【附加15套高考模拟】【全国百强校】四川省南充高级中学2020届高三2月线上月考数学(理)试题含答案

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2020年四川省南充高中高考数学模拟试卷(文科)(2月份) (含答案解析)

2020年四川省南充高中高考数学模拟试卷(文科)(2月份)一、单项选择题（本大题共12小题，共60.0分）1.已知复数z满足(z−i)(−i)=5，则z=()A. 6iB. −6iC. −6D. 62.已知集合A={(x,y)|y=x2}，B={(x,y)|2x−y−1=0}，则A∩B=()A. x=1，y=1B. (1,1)C. {1,1}D. {(1,1)}3.从A，B，C三个同学中选2名代表，则A被选中的概率为()A. 13B. 14C. 12D. 234.已知α∈[π,3π2]，sinα=−35，则tanα=()A. −43B. 43C. −34D. 345.已知双曲线m y2−x2=1( m∈R )与椭圆y25+x2=1有相同的焦点，则该双曲线的渐近线方程为()A. y=±√3xB. y=±√33x C. y=±13x D. y=±3x6.设向量a⃗=(m,1),b⃗ =(1,−3)，且，则m=()A. 3B. −2C. 1或−2D. 1或37.在△ABC中，内角A，B，C所对的边分别是a，b，c，若A=60°，b=16，此三角形的面积S=220√3，则a的值为()A. 7B. 25C. 55D. 498.已知函数f(x)=√3sinx+2cos2x2，则下列说法正确的是()A. 函数f(x)的最小正周期为2π，最大值为2B. 函数f(x)图象的对称轴方程为x=π3+kπ，k∈Z，对称中心为(−π6+kπ,1)，k∈ZC. 函数f(x)的最小正周期为2π，最小值为−2D. 函数f(x)图象的对称轴方程为x=−π6+kπ，k∈Z，对称中心为(π3+kπ,1)，k∈Z9.在底面是正三角形，侧棱与底面垂直的三棱柱ABC−A1B1C1中，若AB=√2BB1，则AB1与C1B所成角的大小为()A. 300B. 450C. 600D. 900 10. 已知f(x)是定义在上的奇函数，满足f(x)=f(2−x).若f(1)=3，则f(1)+f(2)+f(3)+⋯+f(2019)=( ) A. −2019 B. 0 C. 3 D. 201911. 已知点P(a,b)是抛物线x 2=20y 上一点，焦点为F ，|PF|=25，则|ab|=( )A. 100B. 200C. 360D. 40012. 若函数f (x )=e −x +tlnx 有两个极值点，则实数t 的取值范围是( )A. (0,1e )B. (−∞,1e )C. (−1e ,0)D. (1e ,+∞) 二、填空题（本大题共4小题，共20.0分）13. 若实数x,y 满足约束条件{x +2y ⩾0x −y ⩽0x −2y +2⩾0，则z =3x −y 的最小值等于_____．14. 已知函数f(x)=ae x +x +b ，若函数f(x)在(0,f(0))处的切线方程为y =2x +5，则ab 的值为___．15. 台风中心从A 地以每小时20千米的速度向东北方向移动，离台风中心30千米的地区为危险区，城市B 在A 地正东40千米处，则城市B 处在危险区内的时间是______ ．16. 在三棱锥S −ABC 中，SA ⊥平面ABC ，SA =2,AB =1,BC =2√2,AC =√5，则该三棱锥的外接球表面积为________．三、解答题（本大题共7小题，共82.0分）17. 已知数列{a n }满足a 1=1，na n+1−2n(n +1)−(n +1)a n =0，设b n =a n n ，n ∈N ∗．(Ⅰ)证明：{b n }是等差数列；(Ⅱ)求数列{b n2}的前n 项和T n ．18.随着社会发展，淮北市在一天的上下班时段也出现了堵车严重的现象。

四川省南充高级中学2020届高三2月线上月考数学(文)试题含答案

2017级⾼三寒假⾼考数学试卷（⾼）⾼、选择题（本题共12⾼题，每⾼题5分，共60分）1.在复平⾯内，复满，的共轭复数对应的点位于（）A.第⾯象限B.第⾯象限C.第三象限D. 第四象限2.已知集，，的元素个数是（）A.4B.3C.2D.13.质监部⾯对2辆新能源汽⾯和3辆燃油汽⾯进⾯质量检测，现从中任选2辆，则选中的2辆都为燃油汽⾯的概率为（）A.0.3B.0.4C.0.5D.0.64. 若，且，=（）A. B. C. D.5.已知椭圆与双曲线的焦点相同，则双曲线的渐近线⾯程为（）A. B. C. D.6.已知向, ，则A.8B.6C.-6D.-87.在解三⻆形的问题中，其中⾯个⾯较困难的问题是如何由三⻆形的三边直接求出三⻆形的⾯积。

据说这个问题最早由古希腊数学家阿基⾯德解决的，他得到了海伦公式，即,其中.我国南宋著名数学家秦九韶（约1202-1261）在《数学九章》⾯⾯给出了⾯个等价解法，这个解法写成公式就是,这个公式中应该是（）A. B. C. D.8.已知函数，则下列结论正确的是（）A. 的最⾯值为1B. 的最⾯正周期为2C. 的图象关于对称D. 的图象关于对称9.在直三棱中，,则异⾯直与所成⻆的⾯⾯为（）A.30°B.60°C.90°D.120°10.已知函为定义在R上的奇函数是偶函数，且时，则=()A.-3B.-2C.-1D.011.已知O为坐标原点，抛物线上⾯点A到焦点F的距离为6，若P为抛物线C准线上的⾯个动点，则的最⾯值为（）A.4B.C.D.12.已知函数有两个不同的极值点，则实的取值范围为（）A. B. D.⾼、填空题（本题共4⾼题，每⾼题5分，共20分）13.已知实满⾯不等式组,的最⾯值为14.已知函的图像在处的切线⾯程,=15．代号为“狂飙”的台⾯于某⾯晚8点在距港⾯的A码头南偏东60°的400千⾯的海⾯上形成，预计台⾯中⾯将以40千⾯／时的速度向正北⾯向移动，离台⾯中⾯350千⾯的范围都会受到台⾯影响，则A码头从受到台⾯影响到影响结束，将持续多少⾯时16.已知所在的平⾯与矩形所在的平⾯互相垂直，且满，则多⾯的外接球的表⾯积为三、解答题（本题共6⾼题，共70分，请在指定位置写出解答过程）17.（本⾯题满分12分）设数列满⾯，其中.（1）证明：是等⾯数列；，求使成⾯的最⾯⾯然数n 的值.（2）令，设数列的前n 项和为，， 18.（本⾯题满分12分）交通拥堵指数是综合反映道路⾯畅通或拥堵的概念，记交通拥堵指数，其范围为，分别有五个级别畅通基本畅通轻度拥堵中度拥堵严重拥堵．晚⾯峰时段( )，从某市交通指挥中⾯选取了市区个交通路段，依据其交通拥堵指数数据绘制的直⾯图如图所示．（1）⾯分层抽样的⾯法从交通指数在级别路段的个数；的路段中共抽取6个路段，求依次抽取的三个（2）从（1 ）中抽出的6个路段中任取2个，求⾯少有1个路段为轻度拥堵的概率．.本⾯题满分12分）如图1所示，在等腰梯形ABCD 中，，，垂⾯为E ，，将沿EC 折起到的位置，如图2所示，使平⾯平⾯ABCE ．（1）连结BE ，证明平⾯；（2）在上是否存在点G ，使得平面若存在，直接指出点G 的位不必说明理，并求出此时三棱的体积；若不存在，请说明理由．（本⾯题满分12分）如图，已知抛物线的焦点是，准线.抛物线上任意⾯点M 到y 轴的距离⾯到准线的距离少2 （1）写出焦的坐标和准的⾯程；（2）已知，若过点的直线交抛物于不同的两点A 、B（均不.直线PA,PB 分别交准线于点M,N,求证21.（本⾯题满分12分）已知函数为⾯然对数的底数），．（1）当时，求函数的极⾯值；（2）若当时，关的⾯程有且只有⾯个实数解，求实的取值范围．选考题：共10分．请考⾼在第22、23题中任选⾼题作答．如果多做则按所做的第⾼题计分．22．在新中国成周年国庆阅兵庆典中，众多群众在脸上贴着⾯颗红⾯，以此表达对祖国的热爱之情. 在数学中，有多种⾯程都可以表示⾯型曲线，其中有著名的笛卡尔⾯型曲线.如图，在直⻆坐标系中，以原点为极点轴正半轴为极轴建⾯极坐标系。

