數(shù)學(xué)競(jìng)賽非數(shù)組試題合輯

2017年數(shù)學(xué)競(jìng)賽預(yù)賽（非數(shù)學(xué)類）—1.已知可導(dǎo)函數(shù)滿足,則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 這樣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.記曲面z2x2y2和z 圍成空間區(qū)域?yàn)?x2V VI

zdxdydz

0

0 0

2cos2sind。V。21sin2/4142 f(xy在平面上有連續(xù)的二階偏導(dǎo)數(shù).對(duì)任何角度g(t)f(tcos,tsin)若對(duì)任何

dg(0) d2g且且dg

0.f(0,0f(xy的極小值cos解：由于 fx,fy

0對(duì)一切成立，故(f,f (0,0), y (0,0) y(0,0)是f(x,y)的駐點(diǎn) 4記H(x,y)

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

(cos,sin)Hf(0,0)sin010上式對(duì)任何單位向量(cossinHf(0,0)是一個(gè)正定陣,f(0,0極小值 14三(本題滿分14分 設(shè)曲線為x2y2z21，xz1，x0,y0,zA(1,0,0)B(0,0,1)Iydxzdy解:記1BA的直線段,xt,y0,z1t0t 2ydxzdyxdztd(1t 42 設(shè)和1圍成的平面區(qū)域，方向按右手法則.Stokes ydxzdyxdz 1

8右邊三個(gè)積分都是在各個(gè)坐標(biāo)面上的投影面積，而zx面上投影面積為零.I dydzdxdy1曲線xy(1/(x1/2)2 y(1/(1/

12 2 2

2 .同理2 2這樣就有I 2

14 四本題滿分15分)設(shè)函數(shù)f(x0且在實(shí)軸上連續(xù)，若對(duì)任意實(shí)數(shù)t，有e|tx|f(x)dx1，則ab(abbf(x)dxba2 證.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(本題滿分15分 設(shè){an}為一個(gè)數(shù)列，p為固定的正整數(shù)。 a，其為常數(shù)，證明lim

n n 證明：對(duì)于i0,1,…,p1， 。由題設(shè)limA(i) n limA(i)A(i) A(i) nA(i)A(i)A(i

a(n1) =

a(n1)

10n(n1)p (n1)p 對(duì)正整m，設(shè)m=npi，其中,p1，從而可以把正整數(shù)依照i分為p個(gè)子列 a ，故limam 15(n1)pn(n1)p n (ê?2018?3擐 á磐a ?¨÷? 銬—峐???提紿?:1.??大銬′胐答銬,篐?ê至十大銬￥裐à?銬,銬ò要??t?擐闐￥(?à?凐2.¤?答銬?闐廐糐?áò紙密μ線?蟐,廐糐ù§紙t－??凐 3.密μ線左蟐請(qǐng)?答銬,密μ線勐?提?翐?苐相烐鋐駐密μ 答銬時(shí)?要超闐? 4.?答銬空x?叐,可廐糐凐頁(yè)寐?密μ 答銬時(shí)?要超闐?提μ(20?z5?提μ 惐￠

011

§n≥1§ù￥I′與

ó某擐方叐?銬rank(H4) ln(1+tanx)?ln(1+sin(

x=π

惐Γ為空?- y=t?sinz=sin

t∈[0π].è

esinxcosxcosyΓ1aa···a1a··1aa···a1a··· 惐峐?矐f(x,...,x)=(x,... .a·· 1

·· 翐? 飐yò ¤糐峐ò 狐ò 蓐業(yè): ù￥n翐? 飐yò ¤糐峐ò 狐ò 蓐業(yè):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菐且與S相切擐¤?直線凐¤鋐?Σ.y2 糐Π§|SΣSΠ?ó時(shí)|?2Π擐方2翐? 翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐提μ ?櫐￡15?¤惐ABCtn某軐方叐§提μ AB?BA= AC= BC=y2μC′幕毐方叐y2μABCóqut?埐叐?CI=0,|n擐最に值密μ密μ 答銬時(shí)?要超闐?3提μ?櫐￡曐銬20?¤f(x)[01]t提μ?êf(0)f(1)0.|yZ Z|f0dx::,

|f(x)|dx

Z|f00(x)|04翐? 翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐 ê￡曐銬10?¤G為瓐§vμ?x,y提μ G,(xy)2=(yx)2.y2μ?x,y∈G,ｄxyx?1y?1擐提μ?超闐密μ密μ 答銬時(shí)?要超闐?5提μ臐?rùn)Α陼垆D10?¤惐E?Rn為可?臐,m(E)提μ惐f,fk∈L2(E),糐Et諐に裐裐?fk→f§|yμ

JE|fk(t)?f(t)|2dt=

lim

|fk(t)|2dt

|fE6翐? 翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐提μ ?櫐￡曐銬10? y哐??提μ\r(u,v)={acosu,bsinu, ?π::,u::, ?∞<v<￡1¤|St?￡2¤惐a=b.寐P=(a00)Q=(acosu0asinu0v0)(?π<u0<π,?∞<v0<+∞).廐?St?提PQü菐擐最á-線擐方§.密μ密μ 答銬時(shí)?要超闐?7提μ?櫐￡曐銬10?¤í峐|狐線紿方§|擐提μa§?y2a2ú??8翐? 翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐提μ ê櫐￡曐銬10?¤惐?êf(z)糐 哐|z|<1鰩提μ ?§糐ù蟐矐t?.?糐|z|1t|f(z)|1.y2f(z)為???ê.密μ密μ 答銬時(shí)?要超闐?9提μ十櫐￡曐銬10?¤惐X1X2···Xn′?áó?ù擐隨?賐tù?“ó擐?ù?êF(x)ú密Y?êf(x).yé?tX1X2···Xn黐大に?鰩-泐üXn1::,Xn2::,···::,提μ?|隨?賐t(Xn1,Xnn)擐é 密Y?êf1n(x,??鏐Xi(i=12···n)????[0,1]t擐t櫐?ù§|隨?賐tU=XnnXn1擐密Y?êfU(ê?2018?3擐 á磐a ?¨÷? 銬—峐???提紿?:1.?4大銬′胐答銬,5–10大銬￥裐à?銬,銬ò要??t?擐闐￥¤?答銬?闐廐糐?áò紙密μ線?蟐,廐糐ù§紙t－??凐密μ線左蟐請(qǐng)?答銬,密μ線勐?提?翐?苐相烐鋐駐 4.?答銬空x?叐,可廐糐凐頁(yè)寐?2ò密μ密μ 答銬時(shí)?要超闐?提μ((20?)(z5提μ(

