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安仁县第三中学2019-2020学年上学期高二数学12月月考试题含解析

安仁县第三中学2019-2020学年上学期高二数学12月月考试题含解析

安仁县第三中学2019-2020学年上学期高二数学12月月考试题含解析班级__________ 姓名__________ 分数__________一、选择题1.已知定义在区间[0,2]上的函数y=f(x)的图象如图所示,则y=f(2﹣x)的图象为()A.B.C.D.2.某单位综合治理领导小组成员之问的领导关系可以用框图表示,这种框图通常称为()A.程序流程图B.工序流程图C.知识结构图D.组织结构图3.设a=sin145°,b=cos52°,c=tan47°,则a,b,c的大小关系是()A.a<b<c B.c<b<a C.b<a<c D.a<c<b4.阅读如图所示的程序框图,运行相应的程序.若该程序运行后输出的结果不大于20,则输入的整数i的最大值为()A.3 B.4 C.5 D.65.已知函数f(x)=2x﹣+cosx,设x1,x2∈(0,π)(x1≠x2),且f(x1)=f(x2),若x1,x0,x2成等差数列,f′(x)是f(x)的导函数,则()A.f′(x0)<0 B.f′(x0)=0C.f′(x0)>0 D.f′(x0)的符号无法确定6.已知集合A={0,m,m2﹣3m+2},且2∈A,则实数m为()A.2 B.3 C.0或3 D.0,2,3均可7.直线的倾斜角为()A.B.C.D.8.5名运动员争夺3项比赛冠军(每项比赛无并列冠军),获得冠军的可能种数为()A.35B.C.D.539.若f(x)=sin(2x+θ),则“f(x)的图象关于x=对称”是“θ=﹣”的()A.充分不必要条件B.必要不充分条件C.充要条件 D.既不充分又不必要条件10.已知定义域为的偶函数满足对任意的,有,且当时,.若函数在上至少有三个零点,则。

安仁县第三中学校2018-2019学年高二上学期第二次月考试卷数学

安仁县第三中学校2018-2019学年高二上学期第二次月考试卷数学

安仁县第三中学校2018-2019学年高二上学期第二次月考试卷数学班级__________ 姓名__________ 分数__________一、选择题1. 已知抛物线关于x 轴对称,它的顶点在坐标原点O ,并且经过点M (2,y 0).若点M 到该抛物线焦点的距离为3,则|OM|=( )A .B .C .4D .2. 函数()log 1xa f x a x =-有两个不同的零点,则实数的取值范围是( )A .()1,10B .()1,+∞C .()0,1D .()10,+∞3. 将函数f (x )=3sin (2x+θ)(﹣<θ<)的图象向右平移φ(φ>0)个单位长度后得到函数g (x )的图象,若f (x ),g (x )的图象都经过点P (0,),则φ的值不可能是( )A .B .πC .D .4. 平面向量与的夹角为60°,=(2,0),||=1,则|+2|=( )A .B .C .4D .125. 设向量,满足:||=3,||=4, =0.以,,﹣的模为边长构成三角形,则它的边与半径为1的圆的公共点个数最多为( )A .3B .4C .5D .66. 如图,长方形ABCD 中,AB=2,BC=1,半圆的直径为AB .在长方形ABCD 内随机取一点,则该点取自阴影部分的概率是( )A .B .1﹣C .D .1﹣7. 冶炼某种金属可以用旧设备和改造后的新设备,为了检验用这两种设备生产的产品中所含杂质的关系,调查结果如下表所示.杂质高 杂质低 旧设备 37 121 新设备22202根据以上数据,则( )A .含杂质的高低与设备改造有关B .含杂质的高低与设备改造无关C .设备是否改造决定含杂质的高低D .以上答案都不对8. 命题:“∀x >0,都有x 2﹣x ≥0”的否定是( )A .∀x ≤0,都有x 2﹣x >0B .∀x >0,都有x 2﹣x ≤0C .∃x >0,使得x 2﹣x <0D .∃x ≤0,使得x 2﹣x >09. 已知某市两次数学测试的成绩ξ1和ξ2分别服从正态分布ξ1:N 1(90,86)和ξ2:N 2(93,79),则以下结论正确的是( )A .第一次测试的平均分比第二次测试的平均分要高,也比第二次成绩稳定B .第一次测试的平均分比第二次测试的平均分要高,但不如第二次成绩稳定C .第二次测试的平均分比第一次测试的平均分要高,也比第一次成绩稳定D .第二次测试的平均分比第一次测试的平均分要高,但不如第一次成绩稳定10.已知函数f (x )=x 3+(1﹣b )x 2﹣a (b ﹣3)x+b ﹣2的图象过原点,且在原点处的切线斜率是﹣3,则不等式组所确定的平面区域在x 2+y 2=4内的面积为( )A .B .C .πD .2π11.已知复合命题p ∧(¬q )是真命题,则下列命题中也是真命题的是( ) A .(¬p )∨q B .p ∨q C .p ∧q D .(¬p )∧(¬q )12.把函数y=sin (2x ﹣)的图象向右平移个单位得到的函数解析式为( )A .y=sin (2x ﹣)B .y=sin (2x+)C .y=cos2xD .y=﹣sin2x二、填空题13.设,x y 满足条件,1,x y a x y +≥⎧⎨-≤-⎩,若z ax y =-有最小值,则a 的取值范围为 .14.已知()212811f x x x -=-+,则函数()f x 的解析式为_________.15.若函数f (x )=﹣m 在x=1处取得极值,则实数m 的值是 .16.【盐城中学2018届高三上第一次阶段性考试】已知函数f (x )=()210{ 21(0)xxx e x x x +≥++<,若函数y=f (f (x )﹣a )﹣1有三个零点,则a 的取值范围是_____.17.如图:直三棱柱ABC ﹣A ′B ′C ′的体积为V ,点P 、Q 分别在侧棱AA ′和CC ′上,AP=C ′Q ,则四棱锥B ﹣APQC 的体积为 .18.函数f (x )=2a x+1﹣3(a >0,且a ≠1)的图象经过的定点坐标是 .三、解答题19.(本小题满分13分)椭圆C :22221(0)x y a b a b+=>>的左、右焦点分别为1F 、2F ,直线:1l x my =-经过点1F 与椭圆C 交于点M ,点M 在x 轴的上方.当0m =时,1||2MF =.(Ⅰ)求椭圆C 的方程;(Ⅱ)若点N 是椭圆C 上位于x 轴上方的一点, 12//MF NF ,且12123MF F NF F S S ∆∆=,求直线l 的方程.20.(本小题满分13分)如图,已知椭圆22:14x C y +=的上、下顶点分别为,A B ,点P 在椭圆上,且异于点,A B ,直线,AP BP 与直线:2l y =-分别交于点,M N ,(1)设直线,AP BP 的斜率分别为12,k k ,求证:12k k ⋅为定值;(2)求线段MN的长的最小值;(3)当点P运动时,以MN为直径的圆是否经过某定点?请证明你的结论.【命题意图】本题主要考查椭圆的标准方程及性质、直线与椭圆的位置关系,考查考生运算求解能力,分析问题与解决问题的能力,是中档题.21.已知奇函数f(x)=(c∈R).(Ⅰ)求c的值;(Ⅱ)当x∈[2,+∞)时,求f(x)的最小值.22.某单位组织职工开展构建绿色家园活动,在今年3月份参加义务植树活动的职工中,随机抽取M名职工为样本,得到这些职工植树的株数,根据此数据作出了频数与频率统计表和频率分布直方图如图:(1)求出表中M,p及图中a的值;(2)单位决定对参加植树的职工进行表彰,对植树株数在[25,30)区间的职工发放价值800元的奖品,对植树株数在[20,25)区间的职工发放价值600元的奖品,对植树株数在[15,20)区间的职工发放价值400元的奖品,对植树株数在[10,15)区间的职工发放价值200元的奖品,在所取样本中,任意取出2人,并设X 为X E X 合计23.(本小题满分12分)在ABC ∆中,内角C B A ,,的对边为c b a ,,,已知1cos )sin 3(cos 2cos 22=-+C B B A. (I )求角C 的值;(II )若2b =,且ABC ∆的面积取值范围为,求c 的取值范围. 【命题意图】本题考查三角恒等变形、余弦定理、三角形面积公式等基础知识,意在考查基本运算能力.24.为了预防流感,某学校对教室用药熏消毒法进行消毒.已知药物释放过程中,室内每立方米空气中的含药量y(毫克)与时间t(小时)成正比;药物释放完毕后,y与t的函数关系式为1()16t ay-=(a为常数),如图所示.据图中提供的信息,回答下列问题:(1)写出从药物释放开始,每立方米空气中的含药量y(毫克)与时间t(小时)之间的函数关系式;(2)据测定,当空气中每立方米的含药量降低到0.25毫克以下时,学生方可进教室。

