20142015研究生考试试卷

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2014年全国硕士研究生入学统一考试经济类专业学位联考综合能力试题

法抵抗辐射,微力学有望开发一种芯片,它可以免受辐射损害。因为它仅使用精微机械开关,但这种开关比电子开关的开关速度慢,而且一个芯片只包含 12000 个开关。基于上述关于微力学
E.污染仍存在,但其性质发生改变,鲑鱼能忍受这种改变后的污染。 5.由微小硅片构成的电脑芯片通常包含数百万的电子开关,电子开关是如此小以至于它无
销售量。但是,酒是个例外,一种酒的价格上涨常常导致其销量增加,即使酒本身并没有任何改变。 B.许多顾客在决定买哪种酒时是基于书或期刊中关于酒的评论。 C.顾客在商场里选购酒时常常以酒的价格作为评判酒的质量的主要参考依据。以下哪项如果为真,最有助于解释上述所说的例外? A.零售市场上存在极具竞争力的多个品牌的酒。
里创作的要少。
的皮肤癌病例每年都保持相对稳定的数量,即使与 20 年前盛行晒太阳相比,现在特意将自己暴晒于太阳下的成年人要少得多。 B.继续特意在太阳底下暴晒的人比过去太阳浴者吸收更大剂量的有害放射物。 C.来自于太阳以外的紫外线辐射量逐年增加。 D.尽管现在更少的女性特意在太阳下暴晒,但这样做的男性人数显著增长。 E.大多数皮肤癌同患者病症发作前 30 年经常暴露于紫外线下相关。以下每项如果为真都可解释上述统计数字上的差异,除了: A.因为大气层顶层臭氧含量减少,现在更多的人都将无意识地暴露在过量的太阳紫外线下。
统,物理学论文报导这种实验结果时,每个实验室的参与人员也通常是论文作者。如果上述为真,下面哪一项一定为真? A.涉及多个医院病人的临床实验绝不是仅由一个医院的医生实施。 B.涉及多个医院病人的临床实验报告,大多数有多位作者。 C.如果一篇科技论文有多位作者,他们通常来自不同的科研机构。 D.多个实验室的研究人员共同署名的物理学论文,通常报导使用了每个实验室开展的子系统的实验结果。 E.大多数科技论文的作者仅是那些做了论文所报导的实验的科研人员。 2.对一群以前从不吸烟的青少年进行追踪研究,以确定他们是否吸烟及其精神健康状态的变化。一年后,开始吸烟的人患忧郁症的人数是那些不吸烟的人患忧郁症的四倍。因为香烟中的尼古丁令大脑发生化学变化,可能因而影响情绪。所以,吸烟很可能促使青少年患忧郁症。下面哪项如果为真,最能加强上述论证? A. 研究开始时就已患忧郁症的实验参与者与那时候那些没有患忧郁症的实验参与者,一年后吸烟者的比例一样。 B.这项研究没有在参与者中区分偶尔吸烟与烟瘾很大者。 C.研究中没有或者极少的参与者是朋友亲戚关系。 D.在研究进行的一年里,一些参与者开始出现忧郁症而后又恢复正常了。 E.研究人员没有追踪这些青少年的酒精摄入量。 3.康和制药公司主任认为, 卫生部要求开发的疫苗的开发费用该由政府资助 , 因为疫苗市场比任何其他药品公司市场利润都小。为支持上述主张 ,主任给出下列理由: 疫苗的销量小,因为疫苗的使用是一个人一次,而治疗疾病尤其是慢性疾病的药物,对每位病人的使用是多次的。下列哪项如果为真,将最严重地削弱该主任提出的针对疫苗市场的主张的理由? A.疫苗的使用对象比大多数其他药品的使用对象多。 B.疫苗所预防的许多疾病都可以由药物成功治愈。 C.药物公司偶尔销售既非医学药品也非疫苗的产品。 D.除了康和制药公司外,其他制药公司也生产疫苗。 E.疫苗的使用费不是由生产疫苗的制药公司承担。 — 1 —

2014年全国硕士研究生入学统一考试数学一试题及解析(完整精准版).doc

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智能仪器设计基础试卷(仪器仪表工程硕士)

南昌大学研究生 2014～2015 学年第一学期期末考试试卷
课程名称：姓名：考试占用时间：
题号题分得分一 50 二 30
智能仪器设计基础学号： 120 分钟
三 20 四五
适用专业、班级：学院：
仪器仪表工程
资源环境与化工学院开卷
十总分 100 累分人签名
考试形式（开卷或闭卷）：
得分
评阅人
2、简述限幅滤波的基本原理，并给出相关程序。
第 3 页共 4 页
三、综合题(共 20 分)
利用 80C51 单片机和 AD574 设计一个多路数据采集系统，给出相关硬件电路与巡回数据采集程序。 (硬件电路给出 80C51 与 AD574 的连接即可)
位 TCON IE IP TMOD D7 TF1 EA — GATE D6 TR1 — — C/T D5 TF0 — — M1 D4 TR0 ES PS M0 D3 IE1 ET1 PT1 GATE D2 IT1 EX1 PX1 C/T D1 IE0 ET0 PT0 M1 D0 IT0 EX0 PX0 M0
第 1 页共 4 页
3、键盘有哪几种组成方式？各有何特点？
4、与硬件滤波器相比，数字滤波器有何值是如何进行的。
第 2 页共 4 页
二、设计题(每题 15 分，共 30 分) 1、设计一个由 8031 单片机控制的程控增益放大器的接口电路，给出硬件电路和放大倍数计算过程。
六七八九
考生注意事项： 1、本试卷共 4 页，请查看试卷中是否有缺页或破损。如有立即举手报告以便更换。 2、考试结束后，考生不得将试卷、答题纸和草稿纸带出考场。
一、简答题(每题 10 分，共 50 分)
1、简述数据采集系统基本结构形式，并比较其特点。

2014下学期研究生试卷

硕士研究生《知识产权》课程考试试卷专业：全校年级：2014 考试方式：开卷学分：1 考试时间：110分钟一、长江公司委托黄河大学设计了一项锅炉自动检测系统，但在委托合同中没有明确约定该研究成果专利申请权的归属，黄河大学指派罗教授承担这一委托项目，研究生张某参加了该项目的研究工作，撰写了研究报告，大学科研部的赵老师参加了该项目的评审和验收，并提出了一些改进建议。

项目结束后，罗教授就该项目所产生的技术成果申请了专利，发明人署名为罗教授和研究生张某，但是长江公司和黄河大学对此有异议。

长江公司认为其提供了资金和研究需求，专利申请权应当属于自己独有；黄河大学认为是其提供了研究条件、组建项目团队并最后完成该发明创造，这些发明创造应当属于黄河大学的职务发明，发明人应当为罗教授和研究生张某；科研部的赵老师认为其提出了改进建议应当作为发明人署名。

请回答：1.这一发明创造的专利申请权应当属于谁？为什么？（10分）2.谁是发明人？为什么？（10分）二、2010年1月A石油公司的高级工程师王某研制出一种节油装置，完成了该公司的技术攻坚课题，并达到国际领先水平。

2010年3月，王某未经单位同意，在向某国外杂志的投稿论文中透露了该装置的核心技术，该杂志将论文全文刊载，引起A石油公司不满。

同年6月，丙公司依照该杂志的报道很快研制了样品，并作好了批量生产的必要准备。

A石油公司于2010年7月向我国专利局递交专利申请书。

2010年12月丁公司也根据该杂志开始生产该节油装置。

2012年2月A的申请被公布，2013年5月7日国务院专利行政部门授予A石油公司发明专利，2013年7月A石油公司向法院提起诉讼，分别要求丙公司和丁公司停止侵害并赔偿损失。

问：1. 2010年7月A石油公司申请专利时，该项发明还是否具有新颖性？为什么？（15分）2.高级工程师王某享有哪些权利？为什么？（10分）3.如果A石油公司的专利申请文件于2012年2月被专利局在其官方刊物《专利公告》中公布，自2010年12月开始直到2013年7月丁公司一直在生产销售节油装置，丁公司的这一期间的生产销售行为是否都构成侵权？如果并非都构成侵权，那么哪一期间的生产销售行为构成侵权？为什么？（20分）三、甲公司为了研制新产品，成立了由王某、李某、丁某、朱某组成的课题组。

2014年全国硕士研究生入学统一考试数学一试题及解析.doc

2014年全国硕士研究生入学统一考试数学一试题及解析一、选择题：1~8小题，每小题4分，共32分，下列每题给出四个选项中，只有一个选项符合题目要求的，请将所选项的字母填在答题纸指定位置上。

（1）下列曲线中有渐近线的是（A ）sin y x x =+.(B)2sin y x x =+.(C)1sin y x x =+.(D)21sin y x x=+.【解析】1sin()11lim lim lim(1sin )1x x x x f x x a x x x x→∞→∞→∞+===+= 11lim[()]lim[sin ]limsin 0x x x b f x ax x x x x→∞→∞→∞=-=+-==∴y=x 是y=x +1sin x的斜渐近线【答案】C(2)设函数()f x 具有2阶导数，()()()()011g x f x f x =-+,则在区间[0，1]上（） (A)当0f x '≥（）时，()()f x g x ≥. (B)当0f x '≥（）时，()()f x g x ≤ (C)当0f x '≥（）时，()()f x g x ≥.(D)当0f '≥时，()()f x g x ≤【解析】当() 0f x "≥时，()f x 是凹函数而()g x 是连接()()0,0f 与()1,1f （）的直线段，如右图故()() f x g x ≤ 【答案】D(3)设(),f x y是连续函数，则110(,)ydy f x y -=⎰⎰（A）11110(,)(,)x dx f x y dy dx f x y dy --+⎰⎰⎰.（B）1101(,)(,)xdx f x y dy dx f x y dy --+⎰⎰⎰⎰.（C ）112cos sin 02(cos ,sin )(cos ,sin ).d f r r dr d f r r dr ππθθπθθθθθθ++⎰⎰⎰⎰（D ）112cos sin 02(cos ,sin )(cos ,sin ).d f r r rdr d f r r rdr ππθθπθθθθθθ++⎰⎰⎰⎰【解析】积分区域如图 0≤y ≤1.1x y ≤≤-用极坐标表示，即：D 1:,012r πθπ≤≤≤≤ D 2: 10,02cos sin r πθθθ≤≤≤≤+【答案】D （4）若{}2211,(cos sin )(cos sin )mina b Rx a x b x dx x a x b x dxππππ--∈--=--⎰⎰，则11cos sin a x b x +=（A ）2sin x π.(B)2cos x .(C) 2sin x π. (D)2cos x π. 【解析】令2(,)(cos sin )Z a b x a x b x dx ππ-=--⎰2(cos sin )(cos )0(1)2(cos sin )(sin )0(2)a b Z x a x b x x dx Z x a x b x x dx ππππ--⎧'=---=⎪⎨'=---=⎪⎩⎰⎰由（1）得 202cos 0axdx π=⎰故10,0a a ==由（2）得 0120sin 22sin x xdx b b xdxππ===⎰⎰【答案】A(5)行列式00000000a b abc d c d= （A ）（ad-bc ）2（B ）-（ad-bc ）2。

2014年全国硕士研究生入学统一考试数学一试题

2014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）下列曲线中有渐近线的是（）（A ）sin y x x =+ （B ）2sin y x x =+ （C ）1siny x x =+ （D ）21sin y x x=+ （2）设函数()f x 具有2阶导数，()(0)(1)(1)g x f x f x =-+，则在区间[0,1]上（）（A ）当()0f x '≥时，()()f x g x ≥ （B ）当()0f x '≥时，()()f x g x ≤ （C ）当()0f x ''≥时，()()f x g x ≥ （D ）当()0f x ''≥时，()()f x g x ≤ （3）设(,)f x y 是连续函数，则21101(,)yy dy f x y dx ---=⎰⎰（）（A ）21110010(,)(,)x x dx f x y dy dx f x y dy ---+⎰⎰⎰⎰（B ）211011(,)(,)xx dx f x y dy dx f x y dy ----+⎰⎰⎰⎰（C ）112cos sin 02(cos ,sin )(cos ,sin )d f r r dr d f r r dr ππθθπθθθθθθ++⎰⎰⎰⎰（D ）112cos sin 02(cos ,sin )(cos ,sin )d f r r rdr d f r r rdr ππθθπθθθθθθ++⎰⎰⎰⎰（4）若{}2211,(cos sin )min(cos sin )a b Rx a x b x dx x a x b x dx ππππ--∈--=--⎰⎰，则11cos sin a x b x +=（）（A ）2sin x （B ）2cos x （C ）2sin x π （D ）2cos x π（5）行列式00000000a b abc d c d=（）（A ）2()ad bc - （B ）2()ad bc -- （C ）2222a dbc - （D ）2222b c a d -（6）设123,,ααα均为3维向量，则对任意常数,k l ，向量组1323,k l αααα++线性无关是向量组123,,ααα线性无关的（）（A ）必要非充分条件（B ）充分非必要条件（C ）充分必要条件（D ）既非充分也非必要条件（7）设随机事件A 与B 相互独立，且3.0)(,5.0)(=-=B A P B P ，则=-)(A B P （）（A ）0.1 (B)0.2 (C)0.3 (D)0.4（8）设连续型随机变量1X 与2X 相互独立且方差均存在，1X 与2X 的概率密度分别为1()f x 与2()f x ，随机变量1Y 的概率密度为)]()([21)(211y f y f y f Y +=，随机变量)(21212X X Y +=，则（A ）2121,DY DY EY EY >> （B ）2121,DY DY EY EY == （C ）2121,DY DY EY EY <= （B ）2121,DY DY EY EY >=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）曲面)sin 1()sin 1(22x y y x z -+-=在点)1,0,1(处的切平面方程为 . （10）设)(x f 是周期为4的可导奇函数，且()2(1)f x x '=-，[0,2]x ∈，则(7)f = .（11）微分方程0)ln (ln =-+'y x y y x 满足条件3)1(e y =的解为y = . （12）设L 是柱面122=+y x 与平面0=+z y 的交线，从z 轴正向往z 轴负向看去为逆时针方向，则曲线积分Lzdx ydz +=⎰ .（13）设二次型3231222132142),,(x x x ax x x x x x f ++-=的负惯性指数为1，则a 的取值范围是 .（14）设总体X 的概率密度为⎪⎩⎪⎨⎧<<=其他,02,32),(2θθθθx xx f ，其中θ是未知参数，n X X X ,,,21 为来自总体X 的简单随机样本，若∑=ni i X c 12为2θ的无偏估计，则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）（本题满分10分）求极限)11ln(])1([lim2112xx dtt e t xtx +--⎰+∞→（16）（本题满分10分）设函数)(x f y =是由方程32260y xy x y +++=确定，求)(x f 的极值. （17）（本题满分10分）设函数)(u f 具有2阶连续导数，)cos (y e f z x=满足22222(4cos )x x z zz e y e x y∂∂+=+∂∂，若0)0(,0)0(='=f f ，求)(u f 的表达式. （18）（本题满分10分）设∑为曲面)1(22≤+=z y x z 的上侧，计算曲面积分dxdy z dzdx y dydz x I )1()1()1(33-+-+-=⎰⎰∑（19）（本题满分10分）设数列}{},{n n b a 满足n n n n n b a a b a cos cos ,20,20=-<<<<ππ，且级数1n n b ∞=∑收敛.（I ）证明：;0lim =∞→n n a（II ）证明：级数∑∞=1n nnb a 收敛. （20）（本题满分11分）设E A ,302111104321⎪⎪⎪⎭⎫ ⎝⎛----=为3阶单位矩阵.（I ）求方程组0=Ax 的一个基础解系；（II ）求满足E AB =的所有矩阵B . （21）（本题满分11分）证明：n 阶矩阵⎪⎪⎪⎪⎪⎭⎫⎝⎛111111111与⎪⎪⎪⎪⎪⎭⎫⎝⎛n 00200100 相似（22）（本题满分11分）设随机变量X 的概率分布为21}2{}1{====X P X P ，在给定i X =的条件下，随机变量Y 服从均匀分布)2,1)(,0(=i i U ，（I ）求Y 的分布函数)(y F Y ；（II ）求EY（23）（本题满分11分）设总体X 的分布函数21,0(;)0,0x e x F x x θθ-⎧⎪-≥=⎨⎪<⎩，其中θ是未知参数且大于零，12,,,n X X X 为来自总体X 的简单随机样本.（1）求EX 与2EX ；（2）求θ的最大似然估计量ˆnθ；（3）是否存在实数a ，使得对任何0ε>，都有{}ˆlim 0nn P a θε→∞-≥=？2017考研新大纲权威解析听3小时直播解析，横扫60+增&改考点。

