【加17套高考模拟卷】广西柳州高级中学2020-2021学年高三2月线上月考数学试题含解析
广西柳州高级中学2020-2021学年高三二月模拟文综地理试题
4．A
5．B
【分析】
本题考查等压线图的综合判断。本题难度低，基础性试题，学生只要掌握等压线图上的风向判断和低压系统对天气状况的影响，结合画图分析更直观。
13．阅读材料，完成下列问题。
材料一峡湾是第四纪冰川侵蚀形成冰川谷，在冰期结束后，海平面上升而形成。下图为某地峡湾图片。
材料二下图为北美洲东北部的拉布拉多半岛，在第四纪冰期曾是北美洲大陆三大冰川中心之一，该半岛沿海地区峡湾众多。
（1）描述峡湾地形的特点。
（2）推测峡湾在世界上（拉布拉多半岛除外）的主要分布区。
材料三卤虫，是一种小型甲壳动物，生活在高盐度水域中，以藻类为食。是水产养殖的优质活体饵料，也是候鸟的食物来源。卤虫咋我国是一种稀缺资源。艾比湖是我国最大卤虫产地。
（1）赛里木湖是位于内陆地区、天山山间盆地的湖泊，但盐度低。简析赛里木湖的主要原因。
（2）说明艾比湖成为我国最大卤虫产地的自然原因。
（3）新疆引进高白鲑鱼投入到赛里木湖而没有投到艾比湖，说明其合理性。
广西柳州高级中学2020-2021学年高三二月模拟文综地理试题
学校:___________姓名：___________班级：___________考号：___________
一、单选题
下图为我国东南丘陵某城市郊区土地利用模式图。读图完成下面小题。
1．村庄选择在山坡上，最主要的目的是
A．便于取水，防御洪涝B．节省宝贵的耕地
公路建设会对野生动物造成哪些影响，公路建设最应该采取什么措施保护野生动物？
广西柳州市高级中学2020-2021学年高三上学期第二次统测理综物理试题
广西柳州市高级中学2020-2021学年高三上学期第二次统测理综物理试题学校:___________姓名：___________班级：___________考号：___________一、单选题1．以下原子物理知识中说法正确的是（）A．比结合能越大的原子，其结合能就越大，原子核越稳定B．根据波尔理论可知：一群原子从n=4能级向低能级跃迁最多能辐射6种频率的光子C．原子核发生衰变要遵守电荷守恒和质量守恒规律D．卢瑟福的α粒子散射实验证实了其提出的核式结构的正确性2．水平地面上有一粗糙斜面体，上端有一光滑轻滑轮，轻滑轮上的一根轻绳连接A、B 两物体，并处于静止状态，如图所示．现用一方向向左水平力F作用于物体B上，并缓慢拉开一小角度，此过程中斜面体与物体A仍然静止．（不计绳的质量和绳与滑轮间的摩擦）则下列说法正确的是A．物体A所受细绳的拉力可能不变B．在缓慢拉开B的过程中，水平力F不变C．物体A所受斜面体的摩擦力一定变小D．斜面体对物体A的作用力的合力可能变小3．有一个竖直固定放置的四分之一光滑圆弧轨道，轨道圆心O到地面的高度为5m，小球从轨道最高点A由静止开始沿着圆弧轨道滑下，从轨道最低点B离开轨道，然后做平抛运动落到水平地面上的C点，B点与C点的水平距离也等于5m，则下列说法正确的是（）A ．根据已知条件可以求出该四分之一圆弧轨道的轨道半径为1mB ．当小球运动到轨道最低点B 时，轨道对它的支持力等于重力的4倍C ．小球做平抛运动落到地面时的速度与水平方向夹角θ的正切值tan θ=1D ．小球在圆弧轨道上运动的过程中，重力对小球的冲量在数值上大于圆弧的支持力对小球的冲量4．如图所示，10匝矩形线圈，在磁感应强度为0.2T 的匀强磁场中，绕垂直磁场的轴OO ’以角速度100πrad/s 匀速转动，线框电阻不计，面积为0.5m 2，线框通过滑环与一理想变压器的原线圈相连，副线圈接有两只灯泡L 1和L 2．已知变压器原、副线圈的匝数比为10：1，开关断开时L 1正常发光，且电流表示数为0.01A ，则：A ．若开关S 闭合，灯泡L 1亮度不变，电流表示数将增大B ．闭合开关后电压表示数为C ．灯泡L 1的额定功率为3.14WD ．若从图示位置开始计时，线框中感应电动势的瞬时值为314sin100πt （V ）5．有一电场强度方向沿x 轴的电场，其电势ϕ随x 的分布满足0sin 0.5(V)x ϕϕπ=，如图所示。
【数学】广西柳州高级中学2020届高三2月线上月考 数学(理)(PDF版)
∴ M N {1, 0,1} ，则其子集的个数为 23 8 个．
2．已知复数
z
2
i
，则
z 1
i
在复平面上对应的点所在象限是（
）
A．第一象限 【答案】D
B．第二象限
C．第三象限
D．第四象限
【解析】∵
z
2i
z
，∴
1 i
2i 1 i
1 2
3 2
i ，在复平面对应的点的坐标为 ( 1 2
,
3) 2
，所
柳州高级中学2020届高三二月月考数学（理科）试题 一、选择题：本大题共 12 小题，每小题 5 分，在每小题给出的四个选项
中，只有一项是符合题目要求的．
1．已知集合 M {x | x2 x 2 0} ， N {1, 0,1, 2} ，则 M N 的子集个数为（ ）
A． 2
B． 4
C． 8
3
6
g(x) 1 sin x 3 cos x sin[2(x )] ，
2
2
6
故需将 g(x) 图像上所有点向左平移 个单位长度得到 f (x) ．
3
9． (1 2x y )8 的展开式中 x2 y2 项的系数是（ ） 2
A． 420
【答案】A
B． 420
C．1680
D． 1680
【解析】展开式中
B．②③④
【答案】D
【解析】由平行公理知①对，
由线面垂直的性质定理知②对，
由线面垂直及面面平行定理知③对，
由面面垂直性质定理知④对．
C．①②④
7．下图是一程序框图，若输入的 A 1 ，则输出的值为（ 2
D．①②③④ ）
2
2020-2021年广西柳州市某校高三(上)第一次模拟考试语文试卷(附答案)
2020-2021年广西柳州市某校高三（上）第一次模拟考试语文试卷一、现代文阅读1.阅读下面文本，完成下列各题。
以易解史，即是以易学思维来认识历史、评论历史，这是中国传统史学的一个显著特点。
班固作为封建正统史家的代表，他撰述《汉书》，“综其行事，旁贯《五经》”，自觉以儒家思想作指导。
而在儒家“六经”中，《周易》在班固心目中占有独特的地位。
他以“六经”为诸子之源，而视《周易》为“六经”之首，“《易》的尊崇地位的确立，班固是立了功的。
”正因此，《汉书》重视以易解史，成为汉代史学以易解史的重要代表。
《汉书》深受《周易》及汉易时代天人一体思维的影响，以易学思维为依据，以历史学的形式对天人关系做出了新的探讨，从中表达了对于社会和谐的向往与追求。
首先，“列人事而因以天时”。
《四库全书》说：“《易》之为书，推天道以明人事者也。
”《汉书》在天人关系上，明确认为人事需要顺应天道。
《律历志上》说：《易》金火相革之卦曰“汤武革命，顺乎天而应乎人”，又曰“治历明时”，所以和人道也。
班固明确认为，“列人事而因以天时”，这是孔子作《春秋》的旨趣，也符合《易》的精神。
这里所引“汤武革命，顺乎天而应乎人”和“治历明时”，分别出自《革卦》的《彖辞》与《象辞》，前者以汤武革命之事发论，肯定其乃顺天应人之举，所以取得成功；后者字面含义是整治历法以明四时之序，意为治理国事需要取象历法。
二者其实都是强调人事需要取法天道，也只有取法天道才能成功。
其次，“财成辅相天地之宜”。
人道仿效、顺从天道是促成人事的先决条件。
如何仿效、顺从天道？《汉书》以《易传》为依据，提出了“财成辅相天地之宜”的思想。
《汉书》的这一思想，集中见于《货殖传》的叙述，文中阐发了“育之以时，而用之有节”的思想，主张要顺应自然节气，养育积蓄万物，以足备功用。
其中“后以财成辅相天地之宜，以左右民”一语出自《泰卦·象辞》，这里“后”指圣人君主，“财”通“裁”，是讲如何调整和节制天下万物与社会需求之间的关系问题。
广西柳州高级中学2020┄2021届高三二月份模拟演练英语试题
第Ⅰ卷注意事项：1.答第I卷前，考生务必将自己的姓名、准考证号填写在答题卡相应的位置上。
2. 选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。
如需改动，用橡皮擦干净后，再选涂其他答案标号。
不能答在试题卷上，否则无效。
第一部分听力（共两节，满分30分）第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What is the time now？A. 8:50.B. 9:50.C. 10:00.2. What is Bobby doing now？A. Having dinner.B. Doing his exercise.C. Having his final test.3. What are the speakers going to do？A. Celebrate Mary’s 17th birthday.B. Celebrate Mary’s 18th birthday.C. Celebrate Mary’s 19th birthday.4. For what is the man going to Hawaii？A. On business.B. For his holiday.C. For warm weather.5. What is the probable relationship between the speakers？A. Teammates.B. Workmates.C. Old friends.第二节（共15小题；每小题1分，满分15分）听下面5段对话或独白。
每段对话或独白后有2至4个小题，从题中做给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听每段对话或独白前，你将有5秒钟的时间阅读各个小题；听完后，各小题将给出5秒钟的作答时间。
2020-2021学年广西柳州高三上物理月考试卷(1)
2020-2021学年广西柳州高三上物理月考试卷一、选择题1. 2019年“山东舰”正式服役，标志着我国进入双航母时代，假设，“山东舰”正在沿直线航行，其质量为m，发动机的输出功率恒为P，所受阻力大小恒为f，某时刻速度为v1、加速度为a1，一段时间t后速度变为v2(v2>v1)，在这段时间内位移为x．下列关系式正确的是（）A.f>Pv1B.x=v2+v12tC.Pt=12mv22−12mv12 D.a1=Pmv1−fm2. 对于灾难搜救，中国拥有海洋、风云、高分、遥感等4个型号近10颗卫星为地面搜救行动提供技术支持．假设“高分一号”卫星与同步卫星、月球绕地球运行的轨道都是圆，它们在空间的位置示意图如图所示．下列有关“高分一号”的说法正确的（）A.其绕地球运行的周期比同步卫星绕地球运行的周期大B.其发射速度可能小于7.9km/sC.其在运行轨道上处于完全失重状态，重力加速度为0D.其绕地球运行的角速度比月球绕地球运行的角速度大3. 如图所示，MN和PQ为竖直方向的两平行足够长的光滑金属导轨，间距为L，电阻不计．导轨所在平面与磁感应强度为B的匀强磁场垂直，两端分别接阻值为2R的电阻R1和电容为C的电容器．质量为m、电阻为R的金属杆ab始终垂直于导轨，并与其保持良好接触．杆ab由静止开始下滑，在下滑过程中最大的速度为v，整个电路消耗的最大电功率为P，则（）A.杆ab的最大速度v等于Pmg B.电容器右极板带负电C.杆ab所受安培力的最大功率为P3D.电容器的最大带电量为CBLv34. 如图所示，有一光滑轨道ABC，AB部分为半径为R的14圆弧，BC部分水平，质量均为m的小球a、b固定在竖直轻杆的两端，轻杆长为R，不计小球大小．开始时a球处在圆弧上端A点，由静止释放小球和轻杆，使其沿光滑轨道下滑，下列说法正确的是（）A.a、b滑到水平轨道上时速度为√2gRB.a球下滑过程中机械能保持不变C.从释放到a、b滑到水平轨道上，整个过程中轻杆对b球做的功为mgR2D.从释放到a、b滑到水平轨道上，整个过程中轻杆对a球做的功为mgR25. 如图所示，半径为R的光滑圆弧轨道ABC固定在竖直平面内，O是圆心，OC竖直，OA水平，B是最低点，A点紧靠一足够长的平台MN，D点位于A点正上方，DA距离为有限值．现于D点无初速度释放一个大小可以忽略的小球，在A点进入圆弧轨道，从C点飞出后做平抛运动并落在平台MN上的P点，不计空气阻力，下列说法正确的是（）A.小球在C点的动能大于A点的动能B.只要D点的高度合适，小球可以落在平台MN上任意一点C.如果DA距离为ℎ，则小球经过B点时对轨道的压力为3mg+2mgℎRD.小球从A运动到B的过程中，重力的功率一直增大6. 如图所示为交流发电机的原理图，其矩形线圈在磁感应强度为0.2T的匀强磁场中，从图示位置开始绕垂直于磁场方向的固定轴OO′匀速转动，转动周期为6.28×10−2s，线圈的匝数为100匝、阻值为10Ω，面积为0.1m2．线圈的两端经集流环与阻值为90Ω的电阻R连接，与R并联的交流电压表为理想电表．（π取3.14），下列说法正确的是（）A.电动势的有效值为200VB.线圈在图示位置磁通量的变化率最大C.交流电压表的示数为90√2VD.经过八分之一周期，通过电阻R的电荷量为√2100C7. 如图所示，真空中有一边长为a的等边三角形ABC，P点是三角形的中心．在A点固定一电荷量为q的负点电荷，在B点固定一电荷量为q的正点电荷，已知静电力常量为k，则下列说法正确的是（）A.C点的电势高于P点电势B.C点的电场强度大小为2kqa2C.某一试探电荷在C点与P点所受静电力大小之比为√39D.将某一试探电荷从C点沿CP连线的方向移动到P点，该过程中静电力不做功8. 如图所示，用长为L的轻绳（轻绳不可伸长）连接的甲、乙两物块（均可视为质点），放置在水平圆盘上，甲、乙连线的延长线过圆盘的圆心O，甲与圆心O的距离也为L，甲物体质量为2m，乙物体的质量为m，与圆盘间的动摩擦因数均为μ，物块与圆盘间的最大静摩擦力等于滑动摩擦力，甲、乙始终相对圆盘静止，则下列说法中正确的是（）A.轻绳最大弹力为μmg2B.圆盘转动的角速度最大为√3μg4LC.轻绳最大弹力为μmgD.圆盘转动的角速度最大为√μg2L 二、实验探究题如图甲所示，用铁架台、弹簧和多个已知质量且质量相等的钩码，探究在弹性限度内弹簧弹力与弹簧伸长量的关系．（1）实验中还需要的测量工具有：________．（2）用纵轴表示钩码质量m，横轴表示弹簧的形变量x，根据实验数据绘制出乙图．由图可知：弹簧的劲度系数k＝________N/m（重力加速度g取9.8m/s2）．（3）如图丙所示，实验中用两根不同的弹簧a和b．画出弹簧弹力F与弹簧长度L的关系图像．则下列说法中正确的是（）．A.a的劲度系数比b的小B.a的原长比b的长C.弹力与弹簧长度成正比D.a的劲度系数比b的大实验室使用的直流电流表是由毫安表改装而成，该电流表的内部电路如图乙所示．（1）某次用0∼3A量程测量时，示数如图丙所示，其读数为________A．（2）该电流表0∼0.6A量程对应的内阻为_______（结果保留两位小数）．（3）毫安表允许通过的最大电流约为__________mA（结果保留两位有效数字）．三、解答题如图所示，两块竖直放置的导体板间存在水平向左的匀强电场，板间距离为d．有一带电量为+q、质量为m 的小球（可视为质点）以水平速度从A孔进入匀强电场，且恰好没有与右板相碰，小球最后从B孔离开匀强电场，若A、B两孔的距离为4d，重力加速度为g，求：（1）两板间的场强大小；（2）小球从A孔进入电场时的速度；（3）从小球进入电场到其速度达到最小值，小球电势能的变化量为多少？如图所示，在光滑的水平面上静止着足够长、质量为2kg的长木板，木板上摆放着可视为质点、质量均为1kg、相距为1m的两木块a和b，且木块a恰在长木板的最左端，两木块与木板间的动摩擦因数均为0.25．现同时给木块a、b方向水平向右、大小分别为1m/s和2m/s的初速度，已知重力加速度g=10m/s2，求：（1）系统由于摩擦增加了多少内能？（2）最后木块a、b间的距离为多少？（3）要使木块b不掉下去，则长木板至少多长？如图所示，左右两个粗细均匀内部光滑的汽缸，其下部由体积可以忽略的细管相连，左汽缸顶部封闭，右汽缸与大气连通，左右两汽缸高度均为H、横截面积之比为S1:S2=1:2，两汽缸除左汽缸顶部导热外其余部分均绝热．两汽缸中各有一个厚度不计的绝热轻质活塞封闭两种理想气体A和B，当大气压为p0、外界和气体温度均为27∘C时处于平衡状态，此时左、右侧活塞均位于汽缸正中间．（1）若外界温度不变，大气压为0.9p0时，求左侧活塞距汽缸顶部的距离；（2）若外界温度不变，大气压仍为p0，利用电热丝加热气体B，使右侧活塞上升H4，求此时气体B的温度．如图所示，ABCD是某种透明材料的截面，AB面为平面，CD面是半径为R的圆弧面，O1O2为对称轴，一束单色光从O1点斜射到AB面上折射后照射到圆弧面上E点时刚好发生全反射，AB面上入射角的正弦值为√33，∠DO2C=120∘，透明材料对单色光的折射率为2√33，光在真空中传播速度为c，求：（1）O1O2与O2E的夹角θ的大小；（2）光在透明材料中传播的时间（不考虑光在BC面的反射）（结果可以用根号表示）．参考答案与试题解析2020-2021学年广西柳州高三上物理月考试卷一、选择题1.【答案】此题暂无答案【考点】机表启透问叶圈恒定功率动能射初的应用【解析】此题暂无解析【解答】此题暂无解答2.【答案】此题暂无答案【考点】万有引力都律及其应还随地体绕某问题【解析】此题暂无解析【解答】此题暂无解答3.【答案】此题暂无答案【考点】单杆常割陆感线闭合电因的欧优定律瞬使功胞【解析】此题暂无解析【解答】此题暂无解答4.【答案】此题暂无答案【考点】机械能于恒的氨断单表体木神种能守恒问题动能射初的应用【解析】此题暂无解析【解答】此题暂无解答5.【答案】此题暂无答案【考点】竖直体内火圆是运主-弹力平抛来动基稳造或及推论的应用系统都械问守迹定律逆应用瞬使功胞【解析】此题暂无解析【解答】此题暂无解答6.【答案】此题暂无答案【考点】交流发根机及其有生铁弦件电流的原理交变电三的有效州交变电流于周期术频率【解析】此题暂无解析【解答】此题暂无解答7.【答案】此题暂无答案【考点】点电因的场斯电场根隐钢与根势能变化的关系电都升叠加【解析】此题暂无解析【解答】此题暂无解答8.【答案】此题暂无答案【考点】水平射内开圆认运铵-摩擦力【解析】此题暂无解析【解答】此题暂无解答二、实验探究题【答案】此题暂无答案【考点】探究来力和脂簧伸情毒关系【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】电流表来装电斗表【解析】此题暂无解析【解答】此题暂无解答三、解答题【答案】此题暂无答案【考点】带电水使在重之场和硝场中的运动【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】板块使以问题摩擦力做体与硫量转化动于深了定乘的综合应用【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】理想气射的状态面程【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】光的来射和脂反射含综合主用【解析】此题暂无解析【解答】此题暂无解答。
2020-2021学年广西柳州高三上数学月考试卷
