2012真题
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2012年普通高等学校招生全国统一考试(辽宁卷)英语本试卷分第一卷(选择题)和第二卷(非选择题)两部分。
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1. Where does this conversation probably take place?A. In the bookstoreB. In a classroomC. In a library2. At what time will the film begin?A. 7:20B. 7:15C. 7:003.What are the two speakers mainly talking about?A. Their friend Jane.B. A weekend trip.C. A radio programme.4. What will the woman probably do?A. Catch a train.B. See the man off.C. Go shopping.4.Why did the woman apologize?A. She made a late delivery.B. She went to the wrong place.C. She couldn’t take the cake back.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。
2012年全国硕士研究生入学统一考试数学一试题一、选择题:1 8小题,每小题4分,共32分.下列每题给出的四个选项中,只有一个选项符合题目要求的,请将所选项前的字母填在答题纸...指定位置上. (1) 曲线221x x y x +=-渐近线的条数 ( )(A) 0 (B) 1 (C) 2 (D) 3(2) 设函数2()(1)(2)()x x nx y x e e e n =--- ,其中n 为正整数,则(0)y '= ( )(A) 1(1)(1)!n n --- (B) (1)(1)!n n -- (C) 1(1)!n n -- (D) (1)!n n - (3) 如果函数(,)f x y 在(0,0)处连续,那么下列命题正确的是 ( )(A) 若极限0(,)limx y f x y x y →→+存在,则(,)f x y 在(0,0)处可微(B) 若极限2200(,)limx y f x y x y→→+存在,则(,)f x y 在(0,0)处可微(C) 若(,)f x y 在(0,0)处可微,则 极限00(,)limx y f x y x y →→+存在(D) 若(,)f x y 在(0,0)处可微,则 极限2200(,)lim x y f x y x y→→+存在(4)设2sin (1,2,3)k xK exdx k π==⎰I 则有 ( )(A)123I I I << (B) 321I I I << (C) 231I I I << (D)213I I I <<(5)设1100C α⎛⎫⎪= ⎪ ⎪⎝⎭,2201C α⎛⎫ ⎪= ⎪ ⎪⎝⎭ ,3311C α⎛⎫ ⎪=- ⎪ ⎪⎝⎭ ,4411C α-⎛⎫ ⎪= ⎪ ⎪⎝⎭ ,其中1234,,,C C C C 为任意常数,则下列向量组线性相关的为( )(A)123,,ααα (B) 124,,ααα (C)134,,ααα (D)234,,ααα(6) 设A 为3阶矩阵,P 为3阶可逆矩阵,且1100010002p AP -⎛⎫⎪= ⎪⎪⎝⎭.若P=(123,,ααα),1223(,,)ααααα=+,则1QAQ -= ( )(A) 100020001⎛⎫ ⎪ ⎪ ⎪⎝⎭(B) 100010002⎛⎫ ⎪ ⎪ ⎪⎝⎭(C) 200010002⎛⎫ ⎪ ⎪ ⎪⎝⎭(D)200020001⎛⎫⎪ ⎪ ⎪⎝⎭(7)设随机变量X 与Y 相互独立,且分别服从参数为1与参数为4的指数分布,则{}p X Y <=( )(A)15(B) 13(C)25(D)45(8)将长度为1m 的木棒随机地截成两段,则两段长度的相关系数为 ( )(A) 1 (B)12(C) 12-(D)1-二、填空题:9 14小题,每小题4分,共24分.请将答案写在答题纸...指定位置上. (9)若函数()f x 满足方程'''()()2()0f x f x f x +-=及''()()2f x f x e +=,则()f x =(10)20x =⎰(11)(2,1,1)()|z grad xy +y=(12)设(){},,1,0,0,0x y z x y z x y z ∑=++=≥≥≥,则2y ds ∑=⎰⎰(13)设X 为三维单位向量,E 为三阶单位矩阵,则矩阵T E XX -的秩为 (14)设A ,B ,C 是随机变量,A 与C 互不相容,()()()11,,23p A B P C p A B C ===三、解答题:15~23小题,共94分.请将解答写在答题纸...指定位置上.解答应写出文字说明、证明过程或演算步骤.(15) 证明21ln cos 1(11)12x xx x x x++≥+-<<-(16)求函数222(,)x y f x y xe +-=的极值(17)求幂级数22044321nn n n xn ∞=+++∑的收敛域及和函数(18) 已知曲线(),:(0),cos 2x f t L t y tπ=⎧≤<⎨=⎩其中函数()f t 具有连续导数,且'(0)0,()0(0).2f f t t π=><<若曲线L的切线与x 轴的交点到切点的距离恒为1,求函数()f t 的表达式,并求此曲线L 与x 轴与y 轴无边界的区域的面积。
2012年全国统一高考数学试卷(理科)(大纲版)一、选择题(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.(5分)复数=()A.2+i B.2﹣i C.1+2i D.1﹣2i2.(5分)已知集合A={1,3,},B={1,m},A∪B=A,则m的值为()A.0或B.0或3C.1或D.1或33.(5分)椭圆的中心在原点,焦距为4,一条准线为x=﹣4,则该椭圆的方程为()A.B.C.D.4.(5分)已知正四棱柱ABCD﹣A1B1C1D1中,AB=2,CC1=2,E为CC1的中点,则直线AC1与平面BED的距离为()A.2B.C.D.15.(5分)已知等差数列{a n}的前n项和为S n,a5=5,S5=15,则数列的前100项和为()A.B.C.D.6.(5分)△ABC中,AB边的高为CD,若=,=,•=0,||=1,||=2,则=()A.B.C.D.7.(5分)已知α为第二象限角,,则cos2α=()A.﹣B.﹣C.D.8.(5分)已知F1、F2为双曲线C:x2﹣y2=2的左、右焦点,点P在C上,|PF1|=2|PF2|,则cos∠F1PF2=()A.B.C.D.9.(5分)已知x=lnπ,y=log52,,则()A.x<y<z B.z<x<y C.z<y<x D.y<z<x 10.(5分)已知函数y=x3﹣3x+c的图象与x轴恰有两个公共点,则c=()A.﹣2或2B.﹣9或3C.﹣1或1D.﹣3或1 11.(5分)将字母a,a,b,b,c,c排成三行两列,要求每行的字母互不相同,每列的字母也互不相同,则不同的排列方法共有()A.12种B.18种C.24种D.36种12.(5分)正方形ABCD的边长为1,点E在边AB上,点F在边BC上,,动点P从E出发沿直线向F运动,每当碰到正方形的边时反弹,反弹时反射角等于入射角,当点P第一次碰到E时,P与正方形的边碰撞的次数为()A.16B.14C.12D.10二、填空题:本大题共4小题,每小题5分,共20分,把答案填在题中横线上.(注意:在试题卷上作答无效)13.(5分)若x,y满足约束条件则z=3x﹣y的最小值为.14.(5分)当函数y=sinx﹣cosx(0≤x<2π)取得最大值时,x=.15.(5分)若的展开式中第3项与第7项的二项式系数相等,则该展开式中的系数为.16.(5分)三棱柱ABC﹣A1B1C1中,底面边长和侧棱长都相等,∠BAA1=∠CAA1=60°,则异面直线AB1与BC1所成角的余弦值为.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.(10分)△ABC的内角A、B、C的对边分别为a、b、c,已知cos(A﹣C)+cosB=1,a=2c,求C.18.(12分)如图,四棱锥P﹣ABCD中,底面ABCD为菱形,PA⊥底面ABCD,,PA=2,E是PC上的一点,PE=2EC.(Ⅰ)证明:PC⊥平面BED;(Ⅱ)设二面角A﹣PB﹣C为90°,求PD与平面PBC所成角的大小.19.(12分)乒乓球比赛规则规定:一局比赛,双方比分在10平前,一方连续发球2次后,对方再连续发球2次,依次轮换.每次发球,胜方得1分,负方得0分.设在甲、乙的比赛中,每次发球,发球方得1分的概率为0.6,各次发球的胜负结果相互独立.甲、乙的一局比赛中,甲先发球.(Ⅰ)求开始第4次发球时,甲、乙的比分为1比2的概率;(Ⅱ)ξ表示开始第4次发球时乙的得分,求ξ的期望.20.(12分)设函数f(x)=ax+cosx,x∈[0,π].(Ⅰ)讨论f(x)的单调性;(Ⅱ)设f(x)≤1+sinx,求a的取值范围.21.(12分)已知抛物线C:y=(x+1)2与圆(r>0)有一个公共点A,且在A处两曲线的切线为同一直线l.(Ⅰ)求r;(Ⅱ)设m,n是异于l且与C及M都相切的两条直线,m,n的交点为D,求D到l的距离.22.(12分)函数f(x)=x2﹣2x﹣3,定义数列{ x n}如下:x1=2,x n+1是过两点P (4,5),Q n(x n,f(x n))的直线PQ n与x轴交点的横坐标.(Ⅰ)证明:2≤x n<x n+1<3;(Ⅱ)求数列{ x n}的通项公式.2012年全国统一高考数学试卷(理科)(大纲版)参考答案与试题解析一、选择题(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.(5分)复数=()A.2+i B.2﹣i C.1+2i D.1﹣2i【考点】A5:复数的运算.【专题】11:计算题.【分析】把的分子分母都乘以分母的共轭复数,得,由此利用复数的代数形式的乘除运算,能求出结果.【解答】解:===1+2i.故选:C.【点评】本题考查复数的代数形式的乘除运算,是基础题.解题时要认真审题,仔细解答.2.(5分)已知集合A={1,3,},B={1,m},A∪B=A,则m的值为()A.0或B.0或3C.1或D.1或3【考点】1C:集合关系中的参数取值问题.【专题】5J:集合.【分析】由题设条件中本题可先由条件A∪B=A得出B⊆A,由此判断出参数m 可能的取值,再进行验证即可得出答案选出正确选项.【解答】解:由题意A∪B=A,即B⊆A,又,B={1,m},∴m=3或m=,解得m=3或m=0及m=1,验证知,m=1不满足集合的互异性,故m=0或m=3即为所求,故选:B.【点评】本题考查集合中参数取值问题,解题的关键是将条件A∪B=A转化为B⊆A,再由集合的包含关系得出参数所可能的取值.3.(5分)椭圆的中心在原点,焦距为4,一条准线为x=﹣4,则该椭圆的方程为()A.B.C.D.【考点】K3:椭圆的标准方程;K4:椭圆的性质.【专题】11:计算题.【分析】确定椭圆的焦点在x轴上,根据焦距为4,一条准线为x=﹣4,求出几何量,即可求得椭圆的方程.【解答】解:由题意,椭圆的焦点在x轴上,且∴c=2,a2=8∴b2=a2﹣c2=4∴椭圆的方程为故选:C.【点评】本题考查椭圆的标准方程,考查椭圆的几何性质,属于基础题.4.(5分)已知正四棱柱ABCD﹣A1B1C1D1中,AB=2,CC1=2,E为CC1的中点,则直线AC1与平面BED的距离为()A.2B.C.D.1【考点】MI:直线与平面所成的角.【专题】11:计算题.【分析】先利用线面平行的判定定理证明直线C1A∥平面BDE,再将线面距离转化为点面距离,最后利用等体积法求点面距离即可【解答】解:如图:连接AC,交BD于O,在三角形CC1A中,易证OE∥C1A,从而C1A∥平面BDE,∴直线AC1与平面BED的距离即为点A到平面BED的距离,设为h,=S△ABD×EC=××2×2×=在三棱锥E﹣ABD中,V E﹣ABD=×2×=2在三棱锥A﹣BDE中,BD=2,BE=,DE=,∴S△EBD∴V A=×S△EBD×h=×2×h=﹣BDE∴h=1故选:D.