中原大学95学年度硕士班入学考试

1995年全国硕士学位研究生入学考试政治试卷参考答案（文科一、单项选择题（每小题1分，共10分）1.C2.A3. B4.A5.D6.B7.C8. C9.D 10.D二、多项选择题（每小题2分，共20分）11.ACDE 12.BC 13.ABD 14.ABCE 15.ABDE 16.DCD 17.AC 18.ACDE 19.AC 20.ABCDE三、简答题21.答案略。

22.答案略。

四、辨析题23.答案要点：偶然性是必然性相对应的哲学范畴。

必然性是由内部根本矛盾决定的事物发展的确定不移的趋势，仍然性是由非根本矛盾和外部因索决定的事物发展的不确定趋势。

二者辩证联系。

不可分割。

偶然性是必然性的补充和表演形式，偶然性的背后隐藏着必然性，必然性通过无数偶然事件为自己开辟道路并得以实现。

（3分）科学探索的任务是通过反复试验揭示事物的发展规律，即通过大量偶然性发现必然性。

偶然性是科学活动中的“机遇”，对科学发现有着重要作用。

把偶然性看作科学的敌人就无法认识必然性，实际也就取消了科学。

（2分）24.答案要点：商品市场价格以生产价格为中心，随供求关系的变化而上下波动是资本主义自由竞争阶段价值规律作用的表现形式。

（1分）资本主义垄断阶段，垄断组织控制了大部分的生产与销售，限制了资本的流动，使价值转化为生产价格的过程发生障碍。

垄断组织凭借其垄断地位规定垄断价格，获得超过平均利润的垄断利润，使垄断价格长期、稳定地背离生产价格以至价值，使价值规律进一步改变了表现形式。

（3分）25.答案要点：任何政府都是由少数人组成的，但政府作为国家政权机关是有阶级性的，是统治阶级的代表。

在资本主义国家，不管是“使用一个人的名义，还是使用许多人的名义”，无论采取何种政体，其政府都是代表资产阶级的利益，“终归是少数人的政府”。

（3分）在社会主义国家，尽管政府也是由少数人组成的，但它却代表了广大人民群众的利益，是为多数人服务的政府。

上述说法混淆了不同国体的本质区别。

1995年全国硕士研究生入学统一考试数学一试题1995年全国硕士研究生入学统一考试数学一试题一、填空题(本题共5个小题,每小题3分,满分15分.) (1) 2sin 0lim(13)xx x →+=______________.(2) 202cos xd x t dt dx =⎰______________.(3) 设()2a b c ⨯⋅=,则[()()]()a b b c c a +⨯+⋅+=______________. (4) 幂级数2112(3)n n nn nx ∞-=+-∑的收敛半径R =______________. (5) 设三阶方阵A 、B 满足关系式：16A BA A BA -=+,且100310041007A ⎛⎫ ⎪⎪ ⎪= ⎪ ⎪ ⎪ ⎪⎝⎭,则B =______________.二、选择题(本题共5个小题,每小题3分,满分15分.)(1) 设有直线3210,:21030x y z L x y z +++=⎧⎨--+=⎩及平面:4230x y z ∏-+-=,则直线L( )(A) 平行于∏ (B) 在∏上 (C) 垂直于∏ (D) 与∏斜交(2) 设在[0,1]上()0f x ''>,则(0)f '、(1)f '、(1)(0)f f -或(0)(1)f f -的大小顺序是()(A) (1)(0)(1)(0)f f f f ''>>- (B) (1)(1)(0)(0)f f f f ''>->(C) (1)(0)(1)(0)f f f f ''->> (D) (1)(0)(1)(0)f f f f ''>-> (3) 设()f x 可导,()()(1|sin |)F x f x x =+,则(0)0f =是()F x 在0x =处可导的 ( )(A) 充分必要条件 (B) 充分条件但非必要条件(C) 必要条件但非充分条件 (D) 既非充分条件又非必要条件(4)设(1)ln 1n n u ⎛=-+ ⎝,则级数( )(A) 1n n u ∞=∑与21nn u ∞=∑都收敛 (B) 1n n u ∞=∑与21n n u ∞=∑都发散(C) 1n n u ∞=∑收敛而21nn u ∞=∑发散 (D) 1n n u ∞=∑发散而21n n u ∞=∑收敛(5) 设111213212223313233a a a A a a a a a a ⎛⎫⎪= ⎪ ⎪⎝⎭,212223111213311132123313a a a B a a a a a a a a a ⎛⎫⎪= ⎪ ⎪+++⎝⎭,1010100001P ⎛⎫⎪= ⎪ ⎪⎝⎭, 2100010101P ⎛⎫⎪= ⎪ ⎪⎝⎭,则必有( )(A) 12APP B = (B) 21AP P B =(C) 12PP A B = (D) 21P P A B =三、(本题共2小题,每小题5分,满分10分.)(1) 设2(,,),(,,)0,sin y u f x y z x e z y x ϕ===,其中f 、ϕ都具有一阶连续偏导数,且0zϕ∂≠∂,求du dx .(2) 设函数()f x 在区间[0,1]上连续,并设10()f x dx A =⎰,求 11()()xdx f x f y dy ⎰⎰.四、(本题共2小题,每小题6分,满分12分.)(1) 计算曲面积分zdS ∑⎰⎰,其中∑为锥面z 在柱体222x y x +≤内的部分.(2) 将函数()1(02)f x x x =-≤≤展开成周期为4的余弦级数.五、(本题满分7分)设曲线L 位于xOy 平面的第一象限内,L 上任一点M 处的切线与y 轴总相交,交点记为A .已知MA OA =,且L 过点33,22⎛⎫⎪⎝⎭,求L 的方程.六、(本题满分8分)设函数(,)Q x y 在xOy 平面上具有一阶连续偏导数,曲线积分2(,)Lxydx Q x y dy +⎰与路径无关,并且对任意t 恒有(,1)(1,)(0,0)(0,0)2(,)2(,)t t xydx Q x y dy xydx Q x y dy +=+⎰⎰,求(,)Q x y .七、(本题满分8分)假设函数()f x 和()g x 在[,]a b 上存在二阶倒数,并且()0g x ''≠,()()()()f a f b g a g b ===,试证：(1) 在开区间(,)a b 内()0g x ≠；(2) 在开区间(,)a b 内至少存在一点ξ,使()()()()f fg g ξξξξ''=''.八、(本题满分7分)设三阶实对称矩阵A 的特征值为11λ=-,231λλ==,对应于1λ的特征向量为1(0,1,1)T ξ=,求A .九、(本题满分6分)设A 是n 阶矩阵,满足T AA E =(E 是n 阶单位阵,T A 是A 的转置矩阵),0A <,求A E +.十、填空题(本题共2小题,每小题3分,满分6分.)(1) 设X 表示10次独立重复射击命中目标的次数,每次射中目标的概率为0.4,则2X 的数学期望2()E X =___________. (2) 设X 和Y 为两个随机变量,且{}30,07P X Y ≥≥=, 4(0)(0)7P X P Y ≥=≥=, 则{}max(,)0P X Y ≥=___________.十一、(本题满分6分)设随机变量X 的概率密度为, 0,()0, 0,x X e x f x x -⎧≥=⎨<⎩求随机变量X Y e =的概率密度()Y f y .1995年全国硕士研究生入学统一考试数学一试题解析一、填空题(本题共5个小题,每小题3分,满分15分.) (1)【答案】6e【解析】这是1∞型未定式求极限,2123sin 3sin 0lim(13)lim(13)x xx xx x x x ⋅⋅→→+=+,令3x t =,则当0x →时,0t →,所以1130lim(13)lim(1)xtx t x t e →→+=+=,故 00266lim6lim6sin sin sin sin 0lim(13)lim x x x x x xxx xx x x eeee →→→→+====.(2)【答案】20224cos 2cos xt dt x x -⎰ 【解析】()220022cos cos x x d d x t dt x t dt dx dx=⎰⎰ ()()20222cos cos 2xt dt x x x =-⋅⎰20224cos 2cos xt dt x x =-⎰.【相关知识点】积分上限函数的求导公式：()()()()()()()()()x x d f t dt f x x f x x dxβαββαα''=-⎰. (3)【答案】4【解析】利用向量运算律有[()()]()a b b c c a +⨯+⋅+[()]()[()]()a b b c a a b c c a =+⨯⋅+++⨯⋅+()()()()a b b b c a a c b c c a =⨯+⨯⋅++⨯+⨯⋅+ (其中0b b ⨯=) ()()()()a b c a b a a c c b c a =⨯⋅+⨯⋅+⨯⋅+⨯⋅ ()()a b c b c a =⨯⋅+⨯⋅ ()()4a b c a b c =⨯⋅+⨯⋅=.(4)【解析】令212(3)n n n nna x -=+-,则当n →∞时,有2(1)1111212211112(3)lim lim 2(3)23(1)311lim ,323(1)3n n n n n n n nn n nn n n n n n n xa a nx n x x n +-+++→∞→∞-+→∞++++-=+-⎡⎤⎛⎫+-⎢⎥⎪⎝⎭+⎢⎥⎣⎦=⋅⋅=⎡⎤⎛⎫+-⎢⎥⎪⎝⎭⎢⎥⎣⎦而当2113x <时,幂级数收敛,即||x <,此幂级数收敛,当2113x >时,即||x >时,此幂级数发散,因此收敛半径为R =(5)【答案】300020001⎛⎫⎪⎪ ⎪⎝⎭【解析】在已知等式16A BA A BA -=+两边右乘以1A -,得16A B E B -=+,即1()6A E B E --=.因为 1300040007A -⎛⎫ ⎪= ⎪ ⎪⎝⎭,所以116()6B A E --=-=1200030006-⎛⎫ ⎪ ⎪ ⎪⎝⎭=300020001⎛⎫⎪⎪ ⎪⎝⎭.二、选择题(本题共5个小题,每小题3分,满分15分.) (1)【答案】(C)【解析】这是讨论直线L 的方向向量与平面∏的法向量的相互关系问题. 直线L 的方向向量132281477(42)2110i jk l i j k i j k ⎛⎫ ⎪==-+-=--+ ⎪ ⎪--⎝⎭,平面∏的法向量42n i j k =-+,l n ,L ⊥∏.应选(C). (2)【答案】(B)【解析】由()0f x ''>可知()f x '在区间[0,1]上为严格单调递增函数,故(1)()(0),(01)f f x f x '''>> <<由微分中值定理,(1)(0)(),(01)f f f ξξ'-=<<.所以(1)(1)(0)()(0)f f f f f ξ'''>-=>,(01)ξ<<故应选择(B). (3)【答案】(A)【解析】由于利用观察法和排除法都很难对本题作出选择,必须分别验证充分条件和必要条件.充分性:因为(0)0f =,所以0000()(1sin )()(0)()()(0)lim lim lim lim (0)x x x x f x x F x F f x f x f f x x x x→→→→+--'====, 由此可得 ()F x 在0x =处可导.必要性:设()F x 在0x =处可导,则()sin f x x ⋅在0x =处可导,由可导的充要条件知0()sin ()sin lim lim x x f x x f x xx x-+→→⋅⋅=. ① 根据重要极限0sin lim1x xx→=,可得00sin sin lim lim 1x x x x x x --→→=-=-,00sin sin lim lim 1x x x xx x++→→==, ② 结合①,②,我们有(0)(0)f f =-,故(0)0f =.应选(A). (4)【答案】(C)【解析】这是讨论1n n u ∞=∑与21n n u ∞=∑敛散性的问题.11(1)ln 1nn n n u ∞∞==⎛=- ⎝∑∑是交错级数,显然ln(1+单调下降趋于零,由莱布尼兹判别法知,该级数收敛.正项级数2211ln 1n n n u ∞∞==⎛= ⎝∑∑中,2221ln 1~n u n ⎛=+= ⎝. 根据正项级数的比较判别法以及11n n ∞=∑发散,21n n u ∞=⇒∑发散.因此,应选(C).【相关知识点】正项级数的比较判别法：设1n n u ∞=∑和1n n v ∞=∑都是正项级数,且lim,nn nv A u →∞=则 ⑴ 当0A <<+∞时,1n n u ∞=∑和1n n v ∞=∑同时收敛或同时发散；⑵ 当0A =时,若1n n u ∞=∑收敛,则1n n v ∞=∑收敛；若1n n v ∞=∑发散,则1n n u ∞=∑发散；⑶ 当A =+∞时,若1n n v ∞=∑收敛,则1n n u ∞=∑收敛；若1n n u ∞=∑发散,则1n n v ∞=∑发散.(5)【答案】(C)【解析】1P 是交换单位矩阵的第一、二行所得初等矩阵,2P 是将单位矩阵的第一行加到第三行所得初等矩阵；而B 是由A 先将第一行加到第三行,然后再交换第一、二行两次初等交换得到的,因此 12PP A B =,故应选(C).三、(本题共2小题,每小题5分,满分10分.)(1)【解析】这实质上已经变成了由方程式确定的隐函数的求导与带抽象函数记号的复合函数求导相结合的问题.先由方程式2(,,)0y x e z ϕ=,其中sin y x =确定()z z x =,并求dz dx. 将方程两边对x 求导得1232cos 0y dzx e x dxϕϕϕ'''⋅+⋅+⋅=, 解得 ()12312cos y dz x e x dx ϕϕϕ''=-⋅+⋅'. ①现再将(,,)u f x y z =对x 求导,其中sin y x =,()z z x =, 可得 123cos du dzf f x f dx dx'''=+⋅+⋅. 将①式代入得()213321cos 12cos y du f f x f dx x e x ϕϕϕ'''=+⋅-⋅''⋅+⋅'. 【相关知识点】多元复合函数求导法则：如果函数(,),(,)u x y v x y ϕψ==都在点(,)x y 具有对x 及对y 的偏导数,函数(,)z f u v =在对应点(,)u v 具有连续偏导数,则复合函数((,),(,))z f x y x y ϕψ=在点(,)x y 的两个偏导数存在,且有12z z u z v u vf f x u x v x x x∂∂∂∂∂∂∂''=+=+∂∂∂∂∂∂∂； 12z z u z v u v f f y u y v y y y∂∂∂∂∂∂∂''=+=+∂∂∂∂∂∂∂. (2)【解析】方法一：用重积分的方法.将累次积分11()()xI dx f x f y dy =⎰⎰表成二重积分()()DI f x f y dxdy =⎰⎰,其中D 如右图所示.交换积分次序1()()yI dy f x f y dx =⎰⎰.由于定积分与积分变量无关,改写成10()()xI dx f y f x dy =⎰⎰.⇒ 1110002()()()()xx I dx f x f y dy dx f x f y dy =+⎰⎰⎰⎰111120()()()().dx f x f y dy f x dx f y dy A ===⎰⎰⎰⎰⇒ 212I A =. 方法二：用分部积分法.注意()1()()xdf y dy f x dx =-⎰,将累次积分I 写成()()()111111212()()()()11().22xxxx xx I f x f y dy dx f y dyd f y dy f y dy A ====-=-=⎰⎰⎰⎰⎰⎰四、(本题共2小题,每小题6分,满分12分.)(1)【解析】将曲面积分I 化为二重积分(,)xyD I f x y dxdy =⎰⎰.首先确定被积函数(,)f x y ==对锥面z =而言==. 其次确定积分区域即∑在xOy 平面的投影区域xy D (见右图),按题意：22:2xy D x y x +≤,即22(1)1x y -+≤.xyD I =⎰⎰.作极坐标变换cos ,sin x r y r θθ==,则:02cos ,22xy D r ππθθ≤≤-≤≤,因此2cos 2cos 322000213I d r rdr r d θππθπθθ-=⋅==⎰(2)【解析】这就是将()f x 作偶延拓后再作周期为4的周期延拓.于是得()f x 的傅氏系数：0(1,2,3,)n b n ==2002200222220222()cos 2(1)cos 222(1)sin sin 2244cos ((1)1)28,21,(21)1,2,3,0,2,l n n n x n a f x dx l x xdxl l n n x d x xdx n n n x n n n k k k n k ππππππππππ==-=-=-==---⎧=-⎪-==⎨⎪=⎩⎰⎰⎰⎰2222000021()(1)(1)022a f x dx x dx x ==-=-=⎰⎰.由于(延拓后)()f x 在[2,2]-分段单调、连续且(1)1f -=.于是()f x 有展开式22181(21)()cos ,[0,2](21)2n n f x x x n ππ∞=-=-∈-∑.五、(本题满分7分)【解析】设点M 的坐标为(,)x y ,则M 处的切线方程为 ()Y y y X x '-=-.令0X =,得Y y xy '=-,切线与y 轴的交点为(0,)A y xy '-.由MA OA =,有y xy '=-.化简后得伯努利方程 212,yy y x x '-=- ()221y y x x'-=-. 令2z y =,方程化为一阶线性方程 ()1z z x x '-=-.解得 ()z x c x =-,即 22y cx x =-,亦即y =又由3322y ⎛⎫⎪⎭= ⎝,得3c =,L 的方程为3)y x <<.六、(本题满分8分)【解析】在平面上LPdx Qdy +⎰与路径无关(其中,P Q 有连续偏导数),⇔P Q y x ∂∂=∂∂,即 2Q x x∂=∂. 对x 积分得 2(,)()Q x y x y ϕ=+,其中()y ϕ待定.代入另一等式得对t ∀, ()()(,1)(1,)(0,0)2(0)2,0()()22t t xydx dy xydx d x y y x y ϕϕ+=+++⎰⎰. ①下面由此等式求()y ϕ.方法一：易求得原函数()()()022222()()2(()()).yyxydx dy ydx dyd x y dd x y x s d x dy y s s y ds ϕϕϕϕ+=+=+=+++⎰⎰于是由①式得 ()()(,1)(1,)2200(0,0)(0,0)()()t t yyx y dsx y d s s sϕϕ+=+⎰⎰.即 120()()tt ds t ds s s ϕϕ+=+⎰⎰,亦即 21()ts t t ds ϕ=+⎰.求导得 )2(1t t ϕ=+,即 ()21t t ϕ=-. 因此 2(,)21Q x y x y =+-.方法二：取特殊的积分路径：对①式左端与右端积分分别取积分路径如下图所示.于是得 ()()120()1()tt dy dy y y ϕϕ+=+⎰⎰.即 12()()tt dy t dy y y ϕϕ+=+⎰⎰,亦即 21()ty t t dy ϕ=+⎰.其余与方法一相同.七、(本题满分8分)【解析】(1)反证法.假设(,)c a b ∃∈,使()0g c =.则由罗尔定理,1(,)a c ξ∃∈与2(,),c b ξ∈使12()()0g g ξξ''==；从而由罗尔定理, 12(,)(,)a b ξξξ∃∈⊂,()0g ξ''=.这与()0g x ''≠矛盾.(2)证明本题的关键问题是：“对谁使用罗尔定理？”换言之,“谁的导数等于零？”这应该从所要证明的结果来考察.由证明的结果可以看出本题即证()()()()f x g x f x g x ''''-在(,)a b 存在零点.方法一：注意到 ()()()()()()()()()f x g x f x g x f x g x f x g x '''''''-=-, 考察()()()()f x g x f x g x ''''-的原函数,令()()()()()x f x g x f x g x ϕ''=-,()x ϕ⇒在[,]a b 可导,()()0a b ϕϕ==.由罗尔定理,(,)a b ξ∃∈,使()0ϕξ'=.即有()()()()0f g f g ξξξξ''''-=,亦即()()()()f fg g ξξξξ''=''.方法二：若不能像前面那样观察到()()()()f x g x f x g x ''''-的原函数,我们也可以用积分来讨论这个问题：[]()()()()(?)()()()()?f x g x f x g x f x g x f x g x dx '''''''''-=⇔-=⎰.[]()()()()()()()()f x g x f x g x dx f x dg x g x df x ''''''-=-⎰⎰⎰()()()()()()()()f x g x g x f x dx f x g x f x g x dx ⎡⎤⎡⎤''''''=---⎣⎦⎣⎦⎰⎰()()()()f x g x f x g x ''=-(取0C =).令()()()()()x f x g x f x g x ϕ''=-,其余与方法一相同.八、(本题满分7分)【解析】设对应于231λλ==的特征向量为123(,,)T x x x ξ=,因为A 为实对称矩阵,且实对称矩阵的不同特征值所对应的特征向量相互正交,故10T ξξ=,即230x x +=.解之得 23(1,0,0),(0,1,1)T T ξξ==-. 于是有 123112233(,,)(,,)A ξξξλξλξλξ=, 所以 1112233123(,,)(,,)A λξλξλξξξξ-=1010010100101101001101101010-⎛⎫⎛⎫⎛⎫ ⎪⎪ ⎪=-=- ⎪⎪ ⎪ ⎪⎪ ⎪----⎝⎭⎝⎭⎝⎭.九、(本题满分6分)【解析】方法一：根据T AA E =有|||||()|||||||||T T A E A AA A E A A E A A A E +=+=+=+=+,移项得 (1||)||0A A E -+=. 因为0A <,故1||0A ->.所以||0A E +=.方法二：因为()T T T T A E A AA A E A E A +=+=+=+, 所以 A E A E A +=+, 即 (1||)||0A A E -+=. 因为0A <,故1||0A ->.所以||0A E +=.十、填空题(本题共2小题,每小题3分,满分6分.)(1)【解析】由题设,因为是独立重复实验,所以X 服从10,0.4n p ==的二项分布.由二项分布的数学期望和方差计算公式,有()4,()(1) 2.4E X np D X np p ===-=,根据方差性质有 22()()[()]18.4E X D X E X =+=. (2)【解析】令{0},{0}A X B Y =<=<,则{max(,)0}1{max(,)0}1{0,0}P X Y P X Y P X Y ≥=-<=-<<.由概率的广义加法公式 ()()()()P A B P A P B P AB =+-,有{max(,)0}1[1()]()()()()P X Y P AB P A B P A p B P AB ≥=--=+=+-4435.7777=+-=十一、(本题满分6分)【解析】方法1：用分布函数法先求Y 的分布函数()Y F y . 当1y ≤时, ()0;Y F y =当1y >时, (){}()X Y F y P Y y P e y =≤=≤{}ln P X y =≤ln ln 011,yy x xe dx e y--==-=-⎰所以由连续型随机变量的概率密度是分布函数的微分,得21, 1,()()0, 1.Y Y y yf y F y y ⎧>⎪'==⎨⎪≤⎩或者直接将ln 0yx e dx -⎰对y 求导数得ln ln 2011.y x y d e dx e dy y y--==⎰ 方法2：用单调函数公式直接求Y 的概率密度.由于x y e =在()0,+∞内单调,其反函数()ln x h y y ==在()1,+∞内可导且其导数为10y x y'=≠,则所求概率密度函数为 ()()()()ln 1,1,,1,0, 1.0, 1.y X Y e y h y f h y y y f y y y -⎧⎧'⋅>⋅>⎪⎪==⎨⎨≤⎪⎪⎩≤⎩21, 1,0, 1.y yy ⎧>⎪=⎨⎪≤⎩ 【相关知识点】对积分上限的函数的求导公式：若()()()()t t F t f x dx βα=⎰,()t α,()t β均一阶可导,则[][]()()()()()F t t f t t f t ββαα'''=⋅-⋅.。

