1989年全国硕士研究生入学统一考试
考研数学三历年真题：1987年至2018年
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(含 31 年共 31 套考研《数学三》历年真题)1987 年—2018 年全国硕士研究生入学统一考试《数学 三》真题试卷及答案
全国硕士研究生入学统一考试《数学三》真题目录（31 套）
1987 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1989 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1990 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1991 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1992 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1993 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1994 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1995 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1996 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1997 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1998 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 1999 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2000 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2001 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2002 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2003 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2004 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2005 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2006 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2007 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2008 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2009 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2010 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2011 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2012 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2013 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2014 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2015 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2016 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2017 年全国硕士研究生入学统一考试《数学三》真题试卷及答案 2018 年全国硕士研究生入学统一考试《数学三》真题试卷及答案
1989考研数二真题及解析
1989考研数二真题及解析1989年全国硕士研究生入学统一考试数学二试题一、填空题(每小题3分,满分21分.把答案填在题中横线上.)(1) 0lim cot 2x x x →=＿＿＿＿＿＿.(2) 0sin t tdt π=⎰＿＿＿＿＿＿.(3) 曲线0(1)(2)x y t t dt=--⎰在点(0,0)处的切线方程是＿＿＿＿＿＿.(4) 设()(1)(2)()f x x x x x n =++⋅⋅+L ,则(0)f '=＿＿＿＿＿＿.(5) 设()f x 是连续函数,且1()2()f x x f t dt=+⎰,则()f x =＿＿＿＿＿＿. (6) 设2,0()sin ,0a bx x f x bxx x⎧+≤⎪=⎨>⎪⎩在0x =处连续,则常数a 与b 应满足的关系是＿＿＿＿＿. (7) 设tan y x y =+,则dy =＿＿＿＿＿＿.二、计算题(每小题4分,满分20分.) (1) 已知arcsin xy e -=求y '.(2) 求2ln dxx x⎰. (3) 求1lim(2sin cos )xx x x →+.(4) 已知2ln(1),arctan ,x t y t ⎧=+⎨=⎩求dy dx及22d y dx . (5) 已知1(2),(2)02f f '==及20()1f x dx =⎰,求12(2)xf x dx''⎰.三、选择题(每小题3分,满分18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x >时,曲线1siny x x=( )(A) 有且仅有水平渐近线 (B) 有且仅有铅直渐近线(C) 既有水平渐近线,也有铅直渐近线 (D) 既无水平渐近线,也无铅直渐近线 (2) 若2350a b -<,则方程532340x ax bx c +++=( )(A) 无实根(B) 有唯一实根(C)有三个不同实根(D) 有五个不同实根(3) 曲线cos ()22y x x ππ=-≤≤与x 轴所围成的图形,绕x轴旋转一周所成的旋转体的体积为( )π(B) π(A)2π (D) 2π(C) 22(4) 设两函数()g x都在x a=处取得极大值,则f x及())(A) 必取极大值(B) 必取极小值(C) 不可能取极值(D) 是否取极值不能确定(5) 微分方程1x''-=+的一个特解应具有形式y y e(式中,a b为常数) ( )(A) x ae b+ (B) x axe b+ (C)x+ (D) x axe bx+ae bx(6) 设()f x在f x在x a=的某个领域内有定义,则()x a=处可导的一个充分条件是( ) (A) 1lim [()()]h h f a f a h→+∞+-存在 (B) 0(2)()lim h f a h f a h h →+-+存在(C) 0()()lim 2h f a h f a h h →+--存在(D) 0()()lim h f a f a h h →--存在四、(本题满分6分)求微分方程2(1)xxy x y e'+-=(0)x <<+∞满足(1)0y =的解.五、(本题满分7分)设0()sin ()()x f x x x t f t dt =--⎰,其中f 为连续函数,求()f x .六、(本题满分7分)证明方程0ln 1cos 2xx xdxe π=--⎰在区间(0,)+∞内有且仅有两个不同实根.七、(本大题满分11分)对函数21x y x +=,填写下表：单调减少区间 单调增加区间 极值点 极值 凹(U )区间 凸(I )区间 拐点 渐近线八、(本题满分10分)设抛物线2y axbx c=++过原点,当01x ≤≤时,0y ≥,又已知该抛物线与x 轴及直线1x =所围图形的面积为13,试确定,,a b c 使此图形绕x 轴旋转一周而成的旋转体的体积V 最小.1989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(每小题3分,满分21分.) (1)【答案】12【解析】这是个0⋅∞型未定式,可将其等价变换成00型,从而利用洛必达法则进行求解. 方法一： 0cos 2lim cot 2lim lim cos 2sin 2sin 2x x x x x x x x x x x→→→==⋅ 0011limlim sin 22cos 22x x x x x →→==洛.方法二： 0cos 2lim cot 2lim sin 2x x x x x x x→→= 0012121lim cos 2lim .2sin 22sin 22x x x x x x x →→=⋅==【相关知识点】0sin lim x xx→是两个重要极限中的一个,0sin lim 1x x x→=. (2)【答案】π【解析】利用分部积分法和牛顿-莱布尼茨公式来求解,sin t tdt π=⎰[]000(cos )cos (cos )td t t t t dt πππ-=---⎰⎰分部法[]00sin (00)t ππππ=++=+-=.(3)【答案】2y x =【解析】要求平面曲线的切线,首先应求出该切线的斜率,即0()f x '.这是一个积分上限函数,满足积分上限函数的求导法则,即(1)(2)y x x '=--.由y '在其定义域内的连续性,可知0(01)(02)2x y ='=--=.所以,所求切线方程为02(0)y x -=-,即2y x =.(4)【答案】!n【解析】方法一：利用函数导数的概念求解,即0()(0)(1)(2)()0(0)limlim x x f x f x x x x n f x x→→-++⋅⋅+-'==Llim(1)(2)()12!x x x x n n n →=++⋅⋅+=⋅⋅⋅=L L .方法二：利用其导数的连续性,由复合函数求导法则可知,()(1)(2)()1(2)()f x x x x n x x x n '=++⋅⋅++⋅⋅+⋅⋅+++L L L (1)(2)(1)1x x x x n ++⋅⋅+-⋅L , 所以 (0)(01)(02)(0)00f n '=++⋅⋅++++L L 12!n n =⋅⋅⋅=L . (5)【答案】1x -【解析】由定积分的性质可知,1()f t dt⎰和变量没有关系,且()f x 是连续函数,故1()f t dt ⎰为一常数,为简化计算和防止混淆,令10()f t dt a=⎰,则有恒等式()2f x x a =+,两边0到1积分得11()(2)f x dx x a dx=+⎰⎰,即[]111112000001(2)222a x a dx xdx a dx x a x ⎡⎤=+=+=+⎢⎥⎣⎦⎰⎰⎰122a=+,解之得12a =-,因此()21f x x a x =+=-. (6)【答案】a b =【解析】如果函数在0x 处连续,则函数在该点处的左右极限与该点处函数值必然相等, 由函数连续性可知(0)(0)0f f a b a -==+⋅=.而 000sin sin sin (0)lim lim lim x x x bx bx bxf b b b x bx bx++++→→→==⋅=⋅=, 如果()f x 在0x =处连续,必有(0)(0)f f -+=,即a b =.(7)【答案】2()dx x y + 【解析】这是个隐函数,按照隐函数求导法,两边微分得2secy dy dx dy⋅=+,所以 222sec 1tan ()dx dx dxdy y y x y ===++,(0x y +≠).二、计算题(每小题4分,满分20分.) (1)【解析】令xu e =,v x =则arcsin arcsin xy eu-==,由复合函数求导法则,222(arcsin )2111v v y u u e v e xuuu''''===⋅=---即 221x xy e xe-'=-【相关知识点】复合函数求导法则：(())y f x ϕ=的导数(())()y f x f x ϕ'''=.(2)【解析】利用不定积分的换元积分法,22ln 1ln ln ln dx d x Cx x x x ==-+⎰⎰.(3)【解析】可将函数转化称为熟悉的形式来求其极限,110lim(2sin cos )lim[1(2sin cos 1)]xxx x x x x x →→+=++-12sin cos 12sin cos 10lim[1(2sin cos 1)]x x x x xx x x +-⋅+-→=++-,令 2sin cos 1x x t +-=,则当0x →时,0t →,则 112sin cos 1lim[1(2sin cos 1)]lim[1]x x tx t x x t +-→→++-=+,这是个比较熟悉的极限,即10lim(1)tt t e→+=.所以 012sin cos 1lim0lim(2sin cos )x x x xx x x x e→+-→+=,而 002sin cos 12cos sin lim lim 21x x x x x x x →→+--=洛,所以 012sin cos 1lim2lim(2sin cos )x x x xxx x x ee →+-→+==.(4)【解析】这是个函数的参数方程,22111221dy dy dt t dx t dx tdt t+===+,2222321111211()()()2222(2)41d y d d dt d t dx t dx dx t dt t dx dt t t tdt t -+==⋅=⋅=⋅=-+.【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'='. (5)【解析】利用定积分的分部积分法求解定积分,111122220000111(2)(2)(2)(2)222x f x dx x df x x f x f x dx '''''⎡⎤==⋅-⎣⎦⎰⎰⎰分部法[]1011(2)0(2)2f xf x dx''=⋅--⎰1011(2)(2)22f xdf x '=-⎰()1100111(2)(2)(2)222f xf x f x dx ⎡⎤'=--⎢⎥⎣⎦⎰1111(2)(2)(2)222f f f x dx '=-+⎰,令2t x =,则11,22x t dx dt ==, 所以 1201(2)()2f x dx f t dt =⎰⎰.把1(2),(2)02f f '==及2()1f x dx =⎰代入上式,得1120111(2)(2)(2)(2)222xf x dx f f f x dx '''=-+⎰⎰21111(2)(2)()2222f f f t dt '=-+⋅⎰1111101022222=⋅-⋅+⋅⋅=.三、选择题(每小题3分,满分18分.) (1)【答案】(A)【解析】函数1sin y x x=只有间断点0x =. 001lim lim sin x x y x x ++→→=,其中1sin x是有界函数.当0x +→时,x 为无穷小,无穷小量和一个有界函数的乘积仍然是无穷小,所以1lim lim sin 0x x y x x++→→==,故函数没有铅直渐近线.01sin1sin lim limlim 11x x x t x y t x t x+→+∞→+∞→=== ,所以1y =为函数的水平渐近线,所以答案为(A). 【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. (2)【答案】(B)【解析】判定方程()0f x =实根的个数,其实就是判定函数()y f x =与x 有几个交点,即对函数图形的描绘的简单应用,令 53()234f x xax bx c=+++,则 42()563f x x ax b'=++.令 2t x =,则422()563563()f x x ax b t at b f t ''=++=++=,其判别式22(6)45312(35)0a b a b ∆=-⋅⋅=-<,所以 2()563f t tat b'=++无实根,即()0f t '>.所以 53()234f x xax bx c=+++在(,)x ∈-∞+∞是严格的单调递增函数.又 53lim ()lim (234)x x f x x ax bx c →-∞→-∞=+++=-∞53lim ()lim (234)x x f x xax bx c →+∞→+∞=+++=+∞所以利用连续函数的介值定理可知,在(,)-∞+∞内至少存在一点0(,)x ∈-∞+∞使得0()0f x =,又因为()y f x =是严格的单调函数,故0x 是唯一的.故()0f x =有唯一实根,应选(B). (3)【答案】(C)【解析】如图cos ()22y x x ππ=-≤≤的图像,则当cos y x=绕x 轴旋转一周,在x 处取微增dx ,则微柱体的体积2cosdV xdxπ=,所以体积V 有222cos V xdxπππ-=⎰222222cos 21cos 22242x dx xd x dxπππππππππ---+==+⎰⎰⎰[][]22222sin 20()422222x x ππππππππππ--=-+=++=.因此选(C). (4)【答案】(D)【解析】题中给出的条件中,除了一处极值点外均未指明函数其它性质,为了判定的方便,可以举出反例而排除.若取2()()()f x g x x a ==--,两者都在x a =处取得极大值0, 而4()()()()F x f x g x x a ==-在x a =处取得极小值,所以(A)、(C)都不正确.若取2()()1()f x g x x a ==--,两者都在x a =处取得极大值1, 而22()()()1()F x f x g x x a ⎡⎤==--⎣⎦在x a =处取得极大值1,所以(B)也不正确,从而选(D). (5)【答案】(B)【解析】微分方程1xy y e ''-=+所对应的齐次微分方程的特征方程为210r-=,它的两个根是121,1r r==-.而形如xy y e ''-=必有特解1xY x ae =⋅；1y y ''-=必有特解1Y b=.由叠加得原方程必有特解xY x ae b=⋅+,应选(B).(6)【答案】(D)【解析】利用导数的概念判定()f x 在x a =处可导的充分条件.(A)等价于0()()lim t f a t f a t →++-存在,所以只能保证函数在x a =右导数存在；(B)、(C)显然是()f x 在x a =处可导的必要条件,而非充分条件, 如 1cos ,00,0x y x x ⎧≠⎪=⎨⎪=⎩在0x =处不连续,因而不可导,但是0001111cos(0)cos(0)cos cos()()lim lim lim 0222h h h f a h f a h h h h h h h h→→→+---+--===,0001111cos()cos(0)cos cos(2)()2222lim lim lim 0h h h f a h f a h h h h h h h h→→→---+-+===均存在； (D)是充分的：00()()()()lim lim x h x h f a x f a f a f a h x h∆=-∆→→+∆---=∆存在0()()()lim h f a f a h f a h →--'⇒=存在,应选(D).四、(本题满分6分)【解析】所给方程为一阶线性非齐次微分方程,先写成标准形式211(1)xy y e x x'+-=,通解为 11(1)(1)21()dxdx x xxy e e e dx C x ---⎰⎰=+⎰211()()x x x x x x e e e dx C e C x x e x=+=+⎰.代入初始条件(1)0y =,得C e=-,所求解为()x xe y e e x=-.【相关知识点】一阶线性非齐次微分方程的标准形式为()()y p x y q x '+=,其通解公式为()()(())p x dxp x dxy e q x e dx C -⎰⎰=+⎰,其中C 为常数.五、(本题满分7分)【解析】先将原式进行等价变换,再求导,试着发现其中的规律, 0()sin ()()sin ()()xx xf x x x t f t dt x x f t dt tf t dt=--=-+⎰⎰⎰,所给方程是含有未知函数及其积分的方程,两边求导,得()cos ()()()cos ()xxf x x f t dt xf x xf x x f t dt'=--+=-⎰⎰,再求导,得()sin ()f x x f x ''=--,即 ()()sin f x f x x ''+=-,这是个简单的二阶常系数非齐次线性微分方程,对应的齐次方程的特征方程为210r+=,此特征方程的根为r i =±,而右边的sin x 可看作sin x e xαβ,0,1,i i αβαβ==±=±为特征根,因此非齐次方程有特解sin cos Y xa x xb x =+.代入方程并比较系数,得10,2a b ==,故cos 2xY x =,所以 12()cos sin cos 2xf x c x c x x=++.又因为(0)0,(0)1f f '==,所以1210,2c c ==,即1()sin cos 22xf x x x=+.六、(本题满分7分)【解析】方法一：判定方程()0f x =等价于判定函数()y f x =与x 的交点个数. 令 0()ln 1cos 2x f x x xdxe π=-+-⎰,其中01cos 2xdxπ-⎰是定积分,为常数,且被积函数1cos2x-在(0,)π非负,故1cos 20xdx π->⎰,为简化计算,令01cos 20xdx k π-=>⎰,即()ln xf x x ke=-+,则其导数11()f x x e'=-,令()0f x '=解得唯一驻点x e =, 即()0,0()0,f x x ef x e x '><<⎧⎨'<<<+∞⎩,所以,x e =是最大点,最大值为()ln 0e f e e k k e=-+=>.又因为00lim ()lim (ln )lim ()lim (ln )x x x x x f x x k e x f x x k e ++→→→+∞→+∞⎧=-+=-∞⎪⎪⎨⎪=-+=-∞⎪⎩,由连续函数的介值定理知在(0,)e 与(,)e +∞各有且仅有一个零点(不相同),故方程0ln 1cos 2xx xdxe π=--⎰在(0,)+∞有且仅有两个不同实根. 方法二：201cos 2sin xdx xdxππ-=⎰⎰,因为当0x π≤≤时,sin 0x ≥,所以]2002sin 2sin 2cos 220xdx xdx x πππ==-=>⎰.其它同方法一.七、(本大题满分11分)【解析】函数21x y x +=的定义域为()(),00,-∞+∞U ,将函数化简为211,y x x=+ 则 32243321126216(1),(2)y y x xx x x x x x '''=--=--=+=+.令0y '=,得2x =-,即2212(1)0,(2,0),12(1)0,(,2)(0,),y x x x y x x x⎧'=-->∈-⎪⎪⎨⎪'=--<∈-∞-+∞⎪⎩U 故2x =-为极小值点.令0y ''=,得3x =-,即3316(2)0,(3,0)(0,),16(2)0,(,3)y x x x y x x x⎧''=+>∈-+∞⎪⎪⎨⎪''=+<∈-∞-⎪⎩U 为凹，，为凸，y ''在3x =-处左右变号,所以23,(3)9x y =--=-为函数的拐点.又 211lim lim(),x x y x x→→=+=∞故0x =是函数的铅直渐近线；211lim lim()0,x x y x x→∞→∞=+=故0y =是函数的水平渐近线.填写表格如下： 单调减少区间 (,2)(0,)-∞-+∞U单调增加区间 (2,0)- 极值点 2x =- 极值 14y =-凹区间 (3,0)(0,)-+∞U 凸区间 (,3)-∞- 拐点 2(3,)9-- 渐近线 0,0x y ==Born to win八、(本题满分10分)【解析】由题知曲线过点(0,0),得0c =,即2y ax bx =+. 如图所示,从x x dx →+的面积dS ydx =,所以 11123200011()32S ydx ax bx dx ax bx ⎡⎤==+=+⎢⎥⎣⎦⎰⎰32a b =+, 由题知 1323a b +=,即223a b -=. 当2y ax bx =+绕x 轴旋转一周,则从x x dx →+的体积2dV y dx π=,所以旋转体积 1254232211222000()()523523a x abx b x a ab b V y dx ax bx dx ππππ⎡⎤==+=++=++⎢⎥⎣⎦⎰⎰,b 用a 代入消去b ,得224(1)(1)5273a a a a V π⎡⎤--=++⎢⎥⎣⎦,这是个含有a 的函数,两边对a 求导得4(1)275dV a da π=+,令其等于0得唯一驻点54a =-,dV da在该处由负变正,此点为极小值点,故体积最小, 这时32b =,故所求函数225342y ax bxc x x =++=-+.。
1989考研英语一真题(含答案解析)
1989年全国硕士研究生入学统一考试英语试题Section I Close TestFor each numbered blank in the following passage there are four choices labeled ［A］, ［B］, ［C］and ［D］. Choose the best one and put your choice in the ANSWER SHEET. Read the whole passage before making your choice. (10 points)①One day drought may be a thing of the past at least in coastal cities.②V ast areas of desert throughout the world may for the first time 1 and provide millions of hectares of land where now nothing grows.③By the end of this century this may not be mere 2 .④Scientists are already looking into the possibility of using some of the available ice in the Arctic and Antarctic.⑤In these regions there are vast ice-caps formed by snow that has fallen over the past 50,000 years.⑥Layer 3 layer of deep snow means that, when melted, the snow water would be pure, not salty as sea-ice would be.⑦There is so much 4 pure water here that it would need only a fraction of it to turn much of the desert or poorly irrigated parts of the world into rich farmland.⑧And what useful packages it would come in!⑨It should be possible to cut off a bit of ice and transport it!⑩Alternatively perhaps a passing iceberg could be 5 . ○11They are always breaking away from the main caps and floating around, pushed by currents, until they eventually melt and are wasted.○12Many icebergs are, of course, far too small to be towed 6 distance, and would melt before they reached a country that needed them anywhere. ○13It would be necessary to locate one that was 7 and that was big enough to provide a good supply of ice when it reached us. ○14Engineers think that an iceberg up to seven miles long and one and a half miles wide could be transported if the tug pulling it was as big as a supertanker! ○15Even then they would cover only twenty miles every day. ○16However, 8 the iceberg was at its destination, more that 7,000 million cubic metres of water could be taken from it! ○17That would probably be more than enough for any medium-sized city even in the hottest summer! ○18But no doubt a use could be found for it. ○199 , scientist say, there would not be too much wastage in such a journey. ○20The larger the iceberg, the slower it melts, even if it is towed through the tropics. ○21This is because when the sun has a bigger area to warm 10 , less heat actually gets into the iceberg. ○22The vast frozen centre would be unaffected. [394 words]1. ［A］come to life ［B］come into existence［C］come into activity ［D］come round2. ［A］speculation ［B］imagination ［C］computation ［D］expectation3. ［A］above ［B］of ［C］upon ［D］over4. ［A］essential ［B］potential ［C］claimable ［D］obtainable5. ［A］seized ［B］snatched ［C］grabbed ［D］captured6. ［A］much ［B］any ［C］some ［D］certain7. ［A］manageable ［B］manipulative ［C］operable ［D］controllable8. ［A］after ［B］while ［C］since ［D］once9. ［A］Apparently ［B］Noticeably ［C］Distinctly ［D］Notably10.［A］round ［B］over ［C］up ［D］throughSection II Reading ComprehensionEach of the two passages below is followed by five questions. For each question there are four answers. Read the passages carefully and choose the best answer to each of the questions. Put your choice in the brackets on the left. (10 points)Text 1A scientist once said: “I have concluded that the earth is being visited by intelligently co ntrolled vehicles from outer space.”If we take this as a reasonable explanation for UFOs (unidentified flying objects), questions immediately come up.“Why don’t they get in touch with us, then? Why don’t they land right on the White House lawn and decla re themselves?” people asked.In reply, scientists say that, while this may be what we want, it may not necessarily be what they want.“The most likely explanation, it seems to me,” said Dr. Mead, “is that they are simply watching what we are up to -- tha t responsible society outside our solar system is keeping an eye on us to see that we don’t set in motion a chain reaction that might have unexpected effects for outside our solar system.”Opinions from other scientists might go like this: “Why should they want to get in touch with us? We may feel we’re more important than we really are! They may want to observe us only and not interfere with the development of our civilization. They may not care if we see them but they also may not care to say ‘hello’.”①Some scientists have also suggested that Earth is a kind of zoo or wildlife reserve. ②Just as we set aside wilderness areas and wildlife reserves to allow animals and growing things to develop naturally while we observe them, so perhaps Earth was set aside ages ago for the same purpose.①Are we being observed by intelligent beings from other civilizations in the universe? ②Are they watching our progress in space travel? ③Do we live in a gigantic “zoo” observed by our “keepers,” but having no communication with them?①Never before in our history have we had to confront ideas like these. ②The simple fact is that we, who have always regarded ourselves as supreme in the universe, may not be so. ③Now we have to recognize that, among the stars in the heavens, there may very well be worlds inhabited by beings who are to us as we are to ants.11. People who ask the question “Why don’tthey get in touch with us... and declarethemselves?” think that ________.[A] there are no such things as UFOs[B] UFOs are visitors from solar system[C] there’s no reason for UFOs sooner or later[D] we are bound to see UFOs sooner or later12. According to Dr. Mead, the attitude ofbeings from outer space toward us is one of________.[A] unfriendliness[B] suspicion[C] superiority[D] hostility13. The tone of the writer is that of ________.[A] doubt[B] warning[C] indifference[D] criticismText 2①The use of the motor is becoming more and more widespread in the twentieth century; as an increasing number of countries develop both technically and economically, so a larger proportion of the world’s population is able to buy and use a car. ②Possessing a car gives a much greater degree of mobility, enabling the driver to move around freely. ③The owner of a car is no longer forced to rely on public transport and is, therefore, not compelled to work locally. ④He can choose from different jobs and probably changes his work more frequently as he is not restricted to a choice within a small radius. ⑤Travelling to work by car is also more comfortable than having to use public transport; the driver can adjust the heating in winter and the air conditioning in the summer to suit his own needs and preference. ⑥There is no irritation caused by waiting for trains, buses or underground trains, standing in long patient queues, or sitting on windy platforms, for as long as half an hour sometimes. ⑦With the building of good, fast motorways long distances can be covered rapidly and pleasantly. ⑧For the first time in this century also, many people are now able to enjoy their leisure time to the full by making trips to the country or seaside at the weekends, instead of being confined to their immediate neighbourhood. ⑨This feeling of independence, and the freedom to go where you please, is perhaps the greatest advantage of the car.①When considering the drawbacks, perhaps pollution is of prime importance. ②As more and more cars are produced and used, so the emission from their exhaust-pipes contains an ever larger volume of poisonous gas. ③Some of the contents of this gas, such as lead, not only pollute the atmosphere but cause actual harm to the health of people. ④Many of the minor illnesses of modern industrial society, headaches, tiredness, and stomach upsets are thought to arise from breathing polluted air; doctors’ surgeries are full of people suffering from illnesses caused by pollution. ⑤It is also becoming increasingly difficult to deal with the problem of traffic in towns; most of the important cities of the world suffer from traffic congestion. ⑥In fact any advantage gained in comfort is often cancelled out in city driving by the frustration caused by traffic jams: endless queues of cars crawling one after another through all the main streets. ⑦As an increasing number of traffic regulation schemes are devised, the poor bewildered driver finds himself diverted and forced into one-way systems which cause even greater delays than the traffic jams they are supposed to prevent. ⑧The mounting cost of petrol and the increased license fees and road tax all add to the driver’s worries. ⑨In fact, he must sometimes wonder if the motor car is such a blessing and not just a menace.14. More and more people can afford to buyand use cars because ________.[A] an increasing number of cars are beingproduced[B] the cost of cars is getting cheaper with thedevelopment of technology[C] lots of countries have become moredeveloped[D] the use of cars has proved to be moreeconomical15. The advantages of having a car are bestexperienced in the driver’s ________.