2018-2019年上海市复旦附中高三上第一次月考
2021届上海复旦附中高三上学期第一次月考数学试题word版含答案
2021届上海复旦附中高三上学期第一次月考数学试题一、填空题（1-6题每题4分，7-12题每题5分）1. 已知全集U R =，集合102x A x x ⎧+⎫=≤⎨⎬-⎩⎭，则集合u A =____________. 2. 已知函数34()log 2f x x ⎛⎫=+ ⎪⎝⎭，则方程1()4f x -=的解x =____________. 3. 已知线性方程组的增广矩阵为11334a --⎛⎫ ⎪⎝⎭，若该线性方程组的解为12-⎛⎫ ⎪⎝⎭，则实数a =____________. 4. 无穷等比数列*{}()n a n N ∈的前n 项的和是n S ，且1lim 2n n S →∞=，则首项1a 的取值范围是____________. 5. 已知虚数z 满足216z z i -=+，则||z =____________.6. 3(1)n x -展开式的二项式系数之和为256，则展开式中x 的系数为____________.7. 已知ABC 的内角A 、B 、C 所对应边的长度分别为a 、b 、c ，若a cb ac a b b --=，则角C 的大小是____________.8. 学校有两个食堂，现有3名学生前往就餐，则三个人不在同一个食堂就餐的概率是____________.9. 若数12345,,,,a a a a a 的标准差为2，则数1234532,32,32,32,32a a a a a -----的方差为____________.10. 如图，在矩形OABC 中，点E 、F 分别在线段AB 、BC 上，且满足3,3AB AE BC CF ==，若(,)OB OE OF R λμλμ=+∈，则λμ+=____________.11. 已知22430()230x x x f x x x x ⎧-+≤=⎨--+>⎩，当[,1]x a a +时不等式()(2)f x a f a x +≥-恒成立，则实数a 的最大值是____________.12. 已知()f x 是定义在R 上的奇函数，当01x ≤≤时，2()f x x =，当0x >时，(1)()(1)f x f x f +=+，若直线y kx =与函数()y f x =的图像恰有11个不同的公共点，则实数k 的取值范围为____________.二、选择题（每题5分）13. 下列四个命题中，为真命题的是（ ）A. 若a b >，则22ac bc >B. 若,a b c b >>，则a c b d ->-C. 若||a b >，则22a b >D. 若a b >，则11a b < 14. 下列函数中，既是奇函数，又在区间(0,)+∞上递增的是（ ）A. ||2x y =B. ln y x =C. 13y x = D. 1y x x=+ 15. 设,a b 是两个单位向量，其夹角为θ，则“63ππθ<<”是“||1a 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16. 设,,x y z 是互不相等的正数，则下列不等式中不恒成立的是（ ）A. 2211x x x x +≥+B. 312x x x x +-+≤+-C. 1||2x y x y -+≥-D. ||||||x y x z y z -≤-+-三、解答题17. （本题满分14分，第1小题8分，第2小题6分）如图，某人打算做一个正四棱锥形的金字塔模型，先用木料搭边框，再用其他材料填充。
最新上海市复旦附中2018-2019学年高一(上)期中数学模拟试卷
2018-2019学年上海市复旦附中高一（上）期中数学模拟试卷一．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．2．函数y=+lg（3﹣x）的定义域为．3．命题“若ab=0，则b=0”的逆否命题是．4．函数y=+2的单调区间是．5．已知f（x）为定义在R上的奇函数，当x≥0时，f（x）=x（1+x），则当x＜0时，f （x）=．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．7．函数的值域为．8．已知a＞0，b＞0，则的最小值为．9．设集合A={1，2，6}，B={2，4}，C={x∈R|﹣1≤x≤5}，则（A∪B）∩C=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．11．若函数f（x）=，则f（5）=．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（x，y）|?A，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）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14．已知函数f（x）=3x+x，g（x）=log3x+x，h（x）=sinx+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＜x115．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+2﹣1三．解答题（共5小题，满分76分）17．（14分）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，且p∧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．18．（14分）已知函数y=f（x）为定义在（﹣∞，0）∪（0，+∞）上的奇函数，且当x＞0时（Ⅰ）试求f（﹣2）的值；（Ⅱ）指出f（x）的单调递增区间；（直接写出结论即可）；（Ⅲ）求出f（x）的零点．19．（14分）已知函数f（x）=|x﹣2|+|x+3|．（1）求不等式f（x）≤15的解集；（2）若﹣x2+a≤f（x）对x∈R恒成立，求a的取值范围．20．（16分）函数f（x）的定义域为D={x|x≠0}，且满足对于任意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．21．（18分）已知函数，a∈R．（1）若函数f（x）是奇函数，求实数a的值；（2）在（1）的条件下，判断函数y=f（x）与函数y=lg2x的图象公共点个数，并说明理由；（3）当x∈[1，2）时，函数y=f（2x）的图象始终在函数y=lg（4﹣2x）的图象上方，求实数a的取值范围．2018-2019学年上海市复旦附中高一（上）期中数学模拟试卷参考答案与试题解析一．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{﹣1，﹣2，2，0} ．【分析】由实数a满足：a2∈{1，4，a}，得到a2=1或a2=4，或a2=a，由此能求出实数a的取值集合．【解答】解：∵实数a满足：a2∈{1，4，a}，∴a2=1或a2=4，或a2=a，解得a=﹣2或a=2或a=﹣1或a=1或a=0，当a=1时，{1，4，1}不成立，当a=﹣1，或a=±2，或a=0时，都成立．∴实数a的取值集合为{﹣1，﹣2，2，0}．故答案为：{﹣1，﹣2，2，0}．【点评】本题考查集合的求法，是基础题，解题时要认真审题，注意集合中元素的性质的合理运用．2．函数y=+lg（3﹣x）的定义域为[﹣2，3）．【分析】由根式内部的代数式大于等于0，对数式的真数大于0联立不等式组求解．【解答】解：由，解得﹣2≤x＜3．∴函数y=+lg（3﹣x）的定义域为：[﹣2，3）．故答案为：[﹣2，3）．【点评】本题考查函数的定义域及其求法，是基础题．3．命题“若ab=0，则b=0”的逆否命题是若b≠0，则ab≠0．【分析】根据命题“若p，则q”的逆否命题是“若￢q，则￢p”，直接写出答案即可．【解答】解：根据原命题与逆否命题的关系，知：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．【点评】本题考查了原命题与它的逆否命题之间的相互转化问题，解题时应明确四种命题之间的关系，是基础题．4．函数y=+2的单调区间是（﹣∞，0）和（0，+∞）．【分析】求出函数的定义域，利用反比例函数的单调性可求得答案．【解答】解：函数y=+2的定义域为（﹣∞，0）∪（0，+∞），由y=在（﹣∞，0）和（0，+∞）上单调递减，知函数y=+2的单调减区间是（﹣∞，0）和（0，+∞），故答案为：（﹣∞，0）和（0，+∞）．【点评】该题考查函数的单调性及单调区间的求解，属基础题，熟练掌握常见基本函数的单调性是解题基础．5．已知f（x）为定义在R上的奇函数，当x≥0时，f（x）=x（1+x），则当x＜0时，f （x）=x（1﹣x）．【分析】根据f（x）是奇函数可得出f（﹣x）=﹣f（x），根据xx≥0时，f（x）=x（1+x），可设x＜0，从而可求出f（﹣x）=﹣x（1﹣x）=﹣f（x）．【解答】解：∵f（x）是R上的奇函数；∴f（﹣x）=﹣f（x）；又x≥0时，f（x）=x（1+x）；∴设x＜0，﹣x＞0，则f（﹣x）=﹣x（1﹣x）=﹣f（x）；∴f（x）=x（1﹣x）．故答案为：x（1﹣x）．【点评】考查奇函数的定义，函数解析式的定义及求法．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{} ．【分析】分类讨论，分别求出等价函数，分别求解其零点个数，然后相加即可．【解答】解：①x＞0时，函数f（x）=sgn（x）﹣2x转化为函数f（x）=1﹣2x，令1﹣2x=0，得x=，即当x＞0时．函数f（x）=sgn（x）﹣2x的零点是；②x=0时，函数f（x）=sgn（x）﹣2x转化为函数f（x）=0，函数f（x）=sgn（x）﹣2x的零点是0；③x＜0时，函数f（x）=sgn（x）﹣2x转化为函数f（x）=﹣1﹣2x，令﹣1﹣2x=0，得x=﹣，即当x＜0时．函数f（x）=sgn（x）﹣2x的零点是﹣；综上函数f（x）=sgn（x）﹣x的零点的集合为：{}．故答案为：{}．【点评】本题主要考查了根的存在性及根的个数判断，考查转化思想，分类讨论思想，是基础题．7．函数的值域为（0，+∞）．【分析】根据8x＞0即可得出8x+1＞1，从而可求出，即得出f（x）的值域．【解答】解：8x＞0；∴8x+1＞1；∴；∴f（x）的值域为（0，+∞）．故答案为：（0，+∞）．【点评】考查函数值域的概念及求法，指数函数的值域，对数函数的单调性．8．已知a＞0，b＞0，则的最小值为4．【分析】构造基本不等式的性质即可求解．【解答】解：由==（a+2b）+，∵a＞0，b＞0，∴（a+2b）+≥2=4，当且仅当a+2b=2时取等号．则的最小值为4．故答案为：4．【点评】本题考查了“构造思想”与基本不等式的性质运用，属于基础题．9．设集合A={1，2，6}，B={2，4}，C={x∈R|﹣1≤x≤5}，则（A∪B）∩C={1，2，4}【分析】根据并集与交集的定义计算即可．【解答】解：集合A={1，2，6}，B={2，4}，∴A∪B={1，2，4，6}，又集合C={x|﹣1≤x≤5，x∈R}，∴（A∪B）∩C={1，2，4}．故答案为：{1，2，4}．【点评】本题考查了交集与并集的运算问题，是基础题．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（﹣∞，2）．【分析】根据y=f（x）是定义在（﹣∞，+∞）上的单调减函数即可由f（x）＜f（2x﹣2）得出x＞2x﹣2，这样即可解出x的取值范围．【解答】解：∵y=f（x）是定义在（﹣∞，+∞）上的单调减函数；∴由f（x）＜f（2x﹣2）得：x＞2x﹣2；∴x＜2；∴x的取值范围为（﹣∞，2）．故答案为：（﹣∞，2）．【点评】考查单调减函数的定义，以及一元一次不等式的解法．11．若函数f（x）=，则f（5）=1．【分析】推导出f（5）=f（3）=f（1），由此能求出结果．【解答】解：∵函数f（x）=，∴f（5）=f（3）=f（1）=1．故答案为：1．【点评】本题考查函数值的求法，考查函数性质等基础知识，考查运算求解能力，考查函数与方程思想，是基础题．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（x，y）|?A，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①③．（请写出所有符合条件的序号）【分析】根据新定义进行计算后判断，弄清开集的定义是解决本题的关键．即所选的集合需要满足存在以该集合内任意点为圆心，任意正实数为半径的圆内部分均在该集合内．初步确定该集合不含边界【解答】解：对于①：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B={（x，y）|＜r}?A，故①不是开集．对于②：A={（x，y）|x+y+2＞0}平面点集A中的任一点（x0，y0），则该点到直线的距离为d，取r=d，则满足B={（x，y）|＜r}?A，故②是开集；对于③：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B={（x，y）|＜r}?A，故该集合不是开集；对于④：A=表示以点（0，）为圆心，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B={（x，y）|＜r}?A，故该集合是开集；故答案为：①③．【点评】本题属于集合的新定义型问题，考查学生即时掌握信息，解决问题的能力．正确理解好集的定义是解决本题的关键二．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【分析】根据绝对值不等式和一元二次不等式的解法求出不等式的等价条件，结合充分条件和必要条件的定义进行判断即可．【解答】解：由|x﹣2|＜1得﹣1＜x﹣2＜1，得1＜x＜3由x2﹣x﹣6＜0得﹣2＜x＜3，即“|x﹣2|＜1”是“ｘ2﹣x﹣6＜0”的充分不必要条件，故选：A．【点评】本题主要考查充分条件和必要条件的判断，结合不等式的解法求出不等式的等价条件是解决本题的关键．14．已知函数f（x）=3x+x，g（x）=log3x+x，h（x）=sinx+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【分析】利用数形结合，画出函数的图象，判断函数的零点的大小即可．【解答】解：函数f（x）=3x+x，g（x）=log3x+x，h（x）=sinx+x的零点依次为x1，x2，x3，在坐标系中画出y=3x，y=log3x，y=sinx与y=﹣x的图象如图：可知x1＜0，x2＞0，x3=0，满足x1＜x3＜x2．故选：B．【点评】本题考查了函数的零点的判定理，数形结合的应用，属于基础题．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【分析】根据新定义运算法则“若x∈M，则∈M”求得集合M的所有元素，然后求其积即可．【解答】解：依题意，得当4∈M时，有=﹣∈M，从而=∈M，=4∈M，于是集合M的元素只有4，﹣，，所有元素之积等于4×（﹣）×=﹣1．故选：C．【点评】本题考查了元素与集合关系的判断．解题的关键是根据“若x∈M，则∈M”得到集合M的所有元素．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【分析】根据奇函数的性质和周期性可得f（x）关于（1，0）对称，求出x∈（﹣1，0]时，函数的解析式，再根据函数的周期性，即可得到函数y=f（x）的图象，再画出y=lg|x|的图象，由图象即可判断．【解答】解：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），可得f（x）为周期为2的奇函数，可得f（﹣x+2）=f（﹣x）=﹣f（x），即有f（x）的图象关于点（1，0）对称，故A错误；当x∈[0，1）时，f（x）=2x﹣1，由x∈[2，3]时，x﹣2∈[0，1]时，可得f（x﹣2）=f（x）=2x﹣2﹣1，故D错误；当x∈[﹣1，0）时，﹣x∈[0，1）时，f（﹣x）=2﹣x﹣1=﹣f（x），即f（x）=1﹣2﹣x，可得f（x）无最小值和最大值，故B错误；画出函数y=f（x）与y=lg|x|的图象，如图所示，结合图象可得函数f（x）无对称轴，f（x）的最大值与最小值之和为0，当x＞0时，y=f（x）与y=lg|x|有个交点，当x＜0y=f（x）与y=lg|x|有5个交点，故方程f（x）﹣lg|x|=0有10个实数根，故C正确．故选：C．【点评】本题考查了函数的奇偶性周期性，对称性，以及函数零点的问题，考查了转化能力和运算能力，属于中档题．三．解答题（共5小题，满分76分）17．（14分）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，且p∧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．【分析】（1）运用真值表判断命题真假即可；（2）运用充分必要条件的判断可解出．【解答】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∧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]．【点评】本题考查充分必要条件和命题真假的判断．18．（14分）已知函数y=f（x）为定义在（﹣∞，0）∪（0，+∞）上的奇函数，且当x＞0时（Ⅰ）试求f（﹣2）的值；（Ⅱ）指出f（x）的单调递增区间；（直接写出结论即可）；（Ⅲ）求出f（x）的零点．【分析】（Ⅰ）利用函数的奇偶性以及函数的解析式真假求解f（﹣2）的值；（Ⅱ）利用函数的奇偶性以及分段函数的解析式写出f（x）的单调递增区间；（Ⅲ）f（x）的零点转化为方程的根，求解即可．【解答】本题满分（12分）解：（I）由已知f（﹣2）=﹣f（2），2∈（0，3]，所以（II）函数y=f（x）为定义在（﹣∞，0）∪（0，+∞）上的奇函数，当x＞0时，，单调递增区间为（﹣∞，﹣3）和（3，+∞）（8分）（III）由，且0＜x≤3，解得又f（x）为奇函数，可得另一个零点为综上：f（x）的零点为和（12分）【点评】本题考查分段函数的应用幂函数的奇偶性以及函数值的求法，函数的零点的求法，考查计算能力．19．（14分）已知函数f（x）=|x﹣2|+|x+3|．（1）求不等式f（x）≤15的解集；（2）若﹣x2+a≤f（x）对x∈R恒成立，求a的取值范围．【分析】（1）化简函数为分段函数，然后求解不等式f（x）≤15的解集；（2）若﹣x2+a≤f（x）对x∈R恒成立，求出函数的最小值，即可求a的取值范围．【解答】解：（1）因为，所以当x＜﹣3时，由f（x）≤15得﹣8≤x＜﹣3；当﹣3≤x≤2时，由f（x）≤15得﹣3≤x＜2；当x＞2时，由f（x）≤15得﹣2＜x≤7．综上，f（x）≤15的解集为[﹣8，7]．（2）由﹣x2+a≤f（x）得a≤x2+f（x），因为f（x）≥|（x﹣2）﹣（x+3）|=5，当且仅当﹣3≤x≤2取等号，所以当﹣3≤x≤2时，f（x）取得最小值5，所以当x=0时，x2+f（x）取得最小值5，故a≤5，取a的取值范围为（﹣∞，5]．【点评】本题考查不等式的解法，函数恒成立条件的应用，考查转化思想以及计算能力．20．（16分）函数f（x）的定义域为D={x|x≠0}，且满足对于任意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．【分析】（1）令x1=x2=1，代入已知条件，即可求解得f（1）．（2）f（x）为偶函数．利用偶函数的定义证明即可．（3）求解f（4×4）=2，由（2）知，f（x）是偶函数，转化f（x﹣1）＜2?f（|x﹣1|）＜f（16）．然后求解即可．【解答】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）f（x）为偶函数．证明：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}．【点评】本题考查抽象函数的应用，函数的奇偶性以及函数的单调性的应用，考查计算能力．21．（18分）已知函数，a∈R．（1）若函数f（x）是奇函数，求实数a的值；（2）在（1）的条件下，判断函数y=f（x）与函数y=lg2x的图象公共点个数，并说明理由；（3）当x∈[1，2）时，函数y=f（2x）的图象始终在函数y=lg（4﹣2x）的图象上方，求实数a的取值范围．【分析】（1）运用奇函数的定义，以及对数的运算性质，结合恒成立思想解方程可得a 的值；（2）求得f（x）的定义域，要求方程解的个数，即求方程在定义域D上的解的个数．构造函数，运用函数零点存在定理，即可得到所求零点个数；（3）要使x∈[1，2）时，函数y=f（2x）的图象始终在函数y=lg（4﹣2x）的图象的上方，必须使在x∈[1，2）上恒成立，令t=2x，则t∈[2，4），上式整理得t2+（a﹣5）t+6﹣a＞0在t∈[2，4）恒成立．由参数分离和基本不等式可得最值，进而得到所求范围．【解答】解：（1）函数为奇函数，所以对于定义域内任意x，都有f（x）+f（﹣x）=0，即，∴，显然x≠1，由于奇函数定义域关于原点对称，所以必有x≠﹣1．上面等式左右两边同时乘以（x﹣1）（x+1）得[a（x﹣1）+2]?[a（x+1）﹣2]=x2﹣1，化简得（a2﹣1）x2﹣（a2﹣4a+3）=0，上式对定义域内任意x恒成立，所以必有，解得a=1；（2）由（1）知a=1，所以，即，由得x＜﹣1或x＞1，所以函数f（x）定义域D=（﹣∞，﹣1）∪（1，+∞），由题意，要求方程解的个数，即求方程在定义域D上的解的个数．令，显然F（x）在区间（﹣∞，﹣1）和（1，+∞）均单调递增，又，，且，，所以函数F（x）在区间和上各有一个零点，即方程在定义域D上有2个解，所以函数y=f（x）与函数y=lg2x的图象有2个公共点；（3）要使x∈[1，2）时，函数y=f（2x）的图象始终在函数y=lg（4﹣2x）的图象的上方，必须使在x∈[1，2）上恒成立，令t=2x，则t∈[2，4），上式整理得t2+（a﹣5）t+6﹣a＞0在t∈[2，4）恒成立．因为t2+（a﹣5）t+6﹣a＞0在t∈[2，4）恒成立．即（t﹣1）a＞﹣t2+5t﹣6，又1≤t﹣1＜3，所以得在t∈[2，4）恒成立，令u=t﹣1，则u∈[1，3），且t=u+1，所以，由基本不等式可知（当且仅当时，等号成立）即，所以，所以a的取值范围是．【点评】本题考查函数的性质和运用，主要是奇偶性的定义和性质，考查函数零点存在定理的应用以及参数分离、构造函数和基本不等式的运用，考查化简整理的运算能力，属于难题．。
2017-2018年上海市交大附中高三上第一次月考
（3）第 1 天的利润最大，为 1
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上海市奉贤中学2018—2019学年高三上学期第一次月考数学试题
绝密★启用前上海市奉贤中学2018—2019学年高三上学期第一次月考数学试题试卷副标题注意事项：1．答题前填写好自己的姓名、班级、考号等信息 2．请将答案正确填写在答题卡上第I 卷（选择题)请点击修改第I 卷的文字说明 一、单选题1．下列命题为真命题的是（ ）A ．若 ，则B ．若 ，则C ．若，则 D ．若 ，则2．设在ABC ∆中,角,A B C ，所对的边分别为,a b c ，, 若cos cos sin b C c B a A +=, 则ABC ∆的形状为 （ ） A ．锐角三角形B ．直角三角形C ．钝角三角形D ．不确定3．某学生对一些对数进行运算，如下图表格所示：现在发觉学生计算中恰好有两次地方出错，那么出错的数据是 （ ） A.()()3,8B.()()4,11C.()()1,3D.()()1,44．已知()i ai y f x x == ，1,22019i =⋅⋅⋅为2019个不同的幂函数，有下列命题：① 函数()()()122019y f x f x f x =++⋅⋅⋅+ 必过定点()1,2019； ② 函数()()()122019y f x f x f x =++⋅⋅⋅+可能过点()1,2018-；③ 若1220192a a a ++⋅⋅⋅+= ，则函数()()()122019y f x f x f x =⋅⋅⋅⋅⋅⋅为偶函数； ④ 对于任意的一组数1a 、2a 、…、2019a ，一定存在各不相同的1009个数1i 、2i 、…、{}2019122019i ∈⋅⋅⋅、、使得()()()122019i i i y f f f x =⋅⋅⋅在()0,∞+上为增函数.其中真命题的个数为（ ） A.1个 B.2个C.3个D.4个第II卷（非选择题)请点击修改第II卷的文字说明二、填空题5．已知集合{}02A x x=<<，()(){}110B x x x=-+>，则A B=______；6．不等式23xx-≥+的解集为______；7．方程14230x x+--=的解是 .8．已知实数x、y，命题“若0x>且0y>，则0x y+>”的逆否命题是______；9．函数()20.3log32y x x=-+的单调增区间是______10．已知()0,aπ∈，1sin cos5αα+=，则cos2=α______；11．若不等式23x t-<的解集为（m，n），则m n+=______；12．设()f x是定义在R上的奇函数，若当0x>时()2sin1f x x x=-+，则()f x的解析式为______；13．函数1()2sin(),[2,4]1f x x xxπ=-∈--的所有零点之和为．14．设P是一个数集，且至少含有两个数，若对任意a、b∈P，都有a+b、a-b、ab、ab ∈P（除数b≠0）则称P是一个数域，例如有理数集Q是数域，有下列命题：①数域必含有0，1两个数；②整数集是数域；③若有理数集Q⊆M，则数集M必为数域；④数域必为无限集．其中正确的命题的序号是．（填上你认为正确的命题的序号）15．已知x、y、z R+∈，230x y z-+=，且20y txz-≥恒成立，则实数t最大值是______；16．当x、y∈（0，1）时，{}1min8,8,8x x y y---的最大值是______.三、解答题17．设集合{}2320A x x x=++=，(){}210B x x m x m=+++=，{}21C x a x a =-≤≤（1）若B A ⊆，求实数m 的值；（2）若C A ⋂为空集，求实数a 的取值范围。
上海市2018-2019学年华一附中高三上学期数学第一次月考
上海市2018-2019学年华一附中高三上学期数学第一次月考一. 填空题1. 已知R 为实数集，2{|20}A x x x =-<，{|1}B x x =≥，则()A C B =R2. 函数3log (1)y x +的定义域为3. 不等式103xx -≥-的解为 4. 若函数()3sin(2)f x x ϕ=+，(0,)ϕπ∈为偶函数，则ϕ= 5. 已知幂函数()f x 过点(2,8)，1()f x -是它的反函数，则11()8f -= 6. 函数3cos()45x π-=，那么sin 2x = 7. 若“2230x x -->”是“x a <”的必要非充分条件，则a 的最大值为 8. 已知函数()2sin f x x ω=（0ω>）在区间[,]43ππ-上的最大值为2，则实数ω的最小 值为9. 函数log (2)1a y x =+-（0a >且1a ≠）图像恒过定点A ，若A 在直线10mx ny ++= 上，其中0mn >，则21m n+的最小值为 10. 已知函数()log a f x x =（0a >且1a ≠）满足(3)(4)f f >，若1()y f x -=是()y f x =的反函数，则关于x 的不等式11(1)1f x-->的解集是11. 已知()f x 是定义在R 上的偶函数，且在区间(,0)-∞上单调递增，若实数a 满足|1|(2)(a f f ->，则a 的取值范围是12. 对于函数()f x 和()g x ，设{|()0}x f x α∈=，{|()0}x g x β∈=，若存在α、β，使得||1αβ-≤，则称()f x 和()g x 互为“零点相邻函数”，若函数2()33x f x x -=+-与2()4g x x ax x =--+互为“零点相邻函数”，则实数a 的取值范围是二. 选择题13. 下列函数中，值域为R 的函数是（ ） A. 21y x =- B. 11x y x +=- C. 12x y -= D. lg(1)y x =- 14. 已知0ab ≠，则“1b a >”是“1ab<”成立的（ ） 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15. 已知两个不相等的实数a 、b 满足以下关系式：2sin cos 04a a πθθ⋅+⋅-=，2sin cos 04b b πθθ⋅+⋅-=，则连接2(,)A a a 、2(,)B b b 两点的直线与圆心在原点的单位圆的位置关系是（ ）A. 相离B. 相切C. 相交D. 不能确定16. 若函数|lg(||1)|||1()sin()||12x x f x a x x π->⎧⎪=⎨≤⎪⎩，关于x 的方程2()(1)()0f x a f x a -++=，给出 下列结论：① 存在实数a ，使得方程有5个不同的实根；② 不存在实数a ，使得方程有9个不同的实根；③ 存在实数a ，使得方程有10个不同的实根；④ 存在实数a ，使得方程有6个不同的实根；其中正确的个数是（ ）A. 4个B. 3个C. 2个D. 1个三. 解答题17. △ABC 中，A 、B 、C 对边分别是a 、b 、c ，1cos 23A =-，c =sin A C =. （1）求a 的值；（2）若角A 为锐角，求b 的值及△ABC 的面积.18. 已知()sin()cos()f x x x ωϕωϕ=+++（0ω>，0||2πϕ<<），(0)0f =，且函数()f x 图像上的任意两条对称轴之间距离的最小值是2π. （1）求()8f π的值和()f x 的单调增区间； （2）将函数()y f x =的图像向右平移6π个单位后，得到函数()y g x =的图像，求函数()g x 在[,]62ππ上的最值，并求取得最值时的x 的值.19. 已知函数131()log ()1axf x x -=-满足(2)1f -=，其中a 为实常数.（1）求a 的值，并判断函数()f x 的奇偶性；（2）若不等式1()()3x f x t >+在[2,3]x ∈上恒成立，求t 的取值范围.20. 设a ∈R ，函数()||3f x x a x x =⋅-+.（1）若3a =，求函数()f x 在区间[0,4]上的最大值；（2）若3a >，写出函数()f x 的单调区间（写出必要的过程，不必证明）；（3）若存在(3,6]a ∈，使得关于x 的方程()()f x t f a =⋅有三个不相等的实数解，求实数t 的取值范围.21. 对于函数()f x ，若在定义域内存在实数0x ，满足00()()f x f x -=-，则称()f x 为“M 类函数”.（1）已知函数()2cos()3f x x π=-，试判断()f x 是否为“M 类函数”？并说明理由；（2）设1()423x x f x m +=-⋅-是定义域R 上的“M 类函数”，求实数m 的取值范围；（3）若22log (2)3()23x mx x f x x ⎧-≥=⎨-<⎩为其定义域上的“M 类函数”，求实数m 取值范围.参考答案一. 填空题1. (0,1)2. (1,2]-3. [1,3)4. 2π 5.12 6. 725- 7. 1- 8. 329. 3+ 10. (0,1) 11. 13(,)2212. [3,4]二. 选择题13. D 14. A 15. C 16. B （①②④）三. 解答题17.（1）a =2）5b =，S =18.（1）()f x x =，()18f π=，增区间[,]44k k ππππ-++，减区间3[,]44k k ππππ++；（2）())3g x x π=-，max 5()()12g x g π==min ()()06g x g π==.19.（1）1a =-，奇函数；（2）参变分离，根据单调性求最值，109t <-. 20.（1）max ()(4)16f x f ==；（2）在3(,)2a +-∞递增，3[,]2a a +递减，(,)a +∞递增； （3）即3()()()2a f a t f a f +<⋅<，存在性问题，分离求最值，解得908t <<. 21.（1）是；（2）(0)0f ≤，∴1m ≥-；（3）(3)2f ≤，∴56m ≥.。
