数学分析1测试题答案.pdf
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数学分析1测试题答案.pdf
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数学分析数学分析1测试题测试题1参考答案参考答案一、填空题(每小题3分,共24分)1.3;022.12.3.04.1,|10,|1aa.5.000(,)lim,lim()nnnnnxUxxxfx=对且都有存在且相等.6.;.二7.,()2nnZ+.8.1.二、计算题(6分,共36分)1.112nnnnnnannn+=时在处连续-(4分)要使f(x)在x=0可导,即要使100()(0)1limlimsin0xxfxfxxx=存在-(7分)故当101,()0fxx=即时在处可导.-(8分)6.32000sin01cos01sin1limlimlim03066xxxxxxxxxx=型型-(8分)7.1ln00coslncossinlimlnlimsinlnxxxxxxxx+=Q型-(4分)=00sincoscossinlimlim11sincosxxxxxxxxxx+=-(7分)所以,1lncosln1sin0limxxxxee+=-(8分)三、证明题(每小题10分,共20分)1.先限制|0|10,xx且则-(2分)211111|0|0|222(11)xxxxx+=当有1111|0|22xxx+-(9分)故0111lim2xxx+=-(10分)2.在12,(0,)2xx对y=sinx应用拉格朗日中值定理,得12(,)xx,使得212121sinsincos()xxxxxx=0,202f=,知0,2x,使得f(x)=0,即1+sinx=2x有根.-(7分)下证唯一性.若方程存在两个根1212,0,2xxxx且,则有2121sinsin2()xxxx=这与前面已证2121sinsin()xxxx相矛盾,故12xx=-(10分)数学分析数学分析1测试题测试题2参考答案参考答案一、填空题(每小题3分,共24分)1.1,1.2.2.3.1e.4.,0;0,0;,=5.00,()0,0|,|()|.xxfxA=对使当时有6.32,;.2xkkZ=+二7.,.kkZ8.2.二、计算题(每小题8分,共56分)1.2nnnnnnbbabb=+Q-(5分)而lim,lim2nnnbbbb=-(7分)limnnnnabb+=(由迫敛性)-(8分)2.()221(),121xuxux=+=+-(4分)221()1121xfxxxx=+-(8分)3.取对数,得ln()coslnsinfxxx=-
(2)两边对x求导,得1()sinlnsincos()fxxxxctgxfx=+-(2分)所以cos()(sin)(cossinlnsin)xfxxxctgxxx=-(8分)4.取对数,得1ln()lnfxxx=,-(2分)两边对x求导,得22111()ln()fxxfxxx=+-(6分)所以,12()(1ln)xfxxx=-(8分)5.因为f(x)在x=0处可导,故在x=0处连续,所以9a=.又2089(9)()lim80xxxfxx+=-(3分)-(2分)01(9)1()lim0xaxbfxxb+=-(6分)而11(0)(0),8,.8ffbb+=有得-(8分)6.32000
(2)20
(1)101limlimlim03066xxxxxxexxxexexxx+=型型-(8分)7.()00limlnsinlimlnsinxxxxxx+=Q-(4分)=00lnsinlimlimcos01sinxxxxxxxx+=-(7分)所以,原式01e=-(8分)三、证明题(每小题10分,共20分)1.1112122(21)nnnn=+当时有1212nn+-(9分)故1lim212nnn=+-(10分)2.令()ln
(1)fxx=+.则f(t)在0,x上连续,在(0,x)内可导.故由拉格朗日中值定理,得(0,)x,使得ln
(1)ln1()0xfx+=-(3分)即ln
(1)11,0.1xxx+=+所以ln
(1).xx+-(6分)显然1,1,2,nnaan+=L-(7分)且2221111112201122212na+=对使当时有4.25.22x.6.2.7.,nnZ8.1()2ab+二、计算题(每小题8分,共48分)1.由lim21nn=及迫敛性知:
当01x时,12()nnnnnxxxxxn=+,这时lim1nnnxx+=-(8分)2.归结原则:
000(,)lim,lim()nnnnnxUxxxfx=对且都有存在且相等.-(3分)取21,lim0()nnnxxn=则,但()cos