2020-2021学年广西柳州高三上数学月考试卷一、选择题1. 已知复数z满足(2−i)z=1+2i(其中i为虚数单位)，则|z|=( )A.−iB.iC.5D.12. 设集合A={x|x2+x−2≤0}，B={x||x|≤2}，则A∩B=( )A.[−2,2]B.[−2,1]C.[−1,1]D.[−1,2]3. 已知x，y满足约束条件{x+y−2≥0,x−y−2≤0,y≤2,则z=2x+y的最小值为( )A.10B.6C.2D.44. 《周髀算经》是我国古老的天文学和数学著作，其书中记载：一年有二十四个节气，每个节气晷长损益相同（晷是按照日影测定时刻的仪器，晷长即为所测影子的长度），夏至、小暑、大暑、立秋、处暑、白露、秋分、寒露、霜降是连续的九个节气，其晷长依次成等差数列，经记录测算，这九个节气的所有晷长之和为49.5尺，夏至、大暑、处暑三个节气晷长之和为10.5尺，则立秋的晷长为( )A.3.5尺B.1.5尺C.4.5尺D.2.5尺5. 已知a=log52，b=ln2，c=23，则a，b，c的大小关系正确的是( )A.b<c<aB.b<a<cC.a<b<cD.a<c<b6. 已知2cos2α=3sinα，则cos2α=( )A.−12B.12C.−19D.197. 函数f(x)=sin x+xcos x+x2在[−π,π]的图像大致为( ) A. B.C. D.8. 如图所示的程序框图，当其运行结果为31时，则图中判断框①处应填入的是()A.i≤5B.i≤3C.i≤6D.i≤49. 某几何体的三视图如图所示，则该几何体外接球的表面积是( )A.6πB.12πC.24πD.16π10. 函数f(x)=2sin(ωx+φ)(ω>0,|φ|<π2)的最小正周期为π，若其图象向右平移π6个单位后得到的函数为奇函数，则函数f(x)的图象( )A.在x=π6处取最大值 B.关于直线x=π3对称C.关于点(π3,0)对称 D.在(−π2,π2)上单调递增11. 已知函数f(x)的定义域为R，且f(x+1)是偶函数，f(x−1)是奇函数，f(x)在[−1,1]上单调递增，则( )A.f(2020)>f(0)>f(2019)B.f(2020)>f(2019)>f(0)C.f(0)>f(2020)>f(2019)D.f(0)>f(2019)>f(2020)12. 已知不等式x +a ln x +1ex ≥x a对x ∈(1, +∞)恒成立，则实数a 的最小值为( )A.−e2 B.−√e C.−e D.−2e二、填空题已知向量a →=(2,−1)，b →=(1,m )，若向量a →+b →与a →垂直，则m =________.在递增等比数列{a n }中，S n 是其前n 项和，若a 2+a 4=5，a 1⋅a 5=4，则S 4=________.已知直线l ：y =4−x 与曲线C ：y =2C 上随机取一点M ，则点M 到直线l 的距离不大于√2的概率为________.已知F 1(−c,0),F 2(c,0)是双曲线C ：x 2a2−y 2b 2=1的左、右焦点，F 1关于双曲线的一条渐近线的对称点为P ，且点P 在抛物线y 2=4cx 上，则双曲线的离心率为________. 三、解答题偏差是指个别测定值与测定的平均值之差，在成绩统计中，我们把某个同学的某科考试成绩与该科班平均分的差叫某科偏差，在某次考试成绩统计中，某老师为了对学生数学偏差x （单位：分）与物理偏差y （单位：分）之间的关系进行分析，随机挑选了8位同学，得到他们的两科成绩偏差数据如下：∑x i 8i=1y i=20×6.5+15×3.5+13×3.5+3×1.5+2×0.5+(−5)×(−0.5)+(−10)×(−2.5)+(−18)×(−3.5)=324∑x 28i=1=202+152+132+32+22+(−5)2+(−10)2+(−18)2=1256参考公式： b ̂=∑x i ni=1y i −nx ¯y¯∑x i 2n i=1−nx¯2, a ̂=y ¯−b ̂x ¯.(1)若x 与y 之间具有线性相关关系，求y 关于x 的线性回归方程；(2)若该次考试数学平均分为120分，物理平均分为91.5分，试由(1)的结论预测数学成绩为128分的同学的物理成绩．在△ABC 中，角A ，B ，C 的对边分别为a ，b ，c ，且(2b −c )(b 2−a 2+c 2)=2abc cos C .(1)求角A 的大小．(2)若∠ABC =π3，D 为△ABC 外一点，BD=2,CD =1，四边形ABDC 的面积是5√34+2，求∠BDC 的大小．在四棱锥P −ABCD 中，AB // CD ，CD =2AB ，AC 与BD 相交于点M ，点N 在线段AP 上，AN =λAP(λ>0)，且MN // 平面PCD ．(1)求实数λ的值；(2)若AB =AD =DP =1,PA =PB =√2，∠BAD =60∘，求点N 到平面PCD 的距离．已知椭圆C ：x 2a2+y 2b 2=1(a >b >0)的离心率为√32，短轴长为2.(1)求椭圆C 的标准方程；(2)设直线l:y =kx +m 与椭圆C 交于M ， N 两点，O 为坐标原点，若k OM ⋅k ON =54，求证：m 2+k 2为定值．已知函数f(x)=ln x+mx+1，g(x)=12mx2−(1+2m)x−1.(1)若f(x)的最大值是0，求函数f(x)的图像在x=e处的切线方程；(2)设函数F(x)=f(x)+g(x)，对于定义域内的x，讨论函数F(x)的极值情况．已知函数f(x)=|x−1|−|x+2|.(1)求不等式f(x)<x的解集；(2)函数f(x)的最大值为M，若正实数a，b，c满足a+4b+9c=13M，求1a+1b+1c的最小值．参考答案与试题解析2020-2021学年广西柳州高三上数学月考试卷一、选择题1.【答案】此题暂无答案【考点】复根的务复验热数术式工乘除运算【解析】此题暂无解析【解答】此题暂无解答2.【答案】此题暂无答案【考点】交集根助运算一元二次正等式的解且绝对常不等至的保法与目明【解析】此题暂无解析【解答】此题暂无解答3.【答案】此题暂无答案【考点】求线性目于函数虫最值【解析】此题暂无解析【解答】此题暂无解答4.【答案】此题暂无答案【考点】等差数常的占n项和等三中弧【解析】此题暂无解析【解答】此题暂无解答5.【答案】此题暂无答案【考点】对数值于小的侧较指数表、对烧式守综合员较【解析】此题暂无解析【解答】此题暂无解答6.【答案】此题暂无答案【考点】同角正角测数解的当本关系二倍角三余弦公最【解析】此题暂无解析【解答】此题暂无解答7.【答案】此题暂无答案【考点】函数因象的优法【解析】此题暂无解析【解答】此题暂无解答8.【答案】此题暂无答案【考点】程正然图【解析】此题暂无解析【解答】此题暂无解答9.【答案】此题暂无答案【考点】由三视于求表械积球内较多面绕球的表体积决体积【解析】此题暂无解析【解答】此题暂无解答10.【答案】此题暂无答案【考点】函数y射Asi过（ω复非φ）的图象变换正弦函较的对盛性正弦函射的单调长正弦射可的图象【解析】此题暂无解析【解答】此题暂无解答11.【答案】此题暂无答案【考点】奇偶性与根调性的助合函数奇明性研性质函数水因期性【解析】此题暂无解析【解答】此题暂无解答12.【答案】此题暂无答案【考点】利验热数技究女数的最值利用都数资究不长式化成立问题利用验我研究务能的单调性【解析】此题暂无解析【解答】此题暂无解答二、填空题【答案】此题暂无答案【考点】数量积常断换个平只存量的垂直关系数量积正率标表达式【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】等比使香的性质等比数使的前n种和等比数表的弹项公式【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】几何概表计声（集长样、角度奇附积、体积有关的几何概型）点到直使的距离之式【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】双曲根气离心率双曲根气渐近线【解析】此题暂无解析【解答】此题暂无解答三、解答题【答案】此题暂无答案【考点】求解线都接归方程最都问乘法【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】余于视理正因归理两角和与表擦正弦公式【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】平三线硬线万蓝比例定理直线与平三平行要性质直线与平正垂直的判然平面与平明垂钾的判定【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】椭圆水明心率椭圆较标准划程圆锥来线中雨配点缺定值问题【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】利用验我研究务能的单调性利用三数定究曲纵上迹点切线方程利验热数技究女数的最值利来恰切研费函数的极值【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】绝对来不等阅基本常等式簧最母问赤中的应用绝对值射角不等开【解析】此题暂无解析【解答】此题暂无解答。
广西柳州高级中学2020┄2021届高三2月线上月考试题英语
第一部分听力（共两节，满分30分）第一节听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
Who is the woman talking with？A. Her brother.B. Carl’s brother.C. Her teacher.How much money did the speakers raise this year？Two hundred dollars.Three hundred dollars.Four hundred dollars.What does the man want to know？Where the sign-up sheet is.When he can take the field trip.Whether the woman will go on the field trip.Which gate should the woman go to？A. F56.B. B5.C. C34What will the man probably do tomorrow？A. Stay alone.B. Go on a school outing.C. Relax with the woman.第二节听下面5段对话或独白。
每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料，回答第6、7题。
6. What does the man want to reserve？A. Business suites.B. Double rooms.C. Single rooms.7. When will the man check in？A. On the 16th.B. On the 19th.C. On the 23rd.听第7段材料，回答第8、9题。
2020-2021学年广西柳州高级中学高三英语模拟试题及答案解析
2020-2021学年广西柳州高级中学高三英语模拟试题及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ACome and enjoy Vivaldi's TheFour Seasonsperformed by live musicians!Tickets△Zone A Sating (Excellent Visibility, $75)△Zone B Seating (Great Visibility, $60)△Zone C Seating (Good Visibility, $45)△Zone D Seating (Restricted Visibility, 30)Zone A and Zone B audiences will get the chance to take pictures with the performers on the stage after the show.Highlights* A beautiful venue bathed in candlelight.*Classical music performance by the Angel Strings quartet*A safe and socially-distanced event, ensuring you are comfortable and at ease.General Info*Dates and times: Various dates, at 6:30 pm and 8:30 pm (select during purchase).*How long: 65 minutes. Doors open 45 minutes before the start time. We recommend you arrive at least 30 minutes before the start of the event, as late entry is not permitted.*Where: Events on Oxlade*Age requirement: Must be 8 years old or older to attend. Anyone under the age of 16 must be accompanied by an adult.*Please note: The 6:30 pm seating will take place during daylight hours outdoors, and the space will not be that dark. In the case of rain, the event will be moved to the indoor area of the venue.DescriptionWhether you're looking for a beautifully unique classical music performance or a romantic candlelit experience, this performance is for you. You don't need to know all things about Vivaldi to enjoy the evening; simply sit back and admire the wonderful atmosphere and the pieces you'll hear.Join our musicians for an evening under the stars, and prepare to be taken into the clouds with Vivaldi' s most treasured masterpieces!1.What can someone with a $45 ticket do?A.Perform on the stage.B.Enjoy good visibility.C.Select a seat in Zone B.D.Take photos with the musicians.2.What should potential audiences keep in mind?A.Arrive at the venue on time.B.Learn about Vivaldi in advance.C.The performance lasts 45 minutes.D.The event will be canceled if it rains.3.What do we know about the 8:30 p.m. performance?A.It welcomes children under the age of 8.B.Its performers differ on different dates.C.Its stage will be decorated with candles.D.It will be shown in the indoor area of the venue.BAt the World Economic Forum last month, President Trump drew claps when he announced the United States would respond to the forum's proposal to plant one trillion(万亿) trees to fight climate change. The trillion-tree idea won wide attention last summer after a study published in the journal Science concluded thatplanting so many trees was “the most effective climate change solution to date”.If only it were true. But it isn't. Planting trees would slow down the planet's warming, but the only thing that will save us and future generations from paying a huge price in dollars, lives and damage to nature is rapid and considerable reductions in carbon release from fossil fuels, to net zero by 2050.Focusing on trees as the big solution to climate change is a dangerous diversion(偏离). Worse still, it takes attention away from those responsible for the carbon release that are pushing us toward disaster. For example, in the Netherlands, you can pay Shell an additional 1 euro cent for each liter of regular gasoline you put in your tank, to plant trees to balance the carbon release from your driving. That's clearly no more than disaster slightly delayed. The only way to stop this planet from overheating is through political, economic, technological and social solutions that end the use of fossil fuels.There is no way that planting trees, even across a global area the size of theUnited States, can absorb the huge amounts of fossil carbon released from industrial societies. Trees do take up carbon from the atmosphere as they grow. But this uptake merely replaces carbon lost when forests were cleared in the first place, usually longago. Regrowing forests where they once grew can undo some damage done in the past, but even a trillion trees can't store enough carbon to head off dramatic climate changes this century.In a sharp counter argument to last summer's Paper in Science, Justin Gillis wrote in the same journal in October that the study's findings were inconsistent with the dynamics of the global carbon cycle. He warned that “the claimthat global tree restoration(复原) is our most effective climate solution is simply scientifically incorrect and dangerously misleading”.4. What do we know about the trillion-tree idea?A. It was published in a journal.B. It was proposed last summer.C. It was put forward by Trump.D. It drew lots of public attention.5. What is paragraph 3 mainly about?A. A drawback of the tree planting strategy.B. An example of balancing carbon release.C. An anecdote of making a purchase at Shell.D. A responsibility for politicians and economists.6. What was Justin Gillis's attitude towards global tree