[A] freedom in choosing his job[B] comfort during the travels[C] enjoyment of his leisure time[D] feeling of self-reliance16. What is considered by the writer as thegreatest menace to the people caused by thewidespread use of motor cars?[A] air pollution[B] traffic jams[C] fatal diseases[D] high costText 3①Manners nowadays in metropolitan cities like London are practically non-existent. ②It is nothing for a big, strong schoolboy to elbow an elderly woman aside in the dash for the last remaining seat on the tube or bus, much less stand up and offer his seat to her, as he ought. ③In fact, it is saddening to note that if a man does offer his seat to an older woman, it is nearly always a Continental man or one from the older generation.①This question of giving up seats in public transport is much argued about by young men, who say that, since women have claimed equality, they no longer deserve to be treated with courtesy and that those who go out to work should take their turn in the rat race like anyone else. ②Women have never claimed to be physically as strong as men. ③Even if it is not agreed, however, that young men should stand up for younger women, the fact remains that courtesy should be shown to the old, the sick and the burdened. ④Are we really so lost to all ideals of unselfishness that we can sit there indifferently reading the paper or a book, saying to ourselves “First come, first served,” while a grey-haired woman, a mother with a young child or a cripple stands? ⑤Yet this is all too often seen.①Conditions in travel are really very hard on everyone, we know, but hardship is surely no excuse. ②Sometimes one wonders what would have been the behaviour of these stout young men in a packed refugee train or a train on its way to a prison-camp during the War. ③Would they have considered it only right and their proper due to keep the best places for themselves then?①Older people, tired and irritable from a day’s work, are not angels, eit her —far from it. ②Many a brisk argument or an insulting quarrel breaks out as the weary queues push and shove each other to get on buses and tubes. ③One cannot commend this, of course, but one does feel there is just a little more excuse.①If cities are to remain pleasant places to live in at all, however, it seems imperative, not only thatcommunications in transport should be improved, but also that communication between human beings should be kept smooth and polite. ②All over cities, it seems that people are too tired and too rushed to be polite. ③Shop assistants won’t bother to assist, taxi drivers growl at each other as they dash dangerously round corners, bus conductor pull the bell before their desperate passengers have had time to get on or off the bus, and so on and so on.④It seems to us that it is up to the young and strong to do their small part to stop such deterioration.17. From what you have read, would youexpect manners to improve among people________?[A] who are physically weak or crippled[B] who once lived in a prison-camp during theWar[C] who live in big modern cities[D] who live only in metropolitan cities18. What is the writer’s opinion concerningcourteous manners towards women?[A] Now that women have claimed equality, theyno longer need to be treated differently from men.[B] It is generally considered old-fashioned foryoung men to give up their seats to young women.[C] “Lady First” should be universally practiced.[D] Special consideration ought to be shown them.19. According to the author communicationbetween human beings would be smoother if________.[A] people were more considerate towards eachother[B] people were not so tired and irritable[C] women were treated with more courtesy[D] public transport could be improved20. What is the possible meaning of the word“deterioration” in the last paragraph?[A] worsening of general situation[B] lowering of moral standards[C] declining of physical constitution[D] spreading of evil conductSection III English-Chinese TranslationTranslate the following passage into Chinese. Only the underlined sentences are to be translated. (20 points)When Jane Matheson started work at Advanced Electronics Inc. 12 years ago, (21) she laboured over a microscope, hand-welding tiny electronic computers and turned out 18 per hour. Now she tends the computerized machinery that turns out high capacity memory chips at the rate of 2,600 per hour. Production is up, profits are up, her income is up and Mrs. Matheson says the work is far less strain on her eyes.But the most significant effect of the changes at AEI was felt by the workers who are no longer there. Before the new computerized equipment was introduced, there were 940 workers at the plant. Now there are 121. (22) A plant follow-up survey showed that one year after the layoffs only 38% of the released workers found new employment at the same or better wages. Nearly half finally settled for lower pay and more than 13% are still out of work. The AEI example is only one of hundreds around the country which forge intelligently ahead into the latest technology, but leave the majority of their workers behind.(23) Its beginnings obscured by unemployment caused by the world economic slow-down, the new technological unemployment may emerge as the great socio-economic challenge of the end of the 20th century. One corporation economist says the growth of “machine job replacement” has been with us since the beginning of the industrial revolution, but never at the pace it is now. The human costs will be astonishing. (24) “It’s humiliating to be done out of your job by a machine and there is no way to fight back, but it is the effort to find a new job that really hurts.” Some workers, like Jane Matheson, are retrained to handle the new equipment, but often a whole new set of skills is required and that means a new, and invariably smaller set of workers. (25) The old workers, trapped by their limited skills, often never regain their old status and employment. Many drift into marginal areas. They feel no pride in their new work. They get badly paid for it and they feel miserable, but still they are luckier than those who never find it.(26) The social costs go far beyond the welfare and unemployment payments made by the government. Unemployment increases the chances of divorce, child abuse, and alcoholism, a new federal survey shows. Some experts say the problem is only temporary... that new technology will eventually create as many jobs as it destroys.(27) But futurologist Hymen Seymour says the astonishing efficiency of the new technology means there will be a simple and direct net reduction in the amount of human labor that needs to be done. “We should treat this as an opportunity to give people more leisure. It may not be easy, but society will have to reach a new unanimity on the division and distribution of labor,” Seymour says. He predicts most people will work only six-hour days and four-day weeks by the end of the century. But the concern of the unemployed is for now. (28) Federally funded training and free back-to-school programs for laid-off workers are under way, but few experts believe they will be able to keep up with the pace of the new technology. For the next few years, for a substantial portion of the workforce, times are going to be very tough indeed.。
1989考研数一真题及解析
1989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) 已知(3)2f '=,则 0(3)(3)lim2h f h f h→--=_______.(2) 设()f x 是连续函数,且1()2()f x x f t dt =+⎰,则()f x =_______.(3) 设平面曲线L 为下半圆周21,y x =--则曲线积分22()Lx y ds +=⎰_______.(4) 向量场22(,,)ln(1)zu x y z xy i ye j x z k =+++在点(1,1,0)P 处的散度divu =_______.(5) 设矩阵300140003A ⎛⎫ ⎪= ⎪ ⎪⎝⎭, 100010001E ⎛⎫ ⎪= ⎪ ⎪⎝⎭,则逆矩阵1(2)A E --=_______.二、选择题(本题共5个小题,每小题3分,满分15分.) (1) 当0x >时,曲线1siny x x= ( ) (A) 有且仅有水平渐近线 (B) 有且仅有铅直渐近线(C) 既有水平渐近线,也有铅直渐近线 (D) 既无水平渐近线,也无铅直渐近线(2) 已知曲面224z x y =--上点P 处的切平面平行于平面2210x y z ++-=,则点P 的坐标是 ( ) (A) (1,-1,2) (B) (-1,1,2) (C) (1,1,2) (D) (-1,-1,2)(3) 设线性无关的函数1y 、2y 、3y 都是二阶非齐次线性方程()()()y p x y q x y f x '''++=的解,1C 、2C 是任意常数,则该非齐次方程的通解是 ( ) (A) 11223C y C y y ++ (B) 1122123()C y C y C C y +-+ (C) 1122123(1)C y C y C C y +--- (D) 1122123(1)C y C y C C y ++-- (4) 设函数2(),01,f x x x =≤<而1()sin ,,nn S x bn x x π∞==-∞<<+∞∑其中102()sin ,1,2,3,n b f x n xdx n π==⎰…,则1()2S -等于 ( )(A) 12-(B) 14- (C) 14 (D) 12(5) 设A 是n 阶矩阵,且A 的行列式||0A =,则A 中 ( )(A) 必有一列元素全为0(B) 必有两列元素对应成比例(C) 必有一列向量是其余列向量的线性组合 (D) 任一列向量是其余列向量的线性组合三、(本题满分15分,每小题5分.)(1) 设(2)(,)z f x y g x xy =-+,其中函数()f t 二阶可导,(,)g u v 具有连续的二阶偏导数,求2z x y∂∂∂. (2) 设曲线积分2()Cxy dx y x dy ϕ+⎰与路径无关,其中()x ϕ具有连续的导数,且(0)0ϕ=,计算(1,1)2(0,0)()xy dx y x dy ϕ+⎰的值.(3) 计算三重积分()x z dV Ω+⎰⎰⎰,其中Ω是由曲面22z x y =+与221z x y =--所围成的区域.四、(本题满分6分.)将函数1()arctan 1xf x x+=-展为x 的幂级数.五、(本题满分7分.)设0()sin ()()xf x x x t f t dt =--⎰,其中f 为连续函数,求()f x .六、(本题满分7分.)证明方程0ln 1cos 2x x xdx e π=--⎰在区间(0,+∞)内有且仅有两个不同实根.七、(本题满分6分.)问λ为何值时,线性方程组131231234226423x x x x x x x x λλλ+ =⎧⎪++=+⎨⎪++=+⎩ 有解,并求出解的一般形式.八、(本题满分8分.)假设λ为n 阶可逆矩阵A 的一个特征值,证明： (1)1λ为1A -的特征值； (2)Aλ为A 的伴随矩阵A *的特征值.九、(本题满分9分.)设半径为R 的球面∑的球心在定球面2222(0)x y z a a ++=>上,问当R 为何值时,球面∑在定球面内部的那部分的面积最大？十、填空题(本题满分6分,每小题2分.)(1) 已知随机事件A 的概率()P A =0.5,随机事件B 的概率()P B =0.6及条件概率()P B A |=0.8,则和事件A B 的概率()P A B =_______.(2) 甲、乙两人独立地对同一目标射击一次,其命中率分别为0.6和0.5.现已知目标被命中,则它是甲射中的概率为_______. (3) 若随机变量ξ在(1,6)上服从均匀分布,则方程210x x ξ++=有实根的概率是______.十一、(本题满分6分.)设随机变量X 与Y 独立,且X 服从均值为1、标准差(均方差)2,而Y 服从标准正态分布.试求随机变量23Z X Y =-+的概率密度函数.1989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】1- 【解析】原式=01(3)(3)1lim (3)122h f h f f h -→--'-=-=--. (2)【答案】1x -【解析】由定积分的性质可知,1()f t dt ⎰和变量没有关系,且()f x 是连续函数,故1()f t dt ⎰为一常数,为简化计算和防止混淆,令10()f t dt a =⎰,则有恒等式()2f x x a =+,两边0到1积分得11()(2)f x dx x a dx =+⎰⎰,即 []111112000001(2)222a x a dx xdx a dx x a x ⎡⎤=+=+=+⎢⎥⎣⎦⎰⎰⎰122a =+,解之得 12a =-,因此()21f x x a x =+=-. (3)【答案】π【解析】方法一：L 的方程又可写成221(0)x y y +=≤,被积分函数在L 上取值,于是原积分=1Lds π=⎰(半径为1的的半圆周长).方法二：写出L 的参数方程,cos sin x ty t=⎧⎨=⎩,(0)t π-≤≤ 则00222222()(cos sin )(sin )cos 1Lx y ds t t t tdt dt πππ--+=+-+=⋅=⎰⎰⎰.(4)【答案】2【解析】直接用散度公式22[()()(ln(1))]z PP divuxy ye x z x y z∂∂∂=+++∂∂∂ 220(1,1,0)22220()10112110z zy e x e z =++⋅=++⋅=+=++.(5)【答案】10011022001⎛⎫ ⎪ ⎪-⎪ ⎪⎝⎭【解析】由于3002001002140020120003002001A E ⎛⎫⎛⎫⎛⎫⎪ ⎪ ⎪-=-= ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭,为求矩阵的逆可有多种办法,可用伴随,可用初等行变换,也可用分块求逆.方法一：如果对(2)A E E -作初等行变换,则由1(2)((2))A E E E A E --→-可以直接得出1(2)A E --.本题中,第一行乘以()1-加到第二行上;再第二行乘以12,有 10010010010010010011120010020110010022001001001001001001⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪⎪ → -→ - ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭, 从而知 110011(2)022001A E -⎛⎫⎪ ⎪-=-⎪ ⎪⎝⎭. 方法二：对于2阶矩阵的伴随矩阵有规律：a b A c d ⎛⎫=⎪⎝⎭,则求A 的伴随矩阵 *a b d b A c d c a *-⎛⎫⎛⎫== ⎪ ⎪-⎝⎭⎝⎭.如果0A ≠,这样111a b d b d b c d c a c a A ad bc ---⎛⎫⎛⎫⎛⎫== ⎪ ⎪⎪---⎝⎭⎝⎭⎝⎭. 再利用分块矩阵求逆的法则：1110000A AB B ---⎛⎫⎛⎫=⎪ ⎪⎝⎭⎝⎭,本题亦可很容易求出110011(2)022001A E -⎛⎫⎪ ⎪-=-⎪ ⎪⎝⎭.二、选择题(本题共5个小题,每小题3分,满分15分.) (1)【答案】(A)【解析】函数1siny x x =只有间断点0x =. 001lim lim sin x x y x x ++→→=,其中1sin x是有界函数,而当0x +→时,x 为无穷小,而无穷小量和一个有界函数的乘积仍然是无穷小, 所以 001lim lim sin 0x x y x x++→→==,故函数没有铅直渐近线.01sin1sin lim limlim 11x x x t x y t x tx+→+∞→+∞→===令, 所以1y =为函数的水平渐近线,所以答案为(A).【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.(2)【答案】(C)【解析】题设为求曲面:(,,)0S F x y z =(其中22(,,)4F x y z z x y =++-)上点P 使S 在该点处的法向量n 与平面2210x y z ++-=的法向量{}02,2,1n =平行.S 在(,,)P x y z 处的法向量{},,2,2,1F F F n x y x y z ⎧⎫∂∂∂==⎨⎬∂∂∂⎩⎭,若0//,n n 则0,n n λλ=为常数,即22,22,1x y λλλ===.即1,1x y ==. 又点(,,)P x y z S ∈,所以2222(,)(1,1)44112x y z x y ==--=--=,故求得(1,1,2)P .因此应选(C).(3)【答案】(D)【解析】由二阶常系数非齐次微分方程解的结构定理可知,1323,y y y y --为方程对应齐次方程的特解,所以方程()()()y p x y q x y f x '''++=的通解为1132233()()y C y y C y y y =-+-+,即1122123(1)y C y C y C C y =++--,故应选D. (4)【答案】(B)【解析】()S x 是函数()f x 先作奇延拓后再作周期为2的周期延拓后的函数的傅式级数的和函数,由于()S x 是奇函数,于是11()()22S S -=-.当12x =时,()f x 连续,由傅式级数的收敛性定理,21111()()()2224S f ===.因此, 11()24S -=-.应选(B).(5)【答案】(C)【解析】本题考查||0A =的充分必要条件,而选项(A) 、(B)、(D)都是充分条件,并不必要.因为对矩阵A 来说,行和列具有等价性,所以单说列或者单说行满足什么条件就构成了||0A =的必要条件,但是不具有任意性,只需要存在一列向量是其余列向量的线性组合.以3阶矩阵为例,若 112123134A ⎛⎫⎪= ⎪ ⎪⎝⎭,条件(A)必有一列元素全为0,(B)必有两列元素对应成比例均不成立,但有||0A =,所以(A)、 (B)不满足题意,不可选.若123124125A ⎛⎫⎪= ⎪ ⎪⎝⎭,则||0A =,但第三列并不是其余两列的线性组合,可见(D)不正确.这样用排除法可知应选(C).三、(本题满分15分,每小题5分.)(1)【解析】由于混合偏导数在连续条件下与求导次序无关,可以先求zx∂∂,也可以先求z y ∂∂.方法一：先求zx∂∂,由复合函数求导法,1212(2)()()2z f x y g x g xy f g yg x x x x∂∂∂∂''''''=-++=++∂∂∂∂, 再对y 求偏导,得212(2)2(2)z f g yg f x y x y y y∂∂∂'''''=++=-∂∂∂∂ 111222122()()()()g x g xy g yg x yg xy y y y y ⎡⎤⎡⎤∂∂∂∂'''''''''+++++⎢⎥⎢⎥∂∂∂∂⎣⎦⎣⎦111222122200f g xg g yg xyg '''''''''''=-+⋅+++⋅+ 212222f xg g xyg '''''''=-+++. 方法二：先求zy∂∂, 122(2)()()z f x y g x g xy f xg y y y y∂∂∂∂'''''=-++=-+∂∂∂∂, 再对x 求偏导数,得222()z z f xg x y y x x∂∂∂''==-+∂∂∂∂∂ 22122(2)()()f x y g xg x xg xy x x x∂∂∂'''''''=--+++∂∂∂221222f g xg xyg '''''''=-+++. 【相关知识点】复合函数求导法则：若(,)u u x y =和(,)v v x y =在点(,)x y 处偏导数存在,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 u x y v x y =在点(,)x y 处的偏导数存在,且,z f u f v z f u f v x u x v x y u y v y∂∂∂∂∂∂∂∂∂∂=+=+∂∂∂∂∂∂∂∂∂∂. (2)【解析】方法一：先求出()x ϕ,再求曲线积分.设(,),(,)P x y Q x y 有连续偏导数,在所给的单连通区域D 上,LPdx Qdy +⎰与路径无关,则在D 上有Q P x y∂∂=∂∂,所以()2,y x xy ϕ'=即2()2,()x x x x C ϕϕ'==+.由(0)ϕ=0,得0C =,即2()x x ϕ=,因此(1,1)(1,1)(1,1)2222222(0,0)(0,0)(0,0)1()2I xy dx y x dy xy dx yx dy y dx x dy ϕ=+=+=+⎰⎰⎰ (1,1)(0,0)(1,1)2222(0,0)111()()222d x y x y ===⎰. 或取特殊路径如图：11222001LI xy dx yx dy dx y dy =+=+⎰⎰⎰1201122y ⎡⎤==⎢⎥⎣⎦. 方法二：不必求出()x ϕ,选取特殊的路径,取积分路径如图,则(1,1)2(0,0)()I xy dx y x dy ϕ=+⎰11011(0)022y dy xdx ϕ=+=+=⎰⎰. (3)【解析】利用三重积分的性质,Ω关于yz 平面对称,x 对x 为奇函数,所以0xdV Ω=⎰⎰⎰,即()x z dV zdV ΩΩ+=⎰⎰⎰⎰⎰⎰.Ω是由球心在原点半径为1的上半球面与顶点在原点、对称轴为z 轴、半顶角为4π的锥面所围成.故可选用球坐标变换,则020014πθπϕρΩ≤≤≤≤≤≤：，，,所以 2cos sin I zdV d d d ρϕρϕρϕθΩΩ==⋅⎰⎰⎰⎰⎰⎰ 2113344000001cos sin 2sin 22d d d d d πππθϕϕϕρρπϕϕρρ==⎰⎰⎰⎰⎰1440011cos 2248πππϕρ⎡⎤⎡⎤=-⋅=⎢⎥⎢⎥⎣⎦⎣⎦.四、(本题满分6分.)【解析】直接展开()f x 相对比较麻烦,可()f x '容易展开,2222211(1)(1)21()1(1)(1)(1)11()1x x f x x x x x x x--+⋅-'=⋅==+--++++-. 由2011(1)(1),(||1)1n nn n n t t t t t t∞==-+-+-+=-<+∑,令2t x =得242222111(1)(1),(1)11nnn n n x x x x x t x ∞===-+-+-+=-<++∑即 221()(1),(||1)1n n n f x x x x ∞='==-<+∑ 所以()()(0)xf x f u du f '=+⎰,22000010(1)arctan(1)104x x nnnn n n u du u du π∞∞==+=-+=+--∑∑⎰⎰ 210(1)421n nn x n π+∞==+-+∑,(||1)x <当1x =±时,式210(1)21n nn x n +∞=-+∑均收敛,而左端1()arctan 1xf x x +=-在1x =处无定义.因此 2101(1)()arctan,[1,1)1421n n n x f x x x x n π∞+=+-==+∈--+∑.五、(本题满分7分.)【解析】先将原式进行等价变换,再求导,试着发现其中的规律, 0()sin ()()sin ()()xx xf x x x t f t dt x x f t dt tf t dt =--=-+⎰⎰⎰,所给方程是含有未知函数及其积分的方程,两边求导,得()cos ()()()cos ()xxf x x f t dt xf x xf x x f t dt '=--+=-⎰⎰,再求导,得()sin ()f x x f x ''=--,即 ()()sin f x f x x ''+=-.这是个简单的二阶常系数非齐次线性微分方程,对应的齐次方程的特征方程为210r +=, 此特征方程的根为r i =±,而右边的sin x 可看作sin xe x αβ,i i αβ±=±为特征根,因此非齐次方程有特解sin cos Y xa x xb x =+.代入方程并比较系数,得10,2a b ==,故cos 2xY x =,所以 12()cos sin cos 2xf x c x c x x =++,又因为(0)0,(0)1f f '==,所以1210,2c c ==,即1()sin cos 22xf x x x =+.六、(本题满分7分.)【解析】方法一：判定方程()0f x =等价于判定函数()y f x =与x 的交点个数.令 0()ln 1cos 2x f x x xdx e π=-+-⎰,其中1cos 2xdx π-⎰是定积分,为常数,且被积函数1cos2x -在(0,)π非负,故1cos 20xdx π->⎰,为简化计算,令01cos 20xdx k π-=>⎰,即()ln xf x x k e=-+,则其导数11()f x x e'=-,令()0f x '=解得唯一驻点x e =, 即 ()0,0()0,f x x ef x e x '><<⎧⎨'<<<+∞⎩,所以x e =是最大点,最大值为()ln 0ef e e k k e=-+=>. 又因为00lim ()lim (ln )lim ()lim (ln )x x x x x f x x k ex f x x k e ++→→→+∞→+∞⎧=-+=-∞⎪⎪⎨⎪=-+=-∞⎪⎩,由连续函数的介值定理知在(0,)e 与(,)e +∞各有且仅有一个零点(不相同),故方程0ln 1cos 2x x xdx e π=--⎰在(0,)+∞有且仅有两个不同实根.方法二：201cos 2sin xdx xdx ππ-=⎰⎰,因为当0x π≤≤时,sin 0x ≥,所以]2002sin 2sin 2cos 220xdx xdx x πππ==-=>⎰,其它同方法一.七、(本题满分6分.)【解析】对方程组的增广矩阵作初等行变换.第一行分别乘以有()4-、()6-加到第二行和第三行上,再第二行乘以()1-加到第三行上, 有1011011014122012320123261423012430001λλλλλλλλλ⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪+→--+→--+ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪+--+-+⎝⎭⎝⎭⎝⎭. 由于方程组有解的充要条件是()()r A r A =,故仅当10λ-+=,即1λ=时,方程组有解.此时秩()()23r A r A n ==<=,符合定理的第二种情况,故方程组有无穷多解.由同解方程组 1323 1,21,x x x x +=⎧⎨-=-⎩令3,x t =解得原方程组的通解1231,21,,x t x t x t =-+⎧⎪=-⎨⎪=⎩ (其中t 为任意常数). 【相关知识点】1.非齐次线性方程组有解的判定定理：设A 是m n ⨯矩阵,线性方程组Ax b =有解的充分必要条件是系数矩阵的秩等于增广矩阵()A A b =的秩,即是()()r A r A =(或者说,b 可由A 的列向量12,,,n ααα线表出,亦等同于12,,,n ααα与12,,,,n b ααα是等价向量组)设A 是m n ⨯矩阵,线性方程组Ax b =,则（1） 有唯一解 ⇔ ()().r A r A n == （2） 有无穷多解 ⇔ ()().r A r A n =< （3） 无解 ⇔ ()1().r A r A +=⇔ b 不能由A 的列向量12,,,n ααα线表出.八、(本题满分8分.)【解析】(1)由λ为A 的特征值可知,存在非零向量α使A αλα=,两端左乘1A -,得1A αλα-=.因为0α≠,故0λ≠,于是有11A ααλ-=.按特征值定义知1λ是1A -的特征值.(2)由于逆矩阵的定义1||A A A *-=,据第(1)问有1||||A A A A ααααλλ**=⇒=,按特征值定义,即||A λ为伴随矩阵A *的特征值.【相关知识点】矩阵特征值与特征向量的定义：设A 是n 阶矩阵,若存在数λ及非零的n 维列向量X 使得AX X λ=成立,则称λ是矩阵A 的特征值,称非零向量X 是矩阵A 的特征向量.九、(本题满分9分.)【解析】由球的对称性,不妨设球面∑的球心是(0,0,)a , 于是∑的方程是2222()x y z a R ++-=.先求∑与球面2222x y z a ++=的交线Γ：2222222222(),22,x y z a R a R z a x y z a ⎧++-=-⎪⇒=⎨++=⎪⎩. 代入上式得Γ的方程 422224R x y R a+=-.它在平面xOy 上的投影曲线4222222,(02),40,R x y b b R R a az ⎧+==-<<⎪⎨⎪=⎩相应的在平面xOy 上围成区域设为xy D ,则球面∑在定球面内部的那部分面积22()1xyx y D S R z z dxdy ''=++⎰⎰.将∑的方程两边分别对,x y 求偏导得,z x z y x z a y z a∂∂=-=-∂-∂-, 所以 2222()11()()xyxyx y D D x y S R z z dxdy dxdy a z a z''=++=++--⎰⎰⎰⎰ 222221()()xyxyD D x y dxdy dxdy a z a z R x y =++=----⎰⎰⎰⎰.利用极坐标变换(02,0)b θπρ≤≤≤≤有222222()xybD S R dxdy d R x yR πθρρ=---⎰⎰⎰⎰极坐标变换2222200()2b R d R R πθρρ=---⎰⎰ 222202()2()b R R R R b R πρπ=--=--代入42224R b R a =-,化简得32()2R S R R aππ=-.这是一个关于R 的函数,求()S R 在(0,2)a 的最大值点,()S R 两边对R 求导,并令()0S R '=,得23()40R S R R a ππ'=-=,得43aR =. 且 4()0,034()0,23S R R a S R a R a ⎧'><<⎪⎪⎨⎪'<<<⎪⎩,故43aR =时()S R 取极大值,也是最大值. 因此,当43aR =时球面∑在定球面内部的那部分面积最大.十、填空题(本题满分6分,每小题2分.) (1)【解析】 方法一：()()()()P A B P A P B P AB =+-()()()(|)0.7P A P B P A P B A =+-=. 方法二：()()()P AB P B P AB =+()()(|)0.60.50.20.7P B P A P B A =+=+⨯=.(2)【解析】设事件A =“甲射中”,B =“乙射中”,依题意,()0.6P A =,()0.5P B =,A 与B 相互独立,()()()0.60.50.3P AB P A P B =⋅=⨯=.因此,有 ()()()()P AB P A P B P AB =+-0.60.50.30.8=+-=. (())()(|)0.75()()P A A B P A P A AB P A B P A B ===.(3)【解析】设事件A =“方程有实根”,而方程210x x ξ++=有实根的充要条件是其判别式240ξ∆=-≥,即{}{}22404A ξξ=-≥=≥.随机变量ξ在(1,6)上服从均匀分布,所以其分布函数为0, 1,1(), 16,611, 6.x x F x x x <⎧⎪-⎪=≤<⎨-⎪≥⎪⎩由分布函数的定义()()P x k F k ≤=,{}{}21210.20.8.P P ξξ≥=-<=-= 而{}20.P ξ≤-=所以由概率的可加性,有{}{}{}2()422P A P P ξξξ=≥=≥+≤-0.800.8=+=.【相关知识点】广义加法公式：()()()()P AB P A P B P AB =+-.条件概率：()(|)()P BA P B A P A =,所以()()(|)()P AB P BA P B A P A ==. 十一、(本题满分6分.)【解析】~(1,2)X N ,~(0,1)Y N ,由独立的正态变量X 与Y 的线性组合仍服从正态分布,且235,EZ EX EY =-+=44219DZ DX DY =+=⨯+=,得 ~(5,9)Z N .代入正态分布的概率密度公式,有Z 的概率密度函数为 2(5)18()32z Z f z π--=.【相关知识点】对于随机变量X 与Y 均服从正态分布,则X 与Y 的线性组合亦服从正态分布.若X 与Y 相互独立,由数学期望和方差的性质,有()()()E aX bY c aE X bE Y c ++=++, 22()()()D aX bY c a D X b D Y ++=+,其中,,a b c 为常数.。
北京大学1989年研究生入学考试试题
北京大学1989年研究生入学考试试题考试科目：综合考试考试时间：招生专业：中文系各专业研究方向：各研究方向一、填空（每格0.5分，共65分）1、汉代传习《诗经》的有、、、四家。
2、《楚辞》是西汉诗集，东汉章句。
3、1900年，敦煌莫高窟被打开，发现了大量珍贵文献和文物。
4、《古文辞类纂》是清人所编，选录到的古文辞赋，将文体分为类。
5、是我国古代第一部训诂书，也就是最早的一部词典。
6、和是宋人所修的两本研究中古音韵的要籍。
7、《文献通考》是代所撰。
8、《昭明文选》是梁编选，唐显庆年间有为作注。
9、古诗十九首大约产生于末年，其中有八首《玉台新咏》题为作，并不可信。
10、北朝乐府民歌在《乐府诗集》中主要收在里。
11、江淹的、是南朝抒情小赋中的名作。
12、诗歌声律的“八病”是提出的。
13、初唐声律的“八病”是提出的。
14、《秦中吟》共首，是唐代诗人所作。
15、“唐宋八大家”的名称来自的。
16．元好问《论诗绝句》说“曹刘坐啸虎生风，四海无人角两雄”。
“曹刘”指是、。
17、“南洪”“北孔”指清代剧作家、。
18、在清代“格调说”的代表人物是，“肌理说”的代表人物是。
19、《红楼梦》的版本，大体上可以分为和两个系统。
20、梁启超提倡的“诗界革命”、和运动，都是为了使文学成为变法维新，启迪民智的工具。
21、在十年代京派的大本营是（人名）主编的《大公报·文艺副刊》和（人名）主编的《文学杂志》。
22、巴金长篇小说《家》中的和老舍长篇小说《四世同堂》中的，都是封建家庭和旧礼教毒害下的悲剧典型。
23、郭沫若描写潜意识活动的小说有和。
24、“未名社四杰”指的韦素园、韦丛芜、和。
25、抗战时期国统区剧作家陈白尘和袁俊（张骏祥）分别写出了歌颂知识分子的正剧和。
26、郭沫若翻译的《少年维特之烦恼》出版后，一时间洋式的书信体，日记体等样式的作品大量涌现，其中有庐隐的、淦女士的。
27、小说《浮躁》的作者是；小说《大淖记事》的作者是。
历年考研数学一真题及答案
三、(本题共3小题,每小题5分,满分15分)
(1)设 其中函数 二阶可导 具有连续二阶偏导数,求
(2)设曲线积分 与路径无关,其中 具有连续的导数,且 计算
的值.
(3)计算三重积分 其中 是由曲面 与 所围成的区域.
四、(本题满分6分)
(1)过点 且与直线 垂直的平面方程是_____________.
(2)设 为非零常数,则 =_____________.
(3)设函数 ,则 =_____________.
(4)积分 的值等于_____________.
(5)已知向量组
则该向量组的秩是_____________.
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)
(3)与两直线
及 都平行且过原点的平面方程为_____________.
(4)设 为取正向的圆周 则曲线积分 = _____________.
(5)已知三维向量空间的基底为 则向量 在此基底下的坐标是_____________.
二、(本题满分8分)
求正的常数 与 使等式 成立.