(完整word版)2014-2015年考研数学二真题及答案解析,推荐文档

2014年全国硕士研究生入学统一考试数学二试题一、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) 当0x +→时，若ln (12)x +α，1(1cos )x -α均是比x 高阶的无穷小，则α的取值范围是( )(A) (2,)+∞(B) (1,2)(C) 1(,1)2(D) 1(0,)2(2) 下列曲线中有渐近线的是 ( )(A) sin y x x =+ (B) 2sin y x x =+ (C) 1siny x x =+(D) 21siny x x=+ (3) 设函数()f x 具有2阶导数，()(0)(1)(1)g x f x f x =-+，则在区间[0,1]上（）(A) 当()0f x '≥时，()()f x g x ≥ (B) 当()0f x '≥时，()()f x g x ≤ (C) 当()0f x ''≥时，()()f x g x ≥(D) 当()0f x ''≥时，()()f x g x ≤(4) 曲线22741x t y t t ⎧=+⎪⎨=++⎪⎩上对应于1t =的点处的曲率半径是（）(A)50(B)100(C)(D)(5) 设函数()arctan f x x =，若()()f x xf '=ξ，则22limx x→=ξ （）(A)1(B)23(C)12(D)13(6) 设函数(,)u x y 在有界闭区域D 上连续，在D 的内部具有2阶连续偏导数，且满足20ux y ∂≠∂∂及22220u ux y∂∂+=∂∂，则（） (A)(,)u x y 的最大值和最小值都在D 的边界上取得 (B) (,)u x y 的最大值和最小值都在D 的内部上取得(C) (,)u x y 的最大值在D 的内部取得，最小值在D 的边界上取得 (D) (,)u x y 的最小值在D 的内部取得，最大值在D 的边界上取得(7) 行列式0000000ab a bcd c d= （）(A) 2()ad bc - (B) 2()ad bc -- (C) 2222a dbc -(D) 2222b c a d -(8) 设123,,ααα均为3维向量，则对任意常数,k l ，向量组1323,k l ++αααα线性无关是向量组123,,ααα线性无关的（）(A) 必要非充分条件 (B) 充分非必要条件(C) 充分必要条件 (D) 既非充分也非必要条件二、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)12125dx x x -∞=++⎰__________.(10) 设()f x 是周期为4的可导奇函数，且()f x '2(1),x =-[0,2]x ∈，则(7)f =__________. (11) 设(,)z z x y =是由方程2274yzex y z +++=确定的函数，则11(,)22dz =__________.(12) 曲线()r r =θ的极坐标方程是r =θ，则L 在点(,)(,)22r =ππθ处的切线的直角坐标方程是__________.(13) 一根长为1的细棒位于x 轴的区间[0,1]上,若其线密度()221x x x =-++ρ,则该细棒的质心坐标x =__________.(14) 设二次型()22123121323,,24f x x x x x ax x x x =-++的负惯性指数为1，则a 的取值范围为_______.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)(本题满分10分)求极限12121lim.1ln 1xt x t e t dt x x →+∞⎡⎤⎛⎫--⎢⎥ ⎪⎢⎥⎝⎭⎣⎦⎛⎫+ ⎪⎝⎭⎰(16)(本题满分10分)已知函数()y y x =满足微分方程221x y y y ''+=-，且()20y =，求()y x 的极大值与极小值.(17)(本题满分10分)设平面区域(){}22,14,0,0,D x y x y x y =≤+≤≥≥计算(sin Dx dxdy x y+⎰⎰.(18)(本题满分10分)设函数()f u 具有二阶连续导数，(e cosy)xz f =满足22222(4e cos )e x xz z z y x y ∂∂+=+∂∂，若'(0)0,(0)0f f ==，求()f u 的表达式.(19)(本题满分10分)设函数(),()f x g x 的区间[a,b]上连续，且()f x 单调增加，0()1g x ≤≤.证明： (I)0(),[,]xag t dt x a x a b ≤≤-∈⎰,(II)()()d ()g()ba a g t dtb aaf x x f x x dx +⎰≤⎰⎰.(20)(本题满分11分)设函数[](x),0,11xf x x=∈+，定义函数列121()(),()(()),f x f x f x f f x ==,L 1()(()),n n f x f f x -=L ，记n S 是由曲线()n y f x =，直线1x =及x 轴所围成平面图形的面积，求极限lim n n nS →∞.(21)(本题满分11分) 已知函数(,)f x y 满足2(1)fy y∂=+∂，且2(,)(1)(2)ln ,f y y y y y =+--求曲线(,)0f x y =所围成的图形绕直线1y =-旋转所成的旋转体的体积. (22)(本题满分11分)设矩阵123401111203A --⎛⎫ ⎪=- ⎪ ⎪-⎝⎭，E 为三阶单位矩阵.(I)求方程组0Ax =的一个基础解系； (II)求满足AB E =的所有矩阵.(23)(本题满分11分)证明n 阶矩阵111111111⎛⎫⎪⎪ ⎪⎪⎝⎭L LM M M M L与00100200n ⎛⎫⎪⎪⎪ ⎪⎝⎭LL M M M M L 相似.2014年全国硕士研究生入学统一考试数学二试题答案一、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) 当0x +→时，若ln (12)x +α，1(1cos )x -α均是比x 高阶的无穷小，则α的取值范围是( )(A) (2,)+∞(B) (1,2)(C) 1(,1)2(D) 1(0,)2【答案】B【解析】由定义 1000ln (12)(2)limlim lim 20x x x x x x x x-→→→+===αααα 所以10->α，故1>α.当0x +→时，211(1cos )~2xx -ααα是比x 的高阶无穷小，所以210->α，即2<α.故选B(2) 下列曲线中有渐近线的是 ( )(A) sin y x x =+ (B) 2sin y x x =+ (C) 1sin y x x =+(D) 21siny x x=+ 【答案】C【解析】关于C 选项：11sinsinlimlim1lim 101x x x x x x x x →∞→∞→∞+=+=+=. 11lim[sin ]limsin 0x x x x x x →∞→∞+-==，所以1sin y x x=+存在斜渐近线y x =. 故选C(3) 设函数()f x 具有2阶导数，()(0)(1)(1)g x f x f x =-+，则在区间[0,1]上（）(A) 当()0f x '≥时，()()f x g x ≥ (B) 当()0f x '≥时，()()f x g x ≤ (C) 当()0f x ''≥时，()()f x g x ≥(D) 当()0f x ''≥时，()()f x g x ≤【答案】D【解析】令()()()(0)(1)(1)()F x g x f x f x f x f x =-=-+-，则(0)(1)0F F ==,()(0)(1)()F x f f f x ''=-+-,()()F x f x ''''=-.若()0f x ''≥，则()0F x ''≤，()F x 在[0,1]上为凸的.又(0)(1)0F F ==，所以当[0,1]x ∈时，()0F x ≥，从而()()g x f x ≥. 故选D.(4) 曲线22741x t y t t ⎧=+⎪⎨=++⎪⎩上对应于1t =的点处的曲率半径是（）(C)(D)【答案】C 【解析】1112'21122432212t t t t t dy t dxtd y dy tdx dx t=====+==-===-()()''33'22211,11y k R kq y ==∴==++ 故选C(5) 设函数()arctan f x x =，若()()f x xf '=ξ，则22limx x→=ξ （）(A)1 (B)23(C)12(D)13【答案】D【解析】因为'2()1()1f x f x ==+ξξ，所以2()()x f x f x -=ξ 22222200011()arctan 11limlimlim lim ()arctan 33x x x x x f x x xx x x f x x x x →→→→---+====ξ故选D.(6) 设函数(,)u x y 在有界闭区域D 上连续，在D 的内部具有2阶连续偏导数，且满足20ux y ∂≠∂∂及22220u ux y∂∂+=∂∂，则（） (A)(,)u x y 的最大值和最小值都在D 的边界上取得 (B) (,)u x y 的最大值和最小值都在D 的内部上取得(C) (,)u x y 的最大值在D 的内部取得，最小值在D 的边界上取得 (D) (,)u x y 的最小值在D 的内部取得，最大值在D 的边界上取得【答案】A【解析】记22222,,,0,,u u uA B C B A C x x y y∂∂∂===≠∂∂∂∂相反数则2=AC-B 0∆<,所以(x,y)u 在D 内无极值，则极值在边界处取得.故选A(7) 行列式0000000ab a bcd c d= ( )(A)2()ad bc - (B)2()ad bc -- (C)2222a d b c - (D)2222b c a d -【答案】B【解析】由行列式的展开定理展开第一列000000000000a b a b a b a b a cd c b c d dcdc d=--()()ad ad bc bc ad bc =--+- 2()ad bc =--.(8) 设123,,a a a 均为三维向量，则对任意常数,k l ,向量组13a ka +,23a la +线性无关是向量组123,,a a a 线性无关的 ( )(A)必要非充分条件 (B)充分非必要条件 (C)充分必要条件(D)既非充分也非必要条件【答案】A 【解析】()()13231231001k l k l ⎛⎫⎪++= ⎪ ⎪⎝⎭ααααααα.)⇐ 记()1323A k l =++αααα，()123B =ααα，1001k l ⎛⎫⎪= ⎪ ⎪⎝⎭C . 若123,,ααα线性无关，则()()()2r A r BC r C ===，故1323,k l ++αααα线性无关.)⇒ 举反例. 令30=α，则12,αα线性无关，但此时123,,ααα却线性相关.综上所述，对任意常数,k l ，向量1323,k l ++αααα线性无关是向量123,,ααα线性无关的必要非充分条件.故选A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)12125dx x x -∞=++⎰__________.【答案】38π【解析】()111221111arctan 252214132428x dx dx x x x -∞-∞-∞+==++++⎡⎤⎛⎫=--= ⎪⎢⎥⎝⎭⎣⎦⎰⎰πππ(10) 设()f x 是周期为4的可导奇函数，且()f x '2(1),x =-[0,2]x ∈，则(7)f =__________. 【答案】1【解析】()()[]'210,2f x x x =-∈，且为偶函数则()()[]'212,0f x x x =--∈-，又()22f x x x c =--+且为奇函数，故=0c()[]222,0f x x x x ∴=--∈-，又()f x Q 的周期为4，()()711f f ∴=-= (11) 设(,)z z x y =是由方程2274yzex y z +++=确定的函数，则11(,)22dz =__________.【答案】1()2dx dy -+ 【解析】对2274yzex y z +++=方程两边同时对,x y 求偏导22210(22)20yzyz z z e y x x z z e z y y y y ∂∂⎧⋅⋅++=⎪∂∂⎪⎨∂∂⎪+++=∂∂⎪⎩当11,22x y ==时,0z = 故1111(,)(,)222211,22z z x y∂∂=-=-∂∂故11(,)22111()()222dzdx dy dx dy =-+-=-+(12) 曲线lim n n nS →∞的极坐标方程是r =θ，则L 在点(,)(,)22r =ππθ处的切线的直角坐标方程是__________. 【答案】22y x =-+ππ【解析】由直角坐标和极坐标的关系 cos cos sin sin x r y r ==⎧⎨==⎩θθθθθθ，于是(),,,22r ⎛⎫=⎪⎝⎭ππθ对应于(),0,,2x y ⎛⎫= ⎪⎝⎭π 切线斜率cos sin cos sin dydy d dx dx d +==-θθθθθθθθ0,22dy dx ⎛⎫⎪⎝⎭∴=-ππ所以切线方程为()202y x -=--ππ即2=2y x -+ππ(13) 一根长为1的细棒位于x 轴的区间[0,1]上,若其线密度()221x x x =-++ρ,则该细棒的质心坐标x =__________. 【答案】1120【解析】质心横坐标()()1010x x dx x x dx=⎰⎰ρρ()()()()31122100042112310005=2133211=2143212x x dx x x dx x x x x x x dx x x x dx x ⎛⎫-++=-++= ⎪⎝⎭⎛⎫-++=-++= ⎪⎝⎭⎰⎰⎰⎰ρρ111112=5203x ∴=(13) 设二次型()22123121323,,24f x x x x x ax x x x =-++的负惯性指数是1，则a 的取值范围_________. 【答案】[]2,2-【解析】配方法：()()()22222123133233,,24f x x x x ax a x x x x =+---+由于二次型负惯性指数为1，所以240a -≥，故22a -≤≤.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)(本题满分10分)求极限12121lim.1ln 1xtx t e t dt x x →+∞⎡⎤⎛⎫--⎢⎥ ⎪⎢⎥⎝⎭⎣⎦⎛⎫+ ⎪⎝⎭⎰【解析】11221122d d (e 1)(e 1)lim lim 11ln(1)xx t t x x t t t t t t x x x x→+∞→+∞⎡⎤⎡⎤----⎢⎥⎢⎥⎣⎦⎣⎦=+⋅⎰⎰12lim[(e 1)]xx x x →+∞=--12000e 1e 11lim lim lim 222t t t xt t t t t t t t +++=→→→---====. (16)(本题满分10分)已知函数()y y x =满足微分方程221x y y y ''+=-，且()20y =，求()y x 的极大值与极小值.【解析】由221x y y y ''+=-，得22(1)1y y x '+=-………………………………………………………①此时上面方程为变量可分离方程，解的通解为331133y y x x c +=-+ 由(2)0y =得23c =又由①可得 221()1x y x y -'=+当()0y x '=时，1x =±，且有：1,()011,()01,()0x y x x y x x y x '<-<'-<<>'><所以()y x 在1x =-处取得极小值，在1x =处取得极大值 (1)0,(1)1y y -==即：()y x 的极大值为1，极小值为0.(17)(本题满分10分)设平面区域(){}22,14,0,0,D x y xy x y =≤+≤≥≥计算(sin Dx dxdy x y+⎰⎰.【解析】D 关于y x =对称，满足轮换对称性，则：D D=⎰⎰12D D I dxdy ∴==⎢⎥⎣⎦⎰⎰1sin(2Ddxdy =⎰⎰π 2201211sin 21()cos 4d r rdrrd r =⋅=-⎰⎰⎰πθππππ22111cos |cos 4r r rdr ⎡⎤=-⋅-⎢⎥⎣⎦⎰ππ211121sin |4r ⎡⎤=-+-⎢⎥⎣⎦ππ34=-(18)(本题满分10分)设函数()f u 具有二阶连续导数，(e cosy)xz f =满足22222(4e cos )e x xz z z y x y∂∂+=+∂∂，若'(0)0,(0)0f f ==，求()f u 的表达式.【解析】由()cos ,xz f e y =()(cos )cos ,(cos )sin x x x x z zf e y e y f e y e y x y∂∂''=⋅=⋅-∂∂ 22(cos )cos cos (cos )cos x x x x xz f e y e y e y f e y e y x∂'''=⋅⋅+⋅∂， ()()()22(cos )sin sin (cos )cos x x x x xz f e y e y e y f e y e y y∂'''=⋅-⋅-+⋅-∂ 由 ()22222+4cos x x z zz e y e x y∂∂=+∂∂，代入得， ()()22cos [4cos cos ]x x x x x f e y e f e y e y e ''⋅=+即()()cos 4cos cos x x x f e y f e y e y ''-=，令cos =,xe y t 得()()4f t f t t ''-=特征方程 240,2-==±λλ 得齐次方程通解2212t t y c e c e -=+设特解*y at b =+，代入方程得1,04a b =-=，特解*14y t =- 则原方程通解为()22121=4t ty f t c e c e t -=+-由()()'00,00f f ==，得1211,1616c c ==-, 则 ()22111=16164u u y f u e e u -=--.(19)(本题满分10分)设函数(),()f x g x 在区间[,]a b 上连续，且()f x 单调增加，0()1g x ≤≤，证明：（I ）0(),[,]xag t dt x a x a b ≤≤-∈⎰,（II ）()()d ()g()ba a g t dtb aaf x x f x x dx +⎰≤⎰⎰.【解析】（I ）由积分中值定理()()(),[,]xag t dt g x a a x =-∈⎰ξξ()01g x ≤≤Q ，()()()0g x a x a ∴≤-≤-ξ()()0xa g t dt x a ∴≤≤-⎰（II ）直接由()01g x ≤≤，得到()()01=x xaag t dt dt x a ≤≤-⎰⎰（II ）令()()()()()ua u a g t dt aaF u f x g x dx f x dx +⎰=-⎰⎰()()()()()()()()()()'uaua F u f u g u f a g t dt g u g u f u f a g t dt =-+⋅⎡⎤=-+⎢⎥⎣⎦⎰⎰由（I ）知()()0u ag t dt u a ≤≤-⎰ ()uaa a g t dt u ∴≤+≤⎰又由于()f x 单增，所以()()()0u af u f ag t dt -+≥⎰()()'0F u F u ∴≥∴，单调不减，()()0F u F a ∴≥=取u b =，得()0F b ≥，即（II ）成立. (20)(本题满分11分)设函数[](x),0,11xf x x=∈+，定义函数列 1211()(),()(()),,()(()),n n f x f x f x f f x f x f f x -===L L ，记n S 是由曲线()n y f x =，直线1x =及x 轴所围成平面图形的面积，求极限lim n n nS →∞.【解析】123(),(),(),,(),112131n x x x xf x f x f x f x x x x nx====++++L 11100011()11n n x x n n S f x dx dx dx nx nx+-∴===++⎰⎰⎰1110200111111ln(1)1dx dx nx n n nx n n =-=-++⎰⎰ 211ln(1)n n n=-+ ln(1)ln(1)1lim 1lim 1lim 1lim 1n n n x x n x nS n x x→∞→∞→∞→∞++∴=-=-=-+101=-= (21)(本题满分11分) 已知函数(,)f x y 满足2(1)fy y∂=+∂，且2(,)(1)(2)ln ,f y y y y y =+--求曲线(,)0f x y =所围成的图形绕直线1y =-旋转所成的旋转体的体积.【解析】因为2(1)fy y∂=+∂,所以2(,)2(),f x y y y x =++ϕ其中()x ϕ为待定函数. 又因为()2(,)(1)2ln ,f y y y y y =+--则()()12ln y y y =--ϕ，从而()()22(,)212ln (1)2ln f x y y y x x y x x =++--=+--.令(,)0,f x y =可得()2(1)2ln y x x +=-，当1y =-时，1x =或2x =，从而所求的体积为()()2221122112ln ln 22V y dx x xdxx xd x =+=-⎛⎫=- ⎪⎝⎭⎰⎰⎰πππ22211221ln (2)222552ln 2(2)2ln 22ln 2.444x x x x dxx x ⎡⎤⎛⎫=--- ⎪⎢⎥⎝⎭⎣⎦⎛⎫=--=-⋅=- ⎪⎝⎭⎰πππππππ(22)(本题满分11分)设矩阵123401111203A --⎛⎫⎪=- ⎪ ⎪-⎝⎭，E 为三阶单位矩阵.(I)求方程组0Ax =的一个基础解系； (II)求满足AB E =的所有矩阵B .【解析】()123410012341000111010011101012030010431101A E ----⎛⎫⎛⎫⎪ ⎪=-→- ⎪ ⎪ ⎪ ⎪---⎝⎭⎝⎭ 123410010012610111010010213100131410013141---⎛⎫⎛⎫ ⎪ ⎪→-→--- ⎪ ⎪ ⎪ ⎪------⎝⎭⎝⎭,(I)0Ax =的基础解系为()1,2,3,1T=-ξ (II)()()()1231,0,0,0,1,0,0,0,1TTTe e e ===1Ax e =的通解为()()111112,1,1,02,12,13,T Tx k k k k k =+--=--+-+ξ 2Ax e =的通解为()()222226,3,4,06,32,43,TTx k k k k k =+--=--+-+ξ 3Ax e =的通解为()()333331,1,1,01,12,13,TTx k k k k k =+-=--++ξ123123123123261123212134313k k k k k k B k k k k k k ----⎛⎫ ⎪-+-++⎪∴= ⎪-+-++ ⎪ ⎪⎝⎭（123,,k k k 为任意常数）(23)(本题满分11分)证明n 阶矩阵111111111⎛⎫⎪⎪ ⎪⎪⎝⎭L LM M M M L 与00100200n ⎛⎫⎪⎪⎪ ⎪⎝⎭LL M M M M L 相似. 【解析】已知()1111A ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭M L L M ，()12001B n ⎛⎫ ⎪ ⎪ ⎪ ⎪⎝⎭LM =，则A 的特征值为n ，0(1n -重).A 属于n λ=的特征向量为(1,1,,1)T L ；()1r A =，故0Ax =基础解系有1n -个线性无关的解向量，即A 属于0λ=有1n -个线性无关的特征向量；故A 相似于对角阵=0n ⎛⎫⎪⎪Λ ⎪ ⎪⎝⎭O .B 的特征值为n ，0(1n -重)，同理B 属于0λ=有1n -个线性无关的特征向量，故B 相似于对角阵Λ.由相似关系的传递性，A 相似于B .2015年全国硕士研究生入学统一考试数学二试题及答案解析一、选择题：（1~8小题,每小题4分，共32分。

