東南-數(shù)學(xué)分析2002答案

一、敘述定義(5分+5分=10分)

fx解：設(shè)0M0E0,則當(dāng)x時(shí)就有f(xMxa時(shí),f(x)不以A為極限解：設(shè)0,E0,使得當(dāng)xa時(shí),f(xA二、計(jì)算（9分×7=63分yln(1x2),0x12 s

1[f'(x)]2dx1 11 21( 2)2dx 2dx2 1)dxln3 1 01 01 1 u

f(xyzg(x2eyz)0,ysinx,且己知f與g偏導(dǎo)0,du g(xez)0,知2xg1dxeg2dyg3dz0,dxxyxz=f1cosxf2(2xg1cosxeyg2)f求lnx2x lnx t 2 2 e解：令tlnx,則xedxedte

x)dx

2tedt=tedttet2te

t

(lnx)22lnx2C求

axxax

(a axxa x

x2[(ln2

11]o(xa

)}[1xlna

x2(ln2

o(x2 x1=Sx2dydzy2dzdxz2dxdy其中Szx2y2zSSP(xyzx2Q(xyz)y2R(xyz)z2xrcosyrsinzr2，且0r1,02SSx2dydzy2dzdxz S 2d12r(rcosrsinr2)dr x2x2y

2(xy 求常數(shù)，使得曲線積 xrL

dx

x y2

dy0,r x2y

z 1,x 0 4 z

F(xyzx2

1,則 2x, 2y,

2x(Xx)2y(Yy)z(Zz)2

,為X

4x24y2z

,Y

4x24y2z4

,Z

4x24y2z,zdX2Y2Z2(4x24y2z2)2( 1)=1116x 16y z x y z 令P(x,y,z) x y

z2

2y2z4

則P22x0,P

2y0,P321z0,x2y2z2

y

z xy1z2

2,16,

三、證明題（6分+7分+7分+7分=27分 sin級(jí) n101解設(shè)f(x)在區(qū)間[02]上具有二階連續(xù)導(dǎo)數(shù)，且對(duì)一切x[0,2]，均有f

f

f

2 f(0)f(2)

f(x)f'(x)(0x)f''()(0x)22f(x)f'(x)(2x)f''()(22f(2)f(0)2f'(x)1[f''()(2x)2f''()x222f'(x)

f(2)f(0)1[f''()(2x)2f''()x212121212f'(x)2

f(2)f(0)1[f''()(2x)2f''()x2]2

f(2) f(0)1212f''()(2x)2 f''()x 11(2x)21x (x1)2 1212 f

證明積分xexydy在(0,0 設(shè)0,m1,m20,并且當(dāng)m1m2時(shí)

d(xy)[exy

emxemxemx(1e(mm)x

x,取2則

xexydye1ee2m1(1eA(1eA為一常數(shù)證明函數(shù)f(x) xlnx在[1,)上一致連續(xù)2lnx2證明：f'(x) 2lnx2xxf''(x)12