历年考研真题解析

徐之明权威解析2013年考研政治试题马原客观题答案解析一、单项选择题马原部分1.有一副对联，上联是“桔子洲，洲旁舟，舟行洲不行”，下联是“天心阁，阁中鸽，鸽飞阁不飞”。

这形象地说明了运动和静止是相互联系的。

静止是（）A.运动的普遍状态B.运动的内在原因C.运动的衡量尺度D.运动的存在方式【答案】C【思路剖析】本题属于案例型考题，即通过一个具体的事例，去探究其中所包含的哲理。

题干所引对联的作者，是毛泽东和周恩来。

1960年5月，毛泽东、周恩来一行视察长沙，工作之余，到江边散步。

遥望橘子洲头，百舸争流，万帆竞发，毛泽东逸兴遄飞，口占一上联：“橘子洲，洲旁舟，舟行洲不行”，此联动静相对，意境悠远，三个断句，两处“顶针”，“洲”和“舟”又是谐音，应对难度极大。

毛泽东对身边的周恩来说：“恩来，我一时江郎才尽，请你来个锦上添花如何？”周恩来才思敏捷，了解长沙，应声对道：“天心阁，阁中鸽，鸽飞阁不飞。

”天心阁系长沙市内一景，与橘子洲相对。

整个对联工整流畅、浑然一体，留下一段佳句和佳话。

从考试答题的角度论，最关键的是抓住题干中的最后一句：“这形象地说明了运动和静止是相互联系的。

静止是（）。

“.运动的普遍状态”是胡编的干扰项，学术上没有这个说法。

故A.项错误。

“.运动的内在原因”是矛盾，故B项为错误。

物质的存在方式是运动；没有“运动的存在方式”这个说法。

C. 项“静止是运动的衡量尺度”是正确的说法。

该说法的出处是马克思、恩格斯的话：“运动应当从它的反面即从静止找到它的量度”。

所包含的深层哲理是在对立中认识同一。

【必背考点】《马原》第2章之“运动和静止”。

【应试对策】本题的难度，主要不是来自考查的内容，而是来自于干扰项的设定。

“运动”、“普遍”、“内在原因”、“存在方式”等，都是我们学习哲学中经常碰到的基本概念。

如果没有理解作为支撑，并且采取临时抱佛脚的突击记忆方式，比较难做出正确的决断。

2.一位机械工程专家讲过这样一件事：“文革”中，他在某地劳动，有一天公社派他去割羊草。

考研（数学一）历年真题试卷汇编1(题后含答案及解析)题型有：1. 选择题 2. 填空题选择题下列每题给出的四个选项中，只有一个选项符合题目要求。

1．(2011)已知当x→0时，函数f(x)=3sin．x=sin 3x与cxk是等价无穷小，则( )A．k=1，c=4．B．k=1，c=4．C．k=3，c=4．D．k=3，c=-4．正确答案：C解析：因为当x→0时，函数f(x)=3sin x=sin 3x与cxk是等价无穷小，所以从而k-1=2，即k=3，于是故应选C．2．(2012)设函数f(x)=(ex-1)(e2x-2).….(enx-n)，其中n为正整数，则f’(0)=( ) A．(-1)n-1(n-1)!．B．(-1)n(n-1)!．C．(-1)n-1n!．D．(-1)nn!．正确答案：A解析：利用导数的定义求f’(0)．故应选A．3．(2012)曲线的渐近线的条数为( )A．0．B．1．C．2．D．3．正确答案：C解析：应同时考虑水平渐近线、铅直渐近线与斜渐近线．因为所以y=1是曲线的水平渐近线，同时说明曲线无斜渐近线．又因为所以x=1是曲线的铅直渐近线，x=-1不是曲线的铅直渐近线．综上所述，应选C．4．(2009)设A，B均为2阶矩阵，A*，B*分别为A，B的伴随矩阵．若｜A｜=2，｜B｜=3，则分块矩阵的伴随矩阵为( )A．B．C．D．正确答案：B解析：本题主要考查分块矩阵的行列式、伴随矩阵的相关公式以及分块矩阵的逆矩阵．由=(-1)2×2｜A｜｜B｜=6知，矩阵可逆，从而故应选B．5．(2006)设A、B为两个随机事件，且P(B)＞0，P(A｜B)=1，则必有( ) A．P(A∪B)＞P(A)．B．P(A∪B)＞P(B)．C．P(A∪B)=P(A)．D．P(A∪B)=-P(B)．正确答案：C解析：本题主要考查乘法公式与加法公式．由已知条件与乘法公式有P(AB)=P(B)P(A｜B)=P(B)，再由加法公式有P(A∪B)=P(A)+P(B)-P(AB)=P(A)．故应选C．6．(2003)设函数f(x)在(-∞，+∞)内连续，其导函数的图形如图1所示，则f(x)有( )A．一个极小值点和两个极大值点．B．两个极小值点和一个极大值点．C．两个极小值点和两个极大值点．D．三个极小值点和一个极大值点．正确答案：C解析：本题主要考查导函数y=f’(x)与函数y=f(x)的图形的关系与一元函数的极值(点)．由于已知函数是抽象函数，无法用推理法及反例排除法解决．考虑用y=f’(x)与y=f(x)的图形之间的关系画出y=f(x)的图形，利用定性分析的方法解决该问题．根据y=f’(x)的图形画出y=f(x)的图形，如图2所示，根据y=f(x)的图形知，f(x)有两个极小值点和两个极大值点．故应选C．7．(2011)函数f(x)=ln｜(x-1)(x-2)(x-3)｜的驻点个数为( )A．0．B．1．C．2．D．3．正确答案：C解析：因为，所以x=1，x=2，x=3是曲线y=f(x)的铅直渐近线．又，由此可画出f(x)=ln｜(x-1)(x-2)(x-3)｜的草图，如图3所示，由图形可知，存在两点x1，x2，使得f’(x1)=f’(x2)=0，即f(x)有两个驻点．故应选C．8．(2006)设函数y=f(x)具有二阶导数，且f’(x)＞0，f’’(x)＞0，△x为自变量x在点x0处的增量，△y与dy分别为f(x)在点x0处对应的增量与微分，若△x＞0，则( )A．0＜dy＜△y．B．0＜△y＜dy．C．△y＜dy＜0．D．dy＜△y＜0．正确答案：A解析：△y=f(x0+△x)-(x0)=f’(ξ)△x (x0＜ξ＜x0+△x)．因为f’’(x)＞0，所以f’(x)单调增加，从而f’(ξ)＞f’(x0)，于是△y=f’(ξ)△x＞f’(x0)△x=dy．又因为f’(x)＞0，所以0＜dy＜△y．故应选A．9．(1999)设两个相互独立的随机变量X和Y分别服从正态分布N(0，1)和N(1，1)，则( )A．P{X+Y≤0}=B．P{X+Y≤1}=C．P{X-Y≤0}=D．P{X-Y≤1}=正确答案：B解析：由于均服从正态分布且相互独立的随机变量的线性组合仍然服从正态分布，所以由正态分布的几何意义知，正态分布的密度函数关于均值左右对称，于是其小于均值的概率为，从而P{X+Y≤1}=故应选B．10．(2002)设函数y=f(x)在(0，+∞)内有界且可导，则( )A．B．C．D．正确答案：B解析：取，因为排除A、C、D．故应选B．11．(2005)以下四个命题中，正确的是( )A．若f’(x)在(0，1)内连续，则f(x)在(0，1)内有界．B．若f(x)在(0，1)内连续，则f(x)在(0，1)内有界．C．若f’(x)在(0，1)内有界，则f(x)在(0，1)内有界．D．若f(x)在(0，1)内有界，则f’(x)在(0，1)内有界．正确答案：C解析：取f’(x)=，在(0，1)内连续，但f(x)=lnx在(0，1)内无界，排除A．取f(x)=，在(0，1)内连续，但f(x)在(0，1)内无界，排除B．取f(x)=，在(0，1)内有界，但f’(x)=在(0，1)内无界，排除D．故应选C．12．(2004)设f’(x)在[a，b]上连续，且f’(a)＞0，f’(b)＜0，则下列结论中错误的是( )A．至少存在一点x0∈(a，b)，使f(x0)＞f(a)．B．至少存在一点x0∈(a，b)，使f(x0)＞f(b)．C．至少存在一点x0∈(a，b)，使f’(x0)=0．D．至少存在一点x0∈(a，b)，使f(x0)=0．正确答案：D解析：取f(x)=2-x2，x∈[-1，1]，则f’(x)=-2x在[a，b]=[-1，1]上连续，且f’(a)=f’(-1)=2＞0，f’(b)=f’(1)=-2＜0，满足已知条件．由f(x)=2-x2的图形可知，在(-1，1)内，f(x)＞1，即对任意x0∈(-1，1)，都有f(x0)≠0，这表明D选项是错误的．故应选D．13．(2001)设f(x)的导数在x=a处连续，又，则( )A．x=a是f(x)的极小值点．B．x=a是f(x)的极大值点．C．(a，f(a))是曲线y=f(x)的拐点．D．x=a不是f(x)的极值点，(a，f(a))也不是曲线y=f(x)的拐点．正确答案：B解析：由f(x)的导数在x=a处连续及=f’(a)=0，即x=a是f(x)的驻点．从而所以x=a是f(x)的极大值点．故应选B．14．(2003)设f(x)为不恒等于零的奇函数，且f’(0)存在，则函数g(x)=( ) A．在x=0处左极限不存在．B．有跳跃间断点x=0．C．在x=0处右极限不存在．D．有可去间断点x=0．正确答案：D解析：因为f(x)为不恒等于零的奇函数，所以f(0)=0，又f’(0)存在．所以故x=0是g(x)的可去间断点．应选D．15．(2005)设函数u(x，y)=φ(x+y)+φ(x+y)+其中函数φ具有二阶导数，ψ具有一阶导数，则必有( )A．B．C．D．正确答案：B解析：取φ(x)=x2，ψ(x)=0，则u(x，y)=(x+y)2+(x-y)2=2x2+2y2．于是由此可知，选项A、C、D都不正确．故应选B．16．(2005)设an＞0，n=1，2，…，若收敛，则下列结论正确的是( ) A．B．C．D．正确答案：D解析：取收敛，但发散，排除A；发散，排除B；发散，排除C．故应选D．17．(2002)设A是m×n矩阵，B是n×m矩阵，则线性方程组(AB)X=0( ) A．当n＞m时仅有零解．B．当n＞m时必有非零解．C．当m＞n时仅有零解．D．当m＞n时必有非零解．正确答案：D解析：(推理法)因为当n＜m时，齐次线性方程组BX=0有非零解，从而线性方程组(AB)X=0有非零解，故应选D．18．(2002)设向量组α1，α2，α3线性无关，向量β1可由α1，α2，α3线性表示，而向量β2不能由α1，α2，α3线性表示，则对任意常数k，必有( )A．α1，α2，α3，kβ1+β2线性无关．B．α1，α2，α3，kβ1+β2线性相关．C．α1，α2，α3，β1+kβ2线性无关．D．α1，α2，α3，β1+kβ2线性相关．正确答案：A解析：因为β2不能由α1，α2，α3线性表示，则α1，α2，α3，β2线性无关．取k=0，由B知，α1，α2，α3，β2线性相关，与α1，α2，α3，β2线性无关矛盾，排除B．取k=0，由C知，α1，α2，α3，β1线性无关，则β1不能由α1，α2，α3线性表示，与已知条件矛盾，排除C．取k=1，由D知，α1，α2，α3．β1+β2线性相关，因为α1，α2，α3线性无关，所以β1+β2可由α1，α2，α3线性表示，而β1可由α1，α2，α3线性表示，于是β2可由α1，α2，α3线性表示，与已知条件矛盾，排除D．故应选A．填空题19．(2000)=_____，正确答案：解析：由定积分的几何意义，表示由直线x=0，x=1，y=0与曲线y=所围成的图形的面积，如图5所示，所以(其中S为单位圆(x-1)2+y2≤1的面积)．20．(2001)(x3+sin2x)cos2xdx=_______．正确答案：解析：21．(2012)设区域D是由曲线y=sinx，x=，y=1围成，则(x5y-1)dxdy=_______.正确答案：-π解析：22．(2008)设D={(x，y)｜x2+y2≤1}，则(x2-y)dxdy=______．正确答案：解析：因为积分区域D关于x轴对称，函数y关于y是奇函数，所以．由轮换对称性以及极坐标下二重积分的计算方法，有23．(2009)设Ω={(x，y，z)｜x2+y2+z2≤1}，则z2dxdydz=_______.正确答案：解析：利用轮换对称性，有再利用球坐标下三重积分的计算有24．(2007)设曲面∑：｜x｜+｜y｜+｜z｜=1，则=_______.正确答案：解析：因为∑关于yOz平面对称，x关于x为奇函数，所以．由轮换对称性，其中S是∑的表面积，记∑在第一卦限部分的面积为S1．如图8所示，则。

