数学竞赛非数组试题合辑

2017年数学竞赛预赛（非数学类）—1.已知可导函数满足,则ff'(x)cosx+f(x)sinx1,f'(x)+f(x)tanxsecxf(x

tanxdxdxc=elncosx

elncosxdxc=cosx dxc cos =cosxtanxc=sinxccosf(01f(xsinxcosx求limsin2

n2nn2 n2n2nn 1cyv1wxxc2wyy 解 wyc(f2wcffccfcfcfcf=c2f2ff 221wxxc2wyy=4f12

f(sin2=f(xf(0)f'(0)x1f"()x2f(sin2x1f"(sin4x 这样limf(sin2x=limf"()sin4x3 esinxsin5I(1sin解:

dx esinxsinxcos sin (vI

2(1v)2dv

(1 2v1dv2(v1)2dv2v1dv2edv 12v1dv2 v

eve

2esinv1+C1sinx+C 4x26.记曲面z2x2y2和z 围成空间区域为4x2V VI

zdxdydz

0

0 0

2cos2sind。V。21sin2/4142 f(xy在平面上有连续的二阶偏导数.对任何角度g(t)f(tcos,tsin)若对任何

dg(0) d2g且且dg

0.f(0,0f(xy的极小值cos解：由于 fx,fy

0对一切成立，故(f,f (0,0), y (0,0) y(0,0)是f(x,y)的驻点 4记H(x,y)

xy yy d2g(0)d cos dt dt(fx,fy)sin

(cos,sin)Hf(0,0)sin010上式对任何单位向量(cossinHf(0,0)是一个正定阵,f(0,0极小值 14三(本题满分14分 设曲线为x2y2z21，xz1，x0,y0,zA(1,0,0)B(0,0,1)Iydxzdy解:记1BA的直线段,xt,y0,z1t0t 2ydxzdyxdztd(1t 42 设和1围成的平面区域，方向按右手法则.Stokes ydxzdyxdz 1

8右边三个积分都是在各个坐标面上的投影面积，而zx面上投影面积为零.I dydzdxdy1曲线xy(1/(x1/2)2 y(1/(1/

12 2 2

2 .同理2 2这样就有I 2

14 四本题满分15分)设函数f(x0且在实轴上连续，若对任意实数t，有e|tx|f(x)dx1，则ab(abbf(x)dxba2 证.由于ab(ab be|tx|f(x)dxe|tx|f(x)dx aa因 bdtbe|tx|f(x)dxba 4aa b

然 adta f(x)dxa|tf(x)a 其 be|tx|dtxetxdtbextdt2eaxexba

bf(x)(2eaxexb)dxba

10 bf(x)dxba1beaxf(x)dxbexbf(x)dx beaxf(x)dxbe|ax|f(x)dx1，和bf(x)exbdx1 15五(本题满分15分 设{an}为一个数列，p为固定的正整数。 a，其为常数，证明lim

n n 证明：对于i0,1,…,p1， 。由题设limA(i) n limA(i)A(i) A(i) nA(i)A(i)A(i

a(n1) =

a(n1)

10n(n1)p (n1)p 对正整m，设m=npi，其中,p1，从而可以把正整数依照i分为p个子列 a ，故limam 15(n1)pn(n1)p n (êÆ2018䁜3擐 Á磐ª ©¨÷© 铐—峐䭜䱜䱜提绐¿:1.䁜䱜大铐´胐答铐,篐䥜Ê至十大铐¥裐À䭜铐,铐Ò要㑜㥜þ¡擐闐¥(õÀÃ凐2.¤䡜答铐Ñ闐廐糐䅜Áò纸密µ线䩜蟐,廐糐Ù§纸þ－ÆÃ凐 3.密µ线左蟐请Õ答铐,密µ线勐Ø提䡜翐¶苐相烐鋐駐密µ 答铐时Ø要超闐䅜 4.㕜答铐空xØ叐,可廐糐凐页寐¡密µ 答铐时Ø要超闐䅜提µ(20©z5©提µ 惐¢

011

§n≥1§Ù¥I´与

Ó某擐方叐©铐rank(H4) ln(1+tanx)−ln(1+sin(

x=π

惐Γ为空䩜­ y=t−sinz=sin

t∈[0π].È

esinxcosxcosyΓ1aa···a1a··1aa···a1a··· 惐峐䑜矐f(x,...,x)=(x,... .a·· 1

·· 翐¶ 飐yÒ ¤糐峐Ò 狐Ò 蓐业: Ù¥n翐¶ 飐yÒ ¤糐峐Ò 狐Ò 蓐业:1提µ峐櫐(15©)提µ¡ S:a2+b2+c2= a,b,c>苐S勐Ü－菐A(x0,y0,z0)§闐A菐且与S相切擐¤䡜直线凐¤鋐¡Σ.y² 糐Π§¦SΣSΠ¶Ó时¦Ñ²Π擐方2翐¶ 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐提µ 䭜櫐£15©¤惐ABCþn某軐方叐§提µ AB−BA= AC= BC=y²µC´幕毐方叐y²µABCÓquþ䭜埐叐䉜CI=0,¦n擐最に值密µ密µ 答铐时Ø要超闐䅜3提µ䱜櫐£曐铐20©¤f(x)[01]þ提µ¼êf(0)f(1)0.¦yZ Z|f0dx::,

|f(x)|dx

Z|f00(x)|04翐¶ 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐 Ê£曐铐10©¤G为瓐§vµ∀x,y提µ G,(xy)2=(yx)2.y²µ∀x,y∈G,ｄxyx−1y−1擐提µØ超闐密µ密µ 答铐时Ø要超闐䅜5提µ臐櫐£曐铐10©¤惐E⊂Rn为可ÿ臐,m(E)提µ惐f,fk∈L2(E),糐Eþ諐に裐裐䡜fk→f§¦yµ

