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2018-2019学年上海高考标准模拟试卷一(word带答案)

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上海市崇明区2018届高三第一次高考模拟考试数学---精校解析Word版

上海市崇明区2018届高三第一次高考模拟考试数学---精校解析Word版

2018年上海市崇明县高考数学一模试卷一、填空题（本大题共有12题，满分54分，其中1-6题每题4分，7-12题每题5分）1. 已知集合，若，则_____．【答案】【解析】因为集合，且，所以，故答案为.2. 抛物线的焦点坐标为_____．【答案】【解析】抛物线的焦点在轴上，且，所以抛物线的焦点坐标为，故答案为.3. 不等式的解是_____．【答案】【解析】由可得，，所以不等式的解是，故答案为. 4. 若复数满足为虚数单位），则_____．【答案】【解析】试题分析：因为，所以本题也可设，因为由复数相等得：考点：复数的四则运算．5. 在代数式的展开式中，一次项的系数是_____．（用数字作答）【答案】【解析】展开式的通项为，令，得，，故答案为.【方法点晴】本题主要考查二项展开式定理的通项与系数，属于简单题. 二项展开式定理的问题也是高考命题热点之一，关于二项式定理的命题方向比较明确，主要从以下几个方面命题：（1）考查二项展开式的通项公式；（可以考查某一项，也可考查某一项的系数）（2）考查各项系数和和各项的二项式系数和；（3）二项展开式定理的应用.6. 若函数的最小正周期是，则_____．【答案】【解析】因为函数的最小正周期是，所以，解得，故答案为.7. 若函数的反函数的图象经过点，则_____．【答案】【解析】函数的反函数的图象经过点，所以，函数的图象经过点，，，故答案为.8. 将一个正方形绕着它的一边所在的直线旋转一周，所得几何体的体积为，则该几何体的侧面积为_____．【答案】【解析】将一个正方形绕着它的一边所在的直线旋转一周，所得圆柱的体积为，设正方体的边长为，则，解得该圆柱的侧面积为，故答案为.9. 已知函数是奇函数，当时，，且，则_____．【答案】【解析】是奇函数，且当时，，，解得，故答案为.10. 若无穷等比数列的各项和为，首项，公比为，且，则_____．【答案】【解析】无穷等比数列的前项和为，首项为，公比，且，，或，或，，故答案为.11. 从5男3女共8名学生中选出队长1人，副队长1人，普通队员2人组成 4人志愿者服务队，要求服务队中至少有 1 名女生，共有_____种不同的选法．（用数字作答）【答案】12. 在中，边上的中垂线分别交于点．若，则_____．【答案】【解析】设，则，，又，即，故答案为.二、选择题（本大题共有4题，满分20分）13. 展开式为的行列式是（）A. B. C. D.【答案】B【解析】,错误；，正确; ，错误；，错误，故选B.14. 设，若，则（）A. B. C. D.【答案】D【解析】，当时，不成立，根据对数函数的定义，可知真数必需大于零，故不成立，由于正弦函数具有周期性和再某个区间上为单调函数，故不能比较，故不成立，根据指数函数的单调性可知，正确，故选D.15. 已知等差数列的公差为，前项和为，则“”是“”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【答案】C【解析】，，充分性成立，若“”则，必要性成立，所以“”是“”的充分必要条件，故选C.【方法点睛】本题通过等差数列前项和的基本量运算，主要考查充分条件与必要条件，属于中档题. 判断充要条件应注意：首先弄清条件和结论分别是什么，然后直接依据定义、定理、性质尝试.对于带有否定性的命题或比较难判断的命题，除借助集合思想化抽象为直观外，还可利用原命题和逆否命题、逆命题和否命题的等价性，转化为判断它的等价命题；对于范围问题也可以转化为包含关系来处理.16. 直线与双曲线的渐近线交于两点，设为双曲线上任一点，若为坐标原点），则下列不等式恒成立的是（）A. B. C. D.【答案】C【解析】由题意，双曲线渐近线方程为，联立直线，解得不妨设，，，为双曲线上的任意一点，，，时等号成立），可得，故选C.【方法点睛】本题主要考查双曲线的的渐近线、向量相等的应用以及平面向量的坐标运算、不等式的性质，属于难题．向量的运算有两种方法，一是几何运算，往往结合平面几何知识和三角函数知识解答，运算法则是：（１）平行四边形法则（平行四边形的对角线分别是两向量的和与差）；（２）三角形法则（两箭头间向量是差，箭头与箭尾间向量是和）；二是坐标运算：建立坐标系转化为解析几何问题解答，解答本题的关键是根据坐标运算．三、解答题（本大题共有5题，满分76分）17. 如图，长方体中，与底面所成的角为.（1）求四棱锥的体积；（2）求异面直线与所成角的大小．【答案】(1) ；（2）.【解析】试题分析：（1）先证明是与底面所成的角，可得，利用棱锥的体积公式可得结果；（2）由，可得是异面直线与所成角（或所成角的补角），利用余弦定理可得结果．试题解析：（1）∵长方体ABCD﹣A1B1C1D1中，AB=BC=2，∴AA1⊥平面ABCD，AC==2，∴∠A1CA是A1C与底面ABCD所成的角，∵A1C与底面ABCD所成的角为60°，∴∠A1CA=60°，∴AA1=AC•tan60°=2•=2，∵S正方形ABCD=AB×BC=2×2=4，∴四棱锥A1﹣ABCD的体积：V===．（2）∵BD∥B1D1，∴∠A1BD是异面直线A1B与B1D1所成角（或所成角的补角）．∵BD=，A1D=A1B==2，∴cos∠A1BD===．∴∠A1BD=arccos．∴异面直线A1B与 B1D1所成角是arccos．..................18. 已知．（1）求的最大值及该函数取得最大值时的值；（2）在中，分别是角所对的边，若，且，求边的值．【答案】(1)，；（2）.【解析】试题分析：（1）跟据二倍角的正弦、余弦公式以及两角和的正弦公式可得，根据正弦函数的图象与性质可得结果；（2）由，得，结合三角形内角的范围可得或，讨论两种情况分别利用余弦定理可求出边的值.试题解析：f（x）=2sinxcosx+2cos2x﹣1=sin2x+cos2x=2sin（2x+）（1）当2x+=时，即x=（k∈Z），f（x）取得最大值为2；（2）由f（）=，即2sin（A+）=可得sin（A+）=∵0＜A＜π∴＜A＜∴A=或∴A=或当A=时，cosA==∵a=，b=，解得：c=4当A=时，cosA==0∵a=，b=，解得：c=2．19. 2016 年崇明区政府投资 8 千万元启动休闲体育新乡村旅游项目．规划从 2017 年起，在今后的若干年内，每年继续投资 2 千万元用于此项目.2016 年该项目的净收入为 5 百万元，并预测在相当长的年份里，每年的净收入均为上一年的基础上增长．记 2016 年为第 1 年，为第 1 年至此后第年的累计利润（注：含第年，累计利润=累计净收入﹣累计投入，单位：千万元），且当为正值时，认为该项目赢利．（1）试求的表达式；（2）根据预测，该项目将从哪一年开始并持续赢利？请说明理由．【答案】(1)；（2）.【解析】试题分析：（1）由题意知，第一年至此后第年的累计投入为（千万元），第年至此后第年的累计净收入为，利用等比数列数列的求和公式可得；（2）由，利用指数函数的单调性即可得出. 试题解析：（1）由题意知，第1年至此后第n（n∈N*）年的累计投入为8+2（n﹣1）=2n+6（千万元），第1年至此后第n（n∈N*）年的累计净收入为+×+×+…+×=（千万元）．∴f（n）=﹣（2n+6）=﹣2n﹣7（千万元）．（2）方法一：∵f（n+1）﹣f（n）=[﹣2（n+1）﹣7]﹣[﹣2n﹣7]=[﹣4]，∴当n≤3时，f（n+1）﹣f（n）＜0，故当n≤4时，f（n）递减；当n≥4时，f（n+1）﹣f（n）＞0，故当n≥4时，f（n）递增．又f（1）=﹣＜0，f（7）=≈5×﹣21=﹣＜0，f（8）=﹣23≈25﹣23=2＞0．∴该项目将从第8年开始并持续赢利．答：该项目将从2023年开始并持续赢利；方法二：设f（x）=﹣2x﹣7（x≥1），则f′（x）=，令f'（x）=0，得=≈=5，∴x≈4．从而当x∈[1，4）时，f'（x）＜0，f（x）递减；当x∈（4，+∞）时，f'（x）＞0，f（x）递增．又f（1）=﹣＜0，f（7）=≈5×﹣21=﹣＜0，f（8）=﹣23≈25﹣23=2＞0．∴该项目将从第8年开始并持续赢利．答：该项目将从2023年开始并持续赢利．20. 在平面直角坐标系中，已知椭圆的两个焦点分别是，直线与椭圆交于两点．（1）若为椭圆短轴上的一个顶点，且是直角三角形，求的值；（2）若，且是以为直角顶点的直角三角形，求与满足的关系；（3）若，且，求证：的面积为定值．【答案】(1)或；（2）；（3）证明见解析.【解析】试题分析：（1）根据为等腰直角三角形，可得，两种情况讨论，可得的值为或；（2）当时，，设，由，即，由韦达定理及平面向量数量积公式可得结果；（3）由可得，结合韦达定理可得，根据以上结论，利用三角形面积公式化简即可得结论.试题解析：（1）∵M为椭圆短轴上的一个顶点，且△MF1F2是直角三角形，∴△MF1F2为等腰直角三角形，∴OF1=OM，当a＞1时，=1，解得a=，当0＜a＜1时，=a，解得a=，（2）当k=1时，y=x+m，设A（x1，y1），（x2，y2），由，即（1+a2）x2+2a2mx+a2m2﹣a2=0，∴x1+x2=﹣，x1x2=，∴y1y2=（x1+m）（x2+m）=x1x2+m（x1+x2）+m2=，∵△OAB是以O为直角顶点的直角三角形，∴•=0，∴x1x2+y1y2=0，∴+=0，∴a2m2﹣a2+m2﹣a2=0∴m2（a2+1）=2a2，（3）证明：当a=2时，x2+4y2=4，设A（x1，y1），（x2，y2），∵k OA•k OB=﹣，∴•=﹣，∴x1x2=﹣4y1y2，由，整理得，（1+4k2）x2+8kmx+4m2﹣4=0．∴x1+x2=，x1x2=，∴y1y2=（kx1+m）（kx2+m）=k2x1x2+km（x1+x2）+m2=++m2=，∴=﹣4×，∴2m2﹣4k2=1，∴|AB|=•=•=2•=∵O到直线y=kx+m的距离d==，∴S△OAB=|AB|d==•==1.【方法点睛】本题主要考查待定系数法求椭圆标准方程、韦达定理以及平面向量数量积公式、圆锥曲线的定值问题，属于难题. 探索圆锥曲线的定值问题常见方法有两种：① 从特殊入手，先根据特殊位置和数值求出定值，再证明这个值与变量无关；② 直接推理、计算，并在计算推理的过程中消去变量，从而得到定值.21. 若存在常数，使得对定义域内的任意，都有成立，则称函数在其定义域上是“利普希兹条件函数”．（1）若函数是“利普希兹条件函数”，求常数的最小值；（2）判断函数是否是“利普希兹条件函数”，若是，请证明，若不是，请说明理由；（3）若是周期为2的“利普希兹条件函数”，证明：对任意的实数,都有．【答案】(1)；（2）不是，理由见解析；（3）证明见解析.【解析】试题分析：（1）不妨设，则恒成立．，从而可得结果；（2）令，则，从而可得函数不是“利普希兹条件函数”；（3）设的最大值为，最小值为，在一个周期，内，利用基本不等式的性质可证明.试题解析：（1）若函数f（x）=，（1≤x≤4）是“k﹣利普希兹条件函数”，则对于定义域[1，4]上任意两个x1，x2（x1≠x2），均有|f（x1）﹣f（x2）|≤k|x1﹣x2|成立，不妨设x1＞x2，则k≥=恒成立．∵1≤x2＜x1≤4，∴＜＜，∴k的最小值为．（2）f（x）=log2x的定义域为（0，+∞），令x1=，x2=，则f（）﹣f（）=log2﹣log2=﹣1﹣（﹣2）=1，而2|x1﹣x2|=，∴f（x1）﹣f（x2）＞2|x1﹣x2|，∴函数f（x）=log2x 不是“2﹣利普希兹条件函数”．（3）设f（x）的最大值为M，最小值为m，在一个周期[0，2]内f（a）=M，f（b）=m，则|f（x1）﹣f（x2）|≤M﹣m=f（a）﹣f（b）≤|a﹣b|．若|a﹣b|≤1，显然有|f（x1）﹣f（x2）|≤|a﹣b|≤1．若|a﹣b|＞1，不妨设a＞b，则0＜b+2﹣a＜1，∴|f（x1）﹣f（x2）|≤M﹣m=f（a）﹣f（b+2）≤|a﹣b﹣2|＜1．综上，|f（x1）﹣f（x2）|≤1．。

2018年上海市高考数学一模试卷(解析卷)

2018年上海市高考数学一模试卷(解析卷)

