2018年上海杨浦区高三一模试题(附答案)

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2018学年杨浦区高三年级一模试卷

2018学年杨浦区高三年级一模试卷2018.12一、填空题1.设全集{}1,2,3,4,5U =，若集合{}3,4,5A =，则u A =ð___________.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()1ln1xf x x +=-的定义域为集合A ，集合(),1B a a =+，且B A ⊆,则实数a 的取值范围为___________.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()()112n n n nT 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、选择题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 ,,2a f b f θθ+⎛⎫== ⎪⎝⎭sin sin cos c f θθθ⎛⎫=⎪+⎝⎭，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17.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。

上海市杨浦区2018年高考物理一模试卷含解析

2018年上海市杨浦区高考物理一模试卷一、选择题I1．关于质点，下列说法中正确的是（）A．质点就是几何点B．质点就是质量很小的点C．质点就是体积很小的点D．质点就是用来代替物体的有质量的点2．关于物体运动的速度和加速度的关系，下列说法正确的是（）A．加速度就是增加的速度B．速度越大，加速度也越大C．速度变化越快，加速度一定越大D．加速度的方向保持不变，速度方向也一定保持不变3．下列说法中，正确的是（）A．有受力物体，就必定有施力物体B．力只能产生在相互接触的物体之间C．施力物体施力在先，受力物体受力在后D．力是一个物体就能产生的，而并不需要其他物体的存在4．简谐运动中反映物体振动强弱的物理量是（）A．周期 B．频率 C．振幅 D．位移5．关于重力，下列说法正确的是（）A．球体的重心一定在球心上B．物体上只有重心处才受到重力作用C．物体向上抛出时受到的重力小于它静止时受到的重力D．同一物体在同一地点，无论运动状态如何，其所受重力都一样大6．关于机械波，下列说法中正确的是（）A．机械波的振幅与波源振动的振幅不相等B．在机械波的传播过程中，离波源越远的质点振动的周期越大C．在波的传播过程中，介质中质点的振动频率等于波源的振动频率D．在波的传播过程中，介质中质点的振动速度等于波的传播速度7．关于电场，下列说法中正确的是（）A．电场并不是客观存在的物质B．描述电场的电场线是客观存在的C．电场对放入其中的电荷有力的作用D．电场对放入其中的电荷没有力的作用8．如图，一圆形金属环与两固定的平行长直导线在同一竖直平面内，环的圆心与两导线距离相等，环的直径小于两导线间距．两导线中通有大小相等、方向向下的恒定电流．若（）A．金属环向上运动，则环上的感应电流方向为顺时针方向B．金属环向下运动，则环上的感应电流方向为顺时针方向C．金属环向左侧直导线靠近，则环上的感应电流方向为逆时针D．金属环向右侧直导线靠近，则环上的感应电流方向为逆时针9．图示为锥形齿轮的传动示意图，大齿轮带动小齿轮转动，大、小齿轮的角速度大小分别为ω1、ω2，两齿轮边缘处的线速度大小分别为v1、v2，则（）A．ω1＜ω2，v1=v2B．ω1＞ω2，v1=v2C．ω1=ω2，v1＞v2D．ω1=ω2，v1＜v2 10．锂电池能量密度高、绿色环保．现用充电宝为一手机锂电池（图甲）充电，等效电路如图乙所示，充电宝的输出电压为U，输出电流为I，该锂电池的内阻为r，则（）A．充电宝输出的电功率为UI+I2rB．电能转化为化学能的功率为UIC．锂电池产生的热功率为I2rD．锂电池产生的热功率为11．如图，直线a、b和c、d是处于匀强电场中的两组平行线，M、N、P、Q是它们的交点，四点处的电势分别为φM，φN，φP，φQ，一电子由M点分别到N点和P点的过程中，电场力所做的负功相等，则（）A．直线a位于某一等势面内，φM＞φQB．直线c位于某一等势面内，φM＞φNC．若电子由M点运动到Q点，电场力做正功D．若电子由P点运动到Q点，电场力做负功12．如图，两个轻环a和b套在位于竖直面内的一段固定圆弧上：一细线穿过两轻环，其两端各系一质量为m的小球，在a和b之间的细线上悬挂一小物块．平衡时，a、b间的距离恰好等于圆弧的半径．不计所有摩擦，小物块的质量为（）A．B．m C．m D．2m二、选择题II13．在匀速圆周运动中，保持不变的物理量是（）A．速度 B．加速度C．角速度D．周期14．如图所示的电路中，E为电源电动势，r为电源内阻，R1和R3均为定值电阻，R2为滑动变阻器．当R2的滑动触点在ab的中点时合上开关S，此时三个电表A1、A2和V的示数分别为I1、I2和U．现将R2的滑动触点向a端移动，则（）A．电源的总功率减小 B．R3消耗的功率增大C．I1增大，I2减小，U增大D．I1减小，I2不变，U减小15．如图，在正电荷Q的电场中有M、N、P、F四点，M、N、P为直角三角形的三个顶点，F为MN的中点，∠M=30°，M、N、P、F四点处的电势分别用φM、φN、φP、φF表示，已知φM=φN、φP=φF，点电荷Q在M、N、P三点所在平面内，则（）A．连接PF的线段一定在同一等势面上B．将正试探电荷从P点搬运到N点，电场力做负功C．将正试探电荷从P点搬运到N点，电势能减少D．点电荷Q一定在MP的连线上三、填空题16．如图所示，S1、S2是两个振幅相等的相干波源，实线和虚线分别表示在某一时刻它们所发出的波的波峰和波谷，其中d点处于图示波峰和波谷的正中间位置．在a、b、c、d四点中，振动加强点有，振动减弱点有，此时恰处于平衡位置的有．17．目前在我国许多省市ETC联网正式启动运行，ETC是电子不停车收费系统的简称．汽车分别通过ETC通道和人工收费通道的流程如图所示．假设汽车以v1=15m/s朝收费站正常沿直线行驶，如果过ETC通道，需要在收费站中心线前l0m处正好匀减速至v2=5m/s，匀速通过中心线后，再匀加速至v1正常行驶；如果过人工收费通道，需要恰好在中心线处匀减速至零，经过20s缴费成功后，再启动汽车匀加速至v1正常行驶．设汽车加速和减速过程中的加速度大小均为l m/s2．汽车过ETC通道时，从开始减速到恢复正常行驶过程中的位移大小为；汽车通过ETC通道比通过人工收费通道节约的时间是秒．四、综合题I18．如图所示为“研究电磁感应现象”的实验装置，部分导线已连接．（1）用笔线代替导线将图中未完成的电路连接好；（2）将线圈A插入线圈B中，合上开关S，能使线圈B中感应电流的磁场方向与线圈A 中原磁场方向相反的实验操作是；A．插入铁芯F B．拔出线圈A C．使变阻器阻值R变小D．断开开关S（3）某同学第一次将滑动变阻器的触头从变阻器的左端快速滑到右端，第二次将滑动变阻器的触头从变阻器的左端慢慢滑到右端，发现电流计的指针摆动的幅度大小不同，第一次比第二次的幅度（填写“大”或“小”），原因是线圈中的（填写“磁通量”或“磁通量的变化”或“磁通量变化率”）第一次比第二次的大．19．某研究性学习小组用图1所示装置来测定当地重力加速度，主要操作如下：①安装实验器材，调节试管夹（小铁球）、光电门和纸杯在同一竖直线上；②打开试管夹，由静止释放小铁球，用光电计时器记录小铁球在两个光电门间的运动时间t，并用刻度尺（图上未画出）测量出两个光电门之间的高度h，计算出小铁球通过两光电门间的平均速度v；③保持光电门1的位置不变，改变光电门2的位置，重复②的操作，测出多组（h，t），计算出对应的平均速度”；④画出v﹣t图象．请根据实验，回答如下问题：（1）设小铁球到达光电门l时的速度为v0，当地的重力加速度为g．则小铁球通过两光电门间平均速度v的表达式为．（用v0、g和t表示）2（3）根据v﹣t图象，可以求得当地重力加速度g=m/s2，小球通过光电门1时的速度为m/s．（以上结果均保留两位有效数字）五、综合题II21．如图所示，在倾角为37°的斜面上，固定着宽L=0.25m的平行金属导轨，在导轨上端接入电源和变阻器．电源电动势E=12V，内电阻r=1.0Ω．一质量m=20g的金属棒ab与两导轨垂直并接触良好，导轨与金属棒的电阻不计，整个装置处于磁感应强度B=0.80T、垂直于斜面向上的匀强磁场中．若金属导轨是光滑的，取g=10m/s2，且已知sin37°=0.60，cos37°=0.80，要保持金属棒静止在导轨上．求：（1）金属棒ab所受的安培力；（2）回路中电流的大小；（3）滑动变阻器接入电路的阻值．22．如图甲所示，用大型货车运输规格相同的圆柱形水泥管道，货车可以装载两层管道，底层管道固定在车厢里，上层管道堆放在底层管道上，如图乙所示．已知水泥管道间的动摩擦因数μ=，货车紧急刹车时的加速度大小为8m/s2．每根钢管道的质量m=1500kg，重力加速度取g=10m/s2，求：（1）货车沿平直路面匀速行驶时，乙图中管A、B之间的弹力大小；（2）如果货车在水平路面上匀速行驶的速度为43.2km/h，要使货车在紧急刹车时上管道不撞上驾驶室，最初堆放时上层管道最前端应该离驾驶室的最小距离．2018年上海市杨浦区高考物理一模试卷参考答案与试题解析一、选择题I1．关于质点，下列说法中正确的是（）A．质点就是几何点B．质点就是质量很小的点C．质点就是体积很小的点D．质点就是用来代替物体的有质量的点【考点】质点的认识．【分析】质点是只计质量、不计大小、形状的一个几何点，是实际物体在一定条件的科学抽象，能否看作质点物体本身无关，要看所研究问题的性质，看物体的形状和大小在所研究的问题中是否可以忽略．【解答】解：质点就是用来代替物体的有质量的点，与几何点不同，能否看作质点物体本身无关，要看所研究问题的性质，看物体的形状和大小在所研究的问题中是否可以忽略，不是说质量很小、体积很小、或密度很小就可看做质点，故ABC错误，D正确．故选：D2．关于物体运动的速度和加速度的关系，下列说法正确的是（）A．加速度就是增加的速度B．速度越大，加速度也越大C．速度变化越快，加速度一定越大D．加速度的方向保持不变，速度方向也一定保持不变【考点】加速度．【分析】加速度等于单位时间内的速度变化量，反映速度变化快慢的物理量，加速度的方向与速度变化量的方向相同，与速度方向无关．【解答】解：A、加速度大小等于单位时间内速度的变化量，不是增加的速度，故A错误；B．速度大，加速度不一定大．比如比较大的速度做匀速直线运动，速度虽然很大，但加速度很小，为0，故B错误；C．加速度是反映速度变化快慢的物理量．加速度大，速度变化快，故C正确；D．加速度的方向不变，速度的方向可能改变，比如平抛匀速运动，加速度方向不变，速度方向时刻改变，故D错误．故选：C．3．下列说法中，正确的是（）A．有受力物体，就必定有施力物体B．力只能产生在相互接触的物体之间C．施力物体施力在先，受力物体受力在后D．力是一个物体就能产生的，而并不需要其他物体的存在【考点】力的概念及其矢量性．【分析】两个物体不接触也可能有力产生．力是一个物体对另一个物体的作用，有受力物体，就必定有施力物体．施力物体施力和受力物体受力是同时的．【解答】解：A．因为力是物体对物体的作用，一个力必然涉及两个物体：一个是施力物体，一个是受力物体，故A正确；B．在空中飞行的物体尽管没有和地球接触，仍受重力作用，故B错误；C．由于力的作用是相互的，物体受力的同时也对施力物体产生力的作用，故C错误；D．由力的概念知，力是物体对物体的作用，所以说“一个物体就能产生力”这种说法是错误的，故D错误故选：A4．简谐运动中反映物体振动强弱的物理量是（）A．周期 B．频率 C．振幅 D．位移【考点】简谐运动的振幅、周期和频率．【分析】能够反映物体做机械振动强弱的物理量是振幅，不是频率，回复力和周期【解答】解：A、B频率和周期表示振动的快慢．故AB错误．C、振幅是振动物体离开平衡位置的最大距离，表示振动的强弱，故C正确．D、位移大小是振动物体离开平衡位置的距离，不表示振动的强弱，故D错误．故选：C5．关于重力，下列说法正确的是（）A．球体的重心一定在球心上B．物体上只有重心处才受到重力作用C．物体向上抛出时受到的重力小于它静止时受到的重力D．同一物体在同一地点，无论运动状态如何，其所受重力都一样大【考点】重力．【分析】物体的重心与物体的形状和质量分布两个因素有关．重心可以在物体上，也可能在物体之外．只有质量分布均匀，形状规则的物体重心在几何中心．【解答】解：A、重心不仅与物体的形状有关，还与质量分布有关，所以球体的重心不一定在球心，故A错误；B、物体上各处都受到重力，故B错误；C、物体的重力与物体的运动状态无关，则知物体向上抛出受到的重力等于它静止时受到的重力，故C错误；D．同一物体在同一地点，重力加速度g相同，物体的重力G=mg相同．故D正确．故选：D．6．关于机械波，下列说法中正确的是（）A．机械波的振幅与波源振动的振幅不相等B．在机械波的传播过程中，离波源越远的质点振动的周期越大C．在波的传播过程中，介质中质点的振动频率等于波源的振动频率D．在波的传播过程中，介质中质点的振动速度等于波的传播速度【考点】机械波．【分析】明确波的形成和传播规律，知道波的振幅和频率、周期由振源决定，而波的传播速度由介质决定．【解答】解：A．波在传播过程中，机械波的振幅与波源振动的振幅是相等的，故A错误B．在波传播中各点的振动周期均与波源的周期相同，故B错误C．每个质点都在重复波源的振动，因此质点的振动频率和波源的振动频率相同，故C正确D．振动速度与波的传播速度是不相同的，故D错误故选：C．7．关于电场，下列说法中正确的是（）A．电场并不是客观存在的物质B．描述电场的电场线是客观存在的C．电场对放入其中的电荷有力的作用D．电场对放入其中的电荷没有力的作用【考点】电场．【分析】电场是物质客观存在的一种形式，电场线是为了形象描述电场而假象的线；电场的性质，对对放入其中的电荷有力的作用．【解答】解；A、电场是物质客观存在的一种形式，故A错误B、电场线是为了形象描述电场而假象的线，故B错误C、电场的性质，对对放入其中的电荷有力的作用，故C正确，D错误故选：C．8．如图，一圆形金属环与两固定的平行长直导线在同一竖直平面内，环的圆心与两导线距离相等，环的直径小于两导线间距．两导线中通有大小相等、方向向下的恒定电流．若（）A．金属环向上运动，则环上的感应电流方向为顺时针方向B．金属环向下运动，则环上的感应电流方向为顺时针方向C．金属环向左侧直导线靠近，则环上的感应电流方向为逆时针D．金属环向右侧直导线靠近，则环上的感应电流方向为逆时针【考点】楞次定律．【分析】通过线圈面积的磁通量发生变化时，则会出现感应电动势，当电路闭合时，则产生感应电流．结合楞次定律可判定感应电流方向．【解答】解：AB、直导线之间的磁场时对称的，圆环在中间时，通过圆环的磁通量为零，金属环上下运动的时候，圆环的磁通量不变，不会有感应电流产生，故AB错误；C、金属环向左侧直导线靠近，则穿过圆环的磁场垂直纸面向外并且增强，根据楞次定律可得，环上的感应电流方向为顺时针，故C错误；D、金属环向右侧直导线靠近，则穿过圆环的磁场垂直纸面向里并且增强，根据楞次定律可得，环上的感应电流方向为逆时针，故D正确；故选：D9．图示为锥形齿轮的传动示意图，大齿轮带动小齿轮转动，大、小齿轮的角速度大小分别为ω1、ω2，两齿轮边缘处的线速度大小分别为v1、v2，则（）A．ω1＜ω2，v1=v2B．ω1＞ω2，v1=v2C．ω1=ω2，v1＞v2D．ω1=ω2，v1＜v2【考点】线速度、角速度和周期、转速．【分析】大齿轮带动小齿轮转动，轮子边缘上的点在相同时间内走过的弧长相同，则线速度大小相等，结合半径大小关系比较角速度的大小关系．【解答】解：大齿轮带动小齿轮转动，轮子边缘上的点在相同时间内走过的弧长相同，则线速度大小相等，即v1=v2，根据知，r1＞r2，则ω1＜ω2．故选：A．10．锂电池能量密度高、绿色环保．现用充电宝为一手机锂电池（图甲）充电，等效电路如图乙所示，充电宝的输出电压为U，输出电流为I，该锂电池的内阻为r，则（）A．充电宝输出的电功率为UI+I2rB．电能转化为化学能的功率为UIC．锂电池产生的热功率为I2rD．锂电池产生的热功率为【考点】电功、电功率．【分析】手机锂电池是非纯电阻电路，其充电和放电过程是电能和化学能转化的过程，给电池充电是将电能转化为化学能储存在电池内，根据能量守恒定律列式求解．【解答】解：A、充电宝输出的电功率：P=UI，故A错误；=UI﹣I2r，故B错BCD、锂电池产生的热功率为P r=I2r，故电能转化为化学能的功率为P出误，C正确，D错误；故选：C11．如图，直线a、b和c、d是处于匀强电场中的两组平行线，M、N、P、Q是它们的交点，四点处的电势分别为φM，φN，φP，φQ，一电子由M点分别到N点和P点的过程中，电场力所做的负功相等，则（）A．直线a位于某一等势面内，φM＞φQB．直线c位于某一等势面内，φM＞φNC．若电子由M点运动到Q点，电场力做正功D．若电子由P点运动到Q点，电场力做负功【考点】电势差与电场强度的关系；电势．【分析】电子由M点分别到N点和P点的过程中，电场力所做的负功相等，说明电势能增加相等，据此分析电势高低．【解答】解：AB、据题，电子由M点分别到N点和P点的过程中，电场力做负功相等，则电势能增加相等，电势降低，则N、P两点的电势相等，d位于同一等势面内，根据匀强电场等势面分布情况知，直线a不是同一等势面，直线c位于某一等势面内，且φM＞φN．故A错误，B正确．C、由上分析知，直线c位于某一等势面内，M、Q的电势相等，若电子由M点运动到Q点电场力不做功，故C错误．D、电子由P点运动到Q点与电子由P点运动到M点电场力做功相等，所以电场力做正功，故D错误．故选：B．12．如图，两个轻环a和b套在位于竖直面内的一段固定圆弧上：一细线穿过两轻环，其两端各系一质量为m的小球，在a和b之间的细线上悬挂一小物块．平衡时，a、b间的距离恰好等于圆弧的半径．不计所有摩擦，小物块的质量为（）A．B．m C．m D．2m【考点】共点力平衡的条件及其应用．【分析】同一根绳子上的张力大小相等，根据ab距离等于圆环半径可知绳所成角度，据此由平衡分析即可．【解答】解：设悬挂小物块的点为O'，圆弧的圆心为O，由于ab=R，所以三角形Oab为等边三角形．由于圆弧对轻环的支持力垂直于半径，所以小球和小物块对轻环的合力方向由轻环指向圆心O，因为小物块和小球对轻环的作用力大小相等，所以aO、bO是∠mab、∠mba的角平分线，所以∠O'Oa=∠maO=∠mbO=30°，所以由几何关系可得∠aO'b=120°，而在一条绳子上的张力大小相等，故有T=mg，小物块受到两条绳子的拉力作用大小相等，夹角为120°，故受到的合力等于mg，因为小物块受到绳子的拉力和重力作用，且处于平衡状态，故拉力的合力等于小物块的重力为mg，所以小物块的质量为m故ABD错误，C正确．故选：C．二、选择题II13．在匀速圆周运动中，保持不变的物理量是（）A．速度 B．加速度C．角速度D．周期【考点】匀速圆周运动．【分析】对于物理量的理解要明确是如何定义的决定因素有哪些，是标量还是矢量，如本题中明确描述匀速圆周运动的各个物理量特点是解本题的关键，尤其是注意标量和矢量的区别．【解答】解：在描述匀速圆周运动的物理量中，线速度、向心加速度、向心力这几个物理量都是矢量，虽然其大小不变但是方向在变，因此这些物理量是变化的；周期、频率、转速是标量，是不变化的，故AB错误，CD正确．故选：CD．14．如图所示的电路中，E为电源电动势，r为电源内阻，R1和R3均为定值电阻，R2为滑动变阻器．当R2的滑动触点在ab的中点时合上开关S，此时三个电表A1、A2和V的示数分别为I1、I2和U．现将R2的滑动触点向a端移动，则（）A．电源的总功率减小 B．R3消耗的功率增大C．I1增大，I2减小，U增大D．I1减小，I2不变，U减小【考点】闭合电路的欧姆定律．【分析】理清电路，确定电压表测的是什么电压，电流表测定的是什么电流，抓住电动势和内阻不变，采用局部→整体→局部的方法，利用闭合电路电路的欧姆定律进行分析；根据分析电源总功率变化情况，由分析消耗的功率【解答】解：的滑动触点向a端移动时，增大，整个电路的总电阻增大，总电流减小，内电压减小，外电压增大，即电压表示数U增大，电压减小，、并联电压增大，通过的电流增大，即示数增大，而总电流I减小，则通过的电流减小，即示数减小，故C正确；D错误；电源的总功率，总电流I减小，电源的总功率减小，故A正确；消耗的功率，总电流减小，R3消耗的功率减小，故B错误；故选：AC15．如图，在正电荷Q的电场中有M、N、P、F四点，M、N、P为直角三角形的三个顶点，F为MN的中点，∠M=30°，M、N、P、F四点处的电势分别用φM、φN、φP、φF表示，已知φM=φN、φP=φF，点电荷Q在M、N、P三点所在平面内，则（）A．连接PF的线段一定在同一等势面上B．将正试探电荷从P点搬运到N点，电场力做负功C．将正试探电荷从P点搬运到N点，电势能减少D．点电荷Q一定在MP的连线上【考点】匀强电场中电势差和电场强度的关系；点电荷的场强；电势能．【分析】点电荷的等势面是一系列的同心圆，对于圆，圆弧上任意两点的连线的中垂线一定通过圆心；找出电荷位置后，根据电势能的变化情况判断电场力做功情况．【解答】解：AD、点电荷的等势面是一系列的同心圆，对于圆、圆弧上任意两点的连线的中垂线一定通过圆心，故场源电荷在MN的中垂线和FP的中垂线的交点上，在MP的连线上，如图所示，故D正确；线段PF是P、F所在等势面（圆）的一个弦，连接PF的线段一定不在同一等势面上，A错误；BC、在正的点电荷的电场中，离场源越远，电势越低，将正试探电荷从P点搬运到N点，电场力做正功，电势能降低，故B错误、C正确；故选：CD．三、填空题16．如图所示，S1、S2是两个振幅相等的相干波源，实线和虚线分别表示在某一时刻它们所发出的波的波峰和波谷，其中d点处于图示波峰和波谷的正中间位置．在a、b、c、d四点中，振动加强点有bcd，振动减弱点有a，此时恰处于平衡位置的有ad．【考点】横波的图象；波长、频率和波速的关系．【分析】两列波干涉时，两列波的波峰与波峰、波谷与波谷相遇处，振动始终加强，波峰与波谷相遇处振动始终减弱．振动加强点的振动等于波单独传播时振幅的之和．【解答】解：此时b质点处是两列波波峰与波峰叠加的地方，c质点处是波谷与波谷叠加的地方，振动是最强的．d处在振动加强的区域，振动也是最强的．即b、c、d质点振动都最强．a处是波峰与波谷相遇处振动最弱，a质点处是两列波波峰与波谷叠加的地方，振动始终是最弱的，根据波的叠加原则可知，此时恰处于平衡位置的有ad．故答案为：bcd；a；ad17．目前在我国许多省市ETC联网正式启动运行，ETC是电子不停车收费系统的简称．汽车分别通过ETC通道和人工收费通道的流程如图所示．假设汽车以v1=15m/s朝收费站正常沿直线行驶，如果过ETC通道，需要在收费站中心线前l0m处正好匀减速至v2=5m/s，匀速通过中心线后，再匀加速至v1正常行驶；如果过人工收费通道，需要恰好在中心线处匀减速至零，经过20s缴费成功后，再启动汽车匀加速至v1正常行驶．设汽车加速和减速过程中的加速度大小均为l m/s2．汽车过ETC通道时，从开始减速到恢复正常行驶过程中的位移大小为210m；汽车通过ETC通道比通过人工收费通道节约的时间是27秒．【考点】匀变速直线运动的位移与时间的关系；匀变速直线运动的速度与时间的关系．【分析】根据匀变速直线运动的速度位移公式求出加速和减速的位移，以及匀速运动的位移大小求出总位移；根据匀变速直线运动的速度时间公式求出匀加速和匀减速运动的时间，结合通过ETC通道和人工收费通道的时间求出节约的时间．【解答】解：汽车过ETC通道：减速过程有：加速过程与减速过程位移相等，则有：解得：x=210m汽车过ETC通道的减速过程有：总时间汽车过人工收费通道有：。