2020年湖南省郴州市安仁县第三中学高二化学联考试卷含解析

2020年湖南省郴州市安仁县第三中学高二化学联考试卷含解析

2020年湖南省郴州市安仁县第三中学高二化学联考试卷含解析一、单选题(本大题共15个小题,每小题4分。

在每小题给出的四个选项中,只有一项符合题目要求,共60分。

)1. 下列各组不属于同分异构体的是A.丙三醇与乙醇B.2—甲基丙烷与正丁烷C.葡萄糖和果糖D.乙酸与甲酸甲酯参考答案:略2. 下列离子在水溶液中能大量共存的一组是A.Na+、CO32-、SO42- B.Fe2+、NO3-、H+C.NH4+、Cl-、OH- D.K+、H+、HCO3-参考答案:略3. 下列两种有机物是某些药物中的有效成分:以下说法正确的是()A.将各1mol的两种物质分别与氢氧化钠溶液充分反应,消耗等量氢氧化钠B.将各1mol的两种物质分别与溴水充分反应,消耗等量溴分子C.将各1mol的两种物质分别与氢气充分反应,消耗等量氢气D.两种有机物中的所有原子均不可能共平面参考答案:B试题分析:A亮菌甲素含有一个酚羟基、一个酯基,酯基水解得到一个酚羟基一个羧基,1mol该物质与NaOH发生反应消耗3molNaOH;对羟基桂皮酸含有一个酚羟基和一个羧基可以与NaOH发生反应,1mol该物质反应消耗2molNaOH,错误。

B.亮菌甲素和对羟基桂皮酸都只含有一个不饱和的碳碳双键可以与溴水发生加成反应。

所以将各1mol的两种物质分别与溴水充分反应,消耗等量溴分子,正确。

C.羧基、酯基的羰基不能发生加成反应,其它双键或苯环都可以与氢气发生加成反应。

所以1mol亮菌甲素与足量的氢气发生加成反应,消耗5molH2;1mol的对羟基桂皮酸与足量的氢气发生加成反应消耗4molH2,两种物质消耗氢气不等量,错误。

D.亮菌甲素含有饱和碳原子,由于饱和碳原子是四面体结构,原子不可能共平面,而对羟基桂皮酸分子可能共平面,错误。

故答案B。

4. 下列化学电池不易造成环境污染的是()A.氢氧燃料电池 B.锌锰电池 C.镍镉电池 D.铅蓄电池参考答案:A略5. 下列实验操作正确的是A.用酸性高锰酸钾溶液鉴别己烷、苯、甲苯B.用水鉴别苯和溴苯C.提取溶解在水中的少量碘:加入裂化汽油振荡D.制取溴苯:将铁屑、液溴、苯充分混合后加热参考答案:B略6. 以乙醇为原料制取乙二醇,需要经过的反应为()A.加成― 消去― 取代 B.消去― 加成―取代C.消去― 取代― 加成 D.取代― 消去― 加成参考答案:B7. 制造火箭外壳材料中的主要元素是A.Fe B.Al C.CD.Si参考答案:B8. 实验室对某有机物有如下实验记录:①在氧气中完全燃烧,只生成CO2和H2O;②完全燃烧1 mol该有机物需消耗氧气112 L(标准状况下)。

安仁县第三高级中学2018-2019学年高二上学期第一次月考试卷物理

安仁县第三高级中学2018-2019学年高二上学期第一次月考试卷物理

安仁县第三高级中学2018-2019学年高二上学期第一次月考试卷物理班级__________ 座号_____ 姓名__________ 分数__________一、选择题1.如图甲、乙两图是电子技术中的常用电路,a、b是各部分电路的输入端,其中输入的交流高频成分用“≋”表示,交流低频成分用“~”表示,直流成分用“—”表示。

关于两图中负载电阻R上得到的电流特征是()A.图甲中R得到的是交流成分B.图甲中R得到的是直流成分C.图乙中R得到的是低频成分D.图乙中R得到的是高频成分2.甲、乙两车在平直公路上同向行驶,其v-t图像如图所示。

已知两车在t=0时并排行驶,则()A. 两车另一次并排行驶的时刻是t=2sB. 在t=3s时,甲车在乙车前7.5mC. 在t=1s时,甲车在乙车后D. 甲、乙车两次并排行驶的位置之间沿公路方向的距离为40m3.矩形线圈绕垂直磁场线的轴匀速转动,对于线圈中产生的交变电流()A.交变电流的周期等于线圈转动周期B.交变电流的频率等于线圈的转速C.线圈每次通过中性面,交变电流改变一次方向D.线圈每次通过中性面,交变电流达到最大值4.甲、乙两车从同一地点沿同一方向出发,下图是甲、乙两车的速度图象,由图可知()A. 甲车的加速度大于乙车的加速度B. 时刻甲、乙两车的速度相等1tC. 时刻甲、乙两车相遇1t D. 0~时刻,甲车的平均速度小于乙车的平均速度1t 5. (2016·山东师大附中高三月考)质量为m 的物体放在水平面上,它与水平面间的动摩擦因数为μ,重力加速度为g 。

用水平力拉物体,运动一段时间后撤去此力,最终物体停止运动。

物体运动的v -t 图象如图所示。

下列说法正确的是()A .水平拉力大小为F =m v 0tB .物体在3t 0时间内位移大小为v 0t 032C .在0~3t 0时间内水平拉力做的功为mv 122D .在0~3t 0时间内物体克服摩擦力做功的平均功率为μmgv 0126. 右图是质谱仪的工作原理示意图。