JE|fk(t)−f(t)|2dt=

lim

|fk(t)|2dt

|fE6翐¶ 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐提µ Ô櫐£曐铐10© ý哐Ρ提µ\r(u,v)={acosu,bsinu, −π::,u::, −∞<v<£1¤¦Sþÿ£2¤惐a=b.寐P=(a00)Q=(acosu0asinu0v0)(−π<u0<π,−∞<v0<+∞).廐ÑSþë提PQü菐擐最á­线擐方§.密µ密µ 答铐时Ø要超闐䅜7提µ䥜櫐£曐铐10©¤í峐¦狐线绐方§|擐提µª§¿y²ª²ÚÂñ8翐¶ 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐提µ Ê櫐£曐铐10©¤惐¼êf(z)糐 哐|z|<1鳐提µ Û§糐Ù蟐矐þë.䉜糐|z|1þ|f(z)|1.y²f(z)为䡜䭜¼ê.密µ密µ 答铐时Ø要超闐䅜9提µ十櫐£曐铐10©¤惐X1X2···Xn´ÕáÓ©Ù擐随Å賐þÙ䡜“Ó擐©Ù¼êF(x)Ú密ݼêf(x).yéÅþX1X2···Xn黐大に㭜鳐­泐üXn1::,Xn2::,···::,提µÇ¦随Å賐þ(Xn1,Xnn)擐é 密ݼêf1n(x,Ç㕜鏐Xi(i=12···n)Ñ䥜«䩜[0,1]þ擐þ櫐©Ù§¦随Å賐þU=XnnXn1擐密ݼêfU(êÆ2018䁜3擐 Á磐ª ©¨÷© 铐—峐䭜䱜䱜提绐¿:1.䁜4大铐´胐答铐,5–10大铐¥裐À䭜铐,铐Ò要㑜㥜þ¡擐闐¥¤䡜答铐Ñ闐廐糐䅜Áò纸密µ线䩜蟐,廐糐Ù§纸þ－ÆÃ凐密µ线左蟐请Õ答铐,密µ线勐Ø提䡜翐¶苐相烐鋐駐 4.㕜答铐空xØ叐,可廐糐凐页寐¡²Ò密µ密µ 答铐时Ø要超闐䅜提µ((20©)(z5提µ(

011

§n≥1§Ù¥I´

Ó某擐方叐©铐rankH4)=3limln(1+tanx)−ln(1+sinx)=13

x=πsint /(3)惐Γ为空䩜­ y=t−sinz=sin

,t∈[0,π].铐竐峐矐­线È esinxcosxcosyΓ）siny +coszdz=

···x ···(4)惐峐䑜矐f(x,...,x)=(x,...,x 擐Ý叐A

翐¶ 飐yÒ ¤糐峐Ò 翐¶ 飐yÒ ¤糐峐Ò 狐Ò 蓐业:a·· 1a ··a1 Ù¥n1aR铐f糐巐に賐换䉜擐鋐飐磐为((n1)a1)y12(a1)y22···(a 答㙜.(1)Hn´m=2n某é方叐§糐巐に方叐P¦P−1HnP=D´1 \ \

\−1 埐方叐©䥜囐§Hn+1

与

相q©惐HnO O ¤䡜諐埐值´λ1λ2···λm§Hn+1擐¤䡜諐埐值´λ11λ1−1λ21λ21···λm+1λm−1©|㭜êÆ臐诐{韐´y²µHn擐¤䡜ØÓ諐埐值为{nn2k|k=01···n}§¿且z个諐埐值n−2k擐代ê­ê为n

©k!(n−4䅜§rankH4244方法二µ©¬Ð绐:|Lagrange¥½䭜可±zlimln(1+tanx)−ln(1+sinx)=limtanx−sinx=lim1−cosx= -

只鋐¦ÑA擐旐Ü值蛐可wAa−1)I擐秩<1.Aa−1)I擐毐空䩜擐维ê�n−1,䥜囐可惐A擐n个諐征值为λ1=1−a,λ2=1−a,···,λn−1=1−a,绐trA=n,拐提λn=(n−1)a+1.燐鏐§f糐巐に賐换䉜擐鋐飐磐为((n−1)a1)y2−(a−1)y2−···−(a− 2提µ 峐櫐(15©)糐空䩜直埐坐鋐㕜䉜§惐䡜ý提µ S:a2+b2+c2= a,b,c>苐S勐Ü－菐A(x0,y0,z0)§闐A菐且与S相切擐¤䡜直线凐¤鋐¡Σ.y² 糐Π§¦SΣSΠ¶Ó时¦Ñ²Π擐方狐µÏA糐S擐勐ܧ 0 0+0−1> éuM(x,yz)∈SΣ§ëAM擐直线駐为lM§Ùëê§ ˜=x+t(x− −∞<t< 密µ 答铐时Ø要超闐䅜密µ 答铐时Ø要超闐䅜(x+t(x−

(y+t(y−

(z+t(z− =

z2+

（

+

+0+0+0−

0x+0y

（

z0+

a2+b2+c2 a2x+b2y+c2 =

(5©Ï为菐M糐ý¥Sþ§x2+y2+z21.¤±þªz

（x0x

y0y+z0

（

z0 1

0+0+0−

+

1− x+ y+

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+z1

–1=a2 b2

c2u´§菐M(x1y1z1)糐²£§䅜£6¤ª§x0y0z0ØΠ:x0x+y0y+z0z−1= þ§蛐䡜MS反之§éuM(x1y1z1)∈SΠ§

(10©铐S糐M菐擐切²

x0xa21x

+y0b2y

+z0c2z

–1=1x

1y

1z−1= ÏA(x0y0z0)菐§ÏM,A擐ë糐菐MÚý¥¡S相切§§糐鋐¡Σþ.拐MS∩燐Ø提y (15©4提µ 䭜櫐£15©¤惐AB,Cþn某軐方叐§且提µ AB−BA= AC= BC=y²µC´幕毐方叐y²µABCÓquþ䭜埐叐䉜CI=0,¦n擐最に值y²Ù¥Ji为諐埐值λié諐擐Jordan¬.éÝ叐B做与C相Ó擐©¬§B=(Bij)k×k.䅜BC=CB可提JiBij=BijJj,ij=12k.ù懐é裐¿õ项ªp䡜 p(Ji)Bij=Bijp(Jj).寐p为Ji擐最にõ项ª§铐提Bijp(Jj)=0.凐iI=j时§p(Jj)密µ 答铐时Ø要超闐䅜䅜AB−BA=C提AiiBii−BiiAii=Ji,i=1,...,k.拐Tr(Ji)=Tr(AiiBii−BiiAii)=0,i=1,2,...,k.䥜囐λi=0,蛐C为幕毐方密µ 答铐时Ø要超闐䅜盐V0={v∈Cn|Cv=0}.é裐¿v∈V0,䅜uC(Av)=A(Cv)=0,Ï䅜AV0⊆V0.Ó䭜§BV0⊆V0.u´ 糐0I=v∈V0Úλ∈C¦提Av=λv毐駐V1={v|Av=λv}⊆V0,䅜AB−BA=C §é裐¿u∈V1,A(Bu)=B(Au)+Cu=λBu.拐BV1⊆V1.䥜囐糐0I=v1∈V1苐µ∈C¦提Bv1=µv1,Ó时䡜Av1=λ1v1,Cv1=0.òv1珐¿为Cn擐－|Ä{v1,v2,...,vn},盐P=(v1vn)§( AP=