2018年上海市高考数学试卷一.填空题（本大题满分54分）本大题共有12题，1-6每题4分，7-12每题5分考生应在答题纸相应编号的空格内直接填写结果，每个空格填对得分，否则一律得零分.1．（4分）设全集U=Z，集合M={1，2}，P={﹣2，﹣1，0，1，2}，则P∩C U M {﹣2，﹣1，0} ．【解答】解：C U M={﹣2，﹣1，0}，故P∩C U M={﹣2，﹣1，0}故答案为：{﹣2，﹣1，0}2．（4分）已知复数（i为虚数单位），则=．【解答】解：复数==，∴=，∴=•==，故答案为．3．（4分）不等式2＞（）3（x﹣1）的解集为（﹣∞，﹣2）∪（3，+∞）．【解答】解：不等式2＞（）3（x﹣1）化为2＞23﹣3x，即x2﹣4x﹣3＞3﹣3x，∴x2﹣x﹣6＞0，解得x＜﹣2或x＞3，∴原不等式的解集为（﹣∞，﹣2）∪（3，+∞）．故答案为：（﹣∞，﹣2）∪（3，+∞）．4．（4分）函数f（x）=sinxcosx+cos2x的最大值为．【解答】解：函数f（x）=sinxcosx+cos2x=sin2x+cos2x+=sin（2x+）+，当2x+=2kπ+，k∈Z，即x=kπ+，k∈Z，函数取得最大值1+=，故答案为：．5．（4分）在平面直角坐标系xOy中，以直线y=±2x为渐近线，且经过椭圆x2+=1右顶点的双曲线的方程是x2﹣=1．【解答】解：设以直线y=±2x为渐近线的双曲线的方程为x2﹣=λ（λ≠0），∵双曲线椭圆x2+=1右顶点（1，0），∴1=λ，∴双曲线方程为：x2﹣=1．故答案为：x2﹣=1．6．（4分）将圆锥的侧面展开后得到一个半径为2的半圆，则此圆锥的体积为．【解答】解：设圆锥的底面半径为r，则2πr=2π，∴r=1．∴圆锥的高h=．∴圆锥的体积V==．故答案为：．7．（5分）设等差数列{a n}的公差d不为0，a1=9d．若a k是a1与a2k的等比中项，则k=4．【解答】解：因为a k是a1与a2k的等比中项，则a k2=a1a2k，[9d+（k﹣1）d]2=9d•[9d+（2k﹣1）d]，又d≠0，则k2﹣2k﹣8=0，k=4或k=﹣2（舍去）．故答案为：4．8．（5分）已知（1+2x）6展开式的二项式系数的最大值为a，系数的最大值为b，则=12．【解答】解：由题意可得a==20，再根据，解得，即≤r≤，∴r=4，此时b=×24=240；∴==12．故答案为：12．9．（5分）同时掷两枚质地均匀的骰子，则两个点数之积不小于4的概率为．【解答】解：同时掷两枚质地均匀的骰子，基本事件总数n=6×6=36，两个点数之积小于4包含的基本事件（a，b）有：（1，1），（1，2），（2，1），（1，3），（3，1），共5个，∴两个点数之积不小于4的概率为p=1﹣=．故答案为：．10．（5分）已知函数f（x）=有三个不同的零点，则实数a的取值范围是[1，+∞）．【解答】解：由题意可知：函数图象的左半部分为单调递增对数函数的部分，函数图象的右半部分为开口向上的抛物线，对称轴为x=，最多两个零点，如上图，要满足题意，必须指数函数的部分向下平移到与x轴相交，由对数函数过点（1，0），故需左移至少1个单位，故a≥1，还需保证抛物线与x轴由两个交点，故最低点＜0，解得a＜0或a＞，综合可得：a≥1，故答案为：[1，+∞）．11．（5分）已知S n为数列{a n}的前n项和，a1=a2=1，平面内三个不共线的向量，，，满足=（a n﹣1+a n+1）+（1﹣a n），n≥2，n∈N*，若A，B，C在同一直线上，则S2018=2．【解答】解：若A，B，C三点共线，则=x+（1﹣x），∴根据条件“平面内三个不共线的向量，，，满足=（a n﹣1+a n+1）+（1﹣a n），n≥2，n∈N*，A，B，C在同一直线上，”得出a n﹣1+a n+1+1﹣a n=1，∴a n﹣1+a n+1=a n，∵S n为数列{a n}的前n项和，a1=a2=1，∴数列{a n}为：1，1，0，﹣1，﹣1，0，1，1，0，﹣1，﹣1，0，…即数列{a n}是以6为周期的周期数列，前6项为1，1，0，﹣1，﹣1，0，∵2018=6×336+2，∴S2018=336×（1+1+0﹣1﹣1+0）+1+1=2．故答案为：2．12．（5分）已知函数f（x）=m（x﹣m）（x+m+2）和g（x）=3x﹣3同时满足以下两个条件：①对任意实数x都有f（x）＜0或g（x）＜0；②总存在x0∈（﹣∞，﹣2），使f（x0）g（x0）＜0成立．则m的取值范围是（﹣3，﹣2）．【解答】解：对于①∵g（x）=3x﹣3，当x＜1时，g（x）＜0，又∵①∀x∈R，f（x）＜0或g（x）＜0∴f（x）=m（x﹣m）（x+m+2）＜0在x≥1时恒成立则由二次函数的性质可知开口只能向下，且二次函数与x轴交点都在（1，0）的左面，即，可得﹣3＜m＜0又∵②x∈（﹣∞，﹣2），f（x）g（x）＜0∴此时g（x）=3x﹣3＜0恒成立∴f（x）=m（x﹣m）（x+m+2）＞0在x∈（﹣∞，﹣2）有成立的可能，则只要﹣2比x1，x2中的较小的根大即可，（i）当﹣1＜m＜0时，较小的根为﹣m﹣2，﹣m﹣2＞﹣2不成立，（ii）当m=﹣1时，两个根同为﹣1＞﹣3，不成立，（iii）当﹣3＜m＜﹣1时，较小的根为m，即m＜﹣2成立．综上可得①②成立时﹣3＜m＜﹣2．故答案为：（﹣3，﹣2）．二.选择题（本大题满分20分）本大题共有4题，每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分.13．（5分）“a＞b”是“（）2＞ab”成立的（）A．充分而不必要条件B．必要而不充分条件C．充要条件D．既不充分又不必要条件【解答】解：由（）2＞ab得＞ab，即a2+2ab+b2＞4ab，则a2﹣2ab+b2＞0，即（a﹣b）2＞0，则a≠b，则“a＞b”是“（）2＞ab”成立的充分不必要条件，故选：A．14．（5分）已知函数f（x）=2sin（x+），若对任意实数x，都有f（x1）≤f （x）≤f（x2），则|x2﹣x1|的最小值是（）A．πB．2πC．2 D．4【解答】解：对于函数f（x）=2sin（x+），若对任意实数x，都有f（x1）≤f（x）≤f（x2），则|x2﹣x1|的最小值为函数f（x）的半个周期，即===2，故选：C．15．（5分）已知和是互相垂直的单位向量，向量满足：，，n∈N*，设θn为和的夹角，则（）A．θn随着n的增大而增大B．θn随着n的增大而减小C．随着n的增大，θn先增大后减小D．随着n的增大，θn先减小后增大【解答】解：分别以和所在的直线为x轴，y轴建立坐标系，则=（1，0），=（0，1），设=（x n，y n），∵，，n∈N*，∴x n=n，y n=2n+1，n∈N*，∴=（n，2n+1），n∈N*，∵θn为和的夹角，∴tanθn===2+∴y=tanθn为减函数，∴θn随着n的增大而减小．故选：B．16．（5分）在平面直角坐标系xOy中，已知两圆C1：x2+y2=12和C2：x2+y2=14，又点A坐标为（3，﹣1），M、N是C1上的动点，Q为C2上的动点，则四边形AMQN能构成矩形的个数为（）A．0个 B．2个 C．4个 D．无数个【解答】解：如图所示，任取圆C2上一点Q，以AQ为直径画圆，交圆C1与M、N两点，则四边形AMQN能构成矩形，由作图知，四边形AMQN能构成矩形的个数为无数个．故选：D．三．解答题（本大题满分76分）本大题共有5题，解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.17．（14分）如图，在四棱锥P﹣ABCD中，底面ABCD是矩形，PA⊥平面ABCD，PA=AD=2AB=2，E是PB的中点．（1）求三棱锥P﹣ABC的体积；（2）求异面直线EC和AD所成的角（结果用反三角函数值表示）．【解答】解：（1）∵PA⊥平面ABCD，底面ABCD是矩形，高PA=2，BC=AD=2，AB=1，==1．∴S△ABC故V P==．﹣ABC（2）∵BC∥AD，∴∠ECB或其补角为异面直线EC和AD所成的角θ，又∵PA⊥平面ABCD，∴PA⊥BC，又BC⊥AB，∴BC⊥平面PAB，∴BC⊥PB，于是在Rt△CEB中，BC=2，BE=PB=，tanθ==，∴异面直线EC和AD所成的角是arctan．18．（14分）已知抛物线C：y2=2px过点P（1，1）．过点（0，）作直线l与抛物线C交于不同的两点M，N，过点M作x轴的垂线分别与直线OP、ON交于点A，B，其中O为原点．（1）求抛物线C的方程，并求其焦点坐标和准线方程；（2）求证：A为线段BM的中点．【解答】解：（1）∵y2=2px过点P（1，1），∴1=2p，解得p=，∴y2=x，∴焦点坐标为（，0），准线为x=﹣，（2）证明：设过点（0，）的直线方程为y=kx+，M（x1，y1），N（x2，y2），∴直线OP为y=x，直线ON为：y=x，由题意知A（x1，x1），B（x1，），由，可得k2x2+（k﹣1）x+=0，∴x1+x2=，x1x2=∴y1+=kx1++=2kx1+=2kx1+=2kx1+（1﹣k）•2x1=2x1，∴A为线段BM的中点．19．（14分）如图，某大型厂区有三个值班室A、B、C．值班室A在值班室B的正北方向2千米处，值班室C在值班室B的正东方向2千米处．（1）保安甲沿CA从值班室出发行至点P处，此时PC=1，求PB的距离；（2）保安甲沿CA从值班室C出发前往值班室A，保安乙沿AB从值班室A出发前往值班室B，甲乙同时出发，甲的速度为1千米/小时，乙的速度为2千米/小时，若甲乙两人通过对讲机联系，对讲机在厂区内的最大通话距离为3千米（含3千米），试问有多长时间两人不能通话？【解答】解：（1）在Rt△ABC中，AB=2，BC=2，所以∠C=30°，在△PBC中PC=1，BC=2，由余弦定理可得BP2=BC2+PC2﹣2BC•PCcos30°=（2）2+1﹣2×2×1×=7，即BP=；（2）在Rt△ABC中，BA=2，BC=2，AC==4，设甲出发后的时间为t小时，则由题意可知0≤t≤4，设甲在线段CA上的位置为点M，则AM=4﹣t，①当0≤t≤1时，设乙在线段AB上的位置为点Q，则AQ=2t，如图所示，在△AMQ中，由余弦定理得MQ2=（4﹣t）2+（2t）2﹣2•2t•（4﹣t）cos60°=7t2﹣16t+7＞9，解得t＜或t＞，所以0≤t≤；②当1≤t≤4时，乙在值班室B处，在△ABM中，由余弦定理得MB2=（4﹣t）2+4﹣2•2t•（4﹣t）cos60°=t2﹣6t+12＞9，解得t＜3﹣或t＞3+，又1≤t≤4，不合题意舍去．综上所述0≤t≤时，甲乙间的距离大于3千米，所以两人不能通话的时间为小时．20．（16分）设集合A，B均为实数集R的子集，记A+B={a+b|a∈A，b∈B}．（1）已知A={0，1，2}，B={﹣1，3}，试用列举法表示A+B；（2）设a1=，当n∈N*且n≥2时，曲线+=的焦距为a n，如果A={a1，a2，…，a n}，B={﹣，﹣，﹣}，设A+B中的所有元素之和为S n，求S n的值；（3）在（2）的条件下，对于满足m+n=3k，且m≠n的任意正整数m，n，k，不等式S m+S n﹣λS k＞0恒成立，求实数λ的最大值．【解答】解：（1）∵A+B={a+b|a∈A，b∈B}；当A={0，1，2}，B={﹣1，3}时，A+B={﹣1，0，1，3，4，5}；（2）曲线+=，即﹣=，在n≥2时表示双曲线，故a n=2=n，∴a1+a2+a3+…+a n=∵B={﹣，﹣，﹣}，∴A+B中的所有元素之和为S n=3（a1+a2+a3+…+a n）+n（﹣﹣﹣）=3•+n （﹣﹣﹣）=n2，（3）∵∴S m+S n﹣λS k＞0恒成立⇔λ＜=恒成立，∵m+n=3k，且m≠n，∴==＞，∴λ≤，故实数λ的最大值为21．（18分）对于定义在[0，+∞）上的函数f（x），若函数y=f（x）﹣（ax+b）满足：①在区间[0，+∞）上单调递减，②存在常数p，使其值域为（0，p]，则称函数g（x）=ax+b是函数f（x）的“逼进函数”．（1）判断函数g（x）=2x+5是不是函数f（x）=，x∈[0，+∞）的“逼进函数”；（2）求证：函数g（x）=x不是函数f（x）=（）x，x∈[0，+∞）的“逼进函数”（3）若g（x）=ax是函数f（x）=x+，x∈[0，+∞）的“逼进函数”，求a 的值．【解答】解：（1）f（x）﹣g（x）=﹣（2x+5）=，可得y=f（x）﹣g（x）在[0，+∞）递减，且x+2≥2，0＜≤，可得存在p=，函数y的值域为（0，]，则函数g（x）=2x+5是函数f（x）=，x∈[0，+∞）的“逼进函数”；（2）证明：f（x）﹣g（x）=（）x﹣x，由y=（）x，y=﹣x在[0，+∞）递减，则函数y=f（x）﹣g（x）在[0，+∞）递减，则函数y=f（x）﹣g（x）在[0，+∞）的最大值为1；由x=1时，y=﹣=0，x=2时，y=﹣1=﹣＜0，则函数y=f（x）﹣g（x）在[0，+∞）的值域为（﹣∞，1]，即有函数g（x）=x不是函数f（x）=（）x，x∈[0，+∞）的“逼进函数”；（3）g（x）=ax是函数f（x）=x+，x∈[0，+∞）的“逼进函数”，可得y=x+﹣ax为[0，+∞）的减函数，可得导数y′=1﹣a+≤0在[0，+∞）恒成立，可得a﹣1≥，由x＞0时，=≤1，则a﹣1≥1，即a≥2；又y=x+﹣ax在[0，+∞）的值域为（0，1]，则＞（a﹣1）x，x=0时，显然成立；x＞0时，a﹣1＜，可得a﹣1≤1，即a≤2．则a=2．。

【2019高考模拟试题 全国卷】2018-2019学年度模拟测试(一)附答案

【2019高考模拟试题 全国卷】2018-2019学年度模拟测试(一)附答案

学年度模拟测答案第2页，总12页A ．2B ．3C ．4D ．55.集合}{220A x x x =--≤，{}10B x x =-<，则A ∪B = ( ). A.}{1x x < B.}{11x x -≤< C.{}2x x ≤ D.{}21x x -≤< 6.下列选项中，可以作为a b >的必要不充分条件的是 A.000,x a x b ∃≤+> B. 000,x a x b ∃<+≥ C. 0,-x x a b ∀≥> D. 0,-x a b x ∀>≥ 7.设31ln ,2,log 313===c b a π ，则a ，b ，c 的大小是 A. a ＞b ＞c B.b ＞a ＞c C. b ＞c ＞a D.a ＞c ＞b8.已知i 为虚数单位，41iz =+，则复数z 的虚部为( ).A.－2iB.2iC.2D. －2 9.已知函数)(x f y =是定义域为R 的偶函数，当0≥x 时，⎪⎪⎩⎪⎪⎨⎧+≤≤=,2>,1)21(,20,4sin 45)(x x x x f x π，若关于x 的方程),(0)()]([2R b a b x af x f ∈=++.有且仅有6个不同的实数根，则实数的a 取值范围是A. 5(,1)2--B. 599(,)(,1)244---- C. 59(,24-- D. 9(,1)4-- 10.已知函数)1＞,2＜||其中)(sin()(ωπϕϕω+=x x f 是R 上的奇函数，在区间4,0[π上具有单调性，且)(x f y =图象的一条对称轴是直线43π=x ，若锐角△ABC 满足21)42(=-B A f ，21)24(-=-B A f ,则)(C f 的值为A. 23B. 21C. 23-D. 21- 11.已知x ，y 满足不等式组22y xx y x ≤⎧⎪+≥⎨⎪≤⎩，则2z x y =+的最大值与最小值的比值为第3页，总12页( ) A.21 B.2 C.23 D.34 12.已知数列{a n }的前n项和为S n , 首项321-=a ,且满足)2(21≥=++n a S S n nn ,则2018S 等于 A. 20172016- B. 20182017- C. 20192018- D. 20202019-答案第4页，总12页第II 卷（非选择题）二、填空题（本题共4道小题，每小题0分，共013.若()12nx +的二项展开式中的第3项的二项式系数为15，则()12nx +中含3x 项的系数为 ． 14.在正四棱柱ABCD -A 1B 1C 1D 1中，AB =12AA =，设四棱柱的外接球的球心为O ，动点P 在正方形ABCD 的边上，射线OP 交球O 的表面于点M P 从点A 出发，沿着A B C D A →→→→运动一次，则点M 经过的路径长为 . 15.在平面直角坐标系xOy 中，点n A (()12 2nn n n+-⋅，)(*n N ∈)，记21221n n n A A A -+∆积为S n ，则1ni i S ==∑ .16.已知函数()1xx e f x e =+，数列{a n }为等比数列，0n a >，10101a =，则32019(ln )(ln )a f a ++= ．三、解答题（本题共6道小题,第1题0分,第2题0题0分,第4题0分,第5题0分,第6题0分,共0S n , 且)(42*∈+=N n n n S n .{b n }的前n 的项和T n .4万元，并且每生产1百台产品需增R (x )（万元）满足,)100(≤≤x x （其中x 是该产品的月产量，单位：百台），x 的函数)(x f y =；=BC =2, ∠B =90°,D 为BC 边上一点，以边AC 为对角线AC 将△ACE 折起，使得平面ACE ⊥平面ABC ,如图答案第6页，总12页试卷答案1.D2.B∵双曲线的中心在坐标原点，顶点在椭圆上，且与抛物线有相同的焦点∴双曲线的顶点在轴上，且半焦距，顶点坐标为∴双曲线的半实轴长为，则双曲线的半虚轴长为∴其渐近线方程为故选B3.D等差数列中，是函数的两个零点，，的前10项和.故选：D.4.B5.C解得集合，所以，故选C。