【附加15套高考模拟试卷】四川省南充市2020届高三二诊测试(数学理)试题含答案

四川省南充市2020届高三二诊测试（数学理）试题一、选择题：本题共12小题，每小题5分，共60分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1．设函数()()212log1f x x=+112x++，则使得()()21f x f x≤-成立的x的取值范围是（）A．(],1-∞B．[)1,+∞C．1,13⎡⎤⎢⎥⎣⎦D．[)1,1,3⎛⎤-∞⋃+∞⎥⎝⎦2．函数()2sin()(0)3f x xπωω=+>的图象在[0,1]上恰有两个最大值点，则ω的取值范围为（）A．[2,4]ππ B．9[2,)2ππC．1325[,)66ππD．25[2,)6ππ3．如图是为了求出满足321000->n n的最小偶数n，那么在和两个空白框中，可以分别填入( )A．1000>A和1=+n n B．1000>A和2=+n nC．1000≤A和1=+n n D．1000≤A和2=+n n4．在ABC∆中，cos cosa Ab B=，则ABC∆的形状为（）A．等腰三角形B．直角三角形C．等腰或直角三角形 D．等腰直角三角形5．在平面直角坐标系中，(4,0),(1,0)A B--，点(,)(0)P a b ab≠满足||2||AP BP=，则2241a b+的最小值为（）A．4 B．3 C．32D．946．已知函数32()(0)g x ax bx cx d a=+++≠的导函数为()f x，且230a b c++=，(0)(1)0,f f>设12,x x 是方程()0f x=的两根，则12x x-的取值范围是（）A．2 [0,)3B．4[0,)9C．12(,)33D．14(,)997．若直线l不平行于平面a，且l a⊄，则A．a内的所有直线与l异面B．a内不存在与l平行的直线C．a内存在唯一的直线与l平行D．a内的直线与l都相交8．我国古代数学典籍《九章算术》第七章“盈不足”中有一问题：“今有蒲生一日，长三尺，莞生一日，长一尺.蒲生日自半.莞生日自倍.问几何日而长等？”（蒲常指一种多年生草本植物，莞指水葱一类的植物）现欲知几日后，莞高超过蒲高一倍.为了解决这个新问题，设计如图所示的程序框图，输入3A=，1a=.那么在①处应填_______和输出i的值为（）A．2?S T> 4 B．2?S T< 4C．2?T S> 3 D．2?T S< 39．如图为我国数学家赵爽（约3世纪初）在为《周髀算经》作注时验证勾股定理的示意图，现在提供5种颜色给其中5个小区域涂色，规定每个区域只涂一种颜色，相邻区域颜色不相同，则不同的涂色方案共有（）种A．120 B．260 C．340 D．42010．已知函数2(1),0()43,0xe xf xx xx+⎧≤⎪=⎨+->⎪⎩，函数()y f x a=-有四个不同的零点，从小到大依次为1234,,,x x x x则1234x x x x++的取值范围为（）A．(]5,3+eB．[4,4)e+ C．[)4+∞，D．(4,4)e+11．已知集合{}*230A x N x x =∈-<，则满足条件B A ⊆的集合B 的个数为（） A ．2B ．3C ．4D ．812．已知函数()()()31ln 3ln 3xx f x x ⎡⎤=-⎢⎥⎢⎥⎣⎦g ，且()20f x ->，则实数x 的取值范围是（） A ．(),2-∞ B ．()2,+∞C ．()(),22,-∞+∞U D ．(),-∞+∞二、填空题：本题共4小题，每小题5分，共20分。

四川省南充高中2020届高三数学(文)第二次月考试卷(含答案)

A． 2 2
B．8
C．9
D．
8．执行如图所示的程序框图，输出的 S 值为（）
高 2017 级数学（文）第 1页共 4 页
A． 3
B． 6
C． 10
D． 15
9．在区间 0，2 上任取两个数，则这两个数之和大于 3 的概率是（）
A． 1 8
B． 1 4
C． 7 8
D． 3 4
10．已知函数
A．V1 2V2
B． V1 2V2
C． V1 V2 163
D． V1 V2 173
12．已知函数 f (x) ln x (x b)2 (b R) .若存在 x [ 1 , 2] ，使得 f (x) x f (x) ，则实数 b 的取值
x
2
范围是（）
A． (, 2)
B． (, 3) 2
2
2
（1）求 B．
（2）若 b 3 ，求 △ABC 面积的最大值．
18．（12 分）某高校共有学生 15000 人，其中男生 10500 人，女生 4500 人，为调查该校学生每周平均体育运动时间的情况，采用分层抽样的方法，收集 300 名学生每周平均体育运动时间的样本数据单位：小时．
Ⅰ 应收集多少位女生的样本数据？ Ⅱ 根据这 300 个样本数据，得到学生每周平均体育运动时间的频率分布直方图如图所示，其中样本数据的分组区间为：[0,2], (2,4] ,(4,6], (6,8], (8,10], (10,12],估计该校学生每周平均体育运动时间超过 4 小时的概率； Ⅲ 在样本数据中，有 60 位女生的每周平均体育运动时间超过 4 小时，请完成每周平均体育运动时间与性别列联表，并判断是否有 95%的把握认为“该校学生的每周平均体育运动时间与性别有关”

四川省南充高级中学20222023学年高三上学期第二次模拟考试英语试题2

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。
听下面一段长对话，回答小题。
6. Why did the woman do better in the test than the man?
南充高中高2020级高三第二次模拟
英语试题
第一部分听力(共两节，满分30分)
做题时，先将答案标在试卷上。录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。
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。
A.Just Dance.B.Poker Face.C.Bad Romance.
第二部分阅读理解（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和D四个选项中，选出最佳选项。
A
Urban gardens, which provide green spaces to grow sustainable food, make new friends, connect with the earth, and much more, are invaluable to munities. Let’s check out our list of 4 inspiring urban gardens in the US.
Baltimore Urban Gardening with Students

四川省南充高级中学2022-2023学年高三上学期第三次模拟考试理综试题2

＋) = c(Cl－)
第Ⅱ卷(共58分)
二、非选择题：包括必考题和选考题两部分。第8～10题为必考题，每个试题考生都必须作答。第11～12题为选考题，考生根据要求作答。
(一)必考题(共3题，43分)
8.三氯氧磷(POCl3)可用作半导体掺杂剂及光导纤维原料，是能推进中国半导体产业链发展壮大的一种重要的化工原料。工业上可以直接氧化PCl3制备POCl3，反应原理为：P4(白磷)+6Cl2 4PCl3，2PCl3+O2 2POCl3。
①恒温恒压下，不能说明上述反应达到平衡状态是________。
A．容器内混合气体的密度保持不变B．CH3OH(g)的体积分数保持不变
C．CO(g)的物质的量浓度保持不变D．v(CO)正= v(CH3OH)逆
②温度为T1时，该反应中CH3OH的反应速率：v(A)正________v(B)逆(填“>”，“=”或“<”)。
2.罗勒烯是一种无环单萜类化合物，主要存在于罗勒油等植物精油中，其结构如图a所示；叔丁基苯是一种色谱分析标准物质，其结构如图b。下列有关罗勒烯和叔丁基苯的的说法错误的是
A.两种物质均难溶于水，易溶于机溶剂
B.罗勒烯分子中所有碳原子可能共平面
C.叔丁基苯的一氯代物有4种
D.罗勒烯与叔丁基苯互为同分异构体
③温度为T1时，上述反应已达到平衡，若此时保持容器体积不变，再通入0.10 molCH3OH和0.10 molCO 混合气体，再次达到平衡，CO的转化率________80%(填“>”，“=”或“<”)。
④温度由T1到T2，经历5 min时，CH3OH的转化率增大的原因是________。
II．乙酸制氢具有重要意义，制氢过程发生如下反应：

四川省南充高级中学2020届高三数学考前模拟考试试题文

南充高中2020届高三考前模拟考试高三数学（文）试卷本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，满分150分，考试时间120分钟。

第Ⅰ卷（选择题，60分）一、选择题：本题共12小题，每小题5分，共60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1{|ln }2A x x =≤，{|210}B x x =-≤，则A B =I （）A ．1(,]2-∞ B ．(,e]-∞C ．(0,e]D ．1(0,]2 2.已知复数()3biz b R i-=∈的实部和虚部相等，则z =( ) A ．2 B ． 3 C ． 22 D ． 32 3.下列命题正确的个数是（）①命题“2000,13x R x x ∃∈+>”的否定是“2,13x R x x ∀∈+≤”；②函数()22cos sin f x ax ax =-的最小正周期为π是“1a =”的必要不充分条件；③22x x ax +≥在[]1,2x ∈上恒成立()()2max min2x xax ⇔+≥在[]1,2x ∈上恒成立；④“平面向量a r 与b r 的夹角是钝角” 的充分必要条件是“0a b <r rg”. A ．1 B ．2 C.3 D ．44．已知点P 的坐标（x ，y ）满足，过点P 的直线l 与圆C ：x 2+y 2=16相交于A ，B两点，则|AB|的最小值为（） A ．B ．C ．D ．5.中国古代数学著作《算法统宗》中有这样一个问题：“三百七十八里关，初行健步不为难，次日脚痛减一半，六朝才得到其关，要见次日行里数，请公仔细算相还．”其意思为：“有一个人走378里路，第一天健步行走，从第二天起脚痛每天走的路程为前一天的一半，走了6天后到达目的地”，请问第3天比第5天多走A ．12里B ．24里C ．36里D ．48里6．设方程2x|lnx|=1有两个不等的实根x 1和x 2，则（） A ．x 1x 2＜0 B ．x 1x 2=1 C ．x 1x 2＞1 D ．0＜x 1x 2＜1 7．某程序框图如右图所示，其中21()g x x x =+，若输出的20162017S =，则判断框内应填入的条件为（） A.2017n < B.2017n ≤ C.2017n > D.2017n ≥6.如图，小正方形的边长为1，粗线画出的是某空间几何体的三视图，则该几何体的体积可能为（）A.38 B ．16 C. 316 D ．329.为得到函数22cos 32y x x =-的图象，只需将函数2sin 21y x =+的图像（）A ．向左平移π12个长度单位 B ．向右平移π12个长度单位 C ．向左平移5π12个长度单位 D ．向右平移5π12个长度单位10.已知双曲线的渐近线方程为x y 21±=，则双曲线的离心率（） A ．23 B ．25 C ．25或5 D ．23或3 11.我国南宋著名数学家秦九韶发现了从三角形三边求三角形面积的“三斜公式”，设ABC ∆三个内角A B C 、、所对的边分别为a b c 、、，面积为S ，则 “三斜求积”公式为222222142a c b S a c ⎡⎤⎛⎫+-⎢⎥=- ⎪⎢⎥⎝⎭⎣⎦若()222sin 4sin 12a C A a c b =+=+，则用“三斜求积”公式求得ABC ∆的面积为（）3612.若存在m ，使得关于x 的方程()()224ln ln 0x a x m ex x m x ⎡⎤++-+-=⎣⎦成立，其中e 为自然对数的底数，则实数a 的取值范围是（） A.(),0-∞ B. 10,2e ⎛⎫ ⎪⎝⎭ C. ()1,0,2e ⎡⎫-∞⋃+∞⎪⎢⎣⎭ D. 1,2e ⎡⎫+∞⎪⎢⎣⎭第Ⅱ卷（非选择题，90分）二、填空题（每题5分，满分20分）13.曲线3()3f x x x =-+在点(1,(1))P f 处的切线方程为_____________．14.若||1a =r ，||2b =r，c a b =+r r r ，且c a ⊥r r ，那么a r 与b r 的夹角为．15.已知1sin 24α=，则2π2cos 4α⎛⎫-= ⎪⎝⎭__________． 16.在正三棱锥V-ABC 内，有一个半球，其底面与正三棱锥的底面重合，且与正三棱锥的三个侧面都相切，若半球的半径为2，则正三棱锥的体积最小时，其底面边长为 .三、解答题（本大题共6小题，共70分.） 17. (本小题满分12分)在各项均不相等的等差数列{}n a 中，已知54=a ，且3a ，5a ，8a 成等比数列，（1）求n a ；（2）设数列{}n a 的前n 项和为n S ，记nn n S a n b ⋅+=23，求数列{}n b 的前n 项和n T .18. (本小题满分12分)某中学高三年级有学生500人，其中男生300人，女生200人。

