
2
()
1
2
2
11 11 1
(1) (1)( 1)
1
22 22 2
11 (1)
22! !
mm
m
tt t t
m
−
− −− − −− −−+
−=−+ ++− +
L
LL
()
1
22 4 2
2
11 11 1
(1) (1)( 1)
1
22 22 2
11 (1)
22! !
mm
m
xx x x
m
−
− −− − −− −−+
−=−+ ++− +
L
LL
故
[]
()
(2004)
1
(2005)
2
2
1002
0
0
arcsin 1 2004!
x
x
xxb
−
=
=
⎡⎤
=− = ⋅
⎢⎥
⎣⎦
, 其中
1002
1002
11 1
( )( 1) ( 1001)
2003!!
22 2
(1)
1002! 2004!!
b
−−− −−
=− =
L
由
[
(2005)
(2008) 3 3
2008
( ) ( ) arcsin ( )
xC x x xhx
′′′
=+
知
()
(2004)
1
(2008) 3 2
2
2008
0
(0) 3! 1 0
x
fCx
−
=
⎡⎤
=− +
⎢⎥
⎣⎦
3
2008
2003!! 2008! 2003!!
3! 2004! 2004!
2004!! 2005! 2004!!
C==
2003!!
2008 2007 2006 2004!
2004!!
=⋅⋅
2
2008 2007 2006 (2003!!)=⋅⋅⋅
4.求
44 2 2
(, ) 4 2 2 2
xy x y x xy y=+−− −
的极值。
解
3
(, ) 16 4 2 2 0
x
fxy x x y=−− =(1)
3
(, ) 4 2 2 2 0
y
fxy y x y=− −=(2)
(2)式乘
2 减(1)得
33
42( 22 ) 0yx−=,即 2yx= ,因此再由(1)式,得
2
1
(, ) 16( ) 0
2
x
fxy xx=−=
,即
12 3
22
0, ,
22
xx x== =−
,解得
1
3
2
1
3
2
2
2
0
,,
2
2
0
1
1
x
x
x
y
y
y
⎧
⎧
=
⎧
=−
=
⎪⎪
⎨⎨ ⎨
=
⎩
⎪⎪
=−
=
⎩
⎩
因此驻点为
22
(0,0),( ,1),( , 1)
22
−−
。易见
2
(, ) 48 4
xx
fxy x=−
, (, ) 22
xy
fxy=− ,
2
(, ) 12 2
yy
fxy y
−
判别式为
222
(, ) (, ) (, ) (48 4)(12 2) 8
xx yy xy
A f xyf xy f xy x y=−=−−−
当
22
(, ) (0,0),( ,1),( ,1)
22
xy=−−
时,判别式分别为 0, 192, 192A
,因