restoration?A. Indifferent.B. Opposed.C. Hesitant.D. Supportive.7. What is the best title for the text?A. Contradictory Ideas on Tree Planting.B. A Trillion Trees Come to the Rescue.C. Planting Trees Won't Save the World.D. The Best Solution to Climate Change.CAccording to statistics published by the BPI (Buying Power Index) a couple of months ago, digital streaming (流媒体) now accounts for 80 percent of the music consumption in the UK. Despite the incredible growth of online streaming platforms like iTunes, Apple Music and Tidal over the past 15 years, a more traditional medium has also seen a return of interest and sales in the music industry. In 2020, almost one in five of all albums purchased in the UK is vinyl (黑胶唱片), and it has once again become the most popular physical musical medium.With digital streaming so easy and convenient, why are so many peopledrawn to traditional records? Some experts claim that vinyl is a physical medium for experiencing music, something tangible (有形的) to hold and own. For most people, having something tangible and interacting with it gives depth to the experience of music. Listening to an album and touching it the way the artist intended can make them feel more connected to the music and the artist. Records are physical products that can be not only displayed but also gifted, shared, traded and passed down through generations.Sound quality is another hot topic. A lot of music lovers feel that the analogue sound (模拟声音) vinyl offers is superior to modern digital audio, particularly with regards to the compressed formats streaming platforms use. There’s a common belief that old-school analogue audio has a warmer, fuller sound than digitised music. For vinyl followers, the very defect traditional recorders often have, such as the familiar crackle (劈啪作响) when the record starts, bring the music to life in a different way.There’s aritualisticaspect to vinyl that a lot of people are drawn to, too. The act of putting a record on—carefully removing the record from the sleeve, placing it on the record player and gently dropping the needle on the right groove (凹槽)—is a more assiduous (一丝不苟的), mindful way of engaging with music. When you’re listening to vinyl, you can’t tap abutton and go about your day while the streaming service provides hours of music. You need to stay close to the record player to move the needle and flip the record over.It’s clear that the vinyl interest is well underway, and vinyl records are truly making a comeback. In an increasingly digital society, there’s something to be said for analogue experiences. Perhaps one of the great things about being alive in the 21st century is our ability to have the best of both worlds—the timeless appeal of physical records alongside the easy access to vast music libraries that streaming offers.8. What are the statistics published by the BPI used to show?A. An increase in music consumption.B. The recovery of music industry.C. A comeback of a physical medium.D. The acceptance of online streaming.9. According to some experts, why does vinyl interest many people?A. It attracts people by its realistic feel.B. It offers simple access to different music.C. It shares a new way to enjoy music.D. It provides people with perfect sound effect.10. The underlined word “ritualistic” in Para.4 means something ______.A. Overlooked by society.B. Updated very frequently.C. Performed as part of a ceremony.D. Kept for a long time without changing.11. How does the writer feel about the future development of music medium?A. Traditional records will get underway.B. The analogue experiences may matter more.C. Vinyl sales will boom with technological advance.D. There should be a good mix of old and new.DIn 2015, a man named Nigel Richards memorized 386, 000 words in the entireFrench Scrabble Dictionaryin just nine weeks. However, he does not speak French. Richards’ impressive feat is a useful example to show how artificial intelligence works — real AI. Both of Richard and AI take in massive amounts of data to achieve goals with unlimited memory and superman accuracy in a certain field.The potential applications for AI are extremely exciting. Because AI canoutperformhumans at routine tasks — provided the task is in one field with a lot of data — it is technically capable of replacing hundreds of millions of white and blue collar jobs in the next 15 years or so.But not every job will be replaced by AI. In fact, four types of jobs are not at risk at all. First, there are creative jobs. AI needs to be given a goal to optimize. It cannot invent, like scientists, novelists and artists can. Second, the complex, strategic jobs — executives, diplomats, economists — go well beyond the AI limitation of single-field and Big Data. Then there are the as-yet-unknown jobs that will be created by AI.Are you worried that these three types of jobs won’t employ as many people as AI will replace? Not to worry, as the fourth type is much larger: jobs where emotions are needed, such as teachers, nannies and doctors. These jobs require compassion, trust and sympathy — which AI does not have. And even if AI tried to fake it, nobody would want a robot telling them they have cancer, or a robot to babysit their children.So there will still be jobs in the age of AI. The key then must be retraining the workforce so people can do them. This must be the responsibility not just of the government, which can provide funds, but also of corporations and those who benefit most.12. What is the main purpose of paragraph 1?A. To introduce the topic.B. To mention Nigel’s feat.C. To stress the importance of good memory.D. To suggest humans go beyond AI in memory.13. Which of the following best explains “outperform” underlined in paragraph 2?A. Be superior toB. Be equal toC. Be similar toD. Be related to14. Which of the following jobs is the most likely to be replaced?A. The writer.B. The shop assistant.C. The babysitter.D. The psychologist.15. What does the text suggest people do about job replacement of AI?A. Limit the application of AI to a certain degree.B. Get more support from the government.C. Apply for the donation from companies.D. Upgrade themselves all the time.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020-2021学年广西柳州高级中学高三英语模拟试题及答案解析
2020-2021学年广西柳州高级中学高三英语模拟试题及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项APersonal Time Off(PTO)is something my wife and I created after having kids. We learned that，over time，a full life can leave little time for personal rest and for reflection，hanging out with friends，or just being“off. ”So，after a number of years，we make a change. If I can persuade you to take your own PTO，then these might help.■Schedule itFirst of all，one of our favorite family sayings is“schedule it. ”Sounds easy enough，but life gets complicated managing full-time work and full-time family. Put yourPTO time on the calendar and you make it a real thing.■Be flexible and strictIf you can schedule PTO at the same time each week，then all the better. Because our calendar can get rather full，flexibility in scheduling becomes a necessity. But being strict in actually taking the time each week is more important. Skipping it once makes it easier to skip again.■Take enough timeMy typical PTO lasts a couple of hours or longer. Sometimes it might be half the day depending on what I’m doing. The goal is to spend enough time away to1et your shoulders drop.■Do what you want to doRemember，PTO time is about personal time to do what you want to do，not what you have to do. PTO time is about relaxation. Grab a friend and get a beer. Work can wait until tomorrow.1.What can be the first step to take the PTO?A.Persuade the family.B.Have a personal rest.C.Ask friends for advice.D.Make a time plan.2.What does the underlined part“let your shoulders drop”probably mean?A.Get you more focused.B.Have you feel relaxed.C.Shake your shoulders often.D.Make you feel more stressed.3.What does the text mainly talk about?A.Tips on how to take your time offB.Skills to manage work and familyC.Advice on how to free yourselfD.Ways of scheduling your workBFor most thinkers since the Greek philosophers, it was self-evident that there is something called human nature, something that constitutes the essence of man. There were various views about what constitutes it, but there was agreement that such an essence exists—that is to say, that there is something by virtue of which man is man. Thus man was defined as a rational(理性的) being, as a social animal, an animal that can make tools, or a symbol-making animal.More recently, this traditional view has begun to be questioned. One reason for this change was the increasing emphasis given to the historical approach to man. An examination of the history of humanity suggested that man in our time is so different from man in previous times that it seemed unrealistic to assume that men in every age have had in common something that