四川省南充高级中学2020届高三2月线上月考物理试题及参考答案

电流的正方向，下列关于线框中感应电流 i 与线框移动距离 x 的
关系图象正确的是：
D. 16h
A
B
C
D
19．如图所示，不可伸⻓的轻绳，绕过光滑定滑轮 C，与质量为 m 的物体 A 连接，A 放在倾
⻆为的光滑斜面上，绳的另一端和套在固定竖直杆上的物体 B 连接，连接物体 B 的绳最初
水平。从当前位置开始，使物体 B 以速度 v 沿杆匀速向下运动，
同。M、N 之间用轻绳连接，绳拉直后与水平面之间不平行。先用水平拉力 F1 作用于 M，
使两木块一起向左做匀速直线运动，此时轻绳中张力为 T1 ；然后改用水平拉力 F2 作用于
N，使两木块一起向右做匀速直线运动，此时轻绳中张力为 T2 则：
A、 F1>F2 ， T1<T2
B、 F1=F2 ， T1<T2
A. 伽利略通过“理想实验”得出“力是维持物体运动的原因”
B. 法拉第发现了由磁场产生电流的条件和规律——电磁感应定律
C. 爱因斯坦质能方程中：高速运动的粒子质量比其静止时的质量（静质量）更小
D. 汤姆生利用阴极射线管发现了电子，并提出了原子的核式结构模型
15．如图所示，两质量相等的木块 M 、N 放在水平面上，两者与水平面间的动摩擦因数相
C、 F1<F2 ， T1>T2
D、 F1=F2 ， T1>T2
16.在光电效应实验中，分别用频率为νa、νb 的单色光 a、b 照射到同种金属上，测得相应的遏
止电压分别为 Ua 和 Ub、光电子的最大初动能分别为 Eka 和 Ekb。h 为普朗克常量。下列说法正
确的是
A. 若νa>νb，则一定有 Ua<Ub

2020年四川省南充高级中学高三英语二模试卷及答案

2020年四川省南充高级中学高三英语二模试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AFour Best Hikes in the WorldThere's nothing like getting out and getting some fresh air on a hike. No matter whether your idea of a hike is a leisure walk or climbing the highest mountain on Earth, we've got you covered. Below are four best hikes inthe world.Torres del Paine W CircuitLocation (位置): Patagonia. ChileDistance: 37 + milesTime: 5~6 daysBest time to go: October to JanuaryThe W Circuit is one of the most recommended hikes you'll find. Not only will you appreciate the diverse landscapes and striking granite pillars (花岗岩柱子), but you'll probably meet some new friends along the way.Grand Canyon Rim - to - Rim HikeLocation: Arizona, the United StatesDistance: 48 milesTime: 1~3 daysBest time to go: May to June, September to OctoberThere's no better way to experience one of the greatest wonders in the world. Located in one of the USA's most beautiful parks, the views are ly appealing. Just make sure you're prepared for the challenge.Trek to PetraLocation: JordanDistance: 47 milesTime: 5~ 6 daysBest time to go: October to AprilTake the road less traveled through the Kingdom of Jordan and experience one of the seven wonders of the world. Hike through canyons, gorges and ridges, and see tombs and temples along the way all while avoidingcrowds of tourists.Yosemite Grand TraverseLocation: California, the United StatesDistance: 60 milesTime: 6~7 daysBest time to go: July to SeptemberKnown for some of the best hiking in the world, Yosemite National Park is famous for its views and huge sequoia (红杉) trees. Praised byNational Geographic, the Yosemite Grand Traverse will take you through waterfalls and green mountaintops.1.Which of the following is the best time for the hike in Patagonia, Chile?A.AprilB.MayC.AugustD.December2.Where should you go for a less crowded hike?A.JordanB.Patagonia, ChileC.Arizona, the United StatesD.California, the United States3.What can you do along the Yosemite Grand Traverse?A.Plant sequoia treesB.Appreciate waterfallsC.Visit local templesD.Climb granite pillarsBSomeday soon an emoji (表情符号)might really save lives.Hiroyuki Komatsu is a Google engineer who suggested adding a series of new emojis to the standard emoji library. It could help those with food allergies (过敏)understand what they are eating anywhere inthe world. Emojis should cover characters representing major food causing allergies. They make people understand what are used in foods even in foreign countries and safely select meals.Emojis are universal because they are chosen and developed by the Unicode Consortium, a non-profit company that oversees, develops and maintains how text is represented. This is in regards to all software products and standards. It's thanks to the Unicode Standard that when you text a friend six pizza emojis, they’ll see those six pizza pieces on their phone. This is true regardless of whether they use an iPhone or an Android.Because emojis are everywhere and visual(视觉的),they could be helpful for restaurants and food packaging designers. They can communicate whether a product is made with common causing-allergy food. But as Komatsu’s advice argues, many of the most common causing-allergy foods are missing or poorly represented by the presentemoji library. For example, there is an emoji for octopus, but nothing for squid. There is a loaf of bread that could symbolize grain, but a picture of wheat could be clearer. The emojis can be more direct when symbolizing foods.It’s not uncommon for the Unicode Consortium to add new emojis to the library: several food-related emojis were put into use last June, including some long-waited food emojis. Apple included support for multiracial emojis in a recent iOS update. An artist even recreated Moby-Dick in emoji characters. Some might be sorry for the continuing death of the written word if Komatsu’s suggestion is accepted, but look on the bright side: if you ever see that happy poop on a box, you’ll know to stay away.4. How will emojis save lives according to the text?A. By showing what the food contains visually.B. By telling the safest places in the world.C. By teaching people how to treat allergies.D. By adding standard emojis about safety.5. What does paragraph 3 mainly tell us?A. Emojis have the same meanings around the world.B. The Unicode Consortium is a non-profit company.C. What emojis represent is different in different places.D. Different mobile operating systems have different emojis.6. What can be the reason for Komatsu’s advice?A. Emojis are easy to mix up.B. Present emojis are not enough.C. Emojis can't interest most users.D. Emojis can't represent foods directly.7. What is the author’s attitude to Komatsu's suggestion?A. Doubtful.B. Worried.C. Supportive.D. Uninterested.CImaginary friends in childhood refer to the invisible beings that a child gives a personality to and plays with for over three months.Crabbycrab（蟹）appeared on a holiday in Norway by running out of my four-year-old son Fisher's ear after a night of tears from an earache. Like other childhood imaginary friends, Crabby should be a sign thatFisher's mind is growing and developing positively. Indeed, research shows that imaginary friends can help develop children's social skills.Research has shown that the positive effects of having imaginary friends as a child continue into adulthood. Adolescents who remember their imaginary playmates have been found to use more activecoping（应对）styles, such as seeking advice from loved ones rather than bottle things up inside. Even adolescents with behavioral problems who had imaginary friends as children have been found to have better coping skills through the teenage years.Scientists thinkthis could be because these teens have been able to adjust themselves to the social world with imagination rather than choose to be involved in relationships with more difficult classmates. It could also be because the imaginary friends help to reduce these adolescents,loneliness.These teens are also more likely to seek out social connections -they tend to turn to others for advice. Current research by Tori Watson is taking this evidence and looking at how adolescents who have imaginary friends as children deal withbullying（欺凌）at school. It is found that teens who remember their imaginary friends are better at dealing with bullying.While we know a lot about childhood imaginary friends such as Crabby Crab and the positive effects they can have, there is still a lot to learn about imaginary friends.8. What is Crabby crab?A. It is a crab Fisher caught inNorway.B. It is Fisher's imaginary friend.C. It is a toy Fisher like much.D. It is a cause of earache.9. Why do children with imaginary friends have better coping skills?A. Imaginary friends help improve their adjustment.B. Having imaginary friends makes them smarter.C. They have rich imagination.D. They are no longer alone.10. What will a child with imaginary friends probably do if he is bullied?A. Escape from the bully.B. Fight with the bully bravely.C. Keep silent about being bullied.D. Ask a parent or a teacher for help.11. What is the author's attitude towards the effect of imaginary friends?A. Concerned.B. Doubtful.C. Optimistic.D. Indifferent.DNostalgia (怀旧) has become increasingly common in our current climate of accelerated, unexpected change. More and more Americans are turning back with longing towhat feels like simpler, sweeter times. They collect cassette tapes, manual typewriters even decades-old video games.Is it a mistake to get too obsessed with the past? Some psychologists warn that too much devotion to the so-called good old days is an escape from reality; it can indicate loneliness or that a person is having a difficult time coping in the present. Psychologist Stephanie Coontz argues that nostalgia distracts us from addressing the problems of modern life and contribute to anxiety, depression , insomnia etc.But new studies suggest that a modest dose of nostalgia is not only harmless, but actually beneficial. They suggest it helps strengthen our sense of identity and makes us feel more optimistic and inspired. It is also a tool for self — discovery and memories are a psychological immune response that is triggered when you want to take a break from negativity. Interestingly, those happy memories can be particularly beneficial both to kids in their teens and to society's elders. Recalling our childhood reminds us of “the times when we were accepted and loved unconditionally," says Krystine Batcho, a psychologist. "That is such a powerfully comforting phenomenon, knowing that there was a time in life when we didn't have to earn our love." Nostalgia can transform even the most ordinary past into legends which warms the heart and the body. Let's not forget that nostalgia has been a source of inspiration to innumerable American writers. Mark Twain recalled his boyhood, writing, "after all these years, I can picture that old time to myself now, just as it was then：The white town drowsing in the sunshine of a summer's morning."So go ahead, daydream a little about your best childhood friend, your first car, a long - gone family pct. As Dr. Sedikidessays，"Nostalgia is ly central to human experience. "But at the same time, keep these words of wisdom from the great inventor Charles Kettering in mind as well："You can't have a better tomorrow if you are thinking about yesterday all the time. "312. What did some psychologists in paragraph 2 probably agree?A. Nostalgia will cause some mental problems.B. Nostalgia makes us devoted to the good old days.C. Nostalgia shows you are trying to get rid of loneliness.D. Nostalgia helps us cope with the difficult time we are going through.13. There are many benefits of nostalgia except ________A. It can enable us to know ourselves better.B. It can bring us some comfort when we recall.C. We are likely to gain attention if we recall the happy childhood.D. We can sometimes break away from negativity with happy memories.14. What will be talked about in the following paragraph?A. The bad influence of too much devotion to nostalgia.B. The reasons why we should avoid nostalgia.C. The bad memories that always stick around you.D. The great changes nostalgia will bring to you.15. What's the best title of the passage?A. We all have a soft spot for nostalgia.B. Nostalgia is actually good for you.C. Don't be carried away by nostalgia.D. There are many times when we like to recall.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年四川省南充高级中学高三英语模拟试题及答案

