上海市徐汇区2019学年度高三(一模)(含答案)

徐汇区2018-2019学年等级考第一次模拟考试试卷物 理考生注意：1．试卷满分100分，考试时间60分钟。

2．本考试分设试卷和答题纸。

试卷包括三部分，第一部分为选择题，第二部分为填空题，第三部分为综合题。

3．答题前，务必在答题纸上填写学校、姓名、考场号和座位号，并将核对后的条形码贴在指定位置上。

作答必须涂或写在答题纸上，在试卷上作答一律不得分。

第一部分的作答必须涂在答题纸上相应的区域，第二、三部分的作答必须写在答题纸上与试卷题号对应的位置。

一、单项选择题（共40分，1至8题每小题3分，9至12题每小题4分。

每小题只有一个正确选项）1. 在国际单位制（SI ）中，下列物理量的单位属于导出单位的是( )(A) 长度 (B) 电流 (C) 电压 (D) 热力学温度2. 收听广播时会听到“上海东方广播电台FM100.7”。

这里的“100.7”指的是电磁波的(A) 频率 (B) 周期 (C) 波长 (D) 波速3. 碳的同位素146 C 的原子核中有(A) 8个质子 (B) 8个电子 (C) 8个中子 (D) 8个核子4. 某放射性元素经过了3个半衰期，发生衰变的原子核约为总数的(A) 1/4 (B) 3/4 (C) 1/8 (D) 7/85. 若单摆的摆长不变，摆球的质量减小，摆球离开平衡位量的最大摆角减小，则单摆振动的(A) 频率不变，振幅不变 (B) 频率不变，振幅改变(C) 频率改变，振幅不变 (D) 频率改变，振幅改变6. 如图，水平桌面上有一个小铁球和一根条形磁铁，现给小铁球一个沿OP 方向的初速度v ，则小铁球的运动轨迹可能是(A) 甲 (B) 乙 (C) 丙 (D) 丁7. 某种气体在不同温度下的气体分子速率分布曲线如图所示，三条曲线所对应的温度分别为T I 、T II 、T III ，则(A) T I > T II > T III (B) T I <T II <T III(C) T I = T II =T III (D) T I < T II , T II > T III8. 如图，空间有两个等量正点电荷，b、c为电荷连线上两点，b为中点，ab连线垂直于bc连线，三点电势分别为φa、φb、φc。