BP=

CP= 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐A1B1C1÷vA1B1B1A1=C1A1C1=C1A1B1C1=C1B1.䅜êÆ臐{蛐可提,A,B,CÓ时相quþ䭜埐叐 (10©§\=E1§A,B,妨惐C 01.铐䡜AC=CA提A a1a2.㹜q䅜䡜BC=CB提B \0 b1b2.u´AB−BA=0§ùAB−BA=C䑜ñ！拐÷vCI=0擐 にn为 (15©5提µ䱜櫐£曐铐20©¤f(x)[01]þ提µ¼êf(0)f(1�0.¦yr r|ft(x)|dx<

|f(x)|dx

r|ftt(x)|0狐 惐M=max|ft(x)|=|ft(x1)|,m=min|ft(x)|=|ft(x0)|.铐r Ir|ftt(x)|dx�

Iftt(x)

=|ft(x1)−ft(x0)|�M−0痐－方¡

1 J10|ft(x)|dx<

dxM拐只鋐y

(5©rm<0

|f(x)| ft(x[01]¥毐菐,m=0䅜时(2)w痐¤á.y糐㽜惐ft(x)[0þÃ毐菐,Ø妨惐ft(x)>0,Ï囐f(x)î格緐飐.䉜¡©ü«情磐裐Ø (10©情磐1f(0)�0䅜时f(x)�0x∈[01])ft(x)=|ft(x)|�m,r|f(x)|dx

rf(x)dx

r(f(x)−f(0))dx+f0 r0 r r0 (f(x)−f(0))dx0

0

ft(ξ)xdx

mxdx=1 拐,(2)¤á (15©情磐2f(0)0䅜时䡜f(1)<0根âf擐緐飐绐,f(x)<0x∈[0r |f(x)|dx=

1f(x)dx

r(f(1)−f(x))dx−frrr rrr |f(1)−f(x)|dx |ft(ξ)|(1−x)dx

m(1−x)dx

1 䅜时(2)也¤á.绐µf(0)f(1�0可Øf(x�0x∈[01].可只Ä情磐1.(20©)6翐¶ 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐 Ê£曐铐10©¤G为瓐§vµ∀x,y提µ G,(xy)2=(yx)2.y²µ∀x,y∈G,,ｄxyx−1y−1擐提µØ超闐y².䡜yx2y=yx2. þx2y=((xy−1)y)2y=(y(xy−1))2y=yxy−1yx=(3©篐yx−1y−1xxy−1x−1.ù可x−1y−1x=x(x−1)2y−1x=xy−1(x−1)2x=/wÑ (6©密µ 答铐时Ø要超闐䅜 最后凐y(xyx−1y−1)2e.密µ 答铐时Ø要超闐䅜(xyx−1y−1)2=xy(x−1y−1x)yx−1y−1==(xyx)(y−1x−1y)x−1y−1=xyx(yx−1y−1)x−1y−1=(xy)2(x−1y−1)2=(xy)2(xy)−2=y矐 (10©7提µ臐櫐£曐铐10©¤惐E⊂Rn为可ÿ臐÷vm(提µ惐ffkL2(E),糐Eþ裐裐䡜fkf§¦y

lim r

|fk(t)|2dt

|f(t)|2dt<E

|fk(t)−f(t)|2dt=Ey²µ䡜y裐寐F⊂Er

r|fk(t)|2dt

|f 䅜FatouÚ |f(t)|2dtF

liminfF<lim

|fk(t)|2dt<lim

I=lim

|fk(t)|2dt

r<limr

r|fk(t)|2dt−lim r

|f(t)|2dt

liminfr |f(t)|2dt

|f(t)|2dt

|f Ï䅜(∗)¤á (4©䅜u|f|2可ȧ裐给E0§糐δ0¦裐寐可ÿF⊂E÷vm(F)δ时 |f(t)|2dt

E (6©䅜叶鏐âŽ䭜§糐Eδ⊂E§¦提m(E\Eδ)<δ§且糐Eδþfk－致磐Âñf毐Ï糐N1§裐寐k�N1§t∈Eδ

(t)−f(t)|<

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r

|f(t)|2dt<Er 拐糐N2§¦提裐寐k�N2

|fk(t)|2dt

|fr

盐N=max(N1N2)§铐凐k�N§ |fk(t)−f密µ密µ 答铐时Ø要超闐䅜 |fk(t)−f(t)|2dt |fk(t)−f m(E)+3(1+

|f(t)|2dt+k

|f翐¶ 飐yÒ ¤糐峐Ò 狐Ò 翐¶ 飐yÒ ¤糐峐Ò 狐Ò 蓐业: (10©9提µÔ櫐£曐铐10© ý哐Ρ提µr(u,v)={acosu,bsinu, −π<u< −∞<v<£1¤¦Sþÿ£2¤惐ab.寐Pa00)Qacosu0asinu0v0(−πu0π,−∞<v0+∞).廐ÑSþë提PQüá§.狐µ£1¤¦Sþÿ§.狐{－µru={−asinu,bcosu, rv={0,0,¤±§S擐 {向þn=ru×|ru×

=(a2sin2u+b2cos2

12{bcosu,asinu, 1惐γ´Sþÿ§Ùëê§u= v= t∈ Ä䡜§䅜u裐¿­¡þ擐直线£㕜鏐糐擐{¤Ñ´ÿ磐线§Sþ擐直竐线£蛐u=~ê¤þ为ÿ磐线.u´只鋐¦÷v㭜件ut(t)I=0擐ÿ磐线.䅜时§可作γ擐ëê賐换¦提§可±㭜wª¼êv=f(u)£u∈[−ππ]¤绐闐«.u´§γ擐向þªëê方§z为r(u)={acosu,bsinu,f u∈[−π, 烐uëêu¦§rt(u)={−asinu,bcosu,f rtt(u)={−acosu,−bsinu,f ¤±§γ擐 切向þT(u)=|rt(u)|−1rt(u)=|rt|−1{−asinu,bcosu,f 㕜鏐s´γ擐䥜§铐䡜

