全国硕士研究生入学统一考试资料答案附后
全国硕士研究生入学统一考试资料答案附后
【经典资料，ＷＯＲＤ文档，可编辑修改】【经典考试资料，答案附后，看后必过，ＷＯＲＤ文档，可修改】
2015年全国硕士研究生入学统一考试
管理类专业学位联考综合能力考试大纲
I.考试性质
综合能力考试是为高等院校和科研院所招收管理类专业学位硕士研究生而设置的具有选择性质的全国联考科目，其目的是科学、公平、有效地测试考生是否具备攻读专业学位必需的基本素质、一般素质和培养潜能，评估的标准是高等学校本科毕业生所能达到的及格或及格以上水平，以利于高等院校和科研院所在专业上择优选拔，确保专业学位硕士研究生的招生质量。
II.考察目标
1.具有运用数学基础知识、基本方法分析和解决问题的能力。
2.具有较强的分析、推理、论证等逻辑思维能力。
2024 年全国硕士研究生招生考试(教育综合 333)真题及答案
2024年全国硕士研究生招生考试(教育综合333)真题试卷一、单项选择题：第1～30小题，每小题2分，共60分。
下列每小题给出的四个选项中，只有一个选项是最符合题目要求的。
1.在第39个教师带到来之际，习近平总书记专门致信全国优秀教师代表座谈会与会老师，充分肯定与会教师为代表的广大教师的贡献，号召全国广大教师以教育家为榜样，弘扬教育家精神，为强国建设、民族复兴做出新的贡献，教育家精神具有丰富的内涵，其中对教师理想信念的要求是（）。
A.心有大我，至诚报国B.启智润心，因材施教C.勤学笃行，求是创新D.言为士则，行为世范【答案】A【解析】A项“心有大我、至诚报国”中含有对国家的情怀，属于理想信念范畴。
B项“启智润心、因材施教”体现教师的教学方法，是一种教学智慧。
C项“勤学笃行、求是创新”体现教师严谨认真的工作态度。
D项“言为士则、行为世范”强调教师应谨言慎行，为学生做好榜样，是教师应具备的道德情操。
2.20世纪60年代以来，许多国家相继实行教育先行战略，支持这种国家发展战略的理论是（）。
A.社会再生产理论B.人力资本理论C.社会平衡器理论D.文化资本理论【答案】B【解析】人力资本理论认为要重视教育投资对人力资本的作用，教育投资能够转化为人力资本，促进社会经济的发展。
教育先行指教育在本国经济能力可承载的情况下，适度地先于其他行业或经济发展的现有状态而超前、提前发展，体现国家对教育的重视。
3.华虚朋的文纳特卡制在共同要素的教学上，采用了个别化的策略，所遵循的人的身心发展的特点是（）。
A.顺序性B.阶段性C.差异性D.不平衡性【答案】C【解析】人身心发展的差异性要求考虑不同受教育者的个别差异性，做到因材施教、有的放矢，符合华虚朋的个别化策略。
4.教育概念有特定的内涵和外延。
下列故事所述的行为行动中，称得上教育的是（）。
A.曹冲称象B.程门立雪C.岳母刺字D.公车上书【答案】C【解析】教育指有目的地培养人的活动。
1999年全国硕士研究生入学统一考试数学一、二、三、四试题完整版附答案解析及评分标准
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四、(本题满分 5 分)
求 I (ex sin y b(x y)) dx (ex cos y ax) dy ，其中 a, b 为正的常数， L 为从点 L
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I (ex sin y b(x y))dx (ex cos y ax)dy L L1
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2000年全国硕士研究生入学统一考试数学一、二、三、四试题完整版附答案解析及评分标准
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2020年全国硕士研究生入学统一考试数学二试题完整版附答案分析及详解
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0
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全国硕士研究生入学统一考试资料答案附后
全国硕士研究生入学统一考试资料答案附后
The document was prepared on January 2, 2021
【经典资料，ＷＯＲＤ文档，可编辑修改】【经典考试资料，答案附后，看后必过，ＷＯＲＤ文档，可修改】
2015年全国硕士研究生入学统一考试
管理类专业学位联考综合能力考试大纲
I.考试性质
综合能力考试是为高等院校和科研院所招收管理类专业学位硕士研究生而设置的具有选择性质的全国联考科目，其目的是科学、公平、有效地测试考生是否具备攻读专业学位必需的基本素质、一般素质和培养潜能，评估的标准是高等学校本科毕业生所能达到的及格或及格以上水平，以利于高等院校和科研院所在专业上择优选拔，确保专业学位硕士研究生的招生质量。
II.考察目标
1.具有运用数学基础知识、基本方法分析和解决问题的能力。
2.具有较强的分析、推理、论证等逻辑思维能力。
3.具有较强的文字材料理解能力、分析能力以及书面表达能力。
III.考试形式和试卷结构
一、试卷满分及考试时间
1.试卷满分为200分，考试时间为180分钟
二、答题方式
答题方式为闭卷、笔试。
不允许使用计算器。
三、试卷内容与题型结构
数学基础 75分，有一下两种题型：问答求解 15小题，每小题3分，共45分条件充分性判断 10小题，每小题3分，共30分
逻辑推理 30小题，每小题2分，共60分
写作 2小题，其中论证有效性分析30分，论说文35分，共65分。
2024年度全国硕士研究生入学考试《中医综合》备考真题库(含答案)
2024年度全国硕士研究生入学考试《中医综合》备考真题库（含答案）学校:________ 班级:________ 姓名:________ 考号:________一、单选题(35题)1.液的分布特点（）。
A.灌注于骨节，脏腑，脑髓B.布散于皮肤，肌肉，孔窍C.渗灌于筋脉，肓膜，肌肉D.敷布于皮肤，脏腑，筋脉2.患者咳吐涎沫，清稀量多，不渴，短气不足以息，乏力，神疲懒言，舌淡苔白，脉弱，治宜选用（）。
A.桂枝甘草汤B.苓桂术甘汤C.甘草干姜汤D.吴茱萸汤3.热裂津亏与血虚不荣均可见的表现（）。
A.裂纹舌B.齿痕舌C.点刺舌D.胖大舌4.下列各项中，不属于白色主病的是（）。
A.气虚证B.阳虚证C.实寒证D.湿淫证5.患者腹部胀痛，有索状物隆起，纳呆，舌淡红苔白腻，弦滑脉（）。
A.食滞痰阻之聚证B.肝气郁结之聚证C.瘀血内结之积证D.气滞血阻之积证6.《素问·水热穴论》“肾者，胃之关”的含义是（）。
A.肾阳帮助脾胃的运化功能B.肾气的固摄主司二便的排泄C.肾为封藏之本，主司精的藏泄D.肾气的蒸腾气化作用主司尿液的生成与排泄7.证见发热重恶寒轻，鼻流黄涕，心悸失眠，舌红少苔脉细数，可见于（）。
A.表寒里热B.表里倶热C.表里倶实D.表里俱虚8.下列药物中，主治属于肾不纳气之虚喘的药物是：（）。
A.檀香B.沉香C.枳壳D.乌药9.金元时期，朱震亨“相火论”的基本观点是（）。
A.阳常有余，阴常不足B.阴常有余，阳常不足C.五志过极，皆能化火D.六气郁久，皆能化火10.气虚导致的便秘应选用（）。
A.六磨汤B.黄芪汤C.麻子仁丸D.济川煎11.宇宙万物化生和变化的根本条件（）。
A.阴阳互根互用B.阴阳相互转化C.阴阳消长平衡D.阴阳交感互藏12.胆郁痰扰和痰火扰神均可出现的症状（）。
A.头晕耳鸣B.狂躁妄动C.胸胁满闷D.失眠多梦13.可用牡蛎不用龙骨的是（）。
A.肝阳上亢，头痛眩晕B.肾虚不固，遗精滑精C.瘰疬痰核，癥瘕积聚D.心神不宁，失眠多梦14.男性，38周岁，左耳两周以来耳鸣如火车鸣响，头目胀痛，目赤心烦，入睡困难，口苦咽干。
1990年全国硕士研究生入学统一考试数学一、二、三、四、五试题完整版附答案及评分标准
1990 年全国硕士研究生入学统一考试数学一、二、三、四、五试题 完整版附答案及评分标准数 学（试卷一）一、填空题：(本题满分15分，每小题3分)(1)过点)1,2,1(-M 且与直线⎪⎩⎪⎨⎧-=-=+-=1432t z t y t x 垂直的平面方程是 x -3y -z +4=0 .(2)设a 为非零常数，则a xx e a x a x 2)(lim =-+∞→.(3)设函数11,0,1)(>≤⎩⎨⎧=x x x f , 则)]([x f f = ___1___． (4)积分dy e dx xy ⎰⎰-2022的值等于4(1)/2e --.(5)已知向量组 1α=(1,2,3,4)，2α=(2,3,4,5)，3α=(3,4,5,6)，4α=(4,5,6,7)，则该向量组的秩是2二、选择题：(本题满分15分，每小题3分) (1)设()f x 是连续函数，且⎰-=x e xdt t f x F )()(则)(x F '等于(A)（A ）)()(x f e f e x x ----(B) )()(x f e f e x x +---(C))()(x f e f e x x ---(D) )()(x f e f e x x +--(2)已知函数()f x 具有任意阶导数，且[]2)()(x f x f =', 则当n 为大于2的正整数时，()f x 的n 阶导数)()(x fn 是(A)(A) 1)]([!＋n x f n (B) 1)]([+n x f n (C) nx f 2)]([ (D) nx f n 2)]([!(3)设α为常数，则级数]1)sin([12nn na n -∑∞=（C ）(A)绝对收敛(B)条件收敛(C)发散(D)收敛性与α的取值有关.(4)已知()f x 在0x =的某个邻域内连续 ，且(0)0f =,2cos 1)(lim0=-→xx f x 则在点0x =处()f x (D)(A)不可导(B)可导,且0)0(≠'f (C)取得极大值(D)取得极小值(5)已知1β和2β是非齐次线性方程组AX = b 的两个不同的解，21,αα是对应导出组AX = 0基础解系，21,k k 为任意常数，则方程组AX = b 的通解（一般解）必是(B)(A) 2)(2121211ββααα-+++k k (B) 2)(2121211ββααα++-+k k (C) 2)(2121211ββββα-+++k k (D) 2)(2121211ββββα++-+k k 三、(本题满分15分，每小题5分)(1)求dx x x ⎰-+102)2()1ln(.解：11200ln(1)1ln(1)(2)2x dx x d x x +=+--⎛⎛⎜⎜⎠⎠110011ln(1)2(1)(2)x dx x x x =+--+-⎛⎜⎠……2分 101111ln 2()ln 232(1)3dx x x =-+=-+⎰.……5分 (2)设(2,sin )z f x y y x =-，其中(,)f u v 具有连续的二阶偏导数，求yx z∂∂∂2.解：2cos z f fy x x u v ∂∂∂=+∂∂∂.……2分 2222222(2sin cos )sin cos cos z f f f fx y x y x x x x y u u v v v∂∂∂∂∂=-+-++∂∂∂∂∂∂∂. ……5分 (3) 求微分方程x e y y y 244-=+'+''的通解(一般解).解：特征方程为2440r r ++=的根为1,22r =-.对应齐次方程的通解为212()x Y C C x e -=+，其中12,C C 为任意常数. ……2分 设原方程的特解为*2()x y x Ax e 2-=，代入原方程得12A =.……4分 因此，原方程的通解为2*2212()()2xx x y x Y y C C x ee --=+=++. ……5分四、(本题满分6分) 求幂级数∑∞=+0)12(n nxn 的收敛域, 并求其和函数.解：因为123limlim 121n n n n a n a n ρ+→∞→∞+===+，所以11R ρ==.显然幂级数(21)nn n x∞=+∑在1x =±时发散，故此幂级数的收敛域为(1,1)-.……2分又0()(21)2nnnn n n S x n x nx x ∞∞∞====+=+∑∑∑012()1n n x x x∞='=+-∑……5分 2221111(1)1(1)x xx x x x +=+=-<<---.……6分五、(本题满分8分) 求曲面积分I=⎰⎰+sdxdy yzdzdx .2其中S 是球面4222=++z y x外侧在0≥z 的部分解：令2214x y S z ⎧+≤=⎨=⎩，其法向量与z 轴的负向相同. 设1S S 和所围成的区域为Ω，则由奥-高公式有12S I yzdzdx dxdy zdxdydz Ω++=⎰⎰⎰⎰⎰. ……2分而221140,228S S x y yzdzdx dxdy dxdy π+≤==-=-⎰⎰⎰⎰⎰⎰.……4分2222cos sin 4zdxdydz d d r r dr ππθϕϕϕπΩ=⋅=⎰⎰⎰⎰⎰⎰.……7分 所以12I π=.……8分六、(本题满分8分)设不恒为常数的函数)(x f 在闭区间[,]a b 上连续，在开区间(,)a b 内可导，且()()f a f b =. 证明：在(,)a b 内至少存在一点ξ, 使0)(>'ξf .证：因()()()f a f b f x =且不恒为常数，故至少存在一点(,)c a b ∈，使得()()()f c f a f b ≠=.于是()()()()f c f a f c f a ><或.……2分现设()()f c f a >，则在[,]a c 上因()f x 满足拉格朗日定理的条件，故至少存在一点(,)(,)a c a b ξ∈⊂，使得1()[()()]0f f c f a c a ξ'=->-. ……6分对于()()f c f a <情形，类似地可证得此结果.……7分七、(本题满分8分) 设四阶矩阵=B ⎪⎪⎪⎪⎪⎭⎫ ⎝⎛---1000110001100011，=C ⎪⎪⎪⎪⎪⎭⎫⎝⎛2000120031204312且矩阵A 满足关系式E C B C E A =''--)(1, 其中E 为四阶单位矩阵, 1-C 表示C 的逆矩阵,C '表示C 的转置矩阵, 将上述关系化简并求矩阵A .解：因11()[()]()A E C B C A C E C B A C B --''''-=-=-,故()A C B E '-=……2分因此 1[()]A C B -'=-11000210032104321-⎛⎫⎪⎪= ⎪⎪⎝⎭……4分1000210012100121⎛⎫⎪-⎪= ⎪-⎪-⎝⎭……6分八、(本题满分8分)求一个正交变换化二次型32312123222184444x x x x x x x x x f -+-++=成标准形.解：二次型的矩阵122244244-⎛⎫⎪=-- ⎪ ⎪-⎝⎭A ……1分由2122||244(9)244λλλλλλ---=---=----A E ，A 的特征值为1230,9λλλ===.……3分对于120λλ==，122122244000244000λ--⎛⎫⎛⎫⎪ ⎪-=--→ ⎪ ⎪⎪ ⎪-⎝⎭⎝⎭A E ，从而可取特征向量1011P ⎛⎫ ⎪= ⎪ ⎪⎝⎭及与1P 正交的另一特征向量2411P ⎛⎫ ⎪= ⎪ ⎪-⎝⎭. ……5分 对于39λ=，822245254099245000λ----⎛⎫⎛⎫ ⎪ ⎪-=---→-- ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭A E ，取特征向量3122P ⎛⎫⎪=- ⎪ ⎪⎝⎭. ……6分将上述相互正交的特征向量单位化，得1231032,,323ξξξ⎛⎫⎛⎫⎪ ⎪ ⎪ ⎪ ⎪===- ⎪ ⎪⎪ ⎪⎝⎭， ……7分故在正交变换1122331032323x y x y x y ⎛⎫ ⎪⎪⎛⎫⎛⎫⎪ ⎪ ⎪=-=⎪ ⎪ ⎪ ⎪ ⎪⎪⎝⎭⎝⎭⎪⎪⎭下，二次型239f y =. ……8分九、(本题满分8分)质点P 沿着以A,B 为直径的半圆周,从点A(1,2)运动到点B(3,4)的过程中受变力→F 作用 （见图），→F 的大小等于点P 与原点O 之间的距离，其方向垂直于线段OP 且于y 轴正向的夹角小于2π.求变力→F 对质点P 所作的功.解：按题意，变力y x =-+F i j .……3分圆弧AB的参数方程是23443x y θππθθ⎧=⎪-≤≤⎨=⎪⎩.……5分 变力F 所作的功ABW ydx xdy =-+⎰434)sin )cos ]d ππθθθθθ-=⎰()21π=-……8分十、填空题：(本题满分6分，每小题2分)(1)已知随机变量X 的概率密度函数f (x )=x e -21, +∞<<∞-x ，则X 的概率分布函数()F x =1212010xx e x ex -⎧<⎨-≥⎩.(2)设随机事件A ，B 及其事件A B 的概率分别为6.0,3.0,4.0和，若_B 表示B 的对立事件，那么积事件B A 的概率3.0)B A (P =(3)已知离散型随机变量X 服从参数为2的泊松分布，则随机变量32Z X =-的数学期望()E Z = 4 .十一、(本题满分6分)设二维变量(X ,Y )在区域 x y x D <<<,10:内服从均匀分布，求关于X 的边缘概率密度函数及随机变量 Z =2X +1的方差D (Z ).解：(,)X Y 的联合概率密度函数是1,01,||,(,)0,x y x f x y <<<⎧=⎨⎩其它，因此关于X 的边缘概率密度函数是2,01()(,)0,X x x f x f x y dy +∞-∞<<⎧==⎨⎩⎰其它. ……2分22D(Z)(21)4[()(())]D X E X E X =+=-()22X X 4()()x f x dx xf x dx +∞+∞-∞-∞⎡⎤=-⎢⎥⎣⎦⎰⎰……4分()21132001424224299x dx x dx ⎡⎤⎛⎫=-=-= ⎪⎢⎥⎝⎭⎣⎦⎰⎰.……6分数 学（试卷二）一、填空题【 同数学一 第一题 】 二、选择题【 同数学一 第二题 】三、(本题满分15分，每小题5分)【 同数学一 第三题 】 四、(本题满分18分，每小题6分) (1)【 同数学一 第四、(1)题 】(2)求微分方程0)ln (ln =-+dx x y xdy x 满足条件1==ex y的特解.解：将原方程化为11,(1)ln y y x x x x'+=≠.……1分 由公式()()()P x dx P x dx y e Q x e dx C -⎛⎫⎰⎰=+ ⎪⎝⎭⎰……3分 得2ln ln 111ln ln 2dx dx x x x xy e e dx C x C x x -⎛⎫⎛⎫⎰⎰=+=+ ⎪ ⎪⎝⎭⎝⎭. ……4分 又由|1x e y ==，可解出12C =，所以方程的特解是11ln 2ln y x x ⎛⎫=+ ⎪⎝⎭.……6分(3)过点(1,0)P 作抛物线2-=x y 的切线与上述抛物线及x 轴围成一平面图形，求此图形绕x 轴旋转一周所成旋转体的体积.解：设所作切线与抛物线相切于点0(x .因00|x x y =='==，故此切线的方程为)y x x =-.……1分又因该切线过点(1,0)P ，所以有03x =. 从而切线的方程为1(1)2y x =-. ……3分 因此，所求旋转体的体积332121(1)(2)4V x dx x dxππ=---⎰⎰……5分 6π=.……6分五、（本题满分8分）【 同数学一第五题 】 六、（本题满分7分）【 同数学一 第六题 】 七、（本题满分6分）【 同数学一 第七题 】 八、（本题满分8分）【 同数学一 第八题 】 九、（本题满分8分）【 同数学一 第九题】数 学（试卷三）一、填空题：(本题满分15分，每小题3分)(1)曲线⎩⎨⎧==ty t x 33sin cos 上对应于6π=t 点处的法线方程是13-=x y .(2)设x e y x tg 1sin 1⋅=，则='y 1tan 221111(sec sin cos )x e x x x x-⋅+.(3)=-⎰11dx x x15/4(4)下列两个积分的大小关系是：dx e dxe x x ⎰⎰----->121233.(5)【 同数学一 第一、(3) 题 】二、选择题：(本题满分15分，每小题3分)(1)已知0)1(lim 2=--+∞→b ax x x x ，其中,a b 常数，则(C)(A)1,1a b ==(B)1,1a b =-=(C)1,1a b ==-(D)1,1a b =-=-(2)设函数)(x f 在),(+∞-∞上连续，则⎰])([dx x f d 等于(B)(A))(x f (B)dxx f )((C)cx f +)((D)dxx f )('(3)【 同数学一 第二、(3) 题 】(4)【 同数学一 第二、(4) 题 】(5)设⎪⎩⎪⎨⎧=≠=0),0(0,)()(x f x x x f x F ，其中()f x 在0x =处可导，(0)0,(0)0f f '≠=，则0x =是()F x 的 （B ）(A)连续点 (B) 第一类间断点 (C) 第二类间断点（Ｄ）连续点或间断点不能由此确定三、(本题满分15分，每小题3分) (1)已知9)(lim =-+∞→xx ax a x ，求常数a . 解：因2(1)lim()lim (1)x x a x x xa x a x e ax a x→∞→∞++==--……3分 故29a e =,ln 3a =.……5分(2)求由2()ln()y x x y x y -=--所确定的函数()y y x =的微分dy .解：对方程两边求微分2()ln()()dx dydy dx dx dy x y x y x y--=--+--, ……3分故2ln(),3ln()2x y xdy dx dy dx x y x y +-==+--或.……5分 (3)求曲线)0(112>+=x xy 的拐点. 解：22223231,2(1)(1)x x y y x x -'''=-=++. ……2分 令0y ''=,解得x =.因在x =的左右邻近"y 变号,故x =是拐点的横坐标.所以曲线的拐点是3)4.……5分 (4)计算 ⎰-dx x x2)1(ln . 解：原式1ln 1xd x =-⎰ln 11(1)x dxxx x =---⎰……2分 10ln 11()11x dxx x x =-+--⎰……4分 ln |1|ln 1x x C x x-=++-.……5分 (5)见【 数学二 第四（2）题 】四、(本题满分9分)在椭圆12222=+by a x 的第一象限部分上求一点Ｐ，使该点处的切线，椭圆及两坐标轴所围图形的面积为最小（其中0,0a b >>）.解：设00(,)P x y 为所求之点，则此点处的切线方程为00221xx yya b+=. ……2分令0x =，得该切线在y 轴上的截距20b y .令0y =，得该切线在x 轴上的截距2a x . ……4分于是所围图形的面积为2200011,(0,)24a b S ab x a x y π=⋅-∈.……6分 求S的最小值时，不妨设00A x y ==22b A a '=. ……7分令0A '=,解得在(0,)a 内唯一驻点0x =……8分由A '在0x =右侧为负，得知0x =A 的极大点，即S 的极小点.所以0x =S 为最小，此时0y =,即为所求之点.……9分 五、(本题满分9分)证明：当0x >时，有不等式 21π>+x arctgx . 解：考虑函数1()arctan ,02f x x x x π=+->.……2分 有2211()0,01f x x x x '=-<>+. ……4分 所以()f x 在(0,)+∞上是单调减少的.……5分 又lim ()0x f x →+∞=……7分知当10,()arctan 02x f x x x π>=+->时. ……8分 即1arctan 2x x π+>. ……9分六、(本题满分9分)设dt t t x f x⎰+=11ln )(, 其中0,x >求 1()().f x f x+解：111ln ()1xt f dt xt =+⎰. 令1t y =，得11ln ()(1)x y f dy x y y =+⎰. ……3分 于是111ln ln ()()(1)(1)x x t t f x f dt dt x t t t +=+++⎰⎰111()ln (1)(1)x tdtt t t =+++⎰……5分 1111()ln 11x tdt t t t =+-++⎰……7分 21ln 1ln 2x t dt x t ==⎰. ……9分七、（本题满分9分）【 同数学二 第四、（3）题 】 八、(本题满分9分)求微分方程ax e y y y =+'+''44之通解，其中a 为实数.解：特征方程为2440r r ++=，特征根为1,22r =-.对应齐次方程的通解为212()x y C C x e -=+ .……2分 当2a ≠-时，设非齐次方程的特解为*()ax y x Ae =， ……3分代入原方程，可得21(2)A a =+,*21()(2)axy x e a =+. 当2a =-时，设非齐次方程的特解为*21()xy x A x e 2-=.代入原方程，得12A =,*21()2x y x x e 2-=.……8分故通解为212222121()2(2)()()()22x axx C C x e e a a y x x y x C C x e a --⎧++≠-⎪+⎪=⎨⎪=++=⎪⎩，当，当.……9分数 学（试卷四）一、填空题：(本题满分15分，每小题3分) (1)极限n →∞=2(2)设函数()f x 有连续的导函数，0)0(=f 且b f =')0(，若函数00,sin )()(=≠⎪⎩⎪⎨⎧+=x x A xx a x f x F 在0x =处连续，则常数A ＝ a + b .(3)曲线2y x =与直线2y x =+所围成的平面图形的面积为 4.5 .(4)若线性方程组⎪⎪⎩⎪⎪⎨⎧=+-=+=+-=+414343232121a x x a x x a x x a x x 有解，则常数4321,,,a a a a 应满足条件04321=+++a a a a (5)一射手对同一目标独立的进行四次射击，若至少命中一次的概率为8180，则射手的命中率为2/3二、选择题：(本题满分15分，每小题3分) (1)设函数x e tgx x x f sin )(⋅⋅=，则)(x f 是 （B ）（A ）偶函数(B)无界函数(C)周期函数(D)单调函数(2)设函数()f x 对任意x 均满足等式(1)()f x a f x +=， 且有b f =')0(,其中,a b 为非零常数，则 (D)（A ）()f x 在1x =处不可导（B ）()f x 在1x =处可导，且a f =')1(（C ）()f x 在1x =处可导,且 f (1)b '= （D ）()f x 在1x =处可导，且 f (1)ab '=. (3)向量组s ααα,,21⋅⋅⋅⋅线性无关的充分条件是(A)s ααα,,21⋅⋅⋅⋅均不为零向量(B) s ααα,,21⋅⋅⋅⋅中任意两个向量的分量不成比例(C) s ααα,,21⋅⋅⋅⋅中任意一个向量均不能由其余1s -个向量线形表示 (D) s ααα,,21⋅⋅⋅⋅中有一部分向量线形无关(4)设A ，B 为两随机事件，且A B ⊂，则下列式子正确的是(A)(A)P (A+B )= P (A )(B)P(AB )=P(A )(C)P (A B )= P (B )(D)P (B -A )=P (B )-P (A )(5)设随机变量X 和Y 相互独立，其概率分布为则下列式子正确的是 (C )(A )X =Y(B ){}0P X Y ==(C ){}P X Y ==21(D ){}1P X Y ==三、(本题满分20分，每小题5分) (1)求函数()I x =dt t t t xe ⎰+-12ln 2在区间[2,e e ]上的最大值.