2019年上海市徐汇区高考数学一模试卷一、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）考生应在答题纸的相应位置直接填写结果.1．（4分）若复数z满足i•z＝1+2i，其中i是虚数单位，则z的实部为．2．（4分）已知全集U＝R，集合A＝{y|y＝x﹣2，x∈R，x≠0}，则∁U A＝．3．（4分）若实数x，y满足xy＝1，则2x2+y2的最小值为．4．（4分）若数列{a n}的通项公式为a n＝（n∈N*），则a n＝．5．（4分）已知双曲线＝1（a＞0，b＞0）的一条渐近线方程是y＝2x，它的一个焦点与抛物线y2＝20x的焦点相同，则此双曲线的方程是．6．（4分）在平面直角坐标系xOy中，直线经过坐标原点，＝（3，1）是l的一个法向量．已知数列{a n}满足：对任意的正整数n，点（a n+1，a n）均在l上，若a2＝6，则a3的值为．7．（5分）已知（2x2﹣）n（n∈N*）的展开式中各项的二项式系数之和为128，则其展开式中含项的系数是．（结果用数值表示）8．（5分）上海市普通高中学业水平等级考成绩共分为五等十一级，各等级换算成分数如表所示：上海某高中2018届高三（1）班选考物理学业水平等级考的学生中，有5人取得A+成绩，其他人的成绩至少是B级及以上，平均分是64分，这个班级选考物理学业水平等级考的人数至少为人．9．（5分）已知函数f（x）是以2为周期的偶函数，当0≤x≤1时，f（x）＝lg（x+1），令函数g（x）＝f（x）（x∈[1，2]），则g（x）的反函数为．10．（5分）已知函数y＝sin x的定义域是[a，b]，值域是[﹣1，]，则b﹣a的最大值是．11．（5分）已知λ∈R，函数f（x）＝，若函数f（x）恰有2个零点，则λ的取值范围是．12．（5分）已知圆M：x2+（y﹣1）2＝1，圆N：x2+（y+1）2＝1．直线l1、l2分别过圆心M、N，且11与圆M相交于A，B两点，12与圆N相交于C，D两点，点P是椭圆＝1上任意一点，则+的最小值为．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.13．（5分）设θ∈R，则“θ＝”是“sinθ＝”的（）A．充分非必要条件B．必要非充分条件C．充要条件D．既非充分也非必要条件14．（5分）魏晋时期数学家刘徽在他的著作《九章算术注》中，称一个正方体内两个互相垂直的内切圆柱所围成的几何体为“牟合方盖”，刘徽通过计算得知正方体的内切球的体积与“牟合方盖”的体积之比应为π：4．若正方体的棱长为2，则“牟合方盖”的体积为（）A．16B．16C．D．15．（5分）对于函数y＝f（x），如果其图象上的任意一点都在平面区域{（x，y）|（y+x）（y ﹣x）≤0}内，则称函数f（x）为“蝶型函数”，已知函数：①y＝sin x；②y＝，下列结论正确的是（）A．①、②均不是“蝶型函数”B．①、②均是“蝶型函数”C．①是“蝶型函数”；②不是“蝶型函数”D．①不是“蝶型函数”：②是“蝶型函数”16．（5分）已知数列{a n}是公差不为0的等差数列，前n项和为S n，若对任意的n∈N*，都有S n≥S3，则的值不可能为（）A．2B．C．D．三、解答题.17．（14分）如图，已知正方体ABCD﹣A′B′C′D′的棱长为1．（1）正方体ABCD﹣A′B′C′D'中哪些棱所在的直线与直线A′B是异面直线？（2）若M，N分别是A'B，BC′的中点，求异面直线MN与BC所成角的大小．18．（14分）已知函数f（x）＝，其中a∈R．（1）解关于x的不等式f（x）≤﹣1；（2）求a的取值范围，使f（x）在区间（0，+∞）上是单调减函数．19．（14分）我国的“洋垃极禁止入境”政策已实施一年多．某沿海地区的海岸线为一段圆弧AB，对应的圆心角∠AOB＝，该地区为打击洋垃圾走私，在海岸线外侧20海里内的海域ABCD对不明船只进行识别查证（如图：其中海域与陆地近似看作在同一平面内）在圆弧的两端点A，B分别建有监测站，A与B之间的直线距离为100海里．（1）求海域ABCD的面积；（2）现海上P点处有一艘不明船只，在A点测得其距A点40海里，在B点测得其距B 点20海里．判断这艘不明船只是否进入了海域ABCD？请说明理由．20．（16分）已知椭圆Γ：＝1（a＞b＞0）的长轴长为2，右顶点到左焦点的距离为+1，直线l：y＝kx+m与椭圆Γ交于A，B两点．（1）求椭圆Γ的方程；（2）若A为椭圆的上项点，M为AB中点，O为坐标原点，连接OM并延长交椭圆Γ于N，，求k的值．（3）若原点O到直线l的距离为1，＝λ，当时，求△OAB的面积S 的范围．21．（18分）已知项数为n0（n0≥4）项的有穷数列{a n}，若同时满足以下三个条件：①a 1＝1，a＝m（m为正整数）；②a i﹣a i﹣1＝0或1，其中i＝2，3，……，n0；③任取数列{a n}中的两项a p，a q（p≠q），剩下的n0﹣2项中一定存在两项a s，a t（s≠t），满足a p+a q＝a s+a t，则称数列{a n}为Ω数列．（1）若数列{a n}是首项为1，公差为1，项数为6项的等差数列，判断数列{a n}是否是Ω数列，并说明理由．（2）当m＝3时，设Ω数列{a n}中1出现d1次，2出现d2次，3出现d3次，其中d1，d2，d3∈N*．求证：d1≥4，d2≥2，d3≥4；（3）当m＝2019时，求Ω数列{a n}中项数n0的最小值．2019年上海市徐汇区高考数学一模试卷参考答案与试题解析一、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）考生应在答题纸的相应位置直接填写结果.1．【解答】解：由i•z＝1+2i，得z＝，∴z的实部为2．故答案为：2．2．【解答】解：A＝（0，+∞）；∴∁U A＝（﹣∞，0]．故答案为：（﹣∞，0]．3．【解答】解：∵xy＝1，∴2x2+y2≥2＝2，（当且仅当2x＝y＝±时，取等），故答案为：2．4．【解答】解：数列{a n}的通项公式为a n＝＝﹣，则a n＝（﹣）＝﹣1．故答案为：﹣1．5．【解答】解：抛物线y2＝20x的焦点为（5，0），则双曲线的焦点在x轴上，双曲线的一条渐近线为y＝2x，可得b＝2a，由题意双曲线的一个焦点与抛物线y2＝20x的焦点相同，可得＝5，解得a＝，b＝2，则双曲线的方程为：．故答案为：．6．【解答】解：直线经过坐标原点，＝（3，1）是l的一个法向量，可得直线l的斜率为﹣3，即有直线l的方程为y＝﹣3x，点（a n+1，a n）均在l上，可得a n＝﹣3a n+1，即有a n+1＝﹣a n，则数列{a n}为公比q为﹣的等比数列，可得a3＝a2q＝6×（﹣）＝﹣2．故答案为：﹣2．7．【解答】解：由题意，2n＝128，得n＝7．∴（2x2﹣）n＝（2x2﹣）7，其二项展开式的通项＝．由14﹣3r＝﹣1，得r＝5．∴展开式中含项的系数是．故答案为：﹣84．8．【解答】解：设取得A成绩的x人，取得B+成绩的y人，取得B成绩的z人，则70×5+67x+64y+61z＝64×（5+x+y+z），即z﹣x＝10，又x，y，z∈N，即当且仅当x＝0，y＝0，z＝10时，5+x+y+z取得最小值15，取得A成绩的0人，取得B+成绩的0人，取得B成绩的10人，这个班级选考物理学业水平等级考的人数至少为15人，故答案为：159．【解答】解：当﹣1≤x≤0时，0≤﹣x≤1，∴f（x）＝f（﹣x）＝lg（﹣x+1），当1≤x≤2时，﹣1≤x﹣2≤0，∴f（x）＝f（x﹣2）＝lg[﹣（x﹣2）+1]＝lg（﹣x+3）．∴g（x）＝lg（﹣x+3）（1≤x≤2），∴﹣x+3＝10g（x），∴x＝3﹣10g（x），故答案为：g﹣1（x）＝3﹣10x，（0≤x≤lg2）10．【解答】解：函数y＝sin x，令≤a≤，要使b﹣a的最大值，可知b的最大值为：b＝，∴b﹣a的最大值为；故答案为：11．【解答】解：根据题意，在同一个坐标系中作出函数y＝x﹣4和y＝x2﹣4x+3的图象，如图：若函数f（x）恰有2个零点，即函数f（x）图象与x轴有且仅有2个交点，则1＜λ≤3或λ＞4，即λ的取值范围是：（1，3]∪（4，+∞）故答案为：（1，3]∪（4，+∞）．12．【解答】解：由题意可得，M（0，1），N（0，﹣1），r M＝r N＝1，＝（）•（）＝＝，＝（）•＝＝﹣1，∵∵P为椭圆上的点，∴＝+﹣2＝2（x2+y2）＝由题意可知，﹣3≤x≤3，∴8≤，故答案为：8．二、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.13．【解答】解：由θ＝，则有sinθ＝，即“θ＝”是“sinθ＝”的充分条件，由sinθ＝，得：θ＝kπ+（﹣1）k，即“θ＝”是“sinθ＝”的不必要条件，即“θ＝”是“sinθ＝”的充分不必要条件．故选：A．14．【解答】解：正方体的棱长为2，则其内切球的半径r＝1，∴正方体的内切球的体积，又由已知，∴．故选：C．15．【解答】解：由y＝sin x，设g（x）＝sin x+x，导数为cos x+1≥0，即有x＞0，g（x）＞0；x＜0时，g（x）＜0；设h（x）＝sin x﹣x，其导数为cos x﹣1≤0，x＞0时，h（x）＜0，x＜0时，h（x）＞0，可得（y+x）（y﹣x）≤0恒成立，即有y＝sin x为“蝶型函数”；由（+x）（﹣x）＝x2﹣1﹣x2＝﹣1＜0，可得y＝为“蝶型函数”．故选：B．16．【解答】解：∵数列{a n}是公差不为0的等差数列，前n项和为S n，对任意的n∈N*，都有S n≥S3，∴，∴，且∴﹣3d≤a1≤﹣2d，∴当＝＝2时，a1＝﹣3d．成立；当＝＝时，a1＝﹣d．成立；当＝＝时，a1＝﹣2d．成立；当＝＝时，a1＝﹣d．不成立．∴的值不可能为．故选：D．三、解答题.17．【解答】解：（1）正方体ABCD﹣A′B′C′D′中，直线A′B是异面直线的棱所在直线有：AD，B′C′，CD，C′D′，DD′，CC′，共6条．（2）M，N分别是A'B，BC′的中点，以D为原点，DA为x轴，DC为y轴，DD′为z轴，建立空间直角坐标系，则A′（1，0，1），B（1，1，0），C′（0，1，1），M（1，，），N（），B（1，1，0），C（0，1，0），＝（﹣，0），＝（﹣1，0，0），设异面直线MN与BC所成角的大小为θ，则cosθ＝＝＝，∴θ＝45°，∴异面直线MN与BC所成角的大小为45°．18．【解答】解：（1）x的不等式f（x）≤﹣1，即为≤﹣1，即为≤0，当a＝﹣1时，解集为{x|x≠﹣2}；当a＞﹣1时，解集为（﹣2，0]；当a＜﹣1时，解集为（﹣∞，﹣2）∪[0，+∞）；（2）f（x）＝＝a+，由f（x）在区间（0，+∞）上是单调减函数，可得﹣2﹣2a＞0，解得a＜﹣1．即a的范围是（﹣∞，﹣1）．19．【解答】解：（1）∵∠AOB＝，在海岸线外侧20海里内的海域ABCD，AB＝100∴AD＝BC＝20，OA＝OB＝AB＝100，∴OD＝OA+AD＝100+20＝120，∴S ABCD＝•π（OD2﹣OA2）＝π（1202﹣1002）＝（平方海里），（2）由题意建立平面直角坐标系，如图所示；由题意知，点P在圆B上，即（x﹣100）2+y2＝7600…①，点P也在圆A上，即（x﹣50）2+＝1600…②；由①②组成方程组，解得或；又区域ABCD内的点满足，由302+＝3600＜10000，∴点（30，30）不在区域ABCD内，由902+＝15600＞14400，∴点（90，50）也不在区域ABCD内；即这艘不明船只没进入了海域ABCD．20．【解答】解：（1）由题意可知，，于是得到，因为右顶点到左焦点的距离为，所以，c＝1，则，因此，椭圆Γ的方程为；（2）当点A为椭圆的上顶点时，点A的坐标为（1，0），则m＝1，直线l的方程为y＝kx+1，将直线l的方程代入椭圆的方程并化简得（2k2+1）x2+4kx＝0，解得，，所以点B的坐标为，由于点M为线段AB的中点，则点M的坐标为，由于，所以，点N的坐标为，将点N的坐标代入椭圆的方程得，化简得，解得；（3）由于点O到直线l的距离为1，则有，所以，m2＝k2+1．设点A（x1，y1）、B（x2，y2），将直线l的方程代入椭圆方程并化简得（2k2+1）x2+4kmx+2m2﹣2＝0，由韦达定理可得，，＝x1x2+（kx1+m）（kx2+m）＝＝＝＝，由于，即，解得，线段AB的长为＝＝＝＝，所以，．因此，△OAB的面积S的取值范围是．21．【解答】解：（1）若数列{a n}：1，2，3，4，5，6是Ω数列，取数列{a n}中的两项1和2，则剩下的4项中不存在两项a s，a t（s≠t），使得1+2＝a s+a t，故数列{a n}不是Ω数列；（2）若d1≤3，对于p＝1，q＝2，若存在2＜s＜t，满足a p+a q＝a s+a t，∵2＜s＜t，于是s≥3，t≥4，故a5≥a2，a t＞a1，从而a s+a t＞a2+a1，矛盾，故d1≥4，同理d3≥4，下面证明d2≥2：若d2＝1，即2出现了1次，不妨设a k＝2，a1+a k＝a s+a t，等式左边是3，等式右边有几种可能，分别是1+1或1+3或3+3，等式两边不相等，矛盾，于是d1≥2；（3）设出现d1次，2出现d2次…，2019出现d2019次，其中d1，d2，…，d2019∈N*，由（2）可知，d1≥4，d2019≥4，且d2≥2，同理d2018≥2，又∵d3，d4…，d2017∈N*，故项数n0＝d1+d2+…+d2019≥2027，下面证明项数n0的最小值是2027：取d1＝4，d2＝2，d3＝d4＝…＝d2017＝1，d2018＝2，d2019＝4，可以得到数列{a n}：1，1，1，1，2，2，3，4…，2016，2017，2018，2019，2019，2019，2019，接下来证明上述数列是Ω数列：若任取的两项分别是1，1，则其余的项中还存在2个1，满足1+1＝1+1，同理，若任取的两项分别是2019，2019也满足要求，若任取的两项分别是1，2，则其余的项中还存在3个1，1个2，满足要求，同理，若任取的两项分别是2018，2019也满足要求，若任取a p＝1，a q≥3，则在其中的项中取a5＝2，a t＝a q﹣1，满足要求，同理，若a p≤2017，a q＝2019也满足要求，若任取的两项a p，a q满足1＜a p≤a q＜2019，则在其余的项中选取a s＝a p﹣1，a t＝a q+1，每个数最多被选取了1次，于是也满足要求，从而，项数n0的最小值是2027．。