=|rt(u)|.u´γÇ向þ

）

rtt|rt(u)|−1+ =rtt|rt|−2−rt|rt|−3· ϧγSþÿÇ (

ds,n,T=|rt|−2(rtt,n,¤±§γ´ÿ(rttnT0.篐䅜(6(9)Ú(10)§䅜ªI−acosu−sinufbcos asin

≡I−asin bcos fþª

ftt(a2sin2u+b2cos2u)=ft(a2−b2)sinucos £i¤ft≡0.铐v=f(u)≡c´~ꧭ线γ´Sþ擐巐棐线§蛐ý 密µ 答铐时Ø要超闐䅜a2+b2= z密µ 答铐时Ø要超闐䅜£ii¤ftI=0.铐(12)棐䅜ut f (a2−b2)sinucos(log|f|)=ft=a2sin2u+b2cos2=12（

(a2−b2)(sin2(a2−b2)sin2u+2 log(a2sin2u+b2cos2u) 21üuu裐竐È©§f1

=

u+

20IcR.篐È©2 2 f=

(a2sin2u+b2cos2u)1 0I=c∈ 翐¶ 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐¤±§Sþÿ£i¤Sþ擐直竐线£v=~ꤶ£ii¤Sþ擐î棐ý哐£u=~ꤶ£iii¤­线(8)§Ù¥¼êf䅜(13)燐½. rýÎвþs1<s< −∞<v<Ù¥s=s(u)´ý r(u)={acosu,bsinu, −π<u<擐䥜¼êÏ为rΡÐ䩜为²¡é諐擐賐换に±­¡þ­线擐䥜Ø賐§囐に£蛐棐å¤rÿ磐线賐为ÿ磐线§¤±® ý哐Ρþ擐ÿ磐线é諐uþã带磐«域¥擐ÿ磐线蛐直线.根â²þ§§§As+Bv+C=痐－方¡§䅜䥜微©ú ds=|rt(u)|du拐

(a2sin2u+b2cos2

）2(s a2sin2u+b2cos2

）2 ý¥þ¤¦ÿ§r ） a2sin2ub2cos2u2duBvC ABØ㕜鏐A0§铐䡜v=~ê§é諐ýÎþ擐î棐ý哐㕜鏐B0§铐r ）a2sin2u+b2cos2u2du=~ê蛐u=~ê§é諐ý哐Îþ擐直竐线㕜鏐AI=0§BI=0§r ）v≡f(u)=

2 cI=(8©£2¤䅜uba§䅜(13)燐½ÿ§(8)zr(u)={acosu,asinu,c1u+ c1,c2∈R,c1I= ϧéu½Q(acosu0asinu0£i¤㕜鏐u00§铐¤¦擐最á为ëPQ擐直竐线ãµ0v£ii¤㕜鏐v00§铐¤¦最áëPQ擐巐棐ý寐䥜ãµu0<u0£u00)½0<u<u0£u0翐¶ 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐£iii¤㕜鏐u0I=0,v0I=0§铐凐ÿ磐线Ï闐P菐时§可惐c2=0.篐盐c1u0=v0§提c1=u−1v0.䥜囐¤¦擐最á­线方§为 r(u)={acosu,asinu,(u−1v Ù¥凐u0<0时§u∈[u0,0]¶凐u0>0时§u∈[0, (10©密µ密µ 答铐时Ø要超闐䅜提µ䥜櫐£曐铐10©¤í峐¦狐线绐方§|擐提µª§¿y²ª²ÚÂñ Än某线绐方§|Ax=b,ù䵜A为n某¢é¡巐½方叐§b为n维壐向þ毐¦狐鷐方§|棐䅜u¦峐䑜êf(x):=1xTAx−bT2¦¯ªxi+1=xi+αidi,i=0,1,2,...Ù¥x0为给½Ð值§xi´x擐竐iÚ鳐代值§di´iÚ䁜裐方向§αi为ÚÀ寐αi¦提f(xi+αidi)达巐最䉜§´提αi

dTi ,iidTiÙ¥ribAxi为í哐毐w痐ív緐íri+1=b−A(xi+αidi)=ri−Ý运Ý{¥§·¦鳐代方向珐䅜A-巐に§蛐iTAdj=0iI=j,¿且ri与rj(iI=j)珐䅜巐に§Ï䅜§可±凐軐“Ý方向di㕜䉜µd0= di+1=ri+1+ i=0,1,2,...䅜di与di+1A-dTAdi+1=dTA(ri+1+βi+1di)= i=0,1,2,... 拐䅜

dT= idTi dT =rTAd=rT(r− )= rT ±

αi

dTri=(ri+βidi−1)Tri=rT Ù¥㭜巐叐dTri=0.ò±þüªβi+1擐闐达ª可 rTrT βi+1 rTiiiu´¦线绐方§|Ax=b擐“Ý运ÝrTd0=r0=b− αi idTixi+1=xi+ ri+1=ri−rTβi+1 di+1=ri+1+irTii=0,1,2,..., (5©·们y²§“ÝÝõnÚ巐狐擐°燐值毐¢þ§Axi+1 αAd䡜密µ 答铐时Ø要超闐䅜/密µ 答铐时Ø要超闐䅜ù

Axn=Ax0+α1Ad1+...+dT(Axn−b)=dT(Ax0−b)+αidT i=0,1,...,n− ·们㵜²þª¢þ

=dTi(b−Ax0)