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学习-----好资料1995年全国硕士研究生入学统一考试数学二试题一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上.)122?)sincos(xy??y＿＿＿＿＿＿,则(1) 设. x???y??2yx的通解为＿＿＿＿＿＿(2) 微分方程.2?x?1?t?t?2处的切线方程为＿＿＿＿＿＿曲线在. (3) ?3y?t??12n??Llim(?)?＿＿＿＿＿＿.(4) 222n?n?1n?n?2n?n?n??n22?x e?xy的渐近线方程为＿＿＿＿＿＿(5) 曲线.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一项符合题目要求,把所选项前的字母填在题后的括号内.)??(x0)(x)?f(??,??)(x)f(x)f(x)有间断点设(1) ,,为连续函数在和内有定义,,且则( )2??)]([x)](x[f (A) 必有间断点 (B) 必有间断点?(x)?)](f[x必有间断点(C) (D)必有间断点f(x)x)?xx?1)(2?yx(轴所围图形的面积可表示为与曲线( )(2)??x(x?1)(2dx??x)dxx(x?1)(2?x) (B)2?x(x?1)(2??x)dx (A) 021?x(x?1)(2?x)dx (D)0112??dx)?xx(x?xx(?1)(2?x)dx??1)(2 (C) 1020x,xx?xf(x)?f(x))f(x),??(??,都有,当则时内可导(3) 设,且对任意,在221121( )??(?x),f?0x,f0(x)?x (B) (A) 对任意对任意f(?x)?f(?x)单调增加单调增加(D) 函数(C) 函数????(0)、f(1)?f(0)f、f0)f[0,1]xf()(x?(1)f(0)?f(1)的大小则在设函数(4) 上,或更多精品文档．学习-----好资料顺序是 ( )????(0)ff(0)?f(1)?f(1)(1)?f?(0)?f(1)?f(0)f (A) (B)????(0)ff(1)?(1)?f?f(0)?f(0)(1)?f?(0)ff(1) (C) (D)f(x)F(x)?f(x)(1?|sinx|)F(x)x?0处可导可导,在若使,,(5) 设则必有 ( )?(0)?f0f(0)?0 (A) (B)??(0)?f00f(0)?f(0)?f?(0) (C) (D)三、(本题共6小题,每小题5分,满分30分.)1?cosxlim.(1) 求x(1?cosx)?0x?2yd yf(y)?exe?f1?(x)fy?y.,其中由方程(2) 设函数,具有二阶导数,且求确定2dx2x2???ln1)?f(x?x?f[ln(x)]dx)(x.求设(3) ,且,22?x1?xarctan,x?0,??(xf)x?02?)f(xx. 设试讨论在处的连续性(4)??0, x?0,?x?1?cost??2t?0?)的弧长(5) 求摆线一拱(.?y?t?sint?v?v,已知阻力与速度成正比((6) 设单位质点在水平面内作直线运动,初速度比例常00?t v0t？并求到此时刻该质点所经过的路程为多少时此质点的速度为. 数为1),问3)分(本题满分8四、2x?t?dte??t)(2xf()的最大值和最小值求函数.五、(本题满分8分)x?0?y ey?x?xxy?p()y的特设的一个解是微分方程,求此微分方程满足条件ln2x?. 解)8(六、本题满分分更多精品文档．学习-----好资料???0MT,MPy?f(x)y L分别为该曲线在点,,的方程为且又如图,设曲线32?)y(1?2??0yM(x),(?yx,y)MP其中已知线段(的长度为处的切线和法线,0000??y0??????y)(x?y),P(. ),试推导出点的坐标表达式00y L??),P(?)y(x,M00T O x??dx)(xf dt?)(fx.分8)七、(本题满分tsin?x,设计算?t?00分)八、(本题满分8)xf(??1lim?x)?0f(xf?(x). ,,且设证明x0x?更多精品文档．好资料学习-----年全国硕士研究生入学统一考试数学二试题解析1995.) 分分,满分15一、填空题(本题共5小题,每小题322sin)x?cos(1x22???2xsin(xsin)【答案】(1)2xx,满足复合函数求导法则【解析】该函数是由两个复合函数的乘积构成,?11???2222???)cos(?cos(xy)sinxsin?????xx??1111222sinx?sin(x)?2x?)?2sin?cos?(1)???cos(2xxxx22sin)cos(x?1x22?)?sin(xsin??2x .2xx?????)x(f(x(f(x))y))?fy?(. 的导数为【相关知识点】复合函数求导法则：x2x?cosx?csiny?c 【答案】(2)212????01?r?0??y??2xyy?y, 对应的齐次方程【解析】微分方程的特征方程为i??rxsincosCx?C.故对应齐次方程的通解为特征根为,1,221?????x2y?y??0YY?ax?b??aY得则,,代入微分方程设非齐次方程的特解,,x?2?b?0?ax,0,b???2,a x?2Y?故所以通解为比较系数得.2xsinx??Ccosx?Cy.21*)xy(【相关知识点】1.二阶线性非齐次方程解的结构：设是二阶线性非齐次方程????Q(x)y?f((x)yx)Y(xy)?P是与之对应的齐次方程的一个特解.*???(x))?yy?Y(x0yy?P?(x)y(?Qx)是非齐次方程的通解.则的通解,2. 二阶常系数线性齐次方程通解的求解方法：对于求解二阶常系数线性齐次方程的通解????Q(x)y?0P(P(x)yx)Q(x)Y(xy)?均是常数中的,可用特征方程法求解：即、,方程2???0?pr?qr?r,r0y??py?qy变为.其特征方程写为,在复数域内解出两个特征根；21分三种情况：rxrx;e?eCy?Crr,两个不相等的实数根(1) ,则通解为212121更多精品文档．好资料学习-----??rx;C??Cxey rr?则通解为两个相等的实数根,(2)???x????.x?yex?CCcossin i?r?CC,其中,一对共轭复根则通解为(3) 1,22121.12121为常数*???)(xy)x)yx?Q(x)y?fy(P?(可用待定3.对于求解二阶线性非齐次方程,的一个特解 ,有结论如下：系数法??xkx*e)Q(xP(x)e,y(x)?xf(x)?如果则二阶常系数线性非齐次方程具有形如mm?)P(xQ(x)k不是特征方程的根、是特征方而按相同次数的多项式是与,其中的特解,mm2.或0、1程的单根或是特征方程的重根依次取?x??])sin(x)cosxx?P(x?f(x)e[P则二阶常系数非齐次线性微分方程如果,nl???)(xy?q(x)yy?f?p(x)的特解可设为?xk(1)(2)*??]cos?xR(y?xxe)sinx[R(x), mm??(2)(1)????nl,m?max)xR(xR()m ki?i?不是特征)与而(按其中是或次多项式,,mm01. 或方程的根、或是特征方程的单根依次取为0?3x?7?y【答案】(3) 【解析】切线的斜率为dy23dytdt3????t. dx2t2dx2?tt?22t?dt2?t0x?7??8?3(x5)y?3yx?5,y?8?2t?..,时故所求切线方程为当. 化简得?)?t(x???)t(dy?.,【相关知识点】参数方程所确定函数的微分法：如果则????)(dxt)(ty??1(4)【答案】2 .令【解析】应用夹逼准则求数列的极限n12a????n222nnn?2n??n?nn?1?n12????a则n222nn??nnn??nnn??n更多精品文档．好资料学习-----11)?n(nn?1?2?2??22n?22nnn?1n?1.??2n?211)n?n(1nn21?2??12????a???, 又n222222nnn?nnn?n?n?nn?11n1??a??,即n22n?211?111n??lim?lim??lima.所以n2n2?222??n????nn1?lima ,由夹逼准则.得即n2??n12n1??lim(??). 2222nn?n??nn??1n2n???n0y?【答案】(5)2x?2e?yx的定义域为全体实数,且【解析】函数2x2?0e?limy?limx,????xx0y?.所以曲线只有一条水平渐近线xx???(x)limf)y?f(x【相关知识点】铅直渐近线：如函数在其间断点则,处有0x?x0x?x是函数的一条铅直渐近线；0a?x)limf(a ay?.为函数的水平渐近线水平渐近线：当),(则为常数??x.)每小题3分,满分15分本题共二、选择题(5小题,(D)【答案】(1)即有限多个在同一点处连续的函数之乘【解析】方法一：反证法,利用连续函数的性质,.积,仍然在该点处连续??)xx)((??)x)?f(x()x(f则,是连续函数无间断点必无间断点设函数,这与,因为)xf()(fx?)(x(D).有间断点矛盾故应选择,.举出反例排除方法二：排除法,?1,x?0,????xf()1,()x设?0,?x1,?更多精品文档．学习-----好资料2???1)]1,[?(f[x(x)]?[f(x)]?1,(D). (A),(B),(C).故应选择则都处处连续,排除(C)(2)【答案】方法一：利用定积分的求面积公式有【解析】22??dx)(2??xx(x?1)x(x?1)(2?x)dx0021??dx)??1)(2dx?xx(??xx(x?1)(2?x)10(C).应选择)x?1)(2?y?x(x方法二：画出曲线即,所求面积为图中两面积之和,的草图21??dxx)(x?1)(2?x(x?1)(2x)dx???x,10(C). 故应选(D)【答案】(3))x(??x)?fxx?x?x??xf(x, ,即则函数时,当,【解析】因为对任意,22221111)x(?)??f?f(?x)x(??f(D).应选择,故是单调增加的.213x?(fx))x)?f(,xx?xf(xx, 都有(A)(B)(C)可令,则对任意当时,,对于2121212?03xf?(0)?,但0?x2?0??x)(?x)?f3(,3x)??f(?x.在其定义域内单调减少,(A)(B)(C). 故排除(B)【答案】(4)???[0,1])(xx)?f0f(【解析】由故上为严格的单调递增函数在区间可知,???1)??x(0)?,(0(1)?f)(x?ff???1)(?),(0?f(1)f(0)?f?由微分中值定理,.所以?????1)(0?(f)??f(0)(1)?f(1)?f?f(0) ,(B). 应选择(A)【答案】(5)??)()(xffx?xx)f(x. 存在且相等在【解析】函数处可导的充分必要条件是与00?0?)x((x)Ffxx)?(x)f(x?f()|sin|F0?x处可导等价于在可导由于,所以,而|x|(x)sinf0x?.可导在更多精品文档．学习-----好资料?(x)?f(x)|sinx|,则令f(x)|sinx|f(x)sinx???(0)?lim?lim?f(0),???xx??0?xx?0?f(x)|sinx|f(x)sinx????(0)?lim???limf(0),??xx???0x?0?x F(x)?f(0)?f(0)f(0)?00x?.故选择(A). 处可导,当且仅当在,即于是要使三、(本题共6小题,每小题5分,满分30分.)x?0sinxx.时,【解析】利用等价无穷小计算,即当(1)2x??x2?22sin?? 11x111?cos2??2limlim???lim??原式. ??2222x?cosx1???xcosx1???0xx?0?0?x x2sin2x?x2??22??(2)【解析】这是一个由复合函数和隐函数所确定的函数.x求导,得方法一：将方程两边对f(y)f(y)y???yy???efe(?xey)?,f(y)e??y, 即f(yy)?(y)?xfee1yy)f(??yexe?.将得代入并化简,?(y?f))x(1x求导,两边再对得 ??????????)y(y)y))?x(??0fx(1?fy())??(1?f(????y22??????))(fy?x(1f(y))x(1????(yf)1y???.22???))x((1?fy?))(x(1?fy1??y代入并化简得将?x(1?f(y))??(fy)1???y??.32???))yfx(1?(2?))f?yx((1x求导.方法二：方程两边先取对数再对lnx?f(y)?y,方程两边取对数得更多精品文档．-----好资料学习1???y)?y?f?(y, 求导得x1???y1f?所以,因为.?))x(1?fy(.以下同方法一?????(x)x?))f(f(y?y(f(x)). 的导数为【相关知识点】复合函数求导法则：?(x)的表达式.由(3)【解析】首先应求出22?1?x1x2lnln??1)?f(x,2211?2x?x?t?12,?tx?1ln?f(t).又得令t?1?(x)?1??lnx)]?f[ln(x,?(x)?1?x?11)?(x?x?(x)?..解得则因此?(x)?1x?1x?12????(1?)dx?x?2lndx?x?1(x)dx??C. x?1x?1??(x)(x)x?xff)(fx存在且与处的导函数连续的充分必要条件是在(4)【解析】函数0??00?)x(f.必与相等02x12?(x)?arctan?f0x?,由于时,当????x12???(x)?lim?limfarctanlimf?(x)?limf(x)??0?, ??42x1?x242x1?x220x?x?0?x?0??0x???1x)f(0)f(f(x)???lim?f(0)?limlimarctan?,2x?0xx20x?0xx?0????(0)f)limf?(xlimf(x)?.所以x??x0?0??(x)f x?0处连续在故.(5)【解析】由弧微分公式得22????22??dt)cos2(1)cos(1sin)x?ds(t?)y(tdt?t??tdt??t,所以更多精品文档．好资料学习-----ttt????22222?????dttsindts?)2(12?2sindt?2?sindt?2cos2220000?2t??8.1)?4cos??4(?1????? 2??0)tv(t)?v((6)【解析】设质点的运动速度为,按牛顿第二定律有阻力为,,由题设)tdv()t??vm(,dt)tdv()??v(t1?m. 即其中质量,dt t?Cev(t)?. 解之得这是简单变量可分离的微分方程,t?ev(t)?vv?v(0).另有初始条件得,00v v t?00e?vln3t?. ,得时,有当此质点的速度为033到此时刻该质点所经过的路程为21??ln3ln3t??t???v1evdt?s??ve???v?. ????0000330??0) 四、(本题满分8分2xt???0x)f?(dte?t)f(x?)(2 ,【解析】对函数两边求导并令得02x?2?0?)e2x(2?fx(x)?,2?x?0,x?.解得驻点??)2,???x??f(fx(x)?0,,严格单调增??)ff(x)?0,?2?x?0,(x?,严格单调减由于??)x?0,0x?2,(ff(x)?,严格单调增???)?2?x??,f)(x?0,xf(,严格单调减?2)(?2),f(f)f(0)(xf(x)f为函数的极小值点,的极大值点所以,为函数且2222t?t?t????e?dt?(2?t)et)edt??e?1?f(2)?(2?,0000t??0edt?f(0)?(2?t),0??????t?t?t???1dt)?)(?)limfx)?(2?tedt?(2?te??exflim(,又000??????xx2?e1(f?2)??)f?f(0)xf()0(x.为函数最大值所以的最小值,为函数更多精品文档．好资料学习-----【相关知识点】积分上限函数的求导公式：d???x?????????????????????x?xfffxtxdt?. dx???x分)五、(本题满分8xx?ey?y?e【解析】把代入所给的一阶线性微分方程,得和xx x??p(x)xee, x?x(x)?xe?p.解得x??x?(xe?x)xyy?线性方程被确定为即,x??1??e1)yy?(.通解为这是一阶线性非齐次微分方程,??x??x??dxe?1)dx?(e?1)(?C?ey?edx????x???e???e?????xx?x?x?x??ee?xe??xx?e??ex???eCdx?e?dx?e?C?edx?Ce??????x e????xx???x x?xx?eee?Ce(e??ee?C)?.1?2?ln2e?lnln20?e?Ce e?C?0y?2.得,再由即ln2x?1x??xe?x ey??e2故所求的特解为.?)x)y?q(yp?(x:的通解公式为【相关知识点】一阶线性非齐次微分方程??dx)(x)?p(xdxp?)dx?eC()q(xe?y C.为常数,其中??232???本题满分8分)六、(?????yy,y,x,,,P要用表示出,也就是说【解析】要求点. ,的坐标0000y?y1MP?有,由00??32?y?1022???(?x?(?y)) , ①002??y0知又由法线的斜率与切线斜率互为负倒数的关系,??x?0y??, ②0??y0更多精品文档．学习-----好资料?????x)(?(??y)y,即把②式,得到代入①消去0002222?????(1y?y)(y)?/③, 0002?1?y????0y?0y??y?,所以③可化为由 ,容易看出,,知曲线是向上凹的00??y02??)y(1?y???00?x??y?y)??(, 且000??y0?y?2??0?x?(1?y),?00??y?0于是得?1?2??).(1?y?y?00???y?0七、(本题满分8分)【解析】方法一：这是一个积分上限函数求定积分,可以考虑用定积分的分部积分法. sinx??xf)(, 由于??xsint0?dt?0f(0)?,有因而由分部积分法和?t?0???????????dxx)())?f?(x)d(x?)?f(x)(x?xf(x)dx?f(0??????2?cos?dx(?x)?xsinxdx?.000sinx????0?x00sint????x???dtdx?dxf(x),可以通过变换积分次序来求解方法二：对于二重积分. ????t??000sintsint????x?????dtdx?dx?dtf(x)dx,????t??t??000D其中???x?t0?x??,0D?(x,t)????.x,t?0(x,t)?t???于是tsinsint??????????????2?sindx?dtdx?tdtdtdx?)(fx.?????t?t??t0000t更多精品文档．学习-----好资料八、(本题满分8分)f(x)?lim1f(0)?0,,所以必有且【解析】由于x0x?f(x)?f(0)f(x)?(0)?lim?flim?1. x?0x0?0x?x证法一：用函数单调性证明不等式.?(x)?f(x)?x,令?????(0)?f(x)(x)?1?(x)?ff. 则???(x)?f0f(x)单调增加由于, ,所以函数????(0)?0,x?(x)?f(x)?f0,??????f(x)?(0)(x)??0,x?0,f????(x))(x0?0x?x的极小值点也是最小值点是由负变正,在所以,??(0)?f(0)?0?f(x)?x?0(x)?,f(x)?x. 即证法二：用泰勒公式.1122???????x(f)()xx??f(fx)?(0)?f(0)x?f.2!212?????0?)f(x0)x?f(. 所以,因为212????(fxxf?x?x)(). 所以2更多精品文档．。

