haihongyuan.com

# 第5届全国大学生数学竞赛预赛(非数学类)参考答案

1. 求极限

lim1?sin?n??

?

.

n

sin??sin?2n??sin

?

?

?

(2分)

n

?1

?sin?

?exp??lim??n??

nln??1? ?

??

?? ?

?exp

??limn?

?n?? ?exp???1n?e4 ?

??

2 证明广义积分

?

sinx

x

dx不是绝对收敛的. (n?1)?

x, 只要证明?an发散. n?

|sinx|

?

?

n?0

1

(n?1)?

?

(n?1)?

|sinx|dx?(n?1)??sinxdx?2

. n?

1?

0(n?1)?

??

??

3. 设函数y?y(x)由x3?3x2y?2y3?2所确定. 求y(x)的极值.

3x2?6xy?3x2y'?6y2y'?0 故 y'?

x(x?2y)

2y2?x2

，令y'?0，得x(x?2y)?0?x?0或x??2y.

(2分) (2分) (2分) (3分) (1分)

(1分)

?x?0 ??y??1和?x??2. (2分) ?y?1?

(2y2?x2)(2x?2xy'?2y)?(x2?2xy)(4yy'?2x)y???(2y2?x2)2

??2?1?0. y??xy?1y'?0x?0y??1y'?0??1?0，

4．

yx?t) (2分)

S333t??x??t?1，?A的坐标为(1,1). (4分) 44

?????xsinx?arctanexdx. 1?cos2xx?xsinx?arctanexsinx?arctanexdx??dx 01?cos2x1?cos2x

x?xsinx?arctanexsinx?arctane?x

??dx??dx (4分) 001?cos2x1?cos2x

?xsinx??(arctanex?arctane?x)dx 01?cos2x?

??

2?0?xsinxdx (2分) 21?cosx

???????2?2?

2?0sinxdx (4分) 1?cos2x3????????arctan(cosx)??08?2? (2分)

?n?1??1?f??收敛. ?n?

x?0 证 由于f(x)在x?0处连续，且lim

f(x)?f(0) f?(0)?lim?0. (２分) x?0x?0

limx?0f?(x)?f?(0)1f(x)f?(x)lim?f??(0). (３分) ?lim?2x?0x?0x2x2(x?0)2

?1?f???n?1???f(0). (２分) limn?02

2n

?1?1?由于级数?2收敛，从而?f??收敛. (3分) ?n?n?1n?1n?

2. ?am

0???(y)?

?b

asinf(x)dx???x??(y)?

0BA??(y)sinydy (3分) ???12sinydy? (2分) mm

?3?x?dydz??2y3?y?dzdx??3z3?z?dxdy.

I?????3x

V2?6y2?9z2?3?dv?3????x2?2y2?3z2?1?dxdydz. (３分) V

V?(x,y,z)|x2?2y2?3z2?1. (３分) 所以V是一个椭球, ?是椭球V的表面时, 积分I最小. ??

?x?u??(x,y,z) 为求该最小值, 作变换

?y?v/.

I?222u?v?w?1?dudvdw. (4分) ????2?v2?w2?1

2?使用球坐标变换, 我们有

I? 22d?d?r?1rsin?dr???

????000?1?. (４分) 15

??x?(u?v)/解.

22223212u?v?r2, 也是取正向 (2分) 22且有 x?y?u?v, ydx?xdy?vdu?udv,

Ia(r)?vdu?udv. (2分)

22a?(u?v)?

?u?cos??2作变换?,

2(?1a

Ia(r)?r2?2?)?0d??(2cos2?/3?2sin2?)a?2a(1)Ja, 其中Ja??0d?, 0?Ja???. (3分) 22a(2cos?/3?2sin?)

2?

J1?

?

d?

2co2s?/?3

?/2

?4?2

2s?in0

dtan?

?4?2

?2/3?2tan0

??

dt

?. (3分) 2

?t22

?0,a?1?

limIa(r)????,a?1. (2分)

r???

??

?2?,a?11?

1?

???1n?2)

n?1?2

??

n，uann?(n?1)(n?2)

,n?1,2,3,. 因为n充分大时

0?a1

1

n?1?

2

??

n

?1??n11x?1?lnn?

1?n?

?1

n

3/2 而 收敛，所以u收敛. n?1n

3/2?n

n?1（2）a1

k?1?

2

??1

k

(k?1,2,....) 1n1???1nS??an

kk?1(k?1)(k?2)??k?1(k?1)(k?2)??k?1?ak?k?1

?ak?n?k?2?? ???a1?a1?????a2?a2?

????a

n?1?an?1?n?23??34?

??n

n?1?????aa??n?1?nn?2?? ?12a11

1?3(a2?a1)?4

(a3?a2)??

1n?1(a?a1nn?1)?n?2

an ???111?1?2?2?3?3?4

??

1?n?(n?1)?

??1n?2an?1?1n?1n?2

an. 因为0?an?1?lnn 所以0?

ann?2?1?lnn1?lnn

n?2

n??n?2?0. 所以limann??n?2?0. 于是 S?nlim??

Sn?1?0?0?1. 证毕。

(3分)

(2分)

(2分)

(2分) (2分) (3分)