2019-2020学年上海市徐汇中学高三英语一模试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AIt’s the time of year when we start hiking. As you pack, remember to bring your smartphone. Whether you’re going on a short walk or a long trip, there are a handful of apps that can help.MapMyHikeThis app tracks where you're hiking so you have a mapof your route at the end of the hike. It can also track other fitness information like the distance traveled, speed, pace, and even calories burned. You can save the data for your hike, so you can always access the route you look as well as track improvements to your workout. GaiaGPSYou don't always have cellphone service when hiking, but you always want to know where you are. The GaiaGPS app provides that information. Download maps of different parts of the world, and access the GaiaGPS app in the middle of even the most remote trails. The GPS function makes using the maps simple, and the app will also point to areas of interest.Backpacking ChecklistOne of the worst things is being way out on a trail only to discover you left behind something important. That's why checklists are the best. This checklist app helps you build a customized(定制的) list of things to take with you. Organize different lists based on trail lengths or requirements. Track all your essential items by weight and where you can find them.WildObsUsing WildObs, you can record your observations of plants and animals and add them to the database. You can ask the community to help you identify something and keep track of everything you've met, and most importantly, you can become a citizen scientist. By recording what you've seen with this app, you're helping scientists keep track of what's happening to the natural world.1. What can you do with MapMyHike?A. Record your walking speed.B. Design a suitable hiking route.C. Locate popular tourist attractions.D. Store the data of your daily activities.2. What is WildObs intended to do?A. To provide survival skills.B. To lead the way.C. To identify wildlife.D. To help make preparations.3. Which app is most useful before hiking?A. GaiaGPS.B. MapMyHike.C. WildObs.D. Backpacking Checklist.BIt is that time of year when people need to lock their cars. It’ s not because there are a lot of criminals running around stealing cars. Rather, it’ s because of the good-hearted neighbors who want to share their harvest. Especially with this year’s large crop, leaving a car unlocked in my neighborhood is an invitation for someone to stuff it full of zucchini(西葫芦).My sister-in-law, Sharon, recently had a good year for tomatoes. She and her family had eaten and canned so many that they began to feel their skin turn slightly red. That ’ s when she decided it was time to share herblessings.She started calling everyone she knew. When that failed, she began to ask everyone in the neighborhood like a politician, eventually finding a neighbor delighted to have the tomatoes. “ Feel free to take whatever you want,”Sharontold her. She felt happy that she could help someone and that the food didn’t go to waste.A few days later,Sharonanswered the door. There was the neighbor, holding some bread. The neighbor smiled pleasantly, “I want to thank you for all of the tomatoes, and I have to admit that I took a few other things and hope you wouldn’t mind.”Sharoncouldn’t think of anything else in her garden that had been worth harvesting and said so. “Oh, but you did,” the neighbor said. “You had some of the prettiest zucchini I’ve ever seen.”Sharonwas confused. Zucchini in her garden? They hadn’ t even planted any zucchini. But her neighbor insisted that there really were bright-green zucchini in her garden. The two of them walked together into the backyard. When the neighbor pointed at the long green vegetables,Sharonsmiled, “ Well, actually, those are cucumbers that we never harvested, because they got too big, soft and bitter for eating or canning.”The neighbor looked atSharon, shock written all over her face. Then she smiled, and held out the bread that she had shared all over the neighborhood, “I brought you a loaf of cucumber bread. I hope you like it.”4. Why does the author suggest that people in the neighborhood should lock their cars?A. They might be stolen by thieves.B. They might be moved away by the police.C. Their neighbors might fill them with their harvest.D. Their neighbors might throw rubbish in them.5. What does the underlined word “blessings” in the second paragraph mean?A. Tomatoes.B. God’s protection.C.Helpful things.D. Best wishes.6. What did the neighbor do inSharon’s garden?A. She harvested tomatoes only.B. She harvested zucchini by accident.C. She took some cucumbers mistakenly.D. She stole something withoutSharon’s permission.7. We can infer from the article that the neighbor’s bread would taste________.A. bitter but tastyB. strange and bitterC. hard and sourD. soft and sweetCWhen I was trying to find a place where to spend my December holidays, I met by chance some cheap flights to Iceland. After checking just a few winter pictures of Iceland, I realized that the country, known as the land of fire and ice, during the cold months of the year could offer me experiences I had never had before.For sure you can’t miss the chance to go to Iceland in winter if your traveling wish list includes at least one of the crazy experiences Iceland can offer. Iceland in the North Atlantic Ocean is a paradise (乐园) for all those who want to see the northern lights, experience cold weather conditions and put themselves in geothermal (地热的) baths while the snow is falling on their head.The best way to move around Iceland is with a rental car. Distances are huge and public transport in winter is not really common out of the major towns. As we wanted to be even more convenient we decided to rent a small camper (野营车). Sleeping and cooking in a camper saved us a lot of driving, money and gave us the chance tobe always in the right place at the right time.There were also no locals and in many cases no tourist facilities (设备). For us, as we slept in a camper, it was easier. But for tourists traveling by normal cars it is necessary to check the opening times ofhotels and restaurants as many of them run just from June to September.It is amazing to experience how the weather is changing in Iceland. However, Icelanders prefer to stay inside their houses. They have even no time to complain about the weather in December. All they care about isChristmas. They love to decorate their houses, sing Christmas songs and eat typical Christmas food.8. Why is Iceland famous as the land of fire and ice?A. Because tourists would like to play with fire on the ice.B. Because it is too dry to easily cause fire to happen.C. Because it is hot inside a house and cold outside.D. Because there exist hot springs and freezing ice.9. What did the author think of the rented camper?A. It was not only practical but also economical.B. It was convenient but cost them more money.C. It provided the best chance to see the new country.D. It was much faster than other public transport.10. What does the last paragraph imply?A. The Icelanders prefer to live with their family.B. The joy of Christmas drives the freezing weather away.C. December is the coldest month of the year.D. The Icelanders are always positive and stay outside.11. What does this passage most probably come from?A. A textbookB. A scientific reportC. A travel magazineD. A news reportDThere is an old Chinese proverb that states “One generation plants the trees; another gets the shade,” and this is how it should be with mothers and daughters. The relationship between a mother and a daughter is sometimes confusing. The relationship can be similar to friendship. However, the mother and daughter relationship has unique characteristics that distinguish it from a friendship. These characteristics include responsibilities and unconditional love, whichprecludemothers and daughters from being best friends.Marina, 27 years old, said, “I love spending time with my mom, but I wouldn’t consider her my best friend. Best friends don’t pay for your wedding. Best friends don’t remind you how they carried you in their body and gave you life! Best friends don’t tell you how wise they are because they have been alive at least 20 years longer than you.” This doesn’t mean that the mother and daughter relationship can’t be very close and satisfying. This generation of mothers and adult daughters has a lot in common, which increases the likelihood of sharedcompanionship. Mothers and daughters have always shared the common experience of being homemakers, responsible for maintaining(保持) and passing on family values and traditions. Today contemporary mothers and daughters also share the experience of work and technology, which may bring them even closer together.Best friends may ormay not continue to be best friends, but for better or worse; the mother and daughter relationship is permanent, even if for some unfortunate reason they aren’t speaking. Sometimes this is not an equal relationship. Daughters don’t always feel responsible for their mother’s emotional well-being. But mothers never stop being mothers, which includes frequently wanting to protect their daughters and often feeling responsible for their happiness. The mother and daughter relationship is a relationship that is not replaceable by any other. Mothers always “trump(胜过)” friends.12. What does the underlined word “preclude” in paragraph 1 probably mean?A. differ.B. benefit.C. prevent.D. change.13. What can we learn from what Marina said?A. Best friends will not spend money on her wedding.B. Best friends will not remind her of important issues in life.C. Her mother is wiser on account of her age.D. Her mother is definitely not her best friend.14. Why can a mother and a daughter build a even closer relationship today?A. Because they share advanced technology with each other.B. Because they work together to support the whole family.C. Because they experience the same values and traditions.D. Because they have common experience in life and work.15. What is the text mainly about?A. How to build a good mother and daughter relationship.B. A mother-daughter relationship is irreplaceable.C. Mothers want to be daughters’ friends.D. A daughter is a mother’s best friend.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2019年上海市徐汇区高考物理一模试卷一、单选题（本大题共12小题，共40.0分）1.在国际单位制（SI）中，下列物理量的单位属于导出单位的是（）A. 长度B. 电流C. 电压D. 热力学温度2.收听广播时会听到：“上海东方广播电台FM100.7”。

这里的“100.7指的是电磁波的（）A. 频率B. 周期C. 波长D. 波速3.碳的同位素614C的原子核中有（）A. 8个质子B. 8个电子C. 8个中子D. 8个核子4.某放射性元素经过了3个半衰期，发生衰变的原子核约为总数的（）A. 14B. 34C. 18D. 785.若单摆的摆长不变，摆球的质量减小，摆球离开平衡位置的最大摆角减小，则单摆振动的（）A. 频率不变，振幅不变B. 频率不变，振幅改变C. 频率改变，振幅不变D. 频率改变，振幅改变6.如图所示，水平桌面上有一个小钢球和一根条形磁铁，现给小钢球一个沿OP方向的初速度v，则小钢球的运动轨迹可能是（）A. 甲B. 乙C. 丙D. 丁7.某种气体在不同温度下的气体分子速率分布曲线如图所示，三条曲线所对应的温度分别为T I、T II、T III，则（）A. T I>T II>T IIIB. T I<T II<T IIIC. T I=T II=T IIID. T I<T II，T II>T III8.如图，空间有两个等量正点电荷，b、c为电荷连线上两点，b为中点，ab连线垂直于bc连线，三点电势分别为φa、φb、φc将一检验负电荷q从a点移到b点，再从b点移到c点，电荷q的电势能为E．则（）A. φa>φb>φc，E先减小后增大B. φa<φb<φc，E先减小后增大C. φa>φb>φc，E始终减小D. φa<φb<φc，E始终减小9.如图，足够大的匀强磁场中有相距较远的两个导体圆环a、b，磁场方向与圆环所在平面垂直。

2019-2020学年上海市徐汇区高三年级一模考试数学试卷2019.12一. 填空题（本大题共12题，1-6每题4分，7-12每题5分，共54分） 1. 已知集合{|2}M x x =>，集合{|1}N x x =≤，则M N =U 【答案】(](),12,-∞+∞【解析】考察集合的并集，易得(](),12,M N =-∞+∞2. 向量(3,4)a =r 在向量(1,0)b =r方向上的投影为【答案】3【解析】向量a →在向量b →方向上的投影为3cos 31a ba ba a a bbθ→→→→→→→→→⋅⋅=⨯===⨯3. 二项式11(31)x -的二项展开式中第3项的二项式系数为 【答案】55【解析】二项式11(31)x -的二项展开式中第3项的二项式系数为21155C =4. 复数1i34i++的共轭复数为 【答案】712525i + 【解析】()()()()1341713434342525i i i i i i i +-+==-++-，所以它的共轭复数为712525i+5. 已知()y f x =是定义在R 上的偶函数，且它在[0,)+∞上单调递增，那么使得(2)()f f a -≤成立的实数a 的取值范围是 【答案】(][),22,-∞-+∞【解析】由题，()y f x =是定义在R 上的偶函数，且它在[0,)+∞上单调递增，则()f x 在(],0-∞上单调递减，(2)()f f a -≤，则2a -≤，解得a 的取值范围是(][),22,-∞-+∞6. 已知函数()arcsin(21)f x x =+，则1()6f π-=【答案】14-【解析】考察反函数性质，令()()arcsin 216f x x π=+=，则1212x +=，解得14x =-。