2018——2019年上海各区高中数学高三数学一模试卷试题汇总

2018——2019年上海各区高中数学高三数学一模试卷试题汇总

第一学期教学质量检测高三数学试卷一、填空题（本大题共有12题，满分54分）只要求直接填写结果，1-6题每个空格填对得4分，7-12题每个空格填对得5分，否则一律得零分. 1. 已知全集R U =，集合(][)12,,=-∞+∞A ，则U=A ______________.()12,2. 抛物线24=y x 的焦点坐标为_________.()10, 3. 不等式2log 1021>x 的解为____________.4(,)+∞4. 已知复数z 满足(1i)4i z +⋅=（i 为虚数单位），则z 的模为_________. 225. 若函数()=y f x 的图像恒过点01(,)，则函数13()-=+y fx 的图像一定经过定点____.()13,6. 已知数列{}n a 为等差数列，其前n 项和为n S .若936=S ，则348++=a a a ________.127. 在△ABC 中，内角,,A B C 的对边是,,a b c .若22)32(b a ⋅+=，c b =，则=A ___.56π 8. 已知圆锥的体积为π33，母线与底面所成角为3π，则该圆锥的表面积为 ．π3 9.已知二项式n的展开式中，前三项的二项式系数之和为37，则展开式中的第五项为________.358x 10. 已知函数()2||1=+-f x x x a 有三个不同的零点，则实数a 的取值范围为_____.(,-∞11. 已知数列{}n a 满足：211007(1)2018(1)++=-++n n n na n a n a *()∈n N ， 且121,2,a a ==若1lim,+→∞=n n na A a 则=A ___________. 100912. 已知函数()2,24161,22-⎧≥⎪+⎪=⎨⎛⎫⎪< ⎪⎪⎝⎭⎩x ax x x f x x ，若对任意的[)12,∈+∞x ，都存在唯一的()2,2∈-∞x ，满足()()12=f x f x ，则实数a 的取值范围为_________. [)2,6∈-a解：当[)12,∈+∞x 时，1211041616x x ⎛⎤∈ ⎥+⎝⎦，.当()2,2∈-∞x 时，（1）若2a ≥，则()11=22x aa xf x --⎛⎫⎛⎫= ⎪⎪⎝⎭⎝⎭在(),2-∞上是单调递增函数，所以()2210,2a f x -⎛⎫⎛⎫∈ ⎪ ⎪ ⎪⎝⎭⎝⎭.若满足题目要求，则21100,162a -⎛⎫⎛⎤⎛⎫⊆ ⎪ ⎪⎥ ⎪⎝⎦⎝⎭⎝⎭，，所以24111,24,62162a a a -⎛⎫⎛⎫>=∴-<< ⎪⎪⎝⎭⎝⎭.又2a ≥，所以[)2,6a ∈. （2）若2a <，则()1,,21=21, 2.2a xx ax ax a f x a x ---⎧⎛⎫<⎪ ⎪⎪⎝⎭⎛⎫=⎨ ⎪⎝⎭⎛⎫⎪≤< ⎪⎪⎝⎭⎩，()f x 在(),a -∞上是单调递增函数，此时()()0,1f x ∈；()f x 在[),2a 上是单调递减函数，此时()21,12a f x -⎛⎤⎛⎫∈ ⎥ ⎪ ⎝⎭⎥⎝⎦.若满足题目要求，则211,2162aa -⎛⎫≤∴≥- ⎪⎝⎭，又2a <，所以[)2,2a ∈-.综上，[)2,6a ∈-.二、选择题(本大题共有4题，满分20分) 每小题都给出四个选项，其中有且只有一个选项是正确的，选对得 5分，否则一律得零分. 13. “14<a ”是“一元二次方程20-+=x x a 有实数解”的（ A ） （A ）充分非必要条件 （B ）充分必要条件（C ）必要非充分条件 （D ）非充分非必要条件 14. 下列命题正确的是（ D ）（A ）如果两条直线垂直于同一条直线，那么这两条直线平行（B ）如果一条直线垂直于一个平面内的两条直线，那么这条直线垂直于这个平面 （C ）如果一条直线平行于一个平面内的一条直线，那么这条直线平行于这个平面 （D ）如果一个平面内的两条相交直线与另一个平面平行，那么这两个平面平行15. 将4位志愿者分配到进博会的3个不同场馆服务，每个场馆至少1人，不同的分配方案有（ B ）种.（A ）72 （B ）36 （ （D ）81 16. 已知点()()1,2,2,0-A B ，P ⋅AP AB 的取值范围为（ A ）（A ）[]1,7 （B ）[]1,7- （C）1,3⎡+⎣ （D）1,3⎡-+⎣三、解答题（本大题共有5题，满分76分）解答下列各题必须写出必要的步骤． 17．（本小题满分14分，第1小题满分7分，第2小题满分7分 已知直三棱柱ABC C B A -111中，︒=∠===9011BAC ,AA AC AB .（1）求异面直线B A 1与11C B 所成角； （2）求点1B 到平面BC A 1的距离．解：（1）在直三棱柱ABC C B A -111中，AB AA ⊥1，AC AA ⊥1,︒=∠===9011BAC ,AA AC AB所以，211===BC C A B A ．…………………………2分因为，11C B //BC ，所以，BC A 1∠为异面直线B A 1与11C B 所成的角或补角．……4分 在BC A 1∆中，因为，211===BC C A B A ，所以，异面直线B A 1与11C B 所成角为3π．…………………………7分 （2）设点1B 到平面BC A 1的距离为h ， 由（1）得23322211=π⋅⨯⨯=∆sin S BC A ，…………………………9分 21112111=⨯⨯=∆B B A S ，…………………………11分 因为，B B A C BC A B V V 1111--=，…………………………12分所以，CA S h S B B A BC A ⋅=⋅∆∆1113131，解得，33=h ． 所以，点1B 到平面BC A 1的距离为33．…………………………14分 或者用空间向量：（1） 设异面直线B A 1与11C B 所成角为θ，如图建系，则()1011-=,,A ，()01111,,C B -=，…………4分A1C CB1B 1A因为，321221π=θ⇒=⋅-==θcos 所以，异面直线B A 1与11C B 所成角为3π．…………7分 （2）设平面BC A 1的法向量为()w ,v ,u n =，则B A n ,BC n 1⊥⊥． 又()011,,-=，()1011-=,,A ，……………9分所以，由⎩⎨⎧=-=+-⇒⎪⎩⎪⎨⎧=⋅=⋅00001w u v u A ，得()111,,n =．…………12分所以，点1B 到平面BC A 1的距离33==d ．…………………………14分 18．（本小题满分14分，第1小题满分7分，第2小题满分7分）已知函数2()cos 2sin f x x x x =-.（1）若角α的终边与单位圆交于点3455(,)P ，求()f α的值； （2）当[,]63ππ∈-x 时，求()f x 的单调递增区间和值域.解：（1）∵角α的终边与单位圆交于点3455(,)P ，∴43sin =,cos =55αα ……2分2243432()cos 2sin 2()55525αααα=-=⨯-⨯=f …4分（2）2()cos 2sin f x x x x =-2cos21x x =+- …………………6分2sin(2)16x π=+- …………………………8分由222262k x k πππππ-≤+≤+得，36k x k ππππ-≤≤+又[,]63x ππ∈-，所以()f x 的单调递增区间是[,]66x ππ∈-； ………………10分∵[,]63x ππ∈-，∴52666x πππ-≤+≤…………………………12分 ∴1sin(2)126x π-≤+≤，()f x 的值域是[2,1]-. ………………14分19．（本小题满分14分，第1小题满分6分，第2小题满分8分） 某游戏厂商对新出品的一款游戏设定了“防沉迷系统”，规则如下：①3小时以内（含3小时）为健康时间，玩家在这段时间内获得的累积经验值．．．．．E （单位：exp ）与游玩时间t （小时）满足关系式：22016E t t a =++；②3到5小时（含5小时）为疲劳时间，玩家在这段时间内获得的经验值为0（即累积经验．．．．值．不变）； ③超过5小时为不健康时间，累积经验值．．．．．开始损失，损失的经验值与不健康时间成正比例关系，比例系数为50.（1）当1a =时，写出累积经验值．．．．．E 与游玩时间t 的函数关系式()E f t =，并求出游玩6小时的累积经验值．．．．．； （2）该游戏厂商把累积经验值．．．．．E 与游玩时间t 的比值称为“玩家愉悦指数”，记作()H t ；若0a >，且该游戏厂商希望在健康时间内，这款游戏的“玩家愉悦指数”不低于24，求实数a的取值范围.解：（1）22016,03()85,3533550,5t t t E f t t t t ⎧++<≤⎪==<≤⎨⎪->⎩ （写对一段得1分，共3分）6t =时，(6)35E =    （6分）（2）03t <≤时，16()=20aH t t t++  （8分） 16()244≥⇒+≥aH t t t①0319[,]4164a ⎧<≤⎪⇒∈⎨⎪⎩     （10分） ②39(,)1616343a a⎧>⎪⇒∈+∞⎨+≥⎪⎩    （12分） 综上，1[,)4a ∈+∞        （14分）20．（本小题满分16分，第1小题满分4分，第2小题满分6分，第3小题满分6分）已知双曲线Γ: 22221(0,0)x y a b a b-=>>的左、右焦点分别是 1F 、2F ，左、右两顶点分别是 1A 、2A ，弦 AB 和CD 所在直线分别平行于x 轴与 y 轴，线段BA 的延长线与线段CD 相交于点 P （如图）．（1）若(2,3)d =是Γ的一条渐近线的一个方向向量，试求Γ的两渐近线的夹角θ；（2）若1PA =，5PB = ，2PC =，4PD =，试求双曲线Γ的方程；（3）在（．．1．）的条件下．．．．．，且124A A =，点C 与双曲线的顶点不重合，直线1CA 和直线2CA 与直线:1l x =分别相交于点M 和N ，试问：以线段MN 为直径的圆是否恒经过定点？若是，请求出定点的坐标；若不是，试说明理由．解：（1）双曲线22221x y a b-=的渐近线方程为：即0bx ay ±=，所以3b a =，…………2分 从而3tan2θ=22tan 2tan 431tan2θθθ==-， 所以arctan 3θ=………………………………………………..4分（2）设 (,)P P P x y ，则由条件知：11()()322P x PB PA PA PB PA =-+=+=，11()()122P y PC PD PC PD PC =+-=-=，即(3,1)P ．…………6分所以(2,1)A ，(3,3)C ，………………………………………………………..…………7分代入双曲线方程知：2751,2781199114222222==⇒⎪⎩⎪⎨⎧=-=-b a ba b a ……9分 127527822=-y x ………………………………………………………………….. 10分 （3）因为124A A =，所以2a =，由（1）知，3b =Γ的方程为: 22143x y -=， 令00(,)C x y ，所以2200143x y -=，010:(2)2y CA y x x =++，令1x =，所以003(1,)2y M x +， 020:(2)2y CA y x x =--，令1x =，所以00(1,2y N x --， …………12分故以MN 为直径的圆的方程为：200003(1)()()022y y x y y x x --+--=+-， 即222000200033(1)()0224y y y x y y x x x -++--=-+-，即22000039(1)()0224y y x y y x x -++--=-+，…………………………………………….14分 若以MN 为直径的圆恒经过定点),(y x于是⎪⎩⎪⎨⎧=±=⇒⎪⎩⎪⎨⎧=-+-=0231049)1(022y x y x y 所以圆过x 轴上两个定点5(,0)2和1(,0)2-……………………………………………16分21．（本题满分18分，第1小题满分4分，第2小题满分6分，第3小题满分8分） 已知平面直角坐标系xOy ，在x 轴的正半轴上，依次取点123,,,n A A A A （*n N ∈），并在第一象限内的抛物线232y x =上依次取点123,,,,n B B B B （*n N ∈），使得1k k kA B A -∆*()k N ∈都为等边三角形，其中0A 为坐标原点，设第n 个三角形的边长为()f n .（1）求(1),(2)f f ，并猜想()f n （不要求证明）； （2）令9()8n a f n =-，记m t 为数列{}n a 中落在区间2(9,9)mm内的项的个数，设数列{}m t 的前m 项和为m S ，试问是否存在实数λ，使得2λ≤m S 对任意*m N ∈恒成立？若存在，求出λ的取值范围；若不存在，说明理由； （3）已知数列{}n b满足：11,2n b b +==数列{}n c 满足：111,n nc c +==求证：1()2n n n b f c π+<<.解：（1）(1)1f =，(2)2f =  （2分） 猜想()f n n =  （2分） （2）98n a n =-  （5分）由21218899899999m mm m n n --<-<⇒+<<+112191,92,,9---∴=++⋅⋅⋅⋅⋅⋅m m m n  （6分）21199m m m t --∴=-  （7分） 352211(91)(99)(99)(99)m m m S --∴=-+-+-+⋅⋅⋅+- 352121(9999)(1999)m m --=+++⋅⋅⋅+-+++⋅⋅⋅+22129(19)(19)91091191980m m m m +---⋅+=-=-- （9分） 2λ≤m S 对任意*m N ∈恒成立min 12()83λλ⇒≤==⇒≤m S S （10分）.（3）1sin,4b π=记1sin ,4n n b πθθ==，则1sin sin 2n n θθ+== *1()2n n n N πθ+⇒=∈  （12分） 1tan ,4c π=记1tan ,4n n c πϕϕ==，则1sec 1tan tan tan 2n n n n ϕϕϕϕ+-==*1()2n n n N πϕ+⇒=∈  （14分） 11sin,tan ,22n n n n b c ππ++∴==当(0,)2x π∈时，sin tan x x x <<可知： 1111sin()tan ,2222n n n n n n b f c ππππ++++=<=<=  （18分）杨浦区2018学年度第一学期高三年级模拟质量调研数学学科试卷 2018.12.18一、填空题（本大题有12题，满分54分，第1——6题每题4分，第7—12题每题5分） 1、设全集{}1,2,3,4,5U =，若集合{}3,4,5A =，则____u=2、已知扇形的半径为6，圆心角为3π，则扇形的面积为_____ 3、已知双曲线221x y -=，则其两条渐近线的夹角为_____ 4、若()na b +展开式的二项式系数之和为8，则____n = 5、若实数,x y 满足221x y +=，则xy 的取值范围是_____6、若圆锥的母线长()5l cm =，高()4h cm =，则这个圆锥的体积等于_______7、在无穷等比数列{}n a 中，()121lim ,2n n a a a →+∞+++=则1a 的取值范围是____8、若函数()1ln 1xf x x+=-的定义域为集合A ，集合(),1B a a =+，且B A ⊆,则实数a 的取值范围__9、在行列式274434651xx--中，第3行第2列的元素的代数余子式记作()f x ，则()1y f x =+的零点是____10、已知复数())12cos 2,cos z x f x i z x x i =+=++，（,x R i ∈虚数单位）在复平面上，设复数12,z z 对应的点分别为12,Z Z ，若1290Z OZ ∠=，其中是坐标原点，则函数()f x 的最小正周期______ 11、当0x a <<时，不等式()22112x a x +≥-恒成立，则实数a 的最大值为______ 12、设d 为等差数列{}n a 的公差，数列{}n b 的前项和n T ，满足()()112nn n n T b n N *+=-∈， 且52d a b ==，若实数{}()23,3k k k m P x a x a k N k *-+∈=<<∈≥，则称m 具有性质k P ，若是n H 数列{}n T 的前n 项和，对任意的n N *∈，21n H -都具有性质k P ，则所有满足条件的k 的值为_____二、选题题（本题共有4题，满分20分，每题5分）13、下列函数中既是奇函数，又在区间[]1,1-上单调递减的是（ ）（A ）()arcsin f x x= （B ）lg y x= （C ）()f x x=-（D ）()cos f x x =14、某象棋俱乐部有队员5人，其中女队员2人，现随机选派2人参加一个象棋比赛，则选出的2人中恰有1人是女队员的概率为 （ ）（A ）310 （B ） 35 （C ） 25 （D ）2315、已知()sin log ,0,2f x x θπθ⎛⎫=∈ ⎪⎝⎭，设sin cos sin ,,2sin cos a f b f c f θθθθθ+⎛⎫⎛⎫===⎪⎪+⎝⎭⎝⎭，则,,a b c 的大小关系是 （A ）a b c ≤≤ （B ）b c a ≤≤ （C ）c b a ≤≤（D ）a b c ≤≤16、已知函数()22x f x m x nx =⋅++，记集合(){}0,A x f x x R ==∈，集合(){}0,B x f x x R ==∈，若A B =，且都不是空集，则m n +的取值范围是（ ） （ A ）[]0,4 （B ）[]1,4- （C ）[]3,5- （D ）[]0,7三、解答题（本大题共有5题，满分76分） 17、（本题满分14分，第1题满分6分，第2小题满分8分）如图，,PA ABCD ⊥平面四边形ABCD 为矩形，1PA PB ==，2AD =，点F 是PB 的中点，点E 在边BC 上移动。