届高三数学（理）第一次月考模拟试卷及答案
届高三数学（理）第一次月考模拟试卷及答案2018届高三数学(理)第一次月考模拟试卷及答案高考数学知识覆盖面广，我们可以通过多做数学模拟试卷来扩展知识面!以下是店铺为你整理的2018届高三数学(理)第一次月考模拟试卷，希望能帮到你。
2018届高三数学(理)第一次月考模拟试卷题目一、选择题(本题共12道小题，每小题5分，共60分)1.已知全集U=R，A={x|x2﹣2x<0}，B={x|x≥1}，则A∪(∁UB)=( )A.(0，+∞)B.(﹣∞，1)C.(﹣∞，2)D.(0，1)2.已知集合A={1，2，3，4}，B={y|y=3x﹣2，x∈A}，则A∩B=()A.{1}B.{4}C.{1，3}D.{1，4}3.在△ABC中，“ >0”是“△ABC为锐角三角形”的( )A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件4.下列说法错误的是( )A.命题“若x2﹣4x+3=0，则x=3”的逆否命题是：“若x≠3，则x2﹣4x+3≠0”B.“x>1”是“|x|>0”的充分不必要条件C.若p且q为假命题，则p、q均为假命题D.命题p：“∃x∈R使得x2+x+1<0”，则¬p：“∀x∈R，均有x2+x+1≥0”5.已知0A.a2>2a>log2aB.2a>a2>log2aC.log2a>a2>2aD.2a>log2a>a26.函数y=loga(x+2)﹣1(a>0，a≠1)的图象恒过定点A，若点A在直线mx+ny+1=0上，其中m>0，n>0，则 + 的最小值为( )A.3+2B.3+2C.7D.117.已知f(x)是定义在R上的偶函数，在[0，+∞)上是增函数，若a=f(sin )，b=f(cos )，c=f(tan )，则( )A.a>b>cB.c>a>bC.b>a>cD.c>b>a8.若函数y=f(x)对x∈R满足f(x+2)=f(x)，且x∈[-1 ，1]时，f(x)=1﹣x2，g(x)= ，则函数h(x)=f(x)﹣g(x)在区间x∈[-5 ，11]内零点的个数为( ) A.8 B.10 C.12 D.149设f(x)是定义在R上的恒不为零的函数，对任意实数x，y∈R，都有f(x)•f(y)=f(x+y)，若a1= ，an=f(n)(n∈N*)，则数列{an}的前n 项和Sn的取值范围是( )A.[ ，2)B.[ ，2]C.[ ，1)D.[ ，1]10.如图所示，点P从点A处出发，按逆时针方向沿边长为a的正三角形ABC运动一周，O为ABC的中心，设点P走过的路程为x，△OAP的面积为f(x)(当A、O、P三点共线时，记面积为0)，则函数f(x)的图象大致为( )A . B.C. D.11.设函数f(x)=(x﹣a)|x﹣a|+b，a，b∈R，则下列叙述中，正确的序号是( )①对任意实数a，b，函数y=f(x)在R上是单调函数;②对任意实数a，b，函数y=f(x)在R上都不是单调函数;③对任意实数a，b，函数y=f(x)的图象都是中心对称图象;④存在实数a，b，使得函数y=f(x)的图象不是中心对称图象.A.①③B.②③C.①④D.③④12.已知函数，如在区间(1，+∞)上存在n(n≥2)个不同的数x1，x2，x3，…，xn，使得比值= =…= 成立，则n的取值集合是( )A.{2，3，4，5}B.{2，3}C.{2，3，5}D.{2，3，4}第II卷(非选择题)二、填空题(本题共4道小题，每小题5分，共20分)13.命题：“∃x∈R，x2﹣x﹣1<0”的否定是 .14.定义在R上的奇函数f(x)以2为周期，则f(1)= .15.设有两个命题，p：x的不等式ax>1(a>0，且a≠1)的解集是{x|x<0};q：函数y=lg(ax2﹣x+a)的定义域为R.如果p∨q为真命题，p∧q为假命题，则实数a的取值范围是 .16.在下列命题中①函数f(x)= 在定义域内为单调递减函数;②已知定义在R上周期为4的函数f(x)满足f(2﹣x)=f(2+x)，则f(x)一定为偶函数;③若f(x)为奇函数，则 f(x)dx=2 f(x)dx(a>0);④已知函数f(x)=ax3+bx2+cx+d(a≠0)，则a+b+c=0是f(x)有极值的充分不必要条件;⑤已知函数f(x)=x﹣sinx，若a+b>0，则f(a)+f(b)>0.其中正确命题的序号为 (写出所有正确命题的序号).三、解答题(本题共7道小题,第1题12分,第2题12分,第3题12分,第4题12分,第5题12分,第6题10分,第7题10分,共70分)17.已知集合A={x|x2﹣4x﹣5≤0}，函数y=ln(x2﹣4)的定义域为B.(Ⅰ)求A∩B;(Ⅱ)若C={x|x≤a﹣1}，且A∪(∁RB)⊆C，求实数a的取值范围.18.已知关于x的不等式ax2﹣3x+2≤0的解集为{x|1≤x≤b}.(1)求实数a，b的值;(2)解关于x的不等式： >0(c为常数).19.已知函数f(x)= 是定义在(﹣1，1)上的奇函数，且f( )= .(1)确定函数f(x)的解析式;(2)证明f(x)在(﹣1，1)上是增函数;(3)解不等式f(t﹣1)+f(t)<0.20.已知关于x的不等式x2﹣(a2+3a+2)x+3a(a2+2)<0(a∈R).(Ⅰ)解该不等式;(Ⅱ)定义区间(m，n)的长度为d=n﹣m，若a∈R，求该不等式解集表示的区间长度的最大值.21.设关于x的方程2x2﹣ax﹣2=0的两根分别为α、β(α<β)，函数(1)证明f(x)在区间(α，β)上是增函数;(2)当a为何值时，f(x)在区间[α，β]上的最大值与最小值之差最小.选做第22或23题，若两题均选做，只计第22题的分。
上海市2018-2019学年行知中学高三上学期数学第一次月考
上海市2018-2019学年行知中学高三上学期数学第一次月考一. 填空题1.函数lg(1)y x =-定义域为2. 已知角α的终边落到射线2y x =（0x ≤）上，求cos α=3. 函数()sin 1f x x x =++（x ∈R ），若()2f a =，则()f a -的值为4. 若1cos 23α=，(,0)2πα∈-，则sin α=5. 已知函数2()321x x f x =⋅+，则11()4f -= 6. 在△ABC 中，角A 、B 、C 所对的边分别为a 、b 、c ，若s i n :s i n :s i n 6:5:4AB C =，则最大角等于 7. 函数()arccos(sin )f x x =（433x ππ<<）的值域是 8. 集合11{|2,}42x A x x =≤≤∈R ，2{|210}B x x tx =-+≤，若A B ⊆，则实数t 的取值 范围是9. 已知()y f x =是定义在R 上的偶函数，且在(,0)-∞上单调递增，若实数a满足|1|(2)(a f f ->，则a 的取值范围是10. 设函数2018sin [0,1]()log (1,)xx f x x x π∈⎧=⎨∈+∞⎩，若满足()()()f a f b f c ==（a 、b 、c 互不相等），则a b c ++的取值范围是11. 设函数2()3f x x ax a =-++，()g x x a =-，若不存在0x ∈R ，使得0()0f x <与0()0g x <同时成立，则实数a 的取值范围是12. 在△ABC 中，3AC =，4AB =，5BC =，P 为角平分线AT 上一点，且在△ABC 内部，则P 到三边距离倒数之和的最小值为二. 选择题13. 对任意实数α、β，下列等式恒成立的是（ ） A. cos cos 2sinsin22αβαβαβ+-+=⋅B. cos cos 2cos cos 22αβαβαβ+--=⋅ C. 2sin cos sin()sin()αβαβαβ⋅=++- D. 2cos sin sin()cos()αβαβαβ⋅=++-14. 已知非零实数a 、b 满足a b >，则下列不等式中成立的是（ ）A. 22a b >B.11a b < C. 22a b ab > D. 22a b b a> 15. 设a 、b 为正数，考察如下两组条件的关系：α：对任意的1x >有1xax b x +>-都成立；β2>α是β的（ ） A. 充分非必要条件 B. 必要非充分条件 C. 充要条件 D. 非充分非必要条件16. 当函数的自变量取值区间与值域区间相同时，我们称这样的区间为该函数的保值区间， 函数的保值区间有(,]m -∞、[,]m n 、[,)n +∞三种形式，以下四个二次函数图像的对称轴是 直线l ，从图像可知，有三个保值区间的函数是（ ）A. B. C. D.三. 解答题17. 已知集合{||2|1}A x x =->，1{|2}2x B x x +=≥-，{|1}C x a x a =<<+. （1）求A B ；（2）若B C =∅，求实数a 的取值范围.18. 已知(sin ,1)m x =，(3cos ,cos 2)2An A x x =（0A >），函数()f x m n =⋅最大值为6. （1）求A ，并求函数()f x m n =⋅的最小正周期； （2）将函数()y f x =的图像向左平移12π个单位，得到函数()y g x =的图像，求()g x 在[0,]2π上的值域.19. 如图所示，ABCD 是一块长为7米的正方形铁皮，其中ATN 是一半径为6米的扇形，已知被腐蚀不能使用，其余部分完好可利用，工人师傅想在未被腐蚀部分截下一个有边落在BC 与CD 上的长方形铁皮PQCR ，其中P 是TN 上一点，设TAP θ∠=，长方形PQCR 的面积为S 平方米.（1）求S 关于θ的函数解析式；（2）设sin cos t θθ+=，求S 关于t 的表达式 及S 的最大值.20. 对于定义在区间[,]a b 的函数()f x ，定义：()min{()|}f x f t a t x ϕ=≤≤（[,]x a b ∈），()m a x {()|}f x f t a t xφ=≤≤（[,]x a b ∈），其中，min{()|}f t x D ∈表示函数()f x 在D上的最小值，max{()|}f t x D ∈表示函数()f x 在D 上的最大值. （1）若()cos f x x =，[0,]x π∈，试写出()f x ϕ、()f x φ的表达式；（2）设0a >且1a ≠，函数2()(3)1x x f x a a a =+-⋅-，1[,1]2x ∈，如果()f x φ与()f x 恰好为同一函数，求a 的取值范围.（3）若存在最小正整数k ，使得()()()f f x x k x a φϕ-≤-对任意的[,]x a b ∈成立，则称函数()f x 为[,]a b 上的“k 阶收缩函数”，已知函数2()f x x =，[1,4]x ∈-，试判断()f x 是否为[1,4]-上的“k 阶收缩函数”，如果是，求出对应的k ，如果不是，请说明理由.21. 已知函数2()log f x x =.（1）若()f x 的反函数是1()f x -，解方程：11(21)3()1f x f x --+=-；（2）设21(2)n n x f -=，是否存在*n ∈N ，使得等式22sin cos 1n n n x x x n ++=成立？ 若存在，求出n 的所有取值，如不存在，说明理由；（3）对于任意,,[,)a b c M ∈+∞，且a b c ≥≥，当a 、b 、c 能作为一个三角形的三边长时，()f a 、()f b 、()f c 也总能作为某个三角形的三边长，试探究M 的最小值.参考答案一. 填空题1. (1,2)2.3. 04.5. 06. 1arccos 87. 5[0,)6π8. 54t ≤-9. 13(,)22 10. (2,2019) 11. 3[3,]5- 12.二. 选择题13. C 14. D 15. A 16. B三. 解答题 17.（1）(,1)(2,)-∞+∞；（2）(,1][5,)a ∈-∞+∞.18.（1）6A =，T π=；（2）()6g x -≤≤.19.（1）4942(sin cos )36sin cos S θθθθ=-++（02πθ≤≤）；（2）2184231S t t =-+（[1t ∈），当t =S 取最大值为67-20.（1）()cos f x x ϕ=，[0,]x π∈，()1f x φ=，[0,]x π∈；（2）(0,1)(1,9]；（3）存在，4k =.21.（1）0，1-；（2）不存在；（3）2.。
上海市2018-2019学年市北高中高三上学期数学第一次月考
上海市2018-2019学年市北高中高三上学期数学第一次月考一. 填空题1. 设:12p x <<，:21x q >，则p 是q 成立的 条件2. 设集合{||1|2}A x x =-<，{|2,[0,2]}x B y u x ==∈，则AB = 3.函数y 的单调递减区间为4. 函数2()2f x x x =-，[0,3]x ∈的最大值为5. 不等式224x x -<的解为 6. 设1()f x -为()21x f x x =+的反函数，则1(2)f -= 7. 若函数62()3log 2a x x f x x x -+≤⎧=⎨+>⎩（0a >且1a ≠）的值域是[4,)+∞，则实数a 的取值 范围是 8. 若不等式2||10x a x a -+->对一切(1,2)x ∈恒成立，则实数a 的最大值为9. 若(1)1f =-，且1(2)1()f x f x +=-，则(2019)f = 10. 已知函数()x f x a b =+（0a >，1a ≠）的定义域和值域都是[1,0]-，则a b +=11. 设函数311()21x x x f x x -<⎧=⎨≥⎩，则满足()(())2f a f f a =的a 的取值范围是 12. 设函数21()ln(1||)1f x x x =+-+，则使得()(21)f x f x >-成立的x 的取值范围是二. 选择题13. 已知函数2()a x f x x+=（0a >），(0,)x b ∈，则下列判断正确的是（ ） A.当b >()f x的最小值为B.当0b <≤()f x的最小值为C.当0b <≤()f x 的最小值为2a b b+ D. 对任意的0b >，()f x的最小值为14. 不等式221x x +>+的解集是（ ） A. (,1)(0,1)-∞- B. (1,0)(1,)-+∞ C. (1,0)(0,1)- D. (,1)(1,)-∞-+∞15. 设集合{|||1,}A x x a x =-<∈R ，{|||2,}B x x b x =->∈R ，若A B ⊆，则实数a 、b 必满足（ ）A. ||3a b +≤B. ||3a b +≥C. ||3a b -≥D. ||3a b -≤16. 已知实数0a >，0b >，对于定义在R 上的函数()f x ，有下述命题：①“()f x 是奇函数”的充要条件是“函数()f x a -的图像关于点(,0)a 对称”；②“()f x 是偶函数”的充要条件是“函数()f x a -的图像关于直线x a =对称”；③“2a 是()f x 的一个周期”的充要条件是“对任意的x ∈R ，都有()()f x a f x -=-”； ④“函数()y f x a =-与()y f b x =-的图像关于y 轴对称”的充要条件是“a b =”； 其中正确命题的序号是（ ）A. ③④B. ②③C. ①④D. ①②三. 解答题17. 设2{|40}A x x x =+=，22{|2(1)10}B x x a x a =+++-=.（1）若AB B =，求a 的值；（2）若A B B =，求a 的取值范围.18. 设函数()|1|||f x x x a =-+-.（1）若1a =-，解不等式()3f x ≥；（2）如果对任意实数x ，()2f x ≥恒成立，求a 的取值范围.19. 工厂以x 千克/小时的速度匀速生产某种产品（生产条件要求110x ≤≤），每小时可获 得利润是3100(51)x x +-元.（1）要使生产该产品2小时获得的利润不低于3000元，求x 的取值范围；（2）要使生产900千克该产品获得利润最大，问：工厂该选取何种生产速度？并求出最大利润.20. 已知函数21()f x ax x=+，其中a 为常数. （1）根据a 的不同取值，判断函数()f x 的奇偶性，并说明理由；（2）若(1,3)a ∈，判断函数()f x 在[1,2]上的单调性，并说明理由.21. 设()f x 是定义域为R 的函数，对任意x ∈R ，都满足：(1)(1)f x f x +=-， (1)(1)f x f x -=+，且当[0,1]x ∈时，2()2f x x x =-.（1）请指出()f x 在区间[1,1]-上的奇偶性、单调区间、零点；（2）试证明()f x 是周期函数，并求其在区间[21,2]k k -（k ∈Z ）上的解析式；（3）方程()f x x a =+有三个不等根，求a 的取值范围.参考答案一. 填空题1. 充分不必要2. (1,3)-3. (,1-∞4. 35. (1,2)-6. 23- 7. (1,2] 8. 2 9.12 10. 32- 11. 2[,)3+∞ 12. 1(,1)3二. 选择题13. A 14. B 15. C 16. D三. 解答题17.（1）1a =；（2）(,1]{1}-∞-.18.（1）33(,][,)22-∞-+∞；（2）(,1][3,)a ∈-∞-+∞. 19.（1）310x ≤≤；（2）6x =时，max 457500y =元.20.（1）当0a =时，()f x 是奇函数；当0x ≠时，()f x 是非奇非偶函数；（2）单调递增.21.（1）偶函数，[0,1]上递减，[1,0]-上递增，零点0x =；（2）证明略，2()2f x x x =+；（3）1(2,2)4k k -（k ∈Z ）.。
2018-2019学年上海市长宁区复旦中学高一(下)第一次月考数学试卷(解析版)
2018-2019学年上海市长宁区复旦中学高一（下）第一次月考数学试卷一、选择题（本大题共5小题，共20.0分）1.已知sin10°=k，则sin70°=（）A. 1-k2B. 1-2k2C. 2k2-1D. 1+2k22.对于等式sin3x=sin2x+sin x，下列说法中正确的是（）A. 对于任意x∈R，等式都成立B. 对于任意x∈R，等式都不成立C. 存在无穷多个x∈R使等式成立D. 等式只对有限个x∈R成立3.已知α为锐角，且2tan（π-α）-3cos（+β）+5=0，tan（π+α）+6sin（π+β）=1，则sinα的值是（）A. B. C. D.4.设θ为第二象限角，若，则sinθ+cosθ=（）A. -1B. 1C.D.5.若，则sinα+cosα的值为（）.A. B. - C. D.二、填空题（本大题共9小题，共36.0分）6.-=______．7.已知角α的终边经过点（-3，4），则cosα= ______；cos2α=______．8.已知α为锐角，，则=______．9.已cosθ=，则co s2θ=______．10.已知，则的值为______．11.已知sinα-cosα=，则sin2α=______．12.已知，其中，则cosα= ______ ．13.已知α为第二象限角，化简= ______ ．14.已知tan（θ-π）=2，则sin2θ+sinθcosθ-2cos2θ= ______ ．三、解答题（本大题共7小题，共94.0分）15.求证=．16.已知（1）求tanα的值；（2）求的值．17.已知cosα=，cos（α-β）=，且0＜β＜α＜．求：（1）tan2α的值；（2）β的大小．18.已知．，其中α、β为锐角，且．（1）求cos（α-β）的值；（2）若，求cosα及cosβ的值．19.阅读下面材料：根据两角和与差的正弦公式，有sin（α+β）=sinαcosβ+cosαsinβ------①sin（α-β）=sinαcosβ-cosαsinβ------②由①+②得sin（α+β）+sin（α-β）=2sinαcosβ------③令α+β=A，α-β=B有α=，β=代入③得 sin A+sin B=2sin cos．类比上述推证方法，根据两角和与差的余弦公式，证明：cos A-cos B=-2sin sin．20.（1）在△ABC中，若2lgtan B=lgtan A+lgtan C，则B的取值范围是______ ．（2）求函数y=7-4sin x cosx+4cos2x-4cos4x的最大值______ ．21.若△ABC的三个内角A，B，C满足cos2A-cos2B=2sin2C，试判断△ABC的形状．（提示：如果需要，也可以直接利用19题阅读材料及结论）-------- 答案及其解析 --------1.答案：B解析：解：∵sin10°=k，∴sin70°=cos20°=1-2sin210°=1-2k2．故选B所求式子利用诱导公式化简，再利用二倍角的余弦函数公式变形，将sin10°的值代入计算即可求出值．此题考查了二倍角的余弦函数公式，以及诱导公式，熟练掌握公式是解本题的关键．2.答案：C解析：解：∵sin3x=sin2x•cos x+cos2x•sin x，要使sin3x=sin2x+sin x成立仅需要cos x=cos2x=1即可即x=2kπ，k∈Z故存在无穷多个x∈R使等式成立故选C由已知中等式sin3x=sin2x+sin x，根据两角和的正弦公式，我们可得当cos x=cos2x=1成立时，等式一定成立，解得x=2kπ，k∈Z，分析四个答案，即可得到正确的结论．本题考查的知识点是三角函数中的恒等变换应用，其中熟练掌握和、差、倍、半角公式及诱导公式、同角三角函数关系公式等，是解答本题的关键．3.答案：C解析：解：∵，tan（π+α）+6sin（π+β）-1=0∴-2tanα+3sinβ+5=0…①tanα-6sinβ-1=0…②①×2+②得tanα=3∵α为锐角，∴sinα=故选C．先根据诱导公式进行化简整理，然后求出tanα，最后根据同角三角函数关系求出sinα即可．本题主要考查了三角函数的化简求值，同时考查了诱导公式和同角三角函数关系，属于基础题．4.答案：C解析：解：∵θ为第二象限角，若＞0，∴θ+为第三象限角，∵tan（θ+）==，+=1，sin（θ+）＜0，求得sin（θ+）=-，故sinθ+cosθ=2sin（θ+）=-，故选：C．由题意可得θ+为第三象限角，利用同角三角函数的基本关系求得sin（θ+）的值，可得sinθ+cosθ=2sin（θ+）的值．本题主要考查两角和差的三角公式，同角三角函数的基本关系，属于基础题．5.答案：C解析：解析：解：∵,∴,∴sinα+cosα=,故选C．【分析】对已知条件利用差角的正弦及二倍角的余弦化简可得，，从而可求答案．本题主要考查了利用两角差的正弦及二倍角的余弦对三角函数式子进行化简、求值，考查了基本公式的简单运用，属于基础试题．6.答案：解析：【分析】此题考查了二倍角的余弦函数公式，以及特殊角的三角函数值，熟练掌握公式是解本题的关键，属于基础题．把所求的式子利用二倍角的余弦函数公式化简，再利用特殊角的三角函数值，即可得到所求式子的值．【解答】解：cos2-sin2=cos（2×）=cos=．故答案为．7.答案：-；-解析：解：∵角α的终边经过点（-3，4），则x=-3，y=4，r=|OP|=5，∴cosα==-cos2α=2cos2α-1=-，故答案为：-；-．由条件利用任意角的三角函数的定义，求得cosα的值，再利用二倍角的余弦公式求得cos2α 的值．本题主要考查任意角的三角函数的定义，二倍角的余弦公式的应用，属于基础题．8.答案：-3解析：解：由α为锐角，cosα=，得到sinα==，所以tanα=2，则tan（+α）===-3．故答案为：-3由α为锐角和cosα的值，利用同角三角函数间的基本关系求出sinα的值，进而求出tanα的值，然后把所求的式子利用两角和的正切函数公式及特殊角的三角函数值化简后，将tanα的值代入即可求出值．此题考查学生灵活运用同角三角函数间的基本关系化简求值，灵活运用两角和与差的正切函数公式及特殊角的三角函数值化简求值，是一道基础题．9.答案：-解析：【分析】本题主要考查二倍角的余弦公式的应用，属于基础题．直接利用二倍角的余弦公式可得cos2θ=2cos2θ-1，运算求得结果．【解答】解：由二倍角的余弦公式可得cos2θ=2cos2θ-1=2×-1=-，故答案为-．10.答案：解析：解：∵，∴=2，解得tan x=；∴tan2x===∴==故答案为：．先利用两角和的正切公式求得tan x的值，从而求得tan2x，即可求得．本题考查了二倍角的正切与两角和的正切公式，体现了方程思想，是个基础题．11.答案：解析：解：由sinα-cosα=，两边平方可得：sin2α+cos2α-2sinαcosα=，化为1-sin2α=，则sin2α=．故答案为：．由sinα-cosα=，两边平方，再利用同角三角函数基本关系式、倍角公式即可得出．本题考查了同角三角函数基本关系式、倍角公式在三角函数化简求值中的应用，考查了转化思想，属于基础题．12.答案：解析：解：∵，∴，又，∴．则cosα=cos[（）-]=cos（）cos+sin（）sin=．故答案为：．由α的范围求得的范围，由平方关系结合已知求得，再由cosα=cos[（）-]展开两角差的余弦得答案．本题考查已知角的三角函数值求未知角的三角函数值，考查了两角和与差的三角函数，关键是“拆角与配角”思想的应用，是中档题．13.答案：-1解析：【分析】本题考查了诱导公式，以及三角函数的化简求值，熟练掌握诱导公式是解本题的关键，属于基础题．原式利用诱导公式化简，整理化简得到结果．【解答】解：∵α为第二象限角，∴sinα＞0，cosα＜0，则原式===-=-1．故答案为-1.14.答案：解析：解：∵tan（θ-π）=2=tanθ，则sin2θ+sinθcosθ-2cos2θ====，先求得tanθ的值，再利用同角三角函数的基本关系，求得要求式子的值．本题主要考查同角三角函数的基本关系的应用，属于基础题．15.答案：解：由====∴左边等于右边．解析：运用同角的基本关系式和诱导公式对左边化简，弦化切思想，即可化简到右边．本题考查三角函数的化简和证明，考查同角的基本关系式的运用，考查运算能力，属于基础题．16.答案：解：（1）∵，∴tanα=-tanα+1（2）法一：由（1）知：，∴或当，时，原式=当，时，原式=综上：原式=法二：原式分子分母同除以cos2α得：原式==解析：（1）直接弦化切，即可求tanα的值；（2）法一：求出sinα，cosα，分类讨论求的值．法二：原式分子分母同除以cos2α，弦化切，即可求的值．本题考查同角三角函数关系，考查学生的转化能力，属于中档题．17.答案：解：，．，．因为cos（α-β）=，所以sin（α-β）=，所以cosβ=cos[α-（α-β）]=cosαcos（α-β）+sinαsin（α-β）=，所以β=．解析：（1）由cosα=，求出sinα，tanα，再求tan2α的值；（2）利用cosβ=cos[α-（α-β）]=cosαcos（α-β）+sinαsin（α-β），可求β的大小．本题考查同角三角函数间的基本关系，两角和与差的余弦函数，利用cosβ=cos[α-（α-β）]=cosαcos（α-β）+sinαsin（α-β）是关键．18.答案：解：（1）∵，，，∴，平方整理，得，解得因此，；（2）∵，∴．∵α为锐角，∴，又∵α-β∈（-，），∴，①当时，．②当时，cosβ=cos[α-（α-β）]=cosα•cos（α-β）+sinα•sin（α-β）=0．又∵β为锐角，∴cosβ=0不符合题意，舍去．因此可得cosβ的值为．解析：（1）根据向量模的公式建立关于α、β的等式，利用同角三角函数的关系算出，进而可得cos（α-β）的值；（2）利用二倍角的余弦公式与同角三角函数的商数关系，算出，从cosβ的值．本题着重考查了向量模的计算公式、两角和与差的余弦公式、同角三角函数的基本关系与二倍角的三角函数公式等知识，属于中档题．19.答案：证明：因为cos（α+β）=cosαcosβ-sinαsinβ，------①cos（α-β）=cosαcosβ+sinαsinβ②①-②得cos（α+β）-cos（α-β）=-2sinαsinβ③…令α+β=A，α-β=B有α=，β=，代入③得cos A-cos B=-2sin sin．解析：通过两角和与差的余弦公式，令α+β=A，α-β=B有α=，β=，即可证明结果．本小题主要考查类比推理，考查两角和与差三角函数公式、二倍角公式、三角函数的恒等变换等基础知识，考查推理论证能力，运算求解能力，考查化归与转化思想等．20.答案：[，）；10解析：解：（1）由题意，得tan2B=tan A tan C，∵tan B=-tan（A+C）=，∴tan B=，∴tan3B-tan B=tan A+tan C≥2=2tan B，∴tan3B≥3tan B，tan B＞0∴tan B≥，∴B∈[，），（2）∵f（x）=7-4sin x cosx+4cos2x-4cos4x=7-2sin2x+4cos2x•sin2x=7-2sin2x+sin22x=（sin2x-1）2+6．当sin2x=-1时，即2x=2kπ-时，即x=kπ-时，k∈Z时，f（x）有最大值．∴f（x）max=10，故答案为：（1）[，），（2）10（1）通过对数的基本运算，推出三角形的角的关系，利用两角和的正切以及三角形的内角和，求出tan B的范围，即可得到B的范围．（2）利用正弦函数的二倍角公式将f（x）=7-4sin x cosx+4cos2x-4cos4x化为：f（x）═（sin2x-1）2+6，即可得到答案本题考查三角函数中的恒等变换的应用，三角函数的化简求值，着重考察正弦函数的二倍角公式及正弦函数的性质，突出二次函数的配方法的考察，属于中档题．21.答案：解：由二倍角公式，cos2A-cos2B=2sin2C可化为1-2sin2A-1+2sin2B=2sin2C，…（2分）即sin2A+sin2C=sin2B．…（3分）设△ABC的三个内角A，B，C所对的边分别为a，b，c，由正弦定理可得a2+c2=b2．…（5分）根据勾股定理的逆定理知△ABC为直角三角形．…（6分）解析：利用二倍角公式以及正弦定理、勾股定理，即可判断三角形的形状．本小题主要考查二倍角公式、三角函数的恒等变换，勾股定理等基础知识，考查推理论证能力，运算求解能力，考查化归与转化思想，属于基础题．第11页，共11页。
