【中山大学2011年考研专业课真题】一元微积分2011

2011年考研数一真题及答案解析一、选择题1、 曲线()()()()4324321----=x x x x y 的拐点是（ ）（A ）（1，0） （B ）（2，0） （C ）（3，0） （D ）（4，0）【答案】C 【考点分析】本题考查拐点的判断。

直接利用判断拐点的必要条件和第二充分条件即可。

【解析】由()()()()4324321----=x x x x y 可知1,2,3,4分别是()()()()23412340y x x x x =----=的一、二、三、四重根，故由导数与原函数之间的关系可知(1)0y '≠，(2)(3)(4)0y y y '''===(2)0y ''≠，(3)(4)0y y ''''==，(3)0,(4)0y y ''''''≠=，故（3，0）是一拐点。

2、 设数列{}n a 单调减少，0lim =∞→n n a ，()∑===n k k n n a S 12,1 无界，则幂级数()11nn n a x ∞=-∑的收敛域为（ ） （A ） (-1，1] （B ） [-1，1) （C ） [0，2) （D ）(0，2]【答案】C 【考点分析】本题考查幂级数的收敛域。

主要涉及到收敛半径的计算和常数项级数收敛性的一些结论，综合性较强。

【解析】()∑===n k k n n a S 12,1 无界，说明幂级数()11nn n a x ∞=-∑的收敛半径1R ≤；{}n a 单调减少，0lim =∞→nn a ，说明级数()11nn n a ∞=-∑收敛，可知幂级数()11nn n a x ∞=-∑的收敛半径1R ≥。