2019-2018届上海市普陀区高考文科数学模拟试卷及答案word版本 (6页)

2019-2018届上海市普陀区高考文科数学模拟试卷及答案word版本 (6页)

本文部分内容来自网络整理，本司不为其真实性负责，如有异议或侵权请及时联系，本司将立即删除！== 本文为word格式，下载后可方便编辑和修改！ ==2018届上海市普陀区高考文科数学模拟试卷及答案数学是一门逻辑性较强的学科，但是每年高考的题型基本上是不变的，我们可以通过多做一些模拟试卷来熟悉里面的题型，以下是小编为你整理的2018届上海市普陀区高考文科数学模拟试卷，希望能帮到你。

2018届上海市普陀区高考文科数学二模拟试卷题目一、选择题：本大题共12小题，每小题5分，在每小题给出的四个选项中，只有一项是符合题目要求的。

1、设集合A={x| }，B={y|y=x2}，则A∩B=( )A.[﹣2，2]B.[0，2]C.[2，+∞)D.{(﹣2，4)，(2，4)}2、已知条件p：关于的不等式有解;条件q：指数函数为减函数，则p成立是q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3、在△ 中，为边的中点，若，，则 ( )A. B. C. D.4、已知等差数列的公差为 ,若成等比数列, 则 ( )A. B. C. D.5、若函数，，，又，，且的最小值为，则的值为( )A. B. C. D.26、指数函数且在上是减函数，则函数在R上的单调性为( )A.单调递增B.单调递减C.在上递增，在上递减 D .在上递减，在上递增7、已知中，，，D为边BC的中点，则 ( )A.3B.4C.5D.68、数列是等差数列，若，且它的前n项和有最大值，那么当取得最小正值时，n等于( )A.17B.16C.15D.149、在△ABC中，若 (tanB+tanC)=tanBtanC﹣1，则cos2A=( )A.﹣B.C.﹣D.10、函数的单调增区间与值域相同，则实数的取值为( )A. B. C. D.11、已知函数，其中 .若对于任意的，都有，则的取值范围是( )A. B. C. D.12、，则O是三角形的( )A.垂心B.外心C.重心D.内心二、填空题：本大题共4小题，每小题5分。

2018-2019高中英语上海高三专题试卷质量检测试卷一含答案考点及解析

2018-2019高中英语上海高三专题试卷质量检测试卷一含答案考点及解析

2018-2019高中英语上海高三专题试卷质量检测试卷一含答案考点及解析班级:___________ 姓名：___________ 分数：___________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一、单项选择1. No matter how healthy you ________ be, you need to consult a doctor from time to time. A．should B．will C．must D．may【答案】D【解析】D句意：不管你多么健康，你都需要时不时地咨询医生。

may表示“(转折前所述情况属实)也许，可能”。

二、完形填空Section B (18 marks)Directions: 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.When I was 8 years old, I once decided to run away from home. With my suitcase 36 and some sandwiches in a bag, I started for the front door and said to Mom, “I’m leaving.”“If you want to 37 , that’s all right,” she said. “But you came into this home without anything and you can leave the same way.” I 38 my suitcase and sandwiches on the floor heavily and started for the door again.“Wait a minute,” Mom said. “I want your 39 back. You didn’t wear anything when you arrived.” This really angered me. I tore my clothes off—shoes, socks, underwear and all—and 40 , “Can I go n ow?” “Yes,” Mom answered, “but once you close that door, don’t expect to come back.”I was so 41 that I slammed (砰地关上) the door and stepped out on the front porch. 42 I realized that I was outside, with nothing on. Then I noticed that down the street, two neighbor girls were walking toward our house. I ran to 43 behind a big tree in our yard at once. After a while, I was 44 the girls had passed by. I dashed to the front door and banged on it loudly.“Who’s there?” I heard.“It’s Billy! Let me in!”The voice behind the 45 answered, “Billy doesn’t live here anymore. He ran away from home.” Glancing behind me to see if anyone else was coming, I begged, “Aw, c’mon, Mom! I’m 46 your son. Let me in!”The door inched open and Mom’s smiling face appeared. “Did you change your 47 about running away?” she asked.“What’s for supper?” I answered. (277 words)2.A．packed B．returned C．cleaned D．repaired3.A．drop out B．go by C．move around D．run away4.A．pressed B．shook C．threw D．pulled5.A．bag B．clothes C．sandwiches D．suitcase6.A．explained B．suggested C．continued D．shouted7.A．angry B．sorry C．frightened D．ashamed8.A．Certainly B．Naturally C．Suddenly D．Possibly9.A．play B．bide C．rest D．wave10.A．sure B．proud C．eager D．curious11.A．house B．tree C．door D．yard12.A．also B．still C．even D．already13.A．conclusion B．promise C．concern D．decision【答案】2.A3.D4.C5.B6.D7.A8.C9.B10.A11.C12.B13.D【解析】文章大意：记叙文：记叙了一个不懂事想离家出走的“我”在离家出走所遭遇的事。

2018-2019年高中地理上海高一高考真卷模拟试卷【3】含答案考点及解析

2018-2019年高中地理上海高一高考真卷模拟试卷【3】含答案考点及解析

2018-2019年高中地理上海高一高考真卷模拟试卷【3】含答案考点及解析班级:___________ 姓名：___________ 分数：___________题号一二三四五总分得分注意事项：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评卷人得分一、选择题1.下列人口迁移对环境影响的说法，正确的是（）A．人口压力小的农村，人口迁出，使劳动力不足，不利于开发利用土地资源B．新中国成立后的人口迁入区得到发展并使综合环境得到较大改善C．大量农村人口涌入城市，改善了城市的生态环境D．人口迁移对迁出地区环境的影响更广泛【答案】B【解析】试题分析：人口压力小的农村，人口迁出，使劳动力不足，对自然环境破坏减少，A错；新中国成立后的人口迁入区得到发展，使综合环境得到发展，B对；大量农村人口涌入城市，破坏了城市的生态环境，C错；人口迁移对迁出地区环境的影响将减少，D错。

选B正确。

考点：主要考查人口迁移对环境的影响。

2.岩石圈是指A．软流层及其以上部分B．地壳和上地幔C．地壳和上地幔顶部D．地壳【答案】C【解析】试题分析：岩石圈是指由各类岩石组成的圈层，其范围指软流层以上的地壳和上地幔顶部。

故选C。

考点：地球的圈层结构3.一气团从一山峰的西坡流向东坡，如图所示，在经过甲、乙海拔相同的两地（）A．水汽含量甲地>乙地B．气温甲地=乙地C．水汽含量甲地<乙地D．气温甲地<乙地【答案】AD【解析】试题分析：甲位于迎风坡，乙位于背风坡，在迎风坡由于水汽逐渐凝结，释放一定的热量，因此垂直递减率偏低，而乙地无水汽凝结，因此垂直递减率偏高，故背风坡气流下沉升温较快，使同海拔的乙地气温比甲地高，而乙地水汽比甲地少。

故选AD。

考点：本题考查地形对气温、大气湿度的影响。

点评：本题属基础知识题目，解答时注意下垫面对降水、气温的影响。

湿润气流遇到山脉等高地阻挡时被迫抬升而气温降低形成的降水。

上海市崇明区2019届高三第一次高考模拟考试数学试题Word版含答案

上海市崇明区2019届高三第一次高考模拟考试数学试题Word版含答案

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上海市2018-2019年高三第十次模拟考试数学(理)试题含答案