2019-2020学年上海复旦附中高三上学期第一次月考数学试题
2019-2020学年上海复旦附中高三上学期第一次月考数学试题一、填空题1. 若集合，集合，则 ．2.—个几何体的主视图、左视图、俯视图都是以为半径的圆，则该几何体的体积是 ．3.已知是虚数单位，则的平方根是 ．4.函数的反函数是 ．5.设满足约束条件，则的最小值是 ．6.如图，四个棱长为1的正方体排成一个正四棱柱，是一条侧棱，是上、下底面上其余十六个点，则的不同值的个数为 ．7.数列满足，其前项和记为，若，那么．8.若是展开式中项的系数，则 ．9.设函数，其中，若，且的最小正周期大于，则 ．10.已知函数，设，若关于的不等式在上恒成立，则的{}23A x x =-<30x B xx -⎧⎫=>⎨⎬⎩⎭A B ⋃=a i 2-()()210f x x x =+<x y 、2330233030x y x y y +-≤⎧⎪-+≥⎨⎪+≥⎩2z x y =+AB ()1,2,,16i P i =()1,2,,16i AB AP i ⋅={}n a ()1213,5n n n a a a n a --=-≥=n n S 89S =100S =n a ()()*2,2,nx n N n x R +∈≥∈2x 2323222lim nx na a a →∞⎛⎫+++= ⎪⎝⎭()()2sin ,f x x x R ωϕ=+∈0,ωϕπ><5112,088f f ππ⎛⎫⎛⎫==⎪ ⎪⎝⎭⎝⎭()f x 2πϕ=()2,12,1x x f x x x x ⎧+<⎪=⎨+≥⎪⎩a R ∈x ()2x f x a ≥+R a取值范围是 ．11.函数绕原点逆时针旋转，每旋转得到一个新的曲线，旋转一周共得到24条曲线（不包括未旋转时的曲线），请问从中任选其二，均不是函数图像的概率是 ． 12.已知两正实数，满足，则的最大值为 ．二、选择题13. 若为实数，则成立的一个充分不必要条件是（ ）A. B. C. D. 14. 已知是空间两条直线，是平面，以下结论正确的是（ ）A. 如果，则一定有；B. 如果，则一定有；C. 如果，则一定有；D. 如果，则一定有。
2019-2020学年复旦附中高三上一次月考
2019-2020学年复旦附中高三上一次月考Ⅱ. Grammar and VocabularySection ADirections: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.Some people sit outside for hours without getting bitten by mosquitoes, but it always seems like you (21) ________ (bite) alive within minutes of stepping outdoors.(22) ________ this is you, you’re not alone. According to Smithsonian Magazine, around 20 percent of people in the world are especially tasty to mosquitoes. What about these people making mosquitoes’ mouths water?A popular myth claims that mosquitoes prefer certain blood types, but the fact is that they simply can’t tell what your blood type is from a faraway place. Jonathon Day, a professor of medical entomology（昆虫学）at the University of Florida in the US, told NBC it’s not complicated. “The two most important reasons a mosquito (23) ________ (attract) to you have to do with sight and smell.”Mosquitoes are especially active in the late afternoon. While flying along, they use their sense of smell to find possible targets. They find victims by smelling the carbon dioxide (CO2) breathed out my humans and animals. That’s (24) ________ you commonly find them in crowded streets and parks.Joop van Loon, an entomologist at Wageningen University in the Netherlands, told Live Science, “Mosquitoes start orienting (25) ________ to carbon dioxide and keep flying upwind (26) ________ they sense higher concentrations.”As a result, people who simply exhale more of the gas over time - generally, larger people - (27) ________ (show) to attract more mosquitoes than others. “This is why kids don’t get bitten as much ... as adults,” US professor Ted Rosen told Science Alert.This love for CO2 can also put pregnant women at (28) ________ (increase) risk for mosquito bites, as they tend to exhale 21 percent more CO2 than people of the same age and size who aren’t pregnant.In addition to carbon dioxide, the color of the clothes you wear also plays a role in attracting mosquitoes.According to Live Science, mosquitoes can lock (29) ________ targets from up to 50 meters away. At this distance, what we wear has huge effect. Due to their vision, people wearing dark colors are more likely to become targets.(30) ________ (sting) by mosquitoes is annoying, but don’t worry. Some simple tips can help ward them off. Scientists recommend that we use insect repellent and wear light-colored clothing.Section BDirections: After reading the passage below, fill in each blank with a proper word given in the box. Each word can be used only once. Note that there is one more word than you need.The U. S. dollar was supposed to be at the end of its rope. Kicking the bucket. A dying symbol of a dying empire. Well, maybe not. The dollar continues to __31__ gloom-and-doom predictions. After a swoon last year, the dollar is again enjoying a major ___32___. The U. S. dollar index, which measures the dollar’s value against other major currencies, is just off an eight-month high.The main reason behind the dollar’s recovery is actually no real surprise at all. There is no ___33___ able to replace the dollar as the world’s No. 1 currency. What makes currencies so fascinating is that their perceived value is always relative to other currencies. Sure, the U. S. budget deficit（赤字）is ___34___, the government’s debt is increasing, and Wall Street is still repairing itself. But the dollar remains the prettiest of a flick of ugly ducklings. Is any other major industrialized economy ___35___ better off than the U. S. ?Not really. Just about the ___36___ developed world is suffering with the same problems. That’s why when investors get nervous, they still rush to the good old dollar. The dollar wins because no one else is really in the game.The euro has been exposed as a ___37___. Only a few month ago, economists tryly believed the euro could ___38___ the dollar as the top reserve currency. Now experts are questioning if the euro has a future at all. The Greek debt crisis has ___39___ that the euro is only as strong as its weakest link.Maybe over the next 20 or 30 years, the dollar will slowly lose the ___40___ status it holds today. That process, however, could well be driven by the appearance of new rivals.Ⅱ. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Culture influences an individual’s health beliefs, behaviors, activities and medical treatment outcomes.___41____ the significant influence of culture upon health and related outcomes, health care ___42___ should be culturally competent in order to provide ___43___ health care to patients. Cultural competency means considering many options and being more careful about making judgments. For example, scars and bruises that suggest abuse in western culture could ___44____ be symbols of accepted healing methods or sacred rituals. ____45___, different parts of the body are considered sacred in different cultures.Culture competency in health-care ___46___ four major challenges for providers. The first is the straightforward challenge of recognizing clinical ___47___ among people of different ethnic and racial groups, e. g., higher __48__ of hypertension in African Americans and of diabetes in certain Native American groups.The second, and far more complicated, challenge is ___49___. This deals with everything from the need for interpreters to nuances of words in various languages. Many patients, even in western cultures, are __50__ to talk with their doctors about such personal matters as sexual activity or chemical use. How do we overcome this __51__ among more restricted cultures?The third challenge is ethics. ____52___ western medicine is among the best in the world, we do not have all the ___53___. Respect for the belief systems of others and the effects of those beliefs on well-being are critically important to competent care.The final challenge involves ___54__. For some patients, authority figures are immediately mistrusted, sometimes for good reason. Having seen or been victims of atrocities（暴行）at the hands of authorities in their homelands, many people are as ___55____ of caregivers themselves as they are of the care.41. A. Because of B. Regardless of C. Thanks to D. In regard to42. A. professors B. signals C. professionals D. assistants.43. A. optimum B. opposite C. optimistic D. optional44. A. truly B. surely C. actually D. really45. A. In fact B. In addition C. In general D. In theory46. A. provides B. reduces C. contends D. holds47. A. similarities B. differences C. traditions D. experiences48. A. danger B. threat C. risk D. problem49. A. contact B. communication C. comprehension D. translation50. ready B. hard C. reluctant D. impossible51. A. challenge B. fault C. result D. phenomenon52. A. When B. While C. As D. Unless53. A. diagnoses B. medicines C. reputation D. trust54. A. loyalty B. honor C. reputation D. trust55. A. conscious B. careful C. worried D. warySection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements . For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)Plastic-Eating WormsHumans produce more than 300 million tons of plastic every year. Almost half of that winds up in landfills （垃圾填埋场）,and up to 12 million tons pollute the oceans. So far there is no effective way to get rid of it, but a new study suggests an answer may lie in the stomachs of some hungry worms.Researchers in Spain and England recently found that the worms of the greater wax moth can break down polyethylene（聚乙烯）, which accounts for 40% of plastics. The team left 100 was worms on a commercial polyethylene shopping bag for 12 hours, and the worms consumed and broke down about 92 milligrams, or almost 3% of it. to confirm that the worms’ chewing alone was not responsible for the polyethylene breakdown, the researchers made some worms into paste（糊状物）and applied it to plastic films. 14 hours later the films had lost 13% of their mass - apparently broken down by enzymes（酶）from the worms’ stomachs. Their findings were published in Current Biology in 2017.Federica Bertocchini, co-author of the study, says the worms’ ability to break down their everyday food - beeswax - also allows them to break down plastic. “Wax is a complex mixture, but the basic bond in polyethylene, the carbon-carbon, is there as well,” she explains. “The wax worm evolved a method or system to break this bond.”Jennifer DeBruyn, a microbiologist at the University of Tennessee, who was not involved in the study, says it is not surprising that such worms can break down polyethylene. But compared with previous studies, she finds the speed of breaking down in this one exciting. The next step, DeBruyn says, will be to identify the cause of the breakdown. Is it an enzyme produced by the worm itself or by its gut microbes（肠道微生物）?Bertocchini agrees and hopes her team’s findings might one day help employ the enzyme to break down plastics in landfills. But she expects using the chemical in some kind of industrial process -- not simply “ millions of worms thrown on top of the plastic.”56. What can we learn about the worms in the study?A. They take plastics ads their everyday food.B. They are newly evolved creatures.C. They can consume plastics.D. They wind up in landfills.57. According to Jennifer DeBruyn, the next step of the study is to ________.A. identify other means of the breakdownB. find out the source of the enzymeC. confirm the research findingsD. increase the breakdown speed58. It can be inferred from the last paragraph that the chemical might ________.A. help to raise wormsB. help make plastic bagsC. be used to clean the oceansD. be produced in factories in future59. What is the main purpose of the passage?A. To explain a study method on worms.B. To introduce the diet of a special worm.C. To present a way to break down plastics.D. To propose new means to keep eco-balance.