研究生入学考试2000到2013年最新最全数学三考试试题2000年全国硕士研究生入学统一考试数学三试题一、填空题二、选择题2001年全国硕士研究生入学统一考试数学三试题一、填空题二、选择题2002年全国硕士研究生入学统一考试数学三试题一、填空题二、选择题2003年考研数学（三）真题一、 填空题（本题共6小题，每小题4分，满分24分. 把答案填在题中横线上）（1）设,0,0,0,1cos )(=≠⎪⎩⎪⎨⎧=x x xx x f 若若λ其导函数在x=0处连续，则λ的取值范围是_____. （2）已知曲线b x a x y +-=233与x 轴相切，则2b 可以通过a 表示为=2b ________.（3）设a>0，，x a x g x f 其他若,10,0,)()(≤≤⎩⎨⎧==而D 表示全平面，则⎰⎰-=Ddxdy x y g x f I )()(=_______.（4）设n 维向量0,),0,,0,(<=a a a TΛα；E 为n 阶单位矩阵，矩阵 TE A αα-=， T aE B αα1+=， 其中A 的逆矩阵为B ，则a=______.（5）设随机变量X 和Y 的相关系数为0.9, 若4.0-=X Z ，则Y 与Z 的相关系数为________.（6）设总体X 服从参数为2的指数分布，n X X X ,,,21Λ为来自总体X 的简单随机样本，则当∞→n 时，∑==n i i n X n Y 121依概率收敛于______.二、选择题（本题共6小题，每小题4分，满分24分. 每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内）（1）设f(x)为不恒等于零的奇函数，且)0(f '存在，则函数xx f x g )()(=(A) 在x=0处左极限不存在. (B) 有跳跃间断点x=0.(C) 在x=0处右极限不存在. (D) 有可去间断点x=0. [ ] （2）设可微函数f(x,y)在点),(00y x 取得极小值，则下列结论正确的是(A) ),(0y x f 在0y y =处的导数等于零. （B ）),(0y x f 在0y y =处的导数大于零. (C) ),(0y x f 在0y y =处的导数小于零. (D) ),(0y x f 在0y y =处的导数不存在. [ ] （3）设2nn n a a p +=，2nn n a a q -=，Λ,2,1=n ，则下列命题正确的是(A) 若∑∞=1n na条件收敛，则∑∞=1n np与∑∞=1n nq都收敛.(B) 若∑∞=1n na绝对收敛，则∑∞=1n np与∑∞=1n nq都收敛.(C) 若∑∞=1n na条件收敛，则∑∞=1n np与∑∞=1n nq敛散性都不定.(D) 若∑∞=1n na绝对收敛，则∑∞=1n np与∑∞=1n nq敛散性都不定. [ ]（4）设三阶矩阵⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=a b b b a b b b a A ，若A 的伴随矩阵的秩为1，则必有 (A) a=b 或a+2b=0. (B) a=b 或a+2b ≠0.(C) a ≠b 且a+2b=0. (D) a ≠b 且a+2b ≠0. [ ] （5）设s ααα,,,21Λ均为n 维向量，下列结论不正确的是(A) 若对于任意一组不全为零的数s k k k ,,,21Λ，都有02211≠+++s s k k k αααΛ，则s ααα,,,21Λ线性无关.(B) 若s ααα,,,21Λ线性相关，则对于任意一组不全为零的数s k k k ,,,21Λ，都有.02211=+++s s k k k αααΛ(C) s ααα,,,21Λ线性无关的充分必要条件是此向量组的秩为s.(D) s ααα,,,21Λ线性无关的必要条件是其中任意两个向量线性无关. [ ] （6）将一枚硬币独立地掷两次，引进事件：1A ={掷第一次出现正面}，2A ={掷第二次出现正面}，3A ={正、反面各出现一次}，4A ={正面出现两次}，则事件(A) 321,,A A A 相互独立. (B) 432,,A A A 相互独立.(C) 321,,A A A 两两独立. (D) 432,,A A A 两两独立. [ ] 三、（本题满分8分） 设).1,21[,)1(1sin 11)(∈--+=x x x x x f πππ 试补充定义f(1)使得f(x)在]1,21[上连续.四 、（本题满分8分）设f(u,v)具有二阶连续偏导数，且满足12222=∂∂+∂∂v f u f ，又)](21,[),(22y x xy f y x g -=,求.2222ygx g ∂∂+∂∂ 五、（本题满分8分） 计算二重积分 .)sin(22)(22dxdy y x e I Dy x +=⎰⎰-+-π其中积分区域D=}.),{(22π≤+y x y x六、（本题满分9分）求幂级数∑∞=<-+12)1(2)1(1n nnx n x 的和函数f(x)及其极值.七、（本题满分9分）设F(x)=f(x)g(x), 其中函数f(x),g(x)在),(+∞-∞内满足以下条件： )()(x g x f ='，)()(x f x g ='，且f(0)=0, .2)()(xe x g xf =+(1) 求F(x)所满足的一阶微分方程； (2) 求出F(x)的表达式. 八、（本题满分8分）设函数f(x)在[0，3]上连续，在（0，3）内可导，且f(0)+f(1)+f(2)=3, f(3)=1.试证必存在)3,0(∈ξ，使.0)(='ξf九、（本题满分13分） 已知齐次线性方程组⎪⎪⎪⎩⎪⎪⎪⎨⎧=+++++=+++++=+++++=+++++,0)(,0)(,0)(,0)(332211332211332211332211nn nn n n n n x b a x a x a x a x a x b a x a x a x a x a x b a x a x a x a x a x b a ΛΛΛΛΛΛΛΛΛΛΛΛΛΛ 其中.01≠∑=ni ia试讨论n a a a ,,,21Λ和b 满足何种关系时，(1) 方程组仅有零解；(2) 方程组有非零解. 在有非零解时，求此方程组的一个基础解系. 十、（本题满分13分） 设二次型)0(222),,(31232221321>+-+==b x bx x x ax AX X x x x f T ，中二次型的矩阵A 的特征值之和为1，特征值之积为-12. (1) 求a,b 的值；(2) 利用正交变换将二次型f 化为标准形，并写出所用的正交变换和对应的正交矩阵.十一、（本题满分13分） 设随机变量X 的概率密度为;],8,1[,0,31)(32其他若∈⎪⎩⎪⎨⎧=x x x fF(x)是X 的分布函数. 求随机变量Y=F(X)的分布函数.十二、（本题满分13分）设随机变量X 与Y 独立，其中X 的概率分布为⎪⎪⎭⎫⎝⎛7.03.021~X ，而Y 的概率密度为f(y)，求随机变量U=X+Y 的概率密度g(u).2004年全国硕士研究生入学统一考试数学三试题一、填空题：本题共6小题，每小题4分，满分24分. 请将答案写在答题纸指定位置上.(1) 若()0sin limcos 5x x xx b e a→-=-，则a =______，b =______.(2) 函数(),f u v 由关系式()(),f xg y y x g y =+⎡⎤⎣⎦确定，其中函数()g y 可微，且()0g y ≠，则2fu v∂=∂∂______. (3) 设()211,,2211,,2x xe x f x x ⎧-≤<⎪⎪=⎨⎪-≥⎪⎩ 则()2121f x dx -=⎰_____.(4) 二次型()()()()222123122331,,f x x x x x x x x x =++-++的秩为______. (5) 设随机变量X 服从参数为λ的指数分布，则{P X >=______.(6) 设总体X 服从正态分布()21,N μσ，总体Y 服从正态分布()22,N μσ，112,,,n X X X L 和212,,,n Y Y Y L 分别是来自总体X 和Y 的简单随机样本，则()()122211122n n i j i j X X Y Y E n n ==⎡⎤-+-⎢⎥⎢⎥=⎢⎥+-⎢⎥⎣⎦∑∑______. 