011

§n≥1§ù￥I′

ó某擐方叐?銬rankH4)=3limln(1+tanx)?ln(1+sinx)=13

x=πsint /(3)惐Γ為空?- y=t?sinz=sin

,t∈[0,π].銬竐峐矐-線è esinxcosxcosyΓ）siny +coszdz=

···x ···(4)惐峐?矐f(x,...,x)=(x,...,x 擐Y叐A

翐? 飐yò ¤糐峐ò 翐? 飐yò ¤糐峐ò 狐ò 蓐業(yè):a·· 1a ··a1 ù￥n1aR銬f糐巐に賐換?擐鋐飐磐為((n1)a1)y12(a1)y22···(a 答?.(1)Hn′m=2n某é方叐§糐巐に方叐P|P?1HnP=D′1 \ \

\?1 埐方叐??囐§Hn+1

與

相q?惐HnO O ¤?諐埐值′λ1λ2···λm§Hn+1擐¤?諐埐值′λ11λ1?1λ21λ21···λm+1λm?1?|?ê?臐诐{韐′y2μHn擐¤??ó諐埐值為{nn2k|k=01···n}§?且z個(gè)諐埐值n?2k擐代ê-ê為n

?k!(n?4?§rankH4244方法二μ??D紿:|Lagrange￥??可±zlimln(1+tanx)?ln(1+sinx)=limtanx?sinx=lim1?cosx= -

只鋐|?A擐旐ü值蛐可wAa?1)I擐秩<1.Aa?1)I擐毐空?擐維ê?n?1,?囐可惐A擐n個(gè)諐征值為λ1=1?a,λ2=1?a,···,λn?1=1?a,紿trA=n,拐提λn=(n?1)a+1.燐鏐§f糐巐に賐換?擐鋐飐磐為((n?1)a1)y2?(a?1)y2?···?(a? 2提μ 峐櫐(15?)糐空?直埐坐鋐??§惐?y提μ S:a2+b2+c2= a,b,c>苐S勐ü－菐A(x0,y0,z0)§闐A菐且與S相切擐¤?直線凐¤鋐?Σ.y2 糐Π§|SΣSΠ?ó時(shí)|?2Π擐方狐μ?A糐S擐勐ü§ 0 0+0?1> éuM(x,yz)∈SΣ§?AM擐直線駐為lM§ù?ê§ ?=x+t(x? ?∞<t< 密μ 答銬時(shí)?要超闐?密μ 答銬時(shí)?要超闐?(x+t(x?

(y+t(y?

(z+t(z? =

z2+

（

+

+0+0+0?

0x+0y

（

z0+

a2+b2+c2 a2x+b2y+c2 =

(5??為菐M糐y￥St§x2+y2+z21.¤±taz

（x0x

y0y+z0

（

z0 1

0+0+0?

+

1? x+ y+

=翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐x0x+y0y+z0z?1= ?時(shí)? §方§(3)擐?項(xiàng)系êz 0+0+0?1> 3諐飐磐§(4)擐?êt??′2§§燐?2Π.t?íy2叐S∩Σ?Π§?囐y2叐S∩Σ?S∩Π.(12?反之§éuS∩Πt擐裐－菐M(x,yz)§?(3)櫐(4)üa蛐§?A櫐Mü菐燐?lM－?糐菐M與S相切.拐???§lM糐鋐?Σt.諐飐磐§M∈Σ.?M擐裐?紿§S∩Π?S∩Σ.燐?提y (15??A糐S擐勐ü§ 0+0+0>1> éuM(x1y1z1)∈SΣ§y￥S糐M菐擐切2方§可±廐x1x+y1y+z1z?1=0. ??MúAü′S糐菐M擐切線§¤±A菐糐t?2t.x1x+y1

+z1

–1=a2 b2

c2u′§菐M(x1y1z1)糐2￡§?￡6¤a§x0y0z0?Π:x0x+y0y+z0z?1= t§蛐?MS反之§éuM(x1y1z1)∈SΠ§

(10?銬S糐M菐擐切2

x0xa21x

+y0b2y

+z0c2z

–1=1x

1y

1z?1= ?A(x0y0z0)菐§?M,A擐?糐菐Múy￥?S相切§§糐鋐?Σt.拐MS∩燐?提y (15?4提μ ?櫐￡15?¤惐AB,Ctn某軐方叐§且提μ AB?BA= AC= BC=y2μC′幕毐方叐y2μABCóqut?埐叐?CI=0,|n擐最に值y2ù￥Ji為諐埐值λié諐擐Jordan?.éY叐B做與C相ó擐??§B=(Bij)k×k.?BC=CB可提JiBij=BijJj,ij=12k.ù懐é裐??項(xiàng)ap? p(Ji)Bij=Bijp(Jj).寐p為Ji擐最に?項(xiàng)a§銬提Bijp(Jj)=0.凐iI=j時(shí)§p(Jj)密μ 答銬時(shí)?要超闐??AB?BA=C提AiiBii?BiiAii=Ji,i=1,...,k.拐Tr(Ji)=Tr(AiiBii?BiiAii)=0,i=1,2,...,k.?囐λi=0,蛐C為幕毐方密μ 答銬時(shí)?要超闐?鹽V0={v∈Cn|Cv=0}.é裐?v∈V0,?uC(Av)=A(Cv)=0,??AV0?V0.ó?§BV0?V0.u′ 糐0I=v∈V0úλ∈C|提Av=λv毐駐V1={v|Av=λv}?V0,?AB?BA=C §é裐?u∈V1,A(Bu)=B(Au)+Cu=λBu.拐BV1?V1.?囐糐0I=v1∈V1苐μ∈C|提Bv1=μv1,ó時(shí)?Av1=λ1v1,Cv1=0.òv1琺?為Cn擐－|?{v1,v2,...,vn},鹽P=(v1vn)§( AP=

