. Each word can be only used once. Note that there is one word more than youneed.A. journeyingB. foundingC. educationalD. marvelous AB.viewAC. undoubtedly AD. thrill BC. existed BD. sights CD. surprisingly ABC. soilSept 27 is World Tourism Day. Of course, travel isn’t a new discovery. Imagine how Italian traveler Marco Polo must have felt when he found himself on Chinese ____29_____, seeing a way of life quite different from anything he’d seen before.And how ___30______ must it have been to listen to Zhang Qian when he returned to China from his journey through Central Asia and West Asia? His brain must have been packed with everything he’d seen and heard, leading to the ____31_____ of the Silk Road.Travel is one of the most exciting experiences a human being can have. Nowadays, more people are traveling than ever before. By train, plane and car, people all around the globe are ____32_____ to places that people didn’t even know ____33_____ a few centuries ago, or only knew from books.Some people have traveled all over the world, and travel is a way of life to them. They perhaps know what to expect before they travel. That’s why the best travel is when it’s for the first time. Imagine a person who has always wanted to travel to the United States. Of course, they’ve probably seen the Statue of Liberty a thousand times on the TV, and the White House, and all the other famous _______34__. But none of that would compare to the ___35______ of looking out of the cabin window as the plane lands, watching the cities and streets of the real America come into_____36____.Although travel is often just for fun, it’s also ____37_____. We may not know that we are getting an education, but we still are.We’re learning every day: new words in a new language, new people, and new ways of life. But this learning takes place in the school of the world, not the classroom. One of the lessons we learn is ____38_____ a moral one. As we get to know foreign places, we come to understand that there are many different ways to live, and that the way we live isn’t necessarily the best way. The British politician Benjamin Disraeli summed this up well when he said,“Travel teaches toleration.”Ⅲ.Reading Comprehension (40%)Section ADirections : For each blank in the following passages there are four words or phrases marked A,B,C and D. Fill in each blank with the word or phrases that best fits the context.( A )Leif Erickson reached North America around the year 1000, but the attempt to explore was started slowly. It would be five centuries __(39)__ other Europeans landed on that continent.Why were Europeans the ones to __(40)__ to the American? The Chinese and Arabs had the _(41)__ and technology to sail across the seas. __(42)__ of them tool regular voyages in the Indian Ocean and the Asian Pacific for trade. But exploration? By the mid-15th century China had followed the closed-door policy to __(43)__ itself from the rest of the world. The Arabs, with access to the minerals and spices (香料) of Africa and the Far East, saw no __(44)__ to journey into the unknown.Europe, ____(45)________ needed gold and silver; its mines could not meet the demand for coinage. Ottoman Turks blocked the routes across the land to Asia. Only the sea held the ___(46)_____of new wealth.With the return of Magellan’s ships in 1522 from its voyage around the world, the belief was __(47)__ that the oceans were interconnected, promising the age of discovery. The English, as well as the Spanish, Portuguese and French, __(48)__ themselves to finding the “river of the west” through North America to the east.39. A. after B. since C .before D .when40. A .push B .pull C .draw D .drive41. A .sources B .resources C .substances D .matters42. A .Neither B .Both C .Any D .None43. A .prevent B .protect C .isolate D .differ44. A .access B .admission C .application D .association45. A .as a matter of fact B . in other wordsC. for one thing D .on the other hand46. A .symbol B .impression C .promise D .reflection47. A .extended B .estimated C .attracted D .accepted48. A .contributed B .devoted C .referred D .connected( B )Those who keep their word become the most important members of an organization. People come to rely on and trust them. They can be____(49)____.___(50)______you make a promise , be sure to keep it. When you keep your promise, no matter how much ___(51)_______it takes, you will be rewarded.Whenever you say no, stand upon that as well. In a way, a no is also a promise. A good, _____(52)____no can be very important in building trust. Agreements are also important. Whenever you enter an agreement, live by it, ______(53)____you are not too happy with the deal after making the deal, you still live by it. In the long run, your integrity will _____(54)____.Victoria was the manager of a supermarket. She set out to do that____(55)_____. She was very careful, however, of not _____(56)_____. When she did, she moved mountains to make sure that she came through. After a while, the stone employee came to trust and respect her like no______(57)_____they had ever had. The teamwork became magical. People followed her example by living up to their word. She ____(58)______what she said, and people appreciated that.49. A. counted B. counted on C. depended D. numbered50.A. As far as B. Though C. Whenever D. However51.A. pain B. ache C. effort D. money52. A. loud B. clean C. sharp D. loose53. A. As B. As if C. Just as D. Even if54. A. set off B. show off C. pay off D. take off55. A. thoroughly B. entirely C. wholly D. totally56. A. lying B. overpromising C. underestimating D. overlooking57. A. manager B. employee C. supermarket D. others58. A. believed B. trusted C. meant D. promisedSection BDirections :Read the following passages. Each passage is followed byquestions 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 )A few years ago, Paul Gerner began to gather a group of architects in Las Vegas to ask them what it would take to design a public school that used 50 percent less energy, cost much less to build and obviously improved student learning. “I think half of them fell off their chairs,” Gerner says.Gerner manages school facilities for Clark County, Nevada, a district roughly the size of Massachusetts. By 2018, 143,000 additional students will enter the already crowded public-education system. Gerner needs 73 new schools to house them. Four architecture teams have nearly finished designing primary school prototypes （原型）; they plan to construct their schools starting in 2009. The district will then assess how well the schools perform, and three winners will copy those designs in 50 to 70 new buildings.Green schools are appearing all over, but in Clark County, which stands out for its vastness(广阔), such aggressive targets are difficult because design requirements like more natural light for students go against the realities of a desert climate. “One of the biggest challenges is getting the right site orientation,” Mark McGinty, a director at SH Architecture, says. His firm recently completed a high school in Las Vegas. ”You have the same building, same set of windows, but if its orientation is incorrect and it faces the sun, it will be really expensive to cool.”Surprisingly, the man responsible for one of the most