2018届上海市杨浦区高三学业质量调研一模历史试题及答案精品推荐

上海市杨浦区2018届高三学业质量调研（一模）历史试题考生注意：1．考试时间120分钟，试卷满分150分。

一、选择题（共75分）以下每小题2分，共60分。

每题只有一个正确选项。

1．伯里克利在演讲中说：“我们的制度之所以被称为民主政治，因为政权掌握在____手中。

解决私人争执的时候，每个人在法律上都是平等的。

”符合空格中的正确一项是A．男性B．公民C．陪审法庭D．公民大会2．春秋战国时期诸子百家的世界观、人生观，为塑造中华民族的性格奠定了基础。

下列排列组合与儒家、道家、法家所崇尚的理想人格相对应的是A．英雄、隐士、君子B．英雄、隐士、君子C．君子、英雄、隐士D．君子、隐士、英雄3．复旦大学校训“博学而笃志，切问而近思”，出自中国某典籍，全句是“博学而笃志，切问而近思，仁在其中矣。

”按语义判断，复旦校训当出自A．《尚书》B．《墨经》C．《老予》D．《论语》5．“开始于沙漠中的一个宗教，在短短的一个世纪内，已发展成为横跨欧亚大陆的强大帝国，到750年，他统治了从比利牛斯山到信德，从摩洛哥到中国边境的广大地区。

”斯塔夫里阿诺斯《全球通史》里所说的这一帝国和宗教是A．罗马基督教B．阿拉伯伊斯兰教C．拜占庭东正教D．奥斯曼伊斯兰教4．西周时“一人跖（踏）来而耕，不过十亩”；战国时“一夫挟五口，治四百亩”。

引起这种变化的最直接原因是A．公田变为私田B．铁器牛耕的推广C．国家走向统一D．封建制度的确立6．普希金赞美俄国一位伟大的人物时说：“他用独断专行的手，勇敢地撒播着文明。

”其中的“他”是A．威廉·奥伦治B．彼得一世C．亚历山大二世D．尼古拉二世7．孟德斯鸠说：“中国是一个专制的国家，它的原则是恐怖的。

在最初的那些朝代，政府的专制精神也许稍微差些：但是今天的情况却正好相反。

”下列最能佐证引文中“稍许差些”的是A．唐朝完善三省六部制B．元朝实行行省制度C．明朝废除丞相制度D．清朝设置军机处8．恩格斯称法国启蒙思想家是“非常革命”的。

杨浦区2018年高三语文第一次联合考试(解析版)

杨浦区2018年高三语文第一次联合考试语文考试时间：____分钟简答题（综合题）（本大题共6小题，每小题____分，共____分。

）阅读下文，完成1-6题。

（17分）从“事件”看艺术的性质高建平 1艺术是什么？俄国大作家列夫·托尔斯泰认为艺术是传达。

他说，在心中唤起曾使自己激动过的情感，并用语言、线条、色彩、声音等媒介将它表现出来，使别人也感受到同样的情感，这就是艺术。

2小说家无心插柳，提出了许多理论家所不能提出的，有感悟性、穿透性的理论。

正像人们使用语言交流思想一样，艺术从本质上讲，具有传达性。

艺术家心有所悟，掌握媒介，生产意义，进行以传达为目的的制作。

于是，一次艺术活动就完成了。

由小见大，由一次看无数次，要寻找艺术的本质特性，可捕捉住这个线索，从这里出发。

3但理论家们又纠结起来，他们重视托尔斯泰的这个说法，又感到不满足。

有人提出，所传达的不只是“情感”，是否也包括“思想”？传达以后，接受者所感受到的情感，与艺术家的情感是否相同？对于前者，人们倾向于认为，可以带着情感来思考，并思考出富有情感的思想。

不带情感的思考，与不带形象的思考一样，在理论上都不能成立。

对于后者，即感受者是否获得与创作者同样的情感，一个世纪以来出现了许多争论。

4在这里，我们还是集中谈论受艺术传达说的启发而产生的一个假设——艺术即事件。

俞伯牙弹琴，钟子期听懂了。

这当然也是传达。

艺术家可能心中有一个倾诉对象，但，艺术家心中可能会有理想的接受者，但毕竟是在众人旁观的情况下做给理想接受者看的一件事情。

5李白写诗赠汪沦，送孟浩然，心中有诗人群体在，而不只是私相授受。

普西金写《致凯恩》，像是写悄悄话，诗人的修养和训练却使他写成了情诗的范例，成为流传最广的诗。

一个人对另一个人说话，有两种情况，一种是私下说，另一种是在公众场合说。

场合不一样，说出来的话也不一样。

由此类比，艺术是公众在场时的传达。

6艺术品总是具有两面性：一面是它的永恒性，艺术具有永恒价值，可以为人们长久地欣赏，并成为对以后的艺术的创作和欣赏起作用的范本。

2018年杨浦区高三化学一模试卷及参考答案

杨浦区2019学年度第一学期高中等级考模拟质量调研高三年级化学学科试卷2019.12考生留意：1．本试卷满分l00分，考试时间60分钟。

2．本考试设试卷和答题纸两局部，试卷包括试题与答题要求；全部答题必需涂（选择题）或写（综合分析题）在答题纸上；做在试卷上一律不得分。

3．答题前，考生务必在答题纸上用钢笔或圆珠笔清晰填写姓名、准考证号，并将核对后的条形码贴在指定位置上。

4．答题纸与试卷在试题编号上是一一对应的，答题时应特殊留意，不能错位。

相对原子质量：H-1 C-12 N-14 O-16 Br-80 Ag-108一、选择题(本题共40分，每小题2分，每小题只有一个正确答案) 1．对291171Ts中数字的叙述正确的是A．291是Ts元素的相对原子质量B．291是Ts元素的质量数C．117是291171Ts 的质子数 D．117是291171Ts的中子数2．漂粉精暴露在潮湿的空气中易失效，其缘由与下列叙述无关的是A．HClO的酸性比H2CO3弱B．Ca(ClO)2能与碳酸反响C．HClO见光易分解 D．Ca(ClO)2是正盐3．石油分馏的原理是利用物质的A．沸点差异 B．密度差异 C．溶解度差异 D．化学性质差异4．离子晶体具有的性质是A．易导热 B．易熔化 C．熔化状态下导电 D．有延展性5．下列有关苯的说法，错误的是A．分子中全部原子共平面B．分子中无碳碳双键、碳碳单键C．不能发生加成反响D．能发生取代反响6．工业上仅以食盐和水为原料，不能得到的产品是A．烧碱B．NaClO溶液 C．盐酸D．氮肥NH4Cl7．两个容积一样的容器中，分别盛有O2和空气，在同温、同压下，两容器中的气体确定具有一样的A．原子数 B．分子数 C．质量D ．密度8．反响物X 生成Z 的能量改变如图所示，该反响的热化学方程式为A ．2X(g) Z(g)＋（a －b ）kJB ．2X(g) Z(g)＋（b －a ）kJC ．2X(g) Z(g)＋（c －a ）kJD ．2X(g) Z(g)＋（b －c ）kJ 9．对NH 3有关性质的描绘，错误的是A ．NH 3极易溶于水B ．NH 3具有复原性C ．NH 3在确定条件下能分解D ．右图中浓盐酸换成浓硫酸时现象一样 10．下列反响，与硫酸型酸雨的形成确定无关的是A ．SO 2+2H 2S 3S+2H 2OB ．SO 2+H 2OH 2SO 3C ．2SO 2+O 22SO 3D ．SO 3+H 2O H 2SO 4 11．在给定条件下，下列物质间的转化不能实现的是A ．Fe FeCl 3B ．FeFeSO 4C ． Fe Fe 3O 4D ．Fe FeCl 2高H 2O 催化白浓浓常FeCl 3点Cl 2常浓硫C 5H 12~C 11H 24 H 2 反响(CH CO 2 H 2 CO 反响② H 2 反响汽12．25℃时，(NH 4)2SO 4溶液显酸性，是因为A ．NH 4+电离出了H +B ．NH 4+与水反响生成H +C ．SO 42―结合水电离的H +D ．SO 42―抑制水电离13．支撑海港码头根底的钢管桩，常用外加电流的阴极爱护法进展防腐，工作原理如图所示，下列有关表述错误的是A ．利用了电解原理B ．电子由电源负极流向钢管桩C ．钢管桩是负极D ．高硅铸铁作阳极14．CO 2催化加氢可制取汽油（主要是C 5~C 11的烃类混合物），转化过程示意图如下（中括号内是汽油中某些成份的球棍模型）：下列说法错误的是 A ．反响①：CO 2+H 2 CO+H 2O B ．反响②中形成了碳碳键和碳氢键C ．图中b 的名称是5–甲基庚烷D ．图中a 和b 互为同系物15．以下是某工厂的消费流程：催化…ba直流钢高硅海- +平衡状v (反响反应速 v (正)v’(正v’(逆平衡状态Ⅱ下列分析正确的是A ．循环运用的物质：NH 3和CO 2B ．母液中只含有：NH 4+和Cl―C ．向母液中通NH 3、加NaCl 时需加热D ．该工厂的产品是纯碱和NH 4Cl16．右图是铝热反响的装置图，对该反响叙述正确的是A ．铝热反响是放热反响，所以常温下可以反响B ．图中a 为Mg 条，b 为KClO 3，c 为Al 2O 3和Fe 的混合物C ．KClO 3是该反响的催化剂D ．利用铝热反响可焊接钢轨，也可冶炼锰等难熔金属 17．确定温度下，恒容密闭容器中发生N 2(g)+3H 2(g)2NH 3(g)反响，反响速率与反响时间的关系如图所示。

2018年上海杨浦区高三一模语文试卷(附答案)

杨浦区2017学年度第一学期高三年级模拟质量调研语文学科试卷（1）（满分150分，时间150分钟）2017年12月一、积累应用（10分）1.根据提示填空（5分）（1），万钟于我何加焉？（孟子《鱼我所欲也》）。

（2），万方多难起登临。

（杜甫《》）。

（3）人类的物质文明无论多么奢华兼顾，都逃不过时间的淘洗，李白在《登金陵凤凰台》中的一联“，”对此发出感慨。

【考点】再现古诗文中的名篇、名句。

【解析】“吴宫花草埋幽径，晋代衣冠成古丘”叹息物是人非【答案】万钟则不辩礼义而受之；花近高楼伤客心，登楼；吴宫花草埋幽径，晋代衣冠成古丘2.按要求选择（1）现代人做事讲求有法可依，有跟有据，以下表述对应的一项是（）（2分）A.导之以德，齐之以礼，有耻且格天行有常，不为尧存，不为桀亡。

B.没有规矩，不成方圆事不目见，耳闻而臆断，其有无可乎？C.礼禁未然之前，法施已然之后道听而途说，德之弃也。

D.国不可无法，有法而不善用法等不宜言，举人不以人废言【考点】在提供的新情境中使用名篇、名句等【解析】题干强调做事讲究规矩，B强调规矩重要性，不可主观臆断【答案】（1）B（2）填入下面一段空白处的词句，最恰当的一项是（）（3分）中国古典诗文的审美通常停留于农耕时代的意象，中国古代文学理论的气、运、镜、味等均是中国人独特情绪的表现。

然而，面对现代社会的复杂经验，以“诗评、文评”为主要内容的中国古代文学理论只能做出相当有限的反应。

A.精美的意象，隽永的意境仍能酣畅淋漓的表达B.古典诗文的朦胧、雅致倒是十分贴切C.精致的古典诗文，已显得力不从心D.再精美的诗文都已经失去言说的能力【考点】根据语境，使用语文知识进行识别、判断、归类等【解析】从“只能做出相当有限的反应”一句可以看出精致的古典诗文，已显得力不从心【答案】（2）C二、阅读（70分）（一）阅读下文，完成3——7题。