因此，幂级数()11nn n a x ∞=-∑的收敛半径1R =，收敛区间为()0,2。

又由于0x =时幂级数收敛，2x =时幂级数发散。

2011年全国硕士研究生入学统一考试数学试题答案和评分参考数 学（一）一．选择题 ( 1 ～ 8小题，每小题4分，共32分.)(1) 曲线234(1)(2)(3)(4)y x x x x =----的拐点是 （C ） (A) (1,0) (B) (2,0) (C) (3,0) (D) (4,0)(2) 设数列{}n a 单调减少，lim 0n n a →∞=，1(1,2,)nn k k S a n ===∑ 无界，则幂级数1(1)nnn a x ∞=-∑的收敛域为 （C ） (A) (1,1]-(B) [1,1)- (C) [0,2) (D) 0,2]((3) 设函数()f x 具有二阶连续导数，且()0,(0)0f x f '>=，则函数()ln ()z f x f y =在点(0,0)处取得极小值的一个充分条件是 （A ）(A) (0)1,(0)0f f ''>> (B) (0)1,(0)0f f ''>< (C) (0)1,(0)0f f ''<> (D) (0)1,(0)0f f ''<<(4) 设4ln sin I xdx π=⎰,40ln cot J xdx π=⎰,40ln cos K xdx π=⎰, 则,,I J K 的大小关系为（B ）(A) I J K <<(B) I K J <<(C) J I K <<(D) K J I <<(5) 设A 为3阶矩阵，将A 的第2列加到第1列得矩阵B ，再交换B 的第2行与第3行得单位矩阵.记1100110001⎛⎫ ⎪= ⎪ ⎪⎝⎭P ，2100001010⎛⎫⎪= ⎪⎪⎝⎭P ，则A = （D ）(A) 12P P (B) 112-P P (C) 21P P (D) 121-P P(6) 设1234(,,,)αααα=A 是4阶矩阵，*A 为A 的伴随矩阵.若(1,0,1,0)T是方程组0x =A的一个基础解系，则*0x =A 的基础解系可为 （D ）(A)13,αα(B)12,αα (C) 123,,ααα(D)234,,ααα(7) 设1()F x 与2()F x 为两个分布函数，其相应的概率密度1()f x 与2()f x 是连续函数，则必为概率密度的是 （D ） (A) 12()()f x f x(B) 212()()f x F x (C) 12()()f x F x (D) 1221()()()()f x F x f x F x +(8) 设随机变量X 与Y 相互独立，且EX 与EY 存在，记max{,}U X Y =，min{,}V X Y =，则()E UV = （B ） (A)EU EV ⋅ (B) EX EY ⋅ (C) EU EY ⋅ (D) EX EV ⋅ 二、填空题：（9～14小题，每小题4分，共24分.） (9) 曲线0tan (0)4xy tdt x π=≤≤⎰的弧长s =ln(12)(10) 微分方程cos x y y e x -'+=满足条件(0)0y =的解为y =sin xe x -(11) 设函数20sin (,)1xyt F x y dt t =+⎛⎜⎠，则2202x y F x==∂=∂ 4 .(12) 设L 是柱面221x y +=与平面z x y =+的交线，从z 轴正向往z 轴负向看去为逆时针方向，则曲线积分22y Lxzdx xdy dz ++=⎰π.(13) 设二次曲面的方程22232224x y z a x y x z y z +++++=经正交变换化为221144y z +=，则a = 1 .(14) 设二维随机变量(,)X Y 服从正态分布22(,;,;0)N μμσσ，则2()E XY =23μσμ+.三、解答题 ( 15 ～ 23小题，共94分. )(15)（本题满分10分）求极限110ln(1)lim xex x x -→+⎛⎫ ⎪⎝⎭.解：记11ln(1)xex y x -+⎛⎫=⎪⎝⎭. 当0x >时，ln[ln(1)]ln ln 1xx xy e +-=-， 00011ln[ln(1)]ln (1)ln(1)lim ln lim lim 11x x x x x x x x xy e +++→→→-+-++==- ……4分 0(1)ln(1)lim (1)ln(1)x x x x x x x +→-++=++20(1)ln(1)lim x x x x x+→-++= 01ln(1)11lim 22x x x +→-+-==-. ……9分当0x <时，ln[ln(1)]ln()ln 1x x x y e -+--=-， 00ln[ln(1)]ln()1lim ln lim 12x x x x x y e --→→-+--==--.综上可知，110ln(1)lim xex x x e-→+⎛⎫=⎪⎝⎭. ……10分(16)（本题满分9分）设函数(,())z f xy yg x =，其中函数f 具有二阶连续偏导数，函数()g x 可导且在1x =处取得极值(1)1g =，求211x y z x y==∂∂∂.解：由题意(1)0g '= ……2分因为12()zyf yg x f x ∂'''=+∂，……4分 21111222122[()]()()[()]zf y xfg x f g x f yg x xf g x f x y∂''''''''''''=+++++∂∂， ……8分 所以211x y z x y==∂∂∂11112(1,1)(1,1)(1,1)f f f '''''=++.……9分(17)（本题满分10分）求方程arctan 0k x x -=不同实根的个数，其中k 为参数.解：令()arctan f x k x x =-，则()f x 是(,)-∞+∞上的奇函数，且221(0)0,()1k x f f x x --'==+. ……3分 当10k -≤即1k ≤时，()0(0)f x x '<≠，()f x 在(,)-∞+∞内单调减少，方程()0f x = 只有一个实根0x =.……5分当10k ->即1k >时，在1)k -内，()0f x '>，()f x 单调增加；在(1,)k -+∞内，()0f x '<，()f x 单调减少，所以(1)f k -是()f x 在(0,)+∞内的最大值. 由于(0)0f =，所以(1)0f k ->. 又arctan lim ()lim (1)x x k xf x x x→+∞→+∞=-=-∞，所以存在(1,)k ξ∈-+∞，使得()0f ξ=. 由()f x 是奇函数及其单调性可知：当1k >时，方程()0f x =有且仅有三个不同实根,0,x x x ξξ=-==. ……10分(18)（本题满分10分）（I ） 证明：对任意的正整数n ，都有111ln(1)1n n n<+<+成立.（II ） 设111ln (1,2,)2n a n n n=+++-= ，证明数列{}n a 收敛. 解：（I ）根据拉格朗日中值定理，存在(,1)n n ξ∈+，使得 11ln(1)ln(1)ln n n n ξ+=+-=，所以1111ln(1)1n n nξ<+=<+.……4分 （II ）当1n ≥时，由（I ）知111ln(1)01n n a a n n+-=-+<+， ……6分且11111ln ln(11)ln(1)ln(1)ln 22n a n n n n=+++->++++++-ln(1)ln 0n n =+->，所以数列{}n a 单调下降且有下界，故{}n a 收敛. ……10分(19)（本题满分11分）已知函数(,)f x y 具有二阶连续偏导数，且(1,)0f y =，(,1)0f x =，(,)Df x y dxdy a =⎰⎰，其中{(,)|01,01}D x y x y =≤≤≤≤，计算二重积分(,)xy DI xyf x y dxdy ''=⎰⎰.解： 因为(1,)0f y =，(,1)0f x =，所以(1,)0y f y '=，(,1)0x f x '=. ……2分 从而11I (,)xy xdx yf x y dy ''=⎰⎰……4分111000(,)|(,)y x y x x yf x y f x y dy dx ==⎡⎤''=-⎢⎥⎣⎦⎰⎰1100(,)x dy xf x y dx '=-⎰⎰ ……7分 111000(,)(,)x x x f x y f x y dx dy ==⎡⎤=--⎢⎥⎣⎦⎰⎰1100(,)dy f x y dx a ==⎰⎰. ……11分(20)（本题满分11分）设向量组1(1,0,1)T a =，2(0,1,1)T a =，3(1,3,5)T a =不能由向量组T1(1,1,1β=），T 2(1,2,3β=），T3(3,4,a β=）线性表示. （I ）求a 的值； （II ）将123,,βββ用123,,ααα线性表示.解：（I ）4个3维向量123,,,i βββα线性相关(1,2,3)i =，若123,,βββ线性无关，则i α 可由123,,βββ线性表示(1,2,3)i =，与题设矛盾. 于是123,,βββ线性相关. ……3分从而 123113|,,|1245013a aβββ==-=，于是5a =此时，1α不能由向量组123,,βββ线性表示.