2004 1998考研数一真题

114原式lim 1x21x1x2 x2 2

1x2

1111

2

111x2 1 11 211 x2

lim

1x1x14x1

lim211x1x1

214 1 111x1x2ox2 11x1x2ox21 11x1x2ox211x1x2ox2 原式

1x2ox2ox2 1

),x

z

f(xy)y(xy) x

f(xy)x

f(xy)y(xy)2z

x2f

xf

1f(xy)x1f(xy)yf(xy)x(xy)y(xy) 1f(xy)1f(xy)yf(xy)(xy)y(x yf(xy)(xy)y(xz1f(xy)y(xy) f(xy)(xy)y(x2 2z

xf yf(xy)(xy)y(xz z xyxf(xy)yxy(xy) 1f

f

yf(xy)(xy)y(xxyyx视为常数就可以了.【答案】 13x24y212(3x24y2)ds12ds 原式 (3x24y2)ds12a yl1为lx0的部分,则有结论：2lfxyds,fxy关于xfx,yds 2lfxyds,fxy关于yfx,yds l

A0,知0(如果0A的特征值A0A

* *AA

* *

*EAE*

1,故A

E

方法2A0A的特征值0(如果0A的特征值A0A1 *

,A的特征值 ；(A)E的特征值为 1 XAXX成立,则称AXA的特AXX,则AkEXAXkXkXAkE的特征值是k.2AA0A11A

1yy(2,)【解析】首先求X,Yf(xyD(x,y)|1xe2,0y1 e2 dxln f(x,y)1f(x,y) X

Xx1xe2f(x0X 1 当1xe2f(x

f(x,y)dy 0

dy 2f

(2)14【解析】为变限所定义的函数求导数,作积分变量代换ux2t2 t:0xu:x20,dudx2t 2tdtdtduxtf(x2t2)dtux2t20tf(u)1 2t 0

f

12

f

1 dx0tf(xt)dt2dx0f 1f(x2)x21f(x2)2xxf(x2

f(xx2x2)xx21x01f(xx01(x2x2)x(x2

x1x f(x)(x2x2)x(1x2 0x(x2x2)x(x21),1

fxf1

(x2x2)x(1x2)0f(

xfxf1x

x(x2x2)x(1x2)x

0f(xx1f(0)

fxf0

(x2x2)x(x21)0

2 f(0)

fxf0

(x2x2)x(1x2)0

2 f(xx0f(xx1f(x只有2个不可导点,故应选f(xxa(x,其中(xxaf(xxa处可导的充要条件是(a)0.【解析】由yyxy

.1

1 令x0得是xlim0lim

lim

x0

lim x0 x01

x01 x0 1 dy

.1分离变量,得dy dx 1两边积分,得

yarctanxCyCearctanx1 1 )

lim(x)l)ccL:x

y

z

x

y

z11a1

b

c

,L2:a

b

c1 2

a3 c3 a1b1c1 a2b2c20故向量组(a1a2b1b2c1c2与(a2a3b2b3c2c3k1k2k1(a1a2b1b2c1c2k2(a2a3b2b3c2c30,这样(a1a2b1b2c1c2与(a2a3b2b3c2c3L1L2的方向向量,由方向向量线性相关,两直线平行,可知L1,L2不平行.xa3yb3zc3a1 b1 c1xa31yb31zc31a1 b1 c1即xa3a1a2yb3b1b2zc3c1c2即a1 b1 c1xa1yb1zc1a2

b2

c2xa11yb11zc11a2 b2 c2即xa1a2a3b3b2b3zc3c2c3即a2 b2 c2L1L2均过点a2a1a3,b2b1b3,c2c1c3,故两直线相交于一点,选(A).

1P

PABPAPB应选

方法1：L与NL

y 代入平面z1(1t)t2(1t)10t1,N(2,1,0)LM(1,0,1作平面Lx1yz, x1即y

z1L与平面N2(1t)(t)2(12t)10t1321N2(,,33NNL:x2y1z 方法2：L在平面L作垂直于平面的平面0,所求投影线就是平面与0的交线.平面0L上的点(1,与不共线的向量l(1,1x z

0x3y2z10L:xy2z1

0x3y2z1x2 xyz

(2y)(2y)2(1(12

消去Sx

2y

(y ,即

2y1P(xy2xy(x4y2,Q(xy)x2(x4y2),A(xy)P(xyQ(xx0上为某二元函数u(xy的梯度PdxQdyx0上函数u(x,y)QPx Q2x(x4y2)x2(x4y2)14x3P2x(x4y2)2xy(x4y2)12y xy2x(x4y2)x2(x4y2)14x32x(x4y2)2xy(x4y2)12y4x(x4y2)(1)01为求u(xyx0半平面内任取一点,比如点(1,0)作为积分路径的起(x,y)2xydxu(x,y) x4 x2x 01x4dx0

x4yy

dy0x4yy

22 x4(1y x2 y yy d

C d 0x4(1y2

x2

0 y

x2

arctan

x2 x2【相关知识点】1.二元可微函数u(xygraduui+uj 2.定理：DP(xy与Q(xyD内连续且有连续的一QP,(x,y)D

存在二元单值可微函数u(x,y)duPdxPi+Qj为某二元函数u(xygraduPi+Qj法求函数u(x,y).【解析】先建立坐标系,取沉放点为原点O,铅直向下作为F浮Bkvdt