2014考研数学(一)真题

2014年全国硕士研究生招生考试数学（一）真题一、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）1．下列曲线有渐近线的是（）。

（A）（B）sin y x x =+2sin y x x =+ （C）1siny x x =+（D）21siny x x =+2．设函数()f x 具有2阶导数，()(0)(1)(1)g x f x f x =-+，则在区间[0上（）。

,1]（A）当时，()0f x '≥()()f x g x ≥ （B）当()0f x '≥时，()()f x g x ≤ （C）当时，()0f x ''≥()()f x g x ≥（D）当()0f x ''≥时，()()f x g x ≤3．设是连续函数，则110(,)ydy f x y dx -=⎰⎰（）。

（A）110010(,)(,)x dx f x y dy dx f x y dy--+⎰⎰⎰（B）11001(,)(,)xdx f x y dy dx f x y dy--+⎰⎰⎰⎰（C）112cos sin 02(cos ,sin )(cos ,sin )d f r r dr d f r r ++⎰⎰⎰⎰ππθθπθθθθθdrθ（D）112cos sin 02(cos ,sin )(cos ,sin )d f r r rdr d f r r ++⎰⎰⎰⎰ππθθπθθθθθrdrθ4．若{}ππ2211-π-π,(cos sin )min(cos sin )a b Rx a x b x dx x a x b x dx ∈--=--⎰⎰，则11cos sin a x b x +=（）。

（A）2sin x（B）2cos x（C）2sin x π（D）2cos x π5．行列式0000000aba bc d c d =（）。

（A）（B）（C）（D）2(ad bc -))2(ad bc --2222a dbc -2222b c a d -6．设123,,ααα均为三维向量，则对任意常数，向量组l k ,132,k 3l αααα++线性无关是向量组123,,ααα线性无关的（）。

2015年全国硕士研究生统一考试《联考综合》试题.。。doc

一、问题求解（本大题15小题，每小题3分，共45分）下列每题给出的五个选项中，只有一项是符合试题要求的，请在答题卡上将所选项的字母涂黑。

1.某精密仪表商降价5%售出了一台仪器，获利5250元，而以七五折售出，商家将亏损1750元，则此商品的成本是（）万元A 2.8 B2.6 C2.4 D2.2 E.22.已知一个班级的考试成绩为：男生平均成绩为90分，女生平均成绩为81分，全班平均成绩为84分，如果a为男生人数，b为女生人数，则下述一定正确的是（）A a＞bB a＝bC a＜bD a≤bE a≥b3 一项工程，甲单独做比甲、乙两人合作多用4天，乙单独做比甲、乙两个合作多用9天，则乙单独做需（）天。

A.12B.15C.18D.20E.224一列匀速行驶的列车，通过450米长的铁桥，从车头上桥到车尾下桥共用33秒：同一列车穿过760米长的隧道，整个车身在隧道内的时间是22秒，则该列车的长度是（）A.320米B.480米C.240米D.266米E.276米5.有X名同学参加了单循环制的围棋比赛，其中有两人各比赛了3场后退出了比赛，且这两名同学间未进行比赛，这样该项比赛共进行了84场，则X值为（）A.30B.25C.20D.15E.186.某宾馆一楼客房比二楼少5间，某旅游团有48人，若全安排在一楼，每间4人，房间不够：每间5人，有房间没有住满，又若安排住二楼，每间3人，房间不够，每间4人，又有房间没有住满，则该宾馆一楼有客房（）间A.7B.8C.9D.10E.117.汽车从甲地开往乙地，若汽车等速行驶2小时后减速20%，则到乙地后会延误1小时：若汽车等速行驶到最后100公里，才减速20%，到乙地只延误20分钟，那么，甲、乙两地距离的公里数是（）A.380B.410C.450D.460E.4708.相同的5个白球和相同的10个黑球排成一行，要求每个白球的左邻必须是黑球，则共有（A）种不同的排法。

9.A.252B.469C.320D.3270E.649010.在某校举办的足球比赛中规定：胜一场得3分：平一场得1分：负一场得0分，某班足球队参加了12场比赛，共得22分，已知这个队只输了2场，则此队胜了（）场A.8B.7C.6D.5E.411.甲花费5万元购买了股票，随后他将这些股票转卖给乙获利10%，不久乙又将这些股票返卖给甲，但乙损失了10%，最后甲按乙卖给他的价格为9折把这些股票卖掉了，不计交易费，甲在上述股票交易中（）A.不盈不亏B.盈利100元C.盈利50元D.亏损100元E.亏损50元11.在1，2，3，4，5，这个五个数字组成的没有重复数字的三位数中，各位数字之和为奇数的共有（）A.24个B.16个C.28个D.14个E.30个12.某城市按以下规定收取每月煤气费，用煤气如果不超过60立方米，按每立方米0.80元收费：如果超过60立方米。