(B)Your True StoriesOn The WingMy husband had passed tragically and unexpectedly the night before. I returned home the next morning with my sister-in-law, my emotional support. We sat in the upstairs loft, sharing stories about a man who’d left us too young. I glanced out the window and noticed a woodpecker on the roof. It appeared to be watching us. A member of a species rarely seen here, the bird sat for almost 20 minutes as we reminisced. I affectionately named it after my late husband. It has been five years since he passed, and a woodpecker continues to appear at my weakest moments.Shannon Rozewicz Like Son, Like FatherCaught in a sudden downpour on the last day of a bike-packing trip, I ducked into the lobby of a nearbysupermarket for cover. As I waited out the storm with my bike and gear, a teenage boy invited me to spend the night with his family. I gladly accepted, and he went to find his parents. While I waited, an older man made me the same offer. I thanked him and said I already had a place to stay. Shortly after, the boy returned with his parents. The man who had approached me was his father.Philip Wood Color Me ProudWhen my granddaughter Bethany was four years old, she visited my home for a few days. I gave her some crayons and pictures for coloring. When I looked down, I saw she had used a crayon to draw purple marks all over her legs. “Bethany, what are you doing?” I asked. “Why, Grandma,” she said, “you have such pretty purple lines up and down your legs, and I wanted mine to look just like yours.” Since then, I have worn my varicose vein with pride, and they got prettier each year.Margie Anderson60. From the stories, we can learn that ________.A. The woodpecker is Shannon’s emotional supportB. Shannon has been crazy for the death of her husbandC. Purple is Granddaughter’s favorite colorD. Grandma used to be proud of her varicose vein61. When Philip Wood says “Like Son, Like Father”, he means the father and the son ________.A. resemble each other in many waysB. are both helpful and considerateC. like to socialize with strangersD. are fond of accommodating strangers62. What is the general tone of the three stores ?A. Warm and optimisticB. Sad but positiveC. Humorous and ironicD. Hopeful and idealistic(C)①What does it say about the future of meat when the country’s largest processor of chicken, pork, and beef buys a stake（股份）in a start-up that aims to “perfectly replace animal protein with plant protein”?②Tyson Foods announced this week that it purchased a 5 percent stake in Beyond Meat, the Southern California-based food-tech start-up that made headlines earlier this year with its veggie burger that reportedly cooks and tastes like real beef.③To be sure, Beyond Meat’s meatless creations have yet to take the countryby storm. Although the 100 percent plant-based burgers have achieved plenty of positive press since they appeared for the first time in May, so far they’re only available at Whole Foods stores in seven states. Even though the company’s “chicken” strips, “beef” pies, and meatless frozen dinners are available nationwide, Beyond Meat is hardlya household name.④That may be what makes the news of Tyson’s investment all the more noteworthy. While the two companies declined to give details about the deal, it’s doubtful that Tyson’s 5 percent stake made much of dent（凹陷）in the meat giant’s coffers（金库）. The company posted $41.4 billion is sales last year; prior to the deal with Tyson, Beyond Meat had reportedly raised $64 million in project capital funding -- about what Tyson earns before lunch on any given day.⑤Tyson is doing pretty great. The company reported record third-quarterearning per share is August and says that it expects overall meat production to increase 2 to 3 percent during the next financial year. But like a big oil company shelling out cash to invest in wind power, Tyson’s toe-in-the-water move to team up with a start-up devoted to bringing more plant-based protein to American dinner tables seems to suggest the meat industry is starting to see which way the winds are blowing.⑥Sales of plant-based protein, which totaled an estimated $5 billion lastyear,continue to pale compared with the market for meat in America-- but vegetarian alternatives to meat are booming, with sales growing at more than double the rate for good products overall. The steady drumbeat of news about the negative health impacts, environmental problems, and animal welfare concerns associated with meat consumption appears to be sinking in. According to a survey released in April, more than half of Americans surveyed said they plan to eat more plant-based food in the coming year.63. Beyond Meat’s veggie burger made headlines probably because ________.A. it makes perfect use of animal proteinB. it uses high tech in the making processC. it tastes as good as a genuine beef burgerD. it represents the diet trend in South California64. Which of the following statements is TRUE regarding the state of Beyond Meat?A. It is the creator of the country’s first 100 percent plant-based burgers.B. It has been well received as its products are available nationwide.C. It is far from being a match to real food processing giants like Tyson.D. It provides high-quality dining experience in selected Whole Foods stores.65. What does the pale in paragraph 6 mean?A. seem unimportantB. seem whiteC. seem weakD. seem faint66. What does the passage mainly talk about?A. Meat will still take over the marker in spite of other alternatives.B. A major American meat company is betting on plant-based protein.C. Tyson and Beyond Meat work together to build a global meat giant.D. Plants have been found to contain protein that does more good to human being.Section CDirections: Read the passage carefully. Fill in each blanks with a proper sentence given in the box. Each sentences can used only once. Note that there are two more sentences than you need.Write and journalist Cristina Odone aroused widespread anger by suggesting that her daughter was being pressured to take science for graduation exams and this was unreasonable for a child with a literary bent. She even claimed that “ ... this focus on STEM [Science, Technology, Engineering and Mathematics] subjects sends a message that makes her and me uncomfortable: doing a man’s work is more impressive than doing a woman’s.”Like many others, ______67________ Taking science to age 16 should simply be seen as part of obtaining a well-rounded education. Furthermore, identifying STEM as a man’s subject leads in part to our serious lack of variety in the scientific workforce. Meanwhile, many male authors and poets might be surprised to learn that literature is “woman’s work”.Novelist Lucy Ellman once wrote, “The purpose of artists is to ask the right questions, even if we don’t find the answers, whereas the aim of science is to prove some silly points.” But proving some silly points might saveyour life, light your home, allow you to surf the web or visit your relatives living far away. _____68______ However, having said that, I am not trying to denigrate the work of the humanities. I do not see this as an either/or situation but it’s all part of being human. I admire and appreciate those who try to express things hard to be described in words, but is simply isn’t my strength. I may wholeheartedly believe that science is vital but that doesn’t mean I think the humanities (or indeed the social sciences) are not. _______69______ That statement is not equal to saying that the humanities should not be properly funded. Somehow, we are constantly being put in opposition, a divide that is damaging to both scientists and non-scientists.______70_______ I would say it is exactly because I am human. I sit her typing listening to a Schubert piano trio. I have been reading EP Thompson’s The Making of the English Working Class to try to understand how our society was and is the shape it is. Scientists may be capable of dealing with the ethic of their work, but they cannot and should not answer the question of whether we should do this in isolation.Ⅱ. Summary WritingDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your won words as far as possible.Where are the bees?Bees are essential to the production of food we eat. Bees make honey, but they also pollinate large areas of crops, such as strawberries, apples and onions. About a third of the food we eat is a result of pollination of the bees. Unfortunately, bees have been disappearing at an alarming rate.In 2006, bee keepers started reporting about something called Colony Collapse Disaster (CCD). The main sign of CCD is the loss of adult honey bees from a hive. In October of 2006, some beekeepers reported that they had lost between 30 and 90 percent of their hives.There were many theories of the disappearance of the bees. But the most convincing one has to do with pesticides and lifestyles of bees today. Nowadays, beekeepers get most of their income not from producing honey but from renting bees to pollinate plants. This means that the life of the typical bee now consists of traveling all around the country to pollinate crops as the seasons change. That means a lot of traveling on trucks, which is very stressful to bees. It is not unusual for up to 30% of the hive to die during transport due to stress. In addition, bees that spend most of their time locked up on trucks are not exposed to what usually live on. Instead, they live on a sweet liquid from corn, usually polluted with pesticides.No one knows from sure the exact reason for the disappearance of bees, but losing bees is very costly forthe economy. The bee pollination services are worth over $8 billion a year. With no bees, pollination will have to be done by hand, which would have effects on the quality of food and increased food prices. We hear a lot about big environmental disasters almost every day. But one of the biggest may just be the loss of that tiny flying insect.Ⅱ. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.72. 我们无意发动贸易站，但同时也为之做好了充分准备。
2019-2020复旦附中(浦东)高三上英语一次月考
Ⅱ. Grammar and V ocabularySection ADirections: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, used one word that best fits each blank.Wasting Food is Still a Big ProblemChina is famous for its delicious food, and there are so many different dishes to try here. But many of us take this granted and throw away the extra food we don’t eat.According to a report released by the United Nations, there were 815 million undernourished （营养不良的）people in the world in 2016, more than one (21) _______ every 10 on Earth. Yet, 1.3 billion tons of food (22) _______ (waste) annually around the world, which is about one-third of all food produced each year, according to the UN’s Food and Agriculture Organization.(23) _______ (get) a better understanding of different countries’ food waste problems, the Economist intelligence Unit recently surveyed 34 nations according to their food system sustainability（持续性）.According to the report, (24) _______ was released this month, France topped the list of food sustainability, followed by Japan and Germany.The performance of these countries is largely related to their policies to deal with food waste. For example, France was the first country to pass laws to ban food waste. In France, it’s illegal for supermarkets to throw away unsold food, and French restaurants (25) _______ provide doggy bags for people’s leftovers.Germany is trying to deal with the problem by reforming expiration dates（保质期）. “We found in our study that many people believed they should throw away products (26) _______ _______ _______ the “best before’ date has expired,” Martin Kranert, chair of Stuttgart University’s waste management department, told Deutsche Welle. “This is not at all case, and such a persisting lack of knowledge is the first thing (27) _______ has to change.”Some countries are still lagging behind when it comes to (28) _______ (prevent) food waste, however. For example, the Untied Arab Emirates wastes the most food, with each of its citizens (29) _______ (throw) away about 1,000 kilograms every year on average.China has been working hard to reduce food waste. In early 2013, the country’s Clean YourPlate Campaign was Launched, urging people not to waste anything on their plate.“(30) _______ consumers are more conscious of the amount of food waste they generate and everyone plays a role in minimizing their personal food waste, then we can do better as a community,” Joyce Chan from the Foodlink Foundation, a Hong Kong charity dedicated to reducing food waste, told China Daily.Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.Thousands of years ago our ancestors established the winding and far-reaching Silk Road, which __31__ and elevated China to be a nation of most outstanding prosperity and advanced civilization across the globe. Now, inheriting her brilliant commercial, cultural and even __32__ heritage, we, as the new generation witnessing the historic turning point of China’s amazing transformation, are benefiting from its enormous achievements over the past decades and more importantly, are being expected to __33__ our own historical responsibility to dedicate ourselves to making our country an even better place, not just ourselves, but for the whole nation and beyond I Why does it have to be us? Why the initiative of One Belt and One Road is of essential importance? My answers are as follows.As is known to all, China plays an increasingly critical role in the current international community. We __34__ the largest population in the world; we have superb geographically diverse and preferable locations for worldwide trading and business __35__ communication and cooperation. And currently we pride ourselves on being the second biggest economy while still feeling __36__ to build a society featuring a more balanced and equitable development to meet people’s growing desire for a more harmonious and prosperous development.The current reality is that a considerable number of neighboring countries __37__ the precious natural and intellectual resources, which is urgently needed in the global market. If we want to step forward and see the prospect of a better-balanced and more-developed country, we are__38__ to build closer connections with our neighbors and prepare ourselves to be more competitive on the global arena and more cooperative with the neighboring countries. Only through the farseeing initiative of the one Belt and One Road can we achieve the magnificent goals and thus yield a win-win even multi-win vision.As president Xi has pointed out: “History has shown that civilization thrives with openness and nations prosper through exchange.” Now it is our __39__ to take on the historical challenges and responsibilities, join in this __40__ and rejuvenating（充满活力）tide, seize the historic opportunity and carry out the One Belt and One Road initiative in a fully-fledged（飞翔的）manner that connects us with the world.