由原函数与反函数自变量与因变量互换，则1164f π-⎛⎫=- ⎪⎝⎭。

7. 已知x ∈R ，条件2:p x x <，条件1:q a x≥（0a >），若p 是q 的充分不必要条件，则实数a 的取值范围是【答案】(]0,1【解析】由题求得，()0,1p =，10,q a ⎛⎤= ⎥⎝⎦，因为p 是q 的充分不必要条件，可知p q ⇒，则实数a 的取值范围是(]0,18. 已知等差数列{}n a 的公差3d =，n S 表示的前n 项和，若数列{}n S 是递增数列，则1a 的取值范围是 【答案】()3,-+∞【解析】 由题意得，数列{}n S 是递增数列，则()12,n n S S n n N *->≥∈，即()1310na a n =+->，即()131a n >-,()2,n n N *≥∈,解得13a >-9. 数字不重复，且个位数字与千位数字之差的绝对值等于2的四位数的个数为 【答案】840【解析】由题意知，个位数字和千位数字在下列个集合中取值：{}9,7、{}8,6、{}7,5、{}6,4、{}5,3、{}4,2、{}3,1、{}2,0。

2019学年第一学期徐汇区学习能力诊断卷 数学学科参考答案及评分标准2019.12一． 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．(](),12,-∞+∞2．3 3．55 4．712525i + 5．(,2][2,)-∞-+∞ 6．14-7．(]0,1 8．()3,-+∞9．840 10 11．311,44⎛⎫- ⎪⎝⎭12．)12,⎡+∞⎣ 二．选择题：（本大题共有4题，满分20分，每题５分）13．B 14．C 15．D 16．D 三． 解答题：（本大题共5题，满分74分） 17．（本题满分14分，第（1）小题6分，第（2）小题8分） 【解】（1）因为1||=OA ，3SA =，所以在SOA Rt ∆中，h SO ==22=，-------------------------2分所以圆锥的体积ππ322312==h r V ． ----------------------------------------4分 侧面展开图扇形的面积12332S ππ=⋅⋅=．-------------------------------------6分（2）解法一：以OP 、OA 、OS 所在射线为x 轴、y 轴、z 轴建立坐标系，则有()(()(1,0,0,,0,1,0,P S A M ，-----------------------------9分于是()(1,0,22,0,PS AM =-=-，------------------------------------------11分设向量PS 与AM 的夹角为θ，则cos9θ==，--------------------13分所以，异面直线AM 与PS 所成角大小为．------------------------------14分 解法二：取线段OP 中点N ，连MN ，则MN SP 且113222MN SP SA ===,AMN ∠（或其补角）为异面直线AM 与PS 的夹角，------------------------------8分1,OA OM AM AN ==∴===-------------------------10分22232cos322AMN⎛⎫+-⎪⎝⎭⎝⎭∠==,----------------------------------------------13分所以，异面直线AM与PS所成角大小为arccos9．------------------------------------14分18．(本题满分14分，第1小题满分6分，第2小题满分8分)【解】（1）由已知，)()(xfxf=-，………………………………………………（2分）即||||axax+=-，…………………………………………………………………（4分）解得0=a．…………………………………………………………………………（6分）（2）⎪⎩⎪⎨⎧<+-≥-+=axaxxaxaxxxf,,)(22，…………………………………………………（8分）当ax≥时，)(xf的最小值为2)(aaf=，当ax<时，)(xf的最小值为4121-=⎪⎭⎫⎝⎛af，…………………………（12分）由于0214122>⎪⎭⎫⎝⎛-=⎪⎭⎫⎝⎛--aaa，所以函数)(xf的最小值为41-a．……（14分）19．(本题满分14分，第1小题满分6分，第2小题满分8分)【解】（1）设xBD=，则由余弦定理022230cos1685xx-+=，…（2分）即039382=+-xx，解得334±=x，…………（4分）8334>+舍去．所以334-=x，这条公路的长约为3.9km．……（6分）（2）在∆ABD中，由正弦定理得ADBABABDAD∠=∠sinsin，…………（9分）所以54sinsin=∠=∠CBDABD，在∆CBD中，)sin(sin BDCCBDDCB∠+∠=∠0.79≈，…………（12分）由正弦定理得44.05sinBDCDDCB=⨯≈∠景点C与景点D之间的距离约为4.0km．（14分）（注：答案为3.9扣1分）20．(本题满分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)【解】（1）设(,),P x y则221,54x y+=A（0,2）,PA===[)2,2y∈-2y∴=-仅在时，max4PA=∴椭圆Γ是“圆椭圆”．-----------4分30a（2）设(,),P x y 则222 1.4x y a +=PA ===------7分[)2,2,y ∈-关于y 的二次函数2224444a y y a --++的对称轴为284y a -=-， 椭圆Γ为“圆椭圆”，故282,4a -≤--即(.a ∈------------------------------10分 （3）a =，椭圆22:1,84x y Γ+=设直线PQ 方程为,y k x =联立椭圆方程得22(21)8,k x +=解得x =不妨设PQ --------------------------------12分则直线PA方程为(2y k x =+，令0,y =得x =则M，同理N于是圆方程为2(0x x y +=，即22880x y kx +--=令220880x x y kx =⎧⎨+--=⎩即0x y =⎧⎪⎨=±⎪⎩圆过定点(0,±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) 【解】（1）1213214321,,,,,,,,,.1121231234-------------------------------4分 （2）显然i im n +，**(,,)i i i N m n N ∈∈的大小是以自然数顺序（从2开始）排列. 于是按题设规则排列，数列{}n a 各项中分子与分母和为2的为第一组，只有一个数；分子与分母和为3的设为第二组，有两个数；分子与分母和为4的设为第三组，有三个数；分子与分母和为1n +的设第n 组，有n 个数.数列{}n a 中小于1的项*(,,)mm n m n N n<∈在数列{}n a 中第(1)m n +-组中的倒数第m 个数，按题设规则排列得数列{}n b 各项依次为：11212312,,,,,,,,233444553412345,,,,,,,5566666将此数列分母相同的各项分为一组，第n 组中各项为123,,1111nn n n n ++++其和123123111112n n n nT n n n n n +++=++++==+++++ 数列{}n T 为等差数列，通项为,2n nT ={}n b 前10项和1012341234S S T T T T +++==+++=12345,2222+++=-------7分设2019b 在第n 组，则有(1)(1)201922n n n n -+<≤可取64,n =646320162⨯=2019b ∴为第64组的第3个数，故201912363123656565S T T T T =+++++++= 123631232222656565+++++++=11665526(163)63226565⨯+⨯+=综上，102019655265,65S S ==------------------------------------------10分（3）{}1,2,3,2018,2019A = 设{}12312,,,,,k k B d d d d d d d =<<*,k N B A ∈⊆121,,k k k k d d d d d d B ----则都不在中，设{}121,,k k k k M d d d d d d -=---则,M A MB φ⊆=所以(1)2019k k +-≤得1010.k ≤-------------------15分k 能否取到1010，我们构造{}1,3,5,72019B =，，则集合A 的子集B 符合题设，所以集合B 中元素个数最大值为1010.-------------------------------------------18分。