(1)nnfxn=-(6分)因为lim()nnfx不存在,所以01limcosxx不存在(由归结原理)-(8分)(注:
也可取,0,lim()lim()nnnnnnxxfxfx但)3.间断点为1x=-
(2)因为11lim()1,lim()1xxfxfx+=,所以x=1为f(x)的跳跃间断点;-(5分)因为11lim()1,lim()1xxfxfx+=,所以x=-1为f(x)的跳跃间断点.4.0,()2xfxx=时0,()1xfx=不可导-(8)5.取对数lnln(ln)yxx=-(2分)两边对x求导,得11()ln(ln)(ln(ln)ln(ln)lnyxxxxxyx=+=+-(6分)所以1()(ln)ln(ln)lnxfxxxx=+-(8分)(注:
也可以将f(x)写成ln(ln)()xxfxe=再求导)6.因为2()2xfxxe=-(3分)222222()
(2)242
(1)xxxxfxxeexexe=+=+-(6分)所以2222()2(12)xdfxxedx=+-(8分)三、证明题(9+9+10=28分)1.22213133(31)nnnn=+当时有22313nn-(3)证:
由0000lim()()0,(),0,0|,xxfxfxfxxx=取当时有00|()()|()fxfxfx-(7分)即有000()()()()fxfxfxfx=对使当时有4.3.5.2x.6.12.7.,2nnZ+8.223aabb+二、计算题(每小题8分,共48分)a)32332nnnn+Q及-(5分)而lim33n=,lim323nn=-(7分)由迫敛性,得lim233nnnn+=-(8分)2.归结原则:
000(,)lim,lim()nnnnnxUxxxfx=对且都有存在且相等.-(3分)取21,lim0()nnnxxn=则,()sin0nfxn=,lim()0nnfx=-(5分)取21,lim022nnnxxn=+则,()sin212nfxn=+=,lim()1nnfx=,所以01limsinxx不存在(由归结原理)-(8分)3.由2123201,2xxxx+=求得两个根,可知20,12320,1,20,1,2xxxxxxx-(6分)故x=1和x=2是f(x)的第一类(跳跃)间断点;-(8分)4.0,()(cos)sinxfxxx=时0,()(1sin)cosxfxxx=不可导-(8)5.取对数2ln()coslnsinfxxx=-(2分)两边对x求导,得21()sincoslnsincos()fxxxxxctgxfx=+-(6分)所以2cos()cos(sin)(cossinlnsin)xfxxxxctgxxx=-(8分)6.因为sin()sin()()(sin()cos()axbaxbfxeaxbaaxbe+=+=+-(3分)22sin()2sin()()cos()sin()axbaxbfxaaxbeaaxbe+=+-(6分)所以222sin()2()cos()sin()axbdfxaaxbaxbedx+=+-(8分)三、证明题(9+9+10=28分)1.先限制11|1|1,02,131xxx+即1|1|1|1|122|1|2xxxxx=当0时有112xxQ在连续当时有0|()()|guguQ在连续对上述,0|,xx当时有001|()()|uufxfx=只要有,即g(f(x)在0x连续.3.不妨设0x为f(x)的极大值点,则存在邻域000(),()()()UxxUxfxfx使对有.故当0xx时,有00()()0fxfxxx。
由f(x)在故当0x可导及极限的不等式性质,得到00000()()()()lim0xxfxfxfxfxxx=,00000()()()()lim0xxfxfxfxfxxx+=,因此0()0fx=.数学分析数学分析1测试题测试题5参考答案参考答案一、填空题(每小题4分,共32分)1.不存在;0.2.0,|()|MxDfxM对都有.3.11;
(2).24xx4.x.5.,;2kkZ+第一.6.00000001()()();()(),()0()yfxfxxxyfxxxfxfx=其中.7.0|xx.8.4.二、计算题(6分,共36分)1.22222111112nnnnnnnnn+=+QL-(3分)而2lim1nnnn=+-(5分)lim1nna=(由迫敛性)-(6分)2.2222ln(sin)0cos0sin1limlimlim
(2)04
(2)sin08sin4
(2)cos8xxxxxxxxxxxx=型型(8分)3.222222121limlim1112xxxxxxtxx+=+令-(3分)=21212111lim1lim