【点评】本题主要考查了线面平行的判定,线面距离与点面距离的转化,三棱锥的体积计算方法,等体积法求点面距离的技巧,属基础题5.(5分)已知等差数列{a n}的前n项和为S n,a5=5,S5=15,则数列的前100项和为()A.B.C.D.【考点】85:等差数列的前n项和;8E:数列的求和.【专题】11:计算题.【分析】由等差数列的通项公式及求和公式,结合已知可求a1,d,进而可求a n,代入可得==,裂项可求和【解答】解:设等差数列的公差为d由题意可得,解方程可得,d=1,a1=1由等差数列的通项公式可得,a n=a1+(n﹣1)d=1+(n﹣1)×1=n∴===1﹣=故选:A.【点评】本题主要考查了等差数列的通项公式及求和公式的应用,及数列求和的裂项求和方法的应用,属于基础试题6.(5分)△ABC中,AB边的高为CD,若=,=,•=0,||=1,||=2,则=()A.B.C.D.【考点】9Y:平面向量的综合题.【分析】由题意可得,CA⊥CB,CD⊥AB,由射影定理可得,AC2=AD•AB可求AD,进而可求,从而可求与的关系,进而可求【解答】解:∵•=0,∴CA⊥CB∵CD⊥AB∵||=1,||=2∴AB=由射影定理可得,AC2=AD•AB∴∴∴==故选:D.【点评】本题主要考查了直角三角形的射影定理的应用,向量的基本运算的应用,向量的数量积的性质的应用.7.(5分)已知α为第二象限角,,则cos2α=()A.﹣B.﹣C.D.【考点】GG:同角三角函数间的基本关系;GS:二倍角的三角函数.【专题】56:三角函数的求值.【分析】由α为第二象限角,可知sinα>0,cosα<0,从而可求得sinα﹣cosα=,利用cos2α=﹣(sinα﹣cosα)(sinα+cosα)可求得cos2α【解答】解:∵sinα+cosα=,两边平方得:1+sin2α=,∴sin2α=﹣,①∴(sinα﹣cosα)2=1﹣sin2α=,∵α为第二象限角,∴sinα>0,cosα<0,∴sinα﹣cosα=,②∴cos2α=﹣(sinα﹣cosα)(sinα+cosα)=(﹣)×=﹣.故选:A.【点评】本题考查同角三角函数间的基本关系,突出二倍角的正弦与余弦的应用,求得sinα﹣cosα=是关键,属于中档题.8.(5分)已知F1、F2为双曲线C:x2﹣y2=2的左、右焦点,点P在C上,|PF1|=2|PF2|,则cos∠F1PF2=()A.B.C.D.【考点】KC:双曲线的性质.【专题】11:计算题.【分析】根据双曲线的定义,结合|PF1|=2|PF2|,利用余弦定理,即可求cos∠F1PF2的值.【解答】解:将双曲线方程x2﹣y2=2化为标准方程﹣=1,则a=,b=,c=2,设|PF1|=2|PF2|=2m,则根据双曲线的定义,|PF1|﹣|PF2|=2a可得m=2,∴|PF1|=4,|PF2|=2,∵|F1F2|=2c=4,∴cos∠F1PF2====.故选:C.【点评】本题考查双曲线的性质,考查双曲线的定义,考查余弦定理的运用,属于中档题.9.(5分)已知x=lnπ,y=log52,,则()A.x<y<z B.z<x<y C.z<y<x D.y<z<x【考点】72:不等式比较大小.【专题】11:计算题;16:压轴题.【分析】利用x=lnπ>1,0<y=log52<,1>z=>,即可得到答案.【解答】解:∵x=lnπ>lne=1,0<log52<log5=,即y∈(0,);1=e0>=>=,即z∈(,1),∴y<z<x.故选:D.【点评】本题考查不等式比较大小,掌握对数函数与指数函数的性质是解决问题的关键,属于基础题.10.(5分)已知函数y=x3﹣3x+c的图象与x轴恰有两个公共点,则c=()A.﹣2或2B.﹣9或3C.﹣1或1D.﹣3或1【考点】53:函数的零点与方程根的关系;6D:利用导数研究函数的极值.【专题】11:计算题.【分析】求导函数,确定函数的单调性,确定函数的极值点,利用函数y=x3﹣3x+c的图象与x轴恰有两个公共点,可得极大值等于0或极小值等于0,由此可求c的值.【解答】解:求导函数可得y′=3(x+1)(x﹣1),令y′>0,可得x>1或x<﹣1;令y′<0,可得﹣1<x<1;∴函数在(﹣∞,﹣1),(1,+∞)上单调增,(﹣1,1)上单调减,∴函数在x=﹣1处取得极大值,在x=1处取得极小值.∵函数y=x3﹣3x+c的图象与x轴恰有两个公共点,∴极大值等于0或极小值等于0.∴1﹣3+c=0或﹣1+3+c=0,∴c=﹣2或2.故选:A.【点评】本题考查导数知识的运用,考查函数的单调性与极值,解题的关键是利用极大值等于0或极小值等于0.11.(5分)将字母a,a,b,b,c,c排成三行两列,要求每行的字母互不相同,每列的字母也互不相同,则不同的排列方法共有()A.12种B.18种C.24种D.36种【考点】D9:排列、组合及简单计数问题.【专题】11:计算题;16:压轴题.【分析】由题意,可按分步原理计数,对列的情况进行讨论比对行讨论更简洁.【解答】解:由题意,可按分步原理计数,首先,对第一列进行排列,第一列为a,b,c的全排列,共有种,再分析第二列的情况,当第一列确定时,第二列第一行只能有2种情况,当第二列一行确定时,第二列第2,3行只能有1种情况;所以排列方法共有:×2×1×1=12种,故选:A.【点评】本题若讨论三行每一行的情况,讨论情况较繁琐,而对两列的情况进行分析会大大简化解答过程.12.(5分)正方形ABCD的边长为1,点E在边AB上,点F在边BC上,,动点P从E出发沿直线向F运动,每当碰到正方形的边时反弹,反弹时反射角等于入射角,当点P第一次碰到E时,P与正方形的边碰撞的次数为()A.16B.14C.12D.10【考点】IG:直线的一般式方程与直线的性质;IQ:与直线关于点、直线对称的直线方程.【专题】13:作图题;16:压轴题.【分析】通过相似三角形,来确定反射后的点的落的位置,结合图象分析反射的次数即可.【解答】解:根据已知中的点E,F的位置,可知第一次碰撞点为F,在反射的过程中,直线是平行的,利用平行关系及三角形的相似可得第二次碰撞点为G,且CG=,第二次碰撞点为H,且DH=,作图,可以得到回到E点时,需要碰撞14次即可.故选:B.【点评】本题主要考查了反射原理与三角形相似知识的运用.通过相似三角形,来确定反射后的点的落的位置,结合图象分析反射的次数即可,属于难题.二、填空题:本大题共4小题,每小题5分,共20分,把答案填在题中横线上.(注意:在试题卷上作答无效)13.(5分)若x,y满足约束条件则z=3x﹣y的最小值为﹣1.【考点】7C:简单线性规划.【专题】11:计算题.【分析】作出不等式组表示的平面区域,由z=3x﹣y可得y=3x﹣z,则﹣z表示直线3x﹣y﹣z=0在y轴上的截距,截距越大z越小,结合图形可求【解答】解:作出不等式组表示的平面区域,如图所示由z=3x﹣y可得y=3x﹣z,则﹣z表示直线3x﹣y﹣z=0在y轴上的截距,截距越大z越小结合图形可知,当直线z=3x﹣y过点C时z最小由可得C(0,1),此时z=﹣1故答案为:﹣1【点评】本题主要考查了线性规划的简单应用,解题的关键是明确目标函数中z 的几何意义,属于基础试题14.(5分)当函数y=sinx﹣cosx(0≤x<2π)取得最大值时,x=.【考点】GP:两角和与差的三角函数;HW:三角函数的最值.【专题】11:计算题;16:压轴题.【分析】利用辅助角公式将y=sinx﹣cosx化为y=2sin(x﹣)(0≤x<2π),即可求得y=sinx﹣cosx(0≤x<2π)取得最大值时x的值.【解答】解:∵y=sinx﹣cosx=2(sinx﹣cosx)=2sin(x﹣).∵0≤x<2π,∴﹣≤x﹣<,∴y max=2,此时x﹣=,∴x=.故答案为:.【点评】本题考查三角函数的最值两与角和与差的正弦函数,着重考查辅助角公式的应用与正弦函数的性质,将y=sinx﹣cosx(0≤x<2π)化为y=2sin (x﹣)(0≤x<2π)是关键,属于中档题.15.(5分)若的展开式中第3项与第7项的二项式系数相等,则该展开式中的系数为56.【考点】DA:二项式定理.【专题】11:计算题;16:压轴题.【分析】根据第2项与第7项的系数相等建立等式,求出n的值,根据通项可求满足条件的系数【解答】解:由题意可得,∴n=8展开式的通项=令8﹣2r=﹣2可得r=5此时系数为=56故答案为:56【点评】本题主要考查了二项式系数的性质,以及系数的求解,解题的关键是根据二项式定理写出通项公式,同时考查了计算能力.16.(5分)三棱柱ABC﹣A1B1C1中,底面边长和侧棱长都相等,∠BAA1=∠CAA1=60°,则异面直线AB1与BC1所成角的余弦值为.【考点】LM:异面直线及其所成的角.【专题】11:计算题;16:压轴题.【分析】先选一组基底,再利用向量加法和减法的三角形法则和平行四边形法则将两条异面直线的方向向量用基底表示,最后利用夹角公式求异面直线AB1与BC1所成角的余弦值即可【解答】解:如图,设=,,,棱长均为1,则=,=,=∵,∴=()•()=﹣++﹣+=﹣++=﹣1++1=1||===||===∴cos<,>===∴异面直线AB1与BC1所成角的余弦值为【点评】本题主要考查了空间向量在解决立体几何问题中的应用,空间向量基本定理,向量数量积运算的性质及夹角公式的应用,有一定的运算量三.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.(10分)△ABC的内角A、B、C的对边分别为a、b、c,已知cos(A﹣C)+cosB=1,a=2c,求C.【考点】GL:三角函数中的恒等变换应用;HP:正弦定理.【专题】11:计算题.【分析】由cos(A﹣C)+cosB=cos(A﹣C)﹣cos(A+C)=1,可得sinAsinC=,由a=2c及正弦定理可得sinA=2sinC,联立可求C【解答】解:由B=π﹣(A+C)可得cosB=﹣cos(A+C)∴cos(A﹣C)+cosB=cos(A﹣C)﹣cos(A+C)=2sinAsinC=1∴sinAsinC=①由a=2c及正弦定理可得sinA=2sinC②①②联立可得,∵0<C<π∴sinC=a=2c即a>c【点评】本题主要考查了两角和与差的余弦公式及正弦定理的应用,属于基础试题18.(12分)如图,四棱锥P﹣ABCD中,底面ABCD为菱形,PA⊥底面ABCD,,PA=2,E是PC上的一点,PE=2EC.(Ⅰ)证明:PC⊥平面BED;(Ⅱ)设二面角A﹣PB﹣C为90°,求PD与平面PBC所成角的大小.【考点】LW:直线与平面垂直;MI:直线与平面所成的角;MM:向量语言表述线面的垂直、平行关系.【专题】11:计算题.【分析】(I)先由已知建立空间直角坐标系,设D(,b,0),从而写出相关点和相关向量的坐标,利用向量垂直的充要条件,证明PC⊥BE,PC⊥DE,从而利用线面垂直的判定定理证明结论即可;(II)先求平面PAB的法向量,再求平面PBC的法向量,利用两平面垂直的性质,即可求得b的值,最后利用空间向量夹角公式即可求得线面角的正弦值,进而求得线面角【解答】解:(I)以A为坐标原点,建立如图空间直角坐标系A﹣xyz,设D(,b,0),则C(2,0,0),P(0,0,2),E(,0,),B(,﹣b,0)∴=(2,0,﹣2),=(,b,),=(,﹣b,)∴•=﹣=0,•=0∴PC⊥BE,PC⊥DE,BE∩DE=E∴PC⊥平面BED(II)=(0,0,2),=(,﹣b,0)设平面PAB的法向量为=(x,y,z),则取=(b,,0)设平面PBC的法向量为=(p,q,r),则取=(1,﹣,)∵平面PAB⊥平面PBC,∴•=b﹣=0.故b=∴=(1,﹣1,),=(﹣,﹣,2)∴cos<,>==设PD与平面PBC所成角为θ,θ∈[0,],则sinθ=∴θ=30°∴PD与平面PBC所成角的大小为30°【点评】本题主要考查了利用空间直角坐标系和空间向量解决立体几何问题的一般方法,线面垂直的判定定理,空间线面角的求法,有一定的运算量,属中档题19.(12分)乒乓球比赛规则规定:一局比赛,双方比分在10平前,一方连续发球2次后,对方再连续发球2次,依次轮换.