……5分（II ）令 123123(,,|,,)αααβββ=A .对A 施以初等行变换1011131002150131240104210115135001102⎛⎫⎛⎫⎪⎪=→ ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭A ， 从而112324βααα=+-，2122βαα=+，31235102βααα=+-. …… 11分(21)（本题满分11分）设A 为3阶实对称矩阵，A 的秩为2，且111100001111-⎛⎫⎛⎫⎪ ⎪= ⎪ ⎪⎪ ⎪-⎝⎭⎝⎭A .（I ）求A 的所有特征值与特征向量； （II ）求矩阵A . 解：（I ）由于A 的秩为2，故0是A 的一个特征值.由题设可得110011⎛⎫⎛⎫ ⎪ ⎪=- ⎪ ⎪ ⎪ ⎪--⎝⎭⎝⎭A ，110011⎛⎫⎛⎫⎪ ⎪= ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭A ， 所以，1-是A 的一个特征值，且属于1-的特征向量为1(1,0,1)T k -，1k 为任意非零常数； 1也是A 的一个特征值,且属于1的特征向量为2(1,0,1)T k .2k 为任意非零常数； ……4分 设123(,,)T x x x 是A 的属于0的特征向量，由于A 为实对称矩阵，则123(1,0,1)0x x x ⎛⎫ ⎪-= ⎪ ⎪⎝⎭，123(1,0,1)0x x x ⎛⎫⎪= ⎪ ⎪⎝⎭，即131300x x x x -=⎧⎨+=⎩. 于是属于0的特征向量为3(0,1,0)T k ，3k 为任意非零常数； ……6分（II ）令 110001110⎛⎫ ⎪= ⎪ ⎪-⎝⎭P ， 则 1100010000--⎛⎫⎪= ⎪⎪⎝⎭P AP ， ……8分1100010000--⎛⎫ ⎪= ⎪ ⎪⎝⎭A P P 1122112211010000010010100000110000010100--⎛⎫⎛⎫⎛⎫⎛⎫ ⎪ ⎪⎪ ⎪== ⎪ ⎪⎪ ⎪ ⎪⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭⎝⎭. ……11分(22)（本题满分11分）设随机变量X 与Y 的概率分布分别为X 0 1 Y -1 0 1 P1/32/3P1/31/31/3且22{}1P X Y ==.（I ）求二维随机变量(,)X Y 的概率分布； （II ）求Z XY =的概率分布； （III ）求X 与Y 的相关系数XY ρ.解：（I ）由22{}1P X Y ==，得22{}0P X Y ≠=，所以{0,1}{0,1}{1,0}0P X Y P X Y P X Y ==-=======.故(,)X Y 的概率分布为X Y-1 0 1 0 0 1/3 0 11/31/3……4分（II ）Z XY =的可能取值为1,0,1-. 由(,)X Y 的概率分布可得Z 的概率分布为……7分 （III ）由X ，Y 及Z 的概率分布得222,,0,,()0393EX DX EY DY EZ E XY ======，所以 (,)0,Cov X Y =0XY ρ=.…… 11分(23)（本题满分11分）设12,,,n X X X 为来自正态总体20(,)N μσ的简单随机样本，其中0μ已知，20σ>未 知.X 和2S 分别表示样本均值和样本方差.（I ）求参数2σ的最大似然估计 2σ； （II ）计算 2E σ 和2D σ. 解：（I ）设12,,,n x x x 为样本观测值，则似然函数20211()2222()(2).ni i nx L eμσσπσ=---∑=，2220211ln ()ln(2)()22n i i n L x σπσμσ==---∑，Z 1-0 1 P1/31/31/3令 2ln 0()d Ld σ=，得 202411()022n i i n x μσσ=-+-=∑， 从而得2σ的最大似然估计 22011()n i i X n σμ==-∑. ……6分（II ）解法1 由于2202122()()nii Xn n μσσσ=-=~χ∑，……8分所以 222E n n σσσ=⋅=， 442222D n n nσσσ=⋅=. ……11分 解法2 222011()n i i E E X n σμσ==-=∑， ……8分4422221001021112()()()n i i X D D X D X D n n n nμσσσμμσ=-=-=-==∑. ……11分数 学（二）一．选择题（1～8小题，每小题4分，共32分）．(1) 已知当0x →时，函数()3sin sin 3f x x x =-与kcx 是等价无穷小，则 (C) (A) 1,4k c ==(B) 1,4k c ==-(C) 3,4k c ==(D) 3,4k c ==-(2) 设函数()f x 在0x =处可导，且(0)0f =，则2330()2()lim x x f x f x x →-= (B) (A) 2(0)f '- (B) (0)f '- (C) (0)f ' (D) 0(3) 函数()ln |(1)(2)(3)|f x x x x =---的驻点个数为 (C) (A) 0(B) 1(C) 2(D) 3(4) 微分方程2(0)x x y y e e λλλλ-''-=+>的特解形式为 (C)(A) ()x x a e e λλ-+ (B) ()x x ax e e λλ-+ (C) ()x xx ae be λλ-+ (D) 2()x x x ae be λλ-+(5) 【 同数学一（3）题 】 (6) 【 同数学一（4）题 】 (7) 【 同数学一（5）题 】 (8) 【 同数学一（6）题 】二、填空题：（9～14小题，每小题4分，共24分.） (9) 1012lim 2xxx →⎛⎫+= ⎪⎝⎭2.(10) 【 同数学一（10）题 】 (11) 【 同数学一（9）题 】(12) 设函数,0,()00,0,x e x f x x λλλ-⎧>=>⎨≤⎩，则()xf x dx +∞-∞=⎰1λ.(13) 设平面区域D 由直线y x =，圆222x y y +=及y 轴所围成，则二重积分Dxyd σ=⎰⎰712.(14) 二次型222123123121323(,,)3222f x x x x x x x x x x x x =+++++，则f 的正惯性指数为 2 .三、解答题 ( 15 ～ 23小题，共94分. ) (15)（本题满分10分）已知函数20ln(1)()xt dt F x x α+=⎰.设0lim ()lim ()0x x F x F x +→+∞→==，试求α的取值范围. 解：因为2201ln(1)ln(1)lim ()limlim x x x x t dt x F x x x ααα-→+∞→+∞→+∞++==⎰ 22112211lim lim(1)(1)x x x x x x αααααα--→+∞→+∞+==--， 由题意lim ()0x F x →+∞=，得1α>. ……5分又因为2201000ln(1)ln(1)lim ()limlim xx x x t dt x F x x x ααα+++-→→→++==⎰231001lim lim x x x x x αααα++--→→==， 由题意 0lim ()0x F x +→=，得3α<. 综上所述，13α<<. ……10分(16)（本题满分11分）设函数()y y x =由参数方程3311331133x t t y t t ⎧=++⎪⎪⎨⎪=+⎪⎩-确定，求()y y x = 的极值和曲线()y y x =的凹凸区间及拐点.解：令22101dy t dx t -==+，得1t =±.当1t =时，53x =； 当1t =-时，1x =-. ……3分令222222344(1)01(1)td y t t dx t t +===++，得0t =，即13x =. ……6分 列表如下：由此可知，函数()y x 的极大值为1(1)|1t y y =--==，极小值为151()|33t y y ===-. 曲线()y y x =的凹区间为1(,)3+∞，凸区间为1(,)3-∞.由于011()|33t y y ===，所以曲线()y y x =的拐点为11(,33）. ……11分(17)（本题满分10分）【 同数学一（16）题 】 (18)（本题满分10分）设函数()y x 具有二阶导数，且曲线:()l y y x =与直线y x =相切于原点. 记α为曲线l 在点(,)x y 处切线的倾角，若d dydx dxα=，求()y x 的表达式. 解：由于tan y α'=，即arctan y α'=，所以 ……2分21d y dx y α''='+.于是有21y y y '''='+，即2(1)y y y ''''=+ ① ……4分令y p '=，则"y p '=，代入①式得2(1)p p p '=+，分离变量得2(1)dpdx p p =+， 两边积分得212ln 2ln 1p x C p=++ ② 由题意(0)1y '=，即当0x =时1p =，代入②式 得112C =，于是有 22121x e xy p e'==- ……7分两边积分得222221()x xx e e e y dx C ==+-，由(0)0y =得24C π=-.