三、(本题满分7分)
(1)设 、 为连续可微函数 求
(1)设 是连续函数,且 则 等于
(A) (B)
(C) (D)
(2)已知函数 具有任意阶导数,且 则当 为大于2的正整数时 的 阶导数 是
(A) (B)
(C) (D)
(3)设 为常数,则级数
(A)绝对收敛(B)条件收敛
(C)发散(D)收敛性与 的取值有关
(4)已知 在 的某个邻域内连续,且 则在点 处
考研数学三真题(1989年)
(D) A 中至少有一行(列)的元素全为 0
(4) 设 A 和 B 均为n n 矩阵,则必有 ( (A) A B A B ) (B) AB BA (D)
(C) AB BA
A B 1 A1 B 1
)
(5) 以 A 表示事件“甲种产品畅销,乙种产品滞销”,则其对立事件 A 为 ( (A) “甲种产品滞销,乙种产品畅销” (C) “甲种产品滞销” (B) “甲、乙两种产品均畅销”
x
4
五、(本题满分 9 分)
已知函数 0 x 1, x, f (x) 2 x, 1 x 2. 试计算下列各题： (1) S 0 f (x)e xdx; (4 分)
0 2
(2) S1 f (x 2)e xdx; (2 分)
2
4
(3) Sn 2 n
2n2
f (x 2n )e xdx (n 2, 3, ); (1 分) (4) S S n .(2 分)
n0
5
六、(本题满分 6 分)
假设函数 f (x) 在 [a, b] 上连续,在 (a, b) 内可导,且 f (x) 0 ,记 F (x) 证明在(a, b) 内, F (x) 0 . 1 x a
x
a
f (t)dt,
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设1 (1,1,1),2 (1, 2, 3),3 (1,3, t) . (1) 问当t 为何值时,向量组1,2,3 线性无关?(3 分) (2) 问当t 为何值时,向量组1,2,3 线性相关?(1 分) (3) 当向量组1,2,3线性相关时,将3表示为1和2 的线性组合.(2 分)
1989数学三真题答案解析(试卷四)
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(2) 问当 t 为何值时，向量组1,2 ,3 线性相关？（1 分）
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四、（本题满分 9 分）
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1989年考研数学一真题及答案解析
1989年全国硕士研究生入学统一考试真题试卷数学(一)试题一、填空题(本题共5个小题,每小题3分,满分15分.) (1) 已知(3)2f '=,则 0(3)(3)lim2h f h f h→--=_______.(2) 设()f x 是连续函数,且1()2()f x x f t dt =+⎰,则()f x =_______.(3) 设平面曲线L 为下半圆周y =则曲线积分22()Lx y ds +=⎰_______.(4) 向量场22(,,)ln(1)zu x y z xy i ye j x z k =+++在点(1,1,0)P 处的散度divu =_______.(5) 设矩阵300140003A ⎛⎫⎪= ⎪⎪⎝⎭, 100010001E ⎛⎫ ⎪= ⎪ ⎪⎝⎭,则逆矩阵1(2)A E --=_______.二、选择题(本题共5个小题,每小题3分,满分15分.) (1) 当0x >时,曲线1siny x x= ( ) (A) 有且仅有水平渐近线 (B) 有且仅有铅直渐近线(C) 既有水平渐近线,也有铅直渐近线 (D) 既无水平渐近线,也无铅直渐近线(2) 已知曲面224z x y =--上点P 处的切平面平行于平面2210x y z ++-=,则点P 的坐标是 ( ) (A) (1,-1,2) (B) (-1,1,2) (C) (1,1,2) (D) (-1,-1,2)(3) 设线性无关的函数1y 、2y 、3y 都是二阶非齐次线性方程()()()y p x y q x y f x '''++=的解,1C 、2C 是任意常数,则该非齐次方程的通解是 ( )(A) 11223C y C y y ++ (B) 1122123()C y C y C C y +-+ (C) 1122123(1)C y C y C C y +--- (D) 1122123(1)C y C y C C y ++--。
1989考研数学三【解析版】【无水印】
1989年全国硕士研究生入学统一考试数学三试题解析一、填空题(本题满分15分,每小题3分.) (1)【答案】1y x =+【解析】对函数2sin y x x =+两边对x 求导,得12cos y sin x x,'=+ 令2x π=得212sincos122x y .πππ='=+=所以该曲线在点122,ππ⎛⎫+ ⎪⎝⎭处的切线的斜率为1,所以 切线方程是122y x ,ππ⎛⎫-+=- ⎪⎝⎭即1y x =+为所求. (2)【答案】[1,1)-【解析】因系数1n n a a +==,从而1lim1,n n n n a a +→∞== 即幂级数的收敛半径1R =,当11x -<<时幂级数绝对收敛. 当1x =-时得交错级数n n ∞=(条件收敛)；当1x =时得正项级数0n ∞=(发散). 于是,幂级数的收敛域是[1,1)-. (3)【答案】1λ≠【解析】n 个方程n 个未知数的齐次方程组0Ax =有非零解的充分必要条件是0A =, 因为此时未知数的个数等于方程的个数,即A 为方阵时,用0A =判定比较方便.而 21110011010(1),111111A λλλλλ-==-=- 所以当0A ≠时1λ≠.所以此题应填:1λ≠. (4)【答案】1,12【解析】由于任何随机变量X 的分布函数()F x 是右连续函数,因此对任何x ,有()(0)F x F x =+.对于2x π=,有()sin,(0) 1.222F A A F πππ==+= 令 ()2F π=(0)2F π+,得到1A =,其中0(0)lim ()x F x F x +→+=.又 666P X P X ,πππ⎧⎫⎧⎫<=-<<⎨⎬⎨⎬⎩⎭⎩⎭因()F x 在6x π=处连续,连续函数在任何一个点上的概率为0,因此06P X .π⎧⎫==⎨⎬⎩⎭所以 666P X P X πππ⎧⎫⎧⎫<=-<≤⎨⎬⎨⎬⎩⎭⎩⎭66F F ππ⎛⎫⎛⎫=-- ⎪ ⎪⎝⎭⎝⎭162sin .π== (5)【答案】19【解析】由切比雪夫不等式2{}DXP X EX εε-≥≤,有221{3}(3)9P X σμσσ-≥≤=.二、选择题(本题满分15分,每小题3分.) (1)【答案】(B)【解析】由洛必达法则有()0002322ln23ln3lim lim lim ln2ln31x x x x x x x f x x x →→→+-+===+. 所以()f x 与x 是同阶但非等价无穷小量. (2)【答案】(C)【解析】由不定积分的概念和性质可知,()()()()df x dx f x dx f x .dx'==⎰⎰()()()f x dx df x f x C,'==+⎰⎰C 为常数.()()d f x dx f x dx.=⎰故应选(C).(3)【答案】(C)【解析】本题考查||0A =的充分必要条件,而选项(A)、(B)、(D)都是充分条件,并不必要.因为对矩阵A 来说,行和列具有等价性,所以单说列或者单说行满足什么条件就构成了||0A =的必要条件,但是不具有任意性,只需要存在一列向量是其余列向量的线性组合.以3阶矩阵为例,若 112123134A ⎛⎫ ⎪= ⎪ ⎪⎝⎭,条件(A)必有一列元素全为0,(B)必有两列元素对应成比例均不成立,但有||0A =,所以(A)、 (B)不满足题意,不可选.若123124125A ⎛⎫⎪= ⎪ ⎪⎝⎭,则||0A =,但第三列并不是其余两列的线性组合,可见(D)不正确.这样用排除法可知应选(C).(4)【答案】(C) 【解析】当行列式的一行(列)是两个数的和时,可把行列式对该行(列)拆开成两个行列式之和,拆开时其它各行(列)均保持不变.对于行列式的这一性质应当正确理解.因此,若要拆开n 阶行列式A B +,则应当是2n 个n 阶行列式的和,所以(A)错误.矩阵的运算是表格的运算,它不同于数字运算,矩阵乘法没有交换律,故(B)不正确.若1010,0102A B ⎡⎤⎡⎤==⎢⎥⎢⎥⎣⎦⎣⎦,则 ()111111020020102,1310301000223A B A B ----⎡⎤⎡⎤⎡⎤⎢⎥⎡⎤⎡⎤⎢⎥⎢⎥+==+=+=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎢⎥⎣⎦⎣⎦⎢⎥⎣⎦. 而且()1A B -+存在时,不一定11,A B --都存在,所以选项(D)是错误的. 由行列式乘法公式AB A B B A BA =⋅=⋅=知(C)正确.注意,行列式是数,故恒有A B B A ⋅=⋅.而矩阵则不行,故(B)不正确. (5)【答案】D【解析】设事件B =“甲种产品畅销”,事件C =“乙种产品滞销”,则 A 事件“甲种产品畅销,乙种产品滞销”可表示为A BC,=则_____A BCBC ===“甲种产品滞销或乙种产品畅销”,应选(D).三、计算题(本题满分15分,每小题5分.) (1)【解析】这是1∞型未定式求极限.设1u x=,则当x →∞时,0u →.于是 1011lim(sin cos )lim(sin cos )xux u u u x x→∞→+=+ 1sin cos 1sin cos 10lim(1sin cos 1)u u u u uu u u +-⋅+-→=++-,令sin cos 1u u t +-=,则0u →时0t →, 所以 11sin cos 1lim(1sin cos 1)lim(1)u u tu t u u t e +-→→++-=+=,所以 01sin cos 1sin cos 1sin cos 1limsin cos 10lim(1sin cos 1)lim u u u u u u u u u uuuu u u u ee→+-+-+-⋅+-→→++-==,由洛必达法则得00sin cos 1cos sin limlim 11u u u u u uu →→+--==,所以 111lim(sin cos )x x e e x x→∞+==.(2)【解析】方法一：先求zx∂∂,再求2z x y ∂∂∂.由复合函数求导法则,z f u f v f fy ,x u x v x u v∂∂∂∂∂∂∂=+=+∂∂∂∂∂∂∂ 故2()z f fy x y y u v∂∂∂∂=+∂∂∂∂∂ 222222f u f v ff u f v y u y u v y v v u y v y ⎛⎫∂∂∂∂∂∂∂∂∂=++++ ⎪∂∂∂∂∂∂∂∂∂∂∂⎝⎭222222f f f f f x y xy u u v v u v v ∂∂∂∂∂=++++∂∂∂∂∂∂∂ 22222()f f f fx y xy u u v v v∂∂∂∂=++++∂∂∂∂∂. 方法二：利用一阶全微分形式不变性,可得1212()()()()dz f d x y f d xy f dx dy f ydx xdy ''''=++=+++1212()()f yf dx f xf dy ''''=+++.于是有 12x z f yf '''=+. 再对y 外求偏导数,即得122111221222()()()xy y y z f y f f f xf y f xf f ''''''''''''''''=++=++++1112222()f x y f xyf f '''''''=++++. 【相关知识点】复合函数求导法则：若(,)u u x y =和(,)v v x y =在点(,)x y 处偏导数存在,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 u x y v x y =在点(,)x y 处的偏导数存在,且,z f u f v z f u f vx u x v x y u y v y∂∂∂∂∂∂∂∂∂∂=+=+∂∂∂∂∂∂∂∂∂∂. (3)【解析】微分方程562xy y y e -'''++=对应的齐次方程560y y y '''++=的特征方程为2560r r ++=,特征根为122,3r r =-=-,故对应齐次微分方程的通解为2312xx C eC e --+.设所给非齐次方程的特解为*()xy x Ae -=,代入方程562xy y y e-'''++=,比较系数,得1A =,故所求方程的通解为231212,,x x x y C e C e e C C ---=++ 为常数.【相关知识点】关于微分方程特解的求法:如果()()xm f x P x e λ=,则二阶常系数非齐次线性微分方程()()()y p x y q x y f x '''++=具有形如*()k xm y x Q x e λ=的特解,其中()m Q x 与()m P x 同次(m 次)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为0、1或2.四、(本题满分9分) 【解析】(1)收益函数2()10,06x R x xP xe x -==≤≤.边际收益函数25(2)x dRMR x e dx-==-.(2)由 25(2)0x dRx e dx-=-=,得2x =.又2222255(4)02x x x d Rx e dx e-===-=-<.因此()R x 在2x =取极大值.又因为极值点惟一,故此极大值必为最大值,最大值为20(2)R e=. 所以,当生产量为2时,收益取最大值,收益最大值为20e .而相应的价格为10e. (3)五、(本题满分9分)【解析】(1)()f x 为分段函数,由定积分的性质, 212001()()()xxx S f x e dx f x e dx f x e dx ---==+⎰⎰⎰1201(2)x x xe dx x e dx --=+-⎰⎰ 121(2)x x xde x de --=-+-⎰⎰12120101(2)x xxxxe edx x e e dx ----⎡⎤⎡⎤=-++--⎣⎦⎣⎦⎰⎰1220111101()x xe e e e e --⎡⎤⎡⎤=+---=-+--+⎣⎦⎣⎦2121e e=-+. (2)用定积分换元法,令2x t -=,则2,x t dx dt =+=,所以 422(2)212(2)()()x t t S f x e dx f t e dt e f t e dt --+--=-==⋅⎰⎰⎰,而 202012()1x S f x e dx e e-==-+⎰, 故 2222102012()(1)t S e f t e dt S e e e e----=⋅==-+⎰. (3) 用定积分换元法,令2x n t -=,则2,x t n dx dt =+=,所以2222(2)220(2)()()n xt n nt n nS f x n e dx f t edt ef t e dt +--+--=-==⋅⎰⎰⎰而 20212()1x S f x e dx e e-==-+⎰, 故 22220212()(1)nt n n n S ef t e dt S e e e e----=⋅==-+⎰. (4)利用以上结果,有2002001nnn n n n S S S e S e ∞∞∞-===⎛⎫=== ⎪⎝⎭∑∑∑()22002221111111e S e S e e e e e--====--+-.六、(本题满分6分) 【解析】对1()()xa F x f t dt x a=-⎰两边对x 求导,得 22()()()()()()()()xxa a f t dtx a f x f t dtf x F x x a x ax a ---'=+=---⎰⎰.证法一：由积分中值定理知,在(,)a x 内存在一点ξ使得()()()xaf t dt f x a ξ=-⎰,所以 22()()()()()()()()()()()()xa x a f x f t dtx a f x f x a f x f F x x a x a x aξξ------'===---⎰. 又因为()0,f x a x ξ'≤<<,故有()()0f x f ξ-≤,所以()0F x '≤. 证法二：令()()()()xag x x a f x f t dt =--⎰,则()()()()()()()g x f x x a f x f x x a f x '''=+--=-.因为,()0x a f x '>≤,所以()0g x '≤, 即()()()()xag x x a f x f t dt =--⎰在(,)a b 上为减函数,所以()()0g x g a ≤=,所以 2()()0()g x F x x a '=≤-.七、(本题满分5分)【解析】方法一：本题可采用一般的解法如下： 由X AX B,=+得()E A X B.-=因为 ()1111002111013213102011E A ,---⎡⎤⎡⎤⎢⎥⎢⎥-=-=-⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦所以 ()102111311321202030115311X E A B .---⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥=-=-=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥---⎣⎦⎣⎦⎣⎦ 方法二：本题还可用由()E A X B -=作初等行变换()()E A B E X -→,此解法优点是少算一次矩阵乘法,可以适当减少计算量.()110111012010253E A B --⎡⎤⎢⎥-=-⎢⎥⎢⎥-⎣⎦, 第一行乘以()1-分别加到第二行和第三行上,再第三行乘以()1-加到第三行上,得110110111100333--⎡⎤⎢⎥-⎢⎥⎢⎥-⎣⎦第三行自乘13,再加到第二行上,第二行再加到第一行上,有100310102000111-⎡⎤⎢⎥⎢⎥⎢⎥-⎣⎦, 所以312011X .-⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦八、(本题满分6分) 【解析】m 个n 维向量12m ,,,ααα线性相关的充分必要条件是齐次方程组.()12120m m x x x ααα⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦有非零解.特别地,n 个n 维向量12,,,n ααα线性相关的充分必要条件是行列式12,,,0n ααα=.由于123111,,123513t tααα==-,故当5t ≠时,向量组123,,ααα线性无关；5t =时向量组123,,ααα线性相关. 当5t =时,设11223x x ααα+=将坐标代入有1212121,23,3 5.x x x x x x +=⎧⎪+=⎨⎪+=⎩解出121, 2.x x =-=即3122ααα=-+.九、(本题满分5分)【解析】(1) 矩阵A 的特征方程为122212221E A λλλλ+---=-+-+, 经过行列式一系列的初等行变换和初等列变换,有122122112034021021E A λλλλλλλλ-------=-+=+++ ()()()234115021λλλλλ+=-=-+=+,故矩阵A 的特征值为：115,,-.(2)由λ为A 的特征值可知,存在非零向量α使A αλα=,两端左乘1A -,得1A αλα-=.因为0α≠,故0λ≠,于是有11Aααλ-=.按特征值定义知1λ是1A -的特征值.由A 的特征值是115,,,-可知1A -的特征值为1115,,.-又因为()11(1)E A ααλ-+=+, 那么1E A -+的特征值是4225,,.【相关知识点】矩阵特征值与特征向量的定义：设A 是n 阶矩阵,若存在数λ及非零的n 维列向量X 使得AX X λ=成立,则称λ是矩阵A 的特征值,称非零向量X 是矩阵A 的特征向量.十 、(本题满分7分)【解析】(1) 由二维连续型随机变量的概率求法,概率等于对应区域上的二重积分()0{}(,)yx y x yP X Y f x y dxdy dy e dx +∞-+<<==⎰⎰⎰⎰y y x e dy e dx +∞--=⎰⎰0()x y y x x e e dy +∞=--==-⎰201(1)2y y y y e e dy e e +∞+∞----⎛⎫=-=-+ ⎪⎝⎭⎰1.2=(2) 由二维连续型随机变量的数学期望定义得()0()(,)x y E XY xyf x y dxdy xye dxdy +∞+∞+∞+∞-+-∞-∞==⎰⎰⎰⎰x y xe dx ye dy +∞+∞--=⎰⎰.因为由分部积分法有y y y y ye dy yde ye e d y +∞+∞+∞----+∞=-=-+⎰⎰⎰yyye e --+∞+∞=--,由洛必达法则,对∞∞型极限,有1lim lim 0yy y y ye e -→∞→∞==.所以有() 1.E XY =十一、(本题满分8分)【解析】以A 表示事件“对X 的观测值大于3”,依题意,X 的概率密度函数为1,25,()30,x f x ⎧≤≤⎪=⎨⎪⎩其它. 因此 5312(){3}.33P A P X dx p =>==⎰设随机变量Y 表示三次独立观测中观测值大于3的次数(即在三次独立试验中事件A 出现的次数).显然, Y 服从参数23,3n p ==的二项分布,因此,所求概率为 {2}{2}{3}P Y P Y P Y ≥==+=223321220()()()33327C =+=.【相关知识点】二项分布的概率计算公式：若(,)Y B n p ~,则{}(1)k kn k n P Y k C p p -==-, 0,1,,k n =.。
1989考研数三真题及解析
1989年全国硕士研究生入学统一考试数学三试题一、填空题(本题满分15分,每小题3分.把答案填在题中横线上.) (1) 曲线2sin y x x =+在点122,ππ⎛⎫+⎪⎝⎭处的切线方程是__ _ .(2)幂级数nn ∞=的收敛域是__ _ . (3) 齐次线性方程组1231231230,0,0x x x x x x x x x λλ++=⎧⎪++=⎨⎪++=⎩ 只有零解,则λ应满足的条件是__ _ . (4) 设随机变量X 的分布函数为()00sin 0212,x ,F x A x,x ,,x ,ππ⎧⎪<⎪⎪=≤≤⎨⎪⎪>⎪⎩则A =__________,6P X π⎧⎫<=⎨⎬⎩⎭ . (5) 设随机变量X 的数学期望()E X μ=,方差2()D X σ=,则由切比雪夫(Chebyshev)不等式,有{3}P X μσ-≥≤__ _ .二、选择题(本题满分15分,每小题3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()232xxf x ,=+-则当0x →时 ( )(A) ()f x 与x 是等价无穷小量 (B) ()f x 与x 是同阶但非等价无穷小量 (C) ()f x 是比x 较高阶的无穷小量 (D) ()f x 是比x 较低阶的无穷小量 (2) 在下列等式中,正确的结果是 ( )(A)()()f x dx f x '=⎰ (B) ()()df x f x =⎰(C)()()df x dx f x dx =⎰(D) ()()d f x dx f x =⎰ (3) 设A 为n 阶方阵且0A =,则 ( ) (A) A 中必有两行(列)的元素对应成比例(B) A 中任意一行(列)向量是其余各行(列)向量的线性组合 (C) A 中必有一行(列)向量是其余各行(列)向量的线性组合 (D) A 中至少有一行(列)的元素全为0(4) 设A 和B 均为n n ⨯矩阵,则必有 ( )(A) A B A B +=+ (B)AB BA = (C) AB BA = (D) ()111A B A B ---+=+(5) 以A 表示事件“甲种产品畅销,乙种产品滞销”,则其对立事件A 为 ( )(A) “甲种产品滞销,乙种产品畅销” (B) “甲、乙两种产品均畅销”(C) “甲种产品滞销” (D) “甲种产品滞销或乙种产品畅销”三、计算题(本题满分15分,每小题5分)(1) 求极限11lim sin cos xx .x x →∞⎛⎫+ ⎪⎝⎭(2) 已知(,),,,z f u v u x y v xy ==+=且(,)f u v 的二阶偏导数都连续.求2zx y∂∂∂.(3) 求微分方程562xy y y e-'''++=的通解.四、(本题满分9分)设某厂家打算生产一批商品投放市场.已知该商品的需求函数为2()10x P P x e -==,且最大需求量为6,其中x 表示需求量,P 表示价格.(1) 求该商品的收益函数和边际收益函数.(2分)(2) 求使收益最大时的产量、最大收益和相应的价格.(4分) (3) 画出收益函数的图形.(3分)五、(本题满分9分)已知函数,01,()2,1 2.x x f x x x ≤≤⎧=⎨-≤≤⎩试计算下列各题：(1) 200();xS f x e dx -=⎰(4分) (2) 412(2);x S f x e dx -=-⎰(2分)(3) 222(2)(2,3,);n xn nS f x n e dx n +-=-=⎰(1分) (4) 0n n S S ∞==∑.(2分)六、(本题满分6分)假设函数()f x 在[,]a b 上连续,在(,)a b 内可导,且()0f x '≤,记1()(),xa F x f t dt x a=-⎰ 证明在(,)a b 内,()0F x '≤.七、(本题满分5分)已知X AX B,=+其中010111101A ,⎡⎤⎢⎥=-⎢⎥⎢⎥--⎣⎦112053B ,-⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦求矩阵X .八、(本题满分6分)设123(1,1,1),(1,2,3),(1,3,)t ααα===. (1) 问当t 为何值时,向量组123,,ααα线性无关?(3分) (2) 问当t 为何值时,向量组123,,ααα线性相关?(1分)(3) 当向量组123,,ααα线性相关时,将3α表示为1α和2α的线性组合.(2分)九、(本题满分5分)设122212221A .-⎡⎤⎢⎥=--⎢⎥⎢⎥--⎣⎦(1)试求矩阵A 的特征值；(2分)(2)利用(1)小题的结果,求矩阵1E A -+的特征值,其中E 是三阶单位矩阵.(3分)十 、(本题满分7分)已知随机变量X 和Y 的联合密度为(),,,(,)0,x y e x y f x y -+⎧<<+∞<<+∞=⎨⎩ 00其它.试求：(1) {}P X Y <；(5分) (2) ()E XY .(2分)十一、(本题满分8分)设随机变量X 在[2,5]上服从均匀分布,现在对X 进行三次独立观测,试求至少有两次观测值大于3的概率.1989年全国硕士研究生入学统一考试数学三试题解析一、填空题(本题满分15分,每小题3分.) (1)【答案】1y x =+【解析】对函数2sin y x x =+两边对x 求导,得12cos y sin x x,'=+ 令2x π=得212sincos122x y .πππ='=+=所以该曲线在点122,ππ⎛⎫+ ⎪⎝⎭处的切线的斜率为1,所以 切线方程是122y x ,ππ⎛⎫-+=- ⎪⎝⎭即1y x =+为所求. (2)【答案】[1,1)-【解析】因系数1n n a a +==,从而1lim1,n n n n a a +→∞== 即幂级数的收敛半径1R =,当11x -<<时幂级数绝对收敛. 当1x =-时得交错级数nn ∞=(条件收敛)；当1x =时得正项级数0n ∞=(发散).于是,幂级数的收敛域是[1,1)-. (3)【答案】1λ≠【解析】n 个方程n 个未知数的齐次方程组0Ax =有非零解的充分必要条件是0A =, 因为此时未知数的个数等于方程的个数,即A 为方阵时,用0A =判定比较方便.而 21110011010(1),111111A λλλλλ-==-=- 所以当0A ≠时1λ≠.所以此题应填:1λ≠.(4)【答案】1,12【解析】由于任何随机变量X 的分布函数()F x 是右连续函数,因此对任何x ,有()(0)F x F x =+.对于2x π=,有()sin,(0) 1.222F A A F πππ==+= 令 ()2F π=(0)2F π+,得到1A =,其中0(0)lim ()x F x F x +→+=.又 666P X P X ,πππ⎧⎫⎧⎫<=-<<⎨⎬⎨⎬⎩⎭⎩⎭因()F x 在6x π=处连续,连续函数在任何一个点上的概率为0,因此06P X .π⎧⎫==⎨⎬⎩⎭所以 666P X P X πππ⎧⎫⎧⎫<=-<≤⎨⎬⎨⎬⎩⎭⎩⎭66F F ππ⎛⎫⎛⎫=-- ⎪ ⎪⎝⎭⎝⎭162sin .π== (5)【答案】19【解析】由切比雪夫不等式2{}DXP X EX εε-≥≤,有221{3}(3)9P X σμσσ-≥≤=.二、选择题(本题满分15分,每小题3分.) (1)【答案】(B)【解析】由洛必达法则有()0002322ln23ln3lim lim lim ln2ln31x x x x x x x f x x x →→→+-+===+. 所以()f x 与x 是同阶但非等价无穷小量. (2)【答案】(C)【解析】由不定积分的概念和性质可知,()()()()df x dx f x dx f x .dx'==⎰⎰()()()f x dx df x f x C,'==+⎰⎰C 为常数.()()d f x dx f x dx.=⎰故应选(C). (3)【答案】(C)【解析】本题考查||0A =的充分必要条件,而选项(A)、(B)、(D)都是充分条件,并不必要.因为对矩阵A 来说,行和列具有等价性,所以单说列或者单说行满足什么条件就构成了||0A =的必要条件,但是不具有任意性,只需要存在一列向量是其余列向量的线性组合.以3阶矩阵为例,若 112123134A ⎛⎫ ⎪= ⎪ ⎪⎝⎭,条件(A)必有一列元素全为0,(B)必有两列元素对应成比例均不成立,但有||0A =,所以(A)、 (B)不满足题意,不可选.若123124125A ⎛⎫⎪= ⎪ ⎪⎝⎭,则||0A =,但第三列并不是其余两列的线性组合,可见(D)不正确.这样用排除法可知应选(C).(4)【答案】(C) 【解析】当行列式的一行(列)是两个数的和时,可把行列式对该行(列)拆开成两个行列式之和,拆开时其它各行(列)均保持不变.对于行列式的这一性质应当正确理解.因此,若要拆开n 阶行列式A B +,则应当是2n个n 阶行列式的和,所以(A)错误.矩阵的运算是表格的运算,它不同于数字运算,矩阵乘法没有交换律,故(B)不正确.若1010,0102A B ⎡⎤⎡⎤==⎢⎥⎢⎥⎣⎦⎣⎦,则()111111020020102,1310301000223A B A B ----⎡⎤⎡⎤⎡⎤⎢⎥⎡⎤⎡⎤⎢⎥⎢⎥+==+=+=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎢⎥⎣⎦⎣⎦⎢⎥⎣⎦. 而且()1A B -+存在时,不一定11,A B --都存在,所以选项(D)是错误的. 由行列式乘法公式AB A B B A BA =⋅=⋅=知(C)正确.注意,行列式是数,故恒有A B B A ⋅=⋅.而矩阵则不行,故(B)不正确.(5)【答案】D【解析】设事件B =“甲种产品畅销”,事件C =“乙种产品滞销”,则 A 事件“甲种产品畅销,乙种产品滞销”可表示为A BC,=则_____A BCBC ===“甲种产品滞销或乙种产品畅销”,应选(D).三、计算题(本题满分15分,每小题5分.) (1)【解析】这是1∞型未定式求极限.设1u x=,则当x →∞时,0u →.于是 1011lim(sin cos )lim(sin cos )xux u u u x x→∞→+=+ 1sin cos 1sin cos 10lim(1sin cos 1)u u u u uu u u +-⋅+-→=++-,令sin cos 1u u t +-=,则0u →时0t →, 所以 11sin cos 1lim(1sin cos 1)lim(1)u u tu t u u t e +-→→++-=+=,所以 01sin cos 1sin cos 1sin cos 1limsin cos 10lim(1sin cos 1)lim u u u u u u u u u uuuu u u u ee→+-+-+-⋅+-→→++-==,由洛必达法则得00sin cos 1cos sin limlim 11u u u u u uu →→+--==,所以 111lim(sin cos )x x e e x x→∞+==.(2)【解析】方法一：先求z x ∂∂,再求2zx y∂∂∂.由复合函数求导法则,z f u f v f fy ,x u x v x u v∂∂∂∂∂∂∂=+=+∂∂∂∂∂∂∂ 故 2()z f fy x y y u v∂∂∂∂=+∂∂∂∂∂222222f u f v ff u f v y u y u v y v v u y v y ⎛⎫∂∂∂∂∂∂∂∂∂=++++ ⎪∂∂∂∂∂∂∂∂∂∂∂⎝⎭222222f f f f f x y xy u u v v u v v ∂∂∂∂∂=++++∂∂∂∂∂∂∂ 22222()f f f fx y xy u u v v v∂∂∂∂=++++∂∂∂∂∂. 方法二：利用一阶全微分形式不变性,可得1212()()()()dz f d x y f d xy f dx dy f ydx xdy ''''=++=+++1212()()f yf dx f xf dy ''''=+++.于是有 12x z f yf '''=+. 再对y 外求偏导数,即得122111221222()()()xy y y z f y f f f xf y f xf f ''''''''''''''''=++=++++1112222()f x y f xyf f '''''''=++++. 【相关知识点】复合函数求导法则：若(,)u u x y =和(,)v v x y =在点(,)x y 处偏导数存在,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 u x y v x y =在点(,)x y 处的偏导数存在,且,z f u f v z f u f vx u x v x y u y v y∂∂∂∂∂∂∂∂∂∂=+=+∂∂∂∂∂∂∂∂∂∂. (3)【解析】微分方程562xy y y e-'''++=对应的齐次方程560y y y '''++=的特征方程为2560r r ++=,特征根为122,3r r =-=-,故对应齐次微分方程的通解为2312xx C eC e --+.设所给非齐次方程的特解为*()xy x Ae -=,代入方程562xy y y e-'''++=,比较系数,得1A =,故所求方程的通解为231212,,x x x y C e C e e C C ---=++ 为常数.【相关知识点】关于微分方程特解的求法:如果()()xm f x P x e λ=,则二阶常系数非齐次线性微分方程()()()y p x y q x y f x '''++=具有形如*()k xm y x Q x e λ=的特解,其中()m Q x 与()m P x 同次(m 次)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为0、1或2.四、(本题满分9分) 【解析】(1)收益函数2()10,06x R x xP xe x -==≤≤.边际收益函数25(2)x dR MR x e dx-==-.(2)由 25(2)0x dRx e dx-=-=,得2x =. 又2222255(4)02x x x d Rx e dx e-===-=-<.因此()R x 在2x =取极大值.又因为极值点惟一,故此极大值必为最大值,最大值为20(2)R e=. 所以,当生产量为2时,收益取最大值,收益最大值为20e .而相应的价格为10e. (3)五、(本题满分9分)【解析】(1)()f x 为分段函数,由定积分的性质, 212001()()()x x x S f x e dx f x e dx f x e dx ---==+⎰⎰⎰1201(2)xx xe dx x e dx --=+-⎰⎰121(2)xx xdex de --=-+-⎰⎰1212011(2)x x x xxe e dx x e e dx ----⎡⎤⎡⎤=-++--⎣⎦⎣⎦⎰⎰1220111101()x xe e e e e --⎡⎤⎡⎤=+---=-+--+⎣⎦⎣⎦2121e e=-+. (2)用定积分换元法,令2x t -=,则2,x t dx dt =+=,所以 422(2)212(2)()()x t t S f x e dx f t e dt e f t e dt --+--=-==⋅⎰⎰⎰,而 20212()1x S f x e dx e e-==-+⎰,故 2222102012()(1)t S e f t e dt S e e e e----=⋅==-+⎰. (3) 用定积分换元法,令2x n t -=,则2,x t n dx dt =+=,所以 2222(2)220(2)()()n x t n n t n nS f x n e dx f t e dt e f t e dt +--+--=-==⋅⎰⎰⎰而 202012()1x S f x e dx e e-==-+⎰, 故 22220212()(1)nt n n n S ef t e dt S e e e e----=⋅==-+⎰. (4)利用以上结果,有2002001nnn n n n S S S eS e ∞∞∞-===⎛⎫=== ⎪⎝⎭∑∑∑()22002221111111e S e S e e e e e--====--+-.