2020年四川省南充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AA medical capsule robot is a small,often pill-sized device that can do planned movement inside the body after being swallowed or surgically inserted. Most models use wireless electronics or magnets or a combination of the two to control the movement of the capsule. Such devices have been equipped with cameras to allow observation and diagnosis, with sensors that “feel,” and even with mechanical needles that administer drugs.But in practice, Biomechatronics engineer Pietro Valdastri has found that developing capsule models from scratch (从头开始) is costly, time-consuming and requires advanced skills. “The problem was we had to do them from scratch every time,” said Valdastri in an interview. “And other research groups were redeveloping those same modules from scratch, which didn’t make sense.”Since most of the capsules have the same parts of components: a microprocessor, communication submodules, an energy source, sensors, and actuators (致动器), Valdastri and his team made the modular platform in which the pieceswork in concertand can be interchanged with ease. They also developed a flexible board on which the component parts are snapped in like Legos. The board can be folded to fit the body of the capsule, down to about 14 mm. Additionally, they compiled (编译) a library of components that designers could choose from, enabling hundreds of different combinations. They arranged it all in a free online system. Designers can take the available designs or adapt them to their specific needs.“Instead of redeveloping all the modules from scratch, people with limited technological experience can use our modules to build their own capsule robots in clinical use and focus on their innovation,” Valdastri said.Now, the team has designed a capsule equipped with a surgical clip to stop internal bleeding. Researchers at Scotland’s Royal Infirmary of Edinburg have also expressed interest in using the system to make a crawling capsule that takes images of the colon(结肠). One research group, led by professors at the Institute of Digestive Disease of the Chinese University of HongKong, is making a swimming capsule equipped with a camera that pushes itself through the stomach.One limitation of Valdastri’s system is that it’s only for designing models. Researchers can confirm their hypotheses (假设) and do first design using the platform, but will need to move to a custom approach to develop their capsules further and make them practical for clinical use.1. According to the passage, Valdastri and his team created the platform to ________.A. adopt the latest technologiesB. make their robots dream come trueC. help build specialized capsule robotsD. do preciser observation and diagnosis2. What does the underlined phrase “work in concert” mean in Para.3?A. Perform live.B. Run independently.C. Act in a cooperative way.D. Carry on step by step.3. What can be learnt from the passage?A. Valdastri’s system can’t provide a complete capsule creation.B. The modular platform is more useful than a custom approach.C. The capsules can move in human’s body automatically.D. It costs more to module the capsules on the board.BMy first week working in a restaurant, one of the servers said something that stuck with me: Everyone should work in a restaurant for at least a year. At the time, I didn't get it, but I took the advice to heart and worked in restaurants on and off for the next eight years. Before realizing it, I mastered many important skills, one of which is communication skill.When I was little, I was so shy that I used to hide behind my mom whenever someone spoke to me. And when I first started in restaurants, I had two personalities: Restaurant Lizzy and Home Lizzy. It was easier to pretend to be a different person while at work, since it was so different from who I actually was. But gradually, the skills I learned working in restaurants helped Home Lizzy come out of her shell in the real world.When you work in a restaurant, you don't have the luxury of hiding behind your parents to avoid talking to people. I'm still 110% an introvert, but restaurant work helped me communicate. Working in a restaurant not only helped me speak clearly, deliberately and directly but also taught me how to talk about almost everything. Some guests don't want their servers to interact too much with them, and that's fine. But some sit at the bar simply to chat with you. You learn how to judge your guests' level of interest in communicating with you, and how to exit a conversation at the appropriate time.My restaurant work is something that I'm most proud of and I know I wouldn't be the person I am today without those eight years of experience. If you're still on the fence about working in a restaurant for that long,start with one year. I doubt that you'll look back.4. What did the writer think of the server's words?A. Impressive.B. Ridiculous.C. Amusing.D. Logical.5. What do we infer from Paragraph 2?A. The writer tried different jobs.B. The writer became more sociable.C. The writer used an invented name.D. The writer had a hard time at work.6. Which of the following best describes the writer's restaurant work?A. Boring.B. Relaxing.C. Worthwhile.D. Unchallenging.7. What message does the writer try to convey in this passage?A. A strong-willed soul can reach his goal.B. Things are difficult before they are easy.C. Communication skills advance your career.D. Restaurantwork helps to achieve a better self.CAmerica---that glorious symbol of multiculturalism, the great melting pot---qualifies as part of the developed world, right? Not quite, if we consider the state of second language learning in schools across the country. According to a 2018 study, Europe often tops the U.S. in foreign language education in school. 92% of European students are learning a foreign language in school. In America it’s only 20%. In more than 20 European countries, studying a foreign language for at least one year is compulsory. In comparison, only 10 states and D.C. require learning a foreign language for graduation.I went to a public school in Pennsylvania that ranked onNewsweek’s list of America’s top high schools in 2015. Foreign language learning was not a graduation requirement. A common response to such stories and statistics is: So what? Why should Americans care about learning another language when English has recently seemed to dominate as the official language of the world?The world is globalizing faster than we can imagine. More than half of the world’s population speaks more than one language. It is also increasingly becoming the need for success in this globalizing world. Having the ability to speak a second language opens up the possibility of travelling and immersing yourself in another part of the world. This means people can feel other cultures, and traditions in places you were previously kept away because you could not communicate in the language.The teaching of foreign languages can be instrumental in bringing about peace and unity in the country welive in. Learning a second language can be an efficient vehicle to help to bridge communities. To speak another language means having access to a universe of different experiences and world views of another community of people.8. Why does the writer mention Europe?A. America makes studying a foreign language compulsory.B. America is the glorious symbol of multiculturalism.C. America isn’t really qualified as the developed world.D. America doesn’t focus on foreign language education9. What is American’s attitude towards education of foreign languages?A. Less concerned.B. More worried.C. Quite doubtful.D. Very supportive.10. What can visiting people do if they can speak the language where they visit?A. Make the world globalizing.B. Achieve success in business.C. Experience people’s real life there.D. Keep away their own culture.11. What is the main idea of the last paragraph?A. Using a second language can bring harmony.B. Communities of people enjoy the great views.C. Speaking another language is reallynecessary.D. Learning a second language can be difficult.DJohn Montefiore's path to graduation from theUniversityofTorontowas a little unusual. He recently completed his bachelor's degree(学士学位) which he started in1995.Montefiore left university in 1996 and tried to tell himself that a degree wasn't necessary for personal success. But it remained unfinished businessuntil he made the decision to return in 2018 at the age 42. At that time, his job development stalled, so he made up his mind to go back to school in order to improve himself and work well in future.The second time around, Montefiore never missed a class and always sat in the front row. He found many courses invaluable and he received the Award of Excellence twice. He found support services played an important role in his successful return. He said, “I hadn't written a paper for years, so I found the college writing centre to be of great use. Before I handed in a paper, they would give me feedback, which was really great. The university hasso much built in to help students succeed and I'm so thankful for that.” But he also met difficulties this time. As a student much older than others, he found it harder to make friends with classmates.However, no matter what had happened before, he finally got his bachelor's degree. He said, “After all these years, I had thought it wouldn't mean anything, but it means a lot. I totally understand the value of education now. It's not necessarily the value that other people see in it, but the value it gives me as a person, for my confidence and my self-esteem. It also helps me prepare well before I re-enter the workplace.”12. Why was Montefiore's graduation unusual?A. He had achieved personal success before.B. He understood the importance of a degree.C. He finished his degree at a much older age.D. He began his university study at a young age.13. What does the underlined word "stalled" in paragraph 2 probably mean?A. Stopped.B. Started.C. Survived.D. Succeeded.14. What can we learn about Montefiore from the third paragraph?A. He got good grades very easily.B. He took his study seriously this timeC. He was thankful for his classmates' helpD. He had difficulty using support services.15. What did Montefiore learn from his experience?A. Confidence is important in one's life.B. Others' opinions on education matter a lot.C. Higher education is a must for personal success.D. Education makes one feel better about himself.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。

四川省南充高级中学2020届高三物理下学期第二次线上月考试题(含解析)