每次发球,胜方得1分,负方得0分.设在甲、乙的比赛中,每次发球,发球方得1分的概率为0.6,各次发球的胜负结果相互独立.甲、乙的一局比赛中,甲先发球.(Ⅰ)求开始第4次发球时,甲、乙的比分为1比2的概率;(Ⅱ)ξ表示开始第4次发球时乙的得分,求ξ的期望.【考点】C8:相互独立事件和相互独立事件的概率乘法公式;CH:离散型随机变量的期望与方差.【专题】15:综合题.【分析】(Ⅰ)记A i表示事件:第1次和第2次这两次发球,甲共得i分,i=0,1,2;A表示事件:第3次发球,甲得1分;B表示事件:开始第4次发球,甲、乙的比分为1比2,则B=A0A+A1,根据P(A)=0.4,P(A0)=0.16,P (A1)=2×0.6×0.4=0.48,即可求得结论;(Ⅱ)P(A2)=0.62=0.36,ξ表示开始第4次发球时乙的得分,可取0,1,2,3,计算相应的概率,即可求得ξ的期望.【解答】解:(Ⅰ)记A i表示事件:第1次和第2次这两次发球,甲共得i分,i=0,1,2;A表示事件:第3次发球,甲得1分;B表示事件:开始第4次发球,甲、乙的比分为1比2,则B=A0A+A1∵P(A)=0.4,P(A0)=0.16,P(A1)=2×0.6×0.4=0.48∴P(B)=0.16×0.4+0.48×(1﹣0.4)=0.352;(Ⅱ)P(A2)=0.62=0.36,ξ表示开始第4次发球时乙的得分,可取0,1,2,3 P(ξ=0)=P(A2A)=0.36×0.4=0.144P(ξ=2)=P(B)=0.352P(ξ=3)=P(A0)=0.16×0.6=0.096P(ξ=1)=1﹣0.144﹣0.352﹣0.096=0.408∴ξ的期望Eξ=1×0.408+2×0.352+3×0.096=1.400.【点评】本题考查相互独立事件的概率,考查离散型随机变量的期望,确定变量的取值,计算相应的概率是关键.20.(12分)设函数f(x)=ax+cosx,x∈[0,π].(Ⅰ)讨论f(x)的单调性;(Ⅱ)设f(x)≤1+sinx,求a的取值范围.【考点】6B:利用导数研究函数的单调性;6E:利用导数研究函数的最值.【专题】15:综合题.【分析】(Ⅰ)求导函数,可得f'(x)=a﹣sinx,x∈[0.π],sinx∈[0,1],对a进行分类讨论,即可确定函数的单调区间;(Ⅱ)由f(x)≤1+sinx得f(π)≤1,aπ﹣1≤1,可得a≤,构造函数g(x)=sinx﹣(0≤x),可得g(x)≥0(0≤x),再考虑:①0≤x;②,即可得到结论.【解答】解:(Ⅰ)求导函数,可得f'(x)=a﹣sinx,x∈[0,π],sinx∈[0,1];当a≤0时,f'(x)≤0恒成立,f(x)单调递减;当a≥1 时,f'(x)≥0恒成立,f(x)单调递增;当0<a<1时,由f'(x)=0得x1=arcsina,x2=π﹣arcsina当x∈[0,x1]时,sinx<a,f'(x)>0,f(x)单调递增当x∈[x1,x2]时,sinx>a,f'(x)<0,f(x)单调递减当x∈[x2,π]时,sinx<a,f'(x)>0,f(x)单调递增;(Ⅱ)由f(x)≤1+sinx得f(π)≤1,aπ﹣1≤1,∴a≤.令g(x)=sinx﹣(0≤x),则g′(x)=cosx﹣当x时,g′(x)>0,当时,g′(x)<0∵,∴g(x)≥0,即(0≤x),当a≤时,有①当0≤x时,,cosx≤1,所以f(x)≤1+sinx;②当时,=1+≤1+sinx综上,a≤.【点评】本题考查导数知识的运用,考查函数的单调性,考查函数的最值,解题的关键是正确求导,确定函数的单调性.21.(12分)已知抛物线C:y=(x+1)2与圆(r>0)有一个公共点A,且在A处两曲线的切线为同一直线l.(Ⅰ)求r;(Ⅱ)设m,n是异于l且与C及M都相切的两条直线,m,n的交点为D,求D到l的距离.【考点】IM:两条直线的交点坐标;IT:点到直线的距离公式;KJ:圆与圆锥曲线的综合.【专题】15:综合题;16:压轴题.【分析】(Ⅰ)设A(x0,(x0+1)2),根据y=(x+1)2,求出l的斜率,圆心M (1,),求得MA的斜率,利用l⊥MA建立方程,求得A的坐标,即可求得r的值;(Ⅱ)设(t,(t+1)2)为C上一点,则在该点处的切线方程为y﹣(t+1)2=2(t+1)(x﹣t),即y=2(t+1)x﹣t2+1,若该直线与圆M相切,则圆心M到该切线的距离为,建立方程,求得t的值,求出相应的切线方程,可得D 的坐标,从而可求D到l的距离.【解答】解:(Ⅰ)设A(x0,(x0+1)2),∵y=(x+1)2,y′=2(x+1)∴l的斜率为k=2(x0+1)当x0=1时,不合题意,所以x0≠1圆心M(1,),MA的斜率.∵l⊥MA,∴2(x0+1)×=﹣1∴x0=0,∴A(0,1),∴r=|MA|=;(Ⅱ)设(t,(t+1)2)为C上一点,则在该点处的切线方程为y﹣(t+1)2=2(t+1)(x﹣t),即y=2(t+1)x﹣t2+1若该直线与圆M相切,则圆心M到该切线的距离为∴∴t2(t2﹣4t﹣6)=0∴t0=0,或t1=2+,t2=2﹣抛物线C在点(t i,(t i+1)2)(i=0,1,2)处的切线分别为l,m,n,其方程分别为y=2x+1①,y=2(t1+1)x﹣②,y=2(t2+1)x﹣③②﹣③:x=代入②可得:y=﹣1∴D(2,﹣1),∴D到l的距离为【点评】本题考查圆与抛物线的综合,考查抛物线的切线方程,考查导数知识的运用,考查点到直线的距离公式的运用,关键是确定切线方程,求得交点坐标.22.(12分)函数f(x)=x2﹣2x﹣3,定义数列{ x n}如下:x1=2,x n+1是过两点P (4,5),Q n(x n,f(x n))的直线PQ n与x轴交点的横坐标.(Ⅰ)证明:2≤x n<x n+1<3;(Ⅱ)求数列{ x n}的通项公式.【考点】8H:数列递推式;8I:数列与函数的综合.【专题】15:综合题;16:压轴题.【分析】(Ⅰ)用数学归纳法证明:①n=1时,x1=2,直线PQ1的方程为,当y=0时,可得;②假设n=k时,结论成立,即2≤x k<x k+1<3,直线PQ k+1的方程为,当y=0时,可得,根据归纳假设2≤x k<x k+1<3,可以证明2≤x k+1<x k+2<3,从而结论成立.(Ⅱ)由(Ⅰ),可得,构造b n=x n﹣3,可得是以﹣为首项,5为公比的等比数列,由此可求数列{ x n}的通项公式.【解答】(Ⅰ)证明:①n=1时,x1=2,直线PQ1的方程为当y=0时,∴,∴2≤x1<x2<3;②假设n=k时,结论成立,即2≤x k<x k+1<3,直线PQ k+1的方程为当y=0时,∴∵2≤x k<x k+1<3,∴<x k+2∴x k+1<x k+2<3∴2≤x k+1即n=k+1时,结论成立由①②可知:2≤x n<x n+1<3;(Ⅱ)由(Ⅰ),可得设b n=x n﹣3,∴∴∴是以﹣为首项,5为公比的等比数列∴∴∴.【点评】本题考查数列的通项公式,考查数列与函数的综合,解题的关键是从函数入手,确定直线方程,求得交点坐标,再利用数列知识解决.。
2012年普通高等学校招生全国统一考试理科综合能力测试本试卷分第I卷(选择题)和第II卷(非选择题)两部分。
第I卷1至4页,第II卷5至11页。
考试结束后,将本试题卷和答题卡一并交回。
第I卷注意事项:1.答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名、准考证号填写清楚,并贴好条形码。
请认真核准条形码上的准考证号、姓名和科目。
2.每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其他答案标号,在试题卷上作答无效.........。
3.第I卷共21小题,每小题6分,共126分。
一下数据可供解题时参考:相对原子质量(原子量):H 1 C 12 N 14 O 16 Na 23 Cl 35.5 K 39 Fe 56 Cu64 Br 80 Ag 108一、选择题:本题共13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.下列关于膝跳反射的叙述,错误的是A.反射活动由一点的刺激引起B.反射活动中兴奋在突触处双向传递C.反射活动的发生需要反射弧结构完整D.发射活动中需要神经递质参与兴奋的传递2.下列关于叶绿体和线粒体的叙述,正确的是A.线粒体和叶绿体均含有少量的DNAB.叶绿体在光下和黑暗中均能合成ATPC.细胞生命活动所需的ATP均来自线粒体D.线粒体基质和叶绿体基质所含酶的种类相同。
3.一块农田中有豌豆、杂草、田鼠和土壤微生物等生物,其中属于竞争关系的是A.田鼠和杂草B.豌豆和杂草C.豌豆和其根中的根瘤菌D.细菌和其细胞内的噬菌体4.下列关于森林群落垂直结构的叙述,错误的是A.群落中的植物具有垂直分层现象B.群落中的动物具有垂直分层现象C.动物在群落中的垂直分层与植物的分层有关D.乔木层的疏密程度不会影响草木层的水平结构5、下列关于细菌的叙述,正确的是A 不同种类细菌的生长均需要相同碳源B 常用液体培养基分离获得细菌单菌落C 细菌大量培养过程中,芽孢形成于细菌生长的调整期D 培养基中含有高浓度NaCl 有利于金黄色葡萄球菌的筛选6 、下列关于化学键的叙述,正确的一项是A 粒子化合物中一定含有离子键B 单质分子中均不存在化学键C 含有极性键的分子一定是极性分子D 含有共价键的化合物一定是共价化合物7 、能正确表示下列反应的离子方程式是A 硫酸铝溶液中加入过量氨水 3Al ++3OH=Al(OH)3 ↓B 碳酸钠溶液中加入澄清石灰水 Ca(OH) 2 +23CO -=CaCO 3 ↓ + 2OH -C 冷的氢氧化钠溶液中通入氯气 Cl 2 + 2OH -=Cl O - + Cl -+ H 2OD 稀硫酸中加入铁粉 2Fe + 6 H += 23Fe + + 3H 2 ↑ 8 、合成氨所需的氢气可用煤和水作原料经多步反映值得,其中的一步反应为CO (g )+ H 2O(g) −−−→←−−−催化剂CO 2(g) + H 2(g) △H <0 反应达到平衡后,为提高CO 的转化率,下列措施中正确的是A 增加压强B 降低温度C 增大CO 的浓度D 更换催化剂9 、反应 A+B →C (△H <0)分两步进行 ① A+B →X (△H >0) ② X →C (△H <0)下列示意图中,能正确表示总反应过程中能量变化的是10 、 元素X 形成的离子与钙离子的核外电子排布相同,且X 的离子半径小于负二级硫的离子半径,X 元素为A AlB PC ArD K11、 ①②③④ 四种金属片两两相连浸入稀硫酸中都可组成原电池 ,①②相连时,外电路电流从②流向① ;①③相连时,③为正极,②④相连时,②有气泡逸出 ;③ ④ 相连时,③ 的质量减少 ,据此判断这四种金属活动性由大到小的顺序是A ①③②④B ①③④②C ③ ④ ②①D ③ ① ②④12.在常压和500℃条件下,等物质的量的A g2 ,F E (OH)3 ,NH 4HCO 3 ,N a HCO 3完全分解,所得气体体积依次是V 1\V 2\V 3\V 4.体积大小顺序正确的是A.V 3>V 2>V 4>V 1B. V 3>V 4>V 2>V 1C.V 3>V 2>V 1>V 4D.V 2>V 3>V 1>V 413.橙花醇具有玫瑰及苹果香气,可作为香料,其结构简式如下下列关于橙花醇的叙述,错误的是A . 既能发生取代反应,也能发生加成反应B.在浓硫酸催化下加热脱水,可以生成不止一种四烯烃C.1mo1橙花醇在氧气中充分燃烧,需消耗470.4氧化(标准状况D.1mo1橙花醇在室温下与溴四氯化碳溶液反应,最多消耗240g溴二,选择题:本题共8题。
2012年普通高等学校招生全国统一考试(新课标全国卷)文科综合历史部分试题24.汉武帝设置十三州刺史以监察地方,并将豪强大族“田宅逾制”作为重要的监察内容,各地财产达300万钱的豪族被迁到长安附近集中居住。
这表明:A.政权的政治与经济支柱是豪强大族B.政治权力与经济势力出现严重分离C.抑制豪强是缓解土地兼并的重要措施D.经济手段是巩固专制集权的主要方式【考点】古代中国土地制度【解析】关键信息:将豪强大族“田宅逾制”作为重要的监察内容。
“被迁到长安附近集中居住”的原因是“各地财产达300万钱”、“田宅逾制”,说明汉武帝的做法是要抑制豪强,其目的是缓解土地兼并,C项正确;“经济支柱”说法错误,A项排除;B项无从反映;题干的是政治手段,D项排除。