六、(本题满分6分)【解析】对1()()xa F x f t dt x a=-⎰两边对x 求导,得 22()()()()()()()()xxa a f t dtx a f x f t dtf x F x x a x ax a ---'=+=---⎰⎰.证法一：由积分中值定理知,在(,)a x 内存在一点ξ使得()()()xaf t dt f x a ξ=-⎰,所以 22()()()()()()()()()()()()xa x a f x f t dtx a f x f x a f x f F x x a x a x aξξ------'===---⎰.又因为()0,f x a x ξ'≤<<,故有()()0f x f ξ-≤,所以()0F x '≤. 证法二：令()()()()xag x x a f x f t dt =--⎰,则()()()()()()()g x f x x a f x f x x a f x '''=+--=-.因为,()0x a f x '>≤,所以()0g x '≤, 即()()()()xag x x a f x f t dt =--⎰在(,)a b 上为减函数,所以()()0g x g a ≤=,所以 2()()0()g x F x x a '=≤-.七、(本题满分5分)【解析】方法一：本题可采用一般的解法如下： 由X AX B,=+得()E A X B.-=因为 ()1111002111013213102011E A ,---⎡⎤⎡⎤⎢⎥⎢⎥-=-=-⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦所以 ()102111311321202030115311X E A B .---⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥=-=-=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥---⎣⎦⎣⎦⎣⎦ 方法二：本题还可用由()E A X B -=作初等行变换()()E A B E X -→,此解法优点是少算一次矩阵乘法,可以适当减少计算量.()110111012010253E A B --⎡⎤⎢⎥-=-⎢⎥⎢⎥-⎣⎦, 第一行乘以()1-分别加到第二行和第三行上,再第三行乘以()1-加到第三行上,得110110111100333--⎡⎤⎢⎥-⎢⎥⎢⎥-⎣⎦第三行自乘13,再加到第二行上,第二行再加到第一行上,有100310102000111-⎡⎤⎢⎥⎢⎥⎢⎥-⎣⎦, 所以312011X .-⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦八、(本题满分6分) 【解析】m 个n 维向量12m ,,,ααα线性相关的充分必要条件是齐次方程组.()12120m m x x x ααα⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦有非零解.特别地,n 个n 维向量12,,,n ααα线性相关的充分必要条件是行列式12,,,0n ααα=.由于123111,,123513t tααα==-,故当5t ≠时,向量组123,,ααα线性无关；5t =时向量组123,,ααα线性相关. 当5t =时,设11223x x ααα+=将坐标代入有1212121,23,3 5.x x x x x x +=⎧⎪+=⎨⎪+=⎩解出121, 2.x x =-=即3122ααα=-+.九、(本题满分5分)【解析】(1) 矩阵A 的特征方程为122212221E A λλλλ+---=-+-+, 经过行列式一系列的初等行变换和初等列变换,有122122112034021021E A λλλλλλλλ-------=-+=+++ ()()()234115021λλλλλ+=-=-+=+,故矩阵A 的特征值为：115,,-.(2)由λ为A 的特征值可知,存在非零向量α使A αλα=,两端左乘1A -,得1A αλα-=.因为0α≠,故0λ≠,于是有11Aααλ-=.按特征值定义知1λ是1A -的特征值.由A 的特征值是115,,,-可知1A -的特征值为1115,,.-又因为 ()11(1)E A ααλ-+=+, 那么1E A -+的特征值是4225,,.【相关知识点】矩阵特征值与特征向量的定义：设A 是n 阶矩阵,若存在数λ及非零的n 维列向量X 使得AX X λ=成立,则称λ是矩阵A 的特征值,称非零向量X 是矩阵A 的特征向量.十 、(本题满分7分)【解析】(1) 由二维连续型随机变量的概率求法,概率等于对应区域上的二重积分()0{}(,)yx y x yP X Y f x y dxdy dy e dx +∞-+<<==⎰⎰⎰⎰yy x e dy e dx +∞--=⎰⎰0()x y y x x e e dy +∞=--==-⎰201(1)2yy y y e e dy e e +∞+∞----⎛⎫=-=-+ ⎪⎝⎭⎰1.2=(2) 由二维连续型随机变量的数学期望定义得()0()(,)x y E XY xyf x y dxdy xye dxdy +∞+∞+∞+∞-+-∞-∞==⎰⎰⎰⎰xy xe dx ye dy +∞+∞--=⎰⎰.因为由分部积分法有yyyy ye dy ydeyee dy +∞+∞+∞----+∞=-=-+⎰⎰⎰yyye e --+∞+∞=--,由洛必达法则,对∞∞型极限,有1lim lim 0yy y y ye e -→∞→∞==.所以有() 1.E XY =十一、(本题满分8分)【解析】以A 表示事件“对X 的观测值大于3”,依题意,X 的概率密度函数为1,25,()30,x f x ⎧≤≤⎪=⎨⎪⎩其它. 因此 5312(){3}.33P A P X dx p =>==⎰设随机变量Y 表示三次独立观测中观测值大于3的次数(即在三次独立试验中事件A 出现的次数).显然, Y 服从参数23,3n p ==的二项分布,因此,所求概率为 {2}{2}{3}P Y P Y P Y ≥==+=223321220()()()33327C =+=.【相关知识点】二项分布的概率计算公式：若(,)Y B n p ~,则{}(1)k kn k n P Y k C p p -==-, 0,1,,k n =.。
历年考研数学一真题及答案
历年考研数学一真题及答案IMB standardization office【IMB 5AB- IMBK 08- IMB 2C】历年考研数学一真题1987-2014（经典珍藏版）1987年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)当x =_____________时,函数2xy x =⋅取得极小值. (2)由曲线ln y x =与两直线e 1y x =+-及0y =所围成的平面图形的面积是_____________. (3)与两直线 1y t=-+及121111x y z +++==都平行且过原点的平面方程为_____________. (4)设L 为取正向的圆周229,x y +=则曲线积分2(22)(4)Lxy y dx x x dy -+-⎰= _____________.(5)已知三维向量空间的基底为123(1,1,0),(1,0,1),(0,1,1),===ααα则向量(2,0,0)=β在此基底下的坐标是_____________.二、(本题满分8分)求正的常数a 与,b使等式201lim 1sin x x bx x →=-⎰成立. 三、(本题满分7分)(1)设f 、g 为连续可微函数,(,),(),u f x xy v g x xy ==+求,.u v x x∂∂∂∂(2)设矩阵A 和B 满足关系式2,+AB =A B 其中301110,014⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦A 求矩阵.B 四、(本题满分8分)求微分方程26(9)1y y a y ''''''+++=的通解,其中常数0.a >五、选择题(本题共4小题,每小题3分,满分12分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内) (1)设2()()lim1,()x af x f a x a →-=--则在x a =处 (A)()f x 的导数存在,且 ()0f a '≠(B)()f x 取得极大值(C)()f x 取得极小值(D)()f x 的导数不存在(2)设()f x 为已知连续函数0,(),s t I t f tx dx =⎰其中0,0,t s >>则I的值 (A)依赖于s 和t(B)依赖于s 、t 和x(C)依赖于t 、x ,不依赖于s(D)依赖于s ,不依赖于t(3)设常数0,k >则级数21(1)nn k n n ∞=+-∑ (A)发散 (B)绝对收敛(C)条件收敛 (D)散敛性与k 的取值有关(4)设A 为n 阶方阵，且A 的行列式||0,a =≠A 而*A 是A 的伴随矩阵，则*||A 等于(A)a (B)1a(C)1n a -(D)na六、（本题满分10分）求幂级数1112n nn x n ∞-=∑的收敛域，并求其和函数. 七、（本题满分10分） 求曲面积分其中∑是由曲线13()0z y f x x ⎧=≤≤⎪=⎨=⎪⎩绕y 轴旋转一周而成的曲面,其法向量与y 轴正向的夹角恒大于.2π八、（本题满分10分）设函数()f x 在闭区间[0,1]上可微,对于[0,1]上的每一个,x 函数()f x 的值都在开区间(0,1)内,且()f x '≠1,证明在(0,1)内有且仅有一个,x 使得().f x x =九、（本题满分8分） 问,a b 为何值时,现线性方程组有唯一解,无解,有无穷多解并求出有无穷多解时的通解.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)设在一次实验中,事件A 发生的概率为,p 现进行n 次独立试验,则A 至少发生一次的概率为____________;而事件A 至多发生一次的概率为____________.(2)有两个箱子,第1个箱子有3个白球,2个红球, 第2个箱子有4个白球,4个红球.现从第1个箱子中随机地取1个球放到第2个箱子里,再从第2个箱子中取出1个球,此球是白球的概率为____________.已知上述从第2个箱子中取出的球是白球,则从第一个箱子中取出的球是白球的概率为____________. (3)已知连续随机变量X的概率密度函数为221()e ,xx f x -+-=则X 的数学期望为____________,X 的方差为____________. 十一、（本题满分6分）设随机变量,X Y 相互独立,其概率密度函数分别为()Xf x = 10 01x ≤≤其它, ()Y f y =e 0y - 00y y >≤, 求2Z X Y=+的概率密度函数.1988年全国硕士研究生入学统一考试数学(一)试卷一、(本题共3小题,每小题5分,满分15分)(1)求幂级数1(3)3nn n x n ∞=-∑的收敛域. (2)设2()e ,[()]1x f x f x x ϕ==-且()0x ϕ≥,求()x ϕ及其定义域.(3)设∑为曲面2221x y z ++=的外侧,计算曲面积分333.I x dydz y dzdx z dxdy ∑=++⎰⎰二、填空题(本题共4小题,每小题3分,满分12分.把答案填在题中横线上)(1)若21()lim (1),txx f t t x→∞=+则()f t '= _____________. (2)设()f x 连续且31(),x f t dt x -=⎰则(7)f =_____________.(3)设周期为2的周期函数,它在区间(1,1]-上定义为()f x = 22x 1001x x -<≤<≤,则的傅里叶()Fourier 级数在1x =处收敛于_____________.(4)设4阶矩阵234234[,,,],[,,,],==A αγγγB βγγγ其中234,,,,αβγγγ均为4维列向量,且已知行列式4,1,==A B 则行列式+A B = _____________.三、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设()f x 可导且01(),2f x '=则0x ∆→时,()f x 在0x 处的微分dy 是 (A)与x ∆等价的无穷小(B)与x ∆同阶的无穷小(C)比x ∆低阶的无穷小(D)比x ∆高阶的无穷小(2)设()y f x =是方程240y y y '''-+=的一个解且00()0,()0,f x f x '>=则函数()f x 在点0x 处(A)取得极大值(B)取得极小值 (C)某邻域内单调增加(D)某邻域内单调减少 (3)设空间区域2222222212:,0,:,0,0,0,x y z R z x y z R x y z Ω++≤≥Ω++≤≥≥≥则(A)124xdv dv ΩΩ=⎰⎰⎰⎰⎰⎰(B)124ydv ydv ΩΩ=⎰⎰⎰⎰⎰⎰(C)124zdv zdv ΩΩ=⎰⎰⎰⎰⎰⎰(D)124xyzdv xyzdv ΩΩ=⎰⎰⎰⎰⎰⎰(4)设幂级数1(1)nn n a x ∞=-∑在1x =-处收敛,则此级数在2x =处(A)条件收敛(B)绝对收敛 (C)发散(D)收敛性不能确定 (5)n 维向量组12,,,(3)s s n ≤≤ααα线性无关的充要条件是(A)存在一组不全为零的数12,,,,s k k k 使11220s s k k k +++≠ααα(B)12,,,s ααα中任意两个向量均线性无关(C)12,,,s ααα中存在一个向量不能用其余向量线性表示 (D)12,,,s ααα中存在一个向量都不能用其余向量线性表示四、(本题满分6分)设()(),x y u yf xg yx=+其中函数f 、g 具有二阶连续导数,求222.u u x y x x y∂∂+∂∂∂ 五、(本题满分8分)设函数()y y x =满足微分方程322e ,xy y y '''-+=其图形在点(0,1)处的切线与曲线21y x x =--在该点处的切线重合,求函数().y y x =六、（本题满分9分）设位于点(0,1)的质点A 对质点M 的引力大小为2(0kk r >为常数,r 为A 质点与M 之间的距离),质点M沿直线y =(2,0)B 运动到(0,0),O 求在此运动过程中质点A 对质点M的引力所作的功.七、（本题满分6分）已知,=AP BP 其中100100000,210,001211⎡⎤⎡⎤⎢⎥⎢⎥==-⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦B P 求5,.A A 八、（本题满分8分）已知矩阵20000101x ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦A 与20000001y ⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦B 相似. (1)求x 与.y(2)求一个满足1-=P AP B 的可逆阵.P 九、（本题满分9分）设函数()f x 在区间[,]a b 上连续,且在(,)a b 内有()0,f x '>证明:在(,)a b 内存在唯一的,ξ使曲线()y f x =与两直线(),y f x a ξ==所围平面图形面积1S 是曲线()y f x =与两直线(),y f x b ξ==所围平面图形面积2S 的3倍.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)设在三次独立试验中,事件A 出现的概率相等,若已知A 至少出现一次的概率等于19,27则事件A 在一次试验中出现的概率是____________.(2)若在区间(0,1)内任取两个数,则事件”两数之和小于65”的概率为____________.(3)设随机变量X 服从均值为10,均方差为的正态分布,已知 则X 落在区间(9.95,10.05)内的概率为____________. 十一、（本题满分6分）设随机变量X 的概率密度函数为21(),(1)X f x x π=-求随机变量1Y =-的概率密度函数().Yfy1989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)已知(3)2,f '=则0(3)(3)lim2h f h f h→--= _____________.(2)设()f x 是连续函数,且10()2(),f x x f t dt =+⎰则()f x =_____________.(3)设平面曲线L为下半圆周y =则曲线积分22()Lx y ds +⎰=_____________.(4)向量场div u 在点(1,1,0)P 处的散度div u =_____________.(5)设矩阵300100140,010,003001⎡⎤⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦A I 则矩阵1(2)--A I =_____________.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)当0x >时,曲线1sin y x x=(A)有且仅有水平渐近线 (B)有且仅有铅直渐近线(C)既有水平渐近线,又有铅直渐近线(D)既无水平渐近线,又无铅直渐近线(2)已知曲面224z x y =--上点P 处的切平面平行于平面2210,x y z ++-=则点的坐标是(A)(1,1,2)-(B)(1,1,2)- (C)(1,1,2)(D)(1,1,2)--(3)设线性无关的函数都是二阶非齐次线性方程的解是任意常数,则该非齐次方程的通解是(A)11223c y c y y ++(B)1122123()c y c y c c y +-+ (C)1122123(1)c y c y c c y +---(D)1122123(1)c y c y c c y ++--(4)设函数2(),01,f x x x =≤<而1()sin ,,n n S x b n x x π∞==-∞<<+∞∑其中12()sin ,1,2,3,,n b f x n xdx n π==⎰则1()2S -等于(A)12-(B)14-(C)14(D)12(5)设A 是n 阶矩阵,且A 的行列式0,=A 则A 中(A)必有一列元素全为0 (B)必有两列元素对应成比例(C)必有一列向量是其余列向量的线性组合 (D)任一列向量是其余列向量的线性组合三、(本题共3小题,每小题5分,满分15分)(1)设(2)(,),z f x y g x xy =-+其中函数()f t 二阶可导,(,)g u v 具有连续二阶偏导数,求2.z x y∂∂∂(2)设曲线积分2()c xy dx y x dy ϕ+⎰与路径无关,其中()x ϕ具有连续的导数,且(0)0,ϕ=计算(1,1)2(0,0)()xy dx y x dy ϕ+⎰的值.(3)计算三重积分(),x z dv Ω+⎰⎰⎰其中Ω是由曲面z =与z =所围成的区域.四、(本题满分6分) 将函数1()arctan1xf x x+=-展为x 的幂级数. 五、(本题满分7分) 设0()sin ()(),xf x x x t f t dt =--⎰其中f 为连续函数,求().f x六、（本题满分7分）证明方程0ln exx π=-⎰在区间(0,)+∞内有且仅有两个不同实根.七、（本题满分6分）问λ为何值时,线性方程组 有解,并求出解的一般形式. 八、（本题满分8分）假设λ为n 阶可逆矩阵A 的一个特征值,证明(1)1λ为1-A 的特征值.(2)λA为A 的伴随矩阵*A 的特征值.九、（本题满分9分）设半径为R 的球面∑的球心在定球面2222(0)x y z a a ++=>上,问当R 为何值时,球面∑在定球面内部的那部分的面积最大十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)已知随机事件A 的概率()0.5,P A =随机事件B 的概率()0.6P B =及条件概率(|)0.8,P B A =则和事件AB的概率()P A B =____________.(2)甲、乙两人独立地对同一目标射击一次,其命中率分别为和,现已知目标被命中,则它是甲射中的概率为____________. (3)若随机变量ξ在(1,6)上服从均匀分布,则方程210x x ξ++=有实根的概率是____________. 十一、（本题满分6分）设随机变量X 与Y 独立,且X 服从均值为1、标准差(均方差)为的正态分布,而Y 服从标准正态分布.试求随机变量23ZX Y =-+的概率密度函数.1990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)过点(1,21)M -且与直线 34y t =- 垂直的平面方程是_____________ .1z t =- (2)设a 为非零常数,则lim()xx x a x a→∞+-=_____________.(3)设函数()f x =111x x ≤>,则[()]f f x =_____________.(4)积分222ey xdx dy -⎰⎰的值等于_____________.(5)已知向量组1234(1,2,3,4),(2,3,4,5),(3,4,5,6),(4,5,6,7),====αααα则该向量组的秩是_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设()f x 是连续函数,且e ()(),xxF x f t dt -=⎰则()F x '等于(A)e (e )()xx f f x ----(B)e(e )()xx f f x ---+(C)e (e )()xx f f x ---(D)e (e )()x xf f x --+ (2)已知函数()f x 具有任意阶导数,且2()[()],f x f x '=则当n 为大于2的正整数时,()f x 的n 阶导数()()n f x 是(A)1![()]n n f x +(B)1[()]n n f x +(C)2[()]n f x(D)2![()]nn f x (3)设a 为常数,则级数21sin()[n na n ∞=-∑ (A)绝对收敛(B)条件收敛 (C)发散(D)收敛性与a 的取值有关(4)已知()f x 在0x =的某个邻域内连续,且0()(0)0,lim 2,1cos x f x f x→==-则在点0x =处()f x(A)不可导(B)可导,且(0)0f '≠ (C)取得极大值(D)取得极小值(5)已知1β、2β是非齐次线性方程组=AX b 的两个不同的解1,α、2α是对应其次线性方程组=AX 0的基础解析1,k 、2k 为任意常数,则方程组=AX b 的通解(一般解)必是 (A)1211212()2k k -+++ββααα(B)1211212()2k k ++-+ββααα (C)1211212()2k k -+++ββαββ(D)1211212()2k k ++-+ββαββ三、(本题共3小题,每小题5分,满分15分) (1)求120ln(1).(2)x dx x +-⎰(2)设(2,sin ),z f x y y x =-其中(,)f u v 具有连续的二阶偏导数,求2.zx y ∂∂∂(3)求微分方程244e xy y y -'''++=的通解(一般解).四、(本题满分6分)求幂级数0(21)n n n x ∞=+∑的收敛域,并求其和函数.五、(本题满分8分) 求曲面积分其中S 是球面2224x y z ++=外侧在0z ≥的部分. 六、（本题满分7分）设不恒为常数的函数()f x 在闭区间[,]a b 上连续,在开区间(,)a b 内可导,且()().f a f b =证明在(,)a b 内至少存在一点,ξ使得()0.f ξ'>七、（本题满分6分） 设四阶矩阵 且矩阵A 满足关系式其中E 为四阶单位矩阵1,-C 表示C 的逆矩阵,'C 表示C 的转置矩阵.将上述关系式化简并求矩阵.A 八、（本题满分8分）求一个正交变换化二次型22212312132344448f x x x x x x x x x =++-+-成标准型.九、（本题满分8分） 质点P 沿着以AB 为直径的半圆周,从点(1,2)A 运动到点(3,4)B 的过程中受变力F 作用(见图).F 的大小等于点P 与原点O 之间的距离,其方向垂直于线段OP 且与y轴正向的夹角小于.2π求变力F对质点P 所作的功.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)已知随机变量X 的概率密度函数 则X 的概率分布函数()F x =____________.(2)设随机事件A 、B 及其和事件的概率分别是、和,若B 表示B 的对立事件,那么积事件AB 的概率()P AB =____________.(3)已知离散型随机变量X 服从参数为2的泊松()Poisson 分布,即22e {},0,1,2,,!k P X k k k -===则随机变量32Z X =-的数学期望()E Z =____________.十一、（本题满分6分）设二维随机变量(,)X Y 在区域:01,D x y x <<<内服从均匀分布,求关于X 的边缘概率密度函数及随机变量21Z X =+的方差().D Z1991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21cos x t y t=+=,则22d ydx =_____________.(2)由方程xyz =(,)z z x y =在点(1,0,1)-处的全微分dz =_____________.(3)已知两条直线的方程是1212321:;:.101211x y z x y zl l ---+-====-则过1l 且平行于2l 的平面方程是_____________.(4)已知当0x →时123,(1)1ax +-与cos 1x -是等价无穷小,则常数a =_____________.(5)设4阶方阵52002100,00120011⎡⎤⎢⎥⎢⎥=⎢⎥-⎢⎥⎣⎦A 则A 的逆阵1-A =_____________. 二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)曲线221e 1ex x y --+=-(A)没有渐近线(B)仅有水平渐近线(C)仅有铅直渐近线(D)既有水平渐近线又有铅直渐近线(2)若连续函数()f x 满足关系式20()()ln 2,2tf x f dt π=+⎰则()f x 等于(A)e ln 2x(B)2eln 2x(C)e ln 2x+(D)2e ln 2x +(3)已知级数12111(1)2,5,n n n n n a a ∞∞--==-==∑∑则级数1nn a∞=∑等于(A)3(B)7 (C)8(D)9(4)设D 是平面xoy 上以(1,1)、(1,1)-和(1,1)--为顶点的三角形区域1,D 是D 在第一象限的部分,则(cos sin )Dxy x y dxdy +⎰⎰等于(A)12cos sin D x ydxdy ⎰⎰(B)12D xydxdy ⎰⎰(C)14(cos sin )D xy x y dxdy +⎰⎰(D)0(5)设n 阶方阵A 、B 、C 满足关系式,=ABC E 其中E 是n 阶单位阵,则必有(A)=ACB E(B)=CBA E (C)=BAC E(D)=BCA E三、(本题共3小题,每小题5分,满分15分)(1)求2lim .x π+→(2)设n 是曲面222236x y z ++=在点(1,1,1)P 处的指向外侧的法向量,求函数u =在点P 处沿方向n 的方向导数.(3)22(),x y z dv Ω++⎰⎰⎰其中Ω是由曲线220y z x ==绕z 轴旋转一周而成的曲面与平面4z =所围城的立体. 四、(本题满分6分)过点(0,0)O 和(,0)A π的曲线族sin (0)y a x a =>中,求一条曲线,L 使沿该曲线O 从到A 的积分3(1)(2)Ly dx x y dy +++⎰的值最小.五、(本题满分8分)将函数()2(11)f x x x =+-≤≤展开成以2为周期的傅里叶级数,并由此求级数211n n∞=∑的和.六、（本题满分7分）设函数()f x 在[0,1]上连续,(0,1)内可导,且1233()(0),f x dx f =⎰证明在(0,1)内存在一点,c 使()0.f c '=七、（本题满分8分）已知1234(1,0,2,3),(1,1,3,5),(1,1,2,1),(1,2,4,8)a a ===-+=+αααα及(1,1,3,5).b =+β (1)a 、b 为何值时,β不能表示成1234,,,αααα的线性组合(2)a 、b 为何值时,β有1234,,,αααα的唯一的线性表示式写出该表示式.八、（本题满分6分）设A 是n 阶正定阵,E 是n 阶单位阵,证明+A E 的行列式大于1. 九、（本题满分8分）在上半平面求一条向上凹的曲线,其上任一点(,)P x y 处的曲率等于此曲线在该点的法线段PQ 长度的倒数(Q 是法线与x 轴的交点),且曲线在点(1,1)处的切线与x 轴平行.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)若随机变量X 服从均值为2、方差为2σ的正态分布,且{24}0.3,P X <<=则{0}P X <=____________.(2)随机地向半圆0y a <<为正常数)内掷一点,点落在半圆内任何区域的概率与区域的面积成正比,则原点和该点的连线与x 轴的夹角小于4π的概率为____________. 十一、（本题满分6分）设二维随机变量(,)X Y 的密度函数为 求随机变量2Z X Y =+的分布函数.1992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)设函数()y y x =由方程e cos()0x y xy ++=确定,则dydx=_____________.(2)函数222ln()u x y z =++在点(1,2,2)M -处的梯度grad Mu=_____________.(3)设()f x = 211x-+ 00x x ππ-<≤<≤,则其以2π为周期的傅里叶级数在点x π=处收敛于_____________.(4)微分方程tan cos y y x x '+=的通解为y =_____________.(5)设111212121212,n n n n n n a b a b a b a b a ba b a b a b a b ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦A 其中0,0,(1,2,,).i ia b i n ≠≠=则矩阵A的秩()r A =_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)当1x →时,函数1211e 1x x x ---的极限(A)等于2(B)等于0 (C)为∞(D)不存在但不为∞(2)级数1(1)(1cos )(n n a n∞=--∑常数0)a >(A)发散(B)条件收敛 (C)绝对收敛(D)收敛性与a 有关(3)在曲线23,,x t y t z t ==-=的所有切线中,与平面24x y z ++=平行的切线(A)只有1条(B)只有2条 (C)至少有3条(D)不存在(4)设32()3,f x x x x =+则使()(0)n f 存在的最高阶数n 为 (A)0(B)1 (C)2(D)3(5)要使12100,121⎛⎫⎛⎫ ⎪ ⎪== ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭ξξ都是线性方程组=AX 0的解,只要系数矩阵A 为(A)[]212-(B)201011-⎡⎤⎢⎥⎣⎦ (C)102011-⎡⎤⎢⎥-⎣⎦(D)011422011-⎡⎤⎢⎥--⎢⎥⎢⎥⎣⎦三、(本题共3小题,每小题5分,满分15分) (1)求0x x →(2)设22(e sin ,),xz f y x y =+其中f 具有二阶连续偏导数,求2.z x y∂∂∂(3)设()f x= 21exx -+ 00x x ≤>,求31(2).f x dx -⎰ 四、(本题满分6分)求微分方程323e x y y y -'''+-=的通解. 五、(本题满分8分)计算曲面积分323232()()(),x az dydz y ax dzdx z ay dxdy ∑+++++⎰⎰其中∑为上半球面z =. 六、（本题满分7分）设()0,(0)0,f x f ''<=证明对任何120,0,x x >>有1212()()().f x x f x f x +<+ 七、（本题满分8分）在变力F yzi zxj xyk =++的作用下,质点由原点沿直线运动到椭球面2222221x y z a b c++=上第一卦限的点(,,),M ξηζ问当ξ、η、ζ取何值时,力F 所做的功W 最大并求出W 的最大值. 八、（本题满分7分）设向量组123,,ααα线性相关,向量组234,,ααα线性无关,问: (1)1α能否由23,αα线性表出证明你的结论. (2)4α能否由123,,ααα线性表出证明你的结论. 九、（本题满分7分）设3阶矩阵A 的特征值为1231,2,3,λλλ===对应的特征向量依次为1231111,2,3,149⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=== ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ξξξ又向量12.3⎛⎫⎪= ⎪ ⎪⎝⎭β (1)将β用123,,ξξξ线性表出. (2)求(n n A β为自然数).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11()()(),()0,()(),46P A P B P C P AB P AC P BC ======则事件A、B 、C 全不发生的概率为____________.(2)设随机变量X 服从参数为1的指数分布,则数学期望2{e }X E X -+=____________.十一、（本题满分6分）设随机变量X 与Y 独立,X 服从正态分布2(,),N Y μσ服从[,]ππ-上的均匀分布,试求Z X Y =+的概率分布密度(计算结果用标准正态分布函数Φ表示,其中22()e)t xx dt --∞Φ=.1993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)函数1()(2(0)x F x dt x =->⎰的单调减少区间为_____________.(2)由曲线 2232120x y z +==绕y 轴旋转一周得到的旋转面在点处的指向外侧的单位法向量为_____________.(3)设函数2()()f x x x x πππ=+-<<的傅里叶级数展开式为1(cos sin ),2n n n a a nx b nx ∞=++∑则其中系数3b 的值为_____________. (4)设数量场u =则div(grad )u =_____________. (5)设n 阶矩阵A 的各行元素之和均为零,且A 的秩为1,n -则线性方程组=AX 0的通解为_____________.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sin 2340()sin(),(),xf x t dtg x x x ==+⎰则当0x →时,()f x 是()g x 的 (A)等价无穷小(B)同价但非等价的无穷小 (C)高阶无穷小(D)低价无穷小(2)双纽线22222()x y x y +=-所围成的区域面积可用定积分表示为 (A)402cos 2d πθθ⎰(B)404cos 2d πθθ⎰(C)2θ(D)2401(cos 2)2d πθθ⎰(3)设有直线1158:121x y z l --+==-与2:l 623x y y z -=+=则1l 与2l 的夹角为(A)6π(B)4π(C)3π(D)2π(4)设曲线积分[()e ]sin ()cos x L f t ydx f x ydy --⎰与路径无关,其中()f x 具有一阶连续导数,且(0)0,f =则()f x 等于(A)e e 2x x--(B)e e 2x x--(C)e e 12x x-+-(D)e e 12x x-+-(5)已知12324,369t ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦Q P 为三阶非零矩阵,且满足0,=PQ 则 (A)6t =时P 的秩必为1 (B)6t =时P 的秩必为2 (C)6t ≠时P 的秩必为1 (D)6t ≠时P 的秩必为2三、(本题共3小题,每小题5分,满分15分) (1)求21lim(sin cos ).x x x x→∞+(2)求.x(3)求微分方程22,x y xy y '+=满足初始条件11x y ==的特解.四、(本题满分6分)计算22,xzdydz yzdzdx z dxdy ∑+-⎰⎰其中∑是由曲面z =与z =所围立体的表面外侧.