progressive green-design competitions has doubts about ideas of eco-friendly buildings. “I don’t believe in the new green religion,” Gerner says. “Some of the building technologies that you get are impractical. I’m interested in those that work.”But he wouldn’t mind if some green features inspire students. He says he hopes to set up green energy systems that allow them to learn about the process of harvesting wind and solar power. “You never know what’s going to start the interest of a child to study math and science,” he says.59. How did the architects react to Gerner’s design requirements?A .They lost balance in excitement.B .They showed strong disbelief.C .They expressed little interest.D .They burst into cheers.60. Which order of steps is followed in carrying out the project?A .Assessment — Prototype — Design — Construction.B .Assessment — Design — Prototype — Construction.C .Design — Assessment — Prototype — Construction.D .Design — Prototype — Assessment — Construction.61 What makes it difficult to build green schools in Clark County?A .The large size.B .Limited facilities.C .The desert climate.D .Poor natural resources.62. What does Gerner think of the ideas of green schools?A .They are questionable.B .They are out of date.C .They are advanced.D .They are practical.(B)Spring is in the air, as is romance. Perhaps you're even thinking of taking the plunge and making a honey trip with your significant other?Before you do, why not take a look at a list of the top three places in the world to propose(求婚)? With suggestions for traditionalists and private types, you just might find the perfect spot to pop the question.New YorkIf you pay attention to romantic comedies, New York is the place for romance. Ever since the Dutch first entered this harbor in 1624 it has become a historic landmark and a must visit dreamland. Every stretch of this island is packed full of fun that will keep everyone busy. Take a leaf out of sleepless in Seattle and head to the top of the Empire State Building at night, or how about run to your loved one on the streets of NYC this New Years even like in when harry Met Sally?ParisSweeping views? Check. Grand old stately home? Check. Luxurious spots to propose? check. pack your passports and say goodbye to your daily routine life-it's less than three hours to Paris from London by Eurostar. Famous for its breathtaking architecture, and many cultural attractions, France's capital is a must-see destination. While we're not suggesting you climb the Eiffel Tower for the big moment. There's a reason why this is the city of love, if you keep your proposal original. So gentleman, take your lady to the Louvre, turn your back on the Mona Lisa and declare she is more beautiful than thefamous work of art.Las VegasDo you want to bundle the proposal and wedding all-in-one? plead to. Las Vegas, get down on one knee, show your love and get married in less time than it takes to order 'honeymoon suit’ and if it doesn't work out. There's still plenty of entertain. Once the playground of the rich and famous, from Elvis Presley to Marilyn Monroe, Las Vegas now attracts millions of visitors by its bright lights and thrills each year. Whether you're to try your luck at the casinos (赌场), or watch a show, there's a wonder to explore in Las Vegas. 63. Which of the following city on the top-3 list would be the ideal place for film fans to propose?A. New YorkB. Paris.C. Las Vegas.D. None.64. Which place is recommended for a proposal in the passage?A. On the top of the Eiffel TowerB. In the Eurostar train from LondonC. Any luxurious spot in PairsD. In front of the Mona Lisa in the Louvre64. As a perfect place for propose, Las Vegas will provide you with all the following EXCEPT.A. a honeymoon suiteB. a package weddingC. a bundle of flowersD. various ways to entertain( C )Do you look happy? angry? Have you ever wondered how you know what another person’s mood is just by looking at his or her face? Studies have shown that one instantaneously (瞬间地) and subconsciously makes a determination of another person’s mood just by glancing at the position and appearance of both the eyebrows and mouth. But what if only the position of the eyebrows was varied , and not the mouth? Would you still be able to determine his or her mood or personality just by seeing the eyebrows?Observe the four faces shown below. Notice how each face has exactly the same shape. the same smile, and the same eyes. The only difference between each of the faces is the position of the eyebrows.Low, flat eyebrows that hang over the eyes indicate tiredness, while an eyebrow that is highest in the middle denotes sadness. Downward slanting eyebrows show anger, while highly arched eyebrows show happiness.Frequently, malposition of the eyebrows, such as eyebrow sagging(下垂) and upper eyelid fullness, can be overlooked when considering facial aging and expression. Furthermore, the heavy eyebrow skin pushed the eyelid down causing a tired appearance.To avoid eyebrow sagging, many people raise their eyebrows to remove the brow and eyelid skin from their visual space. These people eventually develop wrinkles in the forehead due to the constant movement of the muscles that raise the eyebrow. In addition, there are those who are sensitive to light and frequently’ squint(斜视) and squeeze’ their eyes. This effectively pulls the eyebrows down and leads to wrinkles between the eyebrows and at the corners of the eyes.Botox is an effective measure for fine lines of the forehead and lines between the eyebrows. Botox works by freezing the muscle movement for several months. For the purpose of creating a more pleasant appearance, the best corrective measure is brow-lift surgery.66. What can this passage be?A. An advertisementB. A magazine articleC. A business reportD. A scientific report67. A person’s eyebrows that hang over the eyes low and flat can cause a(n) _________appearance.A. sadB. tiredC. happyD. angry68. What is the purpose of the passage?A. To explain the fact that varied eye position can tell a person’s mood.B. To arouse people’s interest in the best way of removing wrinkles in the forehead and between the eyebrows.C. To explain what leads to wrinkles in the forehead and near the eyebrow.D. To enable people to know how to avoid developing fine lines on their faces.69. If the passage continues, what would the writer most likely discuss in the next paragraph?A. Facial agingB. Eye operationC. Brow life surgeryD. Face-lifts.