山东省济宁市学而优教育咨询有限公司2016-2017学年高二下学期期末考试1英语试题(2017-7-10)

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山东省济宁市2016-2017学年高二下学期期末考试语文试题答案不完整含答案

山东省济宁市2016-2017学年高二下学期期末考试语文试题答案不完整含答案

山东省济宁市2016-2017学年高二下学期期末考试语文试题―、现代文阅读。

(35 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。

任何经典,总是活在当下,总是与一切时代同在,回答每一个它的读者所处时代必然会提出的问题,无论它们具体的内容是什么。

对于一个真正有思想能力和发现能力的人来说,所有的经典都是他那个时代的经典。

只有思维能力孱弱、缺乏足够想象力的人,才会把《论语》或《史记》看作是过去时代的书.也没有一个好学深思者,会认为荷马史诗表达的只是虚构的希腊神话,而不是复杂的人类经验。

没有一个真正用思想读书的人,会认为先秦思想家或古希腊哲学家只属于先秦和古希腊,而不是我们的同时代人。

经典与一般著作不同的地方就在于,它们不是单纯的书,而是人类经验不可分割的基本组成部分,与之一起生活、成长。

另一方面,阅读经典是人类成长的基本方式,人类每次总是带着新的经验和新的思想去阅读经典,经典也因而每次都会展现出新的深度和广度。

每次在经典中看到的都是他第一次阅读经典时看到的东西的人,一定是没有思想活力和生命活力的读者。

经典是意义的渊薮,是思想取之不尽的源泉。

经典是无法一览无余的,它随着我们的理解力和领悟力以及我们的问题意识的提高而愈益精深博大。

德国当代哲学家、现代哲学解释学的创始人——伽达默尔认为,理解文本的首要前提是我们先向它提出问题,然后将文本视为对我们问题的回答。

“因为提出问题,就是打开了意义的各种可能性,因而就让有意义的东西进入自己的意见中。

"文本的意义是无穷尽的,因为一代又一代的人会提出不同的问题,以不同的方式去理解,文本因而获得新的意义。

这样的问答逻辑,主观主义的意味是很明显的,它强调的是读者的主动性———文本似乎只能通过被动回答读者或解释者所提的问题,而产生它的意义,却没有看到文本,尤其是经典文本对读者的引导作用。

读者不可能随便提问,他的问题也不可能不围绕着文本提出。

2016-2017学年山东省济宁市高二(下)期末数学试卷(理科)(解析版)

2016-2017学年山东省济宁市高二(下)期末数学试卷(理科)(解析版)

2016-2017学年山东省济宁市高二(下)期末数学试卷(理科)一、选择题(共12小题,每小题5分,满分60分)1.(5分)在复平面内复数z=(i为虚数单位)对应的点在()A.第一象限B.第二象限C.第三象限D.第四象限2.(5分)在用线性回归方程研究四组数据的拟合效果中,分别作出下列四个关于四组数据的残差图,则用线性回归模式拟合效果最佳的是()A.B.C.D.3.(5分)已知向量,且,则x的值为()A.12B.10C.﹣14D.144.(5分)现抛掷两枚骰子,记事件A为“朝上的2个数之和为偶数”,事件B为“朝上的2个数均为偶数”,则P(B|A)=()A.B.C.D.5.(5分)如图阴影部分的面积是()A.e+B.e+﹣1C.e+﹣2D.e﹣6.(5分)设随机变量X,Y满足:Y=3X﹣1,X~B(2,p),若P(X≥1)=,则D(Y)=()A.4B.5C.6D.77.(5分)函数y=x﹣2sin x的图象大致是()A.B.C.D.8.(5分)某班数学课代表给全班同学出了一道证明题,以下四人中只有一人说了真话,只有一人会证明此题.甲:我不会证明.乙:丙会证明.丙:丁会证明.丁:我不会证明.根据以上条件,可以判定会证明此题的人是()A.甲B.乙C.丙D.丁9.(5分)做一个无盖的圆柱形水桶,若要使其体积是64π,且用料最省,则圆柱的底面半径为()A.3B.4C.5D.610.(5分)直三棱柱ABC﹣A1B1C1中,∠BCA=90°,CA=CC1=2CB,则直线BC1与直线AB1所成角的余弦值为()A.B.C.D.11.(5分)某教师有相同的语文参考书3本,相同的数学参考书4本,从中取出4本赠送给4为学生,每位学生1本,则不同的赠送方法共有()A.15种B.20种C.48种D.60种12.(5分)已知函数f(x)=x3+a与函数g(x)=x2﹣2x的图象上恰有三对关于y轴对称的点,则实数a的取值范围是()A.(﹣,)B.(,)C.(﹣,)D.(﹣,﹣)二、填空题(共4小题,每小题5分,满分20分)13.(5分)曲线y=sin x+e x在点(0,1)处的切线方程为.14.(5分)已知(a+2x)(1+)6的展开式的所有项系数的和为192,则展开式中x2项的系数是.15.(5分)如图,已知二面角α﹣l﹣β的大小为60°,其棱上有A,B两点,直线AC,BD 分别在这个二面角的两个半平面内,且都垂直于AB,已知AB=2,AC=3,BD=4,则线段CD的长为.16.(5分)在探究系数一元二次方程的根与系数的关系时,可按下述方法进行:设实系数一元二次方程a2x2+a1x+a0=0…①在复数集C内的根为x1,x2,则方程①可变形为a2(x﹣x1)(x﹣x2)=0,展开得a1x2﹣a2(x1+x2)x+a2x1x2=0,…②比较①②可以得到:类比上述方法,设实系数一元n次方程a n x n+a n﹣1x n﹣1+…+a1x+a0=0(n≥2且n∈N*)在复数集C内的根为x1,x2,…,x n,则这n个根的积x i=.三、解答题(共5小题,满分60分)17.(12分)观察下列等式:﹣1=﹣1;﹣1+3=2;﹣1+3﹣5=﹣3;﹣1+3﹣5+7=4;…(1)照此规律,归纳猜想出第n个等式(2)用数学归纳法证明(1)中的猜想.18.(12分)甲、乙两企业生产同一种型号零件,按规定该型号零件的质量指标值落在[45,75)内为优质品,从两个企业生产的零件中各随机抽出了500件,测量这些零件的质量指标值,得结果如表:甲企业:乙企业:(1)已知甲企业的500件产品质量指标值的样本方差s2=142,该企业生产的零件质量指标值X服从正态分布N(μ,σ2),其中μ近似为质量指标值的样本平均数(注:求时,同一组数据用该区间的中点值作代表),σ2近似为样本方差s2,试根据该企业的抽样数据,估计所生产的零件中,质量指标值不低于71.92的产品的概率(精确到0.001)(2)由以上统计数据完成下面2×2列联表,并问能否在犯错误的概率不超过0.01的前提下,认为“两个分厂生产的零件的质量有差异”附注:参考数据:≈11.92参考公式:P(μ﹣σ<X<μ+σ)=0.6827,P(μ﹣2σ<X<μ+2σ)=0.9545,P(μ﹣3σ<X<μ+3σ)=0.9973.K2=19.(12分)如图,在三棱锥P﹣ABC中,AB⊥BC,P A=PB,E为AC的中点(1)求证:PE⊥AB(2)设平面P AB⊥平面ABC,PB=BC=2,AC=4,求二面角B﹣P A﹣C的平面角的正弦值.20.(12分)在某校歌咏比赛中,甲班、乙班、丙班、丁班均可从A、B、C、D四首不同曲目中任选一首(1)求甲、乙两班选择不同曲目的概率(2)设这四个班级总共选取了X首曲目,求X的分布列及数学期望EX.21.(12分)已知函数f(x)=ax﹣1﹣lnx(a∈R)(1)讨论函数f(x)极值点的个数,并说明理由(2)若∀x>1,xf(x)<ax2﹣ax+a恒成立,求a的最大整数值.[选修4-4:坐标系与参数方程]22.(10分)在直角坐标系xOy中,直线l的参数方程是(t为参数),以坐标原点O为极点,以x轴的正半轴为极轴,建立极坐标系,曲线C的极坐标方程是ρ=2cosθ.(1)求直线l的普通方程和曲线C的直角坐标方程;(2)设点P(m,0),若直线l与曲线C交于A、B两点,且|P A|•|PB|=1,求实数m的值.[选修4-5:不等式选讲]23.已知函数f(x)=|x﹣1|﹣|x|+a.(1)若a=0,求不等式f(x)≥0的解集;(2)若方程f(x)+x=0有三个不同的解,求实数a的取值范围.2016-2017学年山东省济宁市高二(下)期末数学试卷(理科)参考答案与试题解析一、选择题(共12小题,每小题5分,满分60分)1.【解答】解:复数z===对应的点在第二象限.故选:B.2.【解答】解:当残差点比较均匀地落在水平的袋装区域中,说明选用的模型比较合适,这样的带状区域的宽度越窄,说明拟合精度越好,拟合效果越好,对比4个残差图,易知选项C的图对应的袋装区域的宽度越窄.故选:C.3.【解答】解:因为向量,且,属于=﹣8﹣6+x=0,解得x=14;故选:D.4.【解答】解:事件A为“朝上的2个数之和为偶数“所包含的基本事件有:(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(1,3),(3,1),(1,5)、(5,1),(3,5),(5,3),(2,4),(4,2),(2,6),(6,2),(4,6),(6,4)共18个事件AB,所包含的基本事件有:(2,2),(4,4),(6,6),(2,4),(4,2),(2,6),(6,2),(4,6),(6,4)共9个根据条件概率公式P(B|A)==,故选:D.5.【解答】解:利用定积分可得阴影部分的面积S==(e x+e﹣x)=e+﹣2.故选:C.6.【解答】解:∵随机变量X,Y满足:Y=3X﹣1,X~B(2,p),P(X≥1)=,∴P(X=0)=1﹣P(X≥1)==,解得p=,∴X~B(2,),∴D(X)=2×=,∴D(Y)=9E(X)=9×=4.故选:A.7.【解答】解:函数y=x﹣2sin x可知2sin x∈[﹣2,2],当x>2时,y>0,排除选项C,D;当x=时,y=<0,排除选项A.故选:B.8.【解答】解:四人中只有一人说了真话,只有一人会证明此题.丙:丁会证明.丁:我不会证明.所以丙与丁中一定有一个是正确的;若丙说了真话,则甲必是假话,矛盾;若丁说了真话,则甲说的是假话,甲就是会证明的那个人,符合题意;以此类推.易得出答案:A.故选:A.9.【解答】解:设圆柱的底面半径为r,则高h==,则圆柱的表面积S=πr2+2==πr2+≥3=48π.当且仅当,即r=4时,取等号.∴要使其体积是64π,且用料最省,则圆柱的底面半径为4.故选:B.10.【解答】解:如图,∵直三棱柱ABC﹣A1B1C1中,∠BCA=90°,CA=CC1=2CB,∴以C1为原点,C1B1为x轴,C1A1为y轴,C1C为z轴,建立空间直角坐标系,设CA=2,则B(1,0,2),C1(0,0,0),A(0,2,2),B1(1,0,0),=(﹣1,0,﹣2),=(1,﹣2,﹣2),设直线BC1与直线AB1所成角为θ,则cosθ===.故选:D.11.【解答】解:根据题意,按取出4本书的情况不同分4种情况讨论:①、若取出的4本书全部是数学参考书,将其赠送给4位学生,有1种情况,②、若取出的4本书有1本语文参考书,3本数学参考书,需要在4个学生中选取1人,接受语文参考书,剩下的3人接受数学参考书,有C41=4种赠送方法,③、若取出的4本书有2本语文参考书,2本数学参考书,需要在4个学生中选取2人,接受语文参考书,剩下的2人接受数学参考书,有C42=6种赠送方法,④、若取出的4本书有3本语文参考书,1本数学参考书,需要在4个学生中选取3人,接受语文参考书,剩下的1人接受数学参考书,有C43=4种赠送方法,则一共有1+4+6+4=15种赠送方法,故选:A.12.【解答】解:由题意可知f(x)=g(﹣x)有三解,即a=﹣x3+x2+2x有三解,设h(x)=﹣x3+x2+2x,则h′(x)=﹣x2+x+2,令h′(x)=0可得x=﹣1或x=2.∴当x<﹣1或x>2时,h′(x)<0.当﹣1<x<2时,h′(x)>0,∴h(x)在(﹣∞,﹣1)上单调递减,在(﹣1,2)上单调递增,在(2,+∞)上单调递减,∴当x=﹣1时,h(x)取得极小值h(﹣1)=﹣,当x=2时,h(x)取得极大值.∴﹣<a<.故选:C.二、填空题(共4小题,每小题5分,满分20分)13.【解答】解:y=sin x+e x的导数为y′=cos x+e x,在点(0,1)处的切线斜率为k=cos0+e0=2,即有在点(0,1)处的切线方程为y=2x+1.故答案为:y=2x+1.14.【解答】解:令x=1,可得(a+2x)(1+)6的展开式的所有项系数的和为(a+2)•26=192,∴a=1.∴(a+2x)(1+)6=(1+2x)(1+)6,而(1+)6的展开式的通项公式为T r+1=•,故展开式中x2项的系数是+2=45,故答案为:45.15.【解答】解:∵二面角α﹣l﹣β的大小为60°,其棱上有A,B两点,直线AC,BD分别在这个二面角的两个半平面内,且都垂直于AB,AB=2,AC=3,BD=4,∴=,∴=()2=+2+2+2=4+9+16+2||•||cos120°=29﹣12=17,∴||=,即CD的长为.故答案为:.16.【解答】解:考查一元三次方程:①,在复数集C内的根为x1,x2,x3,则方程①可变形为a3(x﹣x1)(x﹣x2)(x﹣x3)=0,展开得②,结合①②可得:,同理考查一元四次方程可得:,据此归纳可得:.故答案为:.三、解答题(共5小题,满分60分)17.【解答】解:(1)观察等式:﹣1=﹣1,﹣1+3=2,﹣1+3﹣5=﹣3,﹣1+3﹣5+7=4,…可得﹣1+3﹣5+…+(﹣1)n(2n﹣1)=(﹣1)n•n.(2)证明:①n=1时,左式=右式=﹣1,等式成立.②假设n=k时,等式成立,即﹣1+3﹣5+…+(﹣1)k(2k﹣1)=(﹣1)k•k,则当n=k+1时,左式=﹣1+3﹣5+…+(﹣1)k(2k﹣1)+(﹣1)k+1(2k+1)=(﹣1)k•k+(﹣1)k+1(2k+1)=(﹣1)k+1(﹣k+2k+1)=(﹣1)k+1(k+1)=右式,即n=k+1时,等式成立.根据①,②,等式对任意的n∈N*均成立.18.【解答】解:(1)计算甲企业数据的平均值为:=×(30×10+40×40+50×115+60×165+70×120+80×45+90×5)=60,∴μ=60,σ2=142,且甲企业产品的质量指标值X服从正态分布X~N(60,142),又σ=≈11.92,则P(60﹣11.92<X<60+11.92)=P(48.08<X<71.92)=0.6826,P(X>71.92)===0.1587≈0.159,估计所生产的零件中,质量指标值不低于71.92的产品的概率为0.159;(2)由以上统计数据填写2×2列联表,计算K2==≈8.772>6.635,对照临界值表得出,在犯错误的概率不超过0.01的前提下认为“两个分厂生产的零件质量有差异”.19.【解答】(1)证明:取AB的中点D,连接PD,∵P A=PB,D为AB中点,∴PD⊥AB.∵D、E分别为AB、AC中点,∴DE∥BC,∵BC⊥AB,∴DE⊥AB,又∵PD∩DE=D,PD,DE⊂平面PDE∴AB⊥平面PDE,∵PE⊂平面PDE,∴PE⊥AB;(2)解:∵平面P AB⊥平面ABC,ED⊥AB,∴ED⊥平面P AB,则PD⊥DE.如图,以D为原点建立空间直角坐标系,由P A=PB=BC=2,AC=4,则A(0,﹣,0),P(0,0,1),E(1,0,0),∴=(0,,1),=(1,,0).设平面P AC的法向量=(x,y,z),则,令z=,得=(,﹣1,)∵DE⊥平面P AB,∴平面P AB的法向量为=(1,0,0),∴cos<>=.∴二面角B﹣P A﹣C的平面角的正弦值为.20.【解答】解:(1)甲班、乙班、丙班、丁班均可从A、B、C、D四首不同曲目中任选一首,∴甲、乙两班选择不同曲目的概率P==;(2)∵这四个班级总共选取了X首曲目,∴X的可能取值为1,2,3,4,P(X=1)==,P(X=2)==,P(X=3)==,p(X=4)==.∴X的分布列为:E(X)=1×+2×+3×+4×=.21.【解答】解:(1)由f(x)=ax﹣lnx﹣1,得f′(x)=a﹣=,当a≤0时,f′(x)<0在(0,+∞)恒成立,∴函数f(x)在(0,+∞)上单调递减,∴f(x)在(0,+∞)上没有极值点;当a>0时,由f′(x)<0,得0<x<,由f′(x)>0,得x>.∴f(x)在(0,)上单调递减,在(,+∞)上单调递增,即f(x)在x=处有极小值.∴当a≤0时,f(x)在(0,+∞)上没有极值点,当a>0时,f(x)在(0,+∞)上有一个极值点;(2)对∀x>1,xf(x)<ax2﹣ax+a恒成立等价于a<对∀x>1恒成立,设函数g(x)=(x>1),则g′(x)=(x>1),令函数φ(x)=x﹣lnx﹣2,则φ′(x)=1﹣(x>1),当x>1时,φ′(x)=1﹣>0,故φ(x)在(1,+∞)递增,又φ(3)=1﹣ln3<0,φ(4)=2﹣ln4>0,故存在x0∈(3,4),使得φ(x0)=0,即g′(x0)=0,且当x∈(1,x0)时,φ(x)<0,即g(x)<0,故g(x)在(1,x0)递减,当x∈(x0,+∞)时,φ(x)>0,即g(x)>0,故g(x)在(x0,+∞)递增,故x∈(1,+∞)时,g(x)有最小值g(x0)=,由φ(x0)=0,得x0﹣lnx0﹣2=0,即lnx0=x0﹣2,故g(x0)==x0,故a<x0,又x0∈(3,4),故实数a的最大整数值是3.[选修4-4:坐标系与参数方程]22.【解答】解:(1)直线l的参数方程是(t为参数),消去参数t可得普通方程:x﹣y﹣m=0.曲线C的极坐标方程是ρ=2cosθ,即ρ2=2ρcosθ,可得直角坐标方程:x2+y2=2x.(2)设点P(m,0),把直线l的参数方程(t为参数)代入圆C的方程:t2+(m﹣)t+m2﹣2m=0,△=﹣4(m2﹣2m)>0,解得﹣1<m<3.∴t1t2=m2﹣2m,∴|P A|•|PB|=|t1t2|=|m2﹣2m|=1,又﹣1<m<3.解得m=1,m=1.∴实数m的值为1,1.[选修4-5:不等式选讲]23.【解答】解:(1)当a=0时,f(x)=|x﹣1|﹣|x|=,所以当x<0时,f(x)=1>0,符合题意;当0≤x<1时,f(x)=1﹣2x≥0,解得0≤x;当x≥1时,f(x)=﹣1<0,不符合题意.综上可得,f(x)≥0的解集为(].(2)设u(x)=|x|﹣|x﹣1|﹣x,y=u(x)的图象和y=a的图象如图所示.易知y=u(x)的图象与y=a的图象有3个交点时,a∈(﹣1,0),所以实数a的取值范围为(﹣1,0).。