（16分）标准化时代的文化乡愁（有删节）（1）用坐标横轴表示诗的完美性，用纵轴表示重要性，确立一个坐标点，计算其所占面积，便可测算诗的伟大指数。

2017-2018上海市杨浦区高三数学一模试卷

2017-2018上海市杨浦区高三数学一模试卷学校:___________姓名：___________班级：___________考号：___________一、填空题1．计算1lim(1)n n→∞-的结果是________ 2．已知集合1,{}2,A m =，{3,4}B =，若{3}A B ⋂=，则实数m =________ 3．已知3cos 5θ=-，则sin()2πθ+=________ 4．若行列式124012x -=，则x = .5．已知一个关于x y 、的二元一次方程组的增广矩阵是112012-⎛⎫⎪⎝⎭，则x y +=_________6．在62()x x -的二项展开式中，常数项的值为________7．若将一颗质地均匀的骰子（一种各面上分别标有1，2，3，4，5，6个点的正方体玩具），先后抛掷两次，则出现向上的点数之和为4的概率是．8．.数列{}n a 的前n 项和为n S ，若点(,)n n S （*n N ∈）在函数的反函数的图像上，则n a =________.9．在ABC ∆中，若sin A 、sin B 、sin C 成等比数列，则角B 的最大值为________ 10．抛物线28y x =-的焦点与双曲线2221x y a -=的左焦点重合，则这条双曲线的两条渐近线的夹角为 .11．已知函数()cos (sin )f x x x x =+x ∈R ，设0a >，若函数()()g x f x α=+为奇函数，则α的值为________12．已知点C 、D 是椭圆2214x y +=上的两个动点，且点(0,2)M ，若MD MC λ=，则实数λ的取值范围为________二、单选题13．在复平面内，复数2i i-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 14．给出下列函数：①2log y x =；②2y x ；③||2x y =；④arcsin y x =.其中图像关于y 轴对称的函数的序号是（）A ．①②B ．②③C ．①③D ．②④ 15．“0t ≥”是“函数()2f x x tx t =+-在(),-∞+∞内存在零点”的（）A ．充分而不必要条件B ．必要而不充分条件C ．充分必要条件D ．既不充分也不必要条件16．设A B C D 、、、是半径为1的球面上的四个不同点，且满足0AB AC ⋅=，0AC AD ⋅=，0AD AB ⋅=，用123S S S 、、分别表示△ABC 、△ACD 、△ABD 的面积，则123S S S 的最大值是（）.A ．12B ．2C ．4D ．8三、解答题17．（2020-2021学年上海市杨浦区高三数学一模）如图所示，用总长为定值l 的篱笆围成长方形的场地，以墙为一边，并用平行于一边的篱笆隔开.（1）设场地面积为y ，垂直于墙的边长为x ，试用解析式将y 表示成x 的函数，并确定这个函数的定义域；（2）怎样围才能使得场地的面积最大？最大面积是多少？18．如图，已知圆锥的侧面积为15π，底面半径OA 和OB 互相垂直，且3OA =，P 是母线BS 的中点.（1）求圆锥的体积；（2）求异面直线SO 与PA 所成角的大小. （结果用反三角函数值表示）19．已知函数()1ln 1x f x x+=-的定义域为集合A ，集合(),1B a a =+，且B A ⊆. （1）求实数a 的取值范围；（2）求证：函数()f x 是奇函数但不是偶函数.20．设直线l 与抛物线2:4y x Ω=相交于不同两点A 、B ，O 为坐标原点.（1）求抛物线Ω的焦点到准线的距离；（2）若直线l 又与圆22:(5)16C x y -+=相切于点M ，且M 为线段AB 的中点，求直线l 的方程；（3）若0OA OB ⋅=，点Q 在线段AB 上，满足OQ AB ⊥，求点Q 的轨迹方程. 21．若数列A ：1a ，2a ，⋅⋅⋅，n a （3n ≥）中*i a N ∈（1i n ≤≤）且对任意的21k n ≤≤-，112k k k a a a +-+>恒成立，则称数列A 为“U -数列”.（1）若数列1，x ，y ，7为“U -数列”，写出所有可能的x 、y ；（2）若“U -数列” A ：1a ，2a ，⋅⋅⋅，n a 中，11a =，2017n a =，求n 的最大值；（3）设0n 为给定的偶数，对所有可能的“U -数列”A ：1a ，2a ，⋅⋅⋅，0n a ，记{}012max ,,,n M a a a =⋅⋅⋅，其中12max{,,,}s x x x ⋅⋅⋅表示1x ，2x ，⋅⋅⋅，s x 这s 个数中最大的数，求M 的最小值.参考答案1．1【解析】11lim(1)1lim 101n n nn →∞→∞-=-=-= 故答案为12．3【解析】∵ 集合{}1,2,A m =，{}3,4B =，且{}3A B ⋂=∴3m =故答案为33．35【解析】 ∵3cos 5θ=- ∴3sin()cos 25πθθ+==- 故答案为354．2 【解析】试题分析：由行列式的定义把方程转化为一般代数式方程即可.ab ad bc cd =-.考点：行列式的定义.5．6【分析】根据关于x y 、的二元一次方程组的增广矩阵，写出方程组，求出方程组的解，即可得到结论.【详解】解：由题意关于x y 、的二元一次方程组的增广矩阵是112012-⎛⎫ ⎪⎝⎭，可得关于x y 、的二元线性方程组22x y y -=⎧⎨=⎩，可得42x y =⎧⎨=⎩，故6x y +=，故答案为：6.【点睛】本题主要考查二元线性方程组的增广矩阵的涵义，计算量小，属于基础题型.6．－160【解析】展开式的通项为6621662()(2)r r r r r r r T C xC x x--+=-=- 令620r -=，得3r = ∴在62x x ⎛⎫- ⎪⎝⎭的二项展开式中，常数项的值为336(2)160C -=- 故答案为160-点睛：求二项展开式有关问题的常见类型及解题策略（1）求展开式中的特定项：可依据条件写出第1r +项，再由特定项的特点求出r 值即可；（2）已知展开式的某项，求特定项的系数：可由某项得出参数项，再由通项写出第1r +项，由特定项得出r 值，最后求出其参数.7．112【分析】分别求出基本事件数，“点数和为4”的种数，再根据概率公式解答即可．【详解】解：所有的基本事件共6636⨯=个，其中，点数和为4的有(1,3)、(2,2)、(3,1)共3个，∴出现向上的点数之和为4的概率是313612=，故答案为：112．【点睛】本小题考查古典概型及其概率计算公式，如果一个事件有n 种可能，而且这些事件的可能性相同，其中事件A 出现m 种结果，那么事件A 的概率P （A ）m n=，属于基础题．8．12n -【解析】解：因为 221log (1)log (1)12212nn n n n n n y x n S S S a -=+∴=+∴+=∴=-∴= 9．3π 【解析】∵在ABC ∆中，sin A 、sin B 、sin C 依次成等比数列，∴2sin sin sin B A C = ，则由正弦定理可得：2b ac = 根据余弦定理得2222221cos 2222a cb ac ac ac ac B ac ac ac +-+--==≥=，当且仅当a c =时取等号∴B 的取值范围为(0,]3π，即角B 的最大值为3π 故答案为3π 10．3π 【解析】试题分析：因为抛物线28y x =-的焦点为(2,0),-所以22212, 3.a a +==所以双曲线2221x y a-=的渐近线方程为y =3π. 考点：双曲线的渐近线考点：11．*()26k k N ππα=-∈【解析】∵()()cos sin f x x x x =∴sin 2cos 2)()sin(2)2223x x f x x π+=+-=+ ∵函数()()g x f x α=+为奇函数 ∴()sin(22)3g x x πα=++为奇函数，则2()3k k Z παπ+=∈∵0a > ∴*()26k k N ππα=-∈ 故答案为*()26k k N ππα=-∈ 12．1[,3]3 【解析】①当直线斜率存在时，设过点()0,2M 的直线方程为2y kx =+，联立方程222{14y kx x y =++=，整理可得22(14)16120k x kx +++=，则22(16)4(14)120k k ∆=-⨯+⨯≥，即234k ≥ 设11(,)C x y ，22(,)D x y ，则1221614k x x k +=-+，1221214x x k ⋅=+ ∵MD MC λ=∴12x x λ= ∴2216(1)14k x k λ+=-+，22212()14x k λ⋅=+，即2222222(1)161464641()114123(14)34k k k k k k λλ++=⨯==⨯+++ ∵234k ≥ ∴2(1)1643λλ+≤<∴133λ<< ②当直线斜率不存在时，则过点()0,2M 的直线方程为0x =，此时(0,1)C ，(0,1)D -，或(0,1)C -，(0,1)D当(0,1)C ，(0,1)D -时，3λ=；当(0,1)C -，(0,1)D 时，13λ= 综上，133λ≤≤ 故答案为1[,3]3 点睛：本题考查解析几何问题和向量的联系，题设中出现MD MC λ=，可以得出12x x λ=，结合韦达定理找到λ与k 之间的关系，再利用0∆≥建立不等关系即可得解，本题要特别注意直线斜率是否存在的问题，避免不分类讨论造成遗漏.13．C【分析】根据复数除法运算法则，求出2i i-的实部和虚部，即可得出结论. 【详解】 22i (2)()12i i i i i ---==---， 2i i-对应点的坐标为(1,2)--，位于第三象限. 故选：C.【点睛】本题考查复数的代数运算以及复数的几何意义，属于基础题.14．B【解析】对于①，2log y x =的定义域为(0,)+∞，定义域关于原点不对称，则函数为非奇非偶函数；对于②，2y x =是偶函数，图象关于y 轴对称，满足条件；对于③，2x y =是偶函数，图象关于y 轴对称，满足条件；对于④，arcsin y x =是奇函数，图象关于y 轴不对称，不满足条件.故选B15．A【解析】函数()2f x x tx t =+-在(),-∞+∞内存在零点，则240t t =+≥，解得0t ≥或4t ≤-. 所以“0t ≥”是“函数()2f x x tx t =+-在(),-∞+∞内存在零点”的充分而不必要条件. 故选A.点睛：解本题的关键是处理二次函数在区间上的零点问题，对于二次函数的研究一般从以几个方面研究：一是，开口；二是，对称轴，主要讨论对称轴与区间的位置关系；三是，判别式，决定于x 轴的交点个数；四是，区间端点值.16．B【解析】设AB a =，AC b =，AD c = ．∵0AB AC ⋅=，0AC AD ⋅=，0AD AB ⋅=，∴AB ，AC ，AD 两两互相垂直，扩展为长方体，它的对角线为球的直径，即222244a b c R ++==．∵1S 、2S 、3S 分别表示ABC ∆、ACD ∆、ABD ∆的面积， ∴()()22212311222S S S ab ac bc a b c ++=++≤++=，当且仅当a b c ==时取等号 ∴123S S S ++的最大值是2，故选B ．点睛：本题考查球的内接多面体及基本不等式求最值问题，能够把几何体扩展为长方体，推知多面体的外接球是同一个球，是解答本题的关键．17．（1）()3y x l x =-，0,3l x ⎛⎫∈ ⎪⎝⎭；（2）6l x =时，2max 12l y =. 【解析】（1）设平行于墙的边长为a ，则篱笆总长3l x a =+，即3a l x =-，∴场地面积()3y x l x =-，0,3l x ⎛⎫∈ ⎪⎝⎭.（2）()222333612l l y x l x x lx x ⎛⎫=-=-+=--+ ⎪⎝⎭，0,3l x ⎛⎫∈ ⎪⎝⎭，∴当且仅当6l x =时，2max 12l y =. 综上，当场地垂直于墙的边长x 为6l 时，最大面积为212l.18．（1）12π.（2）arctan 4. 【解析】试题分析：（1）根据圆锥的侧面积求出5BS =，从而求出4SO =，由此能求出圆锥的体积；（2）取OB 中点H ，连结PH AH 、，由P 是SB 的中点知PH ∥SO ，则APH ∠（或其补角）就是异面直线SO 与PA 所成角，由此能求出异面直线SO 与PA 所成角的大小．试题解析：（1）由题意，15OA SB ππ⋅⋅=得5BS =，故4SO == ，从而体积2211341233V OA SO πππ=⋅⋅=⨯⨯=. （2）如图，取OB 中点H ，连结PH AH 、. 由P 是SB 的中点知PH ∥SO ，则APH ∠（或其补角）就是异面直线SO 与PA 所成角.由SO ⊥平面OAB ⇒ PH ⊥平面OAB ⇒ PH AH ⊥.在OAH ∆中，由OA OB ⊥得AH ==在Rt APH ∆中，90AHP O ∠=，122PH SB ==，AH =则tan AH APH PH ∠==，∴异面直线SO 与PA 所成角的大小 . 19．（1）[1,0]- ;（2）见解析. 【解析】试题分析：（1）由对数的真数大于0，可得集合A ，再由集合的包含关系，可得a 的不等式组，解不等式即可得到所求范围；（2）求得()f x 的定义域，计算()f x -与()f x 比较，即可得到所求结论．试题解析：（1）令101xx+>-，解得11x -<<，所以()1,1A =-，因为B A ⊆，所以111a a ≥-⎧⎨+≤⎩，解得10a -≤≤，即实数a 的取值范围是[]1,0-（2）函数()f x 的定义域()1,1A =-，定义域关于原点对称()()()1ln 1x f x x ---=+- ()1111ln ln ln 111x x x f x x x x -+--⎛⎫===-=- ⎪-++⎝⎭ 而1ln32f ⎛⎫= ⎪⎝⎭，11ln 23f ⎛⎫-= ⎪⎝⎭，所以1122f f ⎛⎫⎛⎫-≠ ⎪ ⎪⎝⎭⎝⎭所以函数()f x 是奇函数但不是偶函数.20．（1）2;（2）1x =，9x =;（3）2240x x y -+= 【解析】试题分析：（1）根据题意，由抛物线的方程分析可得p 的值，即可得答案；（2）根据题意，设直线l 的方程为x my b =+，分0m =与0m ≠两种情况讨论，分析m 的取值，综合可得m 可取的值，将m 的值代入直线的方程即可得答案；（3）设直线:AB x my b =+，设()11,A x y 、()22,B x y ，将直线的方程与抛物线方程联立，结合0OA OB ⋅=，由根与系数的关系分析可得答案．试题解析：（1）∵抛物线Ω的方程为24y x =∴抛物线Ω的焦点到准线的距离为2 （2）设直线:l x my b =+当0m =时，1x =和9x =符合题意;当0m ≠时，()11,A x y 、()22,B x y 的坐标满足方程组24x my by x=+⎧⎨=⎩， ∴2440y my b --=的两根为1y 、2y ，()2160m b ∆=+>，124y y m += ∴2121242x x my b my b m b +=+++=+，∴线段AB 的中点()22,2M m b m + ∵1AB CM k k ⋅=-，1AB k m= ∴2225CM mk m m b ==-+-，得232b m =- ∴()()22161630m b m ∆=+=->，得203m<<∵4r ===∴23m =（舍去）综上所述，直线l 的方程为：1x =，9x = （3）设直线:AB x my b =+，()11,A x y 、()22,B x y 的坐标满足方程组24x my by x =+⎧⎨=⎩，∴2440y my b --=的两根为1y 、2y()2160m b ∆=+>，124y y m +=，124y y b =-∴222121212124044y y OA OB x x y y y y b b ⋅=+=⋅+=-=，得0b =或4b =0b =时，直线AB 过原点，所以()0,0Q ； 4b =时，直线AB 过定点()4,0P设(),Q x y ∵OQ AB ⊥，∴()()22,4,40OQ PQ x y x y x x y ⋅=⋅-=-+=（0x ≠），综上，点Q 的轨迹方程为2240x x y -+=点睛：本题主要考查直线与圆相切，求直线方程，分类讨论，轨迹方程的求法等，属于中档题.注意解决本类问题时，要使用直线和圆相切的性质，设直线时注意分类讨论，严防漏解，求轨迹方程时一般先设出动点坐标，再根据条件建立关于,x y 的关系，化简即可求出轨迹方程.21．（1）12x y =⎧⎨=⎩，13x y =⎧⎨=⎩或24x y =⎧⎨=⎩ ；（2）n 的最大值为65；(3)20288n n -+．【解析】试题分析：（Ⅰ）直接根据“U -数列”的定义，讨论列举法即可求出x ，y ；（Ⅱ）11112k k k k k k k a a a a a a a +-+-+>⇔->-可得()()112201712n n --≤-，解得：6265n -≤≤，故65n ≤，另外，任意的264k ≤≤，1k k b b ->，故数列{}n a 为“U -数列”，此时651012632017a =++++⋅⋅⋅+=，即65n =符合题意；（Ⅲ）利用放缩法()221120012822228m m m a a m m a a n n m m M +++-+-+-+≥≥≥=，即可得结论. 试题解析：：（Ⅰ）12x y =⎧⎨=⎩，13x y =⎧⎨=⎩或24x y =⎧⎨=⎩ （Ⅱ）n 的最大值为65，理由如下一方面，注意到：11112k k k k k k k a a a a a a a +-+-+>⇔->-对任意的11i n ≤≤-，令1i i i b a a +=-，则i b Z ∈且1k k b b ->（21k n ≤≤-），故11k k b b -≥+对任意的21k n ≤≤-恒成立．当11a =，2017n a =时，注意到121110b a a =-≥-=，得()()()1122111i i i i i b b b b b b b b i ---=-+-+⋅⋅⋅+-+≥-（21i n ≤≤-）此时()()112110122122n n a a b b b n n n --=++⋅⋅⋅+≥+++⋅⋅⋅+-=-- 即()()112201712n n --≤-，解得：6265n -≤≤，故65n ≤ 另一方面，取1i b i =-（164i ≤≤），则对任意的264k ≤≤，1k k b b ->，故数列{}n a 为“U -数列”，此时651012632017a =++++⋅⋅⋅+=，即65n =符合题意．综上，n 的最大值为65．（Ⅲ）M 的最小值为200288n n -+，证明如下：当02n m =（2m ≥，*m N ∈）时，一方面：由（★）式，11k k b b +-≥，()()()1121m k k m k m k m k m k k k b b b b b b b b m +++-+-+-+-=-+-+⋅⋅⋅+-≥．此时有：()()()()()()()()121122112111222111m m m m m m m m m m m a a a a b b b b b b b b b b b b m m m m m +++--++--+-+=++⋅⋅⋅+-++⋅⋅⋅+=-+-+⋅⋅⋅+-≥++⋅⋅⋅+=-故()221120012822228m m m a a m m a a n n m m M +++-+-+-+≥≥≥=另一方面，当11b m =-，22b m =-，…，11m b -=-，0m b =，11m b +=，…，211m b m -=-时，()()11111210k k k k k k k k k a a a a a a a b b +-+--+-=---=-=>取1m a =，则11m a +=，123m a a a a >>>⋅⋅⋅>，122m m m a a a ++<<⋅⋅⋅<，且()()11211112m m a a b b b m m -=-++⋅⋅⋅+=-+ ()()2112211112mm m m m a a b b b m m +++-=+++⋅⋅⋅+=-+此时()200122811128mn n M a a m m -+===-+=．综上，M 的最小值为200288n n -+．。

《上海市杨浦区届高三上学期期末质量抽测(一模)(化学)》

杨浦区 2018 学年度高三学科测试化学试卷本试卷分为第I 卷 (第 1—4 页 )和第Ⅱ卷 (第 5— 8 页 )两部分。

全卷共8 页。

满分 l50 分，考试时间l20 分钟。

第I卷 (共66分)考生注意：1．答第I 卷前，考生务必在答题卡上用钢笔或圆珠笔清楚填写学校、姓名、准考据号，并用2B 铅笔正确涂写准考据号。

2．第I 卷 (1-22 小题 )，由机器阅卷，答案一定所有涂写在答题卡上。

考生应将代表正确答案的小方格用2B 铅笔涂黑。

注意试卷题号和答题纸编号一一对应，不可以错位。

答案需要改正时，一定将原选项用橡皮擦去，从头选择。

答案不可以涂写在试卷上，涂写在试卷上一律不给分。

相对原子质量：Fe-56一、选择题 (此题共 10 分，每题 2 分，只有一个正确选项。

)1．卢瑟福的α粒子散射实考证明原子中存在A ．α粒子B ．电子 C．中子 D ．原子核2．我国是世界最大的耗煤国家。

以下对煤综合利用的表达错误的选项是A ．煤的气化是化学变化B ．煤的干馏是物理变化C．煤的液化是化学变化D．煤焦油分馏出苯的同系物是物理变化3．已知： C( 金刚石，固 )C(石墨，固 ) +1.9kJ ，则以下判断正确的选项是A．金刚石转变为石墨可用右图表示B．等质量的石墨比金刚石能量高C．石墨比金刚石稳固D．金刚石转变为石墨没有化学键的断裂与生成能量石墨金刚石反响进度4．以下对化学用语的理解正确的选项是A ．原子构造表示图+624：表示1213CC，也能够表示B ．比率模型：表示二氧化碳分子，也能够表示水分子．．C．电子式．：：表示羟基，也能够表示氢氧根离子O H．．D ．构造简式 (CH 3)2CHOH ：表示 2–丙醇，也能够表示1–丙醇5．以下物质发生变化时，所战胜的粒子间互相作用属于同种种类的是1/11A．液溴和己烷分别受热变为气体B．干冰和氯化铵分别受热变为气体C．硅和铁分别受热融化D．氯化氢和蔗糖分别溶解于水二、选择题 (此题共 36 分，每题 3 分，只有一个正确选项。