αk−1dTiAdk−1=dTi(b−翐¶ 飐翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐\/也Ò´㵜§Axn−b与¤䡜di巐に§i=0,1,...,n−1.䥜囐Axn= (10©提µÊ£曐铐10©¤惐¼êf(z糐ü哐|z|<1鳐狐Û§糐Ù蟐矐þë.䉜糐|z|1þ|f(z)|1.y²f(z)为䡜提µy²µÏ¼êf(z)糐ü哐|z|<1鳐狐Û,|z|=þ|f(z)|=1,铐f(z)糐ü哐|z|<1鳐擐毐菐只䡜䡜限õ个,惐为z1z2zn£毐菐算k个ü毐菐 (2©做賐 z−f(z)= (1<k< 1−zkαk为¢ê.铐fk(zr|z|1に磐韐峐为|fk(z)|1,|z|=1时|f(z)|=fk(zk)=0(1<k< (5©作¼F(z)= f−1(z)f(z)=f 1−zkzk=1 k=1z−Ù¥α= αk.铐¼êF(z)糐 哐|z|<1鳐狐Û§v䡜毐菐§糐|z|=þ|F(z)|= (7©䅜狐Û¼ê擐最大最に旐俐䭜§|z|<1þF(z)=C(~ê)§且|C|=uf(z)= z−zk=C1 z−k=11− k=11−为䡜䭜¼ê (10©提µ 十櫐£10©¤惐X1X2···Xn´ÕáÓ©提µ 擐随ÅþÙ“Ó©Ù¼êF(x)Úݼêf(x).yéÅþX1X2···Xn黐大に㭜鳐­泐üXn1<Xn2<···<Xnn.Ǧ随Å賐þ(Xn1,Xnn)擐é 密ݼêf1n(x,Ç㕜鏐Xi(i=12···n)Ñ䥜«䩜[0,1]þ擐þ櫐©Ù§¦随Å賐þU=XnnXn1擐密ݼêfU狐µ(a)駐(Xn1Xnn)擐éÜ©Ù¼êFXn1Xnn(x䉜xy, FX

(x,y)=P(Xn1<x,Xnn< n1密µ 答铐时Ø要超闐䅜=P(Xnn<y)−P(Xn1>x,X密µ 答铐时Ø要超闐䅜=P(X1<y,X2<y,···,Xn<y)−P(x<X1<y,x<X2<y,···,x<Xn<=[F(y)]n−[F(y)−F䉜x�yFn1Xnn(xyP(Xn1<xXnn<yP(Xnn<yF拐(Xn1Xnn)擐éÜ密ݼêl

(x,y)

n(n−1)[F(y)−F(x)]n−2f(x)f x< Ù

(4©(b䅜uXiÑ[0,1]þþ©Ùl 0<x< x<f(x)

0,Ù

,F(x) 0<x<1,x�u´䅜£a¤提(Xn1Xnn)擐éÜ密ݼêl

(x,y)

n(n−1)(y−x)n−2,0<x<y<1, Ù§.

(6©翐¶ 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐l蛐方蟐駐v=xnnxn1,

u=xnn+

2

,鷐賐换擐v=xnn−

xnn=可烐竐壐ªJ=∂(xn1,xnn)=1§铐䅜(Xn1

nn)擐éÜÝ(UV

擐éÜÝfUV(u,v)=fl

u−v,u+

|Jn(n−1)vn−2,0<v<1,0<u−v<2,0<u+v< Ù铐UXnnXn1擐密ݼrfU(u)

fUV(u,J

vn−2dv=n 0<u<—= n(n21)J02−uvn−2dv2—=

–u)n−1,1<u< Ù蛐方法二蟐¦U擐©Ù¼êFU(u).wU擐寐值范围´[0,¤±§凐u0时§FU(uP(U<u0凐u�2时§FU(u凐0u1时§r

(10©FU(u)=P(U<u)凐1u2时

x+y<uf1n(x,y)dxdyr

r0

rx

n(n−1)(y−x)n−2dy

12FU(u)=P(U<u)

f1n(x,r r

rn(n−1)(y−x)n−2dy

r

n(n−1)(y−0 1

1(2−2

拐UXnnXn1擐密ݼfU(u)

n 0<u<22n(2−u)n−1,1<u<22 Ù

(10©(êÆ2018䁜3擐 Á磐ª ©¨÷© 铐—峐䭜䱜Ê臐䱜提绐¿:1.曐Áò“臐大铐2.¤䡜答铐ÑÁòµ,廐糐Ù§纸þ－Æà 3.密µÕ,密µØ¶4.㕜答铐空xØ叐,可廐糐凐页寐¡²Ò密µ 答铐时Ø要超闐䅜提µ(20©密µ 答铐时Ø要超闐䅜提µ (1)惐¢

011

§n≥1§Ù¥I´与

Ó某擐方叐©铐rank(H4) ln(1+tanx)−ln(1+sin(

x=π

惐Γ为空䩜­ y=t−sinz=sin

t∈[0π.È

esinxcosxcosyΓ1aa···a1a···1aa···a1a··· 惐峐䑜矐f(x,...,x)=(x,... .a·· 1

·· 翐¶ 飐yÒ ¤糐峐Ò 狐Ò 蓐业: Ù¥n翐¶ 飐yÒ ¤糐峐Ò 狐Ò 蓐业:1提µ峐櫐(15©)提µ¡ S:a2+b2+c2= a,b,c>苐S勐Ü－菐A(x0,y0,z0)§闐A菐且与S相切擐¤䡜直线凐¤鋐¡Σ.y² 糐Π§¦SΣSΠ¶Ó时¦Ñ²Π擐方2翐¶ 翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐提µ 䭜櫐£15©¤惐ABCþn某軐方叐§提µ AB−BA= AC= BC=y²µC´幕毐方叐y²µABCÓquþ䭜埐叐䉜CI=0,¦n擐最に值密µ密µ 答铐时Ø要超闐䅜3提µ䱜櫐£曐铐20©¤f(x)[01]þ提µ¼êf(0)f(1)0.¦yZ Z|f0dx::,

|f(x)|dx

Z04提µ Ê£15©¤惐α(12)(1−x)α提µ\\裐ê akxk,n×n¢~êÝ叐A为幕毐Ý叐,I为n 叐.惐Ý阵值¼êG(x)½ÂG(x)

1

ak(xI+ 0::,x<翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐Áyéu1ijn,È gij(xdx翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐0密µ密µ 答铐时Ø要超闐䅜5提µ臐櫐£曐铐15©¤䡜矐ë¼êg(t):R→提µv1g(t2.x(t)(tR)´§¨(t)=g(t)x擐ü¦y 糐~êC2C106(êÆ2018䁜3擐 Á磐ª ©¨÷© 铐—峐䭜䱜Ê臐䱜提绐¿:1.曐Áò“6大铐¤䡜答铐Ñ闐廐糐䅜Áò纸密µ线䩜蟐,廐糐Ù§纸þ一ÆÃ凐密µ线左蟐请Õ答铐,密µ线勐Ø提䡜翐¶苐相烐鋐駐 4.㕜答铐空xØ叐,可廐糐凐页寐¡²Ò密µ密µ 答铐时Ø要超闐䅜提µ(20©z5©提µ 惐¢方叐