所以24x e y π=-. ……10分(19)（本题满分10分）【 同数学一（18）题 】(20)（本题满分11分）一容器的内侧是由图中曲线绕y 轴旋转一周而成的曲面，该曲线由2212()2x y y y +=≥与2211()2x y y +=≤连接而成.（I ）求容器的容积；（II ）若将容器内盛满的水从容器顶部全部抽出，至少需要做多少功？ （长度单位：m ，重力加速度为g 2/m s ，水的密度为3310/kg m .） 解： (I) 由对称性，所求的容积为12212V x dy π-=⎰……3分122192(1)4y dy ππ-=-=⎰，即该容器的容积为94π立方米.……5分（II ）123232211210(1)(2)10[1(1)](2)W y y gdy y y gdy ππ-=--+--⎰⎰－ ……8分132323232112271010(22)(44)8g y y y dy y y y dy g ππ-⎡⎤⨯=--++-+=⎢⎥⎣⎦⎰⎰.即所求的功为32710/8g π⨯焦耳. ……11分(21)（本题满分11分）【 同数学一（19）题 】 (22)（本题满分11分）【 同数学一（20）题 】 (23)（本题满分11分）【 同数学一（21）题 】数 学（三）一．选择题 (1～8小题，每小题4分，共32分) . (1) 【 同数学二（1）题 】 (2) 【 同数学二（2）题 】(3) 设{}n u 是数列，则下列命题正确的是 (A) (A) 若1nn u ∞=∑收敛，则2121()n n n u u ∞-=+∑收敛 (B) 若2121()n n n u u ∞-=+∑收敛，则1n n u ∞=∑收敛 (C) 若1n n u∞=∑收敛，则2121()n n n uu ∞-=-∑收敛 (D) 若2121()n n n u u ∞-=-∑收敛，则1n n u ∞=∑收敛(4) 【 同数学一（4）题 】 (5) 【 同数学一（5）题 】(6) 设A 为43⨯矩阵，123,,ηηη是非齐次线性方程组β=Ax 的3个线性无关的解，12,k k 为任意常数，则β=Ax 的通解为 (C) (A)23121()2k ηηηη++-(B)23121()2k ηηηη-+-(C) 23121231()()2k k ηηηηηη++-+-(D) 23121231()()2k k ηηηηηη-+-+-(7) 【 同数学一（7）题 】(8) 设总体X 服从参数为(0)λλ>的泊松分布，12,,,(2)n X X X n ≥ 为来自该总体的简单随机样本.则对于统计量111n i i T X n ==∑和121111n in i T X X n n -==+-∑，有 (D) (A)1212,ET ET DT DT >> (B) 1212,ET ET DT DT >< (C) 1212,ET ET DT DT <>(D) 1212,ET ET DT DT <<二、填空题：（9～14小题，每小题4分，共24分.） (9) 设0()lim (13)x tt f x x t →=+，则()f x '=3(13)xx e +.(10) 设函数(1)x yxyz =+，则(1,1)|dz =(12ln 2)()dx dy +-.(11) 曲线tan()4y x y e π++=在点(0,0)处的切线方程为2y x=-.(12) 曲线21y x =-2x =及x 轴所围的平面图形绕x 轴旋转所成的旋转体的体积为4/3π. (13) 设二次型123(,,)T f x x x =x Ax 的秩为1，A 的各行元素之和为3，则f 在正交变换Q =x y 下的标准形为213y .(14) 【 同数学一（14）题 】三、解答题 ( 15 ～ 23小题，共94分. ) (15)（本题满分10分） 求极限 012sin 1x x x →+--.解：0012sin 112sin 1x x x x x x →→+--+--= ……2分 0112sin x x→-+= ……4分 0cos 12sin 12sin x x x x →-+=+0sin 12sin x x x →--+= ……8分 12=-.……10分 (16)（本题满分10分）已知函数(,)f u v 具有二阶连续偏导数，(1,1)2f =是(,)f u v 的极值，(,(,))z f x y f x y =+.求2(1,1)z x y∂∂∂.解：121(,(,))(,(,))(,)zf x y f x y f x y f x y f x y x∂'''=+++⋅∂， ……3分 2111212(,(,))(,(,))(,)(,)(,(,))zf x y f x y f x y f x y f x y f x y f x y f x y x y∂''''''''=+++⋅+⋅+∂∂22 12122()(,(,))(,(,))(,)f x y f x y f x y f x y f x y f x y ⎡⎤''''''+++++⋅⎣⎦2 ……7分由题意知，1(1,1)0f '=，2(1,1)0f '=， ……9分从而2(1,1)z x y ∂∂∂11212(2,2)(2,2)(1,1)f f f '''''=+.……10分(17)（本题满分10分） 求不定积分arcsinln x xx+⎛⎜⎠.解：arcsinln 2ln )x xx x x x+=⎰……2分2(arcsin ln )21x x x x x =--⎛⎛⎜⎜⎠⎠ ……6分 2(arcsinln )41x x x x x=+-- ……8分2(arcsin ln )214x x x x x C =+- ……10分(18)（本题满分10分）证明方程44arctan 303x x π-+=恰有两个实根. 证：设 4()4arctan 3f x x x π=-+24(3)(3)()11x x f x x -+'=-=+ ……2分 令()0f x '=，解得驻点123,3x x =-=.由单调性判别法知()f x 在(,3]-∞-上单调减少，在[3,3]-上单调增加，在[3,)+∞上单调减少. ……5分 因为(3)0f -=，且由上述单调性可知(3)f -是()f x 在(3]-∞上的最小值， 所以3x =()f x 在(,3]-∞-上的唯一的零点. ……7分又因为4(3)2(3)03f π=>，且l i m ()x f x →+∞=-∞，所以由连续函数的介值定理知()f x 在(3,)+∞内存在零点，且由()f x 的单调性知零点唯一.综上可知，()f x 在(,)-∞+∞内恰有两个零点，即原方程恰有两个实根. ……10分(19)（本题满分10分）设函数()f x 在区间[0,1]上具有连续导数，(0)1f =，且满足()()ttD D f x y dxdy f t dxdy '+=⎰⎰⎰⎰，其中{(,)|0,0}(01)tD x y y t x x t t =≤≤-≤≤<≤.求()f x 的表达式.解：()()tt t x D f x y dxdy dx f x y dy -''+=+⎰⎰⎰⎰……2分(()())()()ttf t f x dx tf t f t dx =-=-⎰⎰. ……4分又2()()2tD t f t dxdy f t =⎰⎰，由题设有20()()()2t t tf t f t dx f t -=⎰. ……6分两边求导整理得 (2)()2()t f t f t '-=，解得2()(2)Cf t t =-. ……9分代入(0)1f =，得4C =，故24()(01)(2)f x x x =≤≤-. ……10分(20)（本题满分11分）【 同数学一（20）题 】 (21)（本题满分11分）【 同数学一（21）题 】 (22)（本题满分11分）【 同数学一（22）题 】 (23) （本题满分11分）设二维随机变量(,)X Y 服从区域G 上的均匀分布，其中G 是由0x y -=，2x y +=与0y =所围成的三角形区域.(I) 求X 的概率密度()X f x ； (II) 求条件概率密度|(|)X Y f x y . 解：(I) (,)X Y 的概率密度为1,(,)(,)0,x y Gf x y ∈⎧=⎨⎩其他.X 的概率密度为()(,)X f x f x y dy +∞-∞=⎰.（1）当0x <或2x >时，()0X f x =； （2）当01x ≤≤时，0()xX f x dy x ==⎰； （3）当12x <≤时，20()2xXf x dy x -==-⎰；综上所述 ,01()2,120,X x x f x x x ≤≤⎧⎪=-<≤⎨⎪⎩其 他……7分(II) Y 的概率密度为2,01,2(1),01,()(,)0,0,y yY dx y y y f y f x y dx -+∞-∞⎧≤≤-≤≤⎧⎪==⎨⎨⎩⎪⎩⎰⎰=其他其他； 在(01)Y y y =≤<时，X 的条件概率密度为|1,2,(,)2(1)(|)()0,X Y Y y x y f x y y f x y f y ⎧<<-⎪-==⎨⎪⎩其他. ……11分。