2022年湖南省郴州市安仁县第三中学高二物理联考试题带解析

2022年湖南省郴州市安仁县第三中学高二物理联考试题带解析

2022年湖南省郴州市安仁县第三中学高二物理联考试题含解析一、选择题:本题共5小题,每小题3分,共计15分.每小题只有一个选项符合题意1. (多选)如图所示,通有稳恒电流的螺旋管竖直放置,有一铜环R沿螺旋管的轴线加速下落.在下落过程中,环面始终保持水平,铜环先后经过轴线上1、2、3位置时加速度分别为a1、a2、a3,位置2在螺旋管的中心,位置1、3与位置2等距离,则()A.a2=g B.a2<g C.a1=a3<a2 D.a3<a1<a2参考答案:AD2. (多选题)A、B两船的质量均为m,都静止在平静的湖面上,现A船上质量为m的人,以对地水平速度v从A船跳到B船,再从B船跳到A船,经n次跳跃后,人停在B船上,不计水的阻力,则()A.A、B两船速度大小之比为2:3B.A、B(包括人)两船动量大小之比为1:1C.A、B(包括人)两船动能之比为3:2D.A、B(包括人)两船动能之比为1:1参考答案:BC【考点】动量守恒定律.【分析】对系统应用动量守恒定律求出动量之比,然后求出船的速度之比,根据动量守恒定律求出系统总动量.【解答】解:A、最终人在B船上,以系统为研究对象,在整个过程中,以A的速度方向为正方向,由动量守恒定律得:mv A﹣(m+M)v B=0,解得:=,故A错误;B、以人与两船组成的系统为研究对象,人在跳跃过程中总动量守恒,所以A、B两船(包括人)的动量大小之比总是1:1,故B正确;C、两船的动能之比:==,故C正确,D错误;故选:BC.3. 图中为一“滤速器”装置示意图。

a、b为水平放置的平行金属板,一束具有各种不同速率的电子沿水平方向经小孔O进入a、b两板之间。

为了选取具有某种特定速率的电子,可在a、b间加上电压,并沿垂直于纸面的方向加一匀强磁场,使所选电子仍能够沿水平直线OO'运动,由O'射出。

不计重力作用。

可能达到上述目的的办法是A.使a板电势高于b板,磁场方向垂直纸面向里B.使a板电势低于b板,磁场方向垂直纸面向里C.使a板电势高于b板,磁场方向垂直纸面向外D.使a板电势低于b板,磁场方向垂直纸面向外参考答案:AD4. (单选)某人估测一竖直枯井深度,从井口静止释放一石头并开始计时,经2s听到石头落地声,由此可知井深约为(不计声音传播时间,重力加速度g取10m/s2)()A.10m B.20m C.30m D.40m参考答案:B5. 图a是一列简谐横波在t=2s时的波形图,图b是这列波中P点的振动图象,则该波的传播速度和传播方向是()A.v=25m/s,向x轴负方向传播B.v=25m/s,向x轴正方向传播C.v=50m/s,向x轴负方向传播D.v=50m/s,向x轴正方向传播参考答案:C【考点】波长、频率和波速的关系;波的形成和传播;横波的图象.【分析】由b点的振动图象可知t=2s时刻P点的振动方向,则可知波的传播方向;由振动图象可知波的周期,由波动图象可知波长,则由波长、频率和波速的关系可求得波速.【解答】解:由b图可知,2s时刻P点正经过平衡位置向正方向运动,则由波动图象可知波应向x轴的负方向传播;而波的周期为2s,波长为100m,则波速v==m/s=50m/s;故C正确;故选C.二、填空题:本题共8小题,每小题2分,共计16分6. (4分)甲乙两物体做自由落体运动,已知甲物体的重量是乙物体的2倍,而甲物体距地面的高度是乙距地面高度的一半,则甲物体的加速度是乙物体的倍,甲物体下落的时间是乙物体下落时间的倍。