can be called “human nature.” The historical approach was strengthened, particularly in the United States, by studies in the field of cultural anthropology (人类学). The study of primitive peoples has discovered such a diversity of customs, values, feelings, and thoughts that many anthropologists arrived at the concept that man is born as a blank sheet of paper on which each culture writes its text. Another factor contributing to the tendency to deny the assumption of a fixed human nature was that the concept has so often been abused as a shield(盾牌) behind which the most inhuman acts are committed. In the name of human nature, for example, Aristotle and most thinkers up to the eighteenth century defended slavery. Or in order to prove the rationality and necessity of the capitalist form of society, scholars have tried to make a case for acquisitiveness, competitiveness, and selfishness as natural human characters. Popularly, one refers cynically（愤世嫉俗地）to “human nature” in accepting the inevitability of such undesirable human behavior as greed, murder, cheating and lying.Another reason for disbelief about the concept of human nature probably lies in the influence of evolutionary thinking. Once man came to be seen as developing in the process of evolution, the idea of a substance which is contained in his essence seemed untenable. Yet I believe it is precisely from an evolutionary standpoint that we can expect new insight into the problem of the nature of man.4. Most philosophers believed that human nature ________.A. is the quality distinguishing man from other animalsB. consists of competitiveness and selfishnessC. is something partly innate and partly acquiredD. consists of rationality and undesirable behavior5. The traditional view of “human nature” was strongly challenged by ________.A. the emergence of the evolutionary theoryB. the historical approach to manC. new insight into human behaviorD. the philosophical analysis of slavery6. According to the passage, anthropologists believe that human beings ________.A. have some characters in commonB. are born with diverse culturesC. are born without a fixed natureD. change their characters as they grow up7. The author mentioned Aristotle, a great ancient thinker, in order to ________.A. emphasize that he contributed a lot to defining the concept of “human nature”B. show that the concept of “human nature” was used to justify social evilsC. prove that he had a profound influence on the concept of “human nature”D. support the idea that some human characters are inherited.CBeing an Olympian (奥运会选手) demands focus, determination, and a competitive spirit. Plus, representing your country is a lot of pressure. However, two athletes recently showed the world another quality that is definitely worth championing.Qatar's Mutaz Essa Barshim and Italy's Gianmarco Tamberi were competing in the high jump on Sunday when they reached a stalemate (僵局). Both men had managed to jump over a surprising 2.37 meters with no faults along the way. However, after three attempts neither managed the next level of 2.39 meters.An Olympic official suggested ajump-offbetween the two friends and rivals (竞争对手) to determine who would get the gold medal. But Barshim had another plan to reward their efforts.“Can we have two golds?” he asked the official.The official agreed and the two men jumped for joy. This was the first time a gold medal had been shared since 1912. “He is one of my best friends, not only on the track, but outside the track. We work together. This is a dream come true.” shared Barshim.The decision to share the medal was particularly meaningful to Tamberi. The Italian had suffered an ankleinjury that prevented him from competing in the Rio Olympics in 2016, and it nearly put an end to his career altogether. So this year he brought along his cast to this year's Olympics with “Road to Tokyo 2021” to inspire him along the way.For Barshim, the gold has topped off his already impressive medal collection, having received a bronze and a silver medal in 2012 and 2016 respectively.Despite all the glory of receiving a gold medal for their countries, their achievement means so much more. These two individuals, trying to do their very best for their countries, have provided a wonderful example to all those competing in sports. They've summed up exactly what it means to take part in a global event with a generous and compassionate (有同情心的) spirit.8. What happened to Barshim and Tamberi in the competition?A. They ended in a tie.B. They quit the competition.C. They set a new record.D. They ran out of strength.9. What does the underlined word “jump-off” in Paragraph 3 refer to?A. Debate.B. Vote.C. Celebration.D. Extra round.10. Why was the gold medal particularly meaningful to Tamberi?A. It could bring him a lot of money.B. It may make up for his regret in 2016.C. It was a glory for his country.D. It could complete his medal collection.11. What does the author mainly want to convey in the text?A. The importance of sharing.B. The glory of winning gold medals.C. The valuable and special team spirit.D. The considerate and sharing Olympic spirit.DThe China International Search and Rescue Team(CISAR) was formed in 2001 and is now made up of several hundred rescue workers and about 20 police dogs. The team brings help and hope to those whose lives are changed by astorm, flood, earthquake, or any other natural disasters.After long and careful training, the team went on its first international rescue tasks in 2003. That year, the Chinese team helped save lives after earthquakes inAlgeriaandIran. It was the first time that a Chinese team had worked outsideChinawhose members won high praise for bravery and skill.Since then, the CISAR has completed many tasks. The list of people to whom help has been given is long. The team treated more than 3,000 people who were wounded in the 2006 earthquake inIndonesia, helped 2,500 wounded people after the earthquake that hitHaitiin 2010, and spent several months giving aid to over 25,000 people suffering from the 2010 floods inPakistan. On April 26, 2015, a group of 62 people from CISAR went toNepalafter the 8.1 magnitude earthquake that happened there.Rescue workers are trained to find people, treat wounds, and hand out food, water, and other supplies. They have to be able to do work that is difficult under conditions which can be very dangerous. After a disaster, there is usually no electricity or water, and there may be diseases and other dangers. Rescue workers get to save lives, but they must also bury the dead. That means they have to be strong in both body and mind.Rescue workers must have big hearts, too. It takes a lot of love and courage to risk one’s own life to save someone else’s. The members of the CISAR have plenty of both and are always ready to go wherever help is needed.12. What is the function of the numbers in Paragraph 3?A. To advertise for the CISAR.B. To add some basic information.C. To praise Recue Workers’ contributions.D. To stress the dangers Rescue Workers face.13. What is the author’s attitude towards Rescue Workers?A. Hopeful.B. Respectful.C. Curious.D. Supportive.14. What are the last two paragraphs mainly about?A. The duty rescue workers must perform.B. The qualities rescue workers must own.C. The difficulties rescue workers must go through.D. The willingness rescue workers should require.15. What may be the best title of the passage?A. China to the RescueB. How to train CISARC. Welcome to CISARD. Rescue on request第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
广西柳州高级中学2020届高三2月线上月考语文试题 Word版含答案
柳州高级中学2020届高三2月月考语文试题一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题，9分)阅读下面的文字，完成1～3题。
世界需要“中国新思想”乐黛云1937年，林语堂用英文写了一本向西方介绍中国文化的书——《生活的艺术》。
书一出版就引起轰动，持续52个星期位居美国畅销书排行榜第一名，在美国重印40多次，被翻译成10多种不同的文字。
这种现象说明世界需要于中国的，是不同于他们自己原有思想的另一种思想，不同于他们惯常生活方式的另一种生活方式。
但是，百余年来，中西文化对话中，中国面对的往往是强势文化的灌输和覆盖。
今天，是根本改变这种局面的时候了。
对中国文化来说，通过今人的自觉，从深厚的中国文化土壤中生长出“中国新思想”，是根本之道。
当代世界深陷各种冲突之中。
从军事战争到经济战争，从资源争夺到社会斗争，从国际冲突到文化对峙，如何克服冲突、形成合作，是人类一直未能解决的最大问题。
中国先哲认为，个人无法独立生存，人的初始状态就是与父母和他人的关系，这首先就是一种社会合作状态。
因此，社会的基因不是个人，而是人与人的关系。
所谓“礼”，就是讲社会关系中的人如何存在，而非强调抽象的人。
我国学者从中国文化的这个立场出发，参与世界重大问题的思考和讨论，这种出发点已引起国际学术界重视。
中国学者李泽厚在《伦理学纲要》中，更提出了中国文化的“情本体”，引起国内外学术界对中国的“情理”与西方“理性”的讨论，也引发西方文明对自身的反思。
作为欧美对外政策基础的“帝国理论”，曾经带来三个世纪的战争灾难，已经走到尽头。
中国哲学家赵汀阳，几年前出版《天下体系：世界制度哲学导论》，描述了一个不同于帝国理论的、拥有普世正当性的中国的世界秩序模式。
中国古代的“天下”，不同于西方以民族国家为主体的利益博弈模式，而是一个“至大无外”的思考单位，从“天下—国—家”这样由大至小的方向，思考如何实现各个层次的和谐。
从此，“天下体系”作为不同于长期拥有文化霸权的“帝国理论”的另一种可替代的概念和思考方式，逐渐引起世界思想界重视。
2020-2021学年广西柳州高三上数学月考试卷