五、(本题满分7分)求级数20(1)(1)2n nn n n ∞=--+∑的和. 六、(本题共2小题,每小题5分,满分10分)(1)设在[0,)+∞上函数()f x 有连续导数,且()0,(0)0,f x k f '≥><证明()f x 在(0,)+∞内有且仅有一个零点.(2)设,b a e >>证明.ba ab >七、（本题满分8分）已知二次型22212312323(,,)2332(0)f x x x x x x ax x a =+++>通过正交变换化成标准形22212325,fy y y =++求参数a 及所用的正交变换矩阵.八、（本题满分6分）设A 是n m ⨯矩阵,B 是m n ⨯矩阵,其中,n m <I 是n 阶单位矩阵,若,=AB I 证明B 的列向量组线性无关.九、（本题满分6分）设物体A 从点(0,1)出发,以速度大小为常数v 沿y 轴正向运动.物体B 从点(1,0)-与A 同时出发,其速度大小为2,v 方向始终指向,A 试建立物体B 的运动轨迹所满足的微分方程,并写出初始条件.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)一批产品共有10个正品和2个次品,任意抽取两次,每次抽一个,抽出后不再放回,则第二次抽出的是次品的概率为____________.(2)设随机变量X 服从(0,2)上的均匀分布,则随机变量2Y X =在(0,4)内的概率分布密度()Y f y =____________.十一、（本题满分6分）设随机变量X 的概率分布密度为1()e ,.2x f x x -=-∞<<+∞(1)求X 的数学期望EX 和方差.DX(2)求X 与X 的协方差,并问X 与X 是否不相关(3)问X 与X 是否相互独立为什么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)011lim cot ()sin x x xπ→-= _____________.(2)曲面e 23x z xy -+=在点(1,2,0)处的切平面方程为_____________.(3)设e sin ,x x u y -=则2ux y ∂∂∂在点1(2,)π处的值为_____________.(4)设区域D 为222,x y R +≤则2222()Dx y dxdy a b +⎰⎰=_____________.(5)已知11[1,2,3],[1,,],23==αβ设,'=A αβ其中'α是α的转置,则nA =_____________.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4342342222222sin cos ,(sin cos ),(sin cos ),1x M xdx N x x dx P x x x dx x ππππππ---==+=-+⎰⎰⎰则有 (A)N P M <<(B)MP N<<(C)N MP <<(D)P M N<<(2)二元函数(,)f x y 在点00(,)x y 处两个偏导数00(,)x f x y '、00(,)y f x y '存在是(,)f x y 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0,λ>且级数21n n a ∞=∑收敛,则级数1(1)nn ∞=-∑(A)发散(B)条件收敛 (C)绝对收敛(D)收敛性与λ有关 (4)2tan (1cos )lim 2,ln(12)(1)x x a x b x c x d e -→+-=-+-其中220,a c +≠则必有(A)4b d =(B)4b d =- (C)4a c =(D)4a c =-(5)已知向量组1234,,,αααα线性无关,则向量组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)011lim cot ()sin x x xπ→-= _____________.(2)曲面e 23x z xy -+=在点(1,2,0)处的切平面方程为_____________.(3)设e sin ,xx u y -=则2ux y ∂∂∂在点1(2,)π处的值为_____________.(4)设区域D 为222,x y R +≤则2222()Dx y dxdy a b +⎰⎰=_____________.(5)已知11[1,2,3],[1,,],23==αβ设,'=A αβ其中'α是α的转置,则nA =_____________.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4342342222222sin cos ,(sin cos ),(sin cos ),1x M xdx N x x dx P x x x dx x ππππππ---==+=-+⎰⎰⎰则有 (A)N P M <<(B)MP N<<(C)N MP <<(D)P M N<<(2)二元函数(,)f x y 在点00(,)x y 处两个偏导数00(,)x f x y '、00(,)y f x y '存在是(,)f x y 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0,λ>且级数21n n a ∞=∑收敛,则级数1(1)nn ∞=-∑(A)发散(B)条件收敛 (C)绝对收敛(D)收敛性与λ有关 (4)2tan (1cos )lim 2,ln(12)(1)x x a x b x c x d e -→+-=-+-其中220,a c +≠则必有(A)4b d =(B)4b d =- (C)4a c =(D)4a c =-(5)已知向量组1234,,,αααα线性无关,则向量组 (A)12233441,,,++++αααααααα线性无关(B)12233441,,,----αααααααα线性无关 (C)12233441,,,+++-αααααααα线性无关(D)12233441,,,++--αααααααα线性无关 三、(本题共3小题,每小题5分,满分15分)(1)设2221cos()cos()tx ty t t udu==-⎰,求dydx、22d ydx在t=.(2)将函数111()ln arctan412xf x x xx+=+--展开成x的幂级数.(3)求.sin(2)2sindxx x+⎰四、(本题满分6分)计算曲面积分2222,Sxdydz z dxdyx y z+++⎰⎰其中S是由曲面222x y R+=及,(0)z R z R R==->两平面所围成立体表面的外侧.五、(本题满分9分)设()f x具有二阶连续函数,(0)0,(0)1,f f'==且2[()()][()]0xy x y f x y dx f x x y dy'+-++=为一全微分方程,求()f x及此全微分方程的通解.六、(本题满分8分)设()f x在点0x=的某一邻域内具有二阶连续导数,且0()lim0,xf xx→=证明级数11()nfn∞=∑绝对收敛.七、（本题满分6分）已知点A与B的直角坐标分别为(1,0,0)与(0,1,1).线段AB绕x轴旋转一周所成的旋转曲面为.S求由S及两平面0,1z z==所围成的立体体积.八、（本题满分8分）设四元线性齐次方程组(Ⅰ)为1224x xx x+=-=,又已知某线性齐次方程组(Ⅱ)的通解为12(0,1,1,0)(1,2,2,1).k k+-(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A 为n 阶非零方阵*,A 是A 的伴随矩阵,'A 是A 的转置矩阵,当*'=A A 时,证明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A 、B 两个事件满足条件()(),P AB P AB =且(),P A p =则()P B =____________.(2)设相互独立的两个随机变量,X Y 具有同一分布率,且X 的分布率为则随机变量max{,}Z X Y =的分布率为____________. 十一、（本题满分6分）设随机变量X 和Y 分别服从正态分布2(1,3)N 和2(0,4),N 且X 与Y 的相关系数1,2xy ρ=-设,32X Y Z =+ (1)求Z 的数学期望EZ 和DZ 方差.(2)求X 与Z 的相关系数.xz ρ (3)问X 与Y 是否相互独立为什么1995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)2sin 0lim(13)xx x →+=_____________.(2)202cos xd x t dt dx ⎰= _____________.(3)设()2,⨯=a 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16,-=+A BA A BA 且100310,41007⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦A 则B =_____________.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:L 321021030x y z 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(2)设在[0,1]上()0,f x ''>则(0),(1),(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 ''>->(3)设()f x 可导,()()(1sin ),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)1nn u ∞=∑与21nn u ∞=∑都发散(C)1n n u ∞=∑收敛,而21nn u ∞=∑发散(D)1n n u ∞=∑收敛,而21nn u ∞=∑发散(5)设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 则必有(A)12AP P =B(B)21AP P =B (C)12P P 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(,,),(,e ,)0,sin ,y u f x y z x 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(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的余弦函数. 五、(本题满分7分)设曲线L 位于平面xOy 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 在平面xOy 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 上存在二阶导数,并且()0,()()()()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 fg g ξξξξ''='' 八、（本题满分7分）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101,1⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦ξ求.A 九、（本题满分6分）设A 为n 阶矩阵,满足('=AA I I 是n 阶单位矩阵,'A 是A 的转置矩阵),0,<A 求.+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,则2X 的数学期望2()E X =____________.(2)设X 和Y 为两个随机变量,且 则{max(,)0}P X Y ≥=____________. 十一、（本题满分6分） 设随机变量X 的概率密度为()X f x =e 0x- 00x x ≥<, 求随机变量e XY =的概率密度().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2lim()8,x x x a x a→∞+=-则a =_____________.(2)设一平面经过原点及点(6,3,2),-且与平面428x y z -+=垂直,则此平面方程为_____________.(3)微分方程22e x y y y '''-+=的通解为_____________. (4)函数ln(u x =+在点(1,0,1)A 处沿点A 指向点(3,2,2)B -方向的方向导数为_____________.(5)设A 是43⨯矩阵,且A 的秩()2,r =A 而102020,103⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦B 则()r AB =_____________.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2()()x ay dx ydyx y +++为某函数的全微分,a 则等于 (A)-1(B)0 (C)1(D)2(2)设()f x 具有二阶连续导数,且0()(0)0,lim1,x f x f x→'''==则 (A)(0)f 是()f x 的极大值 (B)(0)f 是()f x 的极小值 (C)(0,(0))f 是曲线()y f x =的拐点(D)(0)f 不是()f x 的极值,(0,(0))f 也不是曲线()y f x =的拐点(3)设0(1,2,),n a n >=且1n n a ∞=∑收敛,常数(0,),2πλ∈则级数21(1)(tan)nn n n a nλ∞=-∑(A)绝对收敛(B)条件收敛 (C)发散(D)散敛性与λ有关(4)设有()f x 连续的导数220,(0)0,(0)0,()()(),xf f F x x t f t dt '=≠=-⎰且当0x →时,()F x '与kx 是同阶无穷小,则k 等于(A)1(B)2 (C)3(D)4(5)四阶行列式112233440000000a b a b a b b a 的值等于(A)12341234a a a a b b b b -(B)12341234a a a a b b b b +。
历年考研数学一真题及答案(1987-2013)
历年考研数学一真题1987-2013（经典珍藏版）1987年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上)(1)当x =_____________时,函数2x y x =⋅取得极小值.(2)由曲线ln y x =与两直线e 1y x =+-及0y =所围成的平面图形的面积是_____________.1x = (3)与两直线 1y t =-+2z t =+及121111x y z +++==都平行且过原点的平面方程为_____________.(4)设L为取正向的圆周229,x y +=则曲线积分2(22)(4)Lxy y dx x x dy -+-⎰= _____________.(5)已知三维向量空间的基底为123(1,1,0),(1,0,1),(0,1,1),===ααα则向量(2,0,0)=β在此基底下的坐标是_____________.二、(本题满分8分)求正的常数a 与,b 使等式201lim 1sin x x bx x →=-⎰成立.三、(本题满分7分)(1)设f 、g 为连续可微函数,(,),(),u f x xy v g x xy ==+求,.u v x x∂∂∂∂ (2)设矩阵A和B满足关系式2,+AB =A B 其中301110,014⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦A 求矩阵.B四、(本题满分8分)求微分方程26(9)1y y a y ''''''+++=的通解,其中常数0.a >五、选择题(本题共4小题,每小题3分,满分12分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内) (1)设2()()lim1,()x af x f a x a →-=--则在x a =处 (A)()f x 的导数存在,且()0f a '≠ (B)()f x 取得极大值(C)()f x 取得极小值 (D)()f x 的导数不存在 (2)设()f x 为已知连续函数0,(),st I t f tx dx =⎰其中0,0,t s >>则I 的值(A)依赖于s 和t (B)依赖于s 、t 和x(C)依赖于t 、x ,不依赖于s (D)依赖于s ,不依赖于t(3)设常数0,k >则级数21(1)n n k n n∞=+-∑(A)发散 (B)绝对收敛(C)条件收敛 (D)散敛性与k 的取值有关(4)设A 为n 阶方阵，且A 的行列式||0,a =≠A 而*A 是A 的伴随矩阵，则*||A 等于(A)a (B)1a(C)1n a - (D)na六、（本题满分10分）求幂级数1112n n n x n ∞-=∑的收敛域，并求其和函数.七、（本题满分10分） 求曲面积分2(81)2(1)4,I x y dydz y dzdx yzdxdy ∑=++--⎰⎰其中∑是由曲线13()0z y f x x ⎧=≤≤⎪=⎨=⎪⎩绕y 轴旋转一周而成的曲面,其法向量与y 轴正向的夹角恒大于.2π八、（本题满分10分）设函数()f x 在闭区间[0,1]上可微,对于[0,1]上的每一个,x 函数()f x 的值都在开区间(0,1)内,且()f x '≠1,证明在(0,1)内有且仅有一个,x 使得().f x x =九、（本题满分8分） 问,a b 为何值时,现线性方程组123423423412340221(3)2321x x x x x x x x a x x b x x x ax +++=++=-+--=+++=-有唯一解,无解,有无穷多解?并求出有无穷多解时的通解.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)设在一次实验中,事件A 发生的概率为,p 现进行n 次独立试验,则A 至少发生一次的概率为____________;而事件A 至多发生一次的概率为____________.(2)有两个箱子,第1个箱子有3个白球,2个红球, 第2个箱子有4个白球,4个红球.现从第1个箱子中随机地取1个球放到第2个箱子里,再从第2个箱子中取出1个球,此球是白球的概率为____________.已知上述从第2个箱子中取出的球是白球,则从第一个箱子中取出的球是白球的概率为____________. (3)已知连续随机变量X 的概率密度函数为221(),xx f x-+-=则X 的数学期望为____________,X 的方差为____________.十一、（本题满分6分）设随机变量,X Y 相互独立,其概率密度函数分别为()X f x = 1001x ≤≤其它,()Y f y e 0y - 00y y >≤, 求2Z X Y =+的概率密度函数.1988年全国硕士研究生入学统一考试数学(一)试卷一、(本题共3小题,每小题5分,满分15分)(1)求幂级数1(3)3nnn x n ∞=-∑的收敛域. (2)设2()e ,[()]1x f x f x x ϕ==-且()0x ϕ≥,求()x ϕ及其定义域.(3)设∑为曲面2221x y z ++=的外侧,计算曲面积分333.I x dydz y dzdx z dxdy ∑=++⎰⎰二、填空题(本题共4小题,每小题3分,满分12分.把答案填在题中横线上)(1)若21()lim (1),tx x f t t x→∞=+则()f t '= _____________.(2)设()f x 连续且31(),x f t dt x -=⎰则(7)f =_____________. (3)设周期为2的周期函数,它在区间(1,1]-上定义为()f x =22x1001x x -<≤<≤,则的傅里叶()Fourier 级数在1x =处收敛于_____________.(4)设4阶矩阵234234[,,,],[,,,],==A αγγγB βγγγ其中234,,,,αβγγγ均为4维列向量,且已知行列式4,1,==A B 则行列式+A B = _____________.三、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内) (1)设()f x 可导且01(),2f x '=则0x ∆→时,()f x 在0x 处的微分dy 是(A)与x ∆等价的无穷小 (B)与x∆同阶的无穷小(C)比x ∆低阶的无穷小 (D)比x ∆高阶的无穷小 (2)设()y f x =是方程240y y y '''-+=的一个解且00()0,()0,f x f x '>=则函数()f x 在点0x 处(A)取得极大值 (B)取得极小值(C)某邻域内单调增加 (D)某邻域内单调减少 (3)设空间区域2222222212:,0,:,0,0,0,x y z R z x y z R x y z Ω++≤≥Ω++≤≥≥≥则(A)124xdv dv ΩΩ=⎰⎰⎰⎰⎰⎰(B)124ydv ydv ΩΩ=⎰⎰⎰⎰⎰⎰(C)124zdv zdv ΩΩ=⎰⎰⎰⎰⎰⎰(D)124xyzdv xyzdv ΩΩ=⎰⎰⎰⎰⎰⎰(4)设幂级数1(1)n n n a x ∞=-∑在1x =-处收敛,则此级数在2x =处(A)条件收敛 (B)绝对收敛(C)发散 (D)收敛性不能确定(5)n 维向量组12,,,(3)s s n ≤≤ααα线性无关的充要条件是(A)存在一组不全为零的数12,,,,s k k k 使11220s s k k k +++≠ααα(B)12,,,s ααα中任意两个向量均线性无关(C)12,,,s ααα中存在一个向量不能用其余向量线性表示(D)12,,,s ααα中存在一个向量都不能用其余向量线性表示四、(本题满分6分)设()(),x y u yf xg yx=+其中函数f 、g 具有二阶连续导数,求222.u u x y x x y∂∂+∂∂∂五、(本题满分8分)设函数()y y x =满足微分方程322e ,x y y y '''-+=其图形在点(0,1)处的切线与曲线21y x x =--在该点处的切线重合,求函数().y y x =六、（本题满分9分）设位于点(0,1)的质点A 对质点M 的引力大小为2(0kk r>为常数,r 为A 质点与M 之间的距离),质点M 沿直线y =(2,0)B 运动到(0,0),O 求在此运动过程中质点A 对质点M 的引力所作的功.七、（本题满分6分）已知,=AP BP 其中100100000,210,001211⎡⎤⎡⎤⎢⎥⎢⎥==-⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦B P 求5,.A A八、（本题满分8分）已知矩阵20000101x ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦A 与20000001y ⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦B 相似. (1)求x 与.y (2)求一个满足1-=PAP B的可逆阵.P九、（本题满分9分）设函数()f x 在区间[,]a b 上连续,且在(,)a b 内有()0,f x '>证明:在(,)a b 内存在唯一的,ξ使曲线()y f x =与两直线(),y f x a ξ==所围平面图形面积1S 是曲线()y f x =与两直线(),y f x b ξ==所围平面图形面积2S 的3倍.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)设在三次独立试验中,事件A 出现的概率相等,若已知A 至少出现一次的概率等于19,27则事件A 在一次试验中出现的概率是____________.(2)若在区间(0,1)内任取两个数,则事件”两数之和小于65”的概率为____________.(3)设随机变量X 服从均值为10,均方差为0.02的正态分布,已知22(),(2.5)0.9938,u xx du φφ-==⎰则X 落在区间(9.95,10.05)内的概率为____________.十一、（本题满分6分） 设随机变量X 的概率密度函数为21(),(1)X f x x π=-求随机变量1Y =的概率密度函数().Y f y1989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)已知(3)2,f '=则0(3)(3)lim2h f h f h→--= _____________. (2)设()f x 是连续函数,且10()2(),f x x f t dt =+⎰则()f x =_____________.(3)设平面曲线L为下半圆周y =则曲线积分22()Lxy ds +⎰=_____________.(4)向量场div u在点(1,1,0)P 处的散度div u =_____________.(5)设矩阵300100140,010,003001⎡⎤⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦A I 则矩阵1(2)--A I =_____________.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)当0x >时,曲线1sin y x x=(A)有且仅有水平渐近线 (B)有且仅有铅直渐近线(C)既有水平渐近线,又有铅直渐近线 (D)既无水平渐近线,又无铅直渐近线(2)已知曲面224z x y =--上点P 处的切平面平行于平面2210,x y z ++-=则点的坐标是(A)(1,1,2)- (B)(1,1,2)-(C)(1,1,2) (D)(1,1,2)-- (3)设线性无关的函数都是二阶非齐次线性方程的解是任意常数,则该非齐次方程的通解是(A)11223c y c y y ++ (B)1122123()c y c y c c y +-+(C)1122123(1)c y c y c c y +--- (D)1122123(1)c y c y c c y ++--(4)设函数2(),01,f x x x =≤<而1()sin ,,n n S x b n x x π∞==-∞<<+∞∑其中12()sin ,1,2,3,,n b f x n xdx n π==⎰则1()2S -等于(A)12- (B)14-(C)14(D)12(5)设A 是n 阶矩阵,且A 的行列式0,=A 则A 中 (A)必有一列元素全为0 (B)必有两列元素对应成比例(C)必有一列向量是其余列向量的线性组合 (D)任一列向量是其余列向量的线性组合三、(本题共3小题,每小题5分,满分15分) (1)设(2)(,),z f x y g x xy =-+其中函数()f t 二阶可导,(,)g u v 具有连续二阶偏导数,求2.zx y∂∂∂ (2)设曲线积分2()c xy dx y x dy ϕ+⎰与路径无关,其中()x ϕ具有连续的导数,且(0)0,ϕ=计算(1,1)2(0,0)()xy dx y x dy ϕ+⎰的值.(3)计算三重积分(),x z dv Ω+⎰⎰⎰其中Ω是由曲面z =与z =所围成的区域.四、(本题满分6分)将函数1()arctan 1x f x x+=-展为x 的幂级数.五、(本题满分7分)设0()sin ()(),xf x x x t f t dt =--⎰其中f 为连续函数,求().f x六、（本题满分7分）证明方程0ln exx π=-⎰在区间(0,)+∞内有且仅有两个不同实根.七、（本题满分6分）问λ为何值时,线性方程组13x x λ+= 123422x x x λ++=+ 1236423x x x λ++=+有解,并求出解的一般形式. 八、（本题满分8分）假设λ为n 阶可逆矩阵A 的一个特征值,证明 (1)1λ为1-A 的特征值.(2)λA为A 的伴随矩阵*A 的特征值.九、（本题满分9分）设半径为R 的球面∑的球心在定球面2222(0)x y z a a ++=>上,问当R 为何值时,球面∑在定球面内部的那部分的面积最大?十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)已知随机事件A 的概率()0.5,P A =随机事件B 的概率()0.6P B =及条件概率(|)0.8,P B A =则和事件AB的概率()P AB =____________.(2)甲、乙两人独立地对同一目标射击一次,其命中率分别为0.6和0.5,现已知目标被命中,则它是甲射中的概率为____________.(3)若随机变量ξ在(1,6)上服从均匀分布,则方程210x x ξ++=有实根的概率是____________.十一、（本题满分6分）设随机变量X 与Y 独立,且X 服从均值为1、标准差(均方差)的正态分布,而Y 服从标准正态分布.试求随机变量23Z X Y =-+的概率密度函数.1990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)2x t =-+(1)过点(1,21)M -且与直线 34y t =-垂直的平面方程是_____________.1z t =-(2)设a 为非零常数,则lim()x x x a x a→∞+-=_____________.(3)设函数()f x =111x x ≤>,则[()]f f x =_____________.(4)积分2220e y x dx dy -⎰⎰的值等于_____________. (5)已知向量组1234(1,2,3,4),(2,3,4,5),(3,4,5,6),(4,5,6,7),====αααα则该向量组的秩是_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设()f x 是连续函数,且e ()(),xx F x f t dt -=⎰则()F x '等于(A)e (e )()x x f f x ---- (B)e (e )()x x f f x ---+(C)e (e )()x x f f x ---(D)e (e )()x x f f x --+ (2)已知函数()f x 具有任意阶导数,且2()[()],f x f x '=则当n 为大于2的正整数时,()f x 的n 阶导数()()n f x 是(A)1![()]n n f x + (B)1[()]n n f x +(C)2[()]n f x(D)2![()]n n f x(3)设a 为常数,则级数21sin()[n na n∞=∑ (A)绝对收敛 (B)条件收敛(C)发散 (D)收敛性与a 的取值有关 (4)已知()f x 在0x =的某个邻域内连续,且0()(0)0,lim2,1cos x f x f x→==-则在点0x =处()f x (A)不可导 (B)可导,且(0)0f '≠(C)取得极大值 (D)取得极小值(5)已知1β、2β是非齐次线性方程组=AX b 的两个不同的解1,α、2α是对应其次线性方程组=AX 0的基础解析1,k 、2k 为任意常数,则方程组=AX b 的通解(一般解)必是(A)1211212()2k k -+++ββααα(B)1211212()2k k ++-+ββααα (C)1211212()2k k -+++ββαββ(D)1211212()2k k ++-+ββαββ三、(本题共3小题,每小题5分,满分15分)(1)求120ln(1).(2)x dx x +-⎰(2)设(2,sin ),z f x y y x =-其中(,)f u v 具有连续的二阶偏导数,求2.zx y∂∂∂(3)求微分方程244e x y y y -'''++=的通解(一般解).四、(本题满分6分)求幂级数0(21)n n n x ∞=+∑的收敛域,并求其和函数.五、(本题满分8分) 求曲面积分2SI yzdzdx dxdy =+⎰⎰其中S 是球面2224x y z ++=外侧在0z ≥的部分.六、（本题满分7分）设不恒为常数的函数()f x 在闭区间[,]a b 上连续,在开区间(,)a b 内可导,且()().f a f b =证明在(,)a b 内至少存在一点,ξ使得()0.f ξ'>七、（本题满分6分） 设四阶矩阵1100213401100213,0011002100010002-⎡⎤⎡⎤⎢⎥⎢⎥-⎢⎥⎢⎥==⎢⎥⎢⎥-⎢⎥⎢⎥⎣⎦⎣⎦B C 且矩阵A 满足关系式1()-''-=A E C B C E其中E 为四阶单位矩阵1,-C 表示C 的逆矩阵,'C 表示C 的转置矩阵.将上述关系式化简并求矩阵.A八、（本题满分8分）求一个正交变换化二次型22212312132344448f x x x x x x x x x =++-+-成标准型.九、（本题满分8分）质点P 沿着以AB 为直径的半圆周,从点(1,2)A 运动到点(3,4)B 的过程中受变力F 作用(见图).F 的大小等于点P 与原点O 之间的距离,其方向垂直于线段OP 且与y 轴正向的夹角小于.2π求变力F 对质点P 所作的功.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)已知随机变量X 的概率密度函数1()e ,2xf x x -=-∞<<+∞ 则X 的概率分布函数()F x =____________.(2)设随机事件A 、B 及其和事件的概率分别是0.4、0.3和0.6,若B 表示B 的对立事件,那么积事件AB 的概率()P AB =____________.(3)已知离散型随机变量X 服从参数为2的泊松()Poisson 分布,即22e {},0,1,2,,!k P X k k k -===则随机变量32Z X =-的数学期望()E Z =____________.十一、（本题满分6分）设二维随机变量(,)X Y 在区域:01,D x y x <<<内服从均匀分布,求关于X 的边缘概率密度函数及随机变量21Z X =+的方差().D Z1991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 21cos x t y t=+=,则22d ydx =_____________.(2)由方程xyz +=所确定的函数(,)z z x y =在点(1,0,1)-处的全微分dz =_____________.(3)已知两条直线的方程是1212321:;:.101211x y z x y zl l ---+-====-则过1l 且平行于2l 的平面方程是_____________.(4)已知当0x →时123,(1)1ax +-与cos 1x -是等价无穷小,则常数a =_____________.(5)设4阶方阵52002100,00120011⎡⎤⎢⎥⎢⎥=⎢⎥-⎢⎥⎣⎦A 则A的逆阵1-A =_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)曲线221e 1ex x y --+=-(A)没有渐近线 (B)仅有水平渐近线(C)仅有铅直渐近线 (D)既有水平渐近线又有铅直渐近线 (2)若连续函数()f x 满足关系式20()()ln 2,2tf x f dt π=+⎰则()f x 等于(A)e ln 2x(B)2e ln 2x(C)eln 2x+(D)2eln 2x+(3)已知级数12111(1)2,5,n n n n n a a ∞∞--==-==∑∑则级数1n n a ∞=∑等于(A)3 (B)7 (C)8 (D)9 (4)设D 是平面xoy 上以(1,1)、(1,1)-和(1,1)--为顶点的三角形区域1,D 是D 在第一象限的部分,则(cos sin )Dxy x y dxdy +⎰⎰等于(A)12cos sin D x ydxdy ⎰⎰ (B)12D xydxdy ⎰⎰(C)14(cos sin )D xy x y dxdy +⎰⎰ (D)0(5)设n 阶方阵A 、B 、C 满足关系式,=ABC E 其中E 是n 阶单位阵,则必有(A)=ACB E (B)=CBA E(C)=BAC E (D)=BCA E三、(本题共3小题,每小题5分,满分15分) (1)求20lim ).x π+→(2)设n 是曲面222236x y z ++=在点(1,1,1)P 处的指向外侧的法向量,求函数u =在点P 处沿方向n 的方向导数.(3)22(),x y z dv Ω++⎰⎰⎰其中Ω是由曲线 220yz x ==绕z 轴旋转一周而成的曲面与平面4z =所围城的立体.四、(本题满分6分)过点(0,0)O 和(,0)A π的曲线族sin (0)y a x a =>中,求一条曲线,L 使沿该曲线O 从到A 的积分3(1)(2)Ly dx x y dy +++⎰的值最小.五、(本题满分8分)将函数()2(11)f x x x =+-≤≤展开成以2为周期的傅里叶级数,并由此求级数211n n∞=∑的和.六、（本题满分7分） 设函数()f x 在[0,1]上连续,(0,1)内可导,且1233()(0),f x dx f =⎰证明在(0,1)内存在一点,c 使()0.f c '=七、（本题满分8分） 已知1234(1,0,2,3),(1,1,3,5),(1,1,2,1),(1,2,4,8)a a ===-+=+αααα及(1,1,3,5).b =+β(1)a 、b 为何值时,β不能表示成1234,,,αααα的线性组合?(2)a 、b 为何值时,β有1234,,,αααα的唯一的线性表示式?写出该表示式.