山东省济宁市2016-2017学年高二化学下学期期末考试试题(扫描版)

山东省济宁市2016-2017学年高二化学下学期期末考试试题(扫描版)

2016—2017学年度高二下学期期末考试参考答案17.(1)①2H2(g) + SO2(g)=S(s) + 2H2O (g) ΔH1=+90.4 kJ·mol-1(2分)② a(1分)③ 5 (2分)(2)NO -3e- +2H2O = NO3- + 4H+(2分)NH3(2分)(3)CuS(1分) 6.3×10-13mol·L-1(2分)选修三:18.(14分)(1)B2O3+2NH3=2BN+3H2O(2分)(2)2s22p3(2分)(3)N(1分); +3(1分)(4) sp3(1分); 109028’ (1分); PO43-、CCl4、SiF4(等合理即可)(2分)(5)平面正三角形(2分);配位键(2分)19.(12分)(1)Cr(2分)(2)2(2分)(3) π(2分)(4)B(2分)(5)Cu2+ + 4NH3 = 2+(2分) N、F、H三种元素的电负性:F>N>H,在NF3中,共用电子对偏向F原子,使得N原子上的孤对电子难与Cu2+形成配位键(2分)20.(14分)(1)C、N、O(每空1分,共3分)(2)第四周期Ⅷ族(1分)(3)3:1(2分);金属键(2分)Cu3AuH8(2分)(4) 8(2分)(5) 2π(a3+b3)/3(a+b)3(2分)选修五:18.(1) (2分)(2) 吸收挥发出来的溴蒸气(2分) D中有白雾,试液变红(1分) E中出现淡黄色沉淀(1分)(3) 广口瓶中的水沿导管进入三颈烧瓶中,瓶中液体出现分层(2分)(4) ②将多余的溴反应掉(2分)③干燥(2分)④C(2分)19.(1)(1分)(2)碳碳双键、醛基(2分)(3)保护醛基不被氧化(1分)(4)+2NaOH+2NaBr (2分)(5)+2Ag(NH3)2OH+2Ag+3NH3+H2O (2分)(6)6种(2分)或(2分)20.(1)C7H8 (1分) 3 (2分)(2)对甲基苯乙烯(2分) 14(2分)(3)(2分)(4)(2分)(5)(3分)。

山东省济宁市2016-2017学年高二下学期期末考试物理试题

山东省济宁市2016-2017学年高二下学期期末考试物理试题

一、选择题:共8小题,每小题5分,共40分,在每小题给出的四个选项中,第1~5题只有一个选项符合题目要求,第6~8题有多项符合题目要求,全选对得5分;选对但不全的得3分,有选错或不答的得0分1.关于动量守恒的条件,下列说法正确的是A .只要系统内存在摩擦力,动量不可能守恒B .只要系统内某个物体做加速运动,动量就不守恒C .只要系统所受合外力恒定,动量守恒D .只要系统所受合外力为零,动量守恒2.如图所示为交流电发电机示意图,线圈的AB 边连在金属滑环K 上,CD 边连在滑环L 上,两个电刷E 、F 分别压在两个滑环上,线圈在转动时可以通过滑环和电刷保持与外电路的连接。

下列说法正确的是A .当线圈平面转到中性面的瞬间,穿过线圈的磁通量最小B .当线圈平面转到跟中性面垂直的瞬间,穿过线圈的磁通量最小C .当线圈平面转到中性面的瞬间,线圈中的感应电流最大D .当线圈平面转到跟中性面垂直度的瞬间,线圈中的感应电流最小3.如图所示,氢原子在下列各能级间跃迁;(2)从n=3到n=1;(2)从n=5到n=3;(3)从n=4到n=2.在跃迁过程中辐射的电磁波的波长分别用123λλλ、、表示。

波长123λλλ、、的大小顺序是A .231λλλ<<B .321λλλ<<C .132λλλ<<D .123λλλ<<4.一单色光照射到某金属表面时,有光电子从金属表面逸出,下列说法正确的是A .增大入射光的频率,金属逸出功将减小B .延长入射光照射时间,光电子的最大初动能将增大C .增大入射光的频率,光电子的最大初动能将增大D .增大入射光的频率,光电子逸出所经历的时间将缩短5.2个质子和2个中子结合成氦核的核反应方程为11410222H n He +→,已知质量的质量271.672610H m kg -=⨯,中子的质量271.674910n m kg -=⨯,α粒子的质量271.646710m kg α-=⨯,光速83.010/c m s =⨯,则α粒子的结合能约为A .124.310J -⨯ B .125.210J -⨯ C .104.310J -⨯ D .105.210J -⨯6.如图所示,AB 为固定的光滑圆弧轨道,O 为圆心,AO 水平,BO 竖直,轨道半径为R ,将质量为m 的小球(可视为质点)从A 点由静止释放,小球沿轨道从A 点运动到B 点的过程中,下列判断正确的是A .小球所受合力的冲量水平向右B .小球所受支持力的冲量水平向右C .小球所受合力的冲量大小为D .小球所受重力的冲量大小为零7.如图甲所示,一小型理想变压器原副线圈匝数之比12:10:1n n =,连线柱a 、b 接上一个正弦交变电流,电压随时间变化规律如图乙所示,变压器右侧部分为一火灾报警系统原理图,其中2R 为半导体热敏材料(电阻随温度升高而减小)制成的传感器,1R 为一定值电阻,下列说法中正确的是A .电压表V 的示数为22VB .当传感器2R 所在处出现火灾时,电压表V 的示数减小C .当传感器2R 所在处出现火灾时,电流表A 的示数减小D .当传感器2R 所在处出现火灾,电阻1R 的功率变大8.如图所示,在光滑水平面的左侧固定一竖直挡板,质量为1m 的A 球在水平面上静止放置,质量为2m 的B 球向左运动与A 球发生正碰,B 球碰撞前后的速率之比为3:1,A 球垂直撞向挡板,碰后以原速率返回,两球刚好不发生第二次碰撞,A 、B 系统碰前的总动能为1E ,碰后的总动为2E ,则下列说法正确的是A .12:1:4m m =B .12:4:1m m =C .12:9:5E E =D .12:5:9E E =二、实验题:9.如图所示,某同学用“碰撞实验器”验证动量守恒定律,即研究两个小球在轨道水平部分碰撞前后的动量关系。