2018年上海市杨浦区高考高三数学一模试卷及解析

2018年上海市杨浦区高考数学一模试卷一.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.(4分)计算的结果是.2.(4分)已知集合A＝{1,2,m},B＝{3,4},若A∩B＝{3},则实数m＝.3.(4分)已知,则＝.4.(4分)若行列式,则x＝.5.(4分)已知一个关于x、y的二元一次方程组的增广矩阵是,则x＋y ＝.6.(4分)在的二项展开式中,常数项等于.7.(5分)若将一颗质地均匀的骰子(一种各面上分别标有1,2,3,4,5,6个点的正方体玩具),先后抛掷2次,则出现向上的点数之和为4的概率是.8.(5分)数列{a n}的前n项和为S n,若点(n,S n)(n∈N*)在函数y＝log2(x＋1)的反函数的图象上,则a n＝.9.(5分)在△ABC中,若sinA、sinB、sinC成等比数列,则角B的最大值为.10.(5分)抛物线y2＝﹣8x的焦点与双曲线﹣y2＝1的左焦点重合,则这条双曲线的两条渐近线的夹角为.11.(5分)已知函数,x∈R,设a＞0,若函数g(x)＝f(x ＋α)为奇函数,则α的值为.12.(5分)已知点C、D是椭圆上的两个动点,且点M(0,2),若,则实数λ的取值范围为.二.选择题(本大题共4题,每题5分,共20分)13.(5分)在复平面内,复数对应的点位于()A.第一象限B.第二象限C.第三象限D.第四象限14.(5分)给出下列函数：①y＝log2x；②y＝x2；③y＝2|x|；④y＝arcsinx.其中图象关于y轴对称的函数的序号是()A.①②B.②③C.①③D.②④15.(5分)“t≥0”是“函数f(x)＝x2＋tx﹣t在(﹣∞,＋∞)内存在零点”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16.(5分)设A、B、C、D是半径为1的球面上的四个不同点,且满足•＝0,•＝0,•＝0,用S1、S2、S3分别表示△ABC、△ACD、△ABD的面积,则S1＋S2＋S3的最大值是()A. B.2 C.4 D.8三.解答题(本大题共5题,共14＋14＋14＋16＋18＝76分)17.(14分)如图所示,用总长为定值l的篱笆围成长方形的场地,以墙为一边,并用平行于一边的篱笆隔开.(1)设场地面积为y,垂直于墙的边长为x,试用解析式将y表示成x的函数,并确定这个函数的定义域；(2)怎样围才能使得场地的面积最大？最大面积是多少？18.(14分)如图,已知圆锥的侧面积为15π,底面半径OA和OB互相垂直,且OA＝3,P 是母线BS的中点.(1)求圆锥的体积；(2)求异面直线SO 与PA 所成角的大小.(结果用反三角函数值表示)19.(14分)已知函数的定义域为集合A,集合B ＝(a,a ＋1),且B ⊆A.(1)求实数a 的取值范围；(2)求证：函数f(x)是奇函数但不是偶函数.20.(16分)设直线l 与抛物线Ω：y 2＝4x 相交于不同两点A 、B,O 为坐标原点. (1)求抛物线Ω的焦点到准线的距离；(2)若直线l 又与圆C ：(x ﹣5)2＋y 2＝16相切于点M,且M 为线段AB 的中点,求直线l 的方程； (3)若,点Q 在线段AB 上,满足OQ ⊥AB,求点Q 的轨迹方程.21.(18分)若数列A ：a 1,a 2,…,a n (n ≥3)中(1≤i ≤n)且对任意的2≤k ≤n ﹣1,a k＋1＋a k ﹣1＞2a k 恒成立,则称数列A 为“U ﹣数列”.(1)若数列1,x,y,7为“U ﹣数列”,写出所有可能的x 、y ；(2)若“U ﹣数列”A ：a 1,a 2,…,a n 中,a 1＝1,a n ＝2017,求n 的最大值；(3)设n0为给定的偶数,对所有可能的“U ﹣数列”A ：a 1,a 2,…,,记,其中max {x1,x 2,…,x s }表示x 1,x 2,…,x s 这s 个数中最大的数,求M 的最小值.2018年上海市杨浦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12题,1-6每题4分,7-12每题5分,共54分)1.(4分)计算的结果是1.【试题解答】解：当n→＋∞,→0,∴＝1,故答案为：1.2.(4分)已知集合A＝{1,2,m},B＝{3,4},若A∩B＝{3},则实数m＝3.【试题解答】解：∵集合A＝{1,2,m},B＝{3,4},A∩B＝{3},∴实数m＝3.故答案为：3.3.(4分)已知,则＝﹣.【试题解答】解：∵,∴＝.故答案为：﹣.4.(4分)若行列式,则x＝2.【试题解答】解：∵,∴2×2x﹣1﹣4＝0即x﹣1＝1∴x＝2故答案为：25.(4分)已知一个关于x、y的二元一次方程组的增广矩阵是,则x＋y＝6.【试题解答】解：∵一个关于x、y的二元一次方程组的增广矩阵是,∴由二元线性方程组的增广矩阵可得到二元线性方程组的表达式,解得x＝4,y＝2,∴x＋y＝6.故答案为：6.6.(4分)在的二项展开式中,常数项等于﹣160.【试题解答】解：展开式的通项为T r＝x6﹣r(﹣)r＝(﹣2)r x6﹣2r＋1令6﹣2r＝0可得r＝3常数项为(﹣2)3＝﹣160故答案为：﹣1607.(5分)若将一颗质地均匀的骰子(一种各面上分别标有1,2,3,4,5,6个点的正方体玩具),先后抛掷2次,则出现向上的点数之和为4的概率是.【试题解答】解：基本事件共6×6个,点数和为4的有(1,3)、(2,2)、(3,1)共3个,故P＝＝.故答案为：.8.(5分)数列{a n}的前n项和为S n,若点(n,S n)(n∈N*)在函数y＝log2(x＋1)的反函数的图象上,则a n＝2n﹣1.【试题解答】解：由题意得n＝log2(S n＋1)⇒s n＝2n﹣1.n≥2时,a n＝s n﹣s n﹣1＝2n﹣2n﹣1＝2n﹣1,当n＝1时,a1＝s1＝21﹣1＝1也适合上式,∴数列{a n}的通项公式为a n＝2n﹣1；故答案为：2n﹣19.(5分)在△ABC中,若sinA、sinB、sinC成等比数列,则角B的最大值为.【试题解答】解：∵在△ABC中,sinA、sinB、sinC依次成等比数列,∴sin2B＝sinAsinC,利用正弦定理化简得：b2＝ac,由余弦定理得：cosB＝＝≥＝(当且仅当a＝c时取等号),则B的范围为(0,],即角B的最大值为.故答案为：.10.(5分)抛物线y2＝﹣8x的焦点与双曲线﹣y2＝1的左焦点重合,则这条双曲线的两条渐近线的夹角为.【试题解答】解：∵抛物线y2＝﹣8x的焦点F(﹣2,0)与双曲线﹣y2＝1的左焦点重合,∴a2＋1＝4,解得a＝,∴双曲线的渐近线方程为y＝,∴这条双曲线的两条渐近线的夹角为,故答案为：.11.(5分)已知函数,x∈R,设a＞0,若函数g(x)＝f(x＋α)为奇函数,则α的值为.【试题解答】解：函数,＝,＝s,函数g(x)＝f(x＋α)＝为奇函数,则：(k∈Z),解得：,故答案为：12.(5分)已知点C、D是椭圆上的两个动点,且点M(0,2),若,则实数λ的取值范围为.【试题解答】解：假设CD的斜率存在时,设过点M(0,2)得直线方程为y＝kx＋2,联立方程,整理可得(1＋4k2)x2＋16kx＋12＝0,设C(x1,y1),N(x2,y2),则△＝(16k)2﹣4×(1＋4k2)×12≥0,整理得k2≥,x1＋x2＝﹣,x1x2＝,(*)由,可得,x1＝λx2代入到(*)式整理可得＝＝,由k2≥,可得4≤≤,解可得＜λ＜3且λ≠1,当M和N点重合时,λ＝1,当斜率不存在时,则D(0,1),C(0,﹣1),或D(0,1),C(0,﹣1),则λ＝或λ＝3∴实数λ的取值范围.故答案为：.二.选择题(本大题共4题,每题5分,共20分)13.(5分)在复平面内,复数对应的点位于()A.第一象限B.第二象限C.第三象限D.第四象限【试题解答】解：∵＝,∴复数对应的点的坐标为(﹣1,﹣2),位于第三象限.故选：C.14.(5分)给出下列函数：①y＝log2x；②y＝x2；③y＝2|x|；④y＝arcsinx.其中图象关于y轴对称的函数的序号是()A.①②B.②③C.①③D.②④【试题解答】解：①y＝log2x的定义域为(0,＋∞),定义域关于原点不对称,则函数为非奇非偶函数；②y＝x2；是偶函数,图象关于y轴对称,满足条件.③y＝2|x|是偶函数,图象关于y轴对称,满足条件.④y＝arcsinx是奇函数,图象关于y轴不对称,不满足条件,故选：B.15.(5分)“t≥0”是“函数f(x)＝x2＋tx﹣t在(﹣∞,＋∞)内存在零点”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【试题解答】解：t≥0⇒△＝t2＋4t≥0⇒函数f(x)＝x2＋tx﹣t在(﹣∞,＋∞)内存在零点,函数f(x)＝x2＋tx﹣t在(﹣∞,＋∞)内存在零点⇒△＝t2＋4t≥0⇒t≥0或t≤﹣4.∴“t≥0”是“函数f(x)＝x2＋tx﹣t在(﹣∞,＋∞)内存在零点”的充分非必要条件.故选：A.16.(5分)设A、B、C、D是半径为1的球面上的四个不同点,且满足•＝0,•＝0,•＝0,用S1、S2、S3分别表示△ABC、△ACD、△ABD的面积,则S1＋S2＋S3的最大值是()A. B.2 C.4 D.8【试题解答】解：设AB＝a,AC＝b,AD＝c,因为AB,AC,AD两两互相垂直,扩展为长方体,它的对角线为球的直径,所以a2＋b2＋c2＝4R2＝4所以S△ABC ＋S△ACD＋S△ADB＝(ab＋ac＋bc )≤(a2＋b2＋c2)＝2即最大值为：2故选：B.三.解答题(本大题共5题,共14＋14＋14＋16＋18＝76分)17.(14分)如图所示,用总长为定值l的篱笆围成长方形的场地,以墙为一边,并用平行于一边的篱笆隔开.(1)设场地面积为y,垂直于墙的边长为x,试用解析式将y表示成x的函数,并确定这个函数的定义域；(2)怎样围才能使得场地的面积最大？最大面积是多少？【试题解答】解：(1)设场地面积为y,垂直于墙的边长为x,它的面积y＝x(l﹣3x)；由x＞0,且l﹣3x＞0,可得函数的定义域为(0,l)；(2)y＝x(l﹣3x)＝×3x(1﹣3x)≤×()2＝,当x＝时,这块长方形场地的面积最大,这时的长为l﹣3x＝l,最大面积为.18.(14分)如图,已知圆锥的侧面积为15π,底面半径OA和OB互相垂直,且OA＝3,P 是母线BS的中点.(1)求圆锥的体积；(2)求异面直线SO与PA所成角的大小.(结果用反三角函数值表示)【试题解答】(本题满分(14分),第1小题满分(7分),第2小题满分7分)解：(1)由题意,π•OA•SB＝15π,解得BS＝5,…(2分)故…(4分)从而体积.…(7分)(2)如图,取OB中点H,连结PH、AH.由P是SB的中点知PH∥SO,则∠APH(或其补角)就是异面直线SO与PA所成角.…(10分)∵SO⊥平面OAB,∴PH⊥平面OAB,∴PH⊥AH.在△OAH中,由OA⊥OB,得,…(11分)在Rt△APH中,∠AHP＝90 O,,…(12分)则,∴异面直线SO与PA所成角的大小.…(14分)19.(14分)已知函数的定义域为集合A,集合B＝(a,a＋1),且B⊆A.(1)求实数a的取值范围；(2)求证：函数f(x)是奇函数但不是偶函数.【试题解答】解：(1)令,解得﹣1＜x＜1,所以A＝(﹣1,1),因为B⊆A,所以,解得﹣1≤a≤0,即实数a的取值范围是[﹣1,0]；(2)证明：函数f(x)的定义域A＝(﹣1,1),定义域关于原点对称,f(﹣x)＝ln＝ln()﹣1＝﹣ln＝﹣f(x),而,,所以,所以函数f(x)是奇函数但不是偶函数.20.(16分)设直线l与抛物线Ω：y2＝4x相交于不同两点A、B,O为坐标原点.(1)求抛物线Ω的焦点到准线的距离；(2)若直线l又与圆C：(x﹣5)2＋y2＝16相切于点M,且M为线段AB的中点,求直线l的方程；(3)若,点Q在线段AB上,满足OQ⊥AB,求点Q的轨迹方程.【试题解答】解：(1)根据题意,抛物线Ω的方程为y2＝4x,则p＝2,故抛物线Ω的焦点到准线的距离为2；(2)设直线l：x＝my＋b当m ＝0时,x ＝1和x ＝9符合题意；当m ≠0时,A(x 1,y 1)、B(x 2,y 2)的坐标满足方程组,所以y 2﹣4my ﹣4b ＝0的两根为y 1、y 2. △＝16(m 2＋b)＞0,y 1＋y 2＝4m, 所以,所以线段AB 的中点M(2m 2＋b,2m)因为k AB •k CM ＝﹣1,,所以,得b ＝3﹣2m 2所以△＝16(m 2＋b)＝16(3﹣m 2)＞0,得0＜m 2＜3 因为,所以m 2＝3(舍去)综上所述,直线l 的方程为：x ＝1,x ＝9(3)设直线AB ：x ＝my ＋b,A(x 1,y 1)、B(x 2,y 2)的坐标满足方程组,所以y 2﹣4my ﹣4b ＝0的两根为y 1、y 2 △＝16(m 2＋b)＞0,y 1＋y 2＝4m,y 1y 2＝﹣4b 所以,得b ＝0或b ＝4b ＝0时,直线AB 过原点,所以Q(0,0)； b ＝4时,直线AB 过定点P(4,0) 设Q(x,y),因为OQ ⊥AB, 所以(x ≠0),综上,点Q 的轨迹方程为x 2﹣4x ＋y 2＝021.(18分)若数列A ：a 1,a 2,…,a n (n ≥3)中(1≤i ≤n)且对任意的2≤k ≤n ﹣1,a k＋1＋a k ﹣1＞2a k 恒成立,则称数列A 为“U ﹣数列”.(1)若数列1,x,y,7为“U ﹣数列”,写出所有可能的x 、y ；(2)若“U﹣数列”A：a1,a2,…,a n中,a1＝1,a n＝2017,求n的最大值；(3)设n为给定的偶数,对所有可能的“U﹣数列”A：a1,a2,…,,记,其中max{x1,x2,…,x s}表示x1,x2,…,x s这s个数中最大的数,求M的最小值.【试题解答】解：(1)x＝1时,,所以y＝2或3；x＝2时,,所以y＝4；x≥3时,,无整数解；所以所有可能的x,y为,或.(2)n的最大值为65,理由如下：一方面,注意到：a k＋1＋a k﹣1＞2a k⇔a k＋1﹣a k＞a k﹣a k﹣1.对任意的1≤i≤n﹣1,令b i＝a i＋1﹣a i,则b i∈Z且b k＞b k﹣1(2≤k≤n﹣1),故b k≥b k﹣1＋1对任意的2≤k≤n﹣1恒成立.(*)当a1＝1,a n＝2017时,注意到b1＝a2﹣a1≥1﹣1＝0,得(2≤i≤n﹣1)即b i≥i﹣1,此时a n﹣a1＝(a n﹣a n﹣1)＋(a n﹣1﹣a n﹣2)＋…＋(a2﹣a1)＝b n﹣1＋b n﹣2＋…＋b1≥0＋1＋2＋…＋(n﹣2)＝,(**)即,解得：﹣62≤n≤65,故n≤65.另一方面,为使(**)取到等号,所以取b i＝i﹣1(1≤i≤64),则对任意的2≤k≤64,b k＞b k﹣1,故数列{a n}为“U﹣数列”,此时由(**)式得,所以a65＝2017,即n＝65符合题意. 综上,n的最大值为65.(3)M的最小值为,证明如下：当n0＝2m(m≥2,m∈N*)时,一方面：由(*)式,b k＋1﹣b k≥1,b m＋k﹣b k＝(b m＋k﹣b m＋k﹣1)＋(b m＋k﹣1﹣b m＋k﹣2)＋…＋(b k＋1﹣b k)≥m.此时有：(a1＋a2m)﹣(a m＋a m＋1)＝(a2m﹣a m＋1)﹣(a m﹣a1)＝(b m＋1＋b m＋2＋…＋b2m﹣1)﹣(b1＋b2＋…＋b m﹣1)＝(b m＋1﹣b1)＋(b m＋2﹣b2)＋…＋(b2m＋1﹣b m﹣1)≥m＋m＋…＋m＝m(m﹣1).即(a1＋a2m)≥(a m＋a m＋1)＋m(m﹣1)故因为,所以,另一方面,当b1＝1﹣m,b2＝2﹣m,…,b m﹣1＝﹣1,b m＝0,b m＋1＝1,b2m﹣1＝m﹣1时,a k＋1＋a k﹣1﹣2a k＝(a k＋1﹣a k)﹣(a k﹣a k﹣1)＝b k﹣b k﹣1＝1＞0取a m＝1,则a m＋1＝1,a1＞a2＞a3＞…＞a m,a m＋1＜a m＋2＜…＜a2m,且此时.综上,M的最小值为.。