2018-2019学年第一学期徐汇区学习能力诊断卷高三英语试卷考生注意：1. 考试时间120分钟，试卷满分140分。

2. 本考试设试卷和答题纸两部分。

所有答題必须涂（选择题）或写（非选择题）在答题纸上，做在试卷上一律不得分。

3. 答題前，务必在答題纸上填写准考证号和姓名，并将核对后的条形码貼在指定位置上，在答題纸反面清楚地填写姓名。

I. Listening ComprehensionSection A Short ConversationsDirections: 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. At a special party. B. At a hair-dressing salon.C. At a night club.D. At a fashion show.2. A. Finland. B. Egypt. C. Mexico. D. Zambia.3. A. Camping. B. Travelling. C. Sporting. D. Shopping.4. A. She is also a fan of Argentina.B. She is also working very hard.C. She loves American football so much.D. She works for the World Cup.5. A. She threw something at a truck.B. She threw herself out of window and broke her leg.C. She moved a truck to save a little boy.D. She rushed to a moving truck to save a kid.6. A. They planned to go skiing in the rain.B. They just want to grab the chance.C. They will probably change their mind.D. They’ll go skiing even in the rain.7. A. Lisa likes the messy situation.B. Lisa made the mess.C. He and Lisa are settling a problem.D. Lisa likes the new place.8. A. The lady should stop being patient.B. He can’t understand the lady’s feeling.C. The lady should not blame others.D. Nobody may be interested in her problem.9. A. Certain gift from Hawaii.B. A grand wedding party.C. Two plane tickets to Hawaii.D. A picture of the moon.10. A. They went to see a movie.B. The dancers impressed them both.C. The woman is also a dancer.D. The man invited the lady to the show.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 would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Aircraft design. B. Mathematics.C. Engineering.D. Science.12. A. 40. B. 14.C. 4.D. 0.13. A. She stuck to studying engineering at college.B. She addressed to students at high schools and colleges.C. She tried to persuade women not to do engineering for its hard work.D. She researched defense systems of satellites and rockets.Questions 14 through 16 are based on the following passage.14. A. The South Atlantic Ocean.B. The coast of South America.C. African continent.D. The coast of Angola.15. A. He studies the similarities between ancient and modern animals.B. He discovers the remains of ancient sea animals on the coast.C. He studies the cause of separation of South America and Africa.D. He helps do the arrangement of the ancient animal remains in a museum.16. A. Because the remains were exposed on the coast.B. Because these animals used to live close to each other in one place.C. Because these animals were driven to one place and killed.D. Because these animals were all eaten by one large, fierce ancient sea animal.Questions 17 through 20 are based on the following conversation.17. A. A saving account in a single name.B. A saving account in joint names.C. A checking account in a single name.D. A checking account in joint names.18. A. 4. B. 3 C. 2 D. 119. A. The lady and her brothers or sisters.B. The lady’s parents.C. The lady and her father.D. The lady and her mother.20. A. Identification paper, photograph, a letter of introduction and some money.B. Identification paper, some clarifications, a letter of introduction and some money.C. A letter of introduction, photographs, a check book and some money.D. Driving license, identification paper, photos, and a letter of introduction.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.The Battle of Chancellorsville, one of the most famous battles of the Civil War, took place in Virginia in the spring of 1863. For months, the two armies had been staying on opposite banks of a narrow river. The Confederate（南方联盟）troops were led by perhaps (21) _______ (honored) military tactician（战略家）in American history, General Robert E. Lee. The Union （北方联盟）soldiers were led by “Fighting” Joe Hooker.In appearance, personality, and lifestyle, these men were nearly perfect opposites. Lee, an older man in poor health with a gray beard, had a solemn, measured character. Hooker was a blond, broad-shouldered young man (22) _______ pride over his appearance was but one aspect of his self-centeredness. Whereas Lee was loyal and principled, Hooker was known for his rollicking enjoyment of both women and whiskey.Despite the fact that the Confederacy (23) _______ (win) the last four major battles and the Union soldiers were starving, (24) _______ (exhaust), and demoralized, Hooker proclaimed, “My plans are perfect. And when I start to carry them out, (25) _______ God have mercy on Bobby Lee, for I shall have none.” Why was Hooker so confident? Hooker had used spies, analysts, and even hot air balloons to compile a vast amount of intelligence about Lee’s army. He had already been aware, for example, (26) _______ Lee had only 61,000 men to Hooker’s own 134,000. Supported by his superior numbers, Hooker secretly moved 70,000 of his men fifteen miles up and across the river, and then ordered them to sneak back down to position themselves (27) _______ Lee’s army. In effect, Hooker had cut off the Confederate soldiers in front and behind. They were trapped. Satisfied with his advantage, Hooker became convinced that Lee’s only option was to retreat to Richmond, thus (28) _______ (assure) a Union victory.Yet Lee, despite his disadvantages of both numbers and position, did not retreat. Instead, he moved his troops into position to attack. Union soldiers who tried to warn H ooker that Lee was on the offensive (29) _______ (dismiss) as cowards. Having become convinced that Lee had no choice but (30) _______ (retreat), Hooker began to ignore reality. When Lee’s army attacked the Union soldiers at 5:00 p.m., th ey were eating supper, completely unprepared for battle. They abandoned their rifles andfled as Lee’s troops came shrieking out of the brush, bayonets drawn. Against all odds, Lee won the Battle of Chancellorsville, and Hooker’s forces withdrew in defeat. Section BDirections: Fill in each blank with a proper word chosen from the box. Each word can be used only once. Note that there is one word more than you need.A. inadequateB. repeatedlyC. processD. achieveE. directedF. reactionsG. raising H. eliminate I. characterizedJ. immediate K. mechanismThe human body can tolerate only a small range of temperature, especially when the person is engaged in vigorous activity. Heat (31) _______ usually occur when large amounts of water and/or salt are lost through over sweating following exhausting exercise. When the body becomes overheated and cannot (32) _______ this overheatedness, heat exhaustion and heat stroke are possible.Heat exhaustion is generally (33) _______ by sweaty skin, tiredness, sickness, dizziness, plentiful sweating, and sometimes fainting, resulting from a(n) (34) _______ intake of water and the loss of fluids. First aid treatment for this condition includes having the victim lie down, (35) _______ the feet 8 to 12 inches, applying cool, wet cloths to the skin, and giving the victim sips of salt water (1 teaspoon per glass, half a glass every 15 minutes) over a 1-hour period.Heat stroke is much more serious; it is a(n) (36) _______ life-threatening situation. The characteristics of heat stroke are a high body temperature (which may reach 106° F or more); a rapid pulse; hot, dry skin; and a blocked sweating (37) _______. Victims of this condition may be unconscious, and first-aid measures should be (38) _______ at quickly cooling the body. The victim should be placed in a tub of cold water or (39) _______ sponged with cool water until his or her temperature is sufficiently lowered. Fans or air conditioners will also help with the cooling (40) _______. Care should be taken, however, not to over-chill the victim once the temperature is below 102° F.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.For centuries, time was measured by the position of the sun with the use of sundials. Noon was recognized when the sun was the highest in the sky, and cities would set their clock by this apparent (41) _______ time, even though some cities would often be on a slightly different time. Daylight Saving Time (DST), sometimes called summer time, was (42) _______ to make better use of daylight. Thus, clocks are set forward one hour in the spring to move an hour of daylight from the morning to the evening and then set back one hour in the fall to return to (43) _______ daylight.Benjamin Franklin first conceived the idea of daylight saving during his term as an American delegate in Paris in 1784 and wrote about it (44) _______ in hi s essay, “An Economical Project.” It is said that Franklin awoke early one morning and was surprised to see the sunlight at such an hour. Always the (45) _______, Franklin believed the practice of moving the time could save on the use of candlelight, as candles were expensive at the time.In England, builder William Willett (1857–1915) became a strong supporter for Daylight Saving Time upon noticing blinds(百叶窗) of many houses were (46) _______ on an early sunny morning. Willet believed everyone, including himself, would appreciate longer hours of light in the evenings. In 1909, Sir Robert Pearce (47) _______ a bill in the House of Commons to make it obligatory（义务）to (48) _______ the clocks. A bill was drafted and introduced into Parliament several times but met with great opposition, mostly from farmers. (49) _______, in 1925, it was decided that summer time should begin on the day following the third Saturday in April and close after the first Saturday in October.The U.S. Congress passed the Standard Time Act of 1918 to establish standard time and (50) _______ and set Daylight Saving Time across the continent. This act also devised （制定）five time (51) _______ throughout the United States: Eastern, Central, Mountain, Pacific, and Alaska. The first time zon e was set on “the mean astronomical time of the seventy-fifth degree of longitude west from Greenwich” (England). In 1919, this act was abandoned.President Roosevelt established year-round Daylight Saving Time (also called War Time) from 1942–1945. However, after this period, each state (52) _______ its own DST, which proved to be (53) _______ to television and radio broadcasting and transportation.In 1966, President Lyndon Johnson created the Department of Transportation and signed the Uniform Time Act. As a result, the Department of Transportation was given the responsibility for the time laws. During the oil embargo（禁运）and energy crisis of the 1970s, President Richard Nixon (54) _______ DST through the Daylight Saving Time Energy Act of 1973 to conserve energy further. This law was (55) _______ in 1986, and Daylight Saving Time was reset to begin on the first Sunday in April (to spring ahead) and end on the last Sunday in October (to fall back).41. A. popular B. solar C. particular D. singular 42. A. employed B. evaluated C. distributed D. contributed43. A. fruitful B. full C. beautiful D. normal44. A. negatively B. alternatively C. extensively D. aggressively45. A. journalist B. physicist C. chemist D. economist46. A. closed B. opened C. fixed D. installed47. A. introduced B. restricted C. donated D. deleted48. A. stop B. adjust C. wind D. mend49. A. Permanently B. Eventually C. Unfortunately D. Theoretically50. A. reserve B. persevere C. preserve D. observe51. A. places B. districts C. zones D. territories52. A. interrupted B. tempted C. imported D. adopted53. A. pleasing B. confusing C. convincing D. comforting54. A. extended B.afforded C.abandoned D. defended55. A. assembled B. combined C. abused D. modifiedSection 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.AThe lives of the Ancient Greeks revolved（运转）around Eris, a concept by which they defined the universe. They believed that the world existed in a condition of opposites. If there was good, then there was evil；if there was love, then there was hatred; joy, then sorrow; war, then peace; and so on. The Greeks believed that good Eris occurredwhen one held a balanced outlook on life and coped with problems as they arose. It was a kind of ease of living that came from trying to bring together the great opposing forces in nature. Bad Eris was evident in the violent conditions that ruled men’s lives. Although these things were found in nature and sometimes could not be controlled, it was believed that bad Eris occurred when one ignored a problem, letting it grow larger until it destroyed not only that person, but his family as well. The Ancient Greeks saw Eris as a goddess: Eris, the Goddess of Discord, better known as Trouble.One myth that expresses this concept of bad Eris deals with the marriage of King Peleus and the river goddess Thetis. Zeus, the supreme ruler, learns that Thetis would bear a child strong enough to destroy its father. Not wanting to father his own ruin, Zeus convinces Thetis to marry a human, a mortal（凡人）whose child could never challenge the gods. He promises her, among other things, the greatest wedding in all of Heaven and Earth and allows the couple to invite whomever they please. This is one of the first mixed marriages of Greek Mythology and the lesson learned from it still applies today. They do invite everyone . . . except Eris, the Goddess of Discord. In other words, instead of facing the problems brought on by a mixed marriage, they turn their backs on them. They refused to deal directly with their problems and the result is tragic. In her fury （狂怒）, Eris arrives, ruins the wedding, causes a jealous argument between the three major goddesses over a golden apple, and sets in place the conditions that lead to the Trojan War. The war would take place 20 years in the future, but it would result in the death of the only child of the bride and groom, Achilles. Eris would destroy the parents’ hopes for their future, leaving the couple with no legal heirs (继承人) to the throne. Hence, when we are told, “If you don’t invite trouble, trouble comes,” it means that if we don’t deal with our problems, our problems will deal with us . . . with a revenge! It is easy to see why the Greeks considered many of their myths learning myths, for this one teaches us the best way to defeat that which can destroy us.56. Bad Eris is defined in the passage as _______.A. the violent conditions of lifeB. the problems man encountersC. the evil goddess who has a golden appleD. the murderer of generations57. Zeus married Thetis off because _______.A. he needed to buy the loyalty of a great king of mankindB. he feared the gods would create bad Eris by competing over herC. he feared the Trojan War would be fought over herD. he feared being a father of a boy who would kill him in the future58. Zeus did not fear a child of King Peleus because _______.A. he knew that the child could not climb Mt. Olympus and manage to kill a godB. he knew that the child would be killed in the Trojan War which would happen in 20 yearsC. he knew that no matter how strong a mortal child was, he couldn’t overthrow an immortal godD. he knew that Thetis would always love him above everyone else59. What does the myth in the passage want to tell us?A. Do not consider a mixed marriage.B. Do not anger the gods.C. Do not ignore the problems that arise in life.D. Do not take myths seriously.BThe National Storytelling Youth Olympics is an event where thousands of kids fro m grades 6 to 12 compete against each other by telling stories. It is sponsored by the Master’s Degree Program in Reading and Storytelling at East Tennessee State University. The sole purpose of this event is to promote and encourage both the art and scien ce of storytelling among middle school and high school students. Although this event is competitive, its underlying intent and goal is to provide students across the nation with a reason to practice numerous noncompetitive skills.Those skills include skillful sportsmanship, responsible behavior, and an attitude of respect for others and the storytelling genre. The eventual goal of the National Storytelling Youth Olympics is to encourage every classroom in America to discover (or rediscover) the beauty of storytelling and story performance.The National Storytelling Youth Olympics takes place usually around the first weekend in March. Students from all over the country arrive by bus, plane, or automobile in Johnson City, Tennessee. They usually arrive on Thursday or Friday. Those that arrive on Thursday take advantage of their early arrival by telling stories at local schools. On Friday, an evening meal is prepared for all contestants, coaches, and parents. Games are played, stories are told, and lifetime friendships begin. Saturday is the day of the bigevent. A luncheon（午餐会）is held in the afternoon so contestants can familiarize themselves with the surroundings and do a sound check.The event is divided into three categories separated by grades. Contestants are judged not only by their storytelling performance, but also by the attitude and behavior they display during the entire weekend. A winner is picked from each of the three categories; however, there is an overall winner who is granted the name of Grand Torch Bearer. This person is selected not only by the judges, but also by the contestants. After the winners have been announced, the contestants retreat back to their hotel where a celebratory ice cream party is held; and believe it or not, they tell more stories! This is what the National Storytelling Youth Olympics is all about: developing a love for the art of storytelling.56. Which of the following sets of words best describes the Grand Torch Bearer?A. competitive, ambitious, talentedB. respectful, responsible, skilledC. athletic, determined, creativeD. imaginative, individualistic, pessimistic57. Why would someone MOST LIKELY choose to attend the National Storytelling Youth Olympics?A. To refine storytelling skills while meeting new friends.B. To compete fiercely with the best storytellers in the country.C. To earn money and fame.D. To develop one’s personal skill in lecture only.58. When is dinner prepared for all contestants, coaches, and parents?A. Wednesday.B. Friday.C. Thursday.D. Saturday.CPhilosophy of Education is a label applied to the study of the purpose, process, nature and ideals of education. It can be considered a branch of both philosophy and education. Education can be defined as the teaching and learning of specific skills, and the。