山东省济宁市2016-2017学年高二下学期期末考试生物试题-含答案

山东省济宁市2016-2017学年高二下学期期末考试生物试题-含答案

山东省济宁市2016-2017学年高二下学期期末考试生物试题第Ⅰ卷一、选择题(每小题只有一个选项符合题意,1一30题,每题1分,31—40题,每题2分,共50分)1.下列关于原核生物和真核生物的叙述,正确的是A.原核生物细胞无线粒体,不能进行有氧呼吸B.大肠杆菌细胞分裂过程中,无染色体和纺锤体的变化,通过无丝分裂方式增殖C.真核细胞的遗传物质是DNA,原核细胞的遗传物质是RNAD.只有真核细胞才可以进行减数分裂或有丝分裂2.下列关于高倍镜使用的叙述中,正确的是A.因为藓类的叶片大,在高倍镜下容易找到,所以可以直接使用高倍镜观察B.在低倍镜下找到叶片细胞,即可换高倍镜观察C.为了使高倍镜下的视野亮一些,可使用大光圈或凹面反光镜D.换用高倍镜后,必须先用粗准焦螺旋调焦,再用细准焦螺旋调至物像最清晰3.下列关于细胞结构和功能的叙述,正确的是A.核糖体是蛔虫细胞和水绵细胞惟一共有的无膜细胞器B.蓝藻细胞和小麦细胞都有细胞壁,但其成分有差异C.胰岛B细胞分泌胰岛素体现了细胞膜具有选择透过性D.卵细胞体积大有利于与外界进行物质交换4.近期,禽流感病毒H7N9因致病性强、致死率高而备受人们的关注。

下列说法正确的是A.该病毒无线粒体,只能进行无氧呼吸B.该病毒内的DNA和RNA化学元素组成相同C.被该病毒感染的细胞的淸除属于细胞坏死D.该病毒和颤藻均无发生染色体变异的可能5.酒精是生物实验室常用试剂,下列有关酒精及其作用的叙述正确的是A.绿叶中色素的提取和分离用无水乙醇分离绿叶中的色素B.观察DNA和RNA在细胞中的分布用质量分数为8%的酒精进行水解C.生物组织中脂肪的监定用体积分数为50%酒精洗去浮色D.探究酵母菌细胞呼吸方式:酸性重铬酸钾溶液遇酒精由灰绿色变为橙色6.关于DNA和RNA的叙述,正确的是A.DNA有氢键,RNA没有氢键B.DNA可被吡罗红染液染成红色C.原核细胞中既有DNA,也有RNAD.叶绿体、线粒体和核糖体都含有DNA7.下列关于糖类对应正确的一组是①存在于DNA而不存在于RNA的糖类②植物细胞主要的储能物质③叶绿体中具有而线粒体不具有的糖④存在于肝脏细胞而不存在于动物乳汁的糖类A.脱氧核糖、淀粉、葡萄糖、糖原B.脱氧核糖、淀粉、糖原、蔗糖C.核糖、纤维素、葡萄糖、糖原D.核糖、淀粉、葡萄糖、糖原8.生物体的生命活动离不开水。

山东省济宁市2016-2017学年高二下学期期末考试生物试题-含答案

山东省济宁市2016-2017学年高二下学期期末考试生物试题-含答案

山东省济宁市2016-2017学年高二下学期期末考试生物试题第Ⅰ卷一、选择题(每小题只有一个选项符合题意,1一30题,每题1分,31—40题,每题2分,共50分)1.下列关于原核生物和真核生物的叙述,正确的是A.原核生物细胞无线粒体,不能进行有氧呼吸B.大肠杆菌细胞分裂过程中,无染色体和纺锤体的变化,通过无丝分裂方式增殖C.真核细胞的遗传物质是DNA,原核细胞的遗传物质是RNAD.只有真核细胞才可以进行减数分裂或有丝分裂2.下列关于高倍镜使用的叙述中,正确的是A.因为藓类的叶片大,在高倍镜下容易找到,所以可以直接使用高倍镜观察B.在低倍镜下找到叶片细胞,即可换高倍镜观察C.为了使高倍镜下的视野亮一些,可使用大光圈或凹面反光镜D.换用高倍镜后,必须先用粗准焦螺旋调焦,再用细准焦螺旋调至物像最清晰3.下列关于细胞结构和功能的叙述,正确的是A.核糖体是蛔虫细胞和水绵细胞惟一共有的无膜细胞器B.蓝藻细胞和小麦细胞都有细胞壁,但其成分有差异C.胰岛B细胞分泌胰岛素体现了细胞膜具有选择透过性D.卵细胞体积大有利于与外界进行物质交换4.近期,禽流感病毒H7N9因致病性强、致死率高而备受人们的关注。

下列说法正确的是A.该病毒无线粒体,只能进行无氧呼吸B.该病毒内的DNA和RNA化学元素组成相同C.被该病毒感染的细胞的淸除属于细胞坏死D.该病毒和颤藻均无发生染色体变异的可能5.酒精是生物实验室常用试剂,下列有关酒精及其作用的叙述正确的是A.绿叶中色素的提取和分离用无水乙醇分离绿叶中的色素B.观察DNA和RNA在细胞中的分布用质量分数为8%的酒精进行水解C.生物组织中脂肪的监定用体积分数为50%酒精洗去浮色D.探究酵母菌细胞呼吸方式:酸性重铬酸钾溶液遇酒精由灰绿色变为橙色6.关于DNA和RNA的叙述,正确的是A.DNA有氢键,RNA没有氢键B.DNA可被吡罗红染液染成红色C.原核细胞中既有DNA,也有RNAD.叶绿体、线粒体和核糖体都含有DNA7.下列关于糖类对应正确的一组是①存在于DNA而不存在于RNA的糖类②植物细胞主要的储能物质③叶绿体中具有而线粒体不具有的糖④存在于肝脏细胞而不存在于动物乳汁的糖类A.脱氧核糖、淀粉、葡萄糖、糖原B.脱氧核糖、淀粉、糖原、蔗糖C.核糖、纤维素、葡萄糖、糖原D.核糖、淀粉、葡萄糖、糖原8.生物体的生命活动离不开水。

2016-2017学年山东省济宁市高二(下)期末数学试卷(文科)(解析版)

2016-2017学年山东省济宁市高二(下)期末数学试卷(文科)(解析版)