2014年数学一真题与答案解析

2014年全国硕士研究生入学统一考试数学一2014年全国硕士研究生入学统一考试数学一试题答案一、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) 下列曲线有渐近线的是 ( )(A)sin y x x =+ (B)2sin y x x =+ (C)1sin y x x =+ (D)21sin y x x=+ 【答案】(C)【解析】关于C 选项：11sinsinlimlim1lim 101x x x x x x x x →∞→∞→∞+=+=+=，又 11lim[sin ]lim sin 0x x x x x x →∞→∞+-==，所以1sin y x x=+存在斜渐近线y x =. 故选(C).(2) 设函数()f x 具有二阶导数，()(0)(1)(1)g x f x f x =-+，则在区间[0,1]上 ( )(A) 当()0f x '≥时，()()f x g x ≥ (B) 当()0f x '≥时，()()f x g x ≤ (C) 当()0f x ''≥时，()()f x g x ≥ (D) 当()0f x ''≥时，()()f x g x ≤ 【答案】(D)【解析】令()()()(0)(1)(1)()F x g x f x f x f x f x =-=-+-，则(0)(1)0F F ==,()(0)(1)()F x f f f x ''=-+-,()()F x f x ''''=-.若()0f x ''≥，则()0F x ''≤，()F x 在[0,1]上为凸的.又(0)(1)0F F ==，所以当[0,1]x ∈时，()0F x ≥，从而()()g x f x ≥. 故选(D).2014年全国硕士研究生入学统一考试数学一(3) 设()f x 是连续函数，则110(,)ydy f x y dx -=⎰⎰( )(A) 1100010(,)(,)x dx f x y dy dx f x y dy --+⎰⎰⎰ (B)1101(,)(,)xdx f x y dy dx f x y dy --+⎰⎰⎰⎰(C)112cos sin 02(cos ,sin )(cos ,sin )d f r r dr d f r r dr ++⎰⎰⎰⎰ππθθπθθθθθθ(D)112cos sin 02(cos ,sin )(cos ,sin )d f r r rdr d f r r rdr ++⎰⎰⎰⎰ππθθπθθθθθθ【答案】(D) 【解析】1101101(,)(,)(,)yxdy f x y dx dx f x y dy dx f x y dy ---=+⎰⎰⎰⎰⎰112cos sin 02(cos ,sin )(cos ,sin )d f r r rdr d f r r rdr +=+⎰⎰⎰⎰ππθθπθθθθθθ.故选(D). (4) 若{}ππ2211-π-π,(cos sin )min(cos sin )a b Rx a x b x dx x a x b x dx ∈--=--⎰⎰，则11cos sin a x b x +=32260y xy x y +++= ( )(A) 2sin x (B) 2cos x (C) 2sin x π (D) 2cos x π 【答案】(A) 【解析】2222(cos sin )(sin )2cos (sin )cos x a x b x dx x b x a x x b x a x x dx --⎡⎤--=---+⎣⎦⎰⎰ππππ22222(2sin sin cos )x bx x b x a x dx -=-++⎰ππ2222202(sin cos 2sin )x dx b x a x bx x dx -=++-⎰⎰πππ223124()422223a b b =+⋅-⋅+πππ 2232(4)3a b b =+-+ππ2014年全国硕士研究生入学统一考试数学一2232(2)43a b ⎡⎤=+--+⎣⎦ππ当0,2a b ==时，积分最小. 故选(A).(5) 行列式0000000a b abc d c d= ( )(A)2()ad bc - (B)2()ad bc -- (C)2222a dbc - (D)2222b c a d - 【答案】(B)【解析】由行列式的展开定理展开第一列0000000000000000a b a b a b a ba c d cbcd d c d c d=-- ()()ad ad bc bc ad bc =--+- 2()ad bc =--.故选(B).(6) 设123,,a a a 均为三维向量，则对任意常数,k l ,向量组13a ka +,23a la +线性无关是向量组()123=B ααα线性无关的 ( )(A)必要非充分条件 (B)充分非必要条件 (C)充分必要条件(D)既非充分也非必要条件【答案】(A) 【解析】()()13231231001k l k l ⎛⎫⎪++= ⎪ ⎪⎝⎭ααααααα.)⇐ 记()1323A k l =++αααα，()123B =ααα，A . 若123,,ααα线性无关，则2014年全国硕士研究生入学统一考试数学一()()()2r A r BC r C ===，故()0.3P A B -=线性无关.()P B A -= 举反例. 令30=α，则12,αα线性无关，但此时123,,ααα却线性相关.综上所述，对任意常数402Q p =-，向量p 线性无关是向量D 线性无关的必要非充分条件. 故选(A).(7) 设随机事件A 与B 相互独立，且()0.5P B =，()0.3P A B -=，则()P B A -= ( ) (A)0.1 (B)0.2 (C)0.3 (D)0.4 【答案】(B)【解析】已知a =，A 与()2123121323,,24f x x x x x ax x x x =-++独立，a ，()()()()()()P A B P A P AB P A P A P B -=-=-()0.5()0.5()0.3P A P A P A =-==，则 ()0.6P A =，则()()()()()()0.50.50.60.50.30.2P B A P B P AB P B P A P B -=-=-=-⨯=-=.故选(B).(8) 设连续性随机变量1X 与2X 相互独立，且方差均存在，1X 与2X 的概率密度分别为1()f x 与2()f x ，随机变量1Y 的概率密度为1121()[()()]2Y f y f y f y =+，随机变量2121()2Y X X =+，则( )(A) 12EY EY >，12DY DY > (B) 12EY EY =，12DY DY =(C) 12EY EY =，12DY DY < (D) 12EY EY =，12DY DY > 【答案】(D)【解析】用特殊值法. 不妨设12,(0,1)X X N ，相互独立. 22212221())2y y y Y f y ---==，1(0,1)Y N .2014年全国硕士研究生入学统一考试数学一2121()2Y X X =+，212212111()(()())0,()(()())242E Y E X E X D Y D X D X =+==+=. 12121()()0,()1()2E Y E Y D Y D Y ===>=.故选(D).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) 曲面22(1sin )(1sin )z x y y x =-+-在点(1,0,1)处的切平面方程为__________. 【答案】21x y z --=【解析】由于22(1sin )(1sin )z x y y x =-+-，所以22(1sin )cos x z x y x y '=--⋅，(1,0)2x z '=；2cos 2(1sin )yz x y y x '=-+-，(1,0)1y z '=-. 所以，曲面在点(1,0,1)处的法向量为{2,1,1}n =--. 故切平面方程为2(1)(1)(0)(1)0x y z -+----=，即21x y z --=.(10) 设()f x 是周期为4的可导奇函数，且()f x '2(1),x =-[0,2]x ∈，则(7)f =__________.【答案】1【解析】由于()f x '2(1)x =-，[0,2]x ∈，所以2()(1)f x x C =-+，[0,2]x ∈.又()f x 为奇函数，(0)0f =，代入表达式得1C =-，故2()(1)1f x x =--，[0,2]x ∈.()f x 是以4为周期的奇函数，故2(7)(18)(1)(1)[(11)1]1f f f f =-+=-=-=---=.(11) 微分方程(ln ln )0xy y x y '+-=满足条件3(1)y e =的解为y =__________.2014年全国硕士研究生入学统一考试数学一【答案】21(0)x y xe x +=>【解析】(ln ln )0xy y x y '+-=ln()y y y x x'⇒=. 令yu x=，则y x u =⋅，y xu u ''=+，代入原方程得 ln xu u u u '+=(ln 1)u u u x-'⇒=分离变量得，(ln 1)du dxu u x=-，两边积分可得 ln |ln 1|ln u x C -=+，即ln 1u Cx -=.故ln1y Cx x -=. 代入初值条件3(1)y e =，可得2C =，即ln 21yx x=+. 由上，方程的解为21,(0)x y xe x +=>.(12) 设L 是柱面221x y +=与平面0y z +=的交线，从A 0x =轴正向往z 轴负向看去为逆时针方向，则曲线积分Lzdx ydz +=⎰ __________.【答案】π【解析】由斯托克斯公式，得0Ldydz dzdx dxdyzdx ydz dydz dzdx x y z z y∑∑∂∂∂+==+∂∂∂⎰⎰⎰⎰⎰xyD dydz dzdx =+=⎰⎰π，其中22{(,)|1}xy D x y x y =+≤.(13) 设二次型()22123121323,,24f x x x x x ax x x x =-++的负惯性指数是1，则a 的取值范围_________. 【答案】[]2,2-2014年全国硕士研究生入学统一考试数学一【解析】配方法：()()()22222123133233,,24f x x x x ax a x x x x =+---+由于二次型负惯性指数为1，所以240a -≥，故22a -≤≤.(14) 设总体X 的概率密度为()22,2,;30,xx f x ⎧<<⎪=⎨⎪⎩θθθθ其他,其中θ是未知参数，12,,,n X X X 为来自总体X 的简单样本，若221()nii E cX==∑θ，则c =_________.【答案】25n【解析】 222222()(;)3x E X x f x dx x dx +∞-∞==⋅⎰⎰θθθθ 2422215342x =⋅=θθθθ，222215[]()2ni i n E cX ncE X c ===⋅=∑θθ， 25c n∴=. 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)(本题满分10分)求极限12121lim.1ln 1xtx t e t dt x x →+∞⎡⎤⎛⎫--⎢⎥ ⎪⎢⎥⎝⎭⎣⎦⎛⎫+ ⎪⎝⎭⎰【解析】11221122d d (e 1)(e 1)lim lim 11ln(1)xx t t x x t t t t t t x x x x→+∞→+∞⎡⎤⎡⎤----⎢⎥⎢⎥⎣⎦⎣⎦=+⋅⎰⎰12lim [(e 1)]xx x x →+∞=--2014年全国硕士研究生入学统一考试数学一12000e 1e 11lim lim lim 222t t t xt t t t t t t t +++=→→→---====. (16)(本题满分10分)设函数()y f x =由方程32260y xy x y +++=确定，求()f x 的极值. 【解析】对方程两边直接求导：2223220y y y xyy x y xy '''++++= ①令1x 为极值点，则由极值必要性知：1()0y x '=，代入①式得：2111()2()0y x x y x +=.即1()0y x =或11()2y x x =-. 将其代入原方程知：1()0y x =（舍去），即11()2y x x =-. 代入，有 33311184260x x x -+-+=，∴11x =. 即(1)2y =-，(1)0y '=.对①式两边再求导：22226()322()222220y y y y yy x y xyy yy xy x y y xy ''''''''''''+++++++++=.将(1)2y =-，(1)0y '=代入得：4(1)09y ''=>. ∴()y f x =在1x =处取极小值，(1)2y f ==-.(17)(本题满分10分)设函数()f u 具有二阶连续导数，()cos xz f e y =满足()222224cos .x xz z z e y e x y∂∂+=+∂∂若()()00,00f f '==，求()f u 的表达式.【解析】由()cos ,xz f e y =()(cos )cos ,(cos )sin x x x x z zf e y e y f e y e y x y∂∂''=⋅=⋅-∂∂ 22(cos )cos cos (cos )cos x x x x x zf e y e y e y f e y e y x∂'''=⋅⋅+⋅∂，2014年全国硕士研究生入学统一考试数学一()()()22(cos )sin sin (cos )cos x x x x xz f e y e y e y f e y e y y∂'''=⋅-⋅-+⋅-∂ 由 ()22222+4cos x x z zz e y e x y∂∂=+∂∂，代入得，()()22cos 4[cos cos ]x x x x x f e y e f e y e y e ''⋅=+,即()()cos 4cos 4cos x x x f e y f e y e y ''-=，令cos =,x e y t 得()()44f t f t t ''-=特征方程 240,2-==±λλ 得齐次方程通解2212t t y c e c e -=+ 设特解*y at b =+，代入方程得1,0a b =-=，特解*y t =- 则原方程通解为()2212=tty f t c e c et -=+-由()()'00,00f f==，得1211,44c c ==-, 则()2211=44u uy f u e e u -=-- (18)(本题满分10分)设∑为曲面22z x y =+(z 1)≤的上侧，计算曲面积分33(1)(1)(1)I x dydz y dzdx z dxdy ∑=-+-+-⎰⎰.【解析】∑非闭，补1∑：平面1z =，被22z x y =+所截有限部分下侧，由Gauss 公式，有 133+(1)(1)(1)x dydz y dzdx z dxdy ∑∑--+-+-⎰⎰223(1)3(1)1x y dV Ω⎡⎤=-+-+⎣⎦⎰⎰⎰ 223()667x y dV xdV ydV dV ΩΩΩΩ=+--+⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰⎰2014年全国硕士研究生入学统一考试数学一∑和1∑所围立体为Ω，Ω关于yoz 面和zox 面对称，则0xdV ydV ΩΩ==⎰⎰⎰⎰⎰⎰22221221()x y x y x y dV dxdy dz +Ω+≤+=⎰⎰⎰⎰⎰⎰=21220(1)d r r rdr -⎰⎰πθ461011112()2()46466r r =-=-=πππ22112x y zdV dzdxdy zdz Ω+≤===⎰⎰⎰⎰⎰⎰⎰ππ173746222∑+∑∴-=⋅+⋅=+=⎰⎰πππππ 14∑+∑∴-=⎰⎰π又22111(1)(11)0x y z dxdy dxdy ∑∑+≤=-=--=⎰⎰⎰⎰⎰⎰1114I ∑+∑∑∴=-=-⎰⎰⎰⎰π(19)(本题满分10分)设数列{}{},n n a b 满足02n a <<π，02n b <<π，cos cosb n n n a a -=，且级数1nn b∞=∑收敛.(I) 证明：lim 0n n a →∞=.(II) 证明：级数1nn na b ∞=∑收敛. 【解析】(I )1nn b∞=∑收敛 lim 0n n b →∞∴=cos cos 2sinsin 022sin 02n n n n n n n n n a b a ba ab a b+-=-=->-∴<又424nn a b --<< ππ，042n n a b-∴-<<π2014年全国硕士研究生入学统一考试数学一即：n n a b <又0,n n a b << lim 0n n b →∞= lim 0n n a →∞∴=(II )证明：由(I )2sinsin 22n n n n n a b a ba +-=- 2sin sin 22n n n nn n na b a b a b b +--∴= 222222222n n n nn n n n n n n a b b a b a b b b b b +--≤=<= 又 1n n b ∞=∑收敛 ∴12nn b ∞=∑收敛,1n n na b ∞=∑收敛(20)(本题满分11分)设矩阵123401111203A --⎛⎫ ⎪=- ⎪ ⎪-⎝⎭，E 为三阶单位矩阵.(I)求方程组0Ax =的一个基础解系； (II)求满足AB E =的所有矩阵B .【解析】()123410012341000111010011101012030010431101A E ----⎛⎫⎛⎫⎪ ⎪=-→- ⎪ ⎪ ⎪ ⎪---⎝⎭⎝⎭ 123410010012610111010010213100131410013141---⎛⎫⎛⎫ ⎪ ⎪→-→--- ⎪ ⎪ ⎪ ⎪------⎝⎭⎝⎭, (I)0Ax =的基础解系为()1,2,3,1T=-ξ (II)()()()1231,0,0,0,1,0,0,0,1TTTe e e ===1Ax e =的通解为()()111112,1,1,02,12,13,T Tx k k k k k =+--=--+-+ξ 2Ax e =的通解为()()222226,3,4,06,32,43,TTx k k k k k =+--=--+-+ξ 3Ax e =的通解为()()333331,1,1,01,12,13,TTx k k k k k =+-=--++ξ2014年全国硕士研究生入学统一考试数学一123123123123261123212134313k k k k k k B k k k k k k ----⎛⎫ ⎪-+-++⎪∴= ⎪-+-++ ⎪ ⎪⎝⎭（123,,k k k 为任意常数）(21)(本题满分11分)证明n 阶矩阵111111111⎛⎫ ⎪⎪ ⎪ ⎪⎝⎭与00100200n ⎛⎫⎪ ⎪⎪⎪⎝⎭相似. 【解析】已知()1111A ⎛⎫ ⎪ ⎪= ⎪ ⎪⎝⎭ ，()12001B n ⎛⎫ ⎪ ⎪ ⎪ ⎪⎝⎭=，则A 的特征值为n ，0(1n -重).A 属于n =λ的特征向量为(1,1,,1)T ；()1r A =，故0Ax =基础解系有1n -个线性无关的解向量，即A 属于0=λ有1n -个线性无关的特征向量，故A 相似于对角阵0=0n ⎛⎫ ⎪⎪Λ ⎪ ⎪⎝⎭. B 的特征值为n ，0(1n -重)，同理B 属于0=λ有1n -个线性无关的特征向量，故B 相似于对角阵Λ.由相似关系的传递性，A 相似于B . (22)(本题满分11分)设随机变量X 的概率分布为{}{}112,2P X P X ====在给定X i =的条件下，随机变量Y 服从均匀分布()0,,(1,2)U i i =.（I ）求Y 的分布函数()Y F y ；（II ）求EY .【解析】（I ）设Y 的分布函数为(y)Y F ，则2014年全国硕士研究生入学统一考试数学一{}{}{}{}{}()1|12|2Y F y P Y y P X P Y y X P X P Y y X =≤==≤=+=≤={}{}11|1|222P Y y X P Y y X =≤=+≤= 当0y <时，()0Y F y =；当01y ≤<时，13()(y )224Y y yF y =+=；当12y ≤<时，1()(1)22Y yF y =+；当2y ≥时，()1Y F y =. 所以Y 的分布函数为0,03,014()1(1),12221,2Y y y y F y y y y <⎧⎪⎪≤<⎪=⎨⎪+≤<⎪⎪≥⎩(II) Y 的概率密度为3,01,41(y),12,40,Y y f y ⎧<<⎪⎪⎪=≤<⎨⎪⎪⎪⎩其他.120131()=()d 44Y E Y f y y y dy y dy +∞-∞=+⎰⎰⎰ =31113(41)42424⨯+⨯-=(23)(本题满分11 分)设总体X 的分布函数为21(;)0,0,0,x x x e F x -≥<⎧⎪-=⎨⎪⎩θθ其中θ是未知参数且大于2014年全国硕士研究生入学统一考试数学一零.12,,,n X X X 为来自总体X 的简单随机样本.（I ）求()E X ，2()E X ；（II ）求θ的最大似然估计量nθ；（III ）是否存在实数a ，使得对任何0>ε，都有{}lim 0n n P a →∞-≥=θε？【解析】X 的概率密度为22,0(;)(;)0,xx e x f x F x -⎧⎪>'==⎨⎪⎩θθθθ其它（I ）22()(;)x xE X xf x dx xedx -+∞+∞-∞==⎰⎰θθθ222[]x x x xdexeedx ---+∞+∞+∞=-=--⎰⎰θθθ2x edx -+∞=⎰θ12==22222()(;)x xE X x f x dx x edx -+∞+∞-∞==⎰⎰θθθ222220[2]x x x x dex eexdx ---+∞+∞+∞=-=--⋅⎰⎰θθθ22x xedx -+∞=⎰θθθ=θ2014年全国硕士研究生入学统一考试数学一（II ）似然函数2112,0()(;)0,ix n i ni i i x e x L f x -==⎧⎪∏>=∏==⎨⎪⎩θθθθ其它当0(1,,)i x i n >=⋅⋅⋅时，212()i x nii x L e-==∏θθθ，21ln ()[ln 2ln ]ni i i x L x ==--∑θθθ222211ln ()11[][]0n ni i i i x d L x n d ===-+=-=∑∑θθθθθθ 解得 211n i i x n ==∑θ所以，θ的最大似然估计量为211ˆnni i X n ==∑θ （III ）依题意，问ˆnθ是否为θ的一致估计量. 2211ˆ()()()nni i E E X E X n ====∑θθ 242211ˆ()()[()()]nD D XE X E X n n==-θ 24442()(;)x xE X x f x dx x edx -+∞+∞-∞==⎰⎰θθθ2224430[4]x x x x dex eex dx ---+∞+∞+∞=-=--⋅⎰⎰θθθ2304x x edx -+∞=⎰θ22222022[2]x x x x dex eexdx ---+∞+∞+∞=-=--⋅⎰⎰θθθθθ2014年全国硕士研究生入学统一考试数学一24x xedx -+∞=⎰θ2222()x x ed -+∞=--⎰θθθ22=θ2221ˆ()[2]nD n n∴=-=θθθθ ˆlim ()0n n D →∞=θˆn∴θ为θ的一致估计量 a ∴=θ。

