首屆生數(shù)學競賽非類答案

(xy)ln(1yx dxdy ，其中區(qū)域D由直線xy1與 1xyuxvxvyuvdxdy

dudvdudv

(xy)ln(1yxdxdy

ulnuuln dv ln dv ln( 1u 1u1u2lnuu(ulnuu)011

u2

11111令t ，則u1t2，du2tdt，u212t2t4，u(1u)t2(1t)(110(*)21(12tt0 21(12t2t4)dt2t2t31t5 0 0設(shè)f(x)是連續(xù)函數(shù)且滿足f(x)3x2 2f(x)dx2,則f(x) 0解令A(yù) 2f(x)dx，則f(x)3x2A2022A022

A2)dx82(A2)42AA4f(x)3x2 x曲面z y22平行平面2x2yz0的切平面方程 2x2x解因平面2x2yz0的法向量為(2,2,1)，而曲面z y222(2,2,1zxxzy2y知2zx(x0y0x02zy(x0y02y0，x處的切平面方程是2(x2)2(y1)(z1)0，即曲面z y22x22x2yz0的切平面方程是2x2yz50yy(xxef(yeyln29ff1，d2y ef(y)xf(y)yef(y)eyyln11 f( f(y)y yelnx1因eyln29xef(y)0

f(y)yyyx

x(1f(

f(y) y

x2(1f(

x[1f( f(y)x2[1f(y)]3

x2(1f(

f(y)[1f(y)]2x2[1f(y)]3解法 方程xef(y)eyln29取對數(shù)，得f(y)lnxylnln方程（1）xfyy1x即y x(1f(

方程（2）xfyyfyy)2

（4

f(y)y f(y) 1yx2(1f(y))2 f(y)(1f(y))2y(1f(y))x2(1f(y))2y

f(y)[1f(y)]2x2[1f(y)]3exe2xenx（5分）求極限

x，其中n是給定的正整數(shù)n1e e )xlim(1

exe2xenxn) 故Alimexe2xenxn elimexe2xenx elimex2e2xnenxe12nn1

exe2xenxA )xA

eee2exe2xenx

ln(exe2xenx)ln )xelim ex2e2x 12 ne 2

e x0e故

exe2xenxA )xA

eee（15分）f(x

f(xt)d

fx

A

0g(xg(xx0處的連續(xù)性解由

fAf(xf(0)limf(x)limx

f 1

1

g(x)0f(xt)dtg(0)0f(0)dtf(0)0 x0g(x)x0f(u)duxlimg(x)lim0f(u)dulimf(x)f(0)x0g(x)

f(u)duf x2 1xf xfg(0)limg(x)g(0)limx lim limf(x)

f

x0 limg(x)

xf(u)du

]limf

xf(u)duA x2 x0x2 g(xx0處連續(xù)（15分）Dxy|0x0y}LD的正向邊界， xesinydyyesinxdxxesinydyyesinxdx xesinydyyesinydx52 xesinydyyesinxdx (xesiny) D

DxesinydyyesinxdxD

(esinyesinx(xesiny)(yesinxDD

) D(esinyesinxD (esinyesinx)dxdy(esinyesinx xesinydyyesinxdxxesinydyyesinet

2(1 2(1 t ) 2(1t2 故esinxesinx2sin2x21cos2x5cos 由 xesinydyyesinydx(esinyesinx)dxdy(esinyesinx 知xesinydyyesinydx1(esinyesinx)dxdy1(esinyesinx 2 21(esinyesiny)dxdy1(esinxesinx)dxdy(esinxesinx)dxdy2D 2D (esinxesinx)dx5cos2xdx5 即即xesinydyyesinydx5 （10分）y1xexe2xy2xexexy3xexe2xex是某二階常系數(shù)線性非微分方程的三個解，試求此微分方程.y1xexe2xy2xexexy3xexe2xexybycyf的三個解，則y2y1exe2x和y3y1ex都是二階常系數(shù)線 ybycy的此ybycy0的特征多項式是(2)(1)0，而ybycy0的特2bc 微分方程為yy2y0，由y1y12y1yexxex2e2x，y2exxex

f(xf(xyy2yxex2ex4e2xxexex2e2x2(xexe2x (1yy2yex2xe（10分）yax2bx2lnc過原點.當0x1時,y0,1

3yax2bx2lnc過原點，故c11 a b2 0 0即

bx)dt3x

2x3b2(13V(a)

(ax2bx)2dt

0

22(1a)x)2

14 4a(1xdt

4(1a)x

1x01a21a(1a)4(1 V(a)1a21a(1a)4(1 令V(a)2a1(12a)8(1a)0 得即4a5a5,b3,c e（15分）已知un(x滿足un(xun(xxn1ex(n1,2,,且un(1)en數(shù)項級 un(x)之和

,解un(x)un(x)xn1ex即yyyex(C即xyex nx(Cxnun(x)n nun(1)e(CnC0，un(x)

S(x)un(x) 則S(x)即

n

)S(x)xn1exS(x)

1S(x)S(x)

1S(x)ex(C

1dx)1xx0，得0S(0)C，因此級數(shù)un(xS(x)exln(1（10分）x1時,與xn2等價的無窮大量解f(txt2，則因當0x1t1,f(t)2txt2lnx0t2lnf(t)xt2 x在(0, f(t)dt