2016-2017学年山东省济宁市高二(下)期末数学试卷(文科)一、选择题(共12小题,每小题5分,满分60分)1.(5分)已知集合A={x|y=},B={x|y=log2(x﹣1)},则A∩B=()A.{x|0≤x<3}B.{x|1<x≤2}C.{x|1<x<3}D.{x|x≤2}2.(5分)用反证法证明“a、b∈N*,如果a、b能被2017整除,那么a、b中至少有一个能被2017整除”时,假设的内容是()A.a不能被2017整除B.b不能被2017整除C.a、b都不能被2017整除D.a、b中至多有一个能被2017整除3.(5分)设复数z满足(z﹣2i)(2﹣i)=5,则复数z的共轭复数为()A.2﹣3i B.2+3i C.+i D.﹣i 4.(5分)执行如图所示的程序框图,若输入n的值为5,则输出s的值为()A.2B.4C.7D.115.(5分)设f(x)是定义在R上的奇函数,且f(x)=,则g(f (﹣7))=()A.﹣1B.﹣2C.1D.26.(5分)已知函数f(x)=e x+e﹣x,则y=f′(x)的图象大致为()A.B.C.D.7.(5分)已知函数f(x)=为奇函数,g(x)=lnx﹣2f(x),则函数g(x)的零点所在区间为()A.(0,1)B.(1,2)C.(2,3)D.(3,4)8.(5分)已知函数f(x)=2lnx++(5﹣m)x在(2,3)上单调递增,则m的取值范围为()A.(﹣∞,5+2]B.(﹣∞,8]C.[,+∞)D.(﹣∞,5+2)9.(5分)通过随机询问100名性别不同的高二学生是否爱吃零食,得到如下的列联表:其中K2=.则下列结论正确的是()A.在犯错误的概率不超过0.05的前提下,认为“是否爱吃零食与性别有关”B.在犯错误的概率不超过0.05的前提下,认为“是否爱吃零食与性别无关”C.在犯错误的概率不超过0.025前提下,认为“是否爱吃零食与性别有关”D.在犯错误的概率不超过0.025前提下,认为“是否爱吃零食与性别无关”10.(5分)某班数学课代表给全班同学出了一道证明题,以下四人中只有一人说了真话,只有一人会证明此题.甲:我不会证明.乙:丙会证明.丙:丁会证明.丁:我不会证明.根据以上条件,可以判定会证明此题的人是()A.甲B.乙C.丙D.丁11.(5分)已知定义在实数集R的函数f(x)满足f(1)=4,且f(x)导函数f′(x)<3,则不等式f(lnx)>3lnx+1的解集为()A.(1,+∞)B.(e,+∞)C.(0,1)D.(0,e)12.(5分)已知函数f(x)=,若关于x的方程f(x)=k有三个不同的实根,则实数k的取值范围是()A.(0,)B.(﹣∞,0]C.(﹣∞,)D.[,+∞)二、填空题(共4小题,每小题5分,满分20分)13.(5分)从某高校在校大学生中随机选取5名女大学生,由她们身高和体重的数据得到的回归直线方程为=0.79x﹣73.56,数据列表是:则其中的数据a=.14.(5分)给出下列不等式:1++>1,1+++…+>,1+++…+>2…,则按此规律可猜想第n个不等式为.15.(5分)已知复数z1=﹣3+2i(i为虚数单位),若复数z1,z2在复平面内对应的点关于直线y=﹣x对称,则z2=.16.(5分)已知函数f(x)是定义在R上的偶函数,若对于x≥0,都有f(x+2)=﹣f(x)且当x∈[0,2)时,f(x)=xe x﹣1,则f(﹣2017)+f(2018)=.三、必考题(共5小题,满分60分)17.(12分)已知函数f(x)=3x+λ•3﹣x(λ∈R).(1)若f(x)为偶函数,求实数λ的值;(2)若不等式f(x)≤6在x∈[0,2]上恒成立,求实数λ的取值范围.18.(12分)设a,b,c为三角形ABC的三边,求证:.19.(12分)已知某商品的进货单价为1元/件,商户甲往年以单价2元/件销售该商品时,年销量为1万件,今年拟下调销售单价以提高销量增加收益.据估算,若今年的实际销售单价为x元/件(1≤x≤2),则新增的年销量P=4(2﹣x)2(万件).(1)写出今年商户甲的收益f(x)(单位:万元)与x的函数关系式;(2)商户甲今年采取降低单价提高销量的营销策略,是否能获得比往年更大的收益(即比往年收益更多)?请说明理由.20.(12分)菜农定期使用低害杀虫农药对蔬菜进行喷洒,以防止害虫的危害,但采集上市时蔬菜仍有少量的残留农药,食用时需要用清水清洗干净.如表是用清水x(单位:千克)清洗该蔬菜1千克后,蔬菜上残留的农药y(单位:微克)的统计表:(1)令ω=x2,利用给出的参考数据求出y关于ω的回归方程=ω+(,精确到0.1).参考数据:ωi=55,(ωi﹣)(y i﹣)=﹣751,(ωi﹣)2=374.其中ωi=,=ωi.(2)对于某种残留在蔬菜上的农药,当它的残留量不高于20微克时对人体无害,为了放心食用该蔬菜,请估计至少需要多少千克的清水洗1千克蔬菜?(精确到0.1,参考数据≈2.24).(附:对于一组数据(u1,v1),(u2,v2),…,(u n,v n),其回归直线v=α+βu的斜率和截距的最小二乘估计分别为=,=﹣.21.(12分)已知函数f(x)=xe x﹣1﹣mx2﹣mx,m∈R.(1)当m=0时,求曲线y=f(x)在点(1,f(1))处的切线方程;(2)讨论函数f(x)的单调性并判断有无极值,有极值时求出极值.四、选考题:[选修4-4:坐标系与参数方程](请考生在第22、23题中任选一题作答,如果多做,则按所做的第一题计分)(共1小题,满分10分)22.(10分)在直角坐标系xOy中,直线l的参数方程是(t为参数),以坐标原点O为极点,以x轴的正半轴为极轴,建立极坐标系,曲线C的极坐标方程是ρ=2cosθ.(1)求直线l的普通方程和曲线C的直角坐标方程;(2)设点P(m,0),若直线l与曲线C交于A、B两点,且|P A|•|PB|=1,求实数m的值.[选修4-5:不等式选讲]23.已知函数f(x)=|x﹣1|﹣|x|+a.(1)若a=0,求不等式f(x)≥0的解集;(2)若方程f(x)+x=0有三个不同的解,求实数a的取值范围.2016-2017学年山东省济宁市高二(下)期末数学试卷(文科)参考答案与试题解析一、选择题(共12小题,每小题5分,满分60分)1.【解答】解:由A中y=,得到2﹣x≥0,解得:x≤2,即A={x|x≤2},由B中y=log2(x﹣1),得到x﹣1>0,解得:x>1,即B={x|x>1},则A∩B={x|1<x≤2},故选:B.2.【解答】解:由于反证法是命题的否定的一个运用,故用反证法证明命题时,可以设其否定成立进行推证.命题“a、b∈N*,如果a、b能被2017整除,那么a、b中至少有一个能被2017整除”的否定是“a,b都不能被2017整除”.故选:C.3.【解答】解:由(z﹣2i)(2﹣i)=5,得:z﹣2i===2+i∴z=2+3i.∴复数z的共轭复数为2﹣3i故选:A.4.【解答】解:模拟执行程序框图,可得n=5,i=1,s=1满足条件i≤5,s=1,i=2满足条件i≤5,s=2,i=3满足条件i≤5,s=4,i=4满足条件i≤5,s=7,i=5满足条件i≤5,s=11,i=6不满足条件i≤5,退出循环,输出s的值为11,故选:D.5.【解答】解:函数f(x)是定义在R上的奇函数,且f(x)=,设x<0,则﹣x>0,则f(﹣x)=log2(﹣x+1),∵f(﹣x)=﹣f(x),∴f(x)=﹣f(﹣x)=﹣log2(﹣x+1),∴g(x)=﹣log2(﹣x+1)(x<0),∴f(﹣7)=g(﹣7)=﹣log2(7+1)=﹣3,∴g(﹣3)=﹣log2(3+1)=﹣2,故选:B.6.【解答】解:函数f(x)=e x+e﹣x,则y=f′(x)=e x﹣e﹣x=,因为y=e x是增函数,y=是增函数,所以导函数是增函数.故选:D.7.【解答】解:函数f(x)=为奇函数,可得a=0,g(x)=lnx﹣2f(x)=lnx﹣,g(2)=ln2﹣1<0,g(3)=ln3﹣>0,由零点判定定理可知:g(2)g(3)<0,可知函数的零点在(2,3)之间.故选:C.8.【解答】解:f′(x)=+x+(5﹣m),若f(x)在(2,3)递增,则f′(x)≥0在(2,3)恒成立,即m﹣5≤+x在(2,3)恒成立,令g(x)=x+,x∈(2,3),则g′(x)=1﹣>0,g(x)在(2,3)递增,故g(x)>g(2)=3,故m﹣5≤3,解得:m≤8,故选:B.9.【解答】解:根据题意,有所给的数据;k2=≈4.761>3.841,而4.761<5.024;即在犯错误的概率不超过0.05的前提下,认为“是否爱吃零食与性别有关”;故选:A.10.【解答】解:四人中只有一人说了真话,只有一人会证明此题.丙:丁会证明.丁:我不会证明.所以丙与丁中一定有一个是正确的;若丙说了真话,则甲必是假话,矛盾;若丁说了真话,则甲说的是假话,甲就是会证明的那个人,符合题意;以此类推.易得出答案:A.故选:A.11.【解答】解:设t=lnx,则不等式f(lnx)>3lnx+1等价为f(t)>3t+1,设g(x)=f(x)﹣3x﹣1,则g′(x)=f′(x)﹣3,∵f(x)的导函数f′(x)<3,∴g′(x)=f′(x)﹣3<0,此时函数单调递减,∵f(1)=4,∴g(1)=f(1)﹣3﹣1=0,则当x<1时,g(x)>g(1)=0,即g(x)<0,则此时g(x)=f(x)﹣3x﹣1>0,即不等式f(x)>3x+1的解为x<1,即f(t)>3t+1的解为t<1,由lnx<1,解得0<x<e,即不等式f(lnx)>3lnx+1的解集为(0,e),故选:D.12.【解答】解:当x≥1时,f′(x)=,∴当1≤x≤e时,f′(x)≥0,当x>e时,f′(x)<0,∴f(x)在[1,e)上单调递增,在[e,+∞)上单调递减,∴当x=e时,f(x)取得极大值f(e)=.又f(1)=0,当x>1时,f(x)=>0,当x<1时,f(x)=﹣x3+1为减函数,作出f(x)的大致函数图象如图所示:∴当0<k<时,f(x)=k有3个不同的实数根.故选:A.二、填空题(共4小题,每小题5分,满分20分)13.【解答】解:由表中数据计算=×(49+53+56+58+64)=56,根据回归直线经过样本中心(,),可得56=0.79﹣73.56,解得=164;由=×(155+161+a+157+174)=164,解得a=163,故答案为:163.14.【解答】解:观察不等式中最后一项的分母分别是3、7、15、31…将每个数加1得4、8、16、32可知通项为2n+1则3、7、15、31…的通项为2n+1﹣1不等式右边是首项为1,公差为的等差数列,∴按此规律可猜想第n个不等式为1+++…+>.故答案为1+++…+>.15.【解答】解:复数z1=﹣3+2i在复平面内对应的点为:(﹣3,2),又复数z1,z2在复平面内对应的点关于直线y=﹣x对称,∴z2在复平面内对应的点为:(﹣2,3),∴z2=﹣2+3i.故答案为:﹣2+3i.16.【解答】解:若对于x≥0,都有f(x+2)=﹣f(x),则f(x+4)=﹣f(x+2)=f(x),即当x≥0时,函数是周期为4的周期函数,∵f(x)是定义在R上的偶函数,∴f(﹣2017)+f(2018)=f(2017)+f(2018),f(2017)=f(504×4+1)=f(1)=e﹣1,f(2018)=f(504×4+2)=f(2)=﹣f(0)=﹣1,则f(﹣2017)+f(2018)=f(2017)+f(2018)=e﹣1﹣1=e﹣2,故答案为:e﹣2三、必考题(共5小题,满分60分)17.【解答】解:(1)∵f(x)为偶函数,∴f(﹣x)=f(x).∴3﹣x+λ•3x=3x+λ•3﹣x,即(1﹣λ)(3﹣x﹣3x)=0.又∵3﹣x﹣3x不恒为零,∴1﹣λ=0,即λ=1;(2)由f(x)≤6,得3x+λ•3﹣x≤6,即≤6.令t=3x∈[1,9],原不等式等价于t+≤6在t∈[1,9]恒成立.亦即λ≤﹣t2+6t在t∈[1,9]上恒成立.令g(t)=﹣t2+6t,t∈[1,9].当t=9时,g(t)有最小值g(9)=﹣27.∴λ≤﹣27.18.【解答】证明:要证明:需证明:a(1+b)(1+c)+b(1+a)(1+c)>c(1+a)(1+b)(4分)需证明:a(1+b+c+bc)+b(1+a+c+ac)>c(1+a+b+ab)需证明a+2ab+b+abc>c(8分)∵a,b,c是△ABC的三边∴a>0,b>0,c>0且a+b>c,abc>0,2ab>0∴a+2ab+b+abc>c∴成立.(12分)19.【解答】解:(1)由题意可得:f(x)=[1+4(2﹣x)2](x﹣1),1≤x≤2.(2)甲往年以单价2元/件销售该商品时,年销量为1万件,可得收益为1万元.f′(x)=8(x﹣2)(x﹣1)+1+4(2﹣x)2=12x2﹣40x+33=(2x﹣3)(6x﹣11),可得当x∈时,函数f(x)单调递增;当x∈时,函数f(x)单调递减;当x∈时,函数f(x)单调递增.∴x=时,函数f(x)取得极大值,=1;又f(2)=1.∴当x=或x=2时,函数f(x)取得最大值1(万元).因此商户甲今年采取降低单价提高销量的营销策略,不能获得比往年更大的收益.20.【解答】解:(1)令ω=x2,计算=×ωi=×55=11,=×(58+54+39+29+10)=38;(ωi﹣)(y i﹣)=﹣751,(ωi﹣)2=374,===﹣2.0=﹣=38﹣(﹣2.0)×11=60,∴回归方程为=﹣2.0ω+60;(2)当<20时,﹣2.0x2+60.0<20,解得x>2=2×2.24≈4.5,∴为了放心食用该蔬菜,估计需要用4.5千克的清水清洗一千克蔬菜.21.【解答】解:(1)当m=0时,f(x)=xe x﹣1,f′(x)=e x﹣1(x+1),f′(1)=2,f(1)=1,∴曲线y=f(x)在点(1,f(1))处的切线方程y﹣1=2(x﹣1),即2x﹣y﹣1=0(2)f′(x)=e x﹣1(x+1)﹣mx﹣m=(x+1)(e x﹣1﹣m),①当m≤0时,e x﹣1﹣m>0恒成立,当x>﹣1时,f′(x)>0,当x<﹣1时,f′(x)<0,此时函数f(x)在(﹣∞,﹣1)单调递减,在(﹣1,+∞)单调递增,有极小值f(﹣1)=﹣e﹣2+.②当m>0时,令f′(x)=e x﹣1(x+1)﹣mx﹣m=(x+1)(e x﹣1﹣m)=0,可得1=﹣1,x2=lnm+1当0<m<e﹣2时,x2<x1,x∈(﹣∞,lnm+1)时,f′(x)>0,x∈(lnm+1,﹣1)时,f′(x)<0,x∈(﹣1,+∞)时,f′(x)>0,此时,函数f(x)在(﹣∞,lnm+1),(﹣1,+∞)单调递增,在(lnm+1,﹣1)单调递减,有极大值f(lnm+1)=﹣,有极小值f(﹣1)=﹣e﹣2+.;当m=e﹣2时,函数f(x)在(﹣∞,+∞)单调递增,无极值;当m>e﹣2时,x∈(﹣∞,﹣1)时,f′(x)>0,x∈(﹣1,lnm+1)时,f′(x)<0,x∈(lnm+1,+∞)时,f′(x)>0,此时,函数f(x)在(﹣∞,﹣1),(lnm+1,+∞)单调递增,在(﹣1,lnm+1)单调递减,有极小值f(lnm+1)=﹣,有极大值f(﹣1)=﹣e﹣2+.;四、选考题:[选修4-4:坐标系与参数方程](请考生在第22、23题中任选一题作答,如果多做,则按所做的第一题计分)(共1小题,满分10分)22.【解答】解:(1)直线l的参数方程是(t为参数),消去参数t可得普通方程:x﹣y﹣m=0.曲线C的极坐标方程是ρ=2cosθ,即ρ2=2ρcosθ,可得直角坐标方程:x2+y2=2x.(2)设点P(m,0),把直线l的参数方程(t为参数)代入圆C的方程:t2+(m﹣)t+m2﹣2m=0,△=﹣4(m2﹣2m)>0,解得﹣1<m<3.∴t1t2=m2﹣2m,∴|P A|•|PB|=|t1t2|=|m2﹣2m|=1,又﹣1<m<3.解得m=1,m=1.∴实数m的值为1,1.[选修4-5:不等式选讲]23.【解答】解:(1)当a=0时,f(x)=|x﹣1|﹣|x|=,所以当x<0时,f(x)=1>0,符合题意;当0≤x<1时,f(x)=1﹣2x≥0,解得0≤x;当x≥1时,f(x)=﹣1<0,不符合题意.综上可得,f(x)≥0的解集为(].(2)设u(x)=|x|﹣|x﹣1|﹣x,y=u(x)的图象和y=a的图象如图所示.易知y=u(x)的图象与y=a的图象有3个交点时,a∈(﹣1,0),所以实数a的取值范围为(﹣1,0).。