011

§n≥1§Ù¥I´

Ó某擐方叐©铐rankH4)=3limln(1+tanx)−ln(1+sinx)=13

x=πsint 惐Γ为空䩜­ 䥜t=0巐t=π擐一ã.铐竐峐矐­线Èy=t−sin z=sin2tesinxcosxcosydx−sinydy+coszdz=Γ2

·· ·· 惐峐䑜矐f(x1xnx1x

擐ÝA.

a·· 翐¶ 飐yÒ ¤糐峐Ò 狐Ò 翐¶ 飐yÒ ¤糐峐Ò 狐Ò 蓐业: Ù¥n1a∈R.铐f糐巐に賐换䉜擐鋐飐磐为n1)a1)y2(a1)y2···(a 1 答㙜.(1)Hn´m=2n某é¡方叐§糐巐に方叐P¦提P−1HnP=D´ \ \

埐方叐©䥜囐§Hn+1

与

相q©惐HnO O ¤䡜諐埐值´λ1λ2···λm§Hn+1擐¤䡜諐埐值´λ11λ1−1λ21λ21···λm+1λm−1©|㭜êÆ臐诐{韐´y²µHn擐¤䡜ØÓ諐埐值为{nn2k|k=01···n}§¿且z个諐埐值n−2k擐代ê­ê为n

©k!(n−4䅜§rankH4244方法二µ©¬ÝÐ绐:|Lagrange¥½䭜可±{zlimln(1+tanx)−ln(1+sinx)=limtanx−sinx=lim1−cosx= -

只鋐¦ÑA擐旐Ü值蛐可wAa−1)I擐秩<1.Aa−1)I擐毐空䩜擐维ê�n−1,䥜囐可惐A擐n个諐征值为λ1=1−a,λ2=1−a,···,λn−1=1−a,绐trA=n,拐提λn=(n−1)a+1.燐鏐§f糐巐に賐换䉜擐鋐飐磐为((n−1)a 1)y2−(a−1)y2−···−(a− 2提µ 峐櫐(15©)糐空䩜直埐坐鋐㕜䉜§惐䡜ý提µ S:a2+b2+c2= a,b,c>苐S勐Ü一菐A(x0,y0,z0)§闐A菐且与S相切擐¤䡜直线凐¤鋐¡Σ.y² 糐Π§¦SΣSΠ¶Ó时¦Ñ²Π擐方狐µÏA糐S擐勐ܧ 0 0 0−1> éuM(x,yz)∈SΣ§ëAM擐直线駐为lM§Ùëê§ ˜=x+t(x− −∞<t< 密µ 答铐时Ø要超闐䅜密µ 答铐时Ø要超闐䅜(x+t(x−

(z+t(z−z+ + =1.

z2+

(

+

+0+

+0−

0x+0y

z0+

a2+b2+c2 a2x+b2y+c2 =

(6©Ï为菐M糐ý¥Sþ§x2+y2+z21.¤±þªz

(x0x

y0y+z0

(

z0 1

0+0+0−

+

x+ y+

=翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐x0x+y0y+z0z−1= 䅜时䅜 §方§(3)擐Ä项系êz 0+0+0−1> 3諐飐磐§(4)擐㕜êþØÏ´²§§燐½²Π.þãíy²叐S∩Σ⊂Π§䥜囐y²叐S∩Σ⊂S∩Π.(12©反之§éuS∩Πþ擐裐一菐M(x,yz)§䅜(3)櫐(4)üª蛐§䅜A櫐Mü菐燐½lM一½糐菐M与S相切.拐䅜½Â§lM糐鋐¡Σþ.諐飐磐§M∈Σ.䅜M擐裐¿绐§S∩Π⊂S∩Σ.燐Ø提y (15©ÏA糐S擐勐ܧ 0+0+0>1> éuM(x1y1z1)∈SΣ§ý¥S糐M菐擐切²方§可±廐x1x+y1y+z1z−1=0. ÏëMÚAü´S糐菐M擐切线§¤±A菐糐þã²þ.u´§菐M(x1y1z1)糐²

x1xa2

+y1b2

+z1c2

–1=Π:x0x+y0y+z0z−1= þ§蛐䡜M∈S∩ (12©反之§éuM(x1y1z1)∈SΠ§铐S糐M菐擐切²

x0xa2

+y0b2

+z0c2

–1=x1x+y1y+z1z−1= ÏA(x0y0z0)菐§ÏM,A擐ë糐菐MÚý¥¡S相切§§糐鋐¡Σþ.拐MS∩燐Ø提y (15©4提µ 䭜櫐£15©¤惐ABCþn某軐方叐§提µ AB−BA= AC= BC=y²µC´幕毐方叐y²µABCÓquþ䭜埐叐䉜C/=0,¦n擐最に值y²惐C擐ØÓλ1λkØCäJordan鋐飐矐µC=diag(J1Ù¥Ji为諐埐值λié諐擐Jordan¬.éÝB做与C相Ó©¬§B 䅜BC=CB可提JiBij=BijJj,ij=12k.ù懐éõ项ªp密µ 答铐时Ø要超闐䅜 p(Ji)Bij=Bijp(Jj).寐p为Ji擐最にõ项ª§铐提Bijp(Jj)=0.凐i/=j时§p(Jj)可㱜§䥜囐Bij=0.Ï䅜§B=diag(B11,...,Bkk).Ó䭜§A=diag(A11,...,Akk).䅜AB−BA=C提AiiBii−BiiAii=Ji,i=1,...,k.拐Tr(Ji)=Tr(AiiBii−BiiAii)=0,i=1,2,...,k.䥜囐λi=0,蛐C为幕毐方叐密µ 答铐时Ø要超闐䅜盐V0={v∈Cn|Cv=0}§wV0非空év∈V0䅜uC(Av)=A(Cv0,ÏAV0⊆V0.Ó§BV0⊆V0.u´糐0/v∈V0Úλ∈C¦Avλv毐駐V1={v|Av=λvv∈V0}⊆V0,䅜AB−BA=C§é裐¿u∈V1,A(BuB(AuCu=λBu.BV1⊆V1.䥜囐糐0/v1∈V1苐µ∈C¦Bv1µv1,ÓAv1λ1v1,Cv1=0.òv1Cn擐一|Ä{v1v2vn},盐P=(v1vn)§ AP=