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及答案

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及答案 2018.12.考生注意： 1．答卷前，考生务必在答题纸写上姓名、考号,并将核对后的条形码贴在指定位置上.2．本试卷共有21道题，满分150分，考试时间120分钟．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填写结果.1．设全集{}=1,2,3,4,5U ，若集合{}3,4,5A =，则U A =ð ▲ ．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nb a )(+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=l )(5cm ，高)(4cm h =，则这个圆锥的体积等于▲________()3cm . 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1()ln1xf x x+=-的定义域为集合A ，集合(,1)B a a =+. 且B A ⊆，则实数a 的取值范围为▲________．9. 在行列式中，第3行第2列的元素的代数余子式记作，则的零点是▲________10. 已知复数1cos 2()i z x f x =+，2cos )i z x x =++ (,R x λ∈，i 为虚数单位）．在复平面上，设复数12,z z 对应的点分别为12,Z Z ，若︒=∠9021OZ Z ，其中O 是坐标原点，则函数()f x 的最小正周期 ▲________． 11. 当a x <<0时，不等式2)(1122≥-+x a x 恒成立，则实数a 的最大值为 ▲________． 274434651xx--()f x 1()y f x =+12. 设d 为等差数列}{n a 的公差，数列}{n b 的前n 项和n T ，满足)N ()1(21*∈-=+n b T n n n n ，且25b a d ==. 若实数)3,N }(|{*32≥∈<<=∈+-k k a x a x P m k k k ，则称m 具有性质k P .若n H 是数列}{n T 的前n 项和，对任意的*N ∈n ，12-n H 都具有性质k P ，则所有满足条件的k 的值为▲________.二、选择题（本题共有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. 下列函数中既是奇函数，又在区间[-1,1]上单调递减的是 ………( ). x x f arcsin )(=. lg y x =.()f x x =-.()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x x f θsin log )(=，，设sin cos ,2a f θθ+⎛⎫=⎪⎝⎭b f =，sin 2sin cos c f θθθ⎛⎫=⎪+⎝⎭，则c b a ,,的大小关系是 ………( )()A .b c a ≤≤.()B .a c b ≤≤. ()C .a b c ≤≤.()D .c b a ≤≤.16. 已知函数nx x m x f x ++⋅=22)(，记集合},0)(|{R x x f x A ∈==，集合},0)]([|{R x x f f x B ∈==，若B A =，且都不是空集，则n m +的取值范围是………( )()A . [0,4) ()B . [1,4)- ()C . [3,5]- ()D . [0,7)()A ()B ()C ()D )2,0(πθ∈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）如图，PA ⊥平面ABCD ，四边形ABCD 为矩形，1PA AB ==，2AD =，点F 是PB 的中点，点E 在边BC 上移动.（1）求三棱锥E PAD -的体积；（2）证明：无论点E 在边BC 的何处，都有AF PE ⊥.18. （本题满分14分，第1小题满分7分，第2小题满分7分）在ABC ∆中，角,,A B C 所对的边分别为,,a b c ，且5cos 13B =．（1）若4sin 5A =，求cos C ；（2）若4b =，求证：5-≥⋅BC AB ．19. （本题满分14分，第1小题满分6分，第2小题满分8分）上海某工厂以x 千克/小时的速度匀速生产一种产品，每一小时可获得的利润是)315(xx -+元，其中101≤≤x .（1）要使生产该产品2小时获得的利润不低于30元，求x 的取值范围；（2）要使生产900千克该产品获得的利润最大，问：该厂应选取何种生产速度？并求最大利润.20. （本题满分16分，第1小题满分4分，第2小题满分5分，第3小题满分7分）如图，已知点P 是y 轴左侧（不含y 轴）一点，抛物线x y C 4:2=上存在不同的两点B A ,，满足PB PA ,的中点均在抛物线C 上.（1）求抛物线C 的焦点到准线的距离；（2）设AB 中点为M ，且),(),,(M M P P y x M y x P ，证明：M P y y =；（3）若P 是曲线221(0)4y x x +=<上的动点，求PAB ∆面积的最小值.21. （本题满分18分，第1小题满分4分，第2小题满分5分，第3小题满分9分）记无穷数列{}n a 的前n 项中最大值为n M ，最小值为n m ，令2n nn M m b +=，其中*N ∈n . (1) 若2cos2n n n a π=+，请写出3b 的值； (2) 求证：“数列{}n a 是等差数列”是“数列{}n b 是等差数列”的充要条件；(3) 若对任意n ，有||2018n a <, 且||1n b =，请问：是否存在*K ∈N ，使得对于任意不小于K 的正整数n ，有1n n b b += 成立？请说明理由．青浦区2018学年第一学期高三年级期终学业质量调研测试数学参考答案及评分标准 2018.12说明1．本解答列出试题一种或几种解法，如果考生的解法与所列解法不同，可参照解答中评分标准的精神进行评分．2．评阅试卷，应坚持每题评阅到底，不要因为考生的解答中出现错误而中断对该题的评阅．当考生的解答在某一步出现错误，影响了后续部分，但该步以后的解答未改变这一题的内容和难度时，可视影响程度决定后面部分的给分，但是原则上不应超出后面部分应给分数之半，如果有较严重的概念性错误，就不给分．3．第17题至第21题中右端所注的分数，表示考生正确做到这一步应得的该题分数． 4．给分或扣分均以1分为单位．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. 1．{}1-； 2．“若a b <，则22am bm <”； 3．()2,3-；4．43； 5．12π；67．(0,4)(4,8)； 8．32；9. 80； 10. 14；11．10,2⎛⎤ ⎥⎝⎦；12.1,3⎤⎦.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. A ；14. D ； 15．C ；16. C .三．解答题（本大题满分74分）本大题共有5题，解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.17.(本题满分14分）本题共2小题，第（1）小题6分，第（2）小题8分. 解：（1）在正四棱柱1111ABCD A B C D -中， ∵1AA ⊥平面ABCD ，AD ⊂≠平面ABCD ， ∴1AA AD ⊥，故14AA =， ∴正四棱柱的侧面积为(43)448⨯⨯=，体积为2(3)436⨯=．（2）建立如图的空间直角坐标系O xyz -，由题意可得(0,0,0)D ,(3,3,0)B ,1(3,0,4)A ,(0,0,0)D ,3(,0,2)2E ,1(0,0,4)AA =，3(,3,2)2BE =--，设1AA 与BE 所成角为α，直线BE 与平面ABCD 所成角为θ，则11cos ||||AA BEAA BE α⋅===⋅ 又1AA是平面ABCD 的一个法向量，故sin cos θα==，θ=．所以直线BE 与平面ABCD所成的角为arcsin61．【另法提示：设AD 中点为G ，证EBG ∠即为BE 与平面ABCD 所成的角，然后解直角三角形EBG ，求出EBG ∠】arctan 1518．（本题满分14分）第（1）小题满分8分，第（2）小题满分6分.解：（1）,1,01BP t CP t t ==-≤≤45DAQ θ∠=︒-,1tan(45)1tDQ tθ-=︒-=+, 12111t tCQ t t-=-=++所以211t PQ t +===+ 故221111211t t l CP CQ PQ t t t t t+=++=-++=-++=++ 所以△CPQ 的周长l 是定值2（2）111221ABP ADQ ABCD t t S S S S t ∆∆-=--=--⨯+正方形122(1)221t t=-++≤+当且仅当1t =时，等号成立所以摄像头能捕捉到正方形ABCD 内部区域的面积S至多为22hm19.(本题满分14分）本题共2小题，第（1）小题6分，第（2）小题8分. 解：（1）因为函数()3g x x =是函数()3mf x x x=+在区间[)+∞4，上的弱渐近函数，所以()()1mf xg x x-=≤ ，即m x ≤在区间[)+∞4，上恒成立，即444m m ≤⇒-≤≤（2）()()2f x g x x x -==[)2,+x ∈∞，()()22(f x g x x x ∴-==-A DCBθP Q45令2()()()2(x xh x f x g x x=-===任取122x x≤<，则2212311x x≤-<-≤<120xx<<12()()h x h x⇒>⇒<即函数()()()2(h x f x g x x=-=在区间[)2,+∞上单调递减，所以(()()0,4f x gx-∈-，又([]0,41,1-⊆-，即满足()2g x x=使得对于任意的[)2,x∈+∞有()()1f xg x-≤恒成立，所以函数()2g x x=是函数()f x=在区间[)2,+∞上的弱渐近函数．20.(本题满分16分）本题共3小题，第（1）小题4分，第（2）小题6分，第（3）小题6分.解：（1）242a a=⇒=，又双曲线的渐近线方程为y=，所以bba==双曲线的标准方程是221412x y-=．（2）法一：由题不妨设11()A x，22(,)B x，则1212(,)22x xP+，由P在双曲线上，代入双曲线方程得124x x⋅=；法二：当直线AB的斜率不存在时，显然2x=±，此时124xx⋅=；当直线AB的斜率存在时，设直线AB的方程为(0,y kxt k k=+≠≠则由y kx tAy=+⎛⎧⎪⇒⎨=⎪⎩同理y kx tBy=+⎛⎧⎪⇒⎨=⎪⎩此时223,33kt t P k k ⎛⎫ ⎪--⎝⎭代入双曲线方程得224(3)t k =-，所以212243t x x k ⋅==-（3）①对称中心：原点；对称轴方程：,y y x ==②顶点坐标：3,22⎛⎫⎪ ⎪⎝⎭，322⎛⎫-- ⎪ ⎪⎝⎭；焦点坐标：(，(1,-实轴长：2a =、虚轴长：22b =、焦距：24c =③范围：()0,,2,x y ⎡≠∈-∞+∞⎣④渐近线：0,3x y x ==21.(本题满分18分）本题共3小题，第（1）小题4分，第（2）小题6分，第（3）小题8分.解：（1）因为数列{}n b 是“Γ数列”，且11b =，3k =、4d =、0c =，所以当1n ≥，n *∈N 时，310n b +=，又*2016672N 3=∈，即20170b =， 20182017044b b d =+=+=，20192018448b b d =+=+= （2）因为数列{}n b 是“Γ数列”，且12b =，4k =、2d =、1c =()()()414344341434243434312336n n n n n n n n n n b b cb b b d b b d b b d b d +---------=-=⨯+-=+-=+-==则数列前4n 项中的项43n b -是以2为首项，6为公差的得差数列，易知{}4n b 中删掉含有43n b -的项后按原来的顺序构成一个首项为2公差为2的等差数列，41543()n n S b b b -∴=+++()()()()23467846454442414+n n n n n n b b b b b b b b b b b b -----++++++++++++⎡⎤⎣⎦2(1)3(31)26(3)2212822n n n n n n n n --=+⨯+⨯+⨯=+ 43nn S λ≤⋅，43nn S λ∴≤，设2412833n n n n S n n c +==，则()max n c λ≥，22211112(1)8(1)12824820333n n n n n n n n n n n c c +++++++-++-=-=当1n =时，2248200n n -++>，12c c <；当2n ≥，n *∈N 时，2248200n n -++<，1n n c c +<，∴123c c c <>>，∴()2max 649n c c ==，即()2max 649n c c λ≥==（3）因为{}n b 既是“Γ数列”又是等比数列，设{}n b 的公比为1n nb q b +=，由等比数列的通项公式有1n n b bq -=，当m *∈N 时，21k m k m b b d ++-=，即()11km km km bq bq bq q d +-=-=① 1q =，则0d =，n b b =； ② 1q ≠，则()1kmd qq b=-，则kmq 为常数，则1q =-，k 为偶数，2d b =-，()11n n b b -=-；经检验，满足条件的{}n b 的通项公式为n b b =或()11n n b b -=-.。

2018年上海市杨浦区高考物理一模试卷(解析版)

2018年上海市杨浦区高考物理一模试卷一、选择题（共40分．第1-8小题，每小题3分，第9-12小题，每小题3分．每小题只有一个正确答案．）1．（3分）下列用来定量描述磁场强弱和方向的是（）A．磁感应强度B．磁通量C．安培力D．磁感线2．（3分）20世纪中叶以后，移动电话快速发展．移动电话机（）A．既能发射电磁波，也能接收电磁波B．只能发射电磁波，不能接收电磁波C．不能发射电磁波，只能接收电磁波D．既不能发射电磁波，也不能接收电磁波3．（3分）我国《道路交通安全法》中规定：各种小型车辆前排乘坐的人（包括司机）必须系好安全带．这是因为（）A．系好安全带可以减小惯性B．系好安全带可以防止因人的惯性而造成的伤害C．系好安全带可以防止因车的惯性而造成的伤害D．是否系好安全带对人和车的惯性没有影响4．（3分）电源电动势反映了电源把其它形式的能转化为电能的本领，下列关于电动势的说法中正确的是（）A．电动势是一种非静电力B．电动势越大表明电源储存的电能越多C．电动势就是闭合电路中电源两端的电压D．电动势由电源中非静电力的特性决定，跟其体积、外电路无关5．（3分）如图所示，在两块相同的竖直木板之间，有质量均为m的4块相同的砖，用两个大小均为F的水平力压木板，使砖块静止不动，则第2块砖对第3块砖的摩擦力大小是（）A．0B．mg C．mg D．2mg6．（3分）意大利科学家伽利略在研究物体变速运动规律时，做了著名的“斜面实验”，他测量了铜球在较小倾角斜面上的运动情况，发现铜球做的是匀变速直线运动，且铜球加速度随斜面倾角的增大而增大，于是他对大倾角情况进行了合理的外推，由此得出的结论是（）A．力不是维持物体运动的原因B．力是使物体产生加速度的原因C．自由落体运动是一种匀变速直线运动D．物体都具有保持原来运动状态的属性，即惯性7．（3分）如图所示，虚线a、b、c是电场中的一簇等势线（相邻等势面之间的电势差相等），实线为一α粒子（重力不计）仅在电场力作用下通过该区域时的运动轨迹，P、Q是这条轨迹上的两点，据此可知（）A．a、b、c三个等势面中，a的电势最高B．电子在P点具有的电势能比在Q点具有的电势能小C．α粒子在P点的加速度比Q点的加速度大D．带电质点一定是从P点向Q点运动8．（3分）在如图所示电路中，合上开关S，将滑动变阻器R2的滑动触点向b端移动，则三个电表A1、A2和V的示数I1、I2和U的变化情况是．（）A．I1增大，I2不变，U增大B．I1减小，I2不变，U减小C．I1增大，I2减小，U增大D．I1减小，I2增大，U减小9．（4分）一物体沿竖直方向运动，以竖直向上为正方向，其运动的v﹣t图象如图所示．下列说法正确的是（）A．0～t1时间内物体处于失重状态B．t1～t2时间内物体机械能守恒C．t2～t3时间内物体向下运动D．0～t2时间内物体机械能一直增大10．（4分）利用引力常量G和下列某一组数据，不能计算出地球质量的是（）A．地球的半径及地球表面附近的重力加速度（不考虑地球自转的影响）B．人造卫星在地面附近绕地球做圆周运动的速度及周期C．月球绕地球做圆周运动的周期及月球与地球间的距离D．地球绕太阳做圆周运动的周期及地球与太阳间的距离11．（4分）如图所示，电源E，导线，导电细软绳ab、cd，以及导体棒bc构成闭合回路，导电细软绳ab、cd的a端和d端固定不动，加上恰当的磁场后，当导体棒保持静止时，闭合回路中abcd所在平面与过ad的竖直平面成30°，已知ad和bc等长且都在水平面内，导体棒bc中的电流I＝2A，导体棒的长度L＝0.5m，导体棒的质量m＝0.5kg，g取10m/s2，关于磁场的最小值和方向，说法正确的是（）A．T，竖直向上B．T，竖直向下C．2.5T，由b指向a D．2.5T，由a指向b12．（4分）如图所示，一辆运送沙子的自卸卡车装满沙子，沙粒之间的动摩擦因数为µ1，沙子与车厢底板的动摩擦因数为µ2（µ2＞µ1），车厢的倾角用θ表示，下列说法正确的是（）A．要顺利地卸干净全部沙子，应满足tanθ＜µ2B．要顺利地卸干净全部沙子，应满足sinθ＞µ2C．只卸去部分沙子，车上还留有一部分沙子，应满足µ2＞tanθ＞µ1D．只卸去部分沙子，车上还留有一部分沙子，应满足tanθ＞µ2＞µ1二、填空题（共20分）13．（4分）习惯上规定电荷定向移动的方向为电流的方向（填“正”或“负”）．一次闪电过程中，流动的电荷量约为300C，持续时间约2s，则闪电的平均电流为A．14．（4分）甲、乙两弹簧振子质量相等，其振动图象如图所示，则它们振动频率的大小关系是f甲f乙；在0﹣4s内，甲的加速度为正向最大的时刻是s末。

凯文老师，2018年杨浦区高三高考英语一模答案

2018年杨浦区高三高考英语一模答案（含听力原文）I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The 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. W: Excuse me, Sir, can I borrow these books on medicine?M: Sorry, you are only allowed to have three at a time.Q: Where does the conversation most probably take place? (C)2. W: The parking fee is 10 yuan for per hour. But how much would it be if I park here for anhour and 10 minutes?M: 20 yuan, madam. Because we charge by the number of hours.Q: How much will the woman pay if she parks her car there for 2 hours and 10 minutes? (C) 3. W: How much do you want to spend on the rent?M: Uh, somewhere under $200 a month. I'd prefer to rent a furnished one and I need a parking space.Q: What's the probable relationship between the two speakers? (A)4. W: It's going to be a wonderful day for hiking tomorrow. But the weatherman said it wasgoing to rain later tonight.M: Ah, nonsense. Nothing can spoil our adventure. Those weather reports are never right.Q: What can we learn from the conversation? (D)5. M: I really can't put up with Jane. She is always speaking ill of others.W: So she is, but she seems to have good relations with our boss.Q: What can we learn about Jane? (A)6. M: Are you going to put up lights on the outside?W: Yes, all of my neighbors decorate their houses, and I am making a gift list.Q: What are they talking about? (C)7. W: It's terrible to see so many breaks for advertisements on TV. They should be avoided.M: But I think this is a good chance to give your eyes a rest.Q: What does the man mean? (D)8. M: About 100 million people will travel during the Spring Festival. Most of them will bebehind the wheel!W: But winter weather can make roads dangerous. So take your time and make sure you arrive home safe and sound.Q: What can we learn from the conversation? (B)9. W: Jonny, you should take up a little weight training and you need to go to bed early insteadof watching TV half the night.M: Hey, you're starting to sound like my personal fitness instructor.Q: What does the man mean? (C)10. M: As you know, we Americans cut our kids off when they're 18.W: I wish I could. But we're Greeks. They're ours until they're married.Q: What can we learn from the conversation? (B)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 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.Burns are the most common injuries, but the right, quick treatment for burns can make a big difference. What should you do if someone suffers a burn?First, look at the burn. Is it a small burn? Doctors say a small burn is about 23 millimeters across. Then ask the victim, "Is your burn very painful?'Surprisingly, more serious burns aren't very painful. A small painful burn isn't serious. Y ou can treat it without a doctor. Put the burn under cold running water for several minutes, which is called "flushing", to cool the burned area. Y ou may have to flush the burn for longer if the victim still feels pain. Don't use ice, which is so cold that it can damage the skin.After flushing, cover the burn with a clean bandage or cloth. Y ou can even use a clean plastic bag. Don't use anything that may stick to the burn. The covering will keep the burn clean and help to stop infection.But what if the burn is more serious? If the victim doesn't feel painful, then it is probably more serious. This sounds strange. But burns that take away a person's feeling have damaged the skin and nerves deep down. If the victim's burn is more than 23 millimeters across, it is also serious. Here is what you need to do with large burns:First remove the victim's clothes or jewelry around the burned area. Then treat the burned area with cool running water. Cover the burned area lightly with a clean cloth and get help fast. If you can get the victim to a doctor more quickly, the victim will have a better chance of recovery. Questions:11. In which situation should the victim be sent to the hospital? (C)12. What should be done to treat a burn? (C)13. Why do we use a covering over a burn? (A)Questions 14 through 16 are based on the following passage.A company V olterman has recently released a smart wallet that takes security to the extreme. It comes with a built-in alarm system, a GPS tracker, and even a front facing camera that takes photos of whoever tries to open it, and sends them to the owner.Wallets almost seem obsolete in this digital age, but the smart wallet isn't your average wallet. Apart from the screen, it packs about as much technology as your smart phone, including a built-in camera, a power bank, a GPS tracker, an alarm system, and even Wi-Fi hotspot capabilities. It maylook like a simple leather wallet on the outside, but it's a whole different one on the inside.Perhaps the most unusual feature of the smart wallet is the built-in front-facing camera. When in "lost mode", the wallet will take pictures whenever someone tries to open it, and the wallet sends the pictures to the owner by connecting to the Internet. This feature is listed as optional, so would-be customers concerned about privacy issues needn't be worried."If someone dares to open your wallet without your permission, the little camera will take a picture and send it to your mobile phone," said the founder of V olterman. "We have developed a wallet, which doesn't get lost. Once your wallet is stolen, you will find it easily by receiving the picture of the thief."Questions:14. What does the smart wallet look like on the outside? (B)15. What will happen when the wallet is stolen? (D)16. What can we learn about the smart wallet? (C)Questions 17 through 20 are based on the following conversation.M: Hi, Helene. How are you?W: Hi, David. Well, I'm a little disappointed.M: What's wrong?W: Tonight's concert doesn't end until after Abby's curfew, so she won't be able to go.M: That's too bad.W: Y eah. I won't bend the rules because it's a school night. So what time is your son Jackson's curfew?M: He doesn't have one.W: How does he know when to be home?M: He knows to be home at an appropriate time.W: Doesn't he need more structure?M: We've discussed an appropriate time for him to be home. We trust him to stick to that.W: He's still pretty young. We give Abby strict guidelines.M: Kids certainly need boundaries. But being too strict can discourage them from being responsible.W: Well, teens aren’t always responsible. They can get into a lot of trouble when they're out too late.M: That may be true, but curfews can give parents a false sense of security.W: What do you mean?M: Teens can get into trouble any time of the day.W: But a curfew does reduce the amount of trouble they can get into.M: It might also prevent them from becoming mature. Curfews can cause teens to be too dependent on their parents.W: I don't agree. Abby has a very strict curfew, but she still makes her own decisions.M: Perhaps your Abby is an exception. I know many teens who feel curfews are a punishment. W: Abby understands the curfew is for her benefit. In any case, a curfew works well for us.M: And we're fine without one.Questions:17. Why can’t Abby go to the concert? (A)18. What can we learn about Jackson? (C)19. According to the man, what’s the influence of curfew on children? (D)20. What does Abby think of her curfew? (B)21. that 22. the loudest 23. is encouraged 24. can/may 25. that 26. As 27. making 28. What 29. Because/as 30. covered31-40 D J B G C K A F I H41-55 BADCA BBACA BDACD56-59 DCBA60-62 ABD63-66 C B A D67-70 D A F CFood delivery workers have a higher risk of getting involved in traffic accidents. The large demand for their service, the high pressure from employers and their strong desire to earn more commission lead to their reckless behavior. In some cities, measures like stricter road safety education and more severe punishment have been taken to solve this problem.1.The contest aims at encouraging young people to inherit the Chinese cultural tradition.2.There is no denying that the exposure of personal information may bring about security risks.3.The conflict between people’s d emand for high quality education and the insufficienteducational resources remains to be resolved.4.What surprised us was that the bookworm was ignorant of the news that ShanghaiGreenland Shenhua won the championship of 2017 CFA Cup.。

杨浦区2018届高三一模英语卷(含答案)

杨浦区2017学年度第一学期高三模拟质调研英语学科试卷2017.12本试卷分为第I卷（第1-11页）和第II卷（第12页）两部分。

全卷共12页。

满分140分。

考试时间120分钟。

考生注意：1．答第I卷前，考生务必将条形码粘贴在答题纸的指定区域内。

2.第I卷（1-20小题，31---70小题）由机器阅卷，答案必须全部涂写在答题卡上。

考生应将代表正确答案的小方格用铅笔涂黑。

注意试题题号和答题卡编号一一对应，不能错位。

答案需要更改时，必须将原选项用橡皮擦去，重新选择。

答案写在试卷上一律不给分。

第I卷中的第21-30小题，IV.Summary Writing部分和第II卷的试题，其答案用钢笔或水笔写在答题纸的规定区域内，如用铅笔答题，或写在试卷上则无效。