2020-2021学年广西柳州高三上数学月考试卷一、选择题1. 已知复数z满足(2−i)z=1+2i(其中i为虚数单位)，则|z|=( )A.5B.1C.iD.−i2. 设集合A={x∣x2+x−2≤0}，B={x||x|≤m}，若A∪B={x∣−2≤x≤2}，则实数m=( )A.2B.1C.−1D.−23. 已知x，y满足约束条件{x+y−2≥0,x−y−2≤0,y≤2,则z=2x+y的最小值为( )A.2B.4C.6D.104. 《周髀算经》是我国古老的天文学和数学著作，其书中记载：一年有二十四个节气，每个节气晷长损益相同（晷是按照日影测定时刻的仪器，晷长即为所测影子的长度），夏至、小暑、大暑、立秋、处暑、白露、秋分、寒露、霜降是连续的九个节气，其晷长依次成等差数列，经记录测算，这九个节气的所有晷长之和为49.5尺，夏至、大暑、处暑三个节气晷长之和为10.5尺，则立秋的晷长为( )A.1.5尺B.2.5尺C.3.5尺D.4.5尺5. 下图为四组样本数据的条形图，则对应样本的标准差最大的是( )A. B.C. D.6. 已知2cos2α=3sinα，则cos2α=( ) A.−19B.19C.12D.−127. 函数f(x)=sin x+xcos x+x2在[−π,π]的图像大致为( )A. B.C. D.8. 如图所示的程序框图，当其运行结果为31时，则图中判断框①处应填入的是()A.i≤3B.i≤4C.i≤5D.i≤69. 某几何体的三视图如图所示，则该几何体外接球的表面积是( )A.24πB.16πC.12πD.6π10. 函数f(x)=2sin(ωx+φ)(ω>0,|φ|<π2)的最小正周期为π，若其图象向右平移π6个单位后得到的函数为奇函数，则函数f(x)的图象( )A.关于点(π3,0)对称 B.在(−π2,π2)上单调递增C.关于直线x =π3对称 D.在x =π6处取最大值11. 已知函数f (x )的定义域为R ，且f (x +1)是偶函数，f (x −1)是奇函数，f (x )在[−1,1]上单调递增，则( )A.f (0)>f (2020)>f (2019)B.f (0)>f (2019)>f (2020)C.f (2020)>f (2019)>f (0)D.f (2020)>f (0)>f (2019)12. 已知函数f(x)的导函数为f ′(x)，且对任意的实数x 都有f ′(x)=e −x (2x +3)−f(x)（e 是自然对数的底数），且f(0)=1，若关于x 的不等式f(x)−m <0的解集中恰有两个整数，则实数m 的取值范围是( ) A.[−e 2, 0) B.[−e, 0) C.(−e, 0] D.(−e 2, 0]二、填空题已知向量a →=(2,−1)，b →=(1,m )，若向量a →+b →与a →垂直，则m =________.在递增等比数列{a n }中，S n 是其前n 项和，若a 2+a 4=5，a 1⋅a 5=4，则S 4=________.在(x −1)(2x −1)5展开式中，x 2的系数为________．（结果用数字作答）已知F 1(−c,0),F 2(c,0)是双曲线C ：x 2a 2−y 2b 2=1的左、右焦点，F 1关于双曲线的一条渐近线的对称点为P ，且点P 在抛物线y 2=4cx 上，则双曲线的离心率为________. 三、解答题在△ABC 中，角A ，B ，C 的对边分别为a ，b ，c ，且(2b −c )(b 2−a 2+c 2)=2abc cos C .(1)求角A 的大小．(2)若∠ABC =π3，D 为△ABC 外一点，BD =2,CD =1，四边形ABDC 的面积是5√34+2，求∠BDC 的大小．某试验小组得到6组某植物每日的光照时间x (单位：ℎ)和每日平均增长高度y (单位：mm )的数据，现分别用模型①y ̂=b ̂x +a ̂和模型②y ̂=e m ̂x+n ̂对以上数据进行拟合，得到回归模型，并计算出模型的残差如下表：(模型①和模型②的残差分别为ê1和ê2，残差êi =y i −y ̂i )(1)根据上表的残差数据，应选择哪个模型来刻画该植物每日的光照时间与每日平均增长高度的关系较为合适，简要说明理由；(2)为了优化模型，将(1)中选择的模型残差绝对值最大所对应的一组数据(x ，y)剔除，根据剩余的5组数据，求该模型的回归方程，并预测光照时间为11ℎ时，该植物的平均增长高度．(剔除数据前的参考数据：x ¯=7.5，y ¯=5.9，∑x i 6i=1y i =299.8，∑xi 26i=1=355，z =ln y ，z ¯≈1.41，∑x i 6i=1z i ≈73.10，ln 10.7≈2.37，e4.034≈56.49.) 参考公式：b̂=∑(x i −x ¯)n i=1(y i −y ¯)∑(x i −x ¯)2n i=1=∑x i n i=1y i −nx ¯y¯∑x i 2n i=1−nx¯2，a ̂=y ¯−b ̂x ¯．如图，在以P 为顶点的圆锥中，母线长为√2，底面圆的直径AB 长为2，O 为圆心．C 是圆O 所在平面上一点，且AC 与圆O 相切，连接BC 交圆于点D ．连接PD ，PC ，E 是PC 的中点，连接OE ，ED .(1)求证：平面PBC ⊥平面PAC ；(2)若二面角B −PO −D 的大小为2π3，求平面PAC 与平面DOE 所成锐二面角的余弦值．已知椭圆C:x2a2+y2b2=1(a>b>0)的离心率为√32，短轴长为2．(1)求椭圆C的标准方程；(2)设直线l:y=kx+m与椭圆C交于M，N两点，O为坐标原点，若k OM⋅k ON=54，求证：点(m, k)在定圆上．已知函数f(x)=ln x+mx+1,g(x)=x(e x−1).(1)若f(x)的最大值是0，求函数f(x)的图像在x=e处的切线方程；(2)若对于定义域内任意x，f(x)≤g(x)恒成立，求m的取值范围．已知函数f(x)=|x−1|−|x+2|.(1)求不等式f(x)<x的解集；(2)函数f(x)的最大值为M，若正实数a，b，c满足a+4b+9c=13M，求1a+1b+1c的最小值．参考答案与试题解析2020-2021学年广西柳州高三上数学月考试卷一、选择题1.【答案】B【考点】复数的模复数代数形式的乘除运算【解析】先化简复数z，再求模即可.【解答】解：由题意可得：z=1+2i2−i =(1+2i)(2+i)(2−i)(2+i)=5i5=i，∴|z|=1.故选B.2.【答案】A【考点】并集及其运算集合关系中的参数取值问题【解析】化简集合A，B，再利用并集关系，确定参数范围.【解答】解：∵A={x∣x2+x−2≤0}={x∣−2≤x≤1}，当m<0时，B=⌀，此时不满足A∪B={x∣−2≤x≤2}，当m≥0时，B={x||x|≤m}={x|−m≤x≤m}，由A∪B={x∣−2≤x≤2}，∴m=2.故选A.3.【答案】A【考点】求线性目标函数的最值【解析】本题考查线性规划中的线性目标函数的最值问题，作出平面区域，利用z的几何意义，确定最小值. 【解答】解：作出不等式组{x+y−2≥0,x−y−2≤0,y≤2,所表示的平面区域，由于z=2x+y，则y=−2x+z，∴z表示直线z=2x+y与y轴的截距，∴当直线z=2x+y平移至A(0,2)时，截距最小，∴z的最小值为2×0+2=2.故选A.4.【答案】D【考点】等差数列的前n项和等差中项【解析】此题暂无解析【解答】解：设这九个节气的晷长为等差数列{a n}，公差为d，前n项和为S n，由已知得，a1+a3+a5=10.5⇒3a3=3a1+6d=10.5，S9=9a1+9×82d=9a1+36d=49.5，解得a1=1.5，d=1,故a4=a1+3d=1.5+3=4.5，即立秋的晷长为4.5尺.故选D.5.【答案】D【考点】极差、方差与标准差用样本的频率分布估计总体分布【解析】直接利用条形图的数据分布，推断出数据的分布情况.【解答】解：由条形图可知：四幅图的平均数均为8，从频率分布来看，BC 选项在平均数附近的数目较多，故BC 标准差相对较小；AD 选项比较，A 选项数据相对比较均衡，而D 选项在最大，最小的数据较多，故和平均数的差异大的数据多，就会导致方差过大，故标准差也会最大. 故选D . 6.【答案】 C【考点】同角三角函数间的基本关系 二倍角的余弦公式 【解析】利用同角三角函数基本关系求出sin α=12，再利用二倍角的余弦公式求解即可. 【解答】解：∵ 2cos 2α=3sin α， ∴ 2(1−sin 2α)=3sin α， 解得sin α=−2(舍去)或sin α=12，∴ cos 2α=1−2sin 2α=1−2×(12)2=12.故选C .7.【答案】 D【考点】函数图象的作法 【解析】 此题暂无解析 【解答】 解：∵ f(−x)=sin (−x)+(−x)cos (−x)+(−x)2=−sin x+x cos x+x 2=−f(x)，∴ f(x)为奇函数，可排除A ； 而f(π)=π−1+π2>0，∴ 可排除B 和C . 故选D . 8.【答案】 C【考点】 程序框图 【解析】 此题暂无解析 【解答】解：第一次输出 S =10,i =8； 第二次输出 S =18,i =7 ； 第三次输出 S =25,i =6 ； 第四次输出 S =31,i =5 ； 其运行结果为31.故判断框①处应填入的是i ≤5 故选C . 9.【答案】 D【考点】由三视图求表面积 球内接多面体 球的表面积和体积【解析】根据三视图可知该几何体为四棱锥，且四棱锥的一条侧棱与底面垂直，把四棱锥补成长方体，则长方体的长宽高分别为2，2，4，利用CFT 的对角线为外接球的直径求外接球的半径，代入球的表面积公式计算． 【解答】解：由三视图知：几何体为四棱锥， 把四棱锥补成长方体，则长方体的长宽高分别为2，1，1，∴ 长方体的外接球就是四棱锥的外接球， ∴ 外接球的直径2R =√4+1+1=√6， ∴ R =√62， ∴ 外接球的表面积S =4πR 2=4π×64=6π． 故选D .10.【答案】 A【考点】函数y=Asin （ωx+φ）的图象变换 正弦函数的对称性 正弦函数的单调性 正弦函数的图象【解析】由题意利用函数y ＝A sin (ωx +φ)的图象变换规律，三角函数的奇偶性，求出f(x)的解析式，再根据三角函数图象的对称性，得出结论． 【解答】解：∵ 函数f(x)=2sin (ωx +φ)的最小正周期为T =2πω=π，∴ ω=2，∵函数图象向右平移π6个单位后得到的函数为：y=2sin[2(x−π6)+φ]=2sin(2x−π3+φ)的图象，又∵所得函数为奇函数，|φ|<π2，∴可得φ=π3，∴f(x)=2sin(2x+π3)．对于A，f(π3)＝sin(2π3+π3)=0，则f(x)的图象关于点(π3, 0)对称，故A成立；对于B，由于2kπ−π2≤2x+π3≤2kπ+π2，k∈Z，可得：−5π12+kπ≤x≤π12+kπ，k∈Z，可得f(x)的单调递增区间为[−5π12+kπ, π12+kπ]，k∈Z，故B错误；对于C，由2x+π3=π2+kπ，k∈Z，解得x=π12+12kπ，k∈Z，故C错误；对于D，f(π6)=2sin2π3=√3≠±2，故D错误．故选A.11.【答案】B【考点】奇偶性与单调性的综合函数奇偶性的性质函数的周期性【解析】由题意得到f(x)周期为8，再利用函数的单调性进行比较即可求解. 【解答】解：∵f(x+1)是偶函数，∴f(x+1)图象关于直线x=0对称，∴f(x)图象关于直线x=1对称，又∵f(x−1)是奇函数，∴f(x−1)图象关于点(0,0)中心对称，∴f(x)图象关于点(−1,0)中心对称，由对称性函数具有周期性，∴f(x)周期为：T=4[1−(−1)]=8，∴f(2019)=f(3)=f(−1)，f(2020)=f(4)=f(−2)=−f(0)，又∵f(−1)=0，∴f(0)>0，∴f(0)>f(−1)>−f(0)，∴f(0)>f(2019)>f(2020).故选B.12.【答案】C【考点】利用导数研究函数的单调性函数的零点【解析】利用导数研究其单调性极值与最值并且画出图象即可得出．【解答】解：∵f′(x)=e−x(2x+3)−f(x)，∴e x[f(′x)+f(x)]=2x+3，∴e x f(x)=x2+3x+c，∵f(0)=1，∴1=0+0+c，解得c=1∴f(x)=(x2+3x+1)e−x，∴f′(x)=−(x2+x−2)e−x=−(x−1)(x+2)e−x．令f′(x)=0，解得x=1或x=−2，当x<−2或x>1时，f′(x)<0，函数f(x)单调递减，当−2<x<1时，f′(x)>0，函数f(x)单调递减增，可得：x=1时，函数f(x)取得极大值，x=−2时，函数f(x)取得极小值，∵f(1)=5e，f(−2)=−e2<0，f(−1)=−e，f(0)=1>0，f(−3)=e3>0∴−e<m≤0时，f(x)−m<0的解集中恰有两个整数−1，−2．故m的取值范围是(−e, 0].故选C.二、填空题【答案】 7【考点】数量积判断两个平面向量的垂直关系 数量积的坐标表达式 【解析】先求出a →+b →=(3,m −1)，再利用(a →+b →)⋅b →=2×3−1×(m −1)=0，求解即可. 【解答】解：∵ 向量a →=(2,−1)，b →=(1,m )， ∴ a →+b →=(3,m −1)， ∵ 向量a →+b →与a →垂直，∴ (a →+b →)⋅a →=2×3−1×(m −1)=0， 解得：m =7. 故答案为：7. 【答案】15 【考点】等比数列的性质等比数列的前n 项和 等比数列的通项公式【解析】利用等比数列性质得到a 2=1，a 4=4，再利用等比数列通项求出首项和公比，再利用求和公式即可得到答案.【解答】解：在递增等比数列{a n }中，设公比为q ， ∵ a 1⋅a 5=4， ∴ a 2⋅a 4=4， 又∵ a 2+a 4=5， ∴ a 2=1，a 4=4， ∴ a 1q =1，a 1q 3=4， ∴ a 1=12，q =2， ∴ S 4=12(1−24)1−2=152.故答案为：152. 【答案】50【考点】二项展开式的特定项与特定系数 二项式定理的应用【解析】利用通项即可求得对应的答案. 【解答】解：∵ (2x −1)5展开式的通项为：T r+1=C 5r (2x)5−r (−1)r =(−1)r ×25−r ⋅C 5r x5−r， 又∵ (x −1)(2x −1)5=x(2x −1)5−(2x −1)5，∴ 含x 2的项的系数为1×2×C 54+1×22×C 53=50. 故答案为：50. 【答案】√5+12【考点】双曲线的离心率 双曲线的渐近线【解析】 此题暂无解析 【解答】解：∵ F 1关于双曲线的一条渐近线的对称点为P ， 设F 1P 与该渐近线的交点为M ， ∴ F 1P ⊥OM ，MF 1=bc √a2+b 2=b ，∴ OM =a, PF 2=2OM =2a.在△F 1PF 2中，PF 1=2b,PF 2=2a,tan ∠F 1F 2P =ba ，∴ cos ∠F 1F 2P =ac.作PN ⊥F 1F 2，在△PF 2N 中，NF 2=PF 2⋅cos ∠F 1F 2P .又∵ 点P 在抛物线y 2=4cx 上， ∴ F 1F 2=PF 2+PF 2cos ∠F 1F 2P ， ∴ 2c =2a +2a cos ∠F 1F 2P ，∴e2−e−1=0，∴e=√5+12.故答案为：√5+12.三、解答题【答案】解：(1)∵(2b−c)(b2−a2+c2)=2abc cos C，∴(2b−c)(b2+c2−a2)2bc=a cos C，由余弦定理可得：(2b−c)cos A=a cos C，由正弦定理可得：2sin B cos A−sin C cos A=sin A cos C，∵A+B+C=π，∴2sin B cos A=sin C cos A+cos C sin A=sin(C+A)=sin B．∵sin B≠0．∴cos A=12，由A∈(0,π)，则A=π3．(2)在△BCD中，BD=2，CD=1，由余弦定理得：BC2=12+22−2×1×2cos D=5−4cos D．∵A=∠ABC=π3，∴∠ACB=π3，∴△ABC为等边三角形，∴S△ABC=12×BC2⋅sinπ3=5√34−√3cos D，∵S△BDC=12⋅BD⋅DC⋅sin D=sin D，∴S四边形ABDC =5√34+sin D−√3cos D=5√34+2sin(D−π3)=5√34+2，∴sin(D−π3)=1，∵D∈(0,π)，∴D=5π6．【考点】余弦定理正弦定理两角和与差的正弦公式【解析】【解答】解：(1)∵(2b−c)(b2−a2+c2)=2abc cos C，∴(2b−c)(b2+c2−a2)2bc=a cos C，由余弦定理可得：(2b−c)cos A=a cos C，由正弦定理可得：2sin B cos A−sin C cos A=sin A cos C，∵A+B+C=π，∴2sin B cos A=sin C cos A+cos C sin A=sin(C+A)=sin B．∵sin B≠0．∴cos A=12，由A∈(0,π)，则A=π3．(2)在△BCD中，BD=2，CD=1，由余弦定理得：BC2=12+22−2×1×2cos D=5−4cos D．∵A=∠ABC=π3，∴∠ACB=π3，∴△ABC为等边三角形，∴S△ABC=12×BC2⋅sinπ3=5√34−√3cos D，∵S△BDC=12⋅BD⋅DC⋅sin D=sin D，∴S四边形ABDC=5√34+sin D−√3cos D=5√34+2sin(D−π3)=5√34+2，∴sin(D−π3)=1，∵D∈(0,π)，∴D=5π6．【答案】解：(1)应选择模型①，理由如下：因为模型①每组数据对应的残差绝对值都比模型②的小，残差波动小，残差点比较均匀地落在水平的带状区域内，说明拟合精度高．(2)由(1)知，需剔除第一组数据，得到下表则上表的数据中，x ¯=7.5×6−55=8，y ¯=5.9×6−0.45=7，5x ¯ y ¯=280，5x ¯2=320，∑x i 5i=1y i =299.8−5×0.4=297.8，∑x i 25i=1=355−25=330，所以b ̂=∑x i 5i=1y i −5x ¯ y ¯∑x i 25i=1−5x¯2=297.8−280330−320=17.810=1.78， a ̂=y ¯−b ̂x ¯=7−1.78×8=−7.24，得模型①的回归方程为y =1.78x −7.24，则x =11时，y =1.78×11−7.24=12.34mm ，故光照时间为11ℎ时，该植物的平均增长高度为12.34mm ． 【考点】根据实际问题选择函数类型 求解线性回归方程 【解析】左侧未给出解答解析． 左侧未给出解答解析．【解答】解：(1)应选择模型①，理由如下：因为模型①每组数据对应的残差绝对值都比模型②的小，残差波动小，残差点比较均匀地落在水平的带状区域内，说明拟合精度高．(2)由(1)知，需剔除第一组数据，得到下表则上表的数据中，x ¯=7.5×6−55=8，y ¯=5.9×6−0.45=7，5x ¯ y ¯=280，5x ¯2=320，∑x i 5i=1y i =299.8−5×0.4=297.8，∑x i 25i=1=355−25=330，所以b̂=∑x i 5i=1y i −5x ¯ y¯∑x i 25i=1−5x¯2=297.8−280330−320=17.810=1.78，a ̂=y ¯−b ̂x ¯=7−1.78×8=−7.24，得模型①的回归方程为y =1.78x −7.24， 则x =11时，y =1.78×11−7.24=12.34mm ，故光照时间为11ℎ时，该植物的平均增长高度为12.34mm ． 【答案】(1)证明：AB 是底面圆的直径，AC 与圆切于点A ，所以AC ⊥AB ， 又因为PO ⊥底面ABC ， 所以PO ⊥AC .因为PO ∩AB =O ， 所以AC ⊥面PAB ， 所以AC ⊥PB .又因为在三角形PAB 中，PA =PB =√22AB ， 所以PA ⊥PB .因为PA ∩AC =A ， 所以PB ⊥面PAC . 因为PB ⊂面PBC ，所以平面PBC ⊥平面PAC .(2)解：因为OB ⊥PO ，OD ⊥PO ，所以∠BOD 为二面角B −PO −D 的平面角， 所以∠BOD =2π3，如图建立空间直角坐标系，易知OB =1，则A (0,−1,0)，B (0,1,0), D(√32,−12,0)，C (2√33,−1,0)，P(0,0,1), E (√33,−12,12)， 由(1)可知BP →=(0,−1,1)为平面PAC 法向量． 设平面ODE 的法向量为n →=(x,y,z )， OE →=(√33,−12,12)，OE →⋅n →=0⇒√33x −12y +12z =0，OD →=(√32,−12,0)，OD →⋅n →=0⇒√32x −12y =0，解得：n →=(√3,3,1)， cos θ=|n →⋅BP →||n →||BP →|=2√2×√13=√2613，即平面PAC 与平面DOE 所成锐二面角的余弦值为√2613.【考点】二面角的平面角及求法 用空间向量求平面间的夹角 平面与平面垂直的判定【解析】 此题暂无解析 【解答】(1)证明：AB 是底面圆的直径，AC 与圆切于点A ，所以AC ⊥AB ， 又因为PO ⊥底面ABC ， 所以PO ⊥AC .因为PO ∩AB =O ， 所以AC ⊥面PAB ， 所以AC ⊥PB .又因为在三角形PAB 中，PA =PB =√22AB ， 所以PA ⊥PB .因为PA ∩AC =A ， 所以PB ⊥面PAC . 因为PB ⊂面PBC ，所以平面PBC ⊥平面PAC .(2)解：因为OB ⊥PO ，OD ⊥PO ，所以∠BOD 为二面角B −PO −D 的平面角， 所以∠BOD =2π3，如图建立空间直角坐标系，易知OB =1，则A (0,−1,0)，B (0,1,0), D(√32,−12,0)，C (2√33,−1,0)，P(0,0,1), E (√33,−12,12)， 由(1)可知BP →=(0,−1,1)为平面PAC 法向量． 