BP=

CP= 0Ù¥A1B1C1为n−1某軐方叐且÷vA1B1−B1A1C1A1C1C1A1B1C1C1B1.䅜êÆ臐诐{蛐可提,A,B,CÓ时相quþ䭜埐叐 (10©§E1,妨惐C 01.铐䡜AC=CA提A 0

a1

.㹜qBCCB提B にn为

.u´AB−BA=0§ù与AB−BA=C䑜ñ！拐÷vC/=0擐翐¶ 飐yÒ ¤糐峐Ò翐¶ 飐yÒ ¤糐峐Ò Ò 蓐业:狐5提µ䱜櫐£曐铐20©¤f(x)[01]þ提µ¼êf(0)f(1�0.¦yr r|ft(x)|dx<

|f(x)|dx

r|ftt(x)|0狐 惐M=max|ft(x)|=|ft(x1)|,m=min|ft(x)|=|ft(x0)|.铐r Ir

I|ftt(x)|dx ftt(x)dxI=|ft(x1)−ft(x0)|�M− 痐一方J1|ft(x)|dxMJ1dxM拐只鋐y rm<0

|f(x)| (10©ft(x[01]¥毐菐,m=0䅜时(2)w痐¤á.y㽜惐ft(x)[0þÃ毐菐Ø妨惐ft(x0Ïf(xî格緐飐䉜©ü裐Ø情磐1f(0)�0䅜时f(x)�0x∈[01])ft(x|ft(x)|�m,r1|f(x)|dx

1f(x)dx

1(f(x)−f(0))dx+f0 r0

r 拐(2)¤á

(f(x)−f(0))dx0

ft(ξ)xdx0

mxdx 情磐2f(0)0䅜时䡜f(1)<0根âf擐緐飐绐,f(x)<0x∈[0r |f(x)|dx=

f(x)dx

r(f(1)−f(x))dx−frrr rrr |f(1)−f(x)|dx |ft(ξ)|(1−x)dx

m(1−x)dx

2䅜时(2)也¤á绐µ䅜f(0)f(1)�0,可Ø妨惐f(x)�0,x∈[0,1].可只Ä情磐 (20©6提µ Ê£15©¤惐α(12)(1−x)α提µ 裐ê akxk,n×n¢~êÝ叐A为幕毐Ý叐,I为叐.惐Ý阵值¼êG(x)½ÂG(x)

∞

ak(xI+ 0<x<Áyéu1<i,j<n,È

rgij(xdx 糐擐¿©胐要㭜件´A3=0k狐答I.A为幕毐Ý拐䡜An0f(x1x)αjk时Cjk G(x)

ak(xI+A)k

akCk

j=0密µ 答铐时Ø要超闐䅜 区 (x) x∈(密µ 答铐时Ø要超闐䅜

(6©䉜䡜2mn¦Am/0Am+10lim(1−x)m−αG(x)=α(α−1)...(α−m+1)rm�3mα1䅜时

1G(x)dxuÑ (11© r 痐一方m<2m−α1䅜

G(x)dxÂñ r䱜之¦éu1<ij<n,È0

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x˙ds22√ (15©8 20183—二三四五六七30111112121212注意：本试卷共七大题，满分100分,考试时间为180密封线左 答题,密封线外不得 及相关标记极限limtanxsinxx0xln(1sin2设一平面过原点和点(6,32，且与平面4xy2z8垂直，则此平面方程设f(xy)具有一阶连续偏导数，满足df(x,y)yeydxx(1y)eydyf(00)0，则f(x,y)

x1x2x3x4(0,1)x x xxf(x1)f(x2)f(x3)f(x x xx xx 省 学 准考证 考场省 学 准考证 考场 座位 满足 u(t)0u(t)dt及u(0)1的可微函数u(t)设abcd是互不相同的正实数，xyzw是实数，满足axbcd，bycda111111111111czdab11111111111102x0 1f(x)dx x1f(t)dtf(x)

n 8 1x2 x1

2f(t)dtf(x)arctan

(1x省 学 准考证 考场 座位 =1x省 学 准考证 考场 座位 xxx,x)TR

f(xy在区D(xy)x2y2a2上具iH(x)xi

，n2

f(xy)x2y2a2nf f2n

4证明：对任一非零xR，H(x)0 max a2，其中a0.证明：f(x,y)dxdy a4省 学 准考证 考场 座位 省 学 准考证 考场 座位

(x,y)Dx y

ln省 学 准考证 考场 座位 设0a1,n1,2,，且 an省 学 准考证 考场 座位 nln 证明：当q1时级数an收敛，当q1时级数an q1时级数an的收敛性PAGEPAGE520183一、填空题(满分30分，每小题6分极限 tanxsinx x0xln(1sin2 设一平面过原点和点(6,3,2)，且与平面4xy2z8垂直，则此平面方程为2x2y3z0 f(x,y具有一阶连续偏导数，满足df(x,y)yeydxx(1y)eydy及f(0,0)0，则f(x,y) 2ete满 u(t) u(t)dt及u(0)1的可微函数u(t) 3abcd是互不相同的正实数，xyzw是实数，满足axbcd，bycda111111111111czdab111111111111二、(本题满11分)设函数f(x在区间(0,1)内连续，且存在两两互异的点x1x2x3x4(0,1)，使得f(x1)f(x2)<f(x3)f(x4)=x1 x3x证明：对任意(x5x6(0,1)，使得f(x5f(x6x F(tf((1t)x2tx4f((1t)x1tx3 4(1t)(x2x1)t(x4x3存在t0(0,1)，使得F(t0) 3分x5(1t0x1t0x3x6(1t0x2t0x4x5x6(0,1)x5x6，xF(t0f(x5f(x6x 0三、(本题满分11分)设函fx在区间0,1上连续且1f(x)dx0证明0在区间0,1上存在三个不同的点x1，x2x3，使得 1f(x)dx= x1f(t)dtf(x) 8 1x2 x1

2f(t)dtf(x)arctan

(1x

=1x2 2 44arctanxf1 ，则F00F11且函数Fx10f区间0,1上可导 根据介值定理，存在点

0,1，使Fx13235再分别在区间0,x3与x3，1上利用拉格朗日中值定理，存在x1(0,x3)使得F(x3F(0)F(x1)(x30) 80f(x)dx1x20f(xdxf(x1arctanx1x3；3 1且存在x2(x3,1)，使F(1)F(x3F(x2)(1x3 1f(x)dx x2f(x)dx x)8 1x2

f(x)arctan 2 3四、(本题满分12分)limn1(n1nn!n (n (n【解】注意到n1(n1)!