二、选择题：本题共8小题，每小题4分，满分24分. 在每小题给出的四个选项中，只有一项符合题目要求，请把所选项前的字母填在答题纸指定位置上.(7) 函数()()()()2sin 212x x f x x x x -=--在下列哪个区间内有界.（A ）()1,0- （B ）()0,1 （C ）()1,2 （D ）()2,3(8) 设()f x 在(),-∞+∞内有定义，且()lim x f x a →∞=，()1,0,0,0,fx g x x x ⎧⎛⎫≠⎪ ⎪=⎝⎭⎨⎪=⎩则（A ）0x =必是()g x 的第一类间断点 （B ）0x =必是()g x 的第二类间断点 （C ）0x =必是()g x 的连续点 （D ）()g x 在点0x =处的连续性与a 的值有关.(9) 设()()1f x x x =-，则（A ）0x =是()f x 的极值点，但()0,0不是曲线()y f x =的拐点 （B ）0x =不是()f x 的极值点，但()0,0是曲线()y f x =的拐点 （C ）0x =是()f x 的极值点，且()0,0是曲线()y f x =的拐点 （D ）0x =不是()f x 的极值点，()0,0也不是曲线()y f x =的拐点 (10) 设有以下命题： ① 若()2121n n n uu ∞-=+∑收敛，则1n n u ∞=∑收敛② 若1nn u∞=∑收敛，则10001n n u∞+=∑收敛③ 若1lim1n n nu u +→∞>，则1n n u ∞=∑发散 ④ 若()1nn n uv ∞=+∑收敛，则1n n a ∞=∑，1n n v ∞=∑都收敛则以上命题中正确的是（A ）①② （B ）②③ （C ）③④ （D ）①④(11) 设()f x '在[],a b 上连续，且()()0,0f a f b ''><，则下列结论中错误的是 （A ）至少存在一点()0,x a b ∈，使得()()0f x f a > （B ）至少存在一点()0,x a b ∈，使得()()0f x f b > （C ）至少存在一点()0,x a b ∈，使得()00f x '= （D ）至少存在一点()0,x a b ∈，使得()00f x = (12) 设n 阶矩阵A 与B 等价，则必有（A ）当()0A a a =≠时，B a = （B ）当()0A a a =≠时，B a =- （C ）当0A ≠时，0B = （D ）当0A =时，0B =(13) 设n 阶矩阵A 的伴随矩阵*0A ≠，若1234,,,ξξξξ是非齐次线性方程组Ax b =的互不相等的解，则对应的齐次线性方程组0Ax =的基础解系（A ）不存在 （B ）仅含一个非零解向量 （C ）含有两个线性无关的解向量 （D ）含有三个线性无关的解向量(14) 设随机变量X 服从正态分布()0,1N ，对给定的()0,1α∈，数n u 满足{}P X u αα>=，若{}P X x α<=，则x 等于（A ）2u α （B ）12uα-（C ）12u α- （D ）1u α-三、解答题：本题共9小题，满分94分. 请将解答写在答题纸指定的位置上. 解答应写出文字说明、证明过程或演算步骤.（15）（本题满分8分）求22201cos lim sin x x x x →⎛⎫-⎪⎝⎭.（16）（本题满分8分）求)Dy d σ⎰⎰，其中D 是由圆224x y +=和()2211x y ++=所围成的平面区域（如图）.（17）（本题满分8分）设()(),f x g x 在[],a b 上连续，且满足()()xxa a f t dt g t dt ≥⎰⎰，[),x ab ∈，()()bb aaf t dtg t dt =⎰⎰证明：()()bbaaxf x dx xg x dx ≤⎰⎰.（18）（本题满分9分）设某商品的需求函数为1005Q P =-，其中价格()0,20P ∈，Q 为需求量. （Ⅰ）求需求量对价格的弹性()0d d E E >；（Ⅱ）推导()1d dRQ E dP=-（其中R 为收益），并用弹性d E 说明价格在何范围内变化时，降低价格反而使收益增加.（19）（本题满分9分）设级数()468242462468x x x x +++-∞<<+∞⋅⋅⋅⋅⋅⋅L 的和函数为()S x .求： （Ⅰ）()S x 所满足的一阶微分方程； （Ⅱ）()S x 的表达式.（20）（本题满分13分）设()()()1231,2,0,1,2,3,1,2,2TTTa ab a b ααα==+-=---+，()1,3,3Tβ=-. 试讨论当,a b 为何值时，（Ⅰ）β不能由123,,ααα线性表示；（Ⅱ）β可由123,,ααα唯一地线性表示，并求出表示式；（Ⅲ）β可由123,,ααα线性表示，但表示式不唯一，并求出表示式.（21）（本题满分13分）设n 阶矩阵111b b bb A bb ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦L L M M M L. （Ⅰ）求A 的特征值和特征向量；（Ⅱ）求可逆矩阵P ，使得1P AP -为对角矩阵.（22）（本题满分13分）设,A B 为两个随机事件，且()()()111,,432P A P B A P A B ===，令 1,0,.A X A ⎧=⎨⎩发生，不发生 1,0,.B Y B ⎧=⎨⎩发生，不发生求：（Ⅰ）二维随机变量(),X Y 的概率分布； （Ⅱ）X 与Y 的相关系数XY ρ； （Ⅲ）22Z X Y =+的概率分布.（23）（本题满分13分） 设随机变量X 的分布函数为()1,,;,0,.x F x x x βαααβα⎧⎛⎫->⎪ ⎪=⎨⎝⎭⎪≤⎩其中参数0,1αβ>>. 设12,,,n X X X L 为来自总体X 的简单随机样本. （Ⅰ）当1α=时，求未知参数β的矩估计量； （Ⅱ）当1α=时，求未知参数β的最大似然估计量； （Ⅲ）当2β=时，求未知参数α的最大似然估计量.2005年全国硕士研究生入学统一考试数学三试题一、填空题：本题共6小题，每小题4分，满分24分. 请将答案写在答题纸指定位置上.(1) 极限22lim sin1x xx x →∞=+______. (2) 微分方程0xy y '+=满足初始条件()12y =的特解为______. (3) 设二元函数()()1ln 1x yz xex y +=+++，则()1,0dz =______.(4) 设行向量组()()()()2,1,1,1,2,1,,,3,2,1,,4,3,2,1a a a 线性相关，且1a ≠，则a =______.(5) 从数1,2,3,4中任取一个数，记为X ，再从1,,X L 中任取一个数，记为Y ，则{}2P Y ==______.(6) 设二维随机变量(),X Y 的概率分布为若随机事件{}0X =与{}1X Y +=相互独立，则a =______，b =______.二、选择题：本题共8小题，每小题4分，满分24分. 在每小题给出的四个选项中，只有一项符合题目要求，请把所选项前的字母填在答题纸指定位置上.(7) 当a 取下列哪个值时，函数()322912f x x x x a =-+-恰有两个不同的零点.（A ）2 （B ）4 （C ）6 （D ）8(8) 设()()22222123,cos ,cos DDDI I x y d I x y d σσσ==+=+⎰⎰⎰⎰⎰⎰，其中(){}22,1D x y xy =+≤，则（A ）321I I I >> （B ）123I I I >> （C ）213I I I >> （D ）312I I I >> (9) 设0,1,2,,n a n >=L 若1nn a∞=∑发散，()111n n n a ∞-=-∑收敛，则下列结论正确的是（A ）211n n a∞-=∑收敛，21nn a∞=∑发散 （B ）21nn a∞=∑收敛，211n n a∞-=∑发散（C ）()2121n n n aa ∞-=+∑收敛 （D ）()2121n n n a a ∞-=-∑收敛(10) 设()sin cos f x x x x =+，下列命题中正确的是 （A ）()0f 是极大值，2f π⎛⎫⎪⎝⎭是极小值 （B ）()0f 是极小值，2f π⎛⎫⎪⎝⎭是极大值 （C ）()0f 是极大值，2f π⎛⎫⎪⎝⎭也是极大值 （D ）()0f 是极小值，2f π⎛⎫⎪⎝⎭也是极小值 (11) 以下四个命题中，正确的是（A ）若()f x '在()0,1内连续，则()f x 在()0,1内有界 （B ）若()f x 在()0,1内连续，则()f x 在()0,1内有界 （C ）若()f x '在()0,1内有界，则()f x 在()0,1内有界 （D ）若()f x 在()0,1内有界，则()f x '在()0,1内有界 (12) 设矩阵()33ijA a ⨯=满足*T A A =，其中*A 为A 的伴随矩阵，TA 为A 的转置矩阵.