CD．突然松手，左边绳子的拉力瞬间减为0,撞锤在重力和右边吊绳的作用下做圆周运动，速度为0,说明向心方向（沿吊绳方向)合力为0，根据几何关系可知此时两段绳子的夹角为直角,水平方向和吊绳的夹角为，则重力在圆周运动的切向(垂直吊绳方向）提供加速度，根据牛顿第二定律
C错误，D正确。
故选D。
6。我国已掌握“半弹道跳跃式高速再入返回技术”,为实现“嫦娥”飞船月地返回任务奠定基础。如图虚线为地球大气层边界，返回器与服务舱分离后,从a点无动力滑入大气层，然后经b点从c点“跳”出，再经d点从e点“跃入”实现多次减速，可避免损坏返回器。d点为轨迹的最高点，与地心的距离为R，返回器在d点时的速度大小为v，地球质量为M,引力常量为G。则返回器
（3）用另一电源E0(电动势4 V，内阻1.00 Ω）和题给器材连接成图(b)所示的电路,调节滑动变阻器R的阻值，可以改变小灯泡的实际功率．闭合开关S，在R的变化范围内，小灯泡的最小功率为 ________W，最大功率为 ________W（结果均保留2位小数)
【答案】（1). 实验电路原理如图所示．（2). 增大（3）。增大（4)。 0.39 （5）. 1.17
故动能不相等,故B错误；
C．在d点时合力等于万有引力，即
所以加速度大小
故C正确；
D．在d点时万有引力大于所需的向心力，做近心运动，故速度大小
故D正确。
故选CD。
7。如图所示，在竖直平面内半径为R的四分之一圆弧轨道AB、水平轨道BC与斜面轨道CD平滑连接在一起(轨道为光滑凹槽），斜面轨道足够长.在圆弧轨道上静止着N个半径为r（r≪R）的完全相同小球,小球恰好将圆弧轨道铺满，从最高点A到最低点B依次标记为1、2、3……N.现将圆弧轨道末端B处的阻挡物拿走，N个小球由静止开始沿轨道运动,不计摩擦与空气阻力，下列说法正确的是（）