(1)1ttttettt+=+=-(6分)4.1222212212sec23sec2(4)22213213arcsinarcsinxxxxyxtgxtgxxxx=+=+-(6分)5.取对数,得2lnlnlnlnyxxx=-(1分)两边对x求导,得12lnyxyx=-(5分)所以lnln12ln2lnxxyxxxxx=-(6分)6.(cos)cossincossin(sin)sincossincosttttttdyetetetttdxetetettt=+-(3分)222223cossin(sincos)(cossin)2sincos(sin)(sincos)(sincos)(sincos)tttttdyddyttttttdxdxdxetttettett+=+三、证明题(每小题8分,共32分)1.22221|12|21|1|2xxxxxxxx+=-(3分)先限制1130|1|,1,222xxx=当有212xx+,按照定义知,0,NNaa使得.又由na的递增性知,当nN时都有Nnaaa.另一方面,由于a是na的一个上界,故对一切na都有naa,因而更有.naa+当时有所以limsupnnnaaa=举例:
11nan=为递增且有上界数列,11lim11sup1nnn=3.
(1)22()2
(2)xxxfxxexexxe=+=+在0,3连续,故有界.
(2)对于2121212,0,3,(),xxxxxfxxexx0,使得12,|()|,()|xxxfxMfM都有从而|.于是,对12210,0,3,|,xxxxM=只要且就有21|()()|fxfxM=即2()xfxxe=在0,3上致收敛.4.1222111lim11lim()limlim2111xxxxxxxxxxxxxx+=+=+110
(1)11kkkkknnnnn+=+Q,而1lim0knn=,故由迫敛性,得()()lim10kknnn+=数学分析数学分析1测试题测试题6参考答案参考答案一、填空题(每小题4分,共32分)1.2;不存在.2.121212,|()|()xxDxxfxfx=若对只要且有7.0|xx.8.1.二、计算题(6分,共36分)1.3nnnnnnnbbabcb=+Q-(3分)而lim,lim3nnnbbbb=-(5分)limnnnnnabcb+=(由迫敛性)-(6分)2.1,0txtx=+令则有-(1分)10010lim
(1)limlim10ttxxttexeet+=型-(6分)3.1000ln(1sin)0coslimln(1sin)lim(lim101sinxxxxxxxxx+=+型)=-(5分)11ln(1sin)200lim(1sin)limxxxxxee+=-(6分)4.22232cos1cos3333212sin12sinarc1()(9)arcxxxyxxtgxtgxxx=+=+-(6分)5.取对数,得lncoslnsinyxx=-(1分)两边对x求导,得1cos(cos)lnsincos(lnsin)sinlnsincossinxyxxxxxxxyx=+=+-(5分)所以cos(cossinlnsin)(sin)xyxctgxxxx=-(6分)6.322
(2)23
(2)22dytttdxttt=-(3分)22222233232212
(1)2(23)362
(2)(22)4
(1)ttdyddytttttdxdxdxtttt+=三、证明题(每小题8分,共32分)1.先限制11|1|1,02,131xxx+即1|1|1|1|122|1|2xxxxx=当0时有112xx0,即k1时f(x)在x=0处可导,且(0)0f=.-(4分)
(2)当0x时,12111()(sin)sincoskkkfxxkxxxxx=.所以,当k2时0lim()0(0),()0.xfxffxx=这时在处连续-(8分)3.对12120,1,3,|,6xxxx=取则当且时有2221121212|6|6xxxxxxxx=+=故2()fxx=在1,3上一致收敛.上述证明在区间(1,3)上仍成立,故f(x)在(1,3)上也一致收敛.4.由题设知
(2)(0)ff=.不妨设f(x)不恒为常数.令()()()Fxfxfx=+,则(0)()(0),()
(2)()FffFff=故F(x)在0,上连续,且(0)()0FF对且时有.3.例如:
2212,1mmama=.4.1e+;因为()1(),fxgx=+而1与g(x)都在xa时收敛.5.,1mn.6.必要;充分.7.若0()0fx,则在0x的某邻域0()Ux内,0()()fxfx与同号.8.0001()()()yfxxxfx=.二、计算题(每小题8分,共48分)1.