【答案】C25.许仙与白娘子自由相恋因法海和尚作梗终成悲剧,菩萨化身的济公游戏人间维持正义。
这些在宋代杭州流传的故事,反映出当时A.对僧人爱憎交加的社会心态B.民间思想借助戏剧广泛传播C.中国文化的地域性浓厚D.市民阶层的价值取向【考点】古代中国的科学技术与文学艺术:京剧等剧种的产生和发展【解析】题干中的故事具有世俗化的特征,这是因宋代商业经济的发展,市民阶层的兴起,D项正确;ABC项都无法从题干中反映。
【答案】D26.明后期松江人何良俊记述:“(正德)以前,百姓十一在官,十九在田……今去农而改业为工商者三倍于前矣。
昔日原无游手之人,今去农而游手趁食(谋生)者又十之二三也。
大抵以十分百姓言之,已六七分去农。
“据此可知A.工商业的发展造成农业的衰退B.工商业的发展导致了社会结构的变动C.财富分配不均引起贫富分化加剧D.无业游民增加促成了工商业的发展【考点】古代中国经济的发展【解析】关键信息:去农而改业为工商者三倍于前矣、今去农而游手趁食(谋生)者又十之二三也、已六七分去农。
从题干中反映了明后期松江人绝大部分弃农而从事工商业活动的经济现象,B项正确;ACD项题干无从反映。
2012 年国家公务员考试行政能力测试真题第三部分数量关系(共 15 题,参考时限 15 分钟)在这部分试题中,每道题呈现一段表述数字关系的文字,要求你迅速、准确地计算出答案。
请开始答题:66.有 300 名求职者参加高端人才专场招聘会,其中软件设计类、市场营销类、财务管理类和人力资源管理类分别有 100 、80 、70 和 50 人。
问至少有多少人找到工作,才能保证一定有 70 名找到工作的人专业相同? ()A. 71B. 119C. 258D. 27767. 甲乙二人协商共同投资,甲从乙处取了 15000 元,并以两人名义进行了 25000 元的投资,但由于决策失误,只收回 10000 元。
甲由于过失在己,愿意主动承担 2/3 的损失。
问收回的投资中,乙将分得多少钱? ()A. 10000 元B. 9000 元C. 6000 元D. 5000 元68. 某儿童艺术培训中心有 5 名钢琴教师和 6 名拉丁舞教师,培训中心将所有钢琴学员和拉丁舞学员共 76 名分别平均地分给各个老师带领,刚好能够分完,且每位老师所带的学生数量都是质数。
后来由于学生人数减少,培训中心只保留了 4 名钢琴教师和 3 名拉丁舞教师,但每名教师所带的学生数量不变,那么目前培训中心还剩下学员多少人?A. 36B. 37C. 39D. 4169.一只装有动力桨的船,其单靠人工划船顺流而下的速度是水速的 3 倍。
现该船靠人工划动从 A 地顺流到达 B 地,原路返回时只开足动力桨行驶,用时比来时少2/5 。
问船在静水中开足动力浆行驶的速度是人工划船速度的多少倍? ()A.2B.3C. 4D. 570.有 5 对夫妇参加一场婚宴,他们被安排在一张 10 个座位的圆桌就餐,但是婚礼操办者并不知道他们彼此之间的关系,只是随机安排座位。
问 5 对夫妇恰好都被安排在一起相邻而坐的概率是多少 ? ()A.在1‰到5‰之间B.在5‰到 1%之间C.超过 1%D.不超过1‰71. 2010 年某种货物的进口价格是 15 元/公斤, 2011 年该货物的进口量增加了一半,进口金额增加了 20%。
2012年普通高等学校招生全国统一考试(天津卷)数 学 (理工类)本试卷分为第Ⅰ卷(选择题)和第Ⅱ(非选择题)两部分,共150分,考试用时120分钟。
第Ⅰ卷1至2页,第Ⅱ卷3至5页。
答卷前,考生务必将自己的姓名、准考证号填写在答题卡上,并在规定位置粘贴考试用条形码。
答卷时,考生务必将答案涂写在答题卡上,答在试卷上的无效。
考试结束后,将本试卷和答题卡一并交回。
祝各位考生考试顺利!第Ⅰ卷注意事项:本卷共8小题,每小题5分,共40分.一、选择题: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)i 是虚数单位,复数ii+-37= (A ) 2 + i (B )2 – i (C )-2 + i (D )-2 – i【解析】复数i ii i i i i i -=-=+---=+-2101020)3)(3()3)(7(37,选B. 【答案】B(2)设,R ∈ϕ则“0=ϕ”是“))(cos()(R x x x f ∈+=ϕ为偶函数”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 )既不充分与不必要条件【解析】函数)cos()(ϕ+=x x f 若为偶函数,则有Z k k ∈=,πϕ,所以“0=ϕ”是“)cos()(ϕ+=x x f 为偶函数”的充分不必要条件,选A.【答案】A(3)阅读右边的程序框图,运行相应的程序,当输入x 的值为-25时,输出x 的值为(A )-1 (B )1 (C )3 (D )9【解析】第一次循环,415125=-=--=x ,第二次循环11214=-=-=x ,第三次循环不满足条件输出3112=+⨯=x ,选C.【答案】C(4)函数22)(3-+=x x f x在区间(0,1)内的零点个数是 (A )0 (B )1 (C )2 (D )3【解析】因为函数22)(3-+=x x f x的导数为032ln 2)('2≥+=x x f x,所以函数22)(3-+=x x f x 单调递增,又0121)0(<-=-=f ,01212)1(>=-+=f ,所以根据根的存在定理可知在区间)1,0(内函数的零点个数为1个,选B. 【答案】B(5)在52)12(xx -的二项展开式中,x 的系数为(A )10 (B )-10 (C )40 (D )-40【解析】二项展开式的通项为k k k k k k kk x C xx C T )1(2)1()2(310555251-=-=---+,令1310=-k ,解得3,93==k k ,所以x x C T 40)1(232354-=-=,所以x 的系数为40-,选D.【答案】D(6)在ABC ∆中,内角A ,B ,C 所对的边分别是c b a ,,,已知8b=5c ,C=2B ,则cosC=(A )257 (B )257- (C )257± (D )2524【解析】因为B C 2=,所以B B B C cos sin 2)2sin(sin ==,根据正弦定理有BbC c sin sin =,所以58sin sin ==B C b c ,所以545821sin 2sin cos =⨯==B C B 。
2012年中央、国家机关公务员录用考试《行政职业能力测验》试卷第一部分常识判断(共25题,参考时限15分钟)根据题目要求,在四个选项中选出一个最恰当的答案。
请开始答题:1.下列哪项是我国在“十二五”开局之年取得的科技成就()。
A. 我国首辆高速磁浮国产化样车交付使用,标志着我国已经具备了磁浮车辆国产化设计,整车集成和制造能力B. 我国具有自产自识产权的戊型肝炎疫苗研制成功,标志着我国在戊型肝炎疫苗研制上已经处于世界领先地位C. 我国第一台自行设计,自主集成研制的“蛟龙号”载人潜水器3000米级海试取得成功,标志着我国成为世界上掌握3500米以上大深度载人深潜技术的国家之一D. 国内首套具有自主知识产权的“机载SAR测图系统”研制成功,至此,我国可成功实现全天时、全天候从万米高空获取高分辨率测绘数据,及时动态监测地理国情2.下列关于推进“十二五”期间资源节约和环境保护的表述,不正确的是()。
A. 我国耕地资源有限,要加大耕地保护工作的力度B. 提高森林蓄积量和覆盖率是“十二五”期间的重要任务C. 提高化石能源消费的比重,以降低能耗总值和排放水平D. 坚持保护优先和自然修复为主,加大生态保护和建设力度3.在历史上中国共产党曾提出:①讲学心、讲政治、讲正气;②知识青年到农村去;③枪杆子里面出政权;④科学技术是第一生产力。
按时间先后顺序排列正确的是()。
A. ②③④①B. ③②④①C. ②①③④D. ③②①④4.下列关于我国民产党和无党派人士的说法,不正确的是()。
A. 工商联不属于民产党派B. 民产党派是参政党,不是在野党C. 无党派人士是指既不参加中国共产党也不参加民主党派的普通群众D. 中国共产党与民产党派将长期共存,互相监督,肝胆相照、荣辱与共5.关于欧洲主权债务危机的原因,下列说法不正确的是()。
A. 欧元升值B. 欧元区经济低迷C. 巨额财政赤字D. 财政政策与货币政策的不协调6.为了抑制通货膨胀,国家宏观调控部门可以采取的措施是()。
Text 1Come on –Everybody’s doing it. That whispered message, half invitation and half forcing, is what most of us think of when we hear the words peer pressure. It usually leads to no good-drinking, drugs and casual sex. But in her new book Join the Club, Tina Rosenberg contends that peer pressure can also be a positive force through what she calls the social cure, in which organizations and officials use the power of group dynamics to help individuals improve their lives and possibly the word.Rosenberg, the recipient of a Pulitzer Prize, offers a host of example of the social cure in action: In South Carolina, a state-sponsored antismoking program called Rage Against the Haze sets out to make cigarettes uncool. In South Africa, an HIV-prevention initiative known as LoveLife recruits young people to promote safe sex among their peers.The idea seems promising,and Rosenberg is a perceptive observer. Her critique of the lameness of many pubic-health campaigns is spot-on: they fail to mobilize peer pressure for healthy habits, and they demonstrate a seriously flawed understanding of psychology.‖ Dare to be differ ent, please don’t smoke!