解：由222ln ln ()0,[,]21(1)x x I x x e e x x x '==>∈-+-, ……1分可知()I x 在2[,]e e 上单调增加,故222ln max ()(1)e e x e e t I x dt t ≤≤==-⎛⎜⎠21ln 1e e tdt --⎛⎜⎠22ln 1111e e e e t dt t t t =-+⋅--⎛⎜⎠……3分 22121ln11e e t e e t -=-+--11ln ln(1)11e e e e e e+=+=+-++. ……5分(2)计算2y Dxe dxdy -⎰⎰，其中D 是曲线24y x =和29y x =在第一象限所围成的区域.解：原式2302yy y edy xdx+∞-=⎰⎰……2分 20111()249y y y e dy +∞-=-⎰……3分 205572144y ye dy +∞-==⎰.……5分(3)求级数的∑∞=-12)3(n nn x 收敛域. 解：21n a n=,121(1)n a n +=+，212lim lim 1(1)n n n n a n a n +→∞→∞==+, ……2分 因此当131x -<-<，即24x <<级数收敛. ……3分当2x =时，得交错级数211(1)n n n ∞=-∑；当4x =时，得级数211n n∞=∑，二者都收敛，于是原级数的收敛域为[2,4].……5分(4)求微分方程x e x x y y sin )(ln cos -=+'的通解解：cos cos sin (ln )xdxxdx x y e x e e dx C --⎰⎰=⋅⋅+⎰……3分 sin (ln )x e xdx C -=+⎰……4分 sin (ln )x e x x x C -=-+.……5分四、(本题满分9分)某公司可通过电台和报纸两种方式做销售某种商品广告，根据统计资料，销售收入R （万 元）与电台广告费用1x (万元) 及报纸广告费用2x (万元) 之间的关系有如下经验公式：222121211028321415x x x x x x R ---++=. （1）在广告费用不限的情况下, 求最优广告策略；（2）若提供的广告费用为1.5 万元, 求相应的最优广告策略.解：(1) 利润函数为22121212121514328210()x x x x x x x x π=++----+221212121513318210x x x x x x =++---……1分 由12121248130,820310x x x x x x ππ∂∂=--+==--+=∂∂……2分 解得10.75x =(万元),2 1.25x =(万元). 因利润函数12(,)x x ππ=在(0.75,1.25)处的二阶偏导数为:2222211224,8,20A B C x x x x πππ∂∂∂==-==-==-∂∂∂∂. ……3分 故有26480160,40B AC A -=-=-<=-<，……4分 所以函数12(,)x x ππ=在(0.75,1.25)处达到极大值，亦即最大值.……5分(2)若广告费用为1.5万元，则只需求利润12(,)x x ππ=在12 1.5x x +=时的条件极值.拉格朗日函数为221212121212(,,)1513318210( 1.5)L x x x x x x x x x x λλ=++---++-……7分令120,0,0L L L x x λ∂∂∂===∂∂∂，有121212481308203101.50x x x x x x λλ--++=⎧⎪--++=⎨⎪+-=⎩……8分由此可得10x =，2 1.5x =，即将广告费1.5万元全部用于报纸广告，可使利润最大.……9分五、(本题满分6分)设)(x f 在闭区间[0,c]上连续,其导数)(x f '在开区间(0,)c 内存在且单调减少.(0)0f =，试应用拉格郎日中值定理证明不等式()()()f a b f a f b +≤+，其中常 数,a b 满足条件c b a b a ≤+≤≤≤0.证：当0a =时,(0)0f =有()()()()f a b f b f a f b +==+. ……1分当0a >时，在[0,]a 和[,]b a b +上分别应用拉格朗日定理,有()11()(0)()(),0,0f a f f a f a a aξξ-'==∈-；……3分 ()22()()()()(),,()f a b f b f a b f b f b a b a b b aξξ+-+-'==∈++-.……4分 显然120a b a b c ξξ<<≤<<+≤. 因()f x '在[0,]c 上单调减少，故21()()f f ξξ''≤.从而有()()()f a b f b f a a a+-≤.……5分 故由0a >，有()()()f a b f a f b +≤+. ……6分六、(本题满分8分)已知线性方程组 1234512345234512345323022654332x x x x x ax x x x x x x x x bx x x x x ++++=⎧⎪+++-=⎪⎨+++=⎪⎪+++-=⎩(1)问,a b 为何值时，方程组有解?(2)方程组有解时，求出方程组的导出组的一个基础解系；(3)方程组有解时, 求出方程组的全部解.解：(1) 考虑方程组的增广矩阵1111111111321130012263012260000035433120000022a aa A bb a a ⎛⎫⎛⎫ ⎪⎪- ⎪ ⎪=→ ⎪ ⎪- ⎪⎪--⎝⎭⎝⎭……2分当30b a -=且220a -=，即13a b ==且时，方程组的系数矩阵与增广矩阵之秩相等，故1,3a b ==时，方程组有解.……3分(2)当1,3a b ==时，有11111101152012263012263000000000000000000000000a a A ----⎛⎫⎛⎫⎪⎪⎪ ⎪→→ ⎪ ⎪⎪ ⎪⎝⎭⎝⎭，因此，原方程组的同解方程组为13452345522263x x x x x x x x ---=-⎧⎨+++=⎩，故导出组的基础解系为123115226,,100010001v v v ⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪--- ⎪ ⎪ ⎪⎪ ⎪ ⎪=== ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭. ……6分(3)令3450x x x ===，得原方程组的特解23000u -⎛⎫ ⎪ ⎪⎪= ⎪ ⎪ ⎪⎝⎭，于是原方程组的全部解为1231234521153226010000100001x x u x c c c x x -⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪ ⎪ ⎪--- ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪==+++ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭，其中123,,c c c 为任意常数.……8分 七、(本题满分5分)已知对于n 阶方阵A ，存在自然数k ，使得0=kA ，试证明矩阵E A -可逆，并写出 其逆矩阵的表达式（E 为n 阶单位阵）.解：由0kA =及1k k E A E A A E A --+++=-()() ，得1k E A E A A E--+++=()() ……3分 可知E A -可逆，且有11()k E A E A A ---=+++ .……5分八、(本题满分6)设A 为n 阶矩阵，1λ和2λ是A 的两个不同的特征值，21,x x 是分别属于1λ和2λ的特征向量，试证明：21x x +不是A 的特征向量.解：因11122212,,Ax x Ax x λλλλ==≠,故12121122()A x x Ax Ax x x λλ+=+=+……2分 设21x x +是A 的特征向量，则1212()()A x x x x λ+=+，即112212()x x x x λλλ+=+， 于是有1122()()0x x λλλλ-+-=.……4分由于12,x x 属于不同的特征值，所以12,x x 线性无关，故有120,0λλλλ-=-=，即12λλ=， 这与假设矛盾，因此21x x +不是A 的特征向量.……6分九、(本题满分4分)从0,1,2,…,9等十个数字中任意选出三个不同的数字,试求下列事件的概率：=1A { 三个数字中不含0和5 } ；=2A { 三个数字中含0但不含5 }解：3813107()15C P A C ==……2分 33982310214()15C C P A C -==. ……4分十、(本题满分5分)一电子仪器由两个部件构成，以X 和Y 分别表示两个部件的寿命(单位:千小时)，已知X 和Y 的联合分布函数为：⎩⎨⎧≥≥+--=+---它其00,01),()(5.05.05.0y x e e e y x F y x y x .(1)问X 和Y 是否独立?(2)求两个部件的寿命都超过100小时的概率α.解 X 的分布函数1()F x 和Y 的分布函数2()F y 分别为：0.511,0;()(,)0,0x e x F x F x x -⎧-≥=+∞=⎨<⎩若若，0.521,0;()(,)0,0y e y F y F y y -⎧-≥=+∞=⎨<⎩若若……2分 显然12(,)()()F x y F x F y =，故X 和Y 独立，……3分 于是{0.1,0.1}{0.1}{0.1}P X Y P X P Y α=>>=>⋅>……4分 0.050.050.112[1(0.1)][1(0.1)]F F e e e ---=-⋅-=⋅=.……5分十一、(本题满分7分)某地抽样调查结果表明，考生的外语成绩(百分制)近似服从正态分布，平均成绩为72 分，96分以上的占考生总数的2.3 %，试求考生的外语成绩在60分至84分之间的概率.[附表] （表中)(x Φ是标准正态分布函数）解：设X 为考生的外语成绩，由题设知2~(,)X N μσ，其中72μ=. ……1分由条件知{96}0.023P X ≥=，即9672{}0.023X P μσσ--≥=，亦即24()0.977σΦ=，由()x Φ的数值表，可见242σ=.因此12σ=.这样2~(72,12)X N .……4分所求概率为60728472{6084}{}{11}1212X X P X P P μμσσ----≤≤=≤≤=-≤≤(1)(1)2(1)120.84110.682=Φ-Φ-=Φ-=⨯-=.……7分数 学（试卷五）一、填空题 (本题满分15分，每小题3分) (1)【 同数学四 第一、(1) 题 】(2)【 同数学四 第一、(2) 题 】(3)【 同数学四 第一、(3) 题 】(4)【 同数学四 第一、(4) 题 】(5)已知随机变量(3,1),(2,1)X N Y N - ,且,X Y 相互独立，设随机变量27Z X Y =-+,则Z ～ N (0,5) .二、选择题 (本题满分15分，每小题3分) (1)【 同数学四 第二、(1) 题 】(2)【 同数学四 第二、(2) 题 】(3)【 同数学四 第二、(1) 题 】(4)设A 为n 阶可逆矩阵,*A 是A 的伴随矩阵，则*A =(A)(A) 1-n A(B) A (C) nA(D) 1-A(5)已知随机变量X 服从二项分布，且EX=2.4，DX=1.44，则二项分布的参数n ，p 的值为 (B )(A )n = 4，p = 0.6(B )n = 6，p = 0.4(C )n = 8，p = 0.3(D )n = 24，p = 0.1三、(本题满分20分，每小题5分) (1)求极限dte t x x t x x 22)1(1lim20-∞→⎰+解：原式22222202(1)(1)limlim(12)xt x x x x x t e dt x e xex e→∞→∞++==+⎰……3分22(1)1lim (12)2x x x →∞+==+. ……5分(2)求不定积分dx x x x ⎰34sin 2cos . 解 443333cos cos cos1222sin 88sin cos sin 222x x x x x x dx dx dx x x x x ==⎰⎰⎰……2分3211sin sin sin 42282x x x x d xd --==-⎛⎛⎜⎜⎠⎠……3分 22111sin 828sin 2x x dx x-=-+⎛⎜⎜⎠……4分 21cot 428sin 2x x C x -=-+211csc cot 8242x xx C =--+.……5分 (3)设)(22y z y z x ϕ=+，其中ϕ为可微函数，求 yz∂∂.解 将原式两边同时对y 求偏导，得2112()()()z z z z z y z y y y y y yϕϕ∂∂'=+-∂∂ ……3分 解出z y ∂∂，得 ()()()()2()2()z z z z z y z zy y yy y zzyz yz y yyϕϕϕϕϕϕ''--∂==∂''--. ……5分(4)【 同数学四 第三、(2) 题 】四、（本题满分9分）【 同数学四 第四题 】五、(本题满分6分)证明不等式1ln(()x x x +≥-∞<<+∞证：记()1ln(f x x x =++()ln(ln(f x x x x '=+=.……2分 令()0f x '=，知0x =为驻点.由()0f x ''=>……4分可知0x =为极小值点，亦即最小值点.()f x 的最小值为(0)0f =，于是，对于一切(,)x ∈-∞+∞,有()0f x ≥，即1ln(()x x x +≥-∞<<+∞. ……6分六、(本题满分4分)设A 为1010⨯矩阵⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡00001010000 (0010000010)10,计算行列式E A λ-，其中E 为10阶单位矩阵，λ为常数.解：1010000100().......................00011000A E λλλλλ---=--按第一列展开……1分101000100000100100010..............................................00010001101λλλλλλλ----－－－＝－……2分9101010()()1010λλλ=---=-.……4分七、(本题满分5分)设方阵A 满足条件TA A E =，其中TA 是A 的转置矩阵，E 为单位阵.试证明所对应的 特征值的绝对值等于1.证：设x 是A 的实特征向量，其所对应的特征值为λ，则Ax x λ=，即T T Tx A x λ=，于是有2T T T x A Ax x x λ=，即2T Tx x x x λ=，2(1)0T x x λ-=.……3分 因为x 为实特征向量，故0Tx x >，所以得210λ-=，即||1λ=.……5分八、（本题满分8分）【 同数学四 第六题 】九、（本题满分5分）【 同数学四 第九题 分值不同 】 十、(本题满分6分)甲乙两人独立地各进行两次射击，假设甲的命中率为0.2，乙的为0.5，以X 和Y 分别表示甲和乙的命中次数，试求X 和Y 联合概率分布.解：X Y 和都服从二项分布，参数相应为(2,0.2)和(2,0.5).因此X Y 和的概率分布分别为：0120.640.320.04X ⎛⎫⎪⎝⎭，0120.250.50.25Y ⎛⎫ ⎪⎝⎭ ……3分故由独立性，知X Y 和的联合分布为6分十一、(本题满分7分)【 同数学四第十一题 】。
2024年全国硕士研究生招生考试试题及答案(管理类综合能力)
2024年全国硕士研究生招生考试试题及答案(管理类综合能力)2024年全国硕士研究生招生考试试题及参考答案(管理类综合能力)(科目代码：199)一、问题求解:第1～15小题,每小题3分,共45分.下列每题给出的A、B、C、D、E五个选项中,只有一项是符合试题要求的.1. 甲股票上涨后的价格与乙股票下跌后的价格相等，则甲、乙股票的原价格之比为( )A. B. C. D. E.【答案】E2. 将3张写有不同数字的卡片随机排成一排，数字面朝下，翻开左边和中间的2张卡片，如果中间卡片上的数字大，那么取中间的卡片。
否则取右边的卡片，则取出的卡片上的数字最大的概率为( )A. B. C. D. E.【答案】C3. 甲乙两人参加健步走活动，第一天两人走的步数相同，此后甲每一天都比前一天多走100步，乙每天走的步数保持不变.若乙前7天走的总步数与甲前6天走的总步数相同，则甲第7天走了( )A.步B.步C. 步D.步E.步【答案】D4. 函数的最小值为( )A. B. C. D. E.【答案】B5. 已知点，若四边形为平行四边形，则( )A. B. C. D. E.【答案】B6. 已知等差数列满足,且,则公差为( )A. B. C. D. E.【答案】C7. 已知都是正整数,若,则的取值方法有( )A.种B.种C.种D.种E.种【答案】C8. 如图1，正三角形的边长为，以为圆心，以为半径做圆弧，再分别以为圆心，以为半径作圆弧，则阴影部分的面积为A. B. C.D. E.图1【答案】B9. 在雨季，某水库的蓄水量已超警戒水位，同时上游来水均匀注入水库，需要及时泄洪.若开个泄洪闸，则水库的蓄水量降到安全水位需要天;若开个泄洪闸，则水库的蓄水量降到安全水位需要天，若开个泄洪闸，则水库的蓄水量降到安全水位需要( )A.天B.天C.天D. 天E.天【答案】B10. 如图2，在三角形点阵中，第行及其上方所有点的个数之和记为，如.已知是平方数且,则( )A. B. C. D. E.图2【答案】C11. 如图3，在边长为2的正三角形材料中截减出一个半圆形工件，半圆的直径在三角形一条边上，则这个半圆的面积最大为( )A. B. C. D. E.图3【答案】A12. 甲，乙两码头相距100千米，一艘轮船从甲地顺流而下到达乙地用了4小时，返回时游轮的静水速度增加了25%用了5小时，则航道的水流速度为( )A. B. C. D.E.【答案】D13. 如图4，圆柱形容器的底面半径是，将半径为的铁球放入容器后，液面的高度为，液面原来的高度为( )A. B. C. D. E.图4【答案】E14. 有4种不同的颜色，甲乙两人各自随机选2种,则两个所选颜色完全相同的概率为( )A. B. C. D. E.【答案】A15. 设非负实数满足,则的最大值为( )A. B. C. D. E.【答案】E二、条件充分性判断(第16～25题,每小题3分,共30分,要求判断每题给出的条件(1)与条件(2)能否充分支持题干中所陈述的结论.A,B,C,D,E五个选项为判断结果,请选择项符合试题要求的判断.请在答题卡上将所选项的字母涂黑.)A.条件(1)充分,但条件(2)不充分B.条件(2)充分,但条件(1)不充分C.条件(1)和(2)单独都不充分,但条件(1)和条件(2)联合起来充分D.条件(1)充分,条件(2)也充分E.条件(1)和(2)单独都不充分,条件(1)和条件(2)联合起来也不充分16. 已知袋中装有红、黑、白三种颜色的球若干个,随机取出1球，则该球是白球的概率大于.(1) 红球数最少. (2) 黑球数不到一半.【答案】C17. 已知是正整数，则除以余.(1) 除以余.(2) 除以余.【答案】D18. 设二次函数，则能确定.(1) 曲线关于直线对称.(2) 曲线与直线相切.【答案】C19. 设为实数，则.(1) .(2) .【答案】A20. 设为实数，，则.(1)(2)【答案】C21. 设为正实数，则能确定.(1)(2)【答案】B22. 兔窝在兔子正北,狼在兔子正西，兔子和狼同时奔跑兔窝，则兔先到兔窝.(1) 兔子的速度是狼的(2) 兔子的速度是狼的.【答案】A23. 设.为实数，则确定.(1) .(2) .【答案】D24. 设曲线与轴有三个不同交点则.(1) 点的坐标为(2)【答案】C25. 设为等比数列，是的前项和，则确定的公比.(1)(2)【答案】E三、逻辑推理：第26-55小题，每小题2分，共60分。
考研精品文档 2010年全国硕士研究生入学统一考试英语试题附答案详解(试题一)
2010年全国硕士研究生入学统一考试英语试题附答案详解(试题一)20__a plateau and then slackening off. This suggests that the alleged” Hawthor ne effect “ is hard topin down.1. [A] affected [B] achieved [C] extracted [D] restored2. [A] at [B]up [C] with [D] off3. [A]truth [B]sight [C] act [D] proof4. [A] controversial [B] perplexing [C]mischievous [D] ambiguous5. [A]requirements [B]explanations [C] accounts [D] assessments6. [A] conclude [B] matter [C] indicate [D] work7. [A] as far as [B] for fear that [C] in case that [D] so long as8. [A] awareness [B] expectation [C] sentiment [D] illusion9. [A] suitable [B] excessive [C] enough [D] abundant10. [A] about [B] for [C] on [D] by11. [A] compared [B]shown [C] subjected [D] conveyed12. [A] contrary to [B] consistent with [C] parallel with [D] pealliar to13. [A] evidence [B]guidance [C]implication [D]source14. [A] disputable [B]enlightening [C]reliable [D]misleading15. [A] In contrast [B] For example [C] In consequence [D] As usual16. [A] duly [B]accidentally [C] unpredictably [D] suddenly17. [A]failed [B]ceased [C]started [D]continued20. [A]breaking [B]climbing [C]surpassing [D]hitingSection II Reading ComprehensionPart ADirections:Read the following four texts. Answer the questions below each text by choosi ng [A], [B], [C]or [D]. Mark your answers on ANSWER SHEET 1. (40 points) Text 1Of all the changes that have taken place in English-language newspapers duri ng the pastquarter-century, perhaps the m ost far-reaching has been the ine xorable decline in the scope and seriousness of their arts coverage.It is difficult to the point of impossibility for the average reader under the a ge of forty toimagine a time when high-quality arts criticism could be found i n most big-city newspapers. Yeta considerable number of the most significa nt c ollections of criticism published in the 20thcentury consisted in large part of new spaper reviews. To read such books today is to marvel at the fact thattheir learned contents were once deemed suitable for publication in general-circulation dailies.We are even farther rem oved from the unfocused newspaper review spublis hed in Englandbetween the turn of t he 20th century and the eve of World War Ⅱ, at a time when newsprintwas dirt-c heap and stylish arts crit icism was consi dered an ornament to the publicat ions inwhich it appe ared. In those far-off days, it was taken for granted tha t the cri tics of majorpapers woul dwri te in detail and at length about the even ts they covered. Theirs was a seriousbusiness, and even those reviewers who wore their learning lightly, like George Bern ard Shawand Ernest Newman, co uld be trus ted to know what they were a bout. These men believed injournal ism as a calling, and were proud to be published in the daily press.