2019年上海徐汇区高三一模语文试卷一、积累应用（l0分）1.按要求填空。

（5分）（1）锲而不舍，________________。

（__________《劝学》）（2）柳永的词《八声甘州》“________________，________________，________________”三句描写，以“渐”字领格，勾勒出一幅深秋雨后的萧索图景。

2.阅读新闻周刊记者采访金庸先生的片断，按要求选择。

（5分）记者：金庸先生，您辞去浙大人文学院院长和博导的职务，在81岁高龄远赴剑桥攻读历史学硕士，________________？金庸：原先王朔批评我的小说太俗了，我并不在乎，我写通俗小说，俗是免不了的。

现在有人公开批评我学问不好，我就相当地重视了。

学问不好是事实，唯一补救的办法是令自己的学问好一点，不是好过别人，而是今天好过昨天的自己。

（1）依据金庸的回答推断出的提问句，填入空格处恰当的一项是（）。

（2分）A.与有学者批评您忙于院长事务、没精力做学问有关吗？B.与王朔先生批评您眼界太窄、您的武侠小说太俗有关吗？C.与有媒体批评您不甘寂寞、借高龄求学来炒作有关吗？D.是因为有学者批评您学问不够，您想拿学位来证明自己吗？（2）下列各组句子中，能代表金庸先生态度的一组是（）。

（3分）①世事洞明皆学问，人情练达即文章。

②虚怀若谷，方能容纳百川。

③知之为知之，不知为不知，是知也。

④古人学问无遗力，少壮工夫老始成。

A.①②B.②③C.③④D.①④二、阅读（70分）（一）阅读下文，完成第3—7题。

（16分）新媒体传播与城市文脉保护孙玮（1）二维码在2017年夏季的上海吸引了世界的目光。

这个渗透在我们日常生活中的移动网络小端口，竟然成了人类两百万年浩瀚历史的未来象征，“‘大英博物馆百物展：浓缩的世界史’上海展”的第101件展品，是一枚由100组文物构图而成的“二维码”。

上海选中二维码作为本地化且能够代表人类未来的展品，让不少人质疑：二维码这样的新媒体，只不过是个传递信息、再现现实的工具，如何可能成为未来的象征？何以能够承担保护城市文脉的功能？（2）论及新媒体与城市文脉的关系，不少人存在负面认知：在抹平疆界的全球化狂潮中，新媒体是切断文脉的助纣为虐者。