1995年全国硕士研究生考试英语试卷及答案PartⅠSection A:1. Between 1897 and 1919 at least 29 motion pictures in which artificial beings were portrayed_____.A. had producedB. have been producedC. would have producedD. had been produced2. There ought to be less anxiety over the perceived risk of getting cancer than ___ in the public mind todayA. existsB. existC. existingD. existed3. The professor can hardly find sufficient grounds _____ his argument in favor of the new theory.A. which to base onB. on which to baseC. to base on whichD. which to be based on4. ________ can help but be fascinated by the world into which he is taken by the science fiction.A. EverybodyB. AnybodyC. SomebodyD. Nobody5. How many of us ___, say, a meeting that is irrelevant to us would be interested in the discussion?A. attendedB. AttendingC. to attendD. have attended6. Hydrogen is the fundamental element of the universe ____ it provides he building blocs from which the other elements are produced.A. so thatB. but thatC. in thatD. provided that7. We are taught that a business letter should be written in a formal style ____ in a personal one.A. rather thanB. Other thanC. better thanD. less than8. ______ is generally accepted, economical growth is determined by the smooth development of production.A. WhatB. ThatC. ItD. As9. It is believed that today's pop music can serve as a creative force ____ stimulating the thinking of its listeners.A. byB. withC. atD. on10. Just as the soil is a part of the earth, _____ the atmosphere.A. as it isB. the same asC. so isD. and so isSection B(改错):ll . The conveniences that Americans desire reflecting not so much a leisurely lifestyle as a busy lifestyle in which even minutes of time are too valuable to be wasted.A BC D12. In debating one must conect the opponent's facts, deny the relevance of his proof, or deny that what he presents as proof, unless relevant , is sufficient.A B C D13 . We are not conscious of the extent of which provides the psychological satisfaction that canmake the difference between a full and an empty life.A B C D14. The Portuguese give a great deal of credit to one man for having promoted sea travel, that manwas Prince Henry the navigator, who lived in the 15th century.A B C D15 . Accounts of scientific experiments are generally correct for those write about science arecareful in checking the accuracy of their reports.A B C D16. whenever we hear of a natural disaster, even in a distant part of the world, we feel sympathyfor the people to have affected.A B CD17 . It is perhaps not an exaggeration to say that we shall soon be trusting our health, wealth andhappiness to elements with whom very names the general public are unfamiliar.A B C D18. The speaker claimed that no other modern nation devotes so small a portion of its wealth topublic assistance and health than the United States does.A B C D19 . There are those who consider it questionable that these defence-linked research projects willaccount for an improvement in the standard of living or, alternately, to do much to protect our diminishing resources.A B C D20. If individuals are awakend each time as they begin a dream phase of sleep, they are likely tobecome irritable even though their total amount of sleep has been sufficient.A B C DSection C:21. In that country, guests tend to feel they are not highly ___ if the invitation to a dinner party is extended only three or four days before the party date.A. admiredB. regardedC. expectedD. worshipped22. A _____ of the long report by the budget committed was submitted to the mayor for approval.A. shorthandB. schemeC. scheduleD. sketch23. A man has to make ____ for his old age by putting aside enough money to live on when old.A. supplyB. assuranceC. provisionD. adjustment24. The newly-built Science Building seems _____ enough to last a hundred years.A. spaciousB. sophisticatedC. substantialD. steady25. It is well-known that the retired workers in our country are ___ free medical care.A. entitled toB. involved inC. associated withD. assigned to26. The farmers were more anxious for rain than the people in the city because they had more at____.A. dangerB. stakeC. lossD. threat27. I felt ____ to death because I could make nothing of the chairman's speech.A. fatiguedB. tiredC. exhaustedD. bored28. When the engine would nto start, the mechanic inspected all the parts to find what was at ___.A. wrongB. troubleC. faultD. difficulty29. Your advice would be ____ valuable to him, who is at present at his wit's end.A. exceedinglyB. excessivelyC. extensivelyD. exclusively30. He failed to carry out some of the provisions of the contract, and now he has to _____ theconsequences.A. answer forB. run intoC. abide byD. step into31. The river is already _____ its bans because of excessive rainfall; and the city is threatenedwith a likely flood.A. parallel toB. level inC. flat onD. flush with32. People _____ that vertical flight transports would carry millions of passengers as do theairliners of today.A. convincedB. anticipatedC. resolvedD. assured33. In spite of the wide range of reading material specially written or _____ for language learningpurposes, there is yet no comprehensive systematic programmed for the reading skills.A. adaptedB. acknowledgedC. assembledD. appointed34. The mother said she would ____ her son washing the dished If he could finish his assignmentbefore supper.A. let downB. let aloneC. let offD. let out35. We should always keep in mind that _____ decisions often lead to bitter regrets.A. urgentB. hastyC. instantD. prompt36. John complained to the bookseller that there were several pages ____ in the dictionary.A. missingB. losing C dropping D. leaking37. In the past, most foresters have been men, but today, the number of women ____ this field isclimbing.A. engagingB. devotingC. registeringD. pursuing38. The supervisor didn't have time so far to go into it _____, but he gave us an idea about hisplan.A. at handB. in turnC. in conclusionD. at length39. Their demand for a pay raise has not the slightest ____ of being met.A. prospectB. predictionC. prosperityD. permission40. It's usually the case that people seldom behave in a _____ way when in a furious state.A. stableB. rationalC. legalD. crediblePart Two:Sleep is divided into periods of so-called REM sleep, characterized by rapid eye movements and dreaming, and longer periods of non-REM sleep. 41 kind of sleep is at all well-understood , but REM sleep is 42 to serve some restorative function of the brain. The purpose of non-REM sleep is even more 43 .The new experiments, such as these 44 for the first time at a recent meeting of the Society for Sleep Research in Minneapolis, suggest fascinating explanations 45 of non-REM sleep .For example, it has long been known that total sleep 46 is 1OO percent fatal to rats, yet ,47 exanlination of the dead bodies , the animals look completely normal . A researcher has now 48 the mystery of why the aninlals die. The rats 49 bacterial infections of the blood ,50 their immune systems——the self-protecting mechanism against disease——had crashed.41 . (A)Either (B)Ndther (C)Each (D)Any42 . (A) intended ( B)required (C) assumed (D) inferred43 . (A) subtle (B)obvious (C)mysterious (D)doubtful44 . (A) maintained ( B) described (C)settled (D)afforded45. (A)in the light (B)by virtue(C)with the exception (D)for the purpose46 . (A) reduction ( B) destruction (C) deprivation (D) restriction47. (A)upon (B)by (C)through (D)with48. (A)paid attention to (B)caught sight of(C)laid emphasis on (D)cast light on49 . (A) develop (B)produce (C)stimulate (D)induce50. (A)if (B)as if (C)only if (D)if onlyPart ⅢReading ComprehensinnPassage lMoney spent on advertising is money spent as well as any I know of. It serves directly to assist a rapid distribotion of goods at reasonable price, thereby establishing a firm home market and so making it possible to provide for export at competitive prices. By drawing attention to new ideas it helps enormously to raise standards of living. By helping to increase demand it ensures an increased need for labour, and is therefore an effective way to fight unemployment. It lowers the costs of many services: without advertisements your daily newspaper would cost four times as much, the price of your television licence would need to be doubled, and travel by bus or tube would cost 20 per cent more.And perhaps most important of all, advertising provides a guarantee of reasonable value in the products and services you buy. Apart from the fact that twenty-seven acts of Parliament govern the terms of advertising, no regular advertiser dare promote a product that fails to live up to the promise of his advertisements. He might fool some people for a little while through misleading advertising. He will not do so for long, for mercifully the public has the good sense not to buy the inferior article more than once. If you see an article consistently advertised, it is the surest proof I know that the article does what is claimed for it , and that it represents good value.Advertising does more for the material benefit of the community than any other force I can think of.There is one more point I feel I ought to touch on. Recently I heard a well-known television personality declare that he was against advertising because it persuades rather than informs. He was drawing excessively fine distinctions. Of course advertising seeks to persuade.If its message were confined merely to information-and that in itself would be difficult if not impossible to achieve, for even a detail such as the choice of the colour of a shirt is subtly persuasive——advertising would be so boring that no one would pay any attention. But perhaps that is what the well-known television personality wants.51 . By the first sentence of the passage the author means that__.(A) he is fairly familiar with the cost of advertising(B) everybody knows well that advertising is money consuming(C) advertising costs money like everything else(D) it is worthwhile to spend money on advertising52. In the passage, which of the following is NOT included in the advantages of advertising?(A) Securing greater fame.(B) Providing more jobs.(C) Enhancing living standards.(D) Reducing newspaper cost.53 . The author deems that the well-known TV personality is_.(A) very precise in passing his judgement on advertising(B) interested in nothing but the buyers' attention(C) correct in telling the difference between persuasion and information(D) obviously partial in his views on advertising54. In the author's opinton,__.(A) advertising can seldom bring material benefit to man by providing(B) advertising informs people of new ideas rather than wins them over(C) there is nothing wrong with advertising in persuading the buyer(D) the buyer is not interested in getting information from an advenisementPassage 2There are two basic ways to see growth: one as a product, the other as a process. People have generally viewed personal growth as an external result or product that can easily be identified and measured. The worker who gets a promotion, the student whose grades improve, the foreigner who learns a new language-all these are examples of people who have measurable results to show for their efforts.By contrast, the process of personal growth is much more difficult to determine, since by definition it is a journey and not the specific signposts or landmarks along the way.The process is not the road itsetf, but rather the attitudes and feellings people have, their caution or courge, as they encounter new experiences and unexpected obstacles. In this process ,the journey never really ends; there are always new ways to experience the world, new ideas to try, new challenges to accept .In order to grow, to travel new roads, people need to have a willingness to take risks, to confront the unknown, and to accept the possibility that they may "fail"at first. How we see our-selves as we try a new way of being is essential to our abitity to grow. Do we perceive ourselves as quick and curious? If so, then we tend to take more chances and to be more open to unfamiliar experiences. Do we think we're shy and indecisive? Then our sense of timidity can cause us to hesitate, to move slowly, and not to take a step until we know the ground is safe. Do we thiQk we're slow to adapt to change or that we' re not smart enough to cope with a new challenge? Then we are likely to take a more passive role or not try at all.These feelings of insecurity and self-doubt are both unavoidable and necessary if we are to change and grow. If we do not confront and overcome these internal fears and doubts, if we protect ourselves too much, then we cease to grow. We become trapped inside a shell of our own making .55 . A person is generally believed to achieve personal growth then__.(A) he has given up his smoking habit(B) he has made great efforts in his work(C) he is keen on leaming anything new(D) he has tried to determine where he is on his journey56. In the author' s eyes, one who views personal growth as a process would__.(A) succeed in climbing up the social ladder(B) judge his ability to glow from his own achievements(C) face difficulties and take up challenges(D) aim high and reach his goal each time57. When the author says "a new way of being" (line 3, para. 3) he is referring to__.(A) a new approach to experiencing the world(B) a new way of taking risks(C) a new method of perceiving ourselves(D) a new system of adaptation to change58. For personal growth ,the author advocates all of the following except_.(A) curiosity about more chances(B) promptness in self-adaptation(C) open-mindedness to new experiences(D) avoidance of intemal fears and doubtsPassage 3In such a changing , complex society formerly simple solutions to informational needs become complicated. Many of life' s problems which were solved by asking family members, friends or colleagues are beyond the capability of the extended family to resolve. Where to turn for expert information and how to determine which expert advice to accept are qaestions facing many people today.In addition to this, there is the growing mobility of people since World War Ⅱ. As families move away from their stable community, their friends of many years, their extended family relationships, the informal flow of information is cut off, and with it the confidence that information will be available when needed and will be trustworthy and reliable. The almost unconscious flow of information about the simplest aspects of living can be cut off. Thus, things once learned subconsciously through the casual communications of the extended family must be consciously learned .Adding to societal changes today is an enormous stockpile of information. The individual now has more information available than any generation, and the task of finding that one piece of information relevant to his or her specific problem is complicated , time-consuming and sometimes even overwhelming .Coupled with the growing quantity of information is the development of technologies which enable the storage and delivery of more information with greater speed to more locations than has ever been possible before. Computer technology makes it possible to store vast amounts of data in machine-readable files, and to program computers to locate specific information . Telecommunications developments enable the sending of messages via television, radio, and very shortly, electronic mail to bombard people with multitudes of messages. Satellites have extended the power of communications to report events at the instant of occurrence. Expertise can be shared world wide through teleconferencing , and problems in dispute can be settled without the parttcipants leaving their homes and/or jobs to travel to a distant conference site. Technology has facilitated the sharing of information and the storage and delivery of information, thus making more information available to more people.In this world of change and complexity , the need for infomtatian is of greatest importance.Those people who have accurate , reliable up-to-date information to solve the day-to-day problems,the critical problems of their business, social and family life, will survive and succeed. "Knowledge is power" may well be the truest saying and access to information may be the most critical requirement of all people.59. The word "it" (line 3, para. 2) most probably refers to__.(A) the lack of stable communities(B) the breakdown of informal information channels(C) the increased mobility of families(D) the growing number of people moving from place to place60. The main problem people may encounter today arises form the fact that__.(A) they have to learn new things consciously(B) they lack the confidence of securing reliable and trustworthy information(C) they have difficulty obtaining the needed informatton readily(D) they can hardly carry out casual communications with an extended family.61 . From the passage we can infer that__.(A) electronic mail will soon play a dominant role in transmitting messages(B) it will become more difficult for people to keep secrets in an information era(C) people will spend less time holding meetings or conferences(D) events will be reported on the spot mainly through satellites62. We can learn from the last paragraph that __.(A) it is necessary to obtain as much(B) people should make the best use of the information(C) we shoutd realize the importance of accumulating information .(D) it is of vital importance to acquire needed information efficientlyPassage 4Personality is to a large extent inherent——A-type parents usually bring about A-type offspring. But the environment must also have a profound effect, since if competition is important to the parents, it is likely to become a major factor in the lives of their children.One place where children soak up A-characteristics is school , which is, by its very nature, a highly competitive institution. Too many schools adopt the 'win at all costs' moral standard and measure their success by sporting achievements. The current passion for making children compete against their classmates or against the clock produces a two-layer system , in which competitive Atypes seem in some way better than their B-type fellows. Being too keen to win can have dangerous consequences: remember that Pheidippides , the first marathon runner , dropped dead seconds after saying: ' Rejoice, we conquer! 'By far the worst form of competition in schools is the disproportionate emphasis on examinations. It is a rare school that allows pupils to concentrate on those things they do well. The merits of competition by examination are somewhat questionable , but competition in the certain knowledge of failure is positively harmful.Obviously, it is neither practical nor desirable that all A-youngsters change into B' s. The world needs A types, and schools have an important duty to try to fit a child' s personality to his possible future employment . It is top management .If the preoccupation of schools with academic work was lessened, more time might be spent teaching children surer values. Perhaps selection for the caring professions , especially medicine,could be made less by good grades in chemistry and more by such considerations as sensitivity and sympathy. It is surely a mistake to choose our doctors exclusively from A-type stock. B's are important and should be encouraged.63 . According to the passage , A-type individuals are usually__.(A) impatient (B) considerate (C) aggressive (D) agreeable64. The author is strongly opposed to the practice of examinations at schoois because__.(A) the pressure is too great on the students(B) some students are bound to fail(C) failure rates are too high(D) the results of exarninations are doubtful65 . The selection of medical professionals are currentiy based on__.(A) candidates' sensitivity(B) academic acbievements(C) competitive spirit(D) surer values66. From the passage we can draw the oonclusion that__.(A) the personality of a child is well established at birth(B) family innuence dominates the shaping of one' s characteristics(C) the development of one' s personality is due to multiple factors(D) B-type characteristics can find no place in competitive societyPassage 5That experiences influence subsequent behaviour is evidence of an obvious but nevertheless remarkable activity called remembering. Learning could not occur without the function popularly named memory.Constant practice has such as effect on memory as to lead to skillful performance on the piano, to recitation of a poem, and even to reading and understanding these words. So-called intelligent behaviour demands memory , remembering being a primary requirement for reasoning. The ability to solve any problem or even to recognize that a problem exists depends on memory. Typically, the decision to cross a street is based on remembering many earlier experiences .Practice (or review) tends to build and maintain memory for a task or for any learned material. Over a period of no practice what has been learned tends to be forgotten; and the adaptive consquences may not seem obvious. Yet, dramatic instances of sudden forgetting can seem to be adaptive. In this sense, the ability to forget can be intffpreted to have survived through a process of natural selection in animals.Inded, when one's memory of an emotionally painful experience lead to serious anxiety, forgetting may produoe relief. Nevertheless, an evolutionary interpretation might make it difficult to understand how the commonly gradual process of forgetting survived natural selection.In thinking about the evolution of memory together with all its possible aspects,it is helpful to consider what would happen if memories failed to fade. Forgetting clearly aids orientation in time, since old memories weaken and the new tend to stand out,providing clues for inferring duration. Without fotgetting, adaptive ability would suffer, for example ,learned behaviour that might have been correct a decade ago may no longer be. Cases are recorded of people who (by or-dinary standards) forgot so little that their everyday activities were full of confusion. This forgetting seems to serve that survival of the individual and the species.Another line of thought assumes a memory storage system of limited capacity that provides adaptive flexibility specifically through forgetting. In this view, continual adjustments are made between learning or memory storage ( input) and forgetting (output) . Indeed, there is evidence that the rate at which individuals forget is directly related to how much they have learned. Such data offers gross support of contemporary models of memory that assume an input-output balance.67. From the evolutionary point of view,__.(A) forgetting for lack of practice tends to be obviously inadaptive .(B) if a person gets very forgetful all of a sudden he must be very adaptive(C) the gradual process of forgetting is an indication of an individual' s adaptability(D) sudden forgetting may bring about adaptive consequences68. According to the passage, if a person never forgot ,__.(A) he would survive best(B) he would have a lot of trouble(C) his ability to learn would be enhanced(D) the evolution of memory would stop69. From the last paragraph we know that__.(A) forgetfulness is a response to learning(B) the memory storage system is an exactly balanced input-output systenl(C) memory is a compensation for forgetting(D) the capacity of a memory storage system is limited because forgetting occurs70. In this article, the author tries to interpret the function of__.(A) remembering (B) forgetting (C) adapting (D) experiencingPart ⅣEnglish-Chinese TranslationThe standardized educational or psychological test that are widely used to aid in selecting, classifying, assigning, or promoting students, employees, and military personnel have been the target of recent attacks in books, magazines, the daily press, and even in congress. 71 )The target is wrong, for in attacking the tests, critics divert attention form the fault that lies with ill-informed or incompetent users. The tests themselves are merely tools , with characteristics that can be measured with reasonable precision under specified conditions. Whether the results will be valuable , meaningless, or even misleading depends partly upon the tool itself but largely upon the user .All informed predictions of future performance are based upon some knowledge of relevant past performance: school grades, research productivity, sales records, or whatever is appropriate.72 )How well the predictions will be validated by later performance depends upon the amount , reliability , and appropriateness of the information used and on the skill and wisdom with which it is interpreted. Anyone who keeps careful score knows that the information available is always incomplete and that the predictions are always subject to error.Standardized tests should be considered in this context. They provide a quick, objective method of getting some kinds of information about what a person learned , the sktlls he has developed, or the kind of person he is. The information so obtained has, qualitatively, the same advantages and shortcomings as other kinds of information. 73)Whether to use tests. other kinds of information, or both in a particular situation depends, therefore, upon the evidence from experience concerning comparative validity and upon such factors as cost and availability.74)In general,the tests work most effectivelv when the qualities to be measured can be most precisely defined and least effectively when what is to be messured or predicted cannot be well defined. Properly used, they provide a rapid means of getting comparable information about many people Sometimes they identify students whose high potential has not been previously recognized, but there are many things they do not do. 75)For example, they do not compensate for gross social inequality, and thus do not tell how able an underprivileged youngster might have been had he grown up under more favorable circumstances.Part ⅤWrlting (15 points)DIRECTIONS :A. Title: THE "PROJECT HOPE"B. Time limit : 40 minutesC. Word limit : 120 - 150 words (not including the given opening sentence)D. Your composition should be based on the OUTLINE below and should start with thegiven opening sentence : "Education plays a very important role in the modernization ofour country " .E. Your composition must be written neatly on the ANSWER SHEET.OUTLEVE:1. Present sluation2. Necessity of the project3. My suggestion参考答案：1. D2. A3. B4. D5. B6. C7. A8. D9. A 10.C11. A, renect 12. D, if13. B, to which 14. D, being15. C, writing about 或who write about 16. D, affected17. C, whose 18. C, as19. C, do much 20. B, each time21. B 22. D 23. C 24. C 25. A 26. B 27. D 28. C 29. A 30. A31. D 32. B 33. A 34. C 35. B 36. A 37. D 38. D 39. A 40. B41. B 42. C 43. C 44. B 45. D 46. C 47. A 48. D 49. A 50. B51. D 52. A 53. D 54. C 55. A 56. C 57. A 58. D 59. B 60. C61. A 62. D 63. C 64. B 65. B 66. C 67. D 68. B 69. A 70. B71．把标准化测试作为抨击目标是错误的，因为在抨击这类测试时，批评者不考虑其弊病来自人们对测试不甚了解或使用不当。