“So few authors havebrains enough or literary gift enough to keep their own end up in journalism, ”Newman wrote,“that I am tempted to define…journalism' as …a term of cont empt appl ied by writers who are notread to writers who are'. ”Unfortunately, these critics are virtually forgotten. Neville Cardus, who wrote for the ManchesterGuardian from 1917 until shortly before his death in 1975, is now known solely as a writer ofessays ont he game of cricket. During his l i fetime, though, he was also one of England'sforemost classical-music critics, and a stylist so widely admired that hisAutobiography（1947）became a best-seller. He was knighted in 1967, the first music critic tobe so ho nored. Yet on ly one of his books is now in print, and his vast body of writi ngs onmusic is unknown save to specialists.Is there any chance that Cardus's criticism will enjoy a revi val? The prospect seems remote.Jour nalistic tastes had changed long before his death, and po stmodern reader shave little usefor the ric hly upholstered Vicwardian pros e in which he specialized. Moreover, the amateur tradition in music criticism has been in headlong retreat.21. It is indicated in Paragraphs 1 and 2 that[A] arts criticism has disappeared from big-city newspapers.[B] English-language newspapers used to carry more arts reviews.[C] high-quality newspapers retain a large body of readers.[D] young readers doubt the suitability of criticism on dailies.22. Newspaper reviews in England before world warⅡwere characterized by[A] free themes.[B] casual style.[C] elaborate layout.[D] radical viewpoints.23. which of the following would Shaw and Newman most probably agree on ？[A] It is writers' duty to fulfill journalistic goals.[B] It is contemptible for writers to be journalists.[C] Writers are likely to be tempted into journalism.[D] Not all writers are capable of journalistic writing.24. What can be learned about Cardus according to the last two paragraphs?[A] His music criticism may not appeal to readers today.[B] His reputation as a music critic has long been in dispute.[C]His style caters largely to modern specialists.[D]His writings fail to follow the amateur tradition.25. What would be the best title for the text?[A] Newspapers of the Good Old Days.[B] The lost Horizon in Newspapers.[C] Mournful Decline of Journalism.[D] Prominent Critics in Memory.Text 2Over the past decade, thousands of patents have been granted for what are called businessmethods. received one for its “one-click” online p ayment system. Merrill Lynch got legal protection for an asset allocation str ategy. One inventor patented a technique for liftinga box.Now the nation's top patent court appears completely ready to scale back on business-methodpatents, which have been controversial ever since they we re first authorized 10 years ago. Ina move that has intellectual-property la wyers abuzz the U.S. court of Appeals for the federalcircuit said it would use a particular case to conduct a broad review of business-methodpatents. In r e Bilski , as the case is known , is “a very big deal”, says Dennis'D. Crouch of theUniversity of Missouri School of law. It “has the potential to eliminate an entire class ofpatents.”Curbs on business-method claims would be a dramatic about-face, because i t was the federalcircuit itself that introduced such patents with is 1998 decisi on in the so-called state StreetBank case, approving a patent on a way of po oling mutual-fund assets. That ruling producedan explosion in business-method patent filings, initially by emerging internet companiestrying to stake out exclusive pinhts to specific types of online transactions. Later, move es tablished companies raced to add such patents to their files, if only as a defe nsive moveagainst rivals that might beat them to the punch. In 2005, IBM n oted in a court filing that ithad been issued more than 300 business-method patents despite the fact that it questionedthe legal basis for granting them. S imilarly, some Wall Street investment films armedthemselves with patents for financial products, even as they took positions in court cases opposing the pr actice.The Bilski case involves a claimed patent on a method for hedging risk in the energy market.The Federal circuit issued an unusual order stating that the c ase would be heard by all 12 ofthe court's judges, rather than a typical panel of three, and that one issue it wants to evaluate is whether it should”recon sider” its state street Bank ruling.The Federal Circuit's action comes in the wake of a series of recent decisions by the supreme Count that has narrowed the scope of protections for patent holders. Last April, for examplethe justices signaled that too many patents w ere being upheld for “inventions” that are obvious. The judges on the Federal circuit are “reacting to the anti_ patent trend at the supreme court” ,says H arole C.wegner, a partend attorney and professor at aeorgeWashington Univ ersity Law School.26. Business-method patents have recently aroused concern because of[A] their limited value to business[B] their connection with asset allocation[C] the possible restriction on their granting[D] the controversy over authorization27. Which of the following is true of the Bilski case?[A] Its ruling complies with the court decisions[B] It involves a very big business transaction[C] It has been dismissed by the Federal Circuit[D] It may change the legal practices in the U.S.28. The word “about-face” (Line 1, Paro 3) most probably means[A] loss of good will[B] increase of hostility[C] change of attitude[D] enhancement of dignity29. We learn from the last two paragraphs that business-method patents[A] are immune to legal challenges[B] are often unnecessarily issued[C] lower the esteem for patent holders[D] increase the incidence of risks30. Which of the following would be the subject of the text?[A] A looming threat to business-method patents[B] Protection for business-method patent holders[C] A legal case regarding business-method patents[D] A prevailing trend against business-method patentsText 3In his book The Tipping Point, Malcolm Aladuell argues that social epidemics are driven in largepart by the acting of a tiny minority of special individuals, often called influentials, who areunusually informed, persuasive, or well-co nnected. The idea is intuitively compelling, but itdoesn't explain how ideas actually spread.The supposed importance of influentials derives from a plausible sounding b ut largely untested theory called the “two step flow of communication”: Inf ormation flows from themedia to the influentials and from them to everyone e lse. Marketers have embraced the two-step flow because it suggests that if th ey can just find and influence the influentials, thoseselected people will do mo st of the work for them. The theory also seems to explain the suddenand une xpected popularity of certain looks, brands, or neighborhoods. In many su ch cases,a cursory search for causes finds that some small group of people w as wearing, promoting, ordeveloping whatever it is before anyone else paid at tention. Anecdotal evidence of this kind fitsnicely with the idea that only cert ain special people can drive trendsIn their recent work, however, some researchers have come up with the findi ng that influentialshave far less impact on social epidemics than is generally supposed. In fact, they don't seemto be required of all.The researchers' argument stems from a simple observing about social influe nce, with the exception of a few celebrities like Oprah Winfrey-whose outsize presence is primarily a function of media, not interpersonal, influence-even t he most influential members of a population simply don't interact with that many others. Yet it is precisely these non-celebrity influentials who, accordin g to the two-step-flow theory, are supposed to drive social epidemicsby influencing their friends and colleagues directly. For a social epidemic to occur, however,each person so affected, must then influence his or her own acquai ntances, who must in turninfluence theirs, and so on; and just how many oth ers pay attention to each of these peoplehas little to do with the initial influe ntial. If people in the network just two degrees removedfrom the initial infl uential prove resistant, for example from the initial influential prove resis tant, for example the cascade of change won't propagate very far or affect many people.Building on the basic truth about interpersonal influence, the researchers stud ied the dynamicsof populations manipulating a number of variables relating of populations, manipulating anumber of variables relating to people's ability to influence others and their tendency to beinfluenced. Our work shows that th e principal requirement for what we call “global cascades”-the widespread propagation of influence through networks - is the presence not of a fewinfl uentials but, rather, of a critical mass of easily influenced people, each of w hom adopts, say,a look or a brand after being exposed to a single adopting neighbor. Regardless of how influential an individual is locally, he or she can exert global influence only if this critical mass is available to propagate a chain reaction.31.By citing the book The Tipping Point, the author intends to[A]analyze the consequences of social epidemics[B]discuss influentials' function in spreading ideas[C]exemplify people's intuitive response to social epidemics[D]describe the essential characteristics of influentials.32.The author suggests that the “two-step-flow theory”[A]serves as a solution to marketing problems[B]has helped explain certain prevalent trends[C]has won support from influentials[D]requires solid evidence for its validity33.what the researchers have observed recently shows that[A] the power of influence goes with social interactions[B] interpersonal links can be enhanced through the media[C] influentials have more channels to reach the public[D] most celebrities enjoy wide media attention34.The underlined phrase“these people” in paragraph 4 refers to the ones w ho[A] stay outside the network of social influence[B] have little contact with the source of influence[C] are influenced and then influence others[D] are influenced by the initial influential35.what is the essential element in the dynamics of social influence?[A]The eagerness to be accepted[B]The impulse to influence others[C]The readiness to be influenced[D]The inclination to rely on othersText 4Bankers have been blaming themselves for their troubles in public. Behind th e scenes, theyhave been taking aim at someone else: the accounting standar d-setters. Their rules, moan thebanks, have forced them to report enormou s losses, and it's just not fair. These rules saythey must value some assets at the price a third party would pay, not the price managers andregulators woul d like them to fetch.Unfortunately, banks' lobbying now seems to be working. The details may be unknowable, butthe independence of standard-setters, essential to the pro per functioning of capital markets,is being compromised. And, unless banks c arry toxic assets at prices that attract buyers,reviving the banking system wil l be difficult.After a bruising encounter with Congress, America's Financial Accounting Sta ndards Board(FASB) rushed through rule changes. These gave banks more fre edom to use models to valueilliquid assets and more flexibility in recognizing losses on long-term assets in their income statement. Bob Herz, the FASB's chairman, cried out against those who “question ourmotives.” Yet bank share s rose and the changes enhance what one lobby group politely calls“the use of judgment by management.”European ministers instantly demanded that the International Accounting Sta ndards Board(IASB) do likewise. The IASB says it does not want to act withou t overall planning, but the pressure to fold when it completes it reconstruc tion of rules later this year is strong. CharlieMcCreevy, a European commissio ner, warned the IASB that it did “not live in a political vacuum”but “in the re al word” and that Europe could yet develop different rules.It was banks that were on the wrong planet, with accounts that vastly overv alued assets.Today they argue that market prices overstate losses, becausethey largely reflect the temporary illiquidity of markets, not the likely exten t of bad debts. The truth will not be knownfor years. But bank's shares trade below their book value, suggesting that investors are skeptical. And dead ma rkets partly reflect the paralysis of banks which will not sell assets forfear of booking losses, yet are reluctant to buy all those supposed bargains.To get the system working again, losses must be recognized and dealt with. A merica's new planto buy up toxic assets will not work unless banks mark ass ets to levels which buyers find attractive. Successful markets require indepe ndent and even combative standard-setters.The FASB and IASB have been exactly that, cleaning up rules on stock options and pensions,for example, ag ainst hostility form special interests. But by giving in to critics now they are i nviting pressure to make more concessions.36. Bankers complained that they were forced to[A] follow unfavorable asset evaluation rules[B]collect payments from third parties[C]cooperate with the price managers[D]reevaluate some of their assets.37.According to the author , the rule changes of the FASB may result in[A]the diminishing role of management[B]the revival of the banking system[C]the banks' long-term asset losses[D]the weakening of its independence38.According to Paragraph 4, McCreevy objects to the IASB's attempt to[A]keep away from political influences.