……○………学校:_______……○………绝密★启用前2019年上海市徐汇区高考一模（文科)数学试题注意事项：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．已知向量a 与b 不共线，且0a b =≠vv ，则下列结论中正确的是（ ）A ．向量a b +v v 与a b -v v垂直B ．向量a b -v v 与a v垂直C ．向量a b +v v 与a v 垂直D ．向量a b +v v 与a b -v v共线2．若,a b 为实数，则“01ab <<”是“1b a<”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3．设x 、y 均是实数，i 是虚数单位，复数（x ﹣2y ）+（5﹣2x ﹣y ）i 的实部大于0，虚部不小于0，则复数z=x+yi 在复平面上的点集用阴影表示为图中的（ ）A ．B ．C ．D ．4．设函数()y f x =的定义域为D ，若对于任意1x 、2x D ∈，当122x x a +=时，恒有()()122f x f x b +=，则称点(),a b 为函数()y f x =图象的对称中心．研究函数()sin 3f x x x π=+-的某一个对称中心，并利用对称中心的上述定义，可得到1234030403120162016201620162016f f f f f ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫+++⋅⋅⋅++ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭的值为（ ）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．设抛物线的顶点在原点，准线方程为2x=-，则抛物线的标准方程是__________. 6．方程log2(3x−5)=2的解是_______________7．设3nna-=（*n N∈）则数列{}n a的各项和为________8．函数()()sin24f x x x Rπ⎛⎫=-∈⎪⎝⎭的单调递增区间是________________.9．若函数()f x的图象与对数函数4logy x=的图象关于直线x+y=0对称，则()f x的解析式为()f x=______．10．函数()24f x x x a=--有四个零点，则a的取值范围是________．11．设x、y R+∈且191x y+=，则x y+的最小值为___________.12．若三条直线30ax y++=，20x y++=和210x y-+=相交于一点，则行列式111a的值为________________.13．在ABC∆中，边2BC=，AB=C的取值范围是________________. 14．已知四面体ABCD的外接球球心O在棱CD上，CD=2，则A、B两点在四面体ABCD的外接球上的球面距离是________.15．()()322134x x x+++展开后各项系数的和等于______．16．已知函数()21f x x=-的定义域为D，值域为{}0,1，则这样的集合D最多有______个.17．正四面体的四个面上分别写有数字0，1，2，3把两个这样的四面体抛在桌面上，则露在外面的6个数字之和恰好是9的概率为______．18．设1x，2x是实系数一元二次方程20ax bx c++=的两个根，若1x是虚数，212xx是实数，则24816321111112222221x x x x x x S x x x x x x ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫=++++++= ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭______．三、解答题19．在三棱锥S ABC-中，,,SA AB SA AC AC BC⊥⊥⊥且2,AC BC SB ===（Ⅰ）求证：SC BC ⊥; （Ⅱ）求三棱锥的体积S ABC V -.20．已知函数()2sin 2sin 2cos2f x x x x =-.（1）化简函数()f x 的表达式，并求函数()f x 的最小正周期； （2）若点()00,A x y 是()y f x =图象的对称中心，且00,2x π⎡⎤∈⎢⎥⎣⎦，求点A 的坐标. 21．已知实数x 满足2411033903x x ---⋅+≤且()22log log 22x f x =⋅ （1）求实数x 的取值范围.（2）求()f x 的最大值和最小值，并求出此时x 的值. 22．数列{}n a 满足15a =，且()1231111122,n nn n N a a a a a *-++++=≥∈L . （1）求2a 、3a 、4a ；SABC订…………○…内※※答※※题※※订…………○…（2）求数列{}n a 的通项公式； （3）令112nn na b a =-，求数列{}n b 的最大值与最小值.23．某地拟建造一座大型体育馆，其设计方案侧面的外轮廓如图所示，曲线AB 是以点E 为圆心的圆的一部分，其中()()0,025E t t <≤；曲线BC 是抛物线()2500y ax a =-+>的一部分；CD AD ⊥，且CD 恰好等于圆E 的半径.假定拟建体育馆的高50OB =（单位：米，下同）.（1）若20t =，149a =，求CD 、AD 的长度； （2）若要求体育馆侧面的最大宽度DF 不超过75米，求a 的取值范围； （3）若125a =，求AD 的最大值.参考答案1．A 【解析】 【分析】通过计算向量数量积确定是否具有垂直关系. 【详解】因为a b =r r ，()()22220a b a b a b a b +⋅-=-=-=r r r r r r r r ,所以向量a b +r r 与a b -r r 垂直.当(1,0)a =r ，(0,1)b =r 时0a b =≠r r ，但向量a b -r r与a r 不垂直、向量a b +r r 与a r 不垂直、向量a b +r r 与a b -r r不共线故选：A. 【点睛】本题考查利用向量的数量积运算判定向量的垂直关系，属于基础题. 2．D 【解析】 【详解】若“0＜ab ＜1”，当a ，b 均小于0时，b ＞1a 即“0＜ab ＜1”⇒“b ＜1a”为假命题； 若“b ＜1a 当a ＜0时，ab ＞1，即“b ＜1a”⇒“0＜ab ＜1”为假命题，综上“0＜ab ＜1”是“b ＜1a”的既不充分也不必要条件，故选D 3．A 【解析】试题分析：由复数（x ﹣2y ）+（5﹣2x ﹣y ）i 的实部大于0，虚部不小于0，可得，利用线性规划的知识可得可行域即可．解：∵复数（x ﹣2y ）+（5﹣2x ﹣y ）i 的实部大于0，虚部不小于0， ∴，由线性规划的知识可得：可行域为直线x=2y 的右下方和直线的左下方，因此为A ． 故选A ．考点：复数的代数表示法及其几何意义． 4．C 【解析】 【分析】利用函数对称中心的性质可知，当122x x +=时，恒有()()124f x f x +=-，由此求出结果． 【详解】∵()sin 3f x x x π=+-，∴当1x =时，()11sin 32f π=+-=-，∴根据对称中心的定义，可得当122x x +=时，恒有()()124f x f x +=-， ∴1234030403120162016201620162016f f f f f ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫+++⋅⋅⋅++⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭1403120162015201620162016ff f ⎡⎤⎛⎫⎛⎫⎛⎫=++ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦ ()201542=⨯-- 8062=-．故选：C ． 【点睛】本题考查函数值的求法，是基础题，解题时要认真审题，注意函数性质的合理运用． 5．28y x = 【解析】 【分析】根据抛物线准线方程可求出p ，再根据准线方程设出抛物线的标准方程，代入p 值即可. 【详解】 由题意可知：22p=，4p ∴=且抛物线的标准方程的焦点在x 轴的正半轴上 故可设抛物线的标准方程为：22y px =将p 代入可得28y x =.故答案为：28y x =. 【点睛】本题考查了根据抛物线准线的方程求抛物线标准方程，属于基础题. 6．x ＝2 【解析】 【分析】把对数方程转化为指数方程，解方程即可. 【详解】解：由方程log 2(3x −5)=2可得 3x ﹣5＝4，即3x ＝32，解得x ＝2， 故答案为 x ＝2． 【点睛】本题主要考查对数方程的解法，指数对数的运算性质应用，属于基础题． 7．12【解析】 【分析】根据无穷等比数列的各项和的计算方法，即可求解，得到答案. 【详解】由题意，数列{}n a 的通项公式为13()3nn n a -==，且113a =， 所以数列{}n a 的各项和为111311213a q ==--.故答案为：12.【点睛】本题主要考查了无穷等比数列的各项和的求解，其中解答中熟记无穷等比数列的各项和的计算方法是解答的关键，着重考查了推理与运算能力，属于基础题.8．()3,88k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦【解析】 【分析】解不等式()222242k x k k Z πππππ-+≤-≤+∈，即可求出函数()y f x =的单调递增区间. 【详解】 解不等式()222242k x k k Z πππππ-+≤-≤+∈，得()388k x k k Z ππππ-≤≤+∈， 因此，函数()y f x =的单调递增区间为()3,88k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦. 故答案为：()3,88k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦. 【点睛】本题考查正弦型函数单调区间的求解，熟悉正弦函数的单调性是解题的关键，考查计算能力，属于基础题. 9．4x y -=- 【解析】 【分析】先设()f x 上任意一点(),x y ，求出这个点关于0x y +=的对称点，则根据题意该对称点在函数4log y x =的图象上，满足函数4log y x =的解析式，从而可求出点(),x y 的轨迹方程. 【详解】设函数()f x 的图象上任意一点(),x y ，则点(),x y 关于0x y +=的对称点(),x y ''在对数函数4log y x =的图象．由题意知1022y yx xx x y y -⎧=⎪⎪-⎨++⎪+='''⎩'⎪，解得x y '=-，y x '=-又∵点(),x y ''在对数函数4log y x =的图象 ∴()4log x y -=-∴4xy --=，∴4x y -=-；故答案为：4xy -=-．【点睛】本题考查利用函数的图象与性质，求函数的解析式．解题的关键是求出点关于某条直线的对称点，属于基础题． 10．(0,4) 【解析】 【分析】()24f x x x a =--有四个零点则240x x a --=,即24x x a -=有四个根,故画出24y x x =-的图像,与y a =有四个交点即可.【详解】由题240x x a --=即24x x a -=有四个根,画出24y x x =-的图像有当2x =时24224y =⨯-=,故a 的取值范围是(0,4) 故答案为(0,4) 【点睛】本题主要考查了绝对值函数的画法以及数形结合的思想,属于基础题型. 11．16 【解析】 【分析】把代数式变形为1()x y ⋅+，用191x y=+进行代换，最后利用基本不等式求解即可. 【详解】191x y+=Q， ,x y R +∈，199()9()101016x y x y y x x y x y x y x y x y ⎛⎫++∴+=+⋅+=+=++≥+= ⎪⎝⎭（当且仅当9y xx y=时，取等号，即4x =，12y =时取等号）. 故答案为：16 【点睛】本题考查了基本不等式的应用，考查了数学运算能力. 12．1 【解析】 【分析】先由三条直线30ax y ++=，20x y ++=和210x y -+=相交于一点，求出a ，再由二阶行列式的计算法则可计算出行列式111a 的值.【详解】 联立20210x y x y ++=⎧⎨-+=⎩，解得11x y =-⎧⎨=-⎩，由于三条直线30ax y ++=，20x y ++=和210x y -+=相交于一点， 所以，直线30ax y ++=过点()1,1--，则130a --+=，解得2a =，因此，212111111=⨯-⨯=.故答案为：1. 【点睛】本题考查三线共点问题的求解，同时也考查了二阶行列式的计算，是基础题，解题时要认真审题，注意二阶行列式计算法则的合理运用． 13．0,3π⎛⎤ ⎥⎝⎦【解析】 【分析】利用余弦定理构建方程，利用判别式可得不等式，从而可求角C 的取值范围． 【详解】由题意，设AC b =，由余弦定理得2222cos AB AC BC AC BC C =+-⋅⋅， 即2344cos b b C =+-，即24cos 10b b C -+=，216cos 40C ∴∆=-≥，1cos 2C ∴≥或1cos 2C ≤-， AB BC <Q ，C ∴不可能为钝角，则1cos 2C ≥，又0C >，03C π∴<≤.因此，角C 的取值范围是0,3π⎛⎤⎥⎝⎦.故答案为：0,3π⎛⎤ ⎥⎝⎦.