2014-2015年考研数学二真题及答案解析

线性无关的（）
(A)必要非充分条件(B)充分非必要条件
(C)充分必要条件(D)既非充分也非必要条件
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) __________.
(10)设是周期为的可导奇函数，且，则 __________.
(11)设是由方程确定的函数，则 __________.
2014
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)当时，若，均是比高阶的无穷小，则的取值范围是( )
(A) (B) (C) (D)
(2)下列曲线中有渐近线的是( )
(A) (B)
(C) (D)
二、填空题：( )小题，每小题4分，共24分。
(9)设则
【答案】48
【解析】由参数式求导法
再由复合函数求导法则得
=
,
综上所述，本题正确答案是48。
【考点】高等数学-一元函数微分学-复合函数求导
(10)函数处的n阶导数
【答案】
【解析】
解法1用求函数乘积的阶导数的莱布尼茨公式
其中注意 ,于是
因此
【解析】
(21)(本题满分11分)
已知函数满足，且求曲线所围成的图形绕直线旋转所成的旋转体的体积.
【解析】因为 ,所以其中为待定函数.
又因为则，从而
.
令可得，当时，或，从而所求的体积为
(22)(本题满分11分)
设矩阵，为三阶单位矩阵.
(I)求方程组的一个基础解系；
(II)求满足的所有矩阵 .

2014年考研数学二真题及答案解析

(9)
1
x2
1 2x
5
dx
__________.
【答案】 3 8
【解析】
1
x2
1 2x
5
dx
1
x
1
12
4
dx
1 2
arctan
x 1 2
1
1 2
4
2
3 8
(10) 设 f (x) 是周期为 4 的可导奇函数，且 f (x) 2(x 1), x [0, 2] ，则 f (7) __________.
(A)
50
【答案】C
10
(B)
100
(C)10 10
5
(D) 5 10
【解析】
2014 年全国硕士研究生入学统一考试数学二
dy dx
t1
2t 2t
4
t 1
3
d2y dx2
t1
dy ' dx
2
t1
t2 2t
1
t 1
k
y '' 1 y'2
3 2
1
3
1 q2
, R
1 k
10
10
故选 C
) 举反例. 令3 0 ，则1,2 线性无关，但此时1,2 ,3 却线性相关.
综上所述，对任意常数 k, l ，向量1 k3 ,2 l3 线性无关是向量1,2 ,3 线性无关的必
要非充分条件.
故选 A
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.
lim x0
1
1
1 x
2
3x2
1 3

2014年全国硕士研究生入学统一考试数学三及参考答案

2014年全国硕士研究生入学统一考试数学三试题一、选择题：1～8小题，每小题4分，共32分，下列每小题给出的四个选项中，只有一项符合题目要求的，请将所选项前的字母填在答题纸．．．指定位置上.（1）设lim ,n a a =且0,a ≠则当n 充分大时有（）（A ）2n a a >（B ）2n a a <（C ）1n a a n >- （D ）1n a a n<+（2）下列曲线有渐近线的是（）（A ）sin y x x =+ （B ）2sin y x x =+ （C ）1siny x x=+ （D ）21siny x x=+ （3）设23(x)a P bx cx dx =+++ ，当0x → 时，若(x)tanx P - 是比x 3高阶的无穷小，则下列试题中错误的是（A ）0a =（B ）1b = （C ）0c = （D ）16d =（4）设函数()f x 具有二阶导数，()(0)(1)(1)g x f x f x =-+，则在区间[0,1]上（）（A ）当'()0f x ≥时，()()f x g x ≥ （B ）当'()0f x ≥时，()()f x g x ≤ （C ）当'()0f x ≤时，()()f x g x ≥ （D ）当'()0f x ≤时，()()f x g x ≥（5）行列式0000000aba b c dc d= （A ）2()ad bc - （B ）2()ad bc -- （C ）2222a d b c - （D ）2222b c a d -（6）设123,,a a a 均为3维向量，则对任意常数,k l ，向量组1323,k l αααα++线性无关是向量组123,,ααα线性无关的（A ）必要非充分条件（B ）充分非必要条件（C ）充分必要条件（D ）既非充分也非必要条件（7）设随机事件A 与B 相互独立，且P （B ）=0.5，P(A-B)=0.3，求P （B-A ）=（）（A ）0.1 （B ）0.2 （C ）0.3 （D ）0.4（8）设123,,X X X 为来自正态总体2(0,)N σ的简单随机样本，服从的分布为（A ）F （1,1）（B ）F （2,1）（C ）t(1）（D ）t(2)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）设某商品的需求函数为402Q P =-（P 为商品价格），则该商品的边际收益为_________。