第I卷（共100分）I.Listening ComprehensionSection ADirections:In Section A,you will hear ten short conversations between two speakers.At the end of each conversation,a question will be asked about what was said.The 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.In a professor's office. B.In a second-hand book shop.C.In a library.D.In a hospital.2. A.10yuan. B.20yuan.C.30yuan.D.50yuan.3. A.House agent and customer. B.Shop assistant and customer.C.Car mechanic and car owner.D.Employer and employee.4. A.The man doesn't believe what the woman says.B.The weather report spoils the man's good mood.C.They will cancel the hiking due to the bad weather.D.The man thinks it unnecessary to give up the adventure.5. A.She always talks bad about her colleagues.B.She has a good reputation among her colleagues.C.She is good at handling complicated relationships.D.She has good relations with her colleagues and boss.6. A.Harmony in a community. B.Safety in the neighborhood.C.Preparation for Christmas.D.Ways to save electricity.7. A.Watching advertisements may help ease eyestrain（眼疲劳）.B.It's a great chance to break the habit of watching TV.C.The advertisements are long enough for her to have a nap.D.Focusing eyes on the screen for a long time is harmful to eyes.8. A.The man decides to go home by rail.B.Most people travel by car during the festival.C.Most people arrive beyond the scheduled time.D.The man will have a sound sleep on the bus.9. A.He is not a bit overweight.B.He likes his fitness instructor.C.She has set too many rules for him.D.She should talk with his personal trainer.10.A.Greeks are not allowed to get married before18.B.Greek kids are not as independent as American kids.C.American parents don’t pay for children's wedding.D.Greek parents will take care of children until they are18.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 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.Questions11through13are based on the following passage.11. A.The burn is20millimeters across.B.The burn is small but very painful.C.The burn takes away the victim's feeling.D.The burn is small but the skin is damaged.e a clean plastic bag to keep warm.B.Bind up the burn with bandage or cloth.C.Treat the burned area with cold running water.D.Flush（冲洗）the burn with ice water for several minutes.13.A.To avoid infection. B.To ease pain.C.To speed recovery.D.To reduce stickiness.Questions14through16are based on the following passage.14.A.A cell phone. B.A leather wallet.C.A mini camera.D.An alarm clock.15.A.The wallet will sound an alarm.B.It will track the thief with GPS system.C.It will contact the bank to block balance.D.Its owner will receive a picture of the thief.16.A.It's out-dated in this digital age.B.It can text messages automatically.C.It is a multifunctional wallet.D.It is unique in appearance and function.Questions17through20are based on the following conversation.17. A.The concert is beyond her curfew(宵禁).B.She can’t go out on school night.C.Her mother is not available.D.She doesn’t like the band.18.A.His parents set a strict rule for him.B.His parents don’t care when he is back.C.He is self-disciplined and trustworthy.D.He envies those who have curfews.19.A.Promoting maturity. B.Giving sense of security.C.Improving sense of responsibility.D.Discouraging independence.20.A.It’s a severe punishment. B.It’s for her good.C.It’s a ridiculous practice.D.It’s an exceptional case.II.Grammar and VocabularySection ADirections:After reading the passage below,fill in the blanks to make the passages coherent and grammatically correct.For the blanks with a given word,fill in each blank with the proper form of the given word;for the other blanks,use one word that best fits each blank.It’s interesting when you think about how Japan is a nation(21)______appreciates the virtues of silence and good manners,and yet when it comes to eating noodles,Japanese people can be(22)______ (loud)in the world.According to lifestyle website grapee.jp,slurping(发出"哧溜"声)when eating noodles(23)______ (encourage)in Japanese culture.It’s believed that taking air into your mouth(24)______enhance the flavor of the noodles,and that it helps cool down the noodles.It’s also considered to be a way to show appreciation for the dish.Sometimes,just making the noise alone seems to make the noodles more enjoyable.It wasn’t until a new expression–“noodle harassment（骚扰）”--came out last year on social media(25) ______Japanese people started to realize that the slurping noise is making some foreign visitors uncomfortable.(26)______a response,Japanese instant noodle maker Nissin introduced a so-called noise-canceling fork last month.The fork,which looks like an electric toothbrush,is connected wirelessly to a smart phone. When the person using the fork starts to slurp,the fork sends a signal to the person’s phone,(27)______ (make)it play a sound to mask the slurping noise.But is it really necessary?Dining traditions do vary.(28)______is considered to be proper table manners in one country is likely to be seen as rude in another.In India,people eat with their hands(29) ______they think in this way they build a connection with the food.However,people who are used to eating with forks might find it uncomfortable to get their hands(30)______(cover)in oil and bits of food. But this eating method is part of Indian's culture,just like Japan's slurping is part of its own.“So,if your are eating noodles,whether that’s ramen,uudon,or soba,please slurp,”wrote reporter Brian Ashcraft on blog Kotaku.“If anyone gets annoyed while you are doing that,pay them no mind because they're missing the point entirely.”Section BDirections:Fill in each blank with a proper word chosen from the box.Each word can only be used once. Note that there is one word more than you need.MeTro is simple to apply,can be easily stored,and works closely with natural__37__to heal a wound. What’s more,it degrades without leaving any kind of poisonous leftovers in the body.For now the trials are__38__to animal models.But human trials are in the works,and the results to date are incredibly__39__.If the MeTro can be further developed into a__40__product,it could become an essential part of a first responder’s toolkit.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.Competition is good for businesses.In the world of navigation（导航）systems,however,competition is also a necessity---it may not be wise to rely on foreign systems for positioning and tracking services. Now,___41___remarkable accuracy and reliability,China’s BeiDou system has made its presence felt.The BeiDou project was set up in1994.The first BeiDou satellite was not launched until2000.Now, ___42___,there are already more than20BeiDou satellites in orbit(轨道).They form a___43___network that provides positioning,navigation and timing services for China and several other Asian countries.This“home-grown"system is now___44___a major upgrade.Earlier this month,two BeiDou-3 satellites,the first of China’s most powerful___45___of navigation satellites,were launched into space.The launch marks the beginning of the global___46___of the BeiDou navigation system.Over the next three years,China plans to send up30more BeiDou-3satellites;The expanded navigation system will___47___ create a network that is able to support military and civilian applications around the world.Scientists involved in the project said the new system would give civilian users an accuracy of2.5 meters to five meters,overtaking that of the___48___positioning technologies.BeiDou’s chief designer said the new satellites would be able to__49___which lane a car is using on a motorway and__50___the swing of a building in high winds.It will also be able to guide fire trucks to the nearest water hydrant(消防栓).The Chinese military,meanwhile,will be able to use coded signals for millimeter（毫米）___51___.China is only the third country in the world to develop a navigation system on its own,after the United States(GPS)and Russia(GLONASS).Developing BeiDou is a necessity.The system__52___national security by ending a reliance on foreign systems.Moreover,it enhances China’s international reputation for technological___53___.For most of us,the benefits of the new satellite system will be felt in a couple of years when more phones are___54___with BeiDou chips(芯片).Many smartphones today still use GPS and GLONASS. That’ll soon change with the development of BeiDou.One product manager___55___most smartphones to be able to receive BeiDou signals.He says:“In three years’time,people may still say‘I’m using GPS’,but in fact,their phone is tune in to BeiDou.”41.A.dominating B.boasting C.shifting D.inputting42.A.however B.afterwards C.moreover D.therefore43.A.continental B.local C.domestic D.regional44.A.enduring B.encountering C.undergoing D.processing45.A.generation rmation C.examination D.revolution46.A.extension B.expansion C.interaction D.invasion47.A.objectively B.eventually C.sufficiently D.essentially48.A.existing B.progressing C.upcoming D.everlasting49.A.explore B.investigate C.spot D.remind50.A.detect B.prevent C.protect D.adjust51.A.privacy B.accuracy C.fluency D.currency52.A.convinces B.insures C.highlights D.strengthens53.A.innovation B.consumption C.emission D.exhibition54.A.decorated B.furnished C.equipped D.connected55.A.respects B.instructs C.inspects D.expectsSection 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)A Swedish power plant is taking reuse and recycle to the next level by burning unusable clothing instead of coal,Bloomberg reports.Retail giant Hennes&Mauritz,more commonly known as H&M,is helping the utility transition away from coal through its moldy(发霉的)or otherwise unsalable clothing.The multi-fuel power and heating station in Västerås,central Sweden,is planning to be completely fossil-fuel free by2020.It’s the largest station of its kind and Sweden claims it’s one of Europe’s cleanest. To kick its coal habit,the station is turning instead to other burnable materials including recycled wood, rubbish and yes,clothes.“Our goal is to use only renewable and recycled fuels,”Jens Neren,head of fuel supplies at the utility company which owns and operates the Västerås plant,told Bloomberg.Johanna Dahl,head of communications for H&M in Sweden,told Bloomberg that the company allows only the burning of clothes which are no longer safe to use.“It is our legal obligation to make sure that clothes that contain mold or do not meet the requirements of our strict restriction on chemicals are destroyed,”she said.The Västerås plant has burned around15tons of old H&M clothes so far this year,compared with about400,000tons of rubbish,Neren told Bloomberg.Sweden has one of the world’s greener energy generating systems,and has invested in bioenergy,solar power and electric buses.In2015,the Scandinavian country announced an ambitious aim to become one of the first nations in the world to end its dependence on fossil fuels.According to the Swedish government,the country has already heavily reduced its dependence on oil,which accounted for75%of the energy supply in 1970,and now makes up a20%share.56.Which of the following can serve as fuel in the Västerås plant?A.Fashionable coats in H&M chain store.B.Old TV sets deserted as rubbish.C.Wooden furniture in second-hand shop.D.H&M clothes unsuitable for sale.57.The underlined word in the last paragraph“generating”is closest in meaning to______.A.eliminatingB.adjustingC.producingD.circulating58.What can we learn from the passage?A.The Swedish government discourages the development of bioenergy.B.Clothes only take up a small proportion of the burning material.C.Sweden’s fossil-fuel free plan is almost accomplished by now.D.Sweden has an ambition to be the cleanest country in the world.59.What is the main idea of the passage?A.A Swedish power plant is burning unusable H&M clothes for fuel.B.The Swedish government aims high and is taking effective action.C.H&M is looking for a new way to strengthen its position in fashion.D.Coal and oil are no longer regarded as the primary fuels in Sweden.(B)60.Before an adventure,a trekker should ______.A.tell the park officials his destination and time scheduleB.pack up some jungle fruit juice and pre-cooked mealsC.consult a local guide about the most adventurous routeD.have his fitness level assessed at the touristcenter61.Which of the following is NOT suitable for a rainforest trekking?A.Long-sleeved cotton shirts.B.Tight sports shorts.C.Hiking boots.D.A wide brimmed hat.62.If a trekker starts out at dawn,he may______.A.escape being caught in the rainB.sight scared wildlifeC.enjoy the heat of the tropical sunD.see animals seeking food(C)The largest genetic study of mosquitoes has found their ability to resist insecticides is evolving rapidly and spreading across Africa,putting millions of people at higher risk of contracting malaria （疟疾）.British scientists who led the work said mosquitoes'growing resistance to control tools such as insecticide-treated bed nets and insecticide spraying,which have helped cut malaria cases since2000,now threatens“to disturb malaria control”in Africa.“Our study highlights the severe challenges facing public efforts to control mosquitoes and to manage and limit insecticide resistance,”said Martin Donnelly of the Liverpool School of Tropical Medicine,who worked on the study with a team from Britain’s Wellcome Trust Sanger Institute.Latest World Health Organization(WHO)data show that216million people were infected last year with the malaria parasite(寄生虫),which is transmitted by blood-sucking Anopheles mosquitoes.The disease killed445,000people in2016,and the majority of them were children in sub-Saharan Africa.To understand how mosquitoes are evolving,the researchers sequenced the DNA of765wild Anopheles mosquitoes taken from15locations across eight African countries.Their work,published in the journal Nature on Wednesday,created the largest data resource on natural genetic variation for any species of insect.Analyzing the data,the scientists found that the Anopheles gambiae mosquitoes（冈比亚疟蚊）were extremely genetically diverse（多样化的）compared with most other animal species.This high genetic diversity enables rapid evolution,they said,and helps to explain how mosquitoes develop insecticide resistance so quickly.The data also showed the rapid evolution of insecticide resistance appeared to be due to many previously unknown genetic variants（变体）within certain genes.The scientists said these genetic variants for insecticide resistance were not only emerging independently in different parts of Africa,but were also being spread across the continent by mosquito migration.Michael Chew,an expert at Britain’s Wellcome Trust global health charity which helped fund the research,said the finds underlined the importance of pushing scientific research ahead to control malaria.Global efforts to control malaria through effective vaccine,insecticides and the best drug combinations require urgent,united action by scientists,drug companies,governments and the WHO.63.Which of the following is scientists’headache?A.The number of mosquitoes in Africa is growing rapidly.B.Some genetic variants of mosquitoes are still unknown.C.The existing insecticides aren’t as effective as they used to be.lions of African people have resistance to medicines for malaria.64.Malaria cases can be cut by______.A.threatening drug companiesB.spraying insecticidesC.limiting blood donationD.transmitting data65.What CANNOT be concluded from the passage?A.Children are more likely to be bit by mosquitoes.B.Many previously unknown variants are found in the study.C.The mosquito migration contributes to the spread of variants.D.Anopheles mosquitoes have great genetic diversity.66.Which is FALSE about the genetic study of mosquitoes?A.It created the largest data on natural genetic variation for any insect species.B.It found the possible causes for the rapid evolution of insecticide resistance.C.It discovered where the genetic variants emerged and how they were spread.D.It highlighted the public efforts and appealed to limit the use of insecticides.Section CDirections:Read the following passage.Fill in each blank with a proper sentence given in the box.Each sentence can be used only once.Note that there are two more sentences than you need.A.But despite its popularity,experts warn that biometrics might not be as secure aswe’d imagined.B.Security experts don’t think it absolutely necessary to use biometric technology.C.But using multiple security measures is the best defense.D.Now,this type of technology might not be far away.E.If a person’s biometric information is stolen,that could have extremely seriousconsequences for that individualF.The chance that a random person could look at your iPhone X and unlock it withhis face is about one in a million.Imagine you're standing in line to buy an afterschool snack at a store.You step up to the counter and the cashier scans your food.Next,you have to pay.But instead of scanning a QR code with your smartphone, you just hold out your hand so the cashier can scan your fingerprint.Or,a camera scans your face,your eyes or even your ear.__________67__________As technology companies move away from traditional password,biometric （生物识别）security,which includes fingerprint,face and voice ID,is becoming increasingly popular.In2013,Apple introduced the iPhone5s,one of the first smartphones with a fingerprint scanner.Since then,using one’s fingerprint to unlock a phone and make mobile payments has become commonplace, bringing convenience to our lives.And since last year,Samsung has featured eye-scanning technology in its top smartphones,while Apple’s new iPhone X can even scan a user’s face.__________68__________“Biometrics,ideally,are good,”John Michener,a biometric expert,told tech website Inverse.“In practice,not so much.”When introducing the new iPhone’s Face ID feature at Apple’s Keynote Event in September,Phil Schiller,Apple’s senior vice president,said,“__________69__________”But it’s already been done.In a video posted on community website Reddit on Nov3,two brothers showed how they were each able to unlock the same iPhone X using their own face,Quartz reported.And they aren’t even twins.“We may expect too much from biometrics,”Anil Jain,a computer science professor at Michigan State University,told CBS news.“No security systems are perfect.”Earlier this year,Jain found a way to trick biometric ing a printed copy of a thumbprint,she was able to unlock a dead person's smartphone for police.“It’s good to see biometrics being used more,”Jain told CBS News,“because it adds another factor for security.__________70__________”IV.Summary WritingDirections:Read the following passage.Summarize the main idea and the main point(s)of the passage in no more e your own words as far as possible.It’s a common sight to see food delivery workers riding electric bikes through big cities in China.Most of them seem to be in a hurry,as they run red lights to deliver their meals in time.However,such reckless （鲁莽的）behavior often causes serious problems.In the first half of this year,food delivery drivers had76traffic accidents in Shanghai alone,according to the Shanghai Public Security Bureau.This means that on average,there is a food delivery worker that gets hurt or even dies on the road in Shanghai every2.5days.Other cities also share similar problems.In Nanjing,three people died and2,473were injured in road accidents related to food delivery workers in the same period,according to the Ministry of Public Security.The rise of reckless behavior among food delivery workers is closely related to the growing demand for their service,reported People’s Daily.About150million people in China use food delivery services, according to China Radio International(CRI).Such a big market has led to a large demand for food delivery workers,with some companies offering high salaries to attract new workers.However,food delivery workers are often under high pressure from their employers.They face company fines of20yuan for delivering food late and upwards of200yuan for receiving complaints, reported CRI.Moreover,the more orders they take,the more commission（佣金）they can earn,leading to some workers checking their mobile phones for new orders while they're riding their bikes.While most companies have measures requiring delivery workers to follow traffic rules,“there remains a problem of whether these requirements and rules for delivery workers are truly entering their ears,brains and hearts,”Wang Liang,deputy head of the Traffic Police Security Bureau,told news website The Paper.To solve the problem,some cities have taken action.Shanghai has asked companies to train their workers on traffic rules and safety.Now in Shenzhen,if a delivery worker gets caught breaking traffic rules more than twice,he or she will be banned from driving food delivery vehicles for a whole year.第II卷（共40分）I.TranslationDirections:Translate the following sentences into English,using the words given in the brackets.1.这个比赛旨在鼓励年轻人继承中国文化的传统。