四川省南充高级中学2020-2021学年高三上学期第二次月考理科数学试题

南充高中2018级高三第二次月考数学试题（理科）第Ⅰ卷（选择题）一、选择题：本大题共12小题.在每小题给出的四个选项中，只有一项是符合题目要求的.1. 设{}2|log (2)A x y x ==-,{}2|9=≥B x x ，则R A C B =（）A. ()2,+∞B. [)2,3C. ()3,+∞D. ()2,3D求对数函数的定义域求得集合A ，解一元二次不等式求得集合B ，求得集合B 的补集后与集合A 求交集，由此得出正确选项.【详解】对于集合A ，20,2x x ->>，对于集合B ，()()29330x x x -=+-≥，解得3x ≤-或3x ≥，故()3,3R C B =-，所以()()2,3R A C B ⋂=，故选D. 2. 若复数z 满足23z z i +=-，其中i 为虚数单位，则||z =（）A. 2B.C.D. 3C设复数(,)z x yi x y R =+∈，利用相等，求得1,1x y ==-，进而可求复数的模.设复数(,)z x yi x y R =+∈，则22233z z x yi x yi x yi i +=++-=+=-，则1,1x y ==-，所以1z i =-，所以z = C. 3. 设、a b 均为单位向量，则“22-=+a b a b ”是“⊥a b ”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C根据22||=a a ，可化简22-=+a b a b 22224444+-=++a b ab a b ab ，又、a b 均为单位向量，可得0=ab ，即可分析出结果.因为、a b 均为单位向量，所以221,1==a b ，由22-=+a b a b 可得：()()2222-=+a b a b ，即22224444+-=++a b ab a b ab ，所以5454-=+ab ab ，即0=ab ，所以⊥a b ，因此“22-=+a b a b ”是“⊥a b ”的充分必要条件，故选C.4. 若01a b <<<，b x a =，a y b =，b z b =，则x ，y ，z 的大小关系为（） A. x z y << B. y x z <<C. y z x <<D. z y x <<A根据指数函数x y b =以及幂函数b y x =的单调性比较出,,x y z 之间的大小关系. 因为x y b =在0,上单调递减，所以a b b b >，即y z >，又因为b y x =在0,上单调递增，所以b b a b <，即x z <，所以x z y <<，故选：A.5. 我国南宋著名数学家秦九韶发现了三角形三边求三角形面积的“三斜求积公式”，设ABC 三个内角A ，B ，C 所对的边分别为a ，b ，c ，面积为S ，则“三斜求积公式”为S =.若2sin 24sin a C A =，()()()2sin sin 27sin a C B c b a A -+=-，则用“三斜求积公式”求得的S =( )A. 4B.4C.4D.4D根据正弦定理：由a 2sinC=4sinA 得ac=24，则由a （sinC ﹣sinB ）（c+b ）=（27﹣a 2）sinA 得a 2+c 2﹣b 2=27，利用公式可得结论．由224a sinC sinA = 可得224,24a c a ac =∴=，由()()()227a sinC sinB c b a sinA -+=- 可得()()()227a c b c b a a -+=-，整理计算有：22227a c b +-=，结合三角形面积公式可得：S ===.故选D .6. 一个几何体的三视图如图所示，若这个几何体的体积为积为（）A. 36πB. 64πC. 81πD. 100πC首先把三视图转换为几何体，进一步利用几何体的体积公式求出四棱锥体的外接球的半径，最后求出球的表面积．解：根据几何体的三视图可以得到该几何体为四棱锥体，如图所示：该四棱锥的底面是长方形，长为6，宽为5，四棱锥的高即为PD所以1562053V h =⨯⨯⨯=解得25h =设四棱锥的外接球的半径为r ，所以()(222225625r =++，解得92r =，所以294812S ππ⎛⎫=⨯= ⎪⎝⎭球，故选：C7. 设等差数列{}n a 的前n 项和为n S ，若632a a =，则115S S =( ) A. 115B.522C.1110 D.225D等差数列中，1116611111155153311()11211()11222=.5()5()52552a a a a S a a a a S a a a a +⨯+===⨯=++⨯ 本题选择D 选项.8. 已知函数()sin()(0,0,)f x A x A ωϕωϕπ=+>><，其部分图象如图所示，则()f x 的解析式为（）A. 1()3sin 26f x x π⎛⎫=+ ⎪⎝⎭B. 1()3sin 26f x x π⎛⎫=- ⎪⎝⎭C. 15()3sin 26f x x π⎛⎫=+⎪⎝⎭D. 1()3sin 26f x x π⎛⎫=- ⎪⎝⎭或15()3sin 26f x x π⎛⎫=+⎪⎝⎭B根据图象可得A ，T ，根据T 求出ω，根据最高点求出ϕ，即可得解. 由图可知2A =，13433T πππ=-=，所以22142T ππωπ===，又当43433T x ππππ=+=+=时，()f x 取得最大值，所以14sin()123πϕ⨯+=，所以232ππϕπ+=+k ，k Z ∈，所以6k ϕπ=π-，k Z ∈，因为||2ϕπ<，所以6πϕ=-.所以1()3sin 26f x x π⎛⎫=- ⎪⎝⎭.故选：B.9. 函数6(3)3,7(),7x a x x f x a x ---≤⎧=⎨>⎩，若数列{}n a 满足()n a f n =，*n N ∈，且{}n a 是递增数列，则实数a 的取值范围是（）A. 9,34⎡⎫⎪⎢⎣⎭B. 9,34⎛⎫ ⎪⎝⎭C. ()1,3D. ()2,3D根据题意可知分段函数为增函数，且()()87f f >，列出不等式组()86301373a a a a -⎧->⎪>⎨⎪>-⨯-⎩，解不等式组即可求解.由题意可知分段函数为增函数，且()()87f f >，即()86301373a a a a -⎧->⎪>⎨⎪>-⨯-⎩，解得23a <<，故实数a 的取值范围是()2,3.故选：D10. 已知321()(4)(0,0)3f x x ax b x a b =++->>在1x =处取得极值，则11a b +的最小值是（）A. 2B. 2C.2D. 1+D求导()2'24f x x ax b =++-，根据极值点得到23a b +=，()1111123a b a b a b ⎛⎫+=++ ⎪⎝⎭，展开利用均值不等式计算得到答案.()()32143f x x ax b x =++-，故()2'24f x x ax b =++-，根据题意()'11240f a b =++-=，即23a b +=，经检验()f x 在1x =处取得极值.()()1111112123313333b a a b a b a b a b ⎛⎫⎛⎫+=++=++≥=+ ⎪ ⎪⎝⎭⎝⎭，当且仅当2b a a b ==，即632a b -==时，等号成立.故选：D .11. 已知函数()1f x -，x ∈R 是偶函数，且函数()f x 满足()()2f x f x =--，当[]1,1x ∈-时，()1f x x ，则()2020f =（） A. -2 B. -1C. 0D. 1D根据函数()1f x -，x ∈R 是偶函数，以及()()2f x f x =--求出函数周期，再由已知区间的解析式，即可求出结果.因为函数()1f x -，x ∈R 是偶函数，所以()()11f x f x --=-，则()()2f x f x --=，又()()2f x f x =--，所以()()22f x f x --=--，则()()22f x f x -=-+，因此()()4f x f x =-+，所以()()()84f x f x f x +=-+=，即函数()f x 是以8为周期的周期函数，因此()()()()20204+252840f f f f =⨯==-，又当[]1,1x ∈-时，()1f x x ，所以()01f =-，因此()()202001f f =-=.故选：D.12. 已知函数(),0,ln ,0xe xf x x x ⎧≤⎪=⎨>⎪⎩(e 为自然对数的底数)，则函数21()(())()1F x f f x f x e =--的零点个数为（） A. 6 B. 5 C. 4 D. 3A利用换元法将原问题转化为两个函数交点个数的问题，然后结合函数的性质确定交点的范围，最后利用交点的范围即可确定原函数零点的个数．解：()f x t =，则由()0F x =得21()1f t t e=+．作出()y f x =的函数图象如图所示：设直线11y k x =+与曲线x y e =相切，切点为0(x ，0)y ，则010101x x e k e k x ⎧=⎪⎨=+⎪⎩，解得00x =，11k =．设直线21y k x =+与曲线ln y x =相切，切点为1(x ，1)y ，则2121117k x k x ⎧=⎪⎨⎪+=⎩，解得21221,x e k e ==．故直线211y t e =+与()f t 的图象有4个交点，不妨设4个交点横坐标为12t t ，，3t ，4t ，且1234t t t t <<<，由图象可知212340,0,01,t t t t e <=<<=，由()f x 的函数图象可知1()f x t =无解，2()f x t =有一个解，3()f x t =有三个解，4()f x t =有两个解，()F x ∴有6个零点．故选：A ．第Ⅱ卷（非选择题）二、填空题. 13. 已知1sin cos 5θθ+=，(0,)θπ∈，则tan θ=________. 43- 分析】把已知等式两边平方，求出sin cos θθ的值，再利用完全平方公式求出sin cos θθ-的值，联立求解再结合同角三角函数间的基本关系可求得tan θ的值．已知1sin cos 5θθ+=，平方得()2221sin cos sin cos 2sin cos 25θθθθθθ+=++=，得12sin cos 25θθ=-， ∴()222sin cos sin cos 2sin cos 125252449θθθθθθ-=+-=+=，(0,)θπ∈，sin 0,cos 0θθ><， 7sin cos 5θθ∴-=，7ta sin cos 1sin cos n 571t n 51a θθθθθθ=-=-+=+，解得4tan 3θ=-. 故答案为：43-14. 函数43()2f x x x =-的图象在点(1,(1))f 处的切线方程为________.210x y +-=利用导数求出切线的斜率，求出切点，即得解. 由题得32()46f x x x '=-，所以切线的斜率为(1)462k f '==-=-，因为(1)1f =-，所以切点为(1.1)-，所以切线方程为12(1)y x +=--，即210x y +-=. 故答案为：210x y +-=15. 给出下列命题：①函数5sin 22y x π⎛⎫=-⎪⎝⎭是偶函数； ②方程8x π=是函数5sin(2)4y x π=+的图象的一条对称轴方程； ③在锐角ABC 中，sin sin cos cos A B A B >；④函数1()sin 33f x x π⎛⎫=++ ⎪⎝⎭的最小正周期为π；⑤函数()tan 213f x x π⎛⎫=++ ⎪⎝⎭的对称中心是,126k ππ⎛⎫- ⎪⎝⎭，k Z ∈，其中正确命题的序号是________. ①②③由诱导公式化简得函数cos 2y x =，判断①正确；求出函数5sin(2)4y x π=+的图象的对称轴328k x ππ=-（k Z ∈），当1k =时，8x π=，判断②正确；在锐角ABC 中，由cos 0C -<化简得到sin sin cos cos A B A B >，判断③正确；直接求出函数1()sin 33f x x π⎛⎫=++ ⎪⎝⎭的最小正周期为2π，判断④错误；直接求出函数()tan 213f x x π⎛⎫=++ ⎪⎝⎭的对称中心是,164k ππ⎛⎫- ⎪⎝⎭，判断⑤错误.①因为函数5sin 2cos(2)2y x x π⎛⎫=-= ⎪⎝⎭，所以函数5sin 22y x π⎛⎫=- ⎪⎝⎭是偶函数，故①正确；②因为函数5sin(2)4y x π=+，所以函数图象的对称轴5242x k πππ+=+（k Z ∈），即328k x ππ=-（k Z ∈），当1k =时，8x π=，故②正确；③在锐角ABC 中，cos()cos()cos 0A B C C π+=-=-<，即cos cos sin sin 0A B A B -<，所以sin sin cos cos A B A B >，故③正确；④函数1()sin 33f x x π⎛⎫=++ ⎪⎝⎭的最小正周期为2π，故④错误；⑤令232k x ππ+=，解得46k x ππ=-，所以函数()tan 213f x x π⎛⎫=++ ⎪⎝⎭的对称中心是,164k ππ⎛⎫- ⎪⎝⎭，故⑤错误. 故答案为：①②③16. 已知函数()()(0)xf x e aln ax a a a =--+>，若关于x 的不等式()0f x >恒成立，则实数a 的取值范围为______()20,e将不等式()0f x >恒成立转化为()1x e ln ax a a +>-在(1,)+∞上恒成立,进一步转化为1xe xa +>恒成立，即1xe a x <-恒成立.再构造函数,利用导数求最值可解决.易求得函数()f x 的定义域为(1,)+∞ ，由()()0xf x e aln ax a a =--+>，得()1xe ln ax a a+>-，因为函数1xey a=+与函数()y ln ax a =-互为反函数，其图象关于直线y x =对称，所以要使得()0f x >恒成立，只需1xe x a+>恒成立，即1x e a x <-恒成立，设()1xeg x x =-，则()()()221x e x g x x '-=-，()g x 在()1,2上递减，在()2，∞+递增，可知当2x =时，()g x 取得最小值2e ，所以2a e <，又因为0a >，所以a 的取值范围是()20,e .三、解答题：解答应写出文字说明、证明过程或演算步骤.17. 一块成凸四边形ABCD 麦田，如图所示.为了分割麦田，将BD 连接，经测量已知2AB BC CD ===，23AD =（13cos A C -的值；（2）记ABD △与BCD 的面积分别为1S 和2S ，为了更好地规划麦田，请你求出2212S S +的最大值.（1）1；（2）14.（1）根据题中条件，在ABD △中，由余弦定理，得到1683cos BD A =-BCD 中，由余弦定理得到88cos BD C =-（2）根据三角形面积公式，由题中条件，得到2211212cos S A =-，22244cos S C =-，得出2221218cos 142S S C ⎛⎫+=-++ ⎪⎝⎭，进而可得出结果.