1995年全国硕士研究生入学统一考试英语试题Section ⅠUse of EnglishSleep is divided into periods of so-called REM sleep, characterized by rapid eye movements and dreaming, and longer periods of non-REM sleep. 1 kind of sleep is at all well-understood, but REM sleep is 2 to serve some restorative function of the brain. The purpose of non-REM sleep is even more 3 . The new experiments, such as these 4 for the first time at a recent meeting of the Society for Sleep Research in Minneapolis, suggest fascinating explanations 5 of non-REM sleep.For example, it has long been known that total sleep 6 is 100 percent fatal to rats, yet, 7 _examinations of the dead bodies, the animals look completely normal. A researcher has now8 the mystery of why the animals die. The rats 9 bacterial infections of the blood, 10 their immune systems—the self-protecting mechanisrn against disease—had crashed.1. [A] Either [B] Neither [C] Each [D] Any2. [A] intended [B] required [C] assumed [D] inferred3. [A] subtle [B] obvious [C] mysterious [D] doubtful4. [A] maintained [B] described [C] settled [D] afforded5. [A] in the light [B] by virtue [C] with the exception [D] for the purpose6. [A] reduction [B] destruction [C] deprivation [D] restriction7. [A] upon [B] by [C] through [D] with8. [A] paid attention to [B] caught sight of [C] laid emphasis on [D] cast light on9. [A] develop [B] produce [C] stimulate [D] induce10. [A] if [B] as if [C] only if [D] if only一、文章结构总体分析睡眠分为浅睡阶段和较长时间的深睡阶段。

1995年全国硕士研究生入学统一考试数学二试题一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上.) (1) 设221cos()siny x x=,则y '=＿＿＿＿＿＿. (2) 微分方程2y y x ''+=-的通解为＿＿＿＿＿＿.(3) 曲线231x ty t⎧=+⎪⎨=⎪⎩在2t =处的切线方程为＿＿＿＿＿＿.(4) 22212lim()12n nn n n n n n n→∞+++=++++++＿＿＿＿＿＿. (5) 曲线22x y x e -=的渐近线方程为＿＿＿＿＿＿.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一项符合题目要求,把所选项前的字母填在题后的括号内.)(1) 设()f x 和()x ϕ在(,)-∞+∞内有定义,()f x 为连续函数,且()0f x ≠,()x ϕ有间断点,则 ( ) (A) [()]f x ϕ必有间断点 (B) 2[()]x ϕ必有间断点 (C) [()]f x ϕ必有间断点 (D)()()x f x ϕ必有间断点 (2) 曲线(1)(2)y x x x =--与x 轴所围图形的面积可表示为 ( )(A) 2(1)(2)x x x dx ---⎰(B)1201(1)(2)(1)(2)x x x dx x x x dx -----⎰⎰(C) 121(1)(2)(1)(2)x x x dx x x x dx ---+--⎰⎰(D)2(1)(2)x x x dx --⎰(3) 设()f x 在(,)-∞+∞内可导,且对任意12,x x ,当12x x >时,都有12()()f x f x >,则( )(A) 对任意,()0x f x '> (B) 对任意,()0x f x '-≤ (C) 函数()f x -单调增加 (D) 函数()f x --单调增加(4) 设函数()f x 在[0,1]上()0f x ''>,则(1)(0)(1)(0)f f f f ''-、、或(0)(1)f f -的大小顺序是 ( ) (A) (1)(0)(1)(0)f f f f ''>>- (B) (1)(1)(0)(0)f f f f ''>-> (C) (1)(0)(1)(0)f f f f ''->> (D) (1)(0)(1)(0)f f f f ''>-> (5) 设()f x 可导,()()(1|sin |)F x f x x =+,若使()F x 在0x =处可导,则必有 ( )(A) (0)0f = (B) (0)0f '= (C) (0)(0)0f f '+= (D) (0)(0)0f f '-=三、(本题共6小题,每小题5分,满分30分.) (1)求0lim x +→(2) 设函数()y y x =由方程()f y yxee =确定,其中f 具有二阶导数,且1f '≠,求22d ydx.(3) 设222(1)ln 2x f x x -=-,且[()]ln f x x ϕ=,求()x dx ϕ⎰.(4) 设21arctan ,0,() 0, 0,x x f x x x ⎧≠⎪=⎨⎪=⎩试讨论()f x '在0x =处的连续性. (5) 求摆线1cos sin x ty t t =-⎧⎨=-⎩一拱(02t π≤≤)的弧长.(6) 设单位质点在水平面内作直线运动,初速度00t v v ==,已知阻力与速度成正比(比例常数为1),问t 为多少时此质点的速度为03v ？并求到此时刻该质点所经过的路程.四、(本题满分8分)求函数2()(2)x t f x t e dt -=-⎰的最大值和最小值.五、(本题满分8分)设xy e =是微分方程()xy p x y x '+=的一个解,求此微分方程满足条件ln20x y ==的特解.六、(本题满分8分)如图,设曲线L 的方程为()y f x =,且0y ''>,又,MT MP 分别为该曲线在点00(,)M x y 处的切线和法线,已知线段MP 的长度为3220(1)y y '+''(其中00(),y y x ''= 00()y y x ''''=),试推导出点(,)P ξη的坐标表达式.七、(本题满分8分)设0sin ()xtf x dt tπ=-⎰,计算0()f x dx π⎰.八、(本题满分8分)设0()lim1x f x x→=,且()0f x ''>,证明()f x x ≥.1995年全国硕士研究生入学统一考试数学二试题解析一、填空题(本题共5小题,每小题3分,满分15分.) (1)【答案】22222cos()sin12sin()sin x x x x xx ⋅-⋅-【解析】该函数是由两个复合函数的乘积构成,满足复合函数求导法则,222211cos()sin cos()sin y x x x x '⎡⎤''⎡⎤=+⎣⎦⎢⎥⎣⎦22221111sin()2sincos()2sin cos (1)x x x x x x x=-⋅⋅+⋅⋅⋅- 22222cos()sin 12sin()sin x x x x x x⋅=-⋅-. 【相关知识点】复合函数求导法则：(())y f x ϕ=的导数为(())()y f x f x ϕ'''=. (2)【答案】12cos sin 2y c x c x x =+-【解析】微分方程2y y x ''+=-对应的齐次方程0y y ''+=的特征方程为210r +=, 特征根为1,2r i =±,故对应齐次方程的通解为12cos sin C x C x +.设非齐次方程的特解Y ax b =+,则Y a '=,0Y ''=,代入微分方程2y y x ''+=-,得02ax b x ++=-,比较系数得2,0,a b =-=故2Y x =-.所以通解为12cos sin 2y C x C x x =+-.【相关知识点】1.二阶线性非齐次方程解的结构：设*()y x 是二阶线性非齐次方程()()()y P x y Q x y f x '''++=的一个特解.()Y x 是与之对应的齐次方程 ()()0y P x y Q x y '''++=的通解,则*()()y Y x y x =+是非齐次方程的通解.2. 二阶常系数线性齐次方程通解的求解方法：对于求解二阶常系数线性齐次方程的通解()Y x ,可用特征方程法求解：即()()0y P x y Q x y '''++=中的()P x 、()Q x 均是常数,方程变为0y py qy '''++=.其特征方程写为20r pr q ++=,在复数域内解出两个特征根12,r r ； 分三种情况：(1) 两个不相等的实数根12,r r ,则通解为1212;rx r x y C eC e =+(2) 两个相等的实数根12r r =,则通解为()112;rxy C C x e =+(3) 一对共轭复根1,2r i αβ=±,则通解为()12cos sin .x y e C x C x αββ=+其中12,C C 为常数.3.对于求解二阶线性非齐次方程()()()y P x y Q x y f x '''++=的一个特解*()y x ,可用待定系数法,有结论如下：如果()(),x m f x P x e λ=则二阶常系数线性非齐次方程具有形如*()()k xm y x x Q x e λ=的特解,其中()m Q x 是与()m P x 相同次数的多项式,而k 按λ不是特征方程的根、是特征方程的单根或是特征方程的重根依次取0、1或2.如果()[()cos ()sin ]xl n f x e P x x P x x λωω=+,则二阶常系数非齐次线性微分方程()()()y p x y q x y f x '''++=的特解可设为*(1)(2)[()cos ()sin ]k x m m y x e R x x R x x λωω=+,其中(1)()m R x 与(2)()m R x 是m 次多项式,{}max ,m l n =,而k 按i λω+(或i λω-)不是特征方程的根、或是特征方程的单根依次取为0或1. (3)【答案】370y x -+= 【解析】切线的斜率为2222233322t t t t dy dy t dt t dx dx tdt========. 当2t =时,5,8x y ==.故所求切线方程为 83(5)y x -=-.化简得 370y x -+=.【相关知识点】参数方程所确定函数的微分法：如果()()x t y t φϕ=⎧⎨=⎩,则()()dy t dx t ϕφ'='.(4)【答案】12【解析】应用夹逼准则求数列的极限.令2221212n na n n n n n n n=+++++++++ 则 22212n na n n n n n n n n n>+++++++++221(1)1222211.22n n n n n n n n n ++++==+++=⋅+ 又 222221(1)1212122n n n n n a n n n n n n n n n n ++++<+++===+++++, 即 111222n n a n +⋅<<+,所以 11111lim lim lim 22222n n n n n a n →∞→∞→∞+⋅=<<=+.由夹逼准则,得1lim 2n n a →∞=.即222121lim()122n n n n n n n n n →∞+++=++++++. (5)【答案】0y =【解析】函数22x y x e -=的定义域为全体实数,且22lim lim 0x x x y x e -→∞→∞==,所以曲线只有一条水平渐近线0y =.【相关知识点】铅直渐近线：如函数()y f x =在其间断点0x x =处有0lim ()x x f x →=∞,则0x x =是函数的一条铅直渐近线；水平渐近线：当lim ()x f x a →∞=(a 为常数),则y a =为函数的水平渐近线.二、选择题(本题共5小题,每小题3分,满分15分.) (1)【答案】(D)【解析】方法一：反证法,利用连续函数的性质,即有限多个在同一点处连续的函数之乘积,仍然在该点处连续.设函数()()x f x ϕ无间断点,因为()f x 是连续函数,则()()()()x x f x f x ϕϕ=⋅必无间断点,这与()x ϕ有间断点矛盾,故应选择(D).方法二：排除法,举出反例排除.设 1,0,()1,()1,0,x f x x x ϕ-<⎧≡=⎨≥⎩则2[()]1,[()]1,[()]1f x f x x ϕϕϕ≡≡≡都处处连续,排除(A),(B),(C).故应选择(D). (2)【答案】(C)【解析】方法一：利用定积分的求面积公式有22(1)(2)(1)(2)x x x dx x x x dx --=--⎰⎰121(1)(2)(1)(2)x x x dx x x x dx =---+--⎰⎰应选择(C).方法二：画出曲线(1)(2)y x x x =--的草图,所求面积为图中两面积之和,即121(1)(2)(1)(2)x x x dx x x x dx ---+--⎰⎰,故应选(C). (3)【答案】(D)【解析】因为对任意12,x x ,当12x x >时,12x x -<-,则函数12()()f x f x -<-,即12()()f x f x -->--,故()f x --是单调增加的.应选择(D).对于(A)(B)(C)可令3()f x x =,则对任意12,x x ,当12x x >时,都有12()()f x f x >, 但 20(0)30x f x='==,2()3()0f x x '-=-≥,3()f x x -=-,在其定义域内单调减少.故排除(A)(B)(C). (4)【答案】(B)【解析】由()0f x ''>可知()f x '在区间[0,1]上为严格的单调递增函数,故(1)()(0),(01)f f x f x '''>> <<由微分中值定理,(1)(0)(),(01)f f f ξξ'-=<<.所以(1)(1)(0)()(0)f f f f f ξ'''>-=>,(01)ξ<<应选择(B). (5)【答案】(A)【解析】函数()f x 在0x x =处可导的充分必要条件是0()f x -'与0()f x +'存在且相等. 由于()()()|sin |F x f x f x x =+,而()f x 可导,所以()F x 在0x =处可导等价于()|sin |f x x 在0x =可导.令()()|sin |x f x x ϕ=,则0000()|sin |()sin (0)lim lim (0),()|sin |()sin (0)lim lim (0),x x x x f x x f x x f x xf x x f x x f x x ϕϕ++--+→→-→→⎧'===⎪⎪⎨⎪'==-=-⎪⎩于是要使()F x 在0x =处可导,当且仅当(0)(0)f f -=,即(0)0f =.故选择(A).三、(本题共6小题,每小题5分,满分30分.)(1)【解析】利用等价无穷小计算,即当0x →时,sin xx .原式22200022sin 1112lim lim lim 2222x x x x x x +++→→→⎛⎫⋅ ⎪⎝⎭====⋅⎝⎭. (2)【解析】这是一个由复合函数和隐函数所确定的函数.方法一：将方程两边对x 求导,得()()()f y f y y e xe f y y e y '''+⋅⋅=⋅,即 ()()()f y y f y e y e xf y e'='-, 将()f y y xee =代入并化简,得 1(1())y x f y '='-.两边再对x 求导,得[][][][]220(1())(1())(())(1())(1())x f y f y x f y y y x f y x f y ''''''----+-⋅''==''--[]221()(1())(1())y f y x f y x f y '''=-+'-'-. 将1(1())y x f y '='-代入并化简得 []3221()(1())(1())f y y x f y x f y ''''=-+'-'-. 方法二：方程两边先取对数再对x 求导.方程两边取对数得 ln ()x f y y +=,求导得1()f y y y x'''+⋅=, 因为1f '≠,所以 1(1())y x f y '='-.以下同方法一.【相关知识点】复合函数求导法则：(())y f x ϕ=的导数为(())()y f x f x ϕ'''=. (3)【解析】首先应求出()x ϕ的表达式.由2222211(1)ln ln 211x x f x x x -+-==---,令21,x t -=得1()ln1t f t t +=-.又 ()1[()]lnln ()1x f x x x ϕϕϕ+==-,则()1()1x x x ϕϕ+=-.解得1()1x x x ϕ+=-.因此12()(1)2ln 111x x dx dx dx x x C x x ϕ+==+=+-+--⎰⎰⎰. (4)【解析】函数()f x 在0x x =处的导函数连续的充分必要条件是0()f x -'与0()f x +'存在且必与0()f x '相等.当0x ≠时,22412()arctan 1x f x x x '=-+,由于224000012lim ()lim ()lim ()lim arctan 0122x x x x x f x f x f x x x ππ→-→+→→⎡⎤'''===-=-=⎢⎥+⎣⎦, 2000()(0)()1(0)limlim limarctan 02x x x f x f f x f x x x π→→→-'====-, 所以 00lim ()lim ()(0)x x f x f x f →-→+'''==.故()f x '在0x =处连续. (5)【解析】由弧微分公式得ds ===,所以2222000022sin2sin224cos4(11)8.2t t s dt dttπππππ====⎡⎤=-=---=⎢⎥⎣⎦⎰⎰⎰⎰(6)【解析】设质点的运动速度为()v t,由题设,阻力为()v t-,按牛顿第二定律有()()dv tm v tdt=-,其中质量1m=,即()()dv tv tdt=-.这是简单变量可分离的微分方程,解之得()tv t Ce-=.另有初始条件(0)v v=,得()tv t v e-=.当此质点的速度为03v时,有3tvve-=,得ln3t=.到此时刻该质点所经过的路程为ln3ln3000012133t ts v e dt v e v v--⎛⎫⎡⎤==-=--=⎪⎣⎦⎝⎭⎰.四、(本题满分8分)【解析】对函数2()(2)x tf x t e dt-=-⎰两边求导并令()0f x'=,得22()2(2)0xf x x x e-'=-=,解得驻点0,x x==由于()0,()()0,0,()()0,0()()0,,()f x x f xf x x f xf x x f xf x x f x'>-∞<<'<<<'><<'<<<+∞⎧⎪⎪⎨⎪⎪⎩严格单调增,严格单调减,严格单调增,严格单调减,所以(f f为函数()f x的极大值点,(0)f为函数()f x的极小值点,且222200((2)(2)1t t tf t e dt t e e dt e----=-=---=+⎰⎰,(0)(2)0tf t e dt-=-=⎰,又00lim()li(2)(2)1m()x xt t tf x f t e dt t e e dtx+∞→-∞→+∞+∞+∞----=---===⎰⎰, 所以2(1f e-=+为函数()f x最大值,(0)0f=为函数()f x的最小值.【相关知识点】积分上限函数的求导公式：()()()()()()()()()x x d f t dt f x x f x x dxβαββαα''=-⎰.五、(本题满分8分)【解析】把xy e =和xy e '=代入所给的一阶线性微分方程,得()x x xe p x e x +=,解得 ()xp x xex -=-.线性方程被确定为()xxy xe x y x -'+-=,即(1)1x y e y -'+-=.这是一阶线性非齐次微分方程,通解为 (1)(1)x x e dxe dx y e e dx C -----⎛⎫⎰⎰=+ ⎪⎝⎭⎰ ()()xxxxx x e exex e x e x e x e eedx C e dx C e e dx C e -------+--++-⎛⎫⎛⎫'=+=+=+ ⎪ ⎪ ⎪⎝⎭⎝⎭⎰⎰⎰ ()xx xe xe x exe e C e Ce ---+-+=+=+.再由 ln20x y==得ln 2ln 2ln 20e e Ce-++=,即12C e -=-.故所求的特解为 12x e x xy e e-+-=-.【相关知识点】一阶线性非齐次微分方程()()y p x y q x '+=的通解公式为:()()(())p x dx p x dxy e q x e dx C -⎰⎰=+⎰,其中C 为常数.六、(本题满分8分)【解析】要求点P 的坐标,也就是说,要用0000,,,,x y y y '''表示出,ξη. 由()32201MP y y '''=+,有 ()3202200201()()y x y y ξη'+-+-='', ①又由法线的斜率与切线斜率互为负倒数的关系,知x y y ξη-'=--, ②把②式,即000()()x y y ξη'-=--代入①消去,ξ得到 2222000()(1)/y y y η'''-=+, ③ 由0y ''>,知曲线是向上凹的,容易看出0y η>,所以③可化为 20001y y y η'+-='',且 2000000(1)()y y x y y y ξη''+'-=--=-'',于是得 200002000(1),1(1).y x y y y y y ξη'⎧'=-+⎪''⎪⎨⎪'=++⎪''⎩七、(本题满分8分)【解析】方法一：这是一个积分上限函数求定积分,可以考虑用定积分的分部积分法.由于 sin ()xf x xπ'=-, 因而由分部积分法和00sin (0)0tf dt t π==-⎰,有()()()()()()()f x dx f x d x f x x f x x dx πππππππ'=-=-+-⎰⎰⎰[]000sin ()sin cos 2x x dx xdx x xπππππ=-==-=-⎰⎰.方法二：对于二重积分0sin ()x t f x dx dt dx t πππ⎛⎫= ⎪-⎝⎭⎰⎰⎰,可以通过变换积分次序来求解.0sin sin ()x Dt t f x dx dt dx dtdx t t ππππ⎛⎫== ⎪--⎝⎭⎰⎰⎰⎰⎰, 其中{}{}(,)0,0(,)0,.D x t x t x x t t t x πππ=≤≤≤≤=≤≤≤≤于是00sin sin ()sin 2t t t t f x dx dx dt dt dx tdt t t ππππππππ⎛⎫==== ⎪--⎝⎭⎰⎰⎰⎰⎰⎰.八、(本题满分8分) 【解析】由于 0()lim1x f x x→=,所以必有(0)0f =,且0()(0)()(0)limlim 10x x f x f f x f x x→→-'===-. 证法一：用函数单调性证明不等式. 令 ()()x f x x ϕ=-,则 ()()1()(0)x f x f x f ϕ''''=-=-. 由于()0f x ''>,所以函数()f x '单调增加,()()(0)0,0,()()(0)0,0,x f x f x x f x f x ϕϕ'''=->>⎧⎨'''=-<<⎩()x ϕ'在0x =由负变正,所以0x =是()x ϕ的极小值点也是最小值点,()()(0)(0)00x f x x f ϕϕ=-≥=-=,即()f x x ≥. 证法二：用泰勒公式.2211()(0)(0)()()2!2f x f f x f x x f x ξξ'''''=++=+. 因为()0f x ''>,所以 21()02f x ξ''≥. 所以 21()()2f x x f x x ξ''=+≥.。