[B]evade the pressure from their peers.[C]act on their own in rule-setting.[D]take gradual measures in reform.39.The author thinks the banks were “on the wrong planet”in that they[A]misinterpreted market price indicators[B]exaggerated the real value of their assets[C]neglected the likely existence of bad debts.[D]denied booking losses in their sale of assets.40.The author's attitude towards standard-setters is one of[A]satisfaction.[B]skepticism.[C]objectiveness[D]sympathyPart BDirections:For Questions 41-45, choose the most suitable paragraphs from the list A-G and fill them intothe numbered boxes to form a coherent text. Paragraph E h as been correctly placed. There isone paragraph which dose not fit in with th e text. Mark your answers on ANSWER SHEET1. (10points)[A] The first and more important is the consumer's growing preference for e ating out; theconsumption of food and drink in places other than homes has ri sen from about 32 percent oftotal consumption in 1995 to 35 percent in 2000 and is expected to approach 38 percent by2005. This development is boosti ng wholesale demand from the food service segment by 4 to5 percent a year across Europe, compared with growth in retail demand of 1 to 2 percent.Me anwhile, as the recession is looming large, people are getting anxious. They t end to keep atighter hold on their purse and consider eating at home a realis tic alternative.[B] Retail sales of food and drink in Europe's largest markets are at a standsti ll, leavingEuropean grocery retailers hungry for opportunities to grow. Most le ading retailers have alreadytried e-commerce, with limited success, and ex pansion abroad. But almost all have ignoredthe big, profitable opportunity in their own backyard: the wholesale food and drink trade,which appears to b e just the kind of market retailers need.[C] Will such variations bring about a change in the overall structure of the f ood and drinkmarket? Definitely not. The functioning of the market is based o n flexible trends dominated by potential buyers. In other words, it is up to t he buyer, rather than the seller, to decide what tobuy .At any rate, this chang e will ultimately be acclaimed by an ever-growing number of both domesti c and international consumers, regardless of how long the current consumer pattern will take hold.[D] All in all, this clearly seems to be a market in which big retailers could pro fitably apply their scale, existing infrastructure and proven skills in the manag ement of product ranges, logistics,and marketing intelligence. Retailers tha t master the intricacies of wholesaling in Europe maywell expect to rake in su bstantial profits thereby. At least, that is how it looks as a whole.Closer ins pection reveals important differences among the biggest national markets, e speciallyin their customer segments and wholesale structures, as well as the competitive dynamics ofindividual food and drink categories. Big retailers mu st understand these differences beforethey can identify the segments of Eu ropean wholesaling in which their particular abilities mightunseat smaller but entrenched competitors. New skills and unfamiliar business models arenee ded too.[E] Despite variations in detail, wholesale markets in the countries that have been closelyexamined-France, Germany, Italy, and Spain-are made out of the same building blocks. Demandcomes mainly from two sources: independen t mom-and-pop grocery stores which, unlike large retail chains, are two small to buy straight from producers, and food service operators that cater to cons umers when they don't eat at home. Such food service operators range from snack machines to large institutional catering ventures, but most of these b usinesses areknown in the trade as “horeca”: hotels, restaurants, and cafes. Overall, Europe's wholesalemarket for food and drink is growing at the same sluggish pace as the retail market, but thefigures, when added together, m ask two opposing trends.[F] For example, wholesale food and drink sales come to $268 billion in Franc e, Germany, Italy,Spain, and the United Kingdom in 2000-more than 40 perc ent of retail sales. Moreover,average overall margins are higher in wholesal e than in retail; wholesale demand from the foodservice sector is growing qui ckly as more Europeans eat out more often; and changes in the competitive dynamics of this fragmented industry are at last making it feasible for whole salersto consolidate.[G] However, none of these requirements should deter large retailers (and e ven some largegood producers and existing wholesalers) from trying their ha nd, for those that master theintricacies of wholesaling in Europe stand to rea p considerable gains.Part CDirections:Read the following text carefullyand then translate the underlined segments i nto Chinese. Yourtranslation should bewritten clearly on ANSWER SHEET 2.(10 points)One basic weakness in a conservation system based wholly on econom ic motives is thatmost members of the land community have no economic valu e. Yet these creatures aremembers of the biotic community and, if its stabil ity depends on its integrity, they are entitled to continuance.When one of these noneconomic categories is threatened and, if we happen t o love it, we invent excuses to give it economic importance. At the beginning of the century songbirds weresupposed to be disappearing.(46)Scientists ju m ped to the rescue with some distinctly shaky evidence to the effect that ins ects would eat us up if birds failed to control them. The evidence had to be e conomic in order to be valid.It is painful to read these roundabout accounts today. We have no land ethic yet,(47) but wehave at least drawn nearer the point of adm itting that birds should continue as a matter of intrinsic right, regardless of the presence or absen ce of economic advantage to us.A parallel situation exists in respect of predatory mammals and fish-eating bi rds.(48) Timewas when biologists som ewhat overworked the evidence that these creatures preserve thehealth of game by killing the physically weak, or tha t they prey only on “worthless”species.Here again, the evidence had to b e economic in order to be valid. It is only in recent years thatwe hear the mor e honest argumentthat predators are members of the community, and that nospecial interest has the right to exte rminate the m f or the sake of a benef it, real or fancied, toitself.Some species of tree have been“read out of the party” by economics-minded foresters becausethey grow too slowly, or have too low a sale value to pay as timber crops.(49) In Europe,where forestry is ecologically more advanced, the noncommercial tree species are recognizedas members of native forest c ommunity, to be preserved as such, within reason. Moreover,some have be en found to have a valuable function in building up soil fertility. The interd ependence of the forest and its constit uent tree species, ground flora, and fauna istaken for granted.To sum up: a systemof conservation based solely on economic self-interest i s hopelesslylopsided.(50) It tends to ignore , and thus eventually to eliminate, many elements i n theland community that lack commercial value,but that are essential to its healthy functioning. Itassumes, falsely, that the economic parts of the bioti c clock will function without theuneconomic parts.Section III WritingPart A51. Directions:You are supposed to write for the Postgraduates' Associ ation a notice to recr uit volunteers foran international c onference on globalization. The notice sho uld include the basic qualificationsofapplicants and other information which yo u think is relevant.You should write about 100 words on ANSWER SHEET 2.Do not sign your own nam e at the end of the notice. Use "postgraduates' Ass ociation" instead. （10 points）Part B52. Directions:Write an essay of 160-200 words based on the following drawing. In your ess ay, you should1) describe the drawing briefly,2) explain its intended meaning, and3) give your comments.You should write neatly on ANSWER SHEET 2. （20 points）Section I: Use of English (10 points)1. A2. B3. C4. B5. C6. B7. D8. A9. C10. D11. C12. A13. A14. D15. B16. A17. D18. C19. B20. DSection II: Reading Comprehension (60 points) Part A (40 points)21. B22. A23. D24. A25. B26. C27. D28. C29. B30. A31. B32. D33. A34. C35. C36. A37. D38. C39. B40. DPart B (10 points)41. B42. F43. D44. G45. APart C (10 points)46 . 科学家们赶紧拿出某些明显站不住脚的证据来补救，大致说的是如果鸟儿不能控制昆虫数量的话，昆虫就会把我们吃光。
2000年全国硕士研究生入学统一考试日语试题(附答案)
2000年全国硕士研究生入学统一考试日语试题（附答案）Ⅰ文字と語彙（１５点）（一）次の文ののある漢字の読み仮名はどれであるか、それぞれ選択肢abcdの中から最も適切なものを一つ選び、記号で答えなさい。
（５点）1.強い人に負けたので、別に悔しいとは思わない。
aむなbくやcさびdいや2.そんなに強引にやろうたって無理ですよ。
aつよひきbきょういんcつよびきdごういん3.社長は一切の事務を長男に任せて引退した。
aいっせいbいっせつcいっさいdいっせ4.マッチやナイフを弄ぶのは危険なことだ。
aいじあそbもてあそcまちあそdもたあそ5.彼の発言は私たちがその問題を解決する上で示唆的であった。
aじさbしすうcしさdしさつ6.戦争であの町はすっかり廃れてしまいました。
aすたbこわcはいdすか7.私に言わせると、彼の判断は偏っていると思う。
aかたよbまがcかたどdかたま8.その話はいま彼らの間では禁物になっている。
aきんぶつbきんものcきんじものdきんもつ9.ふだん無口な彼も酒を飲むと、意外に思われるほどしゃべりだすのである。
aぶこうbむくちcぶくちdむぐち10.静脈注射をすれば、直るのもはやい。
aせいみゃくbしょうみゃくcじょうみゃくdせいまい（二）次の文のをつけた言葉の赤色の部分はどんな漢字を書くか、それぞれ選択肢abcdの中から同じ漢字が使われるものを一つ選び、記号で答えなさい。
（５点）11.そくたつで送ったので、間に合いました。
a 終日仕事にそくばくされているから旅行はなかなかできない。
b そのことは松尾さんに頼んであたってみたが、そくざに断られた。
c 何一つふそくのない家に生まれたのだから、貧乏の味を知らないd 事態はきゅうそくに収拾に向かうほかなかった。
12.何ともいえないきょうふにおそわれる。
a この問題にはきょうつう性があると思う。
b わざわざおいでいただいてきょうしゅくです。
c 世界的なふきょうが漸次回復しつつある。
1987年全国硕士研究生入学统一考试数学一、二、三、四、五试题完整版附答案及评分标准