若111213,,a a a 为三个相等的正数，则11a 为（A ）3 （B ）3 （C ）13（D (13) 设12,λλ是矩阵A 的两个不同的特征值，对应的特征向量分别为12,αα，则()112,A ααα+线性无关的充分必要条件是（A ）10λ= （B ）20λ= （C ）10λ≠ （D ）20λ≠ (14)（注：该题已经不在数三考纲范围内）三、解答题：本题共9小题，满分94分. 请将解答写在答题纸指定的位置上. 解答应写出文字说明、证明过程或演算步骤.（15）（本题满分8分）求011lim 1x x x e x -→+⎛⎫- ⎪-⎝⎭.（16）（本题满分8分）设()f u 具有二阶连续导数，且(),y x g x y f yfx y ⎛⎫⎛⎫=+ ⎪ ⎪⎝⎭⎝⎭，求222222g g x y x y ∂∂-∂∂.（17）（本题满分9分） 计算二重积分221Dx y d σ+-⎰⎰，其中(){},01,01D x y x y =≤≤≤≤.（18）（本题满分9分） 求幂级数211121n n x n ∞=⎛⎫-⎪+⎝⎭∑在区间()1,1-内的和函数()S x .（19）（本题满分8分）设()(),f x g x 在[]0,1上的导数连续，且()()()00,0,0f f x g x ''=≥≥.证明：对任何[]0,1α∈，有()()()()()()11ag x f x dx f x g x dx f a g ''+≥⎰⎰（20）（本题满分13分） 已知齐次线性方程组（ⅰ）123123123230,2350,0,x x x x x x x x ax ++=⎧⎪++=⎨⎪++=⎩ 和 （ⅱ）()12321230,210,x bx cx x b x c x ++=⎧⎪⎨+++=⎪⎩ 同解，求,,a b c 的值.（21）（本题满分13分） 设T AC D C B ⎛⎫= ⎪⎝⎭为正定矩阵，其中,A B 分别为m 阶，n 阶对称矩阵，C 为m n ⨯阶矩阵.（Ⅰ）计算T P DP ，其中1mn E A C P OE -⎛⎫-=⎪⎝⎭； （Ⅱ）利用（Ⅰ）的结果判断矩阵1T B C A C --是否为正定矩阵，并证明你的结论.（22）（本题满分13分）设二维随机变量(),X Y 的概率密度为()0,01,02,,1,x y x f x y <<<<⎧=⎨⎩其它. 求：（Ⅰ）(),X Y 的边缘概率密度()(),X Y f x f y ； （Ⅱ）2Z X Y =-的概率密度()Z f z ； （Ⅲ）1122P Y X ⎧⎫≤≤⎨⎬⎩⎭.（23）（本题满分13分）设()12,,,2n X X X n >L 为来自总体()20,N σ的简单随机样本，其样本均值为X ，记,1,2,,i i Y X X i n =-=L .（Ⅰ）求i Y 的方差,1,2,,i DY i n =L ； （Ⅱ）求1Y 与n Y 的协方差()1,n Cov Y Y ；（Ⅲ）若()21n c Y Y +是2σ的无偏估计量，求常数c .2006年全国硕士研究生入学统一考试数学三试题一、填空题：1－6小题，每小题4分，共24分. 把答案填在题中横线上. (1) ()11lim ______.nn n n -→∞+⎛⎫=⎪⎝⎭(2) 设函数()f x 在2x =的某邻域内可导，且()()ef x f x '=，()21f =，则()2____.f '''=(3) 设函数()f u 可微，且()102f '=，则()224z f x y =-在点(1,2)处的全微分()1,2d _____.z=(4) 设矩阵2112A ⎛⎫=⎪-⎝⎭，E 为2阶单位矩阵，矩阵B 满足2BA B E =+，则=B . (5)设随机变量X Y 与相互独立，且均服从区间[]0,3上的均匀分布，则{}{}max ,1P X Y ≤=_______.(6) 设总体X 的概率密度为()()121,,,,2xn f x e x X X X -=-∞<<+∞L 为总体X 的简单随机样本，其样本方差为2S ，则2____.ES =二、选择题：7－14小题，每小题4分，共32分. 每小题给出的四个选项中，只有一项符合题目要求，把所选项前的字母填在题后的括号内.(7) 设函数()y f x =具有二阶导数，且()0,()0f x f x '''>>，x ∆为自变量x 在点0x 处的增量，d y y ∆与分别为()f x 在点0x 处对应的增量与微分，若0x ∆>，则（）(A) 0d y y <<∆. (B) 0d y y <∆<.(C) d 0y y ∆<<. (D) d 0y y <∆< .(8) 设函数()f x 在0x =处连续，且()22lim1h f h h →=，则（）(A) ()()000f f -'=且存在 (B) ()()010f f -'=且存在 (C) ()()000f f +'=且存在 (D)()()010f f +'=且存在 (9) 若级数1nn a∞=∑收敛，则级数（）(A)1nn a∞=∑收敛 . （B ）1(1)nn n a ∞=-∑收敛.(C)11n n n a a∞+=∑收敛. (D)112n n n a a ∞+=+∑收敛. (10) 设非齐次线性微分方程()()y P x y Q x '+=有两个不同的解12(),(),y x y x C 为任意常数，则该方程的通解是（）(A) []12()()C y x y x -. (B) []112()()()y x C y x y x +-. (C) []12()()C y x y x +. (D) []112()()()y x C y x y x ++ (11) 设(,)(,)f x y x y ϕ与均为可微函数，且(,)0y x y ϕ'≠，已知00(,)x y 是(,)f x y 在约束条件(,)0x y ϕ=下的一个极值点，下列选项正确的是（）(A) 若00(,)0x f x y '=，则00(,)0y f x y '=. (B) 若00(,)0x f x y '=，则00(,)0y f x y '≠. (C) 若00(,)0x f x y '≠，则00(,)0y f x y '=.(D) 若00(,)0x f x y '≠，则00(,)0y f x y '≠. (12) 设12,,,s αααL 均为n 维列向量，A 为m n ⨯矩阵，下列选项正确的是（） (A) 若12,,,s αααL 线性相关，则12,,,s A A A αααL 线性相关. (B) 若12,,,s αααL 线性相关，则12,,,s A A A αααL 线性无关. (C) 若12,,,s αααL 线性无关，则12,,,s A A A αααL 线性相关.(D) 若12,,,s αααL 线性无关，则12,,,s A A A αααL 线性无关.(13) 设A 为3阶矩阵，将A 的第2行加到第1行得B ，再将B 的第1列的1-倍加到第2列得C ，记110010001P ⎛⎫ ⎪= ⎪ ⎪⎝⎭，则（）(A) 1C P AP -=. (B) 1C PAP -=.(C) T C P AP =. (D) T C PAP =.(14) 设随机变量X 服从正态分布211(,)N μσ，随机变量Y 服从正态分布222(,)N μσ，且{}{}1211P X P Y μμ-<>-<则必有（）(A) 12σσ< (B) 12σσ> (C) 12μμ< (D) 12μμ>三、解答题：15－23小题，共94分. 解答应写出文字说明、证明过程或演算步骤. （15）（本题满分7分）设()1sin,,0,01arctan xy y yf x y x y xy xπ-=->>+，求： (Ⅰ)()()lim ,y g x f x y →+∞=；(Ⅱ)()0lim x g x +→。