2016-2017学年山东省济宁市高二下学期期末考试数学(理)试题(解析版)

2016-2017学年山东省济宁市高二下学期期末考试数学(理)试题(解析版)

2016-2017学年山东省济宁市高二下学期期末考试数学(理)试题一、选择题1.在复平面内,复数34i1iz+=-(i是虚数单位)对应的点在()A. 第一象限B. 第二象限C. 第三象限D. 第四象限【答案】B【解析】复数()()()()341341711122i iiz ii i i+++===-+--+,其所对应的点17,22⎛⎫-⎪⎝⎭位于第二象限;本题选择B选项.2.在用线性回归方程研究四组数据的拟合效果中,分别作出下列四个关于四组数据的残差图,则用线性回归模式拟合效果最佳的是()A.B.C.D.【答案】C【解析】显然从残差图中可以看出,C中各点比较均匀的分布在一条直线附近,说明C对应的回归直线拟合度较好,故选C.3.已知向量()2,3,1a =- , ()4,2,b x =- ,且a b ⊥,则x 的值为( )A. 12B. 10C. 14-D. 14 【答案】D【解析】由向量垂直的充要条件有: ()()24320x ⨯-+-⨯+=,解得: 14x =. 本题选择D 选项.4.现抛掷两枚骰子,记事件A 为“朝上的2个数之和为偶数”,事件B 为“朝上的2个数均为偶数”,则(|)P B A =( ) A.18 B. 14 C. 25 D. 12【答案】D【解析】解:事件AB 的事件包括:()()()()()()()()()2,2,,2,4,2,6,4,2,4,4,4,6,6,2,6,4,6,6,事件A 包括:()()()()()()()()()()()()()()()()()()1,1,1,3,1,5,2,2,,2,4,2,6,3,1,3,3,3,5,4,2,4,4,4,6,5,1,5,3,5,5,6,2,6,4,6,6,由题意可得: ()()918,3636p AB P A == , 由条件概率公式可得: ()()1(|)2P AB P B A P A ==. 本题选择D 选项.5.如图,阴影部分面积是( )A. 1e e +B. 1e 1e +-C. 1e 2e +-D. 1e e- 【答案】C【解析】由定积分的定义可得,阴影部分的面积为()()1101|2xx x x ee dx e e e e---=+=+-⎰.本题选择C 选项.点睛:利用定积分求曲线围成图形的面积的步骤:(1)画出图形;(2)确定被积函数;(3)确定积分的上、下限,并求出交点坐标;(4)运用微积分基本定理计算定积分,求出平面图形的面积.求解时,注意要把定积分与利用定积分计算的曲线围成图形的面积区别开:定积分是一个数值(极限值),可为正,可为负,也可为零,而平面图形的面积在一般意义上总为正. 6.设随机变量X , Y 满足: 31Y X =-, ()2,X B p ~,若()519P X ≥=,则()D Y =( )A. 4B. 5C. 6D. 7 【答案】A【解析】由题意可得: ()()()225110119P X P X C p ≥=-==--=, 解得: 13p =,则: ()()()()212412,34339D X np p D Y D X =-=⨯⨯===。

山东省济宁市2016-2017学年高二下学期期末考试英语试题及答案

山东省济宁市2016-2017学年高二下学期期末考试英语试题及答案

山东省济宁市2016-2017学年高二下学期期末考试英语试题第I卷(共100分)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。

例:How much is the shirt?A. £ 19.15B. £ 9.18C. £ 9.15答案是C。

1. What are the speakers talking about?A. Weather.B. School report.C. Homework.2. Why does Mr. Jones want to go to Australia?A. To have a rest there.B. To find a job there.C. To travel there.3. What does the woman mean?A. She won’t go to the zoo.B. She will go to the zoo.C. She doesn’t like the zoo.4. What is the man most probably going to do next?A. Say sorry to the woman.B. Buy the woman another necklace.C. Help the woman look for the necklace.5. What is the man trying to do?A. Take a bath.B. Clean his room.C. Book a hotel room. 第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

山东省济宁市2016-2017学年高二下学期期末考试文数试题(解析版)

山东省济宁市2016-2017学年高二下学期期末考试文数试题(解析版)