( D )The past ages of man have all been carefully labeled by anthropologists. Descriptions like ‘Palaeolithic Man’, ‘Neolithic Man’, etc., neatly sum up whole periods. When the time comes for anthropologists to turn their attention to the twentieth century, they will surely choose the label ‘Legless Man’. Histories of the time will go something like this: ‘in the twentieth century, people forgot how to use their legs. Men and women moved about in cars, buses and trains from a very early age. There were lifts and escalators in all large buildings to prevent people from walking. This situation was forced upon earth dwellers of that time because of miles each day. But the surprising thing is that they didn’t use their legs even when they went on holiday. They built cable railways, ski-lifts and roads to the top of every huge mountain. All the beauty spots on earth were ruined by the presence of large car parks.’The future history books might also record that we were deprived of the use of our eyes. In our hurry to get from one place to another, we failed to see anything on the way. Air travel gives you a bird’s-eye view of the world—or even less if the wing of the aircraft happens to get in your way. When you travel by car or train a blurred (="not" clear) image of the countryside constantly smears the windows. Car drivers, in particular, are forever obsessed with the urge to go on and on: they never want to stop. Is it the lure of the great motorways, or what? And as for sea travel, it hardly deserves mention. It is perfectly summed up in the words of the old song: ‘I joined the navy to see the world, and what did I see? I saw the sea.’ The typical twentieth-century traveler is the man who always says ‘I’ve been there. ’You mention the remotest, most evocative (引起记忆的) place-names in the world like El Dorado, Kabul, Irkutsk and someone is bound to say ‘I’ve been there’—meaning, ‘I drove through it at 100 miles an hour on the way to somewhere else.’When you travel at high speeds, the present means nothing: you live mainly in the future because you spend most of your time looking forward to arriving at some other place. But actual arrival, when it is achieved, is meaningless.You want to move on again. By traveling like this, y ou suspend all experience; the present ceases to be a reality: you might just as well be dead. The traveler on foot, on the other hand, lives constantly in the present. For him traveling and arriving are one and the same thing: he arrives somewhere with every step he makes. He experiences the present moment with his eyes, his ears and the whole of his body. At the end of his journey he feels a delicious physical weariness. He knows that sound. Satisfying sleep will be his: the just reward of all true travellers.70. Anthropologists label nowadays’ men ‘Legless’ because _________.A. people forget how to use his legs.B. people prefer cars, buses and trains.C. lifts and escalators prevent people from walking.D. there are a lot of transportation devices.71.Why does the author say ‘we are deprived of the use of our eyes’?A. People won’t use their eyes.B. In traveling at high speeds, eyes become useless.C. People can’t see anything on his way of travel.D. People want to sleep during travelling.72. Travelling at high speed means _________.A. people’s focus on the futureB. a pleasureC. satisfying drivers’ great thrillD. a necessity of life73.What's the best title of the passage?A. More haste, less speedB. Modern means of transportation make the world a small placeC. Eyes open and mind broadenD. The only way to travel is on foot.( 以下各题请务必做在答卷纸上)Ⅰ. Grammar (8%)\One of the main points of traveling is to relax and take a break from your normal daily life. (1)__________the truth is, we’re not always free to do what we like when travelling to a foreign country, and a US tourist learned that the hard way.On Aug 12, the unnamed 41-year-old man was beaten by a passerby after he was seen giving Nazi salutes(纳粹礼) again and again on a street inDresden, Germany.Ever since the end of World WarⅡ, Germany has strict laws (2)_________(forbid) the Nazi salute, as well as other symbols of Nazism.In fact, most countries have their own taboos. If you plan to travel overseas, it’s best to get familiar with these taboos(3)______________you start touring local sites. Below, TEENS gives some examples.SingaporeYou can get (4) ________(fine) fore a lot things in Singapore, including feeding birds, spitting , urinating (小便)in public, smoking in public , not flushing a public toilet after you use it, and eating or drinking on buses or trains.JapanYou’re not supposed to wear your shoes in someone’s house, but you’re not supposed to take your shoes off (5)________the house either. Instead, there’s a small areas inside the door called a “ genkan” which is (6)__________your shoes should go. If you’re still not sure where that is , pay attention to what other people do and do the same.FranceFrench people don’t like (7)________when you talk about money. It’s OK if you say that you want to quit a job because you (8)___________(pay) little money, but you should never say the exact amount. Money is a “ dirty” topic in France.Ⅱ. Recitation (4%)In western cultures, (1)_______eye contact in conversation is necessary. As a matter of fact, a westerner might consider a lack of eye contact as a lack of interest. In Spain, Italy and Greece, where people stand close together talking to each other, eye contact is more frequent and lasts longer.In many Asian cultures, people avoid eye contact to show respect. It is done when talking with anyone in (2)__________or with anyone older.Habits like this can cause problems when people do not understand them. For example, an Asian person might close his or her eyes in (3) ___________ or look down while listening to a speaker. A Western speaker might thing the person is not interested.Eye contact is a (4) ___________ thing. A lack of eye contact may be considered impolite. But if you stare at others, it is also considered rude and should be avoided.Ⅲ. Translation (20%)1. 在炎热的夏日，将食物放在冰箱被认为是保藏食物最行之有效的方法。

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。

2018-2019学年上海市复旦附中高三（上）期中数学试卷试题数：21.满分：01.（填空题.4分）n→∞(1+12019n)2019=___ ．2.（填空题.4分）若复数z满足z•（1+i）=（1-i）2（i为虚数单位）.则|z|=___ ．3.（填空题.4分）已知向量a⃗ =（-2.1）. b⃗⃗ =（3.-1）. c⃗ =（2.y）.且（a⃗ - b⃗⃗）⊥ c⃗ .则y=___ ．4.（填空题.4分）若集合A={x|y=lg（x- 12）}.B={y|y=arcsinx}.则A∩B=___．5.（填空题.4分）在(x−2x2)6的二项展开式中.x3的系数是___ ．6.（填空题.4分）在△ABC中.角A.B.C所对的边分别是a.b.c.若a=3.b= √6，A=π3.则角C的大小为___ ．7.（填空题.4分）若圆锥侧面积为20π.且母线与底面所成角正切为34.则该圆锥的体积为___ ．8.（填空题.4分）若无穷等比数列{a n}的前n项和为S n.首项a1=1.a2=a- 32 .且n→∞S n =a.则a=___ ．9.（填空题.4分）某学生选择物理、化学、地理这三门学科参加等级考.已知每门学科考A+得70分.考A得67分.考B+得64分.该生每门学科均不低于64分.则其总分至少为207分的概率为___ ．10.（填空题.4分）已知a.b∈R.且4≤a2+b2≤25.则a2+b2+ab的取值范围是___ ．11.（填空题.6分）已知函数f（x）=a sinx+bcos x+c的图象经过A（0.5）.B（π2.5）两点.当x∈[0. π2]时.|f（x）|≤10.则实数c的取值范围是___ ．12.（填空题.6分）定义在[0.+∞）上的函数f（x）满足：① 当x∈[1.2）时. f(x)=1 2sin(2πx+π2)；② 对任意x∈[0.+∞）都有f（2x）=2f（x）．设关于x的函数F（x）=f（x）-a的零点从小到大依次为x1.x2.x3.…x n.….若a∈(12，1) .则x1+x2+…+x2n=___ ．13.（单选题.5分）若平面中两条直线l1.l2的方向向量分别是a⃗ . b⃗⃗ .则l1 || l2是a⃗ || b⃗⃗的（）A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件14.（单选题.5分）若定义域为R的奇函数y=f（x）有反函数y=f-1（x）.那么必在函数y=f-1（x+1）图象上的点是（）A.（-f（t-1）.-t）B.（-f（t+1）.-t）C.（-f（t）-1.-t）D.（-f（t）+1.-t）15.（单选题.5分）已知数列{a n}的通项公式为a n= 1n(n+1)（n∈N*.其前n项和S n= 910.则双曲线x2 n+1 - y2n=1的渐近线方程为（）A. y=±2√23xB. y=±3√24xC. y=±3√1010xD. y=±√103x16.（单选题.5分）已知定义在[0.+∞）上的函数f（x）满足f（x+2）=f（x）+x.且当x∈[0.2）时.f（x）=x-8.则f（93）=（）A.2019B.2109C.2190D.290117.（问答题.14分）已知递增的等差数列{a n}的首项a1=1.且a1、a2、a4成等比数列．（1）求数列{a n}的通项公式a n；（2）设数列{b n}满足b n=2a n+（-1）n a n.T n为数列{b n}的前n项和.求T2n．18.（问答题.14分）已知函数f（x）=2sinxcosx+2 √3 cos2x- √3 .x∈R．（1）求函数y=f（-3x）+1的最小正周期和单调递减区间；（2）已知△ABC中的三个内角A.B.C所对的边分别为a.b.c.若锐角A满足f（A2 - π6）= √3 .且a=7.sinB+sinC= 13√314.求b.c的长．19.（问答题.14分）某饮料生产企业为了占有更多的市场份额.拟在2013年度进行一系列促销活动.经过市场调查和测算.饮料的年销售量x 万件与年促销费t 万元间满足x= 3t+1t+1 .已知2013年生产饮料的设备折旧、维修等固定费用为3万元.每生产1万件饮料需再投入32万元的生产费用.若将每件饮料的售价定为其生产成本的150%与平均每件促销费的一半之和.则该年生产的饮料正好能销售完．（1）将2013年的利润y （万元）表示为促销费t （万元）的函数； （2）该企业2013年的年促销费投入多少万元时.企业的年利润最大？ （注：利润=销售收入-生产成本-促销费.生产成本=固定费用+生产费用）20.（问答题.16分）设数列{a n }的前n 项和为S n .对任意n∈N *.点 (n ，S nn ) 都在函数 f (x )=x +a n2x的图象上． （1）求a 1.a 2.a 3.归纳数列{a n }的通项公式（不必证明）．（2）将数列{a n }依次按1项、2项、3项、4项、5项循环地分为（a 1）.（a 2.a 3）.（a 4.a 5.a 6）.（a 7.a 8.a 9.a 10）.（a 11.a 12.a 13.a 14.a 15）.（a 16）.（a 17.a 18）.（a 19.a 20.a 21）.….分别计算各个括号内各数之和.设由这些和按原来括号的前后顺序构成的数列为{b n }.求b 6+b 100的值． （3）设A n 为数列 {a n −1a n} 的前n 项积.若不等式 A n √a n +1＜f (a )−a n +32a对一切n∈N *都成立.其中a ＞0.求a 的取值范围．21.（问答题.18分）已知函数h （x ）= mx+n 3x 2+27 为奇函数.k （x ）=（ 13 ）|x-m|.其中m 、n∈R ． （1）若函数h （x ）的图象过点A （1.1）.求实数m 和n 的值；（2）若m=3.试判断函数f （x ）= 1ℎ(x ) + 1k (x ) 在x∈[3.+∞）上的单调性并证明； （3）设函数 g (x )={ℎ(x )，x ≥39k (x )，x ＜3若对每一个不小于3的实数x 1.都恰有一个小于3的实数x 2.使得g （x 1）=g （x 2）成立.求实数m 的取值范围．2018-2019学年上海市复旦附中高三（上）期中数学试卷参考答案与试题解析试题数：21.满分：01.（填空题.4分）n→∞(1+12019n )2019=___ ．【正确答案】：[1]1【解析】：利用数列的极限的运算法则.化简求解即可．【解答】：解：n→∞(1+12019n )2019 = lim n→∞(1+C 20191•12019n+C 20192(12019n)2+⋯+C 20192019(12019n )2019) =1．故答案为：1．【点评】：本题考查数列的极限的运算法则的应用.考查计算能力．2.（填空题.4分）若复数z 满足z•（1+i ）=（1-i ）2（i 为虚数单位）.则|z|=___ ． 【正确答案】：[1] √2【解析】：把已知等式变形.再由复数代数形式的乘除运算化简.利用复数模的计算公式求解．【解答】：解：由z•（1+i ）=（1-i ）2=-2i. 得z=−2i1+i=−2i (1−i )(1+i )(1−i )=−1−i .∴|z|= √2 ． 故答案为： √2 ．【点评】：本题考查复数代数形式的乘除运算.考查复数模的求法.是基础的计算题． 3.（填空题.4分）已知向量 a ⃗ =（-2.1）. b ⃗⃗ =（3.-1）. c ⃗ =（2.y ）.且（ a ⃗ - b ⃗⃗ ）⊥ c ⃗ .则y=___ ． 【正确答案】：[1]5【解析】：可求出 a ⃗−b ⃗⃗=(−5，2) .根据 (a ⃗−b ⃗⃗)⊥c ⃗ 即可得出 (a ⃗−b ⃗⃗)•c ⃗=0 .进行数量积的坐标运算即可求出y ．【解答】：解：a⃗−b⃗⃗=(−5，2)；∵ (a⃗−b⃗⃗)⊥c⃗；∴ (a⃗−b⃗⃗)•c⃗=−10+2y=0；∴y=5．故答案为：5．【点评】：考查向量坐标的减法和数量积运算.以及向量垂直的充要条件．4.（填空题.4分）若集合A={x|y=lg（x- 12）}.B={y|y=arcsinx}.则A∩B=___ ．【正确答案】：[1]（12，π2]【解析】：先求出集合A.B.由此能求出A∩B．【解答】：解：∵集合A={x|y=lg（x- 12）}={x|x＞12}.B={y|y=arcsinx}={x|- π2≤x≤ π2}.∴A∩B={x| 12＜x≤π2）=（12，π2]．故答案为：（12，π2]．【点评】：本题考查交集的求法.考查交集定义等基础知识.考查运算求解能力.考查函数与方程思想.是基础题．5.（填空题.4分）在(x−2x2)6的二项展开式中.x3的系数是___ ．【正确答案】：[1]-12【解析】：在二项展开式的通项公式中.令x的幂指数等于3.求出r的值.即可得到x3的系数．【解答】：解：在(x−2x2)6的二项展开式中.它的通项公式为 T r+1= C6r•（-2）r•x6-3r.令6-3r=3.求得r=1.故x3的系数为C61•（-2）=-12.故答案为：-12．【点评】：本题主要考查二项式定理的应用.二项展开式的通项公式.二项式系数的性质.属于基础题．6.（填空题.4分）在△ABC中.角A.B.C所对的边分别是a.b.c.若a=3.b= √6，A=π3.则角C的大小为___ ．【正确答案】：[1] 5π12【解析】：先由正弦定理求出sinB.然后通过a＞b判断出B为锐角.求出B.最后利用三角形内角和为π.求出C．【解答】：解：在三角形ABC中.由正弦定理得：asinA = bsinB.即3sinπ3= √6sinB.解得：sinB= √22.又b＜a.∴B＜A.∴B= π4 .∴C=π- π3- π4= 5π12.故答案为：5π12【点评】：本题考查了正弦定理．属基础题．7.（填空题.4分）若圆锥侧面积为20π.且母线与底面所成角正切为34.则该圆锥的体积为___ ．【正确答案】：[1]16π【解析】：根据圆锥的侧面积和圆锥的母线长求得圆锥的弧长.利用圆锥的侧面展开扇形的弧长等于圆锥的底面周长求得圆锥的底面半径即可．【解答】：解：∵设圆锥的母线长是l.底面半径为r.母线与底面所成角正切为34 .即余弦为45．可得rl = 45①∵侧面积是20π.∴πrl=20π. ②由① ② 解得：l=5.r=4.故圆锥的高h= √l2−r2 = √25−16 =3.则该圆锥的体积为：13×πr2×3=16π.故答案为：16π．【点评】：本题考查了圆锥的有关计算.解题的关键是正确的进行圆锥与扇形的转化．8.（填空题.4分）若无穷等比数列{a n}的前n项和为S n.首项a1=1.a2=a- 32 .且n→∞S n =a.则a=___ ．【正确答案】：[1]2【解析】：由已知得11−(a−1.5)=a.由此能求出a．【解答】：解：∵无穷等比数列{a n}的前n项和为S n.首项为1.公比为a-1.5.∴S n= 1−(a−1．5)n1−(a−1.5).∵n→∞S n =a.∴ 11−(a−1.5)=a.解得a=2．故答案为：2．【点评】：本题考查等比数列的公比的求法.是中档题.解题时要认真审题.注意等比数列的性质的合理运用．9.（填空题.4分）某学生选择物理、化学、地理这三门学科参加等级考.已知每门学科考A+得70分.考A得67分.考B+得64分.该生每门学科均不低于64分.则其总分至少为207分的概率为___ ．【正确答案】：[1] 427【解析】：先求出基本事件总数n=3×3×3=27.其总分至少为207分包含的基本事件个数m= C33+C32C11 =4.由此能求出其总分至少为207分的概率．【解答】：解：某学生选择物理、化学、地理这三门学科参加等级考.每门学科考A+得70分.考A得67分.考B+得64分.该生每门学科均不低于64分.基本事件总数n=3×3×3=27.其总分至少为207分包含的基本事件个数：m= C33+C32C11 =4.∴则其总分至少为207分的概率p= mn =427．故答案为：427．【点评】：本题考查概率的求法.考查古典概型、排列组合等基础知识.考查运算求解能力.是基础题．10.（填空题.4分）已知a.b∈R.且4≤a2+b2≤25.则a2+b2+ab的取值范围是___ ．【正确答案】：[1][2. 752]【解析】：由题意可令a=rcosα.b=rsinα（2≤r≤5）.代入a2+b2+ab.结合三角函数的性质可求．【解答】：解：∵4≤a2+b2≤25.令a=rcosα.b=rsinα（2≤r≤5）.则a2+b2+ab=（rcosα）2+（rsinα）2+r2cosαsinα=r2（1+sinαcosα）=r2（1+ 12sin2α）.∵ 1 2≤1+ 12sin2α≤ 32.∴2≤ 12r2≤ r2（1+ 12sin2α）≤3r22≤752.故答案为：[2. 752]．【点评】：本题以不等式为载体.主要考查了不等式的性质及三角函数性质的简单应用．11.（填空题.6分）已知函数f（x）=a sinx+bcos x+c的图象经过A（0.5）.B（π2.5）两点.当x∈[0. π2]时.|f（x）|≤10.则实数c的取值范围是___ ．【正确答案】：[1][20-5 √2 .20+15 √2 ]【解析】：先求出a=b=5-c.再化简函数f（x）=（5-c）√2 sin（x+ π4）+c.由x的范围和正弦函数的图象与性质.求出f（x）的最值.由条件和恒成立列出不等式.求出实数a的取值范围．【解答】：解：∵已知函数f（x）=a sinx+bcos x+c的图象经过A（0.5）.B（π2.5）两点.∴{0+b+c=5a+0+c=5.∴a=b=5-c．∴函数f（x）=a sinx+bcos x+c=（5-c）（sinx+cosx）+c=（5-c）√2 sin（x+ π4）+c.当x∈[0. π2 ]时.x+ π4∈[ π4. 3π4].sin（x+ π4）∈[ √22.1].∴ √2 sin（x+ π4）∈[.1 √2 ].当sin（x+ π4）=1时.f（x）= √2（5-c）+c；当sin（x+ π4）= √22时.f（x）=5．∵当x∈[0. π2]时.|f（x）|≤10恒成立.∴| √2（5-c）+c|≤10.∴-10≤（1- √2）c+5 √2≤10.求得-5 √2≤c≤20+15 √2 .则实数c的取值范围是[-5 √2 .20+15 √2 ].故答案为：[-5 √2 .20+15 √2 ]．【点评】：本题考查正弦函数的图象与性质.三角恒等变换中的公式.函数的单调性.以及恒成立与存在性问题的转化.考查转化思想.换元法、分离常数法.化简、变形能力.属于难题．12.（填空题.6分）定义在[0.+∞）上的函数f（x）满足：① 当x∈[1.2）时. f(x)=1 2sin(2πx+π2)；② 对任意x∈[0.+∞）都有f（2x）=2f（x）．设关于x的函数F（x）=f（x）-a的零点从小到大依次为x1.x2.x3.…x n.….若a∈(12，1) .则x1+x2+…+x2n=___ ．【正确答案】：[1]6×（2n-1）【解析】：① 当x∈[1.2）时. f(x)=12sin(2πx+π2)；② 对任意x∈[0.+∞）都有f（2x）=2f（x）．x∈[2.4）时. 12x∈[1.2）．可得f（x）=2f（12x）=sin (πx+π2) .…….画出图象.设关于x的函数F（x）=f（x）-a的零点从小到大依次为x1.x2.x3.…x n.….同理.则a∈(12，1) .F （x）=f（x）-a在区间（2.3）和（3.4）上各有1个零点.分别为x1.x2.且满足x1+x2=2×3=6.依此类推：x3+x4=2×6=12.x5+x6=2×12=24….x2n-1+x2n=2×3×2n-1．再利用等比数列的求和公式即可得出．【解答】：解：① 当x∈[1.2）时. f(x)=12sin(2πx+π2)；② 对任意x∈[0.+∞）都有f（2x）=2f（x）．x∈[2.4）时. 12x∈[1.2）．∴f（x）=2f（12x）=sin (πx+π2) .…….画出图象.设关于x的函数F（x）=f（x）-a的零点从小到大依次为x1.x2.x3.…x n.….同理.则a∈(12，1) .F（x）=f（x）-a在区间（2.3）和（3.4）上各有1个零点. 分别为x1.x2.且满足x1+x2=2×3=6.依此类推：x3+x4=2×6=12.x5+x6=2×12=24….x2n-1+x2n=2×3×2n-1．∴当a∈(12，1) .时.x1+x2+…+x2n-1+x2n=6×（1+2+22+…+2n-1）=6× 1−2n=6×（2n-1）.1−2故答案为：6×（2n-1）．【点评】：本题考查了三角函数的图象与性质、等比数列的通项公式与求和公式.考查了数形结合方法、推理能力与计算能力.属于中档题．13.（单选题.5分）若平面中两条直线l1.l2的方向向量分别是a⃗ . b⃗⃗ .则l1 || l2是a⃗ || b⃗⃗的（）A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件【正确答案】：A【解析】：平面中两条直线l1.l2的方向向量分别是a⃗ . b⃗⃗ .可得l1 || l2⇒ a⃗ || b⃗⃗ .反之不成立.可能重合．【解答】：解：平面中两条直线l1.l2的方向向量分别是a⃗ . b⃗⃗ .则l1 || l2⇒ a⃗ || b⃗⃗ .反之不成立.可能重合．∴l1 || l2是a⃗ || b⃗⃗的充分不必要条件．故选：A．【点评】：本题考查了线面位置关系、平面向量的性质、简易逻辑的判定方法.考查了推理能力与计算能力.属于基础题．14.（单选题.5分）若定义域为R的奇函数y=f（x）有反函数y=f-1（x）.那么必在函数y=f-1（x+1）图象上的点是（）A.（-f（t-1）.-t）B.（-f（t+1）.-t）C.（-f（t）-1.-t）D.（-f（t）+1.-t）【正确答案】：C【解析】：由f（-t）=-f（t）得f-1（-f（t））=-t.再由函数图象的平移规律得出答案．【解答】：解；∵f（x）定义在R上的奇函数.∴f（-t）=-f（t）.∴f-1（-f（t））=-t.即（-f（t）.-t）在y=f-1（x）的图象上.∵y=f-1（x+1）图象是由y=f-1（x）的图象向左平移1个单位得到的.∴（-f（t）-1.-t）在y=f-1（x+1）图象上．故选：C．【点评】：本题考查了奇函数、反函数的性质及函数图象变换.利用互为反函数的函数图象关系是关键．15.（单选题.5分）已知数列{a n}的通项公式为a n= 1n(n+1)（n∈N*.其前n项和S n= 910.则双曲线x2 n+1 - y2n=1的渐近线方程为（）A. y=±2√23xB. y=±3√24xC. y=±3√1010xD. y=±√103x【正确答案】：C【解析】：根据数列{a n}的通项利用裂项求和算出S n.代入题中解出n=9.可得双曲线的方程为x2 10−y29=1 .再用双曲线的渐近线方程的公式即可算出该双曲线的渐近线方程．【解答】：解：∵数列{a n}的通项公式为a n=1n(n+1)(n∈N∗) .∴ a n=1n −1n+1.可得S n=(1−12)+(12−13)+⋯+(1n−1−1n)+(1n−1n+1)=910即1- 1n+1 = 910.解之得n=9．∴双曲线的方程为x210−y29=1 .得a= √10 .b=3因此该双曲线的渐近线方程为y= ±ba x .即y=±3√1010x．故选：C．【点评】：本题给出数列的前n项和.求项数n并求与之有关的双曲线渐近线方程．着重考查了数列的通项与求和、双曲线的标准方程与简单几何性质等知识.属于中档题．16.（单选题.5分）已知定义在[0.+∞）上的函数f（x）满足f（x+2）=f（x）+x.且当x∈[0.2）时.f（x）=x-8.则f（93）=（）A.2019B.2109C.2190D.2901【正确答案】：B【解析】：有f（x+2）=f（x）+x得f（x+2）-f（x）=x.利用累加法进行求解即可得到结论．【解答】：解：由f（x+2）=f（x）+x得f（x+2）-f（x）=x.则f（3）-f（1）=1.f（5）-f（3）=3.f（7）-f（5）=5.….f（93）-f（91）=91.=2116.两边同时相加得f（93）-f（1）=1+3+5+…+91= (1+91)×462∴f（93）=f（1）+2116.∵当x∈[0.2）时.f（x）=x-8.∴f（1）=-7.则f（93）=f（1）+2116=-7+2116=2109.故选：B．【点评】：本题主要考查函数值的计算.根据条件.利用累加法进行求解是解决本题的关键．17.（问答题.14分）已知递增的等差数列{a n}的首项a1=1.且a1、a2、a4成等比数列．（1）求数列{a n}的通项公式a n；（2）设数列{b n}满足b n=2a n+（-1）n a n.T n为数列{b n}的前n项和.求T2n．【正确答案】：【解析】：（1）由已知列式求得公差.代入等差数列的通项公式得答案；（2）把数列{a n}的通项公式代入b n=2a n+（-1）n a n.分组后利用等差数列与等比数列前n项和公式求解．【解答】：解：（1）由题意可得.d＞0.a1=1.且a1a4=a22 .即a1(a1+3d)=(a1+d)2 .解得d=1．∴a n=a1+（n-1）d=n；（2）b n=2a n+（-1）n a n=2n+（-1）n n．则T2n=(21+22+⋯+22n)+ [-1+2-3+4-…-（2n-1）+2n]= 2(1−22n)1−2+n =22n+1+n-2．【点评】：本题考查等差数列的通项公式.考查等比数列的性质.