2011年考研政治真题一.单项选择题：1-16小题，每小题1分，共16分，下列每题给出的四个选项中，只有一个选项是符合题目要求的，请在答题卡上将选项的字母填写。

1. 著名数学家华罗庚在一次报告中以“一支粉笔多长为好”为题来讲解他的优选法，对此，他解释道：“每支粉笔都要丢掉一段一定长的粉笔头，单就这一点来说：愈长愈好。

但太长了，使用起来很不方便，而且容易折断，每断一次，必然多浪费一个粉笔头，反而不合适。

因而就出现了粉笔多长最合适的问题—这就是一个优选问题。

”所谓优选问题，从辩证的角度看，就是要（）A.注重量的积累B.保持物质的稳定性C.坚持适度原则D.全面考虑事物属性的多样性2. 社会存在是指社会的物质生活条件，它有多方面的内容其中最能集中体现人类社会性质的是（）A.社会形态B．地理环境C．人口因素．D．生产方式3. 马克思把商品转换成货币称为“商品的惊险的跳跃”，“这个跳跃如果不成熟，坏的不是商品，但一定是商品占有者”。

这是因为只有商品变为货币（）A.货币才能转化为资本B.价值才能转化为使用价值C.抽象劳动才能转化为具体劳动D.私人劳动才能转化为社会劳动4．邓小平指出：“社会主义究竟是个什么样子，苏联搞了很多年，也并没有完全搞清楚，可能列宁的思路比较好，搞了个新经济政策，但是后来苏联模式僵化了”。

列宁新经济政策关于社会主义的思路之所以“比较好”是因为（）A.提出了比较系统的社会主义建设纲领B.根据俄国的实际情况来探索社会主义建设的道路C.为俄国找到一种比较成熟的社会主义发展模式D.按照马克思恩格斯关于未来的设想来建设社会主义5．1927年大革命失败后，党的工作重心开始转向农村，在农村建立革命根据地，则革命根据地能够在中国长期存在和发展的根本原因是（）A.中国是一个政治、经济、文化发展不平衡的半殖民地半封建社会B.良好的群众基础和革命形势的继续向前发展C.相当力量正式红军的存在D.党的领导及其正确的政策6. 社会主义初级阶段基本经济制度，既包括公有制经济，也包括非公有制经济，公有制经济纳入社会主义初级阶段基本经济制度中，是因为非公有制经济（）A.是社会主义物质的经济成分B.是社会主义经济的重要组成部分C.是为社会主义服务的经济成分D.在社会主义初级阶段不占主体地位7. 党的十七大通过的党章把“和谐”与“富强”、“民主”、“文明”一起作为社会主义现代化建设的目标写入了社会主义初级阶段的基本路线，其原因在于社会和谐是（）A.中国特色社会主义的本质属性B.中国传统文化的价值取向C.经济建设的内在要求D.解决收入分配差距的重要途径8. 深化文化体制改革，要坚持公益性文化事业和经营性文化产业协调发展，发展营业性文化产业的根本任务是( )A.繁荣文化市场，满足人们群众多方面．多层次．多样化的文化需求B.保证人们群众基本的文化权益C.构建覆盖全社会的比较完备的公共文化服务体系D.加快文化产业基地和区域性特色文化产业群建设9．从1840年到1919年的80年间，中国人民对外来的侵略进行了英勇顽强的反抗，但历次的反侵略战争，都是以中国的失败，中国政府被迫签订丧权辱国条约而告终，从中国内部分析，其根本原因是( )A. 军事战略的错误B.社会制度的腐败C. 经济技术的落后D.思想观念的保守10．1953年9月，彭德怀在一份报告中说，抗美援朝战争的胜利证明：西方侵略者几百年来只要在东方一个海岸上架起几尊大炮就可霸占一个国家的时代一去不复返了。

2021考研数学二历年真题及详解一、选择题（本题共10小题，每小题5分，共50分。

每小题给出的四个选项中，只有一个选项是符合题目要求，把所选选项前的字母填在答题卡指定位置上）1.当x→0时，是x7的（）。

A．低阶无穷小B．等价无穷小C．高阶无穷小D．同阶但非等价无穷小答案【答案】C【考点】常用等价无穷小；【解析】因为当x→0时，，所以是x7的高阶无穷小，故选C项。

2.函数，在x＝0处（）。

A．连续且取极大值B．连续且取极小值C．可导且导数为0D．可导且导数不为0答案【答案】D【考点】连续和可导的定义；【解析】因为故f（x）在x＝0处连续。

因为故f′（0）＝1/2，故选D项。

3.有一圆柱体，底面半径与高随时间变化的速率分别为2cm/s，－3cm/s，当底面半径为10cm，高为5cm时，圆柱体的体积与表面积随时间变化的速率分别为（）。

A．125πcm3/s，40πcm3/sB．125πcm3/s，－40πcm3/sC．－100πcm3/s，40πcm3/sD．－100πcm3/s，－40πcm3/s答案【答案】C【考点】复合函数求导；【解析】由题意知，dr/dt＝2，dh/dt＝－3，有V＝πr2h，S＝2πrh＋2πr2，则当r＝10，h ＝5时，dV/dt＝－100π，dS/dt＝40π，故选C项。

4.设函数f（x）＝ax－blnx（a＞0）有2个零点，则b/a的取值范围为（）。

A．（e，＋∞）B．（0，e）C．（0，1/e）D．（1/e，＋∞）答案【答案】A【考点】函数单调性及极值；【解析】函数求导得f′（x）＝a－b/x，令f′（x）＝0，则有驻点x＝b/a，得：在区间（b/a，＋∞）上，f′（x）＞0，f（x）单增；在区间（－∞，b/a）上，f′（x）＜0，f（x）单减。