（1）在ABD △中，因为2AB =，23AD =由余弦定理，可得2222cos 1683BD AB AD AB AD A A =+-⋅=-，所以1683cos BD A =- 在BCD 中，2BC CD ==，由余弦定理，可得2222cos 88cos BD AC CD AC CD C C =+-⋅=-，所以88cos BD C =-1683cos 88cos A C -=-3cos 1A C -=， 3cos A C -为一个定值1.（2）依题意，可得22211sin 1212cos 2S AB AD A A ⎛⎫=⋅⋅=- ⎪⎝⎭，22221sin 44cos 2S BC CD C C ⎛⎫=⋅⋅=- ⎪⎝⎭，3cos 1A C =+，所以()222222121212cos 44cos 164cos 14cos S S A C C C +=-+-=-+-2218cos 8cos 128cos 142C C C ⎛⎫=--+=-++ ⎪⎝⎭，由三角形的性质可得，AD AB BD BC CD -<<+，即2324BD -<<，即()88cos 232,4BD C =--∈，解得1cos 31C -<<-，所以221214S S +≤，当1cos 2C =-时取等号，即2212S S +的最大值为14.18. 如图，等腰梯形ABCD 中，//AB CD ，1AD AB BC ===，2CD =，E 为CD 中点，以AE 为折痕把ADE 折起，使点D 到达点P 的位置（P ∉平面ABCE ）（1）证明：AE PB ⊥；（2）若线段PC 的长为102，求二面角A PE C --的余弦值. （1）证明见解析；（2）55-. （1）先证明OP AE ⊥，OB AE ⊥，再证明AE ⊥平面POB ，最后证明AE PB ⊥；（2）先证明PO OC ⊥，再证明PO ⊥平面ABCE ，建立直角坐标系，标记各点坐标，求平面PCE 的一个法向量1(3,1,1)n =-，平面P AE 的一个法向量2(0,1,0)n =，最后求二面角A PE C --的余弦值.解：（1）在等腰梯形ABCD 中，连接BD ，交AE 于点O ，如图 ∵ABCE ，AB CE =，∴四边形ABCE 为平行四边形，∴AE BC AD DE ===，∴ADE 为等边三角形，∴在等腰梯形ABCD 中，3C ADE π∠=∠=，23DAB ABC π∠=∠=，∴在等腰ADB △中，6ADB ABD π∠=∠= ∴2362DBC πππ∠=-=，即BD BC ⊥，∴BD AE ⊥，翻折后可得：OP AE ⊥，OB AE ⊥，又∵OP ⊂平面OB ，OB ⊂平面POB ，OP OB O =， ∴AE ⊥平面POB ，∵PB ⊂平面POB ，∴AE PB ⊥；（2）由（1）知32DO PO ==，连接OC 在OEC △中，由余弦定理可得72OC =. 在POC△中有222PC PO OC =+，可知PO OC ⊥，又PO AE ⊥，OC AE O PO =⇒⊥平面ABCE ，则以O 为原点，OE 为x 轴，OB 为y 轴，OP 为乙轴，建立空间直角坐标系，由题意得，各点坐标为，30,0,2P ⎛⎫ ⎪ ⎪⎝⎭，1,0,02E ⎛⎫⎪⎝⎭，31,,02C ⎛⎫ ⎪ ⎪⎝⎭，∴13,0,22PE ⎛⎫=- ⎪ ⎪⎝⎭，13,,022EC ⎛⎫= ⎪ ⎪⎝⎭. 设平面PCE 的一个法向量为1(,,)n x y z =，则1100PE n EC n ⎧⋅=⎪⎨⋅=⎪⎩，∴13021302x z x y ⎧-=⎪⎪⎨⎪+=⎪⎩. 设3x =，则1y =，1z =，∴1(3,1,1)n =-，由题意得平面P AE 的一个法向量2(0,1,0)n =，设二面角A EP C --为α，12125|cos |5n n n n α⋅===. 易知二面角A EP C --为钝角，所以5cos α=-.（射影面积法也可）19. 有一名高二学生盼望2020年进入某名牌大学学习，假设该名牌大学有以下条件之一均可录取：①2020年2月通过考试进入国家数学奥赛集训队（集训队从2019年10月省数学竞赛一等奖中选拔）：②2020年3月自主招生考试通过并且达到2020年6月高考重点分数线，③2020年6月高考达到该校录取分数线（该校录取分数线高于重点线），该学生具备参加省数学竞赛、自主招生和高考的资格且估计自己通过各种考试的概率如下表若该学生数学竞赛获省一等奖，则该学生估计进入国家集训队的概率是0.2.若进入国家集训队，则提前录取，若未被录取，则再按②、③顺序依次录取：前面已经被录取后，不得参加后面的考试或录取.（注：自主招生考试通过且高考达重点线才能录取）（Ⅰ）求该学生参加自主招生考试的概率；（Ⅱ）求该学生参加考试的次数X 的分布列及数学期望；（Ⅲ）求该学生被该校录取的概率.（Ⅰ）0.9.（Ⅱ）分布列见解析；数学期望3.3；（Ⅲ）0.838（Ⅰ）设该生参加省数学竞赛获一等奖、参加国家集训队时间分别为A ，B 则1()()P P A P AB =+，然后利用互斥事件的概率公式进行求解；（Ⅱ）X 的可能取值为2，3，4，然后分别求出相应的概率，列出分布列，根据数学期望公式进行求解即可；（Ⅲ）设自主招生通过并且高考达重点线录取、自主招生未通过且高考达该校线录取的事件分别为C 、D ，该学生被该校录取的事件分为三种事件，AB 、C 、D ，分别求出对应的概率，最后相加即可．解：（Ⅰ）设该学生参加省数学竞赛获一等奖、参加国家集训队的事件分别为A ，B ，则()0.5P A =，()0.2P B =，1()()P P A P AB =+10.50.5(10.2)0.9=-+⨯-=. 即该学生参加自主招生考试的概率为0.9.（Ⅱ）该该学生参加考试的次数X 的可能取值为2，3，4(2)()()0.50.20.1P X P A P B ===⨯=； (3)()10.50.5P X P A ===-=； (4)()()0.50.80.4P X P A P B ===⨯=.所以X 的分布列为()20.130.540.4 3.3E X =⨯+⨯+⨯=.（Ⅲ）设该学生自主招生通过并且高考达到重点分数线录取，自主招生未通过但高考达到该校录取分数线录取的事件分别为C ，D .()0.1P AB =，()0.90.60.90.486P C =⨯⨯=，()0.90.40.70.252P D =⨯⨯=，所以该学生被该校录取的概率为2()()()0.838P P AB P C P D =++=.20. 在平面直角坐标系xOy 中，已知椭圆2222:1x y C a b+=0a b >>的右焦点为F ，上顶点为B ，30OBF ∠=︒，点2A ⎛⎫⎪ ⎪⎝⎭在椭圆C 上. （1）求椭圆C 的标准方程；（2）动直线l 与椭圆C 相交于P 、Q 两点，与x 轴相交于点M ，与y 轴的正半轴相交于点N ，T 为线段PQ 的中点，若744OP OQ OT OM OT ON ⋅-⋅-⋅为定值n ，请判断直线l 是否过定点，求实数n 的值，并说明理由.（1）22143x y +=；（2）直线l 过定点()0,1，21n =-，理由见解析.（1）设点F 的坐标为(),0c .由30OBF ∠=︒，可得2a c =，b =，故椭圆C 的标准方程为2222143x y c c +=，把点2A ⎛ ⎝⎭代入，求出c ，即得椭圆C 的标准方程；（2）由题意可设直线l 的方程为y kx m =+()0m >，()()1122,,,P x y Q x y ，则(),00,,mkN m M ⎛⎫- ⎪⎝⎭.由22143x y y kx m ⎧+=⎪⎨⎪=+⎩，消去y ，韦达定理可得12121212,,,x x x x y y y y ++.由()21212278427434OP OQ OT OM OT m x x y y ON k ⋅-⋅-⋅=+-+，可得()222214443k m n k ⎡⎤-+-⎣⎦=+为定值，故243m -=，即求,m n ，即得直线l 过定点. （1）设点F 的坐标为(),0c .由OF c =，OB b =，BF a =，30OBF ∠=︒，可得2a c =，b =.∴椭圆C 的标准方程为2222143x y c c +=，点2A ⎛ ⎝⎭在椭圆C 上， 2211122c c∴+=，1c ∴=，故椭圆C 的标准方程为22143x y +=.（2）由题意可知直线l 的斜率存在且不为0，设直线l 的方程为y kx m =+()0m >，设()()1122,,,P x y Q x y ，则(),00,,mkN m M ⎛⎫- ⎪⎝⎭.由22143x y y kx m ⎧+=⎪⎨⎪=+⎩，消去y ，整理可得()2224384120k x kmx m +++-=，则122843km x x k +=-+，212241243m x x k -=+.由()()()222222644434121612390k m k m k m ∆=-+-=-+>，可得22430k m -+>.()()()21212122286224343k m my y kx m kx m k x x m m k k ∴+=+++=++=-+=++，()()()2212121212y y kx m kx m k x x km x x m =++=+++ ()22222222224128312434343k m k m m k m k k k --=-+=+++， 22121227121243m k x x y y k --∴+=+，2243,4343km m T k k ⎛⎫- ⎪++⎝⎭. ,m OM ON m k ⎛⎫∴+=- ⎪⎝⎭，()222222437434343m m m OT OM ON k k k ⋅+=+=+++， ()74474OP OQ OT OM OT ON OP OQ OT OM ON ∴⋅-⋅-⋅=⋅-⋅+()21212228743m x x y y k =+-=+()()2222222277121273121228434343m k m k m k k k -----=+++ ()()22222271231221444343k m k m k k ⎡⎤⎡⎤-+--+-⎣⎦⎣⎦==++，若744OP OQ OT OM OT ON ⋅-⋅-⋅为定值，则必有243m -=，解得1m =±，0m >，1m ∴=，21n ∴=-. 故直线l 过定点()0,1，21n =-.21. 已知函数()()22ln f x x x ax a R =+-∈有两个极值点1x ，2x ，其中12x x <.（Ⅰ）求实数a 的取值范围；（Ⅱ）当a ≥()()12f x f x -的最小值. （Ⅰ）()4,+∞；（Ⅱ）12e e--（Ⅰ）求出函数的导函数()22'2x x f x ax -+=，由题意可得程2220x ax -+=有两个不相等的正根1x ，2x ，利用根与系数的关系即可求解. （2）结合（Ⅰ）可得()()12f x f x -2111222ln x x x x x x =-+，令()1201xt t x =<<，不妨设()12ln t h t t t=-+，求出函数()h t的单调性，结合a ≥，求出()h t 的最小值即可.（Ⅰ）依题意得()f x 的定义域为()0,∞+，()22'2x x f x ax -+=，因为函数()f x 有两个极值点1x ，2x ，12x x <，所以方程2220x ax -+=有两个不相等的正根1x ，2x ，12x x <，所以21212160021a a x x x x ⎧∆=->⎪⎪+=>⎨⎪⋅=⎪⎩，解得4a >，此时()f x 在()10,x 和()2,x +∞上单调递增，在()12,x x 上单调递减，所以实数a 的取值范围是()4,+∞.（Ⅱ）因为1x ，2x 是方程2220x ax -+=的两个根，所以122ax x +=，121=x x ，因为211220x ax -+=，222220x ax -+=，所以21122ax x =+，22222ax x =+，所以()()()()22121112222ln 2ln f x f x x x ax x x ax -=+--+- ()()22221112222ln 222ln 22x x x x x x ⎡⎤⎡⎤=+-+-+-+⎣⎦⎣⎦ 2221122ln 2ln x x x x =-+-222111222ln x x x x x x -=+ 2111222ln x x x x x x =-+. 令()1201x t t x =<<，()12ln t h t t t=-+，则 ()()22222112210'1t t t t t h t t t ---+-=--+==<，即()h t 在()0,1上单调递减.因为a ≥122a x x +=≥，所以()221212x x x x +≥，即22121212212x x x x e x x e ++≥++，所以12211x x e x x e +≥+，即11t e t e+≥+，所以()10t e t e ⎛⎫--≥ ⎪⎝⎭，01t <<，所以10t e<≤. 因为()h t 在10,e ⎛⎤⎥⎝⎦上单调递减，所以()h t 的最小值为112h e e e ⎛⎫=-- ⎪⎝⎭，即()()12f x f x -的最小值为12e e--.选做题：22. 在极坐标系中，曲线C的极坐标方程为π,22sin 6π1,π.2θθρθ≤<⎪+⎪ ⎪=⎨⎝⎭⎪⎪≤≤⎪⎩ （1）求曲线C 与极轴所在直线围成图形的面积；（2）设曲线C 与曲线1sin 2ρθ=交于A ，B 两点，求AB . （1）1π4+；（2）（1）利用互化公式，将曲线C 的极坐标方程化为直角坐标方程，得出曲线C 与极轴所在直线围成的图形是一个半径为1的14圆周及一个两直角边分别为1面积；（2）联立方程组，分别求出A 和B 的坐标，即可求出AB .解：（1）由于C的极坐标方程为π,22sin 6π1,π.2θθρθ≤<⎪+⎪ ⎪=⎨⎝⎭⎪⎪≤≤⎪⎩，根据互化公式得，曲线C 的直角坐标方程为：当0x <≤时，0x +-=，当10x -≤≤时，221x y +=，则曲线C 与极轴所在直线围成的图形，是一个半径为1的14圆周及一个两直角边分别为1的直角三角形，∴围成图形的面积1π4S =.（2）由11sin 2ρρθ=⎧⎪⎨=⎪⎩得5π1,6A ⎛⎫⎪⎝⎭，其直角坐标为321⎛⎫ ⎪ ⎪⎝⎭， 1sin 2ρθ=化直角坐标方程为12y =， 3π2sin 6ρθ=⎛⎫+ ⎪⎝⎭化直角坐标方程为33x y += ∴312B ⎫⎪⎝⎭， ∴33322AB ==. 23. 已知a ，b 为正数，且满足1a b +=．(1)求证：11(1)(1)9a b ++；(2)求证：1125()()4a b a b ++．（1）证明见解析（2）证明见解析（1）把a +b ＝1代入，用柯西不等式证明；（2）根据基本不等式求出ab 的范围，再化简所求结论，根据对勾函数的最值，求出即可．已知a ，b 为正数，且满足a +b ＝1，（1）（11a +）（11b +）＝111a b a b ab ++++=122a b++，（22a b+）（a +b ）≥222＝8，故11119a b ⎛⎫⎛⎫++≥ ⎪⎪⎝⎭⎝⎭；（2）∵a +b ＝1，a ＞0，b ＞0，∴根据基本不等式1＝a+b0＜ab14≤，（a1a+）（b1b+）222222111a b a b a ba b ab+++++=⋅=≥ab12ab++，令t＝ab∈（0，14]，y＝t1t+递减，所以117444miny=+=，故（a1a+）（b1b+）≥2172544+=．18。