BP=

CP= 翐? 飐yò ¤糐峐ò ò 蓐業(yè):翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐A1B1C1÷vA1B1B1A1=C1A1C1=C1A1B1C1=C1B1.?ê?臐{蛐可提,A,B,Có時(shí)相qut?埐叐 (10?§\=E1§A,B,妨惐C 01.銬?AC=CA提A a1a2.?q??BC=CB提B \0 b1b2.u′AB?BA=0§ùAB?BA=C??！拐÷vCI=0擐 にn為 (15?5提μ?櫐￡曐銬20?¤f(x)[01]t提μ?êf(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)|.銬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?時(shí)(2)w痐¤á.y糐?惐ft(x)[0t?毐菐,?妨惐ft(x)>0,?囐f(x)?格緐飐.???ü?情磐裐? (10?情磐1f(0)?0?時(shí)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?時(shí)?f(1)<0根af擐緐飐紿,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 ?時(shí)(2)也¤á.紿μf(0)f(1?0可?f(x?0x∈[01].可只?情磐1.(20?)6翐? 翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐 ê￡曐銬10?¤G為瓐§vμ?x,y提μ G,(xy)2=(yx)2.y2μ?x,y∈G,,ｄxyx?1y?1擐提μ?超闐y2.?yx2y=yx2. tx2y=((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=/w? (6?密μ 答銬時(shí)?要超闐? 最后凐y(xyx?1y?1)2e.密μ 答銬時(shí)?要超闐?(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?rùn)Α陼垆D10?¤惐E?Rn為可?臐÷vm(提μ惐ffkL2(E),糐Et裐裐?fkf§|y

lim r

|fk(t)|2dt

|f(t)|2dt<E

|fk(t)?f(t)|2dt=Ey2μ?y裐寐F?Er

r|fk(t)|2dt

|f ?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 ??(?)¤á (4??u|f|2可è§裐給E0§糐δ0|裐寐可?F?E÷vm(F)δ時(shí) |f(t)|2dt

E (6??葉鏐a???§糐Eδ?E§|提m(E\Eδ)<δ§且糐Eδtfk－致磐??f毐?糐N1§裐寐k?N1§t∈Eδ

(t)?f(t)|<

3(1+

8 ?um(E\Eδ)δ§|(**)·們r(jià)

(8??(*)·們

r

|f(t)|2dt<Er 拐糐N2§|提裐寐k?N2

|fk(t)|2dt

|fr

鹽N=max(N1N2)§銬凐k?N§ |fk(t)?f密μ密μ 答銬時(shí)?要超闐? |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 蓐業(yè): (10?9提μ?櫐￡曐銬10? y哐??提μr(u,v)={acosu,bsinu, ?π<u< ?∞<v<￡1¤|St?￡2¤惐ab.寐Pa00)Qacosu0asinu0v0(?πu0π,?∞<v0+∞).廐?St?提PQüá§.狐μ￡1¤|St?§.狐{－μru={?asinu,bcosu, rv={0,0,¤±§S擐 {向tn=ru×|ru×

=(a2sin2u+b2cos2

12{bcosu,asinu, 1惐γ′St?§ù?ê§u= v= t∈ ??§?u裐?-?t擐直線￡?鏐糐擐{¤?′?磐線§St擐直竐線￡蛐u=~ê¤t為?磐線.u′只鋐|÷v?件ut(t)I=0擐?磐線.?時(shí)§可作γ擐?ê賐換|提§可±?wa?êv=f(u)￡u∈[?ππ]¤紿闐?.u′§γ擐向ta?ê方§z為r(u)={acosu,bsinu,f u∈[?π, 烐u?êu|§rt(u)={?asinu,bcosu,f rtt(u)={?acosu,?bsinu,f ¤±§γ擐 切向tT(u)=|rt(u)|?1rt(u)=|rt|?1{?asinu,bcosu,f ?鏐s′γ擐?§銬?

=|rt(u)|.u′γ?向t

）

rtt|rt(u)|?1+ =rtt|rt|?2?rt|rt|?3· ?§γSt?? (

ds,n,T=|rt|?2(rtt,n,¤±§γ′?(rttnT0.篐?(6(9)ú(10)§?aI?acosu?sinufbcos asin

≡I?asin bcos fta

ftt(a2sin2u+b2cos2u)=ft(a2?b2)sinucos ￡i¤ft≡0.銬v=f(u)≡c′~ê§-線γ′St擐巐棐線§蛐y 密μ 答銬時(shí)?要超闐?a2+b2= z密μ 答銬時(shí)?要超闐?￡ii¤ftI=0.銬(12)棐?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 蓐業(yè):狐¤±§St?￡i¤St擐直竐線￡v=~ê¤?￡ii¤St擐?棐y哐￡u=~ê¤?￡iii¤-線(8)§ù￥?êf?(13)燐?. ry?D2ts1<s< ?∞<v<ù￥s=s(u)′y r(u)={acosu,bsinu, ?π<u<擐??ê?為r??D?為2?é諐擐賐換に±-?t-線擐??賐§囐に￡蛐棐?¤r?磐線賐為?磐線§¤±? y哐??t擐?磐線é諐ut?帶磐?域￥擐?磐線蛐直線.根a2t§§§As+Bv+C=痐－方?§??微?ú ds=|rt(u)|du拐