‖ pleads one billboard campaign aimed at reducing smoking among teenagers-teenagers, who desire nothing more than fitting in. Rosenberg argues convincingly that public-health advocates ought to take a page from advertisers, so skilled at applying peer pressure.But on the general effectiveness of the social cure, Rosenberg is less persuasive. Join the Club is filled with too much irrelevant detail and not enough exploration of the social and biological factors that make peer pressure so powerful. The most glaring flaw of the social cure as it’s presented here is that it doesn’t work very well for very long. Rage Against the Haze failed once state funding was cut. Evidence that the LoveLife program produces lasting changes is limited and mixed.There’s no doubt that our peer groups exert enormous influence on our behavior. An emerging body of research shows that positive health habits-as well as negative ones-spread through networks of friends via social communication. This is a subtle form of peer pressure: we unconsciously imitate the behavior we see every day.Far less certain, however, is how successfully experts and bureaucrats can select our peer groups and steer their activities in virtuous directions. It’s like t he teacher who breaks up the troublemakers in the back row by pairing them with better-behaved classmates. The tactic never really works. And that’s the problem with a social cure engineered from the outside: in the real world, as in school, we insist on choosing our own friends.21. According to the first paragraph, peer pressure often emerges as[A] a supplement to the social cure[B] a stimulus to group dynamics[C] an obstacle to school progress[D] a cause of undesirable behaviors22. Rosenberg holds that public advocates should[A] recruit professional advertisers[B] learn from advertisers’ experience[C] stay away from commercial advertisers[D] recognize the limitations of advertisements23. In the autho r’s view, Rosenberg’s book fails to[A] adequately probe social and biological factors[B] effectively evade the flaws of the social cure[C] illustrate the functions of state funding[D]produce a long-lasting social effect24. Paragraph 5shows that our imitation of behaviors[A] is harmful to our networks of friends[B] will mislead behavioral studies[C] occurs without our realizing it[D] can produce negative health habits25. The author suggests in the last paragraph that the effect of peer pressure is[A] harmful[B] desirable[C] profound[D] questionableText 2A deal is a deal-except, apparently ,when Entergy is involved. The company, a major energy supplier in New England, provoked justified outrage in Vermont last week when it announced it was reneging on a longstanding commitment to abide by the strict nuclearregulations.Instead, the company has done precisely what it had long promised it would not challenge the constitutio nality of Vermont’s rules in the federal court, as part of a desperate effort to keep its Vermont Yankee nuclear power plant run ning. It’s a stunningmove.The conflict has been surfacing since 2002, when the corporation bought Vermont’s only nuclear power plant, an aging reactor in Vernon. As a condition of receiving state approval for the sale, the company agreed to seek permission from state regulators to operate past 2012. In 2006, the state went a step further, requiring that any extension of the p lant’s license be subject to Vermont legislature’s approval. Then, too, the companywent along.Either Entergy never really intended to live by those commitments, or it simply didn’t foresee what would happen next. A string of accidents, including the partial collapse of a cooling tower in 207 and the discovery of an underground pipe system leakage, raised serious questions about both Vermont Yankee’s safety and Entergy’s management–especially after the company made misleading statements about the pipe. Enraged by Entergy’s behavior, the Vermont Senate voted 26 to 4 last year aga inst allowing anextension.Now the company is suddenly claiming that the 2002 agreement is invalid because of the 2006 legislation, and that only the federal government has regulatory power over nuclear issues. The legal issues in the case are obscure: whereas the Supreme Court has ruled that states do have some regulatory authority over nuclear power, legal scholars say that Vermont case will offer a precedent-setting test of how far those powers extend. Certainly, there are valid concerns about the patchwork regulations that could result if every state sets its own rules. But had Entergy kept its word, that debate would be besidethe point.The company seems to have concluded that its reputation in Vermont is already so damaged that it has noting left to lose by going to war with the state. But there should be consequences. Permission to run a nuclear plant is a poblic trust. Entergy runs 11 other reactors in the United States, including Pilgrim Nuclear station in Plymouth. Pledging to run Pilgrim safely, the company has applied for federal permission to keep it open for another 20 years. But as the Nuclear Regulatory Commission (NRC) reviews the company’s application, it should keep it mind what promises from Entergy are worth.26. The phrase ―reneging on‖(Line 3.para.1) is closest in me aning to[A] condemning. [B] reaffirming. [C] dishonoring. [D] securing.27. By entering into the 2002 agreement, Entergy intended to[A] obtain protection from Vermont regulators.[B] seek favor from the federal legislature.[C] acquire an extension of its business license .[D] get permission to purchase a power plant.28. According to Paragraph 4, Entergy seems to have problems with its[A] managerial practices.[B] technical innovativeness.[C] financial goals.