1987 年全国硕士研究生入学统一考试数学一、二、三、四、五试题 完整版附答案及评分标准数 学（试卷一）一、填空题（每小题3分，满分15分. 只写答案不写解题过程）(1)与两直线 112x y t z t =⎧⎪=-+⎨⎪=+⎩及 121121x y z ++-== 都平行，且过原点的平面方程是 50x y -+=(2)当x =1/ln 2-；时，函数2xy x =取得极小值.(3)由ln y x =与两直线(1)y e x =+-及0y =围成图形的面积= 3 / 2 (4)设L 为取正向的圆周922=+y x ，则曲线积分dy x xdx y xy L)4()22(2-+-⎰的值是π18-.(5)已知三维线性空间的一组基底)1,1,0(,)1,0,1(,)0,1,1(321===ααα，则向量α＝(2, 0, 0)在上述基底下的坐标是 ( 1 , 1 , -1 )二、（本题满分8分）求正的常数a 与b ，使式1sin 1lim220=+-⎰→dt ta t x bx x x 成立. 解：假若1b ≠，则根据洛必达法则有2200011lim lim(01sin cos x x x bx x b x →→==≠--⎰，与题设矛盾，于是1b =.此时2222100002111lim lim(lim(sin 1cos x x x x bx x x x →→→===--⎰，即1=，因此4a =.三、（本题满分7分）(1)设函数,f g 连续可微，(,),()u f x xy v g x xy ==+，求,.u vx x∂∂∂∂解：1212()u x xy f f f y f x x x ∂∂∂''''=⋅+⋅=+⋅∂∂∂；()(1)v x xy g y g x x∂∂+''=⋅=+⋅∂∂.(2)设矩阵A 和B 满足2AB A B =+，其中A =301110014⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦，求矩阵B .解：因2AB A B =+，故2AB B A -=，即(2)A E B A -=，故1(2)B A E A -=-=522432223--⎛⎫⎪-- ⎪ ⎪-⎝⎭.四、（本题满分8分）求微分方程26(9)1y y a y ''''''+++=的通解.其中常数0a >.解：由特征方程3222(9)0r r a r +++=，知其特征根根为12,30,3r r ai ==-±. 故对应齐次方程的通解为33123cos sin x x y C C e x C e x --=++ ，其中123,,C C C 为任意常数.设原方程的特解为*()y x Ax =，代入原方程可得A =219a+. 因此，原方程的通解为*33123()cos sin x x y x y y C C e x C e x --=+=+++219a+x . 五、选择题（每小题3分，满分12分） (1)设常数0k >，则级数21)1(n nk n n+-∑∞= （C ）(A)发散(B)绝对收敛(C)条件收敛(D)收敛与发散与k 的值有关.(2)设)(x f 为已知连续函数，⎰=t s dx tx f t I 0)(，0,0s t >>，则I 的值（D ）(A)依赖于s 和t (B)依赖于s 、t 、x(C)依赖于t 和x , 不依赖于s (D)依赖于s , 不依赖于t (3)设1)()()(lim 2-=--→a x a f x f a x ，则在点x a =处(B)(A)()f x 导数存在,0)(≠'a f (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.(4)设A 为n 阶方阵, 且0≠=a A , 而*A 是A 的伴随矩阵，则*A =(C)(A)a(B)a/1(C) 1-n a (D) n a六、（本题满分10分） 求幂级数1121+∞=∑n n n x n 的收敛域，并求其和函数. 解：记112n n n u x n +=，有1112lim lim (1)22n nn n n n n n x u x n u n x +++→∞→∞=⋅=+，令12x <，知原级数在开区间(2,2)-内每一点都收敛.又当2x =-时，原级数=111111(2)2(1)2n n n n n n n ∞∞++==-=-∑∑，故由莱布尼兹判别法知其收敛；而当2x =时，原级数=11111122(1)2n n n n n n n ∞∞++===-∑∑，显然发散，故幂级数的收敛域为)2,2[-. 又记111111()()()22n n n n n x S x x x xS x n n ∞∞+=====∑∑，其中111()()2n n xS x n ∞==∑，有1111()()21/2n n x S x x ∞-='==-∑，于是102()2ln()1/22x dx S x x x ==--⎰，因此幂级数的和函数为2()2ln 2S x x x=-，[2,2)x ∈-.七、（本题满分10分） 计算曲面积分2(81)2(1)4SI x y dydz y dzdx yzdxdy =++--⎰⎰，其中s 是曲线 )31(01≤≤⎩⎨⎧=-=y x y z 绕Y 轴旋转一周所形成的曲面，它的法向量与Y 轴正向的夹角恒大于/2π．解：S 的方程为221y x z =++，记1S ：223,()y x z =+，知1S S +为封闭曲面，设其 方向取外侧，所围区域为Ω，则由高斯公式，有12(81)2(1)4S S I x y dydz y dzdx yzdxdy +=++--⎰⎰12(81)2(1)4S x y dydz y dzdx yzdxdy-++--⎰⎰12102(1)0S dv y dydz Ω=⋅---+⎰⎰⎰⎰⎰=3212(13)yz xD D dy dzdx dzdx--⎰⎰⎰⎰⎰31(1)16234y dy ππ=-+⋅⋅=⎰.八、（本题满分10分）设函数)(x f 在闭区间[0,1]上可微，对于[0,1]上的每个x ，函数的值都在开区间(0,1)内，且1)(≠'x f .证明 在(0,1)内有且仅有一个x ，使()f x x =．证：令()()h t f t t =-，知()h t 在闭区间[0,1]上连续，又由题设知0()1f x <<，于是 有(0)(0)00,(1)(1)10h f h f =->=-<. 故由零点定理，在(0,1)内有x ，使()f x x =.假若)(x f 在开区间(0,1)内有两个不同的点1x 和2x ，使得11()f x x =，22()f x x =， 不妨设12x x <，则易见)(x f 在闭区间[0,1]上连续，在(0,1)内可导，故由拉格朗日定理知，(0,1)ξ∃∈，使得2121()()()f x f x f x x ξ-'=-，即()1f ξ'=.此与1)(≠'x f 矛盾！故在(0,1)内使()f x x =的x 只能有一个.九、（本题满分8分）问,a b 为何值时，线性方程组123423423412340221(3)2321x x x x x x x x a x x b x x x ax +++=⎧⎪++=⎪⎨-+--=⎪⎪+++=-⎩有唯一解？无解？有无穷多解？ 并求出无穷多解时的通解.解：对方程组的增广矩阵进行初等变换，得11110111100122101221()013200101321100010A A b a b a b a a ⎛⎫⎛⎫⎪ ⎪⎪ ⎪==→ ⎪ ⎪----+ ⎪ ⎪--⎝⎭⎝⎭○1 当1≠a 时，系数行列式2(1)0A a =-≠，故由克拉姆法则，原方程组有唯一解；○2 当1a =，且1b ≠-时， ()3,()2r A r A ==， ()()r A r A ≠，故原方程组无解；○3 当1a =，且1b =-时， ()()24r A r A ==<，故原方程组有无穷的解. 此时显然有 11110101110122101221()00000000000000000000A A b ---⎛⎫⎛⎫⎪⎪⎪ ⎪=→→⎪ ⎪⎪⎪⎝⎭⎝⎭可见其通解为：12(1,1,0,0)(1,2,1,0)(1,2,0,1)T T T x c c =-+-+-，其中12,c c 为任意常数.十、填空题（每小题2分，满分6分）(1)在一次试验中事件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np )1(1--；而事件A 至多发生一次的概率为1)1]()1(1[---+n p p n .(2)三个箱子，第一个箱子有4个黑球1个白球，第二个箱子中有3个白球3个黑球，第三个箱子中有3个黑球5五个白球，现随机地取一个箱子，再从这个箱子中取一个球，这个球为白球的概率为53/120，已知取出的是白球，此球属于第二箱的概率是20/53.(3)已知连续随机变量X 的密度为1221)(-+-=x xe xf π，则X 的数学期望为 1 ；X 的方差为 1/2 .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⎩⎨⎧≤≤=它其0101)(x x f X ；⎩⎨⎧≤>=-00)(y y e y f y Y ，求随机变量Z =2X +Y 的概率密度函数()z f z .解：由题设，(,)X Y 的联合密度为01,0(,)()()0y X Y e x y f x y f x f y -⎧≤≤>==⎨⎩其 它， 故Z 的分布函数2()()(2)(,)z x y zF z P Z z P X Y z f x y dxdy +≤=≤=+≤=⎰⎰，○1 当0z <时，2()00z x y zF z dxdy +≤==⎰⎰，此时()00z f z '==；○2 当02z ≤≤时，200001()22z yzz z y y yz z F z dy e dx e dy ye dy ----==-⎰⎰⎰⎰，此时 011()()(1)22z y z z z f z F z e dy e -'===-⎰；○3 当2z >时，121220001()(1)1(1)2z x y x z zz F z dx e dy e dx e e -----==-=--⎰⎰⎰，此时 21()()(1)2zz z f z F z e e -'==-综上所述，Z =2X +Y 的概率密度函数为()z f z =122120(1)02(1)2zz z e z e e z ---<⎧⎪-≤≤⎨⎪->⎩数 学（试卷二）一、（本题满分15分）【 同数学Ⅰ、第一题 】 二、（本题满分14分） (1)（6分）计算定积分2||2(||).x x x e dx --+⎰解：因||x xe-是奇函数，||||x x e -是偶函数，故原式=22||202||226.x x x e dx xe dx e --==-⎰⎰(2)（8分）【 同数学Ⅰ、第二题 】三、（本题满分7分）设函数(,,),yz f u x y u xe ==，其中f 有二阶连续偏导数，求 2.z x y∂∂∂解：121yz u f f f e f x x∂∂''''=⋅+=⋅+∂∂，2111312123()y y y y z f xe f e e f f xe f x y ∂'''''''''=⋅++⋅+⋅+∂∂. 四、（本题满分8分）【同数学Ⅰ、第四题 】 五、（本题满分12分）【 同数学Ⅰ、第五题 】 六、（本题满分10分）【 同数学Ⅰ、第六题 】 七、（本题满分10分）【 同数学Ⅰ、第七题 】 八、（本题满分10分）【 同数学Ⅰ、第八题 】 九、（本题满分8分）【 同数学Ⅰ、第九题 】 十、（本题满分6分）设12,λλ为n 阶方阵A 的特征值，12λλ≠，而21,x x 分别为对应的特征向量，试证明：21x x +不是A 的特征向量.证：假若21x x +是A 的特征向量，设其对应的特征值为3λ，则有12312()()A x x x x λ+=+， 即123132Ax Ax x x λλ+=+. 又由题设条件知111Ax x λ=，222Ax x λ=，故有131232()()0x x λλλλ-+-=.因21,x x 是属于不同特征值的特征向量，所以21,x x 线性无关， 从而13λλ=，且13λλ=，此与12λλ≠矛盾！因此21x x +不是A 的特征向量.数 学（试卷三）一、填空题（每小题2分，满分10分. 把答案填在题中横线上） (1)设)1ln(ax y +=, 其中a 为非零常数，则22)1(,1ax a y ax ay +-=''+='.(2)曲线y arctgx =在横坐标为1点处的切线方程是4221-+=πx y ； 法线方程是4/)8(2++-=πx y .(3)积分中值定理的条件是()[,]f x a b 在闭区间上连续，结论是[,],()()()baa b f x dx f b a ξξ∃∈=-⎰使得(4) 32()1nn n lin e n -→∞-=+.(5)⎰='dx x f )(c x f +)(；⎰'badx x f )2(＝)2(21)2(21a f b f -. 二、（本题满分6分） 求极限 011lim()1x x xe →--解：200000111111lim()lim lim lim lim 1(1)222x x x x x x x x x x e x e x e x x e x e x x x →→→→→------=====--. 三、（本题满分7分）设⎩⎨⎧-=-=)cos 1(5)sin (5t y t t x ，求 22,.dy d y dx dx 解：因5sin ,55cos dy dx t t dt dt ==-，5sin )sin 5(1cos 1cos dy t t dx t t ==--（0+），故t tdx dy cos 1sin -=，且222sin 1()1cos 5(1cos )d y d t dtdx dt t dx t =⋅=---四、（本题满分8分） 计算定积分⎰1arcsin xdx x .解：2211121000111arcsin arcsin 2242x xdx x x π=-=-⎰⎰⎰，令sin x t =，有22120sin cos cos 4t tdt t ππ==⎰⎰，因此101arcsin 4248x xdx πππ=-⋅=⎰. 五、（本题满分8分）设D 是曲线sin 1y x =+与三条直线0x =,π=x ,0y =围成的曲边梯形.求D 绕x 轴旋 转一周所生成的旋转体的体积.解：223(sin 1)42V x dx ππππ=+=+⎰. 六、证明题（本题满分10分）(1)（5分）若()f x 在(,)a b 内可导，且导数)(x f '恒大于零，则()f x 在(,)a b 内单调增加. 证：12,(,)x x a b ∀∈，不妨设12x x <，则()f x 在12[,]x x 上连续，在12(,)x x 内可导，故由拉格朗日中值定理，12(,)(,)x x a b ξ∃∈⊂，使得2121()()()()f x f x f x x ξ'-=-. 由于)(x f '在(,)a b 内恒大于零，所以()0f ξ'>，又210x x ->，因此21()()0f x f x ->， 即21()()f x f x >，表明()f x 在(,)a b 内单调增加.(2)（5分）若()g x 在x c =处二阶导数存在，且0)(='c g ,0)(<''c g ，则()g c 为()g x 的一个极大值.证：因()()()lim 0x c g x g c g c x c →''-''=<-，而0)(='c g ，故()lim 0x c g x x c→'<-.由极限的保号性，0δ∃>，当(,)x c c δ∈-时，有()0g x x c '<-，即()0g x '>，从而()g x 在(,)c c δ-单增；当(,)x c c δ∈+时，有()0g x x c'<-，即()0g x '<，从而()g x 在(,)c c δ-单减.又由0)(='c g 知，x c =是()g x 的驻点，因此()g c 为()g x 的一个极大值.七、（本题满分10分）计算不定积分⎰+x b x a dx2222cos sin ( 其中,a b 为不全为零的非负数 )解：① 当0a =时，原式=22211sec tan xdx x c b b =+⎰；②当0b =时， 原式=22211c cot cs xdx x c a a=-+⎰；③当0ab ≠时，原式=22222(tan )sec 11arctan(tan )tan (tan )1ad x xdx a b x c a a x b ab ab bx b==+++⎰⎰.八、（本题满分15分） (1)（7分）求微分方程y x dxdyx-=，满足条件0|2==x y 的解． 解：原方程即11dy y dx x+=，故其通解为11211()()2dx dx xx y e e dx c x c x -⎰⎰=+=+⎰.因0|2==x y ，所以1c =-.于是所求初值问题的解为xx y 12-=.(2)（8分）求微分方程 x e x y y y =+'+''2的通解.解：由特征方程2210r r ++=，知其特征根根为1,21r =-.故对应齐次方程的通解为12()x y C C x e -=+ ，其中12,C C 为任意常数.设原方程的特解为*()()x y x e ax b =+，代入原方程可得a =14，b =-14. 因此，原方程的通解为*212()()x y x y y C C x e -=+=++ 14(1)x x e -. 九、选择题（每小题4分，满分16分） (1).+∞<<∞=x ex x x f x-,sin )(cos 是（D ）（A ）有界函数 （B ）单调函数 （C ）周期函数 （D ）偶函数(2). 函数()sin f x x x - (D)（A ）当∞→x 时为无穷大 （B ）当∞→x 时有极限 （C ）在),(+∞-∞内有界（D ）在),(+∞-∞内无界(3)设()f x 在x a =处可导，则xx a f x a f x )()(lim 0--+→等于(B)（A ）)(a f '（B ）)(2a f '（C ）0（D ）)2(a f '(4)【 同数学Ⅰ、第五（2）题 】十、（本题满分10分）在第一象限内，求曲线12+-=x y 上的一点，使该点处切线与所给曲线及两坐标围成的面积为最小，并求此最小面积.解：设切点的横坐标为a ，则切线方程为2(1)2()y a a x a --=--，即221y ax a =-++故所围面积2312201112(1)(1)224243a a a s a x dx a a +=+--+=++-⎰. 令0s '=得驻点a =.由于0a s ''>，故所求点的坐标为2)3，其最小值为a s =23.数 学（试卷四）一、判断题（每小题答对得2分，答错得-1分，不答得0分，全题最低0分） (1) 10lim xx e →=∞（ ⨯ ） (2)4sin 0x xdx ππ-=⎰（ √）(3)若级数1nn a∞=∑与1nn b∞=∑均发散，则级数1()nn n ab ∞=+∑必发散（ ⨯ ）(4)假设D 是矩阵A 的r 阶子式，且含D 的一切1r +阶子式都等于0，那么矩阵A 的一切1r +阶子式都等于0（ √） (5)连续型随机变量取任何给定实数值的概率都等于0（ √）二、选择题（每小题2分，满分10分.）(1)下列函数在其定义域内连续的是(A)（A ）()ln sin f x x x =+（B ）⎩⎨⎧>≤=0cos 0sin )(x xx xx f （C ）⎪⎩⎪⎨⎧>-=<+=010001)(x x x x x x f （D ）⎪⎩⎪⎨⎧=≠=0001)(x x xx f (2)若函数f(x)在区间(,)a b 内可导，21,x x 是区间内任意两点，且21x x <，则至少存一点ξ，使得（C ）(A)()()()(),f b f a f b a a b ξξ'-=-<<. (B) 111()()()(),f b f x f b x x b ξξ'-=-<<.(C) 212112()()()(),f x f x f x x x x ξξ'-=-<<. (D) 222()()()(),f x f a f x a a x ξξ'-=-<<. (3)下列广义积分收敛的是（C ）（A ）dx xxe⎰∞+ln （B ）⎰∞+exx dx ln （C ）⎰+∞ex x dx 2)(ln （D ）⎰∞+exx dx ln (4)设A 是n 阶方阵，其秩r < n ， 那么在A 的n 个行向量中(A)(A)必有r 个行向量线性无关(B)任意r 个行向量线性无关(C)任意r 个行向量都构成极大线性无关向量组(D)任意一个行向量都可以由其它r 个行向量线性表示(5)若二事件A 和B 同时出现的概率P( A B ) = 0 , 则(C)(A)A 和B 互不相容（互斥）(B)AB 是不可能事件(C)AB 未必是不可能事件(D)P (A )=0或P (B )=0三、计算下列各题（每小题4分，满分16分） (1)求极限xxx xe 10)1(lim +→.解：因 1ln(1)(1)x xe x xxxe e ++=， 而ln(1)x x xe xe x+ （当0x →）， 故 000ln(1)lim lim lim 1x x x x x x xe xe e xx →→→+===， 从而 10lim(1)x xx xe e →+=.(2)已知1111ln 22++-+=x x y ， 求y '.解：1)1)y =-，y '=-=212xx +. (3)已知y x yx arctg z -+=，求dz .解：222()()()()()()1()1()x y x y dx dy x y dx dy d x y x y dz x y x y x y x y+-+-+---==++++--22ydx xdy x y -+=+(4)求不定积分dx ex ⎰-12.解：t =，有1)t t t t t e tdt te e dt te e c c==-=-+=+⎰⎰⎰四、（本题满分10分）考虑函数sin y x = )2/0(π≤≤x ，问：(1)t 取何值时，图中阴影部分的面积1s 与2s 之和21s s s +=最小？(2 ) t 取何值时，21s s s +=最大？解：因10sin sin sin cos 1ts t t xdx t t t =-=+-⎰，22sin ()sin cos sin sin 22t s xdx t t t t t t πππ=--=+-⎰，故122sin 2cos sin 12s s s t t t t π=+=+--,(0)2t π≤≤.令0s '=，得s 在(0,)2π内的驻点4t π=.而()14s π=，()122s ππ=-，(0)1s =，因此 4t π=时，s 最小；0t =时，s 最大.五、（本题满分6分）将函数231)(2+-=x x x f 展成x 的级数，并指出收敛区间. 解：因111111()(2)(1)121212f x xx x x x x ==-=-⋅------，而011nn x x ∞==-∑，(1,1)x ∈-， 且0011()2212n n n n n x x x ∞∞====-∑∑，(2,2)x ∈-，故1100111()(1)222nn n n n n n n f x x x x ∞∞∞+====+=+∑∑∑，其收敛区间为(1,1)-.六、（本题满分5分） 计算二重积分2x De dxdy ⎰⎰，其中D 是第一象限中由直线y x =和3x y =围成的封闭区域.解：联立y x =和3x y =，可解得两曲线交点的横坐标0x =和1x =，于是22231130()12xx x x Dxe e dxdy dx e dy x x e dx ==-=-⎰⎰⎰⎰⎰七、（本题满分6分）已知某商品的需求量x 对价格P 的弹性为 33p -=η，而市场对商品的最大需求量为1 （万件），求需求函数.解：由弹性的定义，有33p dx p x dp =-，即23dxp dp x=-， 于是有 3px ce -=，c 为待定常数.由题意 0p =时，1x =，故1c =，因此3p x e -=.八、（本题满分8分）解线性方程组 ⎪⎪⎩⎪⎪⎨⎧=-+=++-=-+-=-+-337713343424313214314321x x x x x x x x x x x x x 【123431820160x x k x x -⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪- ⎪ ⎪ ⎪=+ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，k 为任意常数】 解：对方程组的增广矩阵进行初等行变换，有2143410103101130120831101000167073300000---⎛⎫⎛⎫⎪⎪---- ⎪ ⎪→→⎪⎪⎪⎪-⎝⎭⎝⎭故原方程组与下方程组同解：132343826x x x x x =-⎧⎪=-+⎨⎪=⎩，令30x =，可得原方程组的特解(3,8,0,6)T β=-.又显然原方程组的导出组与下方程组同解：1323420x x x x x =-⎧⎪=⎨⎪=⎩，令31x =，可得导出组的基础解系(1,2,1,0)T η=-. 因此原方程组的通解为：1234(,,,)(3,8,0,6)(1,2,1,0)T T x x x x k =-+-，其中k 为任意常数.九、（本题满分7分）设矩阵A 和B 满足2AB A B =+，求矩阵B ，其中A =423110123⎡⎤⎢⎥⎢⎥⎢⎥-⎣⎦.解：因2AB A B =+，故2AB B A -=，即(2)A E B A -=，故1(2)B A E A -=-=3862962129--⎛⎫⎪-- ⎪ ⎪-⎝⎭十、（本题满分6分） 求矩阵A =312014101--⎡⎤⎢⎥-⎢⎥⎢⎥-⎣⎦的实特征值及对应的特征向量.解：令0E A λ-=，即2(1)(45)0λλλ-++=,可见矩阵A 只有一个实特征值1λ=.易见，线性方程组()0E A X λ-=的基础解系为(0,2,1)T ，故A 对应于实特征值1λ=的特征向量为(0,2,1)T k ，（其中k 为非零任意常数）.十一、（每小题4分，满分8分）(1)已知随机变量X 的概率分布为(1)0.2,(2)0.3,(3)0.5P X P X P X ======，试写出X 的分布函数()F x .解：X 的分布函数为()F x =0,0.2,0.5,1,⎧⎪⎪⎨⎪⎪⎩332211≥<≤<≤<x x x x . (2)已知随机变量Y 的概率密度为000)(2222<≥⎪⎩⎪⎨⎧=-y y e y f a y a y , 求随机变量YZ 1=的数学期望EZ .解：222222200111()()y y a a y EZ E f y dy edy dy Yy y a --+∞+∞+∞-∞===⋅==⎰⎰⎰. 十二、（本题满分8分）设有两箱同种零件.第一箱内装50件，其中10件一等品；第二箱内装有30件，其中18件一等品.现从两箱中随机挑出一箱，然后从该箱中先后随机取出两个零件（取出的零件均不放回），试求:(1)先取出的零件是一等品的概率p ；(2)在先取出的零件是一等品的条件下，第二次取出的零件仍然是一等品的条件概率q .解：设i B ={取出的零件为第i 箱中的}，j A ={第j 次取出的是一等品}，,1,2i j =， 显然12,B B 为正概完备事件组，故全概公式得(1) 11112121101182()()()()()2502305p P A P B P A B P B P A B ==+=⋅+⋅=；(2) 1211212122110911817276()()()()()25049230291421P A A P B P A A B P B P A A B ⨯⨯=+=⋅+⋅=⨯⨯， 于是，由贝叶斯公式得q =12211()690()0.48557()1421P A A q P A A P A ===≈.数 学（试卷五）一、判断题（每小题答对得2分，答错得-1分，不答得0分，全题最低0分） (1)【 同数学Ⅳ 第一（1）题 】(2)【 同数学Ⅳ 第一（2）题 】(3)若函数()f x 在区间(,)a b 严格单增，则对区间(,)a b 内任何一点x 有()0f x '>. （ ⨯ ） (4)若A 为n 阶方阵，k 为常数，而A 和kA 为A 和kA 的行列式，则kA k A =. （ ⨯ ） (5)【 同数学Ⅳ 第一（5）题 】二、选择题（每小题2分，满分10分）(1)【 同数学Ⅳ 第二（1）题 】(2)【 同数学Ⅳ 第二（2）题 】(3)【 同数学Ⅳ 第二（3）题 】(4)【 同数学Ⅳ 第二（4）题 】(5)对于任二事件A 和B ，有()P A B -=(C)(A)()()P A P B -(B)()()()P A P B P AB -+(C)()()P A P AB -(D))()()(B A P B P A P --三、计算下列各题（每小题4分，满分20分）(1)求极限1ln(1)limx x arctgx→+∞+. 解：11ln(1)lim ln(1)0lim0lim /2x x x x x arctgx arctgx π→+∞→+∞→+∞++===(2)【 同数学Ⅳ 第三（2）题 】(3)【 同数学Ⅳ 第三（3）题 】(4)计算定积分dxex ⎰-12112解：t =，有111111021tt t te tdt tee dt e e ==-=-=⎰⎰⎰(5)求不定积分⎰++5224x x xdx.解：22422221(1)11arctan 252(1)242xdx d x x c x x x ++==+++++⎰⎰. 四、（本题满分10分）考虑函数2y x =，10≤≤x ，问：(1)t 取何值时，图中阴影部分的面积(与数学Ⅳ第四题类似)1s 与2s 之和21s s s +=最小？ (2 ) t 取何值时，21s s s +=最大？解：132223212041(1)33tts s s t x dx x dx t t t t =+=-+--=-+⎰⎰，(01)t ≤≤令0s '=，得(0,1)内的驻点12t =. 而11()24s =，1(0)3s =，2(1)3s =，因此 12t =时，s 最小；1t =时，s 最大.五、（本题满分5分）【 同数学Ⅳ 第六题 】 六、（本题满分8分）设某产品的总成本函数为21()40032C x x x =++，而需求函数为xp 100=，其中x 为产量（假定等于需求量）,p 为价格. 试求：（1）边际成本； （2）边际收益； （3）边际利润； （4）收益的价格弹性.解：（1）边际成本：()3MC C x x '==+；（2）收益函数：()R x p x =⋅=()MR R x'==；（3）利润函数：21()()()40032L x R x C x x x =-=--， 边际利润：()3ML L x x'==--；（4）收益的价格函数：2(100)()R x p==，收益的价格弹性：2222(100)1(100)p dR p R dp p =-⋅=-. 七、（本题满分8分）【 同数学Ⅳ 第八题 】 八、（本题满分7分）【 同数学Ⅳ 第九题 】 九、（本题满分6分）【 同数学Ⅳ 第十题】十、（本题满分8分）已知随机变量X 的概率分布为(1)0.2,(2)0.3,(3)0.5P X P X P X ======， 试写出X 的分布函数()F x ，并求X 的数学期望与方差.解：X 的分布函数为()F x =0,0.2,0.5,1,⎧⎪⎪⎨⎪⎪⎩332211≥<≤<≤<x x x x ， 10.220.330.5 2.3EX =⨯+⨯+⨯=；222210.220.330.5 5.9EX =⨯+⨯+⨯=222() 5.9 2.30.61DX EX EX =-=-=十一、（本题满分8分）【 同数学Ⅳ 第十二题】。