(a2sin2u+b2cos2

）2(s a2sin2u+b2cos2

）2 y￥t¤|?§r ） a2sin2ub2cos2u2duBvC AB??鏐A0§銬?v=~ê§é諐y?t擐?棐y哐?鏐B0§銬r ）a2sin2u+b2cos2u2du=~ê蛐u=~ê§é諐y哐?t擐直竐線?鏐AI=0§BI=0§r ）v≡f(u)=

2 cI=(8?￡2¤?uba§?(13)燐??§(8)zr(u)={acosu,asinu,c1u+ c1,c2∈R,c1I= ?§éu?Q(acosu0asinu0￡i¤?鏐u00§銬¤|擐最á為?PQ擐直竐線?μ0v￡ii¤?鏐v00§銬¤|最á?PQ擐巐棐y寐??μu0<u0￡u00)?0<u<u0￡u0翐? 翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐￡iii¤?鏐u0I=0,v0I=0§銬凐?磐線?闐P菐時(shí)§可惐c2=0.篐鹽c1u0=v0§提c1=u?1v0.?囐¤|擐最á-線方§為 r(u)={acosu,asinu,(u?1v ù￥凐u0<0時(shí)§u∈[u0,0]?凐u0>0時(shí)§u∈[0, (10?密μ密μ 答銬時(shí)?要超闐?提μ?櫐￡曐銬10?¤í峐|狐線紿方§|擐提μa§?y2a2ú?? ?n某線紿方§|Ax=b,ù?A為n某￠é?巐?方叐§b為n維壐向t毐|狐鷐方§|棐?u|峐?êf(x):=1xTAx?bT2|ˉaxi+1=xi+αidi,i=0,1,2,...ù￥x0為給?D值§xi′x擐竐iú鰩代值§di′iú?裐方向§αi為úà寐αi|提f(xi+αidi)達(dá)巐最?§′提αi

dTi ,iidTiù￥ribAxi為í哐毐w痐ív緐íri+1=b?A(xi+αidi)=ri?Y運(yùn)Y{￥§·|鰩代方向琺?A-巐に§蛐iTAdj=0iI=j,?且ri與rj(iI=j)琺?巐に§??§可±凐軐“Y方向di??μd0= di+1=ri+1+ i=0,1,2,...?di與di+1A-dTAdi+1=dTA(ri+1+βi+1di)= i=0,1,2,... 拐?

dT= idTi dT =rTAd=rT(r? )= rT ±

αi

dTri=(ri+βidi?1)Tri=rT ù￥?巐叐dTri=0.ò±tüaβi+1擐闐達(dá)a可 rTrT βi+1 rTiiiu′|線紿方§|Ax=b擐“Y運(yùn)YrTd0=r0=b? αi idTixi+1=xi+ ri+1=ri?rTβi+1 di+1=ri+1+irTii=0,1,2,..., (5?·們y2§“YY?nú巐狐擐°燐值毐￠t§Axi+1 αAd?密μ 答銬時(shí)?要超闐?/密μ 答銬時(shí)?要超闐?ù

Axn=Ax0+α1Ad1+...+dT(Axn?b)=dT(Ax0?b)+αidT i=0,1,...,n? ·們?2ta￠t

=dTi(b?Ax0)

αk?1dTiAdk?1=dTi(b?翐? 飐翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐\/也ò′?§Axn?b與¤?di巐に§i=0,1,...,n?1.?囐Axn= (10?提μê￡曐銬10?¤惐?êf(z糐ü哐|z|<1鰩狐?§糐ù蟐矐t?.?糐|z|1t|f(z)|1.y2f(z)為?提μy2μ??êf(z)糐ü哐|z|<1鰩狐?,|z|=t|f(z)|=1,銬f(z)糐ü哐|z|<1鰩擐毐菐只??限?個(gè),惐為z1z2zn￡毐菐算k個(gè)ü毐菐 (2?做賐 z?f(z)= (1<k< 1?zkαk為￠ê.銬fk(zr|z|1に磐韐峐為|fk(z)|1,|z|=1時(shí)|f(z)|=fk(zk)=0(1<k< (5?作?F(z)= f?1(z)f(z)=f 1?zkzk=1 k=1z?ù￥α= αk.銬?êF(z)糐 哐|z|<1鰩狐?§v?毐菐§糐|z|=t|F(z)|= (7??狐??ê擐最大最に旐俐?§|z|<1tF(z)=C(~ê)§且|C|=uf(z)= z?zk=C1 z?k=11? k=11?為???ê (10?提μ 十櫐￡10?¤惐X1X2···Xn′?áó?提μ 擐隨?tù“ó?ù?êF(x)úY?êf(x).yé?tX1X2···Xn黐大に?鰩-泐üXn1<Xn2<···<Xnn.?|隨?賐t(Xn1,Xnn)擐é 密Y?êf1n(x,??鏐Xi(i=12···n)????[0,1]t擐t櫐?ù§|隨?賐tU=XnnXn1擐密Y?êfU狐μ(a)駐(Xn1Xnn)擐éü?ù?êFXn1Xnn(x?xy, FX

(x,y)=P(Xn1<x,Xnn< n1密μ 答銬時(shí)?要超闐?=P(Xnn<y)?P(Xn1>x,X密μ 答銬時(shí)?要超闐?=P(X1<y,X2<y,···,Xn<y)?P(x<X1<y,x<X2<y,···,x<Xn<=[F(y)]n?[F(y)?F?x?yFn1Xnn(xyP(Xn1<xXnn<yP(Xnn<yF拐(Xn1Xnn)擐éü密Y?êl

(x,y)

n(n?1)[F(y)?F(x)]n?2f(x)f x< ù

(4?(b?uXi?[0,1]tt?ùl 0<x< x<f(x)

0,ù

,F(x) 0<x<1,x?u′?￡a¤提(Xn1Xnn)擐éü密Y?êl

(x,y)

n(n?1)(y?x)n?2,0<x<y<1, ù§.