上海市2018-2019年高三第十次模拟考试数学(理)试题含答案

高三十模考试 数学试卷（理科）一、选择题（每小题5分，共60分.下列每小题所给选项只有一项符合题意，请将正确答案的序号填涂在答题卡上）1.设集合2{|log (2)}A x y x ==-，2{|320}B x x x =-+<，则A C B =（ ） A ．(,1)-∞ B ．(,1]-∞ C ．(2,)+∞ D ．[2,)+∞2.在复平面内，复数2332iz i-++对应的点的坐标为(2,2)-，则z 在复平面内对应的点位于（ ） A ．第一象限 B ．第二象限 C ．第三象限 D ．第四象限3.已知ABC ∆中，sin 2sin cos 0A B C +=c =，则tan A 的值是（ ）A ．3B ．3C ．34.设{(,)|0,01}A x y x m y =<<<<，s 为(1)n e +的展开式的第一项（e 为自然对数的底数），m ，若任取(,)a b A ∈，则满足1ab >的概率是（ ） A ．2e B ．2e C ．2e e -D ．1e e- 5.函数4lg x x y x=的图象大致是（ ）A ．B ．C ．D ． 6.已知一个简单几何体的三视图如图所示，若该几何体的体积为2448π+，则该几何体的表面积为（ ）A ．2448π+B ．2490π++C ．4848π+D ．2466π++7.已知11717a =，16log b =17log c =，则a ，b ，c 的大小关系为（ ） A ．a b c >> B ．a c b >> C ．b a c >> D ．c b a >> 8.执行如下程序框图，则输出结果为（ ）A ．20200B ．5268.5-C ．5050D ．5151-9.如图，设椭圆E ：22221(0)x y a b a b+=>>的右顶点为A ，右焦点为F ，B 为椭圆在第二象限上的点，直线BO 交椭圆E 于点C ，若直线BF 平分线段AC 于M ，则椭圆E 的离心率是（ ） A ．12 B ．23 C ．13 D ．1410.设函数()f x 为定义域为R 的奇函数，且()(2)f x f x =-，当[0,1]x ∈时，()sin f x x =，则函数()cos()()g x x f x π=-在区间59[,]22-上的所有零点的和为（ ） A ．6 B ．7 C ．13 D ．14 11.已知函数2()sin 20191x f x x =++，其中'()f x 为函数()f x 的导数，求(2018)(2018)f f +-'(2019)'(2019)f f ++-=（ ）A ．2B ．2019C ．2018D ．012.已知直线l ：1()y ax a a R =+-∈，若存在实数a 使得一条曲线与直线l 有两个不同的交点，且以这两个交点为端点的线段长度恰好等于a ，则称此曲线为直线l 的“绝对曲线”.下面给出的四条曲线方程：①21y x =--；②22(1)(1)1x y -+-=；③2234x y +=；④24y x =. 其中直线l 的“绝对曲线”的条数为（ ） A ．1 B ．2 C ．3 D ．4二、填空题：（本大题共4小题，每题5分，共20分）13.已知实数x ，y 满足2202401x y x y y x +-≥⎧⎪+-≤⎨⎪≤+⎩，且341x y m x ++=+，则实数m 的取值范围．14.双曲线22221x y a b-=的左右焦点分别为1F 、2F ，P 是双曲线右支上一点，I 为12PF F ∆的内心，PI 交x 轴于Q 点，若12FQ PF =，且:2:1PI IQ =，则双曲线的离心率e 的值为． 15.若平面向量1e ，2e 满足11232e e e =+=，则1e 在2e 方向上投影的最大值是．16.观察下列各式：311=； 3235=+； 337911=++； 3413151719=+++；……若3*()m m N ∈按上述规律展开后，发现等式右边含有“2017”这个数，则m 的值为．三、解答题：（本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.第17～21为必考题，每个试题考生都必须作答.第22、23题为选考题，考生根据要求作答）17.已知等差数列{}n a 中，公差0d ≠，735S =，且2a ，5a ，11a 成等比数列. （1）求数列{}n a 的通项公式；（2）若n T 为数列11{}n n a a +的前n 项和，且存在*n N ∈，使得10n n T a λ+-≥成立，求实数λ的取值范围.18.为了解学生寒假期间学习情况，学校对某班男、女学生学习时间进行调查，学习时间按整小时统计，调查结果绘成折线图如下：（1）已知该校有400名学生，试估计全校学生中，每天学习不足4小时的人数.（2）若从学习时间不少于4小时的学生中选取4人，设选到的男生人数为X ，求随机变量X 的分布列.（3）试比较男生学习时间的方差21S 与女生学习时间方差22S 的大小.（只需写出结论）19.如图所示，四棱锥P ABCD -的底面为矩形，已知1PA PB PC BC ====，AB =，过底面对角线AC 作与PB 平行的平面交PD 于E.（1）试判定点E 的位置，并加以证明； （2）求二面角E AC D --的余弦值.20.在平面直角坐标平面中，ABC ∆的两个顶点为(0,1)B -，(0,1)C ，平面内两点P 、Q 同时满足：①0PA PB PC ++= ；②QA QB QC == ；③//PQ BC .（1）求顶点A 的轨迹E 的方程；（2）过点F 作两条互相垂直的直线1l ，2l ，直线1l ，2l 与A 的轨迹E 相交弦分别为11A B ，22A B ，设弦11A B ，22A B 的中点分别为M ，N .①求四边形1212A A B B 的面积S 的最小值；②试问：直线MN 是否恒过一个定点？若过定点，请求出该定点，若不过定点，请说明理由. 21.已知函数ln(1)()1x f x ax +=+.（1）当1a =，求函数()y f x =的图象在0x =处的切线方程； （2）若函数()f x 在(0,1)上单调递增，求实数a 的取值范围； （3）已知x ，y ，z 均为正实数，且1x y z ++=，求证(31)ln(1)(31)ln(1)11x x y y x y -+-++--(31)ln(1)01z z z -++≤-. 请考生在22、23题中任选一题作答，如果多做，则按所做的第一题记分.22.[选修4-4：坐标系与参数方程] 在极坐标系中，曲线1C 的极坐标方程是244cos 3sin ρθθ=+，以极点为原点O ，极轴为x 轴正半轴（两坐标系取相同的单位长度）的直角坐标系xOy 中，曲线2C 的参数方程为：cos sin x y θθ=⎧⎨=⎩（θ为参数）. （1）求曲线1C 的直角坐标方程与曲线2C 的普通方程；（2）将曲线2C 经过伸缩变换''2x y y⎧=⎪⎨=⎪⎩后得到曲线3C ，若M ，N 分别是曲线1C 和曲线3C 上的动点，求MN 的最小值. 23.[选修4-5：不等式选讲] 已知()21()f x x a x a R =--+∈. （1）当1a =时，解不等式()2f x >.（2）若不等式21()12f x x x a +++>-对x R ∈恒成立，求实数a 的取值范围.十模数学答案（理）一、选择题1-5: BDACD 6-10: DACCA 11、12：AC二、填空题13. [2,7] 14.3215. 3- 16. 45 三、解答题17.解：（1）由题意可得12111767352(4)()(10)a d a d a d a d ⨯⎧+=⎪⎨⎪+=++⎩，即121352a d d a d +=⎧⎨=⎩. 又因为0d ≠，所以121a d =⎧⎨=⎩.所以1n a n =+.（2）因为111(1)(2)n n a a n n +=++1112n n =-++，所以11112334n T =-+-1112n n +⋅⋅⋅+-++11222(2)n n n =-=++. 因为存在*n N ∈，使得10n n T a λ+-≥成立，所以存在*n N ∈，使得(2)02(2)nn n λ-+≥+成立，即存在*n N ∈，使得22(2)nn λ≤+成立.又2142(2)2(4)n n n n =+++，114162(4)n n≤++（当且仅当2n =时取等号）， 所以116λ≤.即实数λ的取值范围是1(,]16-∞.18.解：（1）由折线图可得共抽取了20人，其中男生中学习时间不足4小时的有8人，女生中学习时间不足4小时的有4人.∴可估计全校中每天学习不足4小时的人数为：1240024020⨯=人. （2）学习时间不少于4本的学生共8人，其中男学生人数为4人，故X 的所有可能取值为0，1，2，3，4.由题意可得4448(0)C P X C ==170=；134448(1)C CP X C ==1687035==； 224448(2)C C P X C ==36187035==； 314448(3)C CP X C ==1687035==； 4448(4)C P X C ==170=. 所以随机变量X 的分布列为∴均值017070EX =⨯+⨯237070+⨯+⨯4270+⨯=. （3）由折线图可得2212s s >. 19.解：（1）E 为PD 的中点，证明如下：连接OE ，因为//PB 平面AEC ，平面PBD 平面AEC OE =，PB ⊄平面AEC ，所以//OE PB ，又O 为BD 的中点，所以E 为PD 的中点.（2）连接PO ，因为四边形ABCD 为矩形，所以OA OC =.因为PA PC =，所以PO AC ⊥.同理，得PO BD ⊥，所以PO ⊥平面ABCD ，以O 为原点，OP 为z 轴，过O 平行于AD 的直线为x 轴，过O 平行于CD 的直线为y 轴建立空间直角坐标系（如图所示）.易知1(,22A -，1(,22B ，1(,,0)22C -，1(,22D --，1(0,0,)2P ，11(,,)444E -，则11(,)444EA =-- ，1(,,0)22OA =- .显然，OP 是平面ACD 的一个法向量.设1(,,)n x y z =是平面ACE 的一个法向量，则1100n EA n OA ⎧⋅=⎪⎨⋅=⎪⎩，即11044102x y z x y ⎧-=⎪⎪⎨⎪=⎪⎩，取1y =，则1n =，所以1cos ,n OP <> 11n OP n OP⋅=11=， 所以二面角E AC D --的余弦值为11. 20.（1）221(0)3x y x +=≠；（2）①S 的最小值的32，②直线MN恒过定点4⎛⎫ ⎪ ⎪⎝⎭. 试题解析：（1）∵2PA PB PO +=， ∴由①知2PC PO =- ，∴P 为ABC ∆的重心. 设(,)A x y ，则,33x y P ⎛⎫⎪⎝⎭，由②知Q 是ABC ∆的外心， ∴Q 在x 轴上由③知,03x Q ⎛⎫ ⎪⎝⎭，由Q C Q A ==化简整理得：221(0)3x y x +=≠. （2）解：F 恰为2213x y +=的右焦点， ①当直线1l ，2l 的斜率存且不为0时，设直线1l的方程为my x =由22330my x x y ⎧=⎪⎨+-=⎪⎩22(3)10m y ⇒++-=，设111(,)A x y ，122(,)B x y，则1223y y m -+=+，12213y y m -=+，①根据焦半径公式得1112)A B x x =+，又1212x x my my +=12()m y y =++223m -=++23m =+，所以1123A B m =+=22221113m A B m ⎫+⎪⎝⎭=+221)31m m +=+， 则2222(1)6(3)(31)m S m m +=++2222(1)64(1)2m m +≥⎛⎫+ ⎪⎝⎭32=， 当22331m m +=+，即1m =±时取等号.②根据中点坐标公式得22,33M m m ⎛⎫ ⎪ ⎪++⎝⎭，同理可求得222,3131N m m ⎛⎫⎪ ⎪++⎝⎭，则直线MN的斜率为MNk =243(1)m m =-， ∴直线MN的方程为y243(1)m x m ⎛=- -⎝⎭，整理化简得()4334ym x m +()263490ym x m y ++-=，令0y =，解得x =∴直线MN恒过定点⎫⎪⎪⎝⎭.②当直线1l ，2l 有一条直线斜率不存在时，另一条斜率一定为0，直线MN 即为x轴，过点⎫⎪⎪⎝⎭. 综上，S 的最小值的32，直线MN恒过定点4⎛⎫ ⎪ ⎪⎝⎭. 21.（1）当1a =时，ln(1)()1x f x x +=+则(0)0f =，21ln(1)'()(1)x f x x -+=+则'(0)1f =， ∴函数()y f x =的图象在0x =时的切线方程为y x =.（2）∵函数()f x 在(0,1)上单调递增，∴10ax +=在(0,1)上无解， 当0a ≥时，10ax +=在(0,1)上无解满足，当0a <时，只需1010a a +≥⇒-≤<，∴1a ≥-①21ln(1)1'()(1)ax a x x f x ax +-++=+， ∵函数()f x 在(0,1)上单调递增，∴'()0f x ≥在(0,1)上恒成立， 即[](1)ln(1)1a x x x ++-≤在(0,1)上恒成立. 设()(1)ln(1)x x x ϕ=++'()ln(1)(1)x x x x ϕ-=+++11ln(1)1x x ⋅-=++， ∵(0,1)x ∈，∴'()0x ϕ>，则()x ϕ在(0,1)上单调递增， ∴()x ϕ在(0,1)上的值域为(0,2ln 21)-.∴1(1)ln(1)a x x x≤++-在(0,1)上恒成立，则12ln 21a ≤-②综合①②得实数a 的取值范围为11,2ln 21⎡⎤-⎢⎥-⎣⎦. （3）由（2）知，当1a =-时，ln(1)()1x f x x+=-在(0,1)上单调递增，于是当103x <≤时，ln(1)()1x f x x +=-134()ln 323f ≤=， 当113x ≤<时，ln(1)()1x f x x +=-134()ln 323f ≥=， ∴(31)()x f x -34(31)ln 23x ≥-⋅，即(31)ln(1)1x x x -+-33(31)ln 24x ≤-⋅， 同理有(31)ln(1)1y y y -+-33(31)ln 24y ≤-⋅，(31)ln(z 1)1z z -+-33(31)ln 24z ≤-⋅，三式相加得(31)ln(1)1x x x -+-(31)ln(1)1y y y -++-(31)ln(z 1)01z z -++≤-.22.解：（1）∵1C 的极坐标方程是244cos 3sin ρθθ=+，∴4cos 3sin 24ρθρθ+=，整理得43240x y +-=，∴1C 的直角坐标方程为43240x y +-=.曲线2C ：cos sin x y θθ=⎧⎨=⎩，∴221x y +=，故2C 的普通方程为221x y +=. （2）将曲线2C经过伸缩变换''2x y y⎧=⎪⎨=⎪⎩后得到曲线3C 的方程为22''184x y +=，则曲线3C 的参数方程为y 2sin x αα⎧=⎪⎨=⎪⎩（α为参数）.设(),2sin N αα，则点N 到曲线1C 的距离为d===(tan ϕ=. 当()sin 1αϕ+=时，dMN23.解：（1）当1a =时，等式()2f x >，即2112x x --+>，等价于11212x x x <-⎧⎨-++>⎩或1121212x x x ⎧-≤≤⎪⎨⎪--->⎩或122112x x x ⎧>⎪⎨⎪--->⎩， 解得23x <-或4x >， 所以原不等式的解集为2(,)(4,)3-∞-+∞ ；（2）设()()1g x f x x x =+-+2x a x =-+，则,2()3,2a a x x f x ax a x ⎧-≤⎪⎪=⎨⎪->⎪⎩，则()f x 在(,)2a-∞上是减函数，在(,)2a +∞上是增函数， ∴当2a x =时，()f x 取最小值且最小值为()22a a f =， ∴2122a a >-，解得112a -<<，∴实数a 的取值范围为1(,1)2-.。

【精品】上海市近两年(2018,2019)高考数学试卷汇总以及答案(word解析版)

【精品】上海市近两年(2018,2019)高考数学试卷汇总以及答案(word解析版)

2018年普通高等学校招生全国统一考试(上海卷)数 学一、填空题（本大题共有12题，满分54分第1-6题每题4分，第7-12题每题5分）1.行列式的值为 。

2.双曲线的渐近线方程为 。

3.在（1+x ）7的二项展开式中，x ²项的系数为 。

（结果用数值表示）4.设常数，函数f (x )=log 2(x +a )，若f (x )的反函数的图像经过点(3,1)，则a= 。

5.已知复数z 满足（i 是虚数单位），则∣z ∣= 。

6.记等差数列的前几项和为S n ，若a 3=0，a 8+a 7=14，则S 7= 。

7.已知α∈{－2,－1,－,,1,2,3}，若幂函数为奇函数，且在(0,+∞)上递减，则α=_____ 8.在平面直角坐标系中，已知点A （-1，0），B （2，0），E ，F 是y 轴上的两个动点，且||=2，则的最小值为______9.有编号互不相同的五个砝码，其中5克、3克、1克砝码各一个，2克砝码两个，从中随机选取三个，则这三个砝码的总质量为9克的概率是______（结果用最简分数表示）10.设等比数列{a n }的通项公式为a n =q ⁿ+1（n ∈N*），前n 项和为S n 。

若，则q=____________ 11.已知常数a >0，函数的图像经过点、，若，则a =__________12.已知实数x ₁、x ₂、y ₁、y ₂满足：，，+的最大值为__________ 二、选择题（本大题共有4题，满分20分，每题5分）每题有且只有一个正确选项.考生应在答题纸的相应位置，将代表正确选项的小方格涂黑.41252214x y -=a R ∈117i z i +=-（）{} n a 2121()n f x x =EF ⋅1Sn 1lim 2n n a →∞+=222()(2)f x ax =+65p p ⎛⎫ ⎪⎝⎭，15Q q ⎛⎫- ⎪⎝⎭，236p q pq +=²²1x y +=₁₁²²1x y +=₂₂212x x y y +=₁₂₁13.设P 是椭圆+=1上的动点，则P 到该椭圆的两个焦点的距离之和为（ ） （A ）2 （B ）2 （C ）2 （D ）414.已知，则“”是“”的（ ） （A ）充分非必要条件（B ）必要非充分条件 （C ）充要条件（D ）既非充分又非必要条件15.《九章算术》中，称底面为矩形而有一侧棱垂直于底面的四棱锥为阳马.设AA ₁是正六棱柱的一条侧棱，如图，若阳马以该正六棱柱的顶点为顶点，以AA ₁为底面矩形的一边，则这样的阳马的个数是（ ）（A ）4 （B ）8 （C ）12 （D ）1616.设D 是含数1的有限实数集，是定义在D 上的函数，若的图像绕原点逆时针旋转后与原图像重合，则在以下各项中，的可能取值只能是（ ）（A （B （C （D ）0 三、解答题（本大题共有5题，满分76分）解答下列各题必须在答题纸的相应位置写出必要的步骤.17.（本题满分14分，第1小题满分6分，第2小题满分8分）已知圆锥的顶点为P ，底面圆心为O ，半径为2（1）设圆锥的母线长为4，求圆锥的体积；（2）设PO =4，OA ，OB 是底面半径，且∠AOB =90°，M 为线段AB 的中点，如图，求异面直线PM 与OB 所成的角的大小.²5x ²3y 2352a R 1a ﹥1a1﹤f x （）f x （）π61f （）18.（本题满分14分，第1小题满分6分，第2小题满分8分）设常数，函数 （1）若为偶函数，求a 的值；（2）若，求方程上的解。