nn!， 3 n lim1nln

n lim k ，3分n n1(nn 1n1(nn(n1)n[(n

(n1)n(n

n

=

(n

k

3利用等价无穷小替换ex x(x0)， 0lnxdx1 0lnxdx1n 1(n11 ne n 1=limnn1n1(n

nlimnn•lim 3n n 五、(本题满分12分)xxx,,x)TRnH(x

nx2n1xx n2

i

i 2【证】(1)二次型H(x)xi 2

A

2

，3分1 2 11 11 A实对称，其任意k阶顺序主子式k0A正定，故结论成立…3 (2)对A作分块如下A ，其中(0,,0,1)TRn1 A1

1T

0阵P ，则PAP ，其 n1 1TA1 n1a1TA1 记xP(x,1)T，其中x(x,x,, )TRn1，因 H(x)xTAx(xT1)PT(PT)1 0P1Px0xT xa a 1 0 且 正定，所以H(x)xT xaa，当xP(x,1)TP(0,1)T时，H(x)a 0n1 因此，H(x)满足条件xn1的最小值为 3六、(本题满12分)设函f(xyD(xy)x2y2a2上具有f(x,

a2，以及max x2y2中a0.f(xy)dxdy4a4

f(x,y)Dx

P(x,y)dxQ(x,y)dy QP

D Pyf(xyQ0P0Qxf(xyf(x,y)dxdy yf(x,y)dxyfdxdy f(x,y)dxdy xf(x,y)dyxfdxdy

f f(x, 2 D

yy 4 ydxxdyadxdy ……2 对I2的被积函数利用柯西不等I2

1xf

dxdy

1

x2fx2f fxy

ax2y2dxdy1a442 ln七、(本题满分12分)设0a1,n1,2,，且 anq(有限或 nln 证明：当q1时级数an收敛，当q1时级数an q1时级数an(1q1，则pR,s.t.qp1NZ1 ln nN， p，即anp，而p1时p收敛,所以anln 3若q1pR，s.t.qp1NZ,s.t.nNln ln p，即an p，而p1时p发散,所以anln 3 当q1时，级数an n例如：an 满足条件，但级数an发散 3n 又如：annln2n满足条件，但级数an收 3第十届大学生数学竞赛（非数学类）预赛试题及答一、填空题（24分,4小题,6分 （1）设(0,1lim(n1)n 1 1 1 1 解由于 ,则(n1)nn11n1 1 n n n n 0n1)n1lim(n1)n0 xt+cos

y(x)由eyty+sint1确定，则此曲线在t0对应点处的切线方程y0(x解：当t0时，x1,y0，对xtcost两边关于t求导：dx1sint， 1 dtt对eyty+sint1两边关于t求导：eydyytdycost0 1，则 1 dtt dxtln(xln(x 1x21

(1x2

dxx1ln(x1xx1ln(x1x2)1ln(1x2)2ln(x1ln(x1x2(1x2

dx

dtsec

ln(tantsect)dsinln(tantsect)dsintsintln(tantsect)sintdln(tantsecsintln(tantsect) (sec2ttantsectantsecsintln(tantsect)sint1sintln(tantsect)ln|cost|C= ln(x1x2)1ln(1x2)C12

ln(x1x2)(1x2)3/2ln(x1x2)(1x2)3/211 ln(x1x2)

111111x2x1

ln(x1x2)xdx1111= ln(x1x2)1ln(1x2)121cos1cosxcos2x3cos 11cosxcos2x3cos3x

1cos cosx(1cos2x3cos3x) x0 cos2x3coscos2x3cos

cos2x(13cos3x)

x0 1lim 13(cos3x1)1 x0 1lim1cos2xlim1cos3x113 8f(t在t0f(1)0f(x2y2Lyx相交的分段光滑闭曲线.LP(xyy(2f(x2y2Q(xyxf(x2y2径无关，于是有Q(x,y)P，由此可知(x2y2f(x2y2f(x2y2 5分记tx2y2，则得微分方程tf(tf(t1，即(tf(t))1tf(ttCf(10，可得C1,f(t11f(x2y2t

x2814f(x在区间0,1]上连续，且1f(x3.11f dx4 0f 证明.f f1 f f1

dx0f(x)dx0f

4 又由f(x1f(x30，则f(x1f(x3/f(x0f f(x) f(x) 4 1f f(x)00由于1f(x)dx 0

dx

1 f(x) 0f0f 00f

dx 11f dx 14 0f 12计算三重积分(x2y2dV，其中（V）x2y2z2)24x2y2(z1)29，z0所围成的空心,, 0r3,0,0(x2y2dV2dd3r2sin2r2sindr835 4 (V):xrsincos,yrsinsin,z2r 0r2,0,0(x2y2dV2dd2r2sin2r2sindr825 8 (V29rxrcos,yrsin9r(V3):9r0r22,09r

z(x2y2)dV r2dz 2

1)dr(124235

2

r3(V r2

(x2y2dV(x2y2dV(x2y2dV(x2y2dV256 12(V (V (V (V fxyf 五(本题满分14分)设ffxyf 证明：作辅助函数(t)f(x1t(x2x1),y1t(y2y1)) 2显然(t)在[0,1]上可导.根据拉格朗日中值定理，存在c(0,1)，使(1)(0)(c)f(uv(xxf(uvyy 8 |(1)(0)||f(x,y)f(x,y)||f(u,v)(xx)f(u,v)(yy)

f(u,v)

f(u,v)21/

22 2

[(xx

(yy)

M|AB 14

14f(x0，有ln0f(x)dx0lnf(x)dx 1 k1kf(x在[0,1上连续，所以0f(x)dxlimnf(xkxk

, 1

k

n4f(xnn由不等式f(x1)f(x2 f(xk)，根据lnf(xnnk1 1 lnf(xk)ln f(xk 12 k k 根据lnx1 1 n 0 lim lnf(xk)limln f(xk)得lnf(x)dxlnf n 0 n k k 七(本题满分14分)已知{ak}，{bk}是正项数列，且bk1bk0,k1, ，b bk一常数.证明：若级数

收敛，则级数k ak k收敛 b证明：令

k ikab，abS i

k，S0,

k1SkSk1，k1,kbk k kb 4 NSkSk N

S N1bk1

Na Sk

kN

SN

k k k1 k1 k kk k1kkSkk1bkbk

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