即f（b/a）为函数f（x）的极小值，若f（x）有2个零点，则f（b/a）＝a·b/a－bln（b/a）＜0，从而ln（b/a）＞1，可得b/a＞e，故选A项。

历年考研数一真题及答案【篇一：历年考研数学一真题及答案(1987-2013)】ss=txt>数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)?=_____________.(2)曲面x2?2y2?3z2?21在点(1,?2,?2)的法线方程为_____________.(3)微分方程xy???3y??0的通解为_____________.?121?(4)已知方程组??23a?2???x1??1?x???3??1a?2???2无解,则a= ???????x3????0??_____________.(5)设两个相互独立的事件a和b都不发生的概率为19,a发生b不发生的概率与b发生a不发生的概率相等,则p(a)=_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内) (1)设f(x)、g(x)是恒大于零的可导函数,且f?(x)g(x)?f(x)g?(x)?0,则当a?x?b时,有(a)f(x)g(b)?f(b)g(x)(b)f(x)g(a)?f(a)g(x)(c)f(x)g(x)?f(b)g(b)(d)f(x)g(x)?f(a)g(a)(2)设s:x2?y2?z2?a2(z?0),s1为s在第一卦限中的部分,则有(a)??xds?4s??xdss1(b)??yds?4??xdsss1(c)??zds?4??xdsss1(d)??xyzds?4??xyzdsss1(3)设级数??un收敛,则必收敛的级数为n?1(a)??(?1)nun (b)??u2nn?1nn?1(c)??(u2n?1?u2n)n?1(d)??(un?un?1)n?1(a)e(x)?e(y)(b)e(x2)?[e(x)]2?e(y2)?[e(y)]2(c)e(x2)?e(y2) (d)e(x2)?[e(x)]2?e(y2)?[e(y)]2三、(本题满分6分) 1求lim(2?exx??4?sinx).1?exx四、(本题满分5分) 设z?f(xy,xy)?g(xy),其中f具有二阶连续偏导数,g具有二阶连续导数,求?2z?x?y.五、(本题满分6分) 计算曲线积分i??xdy?ydxl4x2?y2,其中l是以点(1,0)为中心,r为半径的圆周(r?1),取逆时针方向.六、(本题满分7分)设对于半空间x?0内任意的光滑有向封闭曲面s,都有??xf(x)dydz?xyf(x)dzdx?e2xzdxdy?0,其中函数f(x)在s(0,??)内具有连续的一阶导数,且xlim?0?f(x)?1,求f(x).七、(本题满分6分)求幂级数??1xnn?13n?(?2)nn的收敛区间,并讨论该区间端点处的收敛性.八、(本题满分7分)设有一半径为r的球体,p0是此球的表面上的一个定点,球体上任一点的密度与该点到p0距离的平方成正比(比例常数k?0),求球体的重心位置.九、(本题满分6分) 设函数f(x)在[0,?]上连续,且???f(x)dx?0,?0f(x)cosxdx?0.试证:在(0,?)内至少存在两个不同的点?1,?2,使f(?1)?f(?2)?0.十、(本题满分6分)??1000?000? 设矩阵a的伴随矩阵a*??1??1010??,且?0?308??aba?1?ba?1?3e,其中e为4阶单位矩阵,求矩阵b.十一、(本题满分8分)某适应性生产线每年1月份进行熟练工与非熟练工的人数统计,然后将16熟练工支援其他生产部门,其缺额由招收新的非熟练工补齐.新、老非熟练工经过培训及实践至年终考核有25成为熟练工.设第n年1月份统计的熟练工与非熟练工所占百分比分别为xn和yn,记成向量??xn?y??. ?n(1)求??xn?1?与??xn?的关系式并写成矩阵形?y?n?1??y?n?式:??xn?1??xn?y??a???. n?1??yn??1??是a的两个线性无关的特征向量,并求出相应的特征值.?1?(3)当??x1??2?时,求??y?????xn?1??. 1???1??yn?1??2??十二、(本题满分8分)某流水线上每个产品不合格的概率为p(0?p?1),各产品合格与否相对独立,当出现1个不合格产品时即停机检修.设开机后第1次停机时已生产了的产品个数为x,求x的数学期望e(x)和方差d(x).十三、(本题满分6分) 设某种元件的使用寿命x的概率密度为?2e?2(x??)x??f(x;?)??x???0x1,x2,,其中??0为未知参数.又设,xn是x的一组样本观测值,求参数?的最大似然估计值.2001年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)设y?ex(asinx?bcosx)(a,b为任意常数)为某二阶常系数线性齐次微分方程的通解,则该方程为_____________. (2)r?x2?y2?z2,则div(gradr)(1,?2,2)=_____________.(3)交换二次积分的积分次序:?01?y?1dy?2f(x,y)dx＝_____________. (4)设a2?a?4e?o,则(a?2e)?1= _____________.(5)d(x)?2,则根据车贝晓夫不等式有估计p{x?e(x)?2}? _____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内) (1)设函数f(x)在定义域内可导,y?f(x)的图形如右图所示,则y?f?(x)的图形为(a)(b)(c)【篇二：2000年-2016年考研数学一历年真题完整版(word版)】ss=txt>数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)?=_____________.(2)曲面x2?2y2?3z2?21在点(1,?2,?2)的法线方程为_____________. (3)微分方程xy???3y??0的通解为_____________.1??x1??1??12??????(4)已知方程组23a?2x2?3无解,则a=_____________. ????????1a?2????x3????0??(5)设两个相互独立的事件a和b都不发生的概率为生的概率相等,则p(a)=_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设f(x)、g(x)是恒大于零的可导函数,且f?(x)g(x)?f(x)g?(x)?0,则当a?x?b时,有 (a)f(x)g(b)?f(b)g(x) (c)f(x)g(x)?f(b)g(b)(b)f(x)g(a)?f(a)g(x) (d)f(x)g(x)?f(a)g(a)1,a发生b不发生的概率与b发生a不发9(2)设s:x2?y2?z2?a2(z?0),s1为s在第一卦限中的部分,则有 (a)(c) ??xds?4??xdsss1(b)(d)??yds?4??xdsss1ss1??zds?4??xdsss1??xyzds?4??xyzds(3)设级数?un?1?n收敛,则必收敛的级数为u(a)?(?1)nnn?1n?(b)?un?1?2n(c)?(un?1?2n?1?u2n)(d)?(un?1?n?un?1)(5)设二维随机变量(x,y)服从二维正态分布,则随机变量??x?y 与 ??x?y不相关的充分必要条件为(a)e(x)?e(y)(c)e(x2)?e(y2)三、(本题满分6分)(d)e(x2)?[e(x)]2?e(y2)?[e(y)]2(b)e(x2)?[e(x)]2?e(y2)?[e(y)]2求lim(x??2?e1?e1x4x?sinx). x四、(本题满分5分)xx?2z设z?f(xy,)?g(),其中f具有二阶连续偏导数,g具有二阶连续导数,求. yy?x?y五、(本题满分6分)计算曲线积分i?xdy?ydx??l4x2?y2,其中l是以点(1,0)为中心,r为半径的圆周(r?1),取逆时针方向.六、(本题满分7分)设对于半空间x?0内任意的光滑有向封闭曲面s,都有???xsx?0?(f)x?dyd(z)x?2xyfex?dzd0x,f(x)在z(0,d??x)内具有连续的一阶导数dy其中函数,且limf(x)?1,求f(x).七、(本题满分6分)八、(本题满分7分)1xn求幂级数?n的收敛区间,并讨论该区间端点处的收敛性. n3?(?2)nn?1?设有一半径为r的球体,p0是此球的表面上的一个定点,球体上任一点的密度与该点到p0距离的平方成正比(比例常数k?0),求球体的重心位置.九、(本题满分6分)设函数f(x)在[0,?]上连续,且??f(x)dx?0,?f(x)cosxdx?0.试证:在(0,?)内至少存在两?个不同的点?1,?2,使f(?1)?f(?2)?0.十、(本题满分6分)?10?01*?设矩阵a的伴随矩阵a??10??0?300100?0??,?1?1且aba?ba?3e,其中e为4阶单位矩阵,求0??8?矩阵b.十一、(本题满分8分)1熟练工支援其他生产部62门,其缺额由招收新的非熟练工补齐.新、老非熟练工经过培训及实践至年终考核有成为熟练工.设第5某适应性生产线每年1月份进行熟练工与非熟练工的人数统计,然后将n年1月份统计的熟练工与非熟练工所占百分比分别为xn和yn,记成向量??xn?1??xn??xn?1??xn?与的关系式并写成矩阵形式:?a???????.?yn?1??yn??yn?1??yn??xn??. ?yn?(1)求??4???1??1??1??1??x1??2??xn?1?(3)当?????时,求??.y1y?1????n?1????2?十二、(本题满分8分)某流水线上每个产品不合格的概率为p(0?p?1),各产品合格与否相对独立,当出现1个不合格产品时即停机检修.设开机后第1次停机时已生产了的产品个数为x,求x的数学期望e(x)和方差d(x).十三、(本题满分6分)?2e?2(x??)x??设某种元件的使用寿命x的概率密度为f(x;?)??,其中??0为未知参数.又设x???0x1,x2,?,xn是x的一组样本观测值,求参数?的最大似然估计值.2001年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)设y?ex(asinx?bcosx)(a,b为任意常数)为某二阶常系数线性齐次微分方程的通解,则该方程为_____________.(2)r?x2?y2?z2,则div(gradr)(1,?2,2)= _____________.(3)交换二次积分的积分次序:?0?1dy?1?y2f(x,y)dx＝_____________.2(4)设a?a?4e?o,则(a?2e)?1= _____________.(5)d(x)?2,则根据车贝晓夫不等式有估计p{x?e(x)?2}?_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设函数f(x)在定义域内可导,y?f(x)的图形如右图所示,则y?f?(x)的图形为(a) (b)(c) (d)(2)设f(x,y)在点(0,0)的附近有定义,且fx?(0,0)?3,fy?(0,0)?1则(a)dz|(0,0)?3dx?dy(b)曲面z?f(x,y)在(0,0,f(0,0))处的法向量为{3,1,1}(c)曲线z?f(x,y)在(0,0,f(0,0))处的切向量为{1,0,3}y?0z?f(x,y)(d)曲线在(0,0,f(0,0))处的切向量为{3,0,1}y?0(3)设f(0)?0则f(x)在x=0处可导?f(1?cosh)(a)lim存在2h?0h(c)limh?0f(1?eh)(b) lim存在h?0h(d)limh?0f(h?sinh)存在h2111111111??4??1?0,b???01???1??00000000f(2h)?f(h)存在h?1?(4)设a??1?1??10??0?,则a与b 0??0?(a)合同且相似 (c)不合同但相似(b)合同但不相似 (d)不合同且不相似(5)将一枚硬币重复掷n次,以x和y分别表示正面向上和反面向上的次数, 则x和y相关系数为(a) -1 (c)(b)0 (d)11 2三、(本题满分6分)arctanex. 求?e2x四、(本题满分6分)【篇三：历年考研数学一真题及答案(1987-2015)】1987-2014 （经典珍藏版）1987年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)当x=_____________时,函数y?x?2x取得极小值.(2)由曲线y?lnx与两直线y?e?1?x及y?0所围成的平面图形的面积是_____________.1?x(3)与两直线y??1?tz?2?t及x?1y?2z?11?1?1都平行且过原点的平面方程为_____________.(4)设l为取正向的圆周x2?y2?9,则曲线积分??l(2xy?2y)dx?(x2?4x)dy= _____________.(5)已知三维向量空间的基底为此基底下的坐标是_____________.二、(本题满分8分) 求正的常数a与b,使等式lim1x2x?0bx?sinx?0?1成立.三、(本题满分7分)1(1)设f、g为连续可微函数,u?f(x,xy),v?g(x?xy),求?u?x,?v?x. (2)设矩阵a和b满足关系式ab=a?2b,其中?301?a???110?,求矩阵 ?4?b.?01??四、(本题满分8分)求微分方程y????6y???(9?a2)y??1的通解,其中常数a?0.五、选择题(本题共4小题,每小题3分,满分12分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设limf(x)?f(a)x?a(x?a)2??1,则在x?a处(a)f(x)的导数存在,且f?(a)?0 (b)f(x)取得极大值(c)f(x)取得极小值 (d)f(x)的导数不存在 (2)设f(x)为已知连续函数s,i?t?t0f(tx)dx,其中t?0,s?0,则i的值(a)依赖于s和t (b)依赖于s、t和x(c)依赖于t、x,不依赖于s (d)依赖于s,不依赖于t (3)设常数?k?0,则级数?(?1)nk?nn2n?1(a)发散(b)绝对收敛2(c)条件收敛(d)散敛性与k的取值有关(4)设a为n阶方阵，且a的行列式|a|?a?0,而a*六、（本题满分10分）求幂级数?a1n?1的收敛域，并求其和函数. xnn?2n?1?是a的伴随矩阵，则|a*|等于(a)a (b)1 (c)an?1七、（本题满分10分）求曲面积分i???x(8y?1)dydz?2(1?y2)dzdx?4yzdxdy,?(d)an??z?1?y?3f(x)?其中?是由曲线绕y轴旋转一周而成的曲面,其法向量与y轴正向的夹角恒大于?. ?2x?0??八、（本题满分10分）设函数f(x)在闭区间[0,1]上可微,对于[0,1]上的每一个x,函数f(x)的值都在开区间(0,1)内,且f?(x)?1,证明在(0,1)内有且仅有一个x,使得f(x)?x.九、（本题满分8分）3问a,b为何值时,现线性方程组?x2?x3?x4?02?2x3?2x4?1x2?(a?3)x3?2x4?bx1?2x2?x3?ax4?? 1有唯一解,无解,有无穷多解?并求出有无穷多解时的通解.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)设在一次实验中,事件a发生的概率为p,现进行n次独立试验,则a至少发生一次的概率为____________;而事件a至多发生一次的概率为____________.(2)有两个箱子,第1个箱子有3个白球,2个红球, 第2个箱子有4个白球,4个红球.现从第1个箱子中随机地取1个球放到第2个箱子里,再从第2个箱子中取出1个球,此球是白球的概率为____________.已知上述从第2个箱子中取出的球是白球,则从第一个箱子中取出的球是白球的概率为____________. (3)已知连续随机变量____________.4x的概率密度函数为f(x)??x2?2x?1,则x的数学期望为____________,x的方差为十一、（本题满分6分）设随机变量x,y相互独立,其概率密度函数分别为fx(x)?10?x?1,fy(y)? y?0,求z?2x?y的概率密度函数.?y其它y?05。