八、（本题满分6分）设A 是n 阶正定阵,E 是n 阶单位阵,证明+A E 的行列式大于1.九、（本题满分8分）在上半平面求一条向上凹的曲线,其上任一点(,)P x y 处的曲率等于此曲线在该点的法线段PQ 长度的倒数(Q是法线与x 轴的交点),且曲线在点(1,1)处的切线与x 轴平行.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)若随机变量X 服从均值为2、方差为2σ的正态分布,且{24}0.3,P X <<=则{0}P X <=____________.(2)随机地向半圆0y a <<为正常数)内掷一点,点落在半圆内任何区域的概率与区域的面积成正比,则原点和该点的连线与x 轴的夹角小于4π的概率为____________.十一、（本题满分6分）设二维随机变量(,)X Y 的密度函数为(,)f x y =(2)2e 0,00 x y x y -+>>其它求随机变量2Z X Y =+的分布函数.1992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)设函数()y y x =由方程e cos()0x yxy ++=确定,则dydx=_____________.(2)函数222ln()u x y z =++在点(1,2,2)M -处的梯度grad Mu=_____________.(3)设()f x =211x-+ 00x x ππ-<≤<≤,则其以2π为周期的傅里叶级数在点x π处收敛于_____________. (4)微分方程tan cos y y x x'+=的通解为y=_____________.(5)设111212121212,n n n n n n a b a b a b a b a b a b a b a b a b ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦A 其中0,0,(1,2,,).i i a b i n ≠≠=则矩阵A的秩()r A =_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)当1x →时,函数1211e 1x x x ---的极限(A)等于2 (B)等于0 (C)为∞ (D)不存在但不为∞(2)级数1(1)(1cos )(n n a n∞=--∑常数0)a >(A)发散 (B)条件收敛(C)绝对收敛 (D)收敛性与a 有关(3)在曲线23,,x t y t z t ==-=的所有切线中,与平面24x y z ++=平行的切线(A)只有1条 (B)只有2条(C)至少有3条 (D)不存在(4)设32()3,f x x x x =+则使()(0)n f 存在的最高阶数n 为 (A)0 (B)1(C)2 (D)3(5)要使12100,121⎛⎫⎛⎫ ⎪ ⎪== ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭ξξ都是线性方程组=AX 0的解,只要系数矩阵A 为(A)[]212- (B)201011-⎡⎤⎢⎥⎣⎦(C)102011-⎡⎤⎢⎥-⎣⎦(D)011422011-⎡⎤⎢⎥--⎢⎥⎢⎥⎣⎦三、(本题共3小题,每小题5分,满分15分) (1)求x x →(2)设22(e sin ,),x z f y x y =+其中f 具有二阶连续偏导数,求2.zx y∂∂∂ (3)设()f x = 21exx -+ 00x x ≤>,求31(2).f x dx -⎰四、(本题满分6分)求微分方程323e x y y y -'''+-=的通解.五、(本题满分8分) 计算曲面积分323232()()(),xaz dydz y ax dzdx z ay dxdy ∑+++++⎰⎰其中∑为上半球面z =.六、（本题满分7分） 设()0,(0)0,f x f ''<=证明对任何120,0,x x >>有1212()()().f x x f x f x +<+七、（本题满分8分） 在变力F yzizxj xyk=++的作用下,质点由原点沿直线运动到椭球面2222221x y z a b c++=上第一卦限的点(,,),M ξηζ问当ξ、η、ζ取何值时,力F 所做的功W 最大?并求出W 的最大值.八、（本题满分7分）设向量组123,,ααα线性相关,向量组234,,ααα线性无关,问:(1)1α能否由23,αα线性表出?证明你的结论. (2)4α能否由123,,ααα线性表出?证明你的结论.九、（本题满分7分）设3阶矩阵A 的特征值为1231,2,3,λλλ===对应的特征向量依次为1231111,2,3,149⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=== ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ξξξ又向量12.3⎛⎫⎪= ⎪ ⎪⎝⎭β (1)将β用123,,ξξξ线性表出. (2)求(n n A β为自然数).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11()()(),()0,()(),46P A P B P C P AB P AC P BC ======则事件A、B 、C 全不发生的概率为____________.(2)设随机变量X 服从参数为1的指数分布,则数学期望2{e }X E X -+=____________.十一、（本题满分6分）设随机变量X 与Y 独立,X 服从正态分布2(,),N Y μσ服从[,]ππ-上的均匀分布,试求Z X Y =+的概率分布密度(计算结果用标准正态分布函数Φ表示,其中22()e)t xx dt --∞Φ=.1993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)函数1()(2(0)x F x dt x =->⎰的单调减少区间为_____________.(2)2232120x y z +==绕y 轴旋转一周得到的旋转面在点处的指向外侧的单位法向量为_____________.(3)设函数2()()f x x x x πππ=+-<<的傅里叶级数展开式为1(cos sin ),2n n n a a nx b nx ∞=++∑则其中系数3b 的值为_____________. (4)设数量场u =则div(grad )u =_____________.(5)设n 阶矩阵A 的各行元素之和均为零,且A 的秩为1,n -则线性方程组=AX 0的通解为_____________.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sin 2340()sin(),(),xf x t dtg x x x ==+⎰则当0x →时,()f x 是()g x 的(A)等价无穷小 (B)同价但非等价的无穷小(C)高阶无穷小 (D)低价无穷小(2)双纽线22222()x y x y +=-所围成的区域面积可用定积分表示为(A)402cos 2d πθθ⎰ (B)404cos 2d πθθ⎰(C)2θ(D)2401(cos 2)2d πθθ⎰(3)设有直线1158:121x y z l --+==-与2:l 623x y y z -=+=则1l 与2l 的夹角为(A)6π(B)4π(C)3π(D)2π(4)设曲线积分[()e ]sin ()cos xL f t ydx f x ydy --⎰与路径无关,其中()f x 具有一阶连续导数,且(0)0,f =则()f x 等于(A)e e 2x x--(B)e e 2x x--(C)e e 12x x-+-(D)e e 12x x-+-(5)已知12324,369t ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦Q P 为三阶非零矩阵,且满足0,=PQ 则(A)6t =时P 的秩必为1 (B)6t =时P的秩必为2(C)6t ≠时P 的秩必为1 (D)6t ≠时P的秩必为2三、(本题共3小题,每小题5分,满分15分)(1)求21lim(sin cos ).x x x x→∞+(2)求.x(3)求微分方程22,x y xy y '+=满足初始条件11x y ==的特解.四、(本题满分6分)计算22,xzdydz yzdzdx z dxdy ∑+-⎰⎰其中∑是由曲面z =与z =.五、(本题满分7分)求级数20(1)(1)2n nn n n ∞=--+∑的和.六、(本题共2小题,每小题5分,满分10分) (1)设在[0,)+∞上函数()f x 有连续导数,且()0,(0)0,f x k f '≥><证明()f x 在(0,)+∞内有且仅有一个零点.(2)设,b a e >>证明.ba ab >七、（本题满分8分） 已知二次型22212312323(,,)2332(0)f x x x x x x ax x a =+++>通过正交变换化成标准形22212325,f y y y =++求参数a 及所用的正交变换矩阵.八、（本题满分6分）设A 是n m ⨯矩阵,B 是m n ⨯矩阵,其中,n m <I 是n 阶单位矩阵,若,=AB I 证明B 的列向量组线性无关.九、（本题满分6分）设物体A 从点(0,1)出发,以速度大小为常数v 沿y 轴正向运动.物体B 从点(1,0)-与A 同时出发,其速度大小为2,v 方向始终指向,A 试建立物体B 的运动轨迹所满足的微分方程,并写出初始条件.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)一批产品共有10个正品和2个次品,任意抽取两次,每次抽一个,抽出后不再放回,则第二次抽出的是次品的概率为____________.(2)设随机变量X 服从(0,2)上的均匀分布,则随机变量2Y X =在(0,4)内的概率分布密度()Y f y =____________.十一、（本题满分6分） 设随机变量X的概率分布密度为1()e ,.2xf x x -=-∞<<+∞ (1)求X 的数学期望EX 和方差.DX(2)求X 与X 的协方差,并问X 与X 是否不相关? (3)问X 与X 是否相互独立?为什么?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)011lim cot ()sin x x xπ→-= _____________.(2)曲面e 23x z xy -+=在点(1,2,0)处的切平面方程为_____________. (3)设e sin ,xxu y-=则2u x y ∂∂∂在点1(2,)π处的值为_____________.(4)设区域D为222,x y R +≤则2222()Dx y dxdy a b +⎰⎰=_____________. (5)已知11[1,2,3],[1,,],23==αβ设,'=A αβ其中'α是α的转置,则nA =_____________.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4342342222222sin cos ,(sin cos ),(sin cos ),1x M xdx N x x dx P x x x dx x ππππππ---==+=-+⎰⎰⎰则有(A)N P M << (B)MP N<<(C)N MP <<(D)P MN<<(2)二元函数(,)f x y 在点00(,)x y 处两个偏导数00(,)x f x y '、00(,)y f x y '存在是(,)f x y 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0,λ>且级数21n n a ∞=∑收敛,则级数1(1)nn ∞=-∑(A)发散 (B)条件收敛(C)绝对收敛 (D)收敛性与λ有关 (4)2tan (1cos )lim2,ln(12)(1)x x a x b x c x d e-→+-=-+-其中220,a c +≠则必有(A)4b d = (B)4b d =- (C)4a c = (D)4a c =- (5)已知向量组1234,,,αααα线性无关,则向量组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)011lim cot ()sin x x xπ→-= _____________.(2)曲面e 23x z xy -+=在点(1,2,0)处的切平面方程为_____________. (3)设e sin ,xxu y-=则2u x y ∂∂∂在点1(2,)π处的值为_____________. (4)设区域D为222,x y R +≤则2222()Dx y dxdy a b +⎰⎰=_____________. (5)已知11[1,2,3],[1,,],23==αβ设,'=A αβ其中'α是α的转置,则nA =_____________.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4342342222222sin cos ,(sin cos ),(sin cos ),1x M xdx N x x dx P x x x dx x ππππππ---==+=-+⎰⎰⎰则有(A)N P M << (B)MP N<<(C)N MP <<(D)P MN<<(2)二元函数(,)f x y 在点00(,)x y 处两个偏导数00(,)x f x y '、00(,)y f x y '存在是(,)f x y 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0,λ>且级数21n n a ∞=∑收敛,则级数1(1)nn ∞=-∑(A)发散 (B)条件收敛(C)绝对收敛 (D)收敛性与λ有关 (4)2tan (1cos )lim2,ln(12)(1)x x a x b x c x d e -→+-=-+-其中220,a c +≠则必有(A)4b d = (B)4b d =-(C)4a c = (D)4a c =- (5)已知向量组1234,,,αααα线性无关,则向量组 (A)12233441,,,++++αααααααα线性无关 (B)12233441,,,----αααααααα线性无关 (C)12233441,,,+++-αααααααα线性无关 (D)12233441,,,++--αααααααα线性无关三、(本题共3小题,每小题5分,满分15分)(1)设 2221cos()cos()tx t y t t udu ==-⎰,求dy dx 、22d y dx在t =的值.(2)将函数111()ln arctan 412x f x x x x +=+--展开成x 的幂级数.(3)求.sin(2)2sin dxx x+⎰四、(本题满分6分)计算曲面积分2222,Sxdydz z dxdyx y z +++⎰⎰其中S是由曲面222x y R +=及,(0)z R z R R ==->两平面所围成立体表面的外侧.五、(本题满分9分) 设()f x 具有二阶连续函数,(0)0,(0)1,f f '==且2[()()][()]0xy x y f x y dx f x x y dy '+-++=为一全微分方程,求()f x 及此全微分方程的通解.六、(本题满分8分)设()f x 在点0x =的某一邻域内具有二阶连续导数,且()lim0,x f x x →=证明级数11()n f n ∞=∑绝对收敛.七、（本题满分6分）已知点A 与B 的直角坐标分别为(1,0,0)与(0,1,1).线段AB绕x 轴旋转一周所成的旋转曲面为.S 求由S 及两平面0,1z z ==所围成的立体体积.八、（本题满分8分）设四元线性齐次方程组(Ⅰ)为122400x x x x +=-=,又已知某线性齐次方程组(Ⅱ)的通解为12(0,1,1,0)(1,2,2,1).k k +-(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A 为n 阶非零方阵*,A 是A 的伴随矩阵,'A 是A 的转置矩阵,当*'=AA 时,证明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A 、B 两个事件满足条件()(),P AB P AB =且(),P A p =则()P B =____________.(2)设相互独立的两个随机变量,X Y 具有同一分布率,且X 的分布率为则随机变量max{,}Z X Y =的分布率为____________.十一、（本题满分6分） 设随机变量X和Y 分别服从正态分布2(1,3)N 和2(0,4),N 且X 与Y 的相关系数1,2xy ρ=-设,32X Y Z =+(1)求Z 的数学期望EZ 和DZ 方差. (2)求X 与Z 的相关系数.xz ρ (3)问X 与Y 是否相互独立?为什么?1995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)2sin 0lim(13)xx x →+=_____________.(2)202cos x d x t dt dx ⎰= _____________.(3)设()2,⨯=a 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16,-=+A BA A BA 且100310,41007⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦A 则B =_____________.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:L321021030x y z 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(2)设在[0,1]上()0,f x ''>则(0),(1),(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 ''>->(3)设()f x 可导,()()(1sin ),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21nn u ∞=∑都发散(C)1n n u ∞=∑收敛,而21nn u ∞=∑发散 (D)1n n u ∞=∑收敛,而21nn u ∞=∑发散(5)设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 则必有(A)12AP P =B (B)21AP P =B (C)12P P 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(,,),(,e ,)0,sin ,y u f x y z x z y x ϕ===其中,f ϕ都具有一阶连续偏导数,且0.zϕ∂≠∂求.du dx(2)设函数()f x 在区间[0,1]上连续,并设1(),f x dx A =⎰求110()().x dx 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的余弦函数.五、(本题满分7分)设曲线L 位于平面xOy 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 在平面xOy 上具有一阶连续偏导数,曲线积分2(,)L xydx 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 上存在二阶导数,并且()0,()()()()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 fg g ξξξξ''=''八、（本题满分7分）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101,1⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦ξ求.A九、（本题满分6分）设A 为n 阶矩阵,满足('=AA I I 是n 阶单位矩阵,'A 是A 的转置矩阵),0,<A 求.+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34{0,0},{0}{0},77P X Y P X P Y ≥≥=≥=≥= 则{max(,)0}P X Y ≥=____________.十一、（本题满分6分） 设随机变量X 的概率密度为()X f x = e 0x- 00x x ≥<,求随机变量e XY =的概率密度().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2lim()8,x x x a x a →∞+=-则a =_____________.(2)设一平面经过原点及点(6,3,2),-且与平面428x y z -+=垂直,则此平面方程为_____________.(3)微分方程22e x y y y '''-+=的通解为_____________. (4)函数ln(u x =在点(1,0,1)A 处沿点A 指向点(3,2,2)B -方向的方向导数为_____________.(5)设A 是43⨯矩阵,且A 的秩()2,r =A 而102020,103⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦B 则()r AB =_____________.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2()()x ay dx ydyx y +++为某函数的全微分,a 则等于 (A)-1 (B)0(C)1 (D)2 (2)设()f x 具有二阶连续导数,且()(0)0,lim1,x f x f x→'''==则(A)(0)f 是()f x 的极大值 (B)(0)f 是()f x 的极小值 (C)(0,(0))f 是曲线()y f x =的拐点(D)(0)f 不是()f x 的极值,(0,(0))f 也不是曲线()y f x =的拐点(3)设0(1,2,),n a n >=且1n n a ∞=∑收敛,常数(0,),2πλ∈则级数21(1)(tan )n n n n a nλ∞=-∑(A)绝对收敛 (B)条件收敛(C)发散 (D)散敛性与λ有关 (4)设有()f x 连续的导数220,(0)0,(0)0,()()(),xf f F x x t f t dt '=≠=-⎰且当0x →时,()F x '与k x 是同阶无穷小,则k 等于(A)1 (B)2 (C)3 (D)4(5)四阶行列式112233440000000a b a b a b b a 的值等于(A)12341234a a a a b b b b - (B)12341234a a a a b b b b +(C)12123434()()a a b b a a b b -- (D)23231414()()a a b b a a b b --三、(本题共2小题,每小题5分,满分10分) (1)求心形线(1cos )r a θ=+的全长,其中0a >是常数.(2)设1110,1,2,),n x x n +===试证数列{}n x 极限存在,并求此极限.四、(本题共2小题,每小题6分,满分12分) (1)计算曲面积分(2),Sx z dydz zdxdy ++⎰⎰其中S 为有向曲面22(01),z x y x =+≤≤其法向量与z 轴正向的夹角为锐角.(2)设变换 2u x y v x ay =-=+可把方程2222260z z zx x y y∂∂∂+-=∂∂∂∂简化为20,zu v∂=∂∂求常数.a五、(本题满分7分) 求级数211(1)2nn n ∞=-∑的和.六、(本题满分7分) 设对任意0,x >曲线()y f x =上点(,())x f x 处的切线在y轴上的截距等于01(),x f t dt x⎰求()f x 的一般表达式.。
89--98年考研数学一
1987年全国硕士研究生入学统一考试数学(一)试卷一、填空题(本题共5小题,每小题3分,满分15分.把答案填在题中横线上) (1)当x =_____________时,函数2xy x =⋅取得极小值.(2)由曲线ln y x =与两直线e 1y x =+-及0y =所围成的平面图形的面积是_____________.1x =(3)与两直线 1y t =-+及121111x y z +++==都平行且过原点的平面方程为_____________.2z t =+(4)设L 为取正向的圆周229,x y +=则曲线积分2(22)(4)Lxy y dx xx dy -+-⎰ =_____________.(5)已知三维向量空间的基底为123(1,1,0),(1,0,1),(0,1,1),===ααα则向量(2,0,0=β在此基底下的坐标是_____________.二、(本题满分8分)求正的常数a 与,b 使等式22001lim1sin x x t dt bx x a t→=-+⎰成立.三、(本题满分7分)(1)设f 、g 为连续可微函数,(,),(),u f x xy v g x xy ==+求,.u v x x∂∂∂∂ (2)设矩阵A 和B 满足关系式2,+AB =A B 其中301110,014⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦A 求矩阵.B四、(本题满分8分)求微分方程26(9)1y y a y ''''''+++=的通解,其中常数0.a >五、选择题(本题共4小题,每小题3分,满分12分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设2()()lim1,()x af x f a x a →-=--则在x a =处 (A)()f x 的导数存在,且()0f a '≠ (B)()f x 取得极大值 (C)()f x 取得极小值(D)()f x 的导数不存在(2)设()f x 为已知连续函数0,(),s t I t f tx dx =⎰其中0,0,t s >>则I 的值(A)依赖于s 和t (B)依赖于s 、t 和x (C)依赖于t 、x ,不依赖于s(D)依赖于s ,不依赖于t(3)设常数0,k >则级数21(1)nn k nn ∞=+-∑ (A)发散 (B)绝对收敛 (C)条件收敛(D)散敛性与k 的取值有关(4)设A 为n 阶方阵，且A 的行列式||0,a =≠A 而*A 是A 的伴随矩阵，则*||A 等于(A)a(B)1a(C)1n a -(D)n a六、（本题满分10分） 求幂级数1112n nn x n ∞-=∑ 的收敛域，并求其和函数.七、（本题满分10分） 求曲面积分2(81)2(1)4,I x y dydz y dzdx yzdxdy ∑=++--⎰⎰其中∑是由曲线113()0 z y y f x x ⎧=-≤≤⎪=⎨=⎪⎩绕y 轴旋转一周而成的曲面,其法向量与y轴正向的夹角恒大于.2π八、（本题满分10分）设函数()f x 在闭区间[0,1]上可微,对于[0,1]上的每一个,x 函数()f x 的值都在开区间(0,1)内,且()f x '≠1,证明在(0,1)内有且仅有一个,x 使得().f x x =九、（本题满分8分）问,a b 为何值时,现线性方程组123423423412340221(3)2321x x x x x x x x a x x b x x x ax +++=++=-+--=+++=-有唯一解,无解,有无穷多解?并求出有无穷多解时的通解.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)设在一次实验中,事件A 发生的概率为,p 现进行n 次独立试验,则A 至少发生一次的概率为____________;而事件A 至多发生一次的概率为____________.(2)有两个箱子,第1个箱子有3个白球,2个红球, 第2个箱子有4个白球,4个红球.现从第1个箱子中随机地取1个球放到第2个箱子里,再从第2个箱子中取出1个球,此球是白球的概率为____________.已知上述从第2个箱子中取出的球是白球,则从第一个箱子中取出的球是白球的概率为____________.(3)已知连续随机变量X 的概率密度函数为2211()e ,xx f x π-+-=则X 的数学期望为____________,X 的方差为____________.十一、（本题满分6分）设随机变量,X Y 相互独立,其概率密度函数分别为()X f x = 10 01x ≤≤其它,()Y f y = e 0y - 00y y >≤,求2Z X Y =+的概率密度函数.1988年全国硕士研究生入学统一考试数学(一)试卷一、(本题共3小题,每小题5分,满分15分)(1)求幂级数1(3)3nnn x n ∞=-∑的收敛域. (2)设2()e ,[()]1x f x f x x ϕ==-且()0x ϕ≥,求()x ϕ及其定义域. (3)设∑为曲面2221x y z ++=的外侧,计算曲面积分333.I x dydz y dzdx z dxdy ∑=++⎰⎰二、填空题(本题共4小题,每小题3分,满分12分.把答案填在题中横线上) (1)若21()lim (1),tx x f t t x→∞=+则()f t '= _____________.(2)设()f x 连续且31(),x f t dt x -=⎰则(7)f =_____________.(3)设周期为2的周期函数,它在区间(1,1]-上定义为()f x = 22x1001x x -<≤<≤,则的傅里叶()Fourier 级数在1x =处收敛于_____________.(4)设4阶矩阵234234[,,,],[,,,],==A αγγγB βγγγ其中234,,,,αβγγγ均为4维列向量,且已知行列式4,1,==A B 则行列式+A B = _____________.三、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设()f x 可导且01(),2f x '=则0x ∆→时,()f x 在0x 处的微分dy 是 (A)与x ∆等价的无穷小 (B)与x ∆同阶的无穷小 (C)比x ∆低阶的无穷小(D)比x ∆高阶的无穷小(2)设()y f x =是方程240y y y '''-+=的一个解且00()0,()0,f x f x '>=则函数()f x 在点0x 处(A)取得极大值(B)取得极小值 (C)某邻域内单调增加(D)某邻域内单调减少(3)设空间区域2222222212:,0,:,0,0,0,x y z R z x y z R x y z Ω++≤≥Ω++≤≥≥≥则(A)124xdv dv ΩΩ=⎰⎰⎰⎰⎰⎰(B)124ydv ydv ΩΩ=⎰⎰⎰⎰⎰⎰(C)124zdv zdv ΩΩ=⎰⎰⎰⎰⎰⎰(D)124xyzdv xyzdv ΩΩ=⎰⎰⎰⎰⎰⎰(4)设幂级数1(1)nn n a x ∞=-∑在1x =-处收敛,则此级数在2x =处 (A)条件收敛(B)绝对收敛(C)发散(D)收敛性不能确定(5)n 维向量组12,,,(3)s s n ≤≤ααα 线性无关的充要条件是(A)存在一组不全为零的数12,,,,s k k k 使11220s s k k k +++≠ααα (B)12,,,s ααα 中任意两个向量均线性无关(C)12,,,s ααα 中存在一个向量不能用其余向量线性表示 (D)12,,,s ααα 中存在一个向量都不能用其余向量线性表示四、(本题满分6分)设()(),x yu yf xg y x=+其中函数f 、g 具有二阶连续导数,求222.u u x yx x y ∂∂+∂∂∂五、(本题满分8分)设函数()y y x =满足微分方程322e ,xy y y '''-+=其图形在点(0,1)处的切线与曲线21y x x =--在该点处的切线重合,求函数().y y x =六、（本题满分9分）设位于点(0,1)的质点A 对质点M 的引力大小为2(0kk r>为常数,r 为A 质点与M 之间的距离),质点M 沿直线22y x x =-自(2,0)B 运动到(0,0),O 求在此运动过程中质点A 对质点M 的引力所作的功.七、（本题满分6分）已知,=AP BP 其中100100000,210,001211⎡⎤⎡⎤⎢⎥⎢⎥==-⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦B P 求5,.A A八、（本题满分8分）已知矩阵20000101x ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦A 与20000001y ⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦B 相似.(1)求x 与.y(2)求一个满足1-=P AP B 的可逆阵.P九、（本题满分9分）设函数()f x 在区间[,]a b 上连续,且在(,)a b 内有()0,f x '>证明:在(,)a b 内存在唯一的,ξ使曲线()y f x =与两直线(),y f x a ξ==所围平面图形面积1S 是曲线()y f x =与两直线(),y f x b ξ==所围平面图形面积2S 的3倍.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)设在三次独立试验中,事件A 出现的概率相等,若已知A 至少出现一次的概率等于19,27则事件A 在一次试验中出现的概率是____________. (2)若在区间(0,1)内任取两个数,则事件”两数之和小于65”的概率为____________. (3)设随机变量X 服从均值为10,均方差为0.02的正态分布,已知221()e ,(2.5)0.9938,2u xx du φφπ--∞==⎰则X 落在区间(9.95,10.05)内的概率为____________.十一、（本题满分6分）设随机变量X 的概率密度函数为21(),(1)X f x x π=-求随机变量31Y X =-的概率密度函数().Y f y1989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)已知(3)2,f '=则0(3)(3)lim2h f h f h→--= _____________.(2)设()f x 是连续函数,且1()2(),f x x f t dt =+⎰则()f x =_____________.(3)设平面曲线L 为下半圆周21,y x =--则曲线积分22()Lx y ds +⎰=_____________.(4)向量场div u 在点(1,1,0)P 处的散度div u =_____________.(5)设矩阵300100140,010,003001⎡⎤⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦A I 则矩阵1(2)--A I =_____________.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)当0x >时,曲线1siny x x= (A)有且仅有水平渐近线 (B)有且仅有铅直渐近线(C)既有水平渐近线,又有铅直渐近线(D)既无水平渐近线,又无铅直渐近线(2)已知曲面224z x y =--上点P 处的切平面平行于平面2210,x y z ++-=则点的坐标是(A)(1,1,2)- (B)(1,1,2)- (C)(1,1,2)(D)(1,1,2)--(3)设线性无关的函数都是二阶非齐次线性方程的解是任意常数,则该非齐次方程的通解是(A)11223c y c y y ++(B)1122123()c y c y c c y +-+(C)1122123(1)c y c y c c y +---(D)1122123(1)c y c y c c y ++--(4)设函数2(),01,f x x x =≤<而1()sin ,,nn S x bn x x π∞==-∞<<+∞∑其中102()sin ,1,2,3,,n b f x n xdx n π==⎰ 则1()2S -等于(A)12- (B)14-(C)14 (D)12(5)设A 是n 阶矩阵,且A 的行列式0,=A 则A 中 (A)必有一列元素全为0(B)必有两列元素对应成比例(C)必有一列向量是其余列向量的线性组合(D)任一列向量是其余列向量的线性组合三、(本题共3小题,每小题5分,满分15分)(1)设(2)(,),z f x y g x xy =-+其中函数()f t 二阶可导,(,)g u v 具有连续二阶偏导数,求2.zx y∂∂∂(2)设曲线积分2()cxy dx y x dy ϕ+⎰与路径无关,其中()x ϕ具有连续的导数,且(0)0,ϕ=计算(1,1)2(0,0)()xy dx y x dy ϕ+⎰的值.(3)计算三重积分(),x z dv Ω+⎰⎰⎰其中Ω是由曲面22z x y =+与221z x y =--所围成的区域.四、(本题满分6分)将函数1()arctan 1xf x x+=-展为x 的幂级数.五、(本题满分7分) 设0()sin ()(),xf x x x t f t dt =--⎰其中f 为连续函数,求().f x六、（本题满分7分） 证明方程0ln 1cos 2e x x xdx π=--⎰在区间(0,)+∞内有且仅有两个不同实根. 