安仁县第三中学校2019-2020学年上学期高二数学12月月考试题含解析

安仁县第三中学校2019-2020学年上学期高二数学12月月考试题含解析

安仁县第三中学校2019-2020学年上学期高二数学12月月考试题含解析班级__________ 姓名__________ 分数__________一、选择题1.如图,AB是半圆O的直径,AB=2,点P从A点沿半圆弧运动至B点,设∠AOP=x,将动点P到A,B 两点的距离之和表示为x的函数f(x),则y=f(x)的图象大致为()2.阅读下面的程序框图,则输出的S=()A.14 B.20 C.30 D.553. 已知集合A={0,1,2},则集合B={x ﹣y|x ∈A ,y ∈A}中元素的个数是( ) A .1B .3C .5D .94. 已知f (x )为偶函数,且f (x+2)=﹣f (x ),当﹣2≤x ≤0时,f (x )=2x ;若n ∈N *,a n =f (n ),则a 2017等于( )A .2017B .﹣8C .D .5. 设△ABC 的三边长分别为a 、b 、c ,△ABC 的面积为S ,内切圆半径为r ,则,类比这个结论可知:四面体S ﹣ABC 的四个面的面积分别为S 1、S 2、S 3、S 4,内切球半径为r ,四面体S ﹣ABC 的体积为V ,则r=( )A .B .C .D .6. 过点P (﹣2,2)作直线l ,使直线l 与两坐标轴在第二象限内围成的三角形面积为8,这样的直线l 一共有( )A .3条B .2条C .1条D .0条7. 下列命题正确的是( )A .已知实数,a b ,则“a b >”是“22a b >”的必要不充分条件B .“存在0x R ∈,使得2010x -<”的否定是“对任意x R ∈,均有210x ->”C .函数131()()2xf x x =-的零点在区间11(,)32内D .设,m n 是两条直线,,αβ是空间中两个平面,若,m n αβ⊂⊂,m n ⊥则αβ⊥ 8. 设f (x )=e x +x ﹣4,则函数f (x )的零点所在区间为( ) A .(﹣1,0) B .(0,1) C .(1,2) D .(2,3) 9. 观察下列各式:a+b=1,a 2+b 2=3,a 3+b 3=4,a 4+b 4=7,a 5+b 5=11,…,则a 10+b 10=( )A .28B .76C .123D .19910.如图表示的是四个幂函数在同一坐标系中第一象限内的图象,则幂函数y=x 的图象是( )A .①B .②C .③D .④11.已知集合A ,B ,C 中,A ⊆B ,A ⊆C ,若B={0,1,2,3},C={0,2,4},则A 的子集最多有( ) A .2个 B .4个 C .6个 D .8个12.已知锐角△ABC 的内角A ,B ,C 的对边分别为a ,b ,c ,23cos 2A+cos2A=0,a=7,c=6,则b=( ) A .10B .9C .8D .5二、填空题13在这段时间内,该车每100千米平均耗油量为 升. 14.已知直线5x+12y+m=0与圆x 2﹣2x+y 2=0相切,则m= . 15.【泰州中学2018届高三10月月考】设函数()()21xf x ex ax a =--+,其中1a <,若存在唯一的整数0x ,使得()00f x <,则a 的取值范围是16.抛物线y 2=4x 上一点M 与该抛物线的焦点F 的距离|MF|=4,则点M 的横坐标x= .17.椭圆+=1上的点到直线l :x ﹣2y ﹣12=0的最大距离为 .18.函数y=1﹣(x ∈R )的最大值与最小值的和为 2 .三、解答题19.已知,且.(1)求sin α,cos α的值;(2)若,求sin β的值.20.数列{}n a 中,18a =,42a =,且满足*2120()n n n a a a n N ++-+=∈.(1)求数列{}n a 的通项公式; (2)设12||||||n n S a a a =++,求n S .21.已知定义在区间(0,+∞)上的函数f (x )满足f ()=f (x 1)﹣f (x 2).(1)求f (1)的值;(2)若当x >1时,有f (x )<0.求证:f (x )为单调递减函数;(3)在(2)的条件下,若f (5)=﹣1,求f (x )在[3,25]上的最小值.22.已知曲线21()f x e x ax=+(0x ≠,0a ≠)在1x =处的切线与直线2(1)20160e x y --+= 平行.(1)讨论()y f x =的单调性;(2)若()ln kf s t t ≥在(0,)s ∈+∞,(1,]t e ∈上恒成立,求实数的取值范围.23.(本小题满分12分)某旅行社组织了100人旅游散团,其年龄均在[10,60]岁间,旅游途中导游发现该旅游散团人人都会使用微信,所有团员的年龄结构按[10,20),[20,30),[30,40),[40,50),[50,60]分成5组,分别记为,,,,A B C D E ,其频率分布直方图如下图所示.(Ⅰ)根据频率分布直方图,估计该旅游散团团员的平均年龄;(Ⅱ)该团导游首先在,,C D E 三组中用分层抽样的方法抽取了6名团员负责全团协调,然后从这6名团员中随机选出2名团员为主要协调负责人,求选出的2名团员均来自C 组的概率.24.(本题满分12分)已知数列}{n a 的前n 项和为n S ,233-=n n a S (+∈N n ). (1)求数列}{n a 的通项公式;(2)若数列}{n b 满足143log +=⋅n n n a b a ,记n n b b b b T ++++= 321,求证:27<n T (+∈N n ). 【命题意图】本题考查了利用递推关系求通项公式的技巧,同时也考查了用错位相减法求数列的前n 项和.重点突出运算、论证、化归能力的考查,属于中档难度.安仁县第三中学校2019-2020学年上学期高二数学12月月考试题含解析(参考答案)一、选择题1.【答案】【解析】选B.取AP的中点M,则P A=2AM=2OA sin∠AOM=2sin x2,PB=2OM=2OA·cos∠AOM=2cos x2,∴y=f(x)=P A+PB=2sin x2+2cos x2=22sin(x2+π4),x∈[0,π],根据解析式可知,只有B选项符合要求,故选B.2.【答案】C【解析】解:∵S1=0,i1=1;S2=1,i2=2;S3=5,i3=3;S4=14,i4=4;S5=30,i=5>4退出循环,故答案为C.【点评】本题考查程序框图的运算,通过对框图的分析,得出运算过程,按照运算结果进行判断结果,属于基础题.3.【答案】C【解析】解:∵A={0,1,2},B={x﹣y|x∈A,y∈A},∴当x=0,y分别取0,1,2时,x﹣y的值分别为0,﹣1,﹣2;当x=1,y分别取0,1,2时,x﹣y的值分别为1,0,﹣1;当x=2,y分别取0,1,2时,x﹣y的值分别为2,1,0;∴B={﹣2,﹣1,0,1,2},∴集合B={x﹣y|x∈A,y∈A}中元素的个数是5个.故选C.4.【答案】D【解析】解:∵f(x+2)=﹣f(x),∴f(x+4)=﹣f(x+2)=f(x),即f(x+4)=f(x),即函数的周期是4.∴a2017=f(2017)=f(504×4+1)=f(1),∵f(x)为偶函数,当﹣2≤x≤0时,f(x)=2x,∴f(1)=f(﹣1)=,∴a2017=f(1)=,故选:D.【点评】本题主要考查函数值的计算,利用函数奇偶性和周期性之间的关系是解决本题的关键.5.【答案】C【解析】解:设四面体的内切球的球心为O,则球心O到四个面的距离都是R,所以四面体的体积等于以O为顶点,分别以四个面为底面的4个三棱锥体积的和.则四面体的体积为∴R=故选C.【点评】类比推理是指依据两类数学对象的相似性,将已知的一类数学对象的性质类比迁移到另一类数学对象上去.一般步骤:①找出两类事物之间的相似性或者一致性.②用一类事物的性质去推测另一类事物的性质,得出一个明确的命题(或猜想).6.【答案】C【解析】解:假设存在过点P(﹣2,2)的直线l,使它与两坐标轴围成的三角形的面积为8,设直线l的方程为:,则. 即2a ﹣2b=ab直线l 与两坐标轴在第二象限内围成的三角形面积S=﹣ab=8,即ab=﹣16,联立,解得:a=﹣4,b=4.∴直线l 的方程为:,即x ﹣y+4=0, 即这样的直线有且只有一条,故选:C【点评】本题考查了直线的截距式、三角形的面积计算公式,属于基础题.7. 【答案】C 【解析】考点:1.不等式性质;2.命题的否定;3.异面垂直;4.零点;5.充要条件.【方法点睛】本题主要考查不等式性质,命题的否定,异面垂直,零点,充要条件.