2011年中山大学英语专业（基础英语）真题试卷(总分：78.00，做题时间：90分钟)一、阅读理解(总题数：3，分数：40.00)For an infant just beginning to interact with the surrounding world, it is imperative that he quickly become proficient in his native language. While developing a vocabulary and the ability to communicate using it are obviously important steps in this process, an infant must first be able to learn from the various streams of audible communication around him. To that end, during the course of even the first few months of development, an infant will begin to absorb the rhythmic patterns and sequences of sounds that characterize his language, and will begin to differentiate between the meanings of various pitch and stress changes. However, it is important to recognize that such learning does not take place in a vacuum. Infants must confront these language acquisition challenges in an environment where, quite frequently, several streams of communication or noise are occurring simultaneously. In other words, infants must not only learn how to segment individual speech streams into their component words, but they must also be able to distinguish between concurrent streams of sound. Consider, for example, an infant being spoken to by his mother. Before he can leam from the slight differences of his mother"s speech, he must first separate that speech from the sounds of the dishwasher, the family dog, the bus stopping on the street outside, and, quite possibly, background noise in the form of speech; a newscaster on the television down the hall or siblings playing in an adjacent room. How exactly do infants wade through such a murky conglomeration of audible stimuli? While most infants are capable of separating out two different voices despite the presence of additional, competing streams of sound, this capability is predicated upon several specific conditions. First, infants are better able to learn from a particular speech stream when that voice is louder than any of. the competing streams of background speech; when two voices are of equal amplitude, infants typically demonstrate little preference towards one stream or the other. Most likely, equally loud competing voice streams, for the infant, become combined into a single stream that necessarily contains unfamiliar patterns and sounds that can quite easily induce confusion. Secondly, an infant is more likely to attend to a particular voice stream if it is perceived as more familiar than another stream. When an infant, for example, is presented with a voice stream spoken by his mother and a background stream delivered by an unfamiliar voice, usually he can easily separate out her voice from the distraction of the background stream. By using these simple yet important cues an infant can become quite adept at concentrating on a single stream of communication and, therefore, capable of more quickly learning the invaluable characteristics and rules of his native language.（分数：10.00）(1).Which of the following best conveys the main idea of Paragraph 1 ?（分数：2.00）A.Infants are fully aware of their environments.B.Infants have natural talent to develop vocabulary.C.Infants are able to take in information from the environment.D.Infants like rhythmic patterns and sequences of sounds.(2).The phrase "predicated upon several specific conditions"(Para. 4)is used by the author to suggest that______.（分数：2.00）A.most infants have trouble separating out simultaneous streams of speechB.infants can only learn when they are comfortable in their surroundingsC.only in rare instances do these required conditions occurD.infants are not always able to learn from their surrounding environment(3).The author uses the word "necessarily"(Line 4 of Para. 5)in order to suggest that______.（分数：2.00）A.an individual stream understandably changes character when mixed with anotherB.even adults can have trouble distinguishing between streams of equal volumeC.infants always combine separate streams into a single soundD.it is inevitable that two streams of speech are more confusing than one(4).Before an infant can learn from the slight differences of his mother"s speech, he must first______.（分数：2.00）A.understand his father"s communication streamB.be able to distinguish between his mother"s voice from that of the background noiseC.absorb the sounds of dishwasher and petsD.learn something about his language from the television voice(5).The example in the last paragraph is used to illustrate how______.（分数：2.00）A.an infant who spends little time with his parents would probably have trouble with language acquisitionB.an infant in constant vocal interaction with his parents could experience accelerated language acquisitionC.the complexity of an infant"s native language is not a factor in determining whether that language will be easily acquiredD.infants with particularly attentive parents are more likely to acquire language skills more quicklyWhen I accepted a volunteer position as a social worker at a domestic violence shelter in a developing nation, I imagined the position for which my university experience had prepared me.I envisioned conducting intake interviews and traipsing around from organization to organization seeking the legal, psychological, and financial support that the women would need to rebuild their lives. When I arrived, I felt as if I already had months of experience, experience garnered in the hypothetical situations I had invented and subsequently resolved single-handedly and seamlessly. I felt thoroughly prepared to tackle head-on the situation I assumed was waiting for me. I arrived full of zeal, knocking at the shelter"s door. Within moments, my reality made a sharp break from that which I had anticipated. The coordinator explained that the shelter"s need for financial self-sufficiency had become obvious and acute. To address this, the center was planning to open a bakery. I immediately enthused about the project, making many references to the small enterprise case studies I had researched at the university. In response to my impassioned reply, the coordinator declared me in charge of the bakery and left in order to " get out of my way. " At that moment, I was as prepared to bake bread as I was to run for political office. The bigger problem, however, was that I was completely unfamiliar with the for-profit business models necessary to run the bakery. I was out of my depth in a foreign river with only my coordinator"s confidence to keep me afloat. They say that necessity is the mother of invention. I soon found that it is also the mother of initiative. I began finding recipes and appropriating the expertise of friends. With their help making bread, balancing books, printing pamphlets and making contacts, the bakery was soon running smoothly and successfully. After a short time it became a significant source of income for the house. In addition to funds, baking bread provided a natural environment in which to work with and get to know the women of the shelter. Kneading dough side by side, I shared in the camaraderie of the kitchen, treated to stories about their children and the towns and jobs they had had to leave behind to ensure their safety. Baking helped me develop strong relationships with the women and advanced my understanding of their situations. It also improved the women"s self-esteem. Their ability to master a new skill gave them confidence in themselves, and the fact that the bakery contributed to the upkeep of the house gave the women, many of them newly single, a sense of pride and the conviction that they had the capability to support themselves. Baking gave me the opportunity to work in a capacity I had not at all anticipated, but one that proved very successful. I became a more sensitive and skillful social worker, capable of makinga mean seven-grain loaf. Learning to bake gave me as much newfound self-confidence as it gave the women, and I found that sometimes quality social work can be as simple as kneading dough.（分数：14.00）(1).The primary purpose of the passage is to show how the author______.（分数：2.00）A.was shocked by the discrepancy between her earlier ideas about her work and the reality she facedB.discovered a talent her overly-focused mind had never allowed her to exploreC.broadened how she defined the scope of her workD.developed her abilities to orchestrate a for-profit business enterprise(2).In Line 5 of Para. 1 "garnered" most nearly means______.（分数：2.00）A.exchangedB.collectedC.requiredD.enriched(3).The statement that the author arrived "full of zeal"(Line 1 of Para. 2)indicates that she was______.（分数：2.00）A.anxious and insecureB.eager and interestedC.confident but uninformedD.cheerful but exhausted(4).The author was initially enthusiastic about the idea of the bakery because she______.（分数：2.00）A.considered it from a theoretical point of viewB.hoped to obtain a leadership position in the bakeryC.wanted to demonstrate her baking knowledge to her new coordinatorD.believed it would be a good way to build the women"s self-esteem(5).The comparison in Lines 6 -7 of Para. 2("At that moment...political office")demonstrates the author"s belief that______.（分数：2.00）A.it was unfair of the coordinator to ask the author to run the bakeryB.social workers should not be involved in either baking or politicsC.she was unqualified for a job baking breadD.similar skills were involved in both baking and politics(6).Lines 7 -8 of Para. 2("The bigger...bakery")suggest that the author believed that______.（分数：2.00）A.learning the necessary business practices would be a more daunting challenge than learning to bake breadB.good business practices are more important to running a successful bakery than is the quality of the breadC.her coordinator"s confidence in for-profit business models was misplacedD.for-profit business models are significantly more complex than the non-profit models with which she was familiar(7).The last sentence("Learning...dough")indicates that the author______.