2018年上海市杨浦区高考数学一模试卷

2018年上海市杨浦区高考数学一模试卷一.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．（4分）计算的结果是．2．（4分）已知集合A={1，2，m}，B={3，4}，若A∩B={3}，则实数m=．3．（4分）已知，则=．4．（4分）若行列式，则x=．5．（4分）已知一个关于x、y的二元一次方程组的增广矩阵是，则x+y=．6．（4分）在的二项展开式中，常数项等于．7．（5分）若将一颗质地均匀的骰子（一种各面上分别标有1，2，3，4，5，6个点的正方体玩具），先后抛掷2次，则出现向上的点数之和为4的概率是．8．（5分）数列{a n}的前n项和为S n，若点（n，S n）（n∈N*）在函数y=log2（x+1）的反函数的图象上，则a n=．9．（5分）在△ABC中，若sinA、sinB、sinC成等比数列，则角B的最大值为．10．（5分）抛物线y2=﹣8x的焦点与双曲线﹣y2=1的左焦点重合，则这条双曲线的两条渐近线的夹角为．11．（5分）已知函数，x∈R，设a＞0，若函数g （x）=f（x+α）为奇函数，则α的值为．12．（5分）已知点C、D是椭圆上的两个动点，且点M（0，2），若，则实数λ的取值范围为．二.选择题（本大题共4题，每题5分，共20分）13．（5分）在复平面内，复数对应的点位于（）A．第一象限B．第二象限C．第三象限D．第四象限14．（5分）给出下列函数：①y=log2x；②y=x2；③y=2|x|；④y=arcsinx．其中图象关于y轴对称的函数的序号是（）A．①②B．②③C．①③D．②④15．（5分）“t≥0”是“函数f（x）=x2+tx﹣t在（﹣∞，+∞）内存在零点”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16．（5分）设A、B、C、D是半径为1的球面上的四个不同点，且满足•=0，•=0，•=0，用S1、S2、S3分别表示△ABC、△ACD、△ABD的面积，则S1+S2+S3的最大值是（）A．B．2 C．4 D．8三.解答题（本大题共5题，共14+14+14+16+18=76分）17．（14分）如图所示，用总长为定值l的篱笆围成长方形的场地，以墙为一边，并用平行于一边的篱笆隔开．（1）设场地面积为y，垂直于墙的边长为x，试用解析式将y表示成x的函数，并确定这个函数的定义域；（2）怎样围才能使得场地的面积最大？最大面积是多少？18．（14分）如图，已知圆锥的侧面积为15π，底面半径OA和OB互相垂直，且OA=3，P是母线BS的中点．（1）求圆锥的体积；（2）求异面直线SO与PA所成角的大小．（结果用反三角函数值表示）19．（14分）已知函数的定义域为集合A，集合B=（a，a+1），且B ⊆A．（1）求实数a的取值范围；（2）求证：函数f（x）是奇函数但不是偶函数．20．（16分）设直线l与抛物线Ω：y2=4x相交于不同两点A、B，O为坐标原点．（1）求抛物线Ω的焦点到准线的距离；（2）若直线l又与圆C：（x﹣5）2+y2=16相切于点M，且M为线段AB的中点，求直线l的方程；（3）若，点Q在线段AB上，满足OQ⊥AB，求点Q的轨迹方程．21．（18分）若数列A：a1，a2，…，a n（n≥3）中（1≤i≤n）且对任意的2≤k≤n﹣1，a k+1+a k﹣1＞2a k恒成立，则称数列A为“U﹣数列”．（1）若数列1，x，y，7为“U﹣数列”，写出所有可能的x、y；（2）若“U﹣数列”A：a1，a2，…，a n中，a1=1，a n=2017，求n的最大值；（3）设n为给定的偶数，对所有可能的“U﹣数列”A：a1，a2，…，，记，其中max{x1，x2，…，x s}表示x1，x2，…，x s这s 个数中最大的数，求M的最小值．2018年上海市杨浦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12题，1-6每题4分，7-12每题5分，共54分）1．（4分）计算的结果是1．【解答】解：当n→+∞，→0，∴=1，故答案为：1．2．（4分）已知集合A={1，2，m}，B={3，4}，若A∩B={3}，则实数m=3．【解答】解：∵集合A={1，2，m}，B={3，4}，A∩B={3}，∴实数m=3．故答案为：3．3．（4分）已知，则=﹣．【解答】解：∵，∴=．故答案为：﹣．4．（4分）若行列式，则x=2．【解答】解：∵，∴2×2x﹣1﹣4=0即x﹣1=1∴x=2故答案为：25．（4分）已知一个关于x、y的二元一次方程组的增广矩阵是，则x+y= 6．【解答】解：∵一个关于x、y的二元一次方程组的增广矩阵是，∴由二元线性方程组的增广矩阵可得到二元线性方程组的表达式，解得x=4，y=2，∴x+y=6．故答案为：6．6．（4分）在的二项展开式中，常数项等于﹣160．=x6﹣r（﹣）r=（﹣2）r x6﹣2r【解答】解：展开式的通项为T r+1令6﹣2r=0可得r=3常数项为（﹣2）3=﹣160故答案为：﹣1607．（5分）若将一颗质地均匀的骰子（一种各面上分别标有1，2，3，4，5，6个点的正方体玩具），先后抛掷2次，则出现向上的点数之和为4的概率是．【解答】解：基本事件共6×6个，点数和为4的有（1，3）、（2，2）、（3，1）共3个，故P==．故答案为：．8．（5分）数列{a n}的前n项和为S n，若点（n，S n）（n∈N*）在函数y=log2（x+1）的反函数的图象上，则a n=2n﹣1．【解答】解：由题意得n=log2（S n+1）⇒s n=2n﹣1．n≥2时，a n=s n﹣s n﹣1=2n﹣2n﹣1=2n﹣1，当n=1时，a1=s1=21﹣1=1也适合上式，∴数列{a n}的通项公式为a n=2n﹣1；故答案为：2n﹣19．（5分）在△ABC中，若sinA、sinB、sinC成等比数列，则角B的最大值为．【解答】解：∵在△ABC中，sinA、sinB、sinC依次成等比数列，∴sin2B=sinAsinC，利用正弦定理化简得：b2=ac，由余弦定理得：cosB==≥=（当且仅当a=c时取等号），则B的范围为（0，]，即角B的最大值为．故答案为：．10．（5分）抛物线y2=﹣8x的焦点与双曲线﹣y2=1的左焦点重合，则这条双曲线的两条渐近线的夹角为．【解答】解：∵抛物线y2=﹣8x的焦点F（﹣2，0）与双曲线﹣y2=1的左焦点重合，∴a2+1=4，解得a=，∴双曲线的渐近线方程为y=，∴这条双曲线的两条渐近线的夹角为，故答案为：．11．（5分）已知函数，x∈R，设a＞0，若函数g（x）=f（x+α）为奇函数，则α的值为．【解答】解：函数，=，=s，函数g（x）=f（x+α）=为奇函数，则：（k∈Z），解得：，故答案为：12．（5分）已知点C、D是椭圆上的两个动点，且点M（0，2），若，则实数λ的取值范围为．【解答】解：假设CD的斜率存在时，设过点M（0，2）得直线方程为y=kx+2，联立方程，整理可得（1+4k2）x2+16kx+12=0，设C（x1，y1），N（x2，y2），则△=（16k）2﹣4×（1+4k2）×12≥0，整理得k2≥，x1+x2=﹣，x1x2=，（*）由，可得，x1=λx2代入到（*）式整理可得==，由k2≥，可得4≤≤，解可得＜λ＜3且λ≠1，当M和N点重合时，λ=1，当斜率不存在时，则D（0，1），C（0，﹣1），或D（0，1），C（0，﹣1），则λ=或λ=3∴实数λ的取值范围．故答案为：．二.选择题（本大题共4题，每题5分，共20分）13．（5分）在复平面内，复数对应的点位于（）A．第一象限B．第二象限C．第三象限D．第四象限【解答】解：∵=，∴复数对应的点的坐标为（﹣1，﹣2），位于第三象限．故选：C．14．（5分）给出下列函数：①y=log2x；②y=x2；③y=2|x|；④y=arcsinx．其中图象关于y轴对称的函数的序号是（）A．①②B．②③C．①③D．②④【解答】解：①y=log2x的定义域为（0，+∞），定义域关于原点不对称，则函数为非奇非偶函数；②y=x2；是偶函数，图象关于y轴对称，满足条件．③y=2|x|是偶函数，图象关于y轴对称，满足条件．④y=arcsinx是奇函数，图象关于y轴不对称，不满足条件，故选：B．15．（5分）“t≥0”是“函数f（x）=x2+tx﹣t在（﹣∞，+∞）内存在零点”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【解答】解：t≥0⇒△=t2+4t≥0⇒函数f（x）=x2+tx﹣t在（﹣∞，+∞）内存在零点，函数f（x）=x2+tx﹣t在（﹣∞，+∞）内存在零点⇒△=t2+4t≥0⇒t≥0或t≤﹣4．∴“t≥0”是“函数f（x）=x2+tx﹣t在（﹣∞，+∞）内存在零点”的充分非必要条件．故选：A．16．（5分）设A、B、C、D是半径为1的球面上的四个不同点，且满足•=0，•=0，•=0，用S1、S2、S3分别表示△ABC、△ACD、△ABD的面积，则S1+S2+S3的最大值是（）A．B．2 C．4 D．8【解答】解：设AB=a，AC=b，AD=c，因为AB，AC，AD两两互相垂直，扩展为长方体，它的对角线为球的直径，所以a2+b2+c2=4R2=4所以S△ABC +S△ACD+S△ADB=（ab+ac+bc ）≤（a2+b2+c2）=2即最大值为：2故选：B．三.解答题（本大题共5题，共14+14+14+16+18=76分）17．（14分）如图所示，用总长为定值l的篱笆围成长方形的场地，以墙为一边，并用平行于一边的篱笆隔开．（1）设场地面积为y，垂直于墙的边长为x，试用解析式将y表示成x的函数，并确定这个函数的定义域；（2）怎样围才能使得场地的面积最大？最大面积是多少？【解答】解：（1）设场地面积为y，垂直于墙的边长为x，它的面积y=x（l﹣3x）；由x＞0，且l﹣3x＞0，可得函数的定义域为（0，l）；（2）y=x（l﹣3x）=×3x（1﹣3x）≤×（）2=，当x=时，这块长方形场地的面积最大，这时的长为l﹣3x=l，最大面积为．18．（14分）如图，已知圆锥的侧面积为15π，底面半径OA和OB互相垂直，且OA=3，P是母线BS的中点．（1）求圆锥的体积；（2）求异面直线SO与PA所成角的大小．（结果用反三角函数值表示）【解答】（本题满分（14分），第1小题满分（7分），第2小题满分7分）解：（1）由题意，π•OA•SB=15π，解得BS=5，…（2分）故…（4分）从而体积．…（7分）（2）如图，取OB中点H，连结PH、AH．由P是SB的中点知PH∥SO，则∠APH（或其补角）就是异面直线SO与PA所成角．…（10分）∵SO⊥平面OAB，∴PH⊥平面OAB，∴PH⊥AH．在△OAH中，由OA⊥OB，得，…（11分）在Rt△APH中，∠AHP=90 O，，…（12分）则，∴异面直线SO与PA所成角的大小．…（14分）19．（14分）已知函数的定义域为集合A，集合B=（a，a+1），且B ⊆A．（1）求实数a的取值范围；（2）求证：函数f（x）是奇函数但不是偶函数．【解答】解：（1）令，解得﹣1＜x＜1，所以A=（﹣1，1），因为B⊆A，所以，解得﹣1≤a≤0，即实数a的取值范围是[﹣1，0]；（2）证明：函数f（x）的定义域A=（﹣1，1），定义域关于原点对称，f（﹣x）=ln=ln（）﹣1=﹣ln=﹣f（x），而，，所以，所以函数f（x）是奇函数但不是偶函数．20．（16分）设直线l与抛物线Ω：y2=4x相交于不同两点A、B，O为坐标原点．（1）求抛物线Ω的焦点到准线的距离；（2）若直线l又与圆C：（x﹣5）2+y2=16相切于点M，且M为线段AB的中点，求直线l的方程；（3）若，点Q在线段AB上，满足OQ⊥AB，求点Q的轨迹方程．【解答】解：（1）根据题意，抛物线Ω的方程为y2=4x，则p=2，故抛物线Ω的焦点到准线的距离为2；（2）设直线l：x=my+b当m=0时，x=1和x=9符合题意；当m≠0时，A（x1，y1）、B（x2，y2）的坐标满足方程组，所以y2﹣4my﹣4b=0的两根为y1、y2．△=16（m2+b）＞0，y1+y2=4m，所以，所以线段AB的中点M（2m2+b，2m）因为k AB•k CM=﹣1，，所以，得b=3﹣2m2所以△=16（m2+b）=16（3﹣m2）＞0，得0＜m2＜3因为，所以m2=3（舍去）综上所述，直线l的方程为：x=1，x=9（3）设直线AB：x=my+b，A（x1，y1）、B（x2，y2）的坐标满足方程组，所以y2﹣4my﹣4b=0的两根为y1、y2△=16（m2+b）＞0，y1+y2=4m，y1y2=﹣4b所以，得b=0或b=4b=0时，直线AB过原点，所以Q（0，0）；b=4时，直线AB过定点P（4，0）设Q（x，y），因为OQ⊥AB，所以（x≠0），综上，点Q的轨迹方程为x2﹣4x+y2=021．（18分）若数列A：a1，a2，…，a n（n≥3）中（1≤i≤n）且对任意的2≤k≤n﹣1，a k+1+a k﹣1＞2a k恒成立，则称数列A为“U﹣数列”．（1）若数列1，x，y，7为“U﹣数列”，写出所有可能的x、y；（2）若“U﹣数列”A：a1，a2，…，a n中，a1=1，a n=2017，求n的最大值；（3）设n为给定的偶数，对所有可能的“U﹣数列”A：a1，a2，…，，记，其中max{x1，x2，…，x s}表示x1，x2，…，x s这s 个数中最大的数，求M的最小值．【解答】解：（1）x=1时，，所以y=2或3；x=2时，，所以y=4；x≥3时，，无整数解；所以所有可能的x，y为，或．（2）n的最大值为65，理由如下：一方面，注意到：a k+1+a k﹣1＞2a k⇔a k+1﹣a k＞a k﹣a k﹣1．对任意的1≤i≤n﹣1，令b i=a i+1﹣a i，则b i∈Z且b k＞b k﹣1（2≤k≤n﹣1），故b k≥b k﹣1+1对任意的2≤k≤n﹣1恒成立．（*）当a1=1，a n=2017时，注意到b1=a2﹣a1≥1﹣1=0，得（2≤i≤n﹣1）即b i≥i﹣1，此时a n﹣a1=（a n﹣a n﹣1）+（a n﹣1﹣a n﹣2）+…+（a2﹣a1）=b n﹣1+b n﹣2+…+b1≥0+1+2+…+（n﹣2）=，（**）即，解得：﹣62≤n≤65，故n≤65．另一方面，为使（**）取到等号，所以取b i=i﹣1（1≤i≤64），则对任意的2≤k≤64，b k＞b k﹣1，故数列{a n}为“U﹣数列”，此时由（**）式得，所以a65=2017，即n=65符合题意．综上，n的最大值为65．（3）M 的最小值为，证明如下：当n 0=2m （m ≥2，m ∈N *）时，一方面：由（*）式，b k +1﹣b k ≥1，b m +k ﹣b k =（b m +k ﹣b m +k ﹣1）+（b m +k ﹣1﹣b m +k ﹣2）+…+（b k +1﹣b k ）≥m ．此时有：（a 1+a 2m ）﹣（a m +a m +1）=（a 2m ﹣a m +1）﹣（a m ﹣a 1） =（b m +1+b m +2+…+b 2m ﹣1）﹣（b 1+b 2+…+b m ﹣1） =（b m +1﹣b 1）+（b m +2﹣b 2）+…+（b 2m +1﹣b m ﹣1） ≥m +m +…+m=m （m ﹣1）．即（a 1+a 2m ）≥（a m +a m +1）+m （m ﹣1）故因为，所以，另一方面，当b 1=1﹣m ，b 2=2﹣m ，…，b m ﹣1=﹣1，b m =0，b m +1=1，b 2m ﹣1=m ﹣1时，a k +1+a k ﹣1﹣2a k =（a k +1﹣a k ）﹣（a k ﹣a k ﹣1）=b k ﹣b k ﹣1=1＞0 取a m =1，则a m +1=1，a 1＞a 2＞a 3＞…＞a m ，a m +1＜a m +2＜…＜a 2m ，且此时．综上，M 的最小值为．。

上海市杨浦区2018年高三物理一模试卷

杨浦区2017学年度第一学期高中等级考模拟质量调研高三年级物理学科试卷一、选择题（共40分。

第1-8小题，每小题3分，第9-12小题，每小题4分。

每小题只有一个正确答案。

） 1．下列用来定量描述磁场强弱和方向的是（）（A ）磁感应强度（B ）磁通量（C ）安培力（D ）磁感线2．20世纪中叶以后，移动电话快速发展。

移动电话机（）（A ）既能发射电磁波，也能接收电磁波（B ）只能发射电磁波，不能接收电磁波（C ）不能发射电磁波，只能接收电磁波（D ）既不能发射电磁波，也不能接收电磁波3．我国《道路交通安全法》中规定：各种小型车辆前排乘坐的人（包括司机）必须系好安全带．这是因为（）（A ）系好安全带可以减小惯性（B ）系好安全带可以防止因人的惯性而造成的伤害（C ）系好安全带可以防止因车的惯性而造成的伤害（D ）是否系好安全带对人和车的惯性没有影响4．电源电动势反映了电源把其它形式的能转化为电能的本领，下列关于电动势的说法中正确的是（）（A ）电动势是一种非静电力（B ）电动势越大表明电源储存的电能越多（C ）电动势就是闭合电路中电源两端的电压（D ）电动势由电源中非静电力的特性决定，跟其体积、外电路无关5．如图所示，在两块相同的竖直木板之间，有质量均为m 的4块相同的砖，用两个大小均为F 的水平力压木板，使砖块静止不动，则第2块砖对第3块砖的摩擦力大小是（）（A ）0（B ）mg（C ）12mg（D ）2mg6．意大利科学家伽利略在研究物体变速运动规律时，做了著名的“斜面实验”，他测量了铜球在较小倾角斜面上的运动情况，发现铜球做的是匀变速直线运动，且铜球加速度随斜面倾角的增大而增大，于是他对大倾角情况进行了合理的外推，由此得出的结论是（）（A ）力不是维持物体运动的原因（B ）力是使物体产生加速度的原因（C ）自由落体运动是一种匀变速直线运动（D ）物体都具有保持原来运动状态的属性，即惯性7．如图所示，虚线a 、b 、c 是电场中的一簇等势线（相邻等势面之间的电势差相等），实线为一α粒子（重力不计）仅在电场力作用下通过该区域时的运动轨迹，P 、Q 是这条轨迹上的两点，据此可知（）（A ）a 、b 、c 三个等势面中，a 的电势最高（B ）电子在P 点具有的电势能比在Q 点具有的电势能小（C ）α粒子在P 点的加速度比Q 点的加速度大（D ）带电质点一定是从P 点向Q 点运动8．在如图所示电路中，合上开关S ，将滑动变阻器R 2的滑动触点向b 端移动，则三个电表A 1、A 2和V 的示数I 1、I 2和U 的变化情况是（）（A ）I 1增大，I 2不变，U 增大（B ）I 1减小，I 2不变，U 减小（C ）I 1增大，I 2减小，U 增大（D ）I 1减小，I 2增大，U 减小9．一物体沿竖直方向运动，以竖直向上为正方向，其运动的v ﹣t 图象如图所示．对其运动过程正确描述的是（）（A ）0～t 1时间内物体处于失重状态（B ）t 1～t 2时间内物体机械能守恒（C ）t 2～t 3时间内物体向下运动（D ）0～t 2时间内物体机械能一直增大10．利用引力常量G 和下列某一组数据，不能计算出地球质量的是（）（A ）地球的半径及地球表面附近的重力加速度（不考虑地球自转的影响）（B ）人造卫星在地面附近绕地球做圆周运动的速度及周期（C ）月球绕地球做圆周运动的周期及月球与地球间的距离（D ）地球绕太阳做圆周运动的周期及地球与太阳间的距离11．如图所示，电源E ，导线，导电细软绳ab 、cd ，以及导体棒bc 构成闭合回路，导电细软绳ab 、cd 的a 端和d 端固定不动，加上恰当的磁场后，当导体棒保持静止时，闭合回路中abcd 所在平面与过ad 的竖直平面成30°，已知ad 和bc 等长且都在水平面内，导体棒bc 中的电流I =2A ，导体棒的长度L =0.5m ，导体棒的质量m =0.5kg ，g 取10m/s 2，关于磁场的最小值和方向，说法正确的是（）（A ）533 T ，竖直向上（B ）533 T ，竖直向下（C ）2.5T ，由b 指向a（D ）2.5T ，由a 指向b12．如图，一辆运送沙子的自卸卡车装满沙子，沙粒之间的动摩擦因数为μ1，沙子与车厢底部材料的动摩擦因数为μ2，车厢的倾角用θ表示（已知μ2＞μ1），下列说法正确的是（）（A）要顺利地卸干净全部沙子，应满足tanθ＜μ2（B）要顺利地卸干净全部沙子，应满足sinθ＞μ2（C）只卸去部分沙子，车上还留有一部分沙子，应满足μ2＞tanθ＞μ1（D）只卸去部分沙子，车上还留有一部分沙子，应满足μ2＞μ1＞tanθ二、填空题（共20分）13．习惯上规定____电荷定向移动的方向为电流的方向（填“正”或“负”）。

杨浦区2018届高三一模英语卷(含答案)

杨浦区2017学年度第一学期高三模拟质调研英语学科试卷2017.12本试卷分为第I卷（第1-11页）和第II卷（第12页）两部分。

全卷共12页。

满分140分。

考试时间120分钟。

考生注意：1．答第I卷前，考生务必将条形码粘贴在答题纸的指定区域内。

2.第I卷（1-20小题，31---70小题）由机器阅卷，答案必须全部涂写在答题卡上。

考生应将代表正确答案的小方格用铅笔涂黑。

注意试题题号和答题卡编号一一对应，不能错位。

答案需要更改时，必须将原选项用橡皮擦去，重新选择。

答案写在试卷上一律不给分。

第I卷中的第21-30小题，IV.Summary Writing部分和第II卷的试题，其答案用钢笔或水笔写在答题纸的规定区域内，如用铅笔答题，或写在试卷上则无效。

2018年上海市杨浦区高三一模语文试题(解析版)

2018年上海市杨浦区高三上学期期末（一模）语文试题1.根据提示填空（1）________________，万钟于我何加焉？（孟子《鱼我所欲也》）。

（2）________________，万方多难起登临。

（杜甫《__________》）。

（3）人类的物质文明无论多么奢华兼顾，都逃不过时间的淘洗，李白在《登金陵凤凰台》中的一联“________________，________________”对此发出感慨。

【答案】(1). 万钟则不辩礼义而受之(2). 花近高楼伤客心(3). 登楼(4). 吴宫花草埋幽径(5). 晋代衣冠成古丘【解析】【详解】此题考查学生对文学常识和名句名篇的识记能力。

在默写名句名篇时，要做到“三清”“三不”：“三清”就是卷面清洁，字迹清楚，笔画清晰；“三不”就是不添字，不漏字，不误写。

高考所选择的名句大多出自课内所学篇目，只有少数来自课外，该如何识记这些课内的名句，需要一定的技巧。

高考所考的名句，字不一定有多复杂，但一定很有迷惑性，考生经常在这些“浅水滩”“翻船”，原因是光背不写，光记不辨，不知道联系句意和古文的特性来区分。

所以在平时对名句记忆时要立足于对诗句的理解，在理解的基础上记忆既准确还不容易写错字，然后要立足于“写”，不要背下来但得不到分数。

本题要注意如下字词：钟、辩、礼义、幽径、丘。

2.现代人做事讲求有法可依，有根有据，以下表述对应的一项是（）A. 导之以德，齐之以礼，有耻且格天行有常，不为尧存，不为桀亡。

B. 没有规矩，不成方圆事不目见，耳闻而臆断，其有无可乎？C. 礼禁未然之前，法施已然之后道听而途说，德之弃也。

D. 国不可无法，有法而不善与无法等君子不以言举人，不以言废人【答案】B【解析】【详解】本题考查学生在提供的新情境中使用名篇、名句的能力。

解答此类试题，首先要审清楚题干的要求，如本题“现代人做事讲求有法可依，有根有据，以下表述对应的一项”，然后理解每个选项的意思，再选出符合题干要求的选项。

2018-2019学年杨浦区高三年级英语一模测试卷（含答案）

杨浦区2018学年度第一学期高三模拟质量调研英语学科试卷2018.12考生注意:1.考试时间120分钟,试卷满分140分；2.本次考试设试卷和答题纸两部分.所有答题必须涂(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务必在答题纸上填写准考证号和姓名,井将核对后的条形码贴在指定位置上。

