vip專屬數(shù)學(xué)階段測(cè)評(píng)課6 9月試題sx7 3 answer

,x,y0,1、二元函數(shù)f(x,y)x2 在點(diǎn)0,0處（ x,y0, fxy在0,0

f(0,0)df(x, ,f(0,0)df(0, f(x0)0xf(0y0yfxy在0,0

f(0,

f(0,

0limf(x,y)

1f(0,

x0x2 2f(xyxyx2y2xy成立，則f(xy)f(xy)（ （A）2x2 （B）2x2 （C）x （D）xf(xyxyx2y2xy)(xyxyuxyv，所以f(u,v)uv，故f(x,y) f(x,y)f(x,y)yxxy.故應(yīng)選，所以

1xy14I1ln3(xy)dI2(xy)2dI3sin2xy)d，則（DI1I2

I3I21

（C）I1I3 （D）I3I14

xy1Dln(xy)0sin(xy)x ln3xy)dsin2xy)d(xy)2d.應(yīng)選 D4f(xyf(xyxyf(uv)dudvDy0yx2xDf(xy等于（1（A） （B） 18

Df(xy)xyf(uv)dudvD f(x,y)dxdyxydxdyf(u,v)dudv DD

0

xydy1

，dxdyD1

dy 3則f(x,y)dxdy f(x,y)xy k

ak(n1, an(x1)n的收斂域?yàn)椋ǎˋ） （C）[0, an20(nan0(n1x0

a(xn

(1)na.（n

級(jí)數(shù)

a(xn

an(x1)n

故選

k

n(n1,2,"")無界，所以冪級(jí)數(shù)a(x1n

x26

(1)n1nS（3

8

4S123"(1)n2n1(1)n1n 1S12

3"(1)n2n1(1)n1n3 4S1111"(1)n11(1)n13 1[111"(1)n11](1)n1 )1(1)

13

313

(1)n11[1(1)n](1)n1 利用lim1limn0，即知limS13n n n

nxn1 (1x)2 1n1S() n(

(11 )3) 1 3 3

n n(

3 7、設(shè)zz(x,y)是由方程xyz xyzet2dt確定的隱函數(shù)，則dz0

yzex2y2z2222dx1xyexy

xzex2y2z21xyex2y2dxdydzex2y2z2d(xyz)ex2y2z2yzdxxzdyxydz)，(1xyex2y2z2)dz(yzex2y2z21)dx(xzex2y2z21)dy

yzex2y2z2222dx1xyexy

xzex2y2z21xyex2y28f(xyx2(2y2ylny1e f2x(2y2f2x2ylny ff yAf

解得駐點(diǎn)(0,e1 2(21)xx(0, y

2B

1(0,) (0,

(0,e0，C

(0,1)(2x21

1(0,e所以B2AC2e(21)0，A0， 1是f(x,y)的極小值，極小值f(0, f(0,1)1 9f(xyf(xy

xyf(xy)dDD(xy)x2y21f(xyDxy

Af(xy)df(xyD

xyAAf(x,y)d(xyA)dxydA DDD1(xyx0,y0,x2y21xyxy ydy21x(1x2)dxxyd4xyd40 1 A，即A12

1

1.故f(x,y)xy 10、xy

(x

y)d43x

yxyx

y13(xy)d4(xy)d81dx1xxdy81x(1x)dx3xy

11、冪級(jí)數(shù)2n3)n(x【答案】(2

2n(n(n2n1

則R 收斂區(qū)間為(2,112f(x1

x22x3展開為x的冪級(jí)數(shù)，則f(x)

1

n1

1

，(1x4f(x

1( 1x22x 4x x1[ 1 1]1[(1)nxn1 ()n4x 313

4 3n01[(1)n11]xn，(1x1)4n0 13、設(shè)ufxyzzzxyxexyeyzezduFxyzxexyeyzezFx1exFy1ey ；Fz1ez則用隱函數(shù)求導(dǎo)公式 zFxx1exzzy1ey；zuffzffix1exz z zzuffzffy1 z zz

z z duudxudyffx1exzdxffy1eyz zz zz xexyeyzez1xexdx1yeydy1z解 dz

1z 1xexdx1yeydy dufdxfdyfdzfdxfdy

1z

duffx1exzdxffy1eyz zz zz 又由Q1Q210000，即10000500P15500250P210000，得2P1P2FQ2Q1(P5)2125001000P2 FQ2Q2(P5.5)26875500P2 DD

1x21x2

2D(,)01,0 2 21 11 2D原式D

1

2dd2d

1

11

d

(

44

d2 1

2

1

1[

d2

1221[1arcsin21

1](2 16Ix2y24dDx，y0x2，0yDI4x2y2dx2y2 24x2y2d2d4r2rdr2 2 xy4d0dx4x2xy4dy0xy34 22x8dx224x3dx162

16 2 3 3 3 3 2 2 2

31 （I24xdx令x2sin I416

02

tdt2 342n17f(x)f(xxn1ex（n為正整數(shù)f(1)e，求函數(shù)項(xiàng)級(jí)數(shù)n

f(xn f(x)f(x)

(x)e

Cf(1)

，可確定常數(shù)C0，故f(x)

而fn(x)nex S(x)

nS(x)

1xS(xS(0)xS(t)dtx1dtln(1 01S(xln(1xx1x11x1時(shí)，有

(