历年考研数学真题及答案解析

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202012ὃẆᦪÝÞ⚪
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21ூ$᪆௃ᵫ'“XJ)ᙠ(0,0)ᜐABQV᝞>WXᧅPᙠZ[\ᨵ40,0)=᯿(x,y)=0öT)'+VJTO^᪵lim&4`Qᑏᡂlim""úû°),cᓽ᩽▲lim"*?)4(00)Pᙠ,QVfg+/=AZ+APiᢈ+klim,"ஹ*‘-"஺"o஺,cᓽg,2)-40,0)=OAx+0r+஺Q“+s)஺ᵫQRḄuvQV/(xj)ᙠ(0.0)ᜐQR஺ভ)4=/sinxdx(+=123),ᑣᨵD(A)18%(B)/0,A:e(0,^),ᓽ784=j/sin.mrᐵ'A:ᙠ(0")1/ᓫ=>k5ᦪ?ᵫ'1,2,3e(0"),ᑣ4<4<4ᦑ⌱Dூ!ᫀ௃(B)100)(100ஹூ$᪆௃Q=P110AB=T10Dτஹ001>(001>p00ஹ’100)T00wVI00)f1ஹᦑ஺-஺=-110P'XAP110=-1101110=101Jஹ21஺JI2,1°01,(001>100ᦑ⌱(B)o(7)*xyz{x|y}~ᑖᦪ1|ᦪ4Ḅᢣᦪᑖᑣp{x0.y>0ூ$᪆௃=(X/)Ḅὶᔠᭆ᳛/(x,y)=ᑣ{X

22ூ6᪆*(ᑖXJ,ᯠx+y=l.ᓽ1=-x+l,ᦑ(ὅJឋᐵ¥J¦}ᐵᡠ}ᐵᦪ-1ஹ§¨⚪99-14©⚪ª©S4ᑖᐳ24ᑖ¬ᫀᑏᙠ⚪¯ᢣux°(9)K+ᦪ/(x)±²³´f\x)+/(X)-2/(x)=0µ/(x)+/(x)=2e"Z|/(x)=஺ூᫀ௃9ூ6᪆௃9ᱯ¸³´2-2=0,ᱯ¸᪷0=1º=-2,»¼Rᑖ³´½஺)+/)-2/8)=0+rḄ¾6/(x)=Ge+Ce-21.¿ᵫf(x)+/(x)=2ezÀ2Cex-Á3=2/,QVG=LG=°஺2tᦑ஻x)=/(10)x42x-x2dx஺ூᫀ௃91ூ$᪆௃9Âr=x-lÀ|ᒾ2*_gi&=A=ே96\=|(11)grad/Å+Æ©ூᫀ௃9{1A1}ூ$᪆௃9gradE.9ᓝ={□1}(12)*È={(*>2Ê++2=1<20/20/2:0}‘ᑣ('|',2Ì=஺1Iூᫀᩔூ$᪆௃9ᵫ☢ÎᑖḄÏкÑQVIJ/º=o1+(-Ò+(T)EtMy=/1f/dxdy,ᐸ/6D={(xy)|x>0y>0,x+y<1}»ᦑÓÑ=⍝f/Õ(y2dx=ÖA\'2(l-y)

23n(5)*%=-1«=ஹᐸ/ûüýᦪᑣEᑡᔣ{þឋ}ᐵḄJ(ு(A)%,%,(B)%&(C)%,%&(D)%%%ூᫀ(C)01-1-11=q\[=0,%஺3%ឋᐵ஺ᦑ⌱(C)ூ᪆௃ᵫ|(%%)|=0—11fl[(6)43▤▣!"3▤⌮▣!$"%஻=,1!"=(%%%),IR஺=(%+a”a»%)ᑣ஺"஺=()[1)(1]l+—.-11,ᡠd!A^gr-sinx`0,1-x1-x1-x^/(x)>0,e/(0)=0,ᓽ2xln[+8SX-1-8201-x2

24ᡠdxln——-+cosx>--Fl஺1-x21+X14-X214-x*]%1<%<0,ᨵinj-go,g>1ᡠdJὶ%sinxWO,1—x1%j1-Xᦑf(x)>0,ᓽ2xlnᓃ2+cos%—1%l201-x214-x!xln——-+cosx>l+—,-1?-HD|஺1|4(஻+1)+4(஻+1)+32(n+l)+l,.4n:+4n+32(n+l)+lUP2n+l4(»+1)*+4(M+1)+3S(x)=£4n2+4஻+3/2«+l-xw-0஺(=*4'/+4”+3J஺toJo2«+lx=lBj\E+4E+3/“ᦣ».o2n+l4/+4஻+3QlimT11—=8.13512஻+l