Ⅲ. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.AI automates repetitive learning and discovery discovery through data. But AI is different from hardware-driven, robotic automation. Instead of automating __41__ tasks, AI performs frequent, high-volume, computerized tasks reliably and without tiredness. For this type of automation, human __42__ is still essential to set up the system and ask the right questions.AI adds intelligence to existing products. In most cases, AI will not be sold as an individual application. Rather, products you already use will be improved with AI __43__, much like Siri was added as a feature to a new generation of Apple products. Automation, conversational platforms and smart machines can be combined with large amounts of data to improve many technologies at home and in the workplace, from security __44__ to investment analysis.AI adapts through progressive learning algorithms（演算法）to let the data do the programming. AI finds structure and __45__ in data so that the algorithm acquires a skill: The algorithm becomes a classifier or a predicator. So, just as the algorithm can teach itself how to play chess, it can teach itself what product to recommend next online. And the models adapt when given new data. Back propagation（传播、扩展）is an AI __46__ that allows the model to adjust, through training and added data, when the first answer is not quite right.AI __47__ more and deeper data using neural（神经）networks that have many hidden layers. Building a cheating detection system with five hidden layers was almost impossible a few years ago. All that has changed with incredible computer power and big data. You need lots of data to __48__ deep learning models because they learn directly from the data. The more data you can __49__ them, the more accurate they become.AI achieves incredible accuracy through deep neural networks - which was __50__ impossible. For example, your __51__ with Alexa, Google Search and Google Photos are all based on deep learning - and they keep getting more accurate the more we use them. In the medical field, AI techniques from deep learning, image __52__ and object recognition can now be used to find cancer on MRIs with the same accuracy as __53__ trained radiologists.AI gets the most out of data. When algorithms are self-learning, the data itself can become intellectual property. The answers are in the data; you just have to apply AI to get them out. Since the role of the data is now more important than ever before, it can create a competitive __54__. If you have the best data in a competitive industry, even if everyone is __55__ similar techniques, the best data will win.41. A. mental B. manual C. magic D. mysterious42. A. inquiry B. right C. value D. resource43. A. responsibilities B. certificates C. possibilities D. capabilities44. A. consideration B. intention C. intelligence D. consciousness45. A. regularities B. forms C. conclusions D. characteristics46. A. trick B. technique C. feature D. concept47. A. explores B. investigates C. analyzes D. nails48. A. study B. promote C. follow D. train49. A. feed B. install C. collect D. deposit50. A. previously B. hardly C. formally D. almost51. A. connections B. combinations C. interactions D. innovations52. A. classification B. demonstration C. description D. elimination53. A. psychologically B. highly C. technically D. physically54. A. stress B. atmosphere C. advantage D. market55. A. adjusting B. reforming C. creating D. applyingSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)One evening in February 2007, a student named Paula Ceely brought her car to a stop on a remote road in Wales. She got out to open a metal gate that blocked her path. That’s when she heard the whistle sounded by the driver of a train. Her Renault Clio was parked across a railway line. Seconds later, she watched the train drag her car almost a kilometre down the railway tracks.Ceely’s near miss made the news because she blamed it on her GPS. She had never driven the route before. It was dark and raining heavily. Ceely was relying on her GPS, but it made no mention of the crossing. “I put my complete trust in the device and it led me right into the path of a speeding train,” she told the BBC.Who is to blame here? Rick Stevenson, who tells Ceely’s story in his book when Machines Fail Us, points the finger at the limitations of technology. We put our faith in digital devices, he says, but our digital helpers are too often not up to the job. They are filled with small problems. And it’s not just GPS devices: Stevenson takes us on a tour of digital disasters involving everything from mobile phones to wireless keyboards.The problem with his argument in the book is that it’s not clear why he only focuses on digital technology, while there may be a number of other possible causes. A map-maker might have left the crossing off a paper map. Maybe we should blame Ceely for not paying attention. Perhaps the railway authorities are at fault for poor signaling system. Or maybe someone has studied the relative dangers and worked out that there really is something specific wrong with the GPS equipment. But Stevenson doesn’t say.It’s a problem that runs through the book. In a section on cars, Stevenson gives an account of the advanced techniques that criminals use to defeat computer-based locking systems for cars. He offers two independent sets of figures on car theft; both show a small rise in some parts of the country. He says that once again not all new locks have proved reliable. Perhaps, but maybe it’s also due to the shortage of policemen on the streets, or changing social circumstances, or somecombination of these factors.The game between humans and their smart devices is amusing and complex. It is shaped by economics and psychology and the cultures we live in. Somewhere in the mix of those forces there may be a way for a wiser use of technology.If there is such a way, it should involve more than just an awareness of the shortcomings of our machines. After all, we have lived with them for thousands of years. They have probably been fooling us for just as long.56. What did Paula Ceely think was the cause of her accident?A. She was not familiar with the road.B. It was dark and raining heavily then.C. The railway workers falled to give the signal.D. Her GPS device didn’t tell her about the crossing.57. Which of the following would Rick Stevenson most probably agree with?A. Modern technology is what we can’t live without.B. Digital technology often falls short of our expectation.C. Digital devices are more reliable than they used to be .D. GPS error is not the only cause for Ceely’s accident.58. In the writer’s opinion, Stevenson’s arguments is _______.A. one-sidedB. reasonableC. puzzlingD. well-based59. What is the real concern of the writer of this article?A. The major causes of traffic accidents and car thefts.B. The relationship between human and technology.C. The shortcomings of digital devices we use.D. The human unawareness of technical problems.(B)USING YOUR SCORPIO COOKER: USEFUL HINTSFollow these useful hints to obtain the best results when using your new SCORPIO cooker.Choice of burnerUse a large burner to bring liquids to the boil quickly, and use a small one for stewed dishes andsauces.To conserve gas, place the pan centrally over the burner and adjust the flame so that it does not extend past the edges of the pan.Do not boil food too rapidly. A strong boil does not cook any faster but violently shakes up the food, which may then lose its taste.flame too high -- wastes gas flame not past edges of pan -- conserves gasUtensills（器具）All normally available utensils (aluminum, stainless steel, cast iron, ceramic, etc.) may be used on your new gas cooker, but ensure that they are steady, in order to avoid dangerous spill-over of hot liquids.Caution: Large UtentsilsWhen a cooker is installed close to a worktop, ensure that whenever large utensils are used, they are placed so that they do not overhand the side of hotplate, as this may cause scorching or charring（烧焦）of the worktop surface.OvenThere is a gradual change in temperature between the bottom and top of the oven. The first, or bottom, shelf position is the coolest and the fourth, or top, shelf position is the hottest. Because of the temperature change from one shelf position to another, it is possible to cook various dishes which require different temperatures, at the same time. As a rough guide, the temperature change from one shelf to the next is about ten degrees Celsius.Some recipes do not refer directly to temperature but use descriptions such as “slow”, “moderate”, “hot” etc. when using such recipes, the following chart may be taken as a guide:If preheating is required, allow time for the oven to reach the set temperature. The following table may be used as a guide:Note that the oven light (where fitted) is located on the splash back.Storage drawer (where fitted)The storage drawer situated underneath the oven is designed for the storage of pans and utensils. Do not place plastic utensils or flammable material in this drawer. To remove the drawer, withdraw it to the fully open position. Then lift it dear of the stops. To refit the drawer, locate the nylon drawer slides on the slide tracks. Lift the drawer slightly to clear the stops, then slide it to the fully shut position.60. How do you conserve gas when using a Scorpio cooker?A. Make the flame as high as possible.B. Shake up the food.C. Keep the flame not past the pan edges.D. Make a strong boll.61. If you want to stew beef, you should________.A. use a large utensilB. use a small burnerC. cook it festerD. use a utensil made of stainless steel62. What can we put in the storage drawer underneath the oven?A. Pans and plastic utensilsB. Pans and non-flammable materialC. Utensils and flammable material.D. Pans and various dishes(C)When Liam McGee departed as president of Bank of America in August, his explanation was surprisingly straight up. Rather than cloaking his exit in the usual vague excuse, he came right out and said he was leaving “to pursue my goal of running a company.” Broadcasting his ambition was “very much my decision,” McGee says. Within two weeks, he was talking for the first time with the board of Hartford Financial Services Group, which named him CEO and chairman onSeptember 29.McGee says leaving without a position lined up gave him time to reflect on what kinds of company he wanted to run. It also sent a clear message to the outside world about his aspirations. And McGee isn’t alone. In recent weeks the No. 2 executives at Avon and American Express quit with the explanation that they were looking for a CEO post. As boards scrutinize succession plans in response to shareholder pressure, executives who don’t get the nod also may wish to move on. A turbulent business environment also has senior managers cautious of letting vague pronouncements cloud their reputations.As the first signs of recovery begin to take hold, deputy chiefs may be more willing to make the jump without a net. In the third quarter, CEO turnover was down 23% from a year ago as nervous boards stuck with the leaders they had, according to Liberum Research. As the economy picks up, opportunities will abound for aspiring leaders.The decision to quit a senior position to look for a better one is unconventional. For years executives and headhunters have adhered to the rule that the most attractive CEO candidates are the ones who must be poached. Says Korn Ferry, senior partner Dennis Carey: “I can’t think of a single search I’ve done where a board has not instructed me to look at sitting CEOs first.”Those who jumped without a job haven’t always landed in top