浙江省2+2历届高等数学答案及评分标准

2005年高等数学（A）答案及评分标准：

一. 填空题 1. 14 2. 0 3.

???x?y2 4.

12ln11?x?y?x?c

5.

?9 6.2432 7.F(32?z) 8.

1112

二. 选择题 1. C 2.

A 3.

D 4.

B

5.

D 6. B 7. A 8. C

三. 计算题

t?11.解: 原式x?lim1(1?1t)t?0t2tsin 2’

?limt?sintt?0t3 3’

?lim1?costt?03t2 5’

?limsint1t?06t =6 7’

xlimf(x)?lim(11)?lime?x?12.解: 因

x?0x?0x?ex?1x?0(ex?1)x ?limex

?x?1?1f(0)x?0x22??f(x)在x?0 处连续. 1?1?1'(0)?limf(x)?f(0)?limxexf?12x?0xx?0x

xx=lim2(e?1)?2x?x(e?1)x?02x2(ex?1) 1

’ 3’ 1

=lim

2ex?2?x?xe2x3xx?0=limex?1?xe6x2xx?0

??112 5’

f'(x)??1x2?e(exx2?1)2,

?12x?0 7’

3.解: 原式?

??lnxd[(x?1)] 2’

dx??lnxx2??x??1x2 4’

)??lnxx2?1?1d(1x?arcsin1x?c??lnxx2?11?2?11x2 7’

?14.解:

I???xcosD10x0yxd????xeD?y2d? 1’

?y2??dx?xcos2yxd(yx1101y)??dy?xey22dx2 5’

dy]

=?x2sin1dx01?12[?e0y2?102dy??ye?y222?y2

?131313sin1?1212121[?e0?12dy??yd(e0?y2)]

?y221?sin1?[?e0?1?y22dy?(ye2|0??e011dy)]

=

sin1?e2 . 7’

x3n?15.解: 令

?S(x)??n?1(3n?2)!?xg(x)

x?R 1’

其中

?g(x)??x3n?2n?1(3n?2)!3n

x?Rg(0)?0

?g'(x)??xn?0(3n)!

g'(0)?1

?g''(x)??x3n?1n?1(3n?1)!2

?

g''(x)?g'(x)?g(x)?eg(0)?0,g'(0)?1x , 4’

3i?1?23i

????1?0?x2????1?2,??

, 5’

解得:

g(x)?e2(c1cos3x?c2sin3x)?ex223g(0)?0,g'(0)?1

?c1??13,c12?3?xx)?x[e2(?13?3sin3x)?exS(3cos2x323]. ??120?11?6.解:

?11111????342a?2b?3? 2’

???5a?8315????120?11?

??0?1120???0a?10?2b? 4’

?0???000a?1b??a??1,b?R ?可由?1,?2,?3,?4线性表示. 5’??1000a?b?1??a?1?1000?

??0??0010?2b??a?1? 6’

??1b??000a?1??? ??a?b?12ba?1?1?a?1?3?ba?1?4

7’

7.解: ?

(AP)T(AP)?PTA2P 1’

3

7’

A2?5??4??0???0A2450000100??0? ?0??1? 2’

? 的特征值为: 1,1,1,9. 3’

??1的特征向量: (?1,1,0,0)T,(0,0,10)T,(0,0,0,1)T

??9的特征向量: (1,1,0,0)T

??1?100??22????9?P??11?00?A2P??1???? ?22?

PT?1?0010??????1???0001??8.解:

fX(x)???2e?2x,x?0,

?f?1,x?[1,3]Y(y)?? ?0, x?0?2?0, 其它

f,y)??? e?2x, x?(0,??),y?[1,3](X,Y)(x0, ? 其它 P?X?Y?3??P?(X,Y)?D?

23?x???f(x,y)d???dx?e?2xdy D00

2??(3?x)e?2xdx?514 04?4e?. 9.解: Y 1 2 3 X 1 2642 442 42 2 4242

42

642

4

5’

7’

2’

3’ 4’ 6’

7’

3

642

642

642 4’

X 1 2 3 P

四. 1. 解: 由

f'(x)?1xf(x)?3x1242

1242

1842 6’

157E(X)?12?24?5442? 7’

得

2f(x)?x(3x?c) 3’

9510V(c)???f(x)dx???x(3x?c)dx??(0212c23?32c?) 5’

V'(c)??(

?94?9432?2c3令)?0?c??94 6’

V''()?2?3?0 7’

c?为唯一的极小值点,为最小值点

94).

? f(x)?x(3x?2. 解: 由

?1???2??1?112?32?1??11????1???0??0??0??T010?53?4302??3?1?3?0??? 2’

?(II)的通解: c1(5,4,3,0)?c2(2,1,0,?3)T 3’

5

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