(1)2121212
(1)323limlim131
(1)3xxxxxxxxx+=3423233112lim1113332xxxxxx+=+=42233eee=.
(2)223002122cossin0limlim()sin1cossin(1cos)0xxxxxxxx=3002sin2sincos02sin1lim()limsin2sincos(1cos)02sincos2sin4cossin2xxxxxxxxxxxxxxx=+2.2222()2()()2dfxfxfxxdx=2222222222()4()()()2()()2dfxfxfxxfxxxfxfxxdx=+222221()4
(1)
(1)
(1)2
(1).012xdfxffffdx=+=3.由()0fxx=在的连续性,得200lim(89)9,lim()xxxxxaab+=+=,求出9a=由11(00)8,(00),8ffbb+=得4.sinsin(1cos)1cosdyattdxatt=,22232cos(1cos)sin1(1cos)(1cos)dytttdxatat=5.1223,
(1),
(1)2!
yxyxyx=设()1
(1)
(1)!
kkkykx=,则
(1)()1
(1)
(1)()
(1)
(1)!
()
(1)!
kkkkkkyykkxkx+=故()11
(1)
(1)!
(1,2,)nnnynxn=L6.sup1.A=因为:
(1)1,113nZn+=+总有使三、论证题(10+10+8)1.
(1)2001sin0(0)(0)limlim0xxxfxfxxx+=00(0)(0)0limlim0xxfxfxx+=(0)(0)ff+=
(2)在0x时,11()2sincosfxxxx=在0x=当当2.设3()31,()fxxxfx=+则(多项式)在0,1上连续.由(0)1,
(1)1,(0)
(1)0ffff=知,故由介值定理,知(0,1),()0f=使.若()00,1fx=在上有两个相异的实根1201都使得2.若对0000,()0,|()()|xxxfxfx=使当时有.3.21,0()0,0xfxxx=当当4.充分.5.0;43.6.1;1;0.03;2x.7.(2,3)8.21(ln()(ln().fxfxx二、计算题(每小题8分,共48分)1.
(1)1lim
(1)lim1kmmxmxkmxxkexxk+=+=
(2)0001110101limlim()lim()1
(1)01022xxxxxxxxxxxxexeexexexeexee=+2.22211()2()2
(1)xxdfxxfxfdx=Q2121()2()()|2
(1)
(1)xxdfxxfxfxffdx=又2211()()xxdfxdfxdxdx=,2
(1)2
(1)
(1)fff=故
(1)
(1)1)0ff=3.()3fxx=Q在处可导,所以在x=3处连续,故有39ab+=又233()(3)9(3)limlim633xxfxfxfxx+=33()(3)90limlim()330xxfxfaxbaxx+=.由(3)(3),ff+=求出3a=,再代入9axb+=求得9b=.4.22222112,11
(1)xxxxyarctgxyxxx+=+=+Q22222
(1)dxdyx=+5.2312!
(1)
(1)yyxx=现设()1!
(1)kkkyx+=成立,则
(1)()11!
(1)!
()
(1)
(1)kkkkkkyyxx+=故对任意自然数n,都有()1!