1998年全国硕士研究生入学统一考试英语试题真题(附带详细答案解析)
1998年全国硕士研究生入学统一考试英语试题Section I Cloze TestDirections:For each numbered blank in the following passage, there are four choices marked [A], [B], [C], and [D]. Choose the best one and mark your answer on ANSWER SHEET 1 by blackening the corresponding letter in the brackets with a pencil. (10 points)Until recent l y most historians spoke very critically of the Industrial Revolution. They1that in the long run industrialization greatly raised the standard of living for the 2 man. But they insisted that its 3 results during the period from 1750 to 1850 were widespread poverty and misery for the 4 of the English population. 5 contrast, they saw in the preceding hundred years from 1650 to 1750, when England was still a 6 agricultural country, a period of great abundance and prosperity.This view, 7 , is generally thought to be wrong. Specialists 8 history and economics, have 9 two things: that the period from 1650 to 1750 was 10 by great poverty, and that industrialization certainly did not worsen and may have actually improved the conditions for the majority of the populace.1. ［A］admitted ［B］believed ［C］claimed ［D］predicted2. ［A］plain ［B］average ［C］mean ［D］normal3. ［A］momentary ［B］prompt ［C］instant ［D］immediate4. ［A］bulk ［B］host ［C］gross ［D］magnitude5. ［A］On ［B］With ［C］For ［D］By6. ［A］broadly ［B］thoroughly ［C］generally ［D］completely7. ［A］however ［B］meanwhile ［C］therefore ［D］moreover8. ［A］at ［B］in ［C］about ［D］for9. ［A］manifested ［B］approved ［C］shown ［D］speculated10. ［A］noted ［B］impressed ［C］labeled ［D］markedSection ⅡReading ComprehensionDirections:Each of the passages below is followed by some questions. For each question there are four answers marked [A], [B], [C] and [D]. Read the passages carefully and choose the best answer to each of the questions. Then mark your answer on the ANSWER SHEET 1 by blackening the corresponding letter in the brackets. (40 points)Text 1Few creations of big technology capture the imagination like giant dams. Perhaps it is humankind‟s long suffering at the mercy of flood and drought that makes the idea of forcing the waters to do our bidding so fascinating. But to be fascinated is also, sometimes, to be blind. Several giant dam projects threaten to do more harm than good.The lesson from dams is that big is not always beautiful. It doesn‟t help that building a big, powerful dam has become a symbol of achievement for nations and people striving to assert themselves. Egypt‟s leadership in the Arab world was cemented by the Aswan High Dam. Turkey‟s bid for Firs t World status includes the giant Ataturk Dam.But big dams tend not to work as intended. The Aswan Dam, for example, stopped the Nile flooding but deprived Egypt of the fertile silt that floods left -- all in return for a giant reservoir of disease which is now so full of silt that it barely generates electricity.And yet, the myth of controlling the waters persists. This week, in the heart of civilized Europe, Slovaks and Hungarians stopped just short of sending in the troops in their contention over a dam on the Danube. The huge complex will probably have all the usual problems of big dams. But Slovakia is bidding for independence from the Czechs, and now needs a dam to prove itself.Meanwhile, in India, the World Bank has given the go-ahead to the even more wrong-headed Narmada Dam. And the bank has done this even though its advisors say the dam will cause hardship for the powerless and environmental destruction. The benefits are for the powerful, but they are far from guaranteed.Proper, scientific study of the impacts of dams and of the cost and benefits of controlling water can help to resolve these conflicts. Hydroelectric power and flood control and irrigation are possible without building monster dams. But when you are dealing with myths, it is hard to be either proper, or scientific. It is time that the world learned the lessons of Aswan. You don‟t need a dam to be saved.11. The third sentence of Paragraph 1 implies that ________.[A] people would be happy if they shut their eyes to reality[B] the blind could be happier than the sighted[C] over-excited people tend to neglect vital things[D] fascination makes people lose their eyesight12. In Paragraph 5, “the powerless” probably refers to ________.[A] areas short of electricity[B] dams without power stations[C] poor countries around India[D] common people in the Narmada Dam area13. What is the myth concerning giant dams?[A] They bring in more fertile soil.[B] They help defend the country.[C] They strengthen international ties.[D] They have universal control of the waters.14. What the author tries to suggest may best be interpreted as ________.[A] “It‟s no use crying over spilt milk”[B] “More haste, less speed”[C] “Look before you leap”[D] “He who laughs last laughs best”Text 2Well, no gain without pain, they say. But what about pain without gain? Everywhere you go in America, you hear tales of corporate revival. What is harder to establish is whether the productivity revolution that businessmen assume they are presiding over is for real.The official statistics are mildly discouraging. They show that, if you lump manufacturing and services together, productivity has grown on average by 1.2% since 1987. That is somewhat faster than the average during the previous decade. And since 1991, productivity has increased by about 2% a year, which is more than twice the 1978-87 average. The trouble is that part of the recent acceleration is due to the usual rebound that occurs at this point in a business cycle, and so is not conclusive evidence of a revival in the underlying trend. There is, as Robert Rubin, the treasury sec retary, says, a “disjunction” between the mass of business anecdote that points to a leap in productivity and the picture reflected by the statistics.Some of this can be easily explained. New ways of organizing the workplace -- all thatre-engineering and downsizing -- are only one contribution to the overall productivity of an economy, which is driven by many other factors such as joint investment in equipment and machinery, new technology, and investment in education and training. Moreover, most of the changes that companies make are intended to keep them profitable, and this need not always mean increasing productivity: switching to new markets or improving quality can matter just as much.Two other explanations are more speculative. First, some of the business restructuring of recent years may have been ineptly done. Second, even if it was well done, it may have spread much less widely than people suppose.Leonard Schlesinger, a Harvard academic and former chief executive of Au Bong Pain, a rapidly growi ng chain of bakery cafes, says that much “re-engineering” has been crude. In many cases, he believes, the loss of revenue has been greater than the reductions in cost. His colleague, Michael Beer, says that far too many companies have applied re-engineering in a mechanistic fashion, chopping out costs without giving sufficient thought to long-term profitability. BBDO‟s Al Rosenshine is blunter. He dismisses a lot of the work of re-engineering consultants as mere rubbish -- “the worst sort of ambulance chasing.”15. According to the author, the American economic situation is ________.[A] not as good as it seems[B] at its turning point[C] much better than it seems[D] near to complete recovery16. The official statistics on productivity growth ________.[A] exclude the usual rebound in a business cycle[B] fall short of businessmen‟s anticipation[C] meet the expectation of business people[D] fail to reflect the true state of economy17. The author raises the question “what about pain without gain?” because ________.[A] he questions the truth of “no gain without pain”[B] he does not think the productivity revolution works[C] he wonders if the official statistics are misleading[D] he has conclusive evidence for the revival of businesses18. Which of the following statements is NOT mentioned in the passage?[A] Radical reforms are essential for the increase of productivity.[B] New ways of organizing workplaces may help to increase productivity.[C] The reduction of costs is not a sure way to gain long-term profitability.[D] The consultants are a bunch of good-for-nothings.Text 3Science has long had an uneasy relationship with other aspects of culture. Think of Gallileo’s 17th-century trial for his rebelling belief before the Catholic Church or poet William Blake‟s harsh remarks against the mechanistic worldview of Isaac Newton. The schism between science and the humanities has, if anything, deepened in this century.Until recently, the scientific community was so powerful that it could afford to ignore its critics -- but no longer. As funding for science has declined, scientists have attacked “anti-science” in several books, notably Higher Superstition, by Paul R. Gross, a biologist at the University of Virginia, and Norman Levitt, a mathematician at Rutgers University; and The Demon-Haunted World, by Carl Sagan of Cornell University.Defenders of science have also voiced their concerns at meetings such as “The Flight from Science and Reason,” held in New York City in 1995, and “Science in the Age of (Mis) information,” which assembled last June near Buffalo.Anti-science clearly means different things to different people. Gross and Levitt find fault primarily with sociolog ists, philosophers and other academics who have questioned science‟s objectivity. Sagan is more concerned with those who believe in ghosts, creationism and other phenomena that contradict the scientific worldview.A survey of news stories in 1996 reveals that the anti-science tag has been attached to many other groups as well, from authorities who advocated the elimination of the last remaining stocks of smallpox virus to Republicans who advocated decreased funding for basic research.Few would dispute that the term applies to the Unabomber, whose manifesto, published in 1995, scorns science and longs for return to a pre-technological utopia. But surely that does not mean environmentalists concerned about uncontrolled industrial growth are anti-science, as an essay in US News & World Report last May seemed to suggest.The environmentalists, inevitably, respond to such critics. The true enemies of science, argues Paul Ehrlich of Stanford University, a pioneer of environmental studies, are those who question the evidence supporting global warming, the depletion of the ozone layer and other consequences of industrial growth.Indeed, some observers fear that the anti-science epithet is in danger of becoming meaningless. “The term …anti-science‟ can lump together too many, quite different things,” notes Harvard University philosopher Gerald Holton in his 1993 work Science and Anti-Science. “They have in common only one thing that they tend to annoy or threaten those who regard themselves as more enlightened.”19. Th e word “schism” (Line 4, Paragraph 1) in the context probably means ________.