2016~2017学年度第二学期期末考试高二数学(文)试题第Ⅰ卷(共60分)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.已知集合{A x y ==,(){}2log 1B x y x ==-,则A B =I ( )A. {}03x x ≤<B. {}12x x ≤<C. {}13x x <<D. {}2x x ≤【答案】B 【解析】由题意得,集合{}{|2},1A x x B x x =≤=,所以{|12}A B x x =<≤I ,故选B.2.用反证法证明“*,a b N ∈,如果a 、b 能被2017整除,那么,a b 中至少有一个能被2017整除”时,假设的内容是( ) A. a 不能被2017整除 B. b 不能被2017整除C. ,a b 都不能被2017整除D. ,a b 中至多有一个能被2017整除【答案】C 【解析】命题的否定只否结论,即“,a b 中至少有一个能被2017整除”的否定为,a b 都不能被2017整除,故选C. 3.设复数z 满足()()2i 2i 5z --=,则复数z 的共轭复数为( ) A. 23i - B. 23i +C.1011i 33+ D.1011i 33- 【答案】A 【解析】由题意得52i=(2)2232iz i i i =+++=+-,所以23z i =-,故选A. 4.执行如图所示的程序框图,若输入n 的值为5,则输出s 的值为( )A. 2B. 4C. 7D. 11【答案】D 【解析】模拟执行程序框图,可得5,1,1n i s ===, 满足条件5,1,2i s i ≤==; 满足条件5,2,3i s i ≤==; 满足条件5,4,4i s i ≤==; 满足条件5,7,5i s i ≤==, 满足条件5,11,6i s i ≤==,此时不满足条件5i ≤,推出循环,输出s 的值11,故选D.5.设()f x 是定义在R 上的奇函数,且()()()2log 1,0,0x x f x g x x ≥⎧+⎪=⎨<⎪⎩,则()()7g f -=( )A. 1-B. 2-C. 1D. 2【答案】B 【解析】由题意得,函数()f x 是定义在R 上的奇函数,且()()()2log 1,0,0x x f x g x x ≥⎧+⎪=⎨<⎪⎩, 设0x <,则0x ->,则()2log (1)f x x -=-+ 因()()f x f x -=-,所以()()2log (1)f x f x x =--=--+,所以()2log (1)(0)g x x x =--+<, 所以2(7)(7)log (71)3f g -=-=-+=-, 所以2(3)log (31)2g -=-+=-,故选B.6.已知函数()e e xxf x -=+,则()y f x '=的图象大致为( )A. B. C. D.【答案】D 【解析】函数()e e x xf x -=+,则()1x x xxf x e e e e -=-'=-, 因为xy e =是增函数,1x y e=-也是增函数, 所以导函数也是增函数,故选D. 7.已知函数()1f x x a=-为奇函数,()()ln 2g x x f x =-,则函数()g x 的零点所在区间为( ) A. ()0,1 B. ()1,2C. ()2,3D. ()3,4【答案】C 【解析】 函数()1f x x a=-为奇函数,可得0a =, ()2ln 2()ln g x x f x x x=-=-,所以()()22ln 210,3ln 303g g =-=-,由零点的判定定理可知,()()230g g <,可知函数的零点在(2,3)之间,故选C.8.已知函数()()22ln 52x f x x m x =++-在()2,3上单调递增,则m 的取值范围是( )A. (,522⎤-∞+⎦ B. (],8-∞C. 26,3⎡⎫+∞⎪⎢⎣⎭D. (),522-∞+【答案】B 【解析】 由题意得()2(5)f x x m x'=++-, 若()f x 在区间(2,3)递增,则()0f x '≥在(2,3)上恒成立,即25m x x -≤+在(2,3)上恒成立, 令()2,(2,3)g x x x x =+∈,则()2210g x x'=-+>,所以()g x 在(2,3)上是增函数,故()()3g x g x >=, 所以538m m -≤⇒≤,故选B.9.通过随机询问100名性别不同的高二学生是否爱吃零食,得到如下的列联表:其中()()()()()22,.n ad bc K n a b c d a b c d a c b d -==+++++++则下列结论正确的是A. 在犯错误的概率不超过0.05的前提下,认为“是否爱吃零食与性别有关”B. 在犯错误的概率不超过0.05的前提下,认为“是否爱吃零食与性别无关”C. 在犯错误的概率不超过0.025的前提下,认为“是否爱吃零食与性别有关”D. 在犯错误的概率不超过0.025的前提下,认为“是否爱吃零食与性别无关” 【答案】A【解析】由题意得,22100(10302040) 4.762 3.84150503070K ⨯-⨯=≈>⨯⨯⨯,又因为23.841)0.05(P K >=,所以犯错误的概率不超过5%的前提下,认为“是否爱吃零食与性别有关”,故选A.10.数学老师给同学们出了一道证明题,以下四人中只有一人说了真话,只有一人会证明此题,甲:我不会证明;乙:丙会证明;丙:丁会证明;丁:我不会证明.根据以上条件,可以判定会证明此题的人是( ) A. 甲 B. 乙C. 丙D. 丁【答案】A 【解析】四人中只有一人说了真话,只有一人会证明此题,丙:丁会证明;丁:我不会证明,所以丙与丁中有一个是正确的;若丙说了真话,则甲必是假话,矛盾;若丁说了真话,则甲说的是假话,甲就是会证明的那个人,符合题意,以此类推,即可得到甲说真话,故选A.11.已知定义在实数集R 的函数()f x 满足()14f =,且()f x 导函数()'3f x <,则不等式()ln 3ln 1f x x >+的解集为( ) A. ()1+∞, B. ()e +∞,C. ()0,1D. ()0e ,【答案】D 【解析】 试题分析:设,则,所以是上的单调递减函数,又,因此()ln 3ln 1f x x >+可化为,即,故由单调性可知,即,故应选D.考点:导数和函数性质的综合运用.【易错点晴】导数解决函数问题的重要工具,解答本题时通过借助题设提供的有效信息,巧妙地构造函数,然后运用导数这一重要工具对这个函数求导,凭借题设条件得知函数是上的单调递减函数,为下面不等式的求解创造了条件.求解不等式()ln 3ln 1f x x >+时,以为变量建立不等式,最终通过单调性的定义得到了不等式,使得本题巧妙获解.12.已知函数()3ln ,11,1xx fx x x x ≥⎧⎪=⎨⎪-+<⎩,若关于x 的方程()f x k =有三个不同的实根,则实数k 的取值范围是( ) A. 10,e ⎛⎫ ⎪⎝⎭B. (],0-∞C. 1,e ⎛⎫-∞ ⎪⎝⎭D. 1,e ∞⎡⎫+⎪⎢⎣⎭【答案】A 【解析】由题意得当1x ≥时,()21ln xf x x-'=, 所以当1x e ≤≤时,()0f x '≥;当x e >时,()0f x '<, 所以()f x 在[1,e)上单调递增,在(,)e +∞上单调递减, 所以当x e =时,()f x 取得极大值()1f e e=, 又()10f =,当1x >时,()ln 0xf x x=>, 当1x <时,函数()31f x x =--为减函数, 作出()f x 的图象如图所示, 所以当1k e<<0时,()f x k =有3个不同的实数根,故选A.点睛:本题主要考查了函数与方程思想的应用,其中解答中涉及到利用到时研究函数的单调性,以及利用导数求解函数的极值等知识点,着重考查了分类讨论思想和数形结合思想的应用,其中根据导数研究函数的单调性及极值,作出函数的图象,利用数形结合法求解是解答的关键.第Ⅱ卷(共90分)二、填空题(每题5分,满分20分,将答案填在答题纸上)13.从某高校在校大学生中随机选取5名女大学生,由她们身高和体重的数据得到的回归直线方程为ˆ0.7973.56yx =-,数据列表是:则其中的数据a =__________. 【答案】163 【解析】由4953565864565y ++++==,根据回归直线经过样本中心(),x y ,即560.7973.56x =⨯-,得164x =,由1551611671741645a x ++++==, 得163a =,故答案为163. 14.给出下列不等式:111123++> 111312372+++⋯+>111122315+++⋯+>………则按此规律可猜想第n 个不等式为____________ 【答案】()*11111123212n n n +++++⋯+>∈-N 【解析】观察各式左边为的和的形式,项数分别为3,7,15,…,∴可猜想第n 个式子中左边应有2n +1-1项,不等式右边分别写成,,,…,∴猜想第n 个式子中右边应为,按此规律可猜想此类不等式的一般形式为:1+++…+>(n ∈N *).15.已知复数132i z =-+(i 为虚数单位),若复数1z ,2z 在复平面内对应的点关于直线y x =-对称,则2z =__________.【答案】2+3i - 【解析】由题意得,复数132i z =-+在复平面内对应的点为(3,2)-, 又复数12,z z 在复平面内对应的点关于直线y x =-对称, 所以2z 在复平面内对应的点的坐标为(2,3)-,所以复数223z i =-+.16.已知函数()f x 是定义在R 上的偶函数,若对于0x ≥,都有()()2f x f x +=-且当[)0,2x ∈时,()e 1x f x x =-,则()()20172018f f -+=__________.【答案】e 【解析】 【分析】由已知可得函数是以4为周期的周期函数,且()f x 是定义在R 上的偶函数可得()()()()2017201820172018f f f f -+=+,求出()2017f 与()2018f 的值,可得答案.【详解】解:由已知函数()f x 对于0x ≥,都有()()2f x f x +=-,可得()()()42f x f x f x +=-+=, 即当0x ≥时,函数()f x 是以4为周期的周期函数,又函数()f x 是定义在R 上的偶函数,可得()()()()2017201820172018f f f f -+=+, ()()201750441(1)1f f f e =⨯+==-, ()()201850442(2)(0)1f f f f =⨯+==-=,故可得:()()()()201720182017201811f f f f e e -+=+=-+=, 故答案为:e .【点睛】本题主要考查了函数奇偶性与周期性的综合应用,其中根据已知条件得出函数为周期是4的周期函数是解题关键.三、解答题 (本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤.)17.已知函数()33x xf x λ-=+⋅(R λ∈).(1)若()f x 为偶函数,求实数λ的值;(2)若不等式()6f x ≤在[]0,2x ∈上恒成立,求实数λ的取值范围. 【答案】(1)1λ=;(2)(],27-∞-.【解析】试题分析:(1)函数()f x 是定义在R 上的偶函数,所以()()f x f x -=,化简即可求解实数λ的值;(2)由()6f x ≤得363x xλ≤+,分离参数,换元配方求解最小值,即可得到答案.试题解析:(1)函数()f x 是定义在R 上的偶函数,所以()()f x f x -= 即3333x x x x λλ--+⋅=+⋅ 化简得()()1330xxλ---=所以1λ=(2)由()6f x ≤得336x x λ≤-+⋅,即363xxλ≤+()2363xxλ≤-+⋅ ()2339x =--+又02x ≤≤,所以139x ≤≤ 当39x =即2x =时,()2339x --+取最小值27-故实数λ的取值范围是(],27-∞-. 18.设,,a b c 为三角形ABC 的三边,求证:111a b c a b c+>+++ 【答案】见解析 【解析】试题分析:本题用直接法不易找到证明思路,用分析法,要证该不等式成立,因为0,0,0a b c >>>,所以10,10,10a b c +>+>+>,只需证该不等式两边同乘以(1)(1)(1)a b c +++转化成的等价不等式a(1+b)(1+c)+ b(1+a)(1+c)> c(1+a)(1+b)成立,用不等式性质整理为a+2ab+b+abc>c 成立,用不等式性质及三角不等式很容易证明此不等式成立. 试题解析:要证明:需证明: a(1+b)(1+c)+ b(1+a)(1+c)> c(1+a)(1+b) 5分需证明:a(1+b+c+bc)+ b(1+a+c+ac)> c(1+a+b+ab) 需证明a+2ab+b+abc>c 10分 ∵a,b,c 是的三边 ∴a>0,b>0,c>0且a+b>c,abc>0,2ab>0∴a+2ab+b+abc>c∴成立. 14分考点:分析法证明不等式;三角形两边之和大于第三边.19.已知某商品的进货单价为1元/件,商户甲往年以单价2元/件销售该商品时,年销量为1万件.今年拟下调销售单价以提高销量增加收益.据估算,若今年的实际销售单价为x 元/件(12x ≤≤),则新增的年销量()242P x =-(万件).(1)写出今年商户甲的收益()f x (单位:万元)与x 的函数关系式;(2)商户甲今年采取降低单价提高销量的营销策略,是否能获得比往年更大的收益(即比往年收益更多)?请说明理由.【答案】(1)()f x =324203317x x x -+-(12x ≤≤).(2)见解析. 【解析】试题分析:(1)直接根据题意可写成几年的销售量,从而可计算出客户甲的收益;(2)根据(1)总监理的函数,求导,利用导数等于0,求得函数的极大值点和极大值,在求出2x =时的函数值,比较即可得到函数的最大值,进而得到结论. 试题解析:(1)由题意知,今年的年销售量为()2142x +-(万件). 因为每销售一件,商户甲可获利()1x -元, 所以今年商户甲的收益()()()21421f x x x ⎡⎤=+--⎣⎦324203317x x x =-+-(12x ≤≤).(2)由()324203317f x x x x =-+-(12x ≤≤)得()2124033f x x x =-+' ()()23611x x =--,令()0f x '=,解得32x =或116x = 当31,2x ⎛⎫∈ ⎪⎝⎭时,()0f x '>;当311,26x ⎛⎫∈ ⎪⎝⎭时,()0f x '<; 当11,26x ⎛⎫∈ ⎪⎝⎭时,()0f x '>;∴32 x=为极大值点,极大值为312f⎛⎫=⎪⎝⎭∵()21f=,∴当32x=或2时,()f x在区间[]1,2上的最大值为1(万元),而往年的收益为()2111-⨯=(万元),所以商户甲采取降低单价提高销量的营销策略不能获得比往年更大的收益.20.菜农定期使用低害杀虫农药对蔬菜进行喷洒,以防止害虫的危害,但采集上市时蔬菜仍存有少量的残留农药,食用时需要用清水清洗干净,下表是用清水x(单位:千克)清洗该蔬菜1千克后,蔬菜上残留的农药y(单位:微克)的统计表:(1)令2xω=,利用给出的参考数据求出y关于ω的回归方程ˆˆˆy b aω=+.(ˆa,ˆb精确到0.1)参考数据:5155iiω==∑,()()51751i iiy yωω=--=-∑,()521374iiωω=-=∑其中2i ixω=,5115iiωω==∑(2)对于某种残留在蔬菜上的农药,当它的残留量不高于20微克时对人体无害,为了放心食用该蔬菜,请估计至少需用用多少千克的清水清洗1千克蔬菜?(精确到0.15 2.24≈)附:对于一组数据()11,u v,()22,u v,…,(),n nu v,其回归直线v uαβ=+的斜率和截距的最小二乘估计分别为()()()121ˆni iiniiu u v vu uβ==--=-∑∑,ˆˆv uαβ=-.【答案】(1) 2.060.ˆ0yω=-+;(2)见解析.【解析】试题分析:(1)计算,yω,填表即可,在求出回归系数,即可求解回归直线的方程;(2)由(1)求得ˆy的值,令ˆ20y<,即可求解x的取值范围.试题解析:(1)由题意得,11ω=,38y=.()()()51521ˆi ii i i y y b ωωωω==--=-∑∑ 751 2.0374=-≈- 6ˆ0.0ˆay b ω=-= ∴ 2.060.ˆ0yω=-+ (2)由(1)得, 2.060.ˆ0yω=-+ ∴22.06.0ˆ0yx =-+ 当ˆ20y≤时,即22.060.020x ≤-+,解得 4.5x ≥≈ 所以为了放心食用该蔬菜,估计需要用4.5千克的清水清洗1千克蔬菜.点睛:本题主要考查了回归直线方程的求解及综合应用,此类问题的解答中正确处理数据,利用最小二乘法求解回归系数是解答的一个难点和关键,解答中应细心、认真.21.已知函数()121e 2x f x x mx mx -=--,R m ∈, (1)当0m =时,求曲线()y f x =在点()()1,1f 处的切线方程;(2)讨论函数()f x 的单调性并判断有无极值,有极值时求出极值.【答案】(1)210x y --=;(2)见解析.【解析】试题分析:(1)欲求曲线()y f x =在点(1,(1))f 处的切线方程,只需求出斜率()1k f '=和和()1f 的值,即可利用直线的点斜式方程求解切线的方程;(2)求出()()()111e e e 1x x x f x x mx m m x ---=+--=-+',通过讨论m 的取值范围,求出函数的单调区间,从而求出函数的极值即可,可分0,0m m ≤>两种情况,求出函数的单调区间,得出函数的极值. 试题解析:(1)0m =时,()1ex f x x -=,()11e e x x f x x --+'= 所以()11f =,()12f '=因此曲线()y f x =在点()()1,1f 处的切线方程是()121y x -=-即210x y --=(2)()11e e x x f x x mx m --=+--' ()()1e 1x m x -=-+①当0m ≤时,1e 0x m -->恒成立,所以当(),1x ∈-∞-时()0f x '<,()f x 单调递减当()1,x ∈-+∞时,()0f x '>,()f x 单调递增所以当1x =-时,()f x 取极小值()211e 2m f -=-+ ②当0m >时,由()0f x '=得11x =-或21ln x m =+(ⅰ)当12x x <,即2e m ->时由()0f x '>得1x <-或1ln x m >+由()0f x '<得11ln x m -<<+所以()f x 在(),1-∞-上单调递增,在()1,1ln m -+上单调递减,在()1ln ,m ++∞上单调递增,故1x =-时,()f x 取极大值()211e 2m f -=-+,1ln x m =+时,()f x 取极小值()()211ln 1ln 2f m m m +=-+ (ⅱ)当12x x =,即2e m -=时,()0f x ≥'恒成立此时函数()f x 在(),-∞+∞上单调递增,函数()f x 无极值(ⅲ)当12x x >,即20e m -<<时由()0f x '>得1ln x m <+或1x >-由()0f x '<得1ln 1m x +<<-所以()f x 在(),1ln m -∞+上单调递增,在()1ln ,1m +-上单调递减,在()1,-+∞上单调递增,故1ln x m =+时,()f x 取极大值()()211ln 1ln 2f m m m +=-+ 1x =-时,()f x 取极小值()211e 2m f -=-+. 点睛:本题主要考查导数在函数中的综合应用,本题的解答中涉及利用导数的几何意义求解曲线在某点处的切线方程,利用导数研究函数的单调性和极值,求解函数的单调区间,涉及到分类讨论的数学思想的应用,熟记利用导数研究函数的性质是解答的关键,试题有一定的难度,属于中档试题.请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.22.已知过点(,0)P m 的直线l的参数方程是12x m y t ⎧=+⎪⎪⎨⎪=⎪⎩(t 为参数).以平面直角坐标系的原点为极点,x轴的正半轴为极轴,建立极坐标系,曲线C 的极坐标方程为2cos ρθ=.(Ⅰ)求直线l 的普通方程和曲线C 的直角坐标方程;(Ⅱ)若直线l 与曲线C 交于两点,A B ,且||||1PA PB ⋅=,求实数m 的值.【答案】(Ⅰ)212x t m y t ⎧=+⎪⎪⎨⎪=⎪⎩,(t 为参数),222x y x +=(Ⅱ)1m =?1. 【解析】试题分析:(Ⅰ)消去参数t可得x m =+,由2cos ρθ=,得22cos ρρθ=,可得C 的直角坐标方程;(Ⅱ)把{12x m y t =+=(t 为参数),代入222x y x +=,根据参数的几何意义,结合韦达定理得结果. 试题解析:(Ⅰ)直线L的参数方程是{12x m y t =+=,(t 为参数),消去参数t可得x m =+. 由2cos ρθ=,得22cos ρρθ=,可得C 的直角坐标方程:222x y x +=.(Ⅱ)把{12x m y t =+=(t 为参数),代入222x y x +=,得2220t t m m ++-=. 由>0∆,解得13m -<<,2122t t m m =-∴,12·1PA PB t t ==Q ,221m m -=±∴,解得1m =? 1.又满足>0∆,∴实数1m =? 1.考点:参数方程与普通方程的互化;极坐标方程化为直角坐标;23.选修4-5:不等式选讲已知函数()1f x x x a =--+.(1)若0a =,求不等式()0f x ≥的解集;(2)若方程()0f x x +=有三个不同的解,求实数a 的取值范围.【答案】(1)1,2⎛⎤-∞ ⎥⎝⎦;(2)()1,0-. 【解析】试题分析:(1)当0a =,得到函数()f x 的解析式,根据解析式分别求出()0f x ≥的解集即可;(2)由()0f x x +=得1a x x x -=--+,则方程()0f x x +=有三个不同的解等价于函数y a =-的图象和函数1y x x x =--+的图象有三个不同交点,作出函数的图象,根据图象即可求解实数a 的取值范围.试题解析:(1)当0a =,()1f x x x =--= 1,012,011,1x x x x ≤≤⎧⎪-<⎨⎪->⎩,所以当0x <时,()10f x =>,满足题意;当01x ≤<时,()12f x x =-,由()0f x ≥得120x ≥-,得12x ≤,所以102x ≤<; 当1x >时,()10f x =-<,不合题意.综上,不等式()0f x ≥的解集为1,2⎛⎤-∞ ⎥⎝⎦(2)由()0f x x +=得1a x x x -=--+,则方程()0f x x +=有三个不同的解等价于函数y a =-的图象和函数1y x x x =--+的图象有三个不同交点, 因为1y x x x =--+= 1,01,011,1x x x x x x ≤≤+⎧⎪-<⎨⎪-+>⎩画出其图象,如图所示,结合图象可知,函数y a =-的图象和函数1y x x x =--+的图象有三个不同交点时,则有01a <-<即10a -<<,所以实数a 的取值范围为()1,0-.。

