第12-19届北京市大学生数学竞赛全部试题解答

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200010149:00~11:30..209:30120tan(1cos)lim2ln(12)(1)xxaxbxxce−→+−=−+−=a.220zxy∂=∂∂0x=sinzy=0y=sinzx==z.301!nnn∞=+∑=.40(1)nnnax∞=+∑(4,2)−0(3)nnnnax∞=−∑.52111()0xttdtedx∫∫=.611,,2,xxxyyeyeyeπ====+.7{}nx11sin(2)sin11nnxnnn+++11lim1nknkxn→∞=+∑=.8()fx0x=()0cos1lim11fxxxe→−=−(0)f′=.9()fx10()()sin,(0)0ftxdtfxxxf=+∫=0x≠=()fx′.10C2222xyza++=2222xyzb++==(0,0ab))C3(rxdxydyzdz++∫22rxyz2=++.{()fx(0,)+∞()0fx}na.11()()nnnkafkfx==−∑∫dxS222122xyz++=(0z)PS∈ΠSP(,,)xyzρΠ3(,,)SIzxyzdSρ=∫∫.()ufr=0r+∞(1)0,(1)1ff′==222()ufxyz=++2222220uuuxyz∂∂∂++=∂∂∂.()fr(,,)ufxyz=fyxzfffxyz′′′==ru22rxyz=++2).()fx(,−∞+∞0x=0()lim0xfxax→=11(1)()nnfn∞=−∑11()nfn∞=∑..892103.[,()fx]ab()()0fafb==(,)xab∈()0fx≠()4()bafxdxfxb′′≥−∫a0.1()nnfxaxaxa=+++2n≥0(11)kakn=≤≤−ik≠0ia≠()fxn110kkaa−+.200110139:00~11:30..209:3011sin0()00kxxfxxx⎧≠⎪=⎨⎪=⎩0x=k.2y2yx=(01)yaa=ya=2yx==1x=ya.32102()112xxfx1xx⎧≤≤⎪⎪=⎨⎪−≤⎪⎩01()cos2nnasxanxπ∞==+∑x−∞+∞102()cos,0,1,2,nafxnxdxnπ==∫9(2s−)=.4()yfx=(4)(0)dyyydxββ=−()yfx=0(,3)xβ=.511xyx+=−(10)0xy==.64441(1)1dxxx++∫=+.C7()fx(0)0,(0)1ff′==20020()lim(())xxxftdtftdt→∫∫=.8Ω2221xyz++≤222222(xyzdvabcΩ++∫∫∫)=.9(,)fxy,,xyt2(,)(,)ftxtytfxy=0(1,2,2)P−(,)zfxy=(1,2)4xf′−=0P.10(11m≥na)nmx++nx01nna∞=∑=.()fx[0,1](1)(0)(1)(0)0ffff′′====(0,1)ξ∈()()ffξξ′′=.40tan,1nnaxdxnπ=≥∫1{}na221,1nnaann−+=2−3112(1)2(1)nann+−..ABCPP30(1)(1)!nnnnxn∞=−+∑.RP01PPρ=0P0P0rR.v00,vvv.[0.()fx,1]11()()001fxfyedxedy−≥∫∫1.21nnu∞=∑1nnv∞=∑limnnnulv→∞=0l+∞.200110139:00~11:302011sin0()00kxxfxxx⎧≠⎪=⎨⎪=⎩0x=k.21x≥222arctanarcsin1xxx++=.3131knnnkxnk==+∑limnnx→∞=.4D221xy+≤2222()Dxydabσ+∫∫=.5()fx(ufxyz=)3(),uFttxyzxyz∂==∂∂∂=()Ft.60α0xxeedxα+∞−−−∫=.7()yfx=(4)(0)dyyydxββ=−()yfx=0(,3)xβ=.84441(1)1dxxx++∫=+.C9()fx(0)0,(0)1ff′==20020()lim(())xxxftdtftdt→∫∫=.10(11m≥na)nmx++nx01nna∞=∑=.()fx[0,1](1)(0)(1)(0)0ffff′′====(0,1)ξ∈()()ffξξ′′=.xyxy34xy+A40tan,1nnaxdxnπ=≥∫1{}na221,1nnaann−+=2−3112(1)2(1)nann+−..ABCPP30(1)(1)!nnnnxn∞=−+∑.[0.()fx,1]11()()001fxfyedxedy−≥∫∫200210139:00~11:30..2201lim2nnnnab→∞⎛⎞+⎜⎜⎝⎠⎟⎟=0,0ab1,1ab≠≠.2()fx1x=(1)1f′=0(1)(12sin)2(13tan)limxfxfxfxx→+++−−=.3222(1)(1)xaxdxxx++++∫=a.421limknnknknenne→∞=+∑=.522uxyz=++22(1,1,1)M22zxy=+MnMun∂∂=.62111yxdxedy−∫∫=.7()fx(0)0f=222()[()]tFtzfxydxdydzΩ=++∫∫∫222:,0t1xytzΩ+≤≤≤20()limtFtt+→=.80α→22()CydxxdyI2xyxyα−=++∫C2221xyα+=nα=n.9∑2224xyz++=22()xydS∑+∫∫=.1012!(1)!(2)!kkkkk∞=+++++∑=.[0()fx,]a(0,)a()([0,])fxMxa′′≤∈(0)()ffaMa′′+≤.222200sincos11xxdxdxxxππ≤++∫∫.()fu11()()xxIfdydzfdzdxzdxdyyyxy∑=++∫∫.∑2226,8yxzyxz=++=−−22(,)2(0,0)(,)costtfxydxxydyt+=∫.