[A] confrontation[B] dissatisfaction[C] separation[D] contempt20. Paragraphs 2 and 3 are written to ________.[A] discuss the cause of the decline of science‟s power[B] s how the author‟s sympathy with scientists[C] explain the way in which science develops[D] exemplify the division of science and the humanities21. Which of the following is true according to the passage?[A] Environmentalists were blamed for anti-science in an essay.[B] Politicians are not subject to the labeling of anti-science.[C] The “more enlightened” tend to tag others as anti-science.[D] Tagging environmentalists as “anti-science” is justifiable.22. The author‟s attitude toward the issue of “science vs. anti-science” is ________.[A] impartial[B] subjective[C] biased[D] puzzlingText 4Emerging from the 1980 census is the picture of a nation developing more and more regional competition, as population growth in the Northeast and Midwest reaches a near standstill.This development -- and its strong implications for US politics and economy in years ahead -- has enthroned the South as America‟s most densely populated region for the first time in the history of the nation‟s head counting.Altogether, the US population rose in the 1970s by 23.2 million people -- numerically the third-largest growth ever recorded in a single decade. Even so, that gain adds up to only 11.4 percent, lowest in American annual records except for the Depression years.Americans have been migrating south and west in larger numbers since World War II, and the pattern still prevails.Three sun-belt states -- Florida, Texas and California -- together had nearly 10 million more people in 1980 than a decade earlier. Among large cities, San Diego moved from 14th to 8th and San Antonio from 15th to 10th -- with Cleveland and Washington. D. C., dropping out of the top 10.Not all that shift can be attributed to the movement out of the snow belt, census officials say. Nonstop waves of immigrants played a role, too -- and so did bigger crops of babies as yesterday‟s “baby boom” generation reached its child bearing years.Moreover, demographers see the continuing shift south and west as joined by a related but newer phenomenon: More and more, Americans apparently are looking not just for places with more jobs but with fewer people, too. Some instances—■Regionally, the Rocky Mountain states reported the most rapid growth rate -- 37.1 percent since 1970 in a vast area with only 5 percent of the US population.■Among states, Nevada and Arizona grew fastest of all: 63.5 and 53.1 percent respectively. Except for Florida and Texas, the top 10 in rate of growth is composed of Western states with 7.5 million people -- about 9 per square mile.The flight from overcrowdedness affects the migration from snow belt to more bearable climates.Nowhere do 1980 census statistics dramatize more the American search for spacious living than in the Far West. There, California added 3.7 million to its population in the 1970s, more than any other state.In that decade, however, large numbers also migrated from California, mostly to other partsof the West. Often they chose -- and still are choosing -- somewhat colder climates such as Oregon, Idaho and Alaska in order to escape smog, crime and other plagues of urbanization in the Golden State.As a result, California‟s growth rat e dropped during the 1970s, to 18.5 percent -- little more than two thirds the 1960s‟ growth figure and considerably below that of other Western states.23. Discerned from the perplexing picture of population growth the 1980 census provided,America in 1970s ________.[A] enjoyed the lowest net growth of population in history[B] witnessed a southwestern shift of population[C] underwent an unparalleled period of population growth[D] brought to a standstill its pattern of migration since World War II24. The census distinguished itself from previous studies on population movement in that________.[A] it stresses the climatic influence on population distribution[B] it highlights the contribution of continuous waves of immigrants[C] it reveals the Ameri cans‟ new pursuit of spacious living[D] it elaborates the delayed effects of yesterday‟s “baby boom”25. We can see from the available statistics that ________.[A] California was once the most thinly populated area in the whole US[B] the top 10 states in growth rate of population were all located in the West[C] cities with better climates benefited unanimously from migration[D] Arizona ranked second of all states in its growth rate of population26. The word “demographers” (Line 1, Paragraph 8) most probably means ________.[A] people in favor of the trend of democracy[B] advocates of migration between states[C] scientists engaged in the study of population[D] conservatives clinging to old patterns of lifeText 5Scattered around the globe are more than 100 small regions of isolated volcanic activity known to geologists as hot spots. Unlike most of the world‟s volcanoes, they are not always found at the boundaries of the great drifting plates that make up the earth‟s surface; on the contrary, many of them lie deep in the interior of a plate. Most of the hot spots move only slowly, and in some cases the movement of the plates past them has left trails of dead volcanoes. The hot spots and their volcanic trails are milestones that mark the passage of the plates.That the plates are moving is now beyond dispute. Africa and South America, for example, are moving away from each other as new material is injected into the sea floor between them. The complementary coastlines and certain geological features that seem to span the ocean are reminders of where the two continents were once joined. The relative motion of the plates carrying these continents has been constructed in detail, but the motion of one plate with respect to another cannot readily be translated into motion with respect to the earth‟s interior. It is not possible to determine whether both continents are moving in opposite directions or whether one continent is stationary and the other is drifting away from it. Hot spots, anchored in the deeper layers of the earth, provide the measuring instruments needed to resolve the question. From an analysis of thehot-spot population it appears that the African plate is stationary and that it has not moved during the past 30 million years.The significance of hot spots is not confined to their role as a frame of reference. It now appears that they also have an important influence on the geophysical processes that propel the plates across the globe. When a continental plate come to rest over a hot spot, the material rising from deeper layers creates a broad dome. As the dome grows, it develops deep fissures (cracks); in at least a few cases the continent may break entirely along some of these fissures, so that the hot spot initiates the formation of a new ocean. Thus just as earlier theories have explained the mobility of the continents, so hot spots may explain their mutability (inconstancy).27. The author believes that ________.[A] the motion of the plates corresponds to that of the earth‟s interior[B] the geological theory about drifting plates has been proved to be true[C] the hot spots and the plates move slowly in opposite directions[D] the movement of hot spots proves the continents are moving apart28. That Africa and South America were once joined can be deduced from the fact that________.[A] the two continents are still moving in opposite directions[B] they have been found to share certain geological features[C] the African plate has been stable for 30 million years[D] over 100 hot spots are scattered all around the globe29. The hot spot theory may prove useful in explaining ________.[A] the structure of the African plates[B] the revival of dead volcanoes[C] the mobility of the continents[D] the formation of new oceans30. The passage is mainly about ________.[A] the features of volcanic activities[B] the importance of the theory about drifting plates[C] the significance of hot spots in geophysical studies[D] the process of the formation of volcanoesSection IV English-Chinese TranslationDirections:Read the following passage carefully and then translate the underlined sentences into Chinese. Your translation must be written clearly on the ANSWER SHEET 2. (15 points)They were, by far, the largest and most distant objects that scientists had ever detected: a strip of enormous cosmic clouds some 15 billion light-years from earth. 31) But even more important,it was the farthest that scientists had been able to look into the past, for what they were seeing were the patterns and structures that existed 15 billion years ago. That was just about the moment that the universe was born. What the researchers found was at once both amazing and expected: the US National Aeronautics and Space Administration‟s Cosmic Background Explorer satellite -- Cobe -- had discovered landmark evidence that the universe did in fact begin with the primeval explosion that has become known as the Big Bang (the theory that the universe originated in an explosion from a single mass of energy).32) The existence of the giant clouds was virtually required for the Big Bang, first put forward in the 1920s, to maintain its reign as the dominant explanation of the cosmos. According to the theory, the universe burst into being as a submicroscopic, unimaginably dense knot of pure energy that flew outward in all directions, emitting radiation as it went, condensing into particles and then into atoms of gas. Over billions of years, the gas was compressed by gravity into galaxies, stars, plants and eventually, even humans.Cobe is designed to see just the biggest structures, but astronomers would like to see much smaller hot spots as well, the seeds of local objects like clusters and superclusters of galaxies. They shouldn‟t have long to wait. 33) Astrophysicists working with ground-based detectors at the South Pole and balloon-borne instruments are closing in on such structures, and may report their findings soon.34) If the small hot spots look as expected, that will be a triumph for yet another scientific idea, a refinement of the Big Bang called the inflationary universe theory. Inflation says that very early on, the universe expanded in size by more than a trillion trillion trillion trillion fold in much less than a second, propelled by a sort of antigravity. 35) Odd though it sounds, cosmic inflation is a scientifically plausible consequence of some respected ideas in elementary particle physics, and many astrophysicists have been convinced for the better part of a decade that it is true.31. ________32. ________33. ________34. ________35. ________Section V WritingDirections:[A] Study the following cartoon carefully and write an essay in no less than 150 words.[B] Your essay must be written clearly on the ANSWER SHEET 2. (15 points)[C] Your essay should meet the requirements below:1. Write out the messages conveyed by the cartoon.2. Give your commentsn.1998年英语试题答案Part ⅠCloze Test1. A2. B3. D4. A5. D6. D7. A8.B9. C 10. DPart ⅡReading ComprehensionPart APassage 111. C 12. D 13.D 14. CPassage 215.A 16.B 17.B 18.APassage 319.C 20.D 21.A 22.APassage 423.B 24.C 25.D 26.CPassage 527.B 28.B 29.C 30.CPart ⅢEnglish-Chinese Translation31.更为重要的是，这是科学家们能够观测到的最遥远的过去的景象，因为他们看到的是150亿年前宇宙云的形状和结构。
1989年全国硕士研究生入学统一考试数学一、二、三、四、五试题完整版附答案及评分标准
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二、选择题 (本题满分 15 分，每小题 3 分)
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(A) 有且仅有水平渐近线；
(B) 有且仅有铅直渐近线.