1995年全国硕士研究生入学统一考试英语试题Section ⅠUse of EnglishSleep is divided into periods of so-called REM sleep, characterized by rapid eye movements and dreaming, and longer periods of non-REM sleep. 1 kind of sleep is at all well-understood, but REM sleep is 2 to serve some restorative function of the brain. The purpose of non-REM sleep is even more 3 . The new experiments, such as these 4 for the first time at a recent meeting of the Society for Sleep Research in Minneapolis, suggest fascinating explanations 5 of non-REM sleep.For example, it has long been known that total sleep 6 is 100 percent fatal to rats, yet, 7 _examinations of the dead bodies, the animals look completely normal.A researcher has now8 the mystery of why the animals die. The rats 9 bacterial infections of the blood, 10 their immune systems—the self-protecting mechanisrn against disease —had crashed.1. [A] Either [B] Neither [C] Each [D] Any2. [A] intended [B] required [C] assumed [D] inferred3. [A] subtle [B] obvious [C] mysterious [D] doubtful4. [A] maintained [B] described [C] settled [D] afforded5. [A] in the light [B] by virtue [C] with the exception [D] for the purpose6. [A] reduction [B] destruction [C] deprivation [D] restriction7. [A] upon [B] by [C] through [D] with8. [A] paid attention to [B] caught sight of [C] laid emphasis on [D] cast light on9. [A] develop [B] produce [C] stimulate [D] induce10. [A] if [B] as if [C] only if [D] if only一、文章结构总体分析睡眠分为浅睡阶段和较长时间的深睡阶段。

1. 焦點群體研究（focus group）進行過程中會有一主持人，請問下列對主持人的工作敘述何者有誤？(Ａ)主持人必須確定每一位參與成員都加入討論(Ｂ)主持人必須阻止有強烈個人意識者主導討論過程(Ｃ)主持人必須引導討論的過程，加入討論的行列(Ｄ)主持人必須確保討論的重點是放在主題上。

2. 購買行為大致上分為四類，請問消費者屬於高介入者，而且了解品牌之間存在顯著的差異，請問這群消費者的購買行為屬於那一類型？(Ａ)複雜的購買行為(Ｂ)尋求多樣化的購買行為(Ｃ)降低失調的購買行為(Ｄ)習慣性購買行為。

3. 汽車銷售增加1%，造成汽車廠房設備的需求增加4%，這是因為組織市場具備那一個特性？(Ａ)衍生性需求（derived demand）(Ｂ)無彈性需求（inelastic demand）(Ｃ)彈性需求（elastic demand）(Ｄ)變動需求（fluctuating demand）。

4. 下列敘述何者錯誤？(Ａ)工業品分類中的零件並不會完全成為最終產品的一部分(Ｂ)資本項目是指部分成為最終產品之中的物品(Ｃ)資本項目包括設施及設備兩類(Ｄ)以上皆為錯誤的描述。

5. 請問行銷程序的主要步驟為何？ a.行銷機會的分析 b.規畫行銷方案 c.設計行銷策略 d.研究與選擇目標市場 e.組織、執行與控制行銷努力(Ａ)a-b-c-d-e(Ｂ)a-d-b-c-e(Ｃ)a-d-c-b-e(Ｄ)a-c-b-d-e。

6. 迪斯耐樂園廣告訴求為全世界最大的樂園，請問迪斯耐公司採取的定位策略為何？(Ａ)屬性定位(Ｂ)使用者定位(Ｃ)產品類別定位(Ｄ)品質定位。

7. 選擇目標市場區隔之吸引力時，公司必須優先考慮那些因素？(Ａ)區隔市場所需的成本與可能的收益(Ｂ)目標市場的大小(Ｃ)目標市場遠近(Ｄ)以上皆非。

8. 下列何者有誤？(Ａ)服務基本上是無法產生所有權的(Ｂ)麥當勞的作業標準化是採科技化的策略(Ｃ)以服務為主的產品，其蒐尋品質及經驗品質要比以商品為主的產品高(Ｄ)以上皆錯誤。

我們喜愛「拒絕作弊，堅守正直」的你！理工群組二、化學群組二、商學群組二、商設二、景觀二、資管二、財法二、生環三、醫工三科目： 中英語文能力測驗(中文部份) （共 1 頁第 1 頁）一、填空題：（15％）１、寫出「人生自古誰無死，留取丹心照汗青」這兩句詩的作者是＿＿＿＿。

２、《聊齋誌異》的作者是＿＿＿＿。

３、鴻門之宴與垓下之困是《史記》＿＿＿＿本紀中最精彩的兩段。

４、主張「文章合為時而著，歌詩合為事而作」的大文學家是＿＿＿＿。

５、寫過《看海的日子》、《兒子的大玩偶》，被台灣評論界譽為「小人物的代言人」的作家是＿＿＿＿。

二、成語解釋：（15％）１、得隴望蜀２、買櫝還珠３、罄竹難書４、前倨後恭５、抱薪救火三、作文（20％）求學與做人我們喜愛「拒絕作弊，堅守正直」的你！理工群組二、化學群組二、商學群組二、商設二、景觀二、資管二、財法二、生環三、醫工三科目： 中英語文能力測驗(英文部份) （共 3 頁第 1 頁）□可使用計算機，惟僅限不具可程式及多重記憶者 □不可使用計算機Cloze: Find the best answer to complete each passage (50%)When riots erupt in one of the biggest countries of the supposedly stableEuropean Union (EU), it can be (1) for the government concerned.When those riots go on night after night for two weeks, only to continue gettingworse, it starts to become truly alarming. On Sunday November 6th, theeleventh night (2) of unrest in France, the violence intensified and(3) to new areas, despite promises from the government to (4) ;and on Monday night, hours after the prime minister, Dominque de Villepin,had promised “firmness” in dealing with rioters, the streets of many townsand cities were once again (5) ablaze. What started with a fewdisaffected youths throwing rocks and burning cars on the outskirts of Parishas turned into a national social and political crisis.1. (A)strange (B)embarrassing (C)wonderful (D)smooth2. (A)in a line (B)in a thread (C)in a row (D)in a chance3. (A)limited (B)was contained (C)extended (D)spread4. (A)kill it (B)hurt it (C)stop it (D)harm it5. (A)on fire (B)by fire (C) at fire (D)with fireBitTorrent, the fastest-growing (6) service on the Internet, and eDonkeytogether (7) of all P2P traffic. BitTorrent is especially (8) toHollywood because it can download movies in just a few hours. And thesoftware is designed so that downloading a film gets easier as more people tryto (9) it.6. (A)point-to-point (B)peer-to-peer (C)part-to-part (D)party-to-party7. (A)make up the bulk (B)contain little (C)play a trivial role (D)make by8. (A)beneficial (B)helpful (C)familiar (D)frightening9. (A)control (B)access (C)balance (D)presentIs America's gang problem getting worse? The Justice Department says there are now 30,000 gangs with more than 800,000 members. The National Youth Gang Centre (NYGC), which (10) an annual survey that is (11) by the Justice Department, gives lower ones. It (12) that every town of 250,000 people now has a gang problem, but it thinks the tide has turned: the number of rural counties and small cities reporting gang activity droppedconsiderably between 1996 and 2002.10. (A)says (B)hands over (C)conducts (D)controls11. (A)funded (B)fussed (C)fanned (D)found12. (A)conducts (B)includes (C)conceives (D)concedesDespite the lack of evidence that mobile phones can cause explosions, bans remain (13) around the world, though the rules vary widely. Warning signs abound in Britain, America, Canada and Australia. The city of São Paulo, in Brazil, (14) a ban last year. And, earlier this month, a member ofConnecticut's senate (15) making the use of mobile phones in gasstations in that state punishable by a $250 (16) .13. (A)out of bound (B)in place (C)out of hand (D)in life14. (A)introduced (B)integrated (C)violated (D)exported15. (A)produced (B)claimed (C)provided (D)proposed16. (A)fine (B)bonus (C)salary (D)budgetIt is an old chestnut—a car that drives itself—but General Motors, the world's largest car manufacturer, has become the latest company to (17) to be building one. The car uses (18) technology combined with several(19) innovations and, according to the manufacturer, could be inproduction by 2008. But, while the technology takes some of the boring bits out of driving, it (20) an automatic taxi service and, anyway, various legal, technical and social barriers to its introduction remain.17. (A)require (B)crash (C)acquire (D)claim18. (A)outdated (B)updated (C)uploaded (D)downloaded19. (A)existing (B)dying (C)soothing (D)staggering20. (A)lives up to (B)falls right into (C)remains as (D)falls far short ofWeather systems, as the world has recently been reminded, have awesome power. The energy (21) by a large hurricane can exceed the energy consumption of the human race for a whole year, and even an average tornado has a power (22) that of a large power station. If only mankind could (23) that energy, rather than being at its mercy. Louis Michaud, a Canadian engineer who works at a large oil company, believes he has(24) a way to do just that, by (25) artificial whirlwinds that canbe controlled and harnessed. He calls his invention the “atmospheric vortex engine”.21. (A)expressed (B)explained (C)released (D)extended22. (A)similar to (B)benefited from (C)stimulated by (D)familiar with23. (A)stop (B)use (C)reject (D)resist24. (A)divided (B)figure out (C)finished (D)impended25. (A)blowing (B)flapping (C)generating (D)fitting-End-。

1995年全国硕士研究生入学统一考试——政治试题（文科）一、在下列各题的备选答案中，选出其中一个正确答案，特字母标号填入括号内，错选不给分。

1．彻底的唯物主义一元论的根本要求是A．承认世界是多样的统一B．承认物质对意识的根源性C．坚持一切从实际出发D．反对一切形式的二元论2．唯物辩证法的否定之否定规律揭示了事物发展的A．方向和道路B．形式和状态C。

结构和功能D．源泉和动力3．根据认识的发展规律，在认识的“熟知”与“真知”问题上的正确观点是A．熟知即真知B．熟知不等于真知C．熟知起源于真知D．熟知必然转化为真知4．马克思指出：“如果物没有用，那么其中包含的劳动也就没有用，不能算作劳动，因此不形成价值。

”这段话说明A．价值的存在以物的有用性为前提B．价值的存在与物的有用性互为前提C。

只要物是有用的，它就有价值D．物越是有用就越有价值5．列宁说，“没有革命的理论，就不会有革命的运动”，这句话应理解为A．革命运动是由革命理论派生的B．革命理论是革命运动的基础C。

革命理论对革命实践具有最终决定作用D．革命理论对革命实践具有重要指导作用6．中国共产党把抗日游击战争放在战略地位上加以考察的根据是A．中国的抗日战争是一场持久战B．中国是一个处于进步时代的大而弱的国家C. 中国共产党领导的八路军、新四军的存在D．国民党正面战场的溃退，日军占地甚广7．社会主义市场经济体制的基础是A．以间接手段为主的完善的宏观调控体系B．统一、开放、竞争、有序的市场体系C. 以公有制为主体的现代企业制度D．合理的个人收入分配和社会保障制度8．我国实施社会保障的基本目标是A．保证劳动者的充分就业B．使劳动者老有所养C．满足人们最基本的生活需要D．实现共同富裕9．田中角荣内阁对日本外交政策作出了重要调整。