【点睛】本题考查余弦定理的运用，考查解不等式，解题的关键是利用余弦定理构建方程，利用判别式得不等式，属于中等题. 14．23π 【解析】 【分析】根据球心到四个顶点距离相等可推断出O 为CD 的中点，且OA ＝OB ＝OC ＝OD ，进而在△A 0B 中，利用余弦定理求得cos ∠AOB 的值，则∠AOB 可求，进而根据弧长的计算方法求得答案． 【详解】解：球心到四个顶点距离相等，故球心O 在CD 中点，则OA ＝OB ＝OC ＝OD ＝1，再由AB =A 0B 中，利用余弦定理cos ∠AOB 11312112+-==-⨯⨯，则∠AOB 23π=，则弧AB 23π=•123π=． 故答案为：23π．【点睛】本题主要考查了余弦定理的应用、四面体外接球的性质等，考查了学生观察分析和基本的运算能力． 15．28 【解析】 【分析】根据题意，令1x =，代入多项式即可求出展开式中各项系数的和． 【详解】由于()()322134x x x +++展开后含有字母x ，令1x =，则展开式中各项系数的和为： ()()32121131428+⨯+⨯+=； 故答案为：28． 【点睛】本题考查了求多项式展开式的各项系数和的应用问题，解题的关键应该先令1x =，然后即可求出结果，属于基础题． 16．9 【解析】 【分析】根据值域中的几个函数值，结合函数表达式推断出定义域中可能出现的几个x 值，再加以组合即可得到定义域D 的各种情况． 【详解】令()0f x =，可得1x =±；令()1f x =，可得x =因此，定义域D 的可能结果为：{1,-、{-、{1,、{、{1,1,-、{-、{1,-、{1,、{1,1,-，共9种.故答案为：9.【点睛】本题给出二次函数的一个值域，要我们求函数的定义域最多有几个，着重考查了函数的定义与进行简单合情推理等知识，属于基础题． 17．14【解析】 【分析】首先求出基本事件总数4416n =⨯=，再由列举法求出露在外面的6个数字之和恰好是9包含的基本事件个数，再利用古典概型公式，即可求出露在外面的6个数字之和恰好是9的概率． 【详解】正四面体的四个面上分别写有数字0，1，2，3把两个这样的四面体抛在桌面上， 露在外面的6个数字之和包含的基本事件总数4416n =⨯=， 设两个正四面体中压在桌面的数字分别为m ，n ，则露在外面的6个数字之和恰好是9的基本情况有：()0,3，()3,0，()1,2，()2,1，共包含4个基本事件，∴露在外面的6个数字之和恰好是9的概率41164p ==． 故答案为：14． 【点睛】本题考查古典概型公式的应用，是基础题，解题时要认真审题，注意列举法的合理运用． 18．-2 【解析】 【分析】设1i x s t =+（s ，t ∈R ，0t ≠）．则2i x s t =-．则122x x s +=，2212x x s t =+．利用212x x 是实数，可得223s t =．于是122x x s +=，2212x x s t =+．2112210x x x x ⎛⎫++= ⎪⎝⎭，取12x x ω=，则210ωω++=，31ω=．代入化简即可得出． 【详解】设1i x s t =+（s ，t ∈R ，0t ≠）．则2i x s t =-．则122x x s +=，2212x x s t =+．∵()223223122222i 33i i s t x s st s t t x s t s t s t+--==+-++是实数，∴2330s t t -=， ∴223s t =．∴122x x s +=，2212x x s t =+．∴()22221212121242s x x x x x x x x =+=++=，∴122110x x x x ++=， 取12x x ω=， 则210ωω++=， ∴31ω=． 则2481632248163211111122222211x x x x x x S x x x x x x ωωωωωω⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫=++++++=++++++ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭220ωωωω=++++2=-．故答案为：2-． 【点睛】本题考查了实系数一元二次方程的虚根成对定理，考查了复数的概念，考查了复数的性质210ωω++=，属于中档题.19．（Ⅰ）证明过程详见试题解析；（Ⅱ）3. 【解析】试题分析：（Ⅰ）由线线垂直得到线面垂直，再根据直线所在的平面得到线线垂直；（Ⅱ）根据三棱锥的体积公式13S ABC ABC V S h -∆=⋅⋅求之. 试题解析：（Ⅰ）证明：因为090SAB SAC ∠=∠=,所以,SA AB SA AC ⊥⊥. 又因为AB AC A ⋂=,所以SA ⊥平面ABC ,所以SA BC ⊥.又090ACB ∠=,所以AC BC ⊥.所以BC ⊥平面SAC .故SC BC ⊥.（Ⅱ）在ABC ∆中，090,2,ACB AC BC ∠===所以AB =又在SAB ∆中，,SA AB AB SB ⊥==,所以SA =.又因为SA ⊥平面ABC ,所以112323S ABC V -⎛=⨯⨯⨯=⎝. 考点：（Ⅰ）线面垂直的性质定理；（Ⅱ）三棱锥的体积公式.20．（1）()f x 14242x π⎛⎫=-++ ⎪⎝⎭，最小正周期为2π；（2）31,162π⎛⎫ ⎪⎝⎭或71,162π⎛⎫ ⎪⎝⎭. 【解析】 【分析】（1）利用降幂公式、二倍角公式、辅助角公式化简()14242f x x π⎛⎫=-++ ⎪⎝⎭，代入周期公式计算周期；（2）由对称中心的性质可知0sin 404x π⎛⎫+= ⎪⎝⎭，结合00,2x π⎡⎤∈⎢⎥⎣⎦求出0x ，即可得到点A 的坐标． 【详解】（1）()()21cos 4111sin 2sin 2cos 2sin 4sin 4cos 42222x f x x x x x x x -=-=-=-++14242x π⎛⎫=-++ ⎪⎝⎭， 所以，函数()y f x =的最小正周期为242T ππ==；（2）由()14242f x x π⎛⎫=++ ⎪⎝⎭，Q 点()00,A x y 是函数()y f x =图象的对称中心，则0sin 404x π⎛⎫+= ⎪⎝⎭，得()044x k k Z ππ+=∈，()0416k x k Z ππ∴=-∈， 由()04162k k Z πππ≤-≤∈，解得()1944k k Z ≤≤∈，得1k =或2k =， 当1k =时，0316x π=，此时，点A 的坐标为31,162π⎛⎫⎪⎝⎭； 当2k =时，0716x π=，此时，点A 的坐标为71,162π⎛⎫⎪⎝⎭. 综上所述，点A 的坐标为31,162π⎛⎫ ⎪⎝⎭或71,162π⎛⎫⎪⎝⎭. 【点睛】本题考查了三角函数的恒等变换，正弦函数的图象与性质，属于中档题．21．（1）[]2,4 （2）()min 1=8f x - x =; ()max 0f x = 2x =或4x = 【解析】 【分析】(1)由题意求解关于23x -的二次不等式即可确定函数的定义域；(2)由题意，利用换元法，结合二次函数的性质求解函数的最值和函数取得最值时自变量的取值即可. 【详解】 （1）由2411033903x x ---⋅+≤ 可得242310390x x ---⋅+≤ ，即 ()()2223139013924x x x x -----≤∴≤≤∴≤≤故实数x 的取值范围是[]2,4（2）()()()()2222221*********x f x log log log x log x log x log x ⎛⎫=⋅=--=-- ⎪⎝⎭， 令2log x t =，则[]1,2t ∈ ，()()()()21131122228f xg t t t t ⎛⎫∴==--=-- ⎪⎝⎭ ，()g t Q 在31,2⎡⎤⎢⎥⎣⎦上递减，在3,22⎡⎤⎢⎥⎣⎦上递增，()()3128min min f x g t g ⎛⎫∴===- ⎪⎝⎭，此时232log x =解得x = ()()()()120max max f x g t g g ====，此时21log x =或22log x =即2x =或4x =. 【点睛】二次函数、二次方程与二次不等式统称“三个二次”，它们常结合在一起，有关二次函数的问题，数形结合，密切联系图象是探求解题思路的有效方法．一般从：①开口方向；②对称轴位置；③判别式；④端点函数值符号四个方面分析．22．（1）210a =，3203a =，4209a =；（2）25,1210,23n n n a n -=⎧⎪=⎨⎛⎫⋅≥ ⎪⎪⎝⎭⎩；（3）数列{}n b 的最大值为5，最小值为207-. 【解析】 【分析】（1）由题设条件，分别令2n =、3n =、4n =可计算出2a 、3a 、4a 的值； （2）令3n ≥，由123111112n n a a a a a -++++=L 可得出1232111112n n a a a a a --++++=L ，两式作差可得出123n n a a -=，再利用等比数列的通项公式即可得出数列{}n a 的通项公式； （3）先求出数列{}n b 的通项公式，分23112002n -⎛⎫⋅-> ⎪⎝⎭和23112002n -⎛⎫⋅-< ⎪⎝⎭两种情况讨论，利用数列的单调性即可求出数列{}n b 的最大值与最小值. 【详解】（1）Q 数列{}n a 满足15a =，且()1231111122,n nn n N a a a a a *-++++=≥∈L ，当2n =时，则有2125a =，解得210a =；当3n =时，则有31221111351010a a a =+=+=，解得3203a =； 当4n =时，则有4123211111395102020a a a a =++=++=，解得4409a =； （2）当3n ≥时，由123111112n n a a a a a -++++=L 可得出1232111112n n a a a a a --++++=L ， 两式相减得11122n n n a a a --=-，123n n a a -∴=，123n n a a -∴=，且212a a =， 所以，数列{}n a 从第二项起成等比数列，又210a =，所以25,1210,23n n n a n -=⎧⎪=⎨⎛⎫⋅≥ ⎪⎪⎝⎭⎩； （3）225,12103,2112211203n n n n n n a b n a --=⎧⎪⎛⎫⎪⋅⎪ ⎪==⎨⎝⎭≥-⎪⎛⎫⎪-⋅ ⎪⎪⎝⎭⎩Q ， 当2n ≥时，222210103231120112032n n n n b ---⎛⎫⋅ ⎪⎝⎭==⎛⎫⎛⎫-⋅⋅- ⎪ ⎪⎝⎭⎝⎭. 当23n ≤≤时，23112002n -⎛⎫⋅-< ⎪⎝⎭，此时，数列{}n b 单调递减，且3207n b b ≥=-； 当4n ≥时，23112002n -⎛⎫⋅-> ⎪⎝⎭，此时，数列{}n b 单调递减，且44019n b b ≤=. 1440519b b =>=Q ，因此，数列{}n b 的最大值为15b =，最小值为3207b =-. 【点睛】本题考查了等比数列的通项公式、数列的单调性、递推关系，考查了推理能力与计算能力，属于中档题．23．（1）30CD =，AD =（2）1,100⎡⎫+∞⎪⎢⎣⎭；（3）. 【解析】 【分析】（1）由CD OB OE =-可求出CD 的长，在抛物线方程中，令30y =，可求出OD 的长，在圆E 的方程中，令0y =，可求出AO 的长，相加即可得出AD 的长；（2≤恒成立，根据基本不等式解出即可；（3）先求得OD =E 的方程中，令0y =，可得出OA =AD =())025f t t =<≤，将问题转化为求函数()y f t =在(]0,25t ∈上的最大值.法一：令225cos t α=，0,2πα⎡⎫∈⎪⎢⎣⎭，利用三角函数知识可求出()y f t =的最大值；法二：令x =y =，将问题转化为已知()22250,0x y x y +=≥≥，求()52z AD x y ==+的最大值，利用数形结合思想可求出AD 的最大值.【详解】（1）因为圆E 的半径为5030OB OE t -=-=，所以30CD =米，在215049y x =-+中令30y =，得OD =在圆()222:2030E x y +-=中，令0y =得AO =所以AD AO OD =+== （2）由圆E 的半径为50OB OE t -=-，得50.CD t =-在250y ax =-+中，令50y t =-，得OD =50DF OF OD t =+=-由题意知5075t-+≤对(]0,25t ∈≤恒成立.=25t =取得最小值1010≤，解得1100a ≥. 因此，实数a 的取值范围是1,100⎡⎫+∞⎪⎢⎣⎭； （3）当125a =时，OD = 又圆E 的方程为()()22250x y t t +-=-，令0y =，得x =±所以OA =AD = 下求())025f t t =<≤的最大值.方法一：令225cos t α=，0,2πα⎡⎫∈⎪⎢⎣⎭，则()105sin 55cos AD αααϕ==⨯+⨯=+， 其中ϕ是锐角，且1tan 2ϕ=，从而当2παϕ+=时，AD取得最大值方法二：令x =，y =()22250,0x y x y +=≥≥，求()52z AD x y ==+的最大值.当直线25z y x =-+与圆弧()22250,0x y x y +=≥≥相切时，直线25z y x =-+在y 轴上的截距最大，此时z5≤，0z >Q，解得0z <≤因此，z的最大值为答：当5t =米时，AD的最大值为.【点睛】本题考查了直线和圆的位置关系，考查三角函数问题，考查函数不等式恒成立问题，考查了化归与转化思想的应用，是一道难题．答案第17页，总17页。