山东省济宁市2016-2017学年高二地理下学期期末考试试题(含解析)

山东省济宁市2016-2017学年高二地理下学期期末考试试题(含解析)

济宁市2016-2017学年高二下学期期末考试地理试题第Ⅰ卷(共45分)一、选择题:每小题1.5分,共45分。

加那利群岛是西班牙的旅游用地,全年气温在20-25℃之间,降雨天数不到30天,几乎永远是阳光明媚的好天气。

夏半年,海浪在每座小岛的一侧海域呈现出银白色的波纹,犹如行驶在海洋上的。

读图完成下列各题。

1. 加那利群岛夏半年盛行A. 东南风B. 西南风C. 东北风D. 西北风2. 特内里费岛是加那利群岛的主岛,是全球观赏某类自然景观最理想的地点之一,这种自然景观是A. 星空B. 极光C. 荒漠D. 冰川【答案】1. C 2. A【解析】1. 读图可知,图中加纳利群岛位于北纬20°附近大陆西岸,夏季随着太阳直射点北移,东北信风带北移,该地盛行东北风,故选C。

2. 读图,结合上题分析可知,特内里费位于加纳利群岛中,受东北信风和副热带高气压带控制,加之沿岸的寒流影响,几乎永远是阳光明媚的好天气,天气晴朗利于观测星空,A对。

极光主要发生在高纬度地区,B错。

世界荒漠分布广泛,该地不具代表性,C错。

该地纬度低,海拔不高,没有永久冰川分布,D错。

故选A。

下图示意叙利亚地形图据此完成下列各题。

3. 据图分析叙利亚人口最稀疏的地区是A. 东部河谷地区B. 西部沿海地区C. 南部边境地区D. 北部高原地区4. 阿萨德水库的入库水量最大的季节是A. 春季B. 夏季C. 秋季D. 冬季【答案】3. C 4. D【解析】3. 读图可知,叙利亚大部分位于北纬30°-40°之间,西部沿海地区为地中海气候,南部地区受副高控制时间长,降水少,为热带沙漠气候,不利于人口分布,故选C。

4. 阿萨德水库位于幼发拉底河上,河流上游地区为地中海气候,地中海气候冬季多雨,入库水量大,故选D。

印度尼西亚是中国镍铝矿石的主要供应国,2014年1月该国实施矿石出口禁令,只有在该国建有冶炼厂的公司才会取得原矿出口的特别许可证。

【配套K12】山东省济宁市2016-2017学年高二地理下学期期末考试试题(含解析)

【配套K12】山东省济宁市2016-2017学年高二地理下学期期末考试试题(含解析)

济宁市2016-2017学年高二下学期期末考试地理试题第Ⅰ卷(共45分)一、选择题:每小题1.5分,共45分。

加那利群岛是西班牙的旅游用地,全年气温在20-25℃之间,降雨天数不到30天,几乎永远是阳光明媚的好天气。

夏半年,海浪在每座小岛的一侧海域呈现出银白色的波纹,犹如行驶在海洋上的。

读图完成下列各题。

1. 加那利群岛夏半年盛行A. 东南风B. 西南风C. 东北风D. 西北风2. 特内里费岛是加那利群岛的主岛,是全球观赏某类自然景观最理想的地点之一,这种自然景观是A. 星空B. 极光C. 荒漠D. 冰川【答案】1. C 2. A【解析】1. 读图可知,图中加纳利群岛位于北纬20°附近大陆西岸,夏季随着太阳直射点北移,东北信风带北移,该地盛行东北风,故选C。

2. 读图,结合上题分析可知,特内里费位于加纳利群岛中,受东北信风和副热带高气压带控制,加之沿岸的寒流影响,几乎永远是阳光明媚的好天气,天气晴朗利于观测星空,A对。

极光主要发生在高纬度地区,B错。

世界荒漠分布广泛,该地不具代表性,C错。

该地纬度低,海拔不高,没有永久冰川分布,D错。

故选A。

下图示意叙利亚地形图据此完成下列各题。

3. 据图分析叙利亚人口最稀疏的地区是A. 东部河谷地区B. 西部沿海地区C. 南部边境地区D. 北部高原地区4. 阿萨德水库的入库水量最大的季节是A. 春季B. 夏季C. 秋季D. 冬季【答案】3. C 4. D【解析】3. 读图可知,叙利亚大部分位于北纬30°-40°之间,西部沿海地区为地中海气候,南部地区受副高控制时间长,降水少,为热带沙漠气候,不利于人口分布,故选C。

4. 阿萨德水库位于幼发拉底河上,河流上游地区为地中海气候,地中海气候冬季多雨,入库水量大,故选D。

印度尼西亚是中国镍铝矿石的主要供应国,2014年1月该国实施矿石出口禁令,只有在该国建有冶炼厂的公司才会取得原矿出口的特别许可证。

山东省济宁市高二物理下学期期末考试试题(扫描版,无答案)(2021年整理)

山东省济宁市高二物理下学期期末考试试题(扫描版,无答案)(2021年整理)