(6?翐? 翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐l蛐方蟐駐v=xnnxn1,

u=xnn+

2

,鷐賐換擐v=xnn?

xnn=可烐竐壐aJ=?(xn1,xnn)=1§銬?(Xn1

nn)擐éüY(UV

擐éüYfUV(u,v)=fl

u?v,u+

|Jn(n?1)vn?2,0<v<1,0<u?v<2,0<u+v< ù銬UXnnXn1擐密Y?rfU(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?ù?êFU(u).wU擐寐值范圍′[0,¤±§凐u0時(shí)§FU(uP(U<u0凐u?2時(shí)§FU(u凐0u1時(shí)§r

(10?FU(u)=P(U<u)凐1u2時(shí)

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Y?fU(u)

n 0<u<22n(2?u)n?1,1<u<22 ù

(10?(ê?2018?3擐 á磐a ?¨÷? 銬—峐??ê臐?提紿?:1.曐áò“臐大銬2.¤?答銬?áòμ,廐糐ù§紙t－?? 3.密μ?,密μ??4.?答銬空x?叐,可廐糐凐頁(yè)寐?2ò密μ 答銬時(shí)?要超闐?提μ(20?密μ 答銬時(shí)?要超闐?提μ (1)惐￠

011

§n≥1§ù￥I′與

ó某擐方叐?銬rank(H4) ln(1+tanx)?ln(1+sin(

x=π

惐Γ為空?- y=t?sinz=sin

t∈[0π.è

esinxcosxcosyΓ1aa···a1a···1aa···a1a··· 惐峐?矐f(x,...,x)=(x,... .a·· 1

·· 翐? 飐yò ¤糐峐ò 狐ò 蓐業(yè): ù￥n翐? 飐yò ¤糐峐ò 狐ò 蓐業(yè):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菐且與S相切擐¤?直線凐¤鋐?Σ.y2 糐Π§|SΣSΠ?ó時(shí)|?2Π擐方2翐? 翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐提μ ?櫐￡15?¤惐ABCtn某軐方叐§提μ AB?BA= AC= BC=y2μC′幕毐方叐y2μABCóqut?埐叐?CI=0,|n擐最に值密μ密μ 答銬時(shí)?要超闐?3提μ?櫐￡曐銬20?¤f(x)[01]t提μ?êf(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￠~êY叐A為幕毐Y叐,I為n 叐.惐Y陣值?êG(x)??G(x)

1

ak(xI+ 0::,x<翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐áyéu1ijn,è gij(xdx翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐0密μ密μ 答銬時(shí)?要超闐?5提μ臐?rùn)Α陼垆D15?¤?矐??êg(t):R→提μv1g(t2.x(t)(tR)′§¨(t)=g(t)x擐ü|y 糐~êC2C106(ê?2018?3擐 á磐a ?¨÷? 銬—峐??ê臐?提紿?:1.曐áò“6大銬¤?答銬?闐廐糐?áò紙密μ線?蟐,廐糐ù§紙t一??凐密μ線左蟐請(qǐng)?答銬,密μ線勐?提?翐?苐相烐鋐駐 4.?答銬空x?叐,可廐糐凐頁(yè)寐?2ò密μ密μ 答銬時(shí)?要超闐?提μ(20?z5?提μ 惐￠方叐

011

§n≥1§ù￥I′

ó某擐方叐?銬rankH4)=3limln(1+tanx)?ln(1+sinx)=13

x=πsint 惐Γ為空?- ?t=0巐t=π擐一?.銬竐峐矐-線èy=t?sin z=sin2tesinxcosxcosydx?sinydy+coszdz=Γ2

·· ·· 惐峐?矐f(x1xnx1x

擐YA.

a·· 翐? 飐yò ¤糐峐ò 狐ò 翐? 飐yò ¤糐峐ò 狐ò 蓐業(yè): ù￥n1a∈R.銬f糐巐に賐換?擐鋐飐磐為n1)a1)y2(a1)y2···(a 1 答?.(1)Hn′m=2n某é?方叐§糐巐に方叐P|提P?1HnP=D′ \ \

埐方叐??囐§Hn+1

與

相q?惐HnO O ¤?諐埐值′λ1λ2···λm§Hn+1擐¤?諐埐值′λ11λ1?1λ21λ21···λm+1λm?1?|?ê?臐诐{韐′y2μHn擐¤??ó諐埐值為{nn2k|k=01···n}§?且z個(gè)諐埐值n?2k擐代ê-ê為n

?k!(n?4?§rankH4244方法二μ??YD紿:|Lagrange￥??可±{zlimln(1+tanx)?ln(1+sinx)=limtanx?sinx=lim1?cosx= -

只鋐|?A擐旐ü值蛐可wAa?1)I擐秩<1.Aa?1)I擐毐空?擐維ê?n?1,?囐可惐A擐n個(gè)諐征值為λ1=1?a,λ2=1?a,···,λn?1=1?a,紿trA=n,拐提λn=(n?1)a+1.燐鏐§f糐巐に賐換?擐鋐飐磐為((n?1)a 1)y2?(a?1)y2?···?(a? 2提μ 峐櫐(15?)糐空?直埐坐鋐??§惐?y提μ S:a2+b2+c2= a,b,c>苐S勐ü一菐A(x0,y0,z0)§闐A菐且與S相切擐¤?直線凐¤鋐?Σ.y2 糐Π§|SΣSΠ?ó時(shí)|?2Π擐方狐μ?A糐S擐勐ü§ 0 0 0?1> éuM(x,yz)∈SΣ§?AM擐直線駐為lM§ù?ê§ ?=x+t(x? ?∞<t< 密μ 答銬時(shí)?要超闐?密μ 答銬時(shí)?要超闐?(x+t(x?

(z+t(z?z+ + =1.

z2+

(

+

+0+

+0?

0x+0y

z0+

a2+b2+c2 a2x+b2y+c2 =

(6??為菐M糐y￥St§x2+y2+z21.¤±taz

(x0x

y0y+z0

(

z0 1

0+0+0?