2008年全国硕士研究生入学统一考试政治试题三、分析题：34—38小题，每小题10分，共50分。

要求结合所学知识分析材料回答问题。

将答案写在答题纸指定位置上。

34.最近，四川省搞了一次“医患换位体验”活动，让医生以患者的身份挂号、排队、看病、拿药……结果，医生跑前跑后，既受累又受气，一名全程体验了“患者”的医生感慨道：“医生就像拿着个遥控器，把患者指挥得团团转，当患者确实很苦。

”美国医生爱德华·罗森邦行医50年，忽然患上了喉癌，当他重新审视医学、医院和医生时，感慨地说：“站在我病床边和躺在病床上所看到的角度完全不同”。

他后来在《亲尝我自己的药方》一书中写道：“如果我能从头来过的话，我会以完全不同的方式行医，很不幸的是，生命不给人这种重新来过的机会。

”多年前，有位年轻医生患上甲状腺病，当中国医学科学院著名头脑外科专家屠规益为他主刀。

当手术结束时，屠教授低下身来说：“对不起，让您受苦了！”这是屠教授术后经常对病人说的一句话，虽然简短，却让这位年轻医生深感震撼。

著名医学家袁法祖早年从医，曾在老师的带领下，为一名中年妇女进行开腹手术。

术后没几天，那名妇女就去世了，经解剖发现，患者的死亡与手术并无关系，当时，袁法祖的老师轻轻说了句，“她是4个孩子的妈妈”，就是这句简单的话，让袁法祖至今念念不忘，他知道这句话包含了多少情感，懂得了医生的责任有多重大：医生不仅要看到人身上的病，更要看到生病的人。

结合材料回答问题：(1)“医生换位体验”活动中蕴含着何种哲理？(2)从人的本质属性说明为什么“医生不仅要看到人身上的病，更要看到生病的人”。

(3)你在现实生活中遇到类似医患关系的矛盾，按照矛盾辩证法该如何对待和处理？2009年全国硕士研究生入学统一考试政治试题三、分析题：34～38小题，每小题10分，共50分。

要求结合所学知识分析材料并回答问题。

将答案写在答题纸指定位置上。

34. 结合材料回答问题：华佗是我国东汉名医。

考研政治历年真题试卷汇编12(题后含答案及解析) 题型有：1. 单项选择题 2. 多项选择题单项选择题每小题1分，共16分。

下列每题给出的四个选项中，只有一个选项是符合题目要求的。

1．当代资本主义国际垄断组织的主要形式是A．国际卡特尔B．混合联合企业C．跨国公司D．国际康采恩正确答案：C解析：本题考查垄断资本在世界范围的扩张这一知识点的内容，测试考生对当代资本主义国际垄断组织主要形式的理解和把握，同时也考查考生对下列两个问题的区分。

一是战前和战后国际垄断组织的主要形式；二是战后国内和国际垄断组织的主要形式。

回答好这道题在审题过程中要特别注意题干中的两个限定条件。

一是“当代资本主义”二是“国际垄断组织”。

在这两个限定条件下，就排除了选项A、B、D。

因为国际卡特尔是战前国际垄断组织的主要形式。

而战后即当代资本主义国际垄断组织的主要形式是跨国公司；混合联合企业是战后国内垄断组织的主要形式。

因此。

正确答案是C。

通过解析这道题说明，仔细审题是做好选择题的关键。

2．国家垄断资本主义的产生和发展，从根本上说是A．国内市场竞争的结果B．国际竞争激化的结果C．垄断统治加强的结果D．生产社会化和资本主义私人占有制之间矛盾发展的结果正确答案：D解析：本题考查垄断资本主义的发展的这一知识点的内容，相对比较简单。