(,)fxy(,)fxy.223(6)(8)0yxydxxxydy+++=3()yfx()fx.uhO10g.g10xx=20x.(,)zfxy=0fy∂≠∂()C(,)fxyC=()2220yxxxyxyyyxfffffff′′′′′′′′′′−+=.10()!knnzken∞−==∑1.(0),(1)zz(2)z2.k()zk200210139:00~11:302201210lim(cos)xxx→.2431,2()(1)(2)21,2xaxxfxxxx⎧+−≠−⎪=−+⎨⎪=−⎩1x==a.3()fx1x=(1)1f′=0(1)(12sin)2(13tan)limxfxfxfxx→+++−−=.4222(1)(1)xaxdxxx++++∫=a.520022200220020sinsincosxdxxxπ+∫=.621limknnknknenne→∞=+∑=.72111yxdxedy−∫∫=.8a=0a2311aadxx+∫.92220limxxtxetedtx−→+∞∫=.10222()(0)xRyrrR−+=y=V.2xy+=2.2()2Cqqqαβ=++1(4)qpγ=−Cqp,,αβγ.222200sincos11xxdxdxxxππ≤++∫∫.[0()fx,]a(0,)a()([0,])fxMxa′′≤∈(0)()ffaMa′′+≤.01121,1,(2)nnnFFFFFn−−===+≥1113()22nnnF−−≤≤201nnF∞=∑21lnnnF∞=∑3{(,)1,11},()Dxyxyxfx=≤≤−≤≤[,](1)aaa−≥.2[(1)()(1)()]Dyxfxxfxdxdy++−−∫∫1.–42.sinsinxy+3.4.065.2e1(e1)6−6.0yy′′′−=7.18.09.10.2sincosxxx−−551()5ba−.()fx(0,)+∞111(1)()d(1)()dnnnnnnaafnfxxfnfx+++−=+−=+−∫∫x{}na.n1111()()d()()()d0nknnnkkkafkfxxfnfkfxx+===−=+−∑∑∫∫{}na.{}na.{,,}XYZπ122xXyYzZ++=2221(,,)22xyzxyzρ=⎛⎞⎛⎞++⎜⎟⎜⎟⎝⎠⎝⎠S22122xyz=−−222(,,)4xyzxyρ=−−222122zxxxy∂=∂−−222122zyyxy∂=∂−−2222224d1dddd2122xyzzSxyxyxy−−⎛⎞∂∂⎛⎞=++=⎜⎟⎜⎟∂∂⎝⎠⎝⎠−−xy222222400(1)ddd(1)2()22228DxyrIxyrdrπθπ=−−=−=−∫∫∫∫rr.()uxfrxr∂′=∂22222()()uxrfrfrxrr∂−′′′=+∂23x.22223y22()()uyrfrfryrr∂−′′′=+∂2222223()()uzrzfrfrzrr∂−′′′=+∂.2()()0frfrr′′′+=.()()frpr′=20rpp′+=.12Cpr=12()Cfrr′=12()CfrCr=−1()1frr=−.(,,)(sincos,sinsin,cos)ufxyzfrrrϕθϕθϕ==.yxzffftxyz′′′===,,xyzftxftyftz′′′===(sin)sincossin(sin)sincossin()0xyufrfrtxrtyrtxyxyθϕθϕθϕθϕθ∂′′=⋅−+=−+=−+=∂222coscossincossin(cossincossinsincossincos)xyzufrfrfrtrθϕθϕϕθϕϕθϕϕϕϕϕ∂′′′=⋅+⋅−=+−∂=.2(sincossincos)0trϕϕϕϕ−=ur.0()limxfxax→=(0)0,(0)ffa′==.()fx0x=(0)0fa′=0δ(0,)δ()0fx′.N11,()0,()()1nNfffnn∀+1n1lim()0nfn→∞=1(1)()nnNfn∞=−∑11(1)()nnfn∞=−∑.nN111()()(0)(),0ffffnnnξξ′=−=≤≤1n1()lim(0)01nfnfan→∞′==.11()nfn∞=∑.t()htddhCt=.()xttddxkth=k.T+1tT=0x=tT=2x=tT=+23x=.ddhCt=1hCtC=+.0t=0h=10C=.hCt=ddxkth=d(dxAkAttC==.ln(xAtBB=+.tT=0x=tT=+12x=t+2T=3x=ln0ln(1)2ln(2)3ATBATBATB+=⎧⎪++=⎨⎪++=⎩510.6182T−=≈37572255.()d()d()max()bbaaaxbfxxfxxfxf≤≤′′′′≥∫∫x0044()max()(),,baxbafxdxfxfxxababa≤≤′′≥=−−∫b≠()fx0[,]ax0[,]xb01()()()()0fxfafxaξ′−=−002()()()()fbfxfbxξ′−=−221121()d()d()d()()bafxxfxxfxxffξξξξξξ′′′′′′′′≥≥=−∫∫∫=0000000()()()()(fxfxbafxbxxabxxa−−−=−−−−)200()()()4babxxa−−−≤04()d()bafxxfxba′′≥−∫.(1)21021()knnkfxCCxCx−−−+=+++k+0121(1)!(1)!(1)!,,2!kkiknkCkaCaCai−+1ki+−+−+=−==.(1)()kfx−1nk−+()()kfxnk−()()kfx(1)()kfx−.1,kkaa−+1110,0kkaa−+02,0CC(1)()kfx−0x=.()(1)0(0)0,(0)0kkffC−==()()kfx(1)(1)0()(0)0kkfxfC−−=(1)()kfx−0x()()kfx0x=00x(1)0()0kfxC−≥(1)()kfx−1.22.123.384.3

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