2014年全国硕士研究生入学统一考试数学一试题及答案解析

2014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）下列曲线中有渐近线的是（）（A ）sin y x x =+ （B ）2sin y x x =+ （C ）1sin y x x =+ （D ）21sin y x x=+ 【答案】C【考点】函数图形的渐近线【解析】对于选项A ， lim(sin )x x x →∞+ 不存在，因此没有水平渐近线，同理可知，选项A 没有铅直渐近线，而sinxlimlim x x y x x x→∞→∞+=不存在，因此选项A 中的函数没有斜渐近线；对于选项B 和D ，我们同理可知，对应的函数没有渐近线；对于C 选项，1siny x x=+.由于1sin lim lim1x x x yx x x→∞→∞+==，又()1lim 1lim sin0x x y x x →∞→∞-⋅==.所以1sin y x x=+存在斜渐近线y x =.故选C. （2）设函数()f x 具有2阶导数，()(0)(1)(1)g x f x f x =-+，则在区间[0,1]内（）（A ）当()0f x '≥时，()()f x g x ≥ （B ）当()0f x '≥时，()()f x g x ≤ （C ）当()0f x ''≥时，()()f x g x ≥ （D ）当()0f x ''≥时，()()f x g x ≤ 【答案】D【考点】函数图形的凹凸性【解析】令()()()()(0)(1)(1)F x f x g x f x f x f x =-=---有(0)(1)0F F ==，()()(0)(1)F x f x f f ''=+-，()()F x f x ''''=当()0f x ''≥时，()F x 在[0,1]上是凹的，所以()0F x ≤，从而()()f x g x ≤.选D. （3）设(,)f x y 是连续函数，则21101(,)yy dy f x y dx ---=⎰⎰（）（A ）21110010(,)(,)x x dx f x y dy dx f x y dy ---+⎰⎰⎰⎰（B ）211011(,)(,)xx dx f x y dy dx f x y dy ----+⎰⎰⎰⎰（C ）112cos sin 02(cos ,sin )(cos ,sin )d f r r dr d f r r dr ππθθπθθθθθθ++⎰⎰⎰⎰（D ）112cos sin 02(cos ,sin )(cos ,sin )d f r r rdr d f r r rdr ππθθπθθθθθθ++⎰⎰⎰⎰【答案】D【考点】交换累次积分的次序与坐标系的变换【解析】画出积分区域.21101(,)yy dy f x y dx ---=⎰⎰21111(,)+(,)x xdx f x y dy dx f x y dy ---⎰⎰⎰⎰或112cos sin 02(cos ,sin )(cos ,sin )d f r r rdr d f r r rdr ππθθπθθθθθθ++⎰⎰⎰⎰.故选D.（4）若{}2211,(cos sin )min (cos sin )a b Rx a x b x dx x a x b x dx ππππ--∈--=--⎰⎰，则11cos sin a x b x +=（）（A ）2sin x （B ）2cos x （C ）2sin x π （D ）2cos x π 【答案】A【考点】定积分的基本性质【解析】222(cos sin )[2(cos sin )(cos sin )]x a x b x dx xx a x b x a x b x dx ππππ----=-+++⎰⎰22222[2cos 2sin cos 2sin cos sin ]x ax x bx x a x ab x x b x dx ππ-=--+++⎰22222[2sin cos sin ]x bx x a x b x dx ππ-=-++⎰2222202[2sin cos sin ]x bx x a x b x dx π=-++⎰333222222222(2)(4)[(2)4]32233b a b a b b a b ππππππππ=-++=+-+=+--+故当0,2a b ==时，积分最小.故选A.（5）行列式0000000a b abc d cd=（）（A ）2()ad bc - （B ）2()ad bc -- （C ）2222a dbc - （D ）2222b c a d - 【答案】B【考点】行列式展开定理【解析】2141000000(1)0(1)000000000a b a b a b a ba c d cbcd d c d c d++=⨯-+⨯- 3323(1)(1)a b a b a d c b c d c d ++=-⨯⨯--⨯⨯-a b a bad bcc d c d=-+ 2()()a bbc ad ad bc c d=-=--.故选B. （6）设123,,ααα均为3维向量，则对任意常数,k l ，向量组1323,k l αααα++线性无关是向量组123,,ααα线性无关的（）（A ）必要非充分条件（B ）充分非必要条件（C ）充分必要条件（D ）既非充分也非必要条件【答案】A【考点】向量组的线性无关的充要条件【解析】132312310(,)(,,)01k l k l ααααααα⎛⎫ ⎪++= ⎪ ⎪⎝⎭记132312310(,),(,,),01A k l B C k l ααααααα⎛⎫⎪=++== ⎪ ⎪⎝⎭若123,,ααα线性无关，则1323()()()2,r A r BC r C k l αααα===⇒++线性无关. 由1323,k l αααα++线性无关不一定能推出123,,ααα线性无关.如：123100=0=1=0000ααα⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，，，1323,k l αααα++线性无关，但此时123,,ααα线性相关.故选A.（7）设随机事件A 与B 相互独立，且3.0)(,5.0)(=-=B A P B P ，则=-)(A B P （）（A ）0.1 (B)0.2 (C)0.3 (D)0.4 【答案】B【考点】概率的基本公式【解析】()()()()()()P A B P A P AB P A P A P B -=-=- ()0.5()0.5()0.3()0.6P A P A P A P A =-==⇒=.()()()()()()0.50.50.60.2P B A P B P AB P B P A P B -=-=-=-⨯=.故选B.（8）设连续型随机变量21,X X 相互独立，且方差均存在，21,X X 的概率密度分别为)(),(21x f x f ，随机变量1Y 的概率密度为)]()([21)(211y f y f y f Y +=，随机变量)(21212X X Y +=，则（A ）2121,DY DY EY EY >> （B ）2121,DY DY EY EY == （C ）2121,DY DY EY EY <= （D ）2121,DY DY EY EY >= 【答案】D【考点】统计量的数学期望【解析】2121()2Y X X =+，2121211[()]()22EY E X X EX EX =+=+， 2121211[()]()24DY D X X DX DX =+=+.1121()[()()]2Y f y f y f y =+，1121221[()()]()22y EY f y f y dy EX EX EY +∞-∞=+=+=⎰.2222112121[()()]()22y EY f y f y dy EX EX +∞-∞=+=+⎰， 22222111121211()()()24DY EY EY EX EX EX EX =-=+-+ 2222121212122()()24EX EX EX EX EX EX ⎡⎤=+---⋅⎣⎦ 22121212124DX DX EX EX EX EX ⎡⎤=+++-⋅⎣⎦ 221212121()()24DX DX EX EX EX EX ⎡⎤≥+++-⋅⎣⎦ 2121221()4DX DX EX EX DY ⎡⎤=++-≥⎣⎦ 1212,EY EY DY DY ∴=>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）曲面)sin 1()sin 1(22x y y x z -+-=在点)1,0,1(处的切平面方程为【答案】210x y z ---= 【考点】曲面的切平面【解析】22(,,)(1sin )(1sin )F x y z x y y x z =-+--22(1sin )cos x F x y x y '=--⋅，2cos 2(1sin )y F y x y x '=-⋅+-，1z F '=-∴(1,0,1)2x F '=，(1,0,1)1y F '=-，(1,0,1)1z F '=-曲面在点)1,0,1(处的切平面方程为2(1)(1)(0)(1)(1)0x y z -+--+--=，即210x y z ---=（10）设)(x f 是周期为4的可导奇函数，且]2,0[),1(2)(∈-='x x x f ，则=)7(f【答案】1【考点】函数的周期性【解析】由于]2,0[),1(2)(∈-='x x x f ，所以2()(1),[0,2]f x x C x =-+∈又)(x f 是奇函数，(0)0f =，解得1C =-2()(1)1,[0,2]f x x x ∴=--∈)(x f 是以4为周期的奇函数，故2(7)(3)(1)(1)[(11)1]1f f f f ==-=-=---=（11）微分方程0)ln (ln =-+'y x y y x 满足条件3)1(e y =的解为=y【答案】21x y xe+=【考点】变量可分离的微分方程【解析】(ln ln )0ln 0y xxy y x y y x y''+-=⇒+= ① 令yu x=，则y ux =，y u u x ''=+ 代入①，得ln 0u u x u u '+-=即(ln 1)u u u x-'=分离变量，得(ln 1)(ln 1)ln 1du d u dxu u u x-==--两边积分得1ln ln 1ln u x C -=+，即ln 1u Cx -=即ln 1yCx x-= 代入初值条件3)1(e y =，可得2C =，即ln 12yx x-= 整理可得21x y xe +=.（12）设L 是柱面122=+y x 与平面0=+z y 的交线，从z 轴正向往z 轴负向看去为逆时针方向，则曲线积分⎰=+Lydz zdx【答案】π【考点】斯托克斯公式【解析】由斯托克斯公式，得0xyLD dydz dzdx dxdyzdx ydz dydz dzdx dydz dzdx x y z z yπ∑∑∂∂∂+==+=+=∂∂∂⎰⎰⎰⎰⎰⎰⎰其中{}22(,)1xy D x y x y =+≤（13）设二次型3231222132142),,(x x x ax x x x x x f ++-=的负惯性指数为1，则a 的取值范围是【答案】]2,2[-【考点】惯性指数、矩阵的特征值、配方法化二次型为标准形【详解】【解法一】二次型对应的系数矩阵为：O a a ≠⎪⎪⎪⎭⎫⎝⎛-0221001，记特征值为321,,λλλ则0011)(321=+-==++A tr λλλ，即特征值必有正有负，共3种情况；故二次型的负惯性指数为⇔1特征值1负2正或1负1正1零；0402210012≤+-=-⇔a a a，即]2,2[-∈a【解法二】2222222212312132311332233(,,)2424f x x x x x ax x x x x ax x a x x x x a x =-++=++-+- 2222222213233123()(2)(4)(4)x ax x x a x y y a y =+--+-=-+-若负惯性指数为1，则240[2,2]a a -≥⇒∈-（14）设总体X 的概率密度为⎪⎩⎪⎨⎧<<=其他,02,32),(2θθθθx xx f ，其中θ是未知参数，n X X X ,,,21 为来自总体X 的简单随机样本，若∑=ni i X c 12是2θ的无偏估计，则=c【答案】n52【考点】统计量的数字特征【解析】根据题意，有322222112()()()3n ni i i i x E c X c E X ncE X nc dx θθθ=====∑∑⎰4222221523425nc nc x c nθθθθθ=⋅==∴= 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）（本题满分10分）求极限)11ln(])1([lim2112xx dtt e t xtx +--⎰+∞→【考点】函数求极限、变限积分函数求导、等价无穷小、洛必达法则【详解】11221122((1))((1))limlim11ln(1)xxttx x t e t dt t e t dtx x xx→+∞→+∞----=+⋅⎰⎰1122(1)1lim lim (1)1xx x x x e x x e x→+∞→+∞--==-- 2001111lim lim 22t t t t e t e t x t t ++→→---===令（16）（本题满分10分）设函数)(x f y =由方程322+60y xy x y ++=确定，求)(x f 的极值【考点】极值的必要条件【解析】对方程两边直接求导：2223220y y x y xy y xyy '''++++= ① 令0y '=，得2y x =-，或0y =（舍去）将2y x =-代入原方程得 3660x -+= 解得1x =，此时2y =-. 对①式两端再求导，得222(32)2(3)()4()20y xy x y y x y y x y y ''''+++++++=将1x =，2y =-，0y '=代入上式，得 409y ''=>，即4(1)09f ''=> ()y f x ∴=在1x =处取极小值，极小值为(1)2f =-.（17）（本题满分10分）设函数)(u f 具有2阶连续导数，)cos (y e f z x=满足22222(4cos )x xz z z e y e x y∂∂+=+∂∂，若0)0(,0)0(='=f f ，求)(u f 的表达式. 【考点】多元函数求偏导、二阶常系数非齐次线性微分方程【解析】由)cos (y e f z x=，知(cos )cos x x z f e y e y x ∂'=⋅∂，(cos )(sin )x x zf e y e y y∂'=⋅-∂ 22(cos )cos cos (cos )cos x x x x xz f e y e y e y f e y e y x∂'''=⋅⋅+⋅∂， 22(cos )(sin )(sin )(cos )(cos )x x x x xz f e y e y e y f e y e y y∂'''=⋅-⋅-+⋅-∂ 由22222(4cos )x x z zz e y e x y∂∂+=+∂∂，代入得 22(cos )[4(cos )cos ]x x x x x f e y e f e y e y e ''⋅=+即(cos )4(cos )cos x x x f e y f e y e y ''-= 令cos x u e y =，则()4()f u f u u ''-= 特征方程212402,2r r r -=⇒==- 齐次方程通解为2212uu y C eC e -=+设特解*y au b =+，代入方程得1,04a b =-=，特解*14y u =- 原方程的通解为221214uu y C eC e u -=+-由(0)0,(0)0f f '==，得 1211,1616C C ==- 22111()16164u u y f u e e u -∴==--（18）（本题满分10分）设∑为曲面)1(22≤+=z y x z 的上侧，计算曲面积分dxdy z dzdx y dydz x I )1()1()1(33-+-+-=⎰⎰∑【考点】高斯公式【解析】因∑不封闭，添加辅助面2211:1x y z ⎧+≤∑⎨=⎩，方向向上．133(x 1)(y 1)(z 1)dydz dzdx dxdy ∑+∑-+-+-⎰⎰22(3(1)3(1)1)x y dxdydz Ω=-+-+⎰⎰⎰22(3633631)x x y y dxdydz Ω=++++++⎰⎰⎰ 22(337)x y dxdydz Ω=++⎰⎰⎰1220(z)(337)D dz x y dxdy =++⎰⎰⎰1220(37)4zdz d r rdr πθπ=+=⎰⎰⎰（其中(66)0x y dxdydz Ω+=⎰⎰⎰，因为积分区域关于,xoz yoz对称，积分函数(,)66f x y x y =+分别是,y x 的奇函数．）在曲面1∑上，133(1)(1)(1)0x dydz y dzdx z dxdy ∑-+-+-=⎰⎰故33(1)(1)(1)4x dydz y dzdx z dxdy π∑-+-+-=-⎰⎰ ．（19）（本题满分10分）设数列}{},{n n b a 满足n n n n n b a a b a cos cos ,20,20=-<<<<ππ，且级数1n n b ∞=∑收敛.（I ）证明：;0lim =∞→n n a（II ）证明：级数∑∞=1n nnb a 收敛. 【考点】级数敛散性的判别【解析】证明：（I ）cos cos cos cos n n n n n n a a b a a b -=⇒=-0,022n n a b ππ<<<<，cos cos 00n n n n a b a b ∴->⇒<<级数1n n b ∞=∑收敛，∴级数1n n a ∞=∑收敛，lim 0n n a →∞=.（II ）解法1：2sinsin cos cos 22n n n nn n nn nna b a ba ab b b b +---== 02n a π<<，02n b π<<，sin,sin 2222n n n n n n n n a b a b a b a b++--∴≤≤ 222222n n n nn n n nn n a b a b a b a b b b +--⋅-∴≤=222n n n b b b ≤= 02n a π<<，02n b π<<，且级数1nn b∞=∑收敛，∴级数∑∞=1n nnb a 收敛. 解法2：cos cos 1cos n n n nn n na ab b b b b --=≤21cos 1cos 1lim lim 2n n n n n n n b b b b b →∞→∞--== ∵同阶无穷小有相同的敛散性，∴由1n n b ∞=∑⇒ 11cos n n n b b ∞=-∑收敛⇒∑∞=1n n n b a收敛（20）（本题满分11分）设E A ,302111104321⎪⎪⎪⎭⎫⎝⎛----=为3阶单位矩阵.（I ）求方程组0=Ax 的一个基础解系；（II ）求满足E AB =的所有矩阵B .【考点】齐次线性方程组的基础解系、非齐次线性方程组的通解【详解】对矩阵()A E 施以初等行变换1234100()01110101203001A E --⎛⎫ ⎪=- ⎪ ⎪-⎝⎭1205412301021310013141--⎛⎫ ⎪→--- ⎪ ⎪--⎝⎭ 100126101021310013141-⎛⎫ ⎪→--- ⎪ ⎪---⎝⎭（I ）方程组0=Ax 的同解方程组为⎪⎪⎩⎪⎪⎨⎧===-=4443424132x x x x xx x x ，即基础解系为⎪⎪⎪⎪⎪⎭⎫⎝⎛-1321（II ）⎪⎪⎪⎭⎫ ⎝⎛=001Ax 的同解方程组为：⎪⎪⎩⎪⎪⎨⎧+=-=-=+-=01312244434241x x x x x x x x ，即通解为⎪⎪⎪⎪⎪⎭⎫ ⎝⎛--+⎪⎪⎪⎪⎪⎭⎫ ⎝⎛-011213211k⎪⎪⎪⎭⎫ ⎝⎛=010Ax 的同解方程组为：⎪⎪⎩⎪⎪⎨⎧+=-=-=+-=04332644434241x x x x x x x x ，即通解为⎪⎪⎪⎪⎪⎭⎫⎝⎛--+⎪⎪⎪⎪⎪⎭⎫ ⎝⎛-043613212k ⎪⎪⎪⎭⎫ ⎝⎛=100Ax 的同解方程组为：⎪⎪⎩⎪⎪⎨⎧+=+=+=--=01312144434241x x x x x x x x ，即通解为⎪⎪⎪⎪⎪⎭⎫⎝⎛-+⎪⎪⎪⎪⎪⎭⎫ ⎝⎛-011113213k ，123123123123261212321313431k k k k k k B k k k k k k -+-+--⎛⎫⎪--+ ⎪∴= ⎪--+ ⎪⎝⎭，321,,k k k 为任意常数（21）（本题满分11分）证明：n 阶矩阵⎪⎪⎪⎪⎪⎭⎫⎝⎛111111111与⎪⎪⎪⎪⎪⎭⎫⎝⎛n 00200100 相似【考点】矩阵的特征值、相似对角化【详解】设111111111A ⎛⎫ ⎪ ⎪= ⎪⎪⎝⎭，0010020B n ⎛⎫ ⎪ ⎪=⎪ ⎪⎝⎭因为()1r A =，()1r B =所以A 的特征值为：n A tr n n ======-)(,0121λλλλB 的特征值为：n B tr n n =='='=='='-)(,0121λλλλ 关于A 的0特征值，因为1)()()0(==-=-A r A r A E r ，故有1-n 个线性无关的特征向量，即A 必可相似对角化于⎪⎪⎪⎪⎪⎭⎫⎝⎛n 00 同理，关于B 的0特征值，因为1)()()0(==-=-B r B r B E r ，故有1-n 个线性无关的特征向量，即B 必可相似对角化于⎪⎪⎪⎪⎪⎭⎫ ⎝⎛n 00 由相似矩阵的传递性可知，A 与B 相似. （22）（本题满分11分）设随机变量X 的概率分布为21}2{}1{====X P X P ，在给定i X =的条件下，随机变量Y 服从均匀分布)2,1)(,0(=i i U ，（I ）求Y 的分布函数)(y F Y ；（II ）求EY【考点】一维随机变量函数的分布、随机变量的数字特征（期望）【详解】（I ）()()y F y P Y y =≤(1)(1)(2)(2)P X P Y y X P X P Y y X ==≤=+=≤=11(1)(2)22P Y y X P Y y X =≤=+≤= ① 当0y < 时，(y)0Y F =② 当01y ≤<时，1113(y)2224Y F y y y =+⨯= ③ 当12y ≤<时，1111(y)22224Y yF y =+⨯=+④ 当2y ≥时，11(y)122Y F =+=综上：003y 014(y)1122412Y y y F y y y <⎧⎪⎪≤<⎪=⎨⎪+≤<⎪⎪≥⎩（II ）随机变量Y 的概率密度为'30141(y)(y)1240Y Y y f F y ⎧<<⎪⎪⎪==≤<⎨⎪⎪⎪⎩其他12-013131133()4442424Y EY yf y dy ydy ydy +∞∞==+=⨯+⨯=⎰⎰⎰ （23）（本题满分11分）设总体X 的分布函数21,0(;)00x e x F x x θθ-⎧⎪-≥=⎨⎪<⎩，，其中θ是未知参数且大于零，12,,,n X X X 为来自总体X 的简单随机样本.（Ⅰ）求EX 与2EX ；（Ⅱ）求θ的最大似然估计量ˆnθ；（Ⅲ）是否存在实数a ，使得对任何0ε>，都有{}ˆlim 0nn P a θε→∞-≥=？【考点】统计量的数字特征、最大似然估计、估计量的评选标准（无偏性）【解析】（Ⅰ）X 的概率密度为22,0(;)(;)0,0xx e x f x F x x θθθθ-⎧⎪≥'==⎨⎪<⎩222()(;)()x x xE X xf x dx x edx xd eθθθθ--+∞+∞+∞-∞==⋅=-⎰⎰⎰22200012222x x x xeedx edx θθθθπθπ+∞---+∞+∞=-+==⋅=⎰⎰ 22222202()(;)()x x xE X x f x dx x edx x d e θθθθ--+∞+∞+∞-∞==⋅=-⎰⎰⎰2222200222x x x x xx ex edx x edx edx θθθθθθθ+∞----+∞+∞+∞=-+⋅=⋅=⋅=⎰⎰⎰（Ⅱ）设12,,,n x x x 为样本的观测值，似然函数为2112(),0(1,2,,),()(;)0,0ix n n ni i i i i x e x i n L f x x θθθθ-==⎧≥=⎪==⎨⎪<⎩∏∏当0(1,2,,)i x i n ≥= 时，22111122()()()ni i i x nn x nn i i i i L x ex eθθθθθ=--==∑==∏∏两边取对数，得2211112121ln ()lnln lnln nnnni ii ii i i i L n x x n x x θθθθθ=====+-=+-∑∑∑∏两边求导，得221ln ()1nii d L n xd θθθθ==-+∑令ln ()0d L d θθ=，得211n i i x n θ==∑所以，θ的最大似然估计量为211ˆn i i X n θ==∑.（Ⅲ）存在a θ=.因为{}2n X 是独立同分布的随机变量序列，且21EX θ=<+∞，所以根据辛钦大数定律，当n →∞时，211ˆnn i i X n θ==∑依概率收敛于21EX ，即θ. 所以对于任何0ε>都有{}ˆlim 0nn Pθθε→∞-≥=.。