（分数：2.00）cked self-confidence just as much as the women with whom she workedB.found that performing social work is surprisingly easy with no educationC.underestimated her own ability to learn new skillsD.derived a benefit from her work while helping othersThough he would one day be considered an innovator and founding father of the artistic movement known as Impressionism, Claude Monet(1840-1926)began his career as a fairly traditional representational artist. His painting gradually changed, however, as he became interested in lightand how it affects perception—an interest that led him to attempt to paint light itself rather than the objects off of which light reflected. Monet also rejected the tradition of painting in a dedicated studio, and left the confines of his dusty room to paint outside. Many of his friends and fellow artists, including Pisarro, Renoir and Cezanne, were also interested in working alfresco and joined him in painting outdoors. This group, the core of the movement that would later be classified as Impressionism, made it a common practice to paint the same scene many times in a day to explore the changes in the light, using small patches of color rather than the large brush strokes and blended color that had characterized artistic technique until that time. The Impressionists were thus attempting to evoke a mood rather than document a specific scene or event, as had been the aim of earlier painters. This move away from representation was also effected by a technological development, as photography became more affordable and popular. Before the development of photography, painting was the primary means of documenting the marriages, births, and business successes of the wealthy. Photographers soon took over much of this role because photographs were faster, more accurate, and less expensive than paintings. This freed the Impressionists to find new roles for their medium and encouraged the public to think about painting in a new way. It was no longer just a means of recording significant events; it now reflected an artist"s unique vision of a scene or moment. Today, Impressionism enjoys a privileged position with many art historians and critics, although this was certainly not always the case. As the movement was developing, most critics were at best uninterested and often appalled by the work. Even the name of the movement was originally a derisive critique. A critic who, like most of his colleagues, prized realism in paintings, declared the movement "Impressionism" after the name of the painting Impression: A Sunrise, by Monet. The critic considered the Impressionists" works unfinished—only an impression, rather than a complete painting. It is safe to say that such a critic would be in the minority today, however. Impressionist paintings are now some of the most prized works in the art world. Museums and individuals pay huge sums to add these works to their collections, and the reproductions of the artworks are among the most popular fine art posters sold.（分数：16.00）(1).The primary purpose of the passage is to______.（分数：2.00）A.condemn the critics who prevented the Impressionists from exhibiting their workB.contrast Monet"s work with that of Pisarro, Renoir, and CezanneC.describe the primary characteristics of Monet"s paintingsD.explain the origins of Impressionism and Monet"s role in the movement(2).According to the passage, the Impressionists did all of the following EXCEPT______.（分数：2.00）A.paint the same scene at different times of the dayB.paint the light reflected by objectsC.receive acclaim from their contemporariesD.reconsider the role of painting in society(3).In Line 10 of Para. 1, the author most likely mentions "patches of color" to describe______.（分数：2.00）A.the light that the Impressionists encountered when they worked outdoorsB.a shortcoming of traditional paintingsC.a distinguishing characteristic of modern paintersD.an innovative technique used by Impressionist painters(4).The discussion of photography(Para. 2)serves as______.（分数：2.00）A.a description of an innovation that affected the development of ImpressionismB.the most important context in which to understand ImpressionismC.a demonstration of its similarities to paintingD.a demonstration of the public"s dislike of Impressionism(5).The author of the passage would most likely describe the medium of photography as______.（分数：2.00）A.expensiveB.preciseC.falseD.inconsistent(6).In Para. 3 "a derisive critique" most nearly means that the criticism was made in a(n)______way.（分数：2.00）A.carelessB.constructiveC.exaggeratingD.mocking(7).The "critic" mentioned in Line 5 of Para. 3 would most likely agree that______.（分数：2.00）A.Impressionist paintings are inferior because they fail to clearly represent their subjectsB.Impressionism now enjoys a much more prestigious place in the art world than it once didC.Monet"s Impression; A Sunrise was a highly influential workD.the use of photography to document important events freed painters to explore other roles(8).In Line 8 of Para. 3 "prized" most nearly means______.（分数：2.00）A.awardedB.discoveredC.valuedD.decorated二、句子改错(总题数：10，分数：20.00)1.Correct the mistakes in the following sentences: underline the wrong parts and put the correct ones in the brackets. If there is no error, use a √ or write "No error" on the ANSWER SHEET.(10 points)My parents, my younger sister, and me were delighted to see how much my cousin had grown since we last visited his family in the summer.（分数：2.00）__________________________________________________________________________________________ 2.We spent a most enjoyable afternoon sitting on the grass, watching for unusual shaped cloud formation.（分数：2.00）__________________________________________________________________________________________ 3.Beside the dusty road sets a pond, which serves as a breeding ground for several species of the noisiest animals such as fogs.（分数：2.00）__________________________________________________________________________________________ 4.The other students and she felt unprepared when tested on facts not learned in class.（分数：2.00）__________________________________________________________________________________________ 5.Working two jobs is common among struggling actors, the majority of them work in restaurants that allow them flexible hours to audition.（分数：2.00）__________________________________________________________________________________________ 6.Food produced without pesticides poses fewer danger and promotes easier digestion than that produced through traditional agriculture.（分数：2.00）__________________________________________________________________________________________ 7.When Shirin Abadi was awarded the Nobel Peace Prize, many of her colleagues praised her exceptional efforts for democracy and human rights in Iran.（分数：2.00）__________________________________________________________________________________________ 8.Concerned about the coming game on Saturday, each of the team members spent most of the week practicing their plays.（分数：2.00）__________________________________________________________________________________________9.Even the San Francisco Earthquake in the spring of 1906 leveled many buildings, it was the subsequent series of fires that destroyed most of the city.（分数：2.00）__________________________________________________________________________________________ 10.Studies indicate that the environments in schools where there are fewer adults on staff is often not conducive to learning.（分数：2.00）__________________________________________________________________________________________ 写作11.Here"s a description of a company manager"s personal experience in his job: " I"ve been working in and with international companies for more than a decade, often specifically brought in to help solve cross-cultural communication or management challenges, or to fix disfunctional internal corporate cultures. So my ear has become attuned to the " us versus they " clues. They never listen. They just don"t understand. We are right, they are wrong. " The British poet RudyardKipling(1865-1936)also expressed his understanding of cultural differences by means of a poem "We and They". The following box contains the beginning and the end stanzas excerpted from the poem. Now read the following two stanzas, and then write an argumentation of about 400 words onthe topic: " Why are " They" always wrong?". 2.00）__________________________________________________________________________________________四、英译汉(总题数：1，分数：2.00)12.Translate the following passage into Chinese. Write your translation on the ANSWER SHEET.(20 points) The period of Chinese scientific activity did not begin until the first years of the Republic. The older reformers only introduced a book knowledge of the sciences, without fully understanding their intellectual significance, without adequate equipment for laboratory work, and without adequately trained leaders to organize the studies and researches. Most of the textbooks on science were translated by men who admired science most sincerely but who had only a very superficial book knowledge of the subjects in the Japanese schools, and never did real laboratory work or undertook field expeditions. The schools were beginning to have classroom experiments in physics and chemistry, and botanical and zoological specimens; but they were as bookish as the textbooks, and were useless for the training of scientific workers.（分数：2.00）__________________________________________________________________________________________五、汉译英(总题数：1，分数：2.00)13.Translate the following passage into English. Write your translation on the ANSWER SHEET.(20 points) 徐志摩在《吸烟与文化》中深情地写道：“我在康桥的日子可真是幸福，生怕这辈子再也得不到那样甜蜜的机会了。