充要条件的判定一般有①定义法:先分清条件和结论(分清哪个是条件,哪个是结论),然后找推导关系(判断,p q q p ⇒⇒的真假),最后下结论(根据推导关系及定义下结论). ②等价转化法:条件和结论带有否定性词语的命题,常转化为其逆否命题来判断.8. 【答案】C【解析】解:f (x )=e x+x ﹣4, f (﹣1)=e ﹣1﹣1﹣4<0, f (0)=e 0+0﹣4<0, f (1)=e 1+1﹣4<0, f (2)=e 2+2﹣4>0, f (3)=e 3+3﹣4>0,∵f(1)•f(2)<0,∴由零点判定定理可知,函数的零点在(1,2).故选:C.9.【答案】C【解析】解:观察可得各式的值构成数列1,3,4,7,11,…,其规律为从第三项起,每项等于其前相邻两项的和,所求值为数列中的第十项.继续写出此数列为1,3,4,7,11,18,29,47,76,123,…,第十项为123,即a10+b10=123,.故选C.10.【答案】D【解析】解:幂函数y=x为增函数,且增加的速度比价缓慢,只有④符合.故选:D.【点评】本题考查了幂函数的图象与性质,属于基础题.11.【答案】B【解析】解:因为B={0,1,2,3},C={0,2,4},且A⊆B,A⊆C;∴A⊆B∩C={0,2}∴集合A可能为{0,2},即最多有2个元素,故最多有4个子集.故选:B.12.【答案】D【解析】解:∵23cos2A+cos2A=23cos2A+2cos2A﹣1=0,即cos2A=,A为锐角,∴cosA=,又a=7,c=6,根据余弦定理得:a2=b2+c2﹣2bc•cosA,即49=b2+36﹣b,解得:b=5或b=﹣(舍去),则b=5.故选D二、填空题13.【答案】 8 升.【解析】解:由表格信息,得到该车加了48升的汽油,跑了600千米,所以该车每100千米平均耗油量48÷6=8. 故答案是:8.14.【答案】8或﹣18【解析】【分析】根据直线与圆相切的性质可知圆心直线的距离为半径,先把圆的方程整理的标准方程求得圆心和半径,在利用点到直线的距离求得圆心到直线的距离为半径,求得答案.【解答】解:整理圆的方程为(x ﹣1)2++y 2=1 故圆的圆心为(1,0),半径为1 直线与圆相切∴圆心到直线的距离为半径即=1,求得m=8或﹣18故答案为:8或﹣18 15.【答案】【解析】试题分析:设,由题设可知存在唯一的整数0x ,使得在直线的下方.因为,故当时,,函数单调递减;当时,,函数单调递增;故,而当时,,故当且,解之得,应填答案3,12e ⎡⎫⎪⎢⎣⎭. 考点:函数的图象和性质及导数知识的综合运用.【易错点晴】本题以函数存在唯一的整数零点0x ,使得()00f x <为背景,设置了一道求函数解析式中的参数的取值范围问题,目的是考查函数的图象和性质及导数在研究函数的单调性最值等有关知识的综合运用.同时也综合考查学生运用所学知识去分析问题解决问题的能力.求解时先运用等价转化得到数学思想将问题等价转化为存在唯一的整数0x ,使得在直线的下方.然后再借助导数的知识求出函数的最小值,依据题设建立不等式组求出解之得.16.【答案】3.【解析】解:∵抛物线y2=4x=2px,∴p=2,由抛物线定义可知,抛物线上任一点到焦点的距离与到准线的距离是相等的,∴|MF|=4=x+=4,∴x=3,故答案为:3.【点评】活用抛物线的定义是解决抛物线问题最基本的方法.抛物线上的点到焦点的距离,叫焦半径.到焦点的距离常转化为到准线的距离求解.17.【答案】4.【解析】解:由题意,设P(4cosθ,2sinθ)则P到直线的距离为d==,当sin(θ﹣)=1时,d取得最大值为4,故答案为:4.18.【答案】2【解析】解:设f(x)=﹣,则f(x)为奇函数,所以函数f(x)的最大值与最小值互为相反数,即f(x)的最大值与最小值之和为0.将函数f(x)向上平移一个单位得到函数y=1﹣的图象,所以此时函数y=1﹣(x∈R)的最大值与最小值的和为2.故答案为:2.【点评】本题考查了函数奇偶性的应用以及函数图象之间的关系,奇函数的最大值和最小值互为相反数是解决本题的关键.三、解答题19.【答案】【解析】解:(1)将sin+cos=两边平方得:(sin+cos)2=sin2+2sin cos+cos2=1+sinα=,∴sinα=,∵α∈(,π),∴cosα=﹣=﹣;(2)∵α∈(,π),β∈(0,),∴α+β∈(,),∵sin(α+β)=﹣<0,∴α+β∈(π,),∴cos(α+β)=﹣=﹣,则sinβ=sin=sin(α+β)cosα﹣cos(α+β)sinα=﹣×(﹣)﹣(﹣)×=+=.【点评】此题考查了两角和与差的正弦函数公式,以及运用诱导公式化简求值,熟练掌握公式是解本题的关键.20.【答案】(1)102na n=-;(2)229(5)940(5)nn n nSn n n⎧-≤⎪=⎨-+>⎪⎩.【解析】试题分析:(1)由2120n n na a a++-+=,所以{}na是等差数列且18a=,42a=,即可求解数列{}na的通项公式;(2)由(1)令0na=,得5n=,当5n>时,0na<;当5n=时,0na=;当5n<时,0na>,即可分类讨论求解数列nS.当5n ≤时,12||||||n n S a a a =++2129n a a a n n =+++=-∴229(5)940(5)n n n n S n n n ⎧-≤⎪=⎨-+>⎪⎩.1考点:等差数列的通项公式;数列的求和. 21.【答案】【解析】解:(1)令x 1=x 2>0, 代入得f (1)=f (x 1)﹣f (x 1)=0, 故f (1)=0.…(4分)(2)证明:任取x 1,x 2∈(0,+∞),且x 1>x 2,则>1,由于当x >1时,f (x )<0,所以f ()<0,即f (x 1)﹣f (x 2)<0,因此f (x 1)<f (x 2),所以函数f (x )在区间(0,+∞)上是单调递减函数.…(8分) (3)因为f (x )在(0,+∞)上是单调递减函数,所以f (x )在[3,25]上的最小值为f (25).由f()=f (x 1)﹣f (x 2)得, f (5)=f()=f (25)﹣f (5),而f (5)=﹣1,所以f (25)=﹣2.即f (x )在[3,25]上的最小值为﹣2.…(12分)【点评】本题主要考查抽象函数的应用,利用赋值法以及函数单调性的定义是解决本题的关键.22.【答案】(1)()f x 在1(,)e -∞-,1(,)e +∞上单调递增,在1(,0)e -,1(0,)e 上单调递减;(2)1[,)2+∞.【解析】试题解析:(1)由条件可得221'(1)1f e e a=-=-,∴1a =, 由21()f x e x x=+,可得2222211'()e x f x e x x -=-=, 由'()0f x >,可得2210,0,e x x ⎧->⎨≠⎩解得1x e >或1x e <-;由'()0f x <,可得2210,0,e x x ⎧-<⎨≠⎩解得10x e -<<或10x e <<.所以()f x 在1(,)e -∞-,1(,)e +∞上单调递增,在1(,0)e -,1(0,)e上单调递减.(2)令()ln g t t t =,当(0,)s ∈+∞,(1,]t e ∈时,()0f s >,()ln 0g t t t =>,由()ln kf s t t ≥,可得ln ()t tk f s ≥在(0,)x ∈+∞,(1,]t e ∈时恒成立,即max ln ()t t k f s ⎡⎤≥⎢⎥⎣⎦max()()g t f s ⎡⎤=⎢⎥⎣⎦,故只需求出()f s 的最小值和()g t 的最大值. 由(1)可知,()f s 在1(0,)e 上单调递减,在1(,)e+∞上单调递增,故()f s 的最小值为1()2f e e=,由()ln g t t t =可得'()ln 10g t t =+>在区间(1,]e 上恒成立,所以()g t 在(1,]e 上的最大值为()ln g e e e e ==, 所以只需122e k e ≥=, 所以实数的取值范围是1[,)2+∞.考点:1、利用导数研究函数的单调性及求切线斜率;2、不等式恒成立问题.【方法点晴】本题主要考查的是利用导数研究函数的单调性、利用导数研究函数的最值、不等式的恒成立和导数的几何意义,属于难题.利用导数研究函数()f x 的单调性进一步求函数最值的步骤:①确定函数()f x 的定义域;②对()f x 求导;③令()0f x '>,解不等式得的范围就是递增区间;令()0f x '<,解不等式得的范围就是递减区间;④根据单调性求函数()f x 的极值及最值(闭区间上还要注意比较端点处函数值的大小).23.【答案】【解析】【命题意图】本题考查频率分布直方图与平均数、分层抽样、古典概型等基础知识,意在考查审读能力、识图能力、获取数据信息的能力.24.【答案】 【解析】。