t00,vt0 d2 dv 由 v,

dt dy dyv v 分离变量得dy

dv

0 dy

mgB

dvy

Bmm2g Bmm2gmv m

mgBkvBmmg mgB m2gBm m mdv m(mgB)dv k(mgBkv)

v kd(mgBkv k再根据初始条件v|y00kmykv

kmmg mgBkv lnmg 区域封闭后用高斯公式进行计算,但由于被积函数分母中包含(x2y2z212,因此不能立x2y2即加、减辅助面1z

Iaxdydz(za)2dxdy1axdydz(za)2dxdy ax2y2添加辅助面1z

,其侧向下(由于为下半球面z 侧,而高斯公式要求是整个边界区面的外侧,这里我们取辅助面的下侧,和的上侧组成整个I1axdydz(za)2dxdy1 axdydz(za)2dxdya a 1

a( 1的方向向下；另外由曲面片1yoz平面投影面积为零,则axdydz0,而1z0则za2a21 aI (a2(za))dVadxdya 其中为与D为xoyDxy|x2y2a2 I1 dv a 1

3aa 2

rdr zdza2a20a 0 1 I 2a4 2a z 0d

a2a2r1

a

2 a dr (ar)a 1 a2r r44

14

a4 a 40 a 41

4

a4

a 4 2

axdydz(za)2 (xyzI aaxdydz(xyz

xdydz1(za)2dxdya a

1II1

xdydz

a2x2y2dydz

a2x2y2

a2x2y2函数在x取负数. D为yozDyz|y2z2a2z0} I22 aa2r2 a2r2d(a2r22 (a2r2)3

a 0 0

aa2I1(za)2dxdy1aaa2

a2a2x2y 12da(2a2 r2a a2ra2r

(2a2r r3

2a2rdr

ra2r2dr

aaa aa

22 13 r

aaa

02a3

4 02(a42a4a4)a3 11

yz|y2z2a2I

a2P(xyz、Q(xyzR(xyz在 PQRdvPdydzQdzdx z R

xyzdv 这里是cos、cos、cos是在点(x,y,z)处的法向量的sinsin2 sinnxnnn 1是函数sinx在[0,1]区间上的一个积分和.于是可由定积分sinxdx求得极限limxsin n

sin ni

sin n

nnsinnn

sin

sinnnnnn

ni

nn sin 1 limnsinn0sinxdxnnsin

1 1 limn1limn sinnlimnsinnsinxdxn n 1n sinnnn

ni

2NnNyxzlimylimzalimxa

n

nna0 又(1)naa0(否则级数(1)n

1 又正项级数a单调减少,有 ,而0 1,级数( nann

a

aaa

n 方法2：同方法1,可证明limana0.令bn ,

nan

1a

an

)nan设交错级数(1)n1

nunun1,n1, (2)limun(1)n1u收敛,且其和满足0n

(1)n1uu,余项r 反之,若交错级数(1)n1

条件(2)limun0,所以有limun0.(否则级数(1)n1

设un和vn都是正项级数,且lim n 当0Aun和vn A0时,若un收敛,则vn收敛；若vn发散,则un A时,若vn收敛,则un收敛；若un发散,则vn

设un0,则当 un发散，且limun

0【解析】(1)要证x00,1)x0f(x0xf(x)dx；令(xxf(xxf(t)dt0xx(0)0 (1)0(x)dx0xf(x)dx0(xf 分部 0xf(x)dxxxf

0xf(x)dxf(x在[0,1连续(x在[0,1](x在[0,1]连续,在(0,1)x0(0,1),使(x0)(x0)0.由(xxf(xf(xf(xxf(x2f(x0,知(x在(0,1)内单调增,故)1 1接对(x)用零点定理遇到麻烦时,不妨对(x)的原函数使用罗尔定理.在开区间(abf(af(b

1

P1AP 0记B 0 A(b1)2B11 A13 11

a3,b0EA

1 0x1x2x3于是得方程组(0EA)x0的同解方程组为2x 1EE

1 1

1 0 1 0

0 0x2x3于是得方程组(EA)x0的同解方程组为xx 4EA

3 3

4 4 0x1x2x3于是得方程组(4EA)x0的同解方程组为2x4x xx2,解得基础解系为1,2,1)T

)T

,

)T

3 )T 1 62 2 6因此所求正交矩阵为P 6 1 6 2A与BAB 0 0 0000

A2k30

1111 其中,(a,a,, )T,(a,a,, )T,,(a, ,, )T 1,2 2,2 ,( ABT0两边取转置,有ABTTBAT0ATABY0r(BnBY02nr(B2nnn,恰好等A的行向量个数.故A的行向量组是BY0的基础解系,其通解为其中,(a,a,, )T,(a,a,, )T,,(a, ,, )T 1,2 2,2 k1k2kn性质,可以知道XYN(0,1),这样可以简化整题的计算.E(Z)E(X)E(Y)0,D(Z)D(X)D(Y)111 ZXYN(0,1)DXYDZEZ2EZD(Z)EZ2EZ21EZ211

E

z

e2dz

ze20

z2

z2 e2d e2 0 2 0DX

12【相关知识点】1.X与YX与Y的线性组合亦服从正态X与Y