设平面ODE 的法向量为n →=(x,y,z )，OE →=(√33,−12,12)，OE →⋅n →=0⇒√33x −12y +12z =0，OD →=(√32,−12,0)，OD →⋅n →=0⇒√32x −12y =0，解得：n →=(√3,3,1)， cos θ=|n →⋅BP →||n →||BP →|=2×13=√2613， 即平面PAC 与平面DOE 所成锐二面角的余弦值为√2613. 【答案】解：(1)设焦距为2c ， 已知e =ca =√32，2b =2，a 2=b 2+c 2．∴ b =1，a =2， ∴ 椭圆C 的标准方程为x 24+y 2=1．(2)证明：设M(x 1, y 1)，N(x 2, y 2)， 联立{y =kx +m ，x 24+y 2=1，得：(4k 2+1)x 2+8kmx +4m 2−4=0，依题意，Δ=(8km)2−4(4k 2+1)(4m 2−4)>0， 化简得：m 2<4k 2+1①， x 1+x 2=−8km4k 2+1，x 1x 2=4m 2−44k 2+1，y 1y 2=(kx 1+m)(kx 2+m)=k 2x 1x 2+km(x 1+x 2)+m 2， 若k OM ⋅k ON =54，则y 1y 2x 1x 2=54，即4y 1y 2=5x 1x 2，∴ 4k 2x 1x 2+4km(x 1+x 2)+4m 2=5x 1x 2， ∴ (4k 2−5)⋅4(m 2−1)4k 2+1+4km ⋅(−8km4k 2+1)+4m 2=0，即(4k 2−5)(m 2−1)−8k 2m 2+m 2(4k 2+1)=0， 化简得m 2+k 2=54，②由①②得：0≤m 2<65，120<k 2≤54．∴ 点(m, k)在定圆x 2+y 2=54上． 【考点】椭圆的离心率 椭圆的标准方程圆锥曲线中的定点与定值问题 【解析】（1）设焦距为2c ，由已知e =ca =√32，2b ＝2，a 2＝b 2+c 2联立解得．（2）设M(x 1, y 1)，N(x 2, y 2)，直线方程与椭圆方程联立得(4k 2+1)x 2+8kmx +4m 2−4＝0，依题意，△>0，化简得m 2<4k 2+1，若k OM ⋅k ON =54，则y 1y 2x 1x 2=54，即4y 1y 2＝5x 1x 2，利用根与系数的关系代入化简即可得出．【解答】解：(1)设焦距为2c ， 已知e =c a=√32，2b =2，a 2=b 2+c 2．∴ b =1，a =2，∴ 椭圆C 的标准方程为x 24+y 2=1．(2)证明：设M(x 1, y 1)，N(x 2, y 2)， 联立{y =kx +m ，x 24+y 2=1，得：(4k 2+1)x 2+8kmx +4m 2−4=0，依题意，Δ=(8km)2−4(4k 2+1)(4m 2−4)>0， 化简得：m 2<4k 2+1①， x 1+x 2=−8km4k 2+1，x 1x 2=4m 2−44k 2+1，y 1y 2=(kx 1+m)(kx 2+m)=k 2x 1x 2+km(x 1+x 2)+m 2， 若k OM ⋅k ON =54，则y 1y 2x 1x 2=54，即4y 1y 2=5x 1x 2，∴ 4k 2x 1x 2+4km(x 1+x 2)+4m 2=5x 1x 2， ∴ (4k 2−5)⋅4(m 2−1)4k 2+1+4km ⋅(−8km4k 2+1)+4m 2=0，即(4k 2−5)(m 2−1)−8k 2m 2+m 2(4k 2+1)=0， 化简得m 2+k 2=54，②由①②得：0≤m 2<65，120<k 2≤54． ∴ 点(m, k)在定圆x 2+y 2=54上． 【答案】解：(1)可知f (x )的定义域是(0,+∞)， f ′(x )=1x +m ，若m ≥0，则f ′(x )>0，f (x )在定义域内单调递增，无最大值.若m <0，当x ∈(0,−1m ]时，f ′(x )>0，f (x )单调递增；当x ∈(−1m,+∞)时，f ′(x )<0，f (x )单调递减；∴ 当x =−1m 时，f (x )取得最大值，即ln (−1m )=0， ∴ m =−1．又∵ f ′(e )=1e −1，f (e )=2−e ，∴ 函数f (x )在x =e 处的切线方程为：y =(1e −1)x +1． (2)若f (x )≤g (x )恒成立，即m +1≤e x −ln x+1x在(0,+∞)恒成立.设φ(x )=e x −ln x+1x，则φ′(x )=x 2e x +ln xx 2.设Q (x )=x 2e x +ln x ，则Q ′(x )=(x 2+2x )e x +1x >0，∴ Q (x )在其定义域内单调递增，且Q (12)<0，Q (1)>0， ∴ Q (x )有唯一零点x 0．∴ x 02e x 0+ln x 0=0， ∴ x 0e x 0=−ln x 0x 0，等式两边同时取对数得：x 0+ln x 0=ln (−ln x 0)+(−ln x 0)， 易证明函数y =x +ln x 是增函数， ∴ x 0=−ln x 0，即e x 0=1x 0．∵ φ(x )在(0,x 0]单调递减，在(x 0,+∞)上单调递增， ∴ φ(x )≥φ(x 0)=e x 0−ln x 0+1x 0=1x 0−−x 0+1x 0=1．∴ m 的取值范围是(−∞,0]． 【考点】利用导数研究曲线上某点切线方程 利用导数研究函数的单调性 利用导数研究函数的最值 函数恒成立问题利用导数研究不等式恒成立问题 【解析】无无【解答】解：(1)可知f(x)的定义域是(0,+∞)，f′(x)=1x+m，若m≥0，则f′(x)>0，f(x)在定义域内单调递增，无最大值.若m<0，当x∈(0,−1m]时，f′(x)>0，f(x)单调递增；当x∈(−1m,+∞)时，f′(x)<0，f(x)单调递减；∴当x=−1m 时，f(x)取得最大值，即ln(−1m)=0，∴m=−1．又∵f′(e)=1e−1，f(e)=2−e，∴函数f(x)在x=e处的切线方程为：y=(1e−1)x+1．(2)若f(x)≤g(x)恒成立，即m+1≤e x−ln x+1x在(0,+∞)恒成立.设φ(x)=e x−ln x+1x，则φ′(x)=x2e x+ln xx2.设Q(x)=x2e x+ln x，则Q′(x)=(x2+2x)e x+1x>0，∴Q(x)在其定义域内单调递增，且Q(12)<0，Q(1)>0，∴Q(x)有唯一零点x0．∴x02e x0+ln x0=0，∴x0e x0=−ln x0x0，等式两边同时取对数得：x0+ln x0=ln(−ln x0)+(−ln x0)，易证明函数y=x+ln x是增函数，∴x0=−ln x0，即e x0=1x0．∵φ(x)在(0,x0]单调递减，在(x0,+∞)上单调递增，∴φ(x)≥φ(x0)=e x0−ln x0+1x0=1x0−−x0+1x0=1．∴m的取值范围是(−∞,0]．【答案】解：(1)不等式f(x)<x，即|x−1|−|x+2|<x，①当x≥1时，化简得−3<x，解得：x≥1；②当−2<x<1时，化简得−2x−1<x，解得：−13<x<1；③当x≤−2时，化简得3<x，此时无解．综上，所求不等式的解集为{x|x>−13}．(2)∵|x−1|−|x+2|≤|(x−1)−(x+2)|=3，当且仅当x≤−2时等号成立.∴M=3，即a+4b+9c=1，又∵a，b，c>0，∴1a+1b+1c=(1a+1b+1c)(a+4b+9c)≥(1√a⋅√a1√b⋅√4b1√c√9c)2=(1+2+3)2=36，当且仅当1aa=1b4b=1c9c，即a=16，b=112，c=118时取等号，∴1a+1b+1c的最小值为36.【考点】绝对值不等式基本不等式在最值问题中的应用绝对值三角不等式【解析】【解答】解：(1)不等式f(x)<x，即|x−1|−|x+2|<x，①当x≥1时，化简得−3<x，解得：x≥1；②当−2<x<1时，化简得−2x−1<x，解得：−13<x<1；③当x≤−2时，化简得3<x，此时无解．综上，所求不等式的解集为{x|x>−13}．(2)∵|x−1|−|x+2|≤|(x−1)−(x+2)|=3，当且仅当x≤−2时等号成立.∴M=3，即a+4b+9c=1，又∵a，b，c>0，∴1a +1b+1c=(1a+1b+1c)(a+4b+9c)≥(1√a √a1√b√4b1√c√9c)2=(1+2+3)2=36，当且仅当1aa=1b4b=1c9c，即a=16，b=112，c=118时取等号，∴1a +1b+1c的最小值为36.。
2020-2021学年广西柳州高三上物理月考试卷
2020-2021学年广西柳州高三上物理月考试卷一、选择题1. 2020年1月10日，工程院院士黄旭华获得国家最高科学技术奖，他为核潜艇研制和跨越式发展作出了巨大贡献，下列说法正确的是（）A.核聚变和核裂变都能放出核能，故一切核反应都应放出核能B.核潜艇的核反应堆是利用隔棒控制核反应速度C.核潜艇的核反应也叫热核反应D.原子弹中铀或钚的质量可小于临界质量2. 某变压器原、副线圈匝数比为55:8，原线圈所接交流电源电压按图示规律变化，副线圈接有负载．下列判断正确的是（）A.输出电压的最大值为32VB.变压器输入、输出功率之比为8:55C.交流电源频率为50HzD.原、副线圈中电流之比为55:83. 某带电粒子以速度v垂直射入匀强磁场中，粒子做半径为R的匀速圆周运动，若粒子的速度变为2v，则下列说法正确的是（）A.粒子运动轨迹所包围的磁通量变为原来的4倍B.粒子运动的加速度变为原来的4倍C.粒子运动的周期变为原来的12D.粒子运动的半径仍为R4. a、b两质点同时、同地沿同一直线运动，其v−t图像分别如图中a，b所示，t0时刻a，b两质点速度均为v0，在0∼t0时间内，下列说法正确的是（）A.质点b的平均速度大于v02B.质点a一直做匀加速直线运动C.a、b两质点不存在加速度相同的时刻D.t0时刻两质点相遇5. 2020年计划发射40余颗卫星，堪称中国的航天年．已知某卫星在距地面高为ℎ的轨道绕地球做圆周运动，卫星与地心的连线单位时间内扫过的面积为S，且地球半径为R，忽略地球自转的影响，则地球表面的重力加速度为（）A.4S2R(R+ℎ)2B.2S2R(R+ℎ)2C.2S2R2(R+ℎ)D.4S2R2(R+ℎ)6. 如图所示，A、B、C、D、E为楼梯台阶边缘上的五个点，它们在同一竖直面内，且各级台阶都相同．从A 点沿水平方向先后抛出甲、乙两个小球，甲球刚好可以落到B点，乙球刚好可以落到E点，不计空气阻力，则（）A.甲、乙两球的初速度大小之比为1:4B.甲、乙两球的下落时间之比为1:2C.两小球刚好落到台阶时瞬时速度方向不同D.两小球刚好落到台阶时瞬时速度方向相同7. 避雷针，又名防雷针、接闪杆，是用来保护建筑物、高大树木等避免雷击的装置．图示为避雷针上方有雷雨时避雷针附近的电场线分布示意图，电场线关于直线MN对称．虚线ABCD为一带电粒子在电场中仅在电场力作用下从A运动到D的轨迹，则（）A.该粒子在A点时的加速度比在B点时的加速度大B.该粒子带负电C.运动过程中粒子的电势能先增大后减小D.A点的电势比B点的电势高8. 如图所示，细绳一端固定在O点，另一端拴一小球A，现拉起小球A使细绳水平伸直，然后无初速释放，小球A运动到最低点时，细绳的拉力大小为F．则小球A从开始运动到最低点的过程中，下列说法正确的是（）A.若将细绳变长，小球运动到最低点处时细绳拉力大于FB.若将细绳变长，小球运动到最低点处时细绳拉力仍为FC.小球受到的重力和细绳拉力的合力提供向心力D.小球重力的瞬时功率最大时，小球受到的合力方向水平向左二、实验探究题实验室有一直流电流表，其说明书上附的该电流表的内部电路如图甲所示．（1）某次用0∼3A量程测量时，示数如图乙所示，其读数为________A；（2）该电流表0∼0.6A量程对应的内阻为________Ω（结果保留两位有效数字）；（3）若电阻R1断路，则电流表允许通过的最大电流为________mA．（结果保留两位有效数字）某同学用如图所示的装置做验证动量守恒定律的实验，操作步骤如下：Ⅰ．先将光滑斜槽轨道的末端调至水平，在一块平木板表面先后钉上白纸和复写纸，并将该木板立于靠近槽口处，使小球a从斜槽轨道上某固定点处由静止开始滚下，撞到木板在记录纸上留下压痕O．Ⅱ．将木板向右平移适当距离，再使小球a从原固定点由静止释放，撞到木板在记录纸上留下压痕．Ⅲ．把半径相同的小球b(m a<m b)静止放在斜槽轨道水平段的右边缘，让小球a仍从原固定点由静止开始滚下，与小球b相碰后，a球沿轨道反弹后返回，两球撞在木板上留下压痕．将这些压痕标记为A、B、C，已知b球的压痕为C．（1）本实验必须测量的物理量是________．（填选项前的字母）A.记录纸上O点到A、B、C的距离y1、y2、y3B.斜槽轨道末端到木板的水平距离xC.小球a、b的质量m a、m bD.小球a、b的半径r（2）放上被碰小球b，两球相碰后，小球a在图中的压痕点为________（填“A”或“B”）．（3）用本实验中所测得的物理量来验证两球碰撞过程动量守恒，其表达式为________．三、解答题轻质细线吊着一质量m=0.32kg的单匝线圈，总电阻为r=1Ω．边长为l=0.4m的正方形磁场区域对称分布在线圈下边两侧，如图所示．磁场方向垂直于纸面向里，且B随时间t的关系为B=5t+1(T)，从t=0开始经t0时间细线开始松弛，g=10m/s2，求：（1）在前t0时间内线圈中产生的感应电流大小和方向；（2）在前t0时间内流过线圈横截面的电量大小．某地有一倾角为θ=37∘的如图所示的山坡，山坡上面有一质量为m=1kg的石板B，其上下表面与斜坡平行；板上有一质量也为m=1kg小石块A．已知A与B之间的动摩擦因数为μA=0.3，B与斜坡间动摩擦因数为μB= 0.5，斜坡足够长．系统在外力约束下处于静止状态．现在t=0s时刻撤去约束并同时给B施加一个沿斜面向下的冲量I=12.8N⋅s，设最大静摩擦力等于滑动摩擦力，取重力加速度大小g=10m/s2，sinθ=0.6，cosθ=0.8．（1）求t=0s时A的加速度大小．（2）为保证A不从石板B上端滑出，求石板B的最小长度．（3）假设石板B足够长，求从t=0s时开始B对A的冲量大小．如图所示，体积为V0，内壁光滑的圆柱形导热气缸，气缸顶部有一厚度不计的轻质活塞，气缸内壁密封有密度为ρ，温度为3T0，压强为1.5p0的理想气体（p0和T0分别为大气压强和室温），设容器内气体的变化过程都是缓慢的，气体的摩尔质量为M，阿伏加德罗常数为N A．（1）求气缸内气体与外界大气达到平衡时的体积V1．（2）气缸内气体的分子的总个数N．参考答案与试题解析2020-2021学年广西柳州高三上物理月考试卷一、选择题1.【答案】此题暂无答案【考点】裂变反于和聚侧反应核三应实【解析】此题暂无解析【解答】此题暂无解答2.【答案】此题暂无答案【考点】变压器于构造虫原理交变常流的画象和氮角保数表达式【解析】此题暂无解析【解答】此题暂无解答3.【答案】此题暂无答案【考点】带电根气刚匀强句场中的运动规律【解析】此题暂无解析【解答】此题暂无解答4.【答案】此题暂无答案【考点】v-于图像（匀变技偶问运动）【解析】此题暂无解析【解答】此题暂无解答5. 【答案】此题暂无答案【考点】随地体绕某问题【解析】此题暂无解析【解答】此题暂无解答6.【答案】此题暂无答案【考点】平抛来动基稳造或及推论的应用【解析】此题暂无解析【解答】此题暂无解答7.【答案】此题暂无答案【考点】电根中萄大政问题的分析【解析】此题暂无解析【解答】此题暂无解答8.【答案】此题暂无答案【考点】竖直常内的头范运动立轻绳模型【解析】此题暂无解析【解答】此题暂无解答二、实验探究题【答案】此题暂无答案【考点】电流表来装电斗表【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】利用平使运动规香鸟证洗量守恒定律【解析】此题暂无解析【解答】此题暂无解答三、解答题【答案】此题暂无答案【考点】法拉第体磁目应定律安较脱学花下的平衡问题【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】用牛根盖动识和分析斜面体模型板块使以问题动量定表的病本应用【解析】此题暂无解析【解答】此题暂无解答【答案】此题暂无答案【考点】“汽缸常塞陆”模型【解析】此题暂无解析【解答】此题暂无解答。
2020-2021学年广西柳州高级中学高三英语模拟试题及参考答案
2020-2021学年广西柳州高级中学高三英语模拟试题及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ASome young people win attention because of their good looks or their singing ability. A much smaller number gain fame because they have done something important and worthwhile with their abilities. Rishab Jain is among the latter. In 2018, 13-year-oldRishab developed a way to use AI technology to help pancreatic(胰腺的) cancer patients and won the3MYoung Scientist Challenge, a nationwide middle-school science competition, and its $25,000 prize.In the last stage of the contest, Rishab competed againstnine other finalists at the 3M Innovation Center(创新中心) in St.Paul,Minnesota. Leading up to the big meet, each finalist had partnered with a scientist to further develop their inventions.Rishab explains what led him to create his invention. First,a family friend died of cancer. Then Rishab learned about how deadly pancreatic cancer is, and that its low survival rate is due to how difficult it is to treat. "I'm also into programming, so I was learning about AI technology. I decided to try to solve a real-world problem using it."His winnings have been put in further research and in his nonprofit Samyak Science Society, which helps poor children enter the STEM (science, technology, engineering and math) education. Rishab is also raising awareness about pancreatic cancer. These efforts make him quite different from teenagers of his age. Considering becoming a biomedical engineer or a doctor一or both, he has also put some money aside to further his own learning. Almost certainly the doors of higher education will open wide to him before he even knocks.That's an outstanding outlook for one so young. Rashib is committed to helping very sick people in need. He is also providing teenagers of his age with a much-needed model of what kinds of things youth can achieve.1. What can we learn about the 3M Young