安仁县三中2018-2019学年上学期高二数学12月月考试题含解析

安仁县三中2018-2019学年上学期高二数学12月月考试题含解析

安仁县三中2018-2019学年上学期高二数学12月月考试题含解析班级__________ 姓名__________ 分数__________一、选择题1. 已知棱长为1的正方体的俯视图是一个面积为1的正方形,则该正方体的正视图的面积不可能是( ) A .1B.C.D.2. 若不等式1≤a ﹣b ≤2,2≤a+b ≤4,则4a ﹣2b 的取值范围是( )A .[5,10]B .(5,10)C .[3,12]D .(3,12)3. 在△ABC 中,角A ,B ,C 所对的边分别为a ,b ,c ,若(acosB+bcosA )=2csinC ,a+b=8,且△ABC 的面积的最大值为4,则此时△ABC 的形状为( ) A .等腰三角形B .正三角形C .直角三角形D .钝角三角形4. 命题“设a 、b 、c ∈R ,若ac 2>bc 2则a >b ”以及它的逆命题、否命题、逆否命题中,真命题的个数为( ) A .0 B .1 C .2 D .35. 已知圆O 的半径为1,,PA PB 为该圆的两条切线,,A B 为两切点,那么PA PB ∙ 的最小值为A、4- B、3- C、4-+ D、3-+6. 已知22(0)()|log |(0)x x f x x x ⎧≤=⎨>⎩,则方程[()]2f f x =的根的个数是( )A .3个B .4个C .5个D .6个7. 已知三棱柱111ABC A B C - 的侧棱与底面边长都相等,1A 在底面ABC 上的射影为BC 的中点, 则异面直线AB 与1CC 所成的角的余弦值为( )ABD .348.如右图,在长方体中,=11,=7,=12,一质点从顶点A射向点,遇长方体的面反射(反射服从光的反射原理),将次到第次反射点之间的线段记为,,将线段竖直放置在同一水平线上,则大致的图形是()ABCD9. 设奇函数f (x )在(0,+∞)上为增函数,且f (1)=0,则不等式<0的解集为( )A .(﹣1,0)∪(1,+∞)B .(﹣∞,﹣1)∪(0,1)C .(﹣∞,﹣1)∪(1,+∞)D .(﹣1,0)∪(0,1) 10.等比数列{a n }中,a 4=2,a 5=5,则数列{lga n }的前8项和等于( )A .6B .5C .3D .411.将函数)63sin(2)(π+=x x f 的图象向左平移4π个单位,再向上平移3个单位,得到函数)(x g 的图象, 则)(x g 的解析式为( )A .3)43sin(2)(--=πx x g B .3)43sin(2)(++=πx x g C .3)123sin(2)(+-=πx x g D .3)123sin(2)(--=πx x g【命题意图】本题考查三角函数的图象及其平移变换理论,突出了对函数图象变换思想的理解,属于中等难度. 12.设l ,m ,n 表示不同的直线,α,β,γ表示不同的平面,给出下列四个命题:①若m∥l,m⊥α,则l⊥α;②若m∥l,m∥α,则l∥α;③若α∩β=l,β∩γ=m,γ∩α=n,则l∥m∥n;④若α∩β=l,β∩γ=m,γ∩α=n,n∥β,则l∥m.其中正确命题的个数是()A.1 B.2 C.3 D.4二、填空题13.等比数列{a n}的前n项和为S n,已知S3=a1+3a2,则公比q=.14.若“x<a”是“x2﹣2x﹣3≥0”的充分不必要条件,则a的取值范围为.15.已知sinα+cosα=,且<α<,则sinα﹣cosα的值为.16.已知i是虚数单位,复数的模为.17.设全集U=R,集合M={x|2a﹣1<x<4a,a∈R},N={x|1<x<2},若N⊆M,则实数a的取值范围是.18.若函数63e()()32exxbf x xa=-∈R为奇函数,则ab=___________.【命题意图】本题考查函数的奇偶性,意在考查方程思想与计算能力.三、解答题19.函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<)的一段图象如图所示.(1)求f(x)的解析式;(2)求f(x)的单调减区间,并指出f(x)的最大值及取到最大值时x的集合;(3)把f(x)的图象向左至少平移多少个单位,才能使得到的图象对应的函数为偶函数.20.(本小题满分12分)已知等差数列{}n a 的前n 项和为n S ,且990S =,15240S =. (1)求{}n a 的通项公式n a 和前n 项和n S ; (2)设1(1)n n a b n =+,n S 为数列{}n b 的前n 项和,若不等式n S t <对于任意的*n ∈N 恒成立,求实数t 的取值范围.21.已知数列{a n }满足a 1=﹣1,a n+1=(n ∈N *).(Ⅰ)证明:数列{+}是等比数列;(Ⅱ)令b n =,数列{b n }的前n 项和为S n .①证明:b n+1+b n+2+…+b 2n <②证明:当n ≥2时,S n 2>2(++…+)22. (本题满分12分)在如图所示的几何体中,四边形ABCD 为矩形,直线⊥AF 平面ABCD ,AB EF //,12,2====EF AF AB AD ,点P 在棱DF 上.(1)求证:BF AD ⊥;(2)若P 是DF 的中点,求异面直线BE 与CP 所成角的余弦值; (3)若FD FP 31=,求二面角C AP D --的余弦值.23.已知二次函数f (x )的图象过点(0,4),对任意x 满足f (3﹣x )=f (x ),且有最小值是. (1)求f (x )的解析式;(2)求函数h (x )=f (x )﹣(2t ﹣3)x 在区间[0,1]上的最小值,其中t ∈R ;(3)在区间[﹣1,3]上,y=f (x )的图象恒在函数y=2x+m 的图象上方,试确定实数m 的范围.245(Ⅱ)若同一次考试成绩之差的绝对值不超过5分,则称该次考试两人“水平相当”.由上述5次摸底考试成绩统计,任意抽查两次摸底考试,求恰有一次摸底考试两人“水平相当”的概率.安仁县三中2018-2019学年上学期高二数学12月月考试题含解析(参考答案)一、选择题1.【答案】C【解析】解:水平放置的正方体,当正视图为正方形时,其面积最小为1;当正视图为对角面时,其面积最大为.因此满足棱长为1的正方体的俯视图是一个面积为1的正方形,则该正方体的正视图的面积的范围为.因此可知:A,B,D皆有可能,而<1,故C不可能.故选C.【点评】正确求出满足条件的该正方体的正视图的面积的范围为是解题的关键.2.【答案】A【解析】解:令4a﹣2b=x(a﹣b)+y(a+b)即解得:x=3,y=1即4a﹣2b=3(a﹣b)+(a+b)∵1≤a﹣b≤2,2≤a+b≤4,∴3≤3(a﹣b)≤6∴5≤(a﹣b)+3(a+b)≤10故选A【点评】本题考查的知识点是简单的线性规划,其中令4a﹣2b=x(a﹣b)+y(a+b),并求出满足条件的x,y,是解答的关键.3.【答案】A【解析】解:∵(acosB+bcosA)=2csinC,∴(sinAcosB+sinBcosA)=2sin2C,∴sinC=2sin2C,且sinC>0,∴sinC=,∵a+b=8,可得:8≥2,解得:ab≤16,(当且仅当a=b=4成立)∵△ABC的面积的最大值S△ABC=absinC≤=4,∴a=b=4,则此时△ABC 的形状为等腰三角形. 故选:A .4. 【答案】C【解析】解:命题“设a 、b 、c ∈R ,若ac 2>bc 2,则c 2>0,则a >b ”为真命题; 故其逆否命题也为真命题;其逆命题为“设a 、b 、c ∈R ,若a >b ,则ac 2>bc 2”在c=0时不成立,故为假命题 故其否命题也为假命题故原命题及其逆命题、否命题、逆否命题中,真命题的个数为2个 故选C【点评】本题考查的知识点是四种命题的真假判断,不等式的基本性质,其中熟练掌握互为逆否的两个命题真假性相同,是解答的关键.5. 【答案】D.【解析】设PO t =,向量PA 与PB 的夹角为θ,PA PB ==1sin2t θ=,222cos 12sin 12t θθ=-=-,∴222cos (1)(1)(1)PA PB PA PB t t tθ==-->,2223(1)PA PB t t t∴=+->,依不等式PA PB ∴的最小值为3.6. 【答案】C【解析】由[()]2f f x =,设f (A )=2,则f (x )=A,则2log 2x =,则A=4或A=14,作出f (x )的图像,由数型结合,当A=14时3个根,A=4时有两个交点,所以[()]2f f x =的根的个数是5个。