I.Listening ComprehensionSectionADirections:In Section A,you will hear ten short conversations between two speakers.At the end of e ach conversation,a question will be asked about what was said.Theconversations and the questions will be spoken only once, After you hear aconversation and the question about it,read the four poss ible answers on your paper.and decide which one is the best answer to the question you have heard.1. A.Job applicants. B.University students.C.News reporters.D.Public speakers.2. A.$36. B. $60.C. $24.D.$303. A.Nervous. B.Exc ited.C.Surprised.D.Calm.4. A.They can hardly find time to do exercise.B.Both of them are fond of doing exercise.C.Doing exercise is the last thing they'd like to do.D.The woman disagrees to do exercise with the man.5.A.Student and teacher. B Guest and receptionist.C.Customer and shop assistant.D.Daughter and father.6. A.The Student Union. B.The Tutoring Service Center.C.Her professor's office.D.Her tutor's home.7. A.The coverage of newspapers. B.The nature of humans.C.The increase of crime rate.D.The impact of mass media.高三英语第1页共19页8. A.Doctor Green is very busy on Mondays.B.Doctor Green forgot to put the man on the schedule.C.An unexpected patient visited Doctor Green last Monday.D.Patients can usually see Doctor Green on schedule.9. A.She is disappointed with his decision.B.She is crazy about going camping in the tropical jungle.C.She tries to persuade the man not to go with Jerry.D.She suggests the man reconsider his plan.10. A.Because he favors goods in physical stores.B.Because he enjoys the offline shopping experience.C.Because he doesn't have much money.D.Because he doesn't like the shopping festival.Section BDirection:In Section B, you will hear two short passages and one longer conversation, and you wil be asked several 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. A number of giant rats are hidden in the fields or forests.B. V iolent robbers may come out and hurt innocent civilians.C. People may step onto a landmine(地雷) and get injured.D. Other roads have a litter problem, which spoils good mood.12. A. It is a rare species of rat.B. It has a good sense of smell.C. It sacrifices its life to detect dangers.D. It is terrified of explosive device.高三英语第2页共19页13.A. 13 minutes B. 30 minutesC. 40 minutesD. 4 daysQuestions 14 through 16 are based on the following passage.14.A. It bans the sales of salty and unhealthy food.B. Its residents lose altogether 100,000kg by 2020.C. Everyone weighs himself on giant scales in public.D. Local doctors make personalized diets for everyone15. A. Local people share healthy diet with each other.B. Thousands of residents kick football to keep slim.C. Local restaurants use olive oil to replace butter.D. Government officials persevere in physical exercise.16. A. Students are encouraged to walk or ride to schoolB. Students are allowed to ride bicycles in school libraries.C. Parents are offered devices to monitor children's walking time.D. Fruits are provided for those who are addicted to online games.Questions 17 through 20 are based on the following passage.17. A. Neither of them can find a suitable game.B. They are both enthusiastic about games.C. Good games cost a large amount of money.D. They often go Dutch when buying gifts.18. A. It includes fantasy role-playing.高三英语第3页共19页B. It doesn't have an actual ending.C. It is an expensive online game.D. It is something like a racing game.19. A. Players have opportunities to make new friends.B. Players can have face-to-face communication.C. Players may be rewarded with delic ious food.D. Players can start the game at any time they want.20. A. Add variety to the recommended game.B. Look around to find a cheaper online game.C. Persuade the woman to change her mind.D. Try to find an extraordinary online game.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 with the proper form of the given word; for the other blanks, use one word that best fits each blank.A newly discovered tea plant is caffeine-freeThe world loves tea and some 3m tons of tea are consumed every year. Tea can be good for health, as it contains compounds that help to lower cholesterol（胆固醇）and reduce the risk of heart disease. But there is a downside. Tea contains caffeine which, (21) __________ it improves mental alertness, can also cause anxiety, insomnia and other problems.(22) __________ would be agreeable is that a tea plant that provides all the taste and goodness but with little or (23) __________ of the caffeine has been found. Liang Chen and Ji-Qiang Jin of高三英语第4页共19页the Tea Research Institute of the Chinese Academy of Agricultural Sciences have discovered just such a plant (24) __________ (grow) wild in a remote area in Fujian province, southern China. (25) __________ they report in the Journal of Agricultural and Food Chemistry, not only is the tea plantnaturally caffeine-free but it also contains a number of unique medicinal compounds that, the locals believe, offer considerable health benefits.(26) __________ (know)locally as Hongyacha, the newly discovered plant grows only between 700 and 1,000 metres (27) __________ sea level around a handful of Chinese villages.Now the researchers (28) __________ (explore) methods to protect Hongyacha in its natural habitat while further studies are carried out. It can take time – and sometimes it does not work – for new plant varieties (29) __________ (breed) for commercial use. A pair of naturally caffeine-free coffee plants were discovered in 2003, but little progress (30) __________ (report) Tea enthusiasts will be watching Hongyacha with interest. And others will wonder what else is out there.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.Dealing with Difficult RelationshipsEveryone has at least one awkward or ___31___ relationship. It may be with somebody who will ___32___ your energy whenever you are with them. Or worse, it could be someone who always cuts you down. This person may be a family member or even a friend. No matter who it is, it‟s necessary that you learn to set boundaries for yourself. Otherwise this kind of relationship can chip away at your self-esteem.Setting boundaries for difficult relationships starts by ___33___ how you are affected by the relations hip. Do they bring you closer to your goals or pull you farther away? For example, it‟s time to study for tomorrow‟s test. But your friend wants to take you to a party. Here, setting boundaries will help protect your ___34___ goals.高三英语第5页共19页Next, decide how much time you should spend with these people. It‟s easy to overcommit yourself. But it‟s difficult to help others if you forget to protect your own ___35___.How do you know if a relationship is unhealthy, and it‟s time to set boundaries? Here are a few practical questions to ask yourself.1.How does this relationship affect me?Every ___36___ can affect you positively or negatively. For example, someone whopressuresyou to something you‟re not comfortable doing will ____37___ you out. But a friend who c o ns id ers h ow yo u f e e l w ill r esp ec t y o ur ___38___t o try s om et h ing new.2.Why am I in this relationship in the first place?P eople may try to keep you in an unhealthy relationship. By ___39___ you it‟s yourobligation or duty, you forget about your own needs. Sadly, by remaining ___40___ to these people, you forget who you are. Y ou allow them to take advantage of you or even belittle you.Settling boundaries requires taking a long, honest look at yourself. By saying “no” to harmful patterns in relationships, you say “yes” to a healthier you.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.According to German novelist Heinrich Mann (1871-1950), a house without books is like a room without windows. Mann ___41___ pointed out the value of books, but some researchers have found evidence that people with books in their homes really do gain a window on the world.A new study in the journal Social Science Research, publishing in October, suggests that people who grow up with more books are more likely to have educational ___42___, and to achieve more in life, than people who grow up without them. In the study, over 160,000 adultsfrom 31 countries and regions were asked about the number of books there were in their homes when高三英语第6页共19页they were 16 years old. Then they were_ 43_ in literacy, numeracy (计算能力) and information communication technology.The research shows that the number of books for each household 44_ greatlyfrom country to country.45 _, the number of books is27 in Turkey, 143 in the UK and 218 in Estonia. But "the total46_ of home library size on literacy are largeeverywhere” , according to researcher Joann Sikorn of AustralianNational Universityina paper based on the study.The researcher found that people who had lower levels of secondary educationbut had a large number of books at home got a(n)_47_ score as universitygraduates who grew up with only a few books.The Guardian newspaper commented that “bookish adolescence makes for a good deal of educational advantage.""Reading books in a young age can_ 48 shortcomings not only in adultliteracy but also numeracy; its impacts are __49 to additional years of education."Sikora told Science Alert.Apart from the educational benefits, growing up with more books also plays animportant role in adult success."Early_50_ to books in the parental home matters because books are anessential part of routines and practices that 5l_ lifelong cognitive (认知的)abilities", Sikora told Science Alert. These abilities are_ 52_ to future development.Without doubt, the fact that we are moving toward a digital era could_ 53_ theimportance of printed books. For now, 54_, "they still seem to_ 55_ quite alarge positive benefit, which shows no sign of decreasing", researchers wrote in thepaper.41. A. simply B. definitely C. generally D. randomly42 A. backgrounds B. advantages C. requirements D. development43. A. estimated B. examined C. tested D. investigated44 A. changes B. alters C. shifts D. varies45 A By comparison B. As a result C. For instance D. In total高三英语第7页共19页46. A. effects B. expenses C. inputs D. contributions47. A. higher B. lower C. average D. similar48. A. make up for B break up with C. liveup to D. catch up with49 A. senior B. essential C. equal D. familiar50. A. addition B. engagementC. devotion D. exposure51. A. prove B. demonstrate C. enhance D. acquire52. A. relevant B. crucial C. opposite D. compulsory53. A. weaken B. highlight C. overestimate D. reflect54. A. therefore B. however C otherwise D. likewise55. A. enjoy B. obtain C. maintain D. reduceSection BDirections:Read the following three passages. Each passage is followed by severalquestions 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 thepassage you have just read.(A)"Don't tell anyone". We hear these words when someone tells a secret to us. But it can be hard to keep a secret. We're often tempted to “spill the beans”, even if weregret it later According to Asim Shah, professor in the Menninger Department of Psychiatryand Behavioral Sciences at Baylor College of Medicine, US, keeping a secret may well “become a burden". This is because people often have an "obsessive (无法摆脱的)and anxious urge to share it with someone".An earlier study, led by Anita E. Kelly, a scientist at the University of NotreDame, US, suggested that keeping a secret could cause stress. People entrusted with secrets can suffer from depression, anxiety, and body aches, reported the Daily Mail.高三英语第8页共19页But with secrets so often getting out, why do people share them at all? Shahexplained that people often feel that it will help them keep a person as a friend.Another reason people share secrets is guilt over keeping it from someone close tothem. A sense of distrust can develop when people who are close do not share it with each other. "Keeping or sharing secrets often puts people in a position of eithergaining or losing the trust of someone." according to Shah.He added that talkative people could let secrets slip out. But this doesn't meanthat it is a good idea only to share secrets with quiet people. A quiet person may besomeone who keeps everything inside. To tell such a person a secret may cause themstress, and make them talk about the secret.Shah said that to judge whether to tell someone a secret, you'd better put yourselfin their position. Think about how you would feel to be told that you mustn't give the information away. Shah also recommended that if you accidentally give up someone'ssecret you should come clean about it. Let the person know that their secret isn't so secret anymore.56. The underlined phrase "spill the bean" is closest in meaning to________.A. let out a secretB. tell the truthC. eat our wordsD. take gossip57. A person who is entrusted with a secret ________.A. is oft often obsessed with a certain threatB. often would rather not have been told about itC. usually has a strong desire to keep it until deathD. may suffer from physical or psychological problems58. What can we learn from the passage?A. Quiet people are more likely to keep the secrets to themselves.B. Sharing secrets helps establish friendship or get over the sense of guilt.C. It's better to inform others of the chances of your leaking secrets beforehand.D. Putting yourself in others' shoes helps realize the difficulty of keeping secrets.高三英语第9页共19页59. What‟s the best title for the passage?A. Walls have ears.B. Still tongues make wise heads.C. A word spoken is past recalling.D. Every wall has a crack.(B)Does the thought of cooking an entire Christmas dinner for friends and family make you shudder? Our cookery writer Sophie Godwin has created the ultimate simplified menu for a stress-free day. With a range of time-saving techniques and showstopping flavour twists, this is the only festive menu you'll ever need. Embrace the art of the shortcut!Y our timeplan to eat at 2pmView and print our at-a-glance infographic time plan.·11am Take your turkey out of the fridge.·11.30am Heat the oven.Prepare the rosemary butter and rub underneath the turkey skin.·12pm Put the turkey in the oven to roast.·12.15pm Boil the potatoes and root vegetables. Make the harissa & orange marmalade glaze.Chop the ingredients for the stuffing sprouts.·1pm Baste the turkey, pour in the white wine and nestle cabbage wedges around the bird.Return to the oven.高三英语第10页共19页·1.10pm Prepare the separate ingredients for the mackerel starter.·1.25pm Smash the roasties and put them in the oven.·1.30pm Check the turkey—take out to rest if the internal temperature has reached 65C.Return the cabbage to the oven to finish cooking.·1.35pm Toss the roasted roots in the harissa glaze,then put in the oven.·1.40 pm Make the gravy, cook the sprouts and keep warm.·2 pm Sit down to the starter, with everything for your lunch ready to go.·The d essert takes just 10 minutes to assemble ,so you can do it once you‟ve enjoyed the ma in course.Buy your ingredientsWe‟ve pre-loaded your shopping list so that you can quickly buy all your ingredients for this menu in one go. Just click below to be taken to your shopping list, where you can buy all your ingredients from your preferred supermarket, and remove any ingredients as needed, and even print the list, to use direct when shopping.(Serves 6)·Add to shopping ListThe above shopping list is powered by Whisk. Please check your list thoroughly before purchasing. If the same items appear across different recipes they will not automatically combine in your list and you‟ll need to do this manually. Please als o check the products that have been selected via the Whisk program, as you may prefer a different brand from that listed.60.Which section CAN‟T we find on the website?A. Classic cocktailsB. V egetarian Christmas高三英语第11页共19页C. Roast calculatorD. Wealth management61.How long does it take to toast the turkey?A.1 hourB.2 hoursC.1.5 hoursD.2.5 hours62.Which of the following is True?A. The entire Christmas dinner is composed of a roasted turkey and dessert.B. The automatically-generated shopping list is just for user‟s reference.C. Most of the ingredients on the shopping list come from the same brand.D. Y ou can buy all your ingredients online through Whisky by one click.（C）Y ou‟re walking down a quiet street and suddenly you hear some footsteps .Undoubtedly ,it means that there‟s someone around. But have you ever wonder why it occurs to us that it‟s someone else‟s footsteps , not ours?According to a new study published in the journal Nature in September, this phenomenon arises from a function in our brain to ignore the noise we make ourselves.In order to explore how our brain does this, a group of scientists carried out an experiment with mice at Duke University. The research entered on an intuition(直觉)---that we are usually unaware of the sound of our own footsteps ---as a vehicle for understanding larger neural(神经系统的) phenomena; how this behavior reveals the ability to monitor, recognize, and remember the sound of one‟s own movements in r elation to those of their larger environments.In the experiment, research controlled the sounds of a group of mice could hear, reported Science Daily. During the first several days, the mice would hear the same sound each time they took a step. This was just like “running on a tiny piano with each key playing exactly the same note”, sen ior study author Richard Mooney, a professor of neurobiology at Duke University, told Live Science.Scientists found that their auditory cortex (听觉皮层) – the area of the brain that processes sound – became active at first but decreased its response to the sound after two or three minutes高三英语第12页共19页when the mice became familiar with it.“ It‟s almost like they were wearing special headphones that could filter(过滤)out the sound of their own movements.” David Schneider, an assistant professor at the Center for Neutral Sci ence at New Y ork University, told HuffPost.But once the sound changed, their auditory cortex became active again. This suggests that the “sensory filter” in a mouse‟s brain could help it detect new sounds or abnormal noise in the environment easily after tuning out familiar sounds.“For mice, this is really important,” said Schneider. “They are prey animals, so they really need to be able to listen for a cat creeping up on them, even when they‟re walking and making noise.Being able to ignore the sounds of one's own movements is likely important for humans as well. But the ability to predict the sounds of our own actions is also important for more complex human behaviors such as speaking or playing an instrument."When we learn to speak or to play music, we predict what sounds we are going to hear – such as when we prepare to strike keys on a piano–and we compare this to what we actually hear, "explains Schneider. “We use mismatches between expectation and experience to change how we play – and we get bett er over time because our brain is trying to minimize these errors.”63. What can be discovered about mice in the experiment?A. Their brain responds inactively to the familiar soundsB. They are able to detect sounds other animals don‟t notice.C. They cannot identify different sounds except their own footsteps.D. Different areas of their brain are responsible for different sounds.64. What‟s the function of the sensory filter?A. Getting used to abnormal or unfamiliar sounds.B. Ignoring the sounds made by our companions.C. Identifying the sounds from a larger environment高三英语第13页共19页D. Being sensitive to the sounds of our own movement.65. Why can a good symphony conductor immediately recognize it when a wrong note is played?A. He has the ability to match the wrong note with the instrument player.B. He has an intuition that he should ignore the sound of his own movement.C. He has a low expectation and knows where players are likely to make errors.D. He has a good prediction of how each note should be played in the orchestra.66. What can be inferred from thepassage?A. Noise-filtering ability ensures us a quiet and undisturbed environment.B. The ability to ignore familiar noises helps to detect potential dangers.C. The activeness of auditory cortex determines our activity performance.D. Sound-predicting ability seems not so important for humans as for animals.Section CDirections:Read the passage carefully. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.高三英语第14页共19页Building a lasting social relationshipWe all know that friends are special people who we share our lives with, and who share their lives with us in return. But seeking friends and keeping the friendship going are never easy.According to research recently published in the Journal of Social and Personal Relationships, the key is to use "we-talk".Led by University of California psychologist Megan Robins and her colleagues, the researchers reviewed and analyzed 30 different studies involving over 5, 000participants.________67________.The word "we" moves people from an individual position into a partnership, which makes us more interdependent."__________68____________.Word use is a window into what people are thinking and feeling without asking them." Robbins told Science Daily.___________69___________. The primary point is that interdependence may bring about supportive and relationship-centered behaviors and positive perceptions of the partner--especially important in times of stress and disagreement.Contrary to "we-talk", there is "I-talk", which refers to the frequent use of the first-person singular pronouns, such as "I", "me" and "mine", when writing or speaking. Earlier this year, researcher analyzed a set of data that came from 4,7000 people in Germany and the US.___________70___________. As you can see from the two studies, too much "I-talk" can make you feel depressed. But "we-talk" can encourage you to become more positive and create a chain effect of healthy interdependence with others.So next time you are talking to a friend, try using more "we-talk". Y ou may find yourself feeling more positive--and the effect it will have on your friend will be positive as well.高三英语第15页共19页IV. Summary WritingDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.When we wake up feeling sleepy and with dark circles under eyes, many of us of ten think:” I‟m going to go to bed earlier tomorrow!” But however determined we are, chances are that we don‟t stick to our promise. This behavior is called “bedtime procrastination（拖延症）”.According to a study carried out by a group of health psychologists at Utrecht University in the Netherlands in 2014, bedtime procrastination is a common problem. In their study, 53 percent of the 2,400 participants said they didn‟t follow their sleep schedule, delaying it at least twice a week.It‟s found that they delayed bedtime not because they liked to stay up late, but they couldn‟t stop doing other unimportant things, which were keeping them up in the first place.“People who generally have trouble resisting temptations and sticking to their intentions are more like ly to delay going to bed.” Floor Kroese, a psychologist at Utrecht University, further explained to HuffPost.Y et, according to scientists, lack of self-control is not the only thing to blame. Our body clock also plays an important role when it comes to bed time. In order to check the influence of the body clock on sleep, psychologists at Ulm University in Germany studied the sleep patterns of 108 people in February. It was found that those who get up late are more likely to postpone their bedtime than those who wake early every morning.“The intention to go to bed earlier is not enough,” Jane Kuhnel, a psychologist at Ulm University, concluded in Popular Science. “Biological processes need to support this intention.”People always think that bedtime proc rastination isn‟t a big problem, but as Kroese told Popular Science, “the choices we make that affect out sleep could turn out to be pretty important for our health.”V. T ranslationDirections:Translate the following sentences into English, using the words given in the brackets.高三英语第16页共19页72.无论幸福的定义如何变化，人们永远不会放弃追求幸福的权利。

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