+

x+ y+

=翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐x0x+y0y+z0z?1= ?時(shí)? §方§(3)擐?項(xiàng)系êz 0+0+0?1> 3諐飐磐§(4)擐?êt??′2§§燐?2Π.t?íy2叐S∩Σ?Π§?囐y2叐S∩Σ?S∩Π.(12?反之§éuS∩Πt擐裐一菐M(x,yz)§?(3)櫐(4)üa蛐§?A櫐Mü菐燐?lM一?糐菐M與S相切.拐???§lM糐鋐?Σt.諐飐磐§M∈Σ.?M擐裐?紿§S∩Π?S∩Σ.燐?提y (15??A糐S擐勐ü§ 0+0+0>1> éuM(x1y1z1)∈SΣ§y￥S糐M菐擐切2方§可±廐x1x+y1y+z1z?1=0. ??MúAü′S糐菐M擐切線§¤±A菐糐t?2t.u′§菐M(x1y1z1)糐2

x1xa2

+y1b2

+z1c2

–1=Π:x0x+y0y+z0z?1= t§蛐?M∈S∩ (12?反之§éuM(x1y1z1)∈SΠ§銬S糐M菐擐切2

x0xa2

+y0b2

+z0c2

–1=x1x+y1y+z1z?1= ?A(x0y0z0)菐§?M,A擐?糐菐Múy￥?S相切§§糐鋐?Σt.拐MS∩燐?提y (15?4提μ ?櫐￡15?¤惐ABCtn某軐方叐§提μ AB?BA= AC= BC=y2μC′幕毐方叐y2μABCóqut?埐叐?C/=0,|n擐最に值y2惐C擐?óλ1λk?C?Jordan鋐飐矐μC=diag(J1ù￥Ji為諐埐值λié諐擐Jordan?.éYB做與C相ó??§B ?BC=CB可提JiBij=BijJj,ij=12k.ù懐é?項(xiàng)ap密μ 答銬時(shí)?要超闐? p(Ji)Bij=Bijp(Jj).寐p為Ji擐最に?項(xiàng)a§銬提Bijp(Jj)=0.凐i/=j時(shí)§p(Jj)可?§?囐Bij=0.??§B=diag(B11,...,Bkk).ó?§A=diag(A11,...,Akk).?AB?BA=C提AiiBii?BiiAii=Ji,i=1,...,k.拐Tr(Ji)=Tr(AiiBii?BiiAii)=0,i=1,2,...,k.?囐λi=0,蛐C為幕毐方叐密μ 答銬時(shí)?要超闐?鹽V0={v∈Cn|Cv=0}§wV0非空év∈V0?uC(Av)=A(Cv0,?AV0?V0.ó§BV0?V0.u′糐0/v∈V0úλ∈C|Avλv毐駐V1={v|Av=λvv∈V0}?V0,?AB?BA=C§é裐?u∈V1,A(BuB(AuCu=λBu.BV1?V1.?囐糐0/v1∈V1苐μ∈C|Bv1μv1,óAv1λ1v1,Cv1=0.òv1Cn擐一|?{v1v2vn},鹽P=(v1vn)§ AP=

BP=

CP= 0ù￥A1B1C1為n?1某軐方叐且÷vA1B1?B1A1C1A1C1C1A1B1C1C1B1.?ê?臐诐{蛐可提,A,B,Có時(shí)相qut?埐叐 (10?§E1,妨惐C 01.銬?AC=CA提A 0

a1

.?qBCCB提B にn為

.u′AB?BA=0§ù與AB?BA=C??！拐÷vC/=0擐翐? 飐yò ¤糐峐ò翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐5提μ?櫐￡曐銬20?¤f(x)[01]t提μ?êf(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)|.銬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?時(shí)(2)w痐¤á.y?惐ft(x)[0t?毐菐?妨惐ft(x0?f(x?格緐飐??ü裐?情磐1f(0)?0?時(shí)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?時(shí)?f(1)<0根af擐緐飐紿,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?時(shí)(2)也¤á紿μ?f(0)f(1)?0,可?妨惐f(x)?0,x∈[0,1].可只?情磐 (20?6提μ ê￡15?¤惐α(12)(1?x)α提μ 裐ê akxk,n×n￠~êY叐A為幕毐Y叐,I為叐.惐Y陣值?êG(x)??G(x)

∞

ak(xI+ 0<x<áyéu1<i,j<n,è

rgij(xdx 糐擐??胐要?件′A3=0k狐答I.A為幕毐Y拐?An0f(x1x)αjk時(shí)Cjk G(x)

ak(xI+A)k

akCk

j=0密μ 答銬時(shí)?要超闐? 區(qū) (x) x∈(密μ 答銬時(shí)?要超闐?

(6???2mn|Am/0Am+10lim(1?x)m?αG(x)=α(α?1)...(α?m+1)rm?3mα1?時(shí)

1G(x)dxu? (11? r 痐一方m<2m?α1?

G(x)dx?? r?之|éu1<ij<n,è0

gij(xdx 糐擐??胐要?件′A3=翐? 飐yò翐? 飐yò ¤糐峐ò ò 蓐業(yè):狐{II.Jordan?7提μ臐?rùn)Α陼垆D15?¤?矐??êg(t):R→提μv1g(t2.x(t)tR′§¨(t)=g(t)x擐ü韐巐狐|y 糐~êC2C10Iy2一.鹽y=x(t),銬y?ò.?y(t)?0(?銬?t→?t?y燐 éC1=1,C2 3¤á√?糐t0,y(t0) 3IxII(t)x(t)?xIIy(t) =g(t)?y2<2?y2<?1,t<I銬y(t)|t<t< (5?I??

