2006浙江省高等数学(微积分)竞赛试题(解答)
一、 计算题（每小题12分，满分60分）
1、计算limn??n???x?nx???1???n??e??.
????nx?x??x?解:
?lim??n?x?????1??????x?x?x??n????n??n??1???ex???lim????n??ne????1? ???n????????e????????????xn??n??????1?x?x??e????1?x?x??e?limn??nex????1??n????1???limx?n?n??nex
?e?e?????????????n??1?x?x1?x2ex?1lim?n??e?n??x?x2ex?1lim?1?t?t?et?0t
nt?ln(1?t)010?1?t?t?1?t2x?1t2?xelimt?01 ?x2exlimt?(1?t)ln(1?t)t?0(1?t)t2 ?x2exlimt?(1?t)ln(1?t)t?0t2
00?x2exlim1?1?ln(1?t)t?02t
??x2ex2。
2、求48?1?x?xx(1?x8)dx.
解: ?1?x4?x8x4?x82?x4x(1?x8)dx?11?2?x2(1?x8)dx2?1?1?x2x(1?x)4dx
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?14?1?x?x2424x(1?x)dx?214?1?x?x22x(1?x)dx
1??3???1?ABC?1?212????????dx?dx???4?x?1xx?1?4?x?1xx?1????1?31??ln(x?1)?lnx?ln(x?1)??C4?22??38ln(x?1)?2
??114lnx?18ln(x?1)?C.
?ex23、求?0dy?y??ey??x12??dx. ????dx???11解:
?ex2ydy?e?0?y?x??11?10dy?1yex2xdx??10dy?edxy1y2
???10dx?x2x0ex2xdy??dy?edx0yy2
xedx?3x2?10edx??10(1?y)edy?y2?10e?12.
4、求过(1,2,3)且与曲面z?x?(y?z)的所有切平面皆垂直的平面方程.
解:令F(x,y,z)?x?(y?z)3?z
则Fx?(x,y,z)?1,Fy?(x,y,z)?3(y?z)2,F?(x,y,z)??3(y?z)z2?1
令所求平面方程为: 在曲面z?则A?C?0A(x?1)?B(y?2)?C(z?3)?0,
x?(y?z)3上取一点(1,1,1),则切平面的法向量为{1,0,?1},
x?(y?z)?03在曲面z?上取一点(0,2,1),则切平面的法向量为{1,3,?4},
则A?3B?4C解得:
.
x?y?z?6A?B?C即所求平面方程为: .
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二、(15分)设f(x)?e解: 当x?0, 当x?0,
0x?x36,问f(x)?0有几个实根?并说明理由.
e?0?xx36
x3e?0且ex的增长速度要比
36来得快!所以f(x)?0无实根.
三、(满分20
解: 当
??n?分)求??x??n?1??中x20的系数.
333x?1时,
??x??1?n?3 x???x????????1?x??1?x??n?1??1???1?x??x3??3???xn????x?????2?n?0?2n?2
?3x3?2?n(n?1)xn?2
??n?故??x??n?1?中x20的系数为171.
y?z?R222四、(20分) 计算?Cxyds,其中C是球面x2?的交线.
解： ?C(x?y?z)2ds??C(x2?而?C(x?y?z)2ds?0,
与平面x?y?z?0y?z)ds?2?(xy?yz?zx)ds
22C??C(x?y?z)ds?xyds?222?CRds?2?R23,
C?Cyzds??Czxds,
故?Cxyds???R33.
n五、(20分)设a1,a2,?,an为非负实数,试证:?aksinkxk?1?sinx的充分必要
条件为?kakk?1n?1.
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证明：必要性 由于?aksinkxk?1nn?sinx,则?akk?1nsinkxx?sinxx,
x?0
?limx?0?k?1aksinkxxn??k?1kak?limsinxxx?0?1.
nn充分性;要证明?aksinkx?k?1sinx,只需证明:
?ak?1ksinkx?1sinx,这里
sinx?0,若sinx?0,不等式显然成立;
即只需证明: ?akk?1nsinkxsinxsinkxk?1,
而?akk?1nsinkxsinxn??ak?1sinx,?kakk?1n?1
故只要说明: 当ksinkxsinx?k,即sinkx?ksinx,
?1时,显然成立;
?n时,也成立,即sinnx?nsinx假设当k当k;
?n?1时, sin(n?1)x?sin(nx?x)?sinnxcosx?sinxcosnx?sinnx?sinx?(n?1)sinx.
1sinnx?nsinx 六、(15分)求最小的实数c,使得满足?0f(x)dx?1的连续函数f(x)都有
?10f(x)dx?c.
解: ?01f(x)dx??f(x)dx?2?tft()dx0011011?2?ft()dx01?2,
23?43 取y?2x,显然?0f(x)dx?1,而?f(x)dx?1?102xdx?2?,
取y?(n?1)xn,显然?0而?10f(x)dx?1,
f(x)dx??10(n?1)?x?ndx?2?n?1n?2?2,n??,
故最小的实数c?2.
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2007浙江省高等数学(微积分)竞赛试题(解答)
一.计算题（每小题12分，满分60分） 1、求?x9x5dx.
?1解: 95?x5dx?1xdx5?1t
x?15?x5?15?dtt?1u?t?1?1115?u?1udu?5?udu?15?udu
…… 此处隐藏54字 ……
(1?x)x?(1?2x)2xlim(1?x)x?(1?2x)2xx?0sinx?limx?0x
00?lim?1x?0?(1?x)x?1ln(1?x)?1???2x?1ln(1?2x)???x(x?1)x2??(1?2x)???1)??x(2x2x2??? ?00?lim?1?x?(x??1?2x?(2x?1)ln(1?2x)??x?0?(1?x)x1)ln(1?x)??2x?x2(x?1)??(1?2x)???2x2(2x?1)??? ?11?lim(1?x)x?x?(x?1)ln(1?x)??2xx?0??x2(x?1)??lim(1?2x)2x?(2x?1)ln(1?2x)??x?0??2x2(2x?1)???elimx?(x?1)ln(1?x)?(2x?1)ln(1?2x)x2?elim2xx?0x?02x2
00?elim?ln(1?x)ln(1?2x)x?02x?elim?2x?04x
??e?e?e22.
3、求bp的值,使?a(x?p)2007e(x?p)2dx?0.
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