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(2) 已知曲面 z 4 x2 y2 上点 P 处的切平面平行于平面 2x 2 y z 1 0 ，则点 P 的
1989 年全国硕士研究生入学统一考试数学一、二、三、四、五试题 完整版附答案及评分标准 数 学（试卷一）
一、填空题 (本题满分 15 分，每小题 3 分)
(1) 以知 f (3)=2,则 lim f (3 h) f (3) -1
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„„4 分
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2021年全国硕士研究生入学统一考试参考答案
2021年全国硕士研究生入学统一考试参考答案数学（二）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的，把所选选项前的字母填在答题卡制定的位置上.） （1）当0→x 时，dt e x t ⎰-23)1(是7x 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】C【解析】因为当0→x 时，7'02~)1(2)1(623x e x dt e x x t -=⎥⎦⎤⎢⎣⎡-⎰，所以dt e x t ⎰-230)1(是7x 的阶无穷小 ，正确答案为C.（2）函数⎪⎩⎪⎨⎧=≠-=0,10,1)(x x x e x f x ，在0=x 处（ ）（A ）连续且取得极大值 （B ）连续且取得极小值 （C ）可导且导数为0 （D ）可导且导数不为0 【答案】D【解析】因为)0(11lim)(lim 00f xe xf x x x ==-=→→，故)(x f 在0=x 处连续； 因为211lim 011lim 0)0()(lim 2000=--=---=--→→→x x e x x e x f x f x x x x x ，故正确答案为D.（3）有一圆柱底面半径与高随时间变化的速率分别为s cm /2，s cm /3-，当底面半径为cm 10，高为cm 5时，圆柱体的体积与表面积随时间的变化率分别为（ ） （A ）s cm /1253π，s cm /403π （B ）s cm /1253π，s cm /403π- （C ）s cm /1003π-，s cm /403π （D ）s cm /1003π-，s cm /403π- 【答案】C【解析】由题意知，2=dt dr ，3-=dtdh ，又h r V 2π=，222r rh S ππ+=， 则dt dh r dt dr rh dt dV 22ππ+=，dtdrr dt dh r dt dr h dt dS πππ422++=， 当5,10==h r 时，π100-=dt dV ，π40=dtdS，故正确答案为C. （4）设函数x b ax x f ln )(-=（0>a ）有两个零点，则ab的取值范围是（ ）（A ）),(+∞e （B ）),0(e （C ）)1,0(e （D ）),1(+∞e【答案】A【解析】令0ln )(=-=x b ax x f ，xba x f -=)('，令0)('=x f ，有驻点a b x =，0ln <⋅-⋅=⎪⎭⎫⎝⎛a b b a b a a b f ，从而1ln >a b ，可得e a b >，正确答案为A.（5）设函数x x f sec )(=在0=x 处的2次泰勒多项式为21bx ax ++，则（ ）（A ） 21,1-==b a （B ）21,1==b a （C ）21,0-==b a （D ）21,0==b a【答案】D【解析】由)(2)0('')0(')0()(22x o x f x f f x f +++=，知当x x f sec )(=时，10sec )0(==f ，00tan 0sec )0('=⋅=f ，1)sec tan (sec )0(''032=+==x x x x f ，则)(211sec )(22x o x x x f ++==，故正确答案为D. （6）设函数),(y x f 可微，且2)1(),1(+=+x x e x f x，x x x x f ln 2),(22=，则=)1,1(df （ ）（A ）dy dx + （B ）dy dx - （C ）dy （D ）dy - 【答案】C【解析】)1(2)1(),1('),1('221+++=+++x x x e x f e e x f xxx①x x x x x xf x x f 2ln 4),('2),('2221+=+ ①分别将⎩⎨⎧==00y x ，⎩⎨⎧==11y x 代入①①式有 1)1,1(')1,1('21=+f f ， 2)1,1('2)1,1('21=+f f ，联立可得1)1,1('0)1,1('21==f f ，，于是dy dy f dx f df =+=)1,1(')1,1(')1,1(21，故正确答案为C.（7）设函数)(x f 在区间]1,0[上连续，则=⎰dx x f 1)(（ ）（A ）∑=∞→⎪⎭⎫ ⎝⎛-nk n n n k f 121212lim（B ）∑=∞→⎪⎭⎫ ⎝⎛-nk n nn k f 11212lim（C ）∑=∞→⎪⎭⎫ ⎝⎛-nk n n n k f 212121lim（D ）∑=∞→⎪⎭⎫ ⎝⎛nk n nn k f 2122lim 【答案】B【解析】由定积分的定义知，将)1,0(分成n 份，取中间点的函数值，则=⎰dx x f 1)(∑=∞→⎪⎭⎫ ⎝⎛-nk n n n k f 11212lim ，故正确答案为B.（8）二次型2132********)()()(),,(x x x x x x x x x f --+++=的正惯性指数与负惯性指数依次为（ ）（A ）0,2 （B ）1,1 （C ）1,2 （D ）2,1 【答案】B【解析】313221222132322213212222)()()(),,(x x x x x x x x x x x x x x x x f +++=--+++=所以二次型矩阵⎪⎪⎪⎭⎫ ⎝⎛=011121110A ，故特征多项式为λλλλλλλ)3)(1(1112111-+=-------=-A E ，令上式等于零，故特征值为0,3,1-，故该二次型的正惯性指数为1，负惯性指数为1，故答案应选B.（9）设3阶矩阵),,(321ααα=A ，),,(321βββ=B ，若向量组321,,ααα可以由向量组321,,βββ线性表示，则（ ）（A ）0=Ax 的解均是0=Bx 的解 （B ）0=x A T的解均是0=x B T的解 （C ）0=Bx 的解均是0=Ax 的解 （D ）0=x B T的解均是0=x A T的解 【答案】D【解析】令),,(321ααα=A ，),,(321βββ=B ，由题向量组321,,ααα可以由向量组321,,βββ线性表示，即⎪⎪⎭⎫⎝⎛=⇒=T T TA B r B r A B r B r )(),()(,所以0=x B T与0=⎪⎪⎭⎫ ⎝⎛x A B T T 同解，即0=x B T 的解均是0=x A T的解，故选项D 正确. （10）已知矩阵⎪⎪⎪⎭⎫ ⎝⎛----=521112101A ，若下三角可逆矩阵P 和上三角可逆矩阵Q ，使得PAQ 为对角矩阵，则Q P ,可以分别取（ ）（A ）⎪⎪⎪⎭⎫ ⎝⎛100010001 ，⎪⎪⎪⎭⎫ ⎝⎛100310101 （B ）⎪⎪⎪⎭⎫ ⎝⎛--123012001 ，⎪⎪⎪⎭⎫⎝⎛100010001（C ）⎪⎪⎪⎭⎫ ⎝⎛--123012001，⎪⎪⎪⎭⎫ ⎝⎛100310101 （D ）⎪⎪⎪⎭⎫ ⎝⎛131010001，⎪⎪⎪⎭⎫⎝⎛--100210321 【答案】C【解析】()⎪⎪⎪⎭⎫ ⎝⎛----→⎪⎪⎪⎭⎫ ⎝⎛----=101620012310001101100523010112001101,E A),(123000012310001101P F =⎪⎪⎪⎭⎫ ⎝⎛-----→，则=P ⎪⎪⎪⎭⎫ ⎝⎛--123012001； ⎪⎪⎭⎫⎝⎛Λ=⎪⎪⎪⎪⎪⎪⎪⎪⎭⎫⎝⎛→⎪⎪⎪⎪⎪⎪⎪⎪⎭⎫ ⎝⎛--=⎪⎪⎭⎫ ⎝⎛Q E F 100310101000010001100010001000310101，则=Q ⎪⎪⎪⎭⎫⎝⎛100310101.故正确答案为C. 或者：⎪⎪⎪⎭⎫⎝⎛---−−→−⎪⎪⎪⎭⎫ ⎝⎛----−−→−⎪⎪⎪⎭⎫ ⎝⎛----=+-62031010152131010152111210113122r r r r A⎪⎪⎪⎭⎫ ⎝⎛-−−→−⎪⎪⎪⎭⎫ ⎝⎛-−−→−⎪⎪⎪⎭⎫ ⎝⎛--−−→−+++00001000100031000100031010123132332c c c c r r⎪⎪⎪⎭⎫ ⎝⎛−→−-0000100012r ，取⎪⎪⎪⎭⎫ ⎝⎛--=⎪⎪⎪⎭⎫ ⎝⎛-⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎭⎫ ⎝⎛-=123012001100012001101010001120010001100010001P⎪⎪⎪⎭⎫ ⎝⎛=⎪⎪⎪⎭⎫ ⎝⎛⎪⎪⎪⎭⎫ ⎝⎛=100310101100310001100010101Q二、填空题（本题共6小题，每小题5分，共30分.请将答案写在答题纸指定位置上） （11）.__________32=⎰+∞∞--dx x x【答案】3ln 1 【解析】3ln 133ln 1)(332322222=⋅-=--==∞+-+∞-+∞-+∞∞--⎰⎰⎰x x x x x d dx x dx x .（12）设函数)(x y y =由参数方程⎪⎩⎪⎨⎧+-=++=2)1(412te t y t e x t t确定，则._________022==t dx y d【答案】32【解析】由1224++=tt e tte dx dy ，得222)12()24()12)(244(++++++=t t t t t t e e t te e te e dx y d ， 将0=t 代入得.32022==t dxyd（13）设函数),(y x z z =由方程1)2arctan(ln )1(=-++xy z y z x 确定，则.__________)2,0(=∂∂xz【答案】1【解析】方程两边对x 求导，得04121)1(22=+-∂∂+∂∂++yx yx z z y x z x z ，将0=x ，2=y 代入原方程，得1=z ，再将0=x ，2=y ，1=z 代入，得.1)2,0(=∂∂xz（14）已知函数dy yx dx t f x t⎰⎰=11sin )(，则.__________2'=⎪⎭⎫⎝⎛πf 【答案】2cos 222coscos 223ππππππ--⎰du uu.【解析】交换积分次序，有dx yxdy t f t y t⎰⎰-=2sin )(1，从而⎰⎰⎰⎪⎪⎭⎫ ⎝⎛-=-=t ty tdy y y ty dx y x dy t f 11cos cos sin )(2 ⎰⎰⎰⎰-=-=t t t t t ydy y dy uut ydy y dy y t y 13211cos cos cos cos ， t t t t t t t t dy u u t t f t t cos 21cos cos cos 2)('23323-⎪⎪⎭⎫⎝⎛⋅--=⎰，故 2cos 222coscos )2('223πππππππ--=⎰du u u f .（15）微分方程0'''=-y y 的通解为._________=y【答案】⎪⎪⎭⎫ ⎝⎛++=-x C x C ee C y x x23sin 23cos 32211，R C C C ∈321,, 【解析】由特征方程013=-λ，解得11=λ，i 23213,2±=λ，故方程的通解为 ⎪⎪⎭⎫ ⎝⎛++=-x C x C ee C y x x23sin 23cos 32211，R C C C ∈321,,.（16）多项式xxx x x xx f 11211212121)(--=中3x 项的系数为.__________【答案】5-【解析】112122121311211121212111111211211212121)(---------=--=x x x xx xx x xxx x xxx x x x x f ，所以展开式中含3x 项的 有3x -，34x -，即3x 项得系数为5-.三、解答题（本题共6小题，共70分.请将解答写在答题纸指定位置上，解答应写出文字说明、证明过程或演算过程.）（17）（本题满分10分）求极限⎪⎪⎪⎭⎫⎝⎛--+⎰→x e dt e x x t x sin 111lim 002. 【答案】21 【解析】⎪⎭⎫ ⎝⎛--+-=⎪⎪⎪⎭⎫ ⎝⎛--+→→→⎰⎰x e e dt e x e dt e x x x x t x x x t x sin 111lim 1lim sin 111lim 0000022， xe x x e x x e e x e xx x x x x x x x 2cos lim 11sin lim 1sin )1(1sin lim 1lim 020002-+=+-+=-+-+=→→→→ .212112sin lim 10=-=--+=→x x e x（18）（本题满分12分） 已知xx x x f +=1)(，求)(x f 的凹凸性及渐近线.【答案】凹区间),0(),1,(+∞--∞，凸区间)0,1(-，斜渐近线是1-=x y ，1--=x y .【解析】因为⎪⎪⎩⎪⎪⎨⎧≤+->+=0,10,1)(22x xx x xx x f ，故0>x 时，22)1(2)('x x x x f ++=，3)1(2)(''x x f +=， 0<x 时，22)1(2)(''x x x x f +--=，3)1(2)(''x x f +-=， 所以故x x x x f +=1)(凹区间),0(),1,(+∞--∞，凸区间)0,1(-.因为∞=+-→xxx x 1lim1，所以1-=x 是垂直渐近线；因为1)1(lim=++∞→x x x x x ，11lim -=⎪⎪⎭⎫⎝⎛-++∞→x x x x x ，所以1-=x y 为斜渐近线； 因为1)1(lim -=+-∞→x x xx x ，11lim =⎪⎪⎭⎫⎝⎛++-∞→x x x x x ，所以1+-=x y 为斜渐近线.（19）（本题满分12分） 已知函数)(x f 满足C x x dx xx f +-=⎰261)(，L 为曲线)(x f y =（94≤≤x ），L 的弧长为s ，L 绕x 轴旋转一周所形成的面积为A ，求s 和A . 【答案】322=s ，9425π=A . 【解析】C x x dx x x f +-=⎰261)(两边求导，得131)(-=x xx f ，所以212331)(x x x f -=，曲线的弧长32241421'194942=++=+=⎰⎰dx x x dx y s ； 曲面的侧面积为942521)31(2'12942123942πππ=++-=+=⎰⎰dx x x x x dx y y A .（20）（本题满分12分）函数)(x y y =的微分方程66'-=-y xy ，满足10)3(=y ， （1）求)(x y 的表达式；（2）P 为曲线)(x y y =上的一点，曲线)(x y y =在点P 的法线在y 轴上的就截距为y I ，为使y I 最小，求P 的坐标.【答案】（1）31)(6x x y +=；（2）⎪⎭⎫ ⎝⎛±34,1P 时，y I 有最小值.611【解析】（1）由66'-=-y xy ，得xy x y 66'-=-，所以 6666611)6(Cx C x x C dx e x e y dx x dx x +=⎪⎭⎫ ⎝⎛+=⎥⎦⎤⎢⎣⎡+⎰-⎰=⎰-，将10)3(=y 代入，得31=C ，所以31)(6x x y +=.（2）设),(y x P ，则过P 点的切线方程为)(25x X x y Y -=-， 法线方程为)(215x X xy Y --=-， 令0=X ，得462131xx I Y y ++==，为偶函数，故只需要考虑),0(+∞即可. 令022'55=-=x x I y ，解得1=x ， 所以当)1,0(∈x 时，0'<y I ；当),1(+∞∈x 时，0'>y I ；故当1=x 时462131xx I Y y ++==取得极小值，也是最小值，且最小值为611)1(=±y I ，且P 点坐标为⎪⎭⎫ ⎝⎛±34,1P .（21）（本题满分12分）曲线22222)(y x y x -=+（0,0≥≥y x ）与x 轴围成的区域为D ，求.dxdy xy D⎰⎰【答案】481【解析】使用极坐标计算⎰⎰⎰⎰⎰==4022cos 0340cos sin 2cos 41cos sin πθπθθθθθθθd dr r d dxdy xy D.4812cos 4812cos 2cos 161403402=-=-=⎰ππθθθd（22）（本题满分12分）设矩阵仅有两个不同的特征值. 若相似于对角矩阵，求的值，并求可逆矩阵为对角矩阵.【解】由，故或.（1）当时，由于相似于对角矩阵，二重特征根有两个线性无关的特征向量，从而，故.此时，的两个线性无关的特征向量为，,的一个特征向量为.令为对角矩阵.（2）当时，类似的讨论可知. 此时，的两个线性无关的特征向量为，，的一个特征向量为.令为对角矩阵.。
全国硕士研究生入学综合能力考试资料答案附后
【经典资料，ＷＯＲＤ文档，可编辑修改】【经典考试资料，答案附后，看后必过，ＷＯＲＤ文档，可修改】2013年全国硕士研究生入学统一考试管理类专业学位联考综合能力考试大纲
Ⅰ考试性质
综合能力考试是为高等院校和科研院所招收管理类专业学位硕士研究生而
设置的具有选拔性质的全国联考科目，其目的是科学、公平、有效地测试考
生是否具备攻读专业学位所必须的基本素质、一般能力和培养潜能，评价的
标准是高等学校本科毕业生所能达到的及格或及格以上的水平，以利于各高
等院校和科研院所在专业上择优选拔，确保专业学位硕士研究生的招生质
量。
Ⅱ考查目标
1. 具有运用数学基础知识、基本方法分析和解决问题的能力。
2. 具有较强的分析、推理、论证等逻辑思维能力。
3. 具有较强的文字材料理解能力、分析能力以及书面表达能力。
Ⅲ考试形式和试卷结构
一、试卷满分及考试时间
试卷满分为 200 分,考试时间为 180 分钟。
二、答题方式。
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全国硕士研究生入学统一考试资料答案附后
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【经典资料，ＷＯＲＤ文档，可编辑修改】【经典考试资料，答案附后，看后必过，ＷＯＲＤ文档，可修改】
2015年全国硕士研究生入学统一考试
管理类专业学位联考综合能力考试大纲
I.考试性质
综合能力考试是为高等院校和科研院所招收管理类专业学位硕士研究生而设置的具有选择性质的全国联考科目，其目的是科学、公平、有效地测试考生是否具备攻读专业学位必需的基本素质、一般素质和培养潜能，评估的标准是高等学校本科毕业生所能达到的及格或及格以上水平，以利于高等院校和科研院所在专业上择优选拔，确保专业学位硕士研究生的招生质量。
II.考察目标
1.具有运用数学基础知识、基本方法分析和解决问题的能力。
2.具有较强的分析、推理、论证等逻辑思维能力。
3.具有较强的文字材料理解能力、分析能力以及书面表达能力。
III.考试形式和试卷结构。