Scientist Challenge in 2018?A. It was intended to solve medical problems.B. It was a nationwide AI competition for teenagers.C. It offered the finalists an opportunity to work with scientists.D. It allowed the finalists to learn AI technology in the 3M center.2. How did Rishab win the 3M Young Scientist Challenge?A.He showed excellent programming ability.B. He figured out the survival rate of pancreatic cancer.C. He introduced poor children to STEM education.D. He applied AI technology to treating pancreatic cancer.3. Which of the following best describes Rishab?A. Talented and caring.B. Independent and humorous.C. Responsible and patient.D. Polite and inspiring.BAs a rider, Anna Kiesenhofe’s Olympics victory might be a surprise. The winner of the road race at the Tokyo Olympics left the sport at the end of 2017 when she found herself out of contract (合同). She came into Tokyo without a professional team and left as an Olympic champion.The 30-year old began her cycling career in 2014 after running injuries that prevented her from continuing her pursuits of triathlon (铁人三项). She later joined a Catalan team and won the Spanish National Cup in 2016.The then-26 year old signedher first professional contract with Lotto Soudal Ladies for the following season. However, she ended her 2017 campaign in April and did not sign a contract for 2018, eventually taking a year off the bike. In 2019, Kiesenhofer came back to the sport as a rider, winning the Austrian national road race. Despite her results, Kiesenhofer sill had no professional contract while going into the Tokyo Olympics.Kiesenhofer was the first rider to attack in the Olympic road race, eventually forming a breakaway along with Carl Oberholzer, Omer Shapira, Vera Looser and Anna Plichta, which went on to reach a gap of 11 minutes. After Looser and Oberholzer were dropped, Kiesnhofer ataced her two remaining breakaway companions.After Shapira and Plichta were caught by the peloton (主车群), the rest of the riders seemed to believe that they were racing among themselves for Gold, not knowing that Remehofere was still in front. While it might be a misjudgment from the rest of the peloton that allowed Kiesenhofer to keep her lead of more than two minutes, other riders’ mistakes should not detract from the Austrian’s efforts.Off the bike the new Olympic Champion has a PhD in mathematics after studying at the Technical University of Vienne as well as at Cambridge University. She currently works at the University of Lausanne.4. Why did Anna give up triathlon?A. She got injured.B. She lost interest in it.C. She had to attend university.D. She never won a medal.5. Which is the right order of the following events?①She ended her campaign.②She took a year off the bike.③She began her cycling career.④She won the Austrian national road race.⑤She won the Spanish National Cup.A. ③④①②⑤.B. ②③④①⑤.C. ③⑤①②④.D. ④②③①⑤.6. What were the riders of the peloton unaware of at the Tokyo Olympics?A. The road race was so difficult.B. Anna was a new rider.C. They had caught up with Anna.D. Anna took the lead of them.7. What is Anna’s present job?A. A cycling coach.B. A university teacher.C A professional rider. D. A college student.CIt’s become an accepted part of keeping up to date with extended family and friends, but if schoolchildren were in their parents’ shoes, the majority wouldn’t share posts of their sons and daughters online. Over 55% said they would not upload news about, or images of, their children to their social media feeds, according to survey of over 16,000 pupils by Votes for Schools.While some were concerned about being embarrassed or the longevity of content which could remain online indefinitely, others expressed concern about their personal data beingcompromised. One of the pupils surveyed said, “Although our parents mean well, sometimes theconsequenceof a post can be disastrous.”In response to thesurvey, children’s mental health charity Place2Be and law firm Mishcon de Reya have produced three films ahead of the Christmas holidays – the best period for parental oversharing.In one video, about safety online, 10-year-old Adavan said, “If you share anything with your family, you know who’s going to see it. But if you share it publicly, there are millions of people who can see your picture.”Joe Hancock, a security lead at Mishcon de Reya, encouraged parents to share wisely. “Simple steps, such as checking your privacy settings and asking others not to share content of your children on their accounts if they have not updated their privacy settings, are a good start. And, as we found out from the children during filming, having their permission is key,” he said.The study marks a shift away from the usual debate about teaching children to use the Internet safely. Sandra Davis, head of the law firm’s family department, said, “Children are the experts on the real and immediate impactof sharenting（晒娃）– the full extent of which we cannot know yet. We must ensure we listen to children and take their views into consideration now in order to avoid any unintendedconsequences further down the line.”8.What’s most pupils’ attitude towards sharing posts about children?A. Skeptical.B. Unconcerned.C. Tolerant.D. Unwilling.9.What does the author intend to do in Paragraph 2?A. Summarize the previous paragraph.B. Provide some advice for parents.C. Predict the consequence of sharing posts.D. Give reasons for the result of the survey.10.What should parents first pay attention to according to Joe Hancock?A. Safety.B. Consequence.C. Wisdom.D. Health.11.What can we learn from the last paragraph?A. Parents should take children’s opinions into account.B. Parents should teach children to use the Internet safely.C. Children must make sure to listen to their parents.D. Children should avoid unintended consequences of sharenting.DGetting drunk on ice cream used to be the stuff of dreams, but thanks to Will Rogers, inventor and owner of WDS Dessert Stations in Hinkley, Illinois, it has become a delicious reality. The Below Zero icecream machine uses a unique technique to freeze alcohol, which allows you to turn beers, cocktails and even spirits (烈酒) into delicious soft —serve ice cream.Rogers was trying to create a highly — caffeinated espresso ice cream flavor when he realized hecould use the same technique with alcoholic beverages. He started experimenting with various gums and stabilizers commonly used in the ice cream industry and eventually patented something called the NEA gel. It’s this magical concoction (调制品) that allows the alcohol to freeze to a near solid inside the Below Zero ice cream machine.Even though Below Zero changes the texture (质地) of beer, cocktails and even spirits, essentially turning them into soft —serve ice cream, it does not affect the alcohol contentat all. The ABV (酒精度) remains exactly the same, which means you can get drunk on ice — cream just as you would on the same concoctions in liquid form.Will Rogers claims that it takes around 30 minutes for beer to go from liquid to ice cream form, but higheralcohol content drinks take longer. Essentially, the higher the alcohol level, the longer the wait.The American inventor plans to sell Below Zero ice cream machines to bars and breweries wanting to surprise their patrons. Metro reports that machines will sell for about 6,000.12. What’s the name of the machine which can change beer and spirits into ice cream?A. Will RogersB. WDS Dessert StationsC. HinkleyD. Below Zero13. What makes alcohol to freeze to a near solid inside the machine?A. gums.B. stabilizers.C. NEA gel.D. ABV.14. What can we know from the passage?A. The machine can change all liquids into ice cream.B. It takes 20 minutes for beer to change into ice cream.C. The higher the alcohol level, the shorter the wait will be.D. The machine changes the texture of beer, cocktails and even spirits.15. What can we infer from the passage?A. The machine affects the alcohol content.B. You can get drunk if you have ice—creams made from spirits.C. The American inventor doesn’t want to sell themagical machine.D. Bars and breweries will not become potential buyers of the machine.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2020-2021学年广西柳州高级中学高三英语模拟试题及答案
