西北师范大学2005年研究生入学数学分析试题

硕士研究生入学考试数学分析试题七 西北师范大学硕士研究生2005年入学考试《数学分析》试题

1. 设f(x)?x3?x2?1?x??23?2??的反函数为f3????1(x),求极限limfx?3?1(x).

解:由于当x?时,f?(x)?3x2?2x?3x?x?2??2?,因此在?0,??f(x)??上严格增加?33???且连续,所以f?131?31??2?上严格增加且连续.其中. (x)在?,??f??????2727???3?令x3?x2?1?3,即(x?2)(x2?x?2)?0,得x?2. 从而f(2)?3,f?1(3)?2.又因f?1(x)在x?3处连续,故limfx?3?1(x)?f?1(3)?2.

2. 求?sinxcosx1?sinx323dx.

解:

??sinxcosx1?sinx1?sinx222dx??sinxcosx1?sinx121222d(sinx)??sinx(1?sinx)1?sinx222d(sinx)

?121?sinxd(sinx)?2?2?(1?sinx)1?sinx222d(sinx)?12??2?1??d(sinx) 2?2?1?sinx?12sinx?C.

2??1?sin12xd(1?sinx)?222?1d(sinx)?ln(1?sinx)?22223. 计算??Dxydxdy,其中D是由曲线y?x,x?y?1,x?y?4围成,且在x轴上方.

2解:令x?rcos?,y?rsin?,则J?r,D:xy223?2?4???3?23?4,0?r?1,

23???Ddxdy???44d??rcos?rsin?22221rdr???444cos?sin?2d???rdr?1??441?sin?sin?22d??122r213????4433?1??1d???(?cot???)??222?sin??3???43???32????(4??). 2?2?44. 计算??(x?y)dx?(x?4y)dyx?4y22L,其中L是正方形x?y?2的边界,取逆时针方向.

22解法一:设L0表示椭圆x?4y?1,D表示由封闭曲线L和L0所围成的区域.则L0的参数

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方程为x?cost,y?122sint,t?[0,2?].记P(x,y)?x?yx?4y22,Q(x,y)?x?4yx?4y22,则

?Q?x??x?8xy?4y(x?4y)2222??P?y,?(x,y)?D.

且P,Q,

?Q?P都在D上连续,因此根据Green公式得: ,?x?y(x?y)dx?(x?4y)dyx?4y22??(x?y)dx?(x?4y)dyx?4y22L????L0???0dxdy?0,

D从而??2?0(x?y)dx?(x?4y)dyx?4y22L??(x?y)dx?(x?4y)dyx?4y22L0

????11??cost?sint(?sint)?cost?2sintcost??????dt?22?????2?012dt??.

解法二:L?L1?L2?L3?L4,其中L1:y?2?x,0?x?2;L2:y?2?x,?2?x?0;

L3:y??2?x,?2?x?0;L4:y?x?2,0?x?2.

??(x?y)dx?(x?4y)dyL1x?4y22??0225x?105x?16x?165x?6x?4(x?2)22dx??2010?5x5x?16x?160?222dx,

(x?y)dx?(x?4y)dyL2x?4y20222???20dx???5x?65x?16x?16dx

??5x?65x?16x?16dx,

0?22??(x?y)dx?(x?4y)dyL3x?4y22?5?x?25x?16x?165x?6dx??20210?5x5x?16x?16dx,

(x?y)dx?(x?4y)dyL4x?4y22??2025x?16x?16dx,

???(x?y)dx?(x?4y)dyx?4y822L??L1??????L2L3(x?y)dx?(x?4y)dyL4x?4y82022

?2025x?16x?16dx?8?5201x?2165x?165dx??518?16?x????5?25?2dx

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?2arctan5x?84201???2?arctan?arctan2???.

2??注:解法二实际计算较繁. 5. 设?,???0,????2?x?,证明:

sin(???)1?sin(???)?sin?1?sin??sin?1?sin?.

证明:令f(x)?1?x,x?(0,??).则当x?(0,??)时,f?(x)?1(1?x)2?0,因此f(x)在

???(0,??)内严格增加.由于当?,???0,?时,有

?2?0?sin(???)?sin?cos??cos?sin??sin??sin?,

从而有 f(sin?(??sin?(??)?))f(?s?in, i?s即

?1?sin?(??)?1sin?1?sin??sin?sin(???)1?sin(???)si??ns?i?nsin?s?in ?sinsin?1?sin?sin?1?sin???1?sin??sin?sin?sin???,

故?1?sin??1?sin?.

?x2y,?6. 证明函数f(x,y)??x2?y2?0,?x?y?0,x?y?0,2222在点(0,0)连续且偏导数存在,但在此点

不可微.

证明:ⅰ>因为???0,?????0,使得当x?0??,y?0??时,有

xyx?y222?0?x222x?y2y?y?0??,

所以lim(x,y)?(0,0)xyx?y22?0?f(0,0),从而函数f在原点(0,0)连续;

f(0??x,0)?f(0,0)?x0?0?xⅱ>由于fx?(0,0)?lim?xz?x?x?0?lim?x?0?lim?x?0?0,

fy?(0,0)?lim?yz?y?y?0?limf(0,0??y)?f(0,0)?y?y?0?lim0?0?y?y?0?0,

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即函数f在原点(0,0)的两个偏导数都存在; ⅲ>若函数f在原点(0,0)可微,则

?x?y?x??y222?z?dz?f(0??x,0??y)?f(0,0)?fx?(0,0)?x?fx?(0,0)?y?