positions quickly. Ellen Marram quit as chief of Tropicana when the business became part of PepsiCo (PEP) a decade ago, saying she wanted to be a CEO. It was a year before she became head of a tiny internet-based commodities exchange. Robert Willemstad left Citigroup in 2005 with ambitions to be a CEO. He finally took that post at a major financial institution three years later.Many recruiters say the old disgrace is fading for top performers. The financial crisis has made it more acceptable to be between jobs or to leave a bad one. “The traditional rule was it’s safer to stay where you are, but that’s been fundamentally inverted,” says one headhunter. “The people who’ve been hurt the worst are those who’ve stayed too long.”63. When McGee announced his departure, his manner can best be described as being _______.A. arrogant.B. frank.C. self-centered.D. impulsive.64. According to Paragraph 2, senior executives’ quitting may be encouraged by _______.A. their expectation of better financial statusB. their need to reflect on their private lifeC. their strained relations with the boardsD. their pursuit of new career goals65. It can be inferred from the last paragraph that _______.A. top performers used to cling to their postsB. loyalty of top performers is getting out-datedC. top performers care more about reputationsD. it’s safer to stick to the traditional rules66. Which of the following is the best title for the text?A. CEOs; Where to Go?B. CEOs: All the Way Up?C. Top Managers Jump without a NetD. The Only Way Out for Top PerformersSection CDirections: Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentences can be used only once. Note that there are two more sentences than you need.Have you ever heard of Big Data or are you familiar with Data Scientists and Data Engineers? They are probably new job titles, but the core job roles have been around for a while. Traditionally, anyone who analyzed data would be called a “data analyst” and anyone who created backend platforms to support data analysis would be a “Business intelligence (BI) Developer”.______67_______Here’s an overview of the roles of the Data Analyst, BI Developer, Data Scientist and Data Engineer.Data Analysts are experienced data professionals in their organization who can question and process data, provide reports, summarize and visualize data. They have a strong understanding of how to influence existing tools and methods to solve a problem. _____68_____. However, they are not expected to deal with analyzing big data, nor are they typically expected to have the mathematical or research background to develop new algorithms for specific problems.Skills: Data Analysts need to have a baseline understanding of some core skills: statistics, data munging, data visualization, exploratory data analysis, Tools: Microsoft Excel, SPSS, SPSS Modeler, SAS, SAS Miner, SQL, Microsoft Access, Tableau, SSAS.______69_____. And then they collect requirements, design, and build BI and reporting solutions for the company. They have to design, develop and support new and existing data warehouses, ETL packages, dashboards and analytical reports.Additionally, they work with databases, both relational and multidimensional, and should have great SQL development skills to integrate data from different resources. They use all of these skills to meet the enterprise-wide self-service needs. BI Developers are typically not expected to perform data analyses.Data Engineers are the data professionals who prepare the “big data” infrastructure to be analyzed by Data Scientists. _____70______. Then, they write complex queries on that, make sure it is easily accessible, works smoothly, and their goal is optimizing（优化）the performance of their company’s big data ecosystem.They might also run some ETL (Extract, Transform and Load) on top of big datasets and create big data warehouses that can be used for reporting or analysis by data scientists.Ⅲ. Summary WritingDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.What makes a good gift?What makes a good gift? Here’s my story to share with you:This Christmas I was debating what do give my father. My dad is a hard person to buy for because he never wants anything. I pulled out my phone to read a text message from my mom saying that we were leaving for Christmas shopping for him when I came across a message on my phone that I had locked. The message was from my father. My eyes fell on a photo of a flower underneath a poem by William Blake. The flower, standing against the bright blue sky, inspired me. My dad had been reciting those words to me since I was a child. That may even be the reason why I love writing. I decided that those words would be my gift to my father.I called back. I told my mom to go without me and that I already created my gift. I sent the photo of the cream-colored flower to my computer and typed the poem on top of it. As I was arranging the details another poem came to my mind. The poem was written by Edgar Allan Poe; my dad recited it as much as he did the other. I typed that out as well and searched online for a background to the words of it. The poem was focused around dreaming, and after searching I found the perfect picture. The image was painted with blues, greens and purples, twisting together to create the theme and wonder of a dream. As I watched both poems passing through the printer, the white paper coloring with words that shaped my childhood, I felt that this was a gift that my father would truly appreciate.Christmas soon arrived. The minute I saw the look on my dad’s face as he unwrapped those black letters carefully placed in a cheap frame, I knew I had given the perfect gift.第Ⅲ卷Ⅲ. TranslationDirections: Translate the following sentences into English, using the words given in the bracket1. 我一辈子没听说过也没有见过这样可爱的小昆虫。
2018-2019年上海市复旦附中高二上10月月考数学试卷(有答案)
2018-2019年复旦附中高二上10月月考一. 填空题1. 等差数列{}n a 中，已知1510a =，4590a =，则60a =2. 在等比数列{}n a 中，已知11a =，公比q ∈R ，且1q ≠，12345n a a a a a a =，则n =3. 1123lim 23n nnn n ++→∞-=+ 4. 与向量(3,4)a =-共线的单位向量0a =5. 已知点P 分线段12PP 的比为2-，若1(1,2)P ，2(3,1)P -，则点P 的坐标为 6. 已知(0,1)A 、(1,0)B 、(3,)C k ，若ABC ∠为钝角，则k 的取值范围为7. 若数列100，50，20，⋅⋅⋅的各项加上某个数后恰为一等比数列，则此时等比数列的各项和为 8. 《九章算术》是我国古代一部重要的数学著作，书中有如下问题：“今有良马与驽马发长安，至齐，齐去长安三千里，良马初日行一百九十三里，日增一十三里，驽马初日行九十七里，日减半里，良马先至齐，复还迎驽马，问几何日相逢。
”其意为：“现在有良马和驽马同时从长安出发到齐去，已知长安和齐的距离是3000里，良马第一天行193里，之后每天比前一天多行13里，驽马第一天行97里，之后每天比前一天少行0.5里，良马到齐后，返回去迎接驽马，多少天后两马相遇。
”利用我们所学的知识，可知离开长安后的第 天，两马相逢9. △ABC 的外接圆的圆心为O ，半径为2，若2AO A B A C =+，且||||O A A B =，则向量CA 在向量BC上的投影为10. 已知数列{}n a 的前n 项和1(1)n n S n -=-⋅，若对任意的正整数n ，有1()()0n n a p a p +--<恒成立，则实数p 的取值范围是二. 选择题11. 在数列{}n a 中，111111234212n a n n=-+-+⋅⋅⋅+--，则1k a +=（ ） A. 121k a k ++ B. 112224k a k k +-++C. 122k a k ++D. 112122k a k k +-++12. 数列{}n a 的通项公式是1(1)2nn a +-=，则此数列（ ）A. 有极限，其值是整数B. 有极限，其值是分数C. 有两个极限D. lim n n a →∞不存在13. 已知||3b =，如果a 在b 上的投影是32-，那么a b ⋅为（ ） A. 92- B. 92 C. 2 D. 1214. 如图，正方形ABCD 中，M 是BC 的中点， 若AC AM BD λμ=+，则λμ+=（ ） A. 43 B. 53 C. 158D. 2三. 解答题15. 已知||1a =，||2b =，且a 与b 的夹角为120°. （1）求|32|a b -；（2）若(32)()a b ka b -⊥+，求实数k 的值.16. 已知O 为坐标原点，(3,4)OA =-，(6,3)OB =-，(5,3)OC m m =---. （1）若A 、B 、C 三点共线，求m 的值；（2）若△ABC 是以角A 为直角顶点的直角三角形，求m 的值以及此时三角形的面积.17. 已知数列{}n a 的前n 项和为n S ，且585n n S n a =--，*n ∈N . （1）证明：{1}n a -是等比数列；（2）求数列{}n S 的通项公式，并指出n 为何值时，n S 取得最小值，简单说明理由.18. 数列{}n a 中，11a =，当2n ≥时，前n 项和n S 满足21()2n n n S a S =-.（1）求n S 的表达式； （2）设21nn S b n =+，数列{}n b 的前n 项和为n T ，求lim n n T →∞.19. 在数列{}n a 中，已知11a =，22a =，且数列{}n a 的奇数项依次组成公差为1的等差 数列，偶数项依次组成公比为2的等比数列，数列{}n b 满足212n n na b a -=，记数列{}n b 的前 n 项和为n S .（1）写出数列{}n a 的通项公式； （2）求n S ；（3）证明：当6n ≥时，12n S n-<.（Ⅱ卷）1. 在△ABC 中，90C =︒，3CB =，点M 是AB 上的动点（包含端点），则MC CB ⋅的取值范围是2. 如图，等腰直角三角形ABC ，点G 是△ABC 的重心，过点G 作直线与CA 、CB 两边分别交于M 、N 两点，且CM CA λ=，CN CB μ=，则4λμ+的最小值为3. O 是平面α上一定点，A 、B 、C 是平面α上△ABC 的三个顶点，B 、C 分别是边AC 、AB 的对角，以下命题正确的是 （把你以为正确的序号全部写上）（1）动点P 满足OP OA PB PC =++，则△ABC 的外心一定在满足条件的P 点的集合中； （2）动点P 满足()||||AB ACOP OA AB AC λ=++（0λ>），则△ABC 的内心一定在满足条 件的P 点的集合中；（3）动点P 满足()||sin ||sin AB ACOP OA AB B AC Cλ=++（0λ>），则△ABC 的重心一定在满足条件的P 点的集合中； （4）动点P 满足()||cos ||cos AB ACOP OA AB B AC Cλ=++（0λ>），则△ABC 的垂心一定在满足条件的P 点的集合中； （5）动点P 满足()2||cos ||cos OB OC AB ACOP AB B AC Cλ+=++（0λ>），则△ABC 的 外心一定在满足条件的P 点的集合中； 4. 数列{}n a 中，若11a =，112n n n a a ++=（*n ∈N ），则122lim()n n a a a →∞++⋅⋅⋅+= 5. 已知数列{}n a 的首项11a =，对任意*n ∈N ，n a 、1n a +是方程230n x nx b -+=的两实根，则21n b -=6. 已知数列{}n a 满足n n a n k =⋅（*n ∈N ，01k <<），下面说法正确的是（把你认为正确的命题序号全部填上） （1）当12k =时，数列{}n a 为递减数列； （2）当112k <<时，数列{}n a 不一定有最大项； （3）当102k <<时，数列{}n a 为递减数列；（4）当1kk-为正整数时，数列{}n a 必有两项相等的最大项.。
上海市格致中学2018-2019学年高三上学期第一次月考数学试题
绝密★启用前 上海市格致中学2018-2019学年高三上学期第一次月考数学试题 试卷副标题 注意事项： 1．答题前填写好自己的姓名、班级、考号等信息 2．请将答案正确填写在答题卡上 第I 卷（选择题) 请点击修改第I 卷的文字说明 一、单选题 1．已知直线a ，若直线b 同时满足下列条件：①a 与b 异面；②a 与b 成定角；③a 与b 距离为定值d ；则这样的直线b （ ） A.唯一确定 B.有两条 C.有四条 D.有无数条 2．已知函数()f x 满足：()()()f x y f x f y +=⋅并且(1)1f =，那么2222((1))((2))((3))((1010))(1)(3)(5)(2019)f f f f f f f f +++⋅⋅⋅的值为（ ） A.2019 B.1010 C.4038 D.3030 3．对于函数()f x ，若存在实数m ，使得()()f x m f m +-为R 上的奇函数，则称()f x 是位差值为m 的“位差奇函数”，判断下列函数：①()21f x x =+；②2(1)2f x x x =++；③()2x f x =；④3()sin(4f x x π=+中是“位差奇函数”的有（ ）A.1B.2C.3D.4 4．如右图，一个直径为1的小圆沿着直径为2的大圆内壁的逆时针方 向滚动，M 和N 是小圆的一条固定直径的两个端点.那么，当小圆这 样滚过大圆内壁的一周，点M ，N 在大圆内所绘出的图形大致是( )…………○…………线…………○……答※※题※※ …………○…………线…………○……A. B.C.D.第II卷（非选择题)请点击修改第II卷的文字说明二、填空题5．设集合{2,0,1,9}A=，{|2,}B x x a a A==∈，则A B的所有元素之和为________6．已知sin()45πα-=，35(,)44ππα∈，则sinα=________7．若2(1)(2)3lim42na nb nn→∞-++-=+，则a b+=________8．已知()2n1(2x)n N*x-∈的展开式中各项的二项式系数之和为128，则其展开式中含1x项的系数是______.(结果用数值表示)9．已知x,y满足约束条件10xx yx y m-≥⎧⎪-≤⎨⎪+-≤⎩，若1yx+的最大值为2，则m的值为__________．10．已知函数()|21|xf x a=--，若存在实数1x、2x（12x x≠），使得12()()1f x f x==-，则实数a的取值范围为________.11．已知复数z满足i1iz z+=-，则220181z z z+++⋅⋅⋅+=________12．在平面直角坐标系xOy中，直线l经过坐标原点，(3,1)n=是l的一个法向量，已知数列{}n a满足：对任意的正整数n，点1,()n na a+均在l上，若26a=，则12345a a a a a的值为____13．将1,2,3,4,5,6随机排成一列，记为a，b，c，d，e，f，则abc def+是偶数的概率为______14．在菱形ABCD中，||4AB AD-=，||2AB AD+=，13AF AD=，12DE EC=，则BF AE⋅=________15．已知椭圆22221x ya b+=（0a b>>），F为椭圆的右焦点，AB为过椭圆中心O的弦，则△ABF面积的最大值为_____外…………○………○…………※※※※装※※订※※线※※内…………○………○…………()1f π=，()22f π=，则不等式组()1212x f x ≤≤⎧≤≤⎨⎩的解集为______． 三、解答题 17．设()2sin cos cos 4f x x x x π⎛⎫=-+ ⎪⎝⎭. （Ⅰ）求()f x 的单调区间； （Ⅱ）在锐角ABC ∆中，角,,A B C 的对边分别为,,a b c ,若0,12A f a ⎛⎫== ⎪⎝⎭,求ABC∆面积的最大值.18．如图所示，在三棱锥P ABC -中，PD ⊥平面ABC ，且垂足D 在棱AC 上，1AD =，AB BC ==3CD =，PD =（1）证明：△PBC 为直角三角形；（2）求直线AP 与平面PBC 所成角θ的正弦值.19．如图，两条相交线段AB 、PQ 的四个端点都在椭圆22143x y +=上，其中直线AB 的方程为0x m =>，直线PQ 的方程为12y x n =+.（1）若0n =，BAP BAQ ∠=∠，求m 的值；（2）探究：是否存在常数m ，当n 变化时，恒有BAP BAQ ∠=∠？20．已知函数2()ax bx cf x x d ++=+（其中a ，b ，c ，d 是实数常数，x d ≠-）.（1）若0a =，函数()f x 的图象关于点(1,3)-成中心对称，求b ，d 的值； （2）若函数()f x 满足条件（1），且对任意0[3,10]x ∈，总有0()[3,10]f x ∈，求c 的取值范围； （3）若0b =，函数()f x 是奇函数，(1)0f =，3(2)2f -=-，且对任意[1,)x ∈+∞时，不等式()()0f mx mf x +<恒成立，求负实数m 的取值范围． 21．对于无穷数列{}n a ，若对任意*n N ∈，满足212n n n a a a +++≤且n a M ≤（M 是与n 无关的常数），则称数列{}n a 为T 数列. （1）若1(2n n a =-（*n N ∈），判断数列{}n a 是否为T 数列，说明理由； （2）设350()2n n b n =-，求证：数列{}n b 是T 数列，并求常数M 的取值范围； （3）设数列|1|n p c n =-（*n N ∈，1p >），问数列{}n c 是否为T 数列？说明理由.参考答案1．D【解析】【分析】由题设条件，可作出两个平面，两异面直线分别在两个平面上，以保证两异面直线等距离，结合图象，即可求解，得到答案.【详解】由题意，作出如图所示的图形，其中//,,a b αβαβ⊂⊂，且,a b 异面，则平面β与b 平行的线都满足要求，所以这样的直线由无数条.故选D.【点睛】本题主要考查了空间中直线与直线的位置关系，以及异面直线的定义、夹角、距离等概念是解答本题的关键，着重考查了分析问题和解答问题的能力，属于基础题.2．B【解析】【分析】根据()()()f x y f x f y +=⋅，且(1)1f =，令,1x n y ==，可得()()1f n f n +=，求得()()()()123f f f f n ====，即可求解.【详解】 由题意，函数满足()()()f x y f x f y +=⋅，且(1)1f =，令,1x n y ==，可得()()1f n f n +=，即()()()()123f f f f n ====，所以2222222((1))((2))((3))((1010))(1)(2)(1010)1010(1)(3)(5)(2019)f f f f f f f f f f f +++⋅⋅⋅=+++=. 故选：B.【点睛】本题主要考查了抽象函数，以及数列的求和的应用，解答中合理赋值，得到()()1f n f n +=是解答的关键，着重考查了分析问题和解答问题的能力，属于基础题.3．B【解析】【分析】根据题意，结合“位差奇函数”的定义依次分析四个选项中的函数是否是“位差奇函数”，即可求解，得到答案.【详解】对于①中，函数()21f x x =+，则()()2()1(21)2f x m f m x m m x +-=++-+=，则对任意的实数m ，函数()()f x m f m +-是奇函数，即函数()f x 是位差值为任意实数m 的“位差奇函数”；对于②中，函数22()21(1)f x x x x =++=+，则()()22(1)f x m f m x m x +-=++， 设()22(1)h x x m x =++不会是奇函数，所以函数2(1)2f x x x =++不是“位差奇函数”； 对于③中，函数()2x f x =，则()()222(21)x m m m x f x m f m ++-=-=-，对任意实数m ，函数()()f x m f m +-都不是奇函数，所以()2x f x =不是“位差奇函数”；对于④中，函数3()sin()4f x x π=+，则()()33sin()sin()44f x m f m x m m ππ+-=++-+， 取4m π=时，可得()()sin()sin sin f x m f m x x ππ+-=+-=-是奇函数， 所以函数3()sin()4f x x π=+是“位差奇函数”. 故选：B.【点睛】 本题主要考查了函数的新定义，以及函数的奇偶性的判定及应用，其中解答中正确理解函数的新定义，逐项准确判定是解答的关键，着重考查了分析问题和解答问题的能力，属于中档试题.4．A【解析】【详解】如图：如图，取小圆上一点，连接并延长交大圆于点，连接，，则在小圆中，，在大圆中，，根据大圆的半径是小圆半径的 倍，可知的中点是小圆转动一定角度后的圆心，且这个角度恰好是，综上可知小圆在大圆内壁上滚动，圆心转过角后的位置为点，小圆上的点，恰好滚动到大圆上的也就是此时的小圆与大圆的切点。
2019-2020学年上海市复旦附中高三(上)第一次综合测试数学试卷(10月份)