25X=T]^_`2(-1aᦈcaxe(-Ll)rᦪḄᦈ஺rᦪS(x)=£4e+4;+3/N2w+lx(18)(W⚪Xᑖ10ᑖ)L-.\X=L)[0ofo0(0<2)ÔᵨØ᪍Ø᪍Ú¨2ܨ[3^ydx+(x3+x-2y)dy-[3x2ydx+(x3+x-2y)dy[[(Sx2+1-3x2')dxdy--2)*3-SÝSc+4=2-4c402G2(20)(W⚪Xᑖ10ᑖ)r1a00ஹfn01a0-i/=,b001300᎔(1)mM

26(IDឋPßd=bᨵ¬àá!m©md=6Ḅâ஺1a001a0a0001a0ூl᪆௃E(I)\=lx01a+ox(-l)^10=1—<7401a00101a001001ஹ,1a001]a0-101a0-11a0001a001-ajஹ00a1-a—a,]⌕ä2ÜឋPßᨵ¬àá!ᑣᨵ1-/=0-a-4=0,a=-l஺fl-10000-10ஹª^!ÜឋPßåæ▣a:-10-i010-1-1,è%TᓄêᨬìḄ1o1-1001-10[00000>(0000ojrn(°ஹ1-i-1íḄîïð1⚣òPḄᱯ°,ᦑᐸâk;+0ឋPßd=bõᙠ2ö.½Ḅ!ᨵ|N|=0.211ᓽ÷=0A-10=(2-1)2(/+1)=0,22=1ᡈ-1.11Aq]2=1^!oJp01ஹ(21)(W⚪Xᑖ10ᑖ)8▤▣,4=011!/■▣HḄùG!r(47K)=2,$aòúஹT0a.f=xrArAxo1)ma2)maòúûüḄaòú▣!©@aòúᓄ᪗þú!ᑏJÿ
ᣚ஺ூl᪆o1)ᵫ஻(/ಘ=(4)=2,101011=a+l=0=a=-1-1Qa1f=/,஻,᝖=(,)0

27(202ஹᑣ▣8=,0221224,x-20-2\AE-B\=0A-2-2=x(2-2)(A-6)=0—2—2A-48▣Ḅᱯ4=0;4=2;4=6Tஹ!4=0,(4E—B)x=o"Ḅᱯᔣ$/=1!&=2,(¥-3)'=0"Ḅᱯᔣ$(=01!4=6,(4)-8)ᑐ=0"Ḅᱯᔣ$,=2-஻,஻2,(ᓫ/ᓄ:

28(i)rn.প஺=(%,%&)(22)(2⚪4ᑖ10ᑖ)6789
$X,Y:;XYḄᑖ<=᝞ᓛ⊤ᡠB,X012P121/31/6Y012P1/31/31/3XY0124P7/121301/12C(1)D(X=2F)/<2)cov(x-y.y)E2.1rூ$᪆GX012P1/21/31/6XY0124P7/121/301/12(1)D(x=2F)=D(x=o,y=o)+D(x=2,y=i)=L+o=l44(2)cov(^-yr)=cov(^y)-cov(rr)ss5cov(XJ\=EXY-EXEY,ᐸHI=±£¥?=L£T=1Lᵨ2=2,0X=£YL(EY)2=1—±=;33v799DY^EY2-(EY^=EXY=-''333ᡠ:Ncov(x*)=0,cov(yy)=Dr=|,cov(x—y*)=-g,O=o.s(23)(2⚪4ᑖ11ᑖ)P89
$XEyQRSTUᑖVWXY᝱ᑖ0,

29P2='L'N(1)CzḄᭆ᳛ab/3,/)/(2)Pz“ᩭefgZḄhᓫ89᪵2NC᪗Ḅᨬᜧmᯠop$qr(3)stḄuvop$஺ூ᪆y(1)z“~"("})~'~□:2"),UXEyQRSTNᦑ2=L'N(0,5/),ᡠ:NZḄᭆ᳛ab/(z,஻)==e<—8