2014-2015年安大英语考研真题回忆版

2014-2015年安大英语考研真题回忆版2014回忆版(论坛)基础英语：今年与往年不同，练习册上的原题目几乎是没有。

总共有七大板块：Part 1 V ocabulary(30*1') Part 2 Derivation（20*1'）Part 3 Blank Filling（20*1'）Part 4 Explaining Meaning of the Words and the Phrases(10*2') Part5 Paraphrase(5*4') Part 6 Erroring (10*2') Part 7 Reading Comprehension(10*2')选择题总共有30题，前10题据有的同学说是专四真题，反正我只从网上打印了09到13年的真题，答案也是网上给的。

我只确定有一题我有做过。

He __fifty or so when I first met him in a party. A. had been B. should be C. could be D. must have been 我当时选的是D选项，可是记得当时给的参考答案给的是C。

我当时也没细细去查，以为是“or so”的缘故。

说这个例子，主要是想对学妹学弟们说要弄清楚自己做过的每一道题，不然还不如不做得好。

然后词汇这一块，比往年要简单些，比练习册上的词汇练习要简单些。

单词变型我目前确定我有三处错误：（explore）______discussion应该填exploratory”探索性的；He is the only _____in the earthquake（survive）我记得做过类似的题，我记得survivor不存在，哎，就这样我有错了一题；还有一个到现在还不知道：Her mother ______his father three years.(live) 。

2014年全国考研数学一真题

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2014 年全国硕士研究生入学统一考试真题试卷
数学一试题
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2014 年全国硕士研究生入学统一考试真题试卷
数学一试题参考答案
一、选择题：1～8 小题，每小题 4 分，共 32 分，下列每小题给出的四个选项中，只有一项符合题目要求的，请将所选项前的字母填在答．题纸指定位置上. ．．（1）B （8）（D）二、填空题：914 小题，每小题 4 分，共 24 分，请将答案写在答．题纸指定位置上．． . （11） ln 2 x 1 （9） 2 x y z 1 0 （10） f ( 1 ) 1 （2）D （3）D （4）B （5）B （6）A （7）（B）
所以 y(1 ) 2 为极小值。（17）【答案】
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[整理]2014 2015考研英语真题