训练了等差数列与等比数列前n项和的求法.是中档题．18.（问答题.14分）已知函数f（x）=2sinxcosx+2 √3 cos2x- √3 .x∈R．（1）求函数y=f（-3x）+1的最小正周期和单调递减区间；（2）已知△ABC中的三个内角A.B.C所对的边分别为a.b.c.若锐角A满足f（A2 - π6）= √3 .且a=7.sinB+sinC= 13√314.求b.c的长．【正确答案】：【解析】：（1）利用三角恒等变换化简f（x）.求出y=f（-3x）+1的解析式.再求y的最小正周期和单调减区间；（2）根据题意求出A的值.再利用正弦定理和余弦定理求出b、c的值．【解答】：解：（1）∵ f(x)=2sinxcosx+√3(2cos2x−1)=sin2x+ √3 cos2x=2sin（2x+ π3）；…（2分）∴y=f（-3x）+1=2sin（-6x+ π3）+1=-2sin（6x- π3）+1；∴y=f（-3x）+1的最小正周期为T=2π6=π3；…（3分）由2kπ−π2≤6x−π3≤2kπ+π2得：1 3kπ−π36≤x≤13kπ+5π36.k∈Z.∴y=f（-3x）+1的单调递减区间是[1 3kπ−π36，13kπ+5π36] .k∈Z；…（6分）（2）∵ f(A2−π6)=√3 .∴ 2sin(A−π3+π3)=√3 .∴ sinA=√32.…（7分）∵ 0＜A＜π2 .∴ A=π3；由正弦定理得：sinB+sinC=b+casinA .即13√314=b+c7×√32.∴b+c=13；…（9分）由余弦定理a2=b2+c2-2bccosA得：a2=（b+c）2-2bc-2bccosA.即49=169-3bc.∴bc=40；…（11分）解得b=5.c=8或b=8或c=5．…（12分）【点评】：本题考查了正弦定理和余弦定理的应用问题.也考查了三角恒等变换的应用问题.是综合性题目．19.（问答题.14分）某饮料生产企业为了占有更多的市场份额.拟在2013年度进行一系列促销活动.经过市场调查和测算.饮料的年销售量x万件与年促销费t万元间满足x= 3t+1t+1.已知2013年生产饮料的设备折旧、维修等固定费用为3万元.每生产1万件饮料需再投入32万元的生产费用.若将每件饮料的售价定为其生产成本的150%与平均每件促销费的一半之和.则该年生产的饮料正好能销售完．（1）将2013年的利润y（万元）表示为促销费t（万元）的函数；（2）该企业2013年的年促销费投入多少万元时.企业的年利润最大？（注：利润=销售收入-生产成本-促销费.生产成本=固定费用+生产费用）【正确答案】：【解析】：（1）确定饮料的售价.即可通过x 表示出年利润y.化简代入整理即可求出y 万元表示为促销费t 万元的函数；（2）根据已知代入（1）的函数.分别进行化简.利用关于t 的方程必须有两正根建立关系式.可求出最值.即促销费投入多少万元时.企业的年利润最大．【解答】：解：（1）当年销量为x 万件时.成本为3+32x （万元）． 饮料的售价为3+32x x ×150%+ 12 × tx（万元/万件） 所以年利润y=（3+32x x ×150%+ 12 × tx）x-（3+32x+t ）（万元） 把x= 3t+1t+1 代入整理得到y=−t 2+98t+352t+2.其中t≥0．（2）y= −t 2+98t+352t+2 .去分母整理得到：t 2+2（y-49）t+2y-35=0．该关于t 的方程在[0.+∞）上有解． 当2y-35≤0.即y≤17.5时.必有一解． 当2y-35＞0时.该关于t 的方程必须有两正根所以 {4(y −49)2−4(2y −35)≥0−2(y −49)＞02y −35＞0解得：17.5＜y≤42．综上.年利润最大为42万元.此时促销费t=7（万元）． 所以当促销费定在7万元时.企业的年利润最大．【点评】：本小题主要考查函数模型的选择与应用、方程根的分布等基础知识.考查学生分析问题和解决问题的能力.强调对知识的理解和熟练运用.属于中档题．20.（问答题.16分）设数列{a n }的前n 项和为S n .对任意n∈N *.点 (n ，Snn ) 都在函数 f (x )=x +a n2x的图象上． （1）求a 1.a 2.a 3.归纳数列{a n }的通项公式（不必证明）．（2）将数列{a n }依次按1项、2项、3项、4项、5项循环地分为（a 1）.（a 2.a 3）.（a 4.a 5.a 6）.（a 7.a 8.a 9.a 10）.（a 11.a 12.a 13.a 14.a 15）.（a 16）.（a 17.a 18）.（a 19.a 20.a 21）.….分别计算各个括号内各数之和.设由这些和按原来括号的前后顺序构成的数列为{b n }.求b 6+b 100的值． （3）设A n 为数列 {a n −1a n} 的前n 项积.若不等式 A n √a n +1＜f (a )−a n +32a对一切n∈N *都成立.其中a ＞0.求a 的取值范围．【正确答案】：【解析】：（1）求得S n =n 2+ 12 a n .分别令n=1.2.3.进而归纳出数列{a n }的通项公式；（2）写出几个循环数.可得每一次循环记为一组.由每一个循环含有5个括号.故b 100是第20组中第5个括号内的数之和.每一个循环中含有15个数.20个循环具有300个数.计算可得所求和；（3）由题意可得原不等式即为（1- 1a 1 ）（1- 1a 2 ）…（1- 1a n） √2n +1 ＜a- 32a对一切n∈N *都成立.设g （n ）=（1- 1a 1 ）（1- 1a 2 ）…（1- 1a n） √2n +1 .则只需g （n ）max ＜a- 32a.判断数列g （n ）的单调性.可得最大值.解不等式即可得到所求a 的范围．【解答】：解：（1）由点 (n ，S n n ) 都在函数 f (x )=x +an2x 的图象上. 可得 S n n =n+ a n 2n .即S n =n 2+ 12 a n .当n=1时.a 1=1+ 12a 1.即a 1=2. n=2时.a 1+a 2=4+ 12 a 2.可得a 2=4. n=3时.a 1+a 2+a 3=9+ 12 a 3.可得a 3=6. 由此猜想a n =2n.n∈N*； （2）由a n =2n.n∈N*.将数列{a n }依次按1项、2项、3项、4项、5项循环地分为（2）.（4.6）.（8.10.12）.（14.16.18.20）.（22.24.26.28.30）；（32）.（34.36）.（38.40.42）.….每一次循环记为一组.由每一个循环含有5个括号. 故b 100是第20组中第5个括号内的数之和. 每一个循环中含有15个数.20个循环具有300个数.b 100=（2+4+…+600）-（2+4+…+590）=592+594+596+598+600=2980. 又b 6=32.则b 100+b 6=3012； （3）由a n −1a n =1- 1a n.故A n =（1- 1a 1）（1- 1a 2）…（1- 1a n）.则A n √a n +1 =（1- 1a 1）（1- 1a 2）…（1- 1a n） √2n +1 .又f （a ）-a n +32a =a+ a n 2a - a n +32a =a- 32a. A n √a n +1＜f (a )−a n +32a对一切n∈N *都成立.即为（1- 1a 1）（1- 1a 2）…（1- 1a n） √2n +1 ＜a- 32a对一切n∈N *都成立. 设g （n ）=（1- 1a 1）（1- 1a 2）…（1- 1a n） √2n +1 .则只需g （n ）max ＜a- 32a .由g (n+1)g (n ) =（1- 1a n+1 ）• √2n+3√2n+1= √4n 2+8n+3√4n 2+8n+4 1.即g （n+1）＜g （n ）.可得g （n ）递减.即有g （n ）的最大值为g （1）= √32 . 由a- 32a＞ √32.又a ＞0.可得a ＞ √3 . 可得a 的取值范围是（ √3 .+∞）．【点评】：本题考查数列的通项公式的求法.注意运用归纳法.考查新数列的构造和求和.注意分析规律.考查数列不等式恒成立问题解法.注意运用数列的单调性和转化思想.考查化简整理的运算能力.属于难题．21.（问答题.18分）已知函数h （x ）= mx+n3x 2+27 为奇函数.k （x ）=（ 13 ）|x-m|.其中m 、n∈R ． （1）若函数h （x ）的图象过点A （1.1）.求实数m 和n 的值； （2）若m=3.试判断函数f （x ）= 1ℎ(x ) + 1k (x )在x∈[3.+∞）上的单调性并证明；（3）设函数 g (x )={ℎ(x )，x ≥39k (x )，x ＜3若对每一个不小于3的实数x 1.都恰有一个小于3的实数x 2.使得g （x 1）=g （x 2）成立.求实数m 的取值范围．【正确答案】：【解析】：（1）运用奇函数的定义可得n=0.再由h （x ）图象经过点（1.1）.解方程可得m ； （2）f （x ）=x+ 9x +3x-3在[3.+∞）递增．运用单调性的定义.结合因式分解和指数函数的单调性.即可得证；（3）求得当x≥3时.g （x ）=h （x ）=mx 3x 2+27 = m3x+27x；当x ＜3时.g （x ）=9k （x ）=9•（ 13）|x-m|．分别讨论m≤0.0＜m ＜3.m≥3.运用基本不等式和单调性.求得m 的范围．【解答】：解：（1）函数h （x ）= mx+n3x 2+27 为奇函数. 可得h （-x ）=-h （x ）.即 −mx+n 3x 2+27 =- mx+n3x 2+27 .则n=0. 由h （x ）的图象过A （1.1）.可得h （1）=1.即 m+n30=1. 解得m=30.n=0；（2）m=3.可得f （x ）=x+ 9x +3x-3.f （x ）在[3.+∞）递增． 理由：设3≤x 1＜x 2.则f （x 1）-f （x 2）=x 1+ 9x 1+3x 1-3-x 2- 9x 2-3x 2-3=（x 2-x 1）•9−x 1x 2x 1x 2+3x 1-3-3x 2-3. 由3≤x 1＜x 2.可得x 2-x 1＞0.x 1x 2＞9.3x 1-3-3x 2-3＜0. 则f （x 1）-f （x 2）＜0.即f （x 1）＜f （x 2）. 可得f （x ）在[3.+∞）递增；（3）当x≥3时.g （x ）=h （x ）= mx3x 2+27 =m 3x+27x；当x ＜3时.g （x ）=9k （x ）=9•（ 13 ）|x-m|． ① m≤0时.∀x 1≥3时.g （x 1）=h （x 1）=m3x 1+27x 1≤0；∀x 2＜3时.g （x 2）=9k （x 2）=9•（ 13 ）|x 2-m|．＞0不满足条件.舍去； ② 当0＜m ＜3时.∀x 1≥3时.g （x 1）=h （x 1）=m3x 1+27x 1∈（0. m18 ].∀x 2＜3时.|x 2-m|≥0.g （x 2）=9k （x 2）=9•（ 13 ）|x 2-m|∈（0.9]. 由题意可得（0. m 18 ]⊆（0.9].可得 m18 ≤9.即m≤162； 综上可得0＜m ＜3；③ 当m≥3时.∀x 1≥3时.g （x 1）=h （x 1）=m3x 1+27x 1∈（0. m18 ].∀x 2＜3时.|x 2-m|＞m-3≥0.g （x 2）=9k （x 2）=9•（ 13 ）|x 2-m|∈（0.9•（ 13 ）m-3）. 由题意可得（0. m18 ]⊆（0.9•（ 13 ）m-3）.可得 m 18＜35-m .可令H （x ）=35-x - x 18.则H （x ）在R 上递减.H （6）=0. m18 ＜35-m .可得m ＜6.即3≤m ＜6. 综上可得0＜m ＜6．【点评】：本题考查函数的奇偶性和单调性的定义和运用.考查分类讨论思想方法和化简整理的运算能力.属于难题．。

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．。

2017-2018 上海市复旦附中高一上学期期中考试Grammar and Vocabulary: (18%)Section A (8%)Directions: Beneath each of the following sentences there are four choices marked A, B, C and D. Choose the one answer that best completes the sentence.21. The discovery of gold in Australia led thousands to believe that a fortune_________.A, had been making B. was making C. was to be made D. would make22. The president hopes that people will be better off when he quits than when he _________.A. has startedB. startsC. will startD. started23. Travellers to that area can carry disease to their own countries that have never experienced _________.A. themB. itC. themselvesD. itself24. No one can tell the exact number of U.S. pilots and fighter planes ________were lost in the Pearl Harbor air attack.A. whoB. whichC. thatD. what25. ____________, the climbers, who had already conquered many high mountains before, determined to reach the top of the mountain.A. However it was highB. How high it wasC. However high it wasD. However high was it26. They will not allow others to decide the future of their country, _________.A. which is to be knownB. as is to be knownC. as is known to allD. as what is known to all27. We have come to realize that t he brain must “ forget” some pieces of information ________it can remember others.A. whenB. sinceC. so thatD. if28. They are our school’s volunteers _______to help elderly people cross the streets every day.A. whose task it isB. it is whose taskC. to whom is the taskD. whose task wasSection B (10%)Directions:Complete the following passage by using the words in the box. Each wordSept 27 is World Tourism Day. Of course, travel isn’t a new discovery. Imagine how Italian traveler Marco Polo must have felt when he found himself on Chinese ____29_____, seeing a way of life quite different from anything he’d seen before.And how ___30______ must it have been to listen to Zhang Qian when he returned to China from his journey through Central Asia and West Asia? His brain must havebeen packed with everything he’d seen and heard, leading to the ____31_____ of the Silk Road.Travel is one of the most exciting experiences a human being can