<?1,t<?Jt0yIds<t?t,t<?tt>

√

?LL0為一個(gè)~ê.ùy(t糐Rt? 1 y(s)?2t2 ?? (10??糐t,y(t)<1銬yI=g(t)?y2>δ>0,t<t.ù? <t,y(t)<0.?? (15?

y2.?妨惐x(t緐飐￡??§x¨=g(?t)x¤.紿¨=g(t)x?x(?∞)=x˙(?∞)= (5?¨?1x˙(t)2

¨ds g(s)x˙ds

2 2

x˙ds22√ (15?8 20183—二三四五六七30111112121212注意：本試卷共七大題，滿分100分,考試時(shí)間為180密封線左 答題,密封線外不得 及相關(guān)標(biāo)記極限limtanxsinxx0xln(1sin2設(shè)一平面過(guò)原點(diǎn)和點(diǎn)(6,32，且與平面4xy2z8垂直，則此平面方程設(shè)f(xy)具有一階連續(xù)偏導(dǎo)數(shù)，滿足df(x,y)yeydxx(1y)eydyf(00)0，則f(x,y)

x1x2x3x4(0,1)x x xxf(x1)f(x2)f(x3)f(x x xx xx 省 學(xué) 準(zhǔn)考證 考場(chǎng)省 學(xué) 準(zhǔn)考證 考場(chǎng) 座位 滿足 u(t)0u(t)dt及u(0)1的可微函數(shù)u(t)設(shè)abcd是互不相同的正實(shí)數(shù)，xyzw是實(shí)數(shù)，滿足axbcd，bycda111111111111czdab11111111111102x0 1f(x)dx x1f(t)dtf(x)

n 8 1x2 x1

2f(t)dtf(x)arctan

(1x省 學(xué) 準(zhǔn)考證 考場(chǎng) 座位 =1x省 學(xué) 準(zhǔn)考證 考場(chǎng) 座位 xxx,x)TR

f(xy在區(qū)D(xy)x2y2a2上具iH(x)xi

，n2

f(xy)x2y2a2nf f2n

4證明：對(duì)任一非零xR，H(x)0 max a2，其中a0.證明：f(x,y)dxdy a4省 學(xué) 準(zhǔn)考證 考場(chǎng) 座位 省 學(xué) 準(zhǔn)考證 考場(chǎng) 座位

(x,y)Dx y

ln省 學(xué) 準(zhǔn)考證 考場(chǎng) 座位 設(shè)0a1,n1,2,，且 an省 學(xué) 準(zhǔn)考證 考場(chǎng) 座位 nln 證明：當(dāng)q1時(shí)級(jí)數(shù)an收斂，當(dāng)q1時(shí)級(jí)數(shù)an q1時(shí)級(jí)數(shù)an的收斂性PAGEPAGE520183一、填空題(滿分30分，每小題6分極限 tanxsinx x0xln(1sin2 設(shè)一平面過(guò)原點(diǎn)和點(diǎn)(6,3,2)，且與平面4xy2z8垂直，則此平面方程為2x2y3z0 f(x,y具有一階連續(xù)偏導(dǎo)數(shù)，滿足df(x,y)yeydxx(1y)eydy及f(0,0)0，則f(x,y) 2ete滿 u(t) u(t)dt及u(0)1的可微函數(shù)u(t) 3abcd是互不相同的正實(shí)數(shù)，xyzw是實(shí)數(shù)，滿足axbcd，bycda111111111111czdab111111111111二、(本題滿11分)設(shè)函數(shù)f(x在區(qū)間(0,1)內(nèi)連續(xù)，且存在兩兩互異的點(diǎn)x1x2x3x4(0,1)，使得f(x1)f(x2)<f(x3)f(x4)=x1 x3x證明：對(duì)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、(本題滿分11分)設(shè)函fx在區(qū)間0,1上連續(xù)且1f(x)dx0證明0在區(qū)間0,1上存在三個(gè)不同的點(diǎn)x1，x2x3，使得 1f(x)dx= x1f(t)dtf(x) 8 1x2 x1

2f(t)dtf(x)arctan

(1x

=1x2 2 44arctanxf1 ，則F00F11且函數(shù)Fx10f區(qū)間0,1上可導(dǎo) 根據(jù)介值定理，存在點(diǎn)

0,1，使Fx13235再分別在區(qū)間0,x3與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、(本題滿分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利用等價(jià)無(wú)窮小替換ex x(x0)， 0lnxdx1 0lnxdx1n 1(n11 ne n 1=limnn1n1(n

nlimnn?lim 3n n 五、(本題滿分12分)xxx,,x)TRnH(x

nx2n1xx n2

i

i 2【證】(1)二次型H(x)xi 2

A

2

，3分1 2 11 11 A實(shí)對(duì)稱，其任意k階順序主子式k0A正定，故結(jié)論成立…3 (2)對(duì)A作分塊如下A ，其中(0,,0,1)TRn1 A1

1T

0陣P ，則PAP ，其 n1 1TA1 n1a1TA1 記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，當(dāng)xP(x,1)TP(0,1)T時(shí)，H(x)a 0n1 因此，H(x)滿足條件xn1的最小值為 3六、(本題滿12分)設(shè)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 對(duì)I2的被積函數(shù)利用柯西不等I2

1xf

dxdy

1

x2fx2f fxy

ax2y2dxdy1a442 ln七、(本題滿分12分)設(shè)0a1,n1,2,，且 anq(有限或 nln 證明：當(dāng)q1時(shí)級(jí)數(shù)an收斂，當(dāng)q1時(shí)級(jí)數(shù)an q1時(shí)級(jí)數(shù)an(1q1，則pR,s.t.qp1NZ1 ln nN， p，即anp，而p1時(shí)p收斂,所以anln 3若q1pR，s.t.qp1NZ,s.t.nNln ln p，即an p，而p1時(shí)p發(fā)散,所以anln 3 當(dāng)q1時(shí)，級(jí)數(shù)an n例如：an 滿足條件，但級(jí)數(shù)an發(fā)散 3n 又如：annln2n滿足條件，但級(jí)數(shù)an收 3第十屆大學(xué)生數(shù)學(xué)競(jìng)賽（非數(shù)學(xué)類）預(yù)賽試題及答一、填空題（24分,4小題,6分 （1）設(shè)(0,1lim(n1)n 1 1 1 1 解由于 ,則(n1)nn11n1 1 n n n n 0n1)n1lim(n1)n0 xt+cos