七、（本题满分6分）问λ为何值时,线性方程组13x x λ+=123422x x x λ++=+ 1236423x x x λ++=+有解,并求出解的一般形式. 八、（本题满分8分）假设λ为n 阶可逆矩阵A 的一个特征值,证明(1)1λ为1-A 的特征值. (2)λA为A 的伴随矩阵*A 的特征值.九、（本题满分9分）设半径为R 的球面∑的球心在定球面2222(0)x y z a a ++=>上,问当R 为何值时,球面∑在定球面内部的那部分的面积最大?十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)(1)已知随机事件A 的概率()0.5,P A =随机事件B 的概率()0.6P B =及条件概率(|)0.8,P B A =则和事件A B 的概率()P A B =____________. (2)甲、乙两人独立地对同一目标射击一次,其命中率分别为0.6和0.5,现已知目标被命中,则它是甲射中的概率为____________.(3)若随机变量ξ在(1,6)上服从均匀分布,则方程210x x ξ++=有实根的概率是____________.十一、（本题满分6分）设随机变量X 与Y 独立,且X 服从均值为1、标准差(均方差)为2的正态分布,而Y 服从标准正态分布.试求随机变量23Z X Y =-+的概率密度函数.1990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) 2x t =-+ (1)过点(1,21)M -且与直线 34y t =-垂直的平面方程是_____________.1z t =-(2)设a 为非零常数,则lim()xx x a x a→∞+-=_____________.(3)设函数()f x =1011x x ≤>,则[()]f f x =_____________.(4)积分222e y xdx dy -⎰⎰的值等于_____________.(5)已知向量组1234(1,2,3,4),(2,3,4,5),(3,4,5,6),(4,5,6,7),====αααα 则该向量组的秩是_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)设()f x 是连续函数,且e ()(),xxF x f t dt -=⎰则()F x '等于(A)e (e )()xx f f x ----(B)e (e )()xx f f x ---+(C)e(e )()x x f f x ---(D)e(e )()xx f f x --+(2)已知函数()f x 具有任意阶导数,且2()[()],f x f x '=则当n 为大于2的正整数时,()f x 的n 阶导数()()n f x 是(A)1![()]n n f x + (B)1[()]n n f x +(C)2[()]nf x(D)2![()]nn f x(3)设a 为常数,则级数21sin()1[]n na n n∞=-∑ (A)绝对收敛(B)条件收敛 (C)发散(D)收敛性与a 的取值有关(4)已知()f x 在0x =的某个邻域内连续,且0()(0)0,lim2,1cos x f x f x→==-则在点0x =处()f x(A)不可导(B)可导,且(0)0f '≠ (C)取得极大值(D)取得极小值(5)已知1β、2β是非齐次线性方程组=AX b 的两个不同的解1,α、2α是对应其次线性方程组=AX 0的基础解析1,k 、2k 为任意常数,则方程组=AX b 的通解(一般解)必是(A)1211212()2k k -+++ββααα (B)1211212()2k k ++-+ββααα (C)1211212()2k k -+++ββαββ(D)1211212()2k k ++-+ββαββ三、(本题共3小题,每小题5分,满分15分)(1)求120ln(1).(2)x dx x +-⎰(2)设(2,sin ),z f x y y x =-其中(,)f u v 具有连续的二阶偏导数,求2.zx y∂∂∂ (3)求微分方程244e xy y y -'''++=的通解(一般解).四、(本题满分6分) 求幂级数0(21)nn n x∞=+∑的收敛域,并求其和函数.五、(本题满分8分) 求曲面积分2SI yzdzdx dxdy =+⎰⎰其中S 是球面2224x y z ++=外侧在0z ≥的部分.六、（本题满分7分）设不恒为常数的函数()f x 在闭区间[,]a b 上连续,在开区间(,)a b 内可导,且()().f a f b =证明在(,)a b 内至少存在一点,ξ使得()0.f ξ'>七、（本题满分6分） 设四阶矩阵1100213401100213,0011002100010002-⎡⎤⎡⎤⎢⎥⎢⎥-⎢⎥⎢⎥==⎢⎥⎢⎥-⎢⎥⎢⎥⎣⎦⎣⎦B C 且矩阵A 满足关系式1()-''-=A E C B C E其中E 为四阶单位矩阵1,-C 表示C 的逆矩阵,'C 表示C 的转置矩阵.将上述关系式化简并求矩阵.A八、（本题满分8分）求一个正交变换化二次型22212312132344448f x x x x x x x x x =++-+-成标准型.九、（本题满分8分）质点P 沿着以AB 为直径的半圆周,从点(1,2)A 运动到点(3,4)B 的过程中受变力F 作用(见图).F的大小等于点P 与原点O 之间的距离,其方向垂直于线段OP 且与y 轴正向的夹角小于.2π求变力F 对质点P所作的功.十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上) (1)已知随机变量X 的概率密度函数1()e ,2xf x x -=-∞<<+∞则X 的概率分布函数()F x =____________.(2)设随机事件A 、B 及其和事件的概率分别是0.4、0.3和0.6,若B 表示B 的对立事件,那么积事件AB 的概率()P AB =____________.(3)已知离散型随机变量X 服从参数为2的泊松()Poisson 分布,即22e {},0,1,2,,!k P X k k k -=== 则随机变量32Z X =-的数学期望()E Z =____________.十一、（本题满分6分）设二维随机变量(,)X Y 在区域:01,D x y x <<<内服从均匀分布,求关于X 的边缘概率密度函数及随机变量21Z X =+的方差().D Z1991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 21c o s x t y t=+=,则22d y dx =_____________.(2)由方程2222xyz x y z +++=所确定的函数(,)z z x y =在点(1,0,1)-处的全微分dz =_____________.(3)已知两条直线的方程是1212321:;:.101211x y z x y zl l ---+-====-则过1l 且平行于2l 的平面方程是_____________.(4)已知当0x →时123,(1)1ax +-与cos 1x -是等价无穷小,则常数a =_____________.(5)设4阶方阵52002100,00120011⎡⎤⎢⎥⎢⎥=⎢⎥-⎢⎥⎣⎦A 则A 的逆阵1-A =_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)曲线221e 1ex x y --+=- (A)没有渐近线 (B)仅有水平渐近线(C)仅有铅直渐近线(D)既有水平渐近线又有铅直渐近线(2)若连续函数()f x 满足关系式20()()ln 2,2tf x f dt π=+⎰则()f x 等于 (A)e ln 2x (B)2e ln 2x (C)e ln 2x +(D)2e ln 2x +(3)已知级数12111(1)2,5,n n n n n a a ∞∞--==-==∑∑则级数1n n a ∞=∑等于(A)3 (B)7(C)8(D)9(4)设D 是平面xoy 上以(1,1)、(1,1)-和(1,1)--为顶点的三角形区域1,D 是D 在第一象限的部分,则(cos sin )Dxy x y dxdy +⎰⎰等于(A)12cos sin D x ydxdy ⎰⎰(B)12D xydxdy ⎰⎰(C)14(cos sin )D xy x y dxdy +⎰⎰(D)0(5)设n 阶方阵A 、B 、C 满足关系式,=ABC E 其中E 是n 阶单位阵,则必有 (A)=ACB E (B)=CBA E (C)=BAC E (D)=BCA E三、(本题共3小题,每小题5分,满分15分)(1)求20lim(cos ).x x π+→(2)设n是曲面222236x y z ++=在点(1,1,1)P 处的指向外侧的法向量,求函数2268x y u z+=在点P 处沿方向n 的方向导数.(3)22(),xy z dv Ω++⎰⎰⎰其中Ω是由曲线220y z x ==绕z 轴旋转一周而成的曲面与平面4z =所围城的立体.四、(本题满分6分)过点(0,0)O 和(,0)A π的曲线族sin (0)y a x a =>中,求一条曲线,L 使沿该曲线O 从到A 的积分3(1)(2)Ly dx x y dy +++⎰的值最小.五、(本题满分8分)将函数()2(11)f x x x =+-≤≤展开成以2为周期的傅里叶级数,并由此求级数211n n∞=∑的和.六、（本题满分7分）设函数()f x 在[0,1]上连续,(0,1)内可导,且1233()(0),f x dx f =⎰证明在(0,1)内存在一点,c 使()0.f c '=七、（本题满分8分） 已知1234(1,0,2,3),(1,1,3,5)aa ===-+=+αααα及(1,1,3,5).b =+β(1)a 、b 为何值时,β不能表示成1234,,,αααα的线性组合?(2)a 、b 为何值时,β有1234,,,αααα的唯一的线性表示式?写出该表示式. 八、（本题满分6分）设A 是n 阶正定阵,E 是n 阶单位阵,证明+A E 的行列式大于1. 九、（本题满分8分）在上半平面求一条向上凹的曲线,其上任一点(,)P x y 处的曲率等于此曲线在该点的法线段PQ 长度的倒数(Q 是法线与x 轴的交点),且曲线在点(1,1)处的切线与x 轴平行.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)若随机变量X 服从均值为2、方差为2σ的正态分布,且{24}0.3,P X <<=则{0}P X <=____________.(2)随机地向半圆202(y ax x a <<-为正常数)内掷一点,点落在半圆内任何区域的概率与区域的面积成正比,则原点和该点的连线与x 轴的夹角小于4π的概率为____________.十一、（本题满分6分）设二维随机变量(,)X Y 的密度函数为(,)f x y =(2)2e 0,00 x y x y -+>>其它求随机变量2Z X Y =+的分布函数.1992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)设函数()y y x =由方程e cos()0x yxy ++=确定,则dydx=_____________.(2)函数222ln()u x y z =++在点(1,2,2)M -处的梯度grad Mu =_____________.(3)设()f x = 211x-+00x x ππ-<≤<≤,则其以2π为周期的傅里叶级数在点x π=处收敛于_____________.(4)微分方程tan cos y y x x '+=的通解为y =_____________.(5)设111212121212,n n n n n n a b a b a b a b a b a b a b a b a b ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦A 其中0,0,(1,2,,).i i a b i n ≠≠= 则矩阵A 的秩()r A =_____________.二、选择题(本题共5小题,每小题3分,满分15分.每小题给出的四个选项中,只有一个符合题目要求,把所选项前的字母填在题后的括号内)(1)当1x →时,函数1211e 1x x x ---的极限(A)等于2 (B)等于0(C)为∞(D)不存在但不为∞(2)级数1(1)(1cos )(nn a n ∞=--∑常数0)a > (A)发散(B)条件收敛(C)绝对收敛(D)收敛性与a 有关(3)在曲线23,,x t y t z t ==-=的所有切线中,与平面24x y z ++=平行的切线 (A)只有1条 (B)只有2条 (C)至少有3条(D)不存在(4)设32()3,f x x x x =+则使()(0)n f 存在的最高阶数n 为(A)0 (B)1 (C)2 (D)3(5)要使12100,121⎛⎫⎛⎫ ⎪ ⎪== ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭ξξ都是线性方程组=AX 0的解,只要系数矩阵A 为(A)[]212-(B)201011-⎡⎤⎢⎥⎣⎦(C)102011-⎡⎤⎢⎥-⎣⎦(D)011422011-⎡⎤⎢⎥--⎢⎥⎢⎥⎣⎦三、(本题共3小题,每小题5分,满分15分)(1)求2e sin 1lim.11x x x x→----(2)设22(e sin ,),xz f y x y =+其中f 具有二阶连续偏导数,求2.zx y∂∂∂ (3)设()f x = 21ex x -+ 00x x ≤>,求31(2).f x dx -⎰四、(本题满分6分) 求微分方程323exy y y -'''+-=的通解.五、(本题满分8分) 计算曲面积分323232()()(),x az dydz y ax dzdx z ay dxdy ∑+++++⎰⎰其中∑为上半球面222z a x y =--的上侧.六、（本题满分7分）设()0,(0)0,f x f ''<=证明对任何120,0,x x >>有1212()()().f x x f x f x +<+ 七、（本题满分8分）在变力F y z i z x j x =++ 的作用下,质点由原点沿直线运动到椭球面2222221x y z a b c ++=上第一卦限的点(,,),M ξηζ问当ξ、η、ζ取何值时,力F 所做的功W 最大?并求出W 的最大值.八、（本题满分7分）设向量组123,,ααα线性相关,向量组234,,ααα线性无关,问: (1)1α能否由23,αα线性表出?证明你的结论. (2)4α能否由123,,ααα线性表出?证明你的结论. 九、（本题满分7分）设3阶矩阵A 的特征值为1231,2,3,λλλ===对应的特征向量依次为1231111,2,3,149⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪=== ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ξξξ又向量12.3⎛⎫⎪= ⎪ ⎪⎝⎭β(1)将β用123,,ξξξ线性表出. (2)求(nn A β为自然数).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11()()(),()0,()(),46P A P B P C P AB P AC P BC ======则事件A 、B 、C 全不发生的概率为____________.(2)设随机变量X 服从参数为1的指数分布,则数学期望2{e }XE X -+=____________.十一、（本题满分6分）设随机变量X 与Y 独立,X 服从正态分布2(,),N Y μσ服从[,]ππ-上的均匀分布,试求Z X Y =+的概率分布密度(计算结果用标准正态分布函数Φ表示,其中221()e )2t xx d t π--∞Φ=⎰.1993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)函数11()(2)(0)xF x dt x t=->⎰的单调减少区间为_____________.(2)由曲线2232120x y z +==绕y 轴旋转一周得到的旋转面在点(0,3,2)处的指向外侧的单位法向量为_____________.(3)设函数2()()f x x x x πππ=+-<<的傅里叶级数展开式为01(cos sin ),2n n n a a nx b nx ∞=++∑则其中系数3b 的值为_____________. (4)设数量场222ln,u x y z =++则div(grad )u =_____________.(5)设n 阶矩阵A 的各行元素之和均为零,且A 的秩为1,n -则线性方程组=AX 0的通解为_____________.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sin 2340()sin(),(),xf x t dtg x x x ==+⎰则当0x →时,()f x 是()g x 的(A)等价无穷小 (B)同价但非等价的无穷小 (C)高阶无穷小(D)低价无穷小(2)双纽线22222()x y x y +=-所围成的区域面积可用定积分表示为(A)402cos 2d πθθ⎰(B)404cos 2d πθθ⎰(C)402cos 2d πθθ⎰(D)2401(cos 2)2d πθθ⎰(3)设有直线1158:121x y z l --+==-与2:l 623x y y z -=+=则1l 与2l 的夹角为 (A)6π(B)4π(C)3π(D)2π (4)设曲线积分[()e ]sin ()cos x Lf t ydx f x ydy --⎰与路径无关,其中()f x 具有一阶连续导数,且(0)0,f =则()f x 等于(A)e e 2x x --(B)e e 2x x --(C)e e 12x x-+-(D)e e 12x x-+-(5)已知12324,369t ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦Q P 为三阶非零矩阵,且满足0,=PQ 则(A)6t =时P 的秩必为1(B)6t =时P 的秩必为2 (C)6t ≠时P 的秩必为1(D)6t ≠时P 的秩必为2三、(本题共3小题,每小题5分,满分15分)(1)求21lim(sincos ).x x x x→∞+ (2)求e .e 1x xx dx -⎰(3)求微分方程22,x y xy y '+=满足初始条件11x y ==的特解.四、(本题满分6分) 计算22,xzdydz yzdzdx z dxdy ∑+-⎰⎰ 其中∑是由曲面22z x y =+与222z x y =--所围立体的表面外侧.五、(本题满分7分)求级数20(1)(1)2n nn n n ∞=--+∑的和.六、(本题共2小题,每小题5分,满分10分)(1)设在[0,)+∞上函数()f x 有连续导数,且()0,(0)0,f x k f '≥><证明()f x 在(0,)+∞内有且仅有一个零点.(2)设,b a e >>证明.b a a b > 七、（本题满分8分）已知二次型22212312323(,,)2332(0)f x x x x x x ax x a =+++>通过正交变换化成标准形22212325,f y y y =++求参数a 及所用的正交变换矩阵.八、（本题满分6分）设A 是n m ⨯矩阵,B 是m n ⨯矩阵,其中,n m <I 是n 阶单位矩阵,若,=AB I 证明B 的列向量组线性无关. 九、（本题满分6分）设物体A 从点(0,1)出发,以速度大小为常数v 沿y 轴正向运动.物体B 从点(1,0)-与A 同时出发,其速度大小为2,v 方向始终指向,A 试建立物体B 的运动轨迹所满足的微分方程,并写出初始条件.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)一批产品共有10个正品和2个次品,任意抽取两次,每次抽一个,抽出后不再放回,则第二次抽出的是次品的概率为____________.(2)设随机变量X 服从(0,2)上的均匀分布,则随机变量2Y X =在(0,4)内的概率分布密度()Y f y =____________.十一、（本题满分6分）设随机变量X 的概率分布密度为1()e ,.2xf x x -=-∞<<+∞(1)求X 的数学期望EX 和方差.DX(2)求X 与X 的协方差,并问X 与X 是否不相关? (3)问X 与X 是否相互独立?为什么?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)011limcot ()sin x x xπ→-= _____________.(2)曲面e 23xz xy -+=在点(1,2,0)处的切平面方程为_____________.(3)设e sin ,xx u y -=则2u x y ∂∂∂在点1(2,)π处的值为_____________.(4)设区域D 为222,x y R +≤则2222()Dx y dxdy a b +⎰⎰=_____________.(5)已知11[1,2,3],[1,,],23==αβ设,'=A αβ其中'α是α的转置,则nA =_____________.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4342342222222sin cos ,(sin cos ),(sin cos ),1x M xdx N x x dx P x x x dx x ππππππ---==+=-+⎰⎰⎰则有 (A)N P M << (B)M P N << (C)N M P <<(D)P M N <<(2)二元函数(,)f x y 在点00(,)x y 处两个偏导数00(,)x f x y '、00(,)y f x y '存在是(,)f x y 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0,λ>且级数21nn a∞=∑收敛,则级数21(1)n nn a n λ∞=-+∑(A)发散 (B)条件收敛(C)绝对收敛(D)收敛性与λ有关(4)2tan (1cos )lim2,ln(12)(1)x x a x b x c x d e -→+-=-+-其中220,a c +≠则必有(A)4b d = (B)4b d =- (C)4a c =(D)4a c =-(5)已知向量组1234,,,αααα线性无关,则向量组 (A)12233441,,,++++αααααααα线性无关 (B)12233441,,,----αααααααα线性无关(C)12233441,,,+++-αααααααα线性无关(D)12233441,,,++--αααααααα线性无关三、(本题共3小题,每小题5分,满分15分)(1)设2221c o s ()1c o s ()c o s 2t x t y t t udu u==-⎰,求dy dx 、22d ydx在2t π=的值.(2)将函数111()ln arctan 412x f x x x x +=+--展开成x 的幂级数. (3)求.sin(2)2sin dxx x +⎰四、(本题满分6分)计算曲面积分2222,S xdydz z dxdy x y z +++⎰⎰其中S 是由曲面222x y R +=及,(0)z R z R R ==->两平面所围成立体表面的外侧.五、(本题满分9分) 设()f x 具有二阶连续函数,(0)0f f '==且2[()()][()]0xy x y f x y dx f x x y dy '+-++=为一全微分方程,求()f x 及此全微分方程的通解.六、(本题满分8分)设()f x 在点0x =的某一邻域内具有二阶连续导数,且0()lim0,x f x x→=证明级数11()n f n ∞=∑绝对收敛.七、（本题满分6分）已知点A 与B 的直角坐标分别为(1,0,0)与(0,1,1).线段AB 绕x 轴旋转一周所成的旋转曲面为.S 求由S 及两平面0,1z z ==所围成的立体体积.八、（本题满分8分） 设四元线性齐次方程组(Ⅰ)为122400x x x x +=-=,又已知某线性齐次方程组(Ⅱ)的通解为12(0,1,1,0)(1,2,2,1).k k +-(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A 为n 阶非零方阵*,A 是A 的伴随矩阵,'A 是A 的转置矩阵,当*'=A A 时,证明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A 、B 两个事件满足条件()(),P AB P AB =且(),P A p =则()P B =____________.(2)设相互独立的两个随机变量,X Y 具有同一分布率,且X 的分布率为X 0 1P12 12则随机变量max{,}Z X Y =的分布率为____________.十一、（本题满分6分）设随机变量X 和Y 分别服从正态分布2(1,3)N 和2(0,4),N 且X 与Y 的相关系数1,2xy ρ=-设,32X Y Z =+(1)求Z 的数学期望EZ 和DZ 方差.(2)求X 与Z 的相关系数.xz ρ (3)问X 与Y 是否相互独立?为什么?1995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)2sin 0lim(13)xx x →+=_____________.(2)202cos xd x t dt dx ⎰= _____________.(3)设()2,⨯=a 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16,-=+A BA A BA 且100310,41007⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦A 则B =_____________.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:L 321021030x y z 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(2)设在[0,1]上()0,f x ''>则(0),(1),(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 ''>->(3)设()f x 可导,()()(1sin ),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(1)ln(1),nn u n=-+则级数 (A)1nn u∞=∑与21nn u∞=∑都收敛(B)1nn u∞=∑与21nn u∞=∑都发散(C)1nn u∞=∑收敛,而21nn u∞=∑发散 (D)1nn u∞=∑收敛,而21nn u∞=∑发散(5)设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 则必有(A)12AP P =B (B)21AP P =B (C)12P P 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(,,),(,e ,)0,sin ,yu f x y z x 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(2)设函数()f x 在区间[0,1]上连续,并设1(),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22z x y =+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的余弦函数.五、(本题满分7分)设曲线L 位于平面xOy 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 在平面xOy 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 fg g ξξξξ''=''八、（本题满分7分）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101,1⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦ξ求.A 九、（本题满分6分）设A 为n 阶矩阵,满足('=AA I I 是n 阶单位矩阵,'A 是A 的转置矩阵),0,<A 求.+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34{0,0},{0}{0},77P X Y P X P Y ≥≥=≥=≥=则{max(,)0}P X Y ≥=____________.十一、（本题满分6分） 设随机变量X 的概率密度为()X f x = e 0x - 0x x ≥<,求随机变量e X Y =的概率密度().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2lim()8,xx x a x a→∞+=-则a =_____________.(2)设一平面经过原点及点(6,3,2),-且与平面428x y z -+=垂直,则此平面方程为_____________.(3)微分方程22e xy y y '''-+=的通解为_____________. (4)函数22ln()u x y z =++在点(1,0,1)A 处沿点A 指向点(3,2,2)B -方向的方向导数为_____________.(5)设A 是43⨯矩阵,且A 的秩()2r =A 而102020,103⎡⎤⎢⎥=⎢⎥⎢⎥-⎣⎦B 则()r AB =_____________.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2()()x ay dx ydyx y +++为某函数的全微分,a 则等于 (A)-1 (B)0 (C)1(D)2(2)设()f x 具有二阶连续导数,且0()(0)0,lim1,x f x f x→'''==则 (A)(0)f 是()f x 的极大值 (B)(0)f 是()f x 的极小值(C)(0,(0))f 是曲线()y f x =的拐点(D)(0)f 不是()f x 的极值,(0,(0))f 也不是曲线()y f x =的拐点(3)设0(1,2,),n a n >= 且1nn a∞=∑收敛,常数(0,),2πλ∈则级数21(1)(tan )n n n n a n λ∞=-∑(A)绝对收敛(B)条件收敛(C)发散(D)散敛性与λ有关(4)设有()f x 连续的导数220,(0)0,(0)0,()()(),xf f F x x t f t dt '=≠=-⎰且当0x →时,()F x '与k x 是同阶无穷小,则k 等于(A)1 (B)2 (C)3(D)4(5)四阶行列式112233440000000a b a b a b b a 的值等于(A)12341234a a a a b b b b -(B)12341234a a a a b b b b + (C)12123434()()a a b b a a b b --(D)23231414()()a a b b a a b b --三、(本题共2小题,每小题5分,满分10分) (1)求心形线(1cos )r a θ=+的全长,其中0a >是常数.(2)设1110,6(1,2,),n n x x x n +==+= 试证数列{}n x 极限存在,并求此极限.四、(本题共2小题,每小题6分,满分12分)(1)计算曲面积分(2),Sx z dydz zdxdy ++⎰⎰其中S 为有向曲面22(01),z xy x =+≤≤其法向量与z 轴正向的夹角为锐角.(2)设变换 2u x yv x a y =-=+可把方程2222260z z z x x y y ∂∂∂+-=∂∂∂∂简化为20,zu v∂=∂∂求常数.a五、(本题满分7分) 求级数211(1)2nn n ∞=-∑的和.六、(本题满分7分)设对任意0,x >曲线()y f x =上点(,())x f x 处的切线在y 轴上的截距等于1(),xf t dt x ⎰求()f x 的一般表达式. 七、（本题满分8分）设()f x 在[0,1]上具有二阶导数,且满足条件(),(),f x a f x b ''≤≤其中,a b 都是非负常数,c 是(0,1)内任意一点.证明()2.2b fc a '≤+八、（本题满分6分）设,TA =-I ξξ其中I 是n 阶单位矩阵,ξ是n 维非零列向量,Tξ是ξ的转置.证明(1)2=A A 的充分条件是 1.T=ξξ(2)当1T=ξξ时,A 是不可逆矩阵. 九、（本题满分8分）已知二次型222123123121323(,,)55266f x x x x x cx x x x x x x =++-+-的秩为2,(1)求参数c 及此二次型对应矩阵的特征值. (2)指出方程123(,,)1f x x x =表示何种二次曲面.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A 和工厂B 的产品的次品率分别为1%和2%,现从由A 和B 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A 生产的概率是____________.(2)设,ξη是两个相互独立且均服从正态分布21(0,())2N 的随机变量,则随机变量ξη-的数学期望()E ξη-=____________.十一、（本题满分6分）设,ξη是两个相互独立且服从同一分布的两个随机变量,已知ξ的分布率为1(),1,2,3.3P i i ξ===又设max(,),min(,).X Y ξηξη==(1)写出二维随机变量的分布率: (2)。
1989年全国硕士研究生入学统一考试数学一、二、三、四、五试题完整版附答案及评分标准
则逆矩阵 ( A
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1
2 0
0 1
二、选择题 (本题满分 15 分，每小题 3 分)
(1) 当 x 0 时，曲线 y x sin 1 x
(A) 有且仅有水平渐近线；
(B) 有且仅有铅直渐近线.
(C) 既有水平渐近线，也有铅直渐近线； (D) 既无水平渐近线，也无铅直渐近线.
(2) 已知曲面 z 4 x2 y2 上点 P 处的切平面平行于平面 2x 2 y z 1 0 ，则点 P 的
1989 年全国硕士研究生入学统一考试数学一、二、三、四、五试题 完整版附答案及评分标准 数 学（试卷一）
一、填空题 (本题满分 15 分，每小题 3 分)
(1) 以知 f (3)=2,则 lim f (3 h) f (3) -1
h0
2h
1
(2) 设 f (x) 是连续函数，且 f (x) x 2 f (t)dt ，则 f (x) x 1 . 0
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„„2 分
这是二阶常系数非齐次线性微分方程,初始条件为 y |x0 f (0) 0, y |x0 f (0) 1.„„3 分
其对应齐次方程的通解为 y C1 sin x C2 cos x .
„„4 分
设非齐次方程的特解 y* x(asin x bcos x),可得 a 0,b 1 ；于是 y* x cos x . „„5 分
„„2 分
„„5 分 „„6 分
五、(本题满分 7 分)
x
设 f (x) sin x (x t) f (t)dt ,其中 f 为连续函数，求 f (x) . 0
x
x
数学三考研真题(1989-1999年)
假设函数 f ( x) 在 [ a,b] 上连续 , 在 (a,b) 内可导 , 且 f ( x)
证明在 (a,b) 内, F (x)
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七、 ( 本题满分 5 分 )
2
已知 X AX B, 其中 A
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x, 0 x 1, f (x)
2 x, 1 x 2.
试计算下列各题：
(1) S0
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(2)
(3) Sn
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2
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n0
六、 ( 本题满分 6 分 )
x1 x2 x3 0, 只有零解 , 则 应满足的条件是 __ _ .
x1 x2 x3 0
(4) 设随机变量 X 的分布函数为
0,
x 0,
F x Asinx, 0 x , 则 2
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2
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(5) 设随机变量 X 的数学期望 E ( X ) (Chebyshev) 不等式 , 有 P{ X
x
P P ( x) 10e 2 ,
且最大需求量为 6, 其中 x 表示需求量 , P 表示价格 .