[D] business vision29. In the author’s view, the Vermont case will test[A] Entergy’s capacity to fulfill all its promis es.[B] the mature of states’ patchwork regulations.[C] the federal authority over nuclear issues .[D] the limits of states’ power over nuclear issues.30. It can be inferred from the last paragraph that[A] Entergy’s business elsewhe re might be affected.[B] the authority of the NRC will be defied.[C] Entergy will withdraw its Plymouth application.[D] Vermont’s reputation might be damaged.Text 3In the idealized version of how science is done, facts about the world are waiting to be observed and collected by objective researchers who use the scientific method to carry out their work. But in the everyday practice of science, discovery frequently follows an ambiguous and complicated route. We aim to be objective, but we cannot escape the context of our unique life experience. Prior knowledge and interest influence what weexperience, what we think our experiences mean, and the subsequent actions we take.Opportunities for misinterpretation, error, and self-deception abound.Consequently, discovery claims should be thought of as protoscience. Similar to newly staked mining claims, they are full of potential. But it takes collective scrutiny and acceptance to transform a discovery claim into a mature discovery. This is the credibility process, through which the individual researcher’s me, here, now becomes the community’s anyone, anywhere, anytime. Objective knowledge is the go al, not the startingpoint.Once a discovery claim becomes public, the discoverer receives intellectual credit. But, unlike with mining claims, the community takes control of what happens next. Within the complex social structure of the scientific community, researchers make discoveries; editors and reviewers act as gatekeepers by controlling the publication process; other scientists use the new finding to suit their own purposes; and finally, the public (including other scientists) receives the new discovery and possibly accompanying technology. As a discovery claim works it through the community, the interaction and confrontation between shared and competing beliefs about the science and the technology involved transforms an individual’s discovery claim into the community’s credible discovery.Two paradoxes exist throughout this credibility process. First, scientific work tends to focus on some aspect of prevailing Knowledge that is viewed as incomplete or incorrect. Little reward accompanies duplication and confirmation of what is already known and believed. The goal is new-search, not re-search. Not surprisingly, newly published discovery claims and credible discoveries that appear to be important and convincing will always be open to challenge and potential modification or refutation by future researchers.Second, novelty itself frequently provokes disbelief. Nobel Laureate and physiologistAlbert Azent-Gyorgyi once described discovery as ―seeing what everybody has seen and thinking what nobody has thought.‖ But thinking what nobody else has thought and telling others what they have missed may not change their views. Sometimes years are required for truly novel discovery claims to be accepted and appreciated.In the end, credibility ―happens‖ to a discovery claim – a process that corresponds to what philosopher Annette Baier has described as the commons of the mind. ―We reason together, challenge, revise, and complete each other’s reasoning and each other’sconceptions of reason.‖31. According to the first paragraph, the process of discovery is characterized by its[A] uncertainty and complexity.[B] misconception and deceptiveness.[C] logicality and objectivity.[D] systematicness and regularity.32. It can be inferred from Paragraph 2 that credibility process requires[A] strict inspection.[B]shared efforts.[C] individual wisdom.[D]persistent innovation.33.Paragraph 3 shows that a discovery claim becomes credible after it[A] has attracted the attention of the general public.[B]has been examined by the scientific community.[C] has received recognition from editors and reviewers.[D]has been frequently quoted by peer scientists.34. Albert Szent-Gy?rgyi would most likely agree that[A] scientific claims will survive challenges.[B]discoveries today inspire future research.[C] efforts to make discoveries are justified.[D]scientific work calls for a critical mind.35.Which of the following would be the best title of the test?[A] Novelty as an Engine of Scientific Development.[B]Collective Scrutiny in Scientific Discovery.[C] Evolution of Credibility in Doing Science.