2004至2010年浙江省高等数学（微积分）竞赛工科类试题

(2) limun?limvn?lim?n??n??11??1?????

n??n?1n?23n????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：

（1） 通过C，E两点的直线绕z轴旋转所得的旋转曲面方程；

（2） 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?4?y?2xDM(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???z??z?MQ?x?y?NQ???2????2?

?2???2??2222 21

z2?8， 化简得：所求的旋转曲面方程为：x?y?2?222（2）A(0,0,4?)，故过A(0,0,4?)垂直z轴的平面方程为：z?4?

z2?8， 令x?0，解得在坐标面yoz上的曲线方程为：y?22?2图中所求的旋转体的体积为： V??4?0?z????8?dz ?2?2???22z?z?????2?8?dz 0?2??4?2z?4???4?0zdz?32?2 2?2yB32?2128?22??32??.

33xz2x?y?2?82?22x2?yz222四、(20分) 求函数f(x,y,z)?2,在D?{(x,y,z)1?x?y?z?4}的最大22x?y?z值、最小值.

2x(x2?y2?z2)?2x(x2?yz)2xy2?2xz2?2xyz解： fx?(x,y,z)? ?(x2?y2?z2)2(x2?y2?z2)2z(x2?y2?z2)?2y(x2?yz)zx2?z3?2yx2?y2z fy?(x,y,z)? ?22222222(x?y?z)(x?y?z)y(x2?y2?z2)?2z(x2?yz)yx2?y3?2zx2?z2y fz?(x,y,z)? ?22222222(x?y?z)(x?y?z)由于x,y具有轮换对称性,令x?y, x?0或y?z?0 解得驻点: (0,y,y)或(x,0,0)

x2?yz1x2?yz对f(0,y,y)?2?, f(x,0,0)?2?1,

x?y2?z22x?y2?z2在圆周x?y?z?1上,由条件极值得: 令F(x,y,z)?x?yz??(x?y?z?1)

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2222222Fx?(x,y,z)?2x?2?x?0

Fy?(x,y,z)?z?2?y?0 Fz?(x,y,z)?y?2?z?0

F??(x,y,z)?x2?y2?z2?1?0

解得: (0,22222222,),(0,,?),(0,?,?),(0,?,),(1,0,0),(?1,0,0) 22222222221,)?222,

f(0,f(0,221,?)??222,

f(0,?221,?)?222,

f(0,?221,)??,f(1,0,0)?1,f(?1,0,0)?1; 222在圆周x2?y2?z2?4上,由条件极值得: 令F(x,y,z)?x2?yz??(x2?y2?z2?4)

Fx?(x,y,z)?2x?2?x?0

Fy?(x,y,z)?z?2?y?0 Fz?(x,y,z)?y?2?z?0

F??(x,y,z)?x2?y2?z2?4?0

解得: (0,2,2),(0,2,?2),(0,?2,?2),(0,?2,2) ,(2,0,0),(?2,0,0)

111,f(0,2,?2)??,f(0,?2,?2)?, 2221f(0,?2,2)??,f(2,0,0)?1,f(?2,0,0)?1;

2f(0,2,2)?x2?yz222,在f(x,y,z)?2D?{(x,y,z)1?x?y?z?4}的最大值为1,最小值22x?y?z为?1. 2五、(15分)设幂级数数的和函数.

?axnn?0?n的系数满足a0?2,nan?an?1?n?1,n?1,2,3,?,求此幂级

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???证明：S(x)??axnn?1??1n?S?(x)?n?1x?

n?0?nan?1nx?n?1?an?1?(n?1)xnn?1? ??an?nnx??S(x)?n?0?nxn?0??nxn

n?0???而

?nxn?xn?1?x??xn???x????xn??n?0???x?1??xn?0?nxn?0n?0???1?x???(1?x)2, 即: S?(x)?S(x)?x(1?x)2 一阶非齐次线性微分方程---常数变易法,

求S?(x)?S(x)?0的通解: S(x)?cex, 令S(x)?c(x)ex代入S?(x)?S(x)?x(1?x)2得: c(x)ex?c?(x)ex?c(x)ex?x(1?x)2,

即: c(x)??xxe?(1?x)2exdx????1??x?1?x???xe?xdx?1?x??11?x?xe?x??dx ?xe?xxe?x1?x????e?x?dx?1?x?e?x?c 故S?(x)?S(x)?x(1?x)2的通解为: S(x)???xe?x?1?x?e?x?c????ex?11?x?cex, ?由于S(0)?0,解得c??1, 故

?anxn的和函数S(x)?1n?0?x?ex1. ?1???x?? ?1?1?x?????1?x????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(x2?1)f(x2)?f2??x1?x?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(xf(x2x1?x2?1)2)?f???2??,?x1,x2?R,

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