指出A. “西方一员”外交方针B．“等距离外交”方针C．“全方位外交”方针D．“多边自主外交”方针10．1973年在阿尔及尔召开的第四次不结盟国家和政府首脑会议具有重要意义，在不结盟运动史上，它A．正式规定不结盟运动的五项标准B．强烈要求废除帝国主义、殖民主义和新殖民主义C．明确指出发展中国家与发达国家之间的经济差距不断扩大D．首次提出要反对霸权主义二、在下列各题的选项中，至少有一项是符合题意的。

1995年全国硕士研究生入学统一考试理工数学一试题详解及评析一、填空题(1)()2sin 0lim 13xx x →+= .【答】 6e .【详解1】 用第二类重要极限:()()621sin 6sin 30lim 13lim 13.x xxxx x x x e →→⎧⎫+=+=⎨⎬⎩⎭【详解2】 化为指数函数求极限:()()()006ln 1322limln 13lim6sin 3sin 0lim 13.x x x x x x xx x eee →→++→+===(2)202cos xd x t dt dx =∫ . 【答】2224cos 2cos xt dt x x −∫.【详解】()()()2222000222220224cos cos cos cos 2 cos 2cos .x x x xd d x t dt x t dt t dt x x x dx dx t dt x x ==+−=−∫∫∫∫(3)设()2,a b c ×⋅=则()()()a b b c c a +×+⋅+=⎡⎤⎣⎦ . 【答】 4. 【详解】()()()()()()()()()()() 4.a b b c c a a b b c a a b c c a a b c b c a a b c a b c +×+⋅+⎡⎤⎣⎦=+×⋅+++×⋅+⎡⎤⎡⎤⎣⎦⎣⎦=+×+×⋅=×⋅+×⋅=(4)幂级数()21123n nnn nx ∞−=+−∑的收敛半径R ＝ .【答】【详解】 令()21,23n n nn n a x −=+− 则当 121lim13n n na x a +→∞=<时，即23,x <也即x <时，此幂级数收敛， 因此(5)设三阶方阵A、B 满足关系式：-1A BA =6A+BA,且100310,41007⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦A =则=B . 【答】300020001⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦. 【详解】 在已知等式-1A BA =6A+BA 两边右乘以-1A ，得A -1B =6E +B,于是()11120030066030020.006001−−−⎡⎤⎡⎤⎢⎥⎢⎥=−==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦B A E二、选择题 （1）设有直线3210:21030x y z L x y z +++=⎧⎨−−+=⎩及平面:4220,x y z π−+−=则直线L（A ）平行于.π （B ）在π上. （C ）垂直于.π （D ）与π斜交.【 】【答】 应选（C ）.【详解】 直线L 的方向向量s 为{}{}{}1,3,22,1,1013274,2,12110i j ks =×−−==−−−− 与平面π的法向量{}4,2,1n =−平行，应此直线L 垂直于.π （2）设在[]0,1上()''0,fx >则()()()()''0110f f f f −、、或()()01f f −的大小顺序是 (A)()()()()''1010.f ff f >>− (B)()()()()''1100.f f f f >−>(C)()()()()''1010.f f f f−>> (D)()()()()''1010.f f f f >−>【 】【答】 应选（B ）. 【详解】 由()''0,fx >知()'f x 单调增加，又()()()()()'1010 01,f f f ξξ−=−<<根据()()()'''01ff f ξ<<知，()()()()''0101.f f f f <−<可见正确选项为(B).（3）设()f x 可导，()()()1sin ,F x f x x =+则()00f =是()F x 在0x =处可导的（A ）充分必要条件. （B ）充分条件但非必要条件. （C ）必要条件但非充分条件. （D ）既非充分条件又非必要条件.【 】【答】 应选（A ）. 【详解】 因为()()()()()()()()()()()'000'01sin 00lim lim00sin lim 00,x x x F x F f x x f F x x f x f x f x x x f f −−−−→→→−−−==−−⎡⎤=−⎢⎥⎣⎦=− ()()()()()()()()()()()'000'01sin 00lim lim00sin lim 00,x x x F x F f x x f F x x f x f x f x x x f f ++++→→→−+−==−−⎡⎤=+⎢⎥⎣⎦=+ 可见，()'0F 存在⇔()()()()()()()''''00000000.F F ff f f f −+=⇔−=+⇔=因此正确选项为（A ）. (4)设()1ln 1,nn u ⎛=−+⎜⎝则级数 （A ）1n n u ∞=∑与21n n u ∞=∑都收敛. （B ）1n n u ∞=∑与21n n u ∞=∑都发散. （C ）1nn u∞=∑收敛而21nn u∞=∑发散. (D)1nn u∞=∑发散而21nn u∞=∑收敛.【 】【答】 应选（C ）. 【详解】因为ln 1n v ⎛=⎜⎝单调递减 且lim 0,n n v →∞= 由莱布尼茨判别法知级数()111nnn n n u v ∞∞===−∑∑收敛，而22ln 1n u ⎛=+⎜⎝且11n n ∞=∑发散， 因此21nn u∞=∑也发散.故正确选项为（C ）.(5)设111213212223212223*********3233311132123313010,,100,001a a a a a a a a a a a a a a a a a a a a a ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥+++⎣⎦⎣⎦⎣⎦A =B P 2100010101⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦P ，则必有（A ）2.=1AP P B (B).=A B 21P P (C).=12P P A B (D).=21P P A B【 】【答】 应选（C ）. 【详解】1P 是交换单位矩阵的第一、二行所得初等矩阵，2P 是将单位矩阵的第一行加到第三行所得初等矩阵，而B 是由A 先将第一行加到第三行，然后再交换第一、二行两次初等交换得到的，因此.=12P P A B 故正确选项为（C ）.三、(1)设()()2,,,,,0,sin ,yu f x y z x e z y x ϕ===其中f ϕ、都具有一阶连续偏导数，且0,z ϕ∂≠∂求.du dx【详解】 等式(),,u f x y z =两边同时对x 求导，得,du f f dy f dzdx x y dx z dx∂∂∂=+⋅+⋅∂∂∂而cos .dyx dx=又等式()2,,0y x e z ϕ=两边同时对x 求导，得 '''12320,y y dy dzx e e dx dxϕϕϕ⋅+⋅+⋅=解得()''12'312cos ,y dz x e x dx ϕϕϕ=−+ 故()'sin '12'31cos 2cos .x du f f f x x e x dx x y z ϕϕϕ∂∂∂=+−+∂∂∂ （2）设函数()f x 在区间[]0,1上连续，并设()10,f x dx A =∫求()()11.xdx f x f y dy ∫∫【详解1】 交换积分次序，得()()()()()()1111,yxxdx f x f y dy dy f x f y dx dx f y f x dy ==∫∫∫∫∫∫于是()()()()()()()()()()11111000011001100121 21 2x x x dx f x f y dy dx f x f y dy dx f x f y dy dx f x f y dy f x dx f y dy⎡⎤=+⎢⎥⎣⎦==⋅∫∫∫∫∫∫∫∫∫∫21.2A =【详解2】 分部积分，得()()()()()()()()()()()()()()111101101111202122201 111 22|xxxx x xx dx f x f y dy f y dy f x dxf y dy d f t dt A f t dt d f y dy A f t dt A A =⋅==−⎡⎤=+=−=⎢⎥⎣⎦∫∫∫∫∫∫∫∫∫∫∫2.2A四、（1）计算曲面积分,zdS ∑∫∫其中∑为锥面z =在柱体222x y x +≤内的部分.【详解】因为,dS σσ==于是2cos 2322002cos DzdS d r dr d ππθπσθθθ−===∑=∫∫∫（2）将函数()()102f x x x =−≤≤展开成周期为4的余弦级数. 【详解】 因为()()()()()20022002220210,2221cos 1sin 22224 sin 11,1,2,.2n na x dx n x n x a x dx x d n n x dx n n n ππππππ=−==−=−⎡⎤=−=−−=⎣⎦∫∫∫∫" 故 ()()[]221114,0,2.2nn n xf x cosx n ππ∞=−−=∈∑五、设曲线L 位于xOy 平面的第一象限内，L 上任一点M 处的切线与y 轴总相交，焦点记为.A 已知,MA OA =且L 过点33,,22⎛⎞⎜⎟⎝⎠求L 的方程.【详解】 设点M 的坐标为(),x y ，则切线MA 的方程为()'.Y y y X x −=−令0,X =则',Y y xy =−故点A 的坐标为()'0,y xy −. 由,MA OA =有'y xy −=化简后，得'212,yy y x x−=− 令2z y =，得1,dz z x dx x−=− 解得 (),z x x c =−+即 22.y x cx =−+由于所求曲线在第一象限内，故y =再以条件3322y ⎛⎞=⎜⎟⎝⎠代入得 3.c =于是所求曲线方程为()0<x<3.y =六、设函数(),Q x y 在xOy 平面上具有一阶连续偏导数，曲线积分()2,Lxydx Q x y dy +∫与路径无关，并且对任意t 恒有()()()()()(),11,0,00,02,2,,t t xydx Q x y dy xydx Q x y dy +=+∫∫求 (),Q x y .【详解】 由曲线积分与路径无关的条件知()22,xy Q x x y∂∂==∂∂于是有 ()()2,,Q x y x C y =+其中 ()C y 为待定函数.又()()()()()()()()()(),111220,0001,20,0002,,2,1,t t ttxydx Q x y dy t C y dy t C y dy xydx Q x y dy C y dy t C y dy ⎡⎤+=+=+⎣⎦⎡⎤+=+=+⎣⎦∫∫∫∫∫∫由题设知 ()()12,tt C y dy t C y dy +=+∫∫两边对t 求导得()21,t C t =+于是 ()21,C t t =−从而()21,C y y =− 故有 ()2,2 1.Q x y x y =+−七、假设函数()f x 和()g x 在[],a b 上存在二阶导数，并且()()()()()''0,,g x f a f b g a g b ≠===试证：（1） 在开区间(),a b 内()0;g x ≠ （2） 在开区间(),a b 内至少存在一点,ξ使()()()()''''.f f g g ξξξξ=【详解】 （1）用反证法：若存在点(),c a b ∈使(),g c 则对()g x 在[],a c 和[],c b 上分别应用罗尔定理，知存在()1,a c ξ∈和()2,,c b ξ∈使()()''120.gg ξξ==再对()'g x 在[]12,ξξ上应用罗尔定理，知存在()312,,ξξξ∈使()''30.g ξ=这与题设()''0g x ≠矛盾，故在(),a b 内()0.g x ≠（3） 令()()()()()'',F x f x g x g x fx =−则()F x 在[],a b 上连续，在(),a b 内可导，且()()0,F a F b ==根据罗尔定理知，存在(),,a b ξ∈使()'0,F ξ=即有 ()()()()''''0,fg f g ξξξξ−=故得 ()()()()''''.f f g g ξξξξ=八、设三阶实对称矩阵A 的特征值为1231,1,λλλ=−==对应于1λ的特征向量为()10,1,1,Tξ=求A .【详解】 设对应于231λλ==的特征向量为()123,,,Tx x x ξ=根据A 为实对称矩阵的假设知10,T ξξ=即230,x x +=解得()()231,0,0,0,1,1.TTξξ==−于是由 ()()123112233,,,,,ξξξλξλξλξ=A 有()()11122331231,,,,010010100 101101001.101101010λξλξλξξξξ−−⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥=−=−⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥−−−−⎣⎦⎣⎦⎣⎦A =九、设A 是n 阶矩阵，满足T=AA E （E 是n 阶单位阵，TA 是A 的转置矩阵，0,<A 求A +E .【详解】 根据T=AA E 有(),====T T A+E A+AA A E +A A E +A A A+E于是()10.−=AA+E因为10,−>A 故0.=A+E十、填空题（1）设X 表示10次独立重复射击命中目标的次数，每次射中目标的概率为0.4，则2X 的数学期望()2E X= .【答】 18.4.【详解】 由题设知，X 服从10,0.4n p ==的二项分布， 因此有()()()4,1 2.4,E X np D X np p ===−= 故()()()2218.4E X D X E X =+=⎡⎤⎣⎦(2)设X 和Y 为两个随机变量，且{}{}{}340,0,00,77P X Y P X P Y ≥≥=≥=≥=则(){}max ,0P X Y ≥= .【答】 5.7【详解】 令{}{}0,0,A X B Y =<=<则(){}(){}{}max ,01max ,010,0P X Y P X Y P X Y ≥=−<=−<<()()()()(){}{}{}11000,04435.7777P AB P A BP A P B P AB P X P Y P X Y ⎡⎤=−−=+⎣⎦=+−=≥+≥−≥≥=+−=十一、设随机变量X 的概率密度为(),00, 0x X e x f x x −⎧≥=⎨<⎩，求随机变量xY e =的概率密度().Y f y【详解】 根据分布函数的定义，有(){}(){}0, 1ln ,1xY y f y P Y y P e y P X y y <⎧=<=<=⎨<≥⎩ 于是当1y ≥时，(){}ln 0ln .yx Y f y P X y e dx −=<=∫因此所求概率密度函数为()()20,11,1Y Y y dF y f y y dy y <⎧⎪==⎨≥⎪⎩1995年全国硕士研究生入学统一考试理工数学二试题详解及评析一、 填空题（1）设221cos()sin,'_____.y x y x==则 【答】 ()()22221122sin sin sin cos .x x x x x x−− 【详解】 222211'[cos()]'sincos()sin 'y x x x x ⎛⎞=+⋅⎜⎟⎝⎠()()22221122sin sin sin cos .x x x x x x=−− （2）微分方程''2y y x +=−的通解为______. 【答】 122cos sin .y x C x C x =−++.【详解】 相应齐次方程的特征方程为21210,,,i i λλλ+===−其根为由于非齐次项为2,0x λ−=不是特征根，可设非齐次方程的特解为*y A Bx =+，代入原方程解，得0,2A B ==−，因此通解为122cos sin y x C x C x =−++.（3）曲线2312x tt y t⎧=+⎪=⎨=⎪⎩在处的切线方程为______. 【答】 370.x y −−=【详解】 当0025,8,t x y ===时，且23,3,2t dydy dydt t dx dx dx dt==== 可知过曲线2312x tt y t⎧=+⎪=⎨=⎪⎩上对应于的切线斜率为3，切点为点（5，8）.因此切线方程为()835y x −=−或370.x y −−=（4）22212lim _____.12n n n n n n n n n →∞⎛⎞+++=⎜⎟++++++⎝⎠" 【答】1.2【详解】 利用夹逼定理，由22211212,1n i n i nn n n n n i n n =++++++<<++++++∑"" 且()()2222111212limlim ,2111212limlim ,112n n n n n n n n n n n n n n n n n n n n →∞→∞→∞→∞++++==++++++++==++++""知 211lim.2nn i i n n i →∞==++∑ （5）曲线22x y x e −=的渐近线方程为______.【答】 y=0.【详解】 由于222222lim limlim0,2x x x x x x x x x eexe−→∞→∞→∞===所以，y=0水平为渐近线.二、选择题（1）设()()(),f x x ϕ−∞+∞和在内有定义，()f x 为连续函数，且()()0,f x x ϕ≠有间断点，则()()()()()()22. ().. ().A f xB xC f xD x ϕϕϕϕ⎡⎤⎡⎤⎣⎦⎣⎦⎡⎤⎡⎤⎣⎦⎣⎦必有间断点必有间断点必有间断点必有间断点【 】【答】 应选（D ）.【详解】 方法一（用反证法）：若()()x f x ϕ无间断点，即连续，则()()()()x f x x f x ϕϕ⋅=也连续，与已知条件矛盾，所以()()x f x ϕ必有间断点.（A ）、（B ）、（C ）均可举反例说明不成立.方法二： 取()()()()41, 0, 1,,1, 1x x f x x f x x x ϕϕ≥⎧==+⎨−<⎩则符合要求，而()1,f x ϕ=⎡⎤⎣⎦()()21,2x f x ϕϕ==⎡⎤⎣⎦均无间断点，故排除（A ）、（B ）、（C ），应选（D ）. （2）曲线()()12y x x x =−−与x 轴所围图形的面积可表示为()()()()()()()()()()()()()()()21200120020()12.1212.1212.12.A x x x dxB x x x dx x x x dxC x x x dx x x x dxD x x x dx −−−−−−−−−−−+−−−−∫∫∫∫∫∫【 】【答】 应选（C ）.【详解】 曲线()()12y x x x =−−与x 轴的交点为0,1,2x x x ===，因此该曲线与x 轴所围图形的面积可表示为()()()()()()2121121212.x x x dx x x x dx x x x dx −−=−−−+−−∫∫∫（3）设()(),f x −∞+∞在内可导，且对任意1x 、()()21212,x x x f x f x >>当时，都有则 ()()()()()()()(),'0.,'0...A x f x B x f x C f x D f x >≤−−对任意对任意函数单调增加函数-单调增加【 】【答】 应选（D ）.【详解】 因为对任意1x 、()()2121212,,x x x x x f x f x >−<−当时，-<-,则有 即 ()()12f x f x −−>−− ， 故()f x −−是单调增加的.(4) 设函数()[]()()()()()()()0,1''0,'1'01001f x f x f f f f f f >−−在上则、、或的大小顺序是()()()()()()()()()()()()()()()()()()()()'1'010'110'010'1'0'101'0A f f f f B f f f f C f f f f D f f f f >>−>−>−>>>−>【 】【答】 应选(B)。