[D]Challenge to Credibility at the Gate to Science.Text 4If the trade unionist Jimmy Hoffa were alive today, he would probably represent civil ser vant. When Hoffa’s Teamsters were in their prime in 1960, only one in ten American government workers belonged to a union; now 36% do. In 2009 the number of unionists in America’s public sector passed that of their fellow members in the private sector. In Britain, more than half of public-sector workers but only about 15% of private-sector ones areunionized.There are three reasons for the public-sector unions’ thriving. First, they can shut things down without suffering much in the way of consequences. Second, they are mostly bright and well-educated. A quarter of America’s public-sector workers have a university degree. Third, they now dominate left-of-centre politics. Some of their ties go back a long way. Britain’s Labor Party, as its name implies, has long been associated with trade unionism.Its current leader, Ed Miliband, owes his position to votes from public-sector unions.At the state level their influence can be even more fearsome. Mark Baldassare of the Public Policy Institute of Califor nia points out that much of the state’s budget is patrolled by unions. The teachers’ unions keep an eye on schools, the CCPOA on prisons and avariety of labor groups on health care.In many rich countries average wages in the state sector are higher than in the private one. But the real gains come in benefits and work practices. Politicians have repeatedly ―backloaded‖ public-sector pay deals, keeping the pay increases modest but adding to holidays and especially pensions that are already generous.Reform has been vigorously opposed, perhaps most egregiously in education, where charter schools, academies and merit pay all faced drawn-out battles. Even though there is plenty of evidence that the quality of the teachers is the most important variable, teachers’ unions have fought against getting rid of bad on es and promoting good ones. As the cost to everyone else has become clearer, politicians have begun to clamp down. In Wisconsin the unions have rallied thousands of supporters against Scott Walker, the hardline Republican governor. But many within the public sector suffer under the currentsystem, too.John Donahue at Harvard’s Kennedy School points out that the norms of culture in Western civil services suit those who want to stay put but is bad for high achievers. The only American public-sector workers who earn well above $250,000 a year are university sports coaches and the president of the United States. Bankers’ fat pay packets have attracted much criticism, but a public-sector system that does not reward high achieversmay be a much bigger problem for America.36. It can be learned from the first paragraph that[A] Teamsters still have a large body of members.[B] Jimmy Hoffa used to work as a civil servant.[C] unions have enlarged their public-sector membership.[D]the government has improved its relationship with unionists.37. Which of the following is true of Paragraph 2?[A] Public-sector unions are prudent in taking actions.[B] Education is required for public-sector union membership.[C] Labor Party has long been fighting against public-sector unions.[D]Public-sector unions seldom get in trouble for their actions.38. It can be learned from Paragraph 4 that the income in the state sector is[A] illegally secured.[B] indirectly augmented.[C] excessively increased.[D]fairly adjusted.39. The example of the unions in Wisconsin shows that unions[A]often run against the current political system.[B]can change people’s political attitudes.[C]may be a barrier to public-sector reforms.[D]are dominant in the government.40. John Donahue’s attitude towards the public-sector system is one of[A]disapproval.[B]appreciation.[C]tolerance.[D]indifference.Part BDirections:In the following text, some sentences have been removed. For Questions 41-45, choose the most suitable one from the list A-G to fit into each of the numbered blanks. There are two extra choices, which do not fit in any of the blanks. Mark your answers on ANSWER SHEET1.(10 points)Think of those fleeting moments when you look out of an aeroplane window and realisethat you are flying, higher than a bird. Now think of your laptop, thinner than a brown-paper envelope, or your cellphone in the palm of your hand. Take a moment or two to wonder at those marvels. You are the lucky inheritor of a dream come true.The second half of the 20th century saw a collection of geniuses, warriors, entrepreneurs and visionaries labour to create a fabulous machine that could function as a typewriter and printing press, studio and theatre, paintbrush and gallery, piano and radio,the mail as well as the mail carrier. (41The networked computer is an amazing device, the first media machine that serves as the mode of production, means of distribution, site of reception, and place of praise and critique. The computer is the 21st century's culture machine.But for all the reasons there are to celebrate the computer, we must also tread with caution. (42)I call it a secret war for two reasons. First, most people do not realise that there are strong commercial agendas at work to keep them in passive consumption mode. Second, the majority of people who use networked computers to upload are not evenaware of the significance of what they are doing.All animals download, but only a few upload. Beavers build dams and birds make nests. Yet for the most part, the animal kingdom moves through the world downloading. Humans are unique in their capacity to not only make tools but then turn around and use them to create superfluous material goods - paintings, sculpture and architecture - and superfluous experiences - music, literature, religion and philosophy. (43) For all the possibilities of our new culture machines, most people are still stuck in download mode. Even after the advent of widespread social media, a pyramid of production remains, with a small number of people uploading material, a slightly larger group commenting on or modifying that content, and a huge percentage remainingcontent to just consume. (44Television is a one-way tap flowing into our homes. The hardest task that television asks of anyone is to turn the power off after he has turned it on.