山东省济宁市2016-2017学年高二物理下学期期末考试试题(扫描版,无答案)
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2016—2017学年度第二学期期末考试高二英语试题阅读理解AWealth starts with a goal and saving a dollar at a time. Call it the piggy bank strategy. There are lessons in that time-honored coin-saving container.Any huge task seems easier when reduced to baby steps. If you wished to climb a 12,000-foot mountain, and could do it a day at a time, you would only have to climb 33 feet daily to reach the top in a year. If you want to take a really nice trip in 10 years for a special occasion, to collect the $15,000 cost, you have to save $3.93 a day. If you drop that into a piggy bank and then once a year put $1,434 in a savings account at 1% interest rate after-tax, you will have your trip money.When I was a child, my parents gave me a piggy bank to teach me that, if I wanted something, I should save money to buy it. We associate piggy banks with children, but in many countries, the little containers are also popular with adults. Europeans see a piggy bank as a sign of good fortune and wealth. Around the world, many believe a gi ft of a piggy bank on New Year’s Day brings good luck and financial success. Ah, but you have to put something in it.Why is a pig used as a symbol of saving? Why not an elephant bank, which is bigger and holds more coins? In the Middle Ages, before modern banking and credit instruments, people saved money at home, a few coins at a time dropped into a jar or dish. Potters made these inexpensive containers from an orange-colored clay called “pygg,” and folks saved coins in pygg jars. The Middle English wo rd for pig was “pigge”. While the Saxons pronounced pygg, referring to the clay, as “pug”, eventually the two words changed into the same pronunciation, sounding the “i” as in pig or piggy. As the word became less associated with the orange clay and more with the animal, a clever potter fashioned a pygg jar in the shape of a pig, delighting children and adults. The piggy bank was born.Originally you had to break the bank to get to the money, bringing in a sense of seriousness into savings. While piggy banks teach children the wisdom of saving, adults often need to relearn childhood lessons. Think about the things in life that require large amounts of money---college education, weddings, cars, medical care, starting a business, buying a home, and fun stuff like great trips. So when you have money, take off the top 10%, put it aside, save and invest wisely.21. What is the piggy bank strategy?A. Paying 1% income tax at a time.B. Setting a goal before making a travel plan.C. Aiming high even when doing small things.D. Putting aside a little money regularly for future use.22. Why did the writer’s parents give him a piggy bank as a gift?A. To delight him with the latest fashion.B. To encourage him to climb mountains.C. To help him form the habit of saving.D. To teach him English pronunciation.23. What does the underlined word “something”(Paragraph 3) most probably refer to?A. MoneyB. GiftsC. Financial successD. Good luck24. The last paragraph talks about ________.A. the seriousness of educating childrenB. the enjoyment of taking a great tripC. the importance of managing moneyD. the difficulty of starting a businessBStudents who date in middle school have significantly worse study skills. They are four times more likely to drop out of school and report twice as much alcohol and tobacco use as their single classmates, according to a new research from the University of Georgia."Romantic relationships are a trademark of adolescence, but very few studies have examined how adolescents differ in the development of these relationships," said Pamela Orpinas, study author and professor in the College of Public Health and head of the Department of Health Promotion and Behavior.Orpinas followed a group of 624 students over a seven-year period from 6th to 12th grade. Each year, the group of students completed a survey indicating whether they had dated and reported the frequency of different behaviors, including the use of drugs and alcohol. Their teachers completed questionnaires about the students’ academic efforts. He found some students never or hardly ever reported dating from middle to high school, and these studentshad consistently the best study skills according to their teachers. Other students dated infrequently in middle school but increased the frequency of dating in high school."At all points in time, teachers rated the students who reported the lowest frequency of dating as having the best study skills and the students with the highest dating as having the worst study skills,” according to the journal article. Study skills refer to behaviors that lead to academic success such as doing work for extra credit being well organized, finishing homework, working hard and reading assigned chapters."Dating a classmate may have the same emotional effect of dating a co-worker," Orpinas said, "When the couple break up, they have to continue to see each other in class and perhaps witness the ex-partner dating someone else. It is reasonable to think this could be linked to depression and distract attention from studying.”“Dating should not be considered a ceremony of growth in middle school,” Orpinas concluded.25. When doing his study, Orpinas _____.A. followed a group of students of 6th and 12th gradeB. completed a survey and a report each yearC. completed questionnaires about the students’ academic effortsD. found that the students’ study skills have connection with their frequency of dating26. Study skills may include the following behaviors and qualities Except _____.A. being diligentB. being well organizedC. being kind and helpfulD. finishing assigned schoolwork27. Orpinas’ attitude towards dating in m iddle school is _____.A. supportiveB. positiveC. negativeD. indifferentCMotorists who used to listen to the radio or their favorite tunes on CDs may have a new way to entertain themselves, after engineers in Japan developed a musical road surface.A team from the Hokkaido Industrial Research Institute has built a number of “melody roads”, which use cars as tuning forks to play music as they travel.The concept works by using grooves(凹槽). They are cut at very specific intervals in the road surface. The melody road uses the spaces between to create different notes.Depending on how far apart the grooves are, a car moving over them will produce a series of high or low notes, and designers are able to create a distinct tune.Paten d ocuments for the design describe it as notches “formed in a road surface so as to play a melody without producing simple sound or rhythm and reproduce melody-like tones.”There are three musical strips in central and northern Japan—one of which plays the tune of a Japanese pop song. Reports say the system was invented by Shizuo Shinoda. He scraped some markings into a road with a bulldozer before driving over them and found that they helped to produce all kinds of tones.The optimal speed for melody road is 44kph, but people say it is not always easy to get the intended sound.“You need to keep the car windows closed to hear well,” wrote one Japanese blogger. “Driving too fast will sound like playing fast forward, while driving around 12mph [20km/h] has a slow-motion effect, making you almost car-sick.”28. We can learn from the passage that the highness of notes is depended on _______.A. how far the grooves areB. how big the grooves areC. the number of the groovesD. the speed of the car29. The und erlined word “optimal” in the passage might mean _______.A. fastestB. possibleC. bestD. slowest30. In order to hear the music well, you have to ______.A. drive very fastB. drive slowlyC. open the windows wideD. keepthe windows closed31. What’s the best title of the passage?A. A New Type of MusicB. Melody Roads in JapanC. A Musical Road SurfaceD. A New Invention in JapanDMeasles (麻疹), which once killed 450 children each year and disabled even more, was nearly wiped out in the United States 14 years ago by the universal use of the MMR vaccine (疫苗). But the disease is making a comeback, caused by a growing anti-vaccine movement and misinformation that is spreading quickly. Already this year, 115 measles cases have been reported in the USA, compared with 189 for all of last year.The numbers might sound small, but they are the leading edge of a dangerous trend. When vaccination rates are very high, as they still are in the nation as a whole, everyone is protected. This is called “herd immunity”, which protects the people who get hurt easily, including those who can’t be vaccinated for medical reasons, babies too young to get vaccinated and people on whom the vaccine doesn’t work.But herd immunity works only when nearly the whole herd joins in. When some refuse vaccination and seek a free ride, immunity breaks down and everyone is in even bigger danger.That’s exactly what is happening in small neighborhoods around the cou ntry from Orange County, California, where 22 measles cases were reported this month, to Brooklyn, N.Y., where a 17-year-old caused an outbreak last year.The resistance to vaccine has continued for decades, and it is driven by a real but very small risk. Those who refuse to take that risk selfishly make others suffer.Making things worse are state laws that make it too easy to opt out (决定不参加) of what are supposed to be required vaccines for all children entering kindergarten. Seventeen states allow parents to get an exemption (豁免), sometimes just by signing a paper saying they personally object to a vaccine.Now, several states are moving to tighten laws by adding new regulations for opting out. But no one does enough to limit exemptions.Parents ought to be able to opt out only for limited medical or religious reasons. But personal opinions? Not good enough. Everyone enjoys the life-saving benefits vaccines provide, but they’ll exist only as long as everyone shares in the risks.32. The first two paragraphs suggest that ________.A. a small number of measles cases can start a dangerous trendB. the outbreak of measles attracts the public attentionC. anti-vaccine movement has its medical reasonsD. information about measles spreads quickly33. Herd immunity works well when ________.A. exemptions are allowedB. the whole neighborhood is involved inC. several vaccines are used togetherD. new regulations are added to the state laws34. What is the main reason for the comeback of measles?A. The vaccine opt-outs of some people.B. The lack of medical care.C. The features of measles itself.D. The overuse of vaccine.35. What is the purpose of the passage?A. To introduce the idea of exemption.B. To stress the importance of vaccination.C. To discuss methods to cure measles.D. To appeal for equal rights in medical treatment.七选五阅读(一)Tips for Staying SharpIt’s not abnormal to feel out of it from time to time or perhaps you’re feeling a b it sluggish (迟钝的) when it comes to remembering things. If you’re worried about your forgetfulness,try out these five tips to stay sharp.36You’ve probably heard the term “you are what you eat.” If you decide to eat sugary foods that are heavy with fats then expect to feel lazy and have no energy. But if you decide to eat fruits, vegetables and make other healthy eating decisions you’11 find that you’re full of energy.Never stop learningChallenge yourself constantly—whether it is puzzles,reading, cooking or other tasks that will keep your mind working. If your mind is always learning new things and active, you’11 realize that it’s much easier for you to learn new things and to function. 37Stay fitBeing in good health and staying in shape is a big part of staying sharp. 38 If you choose to be lazy then your body is not going to give you the energy that requires remembering things and completing tasks. However, if you put forth the effort into being healthy then your body will expel (释放) the energy you need to achieve your goals.Ensure enough sleepBe sure to get the necessary amount of sleep that your body needs to function on a daily basis. 39 It also helps your body build up an energy reserve so you can improve your ability to focus and avoid distractions.Socialize more40 Social interaction will help you develop multitasking, problem solving and other skills that are necessary in life to keep your mind sharp.A. Feed your brain.B. Avoid eating too much.C. Your body will only give you what you put into it.D. Lack of sleep contributes to tiredness and slow reaction.E. Conversations require individuals to stay aware and active.F. According to studies, sleep helps strengthen and recover your memories.G. Remember your brain is a muscle and if you want to get the most out of it, you must use it.(二)Business is the organized approach to providing customers with the goods and services they want. The word business also refers to an organization that provides these goods and services. Most businesses seek to make a profit—that is, they aim to achieve income that is more than the costs of operating the business. ________ Commonly called nonprofits, these organizations are primarily nongovernmental service providers. ________Business management is a term used to describe the techniques of planning, direction, and control of the operations of a business. ________ One is the establishment of broad basic policies with respect to production; sales; the purchase of equipment, materials and supplies; and accounting. ________ The third relates to the establishment of standards of work in all departments. Direction is concerned primarily with supervision (监管) and guidance by the management in authority. ________A. Control includes the use of records and reports to compare actual work with the set standards for work.B. In this connection there is the difference between top management and operative management.C. Examples of nonprofit business include such organizations as social service agencies and hospitals.D. However, some businesses only seek to earn enough to cover their operating costs.E. The second aspect relates to the application of these policies by departments.F. In the theory of business management, organization has two main aspects.G. Planning in business management has three main aspects.完形填空(一)How much do you laugh and smile during the day? Do you take your life and your illness or injury so 41 that there is no room for joy to fill you? Want a totally free, simple way to lift your 42 and improve your health with no medicine needed? Then laughing and smiling is 43 to you.So laugh your way to happiness. 44 it takes to put a smile on your face is what you should be doing. Research shows that laughing can increase the immune system, 45 the body to stay disease free and fight colds and the flu.If you are facing an illness, having a positive life opinion and a 46 of humor will keep your body open to healing. If you are healthy, laughing will help to make sure you stay that way, and can 47 enjoyment to your work and home life and 48 your daily stress.Certainly, it can be 49 to keep a positive opinion of life all the time. Simply taking the time to 50 on the positive and treat for the good things 51 in your life can help 52 , but if you are struggling with negative emotions that you just can’t seem to 53 , there are tools that can help.There are so many things out there to smile about and 54 you have to do is find 55 . Practice looking for the bright 56 of every situation. Avoid the negative: don’t 57 yourself with your own problems—or 58 others for their “shortcomings”. And don’t pay no attention to the joy in everyday things. Create your own “Laugh for Health”59 – anything to add to your collection and to give you fresh materials that 60 to you.41. A. proudly B. seriously C. carefully D. freely42. A. burdens B. spirits C. loads D. values43. A. beneficial B. changeable C. suitable D. harmful44. A. However B. Whenever C. Whatever D. Whichever45. A. helping B. making C. letting D. causing46. A. taste B. sense C. knowledge D. joy47. A. adjust B. adapt C. apply D. add48. A. improve B. arise C. cancel D. reduce49. A. hard B. awful C. energetic D. helpful50. A. absorb B. put C. focus D. interview51. A. carefully B. thankfully C. particularly D. differently52. A. remarkably B. accidentally C. frequently D. purposefully53. A. seek B. stress C. accept D. overcome54. A. all B. that C. how D. why55. A. it B. that C. this D. one56. A. side B. plan C. aspect D. future57. A. depend B. abandon C. load D. ban58. A. praise B. miss C. approve D. blame59. A. ambition B. summary C. message D. collection60. A. appeal B. apply C. center D. assume(二)I went to a group activity, “Sensitivity Sunday”, which was to make us more 36 the problems faced by disabled people. We were asked to “37 a disability” for several hours one Sunday. Some members, 38 , chose to use wheelchairs. Others wore sound-blocking earplugs (耳塞) or blindfolds (眼罩).Just sitting in the wheelchair was a 39 experience. I had never considered before how 40 it would be to use one. As soon as I sat down, my 41 made the chair begin to roll. Its wheels were not 42 . Then I wondered where to put my 43 . It took me quite a while to get the metal footrest into 44 . I took my first uneasy look at what was to be my only means of 45 for several hours. For disabled people, “adopting a wheelchair” is not a temporary46 .I tried to find a 47 position and thought it might be restful, 48 kind of nice, to be49 around for a while. Looking around, I 50 I would have to handle the thing myself! My hands started to ache as I 51 the heavy metal wheels. I came to know that controlling the 52 of the wheelchair was not going to be a(n) 53 task.My wheelchair experiment was soon 54 . It made a deep impression on me. A fewhours of “disability” gave me only a taste o f the 55 , both physical and mental, that disabled people must overcome.36. A. curious about B. interested in C. aware of D. careful with37. A. cure B. prevent C. adopt D. analyze38. A. instead B. strangely C. as usual D. like me39. A. learning B. working C. satisfying D. relaxing40. A. convenient B. awkward C. boring D. exciting41. A. height B. force C. skill D. weight42. A. locked B. repaired C. powered D. grasped43. A. hands B. feet C. keys D. handles44. A. place B. action C. play D. effect45. A. operation B. communication C. transportation D. production46. A. exploration B. education C. experiment D. entertainment47. A. flexible B. safe C. starting D. comfortable48. A. yet B. just C. still D. even49. A. shown B. pushed C. driven D. guided50. A. realized B. suggested C. agreed D. admitted51. A. lifted B. turned C. pressed D. seized52. A. path B. position C. direction D. way53. A. easy B. heavy C. major D. extra54. A. forgotten B. repeated C. conducted D. finished55. A. weaknesses B. challenges C. anxieties D. illnesses语篇填空(一)Belt and Road Forum for International CooperationBEIJING, May 14 (Xinhua) —Chinese President Xi Jinping on Sunday said, “The Belt and Road Initiative (倡议)________ (be) ‘a project of the century’ that will benefit people across the world.”Xi made the remarks when __________ (deliver) a speech at the opening ceremony of the two-day Belt and Road Forum for International Cooperation. Named after the historic Silk Road,the Belt and Road Initiative __________ (propose) by Xi in 2013 to chart out new fields for international cooperation.“Going through thousands of miles and y ears, the __________ (anciently) silk routes embody (体现) the spirit of peace and cooperation, openness and inclusiveness (包容), mutual learning __________ mutual benefit,” Xi told an audience of more than 1,500 from across the globe. “The Silk Road spirit __________ (become) a great heritage (遗产) of human civilization so far,” he said.A total of 29 foreign __________ (head) of state and government leaders attended the forum, including Russian President Vladimir Putin. Other delegates (代表) include officials, entrepreneurs, financiers and journalists _________ over 130 countries and representatives of key international organizations. The United States sent a delegation _______ (lead) by Matt Pottinger.By all means, the forum, __________ also features a round-table summit of global leaders on Monday, is one of the premium gatherings in today’s world.(二)Tower Bridge is one of London’s most striking ________ (attract) that you shouldn’t miss when visiting the capital. This magnificent piece of Victorian architecture ________ (date) back to the late 1880s, and until the mid-1970s the main bridge ________ (operate) by a steam engine.Today Tower Bridge is a splendid landmark typical of London, ________ (surround) by cafes, and bars where tourists and locals alike can sit and enjoy the beautiful bridge, ________ looks like two small castles over the Thames, just like something out of a fairy tale.It is free to admire Tower Bridge and walk on it ________ any time; it’s striking to look at during the day and magic when lit up at night, ________ don’t miss out on this wonderful scene!If you’d like ________ (step) inside the towers of the famous bridge, you can book Tower Bridge Exhibition tickets. During the exhibition, you can visit the famed engine rooms and impressive towers and find out ________ the bridge came into existence through digital displays. You can also pay ________ visit to the glass room of Tower Bridge on the high-level walkways. There are even yoga classes available on the glass floor of Tower Bridge!短文改错(一)Last summer, to my delight, I got a chance to go to the United Kingdom. I was eager to see how England was like. Although my parents objected to my decision at first, but finally they were persuaded. It was the first time that I have travelled abroad alone. Felt excited, I boarded the plane. Suddenly, however, I found my luggages gone and felt helplessly. It was on that moment that the airhostess came to my assistance. After helping me find it back, she suggested that I attached a label to my suitcase on my next journey. I was thankful to the kind lady for his help! What unforgettable trip!(二)I hardly remember my grandmother. She used to holding me on her knees and sing old songs. I was only four when she passes away. She is just a distant memory for me now.I remember my grandfather very much. He was tall, with broad shoulder and a beard that turned from black toward gray over the years. He had a deep voice, which set himself apart from others in our small town, he was strong and powerful. In a fact, he even scared my classmates away during they came over to play or do homework with me. However, he was the gentlest man I have never known。

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