精品资料1995考研数四真题及解析........................................1995年全国硕士研究生入学统一考试数学四试题一、填空题(本题共5小题,每小题3分,满分15分.)(1) 设1lim xt x x te dt x αα-∞→∞+⎛⎫= ⎪⎝⎭⎰,则常数α= .(2) 设()y z xyf x=,()f u 可导,则xyxz yz ''+= .(3) 设(ln )1f x x '=+,则()f x = .(4) 设100220345A ⎛⎫⎪= ⎪ ⎪⎝⎭,*A 是A 的伴随矩阵,则1()A *-= .(5) 设X 是一个随机变量,其概率密度为则方差()D X = .二、选择题(本题共5小题,每小题3分,满分15分.) (1) 设()f x 为可导函数,且满足条件0(1)(1)lim 12x f f x x→--=-,则曲线()y f x =在点 (1,(1))f 处的切线斜率为( )(A) 2 (B) 1- (C) 12(D) 2- (2) 下列广义积分发散的是( )(A) 111sin dx x -⎰(B) 1dx -⎰(C) 2x edx +∞-⎰ (D) 221ln dx x x+∞⎰(3) 设n 维行向量11(,0,,0,)22α=,矩阵T A E αα=-,2T B E αα=+,其中E 为n 阶单位矩阵,则AB 等于 ( )(A) 0 (B) E - (C) E (D)T E αα+(4) 设矩阵m n A ⨯的秩为()r A m n =<,m E 为m 阶单位矩阵,下述结论中正确的是 ( )(A) A 的任意m 个列向量必线性无关 (B) A 的任意一个m 阶子式不等于零(C) A 通过初等行变换,必可以化为(,0)m E 的形式(D) 非齐次线性方程组Ax b =一定有无穷多组解(5) 设随即变量X 服从正态分布2(,)N μσ,则随σ的增大,概率{}P X μσ-< ( )(A) 单调增大 (B) 单调减少 (C) 保持不变 (D) 增减不定三、(本题满分6分)设2202(1cos ),0()1,01cos ,0xx x x f x x t dt x x ⎧-<⎪⎪==⎨⎪⎪>⎩⎰,试讨论()f x 在0x =处的连续性和可导性.四、(本题满分6分)求不定积分2(arcsin )x dx ⎰.五、(本题满分7分)设()f x 、()g x 在区间[,]a a -(0a >)上连续,()g x 为偶函数,且()f x 满足条件()()f x f x A +-=(A 为常数).(1) 证明0()()()a aaf xg x dx A g x dx -=⎰⎰；(2) 利用(1)的结论计算定积分22sin arctan x x e dx ππ-⎰.六、(本题满分6分)设某产品的需求函数为()Q Q p =,收益函数为R pQ =,其中p 为产品价格,Q 为需求量(产品的产量),()Q p 为单调减函数.如果当价格为0p ,对应产量为0Q 时,边际收益00Q Q dR a dQ ==>,收益对价格的边际效应0p p dRc dp==<,需求对价格的弹性1p E b =>.求0p 和0Q .七、(本题满分5分)设()f x 在区间[,]a b 上连续,在(,)a b 内可导.证明：在(,)a b 内至少存在一点ξ,使()()()()bf b af a f f b aξξξ-'=+-.八、(本题满分9分)求二元函数2(,)(4)z f x y x y x y ==--在由直线6x y +=、x 轴和y 轴所围成的闭区域D 上的极值、最大值与最小值.九、(本题满分8分)对于线性方程组1231231233,2,2.x x x x x x x x x λλλλ++=-⎧⎪++=-⎨⎪++=-⎩ 讨论λ取何值时,方程组无解、有惟一解和有无穷多组解.在方程组有无穷多组解时,试用其导出组的基础解系表示全部解.十、(本题满分8分)设三阶矩阵A 满足(1,2,3)i i A i i αα==,其中列向量1(1,2,2)T α=,2(2,2,1)T α=-, 3(2,1,2)T α=--.试求矩阵A .十一、(本题满分8分)假设一厂家生产的每台仪器,以概率0.70可以直接出厂；以概率0.30需进一步调试,经调试后以概率0.80可以出厂；以概率0.20定为不合格品不能出厂.现该厂新生产了(2)n n ≥台仪器(假设各台仪器的生产过程相互独立).求：(1) 全部能出厂的概率α；(2) 其中恰好有两台不能出厂的概率β； (3) 其中至少有两台不能出厂的概率θ.十二、(本题满分7分)假设随机变量X 服从参数为2的指数分布.证明：21X Y e -=-在区间(0,1)上服从均匀分布.1995年全国硕士研究生入学统一考试数学四试题解析一、填空题(本题共5小题,每小题3分,满分15分.) (1)【答案】2【解析】等式左端是1∞型未定式求极限,11lim lim 1xx x x x e x x ααα⋅→∞→∞+⎛⎫⎛⎫=+= ⎪ ⎪⎝⎭⎝⎭,等式右端是求一个定积分,可以用分部积分法求得. (1)t t t ttte dt tde e dt te e e ααααααα-∞-∞-∞-∞-∞=-=-=-⎰⎰⎰.由题设有(1)e e ααα=-,解得2α=.【相关知识点】分部积分公式：假定()u u x =与()v v x =均具有连续的导函数,则,uv dx uv u vdx ''=-⎰⎰ 或者 .udv uv vdu =-⎰⎰(2)【答案】2y xyf x ⎛⎫⎪⎝⎭【解析】根据复合函数求导法则,可得22x y y y y y y z yf xyf yf f x x x x x x ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫'''=+⋅-=-⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭, 1y y y y y z xf xyf xf yf x x x x x ⎛⎫⎛⎫⎛⎫⎛⎫'''=+⋅=+ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭. 所以 222x y y y y y y yf y f xyf y f yf x x x x x xz yz x x ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫''-++= ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭'⎝⎭⎝'+⎭=.【相关知识点】复合函数求导法则：(())y f x ϕ=的导数为(())()y f x f x ϕ'''=. (3)【答案】x x e C ++【解析】在(ln )1f x x '=+中令ln x t =,则()1t f t e '=+,从而()()1()t t x f t e dt t e C f x x e C =+=++⇒=++⎰.(4)【答案】100122010345⎛⎫ ⎪⎪ ⎪⎝⎭【解析】由AA A E *=,有A A E A *=,故()1AA A-*=. 而 10022010345A ==,所以 ()1100122010345A A A -*⎛⎫⎪== ⎪ ⎪⎝⎭. (5)【答案】16【解析】 011()(1)(1)0EX xf x dx x x dx x x dx +∞-∞-==++-=⎰⎰⎰,01222211()(1)(1)6EX x f x dx x x dx x x dx +∞-∞-==++-=⎰⎰⎰, ()2216DX EX EX =-=. 【相关知识点】连续型随机变量的数学期望和方差的定义：(),EX x f x dx +∞-∞=⋅⎰22()()()D X E X E X =-.二、选择题(本题共5小题,每小题3分,满分15分.) (1)【答案】(D)【解析】因 0(1)(1)(1)(1)(1)limlimx x f x f f x f f x x xx→→+---'==--00(1)(1)lim(1)(1)2lim 2,2x x f f x xf f x x →→--=--==-所以应选(D).(2)【答案】(A)【解析】由计算知111arcsin x π--==⎰,222111ln ln ln 2dx x x x +∞+∞=-=⎰,且泊松积分20x e dx +∞-=⎰故应选(A).注：对于本题选项(A),由于当0x =时sin 0x =,故在积分区间[1,1]-中0x =是瑕点,反常积分111sin dx x-⎰应分解为两个反常积分之和: 101110111sin sin sin dx dx dx x x x --=+⎰⎰⎰,而且111sin dx x -⎰收敛的充要条件是两个反常积分011sin dx x -⎰与101sin dx x⎰都收敛. 由于广义积分 11001ln tan sin 2x dx x ⎛⎫==+∞ ⎪⎝⎭⎰, 即101sin dx x ⎰发散,故111sin dx x-⎰发散. 在此不可误以为1sin x是奇函数,于是1110sin dx x -=⎰,从而得出它是收敛的错误结论. (3)【答案】(C)【解析】利用矩阵乘法的分配律、结合律,有()()222T T T T T T AB E E E αααααααααααα=-+=+--()2T T T E αααααα=+-.由于 120111(,0,,0,)222012ααT ⎛⎫ ⎪ ⎪ ⎪ ⎪== ⎪ ⎪ ⎪ ⎪⎝⎭,所以T T AB E E αααα=+-=.故应选(C). (4)【答案】(D)【解析】()r A m =表示A 中有m 个列向量线性无关.有m 阶子式不等于零,并不是任意的,因此(A)、(B)均不正确.经初等变换可把A 化成标准形,一般应当既有初等行变换也有初等列变换,只有一种不一定能化为标准形.例如010001⎛⎫⎪⎝⎭,只用初等行变换就不能化成2(0)E 的形式,故(C)不正确.关于(D),因为A 为m n ⨯矩阵,且()r A m =,故增广矩阵的秩必为m ,那么()()r A r A m n ==<,所以方程组Ax b =必有无穷多组解,故选 (D). (5)【答案】(C) 【解析】由于2(,),XN μσ将此正态分布标准化,故()0,1X N μσ-,{}()1211X P X P .μμσσ⎧-⎫-<=<=Φ-⎨⎬⎩⎭计算看出概率{}P X μσ-<的值与σ大小无关.所以本题应选(C).三、(本题满分6分)【解析】这是一道讨论分段函数在分界点处的连续性和可导性的问题.一般要用连续性与可导性的定义并借助函数在分界点处的左极限与右极限以及左导数和右导数.222000122(1cos )2lim ()lim lim 1x x x x x f x x x ---→→→⋅-===, 220cos cos lim ()limlim 11xx x x t dt x f x x+++→→→===⎰, 故(00)(00)(0)f f f +=-=,即()f x 在0x =处连续.2004222001cos 1()(0)(0)lim lim 01cos cos 12lim lim lim 0,22x x x xx x x t dt f x f xf x xx t dt x x x x x++++++→→→→→--'==----====⎰⎰2002320002(1cos )1()(0)(0)lim lim02(1cos )2sin 22(cos 1)lim lim lim 0.36x x x x x x f x f x f x xx x x x x x x x------→→→→→---'==-----==== 即(0)(0)0f f +-''==,故()f x 在0x =处可导,且(0)0f '=.四、(本题满分6分)【解析】当被积函数仅仅为反三角函数时,这种积分肯定要用分部积分法. 方法一：22(arcsin )(arcsin )x dx x x =-⎰2(arcsin )2arcsin x x =+⎰2(arcsin )2x x x dx =+-⎰2(arcsin )2.x x x x C =+-+方法二：令arcsin x u =,则2222222(arcsin )cos sin sin 2sin sin 2cos sin 2cos 2cos sin 2cos 2sin .x dx uudu u d uu u u udu u u ud uu u u u udu u u u u u C ===-=+=+-=+-+⎰⎰⎰⎰⎰⎰五、(本题满分7分)【解析】(1)由要证的结论可知,应将左端积分化成[]0,a 上的积分,即00()()()()()()aaaaf xg x dx f x g x dx f x g x dx --=+⎰⎰⎰,再将0()()af xg x dx -⎰作适当的变量代换化为在[]0,a 上的定积分.方法一：由于 00()()()()()()a aaaf xg x dx f x g x dx f x g x dx --=+⎰⎰⎰,在0()()af xg x dx -⎰中令x t =-,则由:0x a -→,得:0t a →,且00()()()()()()()()()a aa af xg x dx f t g t d t f t g t dt f x g x dx -=---=-=-⎰⎰⎰⎰,所以 []00()()()()()()aa aaf xg x dx f x f x g x dx A g x dx -=+-=⎰⎰⎰. 方法二：在()()aaf xg x dx -⎰中令x t =-,则由:x a a -→,得:t a a →-,且()()()()()()()()()aaaaaaaf xg x dx f t g t d t f t g t dt f x g x dx ---=----=-=-⎰⎰⎰⎰.所以 1()()()()()()2aaa aa a f x g x dx f x g x dx f x g x dx ---⎡⎤=+-⎢⎥⎣⎦⎰⎰⎰[]01()()()()().22a aa a aA f x f x g x dx g x dx A g x dx --=+-==⎰⎰⎰(2)令()arctan x f x e =,()sin g x x =,可以验证()f x 和()g x 符合(1)中条件,从而可以用(1)中结果计算题目中的定积分. 方法一：取()arctan x f x e =,()sin g x x =,2a π=.由于()()arctan arctan x x f x f x e e -+-=+满足()22arctan arctan 011x xxxx xe e eee e----'+=+≡++, 故 arctan arctan x x e e A -+=. 令0x =,得2arctan12A A π=⇒=,即()()2f x f x π+-=.于是有2222sin arctan sin sin 222xx e dx x dx xdx πππππππ-===⎰⎰⎰.方法二：取()arctan x f x e =,()sin g x x =,2a π=,于是1()()arctan arctan2x x f x f x e e π+-=+=. (这里利用了对任何0x >,有1arctan arctan 2x x π+=) 以下同方法一.六、(本题满分6分)【解析】本题的关键在于p 和Q 之间存在函数关系,因此R pQ =既可看作p 的函数,也可看作Q 的函数,由此分别求出dR dp 及dR dQ,并将它们与弹性p p dQE Q dp =联系起来,进而求得问题的解.由()Q Q p =是单调减函数知0dQ dp <,从而需求对价格的弹性0p p dQE Q dp=<,这表明题设1p E b =>应理解为1p p E E b =-=>.又由()Q Q p =是单调减函数知存在反函数()p p Q =且1dp dQdQ dp=.由收益R pQ =对Q 求导,有 1(1)p dR dp p p Q p p p dQ dQ dQ E Q dp=+=+=+,从而001(1)Q Q dR p a dQ b ==-=,得01abp b =-.由收益R pQ =对p 求导,有(1)(1)p dR dQ p dQ Q p Q Q E dp dp Q dp=+=+=+, 从而 00(1)p p dR Q b c dp==-=,于是01c Q b=-.七、(本题满分5分)【解析】由于本题中要证的结论中出现了()()f f ξξξ'+,所以应考虑辅助函数()()F x xf x =.因为()()()F x f x xf x ''=+.令()()F x xf x =,则()F x 在[,]a b 上连续,在(,)a b 内可导,满足拉格朗日中值定理条件,从而在(,)a b 内至少存在一点ξ,使()()()F b F a F b aξ-'=-,即 ()()()()bf b af a f f b aξξξ-'=+-.八、(本题满分9分)【解析】首先在区域D 内求驻点.令2(832)0,(42)0,x y f xy x y f x x y '=--=⎧⎪⎨'=--=⎪⎩ 在D 内仅有唯一驻点(2,1).在点(2,1)处,有2(2,1)2(2,1)2(2,1)(2,1)(862)60,(2,1)(834)4,(2,1)28.xxxyyyA f y xy yB f x x xyC f x ''==--=-<''==--=-''==-=-于是2320B AC -=-<,因此点(2,1)是极大值点,且极大值(2,1)4f =.在D 的边界0(06)x y =≤≤和0(06)y x =≤≤上,(,)0f x y =. 在D 的边界6(06)x y x +=<<上,把6y x =-代入(,)f x y 可得22(6)(06)z x x x =-<<.由于004,6(4)04,046,x z x x x x <<<⎧⎪'=-==⎨⎪><<⎩所以点(4,2)是这段边界上z 的最小值点,最小值(4,2)64f =-.综合以上讨论知(,)f x y 在D 的边界上的最大值是0,最小值是(4,2)64f =-. 比较D 内驻点的函数值(2,1)4f =和(,)f x y 在D 的边界上的最大值和最小值可得(,)f x y 在D 上的最大值和最小值分别是max (,)(2,1)4Df x y f ==,min (,)(4,2)64Df x y f ==-.【相关知识点】1.驻点：凡是能使(,)0x f x y =,(,)0y f x y =同时成立的点00(,)x y 称为函数(,)z f x y =的驻点.具有偏导数的函数的极值点必定是驻点.但函数的驻点不一定是极值点.例如,点(0,0)是函数z xy =的驻点,但函数在该点并无极值. 2.判定一个驻点是否是极值点的定理：定理：设函数(,)z f x y =在点00(,)x y 的某邻域内连续且有一阶及二阶连续偏导数,又(,)0x f x y =,(,)0y f x y =,令000000(,),(,),(,)xx xy yy f x y A f x y B f x y C ===,则(,)f x y 在00(,)x y 处是否取得极值的条件如下：(1) 20AC B ->时具有极值,且当0A <时有极大值,当0A >时有极小值； (2) 20AC B -<时没有极值；(3) 20AC B -=时可能有极值,也可能没有极值,还需另作讨论. 3.具有二阶连续偏导数的函数(,)z f x y =的极值的求法如下：第一步：解方程组(,)0x f x y =,(,)0y f x y =,求得一切实数解,即可求得一切驻点.第二步：对于每一个驻点00(,)x y ,求出二阶偏导数的值A 、B 和C . 第三步：定出2AC B -的符号,按上述定理的结论判定00(,)f x y 是不是极值、是最大值还是最小值.九、(本题满分8分)【解析】对增广矩阵作初等行变换,有113112112112112113λλλλλλλλ--⎡⎤⎡⎤⎢⎥⎢⎥-→-⎢⎥⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦()()21121120110011001133002133λλλλλλλλλλλλ--⎡⎤⎡⎤⎢⎥⎢⎥→--→--⎢⎥⎢⎥⎢⎥⎢⎥----+--⎣⎦⎣⎦当1λ≠且2λ≠-时,()()3r A r A ,==方程组有唯一解. 当2λ=-时,()()23r A ,r A ,==方程组无解. 当1λ=时,()()1r A r A ,==方程组有无穷多组解. 其同解方程组为：1232x x x ++=-. 令230x x ==,得到特解(2,0,0)T α=-.令1231,1,0x x x =-==及1231,0,1x x x =-==得到导出组的基础解系()()12110101T T,,,,,.ηη=-=-因此,方程组的通解是1122,k k αηη++其中12k ,k 是任意常数.十、(本题满分8分)【解析】由(1,2,3)i i A i i αα= =知123,,ααα是矩阵A 的不同特征值的特征向量,它们线性无关.利用分块矩阵,有()()12312323A ,,,,.αααααα= 由123,,ααα线性无关,知矩阵()123,,ααα可逆,故()()112312323A ,,,,αααααα-=而 ()1112312212212212219212212,,ααα---⎡⎤⎡⎤⎢⎥⎢⎥=--=-⎢⎥⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦故 72033146122152243221093322621222233A .⎡⎤-⎢⎥-⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥=--⋅-=-⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥--⎢⎥⎣⎦⎣⎦⎢⎥--⎢⎥⎣⎦十一、(本题满分8分)【解析】对于新生产的每台仪器,设事件A 表示“仪器需要进一步调试”,B 表示“仪器能出厂”,则A =“仪器能直接出厂”,AB =“仪器经调试后能出厂”.且B A AB =,A 与AB 互不相容,应用加法公式与乘法公式,且由条件概率公式()(|)()(|)()()P AB P B A P AB P B A P A P A =⇒=⋅, 有 ()()()()070308094P B P A P A P B |A ....=+=+⨯=.设X 为所生产的n 台仪器中能出厂的台数,则X 服从二项分布()094B n,..由二项分布的概率计算公式,可得所求概率为(1) {}0.94n P X n α===；(2) {}22220.940.06;n n P X n C β-==-=⋅⋅(3){}{}{}121110.060.940.94n n P X n P X n P X n n θ-=≤-=-=--==-⨯-⋅【相关知识点】二项分布的概率计算公式：若(,)Y B n p ~,则{}(1)k k n k n P Y k C p p -==-, 0,1,,k n =.十二、(本题满分7分)【解析】要证明21X Y e -=-在区间(0,1)上服从均匀分布,只需证明随机变量Y 的概率密度()1010Y ,y ,f y ,<<⎧=⎨⎩其他，或证明Y 的分布函数为()110001Y ,y ,F y ,y ,y y .≥⎧⎪=<⎨⎪≤<⎩， 方法1：21x y e -=-是()0,+∞上的单调函数,且其反函数为()()1ln 1.2x h y y -==- 在区间(0,1)上()()1021h y .y '=≠-应用单调函数公式法,Y 的概率密度为()()()()010X Y h y f h y ,y ,f y ,⎧'⋅<<⎪=⎨⎪⎩其他，()()212011012100h y e,y ,,y ,y ,,-⎧<<<<⎧⎪-==⎨⎨⎩⎪⎩其他.其他，由计算可知()Y f y 恰是(0,1)上均匀分布的密度函数. 方法2：用分布函数法求Y 的分布函数.当0y ≤时,()0Y F y =；当1y ≥时,()1Y F y =；当01y <<时,()1ln 102y -->,(){}{}()211ln 12X Y F y P Y y P e y P X y -⎧⎫=≤=-≤=≤--⎨⎬⎩⎭()()12ln 121ln 112y X F y e y ⎡⎤---⎢⎥⎣⎦⎡⎤=--=-=⎢⎥⎣⎦.计算可知Y 的分布函数为()110001Y ,y ,F y ,y ,y y .≥⎧⎪=<⎨⎪≤<⎩，上式恰好是区间(0,1)上均匀分布随机变量的分布函数.【相关知识点】一维均匀分布的定义：若连续型X 的概率密度为1, ,()0, a x b f x b a⎧≤≤⎪=-⎨⎪⎩其他,则称X 服从区间[,]a b 上的均匀分布.其分布函数为0, ,(), ,1, x a x a F x a x b b a <⎧⎪-⎪=≤≤⎨-⎪⎪⎩其他.。