(45What counts as meaningful uploading? My definition revolves around the concept of "stickiness" - creations and experiences to which others adhere.[A] Of course, it is precisely these superfluous things that define human culture and ultimately what it is to be human. Downloading and consuming culture requires great skills, but failing to move beyond downloading is to strip oneself of a defining constituent ofhumanity.[B] Applications like , which allow users to combine pictures, words and other media in creative ways and then share them, have the potential to add stickiness by amusing, entertaining and enlightening others.[C] Not only did they develop such a device but by the turn of the millennium they had also managed to embed it in a worldwide system accessed by billions of people every day.[D] This is because the networked computer has sparked a secret war between downloading and uploading - between passive consumption and active creation - whose outcome will shape our collective future in ways we can only begin to imagine.[E] The challenge the computer mounts to television thus bears little similarity to one format being replaced by another in the manner of record players being replaced by CDplayers.[F] One reason for the persistence of this pyramid of production is that for the past half-century, much of the world's media culture has been defined by a single medium - television - and television is defined by downloading.[G]The networked computer offers the first chance in 50 years to reverse the flow, to encourage thoughtful downloading and, even more importantly, meaningful uploading.Part CDirections:Read the following text carefully and then translate the underlined segments into Chinese. Your translation should be written clearly on ANSWER SHEET 2. (10points)Since the days of Aristotle, a search for universal principles has characterized the scientific enterprise. In some ways, this quest for commonalities defines science. Newton’s laws of motion and Darwinian evolution each bind a host of di fferent phenomenainto a single explicatory frame work.(46)In physics, one approach takes this impulse for unification to its extreme, and seeksa theory of everything—a single generative equation for all we see.It is becoming less clear, however, that such a theory would be a simplification, given the dimensions and universes that it might entail, nonetheless, unification of sorts remains a major goal.This tendency in the natural sciences has long been evident in the social sciences too.(47)Here, Darwinism seems to offer justification for it all humans share common origins itseems reasonable to suppose that cultural diversity could also be traced to more constrained beginnings. Just as the bewildering variety of human courtship rituals might all be considered forms of sexual selection, perhaps the world’s languages, music, social and religious customs and even history are governed by universal features. (48)To filter out what is unique from what is shared might enable us to understand how complex cultural behavior arose and what guides it in evolutionary or cognitive terms.That, at least, is the hope. But a comparative study of linguistic traits published online today supplies a reality check. Russell Gray at the University of Auckland and his colleagues consider the evolution of grammars in the light of two previous attempts to finduniversality in language.The most famous of these efforts was initiated by Noam Chomsky, who suggested that humans are born with an innate language—acquisition capacity that dictates a universal grammar. A few generative rules are then sufficient to unfold the entire fundamental structure of a language, which is why children can learn it so quickly.(49)The second, by Joshua Greenberg, takes a more empirical approach to universalityidentifying traits (particularly in word order) shared by many language which are considered to represent biases that result from cognitive constraintsGray and his colleagues have put them to the test by examining four family trees that between them represent more than 2,000 languages.(50)Chomsky’s grammar should show patterns of language change that are independent of the family tree or the pathway tracked through it. Whereas Greenbergian universality predicts strong co-dependencies between particular types of word-order relations. Neither of these patterns is borne out by the analysis, suggesting that the structures of the languages are lire age-specific and notgoverned by universals答案Section ⅠUse of English1-5 BABDC 6-10 BDBAB 11-15 ACCDA 16-20 CACDDSection ⅡReading ComprehensionPart A21.D 22.B 23.A 24.C 25.D 26.C 27.D 28.A 29.D 30.A31.A 32.B 33.B 34.B 35.C 36.C 37.D 38.B 39.C 40.APart B41.C 42.D 43.A 44.F 45.GPart C46.在物理学领域,有一种方法将这种―万物归一的推动力‖推向了极致,它试图探寻到能解释一切的,最底层的公式。