安仁县第三高级中学2018-2019学年高二上学期第一次月考试卷物理

安仁县第三高级中学2018-2019学年高二上学期第一次月考试卷物理

安仁县第三高级中学2018-2019学年高二上学期第一次月考试卷物理班级__________ 座号_____ 姓名__________ 分数__________一、选择题1.如图甲、乙两图是电子技术中的常用电路,a、b是各部分电路的输入端,其中输入的交流高频成分用“≋”表示,交流低频成分用“~”表示,直流成分用“—”表示。

关于两图中负载电阻R上得到的电流特征是()A.图甲中R得到的是交流成分B.图甲中R得到的是直流成分C.图乙中R得到的是低频成分D.图乙中R得到的是高频成分2.甲、乙两车在平直公路上同向行驶,其v-t图像如图所示。

已知两车在t=0时并排行驶,则()A. 两车另一次并排行驶的时刻是t=2sB. 在t=3s时,甲车在乙车前7.5mC. 在t=1s时,甲车在乙车后D. 甲、乙车两次并排行驶的位置之间沿公路方向的距离为40m3.矩形线圈绕垂直磁场线的轴匀速转动,对于线圈中产生的交变电流()A.交变电流的周期等于线圈转动周期B.交变电流的频率等于线圈的转速C.线圈每次通过中性面,交变电流改变一次方向D.线圈每次通过中性面,交变电流达到最大值4.甲、乙两车从同一地点沿同一方向出发,下图是甲、乙两车的速度图象,由图可知()A. 甲车的加速度大于乙车的加速度t时刻甲、乙两车的速度相等B.1C. 1t 时刻甲、乙两车相遇D. 0~1t 时刻,甲车的平均速度小于乙车的平均速度5. (2016·山东师大附中高三月考)质量为m 的物体放在水平面上,它与水平面间的动摩擦因数为μ,重力加速度为g 。

用水平力拉物体,运动一段时间后撤去此力,最终物体停止运动。

物体运动的v -t 图象如图所示。

下列说法正确的是( )A .水平拉力大小为F =m v 0t 0B .物体在3t 0时间内位移大小为32v 0t 0C .在0~3t 0时间内水平拉力做的功为12mv 20D .在0~3t 0时间内物体克服摩擦力做功的平均功率为12μmgv 06. 右图是质谱仪的工作原理示意图。