2011年中山大学学硕真题（药学综合A）一、单选题（每题3分，20题共60分；请选择正确答案的代号写在答题纸上，并标明题号）1．胰岛素等多肽类药物在胃肠道中易受到酶破环而被分解，因此临床恰当地给药方式为（C）A.口服B.皮内注射C.皮下注射D.静脉注射E.肌肉注射2．脂质体静脉注射后易被网状内皮系统的巨噬细胞吞噬，体现剂型的（D）A.不同剂型改变药物的作用性质B.不同剂型改变药物的作用速度C.不同剂型改变药物的毒副作用D.有些剂型可以产生靶向作用E.剂型可以影响疗效3．将碘配成聚维酮碘液应用于临床，其中聚维酮的作用是（A）A.增溶B.助溶C.潜溶D.乳化E.助悬4．在药物制剂生产中最常用的灭菌方法是（A）A.热压灭菌法B.流通蒸汽灭菌法C.煮沸灭菌法D.低温间歇灭菌法E.辐射灭菌法5．处方中能减轻维生素C注射液疼痛的辅料是（B）A.依地酸二钠B.碳酸氢钠C.亚硫酸氢钠D.亚硫酸钠E.二氧化碳6．取供试品2片，置20℃±1℃C的100ml水中，振摇3分钟，应全部崩解并通过二号筛的片剂是（D）A.包衣片B.泡腾片C.咀嚼片D.分散片E.口腔速崩片7．以下属于肠溶包衣材料的是（D）A.HPMCB.HECC.HPCD.HPMCPE.PEG8．胃溃疡病人因感冒头痛发烧应选用哪种解热镇痛药最合适（C）A.阿司匹林B.吲哚美辛C.对乙酰氨基酚D.保泰松E.布洛芬9．抑制细菌细胞壁合成的抗菌药中可除外（E）A.青霉素类B.头孢菌素类C.杆菌肽D.万古霉素类E.利福平10．西地泮临床不用于（B）A.焦虑症B.诱导麻醉C.小儿高热惊厥D.麻醉前用药E.癫痫持续状态11．维生素B6降低左旋多巴的疗效是因为（A）A.增加外周脱羧酶的活性B.减少左旋多巴的吸收C.减少外周脱羧酶的活性D.直接对抗左旋多巴的作用E.增加左旋多巴进入脑内12．下列有关药物的杂质检查项目中，属于特殊杂质检查项目的有（D）A.青霉素中水分的检查B.盐酸普鲁卡因中铁盐的检查C.对乙酰氨基酚中重金属的检查D.维生素E中游离生育酚的检查E.地西洋中氯化物的检查13．色谱法定量分析时采用内标法的优点是（C）A.优化共存组分的分离效果B.消除和减轻拖尾因子C.消除仪器、操作等影响，优化测定的精密度D.内标物易建立E.为了操作方便14．在药品质量标准分析方法验证中，以下叙述错误的（C）A.用于原料药含量测定的方法除不要求定量限和检测限外，其余均有所要求B.用于原料药中杂质限度检查时只要求检测限、专属性和耐用性，其余均无要求C.用于原料药中杂质含量测定时，所有八项指标均有所要求D.用于鉴别试验时只要求专属性和耐用性，其余均无要求E.用于制剂中有效成分的溶出度测定时，除不要求定量限和检测限外，其余均有所要求15．薄层色谱法用于药物鉴别，具有优点，但常需采用对照品是因为（C）A.一些药物的Rf值数据不足B.硅胶G等试剂的性质欠稳定C.为了消除实验因素的影响，使结果可靠D.质量标准的例行要求E.观察方法的需要16．用酸性染料比色法测定生物碱类药物，有机溶剂萃取测定的有机物是（C）A.生物碱盐B.指示剂C.离子对D.游离生物碱E.离子对和指示剂的混合物17．下列药物中属于吸入性全麻药的是（A）A.恩氟烷B.盐酸氯胺酮C.羟丁酸钠D.硫喷妥钠E.利多卡因18．胆碱酯酶抑制剂临床用于治疗（B）A.治疗胃溃疡B.治疗重症肌无力和青光眼C.抗过敏D.麻醉前给药E.以上都不对19．与抗凝血作用有关的维生素是（E）A.维生素D3B.维生系B2铁盐的检查C.维生素CD.维生素K3E.维生素E20、磺酰脲类降血糖药作用机理是（B）A.增加血液中胰岛素的含量B.与胰腺β细胞膜的受体结合从而促进释放胰岛素而降糖C.增强机体对胰岛素的敏感性D.促进脂肪组织摄取葡萄糖，减少葡萄糖经消化道吸收，使血糖下降E.可降低多糖及蔗糖分解生成葡萄糖减少并延缓吸收而降低血糖二、填空题（每题3分，20题共60分：请把答案按顺序写在答题纸上，并标明题号）1．Stoke's定律描述的是混悬粒子的沉降速度及影响因素。

