2004 1998考研數(shù)一真題

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視為常數(shù)就可以了.【答案】 13x24y212(3x24y2)ds12ds 原式 (3x24y2)ds12a yl1為lx0的部分,則有結(jié)論：2lfxyds,fxy關(guān)于xfx,yds 2lfxyds,fxy關(guān)于yfx,yds l

A0,知0(如果0A的特征值A(chǔ)0A

* *AA

* *

*EAE*

1,故A

E

方法2A0A的特征值0(如果0A的特征值A(chǔ)0A1 *

,A的特征值 ；(A)E的特征值為 1 XAXX成立,則稱AXA的特AXX,則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 當(dāng)1xe2f(x

f(x,y)dy 0

dy 2f

(2)14【解析】為變限所定義的函數(shù)求導(dǎo)數(shù),作積分變量代換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個(gè)不可導(dǎo)點(diǎn),故應(yīng)選f(xxa(x,其中(xxaf(xxa處可導(dǎo)的充要條件是(a)0.【解析】由yyxy

.1

1 令x0得是xlim0lim

lim

x0

lim x0 x01

x01 x0 1 dy

.1分離變量,得dy dx 1兩邊積分,得

yarctanxCyCearctanx1 1 )

lim(x)l)ccL:x

y

z

x

y

z11a1

b

c

,L2:a

b

c1 2

a3 c3 a1b1c1 a2b2c20故向量組(a1a2b1b2c1c2與(a2a3b2b3c2c3k1k2k1(a1a2b1b2c1c2k2(a2a3b2b3c2c30,這樣(a1a2b1b2c1c2與(a2a3b2b3c2c3L1L2的方向向量,由方向向量線性相關(guān),兩直線平行,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均過點(diǎn)a2a1a3,b2b1b3,c2c1c3,故兩直線相交于一點(diǎn),選(A).

1P

PABPAPB應(yīng)選

方法1：L與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與平面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,所求投影線就是平面與0的交線.平面0L上的點(diǎn)(1,與不共線的向量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上為某二元函數(shù)u(xy的梯度PdxQdyx0上函數(shù)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為求u(xyx0半平面內(nèi)任取一點(diǎn),比如點(diǎn)(1,0)作為積分路徑的起(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【相關(guān)知識(shí)點(diǎn)】1.二元可微函數(shù)u(xygraduui+uj 2.定理：DP(xy與Q(xyD內(nèi)連續(xù)且有連續(xù)的一QP,(x,y)D

存在二元單值可微函數(shù)u(x,y)duPdxPi+Qj為某二元函數(shù)u(xygraduPi+Qj法求函數(shù)u(x,y).【解析】先建立坐標(biāo)系,取沉放點(diǎn)為原點(diǎn)O,鉛直向下作為F浮Bkvdt

t00,vt0 d2 dv 由 v,

dt dy dyv v 分離變量得dy

dv

0 dy

mgB

dvy

Bmm2g Bmm2gmv m

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

v kd(mgBkv k再根據(jù)初始條件v|y00kmykv

kmmg mgBkv lnmg 區(qū)域封閉后用高斯公式進(jìn)行計(jì)算,但由于被積函數(shù)分母中包含(x2y2z212,因此不能立x2y2即加、減輔助面1z

Iaxdydz(za)2dxdy1axdydz(za)2dxdy ax2y2添加輔助面1z

,其側(cè)向下(由于為下半球面z 側(cè),而高斯公式要求是整個(gè)邊界區(qū)面的外側(cè),這里我們?nèi)≥o助面的下側(cè),和的上側(cè)組成整個(gè)I1axdydz(za)2dxdy1 axdydz(za)2dxdya a 1

a( 1的方向向下；另外由曲面片1yoz平面投影面積為零,則axdydz0,而1z0則za2a21 aI (a2(za))dVadxdya 其中為與D為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函數(shù)在x取負(fù)數(shù). D為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 這里是cos、cos、cos是在點(diǎn)(x,y,z)處的法向量的sinsin2 sinnxnnn 1是函數(shù)sinx在[0,1]區(qū)間上的一個(gè)積分和.于是可由定積分sinxdx求得極限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(否則級數(shù)(1)n

1 又正項(xiàng)級數(shù)a單調(diào)減少,有 ,而0 1,級數(shù)( nann

a

aaa

n 方法2：同方法1,可證明limana0.令bn ,

nan

1a

an

)nan設(shè)交錯(cuò)級數(shù)(1)n1

nunun1,n1, (2)limun(1)n1u收斂,且其和滿足0n

(1)n1uu,余項(xiàng)r 反之,若交錯(cuò)級數(shù)(1)n1

條件(2)limun0,所以有l(wèi)imun0.(否則級數(shù)(1)n1

設(shè)un和vn都是正項(xiàng)級數(shù),且lim n 當(dāng)0Aun和vn A0時(shí),若un收斂,則vn收斂；若vn發(fā)散,則un A時(shí),若vn收斂,則un收斂；若un發(fā)散,則vn

設(shè)un0,則當(dāng) un發(fā)散，且limun

0【解析】(1)要證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連續(xù)(x在[0,1](x在[0,1]連續(xù),在(0,1)x0(0,1),使(x0)(x0)0.由(xxf(xf(xf(xxf(x2f(x0,知(x在(0,1)內(nèi)單調(diào)增,故)1 1接對(x)用零點(diǎn)定理遇到麻煩時(shí),不妨對(x)的原函數(shù)使用羅爾定理.在開區(qū)間(abf(af(b

1

P1AP 0記B 0 A(b1)2B11 A13 11

a3,b0EA

1 0x1x2x3于是得方程組(0EA)x0的同解方程組為2x 1EE

1 1

1 0 1 0

0 0x2x3于是得方程組(EA)x0的同解方程組為xx 4EA

3 3

4 4 0x1x2x3于是得方程組(4EA)x0的同解方程組為2x4x xx2,解得基礎(chǔ)解系為1,2,1)T

)T

,

)T

3 )T 1 62 2 6因此所求正交矩陣為P 6 1 6 2A與BAB 0 0 0000

A2k30

1111 其中,(a,a,, )T,(a,a,, )T,,(a, ,, )T 1,2 2,2 ,( ABT0兩邊取轉(zhuǎn)置,有ABTTBAT0ATABY0r(BnBY02nr(B2nnn,恰好等A的行向量個(gè)數(shù).故A的行向量組是BY0的基礎(chǔ)解系,其通解為其中,(a,a,, )T,(a,a,, )T,,(a, ,, )T 1,2 2,2 k1k2kn性質(zhì),可以知道XYN(0,1),這樣可以簡化整題的計(jì)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【相關(guān)知識(shí)點(diǎn)】1.X與YX與Y的線性組合亦服從正態(tài)X與Y