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安仁三中高二年级培优试卷一一、单选1. He studied very hard all the time because he didn’t want to _____in his studies.A. fall behindB. be left behindC. left behindD. Both A and B2. Beth _______ her coat with red buttons on it at the Lost and Found.A. provedB. claimedC. identifiedD. sorted3. ---- I haven’t found any money though I’ve searched the drawer bottom up.---- Then, I’m afraid there is ______ left.A. nothingB. no oneC. noneD. neither4. Maybe all Americans realize that __________ they expect to be next president will face masses of problems.A. whoB. whoeverC. whomeverD. no matter who5. Tough __________ the task was, we _______ finish it ahead of time and therefore we were highly praised for it.A. though; wouldB. although; had toC. as; were able toD. though; must6. ---How would you like your coffee?--- __________________.A. It’s well doneB. very nice. Thank you.C. One cup. That’s enoughD. The stronger , the better.7. The play, ______ , is very wonderful: there are many interesting characters in it.A. out of the questionB. on the wholeC. out of questionD. under no condition8. Changes ______to the construction laws in Los Angeles have strengthened the city’s buildings and highways , _______them more resistant to quakes.A. making; madeB. being made; to makeC. made; madeD. made; making9. The five firemen the director had_____ the people in the fire were highly praised.A. rescueB. rescuedC. to rescueD. had rescued10. ________ their son away to college, the old couple got more room in the house.A. AsB. ForC. ThroughD. With11. The driver might have had an accident but he________ the car suddenly.A. brakedB. would brakeC. had brakedD. forget12. It’s terrible. The car ________ break down just as we were going on our holiday.A. shouldB. wouldC. couldD. must13. — Have you graduated from collage?— Yes , ________.A. I studied French for two years.B. I have been studying French.C. I have studied French for two years.D. I had studied French for two years.14. ---Would you mind giving him a message?--- ________.But _______,please. I’ll get a pen first.A. Certainly not; just a momentB. NO , not at all ;hold upC. Of course , I don’tD. I’d like to very much ; hold on15. --- He was nearly killed by a car once.--- Oh. when was________ exactly?--- It was in 2002 __________ he was riding a bike on the drive-way.A. this; thatB. that; whenC. that; thatD. this; when16. _____ scores of times , but he still couldn’t understand it .A. Having explainedB. Having been explainedC. Though it was explainedD. It was explained17. -Must I show you how to operate the machine?-Oh, thank you. I __________ how to do it.A.had wondered B.would wonder C.was wondering D.did wonder18. ______ different life today is from ____ it was 20 years ago.A. What a , whatB. How , whatC. What , whatD. What a , how19. We waited and waited. _________ we had been looking forward to .A. Then came the hourB. Then did the hour comeC. Then the hour cameD. The hour then came20. The new government ________ the foreign--owned fields one after another.A. took overB. took outC. took offD. took on二、完形The requirements for high school graduation have just changed in my community. As a result, all students must 36 sixty hours of service learning, 37 they will not receive a diploma. Service learning is academic learning that also helps the community. 38 of service learning include cleaning up a polluted river, working in a soup kitchen, or tutoring a student. 39 a service experience, students must keep a journal(日志)and then write a 40 about what they have learned.Supporters claim that there are many 41 of service learning. Perhaps most importantly, students are forced to think 42 their own interests and become 43 of the needs of others. Students are also able to learn real-life skills that 44 responsibility, problem-solving, and working as part of a team. 45 , students can explore possible careers 46 service learning.For example, if a student wonders what teaching is like, he or she can choose to work in an elementary school classroom a few afternoons each month. 47 there are many benefits, opponents (反对者) 48 problems with the new requirement. First, they 49 that the main reason students go to school is to learn core subjects and skills. Because service learning is time-consuming, students spend 50 time studying the core subjects. Second, they believe that forcing students to work without 51 goes against the law. By requiring service, the school takes away an individual's freedom to choose.In my view, service learning is a great way to 52 to the community, learn new skills, and explore different careers. 53 , I don' t believe you should force people to help others – the 54 to help must come from the heart. I think the best 55 is one that gives students choices: a student should be able to choose sixty hours of independent study or sixty hours of service. Choice encourages both freedom and responsibility, and as young adults, we must learn to handle both wisely.36. A. spend B. gain C .complete D. save37. A. and B. or C. but D. for38. A. Subjects B. ideas C. Procedures D. Examples39. A. With B. Before C . During D. After40. A. diary B, report C . note D. notice41. A. courses B. benefits C . challenges D. features42. A. beyond B. about C. over D. in43. A. careful B. proud C. tired D. aware44. A. possess B. apply C. include D. develop45. A. Gradually B. Finally C . Luckily D. Hopefully46. A. through B. across C. of D. on47. A. So B. Thus C . Since D. While48. A. deal with B. look into C . point out D. take down49. A. argue B. doubt C . overlook D. admit50. A. much B. full C . less D. more51. A. cost B. pay C. care. D. praise52. A. contribute B. appeal C. attend D. belong53. A. Therefore B. Otherwise C. Besides D. However54. A. courage B. desire C. emotion D. spirit55. A. decision B. purpose C. solution D. result三、阅读APeople diet to look more attractive. Fish diet to avoid being beaten up, thrown out of their social group, and getting eaten as a result. That is the fascinating conclusion of the latest research into fish behavior by a team of Australian scientists.The research team have discovered that subordinate fish voluntarily diet to avoid challenging their larger competitors. “In studying gobies we noticed that only the largest two individuals, a male and female, had breeding (繁殖)rights within the group,” explains Marian Wong. “All other group members are non-breeding females, each being 5-10% smaller than its next largest competitor. We wanted to find out how they maintain this precise size separation.”The reason for the size difference was easy to see. Once a subordinate fish grows to within 5-10% of the size of its larger competitor, it causes a fight which usually ends in the smaller goby being driven away from the group. More often than not, the evicted fish is then eaten up. It appeared that the smaller fish were keeping themselves small in order to avoid challenging the boss fish. Whether they did so voluntarily, by restraining how much they ate, was not clear. The research team decided to do an experiment. They tried to fatten up some of the subordinate gobies to see what happened. To their surprise, the gobies simply refused the extra food they were offered, clearly preferring to remain small and avoid fights, over having a feast.The discovery challenges the traditional scientific view of how boss individuals keep their position in a group. Previously it was thought that large individuals simply used their weight and size to threaten their subordinates and take more of the food for themselves, so keeping their competitors small.While the habits of gobies may seem a little mysterious, Dr. Wong explains that understanding the relationships between boss and subordinate animals is important to understanding how hierarchical (等级的)societies remain stable.The research has proved the fact that voluntary dieting is a habit far from exclusive tohumans. “As yet, we lack a complete understanding of how widespread the voluntary reduction of food intake is in nature,” the researchers comment. “Data on human dieting suggests that, while humans generally diet to improve health or increase attractiveness, rarely does it improve long-term health and males regularly prefer females that are fatter than the females’ own ideal.”65. When a goby grows to within 5-10% of the size of its larger competitor, it .A. faces dangerB. has breeding rightsC. eats its competitorD. leaves the group itself66. The underlined words “the evicted fish” in Paragraph 3 refer to .A. the fish beaten upB. the fish found outC. the fish fattened upD. the fish driven away67. The experiment showed that the smaller fish .A. fought over a feastB. went on diet willinglyC. preferred some extra foodD. challenged the boss fish68. What is the text mainly about?A. Fish dieting and human dieting.B. Dieting and health.C. Human dieting.D. Fish dieting.BHave you ever noticed the colour of the water in a river or stream after a heavy rainfall? What do you think caused this change in colour? It is soil that has been washed into the river from the riverbank or from the nearby fields.Components of SoftSoil is made up of a number of layers(层),each having its own distinctive colour and texture.The upper layer is known as the litter.It acts like a blanket, limiting temperature changes and reducing water loss.The topsoil layer is made up of small particles of rock mixed with rotten plant and animal matter called humus(腐殖质),which is black and gives the topsoil its dark colour.This layer is usually rich in nutrients,oxygen,and water.Below the topsoil is the subsoil,a layer that contains more stones mixed with only small amounts of organic matter.This layer is lighter in colour because of the lack of humus.Beneath the soil lies a layer of bedrock.Soil forms from the bottom up.Over time bedrock is attacked by rain, wind,frost, and snow.It is gradually broken down into smaller particles in a process called weathering.Plants begin to grow,and rotten materials enrich the topsoil.Most of the soil in Eastern Canada.for example.Was formed from weathered rock that was exposed when the ice disappeared l2.000 years ago.Water Beneath the SoilSurface water collects and flows above the ground in lakes.ponds.and rivers.Once in the soil or rock,it is called groundwater.Gravity pulls groundwater through the soil in a process called percolation(渗透).Eventually the water reaches a layer called the water table.Under this is bedrock through which water cannot percolate.As water percolates downward,it dissolves organic matter and minerals from the soil and carries them to deeper layers.This causes a serious problem because plants require these nutrients for growth.Soil pHSoil can be acidic.neutral.or basic.The pH of the soil is determined by the nature ofthe rock from which it was formed.and by the nature of the plants that grow and rot in it.The acidity of rain and snow can lower the pH of the groundwater that enters the soil.By burning fossil fuels such as coal,oil and gasoline,humans have been contributing to higher levels of acidity in many soils.When fossil fuels are burned.gases are released into the air and then fall back to earth as acid rain.Acid soil increases出e problem of carrying nutrients to lower soil levels.As nutrients are removed,soil is less fertile.Plants grow more slowly in acidic soil,and also become easily attacked by diseases.67.The layer of soil that provides necessary nutrients for plant growth is called.A.1itter B.topsoil C.humus D.subsoil 68.According to the text.which of the following is NOT true?A.Soil forms from weathered rock on the earth surface.B.The deeper layer of soil is darker in colour than the surface soil.C.Air pollution is partially responsible for acid soil.D.Groundwater tends to carry away nutrients for plant growth.69.We can infer from the passage that the water table lies .A.between the topsoil layer and the subsoil layerB.in the subsoil layer above bedrockC.between the subsoil layer and bedrockD.in the bedrock layer beneath the subsoil70.The underlined word “dissolve” is used to express the idea that organic matter and minerals from soil are .A.rushed away into the riverB.cleaned and purified by waterC.destroyed and carried away by waterD.mixed with water and become part of it四、任务型Communication PrinciplesHow you see yourself can make a great difference in how you communicate.“Every individual exists in a continually changing world of experience of which he (or she) is the center”.Many communication scholars and social scientists believe that people are products of how others treat them and of the messages others send them.But every day we experience the centrality of our selves in communication.A student.for instance,may describe a conflict with a teacher as unfair treatment:“I know my teacher doesn’t like the fact that I don’t agree with his opinions.and that’s why he gave me such a poor grade in that class.”The teacher might say the opposite.Each person may believe that he is correct and that the other person’s view is wrong.The concept of serf originates in communication.Through verbal and nonverbal symbols, a child learns to accept roles in response to the expectations of others.You establish self-image。

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