2014年英语（一）真题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 the ANSWER SHEET.(10 points)As many people hit middle age, they often start to notice that their memory and mental clarity are not what they used to be. We suddenly can’t remember ___1___ we put the keys just a moment ago, or an old acquaintance’s name, or the name of an old band we used to love. As the brain ___2___,we refer to these occurrences as "senior moments." ___3___ seemingly innocent, this loss of mental focus can potentially have a (n) ___4___ impact on our professional, social,and personal ___5___.Neuroscientists, experts who study the nervous system, are increasingly showing that there’s actually a lot that can be done. It ___6___ out that the brain needs exercise in much the same way our muscles do, and the right mental ___7___ can significantly improve our basic cognitive ___8___. Thinking is essentially a ___9___ of making connections in the brain. To a certain extent, our ability to ___10___ in making the connections that drive intelligence is inherited. ___11___,because these connections are made through effort and practice, scientists believe that intelligence can expand and fluctuate ___12___ mental effort.Now, a new Web-based company has taken it a step ___13___ and developed the first "brain training program" designed to actually help people improve and regain their mental ___14___.The Web-based program ___15___ you to systematically improve your memory and attention skills. The program keeps ___16___ of your progress and provides detailed feedback ___17___ your performance and improvement. Most importantly, it ___18___modifies and enhances the games you play to ___19___ on the strengths you are developing—much like a(n) ___20___exercise routine requires you to increase resistance and vary your muscle use.1. [A]where [B]when [C]that [D]why2. [A]improves [B]fades [C]recovers [D]collapses3. [A]If [B]Unless [C]Once [D]While4. [A]uneven [B]limited [C]damaging [D]obscure5. [A]wellbeing [B]environment [C]relationship [D]outlook6. [A]turns [B]finds [C]points [D]figures7. [A]roundabouts [B]responses [C]workouts [D]associations8. [A]genre [B]functions [C]circumstances [D]criterion9. [A]channel [B]condition [C]sequence [D]process10. [A]persist [B]believe [C]excel [D]feature11. [A] Therefore [B] Moreover [C] Otherwise [D] However12. [A]according to [B]regardless of [C]apart from [D]instead of13. [A]back [B]further [C]aside [D]around14. [A]sharpness [B]stability [C]framework [D]flexibility15. [A]forces [B]reminds [C]hurries [D]allows16. [A]hold [B]track [C]order [D]pace17. [A]to [B]with [C]for [D]on18. [A]irregularly [B]habitually [C]constantly [D]unusually19. [A]carry [B]put [C]build [D]take20. [A]risky [B]effective [C]idle [D]familiarSection 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 the ANSWER SHEET. (40 points)Text 1In order to "change lives for the better" and reduce "dependency" George Osborne, Chancellor of the Exchequer, introduced the "upfront work search" scheme. Only if the jobless arrive at the jobcentre with a CV, register for online job search, and start looking for work will they be eligible for benefit and then they should report weekly rather than fortnightly. What could be more reasonable?More apparent reasonableness followed. There will now be a seven-day wait for the jobseeker’s allowance. "Those first few days should be spent looking for work, not looking to sign on." he claimed. "We’re doing these things because we know they help people stay off benefits and help those on benefits get into work faster." Help? Really? On first hearing, this was the socially concerned chancellor, trying to change lives for the better, complete with "reforms" to an obviously indulgent system that demands too little effort from the newly unemployed to find work, and subsidises laziness. What motivated him, we were to understand, was his zeal for "fundamental fairness"—protecting the taxpayer, controlling spending and ensuring that only the most deserving claimants received their benefits.Losing a job is hurting: you don’t skip down to the jobcentre with a song in your heart, delighted at the prospect of doubling your income from the generous state. It is financially terrifying, psychologically embarrassing and you know that support is minimal and extraordinarily hard to get. You are now not wanted; you support is minimal and extraordinarily hard to get. You are now not wanted; you are now excluded from the work environment that offers purpose and structure in your life. Worse, the crucial income to feed yourself and your family and pay the bills hasdisappeared. Ask anyone newly unemployed what they want and the answer is always: a job.But in Osborneland, your first instinct is to fall into dependency — permanent dependency if you can get it —supported by a state only too ready to indulge your falsehood. It is as though 20 years of ever-tougher reforms of the job search and benefit administration system never happened. The principle of British welfare is no longer that you can insure yourself against the risk of unemployment and receive unconditional payments if the disaster happens. Even the very phrase "jobseeker’s allowance" —invented in 1996 —is about redefining the unemployed as a "jobseeker" who had no mandatory right to a benefit he or she has earned through making national insurance contributions. Instead, the claimant receives a time-limited "allowance," conditional on actively seeking a job; no entitlement and no insurance, at £71.70 a week, one of the least generous in the EU.21. George Osborne’s scheme was intended to[A]provide the unemployed with easier access to benefits.[B]encourage jobseekers’ active engagement in job seeking.[C]motivate the unemployed to report voluntarily.[D]guarantee jobseekers’ legitimate right to benefits.22. The phrase, "to sign on" (Line 3,Para. 2) most probably means[A]to check on the availability of jobs at the jobcentre.[B]to accept the government’s restrictions on the allowance.[C]to register for an allowance from the government.[D]to attend a governmental job-training program.23. What prompted the chancellor to develop his scheme?[A]A desire to secure a better life for all.[B]An eagerness to protect the unemployed.[C]An urge to be generous to the claimants.[D]A passion to ensure fairness for taxpayers.24. According to Paragraph 3,being unemployed makes one feel[A]uneasy[B]enraged.[C]insulted.[D]guilty.25. To which of the following would the author most probably agree?[A]The British welfare system indulges jobseekers’ laziness.[B]Osborne’s reforms will reduce the risk of unemployment.[C]The jobseekers’ allowance has met their actual needs.[D]Unemployment benefits should not be made conditional.Text 2All around the world, lawyers generate more hostility than the members of any other profession—with the possible exception of journalism. But there are few places where clients have more grounds for complaint than America.During the decade before the economic crisis, spending on legal services in America grew twice as fast as inflation. The best lawyers made skyscrapers-full of money, tempting ever more students to pile into law schools. But most law graduates never get a big-firm job. Many of them instead become the kind of nuisance-lawsuit filer that makes the tort system a costly nightmare.There are many reasons for this. One is the excessive costs of a legal education. There is just one path for a lawyer in most American states: a four-year undergraduate degree in some unrelated subject, then a three-year law degree at one of 200 law schools authorized by the American Bar Association and an expensive preparation for the bar exam. This leaves today’s average law-school graduate with $100,000 of debt on top of undergraduate debts. Law-school debt means that many cannot afford to go into government or non-profit work, and that they have to work fearsomely hard.Reforming the system would help both lawyers and their customers. Sensible ideas have been around for a long time,but the state-level bodies that govern the profession have been too conservative to implement them. One idea is to allow people to study law as an undergraduate degree. Another is to let students sit for the bar after only two years of law school. If the bar exam is truly a stern enough test for a would-be lawyer, those who can sit it earlier should be allowed todo so. Students who do not need the extra training could cut their debt mountain by a third.The other reason why costs are so high is the restrictive guild-like ownership structure of the business. Except in the District of Columbia, non-lawyers may not own any share of a law firm. This keeps fees high and innovation slow. There is pressure for change from within the profession, but opponents of change among the regulators insist that keeping outsiders out of a law firm isolates lawyers from the pressure to make money rather than serve clients ethically.In fact, allowing non-lawyers to own shares in law firms would reduce costs and improve services to customers, by encouraging law firms to use technology and to employ professional managers to focus on improving firms’ efficiency. After all, other countries, such as Australia and Britain, have started liberalizing their legal professions. America should follow.26.a lot of students take up law as their profession due to[A]the growing demand from clients.[B]the increasing pressure of inflation.[C]the prospect of working in big firms.[D]the attraction of financial rewards.27.Which of the following adds to the costs of legal education in most American states?[A]Higher tuition fees for undergraduate studies.[B]Admissions approval from the bar association.[C]Pursuing a bachel or’s degree in another major.[D]Receiving training by professional associations.28.Hindrance to the reform of the legal system originates from[A]lawyers’ and clients’ strong resistance.[B]the rigid bodies governing the profession.[C]the stem exam for would-be lawyers.[D]non-professionals’ sharp criticism.29.The guild-like ownership structure is considered "restrictive" partly because it[A]bans outsiders’ involvement in the profession.[B]keeps lawyers from holding law-firm shares.[C]aggravates the ethical situation in the trade.[D]prevents lawyers from gaining due profits.30.In this text,the author mainly discusses[A]flawed ownership of America’s law firms and its causes.[B]the factors that help make a successful lawyer in America.[C]a problem in America’s legal profession and solutions to it.[D]the role of undergraduate studies in America’s legal education.Text 3The US$3-million Fundamental physics prize is indeed an interesting experiment,as Alexander Polyakov said when he accepted this year’s award in March. And it is far from the only one of its type. As a News Feature article in Nature discusses,a string of lucrative awards for researchers have joined the Nobel Prizes in recent years. Many,like the Fundamental Physics Prize,are funded from the telephone-number-sized bank accounts of Internet entrepreneurs. These benefactors have succeeded in their chosen fields,they say,and they want to use their wealth to draw attention to those who have succeeded in science.What’s not to like? Quite a lot,according to a handful of scientists quoted in the News Feature. You cannot buy class,as the old saying goes,and these upstart entrepreneurs cannot buy their prizes the prestige of the Nobels,The new awards are an exercise in self-promotion for those behind them,say scientists. They could distort the achievement-based system of peer-review-led research. They could cement the status quo of peer-reviewed research.They do not fund peer-reviewed research. They perpetuate the myth of the lone genius.The goals of the prize-givers seem as scattered as the criticism. Some want to shock,others to draw people into science,or to better reward those who have made their careers in research.As Nature has pointed out before,there are some legitimate concerns about how science prizes—both new and old—are distributed. The Breakthrough Prize in Life Sciences,launched this year,takes an unrepresentative view of what the life sciences include. But the Nobel Foundation’s limit of t hree recipients per prize,each of whom must still be living,has long been outgrown by the collaborative nature of modern research—as will be demonstrated by the inevitable row over who is ignored when it comes to acknowledging the discovery of the Higgs boson. The Nobels were,of course,themselves set up by a very rich individual who had decided what he wanted to do with his own money. Time,rather than intention,has given them legitimacy.As much as some scientists may complain about the new awards,two things seem clear. First,most researchers would accept such a prize if they were offered one. Second,it is surely a good thing that the money and attention come to science rather than go elsewhere,It is fair to criticize and question the mechanism—that is the culture of research,after all—but it is the prize-givers’ money to do with as they please. It is wise to take such gifts with gratitude and grace.31. The Fundamental Physics Prize is seen as[A]a symbol of the entrepreneurs’ wealth.[B]a possible replacement of the Nobel Prizes.[C]an example of bankers’ investments.[D]a handsome reward for researchers.32. The critics think that the new awards will most benefit[A]the profit-oriented scientists.[B]the founders of the new awards.[C]the achievement-based system.[D]peer-review-led research.33. The discovery of the Higgs boson is a typical case which involves[A]controversies over the recipients’ status.[B]the joint effort of modern researchers.[C]legitimate concerns over the new prizes.[D]the demonstration of research findings.34. According to Paragraph 4,which of the following is true of the Nobels?[A]Their endurance has done justice to them.[B]Their legitimacy has long been in dispute.[C]They are the most representative honor.[D]History has never cast doubt on them.35.The author believes that the now awards are[A]acceptable despite the criticism.[B]harmful to the culture of research.[C]subject to undesirable changes.[D]unworthy of public attention.Text 4"The Heart of the Matter," the just-released report by the American Academy of Arts and Sciences (AAAS),deserves praise for affirming the importance of the humanities and social sciences to the prosperity and security of liberal democracy in America. Regrettably,however,the report’s failure to address the true nature of the crisis facing liberal education may cause more harm than good.In 2010,leading congressional Democrats and Republicans sent letters to the AAAS asking that it identify actions that could be taken by "federal,state and local governments,universities,foundations,educators,individual benefactors and others" to "maintain national excellence in humanities and social scientific scholarship and education." In response,the American Academy formed the Commission on the Humanities and Social Sciences. Among the commission’s 51 members are top-tier-university presidents,scholars,lawyers,judges,and business executives,as well as prominent figures from diplomacy,filmmaking,music and journalism.The goals identified in the report are generally admirable. Because representative government presupposes an informed citizenry,the report supports full literacy; stresses the study of history and government,particularly American history and American government; and encourages the use of new digital technologies. To encourage innovation and competition,the report calls for increased investment in research,the crafting of coherent curricula that improve students’ ability to solve problems and communicate effectively in the 21st century,increased funding for teachers and the encouragement of scholars to bring their learning to bear on the great challenges of the day. The report also advocates greater study of foreign languages,international affairs and the expansion of study abroad programs.Unfortunately,despite 2½ years in the making,"The Heart of the Matter" never gets to the heart of the matter: the illiberal nature of liberal education at our leading colleges and universities. The commission ignores that for several decades America's colleges and universities have produced graduates who don’t know the content and character of liberal education and are thus deprived of its benefits. Sadly,the spirit of inquiry once at home on campus has been replaced by the use of the humanities and social sciences as vehicles for publicizing "progressive," or left-liberal propaganda.Today,professors routinely treat the progressive interpretation of history and progressive public policy as theproper subject of study while portraying conservative or classical liberal ideas—such as free markets and self-reliance—as falling outside the boundaries of routine,and sometimes legitimate,intellectual investigation.The AAAS displays great enthusiasm for liberal education. Yet its report may well set back reform by obscuring the depth and breadth of the challenge that Congress asked it to illuminate.36. According to Paragraph 1,what is the author’s attitude toward the AAAS’s report?[A] Critical[B] Appreciative[C] Contemptuous[D] Tolerant37. Influential figures in the Congress required that the AAAS report on how to[A] retain people’s interest in liberal education[B] define the government’s role in education[C] keep a leading position in liberal education[D] safeguard individuals’ rights to education38. According to Paragraph 3,the report suggests[A] an exclusive study of American history[B] a greater emphasis on theoretical subjects[C] the application of emerging technologies[D] funding for the study of foreign languages39. The author implies in Paragraph 5 that professors are[A] supportive of free markets[B] cautious about intellectual investigation[C] conservative about public policy[D] biased against classical liberal ideas40. Which of the following would be the best title for the text?[A] Ways to Grasp "The Heart of the Matter"[B] Illiberal Education and "The Heart of the Matter"[C] The AAAS’s Contribution to Liberal Education[D] Progressive Policy vs. Liberal EducationPart BDirections:The following paragraphs are given in a wrong order. For Questions 41-45,you are required to reorganize these paragraphs into a coherent text by choosing from the list A-G and filling them into the numbered boxes. ParagraphsA and E have been correctly placed Mark your answers on the ANSWER SHEET (10 points)[A] Some archaeological sites have always been easily observable—for example,the Parthenon in Athens,Greece,the pyramids of Giza in Egypt; and the megaliths of Stonehenge in southern England. But these sites are exceptions to the norm. Most archaeological sites have been located by means of careful searching,while many others have been discovered by accident. Olduvai Gorge,an early hominid site in Tanzania,was found by a butterfly hunter who literally fell into its deep valley in 1911. Thousands of Aztec artifacts came to light during the digging of the Mexico City subway in the 1970s.[B]In another case,American archaeologists Rene Million and George Cowgill spent years systematically mapping the entire city of Teotihuacan in the Valley of Mexico near what is now Mexico City. At its peak around AD 600,this city was one of the largest human settlements in the world. The researchers ma pped not only the city’s vast and ornate ceremonial areas,but also hundreds of simpler apartment complexes where common people lived.[C] How do archaeologists know where to find what they are looking for when there is nothing visible on the surface of the ground? Typically,they survey and sample (make test excavations on) large areas of terrain to determine where excavation will yield useful information. Surveys and test samples have also become important for understanding the larger landscapes that contain archaeological sites.[D] Surveys can cover a single large settlement or entire landscapes. In one case,many researchers working around the ancient Maya city of Copan,Honduras,have located hundreds of small rural villages and individual dwellings by using aerial photographs and by making surveys on foot. The resulting settlement maps show how the distribution and density of the rural population around the city changed dramatically between AD 500 and 850,when Copan collapsed.[E] To find their sites,archaeologists today rely heavily on systematic survey methods and a variety of high-technology tools and techniques. Airborne technologies,such as different types of radar and photographic equipment carried by airplanes or spacecraft,allow archaeologists to learn about what lies beneath the ground without digging. Aerial surveys locate general areas of interest or larger buried features,such as ancient buildings or fields.[F] Most archaeological sites,however,are discovered by archaeologists who have set out to look for them. Such searches can take years. British archaeologist Howard Carter knew that the tomb of the Egyptian pharaoh Tutankhamun existed from information found in other sites. Carter sifted through rubble in the Valley of the Kings for seven years before he located the tomb in 1922. In the late 1800s British archaeologist Sir Arthur Evan combed antique dealers’ stores in Athens,Greece. He was searching for tiny engraved seals attributed to the ancient Mycenaean culture that dominated Gre ece from the 1400s to 1200s BC. Evans’s interpretations of these engravings eventually led him to find the Minoan palace at Knossos (Knossós) on the island of Crete,in 1900.[G] Ground surveys allow archaeologists to pinpoint the places where digs will be successful. Most ground surveys involve a lot of walking,looking for surface clues such as small fragments of pottery. They often include acertain amount of digging to test for buried materials at selected points across a landscape. Archaeologists also may locate buried remains by using such technologies as ground radar,magnetic-field recording,and metal detectors. Archaeologists commonly use computers to map sites and the landscapes around sites. Two and three-dimensional maps are helpful tools in planning excavations,illustrating how sites look,and presenting the results of archaeological research.Part CDirections:Read the following text carefully and then translate the underlined segments into Chinese. Your translation should be written neatly on the ANSWER SHEET. (10 points)Music means different things to different people and sometimes even different things to the same person at different moments of his life. It might be poetic,philosophical,sensual,or mathematical,but in any case it must,in my view,have something to do with the soul of the human being. Hence it is metaphysical; but the means of expression is purely and exclusively physical: sound. I believe it is precisely this permanent coexistence of metaphysical message through physical means that is the strength of music. (46)It is also the reason why when we try to describe music with words,all we can do is articulate our reactions to it,and not grasp music itself.Beethoven’s importance in music has been principally defined by the rev olutionary nature of his compositions. He freed music from hitherto prevailing conventions of harmony and structure. Sometimes I feel in his late works a will to break all signs of continuity. The music is abrupt and seemingly disconnected,as in the last piano sonata. In musical expression,he did not feel restrained by the weight of convention. (47)By all accounts he was a freethinking person,and a courageous one,and I find courage an essential quality for the understanding,let alone the performance,of his works.This courageous attitude in fact becomes a requirement for the performers of Beethoven’s music. His compositions demand the performer to show courage,for example in the use of dynamics. (48)Beethoven’s habit of increasing the volume with an intense crescendo and then abruptly following it with a sudden soft passage was only rarely used by composers before him.Beethoven was a deeply political man in the broadest sense of the word. He was not interested in daily politics,but concerned with questions of moral behavior and the larger questions of right and wrong affecting the entire society. (49)Especially significant was his view of freedom,which,for him,was associated with the rights and responsibilities of the individual: he advocated freedom of thought and of personal expression.Beethoven’s music tends to move from chaos to order as if order were an imperative of human existence. For him,order does not result from forgetting or ignoring the disorders that plague our existence; order is a necessary development,an improvement that may lead to the Greek ideal of spiritual elevation. It is not by chance that theFuneral March is not the last movement of the Eroica Symphony,but the second,so that suffering does not have the last word. (50)One could interpret much of the work of Beethoven by saying that suffering is inevitable,but the courage to fight it renders life worth living.46. It is also the reason why when we try to describe music with words,all we can do is articulate our reactions to it,and not grasp music itself.47. By all accounts he was a freethinking person,and a courageous one,and I find courage an essential quality for the understanding,let alone the performance,of his works.48. Beethoven's habit of increasing the volume with an extreme intensity and then abruptly following it with a sudden soft passage was only rarely used by composers before him.49.Especially significant was his view of freedom,which,for him,was associated with the rights and responsibilities of the individual: he advocated freedom of thought and of personal expression.50.One could interpret much of the work of Beethoven by saying that suffering is inevitable,but the courage to fight it renders life worth living.Section Ⅲ WritingPart A51. Directions:Write a letter of about 100 words to the president of your university,suggesting how to improve students’ physical condition.You should include the details you think necessary.You should write neatly on the ANSWER SHEET.Do not sign your own 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 briefly,2) interpret its intended meaning,and3) give your comments.You should write neatly on the ANSWER SHEET(20 points)2015年考研英语(一)真题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. (10 points)Though not biologically related,friends are as ―related‖ as fourth cousins,sharing about 1% of genes. That is _(1)_a study,published from the University of California and Yale University in the Proceedings of the National Academy of Sciences,has__(2)_.The study is a genome-wide analysis conducted _(3)__1,932 unique subjects which __(4)__pairs of unrelated friends and unrelated strangers. The same people were used in both_(5)_.While 1% may seem_(6)_,it is not so to a geneticist. As James Fowler,professor of medical genetics at UC San Diego,says,―Most people do not even _(7)_their fourth cousins but somehow manage to select as friends the people who_(8)_our kin.‖The study_(9)_found that the genes for smell were something shared in friends but not genes for immunity .Why this similarity exists in smell genes is difficult to explain,for now,_(10)_,as the team suggests,it draws us to similar environments but there is more_(11)_it. There could be many mechanisms working together that _(12)_us in choosing genetically similar friends_(13)_‖functional Kinship‖ of being friends with_(14)_!One of the remarkable findings of the study was the similar genes seem to be evolution_(15)_than other genes Studying this could help_(16)_why human evolution picked pace in the last 30,000 years,with social environment being a major_(17)_factor.The findings do not simply explain people’s_(18)_to befriend those of similar_(19)_backgrounds,say the researchers. Though all the subjects were drawn from a population of European extraction,care was taken to_(20)_that all subjects,friends and strangers,were taken from the same population.1. [A] when [B] why [C] how [D] what2. [A] defended [B] concluded [C] withdrawn [D] advised3. [A] for [B] with [C] on [D] by4. [A] compared [B] sought [C] separated [D] connected5. [A] tests [B] objects [C]samples [D] examples6. [A] insignificant [B] unexpected [C]unbelievable [D] incredible7. [A] visit [B] miss [C] seek [D] know8. [A] resemble [B] influence [C] favor [D] surpass9. [A] again [B] also [C] instead [D] thus10. [A] Meanwhile [B] Furthermore [C] Likewise [D] Perhaps11. [A] about [B] to [C]from [D]like。