2019-2020学年上海市复旦附中高三（上）第一次综合测试数学试卷（10月份）一、填空题[第16题每题4分，第7-12每题5分，共54分）1. 已知P：“角α的终边在第一象限”，q：“sinα>0”，则p是q的________条件（填“充分非必要”、“必要非充分”“充要”或“既不充分也不必要”）【答案】充分非必要【考点】充分条件、必要条件、充要条件【解析】由“sinα>0”，可以推出“角α的终边在第一，第二象限以及y轴正半轴“；由此即可判断．【解答】若：“角α的终边在第一象限”，则“sinα>0”成立；所以p是q的充分条件；若“sinα>0”，则“角α的终边在第一，第二象限以及y轴正半轴”；所以p不是q的必要条件；2. 函数f(x)＝x2−1(x<0)的反函数f−1(x)＝________−√x+1（________>−1）．【答案】,x【考点】反函数【解析】求出值域值域为(−1, +∞)，根据得出x=−√y+1，转化变量求解反函数即可．【解答】∵函数f(x)＝x2−1(x<0)，∴值域为(−1, +∞)，y＝x2−1，∴反函数f−1(x)=−√x+1(x>−1)，3. 记不等式x2+x−6<0的解集为集合A，函数y＝lg(x−a)的定义域为集合B．若“x∈A”是“x∈B”的充分条件，则实数a的取值范围为________．【答案】(−∞, −3]【考点】充分条件、必要条件、充要条件【解析】根据条件求出A，B，结合充分条件和必要条件的定义进行求解即可．【解答】由x2+x−6<0得−3<x<2，即A(−3, 2)，由x−a>0，得x>a，即B＝(a, +∞)，若“x∈A”是“x∈B”的充分条件，则A⊆B，即a≤−3，4. 设f(x)＝1g1−axx−1为奇函数，则a＝________．【答案】−1【考点】函数奇偶性的性质与判断【解析】根据题意，由奇函数的定义可得f(x)+f(−x)＝0，即1g1−axx−1+1g1+ax−x−1=lg1−a2x21−x2=0，分析可得a的值，验证f(x)的奇偶性即可得答案．【解答】根据题意，f(x)＝1g1−axx−1为奇函数，则f(x)+f(−x)＝0，即1g1−axx−1+1g1+ax−x−1=lg1−a2x21−x=0，必有1−a 2x21−x2=1，解可得a＝1或−1，若a＝1，f(x)＝1g1−xx−1=lg(−1)，没有意义，舍去；若a＝−1，f(x)＝1g1+xx−1，为奇函数，符合题意；故a＝−1；5. 已知a>1，则不等式a+2a−1的最小值为________．【答案】1+2√2．【考点】基本不等式【解析】由基本不等式可得a+2a−1=a−1+2a−1+1≥1+2√2，检验取等号的条件．【解答】解：∵a>1,∴a−1>0,∴a+2a−1=a−1+2a−1+1≥1+2√2，当且仅当a−1=2a−1，即a=1+√2时等号成立．∴不等式a+2a−1的最小值为1+2√2．故答案为1+2√2．6. 已知集合A＝{−2, −1, 0}，B＝{−1, 0, 1, 2}，则集合{a−b|a∈A, b∈B}的子集个数为________．【答案】64【考点】 子集与真子集 【解析】可以求出集合{a −b|a ∈A, b ∈B}＝{1, 0, −1, −2, −3, −4}，从而子集个数为26． 【解答】a ∈A ，b ∈B ，∴ a −b ＝−1，−2，−3，−4，0，1，∴ 集合{a −b|a ∈A, b ∈B}＝{1, 0, −1, −2, −3, −4}； ∴ {a −b|a ∈A, b ∈B}的子集个数为：26＝64．7. 已知sin α+cos α=−713，α∈(−π2,0)，则tan α＝________．【答案】−125【考点】同角三角函数间的基本关系 【解析】把已知等式两边平方，利用完全平方公式及同角三角函数间的基本关系变形求出2sin αcos α的值，进而判断出sin α−cos α的正负，利用完全平方公式及同角三角函数间的基本关系求出sin α−cos α的值，联立求出sin α与cos α的值，即可确定出tan α的值． 【解答】 把sin α+cos α=−713①，两边平方得：(sin α+cos α)2＝1+2sin αcos α=49169， ∴ 2sin αcos α=−120169，∵ α∈(−π2, 0)，∴ sin α<0，cos α>0，即sin α−cos α<0，∴ (sin α−cos α)2＝1−2sin αcos α=289169，即sin α−cos α=−1713②， 联立①②，解得：sin α=−1213，cos α=513，则tan α=−125．8. 已知正数a ，b 满足b a ＝4，且a +log 2b ＝3，则a +b ＝________．【答案】 4或5 【考点】指数式与对数式的互化 【解析】将b a ＝4等号两边取以2为底的对数，结合已知条件，转化为关于a 和log 2b 的方程，求出a 和log 2b ，即可得到所求． 【解答】 ∵ b a ＝4，∴ log 2b a =log 24，即a log 2b ＝2①，又a +log 2b ＝3②，联立①②得{a =1log 2b =2 或者{a =2log 2b =1 ，即{a =1b =4 或者{a =2b =2 ， ∴ a +b ＝4或者a +b ＝5，9. 已知函数f(x)=2x−1x−1的定义域是(−∞, 0]∪[3, +∞)，则f(x)的值域是________．【答案】[1,2)∪(2,52]【考点】函数的值域及其求法 函数的定义域及其求法 【解析】分x ≤0及x ≥3两种情况，利用不等式的性质求解． 【解答】 f(x)=2(x−1)+1x−1=2+1x−1，当x ≤0时，x −1≤−1,−1≤1x−1<0,1≤2+1x−1<2，当x ≥3时，x −1≥2,0<1x−1≤12,2<2+1x−1≤52， 综上，函数f(x)在(−∞, 0]∪[3, +∞)的值域为[1,2)∪(2,52]．10. 对于函数f(x)，若存在正实数M ，对于任意x ∈(1, +∞)，都有|f(x)|≤M ，则称函数f(x)在(1, +∞)上是有界函数．下列函数：①f(x)=xx−1；②f(x)=xx 2+1；③f(x)=x−1x+1；④f(x)＝x sin x ；其中在(1, +∞)上是有界函数的序号为________． 【答案】 ②③ 【考点】函数的值域及其求法 【解析】分析求出当x ∈(1, +∞)时，给定四个函数的值域，进而判断是否存在正实数M ，对于任意任意x ∈(1, +∞)，都有|f(x)|≤M ，进而得出结论． 【解答】对①，函数f(x)=xx−1=x−1+1x−1=1+1x−1，其在(1, +∞)上为减函数，且值域为(1, +∞)，故不是有界函数； 对②，函数f(x)=x x 2+1=1x+1x(x >1)，由于x +1x >2(x >1)，故0<f(x)<12，则|f(x)|<12，即存在M =12，故是有界函数； 对③，函数f(x)=x+1−2x+1=1−2x+1(x >1)，由于0<1x+1<12，0<2x+1<1，故0<1−2x+1<1，则|f(x)|<1，即存在M＝1，故是有界函数；对④，函数f(x)＝x sin x在(1, +∞)上的值域为R，故不是有界函数．11. 在平面直角坐标系xOy中，已知曲线C1、C2、C3依次为y＝21og2x、y＝log2x、y＝k log2x(k为常数，0<k<1)．曲线C1上的点A在第一象限，过A分别作x轴、y轴的平行线交曲线C2分别于点B、D，过点B作y轴的平行线交曲线C3于点C．若四边形ABCD为矩形，则k的值是________12．【答案】12【考点】对数函数的图象与性质【解析】设A(t, 21og2t)(t>1)，则B(t2, 21og2t)，D(t, log2t)，C(t2, 2k log2t)，则有log2t＝2k log2t，解出即可．【解答】设A(t, 21og2t)(t>1)，由AB平行x轴得B(t2, 21og2t)，由AD平行y轴得D(t, log2t)，又BC平行y轴，∴C的坐标为(t2, 2k log2t)，∵四边形ABCD为矩形，∴有log2t＝2k log2t，由于log2t>0，故2k＝1，即k=12．12. 已知函数g(x)的定义域为R，对任何实数m，n，都有g(m+n)＝g(m)+g(n)−3，且函数f(x)=x√1−x2x2+1+g(x)的最大值为p，最小值为q，则p+q值为________．【答案】6【考点】函数的最值及其几何意义【解析】根据题意，可令m＝n＝0，从而求出g(0)＝3，然后令m＝−n可得出函数g(x)−3是奇函数，并且y=x√1−x2x2+1是奇函数，从而可得出f(x)−3是奇函数，从而得出p−3+q−3＝0，从而可求出p+q的值．【解答】∵g(x)的定义域为R，对任何实数m，n，都有g(m+n)＝g(m)+g(n)−3，∴令m＝n＝0得，g(0)＝g(0)+g(0)−3，∴令m＝−n得，g(0)＝g(−n)+g(n)−3＝3，∴[g(−n)−3]+[g(n)−3]＝0，∴g(x)−3是R上的奇函数，且函数y=x√1−x2x2+1是R上的奇函数，∴f(x)−3是R上的奇函数，根据奇函数最大值和最小值互为相反数得，p−3+q−3＝0，∴p+q＝6．二、选择题（每题5分，共20分）设a∈(0, +∞)，b∈(0, +∞)，则“a<b“是“a−1a <b−1b”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【答案】C【考点】充分条件、必要条件、充要条件【解析】将a−1a −b+1b化简成(a−b)(1+1ab)，由此来判断a，b的大小关系；即可求解．【解答】∵a−1a −b+1b=(a−b)(1+1ab)，a∈(0, +∞)，b∈(0, +∞)，∴ ①若“a<b“，则a−1a −b+1b<0，即a−1a<b−1b；所以具有充分性；②若a−1a <b−1b，则(a−b)(1+1ab)<0，即a<b；所以具有必要性；设三棱柱的侧棱垂直于底面，所有棱长都为a，顶点都在一个球面上，则该球的表面积为( )A.πa2B.73πa2 C.113πa2 D.5πa2【答案】B【考点】球内接多面体【解析】由题意可知上下底面中心连线的中点就是球心，求出球的半径，即可求出球的表面积．【解答】解：根据题意条件可知三棱柱是棱长都为a的正三棱柱，上下底面中心连线的中点就是球心，则其外接球的半径为R=√(a2)2+(a2sin60∘)2=√712a2，球的表面积为S=4π⋅7a 212=73πa2.故选B.函数f(x)的定义域为[−1, 1]，图象如图1所示；函数g(x)的定义域为[−1, 2]，图象如图2所示．A ＝{x|f(g(x))＝0}，B ＝{x|g(f(x))＝0}，则A ∩B 中元素的个数为（ ）A.1B.2C.3D.4 【答案】 C【考点】 交集及其运算函数的图象与图象的变换 【解析】结合图象，分别求出集合A ，B ，再根据交集的定义求出A ∩B ，问题得以解决． 【解答】 由图象可知，若f （g(x)）＝0， 则g(x)＝0或g(x)＝1，由图2知，g(x)＝0时，x ＝0，或x ＝2， g(x)＝1时，x ＝1或x ＝−1 故A ＝{−1, 0, 1, 2}， 若g （f(x)）＝0，由图1知，f(x)＝0，或f(x)＝2（舍去）， 当f(x)＝0时，x ＝−1或0或1， 故B ＝{−1, 0, 1}，所以A ∩B ＝{−1, 0, 1}，则A ∩B 中元素的个数为3个．设函数f(x)的定义域为R ，满足f(x +2)＝2f(x)，且当x ∈(0, 2]时，f(x)=x +1x −94．若对任意x ∈(−∞, m]，都有f(x)≥−23，则m 的取值范围是( )A.(−∞,215] B.(−∞,163]C.(−∞,184]D.(−∞,194]【答案】 D【考点】已知函数的单调性求参数问题 基本不等式在最值问题中的应用 【解析】先利用函数f(x)的单调性，求出其在x ∈(0, 2]时的最值，然后根据递推关系可知， 当图象向右平移2个单位时，最小值变为原来的2倍，即可分析出何时f(x)min ≥−23． 【解答】解：当x ∈(4, 6]时，f min =f(5)=−1(1)所以要对任意x ∈(−∞, m]，都有f(x)≥−23，∵x∈(4, 5)时，函数f(x)递减，x∈(5, 6]时，函数f(x)递增，所以当m最大时，m∈(4, 5)，且f(x)min=f(m)=2f(m−2)=4f(m−4)=4[m−4+1m−4−94]≥−23，解得m≤194，故m的取值范围是(−∞, 194]．故选D.三、解笞题（共76分）三角形ABC中，角A，B，C所对边分别为a，b，c，且√2sin B=√3cos B．(1)若cos A=13，求sin C的值；(2)若b=√7，sin A=3sin C，求三角形ABC的面积．【答案】解：(1)由√2sin B=√3cos B，两边平方得2sin2B=3cos B，即2(1−cos2B)=3cos B，解得：cos B=12或cos B=−2（舍去），又B为三角形内角，∴B=π3，∵cos A=13，且A为三角形内角，∴sin A=√1−cos2A=2√23，则sin C=sin(B+A)=sin(π3+A)=√32cos A+12sin A=√3+2√26；(2)∵sin A=3sin C，由正弦定理可得a=3c，∵cos B=12，b=√7，∴由余弦定理知：b2=a2+c2−2ac cos B，即7=9c2+c2−3c2，解得：c=1，a=3c=3，则S△ABC=12ac sin B=3√34．【考点】余弦定理正弦定理同角三角函数间的基本关系【解析】（1）将已知等式两边平方，利用同角三角函数间基本关系化简求出cos B的值，即可确定出B的度数；（2）利用正弦定理化简sin A＝3sin C，得到a＝3c，利用余弦定理列出关系式，将b，cos B，以及a＝3c代入求出c的值，进而求出a的值，再由sin B的值，利用三角形面积公式即可求出三角形ABC的面积．【解答】解：(1)由√2sin B =√3cos B ，两边平方得2sin 2B =3cos B ， 即2(1−cos 2B)=3cos B ，解得：cos B =12或cos B =−2（舍去）， 又B 为三角形内角， ∴ B =π3，∵ cos A =13，且A 为三角形内角， ∴ sin A =√1−cos 2A =2√23， 则sin C =sin (B +A)=sin (π3+A) =√32cos A +12sin A =√3+2√26； (2) ∵ sin A =3sin C ，由正弦定理可得a =3c ， ∵ cos B =12，b =√7，∴ 由余弦定理知：b 2=a 2+c 2−2ac cos B ，即7=9c 2+c 2−3c 2，解得：c =1，a =3c =3， 则S △ABC =12ac sin B =3√34．如图，四棱锥P −ABCD 中，PA ⊥平面ABCD ，AD // BC ，AB ⊥AD ，BC =2√33，AB ＝1，BD ＝PA ＝2．（1）求异面直线BD 与PC 所成角的余弦值；（2）求二面角A −PD −C 的余弦值． 【答案】∵ PA ⊥平面ABCD ，AB ⊂平面ABCD ，AD ⊂平面ABCD ， ∴ PA ⊥AB ，PA ⊥AD ．又AD ⊥AB ，故分别以AB 、AD 、AP 所在直线为x 轴、y 轴、z 轴建立空间直角坐标系． 根据条件得AD =√3，所以B(1, 0, 0)，D(0, √3, 0)，C(1, 2√33, 0)，P(0, 0, 2)． 从而BD →=(−1, √3, 0)，PC →=(1, 2√33, −2)． 设异面直线BD ，PC 所成角为θ，则cos θ＝|cos <BD →，PC →)>|=|BD →⋅PC →|BD →||PC →||=(−1,√3,0)⋅(1,2√33,−2)2×√193=√5738， 即异面直线BD 与PC 所成角的余弦值为√5738； ∵ AB ⊥平面PAD ，∴ 平面PAD 的一个法向量为AB →=(1, 0, 0)． 设平面PCD 的一个法向量为n →=(x, y, z)， 由n →⊥PC →，n →⊥PD →，PC →=(1, 2√33, −2)，PD →=(0, √3, −2)，得{x +2√33y −2z =0√3y −2z =0，令z ＝3，得n →=(2, 2√3, 3)．设二面角A −PD −C 的大小为φ，且为锐角， 则cos φ＝cos <AB →，n →>=AB →⋅n→|AB →||n →|=(1,0,0)⋅(2,2√3,3)1×5=25，即二面角A −PD −C 的余弦值为25．【考点】二面角的平面角及求法 异面直线及其所成的角 【解析】（1）以AB 、AD 、AP 所在直线为x 轴、y 轴、z 轴建立空间直角坐标系，所求值即为BD →与PC →夹角的余弦值的绝对值，计算即可；（2）所求值即为平面PAD 的一个法向量与平面PCD 的一个法向量的夹角的余弦值的绝对值，计算即可． 【解答】∵ PA ⊥平面ABCD ，AB ⊂平面ABCD ，AD ⊂平面ABCD ， ∴ PA ⊥AB ，PA ⊥AD ．又AD ⊥AB ，故分别以AB 、AD 、AP 所在直线为x 轴、y 轴、z 轴建立空间直角坐标系． 根据条件得AD =√3，所以B(1, 0, 0)，D(0, √3, 0)，C(1, 2√33, 0)，P(0, 0, 2)． 从而BD →=(−1, √3, 0)，PC →=(1, 2√33, −2)． 设异面直线BD ，PC 所成角为θ，则cos θ＝|cos <BD →，PC →)>|=|BD →⋅PC →|BD →||PC →||=(−1,√3,0)⋅(1,2√33,−2)2×√193=√5738， 即异面直线BD 与PC 所成角的余弦值为√5738； ∵ AB ⊥平面PAD ，∴ 平面PAD 的一个法向量为AB →=(1, 0, 0)． 设平面PCD 的一个法向量为n →=(x, y, z)， 由n →⊥PC →，n →⊥PD →，PC →=(1, 2√33, −2)，PD →=(0, √3, −2)，得{x +2√33y −2z =0√3y −2z =0，令z ＝3，得n →=(2, 2√3, 3)．设二面角A −PD −C 的大小为φ，且为锐角， 则cos φ＝cos <AB →，n →>=AB →⋅n→|AB →||n →|=(1,0,0)⋅(2,2√3,3)1×5=25，即二面角A −PD −C 的余弦值为25．某地发生特大地震和海啸，使当地的自来水受到了污染，某部门对水质检测后，决定往水中投放一种药剂来净化水质．已知每投放质量为m 的药剂后，经过x 天该药剂在水中释放的浓度y （毫克/升）满足y ＝mf(x)，其中f(x)={x4+2(0<x ≤4)6x−2(x >4)，当药剂在水中释放的浓度不低于4（毫克/升）时称为有效净化；当药剂在水口释放的浓度不低于4（毫克/升）且不高于10（毫克/升）时称为最佳净化．（1）如果投放的药剂质量为m ＝4，试问自来水达到有效净化一共可持续几天？（2）如果投放的药剂质量为m ，为了使在7天（从投放药剂算起包括7天）之内的自来水达到最佳净化，试确定该投放的药剂质量m 的值． 【答案】因为m ＝4，所以y ＝m ⋅f(x)={x +8(0<x ≤4)24x−2(x >4) ； 所以，当0<x ≤4时，x +8≥4显然成立，当x >4时，24x−2≥4，得4<x ≤8；综上知，0<x ≤8；所以，自来水达到有效净化一共可持续8天．由y＝m⋅f(x)={mx4+2m(0<x≤4)6mx−2(x>4)知，在区间(0, 4]上单调递增，即2m<y≤3m，在区间(4, 7]上单调递减，即6m5≤y<3m，综上知，6m5≤y≤3m；为使4≤y≤10恒成立，只要6m5≥4，且3m≤10即可，即m=103；所以，为了使在7天之内的自来水达到最佳净化，该投放的药剂量应为103．【考点】根据实际问题选择函数类型【解析】（1）由m＝4，且y＝m⋅f(x)，可得药剂在水中释放浓度y的函数；因为函数y是分段函数，在求释放浓度不低于4（即y≥4）时，要分区间去求解．（2）由函数y是分段函数，故分区间讨论函数的单调性，从而求得y的取值范围，即药剂在水中释放浓度的大小；为使最佳净化，即4≤y≤10恒成立，只要使y的取值范围在区间[4, 10]内即可，从而解出m的值．【解答】因为m＝4，所以y＝m⋅f(x)={x+8(0<x≤4)24x−2(x>4)；所以，当0<x≤4时，x+8≥4显然成立，当x>4时，24x−2≥4，得4<x≤8；综上知，0<x≤8；所以，自来水达到有效净化一共可持续8天．由y＝m⋅f(x)={mx4+2m(0<x≤4)6mx−2(x>4)知，在区间(0, 4]上单调递增，即2m<y≤3m，在区间(4, 7]上单调递减，即6m5≤y<3m，综上知，6m5≤y≤3m；为使4≤y≤10恒成立，只要6m5≥4，且3m≤10即可，即m=103；所以，为了使在7天之内的自来水达到最佳净化，该投放的药剂量应为103．已知函数f(x)是定义域为R的奇函数，且当x≤0时，f(x)＝x−log2(1+2−x)+a，其中a是常数．（1）求f(x)(x∈R)的解析式；（2）求实数m的值，使得函数ℎ(x)＝2f（x）+1++m⋅2x−2m，x∈[0, 1]的最小值为14；（3）已知函数g(x)(x≥1)满足：对任何不小于1的实数x，都有f(log2x)＝−1+log2[x⋅g(x)−k2]，其中k为不小于2的正整数常数，求证：g(1)+g(2)+...+g(k−1)>g(k+1)+g(k+2)+...+g(2k−1)．【答案】∵函数f(x)是定义域为R的奇函数，且当x≤0时，f(x)＝x−log2(1+2−x)+a，则有f(0)＝0−log2(1+1)+a＝0，∴a＝1．∴当x≤0时，f(x)＝x−log2(1+2−x)+1，令x>0，则−x<0，f(x)＝−f(−x)＝−[−x−log2(1+2x)+1]＝x+log2(1+2x)−1∴f(x)={x−log2(1+2−x)+1,x≤0x+log2(1+2x)−1,x>0；当x∈[0, 1]时，函数ℎ(x)＝2f（x）+1++m⋅2x−2m＝2x+log2(1+2x)+m⋅2x−2m＝2x（(1+2x)+m⋅2x−2m＝(2x)2+(1+m)⋅2x−2m令2x＝t，t∈[1, 2]，ℎ(x)＝G(t)＝t2+(1+m)t−2m，t∈[1, 2]，①当−1+m2≥2，即m≤−5时，函数ℎ(x)最小值为G(2)＝6≠14，不符合题意；②当−1+m2≤1，即m≥−3时，函数ℎ(x)最小值为G(1)＝2−m=14，解得m=74，不符合题意；③当−5<m<−3时，函数ℎ(x)最小值为G(−1+m2)=−m2−10m−14=14，即m2+10m+2＝0，∵{(−5)2+10×(−5)+2<0(−3)2+10×(−3)+2<，∴方程m2+10m+2＝0在(−5, −3)无解；综上，m=74．∵x≥1，∴log2x≥0，则f(log2x)＝−1+log2[x⋅g(x)−k2]⇔log2x+log2(x+1)−1＝−1+log2[x⋅g(x)−k2]⇔log2(x+1)x＝log2[x⋅g(x)−k2]⇔x(1+x)＝x⋅g(x)−k2⇔g(x)＝x+k2x+1．(x≥1, k≥2, k∈N)．g(x1)−g(x2)=(x1x2−k2)(x1−x2)x1x2，令x1＝k−n，x2＝k+n，n∈N•，n≤k−1，则0<x1<k<x2，且x1x2<k2∴g(x1)>g(x2)，亦即g(k−n)>g(k+n)，∴g(1)>g(2k−1)，g(2)>g(2k−2)，……，g(k−1)>g(k+1)，∴g(1)+g(2)+...+g(k−1)>g(k+1)+g(k+2)+...+g(2k−1)．【考点】函数与方程的综合运用【解析】（1）由f(0)＝0−log2(1+1)+a＝0，可得a＝1．当x≤0时，f(x)＝x−log2(1+2−x)+1，令x>0，则f(x)＝−f(−x)＝−[−x−log2(1+2x)+1]＝x+log2(1+2x)−1即可求解．（2）当x∈[0, 1]时，函数ℎ(x)＝2f（x）+1++m⋅2x−2m＝2x+log2(1+2x)+m⋅2x−2m＝2x（(1+2x)+m⋅2x−2m＝(2x)2+(1+m)⋅2x−2m令2x＝t，t∈[1, 2]，ℎ(x)＝G(t)＝t2+(1+m)t−2m，t∈[1, 2]，分类讨论即可．（3）可得0<x1<k<x2，且x1x2<k2时，则g(x1)>g(x2)，令x1＝k−n，x2＝k+n，n∈N•，n≤k−1，则g(k−n)>g(k+n)，由此即可得证．【解答】∵函数f(x)是定义域为R的奇函数，且当x≤0时，f(x)＝x−log2(1+2−x)+a，则有f(0)＝0−log2(1+1)+a＝0，∴a＝1．∴当x≤0时，f(x)＝x−log2(1+2−x)+1，令x>0，则−x<0，f(x)＝−f(−x)＝−[−x−log2(1+2x)+1]＝x+log2(1+2x)−1∴f(x)={x−log2(1+2−x)+1,x≤0x+log2(1+2x)−1,x>0；当x∈[0, 1]时，函数ℎ(x)＝2f（x）+1++m⋅2x−2m＝2x+log2(1+2x)+m⋅2x−2m＝2x（(1+2x)+m⋅2x−2m＝(2x)2+(1+m)⋅2x−2m令2x＝t，t∈[1, 2]，ℎ(x)＝G(t)＝t2+(1+m)t−2m，t∈[1, 2]，①当−1+m2≥2，即m≤−5时，函数ℎ(x)最小值为G(2)＝6≠14，不符合题意；②当−1+m2≤1，即m≥−3时，函数ℎ(x)最小值为G(1)＝2−m=14，解得m=74，不符合题意；③当−5<m<−3时，函数ℎ(x)最小值为G(−1+m2)=−m2−10m−14=14，即m2+10m+2＝0，∵{(−5)2+10×(−5)+2<0(−3)2+10×(−3)+2<，∴方程m2+10m+2＝0在(−5, −3)无解；综上，m=74．∵x≥1，∴log2x≥0，则f(log2x)＝−1+log2[x⋅g(x)−k2]⇔log2x+log2(x+1)−1＝−1+log2[x⋅g(x)−k2]⇔log2(x+1)x＝log2[x⋅g(x)−k2]⇔x(1+x)＝x⋅g(x)−k2⇔g(x)＝x+k2x+1．(x≥1, k≥2, k∈N)．g(x1)−g(x2)=(x1x2−k2)(x1−x2)x1x2，令x1＝k−n，x2＝k+n，n∈N•，n≤k−1，则0<x1<k<x2，且x1x2<k2∴g(x1)>g(x2)，亦即g(k−n)>g(k+n)，∴g(1)>g(2k−1)，g(2)>g(2k−2)，……，g(k−1)>g(k+1)，∴g(1)+g(2)+...+g(k−1)>g(k+1)+g(k+2)+...+g(2k−1)．若定义在R上，且不恒为零的函数y＝f(x)满足：对于任意实数x和y，总有f(x+y)+ f(x−y)＝2f(x)f(y)恒成立，则称f(x)为“类余弦型”函数．（1）已知f(x)为“类余弦型”函数，且f(1)=54，求f(0)和f(2)的值；（2）证明：函数f(x)为偶函数；（3）若f(x)为“类余弦型”函数，且对于任意非零实数t，总有f(t)>1，设有理数x1，x2满足|x1|<|x2|，判断f(x1)和f(x2)的大小关系，并证明你的结论．【答案】令x＝1，y＝0，得f(1)+f(1)＝2f(1)f(0)，∴f(0)＝1；令x＝y＝1得f(2)+f(0)＝2f2(1)，∴f(2)＝2f2(1)−f(0)，∴f(2)＝2×(54)2−1=178．令x＝0，得f(y)+f(−y)＝2f(0)f(y)＝2f(y)，∴f(−y)＝f(y)，即f(−x)＝f(x)，∴f(x)是偶函数．∵f(x+y)+f(x−y)＝2f(x)f(y)，又∵t≠0时f(t)>1，∴2f(x)f(y)>2f(y)，即f(x+y)−f(y)>f(y)−f(x−y)∴令y＝kx（k为正整数），对任意的k为正整数，有f[(k+1)x]−f(kx)>f(kx)−f[(k−1)x]，则f[(k+1)x]−f(kx)>f(kx)−f[(k−1)x]>...>f(x)−f(0)>0，∴对于k为正整数，总有f[(k+1)x]>f(kx)成立．∴对于m，n为正整数，若n<m，则有f(nx)<f[(n−1)x]<...<f(mx)成立．∵x1，x2为有理数，所以可设|x1|=q1p1，|x2|=q2p2，其中q1，q2是非负整数，p1，p2都是正整数，则|x1|=q1p2p1p2，|x2|=q2p1p1p2，令x=1p1p2，t＝q1p2，s＝p1q2，则t，s为正整数．∵|x1|<|x2|，∴t<s，∴f(tx)<f(sx)，即f(|x1|)<f(|x2|)．∵函数f(x)为偶函数，∴f(|x1|)＝f(x1)，f(|x2|)＝f(x2)，∴f(x1)<f(x2)．【考点】抽象函数及其应用【解析】（1）令x＝1，y＝0计算f(0)，再令x＝y＝1计算f(2)；（2）令x＝0，得f(y)+f(−y)＝2f(0)f(y)＝2f(y)，所以f(−y)＝f(y)，即f(−x)＝f(x)，即可得出f(x)的奇偶性．（3）由t≠0时，f(t)>1，则f(x+y)+f(x−y)＝f(x)f(y)>2f(y)，即f(x+y)−f(y)>f(y)−f(x−y)令y＝kx（k为正整数），对任意的k为正整数，有f[(k+1)x]−f(kx)>f(kx)−f[(k−1)x]，再由递推即可得到对于k为正整数，总有f[(k+1)x]>f(kx)成立，即有n<m，则有f(nx)<f(mx)成立，可设||x1|=q1p2p1p2，|x2|=q2p1p1p2，其中q1，q2是非负整数，p1，p2都是正整数，再由偶函数的结论和前面的结论，即可得到大小．【解答】令x＝1，y＝0，得f(1)+f(1)＝2f(1)f(0)，∴f(0)＝1；令x＝y＝1得f(2)+f(0)＝2f2(1)，∴f(2)＝2f2(1)−f(0)，∴f(2)＝2×(54)2−1=178．令x＝0，得f(y)+f(−y)＝2f(0)f(y)＝2f(y)，∴f(−y)＝f(y)，即f(−x)＝f(x)，∴f(x)是偶函数．∵f(x+y)+f(x−y)＝2f(x)f(y)，又∵t≠0时f(t)>1，∴2f(x)f(y)>2f(y)，即f(x+y)−f(y)>f(y)−f(x−y)∴令y＝kx（k为正整数），对任意的k为正整数，有f[(k+1)x]−f(kx)>f(kx)−f[(k−1)x]，则f[(k+1)x]−f(kx)>f(kx)−f[(k−1)x]>...>f(x)−f(0)>0，∴对于k为正整数，总有f[(k+1)x]>f(kx)成立．∴对于m，n为正整数，若n<m，则有f(nx)<f[(n−1)x]<...<f(mx)成立．∵x1，x2为有理数，所以可设|x1|=q1p1，|x2|=q2p2，其中q1，q2是非负整数，p1，p2都是正整数，则|x1|=q1p2p1p2，|x2|=q2p1p1p2，令x=1p1p2，t＝q1p2，s＝p1q2，则t，s为正整数．∵|x1|<|x2|，∴t<s，∴f(tx)<f(sx)，即f(|x1|)<f(|x2|)．∵函数f(x)为偶函数，∴f(|x1|)＝f(x1)，f(|x2|)＝f(x2)，∴f(x1)<f(x2)．。
2017-2018学年上海市交大附中高三上第一次月考化学试卷
A.