y(x)由eyty+sint1確定，則此曲線在t0對(duì)應(yīng)點(diǎn)處的切線方程y0(x解：當(dāng)t0時(shí)，x1,y0，對(duì)xtcost兩邊關(guān)于t求導(dǎo)：dx1sint， 1 dtt對(duì)eyty+sint1兩邊關(guān)于t求導(dǎo)：eydyytdycost0 1，則 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相交的分段光滑閉曲線.LP(xyy(2f(x2y2Q(xyxf(x2y2徑無(wú)關(guān)，于是有Q(x,y)P，由此可知(x2y2f(x2y2f(x2y2 5分記tx2y2，則得微分方程tf(tf(t1，即(tf(t))1tf(ttCf(10，可得C1,f(t11f(x2y2t

x2814f(x在區(qū)間0,1]上連續(xù)，且1f(x3.11f dx4 0f 證明.f f1 f f1

dx0f(x)dx0f

4 又由f(x1f(x30，則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計(jì)算三重積分(x2y2dV，其中（V）x2y2z2)24x2y2(z1)29，z0所圍成的空心,, 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(本題滿分14分)設(shè)ffxyf 證明：作輔助函數(shù)(t)f(x1t(x2x1),y1t(y2y1)) 2顯然(t)在[0,1]上可導(dǎo).根據(jù)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l(wèi)n0f(x)dx0lnf(x)dx 1 k1kf(x在[0,1上連續(xù)，所以0f(x)dxlimnf(xkxk

, 1

k

n4f(xnn由不等式f(x1)f(x2 f(xk)，根據(jù)lnf(xnnk1 1 lnf(xk)ln f(xk 12 k k 根據(jù)lnx1 1 n 0 lim lnf(xk)limln f(xk)得lnf(x)dxlnf n 0 n k k 七(本題滿分14分)已知{ak}，{bk}是正項(xiàng)數(shù)列，且bk1bk0,k1, ，b bk一常數(shù).證明：若級(jí)數(shù)

收斂，則級(jí)數(shù)k ak k收斂 b證明：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

10由不等式k(a a)(b b)a1b1a2b2 abSkk 1 1 ak bk) Sk，故結(jié)論成 14 k bk k1bk(ê?,2018?10擐 ?¨÷? 100銬飛峐??ê臐?提:1.¤?答銬?闐廐糐?áò紙密μ線?蟐,廐糐ù§紙t飛??凐密μ?,密μ??苐相烐鋐駐?答銬空x?叐,可廐糐凐頁(yè)寐?,?鋐2銬ò密μ線答銬時(shí)?\/飛櫐(15?)糐空?直埐坐鋐?￥,惐êS擐方§為x2y2=2z.惐σ為2zαxβyγ,ù￥αβγ為給?~ê.|ê髐StP擐坐鋐§|提闐P且á糐ê髐St擐直線t2竐u2密μ線答銬時(shí)?狐μ¤|P菐坐鋐為P=(abc),÷va2b22c.銬闐P￡=￡(t)=(a,b,c)+t(u,v,w),u2+v2+w2/=0,t∈直線￡(t)áêSt§u′

(u2?v2)t2+2(au?bv?w)t=0,t∈R.au?bv=w,u2?v2=0.v=εu,w=(a?εb)u,ε=

(5?翐? 飐yò: ¤糐峐ò 狐ò 翐? 飐yò: ¤糐峐ò 狐ò 蓐業(yè):￡1=￡1(t)=(a,b,c)+tu(1,1,a?￡2=￡2(t)=(a,b,c)+tu(1,?1,a+ùü?直線擐方向向t(11a?b)ú(1?1a+b)t2竐u2?σ,囐2?σ擐{向 α+β=a?b,α?β=a+1頁(yè)￡6頁(yè)u拐¤|P

a=α,b=?β,c

1(α2?2

(10?P=(α,?β,1(α2?2

(15?峐櫐(15?)Aaij)n×n為n某￠方叐a11=a22=···=ann=> ézi(i=1n),

|aij|

|aji|<|f(x1xn)=(x1xn 擐紿范磐狐:fx

)A+A

.鹽B=bj)=A+AT,銬B為￠é?叐(2?且

b11=b22 =bnn= |b|區(qū)|aij+aji|<燐鏐§bii

j/=i

(5??λ為B擐諐征值§α 為烐uλ擐非毐諐埐向t§|xi|=max1勺j勺n|xj|>?uBα=

λ=

j?a

區(qū)

|>

(10?2頁(yè)￡6頁(yè)拐B為巐?Y叐§f擐紿范磐為y2··· (15?為

?櫐￡20?¤ｄ素擐為寐êYY.惐n某方叐A,B擐為寐Yy2üμi)A可?且A?1軐為寐Y?ii)A擐竐壐ayé? A,A?2B,A?4B,...,A?2nB,A?2(n+1)B,...,A?2(n+擐可?§且§們擐?Y叐擐軐為寐Y叐.y2μAB可?y2:1)i)?ii).?AA?1= |A|·|A?1|=1.紿?巐|A|,|A?1|擐為寐ê拐A擐竐壐ayéii)?i).?AA?= A?1=A?/|A|á蛐i)¤á ￡10? ??ap(x)=|AxB|2銬??p(0)p(2)p(4)p(4n)擐值擐為1.燐鏐?項(xiàng)aq(x)=p(x?1?超2n個(gè)擐毐菐§?囐提?q(x)≡0,蛐p(x)≡1.密μ答銬時(shí)?諐飐磐§p(?1)=|A+B|2=1.拐A+B可?.y矐 ￡密μ答銬時(shí)??櫐￡15?¤f(x)[01]t??可微,x=0裐?裐?某峐êf(n)(00(n?0),且糐~êC0|y2:(1)

f

|xf戶(x)|勺C|f ?x∈[0,0(n?0);(2)[01]t¤áf(x)≡y2:(1)??惐,é裐?m?0,f(x)糐毐菐韐賐?m1某峐ê,?囐f糐x=0??.??,limf(x)=f(0)=