2018届高考模拟试卷一参考答案 .doc

2018届高考模拟试卷一参考答案 .doc

2018届高考模拟试卷一参考答案一、填空题（本大题共14小题，每小题5分，共70分．请把答案填写在答卷规定的横线上）1.22.四3.284.35.8π 6.a ＞2 7.6π 8.54 9.6π10.3π11.448 12.2 13.24 14.()5333， 二、解答题（本大题共6小题，共90分．解答时应写出文字说明、证明过程或演算步骤） 15．（本小题满分14分）如图，在几何体中，四边形ABCD 为菱形，对角线AC 与BD 的交点为O ，四边形DCEF 为梯形，EF ∥CD ，FB FD =．（1）若2CD EF =，求证：OE ∥平面ADF ； （2）求证：平面ACF ⊥平面ABCD ．【解析】（Ⅰ）证明：取AD 的中点G ，连接OG 、FG ，因为O 为对角线AC 与BD 的交点，则O 为AC 中点， 所以OG ∥CD ，且12OG CD =． 又因为EF ∥CD ，且2CD EF =，所以OG ∥EF ，OG EF =，则四边形OGFE 为平行四边形，----------3分 所以OE ∥FG ．又因为FG ⊂平面ADF ，OE ⊄平面ADF ，OE ∥FG ，所以OE ∥平面ADF ；-------------------------------------------------------------------6分（Ⅱ）证明：因为四边形ABCD 为菱形，所以OC BD ⊥，--------------------------7分又因为FB FD =，O 是BD 的中点，所以OF BD ⊥，------------------8分 又有OFOC O OF =⊂，平面ACF ，OC ⊂平面ACF ,所以BD ⊥平面ACF ，----------------------------------------------12分 又因为BD ⊂平面ABCD ，所以平面ACF ⊥平面ABCD ．----------------------------------------14分16．（本小题满分14分）已知函数()2sin()cos 6f x x x π=-.（1）求函数()f x 的最大值和最小正周期；（2）设ABC ∆的角A B C ，，的对边分别为a b c ，，，且c =，1()2f C =，若sin 2sin B A =，求边a ，b 的值.【解析】（Ⅰ）因为)2()2sin()cos 612cos cos 22cos cos 1cos 2221sin(2)62f x x xx x x x x x x x x ππ=-=-=-+=-=---------------------------------------------------------------------4分当且仅当,3x k k Z ππ=+∈时，max 1()2f x =--------------------------------------6分 最小正周期分别为和22T ππ==.------------------------------------------------7分 （Ⅱ）因为11()sin(2)622f C C π=--=，即sin(2)16C π-=，因为0C π<<，所以 112666C πππ-<-<，于是262C ππ-=，即3C π=.------------------------------10分 因为sin 2sin B A =，由正弦定理得2b a =，-------------------------------------12分 由余弦定理得2222cos3c a b ab π=+-，即2212a b ab +-=，联立22212b aa b ab =⎧⎨+-=⎩，解得24a b =⎧⎨=⎩.-------------------------------------------14分17．（本小题满分14分） 在平面直角坐标系xOy 中，椭圆C ：22221(0)x y a b a b+=>>在椭圆C 上.（1）求椭圆C 的方程；-（2）设P 为椭圆上第一象限内的点，点P 关于原点O 的对称点为A ，点P 关于x 轴的对称点为Q ，设PD PQ λ=，直线AD 与椭圆C 的另一个交点为B ，若PA ⊥PB ，求实数λ的值.【解析】17.解:(1)因为点222，在椭圆C 上，则222112a b+=，------------------------------1分 又椭圆C 的离心率为32，可得32ca，即32ca ， 所以2222223124b acaa a ，代入上式，可得22221a a +=， 解得24a ，故22114ba .所以椭圆C 的方程为2214x y += ...............................................................................................5分 (2)设P (x 0，y 0)，则A (-x 0，-y 0)，Q (x 0，-y 0). 因为=λ,则(0，y D -y 0)=λ(0，-2y 0)，故y D =(1-2λ)y 0.所以点D 的坐标为(x 0，(1-2λ)y 0). ................................................................................................. 7分 设B (x 1，y 1)，221222101010222210101010114414PB BAx x y y y y y y k k x x x x x x x x ...............................9分 又0000121BA ADy y y k k x x x故001441PBBAx k k y .----------------------------------------------------------------------11分又PA ⊥PB ，且0PAx k y ， D QBPxAOy第17题所以1PB PA k k ，即0000141x y x y ，解得34. 所以34.................................................................................................................................... 14分 18．（本小题满分16分） 一块圆柱形木料的底面半径为12cm ，高为32cm ，要将这块木料加工成一只毛笔筒，在木料一端正中间掏去一个小圆柱，使小圆柱与原木料同轴，并且掏取的圆柱体积是原木料体积的三分之一，设小圆柱底面半径为r cm ，高为h cm ，要求笔筒底面的厚度超过2cm ． （1）求r 与h 的关系，并指出r 的取值范围；（2）笔筒成形后进行后续加工，要求笔筒上底圆环面、桶内侧面、外表侧面都喷上油漆，其中上底圆环面、外表侧面喷漆费用均为a （元/ cm 2），桶内侧面喷漆费用为2a （元/cm 2），而桶内底面铺贴金属薄片，其费用是7a （元/ cm 2）（其中a 为正常数）． ①将笔筒的后续加工费用y （元）表示为r 的函数；②求出当r 取何值时，能使笔筒的后续加工费用y 最小，并求出y 的最小值．【解析】（Ⅰ）据题意，221(1232)3r h ππ=⋅⋅，所以23248h r ⨯=，----------------------3分 因为322h ->，所以30h <即2324830r ⨯<，解得r >----------------------------------------------------------5分 又012r <<，所以125r <<；----------------------------------------------------------6分 （Ⅱ）①据题意，笔筒的后续加工费用22272(2)(1221232)y a r a rh a r πππππ=++⋅-⋅+⋅⋅，整理得2226412763248641276y a r a rh a a r a r a rππππππ=++⨯⨯=+⋅+⨯ 232326(152)a r rπ⨯=++，定义域为；----------------------11分 ②由①知，33/22323286(2)12r y a r a r rππ⨯-=-=⋅，令/0y =得8(,12)5r =∈，由表知，当8r =时，y 取极小值即最小值2064a π.------------------------15分答：当8r cm =时，能使笔筒的后续加工费用y 最小，最小值为2064a π元．----16分19．（本小题满分16分）已知数列{}n a 中，首项11a =，2a a =，12()n n n a k a a ++=+对任意正整数n 都成立，数列{}n a 的前n 项和为n S ．（1）若12k =，且18171S =，求实数a 的值； （2）是否存在实数k ，使数列{}n a 是公比不为1的等比数列，且任意相邻三项n a ，1n a +，2n a +按某顺序排列后成等差数列．若存在，求出所有的k 的值；若不存在，请说明理由；（3）若12k =-，求n S （用a ，n 表示）． 【解析】（Ⅰ）当12k =时，由12()n n n a k a a ++=+得121()2n n n a a a ++=+，即211n n n n a a a a +++-=-，所以数列{}n a 为等差数列，--------------------1分 公差为211d a a a =-=-，数列{}n a 的前n 项和为(1)(1)2n n n S n a -=+⋅-，由18171S =得18(181)17118(1)2a -=+⋅-， 解得2a =；---------------------------------------------------------3分（Ⅱ）设数列{}n a 为等比数列，则其公比为21a q a a ==，1n n a a -=，1n n a a +=，12n n a a ++=． 1︒若1n a +为等差中项，则122n n n a a a ++=+即112n n n a a a -+=+，解得1a =，与已知不符，舍去； 2︒若n a 为等差中项，则122n n n a a a ++=+即112n n n a a a -+=+，即220a a +-=，解得2a =-或1a =（舍），此时由12()n n n a k a a ++=+得11()n n n a k aa -+=+即2(1)a k a =+，故2215a k a ==-+；3︒ 若2n a +为等差中项，则212n n n a a a ++=+即112n n n a a a +-=+，即2210a a --=，解得12a =-或1a =（舍），仿2︒得2215a k a ==-+．---------------------------------------------------8分 综上，满足要求的实数k 有且仅有一个，25k =-；---------------------------------9分（Ⅲ）当12k =-时，121()2n n n a a a ++=-+，所以211()n n n n a a a a ++++=-+，于是32n n a a +++=211()n n n n a a a a +++-+=+．----------------------------------------11分1︒ 当n 为偶数时，123456112(1)()()()()()22n n n n n a S a a a a a a a a a a -+=++++++++=+=； ---------------------------------------------------------------------------------13分2︒ 当n 为奇数时，1234511231()()()()2n n n n S a a a a a a a a a a --=+++++++=++ 11211[()]1(1)22n n a a a a --=+⋅-+=-+（2n ≥），当1n =时，也适合该式， 所以11(1),2(1),2n n a n S n a n -⎧-+⎪⎪=⎨+⎪⎪⎩为奇数为偶数．-----------------------------------------------16分20．（本小题满分16分）已知函数1()ln f x a x x=+（0a ≠）. （1）求函数()f x 的单调区间；（2）若存在两条直线1y ax b =+，2y ax b =+（12b b ≠）都是曲线()y f x =的切线，求实数a 的取值范围；（3）若{}|()0(0,1)x f x ⊆≤，求实数a 的取值范围.【解析】（Ⅰ）/2211()a ax f x x x x-=-=（0x >）. 当0a <时，/()0f x <，()f x 的递减区间为(0,)+∞；----------------------------1分 当0a >时，由/()0f x =得1x a=，列表得：所以，函数()f x 的递减区间为1(0,)a ，递增区间为1(,)a+∞；-----------------------4分 （Ⅱ）因为存在两条直线1y ax b =+、2y ax b =+（12b b ≠）都是曲线()y f x =的切线， 所以/()f x a =至少有两个不等的正根，-----------------------------------------------5分 令/21()ax f x a x-==，得210ax ax -+=，记其两个根为1x 、2x （12x x <）， 则2124010a a x x a ⎧∆=->⎪⎨=>⎪⎩，解得4a >，------------------------------------------------------------------------------------7分 而当4a >时，曲线()y f x =在点11(,())x f x 、22(,())x f x 处的切线分别为11()y ax f x ax =+-、22()y ax f x ax =+-，设()()F x f x ax =-（0x >），由2//1222()()1()()a x x x x ax ax F x f x a x x----+-=-==知，当12x x x <<时，/()0F x >即()F x 在区间12[,]x x 上是单调函数，因此12()()F x F x ≠，所以11()y ax f x ax =+-、22()y ax f x ax =+-不重合，即1y ax b =+、2y ax b =+（12b b ≠）是曲线()y f x =的两条不同的切线，故4a >；----------------10分（Ⅲ）当0a <时，函数()f x 是(0,)+∞内的减函数，因为11111()ln()10aaaaf ea e e e---=+=-<，而1(0,1)ae-∉，不符合题意；----------------------------------------------------------12分当0a >时，由（Ⅰ）知()f x 的最小值为1()ln (1ln )f a a a a a a=-+=-.1︒若1()0f a>即0a e <<时，{}|()0(0,1)x f x φ≤=⊆，所以0a e <<符合题意；2︒若1()0f a =即a e =时，{}1|()0(0,1)x f x e ⎧⎫≤=⊆⎨⎬⎩⎭，所以a e =符合题意；3︒若1()0f a <即a e >时，101a <<，而(1)10f =>，函数()f x 在1(,)a+∞内递增，所以当1x ≥时，()0f x >，又因为()f x 的定义域为(0,)+∞，所以{}|()0(0,1)x f x ≤⊆，符合题意.综上，实数a 的取值范围为(0,)+∞.----------------------------------------------16分课题经济生活第六课《投资理财的选择》知识目标能力目标考点1、2：我国的商业银行及其主要业务+ 储蓄存款利息的计算方法考点3：储蓄、债券、股票、商业保险等投资理财方式重点难点比较储蓄、债券、股票、商业保险四种投资理财方式的异同（知道排序）；分析不同的投资行为（把握投资原则）。

2018-2019高中英语上海高三专题试卷联考模拟试卷一含答案考点及解析

2018-2019高中英语上海高三专题试卷联考模拟试卷一含答案考点及解析

2018-2019高中英语上海高三专题试卷联考模拟试卷一含答案考点及解析班级:___________ 姓名：___________ 分数：___________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一、单项选择1.________ all of them are strong candidates，only one will be chosen for the post. (2011·陕西，19)A．Since B．WhileC．If D．As【答案】【解析】B解析句意为：尽管他们都是很有实力的求职者，但是只有一个人会被选中担任这一职位。

since因为，既然；while尽管；if如果；as因为。

根据句意选while。

二、完形填空2.第二部分：英语知识及运用（共两节，满分35分）第一节：完形填空（共10小题；每小题2分, 满分20分）阅读下面短文，从短文后所给各题的四个选项A、B、C和D中选出能填入相应空白处的最佳选项。

Rubber ducks are being used to help scientists understand global warming and melting glaciers. NASA researchers have 21 90 ducks into holes in Greenland's fastest moving glacier, the Jakobshavn Glacier between Greenland and Canada. The 22 have each been marked with the words "science experiment" along with an e-mail address. If they are found scientists will be able to 23 how the water moves through the ice and provide information about the 24 of glaciers. Scientists are still 25 about why glaciers speed up in summer and head towards the sea. One theory is that as the summer sun melts ice on top of the glacier's surface, the water moves to the bottom of the glacier, where it helps to 26 the movement of ice toward the coast. The Jakobshavn Glacier is believed to be the 27 of the iceberg that sank the Titanic in 1912. RobertJones, the experiment organizer, said none of the ducks had been 28 yet. "We haven't heard back but it may take some time until somebody actually finds it and decides to send usa/an 29 that they have found it," he said. "These are places that are 30 so there aren't people walking around."21. A. flown B. buried C. hidden D. dropped22. A. results B. toys C. glaciers D. scientists23. A. remember B. invent C. learn D. control24. A. development B. movement C. growth D. travels25. A. unsure B. excited C. concerned D. ignorant26. A. reduce B. control C. speed D. stop27. A. position B. source C. reason D. result28. A. hurt B. eaten C. missing D. reported29. A. email B. card C. fax D. sign30. A. hidden B. lost C. remote D. quiet【答案】21—25 DBCBA 26—30 CBDAC【解析】略三、阅读理解For many parents, raising a teenager is like fighting a long war, but years go by without any clear winner. Like a border conflict between neighboring countries, the parent-teen war is about boundaries: Where is the line between what I control and what you do?Both sides want peace, but neither feels it has any power to stop the conflict. In part, this is because neither is willing to admit any responsibility for starting it. From the parents’ point of view, the only cause of their fight is their adolescent s’ complete unreasonableness. And of course, the teens see it in exactly the same way, except oppositely. Both feel trapped.In this article, I’ll describe three no-win situations that commonly arise between teens and parents and then suggest some ways out of the trap. The first no-win situation is quarrelsover unimportant things. Examples include the color of the teen’s hair, the cleanliness of the bedroom, the preferred style of clothing, the child’s failure to eat a good breakfast before school or his tendency to sleep until noon on the weekends. Second, blaming. The goal of a blaming battle is to make the other admit that his bad attitude is the reason why everything goes wrong. Third, needing to be right. It doesn’t matter what the topic is—politics, the laws of physics, or the proper way to break an egg—the point of these arguments is to prove that you are right and the other person is wrong, for both wish to be considered an authority—someone who actually knows something—and therefore to command respect. Unfortunately, as long as parents and teens continue to assume that they know more than the other, they’ll continue to fight these battles forever and never make any real progress.3.Why does the author compare the parent-teen war to a border conflict?A．Both can continue for generations.B．Both are about where to draw the line.C．Neither has any clear winner.D．Neither can be put to an end.4.What does the underlined part in Paragraph 2 mean?A．The teens blame their parents for starting the conflict.B．The teens agree with their parents on the cause of the conflict.C．The teens accuse their parents of misleading them.D．The teens tend to have a full understanding of their parents.5.Parents and teens want to be right because they want to ______.A．give orders to the otherB．know more than the otherC．gain respect from the otherD．get the other to behave properly6.What will the author most probably discuss in the paragraph that follows?A．Causes for the parent –teen conflicts.B．Examples of the parent –teen war.C．Solutions for the parent –teen problems.D．Future of the parent-teen relationship.【答案】3.B4.A5.C6.C【解析】试题分析：对许多父母来说，抚养孩子就像打一场持久战，但经过长年累月的战斗，依然分不出胜负。