(1) 求该商品的收益函数和边际收益函数 .(2 分) (2) 求使收益最大时的产量、最大收益和相应的价格 .(4 分 ) (3) 画出收益函数的图形 .(3 分 )
1989考研英语一真题(含答案解析)
1989年全国硕士研究生入学统一考试英语试题Section I Close TestFor each numbered blank in the following passage there are four choices labeled ［A］, ［B］, ［C］and ［D］. Choose the best one and put your choice in the ANSWER SHEET. Read the whole passage before making your choice. (10 points)①One day drought may be a thing of the past at least in coastal cities.②V ast areas of desert throughout the world may for the first time 1 and provide millions of hectares of land where now nothing grows.③By the end of this century this may not be mere 2 .④Scientists are already looking into the possibility of using some of the available ice in the Arctic and Antarctic.⑤In these regions there are vast ice-caps formed by snow that has fallen over the past 50,000 years.⑥Layer 3 layer of deep snow means that, when melted, the snow water would be pure, not salty as sea-ice would be.⑦There is so much 4 pure water here that it would need only a fraction of it to turn much of the desert or poorly irrigated parts of the world into rich farmland.⑧And what useful packages it would come in!⑨It should be possible to cut off a bit of ice and transport it!⑩Alternatively perhaps a passing iceberg could be 5 . ○11They are always breaking away from the main caps and floating around, pushed by currents, until they eventually melt and are wasted.○12Many icebergs are, of course, far too small to be towed 6 distance, and would melt before they reached a country that needed them anywhere. ○13It would be necessary to locate one that was 7 and that was big enough to provide a good supply of ice when it reached us. ○14Engineers think that an iceberg up to seven miles long and one and a half miles wide could be transported if the tug pulling it was as big as a supertanker! ○15Even then they would cover only twenty miles every day. ○16However, 8 the iceberg was at its destination, more that 7,000 million cubic metres of water could be taken from it! ○17That would probably be more than enough for any medium-sized city even in the hottest summer! ○18But no doubt a use could be found for it. ○199 , scientist say, there would not be too much wastage in such a journey. ○20The larger the iceberg, the slower it melts, even if it is towed through the tropics. ○21This is because when the sun has a bigger area to warm 10 , less heat actually gets into the iceberg. ○22The vast frozen centre would be unaffected. [394 words]1. ［A］come to life ［B］come into existence［C］come into activity ［D］come round2. ［A］speculation ［B］imagination ［C］computation ［D］expectation3. ［A］above ［B］of ［C］upon ［D］over4. ［A］essential ［B］potential ［C］claimable ［D］obtainable5. ［A］seized ［B］snatched ［C］grabbed ［D］captured6. ［A］much ［B］any ［C］some ［D］certain7. ［A］manageable ［B］manipulative ［C］operable ［D］controllable8. ［A］after ［B］while ［C］since ［D］once9. ［A］Apparently ［B］Noticeably ［C］Distinctly ［D］Notably10.［A］round ［B］over ［C］up ［D］throughSection II Reading ComprehensionEach of the two passages below is followed by five questions. For each question there are four answers. Readthe passages carefully and choose the best answer to each of the questions. Put your choice in the brackets on the left. (10 points)Text 1A scientist once said: “I have concluded that the earth is being visited by intelligently co ntrolled vehicles from outer space.”If we take this as a reasonable explanation for UFOs (unidentified flying objects), questions immediately come up.“Why don’t they get in touch with us, then? Why don’t they land right on the White House lawn and decla re themselves?” people asked.In reply, scientists say that, while this may be what we want, it may not necessarily be what they want.“The most likely explanation, it seems to me,” said Dr. Mead, “is that they are simply watching what we are up to -- tha t responsible society outside our solar system is keeping an eye on us to see that we don’t set in motion a chain reaction that might have unexpected effects for outside our solar system.”Opinions from other scientists might go like this: “Why should they want to get in touch with us? We may feel we’re more important than we really are! They may want to observe us only and not interfere with the development of our civilization. They may not care if we see them but they also may not care to say ‘hello’.”①Some scientists have also suggested that Earth is a kind of zoo or wildlife reserve. ②Just as we set aside wilderness areas and wildlife reserves to allow animals and growing things to develop naturally while we observe them, so perhaps Earth was set aside ages ago for the same purpose.①Are we being observed by intelligent beings from other civilizations in the universe? ②Are they watching our progress in space travel? ③Do we live in a gigantic “zoo” observed by our “keepers,” but having no communication with them?①Never before in our history have we had to confront ideas like these. ②The simple fact is that we, who have always regarded ourselves as supreme in the universe, may not be so. ③Now we have to recognize that, among the stars in the heavens, there may very well be worlds inhabited by beings who are to us as we are to ants.11. People who ask the question “Why don’tthey get in touch with us... and declarethemselves?” think that ________.[A] there are no such things as UFOs[B] UFOs are visitors from solar system[C] there’s no reason for UFOs sooner or later[D] we are bound to see UFOs sooner or later12. According to Dr. Mead, the attitude ofbeings from outer space toward us is one of________.[A] unfriendliness[B] suspicion[C] superiority[D] hostility13. The tone of the writer is that of ________.[A] doubt[B] warning[C] indifference[D] criticismText 2①The use of the motor is becoming more and more widespread in the twentieth century; as an increasing number of countries develop both technically and economically, so a larger proportion of the world’s population is able to buy and use a car. ②Possessing a car gives a much greater degree of mobility, enabling the driver to move around freely. ③The owner of a car is no longer forced to rely on public transport and is, therefore, not compelled to work locally. ④He can choose from different jobs and probably changes his work more frequently as he is not restricted to a choice within a small radius. ⑤Travelling to work by car is also more comfortable than having to use public transport; the driver can adjust the heating in winter and the air conditioning in the summer to suit his own needs and preference. ⑥There is no irritation caused by waiting for trains, buses or underground trains, standing in long patient queues, or sitting on windy platforms, for as long as half an hour sometimes. ⑦With the building of good, fast motorways long distances can be covered rapidly and pleasantly. ⑧For the first time in this century also, many people are now able to enjoy their leisure time to the full by making trips to the country or seaside at the weekends, instead of being confined to their immediate neighbourhood. ⑨This feeling of independence, and the freedom to go where you please, is perhaps the greatest advantage of the car.①When considering the drawbacks, perhaps pollution is of prime importance. ②As more and more cars are produced and used, so the emission from their exhaust-pipes contains an ever larger volume of poisonous gas. ③Some of the contents of this gas, such as lead, not only pollute the atmosphere but cause actual harm to the health of people. ④Many of the minor illnesses of modern industrial society, headaches, tiredness, and stomach upsets are thought to arise from breathing polluted air; doctors’ surgeries are full of people suffering from illnesses caused by pollution. ⑤It is also becoming increasingly difficult to deal with the problem of traffic in towns; most of the important cities of the world suffer from traffic congestion. ⑥In fact any advantage gained in comfort is often cancelled out in city driving by the frustration caused by traffic jams: endless queues of cars crawling one after another through all the main streets. ⑦As an increasing number of traffic regulation schemes are devised, the poor bewildered driver finds himself diverted and forced into one-way systems which cause even greater delays than the traffic jams they are supposed to prevent. ⑧The mounting cost of petrol and the increased license fees and road tax all add to the driver’s worries. ⑨In fact, he must sometimes wonder if the motor car is such a blessing and not just a menace.14. More and more people can afford to buyand use cars because ________.[A] an increasing number of cars are beingproduced[B] the cost of cars is getting cheaper with thedevelopment of technology[C] lots of countries have become moredeveloped[D] the use of cars has proved to be moreeconomical15. The advantages of having a car are bestexperienced in the driver’s ________.[A] freedom in choosing his job[B] comfort during the travels[C] enjoyment of his leisure time[D] feeling of self-reliance16. What is considered by the writer as thegreatest menace to the people caused by thewidespread use of motor cars?[A] air pollution[B] traffic jams[C] fatal diseases[D] high costText 3①Manners nowadays in metropolitan cities like London are practically non-existent. ②It is nothing for a big, strong schoolboy to elbow an elderly woman aside in the dash for the last remaining seat on the tube or bus, much less stand up and offer his seat to her, as he ought. ③In fact, it is saddening to note that if a man does offer his seat to an older woman, it is nearly always a Continental man or one from the older generation.①This question of giving up seats in public transport is much argued about by young men, who say that, since women have claimed equality, they no longer deserve to be treated with courtesy and that those who go out to work should take their turn in the rat race like anyone else. ②Women have never claimed to be physically as strong as men. ③Even if it is not agreed, however, that young men should stand up for younger women, the fact remains that courtesy should be shown to the old, the sick and the burdened. ④Are we really so lost to all ideals of unselfishness that we can sit there indifferently reading the paper or a book, saying to ourselves “First come, first served,” while a grey-haired woman, a mother with a young child or a cripple stands? ⑤Yet this is all too often seen.①Conditions in travel are really very hard on everyone, we know, but hardship is surely no excuse. ②Sometimes one wonders what would have been the behaviour of these stout young men in a packed refugee train or a train on its way to a prison-camp during the War. ③Would they have considered it only right and their proper due to keep the best places for themselves then?①Older people, tired and irritable from a day’s work, are not angels, eit her —far from it. ②Many a brisk argument or an insulting quarrel breaks out as the weary queues push and shove each other to get on buses and tubes. ③One cannot commend this, of course, but one does feel there is just a little more excuse.①If cities are to remain pleasant places to live in at all, however, it seems imperative, not only that communications in transport should be improved, but also that communication between human beings should be kept smooth and polite. ②All over cities, it seems that people are too tired and too rushed to be polite. ③Shop assistants won’t bother to assist, taxi drivers growl at each other as they dash dangerously round corners, bus conductor pull the bell before their desperate passengers have had time to get on or off the bus, and so on and so on.④It seems to us that it is up to the young and strong to do their small part to stop such deterioration.17. From what you have read, would youexpect manners to improve among people________?[A] who are physically weak or crippled[B] who once lived in a prison-camp during theWar[C] who live in big modern cities[D] who live only in metropolitan cities18. What is the writer’s opinion concerningcourteous manners towards women?[A] Now that women have claimed equality, theyno longer need to be treated differently from men.[B] It is generally considered old-fashioned foryoung men to give up their seats to young women.[C] “Lady First” should be universally practiced.[D] Special consideration ought to be shown them.19. According to the author communicationbetween human beings would be smoother if________.[A] people were more considerate towards eachother[B] people were not so tired and irritable[C] women were treated with more courtesy[D] public transport could be improved20. What is the possible meaning of the word“deterioration” in the last paragraph?[A] worsening of general situation[B] lowering of moral standards[C] declining of physical constitution[D] spreading of evil conductSection III English-Chinese TranslationTranslate the following passage into Chinese. Only the underlined sentences are to be translated. (20 points)When Jane Matheson started work at Advanced Electronics Inc. 12 years ago, (21) she laboured over a microscope, hand-welding tiny electronic computers and turned out 18 per hour. Now she tends the computerized machinery that turns out high capacity memory chips at the rate of 2,600 per hour. Production is up, profits are up,her income is up and Mrs. Matheson says the work is far less strain on her eyes.But the most significant effect of the changes at AEI was felt by the workers who are no longer there. Before the new computerized equipment was introduced, there were 940 workers at the plant. Now there are 121. (22) A plant follow-up survey showed that one year after the layoffs only 38% of the released workers found new employment at the same or better wages. Nearly half finally settled for lower pay and more than 13% are still out of work. The AEI example is only one of hundreds around the country which forge intelligently ahead into the latest technology, but leave the majority of their workers behind.(23) Its beginnings obscured by unemployment caused by the world economic slow-down, the new technological unemployment may emerge as the great socio-economic challenge of the end of the 20th century. One corporation economist says the growth of “machine job replacement” has been with us since the beginning of the industrial revolution, but never at the pace it is now. The human costs will be astonishing. (24) “It’s humiliating to be done out of your job by a machine and there is no way to fight back, but it is the effort to find a new job that really hurts.” Some workers, like Jane Matheson, are retrained to handle the new equipment, but often a whole new set of skills is required and that means a new, and invariably smaller set of workers. (25) The old workers, trapped by their limited skills, often never regain their old status and employment. Many drift into marginal areas. They feel no pride in their new work. They get badly paid for it and they feel miserable, but still they are luckier than those who never find it.(26) The social costs go far beyond the welfare and unemployment payments made by the government. Unemployment increases the chances of divorce, child abuse, and alcoholism, a new federal survey shows. Some experts say the problem is only temporary... that new technology will eventually create as many jobs as it destroys.(27) But futurologist Hymen Seymour says the astonishing efficiency of the new technology means there will be a simple and direct net reduction in the amount of human labor that needs to be done. “We should treat this as an opportunity to give people more leisure. It may not be easy, but society will have to reach a new unanimity on the division and distribution of labor,” Seymour says. He predicts most people will work only six-hour days and four-day weeks by the end of the century. But the concern of the unemployed is for now. (28) Federally funded training and free back-to-school programs for laid-off workers are under way, but few experts believe they will be able to keep up with the pace of the new technology. For the next few years, for a substantial portion of the workforce, times are going to be very tough indeed.考研英语真题。
1989年考研英语试题及答案
1989年考研英语试题及答案1989年全国硕⼠研究⽣⼊学统⼀考试英语试题Section I Structure and VocabularyIn each question, decide which of the four choices given will most suitably complete the sentence if inserted at the place marked. Put your choices in the ANSWER SHEET. (15 points)EXAMPLE:I was caught ________ the rain yesterday.[A] in[B] by[C] with[D] atANSWER: [A]1. Modern man faces dangers completely unknown ________ hispredecessors.[A] for[B] to[C] of（B）[D] by2. The chances of seeing a helicopter in my hometown are one________ a million.[A] for[B] to[C] in（C）[D] against3. ________ we have all the materials ready, we should begin the newtask at once.[A] Since that[B] Since now[C] By now（D）[D] Now that4. We hope the measures to control prices, ________ taken by thegovernment, will succeed.[A] when[B] as[C] since（B）[D] after5. The historical events of that period are arranged ________.[A] in alphabetical order[B] in an alphabetical order[C] in the alphabetical orders（A）[D] in alphabetical orders6. In some markets there may be only one seller. ________ is calleda monopoly.[A] Situation as this[B] Such kind of situation[C] Such a situation（C）[D] A situation of this7. He is ________ to speak the truth.[A] too much of a coward[B] too much a coward[C] so much a coward（A）[D] so much of a coward8. He always gives ________ to his wife’s demands and does whatever she tells him to.[A] up[B] away[C] in（C）[D] out9. It’s ________ in the regulations that you can take 20 kilos of luggage with you.[A] laid upon[B] laid out[C] laid up（D）[D] laid down10. Look at all the corruption that’s going on. It’s time the citywas ________.[A] cleaned out[B] cleaned down[C] cleaned away（D）[D] cleaned up11. Though he did not say so directly, the inspector ________ the manwas guilty.[A] declared[B] implied[C] disclosed（B）[D] said12. The Prime Minister refused to ________ on the rumour that he hadplanned to resign.[A] explain[B] comment[C] remark（B）[D] talk13. I asked the tailor to make a small ________ to my trousersbecause they were too long.[A] change[B] variation[C] revision（D）[D] alteration14. Magnificent views over the countryside have often ________ peopleto write poems.[A] excited[B] inspired[C] induced（B）[D] attracted15. The food was divided ________ according to the age and size ofthe children.[A] equally[B] proportionately[C] sufficiently（B）[D] adequatelySection II Reading ComprehensionEach of the three passages below is followed by some questions. For each question there are four answers. Read the passages carefully and choose the best answer to each of the questions. Put your choice in the ANSWER SHEET. (20 points)Text 1A scientist once said: “I have concluded that the earth is being visited by intelligently controlled vehicles fro m outer space.”If we take this as a reasonable explanation for UFOs (unidentified flying objects), questions immediately come up.“Why don’t they get in touch with us, then? Why don’t they land right on the White House lawn and declare themselves?”people asked.In reply, scientists say that, while this may be what we want, it may not necessarily be what they want.“The most likely explanation, it seems to me,” said Dr. Mead, “is that they are simply watching what we are up to -- that responsible society outside our solar system is keeping an eye on us to see that we don’t set in motion a chain reaction that might have unexpected effects for outside our solar system.”Opinions from other scientists might go like this: “Why should they want to get in touch w ith us? We may feel we’re more important than we really are! They may want to observe us only and not interfere with the development of our civilization. They may not care if we see them but they also may not care to say ‘hello’.”Some scientists have also suggested that Earth is a kind of zoo or wildlife reserve. Just as we set aside wilderness areas and wildlife reserves to allow animals and growing things to develop naturally while we observe them, so perhaps Earth was set aside ages ago for the same purpose.Are we being observed by intelligent beings from other civilizations in the universe? Are they watching our progress in space travel? Do we live in a gigantic “zoo” observed by our “keepers,” but having no communication with them?Never before in our history have we had to confront ideas like these. The simple fact is that we, who have always regarded ourselves as supreme in the universe, may not be so. Now we have to recognize that, among the stars in the heavens, there may very well be worldsinhabited by beings who are to us as we are to ants.16. People who ask the question “Why don’t they get in touch withus... and declare themselves?” think that ________.[A] there are no such things as UFOs[B] UFOs are visitors from solar system[C] there’s no reason for UFOs sooner or later（A）[D] we are bound to see UFOs sooner or later17. According to Dr. Mead, the attitude of beings from outer spacetoward us is one of ________.[A] unfriendliness[B] suspicion[C] superiority（B）[D] hostility18. The tone of the writer is that of ________.[A] doubt[B] warning[C] indifference（D）[D] criticismText 2The use of the motor is becoming more and more widespread in the twentieth century; as an increasing number of countries develop both technically and economically, so a larger proportion of the world’s population is able to buy and use a car. Possessing a car gives a much greater degree of mobility, enabling the driver to move around freely. The owner of a car is no longer forced to rely on public transport and is, therefore, not compelled to work locally. He can choose from different jobsand probably changes his work more frequently as he is not restricted to a choice within a small radius. Travelling to work by car is also more comfortable than having to use public transport; the driver can adjust the heating in winter and the air conditioning in the summer to suit his own needs and preference. There is no irritation caused by waiting for trains, buses or underground trains, standing in long patient queues, or sitting on windy platforms, for as long as half an hour sometimes. With the building of good, fast motorways long distances can be covered rapidly and pleasantly. For the first time in this century also, many people are now able to enjoy their leisure time to the full by makingtrips to the country or seaside at the weekends, instead of being confined to their immediate neighbourhood. This feeling of independence, and the freedom to go where you please, is perhaps the greatest advantage of the car.When considering the drawbacks, perhaps pollution is of prime importance. As more and more cars are produced and used, so the emission from their exhaust-pipes contains an ever larger volume of poisonous gas. Some of the contents of this gas, such as lead, not only pollute the atmosphere but cause actual harm to the health of people. Many of the minor illnesses of modern industrial society, headaches, tiredness, and stomach upsets are thought to arise from breathing polluted air; doctors’surgeries are full of people suffering from illnesses caused by pollution. It is also becoming increasingly difficult to deal with the problem of traffic in towns; most of the important cities of the world suffer from traffic congestion. In fact any advantage gained in comfort is often cancelled out in city driving by the frustration caused by traffic jams: endless queues of cars crawling one after another through all the main streets. As an increasing number of traffic regulation schemes are devised, the poor bewildered driver finds himself diverted and forced into one-way systems which cause even greater delays than the traffic jams they are supposed to prevent. The mounting cost of petrol and the increased license fees and road tax all add to the driver’s worries. In fact, he must sometimes wonder if the motor car is such a blessing and not just a menace.19. More and more people can afford to buy and use cars because________.[A] an increasing number of cars are being produced[B] the cost of cars is getting cheaper with the development oftechnology[C] lots of countries have become more developed（C）[D] the use of cars has proved to be more economical20. The advantages of having a car are best experienced in thedriver’s ________.[A] freedom in choosing his job[B] comfort during the travels[C] enjoyment of his leisure time（D）[D] feeling of self-reliance21. What is considered by the writer as the greatest menace to thepeople caused by the widespread use of motor cars?[A] air pollution[B] traffic jams[C] fatal diseases（A）[D] high costText 3Manners nowadays in metropolitan cities like London are practically non-existent. It is nothing for a big, strong schoolboy to elbow an elderly woman aside in the dash for the last remaining seat on the tube or bus, much less stand up and offer his seat to her, as he ought. In fact, it is saddening to note that if a man does offer his seat to an older woman, it is nearly always a Continental man or one from the older generation.This question of giving up seats in public transport is much argued about by young men, who say that, since women haveclaimed equality, they no longer deserve to be treated with courtesy and that those who go out to work should take their turn in the rat race like anyone else. Women have never claimed to be physically as strong as men. Even if it is not agreed, however, that young men should stand up for younger women, the fact remains that courtesy should be shown to the old, the sick and the burdened. Are we really so lost to all ideals of unselfishness that we can sit there indifferently reading the paper or a book, saying to o urselves “First come, first served,” while a grey-haired woman, a mother with a young child or a cripple stands? Yet this is all too often seen.Conditions in travel are really very hard on everyone, we know, but hardship is surely no excuse. Sometimes one wonders what would have been the behaviour of these stout young men in a packed refugee train or a train on its way to a prison-camp during the War. Would they have considered it only right and their proper due to keep the best places for themselves then?Older people, tired and irritable from a day’s work, are not angels, either -- far from it. Many a brisk argument or an insulting quarrel breaks out as the weary queues push and shove each other to get on buses and tubes. One cannot commend this, of course, but one does feel there is just a little more excuse.If cities are to remain pleasant places to live in at all, however, it seems imperative, not only that communications in transport should be improved, but also that communication between human beings should be kept smooth and polite. All over cities, it seems that people are too tired and too rushed to be polite. Shopassistants won’t bother to assist, taxi drivers growl at each other as they dash dangerously round corners, bus conductor pull the bell before their desperate passengers have had time to get on or off the bus, and so on and so on. It seems to us that it is up to the young and strong to do their small part to stop such deterioration.22. From what you have read, would you expect manners to improveamong people ________?[A] who are physically weak or crippled[B] who once lived in a prison-camp during the War[C] who live in big modern cities（C）[D] who live only in metropolitan cities23. What is the writer’s opinion concerning courteous m annerstowards women?[A] Now that women have claimed equality, they no longer need tobe treated differently from men.[B] It is generally considered old-fashioned for young men togive up their seats to young women.[C] “Lady First” should be universally practiced.（D）[D] Special consideration ought to be shown them.24. According to the author communication between human beings wouldbe smoother if ________.[A] people were more considerate towards each other[B] people were not so tired and irritable[C] women were treated with more courtesy（A）[D] public transport could be improved25. What is the possible meaning of the word “deterioration” in thelast paragraph?[A] worsening of general situation[B] lowering of moral standards[C] declining of physical constitution（B）[D] spreading of evil conductSection III Cloze TestFor each numbered blank in the following passage there are four choices labeled [A], [B], [C] and [D]. Choose the best one and put your choice in the ANSWER SHEET. Read the whole passage before makingyour choice. (10 points)One day drought may be a thing of the past at least in coastal cities. Vast areas of desert throughout the world may for the first time and provide millions of hectares of land where now nothing grows.By the end of this century this may not be mere . Scientists are already looking into the possibility of using some of the available ice in the Arctic and Antarctic. In these regions there are vast ice-caps formed by snow that has fallen over the past 50,000 years. Layer layer of deep snow means that, when melted, the snow water would be pure, not salty as sea-ice would be. There is so much pure water here that it would need only a fraction of it to turn much of the desert or poorly irrigated parts of the world into rich farmland. And what useful packages it would come in! It should be possible to cut off a bit of ice and transport it! Alternatively perhaps a passing iceberg could be . They are always breaking away from the main caps and floating around, pushed by currents, until they eventually melt and are wasted.Many icebergs are, of course, far too small to be towed distance, and would melt before they reached a country that needed them anywhere. It would be necessary to locate one that was and that was big enough to provide a good supply of ice when it reached us. Engineers think that an iceberg up to seven miles long and one and a half miles wide could be transported if the tug pulling it was as big as a supertanker! Even then they would cover only twenty miles every day. However, the iceberg was at its destination, more that 7,000 million cubic metres of water could be taken from it! That would probably be more than enough for any medium-sized city even in the hottest summer! But no doubt a use could be found for it. , scientist say, there would not be too much wastage in such a journey. The larger the iceberg, the slower it melts, even if it is towed through the tropics. This is because when the sun has a bigger area to warm , less heat actually gets into the iceberg. The vast frozen centre would be unaffected.26. [A] come to life[B] come into existence[C] come into activity（A）[D] come round27. [A] speculation[B] imagination[C] computation（A）[D] expectation28. [A] above[B] of[C] upon（C）[D] over29. [A] essential[B] potential[C] claimable（B）[D] obtainable30. [A] seized[B] snatched[C] grabbed（D）[D] captured31. [A] much[B] any[C] some（B）[D] certain32. [A] manageable[B] manipulative[C] operable（A）[D] controllable33. [A] after[B] while[C] since（D）[D] once34. [A] Apparently[B] Noticeably[C] Distinctly（A）[D] Notably35. [A] round[B] over[C] up（C）[D] throughSection IV Error-detection and CorrectionEach of the following sentences has four underlined parts. These parts are labeled [A], [B], [C] and [D]. Identify the part of the sentence that is incorrect and put your choice in the ANSWER SHEET. Then, without altering the meaning of the sentence, write down your correction on the line in the ANSWER SHEET. (10 points)EXAMPLE:You have to hurry up if you want to buy something hardly .ANSWER: [C] anything36. bank keeps cash all its depositors in at one time.（[C] topay）37. provide great variety of and entertainment information.（[B]a）38. If it rain the next few weeks, the will have to be watered ifthey are .（[D] to survive）39. This is the most important respect civilized can bedistinguished primitive .（[A] in which）40. a bad-tempered man, he having his lectures interrupted he weresome obscure candidate an election speech.（[A] Being）41. If you awarded a prize of ten thousand dollars, what would youdo it if you in a day?（[D] to spend it）42. The boy is constantly not to scratch the paint the all, but hegoes on it .（[C] doing）43. The parcel you post must be . Inadequate packing delay, damageloss at your .（[D] expense）44. The radio was of inferior quality I took it and asked for abetter .（[A] such）45. I can listen to Bruckner hours without getting bored, but if youhaven’t much of his music before, you it takes some .（[D] getting used to）Section V Verb FormsFill in the blanks with the appropriate forms of the verbs given the brackets. Put your answers in the ANSWER SHEET. (10 points)EXAMPLE:It is highly desirable that a new president ________ (appoint) for this college.ANSWER: (should) be appointed46. Byron is said ________ (live) on vinegar and potatoes.（to havelived）47. You ________ (leave) a note. It was very inconsiderate of you todo so.（should have left）48. If the horse won today, he ________ (win) thirty races in fiveyears.（would/should have won）49. Upon being questioned he denied ________ (write) the article.（having written）50. I was so sick last night that I felt as if the room ________ (go)round.（were/was going）51. Nowadays people usually prefer driving to ________ (drive).（being driven）52. I hope her health ________ (improve) greatly by the time we comeback next year.（will have improved）53. While we were in London that year, the London Bridge ________(repair).（was being repaired）54. Lots of empty bottles were found under the old man’s bed. Hemust have done nothing but ________ (drink).（drink）55. Ford tried dividing the labour, each worker ________ (assign) aseparate task.（assigned）Section VI Chinese-English TranslationTranslate the following sentences into English. (15 points)56. 请乘客们系好安全带，以防碰伤。
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1989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)已知(3)2,f '=则0
(3)(3)
lim
2h f h f h
→--= _____________.
(2)设()f x 是连续函数,且1
()2
(),f x x f t dt =+⎰
则()f x =_____________.
(3)设平面曲线L
为下半圆周y =则曲线积分
22()L
x y ds +⎰
=_____________.
(4)向量场div u 在点(1,1,0)P 处的散度div u =_____________.
(5)设矩阵300100140,010,003001⎡⎤⎡⎤⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦
A I 则矩阵1(2)--A I =_____________.
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)当0x >时,曲线1sin
y x x
= (A)有且仅有水平渐近线 (B)有且仅有铅直渐近线
(C)既有水平渐近线,又有铅直渐近线
(D)既无水平渐近线,又无铅直渐近
线
(2)已知曲面2
2
4z x y =--上点P 处的切平面平行于平面2210,x y z ++-=则点的坐标是
(A)(1,1,2)- (B)(1,1,2)- (C)(1,1,2)
(D)(1,1,2)--
(3)设线性无关的函数都是二阶非齐次线性方程的解是任意常数,则该非齐次方程的通解是
(A)11223c y c y y ++
(B)1122123()c y c y c c y +-+
(C)1122123(1)c y c y c c y +---
(D)1122123(1)c y c y c c y ++--
(4)设函数2
(),01,f x x x =≤<而1
()sin ,,n n S x b n x x π∞
==
-∞<<+∞∑其中
1
02()sin ,1,2,3,,n b f x n xdx n π==⎰ 则1
()2S -等于
(A)12- (B)14-
(C)14 (D)12
(5)设A 是n 阶矩阵,且A 的行列式0,=A 则A 中 (A)必有一列元素全为0
(B)必有两列元素对应成比例
(C)必有一列向量是其余列向量的线性组合
(D)任一列向量是其余列向量的线性
组合
三、(本题共3小题,每小题5分,满分15分)
(1)设(2)(,),z f x y g x xy =-+其中函数()f t 二阶可导,(,)g u v 具有连续二阶偏导数,
求2.z x y
∂∂∂
(2)设曲线积分
2()c
xy dx y x dy ϕ+⎰
与路径无关,其中()x ϕ具有连续的导数,且(0)0,
ϕ=计算
(1,1)
2(0,0)
()xy dx y x dy ϕ+⎰
的值.
(3)计算三重积分
(),x z dv Ω
+⎰⎰⎰
其中Ω是由曲面z =与z =成的区域.
四、(本题满分6分)
将函数1()arctan 1x
f x x
+=-展为x 的幂级数.
五、(本题满分7分) 设0
()sin ()(),x
f x x x t f t dt =-
-⎰
其中f 为连续函数,求().f x
六、（本题满分7分）
证明方程0
ln e x
x π=
-⎰在区间(0,)+∞内有且仅有两个不同实根. 七、（本题满分6分）
问λ为何值时,线性方程组
13x x λ+=
123422x x x λ++=+ 1236423x x x λ++=+
有解,并求出解的一般形式.
八、（本题满分8分）
假设λ为n 阶可逆矩阵A 的一个特征值,证明 (1)
1
λ
为1
-A 的特征值.
(2)
λ
A
为A 的伴随矩阵*
A 的特征值.
九、（本题满分9分）
设半径为R 的球面∑的球心在定球面2222(0)x y z a a ++=>上,问当R 为何值时,球
面∑在定球面内部的那部分的面积最大?
十、填空题(本题共3小题,每小题2分,满分6分.把答案填在题中横线上)
(1)已知随机事件A 的概率()0.5,P A =随机事件B 的概率()0.6P B =及条件概率
(|)0.8,P B A =则和事件A B 的概率()P A B =____________.
(2)甲、乙两人独立地对同一目标射击一次,其命中率分别为0.6和0.5,现已知目标被命中,则它是甲射中的概率为____________.
(3)若随机变量ξ在(1,6)上服从均匀分布,则方程2
10x x ξ++=有实根的概率是____________.
十一、（本题满分6分）
设随机变量X 与Y 独立,且X 服从均值为1、标准差(均方差),而Y 服从标准正态分布.试求随机变量23Z X Y =-+的概率密度函数.。