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B. 用酸性 KMnO4 溶液可以确定 CH2=CH−CHO 中含有 C=C 双键 C. 在酸性条件下，CH3CO18OC2H5 的水解产物是 CH3COOH 和 C2H518OH D. 乙烯和聚氯乙烯都能使溴的四氯化碳溶液褪色
第1页/共9页
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B. 比例模型：
可以表示甲烷分子，也可以表示四氯化碳分子
C. 氯化铵的电子式为：
D. CO2 的结构式为：O−C−O 4. 能正确表达下列反应的离子方程式为（ ）
A. 用醋酸除去水垢：2H++CaCO3 Ca2++CO2↑+H2O B. 硫化亚铁与浓硫酸混合加热：2H++FeS H2S↑+Fe2+ C. 向硫酸铝溶液中滴加碳酸钠溶液：2Al3++3CO23－ Al2(CO3)3↓ D. 电解饱和食盐水获取烧碱和氯气：2Cl－+2H2O 通电 H2↑+Cl2↑+2OH－
A
FeCl3(aq)
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AlCl3(aq)
Al3+(aq) Cu2+
NaOH 溶液、盐酸 Al 粉
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第3页/共9页
18. 乙酰水杨酸是一种常见的解热镇痛药，结构简式如图所示。下列有关叙述正确的是（ ）
A. 乙酰水杨酸的分子式为 C9H10O4 B. 乙酰水杨酸的含氧官能团有三种
29. 蒸发浓缩、冷却结晶；
2019-2020学年上海市复旦附中高三(上)第一次综合测试数学试卷(10月份)(39)
2019-2020学年上海市复旦附中高三（上）第一次综合测试数学试卷（10月份）（39）一、选择题（本大题共4小题，共20.0分）1.“a≤0”是“函数f(x)=lnx+ax+1x在[1,+∞)上为减函数”的()A. 充分不必要条件B. 充要条件C. 必要不充分条件D. 既不充分也不必要2.一个直三棱柱的每条棱长都是4√3，且每个顶点都在球O的球面上，则球O的表面积为()A. 84πB. 96πC. 112πD. 144π3.若集合A={x|xx−1≤0},B={x|x2<2x}则A∩B=()．A. {x|0<x<1}B. {x|0≤x<1}C. {x|0<x≤1}D. {x|0≤x≤1}4.定义：min{a,b}表示a，b两数中较小的数．例如：min{2,4}=2.已知f(x)=min{−x2,−x−2},g(x)=2x+x+m(m∈R)，若对任意x1∈[−2,0]，存在x2∈[1,2]，都有f(x1)≤g(x2)成立，则m的取值范围为()A. [−4,+∞)B. [−6,+∞)C. [−7,+∞)D. [−10,+∞)二、填空题（本大题共12小题，共48.0分）5.“x>0”是“x≠0”的______条件；(填“充分不必要”、“必要不充分”、“充要”、“非充分非必要”)6.函数f(x)=x2(x>0)的反函数为______．7.已知集合A={y|y=x2−32x+1,x∈[34,2]}，B={x|x+m2≥1}.若“x∈A”是“x∈B”的充分条件，则实数m的取值范围________．8.若f(x)=12x−1+a是奇函数，则a=__________．9.已知a>3，求a+1a−3的最小值______ ．10.若集合A={−1,1},B={0,2}，集合C={z|z=x+y,x∈A,y∈B}，则集合C的真子集个数为__________．11.已知sinα+cosα=12,α∈(0,π)，则1−tanα1+tanα=_______．12.已知|a−8b|+(4b−1)2=0，则log2a b=__________．13.已知函数f(x)的定义域为[0,1]，值域为[1,2]，则f(x+2)的定义域是__________，值域是__________．14.函数f(x)=1x2−2x+3，x∈[0,3]的最大值为______ ．15.已知函数y=log√22(x2+4x+5)，x∈[−1,3]的值域为__________．16.已知函数f(x+1)=x2−2x+1的定义域为[−2,0]，则函数f(x)的最大值为______ ．三、解答题（本大题共5小题，共70.0分）17. 在△ABC 中，角A ，B ，C 所对边的长分别为a ，b ，c ，且a =√5，b =3，sinC =2sinA ．(1)求c 的值；(2)求cos2A 的值和三角形ABC 的面积．18. 如图，在底面为直角梯形的四棱锥P −ABCD 中，AD//BC ，∠ABC =90°，PA ⊥平面ABCD ，PA =3，AD =2，AB =2√3，BC =6．(1)求异面直线BD 与PC 所成角的大小；(2)求二面角P −DC −B 的余弦值．19. 由于浓酸泄漏对河流形成了污染，现决定向河中投入固体碱．1个单位的固体碱在水中逐步溶化，水中的碱浓度y(个浓度单位)与时间x(个时间单位)的关系为y ={−24x+3−x +8, 0≤x ≤322312−12x , 32<x ≤236.只有当河流中碱的浓度不低于1(个浓度单位)时，才能对污染产生有效的抑制作用．(1)如果只投放1个单位的固体碱，则能够维持有效抑制作用的时间有多长？(2)当河中的碱浓度开始下降时，即刻第二次投放1个单位的固体碱，此后，每一时刻河中的碱浓度认为是两次投放的碱在该时刻相应的碱浓度的和，求河中碱浓度可能取得的最大值．20. 已知函数f(x)=log 2(x +2)，g(x)=a ⋅4x −2x+1−a +1．(1)判断函数ℎ(x)=f(x)+f(x −6)的单调性，并说明理由；(2)若对任意的x1，x2∈[1,2]，f(x1)<g(x2)恒成立，求a的取值范围．21.已知函数f(x)是R上的增函数，对任意实数m，n，都有f(m+n)=f(m)+f(n)，且f(4)=2．(Ⅰ)求f(2)的值；(Ⅱ)判断函数f(x)的奇偶性，并证明；(Ⅲ)对任意x∈[0,4]，都有f(x)−f(2a−1)<1，求实数a的取值范围．-------- 答案与解析 --------1.答案：C解析：解：函数f(x)=lnx +ax +1x ，在[1,+∞)在为减函数，∴f′(x)=1x +a −1x 2≤0，在[1,+∞)上恒成立，∴a ≤1x 2−1x， 设g(x)=1x 2−1x ，∴g′(x)=x−2x 3，当x ∈[1,2]，g′(x)<0，g(x)为减函数，当x ∈[2,+∞)，g′(x)>0，g(x)为增函数，∴g(x)min =g(2)=14−12=−14， ∴a ≤−14，∴当a ≤−14，函数f(x)=lnx +ax +1x 在[1,+∞)上为减函数，∴“a ≤0”是“函数f(x)=lnx +ax +1x 在[1,+∞)上为减函数”的必要不充分条件， 故选：C ．先根据导数和函数单调性的关系求出a 的范围，再结合充分条件，必要条件的定义即可判断． 本题考查了导数和函数的单调性的关系和充分必要条件的定义，属于中档题．2.答案：C解析：解：∵一个直三棱柱的每条棱长都是4√3，且每个顶点都在球O 的球面上，∴设此直三棱柱两底面的中心分别为O 1，O 2，则球O 的球心O 为线段O 1O 2的中点， 设球O 的半径为R ，则R 2=(4√32)2+(23×√32×4√3)2=28，∴球O 的表面积S =4πR 2=112π．故选：C ．设此直三棱柱两底面的中心分别为O 1，O 2，则球O 的球心O 为线段O 1O 2的中点，设球O 的半径为R ，利用勾股定理求出R 2，由此能求出球O 的表面积．本题考查三棱柱的外接球的表面积的求法，是较易题．3.答案：A解析：【分析】本题考查了交集及其运算．【解答】解：由已知解得：集合A={x|0≤x<1}，B={x|0<x<2}，所以A∩B={x|0<x<1}．故选：A．4.答案：C解析：【分析】本题考查函数的最值以及不等式恒成立和存在性问题的求解．首先将要求的问题转化为最值问题，然后结合函数图象即可求解．【解答】解：如图，当x1∈[−2,0]时，f(x1)max=f(−1)=−1；当x2∈[1,2]时，g(x2)为增函数，则g(x2)max=g(2)=6+m．由题意知f(x1)max⩽g(x2)max，即−1≤6+m，即m≥−7．故选C．5.答案：充分不必要解析：解：原命题：若“x>0”则“x≠0”，此是个真命题其逆命题：若“x≠0”，则“x>0”，是个假命题，因为当“x≠0”时“x<0”，也可能成立，故不一定得出“x >0”，综上知“x >0”是“x ≠0”的充分不必要条件故答案为：充分不必要．将题设中的命题改写成命题的形式，分别判断它的真假及其逆命题的真假，再依据充分条件，必要条件的定义作出判断得出正确答案本题考查充分条件必要条件的判断，解题的关键是熟练掌握充分条件与必要条件的定义，本题是基本概念考查题，难度较低，在高考中出现的机率较小6.答案：f −1(x)=√x(x >0)解析：解：由f(x)=y =x 2(x >0)解得x =√y ，∴f −1(x)=√x(x >0)故答案为f −1 (x)=√x(x >0)由y =x 2(x >0)解得x =√y(y >0)，再交换x 与y 的位置即得反函数．本题考查了反函数，属基础题．7.答案：m ≥34或m ≤−34解析：【分析】本题主要考查充分条件和必要条件的应用，先化简集合A ，利用充分条件和必要条件的关系进行求值．【解答】解：A ={y |y =x 2−32x +1,x ∈[34,2]}={y| 716≤y ≤2}，B ={x | x +m 2≥1}={x | x ≥1−m 2}，∵“x ∈A ”是“x ∈B ”的充分不必要条件，∴A ∈B 且A ≠B ，∴1−m 2≤716，解得m ≥34或m ≤−34，故m 的取值范围为(−∞,−34]∪[34,+∞)．8.答案：12解析：【分析】本题主要考查函数的奇偶性．【解答】解：因为f(x)是奇函数，所以f(x)=−f(−x)，所以12x −1+a =−12−x −1−a ，解得a =12．故答案为12．9.答案：5解析：解：∵a >3，a +1a−3=a −3+1a−3+3≥2√(a −3)⋅1a−3+3=5， 当且仅当a −3=1a−3即a =4时取等号， ∴a +1a−3的最小值是5， 故答案为：5． a +1a−3=a −3+1a−3+3，利用基本不等式可求函数的最值． 该题考查利用基本不等式求函数的最值问题，属基础题，熟记基本不等式的使用条件及常见不等式变形是解题关键． 10.答案：7 解析：【分析】 本题主要考查了真子集个数的判断，属于基本题型． 求出集合C 中的元素，是解决本题的关键． 【解答】 解：∵A ={−1,1},B ={0,2}， ∴集合C ={z |z =x +y,x ∈A,y ∈B}={−1,1,3}， 则集合C 的真子集个数为7． 故答案为7． 11.答案：−√7解析：【分析】本题考查同角三角函数基本关系的运用，属于基础题．由条件利用同角三角函数的基本关系，以及三角函数在各个象限中的符号求得cosα−sinα的值，可得的值．【解答】解：∵已知sinα+cosα=12，α∈(0,π)，∴1+2sinαcosα=14，求得2sinαcosα=−34，结合α∈(0,π)，可得α为钝角，，则=−√7， 故答案为−√7．12.答案：14解析：【分析】本题考查了对数的运算性质，属于基础题．根据绝对值和偶次方的非负性，得{a −8b =04b −1=0，求出a ，b 的值，然后利用对数的运算性质可得结果． 【解答】解：由|a −8b |+(4b −1)2=0，得{a −8b =04b −1=0, 解得a =2，b =14，所以log 2a b =log 2214=14． 故答案为14． 13.答案：[−2,−1][1,2]解析：∵f(x)的定义域为[0,1]，∴0≤x +2≤1，∴−2≤x ≤−1.即f(x +2)的定义域为[−2,−1]，值域仍然为[1,2]．14.答案：12解析：解：设g(x)=x 2−2x +3=(x −1)2+2，∵在[0,1]单调递减，在[1,3]单调递增，∴g(1)=2，g(3)=2，g(3)=6，∴2≤g(x)≤6，∴函数f(x)=1x 2−2x+3的值域为[16,12]故答案为：12．先利用g(x)=x 2−2x +3=(x −1)2+2，求出值域，再利用f(x)=1g(x)求解．本题考查了二次函数的性质，整体求解函数值域，最值问题，属于容易题．15.答案：[−2log 226,−2]．解析：令t =x 2+4x +5，则y =log √22t ，因为x ∈[−1,3]，所以t ∈[2,26]，y ∈[−2log 226,−2]． 16.答案：9解析：解：因为f(x +1)=x 2−2x +1=(x +1)2−4(x +1)+4，所以函数解析式为f(x)=x 2−4x +4，又因为f(x +1)=x 2−2x +1的定义域为[−2,0]，所以x +1∈[−1,1]，所以f(x)的定义域为[−1,1]，并且f(x)在[−1,1]上是减函数，所以f(x)的最大值为f(−1)=1+4+4=9；故答案为：9．首先求出函数的解析式，然后求二次函数的最值． 本题考查了复合函数的定义域求法、解析式的求法以及二次函数解析式最值求法． 17.答案：解：(Ⅰ)∵a =√5，sinC =2sinA ，∴根据正弦定理得：c =sinC sinAa =2a =2√5； (Ⅱ)∵a =√5，b =3，c =2√5，∴由余弦定理得：cosA =2×3×2√5=2√55， 又A 为三角形的内角， ∴sinA =√55， ∴sin2A =2sinAcosA =45，cos2A =cos 2A −sin 2A =35， 三角形ABC 的面积S =12×3×2√5×√55=3．解析：(1)利用正弦定理得到c =sinCsinA a ，将a 的值及sinC =2sinA 代入，即可求出c 的值；(2)利用余弦定理表示出cos A ，将a ，b 及求出的c 值代入，求出cos A 的值，由A 为三角形的内角，利用同角三角函数间的基本关系求出sin A 的值，进而利用二倍角的正弦函数公式求出sin2A 及cos2A 的值，可求三角形ABC 的面积．此题考查了正弦、余弦定理，二倍角的正弦，同角三角函数间的基本关系，以及特殊角的三角函数值，熟练掌握定理及公式是解本题的关键． 18.答案：解：如图以A 为坐标原点，以直线AB ，AD ，AP 分别为x 轴，y 轴，z 轴建立空间直角坐标系A −xyz ， 则A(0,0,0),B(2√3,0,0,),C(2√3,6,0),D(0,2,0),P(0,0,3)．(1)∵BD ⃗⃗⃗⃗⃗⃗ =(−2√3,2,0),PC⃗⃗⃗⃗⃗ =(2√3,6,−3)， ∴BD ⃗⃗⃗⃗⃗⃗ ⋅PC⃗⃗⃗⃗⃗ =0， 即异面直线BD 与PC 所成的角为π2；(2)由题意得：PA ⊥平面ABCD ，则平面BCD 的一个法向量为n 1⃗⃗⃗⃗ =(0,0,1)，设平面PCD 的一个法向量为n 2⃗⃗⃗⃗ =(x,y,z)，∵DC⃗⃗⃗⃗⃗ =(2√3,4,0),PD ⃗⃗⃗⃗⃗ =(0,2,−3)， ∴{n 2⃗⃗⃗⃗ ⋅DC ⃗⃗⃗⃗⃗ =2√3x +4y =0n 2⃗⃗⃗⃗ ⋅PD ⃗⃗⃗⃗⃗ =2y −3z =0, 解得平面PCD 的一个法向量为n 2⃗⃗⃗⃗ =(2√3,−3,−2)，∴cos <n 1⃗⃗⃗⃗ ，n 2⃗⃗⃗⃗ >=n 1⃗⃗⃗⃗⃗ ⋅n 2⃗⃗⃗⃗⃗ |n 1⃗⃗⃗⃗⃗ ||n 2⃗⃗⃗⃗⃗ |=√12+9+4=−25， 由图可得二面角P −DC −B 为锐二面角， 即二面角P −DC −B 的余弦值为25．解析：本题考查求异面直线的夹角、二面角的余弦值，注意解题方法的积累，属于中档题．(1)以A 为坐标原点，以直线AB ，AD ，AP 分别为x 轴，y 轴，z 轴建立空间直角坐标系A −xyz.通过BD ⃗⃗⃗⃗⃗⃗ ⋅PC ⃗⃗⃗⃗⃗ =0，即得异面直线BD 与PC 所成的角为π2； (2)所求值即为平面BCD 的一个法向量与平面PCD 的一个法向量的夹角的余弦值的绝对值，计算即可．19.答案：解：(1)由题意，{−24x+3−x +8≥10≤x ≤32⇒{1≤x ≤30≤x ≤32⇒1≤x ≤32 {2312−12x ≥132<x ≤236⇒32<x ≤116 综上，得1≤x ≤116即若1个单位的固体碱只投放一次，则能够维持有效抑制作用的时间为116−1=56(2)当0≤x ≤32时，y′=24−(x+3)2(x+3)>0，所以y =−24(x+3)−x +8单调递增 当2<x ≤4时，y =2311−12x 单调递减所以当河中的碱浓度开始下降时，即刻第二次投放1个单位的固体碱，即x >32，由{0≤x −32≤3232<x ≤236得32<x ≤3，当x >3时，第一、二次投放的固体碱的浓度均在下降(或降为0)． 所以最大浓度发生的时间位于区间(32,3]当32<x ≤3时，y =2312−12x +[−24(x−32)+3−(x −32)+8] =−32(x +32)−24x +32+413≤−12+413=53故当且仅当x +32=16x+32,即x =52时，y 有最大值53．解析：(1)利用分段函数解析式，分别列出不等式，解之，即可求得x 的范围，从而可得能够维持有效抑制作用的时间；(2)确定函数y =−24(x+3)−x +8单调递增，当2<x ≤4时，y =2311−12x 单调递减，进而可得函数，利用基本不等式，即可求得最值．本题考查分段函数，考查解不等式，考查函数的单调性，考查利用基本不等式求函数的最值，确定函数的解析式是关键．20.答案：解：(1)因为ℎ(x)=log 2(x +2)+log 2(x −4)，定义域为(4,+∞)，设x 2>x 1>4， ℎ(x 2)−ℎ(x 1)=log 2x 2+2x 1+2+log 2x 2−4x 1−4， ∵x 2>x 1>4，∴x 2+2x1+2>1,x 2−4x 1−4>1， ∴log 2x 2+2x 1+2>0,log 2x 2−4x 1−4>0，即ℎ(x 2)>ℎ(x 1)，所以函数ℎ(x)在定义域上为增函数．(2)依题意有，f(x)max <g(x)在[1,2]上恒成立．因为f(x)在[1,2]上单调递增，所以f(x)max =log 24=2，a ·4x −2x+1−a +1>2在[1,2]上恒成立，令t =2x ∈[2,4]，即at 2−2t −a +1>2，所以a >2t+1t 2−1，设g(t)=2t+1t 2−1，g′(t)=−2t 2−2t−2(t −1)<0，故g max =g(2)=53即a 的取值范围为(53,+∞)．解析：(1)先求出函数ℎ(x)的解析式，利用定义即可判断；(2)根据题意可知，只要f(x)max <g(x)恒成立即可，分参换元，求函数y =2t+1t 2−1的最大值，即可以得到a 的取值范围．本题主要考查函数单调性的证明以及不等式恒成立问题的解法，解题关键是转化分参，属于中档题． 21.答案：解：(Ⅰ)因为对任意实数m ，n 都有f(m +n)=f(m)+f(n)，令m =n =2，所以f(4)=f(2)+f(2)=2，解得f(2)=1；(Ⅱ)函数f(x)为奇函数，证明如下：因为f(m +n)=f(m)+f(n)对任意实数m ，n 都成立，令m =x ，n =−x ，所以f(0)=f(x)+f(−x)．令m =n =0，所以f(0)=f(0)+f(0)，解得f(0)=0，所以f(x)+f(−x)=0，即f(−x)=−f(x)，所以f(x)为奇函数；(Ⅲ)因为对于任意x ∈[0,4]，都有f(x)−f(2a −1)<1，所以f(x)<1+f(2a−1)，即f(x)<f(2)+f(2a−1)．又因为f(2)+f(2a−1)=f(2+2a−1)=f(2a+1)，所以f(x)<f(2a+1)，因为函数f(x)在R是增函数，所以2a+1>x．因为任意x∈[0,4]，都有2a+1>x成立，所以2a+1>(x)max，，由此得2a+1>4，即a>32, +∞)．所以a的取值范围是(32解析：(Ⅰ)可令m=n=2，代入计算可得所求值；(Ⅱ)函数f(x)为奇函数，可令m=n=0，求得f(0)，再令m=x，n=−x，结合奇偶性的定义，即可得到所求结论；(Ⅲ)由条件和f(2)=1，可得f(x)<f(2a+1)，再由单调性和恒成立思想，可得a的范围．本题考查抽象函数的函数值和性质，注意运用定义法和转化思想，考查方程思想和赋值法，化简整理的运算能力，属于中档题．。
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复旦附中2018学年年第⼀一学期⾼高三年年级第⼀一次综合测试
2018.09.28
⼀一、填空题(本⼤大题共有12题,满分54分,第16题每题4分,第7-12题每题5分)
1.已知全集则_______.
2.在复数集内⽅方程的解为__________.
3.⽅方程的解是________.
4.已知AB为抛物线的弦,如果此弦的垂直平分线的⽅方程是,则弦AB所在直线的⽅方程是
_________.
5.函数的递增区间是__________.
6.设,则_________.
7.已知函数是偶函数,则实数的值是_________.
8.正⽅方体的体对⻆角线与⾯面对⻆角线所成的⻆角的集合是__________.
9.某班级有38⼈人,现需要随机抽取2⼈人参加⼀一次问卷调查,那么甲同学选上,⼄乙同学未选上的概率是
________(⽤用分数作答).
10.观察下列列等式:
照此规律律,第个等式可以为____________.
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11.已知⼆二次函数的图像为开⼝口向下的抛物线,且对任意都有:,若向量量
则满⾜足不不等式的实数取值范围为__________.
12.某班共有50名学⽣生,已知以下信息:
①男⽣生共有33⼈人；②⼥女女团员共有7⼈人；③住校的⼥女女⽣生共有9⼈人；④不不住校的团员共有6⼈人；
⑤住校的男团员共有6⼈人；⑥男⽣生中⾮非团员且不不住校的共有8⼈人；⑦⼥女女⽣生中⾮非团员且不不住校的共有3⼈人。
根据以上信息,该班住校⽣生共有_________⼈人.
⼆二、选择题(本⼤大题共有4题,满分20分,每题5分)
13.设集合,集合且,刚实数的取值围=范围是
A.B.C.D.
14.条件甲:函数满⾜足；条件⼄乙:函数是偶函数，则甲是⼄乙的
A.充分⾮非必要条件B.必要⾮非充分条件
C.充要条件D.既⾮非充分也⾮非必要条件
15.关于函数的反函数,正确的是
A.有反函数B.有反函数
C.有反函数D.⽆无反函数
16.定义“正对数”:现有四个命题：
①若则；②若则
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③若则④若则
则所有真命题的序号为
A.①②③B.①②④C.③④D.②③④
三、解答题(本⼤大题满分76分)
17.(本题满分14分)
已知集合.
(1)⽤用列列举法写出集合；
(2)是否存在⾃自然数,使得,若存在,求出的值,并写出此时集合P的元素个数；若不不存在,请说明
理理由。
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18.(本题满分14分)
设函数是由曲线确定的。
(1)写出函数,并判断该函数的奇偶性；
(2)求函数的单调区间并证明其单调性.
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19.(本题满分14分)
中国古代建筑中的窗饰是艺术和技术的统⼀一体,给⼈人以美的享受。如图,左图为⼀一古式窗户，右图为是这扇窗
中的⼀一格,呈⻓长⽅方形,⻓长30cm,宽26cm,其内部窗芯(不不含⻓长⽅方形边框)是⽤用⼀一种条形⽊木料料做成的,由两个菱形和
六根⽀支条构成,整个窗芯关于⻓长⽅方形边框的两条对称轴对称。若菱形的两条对⻆角线⻓长分别为和,
⼀一格窗芯所需条形⽊木料料的⻓长度之和为.
(1)试⽤用表示L；
(2)如果要求六根⽀支条的⻓长度均不不⼩小于2cm,且每个萎形的⾯面积为130,那么要做这样⼀一格中的窗芯⾄至少
需要多⻓长的条⽊木料料(不不计卯榫及其他损耗)?
20.(本题满分16分)
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定义在R上的奇函数的最⼩小正周期为4,当时,.
(1)判断并证明在(0,2)上的单调性,并求在上的解析式；
(2)当为何值时,关于的⽅方程在上有实数解?
21.(本题满分18分)
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已知及.
(1)分别求的定义域,并求的值；
(2)求的最⼩小值并说明理理由；
(3)若,是否存在正数,使得对于任意的以
为边⻓长都可以构成三⻆角形,若存在,求出符合条件的正数的取值范围；若不不存在，
请说明理理由。
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参考答案
1、2、3、4、
5、6、07、8、
9、10、
11、12、24⼈人
13-16、CABD
17、（1）；（2）
18、（1），奇函数；（2）增区间：和，⽆无递减区间
19、（1）；（2）
20、（1）减；；（2）；
21、（1）定义域都为；；（2）；
（3）