(1)nnnyx+=.6.23,33xxQ,故inf3A=.验证如下:
2
(1),3,3inf;xAxx=对有00
(2)0,3,inf3xxA+=对使,因此,inf3A=.三、验证题(8+10+10)1.当1x时,先限制|1|1x,即02x,则有221211|1|2133(21)3xxxxxx=取当时有2212213xxx2.2|,|yxxxyx=当时并当时.4.19,8;5.1;6.3;7.22x8.必要。
二、计算题(6+6+6+7+7+7+7+9=49分)1.由于33nnnnnnnnnbbabcbb=+=lim,lim3,limnnnnnnnnbbbbabcb=+=2.566352525442553
(1)535limlim522
(1)5xxxxxxexexex+=.3.222221cosln(sin)01cos01sin1sinlim()limlim()lim
(2)04
(2)420428xxxxxxxxxxxx=.4.2112(13)(ln(arcsin)(13)(arcsin)22213arcsin2ytgxtgxxtgx=+=+=222223sec1123sec2()2213421342arcsin11arcsinxxxtgxtgxxxxxx+=+5.两边取对数得lnln(ln)xyx=,即lnln(ln()yxx=,两边求导,得111ln(ln()lnyxxyxx=+故1(ln)ln(ln)lnxyxxx=+6.222sin(32)sin(32)2sin(32)2()(sin(32)cos(32)6xxxdydeedxexxdx+=+=+7.(cos)cossincossin(sin)sincossincosttttttdyetetetttdxetetettt=+222cossin(cossin)(sincos)(cossin)(sincos)sincos(sin)(sincos)(sincos)tttdyttddyttttttttdxttdxdxetttetet+=+=3(sincos)(sincos)(cossin)(cossin)(sincos)tttttttttett+=2233(sincos)(cossin)2(sincos)(sincos)ttttttettett+=+当2t=时,200211,01tdyxeydx=,所求的切线方程与法线方程分别为22(),yxeyxe=三、证明题(8+9+10)1.先限制10|1|2x,即131,22xx且则2221|1|1|04|1|12xxxxxx+=,要使210xx只要4|1|x,即|1|4x故取1min,0|1|24x=则当即可,即得证.2.设()ln,(),fxxxabfxab=则在上连续,在(a,b)内可导,故由拉格朗日中值定理,至少存在一点(,)ab,使得lnln1(ln),xbaxabba=由1110,abba知,故有1lnln1babbaa,即lnbabbabba=当且时有|()()|fxfx设,(,1),xxc由于11|()()|sinsin|2sincos22xxxxfxfxxxxxxx+=2|xxxxxxc故取2c=,则当|,(,1),xxxxc且时有|()()|fxfx=只要且有5.2;6.0,|11,|1aa7.3;8.1.二、计算题(6+6+7+7+7+7+9=49分)1.由于1121nnnnnnnnnnnnnn+=L,而lim1lim1nnnn=,所以12lim1nnnnnn+=L.2.34423232223332
(1)323limlim311
(1)3xxxxxxexexex+=3.0000202lim()limlimlim2sin01cossincosxxxxxxxxxxxxeexeeeeeexxxxx+=.4.2(arcsin(ln)1xyxxx=+2222111(ln1)11(ln)xxxxxxxx=+=32222ln111ln
(1)xxxx+5.两边对数得lnsinlnyxx=,两边求导得:
1sincoslnxyxxyx=+,所以sinsin(cosln)xxyxxxx=+6.()22ln()ln()2112ln()costgxtgxdydeetgxdxtgxx=7.222sinsin11cos,1cos(1cos)(1cos)dytddytdydxtdxtdxdxatat=00
(1),2xaya=,所求切线方程和法线方程为
(1),222ayaxayxa=+或
(1),22ayaxayx=+或三、证明题(8+9+10=27分)1.先限制1130|1|,1222xxx即且,则2222333223|32|1|23|1|26|1|14xxxxxxxxxx+=则当时有223xx+故得证.2.设()fxarctgx=,则在0,h上应用拉格朗日中值定理,得201,01arctgharctghh=+故22111h+,进而2211111h+,故21011arctgharctghh+即21harctghhh+3.
(1)归结原则:
设f(x)在0x的某邻域00()Ux有定义,则0lim()xxfx存在的充要条件是:
对任何心0x为极限且含于00()Ux的数列nx,极限()nfx存在且相等.
(2)证明301limsinxx不存在.取3
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