2020-2021学年广西柳州高级中学高三英语模拟试题及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AHubeiProvincehas long been a favorite Chinese tourist destination because of its natural scenery, historic cities, and beautiful mountains. Here are some of the best places to visit inHubeiProvince.YellowCraneTowerYellowCraneToweris known as one of the Three Famous Towers South of Yangtze River, together withYueyangTowerinHunanand Tengwang Pavilion inJiangxi. The history ofYellowCraneTowerdates back to the Three Kingdoms period (220-280) in Chinese history. The tower was used as a watchtower by the King of Wu's army in the beginning.Enshi Grand CanyonEnshi Grand Canyon can beChina's answer to the Grand Canyon inArizona, theUnited Statesin beauty. The canyon runs 108 kilometers and occupies a land area of 3,000 square kilometers. The region where Enshi Grand Canyon is located used to be a vast sea with many limestone deposits 230 million years ago.Shennongjia Scenic AreaIt is famous for its varied plant species as well as mountains. Regarded as the “Lungs of Central China", the forest coverage reaches over 90 percent of the area. Due to the special climate, it is neither too hot in summer nor too cold in winter. Sometimes clouds stretch around mountains, rewarding tourists with unforgettable views. Besides, the area is home to some rare animals such as golden monkeys, white bears and antelope.The Three-Gorge Tribe scenic spotLocated in the area of the Xiling Gorge in the city ofYichang, it has the beauty of landscape paintings. The spot is a “ProtectedCenterof the Popular Culture and Art of the Three Gorges”. Since ancient times, a lot of famous scholars have produced a great number of excellent poems praising the beauty of this place, some of which are carved on the stones along theYangtze River.1. What is special aboutYellowCraneTower?A. It has a long history.B. It offers a scenic view.C. It once served war's purposes.D. It is the best-known tower inChina2. Why is the Grand Canyon inArizonamentioned ?A. To explain its popularity inChina.B. To help readers know more about it.C. To prove the beauty of the Chinese canyon.D. To show differences between the two canyons.3. Which place will a Chinese literature lover probably visit?A. Enshi Grand Canyon.B.YellowCraneTower.C. Shennongjia Scenic Area.D. TheThree-Gorge Tribe scenic spot.BMany cars in advertisements and on exhibition in the United States are red, blue or green, but almost 75 percent of new cars sold in the United States are black, white, silver orgray.Les Jackson is a reporter who writes about cars. He says the color1 s of cars Americans choose do not show dirt. He says that means the owners wash their cars less in order to save money. And he notes some areas that are suffering from water shortages do not permit people to wash their cars often.Dan Benton works for a company called Axalta, which makes supplies for international car makers. He says white cars are often sold more expensive than cars of other color1 s. And he notes that white cars “absorb(吸收)less energy” than cars of other color1 s. This means temperatures inside them are lower in warmer areas. Benton also says research at Monash University in Australia suggests that there is a lower risk of crashes during the day for white cars compared with darker ones.Car buyers in other countries also like white. Jane Harrington works for PPG Industries, a company that makes paint for cars. She said in China, buyers say white makes a small car look bigger.About 11 percent of cars sold in North America are red and 8 percent are blue. Green has become less popular. Benton notes that in the mid-1990s green was the most popular color1 in North America. Today, green is hard to find.Sometime in the future, people may not have to choose the color1 of their cars —— technology may let owners change their cars’ paint color1 anytime.4. What can we learn from Paragraph 2?A. Most Americans don’t like red cars.B. People in America are not allowed to wash their cars.C. Many people prefer to choose white cars in America.D. Americans may consider the cost of cleaning when choosing cars.5. Why do many people choose white cars?A. They are much cheaper than cars of other color1 s..B. They are much safer while crashing.C. They are bigger than cars of other color1 s.D. They are more comfortable inside in warmer areas.6. What do we know from the text?A. Les Jackson is a member of Axalta.B. Most Americans rarely wash their cars.C. PPG Industries mainly produces cars in China.D. Green cars were once popular in North America.7. What does the text mainly tell us?A. Choices of car color1 sB. How to buy a good car.C. Differences of car color1 s.D. Popular car color1 s in history.CCraig Blackburn, a father and car fan, built a Batmobile for his son’s hope for using the vehicle to brighten the lives of sick children. And now he hopes to use it for more than just his sons hope after seeing the childrens reaction to the Batmobile.Based on the number of failures he had seen in car groups, he estimated that only about one in 50 attempted constructions was actually finished and he realized what an incredible opportunity he had.Mr. Blackburn started the project at the beginning of 2018 after hearing a friend in the US was doing the same thing. It started with importing an outer shell overseas, before picking brains of a friend who had a background as a worker in a car factory to gain knowledge of how to build the car. With the help of his friend, Mr. Blackburn built the Batmobile in 18 months with the cost reaching six figures.Mr.Blackburn hoped to add a flamethrower(喷火器)onto the back of the vehicle and said he had thought about building the more recent Batman Tumbler from the series film Dark Knight. Though Mr. Blackburn encountered plenty of difficulties to get over during the construction, in September 2019, the carmade its firstshow at the Carnival of Flowers in Toowoomba, before being used by Blackburn’s son for his hope.“It was great. It was so good to see the kids’ and adults’ excitement at seeing the Batmobile.” Mr. Blackburn said. As a result, the car lovers hope to make the car work on the roads as soon as possible, so he can visit sick children and take them out with his son.8. What is Craig Blackburn’s initial purpose of making the Batmobile?A. To realize his son’s dream.B. To donate it to sick children.C. To pay his respects to the film Dark Knight.D. To show off at the Carnival of Flowers in Toowoomba.9. How did Blackburn feel about the car-making at first?A. Hopeful.B. Confused.C. Impossible.D. Unsure.10. How did Blackburn’s friend help him?A. By making an outer shell for him.B. By offering him financial support.C. By sharing the knowledge of building cars.D. By telling him the background of the car factory.11. What is the car lovers’ expectation of the Batmobile?A. It will be driven soon on the roads.B. It can be displayed around the world.C. It can change the lives of sick children.D. It will appear in the next film about Batman.DWhy do you check social media? Is it to keep up with everything that your friends and family are doing? Is it to find new trendy spots to eat?Regardless of the reason, you may find yourself with different degrees of envy or discomfort after a quick look at your phone. Then you might be suffering from a phenomenon known as “Fear of Missing Out (FOMO). ”While the phenomenon of FOMO can be traced back for centuries, it had never been the issue as it is today, causing widespread discussion and research. This rise in checking social media is naturally connected to the increasing leading position that social media holds over our lives. Every time someone opens their WeChat Moments, Facebook, Twitter... etc. , they are bombarded with the highlight reels of other peoples' lives. A sunnybeach, delicious-looking food, a super cute kitten-they are all uplifting photos, yet they're very likely to bring about more unhappiness than joy. You see, the first thought to come out of your mind may be “Wow, that's so cool/delicious/cute”, but then it takes a hard U-turn. You're thinking: “I wish I were there” or “my life is so boring compared to his/hers.” The more you see, the more likely you are to have these negative feelings.What's worse is the habit many people have of turning to social media in search of happiness when they are feeling down, not realizing that they are just going to end up in a negative cycle of endless disappointment. Montesquieu once said: “If one only wished to be happy, this could be easily accomplished; but we wish to be happier than other people, and this is always difficult, for we believe others to be happier than they are.”FOMO will go hand in hand with dissatisfaction and envy. Appreciate what you already have, because someone else out there in the world would gladly give everything to be you.12. What is the purpose of the questions in paragraph 1?A. To introduce the topic of the passage.B. To explain the function of FOMO.C. To describe the features of FOMO.D. To give the reasons for checking social media.13. How might people feel seeing other's perfect life through social media?A. Joyful.B. Admiring.C. Comfortable.D. Envious.14. What can we conclude from Montesquieu's words in paragraph 3?A. We could turn to social media for happiness.B. We couldn't realise our dream without hard work.C. We couldn't harvest happiness through comparison.D. We could live better than others by showing ourselves online.15. What is the author's attitude towards FOMO?A. Indifferent.B. Objective.C. Doubtful.D. Hopeful.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。