have. Nowadays, more people are traveling than ever before. By train, plane and car, people all around the globe are ____32_____ to places that people didn’t even know ____33_____ a few centuries ago, or only knew from books.Some people have traveled all over the world, and travel is a way of life to them. They per haps know what to expect before they travel. That’s why the best travel is when it’s for the first time. Imagine a person who has always wanted to travel to the United States. Of course, they’ve probably seen the Statue of Liberty a thousand times on the TV, and the White House, and all the other famous _______34__. But none of that would compare to the ___35______ of looking out of the cabin window as the plane lands, watching the cities and streets of the real America come into_____36____.Although travel is often just for fun, it’s also ____37_____. We may not know that we are getting an education, but we still are.We’re learning every day: new words in a new language, new people, and new ways of life. But this learning takes place in the school of the world, not the classroom. One of the lessons we learn is ____38_____ a moral one. As we get to know foreign places, we come to understand that there are many different ways to live, and that the way we live isn’t necessarily the best way. The British politi cian Benjamin Disraeli summed this up well when he said, “Travel teaches toleration.”Ⅲ.Reading Comprehension (40%)Section ADirections : For each blank in the following passages there are four words or phrases marked A,B,C and D. Fill in each blank with the word or phrases that best fits the context.( A )Leif Erickson reached North America around the year 1000, but the attempt to explore was started slowly. It would be five centuries __(39)__ other Europeans landed on that continent.Why were Europeans the ones to __(40)__ to the American? The Chinese and Arabs had the _(41)__ and technology to sail across the seas. __(42)__ of them tool regular voyages in the Indian Ocean and the Asian Pacific for trade. But exploration? By the mid-15th century China had followed the closed-door policy to __(43)__ itself from the rest of the world. The Arabs, with access to the minerals and spices (香料) of Africa and the Far East, saw no __(44)__ to journey into the unknown.Europe, ____(45)________ needed gold and silver; its mines could not meet the demand for coinage. Ottoman Turks blocked the routes across the land to Asia. Only the sea held the ___(46)_____of new wealth.With the return of Magellan’s ships in 1522 from its voyage around the worl d, the belief was __(47)__ that the oceans were interconnected, promising the age ofdiscovery. The English, as well as the Spanish, Portuguese and French, __(48)__ themselves to finding the “river of the west” through North America to the east.39. A. after B. since C .before D .when40. A .push B .pull C .draw D .drive41. A .sources B .resources C .substances D .matters42. A .Neither B .Both C .Any D .None43. A .prevent B .protect C .isolate D .differ44. A .access B .admission C .application D .association45. A .as a matter of fact B . in other wordsC. for one thing D .on the other hand46. A .symbol B .impression C .promise D .reflection47. A .extended B .estimated C .attracted D .accepted48. A .contributed B .devoted C .referred D .connected( B )Those who keep their word become the most important members of an organization. People come to rely on and trust them. They can be ____(49)____.___(50)______you make a promise , be sure to keep it. When you keep your promise, no matter how much ___(51)_______it takes, you will be rewarded.Whenever you say no, stand upon that as well. In a way, a no is also a promise.A good, _____(52)____no can be very important in building trust.Agreements are also important. Whenever you enter an agreement, live by it, ______(53)____you are not too happy with the deal after making the deal, you still live by it. In the long run, your integrity will _____(54)____.Victoria was the manager of a supermarket. She set out to do that ____(55)_____. She was very careful, however, of not _____(56)_____. When she did, she moved mountains to make sure that she came through. After a while, the stone employee came to trust and respect her like no______(57)_____they had ever had. The teamwork became magical. People followed her example by living up to their word. She ____(58)______what she said, and people appreciated that.49. A. counted B. counted on C. depended D. numbered50.A. As far as B. Though C. Whenever D. However51.A. pain B. ache C. effort D. money52. A. loud B. clean C. sharp D. loose53. A. As B. As if C. Just as D. Even if54. A. set off B. show off C. pay off D. take off55. A. thoroughly B. entirely C. wholly D. totally56. A. lying B. overpromising C. underestimating D. overlooking57. A. manager B. employee C. supermarket D. others58. A. believed B. trusted C. meant D. promisedSection BDirections :Read the following passages. Each passage is followed by 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 )A few years ago, Paul Gerner began to gather a group of architects in Las Vegas to ask them what it would take to design a public school that used 50 percent less energy, cost much less to build and obviously improved student learning. “I think half of the m fell off their chairs,” Gerner says.Gerner manages school facilities for Clark County, Nevada, a district roughly the size of Massachusetts. By 2018, 143,000 additional students will enter the already crowded public-education system. Gerner needs 73 new schools to house them. Four architecture teams have nearly finished designing primary school prototypes （原型）; they plan to construct their schools starting in 2009. The district will then assess how well the schools perform, and three winners will copy those designs in 50 to 70 new buildings.Green schools are appearing all over, but in Clark County, which stands out for its vastness(广阔), such aggressive targets are difficult because design requirements like more natural light for students go against the realities of a desert climate. “One of the b iggest challenges is getting the right site orientation,” Mark McGinty, a director at SH Architecture, says. His firm recently completed a high school in Las Vegas. ”You have the same building, same set of windows, but if its orientation is incorrect and it faces the sun, it will be really expensive to cool.”Surprisingly, the man responsible for one of the most progressive green-design competitions has doubts about ideas of eco-friendly buildings. “I don’t beli eve in the new green religion,” Gerner says. “Some of the building technologies that you get are impractical. I’m interested in those that work.” But he wouldn’t mind if some green features inspire students. He says he hopes to set up green energy systems that allow them to learn about the process of harvesting wind and solar power. “You never know what’s going to start the interest of a child to study math and science,” he says.59. How did the architects react to Gerner’s design requirements?A .They lost balance in excitement.B .They showed strong disbelief.C .They expressed little interest.D .They burst into cheers.60. Which order of steps is followed in carrying out the project?A .Assessment — Prototype — Design — Construction.B .Assessment — Design — Prototype — Construction.C .Design — Assessment — Prototype — Construction.D .Design — Prototype — Assessment — Construction.61 What makes it difficult to build green schools in Clark County?A .The large size.B .Limited facilities.C .The desert climate.D .Poor natural resources.62. What does Gerner think of the ideas of green schools?A .They are questionable.B .They are out of date.C .They are advanced.D .They are practical.