2018-2019高中英语上海高三专题试卷测试练习试卷一含答案考点及解析

2018-2019高中英语上海高三专题试卷测试练习试卷一含答案考点及解析

2018-2019高中英语上海高三专题试卷测试练习试卷一含答案考点及解析班级:___________ 姓名：___________ 分数：___________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一、单项选择1.—I don't imagine Russ and his poor dog had a good time onthe deserted island，________？—I have no idea.A．didn't they B．do youC．did they D．hadn't they【答案】C【解析】C[该反意疑问句要与宾语从句保持一致，而且还要注意这是否定转移现象，要把宾语从句看作否定句。

]二、完形填空Joe Simpson and Simon Yates were the first people to climb the West Face of the Siula Grande in the Andes mountains. They reached the top ______, but on their way back conditions were very______. Joe fell and broke his leg. They both knew that if Simon _____ alone, he would probably get back ______. But Simon decided to risk his ______ and try to lower Joe down the mountain on a rope.As they ______ down, the weather got worse. Then another ______ occurred. They couldn’t see or hear each other and, ______, Simon lowered his friend over the edge of a precipice(绝壁). It was ______ for Joe to climb back or for Simon to pull him up. Joe’s ______ was pulling Simon slowly towards the precipice. ______, after more than an hour in the dark and the icy cold, Simon had to ______. In tears, he cut the rope. Joe ______ into a large crevasse(裂缝)in the ice below. He had no food or water and he was in terrible pain. He couldn’t walk, but he ______ to get out of the crevasse and started to ______ towards their camp, nearly ten kilometers ____.Simon had ______ the camp at the foot of the mountain. He thought that Joe must be _____, but he didn’t want to leave ______. Three days later, in the middle of the night, he heard Joe’s voice. He couldn’t ______ it. Joe was there, a fe w meters from their tent, still alive.2.A．hurriedly B．carefully C．successfully D．early3.A．difficult B．similar C．special D．normal4.A．climbed B．worked C．rested D．continued5.A．unwillingly B．safely C．slowly D．regretfully6.A．fortune B．time C．health D．life7.A．lay B．settled C．went D．looked8.A．damage B．storm C．change D．trouble9.A．by mistake B．by chance C．by choice D．by luck10.A．unnecessary B．practical C．important D．impossible11.A．height B．weight C．strength D．equipment12.A．Finally B．Patiently C．Surely D．Quickly13.A．stand back B．take a rest C．hold on D．make a decision14.A．jumped B．fell C．escaped D．backed15.A．managed B．planned C．waited D．hoped16.A．run B．skate C．move D．march17.A．around B．away C．above D．along18.A．headed for B．traveled to C．left for D．returned to19.A．dead B．hurt C．weak D．late20.A．secretly B．tiredly C．immediately D．anxiously21.A．find B．believe C．make D．accept【答案】2.C3.A4.D5.B6.D7.C8.D9.A10.D11.B12.A13.D14.B15.A16.C17.B18.D19.A20.C21.B【解析】试题分析：本文讲诉的是乔·辛普森和西蒙·叶芝成功登上了在安第斯山脉的修拉格兰德西坡而要返回山下时出了状况——乔摔断了腿。

上海徐教院附中2018-2019学年高三英语模拟试卷含解析

上海徐教院附中2018-2019学年高三英语模拟试卷含解析

上海徐教院附中2018-2019学年高三英语模拟试卷含解析一、选择题1. The man tried to pretend that he hadn't stolen the money but his expression ________.A. turned him downB. paid him offC. made him outD. gave him away参考答案：2. ——Nowadays few people live in a place permanently, and population mobility is a______.——Yes. The present world is more on the _______ than ever before.A. tendency; goB. trend; moveC. phenomenon; moveD. fashion; go命题意图：考查名词辨析和固定词组：tendency, trend 趋势；趋向，phenomenon现象，fashion 时尚, 固定词组：on the go 忙个不停，on the move 在移动中参考答案：b略3. —Twenty-five counties in Yunnan province have experienced their record low rainfall in history.—The forecast says rain is unlikely in the coming week.A.to seeB.to have seenC.to be seenD.being seen参考答案：C4. ---- Oh! You bumped me!---- I am sorry, but I ____ to catch the coming bus.A. tryB. triedC. am tryingD. was trying参考答案：D5. With international and education exchanges increasing every year, the number of Chinese students studying abroad has increased rapidly.A. cooperationB. competitionC.construction D. conflict参考答案：选A, 随着国际合作与教育交流每年加强, 故选A。

上海市2018-2019年高考语文仿真模拟试题

上海市2018-2019年高考语文仿真模拟试题

绝密★启用前普通高中学业水平考试语文仿真模拟试题B选择题部分一、选择题（本大题共16小题，每小题3分，共48分。

每小题列出的四个备选项中只有一个是符合题目要求的，不选、多选、错选均不得分）1．下语中加点的字，每对读音都不相同的一组是A．湍．急/喘．息禅．让/嬗．变诽．谤/斐．然B．抹．杀/抹．墙框．架/诓．骗着．想/卓．越C．荫．庇/饮．马干．系/干．流边塞．/茅塞．顿开D．肄．业/酒肆．箴．言/缄．默慑．服/蹑．足沙砾2．下列各项中没有错别字的一项是A．国家加强对造谣他人、损害他人声誉等言行的惩戒力度，实际上是为网络空间的言行划出了底限、明确了责任。

B．一双手的力量固然微弱，但每个好心人、志愿者的爱心与善行汇聚起来，就是赈灾止殇的浩荡洪流。

C．无论我们是品读他的诗文，临摹他的书法，或仅仅是看过他的记录片，都可能会收获知识的增益、文化的熏陶、人格的滋养。

D．校准价值座标，坚定理想信念，守望心中的绿洲，就不难守住风清气正的精神家园。

3．下列句子中加点的词语运用不恰当的一项是A．黄土地区森林不多，因此，耕作技术与中东的刀耕火种大相径庭．．．．。

B．每个人的生命史就是自己的吉光片羽．．．．，这可以是艺术的，也可以不是艺术的。

C．地方电视台跟风入市，砸钱做综艺、搞真人秀，但真正值得一看的节目却寥寥无几．．．．。

D．每个人心中的能量都取之不尽．．．．，只是欠缺了一个自我发现的时机。

4．下列句子中没有语病的一项是A．西北工业大学今年首次将AR虚拟校园平台嵌入高考录取通知书。

该校招办相关人员表示，精心嵌入的AR技术，是一种前沿气质的展现。

B．《将改革进行到底》首次全景式地反映了党的十八大以来新一届中央领导集体带领全体中国人民开启并扎实推进全面深化改革。

C．祁连山的高山冰川融水和山区降雨形成了一条条河流，养育了河西走廊。

祁连山还是我国干旱地区生物多样性集中分布的地区之一，分布有高等植物大约1300多种。

D．中国外交部发布《印度边防部队在中印边界锡金段越界进入中国领土的事实和中国的立场》的长篇重磅文件，说明了事件全方位的来龙去脉。

2018-2019学年上海高考标准模拟试卷二(word有答案).

2018-2019学年上海高考标准模拟试卷二(word有答案).