南充高级中学2020届高三数学2月线上月考试题理含解析

∴A的坐标A（4，4 ）∵坐标原点关于准线的对称点的坐标为B（—4，0），
∴|PO|=|PB｜，∴|PA｜+|PO｜的最小值:|AB｜= ．
故选C．
【点睛】本题主要考查抛物线的相关知识．两条线段之和的最小值的求法，是中档题,解题时要认真审题，注意数形结合思想的合理运用．
12。设函数是定义在上的函数，是函数的导函数,若，，为自然对数的底数，则不等式的解集是
故X的取值为0、1、2、3。
.
所以分布列为:
X
0
1
2
3
P
期望值为：
.
【点睛】本题考查频率分布直方图的特点，根据频率分布直方图求平均值，根据统计表求中位数,求随机变量的分布列和数学期望，属于简单题。
19.如图，三棱锥D—ABC中，，E，F分别为DB，AB的中点，且 .
（1)求证:平面平面ABC;
（2）求二面角D-CE—F的余弦值.
四川省南充高级中学2020届高三数学2月线上月考试题理(含解析)
一、选择题：本大题共12个小题,每小题5分,共60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.在复平面内，复数满足，则的共轭复数对应的点位于
A. 第一象限B。第二象限C。第三象限D。第四象限
【答案】A
【解析】
A. 4B. C。 D。
【答案】C
【解析】
【分析】
由已知条件，结合抛物线性质求出A点坐标，求出坐标原点关于准线的对称点的坐标点B，由|PO｜＝|PB,|知｜PA|+|PO|的最小值为｜AB|，由此能求出结果．
【详解】抛物线y2=8x的准线方程为x=—2,∵|AF｜=6，
∴A到准线的距离为6，即A点的横坐标为4，∵点A在抛物线上,不妨设为第一象限,

2019-2020学年四川省南充高级中学高三英语模拟试卷及答案解析

2019-2020学年四川省南充高级中学高三英语模拟试卷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AF the Art World competition.Prize -The winner of each type will get the chance to display their artworks in a week-long exhibition inChelsea,New York..Eligibility - Open to artists all over the world..Entry Fee(参赛费)-$24 for a maximum of 3 submitted(提交的)photos..Date of Exhibition - From 25thApril to 2ndMay.F the Art World is an international art competition organized by which is quite unusual, compared to other competitions. The artworks are not judged on the basis of creativity and skill. Instead, they're judged by the depth of the subject matter. The subject this year is “A Competition About Change”, where artists can try and show how they'd like to change humanity in a good way.Notes*For this competition, there are three types for submissions, namely: street art, fine art, and digital art.*Each type will have a winner as chosen by the organizers and all the winners will have the wonderful chance to exhibit their works at the Unarthodox Gallery inNew Yorkin a week-long exhibition.*All the winners will also receive 100% of the sale price when any of their artworks are sold!Winners will also be displayed on the website and the entrants' artworks will also be displayed in the opening ceremony(仪式)as well.1.What is special about F the Art World Competition?AIt is free to attend.B.It lasts for over a week.C.It is open to artists inNew York.D.It centers on the depth of the artworks.2.What will the winners get?A.Prize money.B.A chance to visit an exhibition.C.A written judgement of their artworks.D.Money from the sale of their artworks.3.Where will the entrants' artworks be shown?A.In the street.B.In the opening ceremony.C.On the website.D.At the market.BScientists say baby sharks are at risk of being born smaller and without the energy they need to survive because of warming oceans from climate change.Scientists studied epaulette sharks, which live off Australia and New Guinea. They found that warmer conditions sped up the sharks’ growing process. That meant the sharks were born earlier and very tired. The findings could be used in the study of other sharks, including those that give birth to live young.The scientists studied 27 sharks. Some were raised in average summer water temperatures, about 27 degrees Celsius. Others were raised in higher temperatures around 29 degrees Celsius and 31 degrees Celsius. They found that the sharks raised in the warmest temperatures weighed much less than those raised in average temperatures. They also showed reduced energy levels.Epaulette sharks can grow to a length of about one meter. Their name comes from large spots on their bodies that look like decorations on some military clothing.One study this year found that worldwide numbers of oceanie sharks and rays dropped more than 70 percent between 1970 and 2018. Overfishing is a main concern, while climate change and pollution also threaten shark.Carolyn Wheeler is a doctoral student at the University of Massachusetts Boston and with the ARC Centre of Excellence for Coral Reef Studies at James Cook University in Australia. She was the lead author of the epaulette shark study. She said that while all the sharks survived, those raised in warmer temperatures were not strong enough to survive for long in the wild.She added that if the sharks are born smaller than usual “they are probably going to have to start looking for food sooner—and they’re going to have less time to adjust to their surroundings.” The study should serve as a warning to ocean governing agencies that careful management is needed to prevent the loss of more sharks.4. In what aspect do the warmer conditions affect the baby sharks?A. Their food.B. Their body weight.C. Their living habits.D. Their moving route.5. How did the scientists carry out their study?A. By studying former data.B. By tracking sharks in the wild.C. By collecting information about climate change.D. By comparing sharks in waters of different temperatures6. What does paragraph 5 mainly talk about?A. The origin of sharks’ names.B. The sharks’ appearance.C. The threats to sharks’ survival.D. The sharks’ living environment.7. Which of the following can be the best title for the text?A. Scientists Raise Sharks to Deal with ProblemsB. Global Warming Has Reduced Shark PopulationsC. Baby Sharks Struggle to Survive in Warming OceansD. Scientists Are Struggling to Save Sharks from ExtinctionCA young female athlete in thePhilippinesrecently won many gold medals during a sports meet despite not having proper running shoes. Rhea Ballos, an 11-year-old student ofSalvationElementary Schoolin Balason,Iloilo, wasonly wearing bandages around her feet when she competed at the Iloilo Schools Sports Meet.Facebook user Valenzuela posted pictures of the girl with her feet wrapped in bandages bearing the famous Nike logo. Ballos even wrote the word “NIKE” on the sides of her “shoe” to complete the “Nike running shoes” look. The bandages were tightly wrapped around her feet, creating a thin protective layer against the track. While she was actually barefoot during the races, she was still able to defeat her competitors who all more proper footwear intended for running,According to the post, Ballos bagged the top awards in the 400-meter dash, the 800-meter run, and the 1500-meter run in the girls' categories in the inter school sporting event held in Iloilo, central Philippines.When pictures of her “Nike” footwear become popular, Flipinos on social media praised her. Many noted that instead of falling into self-pity, she was even able to make light of the situation by drawing the Nike logo on her “running shoes”. Some of the commenters of Valenzuela's post expressed how the girl deserved to be recognized by Nike and that the brand should actually give her a new pair of real Nike shoes. Others started getting in touch with the American sports brand, as well as local basketball specialty store Titan 22.It did not take long for Titan co-founder and Alaska Aces head coach Jeffrey Cariaso to take notice of Ballos' outstanding achievement. Cariaso immediately made an effort to get in touch with the young track runner. Theseven-time PBA champion has since talked to the student as well as her coaches in an apparent bid to help her out.8. Why did Ballos wear bandages around her feet to compete?A. She couldn't afford to buy shoes.B. She wished to be noticed by Nike.C. She wanted to draw public's attention.D. She thought it fashionable and unique.9. What's people's attitude to Ballos' story?A. Surprised.B. Confused.C. Favorable.D. Doubtful.10. What can we infer from the last paragraph?A. Ballos will be recognized by Nike.B. Ballos will be probably helped by Cariaso.C. Ballos is bound to win more champions.D. Ballos will become a great basketball player.11. Which of the following can best describe Ballos?A. Shy and lucky.B. Kind and brave.C. Clever and outstanding.D. Gifted and optimistic.DJapan has announced emergency Covid measures in Tokyo and three other areas in order to control rising infections (传染病), just three months before the country is set to host the Olympics. The government said the state of emergency — set to last for about two weeks— would be “short and powerful”. Under the measures, bars will be required to close and big sporting events will be held without audience. The government has insisted that the Olympics will go ahead in July.Prime minister Yoshihide Suga announced the measures on Friday, saying they would begin on Sunday and remain in place until 11 May. In addition to to Tokyo, Osaka, Kyoto and Hyogo will be affected. It has marked the third state of emergency in japan since the pandemic began. “I sincerely apologize for causing trouble for many people again,” he said. “It is feared that infection in major cities will spread across the whole country if we take no measures.”Under the rules, major facilities like department stores will close, as well as restaurants, bars, and KTV rooms serving alcohol. Restaurants that do not serve alcohol are being told to close early, and companies are being askedto make arrangements for people to work remotely. Schools will remain open. Besides, the emergency measurescoincide withthe country’s “Golden Week” holiday, which runs from late April to the first week of May and is the busiest travel period of the year. Tokyo governor Yuriko Koike urged residents to start taking precautions immediately. In an effort to discourage people from going out at night. She said neon signs (霓虹灯) would be turned off.The coronavirus toll in Japan has been much lower than that in many other countries, with about 558,000 cases and fewer than 10,000 deaths, according to figures collected by Johns Hopkins University. But there are concerns over the latest rise in infections, with reports of hospital bed shortages in some areas.Tokyo 2020 president Seiko Hashimoto said at a news conference on Friday that organizers were not considering canceling the event. “I hope the coronavirus situation improves with the prevention measures the government, Tokyo, and other regional governments have put into place,” she said. “We, as Tokyo 2020, continue to wish for the swift return to normal and will continue to work closely to make sure a safe and secure Olympics can happen.”12. According to the passage, what does Japanese government advise people to do?A. To watch a live game.B. To work at home.C. To hang out in the evening.D. To study at home.13. What is the aim of bringing in emergency Covid measures?A. To relieve people's fear of the the pandemic.B. To settle the problem of hospital bed shortages.C. To test government’s management of emergency.D. To ensure the smooth running of the upcoming Olympics.14. Which words can best describe the government’s action?A. generous and brilliant.B. costly but useless.C. positive and considerate.D. accidental but efficient.15. What does the underlined phrase “coincide with” in Paragraph 3 mean?A. meet withB. deal withC. make upD. set up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。