第二次世界大战后国家垄断资本的迅速发展，根本原因在于资本主义基本矛盾的不断发展和深化。

这是因为，随着科技革命的发展，生产和资本社会化程度日益提高与资本主义占有关系的矛盾日益激化，产生了一系列矛盾。

这些矛盾单靠私人垄断资本无法解决，要求私人垄断资本与国家政权相结合，凭借国家的力量加以缓解。

国家垄断资本主义获得广泛发展，这是当代资本主义的最重要的新特征。

因此，正确答案是D。

3．邓小平指出：“社会主义究竟是个什么样子，苏联搞了很多年，也并没有完全搞清楚。

可能列宁的思路比较好，搞了个新经济政策。

但是后来苏联的模式僵化了。

毛泽东思想概论1998—2008真题及考试空挡统计（已剔除历年真题与新大纲解析不符部分）说明：★：选择题可能出处；★★★：分析题可能出处，在理解的基础上记忆。

第一章毛泽东思想是马克思主义中国化的理论成果一、毛泽东思想的形成与发展1．马克思主义是指导中国革命和建设的行动指南2．毛泽东思想产生的社会历史条件★3．毛泽东思想发展的历史进程★历年真题10．毛泽东完整地提出新民主主义革命总路线和总政策的著作是（2000年）A．《中国社会各阶级的分析》B．《新民主主义论》C．《在晋绥干部会议上的讲话》D．《论人民民主专政》【答案】C【分析】1948年4月，毛泽东《在晋绥干部会议上的讲话》中指出，无产阶级领导的，人民大众的，方队帝国主义、封建主义和官僚资本主义的革命，这就是中国共产党在当前历史阶段的总路线和总政策。

这是对中国共产党的新民主主义革命总路线的完整表述。

因此选项C正确。

A项中提出的是关于新民主主义革命的基本思想，B项形成了完整的新民主主义革命的理论，标志着毛泽东思想的成熟，D项总结了中国民主革命的历史经验，提出了“人民民主专政”是我们的主要经验、主要纲领。

7．毛泽东首次明确提出“新民主主义革命”这一科学概念的著作是（2004年）A．《〈共产党人〉发刊词》B．《中国革命和中国共产党》C．《新民主主义论》D．《论联合政府》【答案】B【分析】毛泽东在《中国革命和中国共产党》一文中，首次明确提出了“新民主主义革命”这个概念，并从理论和实践的结合上对新民主主义革命的对象、任务、动力、性质和前途等问题作了全面深刻的论述。

因此选项B正确。

选项A提出的是党的建设是一项伟大工程以及党的建设的目标；C项第一次系统论述了新民主主义的政治纳领、经济纳领和文化纲领，论述了新民主主主革命的性质以及所从属的世界革命的范畴；D项主要论述了党的最低纲领和最高纲领及其相互关系，论述了中国共产党的三大工作作风。

25．在土地革命战争后期和抗日战争时期，毛泽东思想得到系统总结和多方面展开而达到成熟。

全国硕士研究生入学统一考试英语（一）Section I Use of EnglishDirections:Read the following text. Choose the best word(s) for each numbered blank and mark A, B, C or D on ANSWER SHEET 1. (10 points)Read the following text. Choose the best word(s) for each numbered blank and mark A, B, C or D on ANSWER SHEET 1. (10 points)The ethical judgments of the Supreme Court justices have become an important issue recently. The court cannot _1_ its legitimacy as guardian of the rule of law _2_ justices behave like politicians. Yet, in several instances, justices acted in ways that _3_ the court’s reputation for being independent and impartial.Justice Antonin Scalia, for example, appeared at political events. That kind of activity makes it less likely that the court’s decisions w ill be _4_ as impartial judgments. Part of the problem is that the justices are not _5_by an ethics code. At the very least, the court should make itself _6_to the code of conduct that _7_to the rest of the federal judiciary.This and other similar cases _8_the question of whether there is still a _9_between the court and politics.The framers of the Constitution envisioned law _10_having authority apart from politics. They gave justices permanent positions _11_they would be free to _12_ those in power and have no need to _13_ political support. Our legal system was designed to set law apart from politics precisely because they are so closely _14_.Constitutional law is political because it results from choices rooted in fundamental social _15_ like liberty and property. When the court deals with social policy decisions, the law it _16_ is inescapably political-which is why decisions split along ideological lines are so easily _17_ as unjust.The justices must _18_ doubts about the court’s legitimacy by makin g themselves _19_ to the code of conduct. That would make rulings more likely to be seen as separate from politics and, _20_, convincing as law.1.[A]emphasize [B]maintain [C]mod ify [D] recognize2.[A]when [B]lest[C]before [D] unless3.[A]restored [B]weakened [C]esta blished [D] eliminated4.[A]challenged [B]compromised [C]suspected[D] accepted5.[A]advanced [B]caught [C ]bound [D]founded6.[A]resistant [B]subject [C]immune [D]prone7.[A]resorts [B]sticks[C]loads [D]applies8.[A]evade [B]raise[C]deny [D]settle9.[A]line [B]barrier[C]similarity [D]conflict10.[A]by [B]as[C]though [D]towards[A]so [B]since[C]provided [D]though12.[A]serve [B]satisfy[C]upset [D]replace13.[A]confirm [B]express [C ]cultivate [D]offer14.[A]guarded [B]followed [C]stud ied [D]tied15.[A]concepts [B]theories [C ]divisions [D]conceptions16.[A]excludes [B]questions [C]shapes[D]controls17.[A]dismissed [B]released [C]ra nked [D]distorted18.[A]suppress [B]exploit [C]address [D]ignore19.[A]accessible [B]amiable [C]agreeable[D]accountable20. [A]by all mesns [B]atall costs [C]in aword [D]as a resultSection II Reading ComprehensionPart ADirections:Read the following four texts. Answer the questions below each text by choosing A, B, C or D. Mark your answers on ANSWER SHEET 1. (40 points)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 different, 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 the teacher who br eaks up the troublemakers in the back row by pairing them with better-behaved classmates. Thetactic 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 author’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 itannounced it was reneging on a longstanding commitment to abide by the strict nuclear regulations.Instead, the company has done precisely what it had long promised it would not challenge the constitutionality of Vermont’s rules in the federal court, as part of a desperate effort to keep its Vermont Yankee nuclear power pla nt running. It’s a stunning move.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 plant’s license be subject to Vermont legislature’s approval. Then, too, the company went along.Either Entergy never really in tended 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 against allowing an extension.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 beside the 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 meaning 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 Ver mont case will test[A] Entergy’s capacity to fulfill all its promises.[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 elsewhere 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 we experience, 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 goal, not the starting point.Once a discovery claim becomes public, the discoverer receives intellectual credit. But, unlike with mining claims, the community takes control of what happens next. Withinthe 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 physiologist Albert 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’s conceptions 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-Györgyi 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 servant. 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 are unionized.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 California points out that much of the state’s budget is patrolled by unions. The teachers’ unions ke ep an eye on schools, the CCPOA on prisons and a variety 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 r epeatedly “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 ones 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 current system, 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 achievers may 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 realise that 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. (41)The 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 even aware 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 createsuperfluous 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 remaining content to just consume. (44) Television 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.(45)What 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 of humanity.[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 CD players.[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. (10 points) Since the days of Aristotle, a search for universal principles has characterized the scientific enterprise. In some ways, this quest forcommonalities defines science. Newton’s laws of motion and Darwinia n evolution each bind a host of different phenomena into a single explicatory frame work.(46)In physics, one approach takes this impulse for unification to its extreme, and seeks a 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 it seems 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 find universality 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 universality identifying 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 not governed by universalsSection III WritingPart A51. Directions:Some internationals students are coming to your university. Write them an1)extend your welcome and2)provide some suggestions for their campus life here.You should write about 100 words on ANSWER SHEET2.Do not sign your name at the end of the letter. Use “Li Ming” instead.Do not write the address(10 points)Part B52. Directions: write an essay of 160-200 words based on the following drawing. In your essay you should1) describe the drawing briefly2) explain its intended meaning, and3) give your commentsYou should write neatly on ANSWER SHEET2.（20 points）1.【答案】B【解析】从空后信息可以看出，这句表达的是“_ _法官表现得像政治家”的情况下，法庭就不能保持其作为法律法规的合法卫士的形象，所以应该选C,maintain“维持，保持”，其他显然语义不通。