2019年全国普通高等学校招生统一考试上海英语模拟试卷（二）I.Listening ComprehensionSection ADirections: In Part A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. For one week. B. For less than a week.C.For two weeks.D. Hard to say.2. A. Go to her sister’s wedding ceremo ny. B. Stay at home.C. Go to George’s birthady party.D. Go to George’s house-warming party.3. A. 50 dollars. B. 40 dollars. C. 60 dollars. D.55 dollars.4. A. She feels bored with the idea. B. She thinks ballet is funny.C. She will not go with the men anywayD. She shows interest in the show.5. A. Snowy. B. Sunny. C. Windy. D. Cloudy.6. A. She hasn’t seen Monet’s paintings for ten yearsB. She hasn’t been to the museum for long.C. She has been interested in Monet’s paintings for ten yearsD. She used to own one of Monet’s paintings.7. A. Father and daughter B. Friends.C. Husband and wife.D. Boss and his employee.8. A. The man is not interested in the game this weekend.B. The man is not interested in the team that will play this weekend.C. The man doesn’t want to mention the game,D. The man is not interested in watching any game.9. A. She would seen Ellen at last.B. She saw Ellen for the last time not long ago.C. She has many people to see before Ellen.D. She wouldn’t like to see Ellen at all.10. A. The woman is sorry for not being able to spend the holiday with the man.B. The man is a bit annoyed because the woman didn’t tell him her plan for the winter holiday.C. The man is sorry about not being able to go to Malaysia.D. The woman is excited about spending the winter holiday in Malaysia without the man.Section BDirections: In Section B, you will hear several longer conversation(s) and short passage(s), and you will be asked several questions on each of the conversation(s) and the passage(s). The conversation(s) and the passage(s) will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one is the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. 70. B. 80. C. 130.D. 15.12. A. On Tuesdays . B. On Wednesdays.C. On Sundays .D. On Mondays.13. A. The English Family club.B. The painting club .C. The sports club.D. The music club.Questions 14 through 16 are based on the following passage.14.A. Because they haven’t as much interest in finding the cure as in space travel.B. Because there are too many kinds of cold viruses for them to identify.C. Because it is not economical to find a cure for each type of cold.D. Because they believe people can recover without treatment.15.A. They reveal the seriousness of the problem.B. They indicate how fast the virus spreads.C. They tell us what kind of medicine to take.D. They show our body id fighting the virus.16.A. It can actually does more harm than good.B. It causes damage to some organ of our body.C. It works better when combined with other remedies.D. It helps ue to recover much sooner.Questions 17 through 20 are based on the following conversation.17. A. It lasts till today. B. It lasted about ten years.C. It’s not mentioned in the conversation.D. It lasts forever.18.A. The commercial success of several boys and girls.B. The funny daily stories that happened to a group of lose friends.C. How people in Manhattan made their living.D. American culture,mainly the coffee culture.19. A. It always received positive reviews from the critics.B. It enjoyed a high rate of watching.C. It was used as a tool for English learning all over the world.D. It became a cultural phenomenon.20. A. They may go to the “Centra Park”for a cup of coffee.B. They may open a new coffee shop together .C. They may start practicing English with each other.D. They may go to the woman’s place to enjoy the show.II. Grammar and VocabularySection ADirections: After reading the passage below, fill m 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.Is This Art?Have you ever looked at a painting and thought “I could do better than that”？ Have you ever seen a film without any story or characters? Or heard a piece of music (21)________ doesn’t quite sound like music? If you can answer “yes” to any of these questions, thechances are that (22) ________ you were looking at, watching or listening to was something “avant-garde”.One of the most famous examples of avant-garde art comes from the world of music. John Cage's piece of music 4,33n consists of 4 minutes and 33 seconds of silence. It was written by Cage, a leading American member of the avant-garde, in 1952. It was divided into three movements,(23) ________ (perform) without a single note being played. (24) ________the composer, the musicis actually the sounds the listener hears while “listening” to the performance. These might include, of course, listeners (25) ________ (ask) each other how they know when the piece ends.Cinema has always had avant-garde directors. Possibly the best known is Andy Warhol.(26) ________better known as a painter, between 1963 and 1968 Warhol made more than 60 films,nearly (27) ________of them experimental. One film, Eat, consists of a man eating a mushroom for 45 minutes, while Sleep shows poet John Giomo sleeping for 6 hours. Empire is 8 hours long and only shows the Empire State Building as the sun (28) ________ (set) at dusk. You could eat a lot of popcorn in 8 hours.Some people love avant-garde art and some hate it. Some believe avant-garde artists are geniuses, while others tend (29) ________ (think) they're pretentious. However, whether you love them or hate them, you will probably have to accept that these people are just no (30)________ (passionate) about their art than Michaelangelo, Beethoven or Orson Welles were in their day.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.How sound and colour influence the taste of foodThe sound people hear while they are eating food can change the way they think it tastes, scientists have discovered. In fact, researchers have also found that changing the colour of a food can influence the flavour ____31____ by consumers. Food manufacturers are now hoping to ____32____the findings in an attempt to make their foods more appealingIt was ____33____ thought that the sense of taste and smell were the only human senses that played a role in experiencing flavour. Professor Charles Spence, a sensory psychologist at Oxford University, is a leading expert in his field. He believes that it is possible to change the flavor of food simply by exciting people's sense of hearing. For example, listening to waves hitting the sea shore can make ____34____ detect seafood flavours.Professor Spence has also discovered that ____35____ changing the colour of a food can influence the way it tastes. He found that by changing a drink from yellow to a deep red, it is possible to make it taste up to 12 per cent sweeter than it really is. He said: This colour has strong ___36_____ with very ripe fruit.’Flavour is not just as simple as the way something tastes, as all the other senses come into ____37____ and some can dominate the way the brain will ____38____ a food. Ice cream activates a part of the brain which is just behind the eyes and is where emotions are____39____ . By melting, it changes its physical ____40____ and creates contrasts that continually keep your senses interested.So next time you are in a restaurant or a supermarket, or just sitting down at home to eat something, think about whether your choices have been influenced by sound or colour. You may be surprised!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.An internet troll is someone who posts insulting comments in an online chat room, social-networking site or blog. Their aim is not to contribute to the discussion, but to____41____people or make them angry.Sometimes the comments are fairly___42_____. For example, a troll might visit a website that is dedicated to fighting global warming and post a comment like 'Global warming is a myth made up by second-rate scientists who just want to further their owncareers.' Some visitors to the website might think that the opinion is genuine andstart___43_____. This is exactly what the troll wants: to start a fight.However, many trolls are far more dangerous and post comments that are deeply hurtful. In 2011,45-year-old Nicola Brookes left a message of support for a reality TV contestant on Facebook. The contestant had received hate mail on Facebook and, feeling_____44___ for the young man, Ms Brookes left a message on his Facebook page, saying simply: ‘Keep your chin up, Frankie, theyMl move on to someone else soon/ They did move on to someone else—Ms Brookes—within minutes. More than 100 cruel messages were left in just 24 hours, and a(n) ___45_____ Facebook page was set up in her name.Racist trolling probably has the highest profile cases. When Premier League footballer Fabrice Muamba collapsed on the field during a match in 2012, racist messages appeared on Twitter taking pleasure in his_____46___and laughing at him. ____47____, Muamba made a complete recovery. The effects of such attacks can be terrible, especially for people who are being targeted precisely because they admitted to a vulnerability (弱点)in the first place.So ____48____ do trolls post messages that cause such distress? Some of them are sad, lonely people with their own problems who___49_____ the attention their posts attract. They behave badly online because they feel_____50___ normal standards of polite and considerate behaviour. They are in a virtual world and do not have to face their victims. They say things that they would never dare to say ____51____. Finally, trolls believe that they can remain unknown and that no one will ever ____52____who they are.That is about to_____53___ , however. When Nicola Brookes first approached the police and made a complaint, they told her there was nothing they could do. So she contacted lawyers and began legal action to force Facebook to disclose the identities of the trolls who had been tormenting (折磨）her. She ____54____the case and is free to pursue private prosecutions(起诉) against those responsible for tormenting her. With trolls facing____55____ and possible prosecution in future, hopefully they will think twice before posting insulting messages.41. A. entertain B. locate C. upset D. welcome42. A. informative B. convincing C. harmless D. unfair43. A. laughing B. arguing C. proving D. applauding44. A. anxiety B. sympathy C. dissatisfaction D. hope45. A. appealing B. complex C. famous D. fake46. A. misfortune B. distress C. feeling D. violence47. A. Occasionally B. Strangely C. Generally D. Thankfully48. A. why B. how C. when D. where49. A. fear B. enjoy C. hold D. give50. A. liberated from B. curious about C. satisfied with D. dependent on51. A. for sure B. on stage C. in person D. with confidence52. A. wonder B. discover C. say D. remember53. A. return B. happen C. collapse D.change54. A. lost B. filed C. won D. forgot55. A. competition B. wealth C. embarrassment D. exposureSection 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)After bouncing my rental car across several miles of red-dirt roads I walked for nearly another mile down the beach to a deserted valley. It was comforting to think that at the very least I was finally out of cell-phone range.However, even on Kauai, Hawaii’s ‘Garden Island’，complete escape wasn’t all that easy to achieve. Noisy helicopters full of tourists flew overhead like so many dragonflies. Every 20 minutes or so the comforting sounds of wind and water were broken by the noise of a speeding tour boat racing to complete another lap around the island. Worst of all, not more than five minutes by car from the resort where I was staying, the Atomic Clock Internet Cafe signaled with promises of instant email.I felt uncomfortable every time I drove by the Atomic Clock Cafe. 1 am a technology reporter for an online magazine—my life is driven and dominated by email. I'm drowned in it,usually 400 or 500 messages a day. The main reason for my visit to Kauai was to unplug, disconnect, log off, and get away from it all. No cell phone, no electronic organiser, no laptop. And definitely, no email.Yes, my plan was to lie on the beach and not check my email. My friends and family were outraged as they could not understand how I could bear to live without email. But they didn't understand. In my job, I am online, permanently. Cyberspace is more familiar to me than my backyard. While I am awake, my email is always on. I don't like to be without it for too long.A few hours away from it, and I start to tremble. I am, however, no stranger to beaches and their relaxing qualities and so I knew, even when arriving well after dark at the comfortable cottage in the town of Waimea, that the island of Kauai gave me a good chance of beating my addiction to electronic devices.Maybe it was full moon lighting the black-sand beach not 10 metres from my door. Or the mango trees casting shadows across the veranda(阳台). Or the driftwood piled in loose heaps foras far as I could see along the shore. Without question, the long, slow sound of the waves rolling in calmed my restless soul, and I found I could, in fact, log off.56. Why did the writer come to Kauai?A. To get away from the modem technology.B. To work for the Atomic Clock Internet Cafe.C. To write reports on technological development.D. To find whether there is an alternative to email.57. What can we learn about the writer?A. He wrote articles about reasons around the world.B. He enjoyed beach activities like boat racing.C. He was eager to work in his backyard.D. He spent much time working online.58. The word “outraged”（paragraph 4) is closest in meaning to “__________”.A. relievedB. shockedC. amusedD. offended59. The writer described the scenery in the last paragraph in order to .A. argue against his friends’ doubt of KauaiB. propose a possible destination of his tripC. highlight the beauty of the beach of KauaiD. show Kauai produced a relaxing atmosphere(B)Read the following two blogs and answer the questions that follow.Our buses drive me to distractionI’ve had it up to here with buses! To be more precise, I am sick to death of the service offered by our local bus company and which dares to call itself Reliabus—a misnomer if ever there was one, since its exhaust-fume-coughing, atmosphere-choking buses are anything but reliable. Nine times out of ten they arrive late, and when they do eventually turn up, there's no guarantee you'll get on. I've lost count of the times I have watched as a full-to-bursting number 26—which I try to catch home from college---sails past the bus stop (another misnomer), leaving weary travellers like myself to wait for at least another 20 minutes until the next one comes along. Unless of course that’s full as well, in which case we have to wait even longer. Why don’t they lay on more buses!It annoys me to think that they are turning people away from travelling by public transport, encouraging them to use their cars and causing traffic jams in our already heavily congested town centre. I wrote them an email and gave them a piece of my mind If you're as fed up as I am, why not do the same? They might just sit up and take notice.Leave me alone!If there’s one thing that gets on my nerves, it’s people who keep trying to s ell me things I don’t want. It drives me mad when Tm in the middle of my dinner or watching a film and some smooth- talking idiot phones up and asks me if Tm interested in new kitchen units, a subscription to a book club or a superfast internet connection. No, I’m not, thank you very much, and if I was, I’d get in touch with you! I very nearly burnt the house down the other day answering one of their ridiculous calls. I forgot I’d left something frying in the kitchen—just got back to the blackened remains in time.It’s even worse outside of the home—sometimes you can’t move for people handing out fliers in the street, advertising computer classes or urging you to buy this, that or the other. They push one into your face, and if you don’t take it, they give y ou a nasty look and you canhear them swearing and muttering under their breath about you as you walk away. Not nice at all.60. Both the two blogs are meant to________.A. address a problemB. ask for sympathyC. make a complaintD. clarify a misunderstanding61. By “another misnomer”（paragraph 1)，the blogger actually means________.A. the bus seldom stops thereB. 20 minutes' wait is too longC. she is often too late to catch the busD. public transport is not a good choice62. What annoys the blogger according to the second blog?A. The bad quality of kitchen appliances.B. The endless phone calls to sell something.C. The terrible experiences of cooking for the family.D. The attitudes that her colleagues have towards her.(C)The goings-on in the consulting room have become more transparent(透明的）recently. Thank goodness. We know more than the lines supplied by the movies in which the therapist knows all and gives wisdom to those who, sitting on a couch, consult with them. Therapists are interested in how the individual, the couple or the family experiences and understands their difficulties. That has to be a starting place. We can be of value if our first port of call is to listen, to gradually feel ourselves into the shoes of the other, to absorb the feelings that are being conveyed and to think and then to say some words.The thinking and talking that I do inside the consulting room is at odds with many features of ordinary conversation. Not that it is mysterious, but it isn't concerned with traditional ways of sharing or identifying. The therapist makes patterns and theorises, but they are also reflecting on the words that are spoken, how they are delivered and how the words, once spoken, affect the speaker and the therapist themselves.Words can g ive voice to previously unknown feelings and thoughts. That’s why it’s called the talking cure. But just as words reveal so, too, can they obscure, and this gets us to the listening and feeling part of the therapy. Whatever and however the words are delivered, they will have an impact on me as a therapist. I might feel hopeless, I might feel energised, I might feel pushed away, I might feel demanded of, I might feel pulled to find solutions.The influence of the other is what makes any relationship possible or impossible. A therapist is trained to reflect on how those who consult with them affect them. As I try to step into the shoes of the other and then out again, my effort is to hold both those experiences, plus an awareness of my ease or discomfort with what I encounter in the relationship.Feelings are the bread and butter of our work in the consulting room. They inform or modify our ideas and they enable us to find an emotional bridge to what can so hurt for the people we are working with. Along with the more commonly thought-about theories and ideas we have about the psyche, they are an essential part of the therapist^ toolkit, certainly for me. The talking cure means talking, yes. It also means the therapist is listening, thinking and feeling.63. In which way is the thinking and talking the writer does different from ordinary conversation?A. It may not be understood by patients.B. It is full of terms used by most therapists.C. It is a good reflection of traditional talking.D. It involves thinking about how people speak.64. The word t4obscureM (paragraph 3) is closest in meaning to________.A. cancelB. clarifyC. confirmD. conceal65. Which of the following is the writer most likely to agree with?A. Patients' influence has been neglected by therapists for too long a time.B. Therapists need to think from their own perspectives as well as patients'.C. It is no easy job for therapists to realize how uncomfortable their patients are.D. Therapists had better push away those negative emotions acquired from patients.66. Which of the following might be the best title of the passage?A. Awareness of feelingsB. It’s good to talk—and listenC. Theories that help therapistsD. What is the point of being a therapistSection CDirections: Complete the following passage by using the sentences given below. Each sentence can be used only once. Note that there are two more sentences than you need.One Dollar a Night in New YorkWhen it comes to finding a place to stay for a night in New York, things don't always come cheap. However, artist Miao Jiaxin, a Shanghai native who moved to New York in 2006, is offering people the chance to stay in his apartment inBrooklyn.______67__________.Guests can easily book Miao's room on the Internet. Nevertheless, although they will be housed in his apartment, it appears to have more in common with a jail cell than a regular bedroom as a cage in the center of the room is where guests will stay.__________68______________ Guests must stay m the cage for three hours each morning. ''From 9 a.m. to 12 p.m., you can't access the Internet, and there are no electronic devices, books, radio, pens or craftwork. You can't talk to anybody. You can't do Yoga or any other exercises. And you can't even sleep, writes Miao. If you break any of those rules above, you will be fined 100 dollars.Meanwhile, the cage is monitored and recorded by two cameras and the activity of guests is filmed for the whole time they stay in the cell.________69________________They can enjoy great views of New York on the roof deck outside the room.The room is inspired by the alienation(疏离感) Miao felt as a new immigrant—feelings he believes are universal. “It’s not for fun. It’s for an experience.___________70_____________’’ said M iao.IV. Summary WritingDirections: Read the following passage. Summarize in no more than 60 words the main idea of the passage and how it is illustrated. Use your own words as far as possible.e-learning: Hazy past—better future?How much of an effect does technology have on students' learning? A significant one, it seems, according to experts. Currently available technologies, the most important of which are computers and the Internet, apparently provide a learning environment in which problem-solving and intellectual enquiry can flourish. The process of learning in the classroom may become significantly more effective as students can deal with information on the computer. Or so the theory goes. My own viewpoint is rather different, Tm afraid.Computers have been around for two decades as part of school equipment. There are, of course, obstacles like costs to overcome, but it's just a matter of time and effort. This is because schools have done what every organization does when it sees an innovation--it applies the innovation to its existing model, which adds cost but doesn't transform the standard classroom. We have, during that period, spent over $60 billion on them, but in my view they seem to have had little or no effect on learning in schools. Content is king and the mode of delivery is irrelevant. If a teacher makes the subject matter interesting, it does not matter what, if any, equipment is used.However, change is on the horizon. I think student-centred learning will become the norm and transform education. Computers will pave the way for far more independent learning. Students who currently don't have access to schools or teachers are now able to get online. They can study from home thanks to the fact that more learning programmes are being written for learners who are forced by their circumstances to be self-sufficient. This would prove especially beneficial in those areas of the world where quality education is limited or extremely expensive. Therefore, in a few years' time we could have a completely different conversation about technology and its impact on learning.V. TranslationDirections: Translate the following sentences into English, using the words given in the brackets. 72. 最近你收到过汤姆的来信吗？（hear)73.没人知道他昨天缺席这个重要会议的原因。

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2019年全国普通高等学校招生统一考试
上海英语模拟试卷（一）
I. Listening Comprehension
Section A
Short Conversations
Directions: In Section A, you will hear ten short conversations between two speakers. At the end
of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.
1. A. A colleague. B. Their boss. C. Their workload. D. A coffee machine.
2. A. At a gift shop. B. At a graduation ceremony.
C. In the office of a travel agency.
D. In a school library.
3. A. The woman can search the box for printing paper.
B. He doesn't know where the office supplies are.
C. They have already run out of printing paper.
D. The office supplies are far from enough.
4. A. The dictionary will be re-printed soon.
B. The printing of the dictionary is just good.
C. He can read the explanations for the woman.
D. He cannot read the explanations well, either.
5. A. The woman cannot figure out what is in his notes.
B. The woman should comment on his handwriting.
C. He is very pleased to be able to help.
D. He did not attend today’s classes.
6. A. The restaurant is a good place for birthday celebration.
B. The new restaurant caught her attention immediately.
C. The man has good taste in choosing the restaurant.
D. The man had better choose another restaurant.
7. A. Design her office without charging any fees.
B. Tell her how much the redecoration might cost.
C. Give her tips on which design academy to choose.
D. Confirm whether her office really needs redecorating.
9. A. He has been looking forward to spring.
B. He has been waiting for the winter sale.
C. He will clean the woman’s boots for spring.
D. He will help the woman put things away.
Section B
Directions: In Section B, you will hear two short passages and one longer conversation, and you will be asked three questions on each of the passages and the conversation. The passages and the conversation will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.
Questions 11 through 13 are based on the following passage.
11. A. The possible future of cars as a means of transportation.
B. The contrast between road travel in the past and now.
C. The development of road travel in the past 100 years.
D. The effects of modem means of transport on the environment.
12. A. It is expensive. B. It is too romantic.
D. It moves too fast. C. It depends on the weather.
13. A. They will disappear due to the damage they have done.
B. They will be thought of as a slower means of transport.
C. They will remain a common means of transport.
D. They will no longer be considered practical.
Questions 14 through 16 are based on the following passage.
14. A. They were both initially limited to the services for depositors.
B. They were both created to provide income for the king.
C. They both required a large staff to administer them.
D. They both grew out of the need to store food.
15. A. It normally requires precious metals.
B. It is likely to begin when people are in debt.
C. It can take place without the existence of coins.
D. It was started to provide the state with an income.
16. A. The future of banking. B. The origin of banking.
C. The necessity of banking.
D. The development of banking. Questions 17 through 20 are based on the following conversation.
17. A. At a travel agency. B. At a design school.
C. At a botanic garden.
D. At an evening school.
18. A. Two weeks before the start of the term.
B. A week before the start of the term.
C. A month before the start of the term.
D. Two months before the start of the term.
19. A. It costs 70 pounds for the term.
B. It has already had enough applicants.
C. It is on Tuesday evenings from 7 to 9.
D. It requires no extra fees other than the tuition.
20. A. Contact the botany tutor. B. Buy some drawing materials.。

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