06至10年浙江省高等数学(竞赛 工科类试题)(3)

??1???1?x?????x???1?x???1?1?x?2 六、(15分)已知f(x)二阶可导,且f(x)?0,f??(x)f(x)??f?(x)?2?0,x?R, (1) 证明:f(x1)f(x2)?f2??x1?x2??2??,?x1,x2?R. (2) 若f(0)?1,证明f(x)?ef?(0)x,x?R.

证明: (1) 要证明f(x?x1?x2?1)f(x2)?f2??2??,?x1,x2?R, 只需证明

12lnf(x)?112lnf(x2)?lnf??1?2x1?1?2x2??,?x1,x2?R,

也即说明F(x)?lnf(x)是凹函数,

?lnf(x)???f?(x)??f(x)f??(x)??f?(x)?2f(x), ?lnf(x)??????f(x)??f(x)???f2(x)?0, 故F(x)?lnf(x)是凹函数, 即证. (2) F(x)?F(0)?F?(0)x?F??(?)2x2 ?lnf(0)?f?(0)f(x)f??(x)??f?(x)?2f(0)x?2f2(x)x2?f?(0)x,

x??即: f(x)?ef?(0)x,x?R.

2008浙江省高等数学(微积分)竞赛试题(解答) ?

一.计算题 1x2x3x1、求lim???0?e?e?e?sinxx??3?.

?11x2x3x?sinxx2x3x???lim?e?e?esinx解: lim

0?e?e?ex???3??x?0???1??3?3???3?ex?e2x?e3x1?ex?e2x?e3x?ex?e2x?3x3?sinxex?e2x?e3x?lim3?1sinxx?0???e?3???limx?0e

11

0x?3e3x?0ex?2e2lim3cosx?x?0e?e2。

2、计算

?1cos(3?x)sin(5?x)dx.

解:

?11cos[(5?x)?(cos(3?x)sin(5?x)dx?3?x)]cos2?cos(3?x)sin(5?x)dx ?1cos(5?x)cos(3?x)?sin(5?x)sin(3?x)]cos2?cos(3?x)sin(5?x)dx

?1?cos(5?x)cos(3?x)sin(cos2???cos(3?x)sin(5?x)dx??5?x)sin(3?x)]cos(3?x)sin(5?x)dx?? ??1?cos(5?x)cos2???sin(5?x)dx??sin(3?x)?cos(3?x)dx? ??1?dsin(5?xcos2???)sin(5?x)??dcos(3?x)cos(3?x)dx?? ??1cos2?lnsin(5?x)?lncos(3?x)??C ?1sin(cos2ln5?x)cos(3?x)?C。 法二：

?1cos(3?x)sin(5?x)dx??2sin2(x?4)?sin2dx

??22tan(x?4)dx，令t?tan(x?4),x?arctant?4

1?tan2(x?4)?sin2??212t?11?t2dt??22t2sin2?2t?sin2dt?sin2?dt1?t2?sin2t2?2sin2t?1?21sin2??dt ??t?1?cos2??1?cos2?sin2????t?sin2?? 12

????1?11?dt ???1?cos21?cos2cos2??t??t??sin2sin2??1?cos2tan(x?4)?1sin2?C。 ?ln1?cos2cos2tan(x?4)?sin23、设f(x)?x3arcsinx，求f(2008)(0). 11?x2解: g(x)?arcsinx,则g?(x)?

t?arcsinx，则f(sint)?tsint3

dnf(sint)?tsint3 ndt??(n)0?Cnsint3??(n)1t?Cnsint3??(n?1)

dnf(sint)1?0?Cnsint3ndtt?0??(n?1)t?0

d(2008)f(sint)3?2008sintdt2008t?0??(2007)t?0

?2008?2007?3t2cost3??(2006)t?0

?2008?2007?2006?6tcost3?9t4sint3??(2005)t?0

被积函数是奇函数, 要积分为零, 当且仅当积分区间对称,即: a?p??b?p, 解得: p??4、计算

a?b. 2?a0dx?emax{bx0b22,a2y2}dy,(a?0,b?0).

dy???emax{bxD2222解:

?a0dx?emax{bx022b22,a2y2},a2y2}d?, 其中D如右图

???emax{bxD122,a2y2}d????emax{bxD222,a2y2}d? by????eayd????ebxd?

D1D2D1bxa0bxa??dy?0bayb0ea2y2dx??dx?ebxdy

0a22D2a

13 aba2y2bab2x2??yedy??xedx b0a01ba2y21ab2x22222?ed(ay)?ed(bx) ??002ab2ab1a2b2?(e?1). ab25、计算??(x?y)dS,其中S为圆柱面x2?y2?4,(0?z?1). Sz解:

2??(x?y)dS?S1(x2?y2)dS???ydS ??2SS ?14dS???ydS ??2SS22??x???x??8????y1??????dydz

??y???z?Dyzy2?8????y1??02dydz?8? 24?yDyz被积函数关于y是奇函数,积分区域关于z对称, 二、(20分)设un?1?oyx12112112?????????, 234563n?23n?13nvn?111u????,求: (1)10;(2) limun.

n??n?1n?23nv10解: (1)un?12??1????? 3k?13k?k?1?3k?2n12112112?1??????????,

234563n?23n?13n3n11n1 vn??????

n?kk?1k?1kk?1k2n ??1???111111111??11????????????????1????? 23456n3n?23n?13n??2n?n12?3n1n1?1un?vn??????????

3k?23k?13k?k?1kk?1kk?1??21?n1 ?????????0

3k?k?1kk?1?3kn 14

?un?1; vv11??1?????

n??n?1n?23n??(2) limun?limvn?lim?n??n????1?1111??lim?????? (图来说明积分上下) n??n12k2n??1?1?1??1??nnn??n12n1?lim? n??nkk?11?n??1dx?ln3. 01?x2三、(满分20分)有一张边长为4?的正方形纸(如图),C、D分别为AA?、BB?的中点，E为DB?的中点，现将纸卷成圆柱形，使A与A?重合，B与B?重合，并将圆柱垂直放在xOy平面上，且B与原点O重合，D若在y轴正向上，求：

（3） 通过C，E两点的直线绕z轴旋转所得的旋转曲面方程；

（4） 此旋转曲面、xOy平面和过A点垂直于z轴的平面所围成的立体体积. 解:

C(0,4,4?)2ACA?zA4?Nx?2y?Q2z?M? LCE :2?2?4?yxDM(x,y,z) 旋转曲面上任意取一点B?z?E(2,2,0)?x0?2??2?Bz??2 ， Q(0,0,z) 则N(x0,y0,z0)的坐标为：?y0?2???z0?z??DEB? 15

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