应是较??lim?x??y高阶的无穷小量.为此考察极限

22?z?dz??0??lim?x?y32,

??0(?x??y)222此极限不存在,事实上由于当?x??y时

??1?1??lim??3??3.

??x?0?222?2?lim?x?0?x?x?23?lim?x?01?3?13, lim?x?0?x?x?23(?x??x)2222222(?x??x)222或当?y?k?x,?x?0时,

?z?dzk?x23223?x?0?0lim???x?0?0lim?k(1?k)23与k有关.

(?x?k?x)因而函数f在原点(0,0)不可微. 7. 设无穷积分???af(x)dx收敛,证明函数F(x)??xaf(t)dt在区间[a,??)上一致连续.

证明: 因为F(x)在[a,??)上连续，且limF(x)存在,所以F(x)在[a,??)上一致连续.

x???证明: ???0,ⅰ>因为?有F(x1)?F(x2)???af(x)dx收敛,所以根据柯西准则知?X?a,使得当x1,x2?X时,

?x2x1f(x)dx??.

ⅱ>因为F(x)在区间[a,X?1]上连续,从而F(x)在区间[a,X?1]上一致连续,因此??1?0,使得?x1,x2?[a,X],当x1?x2??1时,有F(x1)?F(x2)??.

取??min??1,1??0,则?x1,x2?[a,??),当x1?x2??时,

Ⅰ>当x1,x2?[a,X]时,因为x1?x2????1,根据ⅱ有F(x1)?F(x2)??. Ⅱ>当x1,x2?[X,??)时,即x1,x2?X,根据ⅰ有F(x1)?F(x2)??.

Ⅲ>当x1?[a,X],x2?[X,??)时,因为x1?x2??,从而有0?x2?X?x2?x1?1,即

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x2?X?1，因此x1,x2?[a,X?1]，根据ⅱ有F(x1)?F(x2)??.

综上讨论知: ?x1,x2?[a,??),当x1?x2??时,有F(x1)?F(x2)??.因此F(x)在

[a,??)上一致连续.

8. 设函数f(x)在区间[0,1]上可微,且?x?[0,1],有f?(x)?M,证明:对任意的正整数n,

10有

?f(x)dx?1nn?i?1M?i?. f???n?n?n证明: 因为?f(x)dx?0ni11n?i?1?i?f??=??n?i?1nini?1nninini?1n?ni?1nf(x)dx???i?1?i?f??dx ?n???i?1??i??nf(x)?f???dx??i?1??n??n?10??i?1??i??f(x)?f???dx, ??n???inini?1n所以

?f(x)dx?1nn?i?1?i?f????n?nn?i?1i??i??nf(x)?f???dx??i?1??n??n?M?1ndx?Mnnini?1n??i?1?i?f(x)?f??dx

?n?nini?1n???i?1?i?f?(?i)???x?dx??n??xi?1?x??i?xi???i?1ni?1n??i?1dx?Mn,

其中

i?1nin.

9. 设liman?a(有限),且a?0,证明:

n???⑴对??0,级数?ane?nx在[?,??)上一致收敛;

n?1`?⑵级数?ane?nx在(0,??)内非一致收敛;

n?1`?⑶函数f(x)??an?1`ne?nx在(0,??)内连续.

证明: ⑴因为liman?a,所以?M?0,使得an?M(n?1,2,?).

n??从而ane?nx?Me?nx?Me?n?(n?1,2,?),

e?n?又因limn??1n2?limne2n??n??limte2t?t??2t???2????lim?lim??t???t??2t??0, t???e????e??? 178

?而级数?n?11n2?收敛,根据M---判别法知:级数?ane?nx在[?,??)上一致收敛;

n?1`⑵由于liman?a?0,因此?N0,使得当n?N0时,有an?n??a2?0.

因为??0?a6?0,?N,?n?N?N0?N,?x??a2e1n?(0,??),使得

ane?nx??ane?nx???1e?a6??0.

或supane?nx?supx?(0,??)x?(0,??)?a?nxn??an?a?0,n??.

?所以ane?nx在(0,??)内非一致收敛于0,故级数?ane?nx在(0,??)内非一致收敛.

n?1`注1:fn(x) 在(0,??)内非一致收敛于0的定义:??0?0,?N,?n?N,?x??(0,??),使得

fn(x)??0.

注2:fn(x) 在(0,??)内非一致收敛于0?limsupx02n??x?(0,??)fn(x)?0.

?⑶?x0?(0,??),取??,则??0且x0?(?,??).因为?ane?nx在[?,??)上一致收敛,

n?1`?且每一项ane??nx都在[?,??)上连续,所以和函数f(x)??an?1`ne?nx在[?,??)上连续.特别函

?n数f(x)??an?1`e?nx在x0处连续,由x0?(0,??)的任意性知函数f(x)??an?1`ne?nx在

(0,??)内连续.

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