1994年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)= _____________. (2)曲面在点处的切平面方程为_____________.(3)设则在点处的值为_____________.(4)设区域为则=_____________.(5)已知设其中是的转置,则=_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设则有 (A)(B) (C) (D)(2)二元函数在点处两个偏导数、存在是在该点连续的(A)充分条件而非必要条件 (B)必要条件而非充分条件(C)充分必要条件(D)既非充分条件又非必要条件(3)设常数且级数收敛,则级数(A)发散 (B)条件收敛(C)绝对收敛(D)收敛性与有关(4)其中则必有(A) (B) (C) (D)(5)已知向量组线性无关,则向量组011limcot ()sin x x x π→-e 23x z xy -+=(1,2,0)e sin ,xx u y -=2u x y ∂∂∂1(2,)πD 222,x y R +≤2222()Dx y dxdy a b +⎰⎰11[1,2,3],[1,,],23==αβ,'=A αβ'ααn A 4342342222222sin cos ,(sin cos ),(sin cos ),1x M xdx N x x dx P x x x dx x ππππππ---==+=-+⎰⎰⎰N P M <<M P N <<N M P <<P M N <<(,)f x y 00(,)x y 00(,)x f x y '00(,)y f x y '(,)f x y 0,λ>21nn a ∞=∑1(1)nn ∞=-∑λ2tan (1cos )lim2,ln(12)(1)x x a x b x c x d e -→+-=-+-220,a c +≠4b d =4b d =-4a c =4a c =-1234,,,αααα(A)线性无关 (B)线性无关 (C)线性无关(D)线性无关三、(本题共3小题,每小题5分,满分15分) (1)设 ,求、在. (2)将函数展开成的幂级数.(3)求12233441,,,++++αααααααα12233441,,,----αααααααα12233441,,,+++-αααααααα12233441,,,++--αααααααα2221cos()cos()t x t y t t udu==-⎰dy dx 22d y dx t =111()ln arctan 412x f x x x x +=+--x .sin(2)2sin dxx x +⎰四、(本题满分6分)计算曲面积分其中是由曲面及两平面所围成立体表面的外侧.五、(本题满分9分)设具有二阶连续函数且为一全微分方程,求及此全微分方程的通解.2222,Sxdydz z dxdy x y z +++⎰⎰S 222x y R +=,(0)z R z R R ==->()f x ,(0)0,(0)1,f f '==2[()()][()]0xy x y f x y dx f x x y dy '+-++=()f x设在点的某一邻域内具有二阶连续导数,且证明级数绝对收敛.七、（本题满分6分）已知点与的直角坐标分别为与线段绕轴旋转一周所成的旋转曲面为求由及两平面所围成的立体体积.()f x 0x =0()lim 0,x f x x →=11()n f n ∞=∑A B (1,0,0)(0,1,1).AB x .S S 0,1z z ==设四元线性齐次方程组(Ⅰ)为,又已知某线性齐次方程组(Ⅱ)的通解为(1)求线性方程组(Ⅰ)的基础解析. (2)问线性方程组(Ⅰ)和(Ⅱ)是否有非零公共解?若有,则求出所有的非零公共解.若没有,则说明理由.九、（本题满分6分）设为阶非零方阵是的伴随矩阵是的转置矩阵,当时,证明122400x x x x +=-=12(0,1,1,0)(1,2,2,1).k k +-A n *,A A ,'A A *'=A A 0.≠A十、填空题(本题共2小题,每小题3分,满分6分.把答案填在题中横线上)(1)已知、两个事件满足条件且则=____________. (2)设相互独立的两个随机变量具有同一分布率,且的分布率为则随机变量的分布率为____________.十一、（本题满分6分）设随机变量和分别服从正态分布和且与的相关系数设(1)求的数学期望和方差.(2)求与的相关系数 (3)问与是否相互独立?为什么?A B ()(),P AB P AB =(),P A p =()P B ,X Y X max{,}Z X Y =X Y 2(1,3)N 2(0,4),N X Y 1,2xy ρ=-,32X Y Z =+Z EZ DZ X Z .xz ρX Y1995年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)=_____________.(2)= _____________.(3)设则=_____________. (4)幂级数的收敛半径=_____________. (5)设三阶方阵满足关系式且则=_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设有直线,及平面则直线(A)平行于 (B)在上 (C)垂直于(D)与斜交(2)设在上则或的大小顺序是 (A) (B) (C)(D)(3)设可导则是在处可导的 (A)充分必要条件 (B)充分条件但非必要条件(C)必要条件但非充分条件(D)既非充分条件又非必要条件(4)设则级数 2sin 0lim(13)xx x →+202cos x d x t dt dx⎰()2,⨯=a b c g [()()]()+⨯++a b b c c a g 2112(3)n n nn n x ∞-=+-∑R ,A B 16,-=+A BA A BA 100310,41007⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦A B :L 321021030x y z x y z +++=--+=:4220,x y z π-+-=L ππππ[0,1]()0,f x ''>(0),(1),(1)(0)f f f f ''-(0)(1)f f -(1)(0)(1)(0)f f f f ''>>-(1)(1)(0)(0)f f f f ''>->(1)(0)(1)(0)f f f f ''->>(1)(0)(1)(0)f f f f ''>->()f x ,()()(1sin ),F x f x x =+(0)0f =()F x 0x =(1)ln(1n n u =-+(A)与都收敛(B)与都发散(C)收敛,而发散(D)收敛,而发散(5)设则必有 (A) (B) (C)(D)三、(本题共2小题,每小题5分,满分10分)(1)设其中都具有一阶连续偏导数,且求(2)设函数在区间上连续,并设求1n n u ∞=∑21n n u ∞=∑1n n u ∞=∑21n n u ∞=∑1n n u ∞=∑21n n u ∞=∑1n n u ∞=∑21n n u ∞=∑11121311121321222321222312313233313233010100,,100,010,001101a a a a a a a a a a a a a a a a a a ⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥====⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦A B P P 12AP P =B 21AP P =B 12P P A =B 21P P A =B 2(,,),(,e ,)0,sin ,y u f x y z x z y x ϕ===,f ϕ0.zϕ∂≠∂.du dx ()f x [0,1]1(),f x dx A =⎰110()().xdx f x f y dy ⎰⎰四、(本题共2小题,每小题6分,满分12分) (1)计算曲面积分其中为锥面在柱体内的部分. (2)将函数展开成周期为4的余弦函数.五、(本题满分7分)设曲线位于平面的第一象限内上任一点处的切线与轴总相交,交点记为已知且过点求的方程.,zdS ∑⎰⎰∑z 222x y x +≤()1(02)f x x x =-≤≤L xOy ,L M y .A ,MA OA =L 33(,),22L六、(本题满分8分)设函数在平面上具有一阶连续偏导数,曲线积分与路径无关,并且对任意恒有求七、（本题满分8分）假设函数和在上存在二阶导数,并且试证: (1)在开区间内(2)在开区间内至少存在一点使(,)Q x y xOy 2(,)L xydx Q x y dy +⎰t (,1)(1,)(0,0)(0,0)2(,)2(,),t t xydx Q x y dy xydx Q x y dy +=+⎰⎰(,).Q x y ()f x ()g x [,]a b ()0,()()()()0,g x f a f b g a g b ''≠====(,)a b ()0.g x ≠(,)a b ,ξ()().()()f f g g ξξξξ''=''八、（本题满分7分）设三阶实对称矩阵的特征值为对应于的特征向量为求九、（本题满分6分）设为阶矩阵,满足是阶单位矩阵是的转置矩阵求A 1231,1,λλλ=-==1λ101,1⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦ξ.A A n ('=AA I I n ,'A A ),0,<A .+A I十、填空题(本题共2小题,每小题3分,满分6分.把答案填在题中横线上) (1)设表示10次独立重复射击命中目标的次数,每次射中目标的概率为0.4, 则的数学期望=____________.(2)设和为两个随机变量,且则____________.十一、（本题满分6分） 设随机变量的概率密度为,求随机变量的概率密度X 2X 2()E X X Y 34{0,0},{0}{0},77P X Y P X P Y ≥≥=≥=≥={max(,)0}P X Y ≥=X ()X f x =e 0x -00x x ≥<e X Y =().Y f y1996年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)设则=_____________.(2)设一平面经过原点及点且与平面垂直,则此平面方程为_____________.(3)微分方程的通解为_____________.(4)函数在点处沿点指向点方向的方向导数为_____________.(5)设是矩阵,且的秩而则=_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)已知为某函数的全微分,则等于 (A)-1 (B)0 (C)1(D)2(2)设具有二阶连续导数,且则 (A)是的极大值 (B)是的极小值 (C)是曲线的拐点(D)不是的极值也不是曲线的拐点(3)设且收敛,常数则级数(A)绝对收敛 (B)条件收敛2lim()8,xx x a x a→∞+=-a (6,3,2),-428x y z -+=22e x y y y '''-+=ln(u x =(1,0,1)A A (3,2,2)B -A 43⨯A ()2,r =A 102020,103⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦B ()r AB 2()()x ay dx ydyx y +++a ()f x 0()(0)0,lim 1,x f x f x→'''==(0)f ()f x (0)f ()f x (0,(0))f ()y f x =(0)f ()f x ,(0,(0))f ()y f x =0(1,2,),n a n >=L 1n n a ∞=∑(0,),2πλ∈21(1)(tan )n n n n a n λ∞=-∑(C)发散 (D)散敛性与有关(4)设有连续的导数且当时与是同阶无穷小,则等于(A)1 (B)2 (C)3 (D)4(5)四阶行列式的值等于(A)(B) (C)(D)三、(本题共2小题,每小题5分,满分10分) (1)求心形线的全长,其中是常数.(2)设试证数列极限存在,并求此极限.λ()f x 220,(0)0,(0)0,()()(),xf f F x x t f t dt '=≠=-⎰0x →,()F x 'k x k 1122334400000000a b a b a b b a 12341234a a a a b b b b -12341234a a a a b b b b +12123434()()a a b b a a b b --23231414()()a a b b a a b b --(1cos )r a θ=+0a>1110,1,2,),n x x n +===L {}n x四、(本题共2小题,每小题6分,满分12分)(1)计算曲面积分其中为有向曲面其法向量与轴正向的夹角为锐角.(2)设变换 可把方程简化为求常数五、(本题满分7分) 求级数的和.(2),Sx z dydz zdxdy ++⎰⎰S 22(01),z x y x =+≤≤z 2u x y v x ay =-=+2222260z z z x x y y ∂∂∂+-=∂∂∂∂20,zu v∂=∂∂.a 211(1)2nn n ∞=-∑设对任意曲线上点处的切线在轴上的截距等于求的一般表达式.七、（本题满分8分）设在上具有二阶导数,且满足条件其中都是非负常数是内任意一点.证明0,x >()y f x =(,())x f x y 01(),xf t dt x⎰()f x ()f x [0,1](),(),f x a f x b ''≤≤,a b ,c(0,1)()2.2b f c a '≤+设其中是阶单位矩阵是维非零列向量是的转置.证明 (1)的充分条件是 (2)当时是不可逆矩阵. 九、（本题满分8分）已知二次型的秩为2, (1)求参数及此二次型对应矩阵的特征值. (2)指出方程表示何种二次曲面.,T A =-I ξξI n ,ξn ,T ξξ2=A A 1.T =ξξ1T =ξξ,A 222123123121323(,,)55266f x x x x x cx x x x x x x =++-+-c 123(,,)1f x x x =十、填空题(本题共2小题,每小题3分,满分6分.把答案填在题中横线上)(1)设工厂和工厂的产品的次品率分别为1%和2%,现从由和的产品分别占60%和40%的一批产品中随机抽取一件,发现是次品,则该次品属生产的概率是____________.(2)设是两个相互独立且均服从正态分布的随机变量,则随机变量的数学期望=____________.十一、（本题满分6分）设是两个相互独立且服从同一分布的两个随机变量,已知的分布率为又设(1)(2)求随机变量的数学期望A B A B A ,ξη2)N ξη-()E ξη-,ξηξ1(),1,2,3.3P i i ξ===max(,),min(,).X Y ξηξη==X ().E X。

科目： 心理及教育統計學 （共 6 頁第 1 頁）▇可使用計算機，惟僅限不具可程式及多重記憶者 □不可使用計算機＊所有假設考驗，皆使用α = .051.（30%）X Y （1） 計算X 與Y 的相關係數？_______________ （2） 計算以X預測Y的迴歸公式？112 10 （3） 計算估計標準誤？118 6 （4） 當X=120 計算68%Y的期望值所落區間？119 13122 14125 6________________2. （20%）研究檢驗是否女性比男性容易回憶情緒事件。

有5個男性及5個女性被要求看20個情緒圖片，在1週後進行回憶測驗。

（1）作假設考驗（2）計算效果量.---------------------------------------男性 女性74 8787 9064 8076 7785 91--------------------------------------3. （30%）Suppose you are interested in determining whether certain situations producedifferent amounts of stress. You randomly assign 15 students into threegroups of 5 each. The students in group 1 have their corticosterone levelmeasured immediately after returning from vacation. The students in group2 have their corticosterone level measured immediately after they havebeen in class for a week. The students in group 1 have their corticosteronelevel measured immediately before final exam week.科目： 心理及教育統計學 （共 6 頁第 2 頁）▇可使用計算機，惟僅限不具可程式及多重記憶者 □不可使用計算機（1）作 ANOVA分析？（2）視需要作 Post Hoc Comparison?（3）計算 Effect size？Group 1 Group 2 Group3Vacation Class Final ExamX1 X2 X32 10 103 8 137 7 142 5 136 10 154. （20%）. A social psychologist is interested in determining whether there is a relationship between the education level of parents and the number ofchildren they have.（1）作卡方檢定？No. of ChildrenTwo or less More than twoCollege education 53 22 75High school37 38 75education onlyColumn marginal 90 60 150科目： 心理及教育統計學 （共 6 頁第 3 頁）▇可使用計算機，惟僅限不具可程式及多重記憶者 □不可使用計算機科目： 心理及教育統計學 （共 6 頁第 4 頁）▇可使用計算機，惟僅限不具可程式及多重記憶者 □不可使用計算機科目： 心理及教育統計學 （共 6 頁第 5 頁）▇可使用計算機，惟僅限不具可程式及多重記憶者 □不可使用計算機科目： 心理及教育統計學 （共 6 頁第 6 頁）▇可使用計算機，惟僅限不具可程式及多重記憶者 □不可使用計算機。

1995年全国硕士研究生入学统一考试——政治试题（理科）一、下列每题的选项中，有一项是符合题意的，请把所选答案的字母填写在试回答题纸相应括号内。

（每小题1分，共10分）1．马克思主义新世界观创立的关键在于马克思确立了。

A.剩余价值论B.阶级斗争理论C.无产阶级历史使命学说D.科学的实践观2．唯物主义一元论同唯心主义一无论对立的根本点在于A.世界发展动力问题B.意识本质问题C.世界本质问题D.实践本质问题3．实践高于理论的认识，是因为实践具有A.普遍性B.绝对性C.客观实在性D.直接现实性4．商品的价值量由生产商品的社会必要劳动时间决定，它是在一A.同类商品的生产者之间的竞争中实现的B.不同商品的生产者之间的竞争中实现的C.商品的生产者和消费者之间的竞争中实现的D.商品的生产者和销售者之间的竞争中实现的5，在资本主义经济发展过程中，资本集中的直接后果是A.社会总资本急剧增加B.个别资本规模迅速扩大C.绝对剩余价值总量快速增长D.社会就业率明显提高6．中国近代历史上第一次反对国内封建势力和外国侵略势力的斗争是A.三元里人民的斗争B.太平天国农民运动C.义和团农民运动D.辛亥革命7．“可能有这样一些共产党人，他们是不曾被拿枪的敌人征服过的，他们在这些敌人面前不愧英雄的称号；但是经不起人们用糖衣裹着炮弹的攻击，他们在糖衣炮弹面前要打败仗。

我们必须预防这种情况。

”毛泽东提出这一警告是在A.抗日战争胜利之际B.全国革命胜利前夜C.新中国成立之际D.社会主义改造时期8．社会主义经济体制的基础是A.以间接手段为主的完善的宏观调控体系B.统一、开放、竞争、有序的市场体系C.以公有制为主体的现代企业制度D.合理的个人收入分配和社会保障制度9．进入90年代，我国对外开放发展到一个新阶段，其显著特点是A.技术引进从注重硬件发展到注重软件B.从兴办经济特区发展到开放沿海港口城市C.从吸收利用外资发展到扩大国际劳务合作的跨国经营D.多层次、多渠道、全方位开放的格局已经形成10．人民民主是社会主义的本质要求和内在用性，归根结底是因为它A.是社会主义公有制经济在政治上的集中体现B.有利于调动最广大人民群众的积极性C.是社会主义制度优越性的重要体现D.体现了多数人对少数人的专政二、下列每题的选项中，至少有一项足符合回意的，请把所述答案的字母填写在回答相应的括号内。