(B)Spring is in the air, as is romance. Perhaps you're even thinking of taking the plunge and making a honey trip with your significant other?Before you do, why not take a look at a list of the top three places in the world to propose(求婚)? With suggestions for traditionalists and private types, you just might find the perfect spot to pop the question.New YorkIf you pay attention to romantic comedies, New York is the place for romance. Ever since the Dutch first entered this harbor in 1624 it has become a historic landmark and a must visit dreamland. Every stretch of this island is packed full of fun that will keep everyone busy. Take a leaf out of sleepless in Seattle and head to the top of the Empire State Building at night, or how about run to your loved one on the streets of NYC this New Years even like in when harry Met Sally?ParisSweeping views? Check. Grand old stately home? Check. Luxurious spots to propose? check. pack your passports and say goodbye to your daily routine life-it's less than three hours to Paris from London by Eurostar. Famous for its breathtaking architecture, and many cultural attractions, France's capital is a must-see destination. While we're not suggesting you climb the Eiffel Tower for the big moment. There's a reason why this is the city of love, if you keep your proposal original. So gentleman, take your lady to the Louvre, turn your back on the Mona Lisa and declare she is more beautiful than the famous work of art.Las VegasDo you want to bundle the proposal and wedding all-in-one? plead to. Las Vegas, get down on one knee, show your love and get married in less time than it takes to order 'honeymoon suit’ and if it doesn't work out. There's still plenty of entertain. Once the playground of the rich and famous, from Elvis Presley to Marilyn Monroe, Las Vegas now attracts millions of visitors by its bright lights and thrills each year. Whether you're to try your luck at the casinos (赌场), or watch a show, there's a wonder to explore in Las Vegas.63. Which of the following city on the top-3 list would be the ideal place for film fans to propose?A. New YorkB. Paris.C. Las Vegas.D. None.64. Which place is recommended for a proposal in the passage?A. On the top of the Eiffel TowerB. In the Eurostar train from LondonC. Any luxurious spot in PairsD. In front of the Mona Lisa in the Louvre64. As a perfect place for propose, Las Vegas will provide you with all the following EXCEPT.A. a honeymoon suiteB. a package weddingC. a bundle of flowersD. various ways to entertain( C )Do you look happy? angry? Have you ever wondered how you know what another person’s mood is just by looking at his or her face? Studies have shown that one instantaneously (瞬间地) and subconsciously makes a determination of another person’s mood just by glancing at the position and appearance of both the eyebrows and mouth. But what if only the position of the eyebrows was varied , and not the mouth? Would you still be able to determine his or her mood or personality just by seeing the eyebrows?Observe the four faces shown below. Notice how each face has exactly the same shape. the same smile, and the same eyes. The only difference between each of the faces is the position of the eyebrows.Low, flat eyebrows that hang over the eyes indicate tiredness, while an eyebrow that is highest in the middle denotes sadness. Downward slanting eyebrows show anger, while highly arched eyebrows show happiness.Frequently, malposition of the eyebrows, such as eyebrow sagging(下垂) and upper eyelid fullness, can be overlooked when considering facial aging and expression. Furthermore, the heavy eyebrow skin pushed the eyelid down causing a tired appearance.To avoid eyebrow sagging, many people raise their eyebrows to remove the brow and eyelid skin from their visual space. These people eventually develop wrinkles in the forehead due to the constant movement of the muscles that raise the eyebrow. In addition, there are those who are sensitive to light and frequently’ squint(斜视) and squeeze’ their eyes. This effectively pulls the eyebrows down and leadsto wrinkles between the eyebrows and at the corners of the eyes.Botox is an effective measure for fine lines of the forehead and lines between the eyebrows. Botox works by freezing the muscle movement for several months. For the purpose of creating a more pleasant appearance, the best corrective measure is brow-lift surgery.66. What can this passage be?A. An advertisementB. A magazine articleC. A business reportD. A scientific report67. A person’s eyebrows that hang over the eyes low and flat can cause a(n) _________appearance.A. sadB. tiredC. happyD. angry68. What is the purpose of the passage?A. To explain the fact that varied eye position can tell a person’s mood.B. To arouse people’s interest in the best way of removing wrinkles in the forehead and between the eyebrows.C. To explain what leads to wrinkles in the forehead and near the eyebrow.D. To enable people to know how to avoid developing fine lines on their faces.69. If the passage continues, what would the writer most likely discuss in the next paragraph?A. Facial agingB. Eye operationC. Brow life surgeryD. Face-lifts.( D )The past ages of man have all been carefully labeled by anthropologists. Descriptions like ‘Palaeolithic Man’, ‘Neolithic Man’, etc., neatly sum up whole periods. When the time comes for anthropologists to turn their attention to the twentieth century, they will surely choose the label ‘Legless Man’. Histories of the time will go something like this: ‘in the twentieth century, people forgot how to use their legs. Men and women moved about in cars, buses and trains from a very early age. There were lifts and escalators in all large buildings to prevent people from walking. This situation was forced upon earth dwellers of that time because of miles each day. Bu t the surprising thing is that they didn’t use their legs even when they went on holiday. They built cable railways, ski-lifts and roads to the top of every huge mountain. All the beauty spots on earth were ruined by the presence of large car parks.’The future history books might also record that we were deprived of the use of our eyes. In our hurry to get from one place to another, we failed to see anything on the way. Air travel gives you a bird’s-eye view of the world—or even less if the wing of the aircraft happens to get in your way. When you travel by car or train a blurred (="not" clear) image of the countryside constantly smears the windows. Car drivers, in particular, are forever obsessed with the urge to go on and on: they never want to stop. Is it the lure of the great motorways, or what? And as for seatravel, it hardly deserves mention. It is perfectly summed up in the words of the old song: ‘I joined the navy to see the world, and what did I see? I saw the sea.’ The typical twentieth-century tr aveler is the man who always says ‘I’ve been there. ’You mention the remotest, most evocative (引起记忆的) place-names in the world like El Dorado, Kabul, Irkutsk and someone is bound to say ‘I’ve been there’—meaning, ‘I drove through it at 100 miles an hour on the way to somewhere else.’When you travel at high speeds, the present means nothing: you live mainly in the future because you spend most of your time looking forward to arriving at some other place. But actual arrival, when it is achieved, is meaningless. You want to move on again. By traveling like this, you suspend all experience; the present ceases to be a reality: you might just as well be dead. The traveler on foot, on the other hand, lives constantly in the present. For him traveling and arriving are one and the same thing: he arrives somewhere with every step he makes. He experiences the present moment with his eyes, his ears and the whole of his body. At the end of his journey he feels a delicious physical weariness. He knows that sound. Satisfying sleep will be his: the just reward of all true travellers.70. Anthropologists label nowadays’ men ‘Legless’ because _________.A. people forget how to use his legs.B. people prefer cars, buses and trains.C. lifts and escalators prevent people from walking.D. there are a lot of transportation devices.71.Why does the author say ‘we are deprived of the use of our eyes’?A. People won’t use their eyes.B. In traveling at high speeds, eyes become useless.C. People can’t see anything on his way of travel.D. People want to sleep during travelling.72. Travelling at high speed means _________.A. people’s focus on the futureB. a pleasureC. satisfying drivers’ great thrillD. a necessity of life73.What's the best title of the passage?A. More haste, less speedB. Modern means of transportation make the world a small placeC. Eyes open and mind broadenD. The only way to travel is on foot.( 以下各题请务必做在答卷纸上)Ⅰ. Grammar (8%)\One of the main points of traveling is to relax and take a break from your normal daily life. (1)__________the truth is, we’re not always free to do what we like when travelling to a foreign country, and a US tourist learned that the hard way. On Aug 12, the unnamed 41-year-old man was beaten by a passerby after he wasseen giving Nazi salutes(纳粹礼) again and again on a street in Dresden, Germany. Ever since the end of World WarⅡ, Germany has strict laws (2)_________(forbid) the Nazi salute, as well as other symbols of Nazism.In fact, most countries have their own taboos. If you plan to travel overseas, it’s best to get familiar with these taboos(3)______________you start touring local sites. Below, TEENS gives some examples.SingaporeYou can get (4) ________(fine) fore a lot things in Singapore, including feeding birds, spitting , urinating (小便)in public, smoking in public , not flushing a public toilet after you use it, and eating or drinking on buses or trains. JapanYou’re not supposed to wear your shoes in someone’s house, but you’re not supposed to take your shoes off (5)________the house either. Instead, there’s a small areas inside the door called a “ genkan” which is (6)__________your shoes should go. If you’re still not sure where that is , pay attention to what other people do and do the same.FranceFrench people don’t like (7)________when you talk about money. It’s OK if you say that you want to quit a job because you (8)___________(pay) little money, but you should never say the exact amount. Money is a “ dirty” topic in France.Ⅱ. Recitation (4%)In western cultures, (1)_______eye contact in conversation is necessary. As a matter of fact, a westerner might consider a lack of eye contact as a lack of interest. In Spain, Italy and Greece, where people stand close together talking to each other, eye contact is more frequent and lasts longer.In many Asian cultures, people avoid eye contact to show respect. It is done when talking with anyone in (2)__________or with anyone older.Habits like this can cause problems when people do not understand them. For example, an Asian person might close his or her eyes in (3) ___________ or look down while listening to a speaker. A Western speaker might thing the person is not interested.Eye contact is a (4) ___________ thing. A lack of eye contact may be considered impolite. But if you stare at others, it is also considered rude and should be avoided.Ⅲ. Translation (20%)1. 在炎热的夏日，将食物放在冰箱被认为是保藏食物最行之有效的方法。

上海市2018-2019学年复旦附中高三上学期数学期中考试一. 填空题（本大题共有12题，满分54分，第1～6题每题4分，第7～12题每题5分）1. =⎪⎭⎫ ⎝⎛+∞→2019201911lim n n .2. 若复数z 满足()()211i i z -=+⋅（i 为虚数单位），则=z .3. 已知向量(2,1),(3,1),(2,)a b c y →→→=-=-=，且()a b c →→→-⊥，则y = . 4. 若集合1|lg 2A x y x ⎧⎫⎛⎫==-⎨⎬ ⎪⎝⎭⎩⎭，{}x y y B arcsin ==，则=B A I .5. 在622⎪⎭⎫ ⎝⎛-x x 的二项展开式中，3x 的系数是 .6. 在ABC ∆中，角,,A B C 所对的边分别是,,a b c，若3,3a b A π===，则角C 的大小为 .7. 若圆锥侧面积为π20，且母线与底面所成角为3arctan4，则该圆锥的体积为 . 8. 若无穷等比数列{}n a 的前n 项和为n S ，首项11=a ，232a a =-，且a S n n =∞→lim ，则=a.9. 某学生选择物理、化学、地理这三门学科参加等级考，已知每门学科考A +得70分，考A 得67分，考B +得64分，该生每门学科均不低于64分，则其总分至少为207分的概率为 .10. 已知,,a b R ∈且22425a b ≤+≤，则22a b ab ++的取值范围是 . 11. 已知函数()sin cos f x a x b x c =++的图像经过(0,5),(,5)2A B π两点，当[0,]2x π∈时，()10f x ≤，则实数c 的取值范围是 .12. 定义在[)+∞,0上的函数()x f 满足：①当[)2,1∈x 时，()1sin 222f x x ππ⎛⎫=+⎪⎝⎭；②对任 意[)+∞∈,0x 都有()()x f x f 22=.设关于x 的函数a x f x F -=)()(的零点从小到大依次为,,,,,321ΛΛn x x x x 若⎪⎭⎫⎝⎛∈1,21a ，则=+++n x x x 221Λ .二. 选择题（本大题共有4题，满分20分，每题5分）13. 若平面中两条直线12,l l 的方向向量分别是,a b →→，则12//l l 是//a b →→的 （ ） A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件14. 已知定义域为R 的奇函数()y f x =有反函数1()y fx -=，那么必在函数1(1)y f x -=+图像上的点是 （ ）15. 已知数列}{n a 的通项公式为*1()(1)n a n N n n =∈+,其前n 项和910n S =,则双曲线2211x y n n-=+的渐近线方程为 （ ）A.3y x =±B.4y x =±C.10y x =±D.3y x =± 16. 已知定义在[0,)+∞上的函数()f x 满足(2)()f x f x x +=+，且当[0,2)x ∈时,()=x f 8-x ，则(93)f = （ ）A.2019B.2109C.2190D.2901三. 解答题（本大题满分76分）本大题共有5题，解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.17. （本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分． 已知递增的等差数列{}n a 的首项11a =，且1a 、2a 、4a 成等比数列． （1） 求数列{}n a 的通项公式n a ；（2） 设数列{}n b 满足()n nan a b n 12-+=，n T 为数列{}n b 的前n 项和，求n T 2.18. (本题满分14分) 本题共有2个小题，第1小题满分6分，第2小题满分8分.已知函数,3cos 32cos sin 2)(2-+=x x x x f R x ∈.（1）求函数1)3(+-=x f y 的最小正周期和单调递减区间；（2）已知ABC Δ中的三个内角C B A ,,所对的边分别为c b a ,,，若锐角A 满足=⎪⎭⎫⎝⎛-62πA f 3，且7=a ，14313sin sin =+C B ，求c b ,的长.19. (本题满分14分) 本题共有2个小题，第1小题满分6分，第2小题满分8分. 某饮料生产企业为了占有更多的市场份额，拟在2019年进行一系列促销活动.经市场调查和测算，饮料的年销售量x（万件）与年促销费t （万元）之间满足311t x t +=+.已知2019年生产饮料的设备折旧，维修费等固定费用为3万元，每生产1万件饮料需要再投入32万元的生产费用.若将每件饮料的售价定为每件饮料．．．．生产成本的150%与平均每件．．．．促销费．．．的一半之和，则该年生产的饮料正好能销售完. （1）将2019年的利润y （万元）表示为促销费t （万元）的函数； （2）该企业2019年的促销费投入为多少万元时，企业的年利润最大？20. (本题满分16分) 本题共有3个小题，第1小题满分4分，第2小题满分6分，第三小 题满分6分.设数列{}n a 的前n 项和为n S ，对任意*N n ∈，点.（1）求321,,a a a ，归纳数列{}n a 的通项公式（不必证明）．（2）将数列{}n a 依次按1项、2项、3项、4，),,(654a a a ，),,,(10987a a a a ，),,,,(1514131211a a a a a ，)(16a ，（3）设n A 为数列的前n 项积，若不等式 *N n ∈都成立，其中，求a 的取值范围．21. (本题满分18分) 本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小 题满分8分. 已知函数2()327mx n h x x +=+为奇函数，||1()()3x m k x -=，其中m 、.n ∈R . （1）若函数()h x 的图像过点(1,1)A ，求实数m 和n 的值； （2）若3m =，试判断函数11()([3,))()()f x x h x k x =+∈+∞在),3[+∞∈x 上的单调性并证明； （3）设函数()()()⎩⎨⎧<≥=3,93,x x k x x h x g 若对每一个不小于3的实数1x ，都恰有一个小于3的实数2x ，使得12()()g x g x =成立，求实数m 的取值范围.。