2011年中山大学（资产评估硕士）真题试卷(总分：76.00，做题时间：90分钟)一、名词解释(总题数：6，分数：12.00)1.资本资产定价模型（分数：2.00）__________________________________________________________________________________________正确答案：(正确答案：，也即某种证券的期望收益=无风险资产收益率+β系数×风险溢价。

这一表明证券的期望收益与该证券的p系数线性相关的公式被称为资本资产定价模型。

资本资产定价模型是在投资组合理论和资本市场理论基础上形成发展起来的，主要研究证券市场中资产的预期收益率与风险资产之间的关系，以及均衡价格的形成。

)解析：2.每股盈余无差别点（分数：2.00）__________________________________________________________________________________________ 正确答案：(正确答案：每股盈余无差别点是指，每股盈余无差别点是指在两种筹资方式下，采用负债筹资与采用普通股筹资每股盈余相等的点，可用息税前盈余来表示。

可用采用每股盈余无差别点来分析、判断在何种情况下运用负债筹资和调整企业资本结构，借以实现企业资金结构优化。

当盈余能力大于每股盈余无差别点时，企业利用负债筹资较为有利，当盈余能力小于每股盈余无差别点时，不应再增加负债，以发行普通股为宜。

)解析：3.价值类型（分数：2.00）__________________________________________________________________________________________ 正确答案：(正确答案：价值类型是指资产评估结果的价值属性及其表现形式。

不同价值类型从不同角度反映资产评估价值的属性和特征。

不同的价值类型所代表的资产评估价值不仅在性质上是不同的，在数量上往往也存在着较大差异。

中山大学2011年数学分析真题题目一、（每小题6分，共48分） （1） 求极限limx→0√1−x 2−1xtanx； （2） 计算积分∫sinxcosx 1+sin 4xdx π20；（3） 已知∑(−1)na n ∞n=1=A ，∑a 2n−1=B ∞n=1，求级数∑a n ∞n=1的和；（4） 计算∬(2x +43y +z)dS S，其中S 为平面x 2+y 3+z4=1在第一卦限部分； （5）计算∫√x 2+y 2dx +y (xy +ln(x +√x 2+y 2))dy L ，其中L 为曲线y =sinx(0≤x ≤π)按x 增大方向； （6） 判断级数∑n √n−lnn∞是绝对收敛，条件收敛还是发散？（7） 设{x =t 3−3t y =t 2+2t，求二阶导数d 2y dx 2； （8）求数列极限lim n→∞12·34····2n−12n。

二、设f (x,y )=√|xy |，求偏导数ðf ðx ,ðf ðy，指出它们的定义域及连续性，并讨论f (x,y )在点(0,0)处的可微性。

三、设f (x )满足 （1） −∞<a ≤f (x )≤b <+∞（2）|f (x )−f (y )|≤L |x −y |，0<L <1；x,yϵ[a,b]任取x 1ϵ[a,b]，做序列x n+1=12(x n +f (x n ))，n =1,2,…。

求证{x n }收敛，且其极限ξϵ[a,b]满足：f (ξ)=ξ。

四、设正项数列{x n }单调递增，且lim n→∞x n =+∞，证明∑(1−x n x n+1)∞n=1发散。

五、已知P 是∠AOB 内固定点，∠AOP =α,∠BOP =β，线段长度OP̅̅̅̅=L ，过P 的直线交射线OA 和OB 与点X 与Y ，求线段长度乘积PX̅̅̅̅·PY ̅̅̅̅的最小值，说明取最值时X ，Y 的位置。

2011考研数学一真题试卷一选择题1.曲线222)4()3()2)(1(----=x x x x y 拐点 CA （1，0）B （2，0）C （3，0）D （4，0）2设数列{}n a 单调递减，∑=∞→⋯===nk k n n n n a S a 1,2,1(,0lim ）无界，则幂级数∑=-nk nk x a 1)1(的收敛域CA(-1,1] B[-1,1) C[0,2) D(0,2]3.设函数)(x f 具有二阶连续导数，且0)0(,0)(>'>f x f ，则函数)(ln )(y f x f z =在点（0，0）处取得极小值的一个充分条件 AA 0)0(,1)0(>''>f fB 0)0(,1)0(<''>f fC 0)0(,1)0(>''<f fD 0)0(,1)0(<''<f f4.设⎰⎰⎰===444000cos ln ,cot ln ,sin ln πππxdx K xdx J xdx I 的大小关系是、、则K J IBA I<J<KB I<K<JC J<I<KD K<J<I5.设A 为3阶矩阵，将A 的第二列加到第一列得矩阵B ，再交换B的第二行与第一行得单位矩阵。

记,010100001,010********⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=P P 则A=DA 21P PB 211P P -C 12P PD 112P P -6.设),,,(4321αααα=A 是4阶矩阵，*A 是A 的伴随矩阵，若T )0,1,0,1(是方程组0=Ax 的一个基础解系，则0*=x A 的基础解系可为 D A 31,αα B 21,αα C 321,,ααα D 432,,ααα7.设)(),(21x F x F 为两个分布函数，其相应的概率密度)(),(21x f x f 是连续函数，则必为概率密度的是 DA )()(21x f x fB )()(222x F x fC )()(21x F x fD )()()()(1221x F x f x F x f +8.设随机变量X 与Y 相互独立，且EX 与EY 存在，记U=max{x,y},V={x,y},则E(UV)= B A EUEV B EXEY C EUEY D EXEV 二填空题9.曲线)40(tan 0⎰≤≤=xx tdt y π的弧长s= __)21ln(+_____10.微分方程x e y y x c o s -=+'满足条件y(0)=0的解为y=___x e y x sin -=_________ 11.设函数⎰+=xydt tty x F 021sin ),(，则__________22=∂∂=x x F 4 12.设L 是柱面方程为122=+y x 与平面z=x+y 的交线，从z 轴正向往z 轴负向看去为逆时针方向，则曲线积分⎰=++___________22dz y xdy xzdx π13.若二次曲面的方程为42223222=+++++yz xz axy z y x ，经正交变换化为442121=+z y ，则=a _______1________ 三解答题15求极限110))1ln((lim -→+x e x xx 原式=21111)1()1ln(lim)1ln(1)1ln(021]))1ln((1[lim e eexxx x x e x x x xxx e x x x x x x x ===-++-+--+-+-+→→-16设))(,(x yg xy f z =，其中函数f 具有二阶连续偏导数，函数g(x)可导，且在x=1处取得极值g(1)=1,求1,12==∂∂∂y x yx z解由g(x)可导且在x=1处取极值g(1)=1所以0)1(='g)1,1()1,1()1,1()](,()()(,([)](,[)()](,[)](,[1211212111221f f f yx zx yg xy f x g x yg xy f x y x yg xy f yx zx g y x yg xy f y x yg xy f x zx ''+''+'=∂∂∂''+''+'=∂∂∂''+'=∂∂17求方程0arctan =-x x k 不同实根的个数，其中k 为参数。