6¶:OyÒ:¤3:)Œ Ò:;’:
—µ‚‰KžØ‡‡Ld‚
183¥IŒÆ)êÆ¿mýmÁò
(êÆa, 2014c10)
Á/ª:4òÁžm: 150©¨÷©: 100©
KÒ ˜ n o Ê 8 o©
֩ 15 15 15 15 20 20 100
©
5¿: 1.¤k‰KÑL3dÁò’—µ‚m>,3Ù§’þ˜ÆÃ.
2.—µ‚†>žÕ‰K,—µ‚ Øk6¶9ƒ'IP.
3.X‰K˜xØ,Œ3¡,¿I²KÒ.
˜!(K15©)®˜mü^†‚
l1:
x 4
1
=
y 3
2
=
z 8
1
;
l2:
x+ 1
7
=
y+ 1
6
=
z+ 1
1
:
1¤y²l1Úl2É¡¶
2¤¦l1Úl2úR‚IO§¶
3¤¦ël1þ?˜:Úl2þ?˜:‚ã¥:;,˜„§"
(1)yyy²²²µl1þk:r1= (4;3;8)§••þv1= (1; 2;1)"
l2þk:r2= ( 1; 1; 1)§••þv2= (7; 6;1)"
q
(r1 r2; v1; v2) =
5 4 9
1 2 1
7 6 1
6= 0;
l1Úl2É¡" (3©)
(2)l1þ?˜:P1=r1+t1v1†l2þ?˜:P2=r2+t2v2ë‚••
þ
!
P1P2=r2 r1+t2v2 t1v1
= ( 5 + 7t2 t1; 4 6t2+ 2t1; 9 +t2 t1):
úR‚••þ
v=v1v2=
i j k
1 2 1
7 6 1
= (4;6;8):
11£5¤
d
!
P1P2†v²1µ
( 5 + 7t2 t1) : ( 4 6t2+ 2t1) : ( 9 +t2 t1) = 4 : 6 : 8
t1= 1; t2= 0:
:r2+ 0v2= ( 1; 1; 1)3úR‚þ§lúR‚IO§
x+ 1
4
=
y+ 1
6
=
z+ 1
8
:
(9©)
(3)P1=r1+t1v1†P2=r2+t2v2¥:
1
2
(3 +t1+ 7t2;2 2t1 6t2;7 +t1+t2):
Ïd¥:;,˜‡²¡§²¡{•þ
v=v1v2= (4;6;8):
q:
1
2
(3;2;7)3²¡þ§;,§
4x+ 6y+ 8z 40 = 0:
(15©)
!(K15©)f2C[0;1]´šKî‚üNO¼ê"
1¤y²µé?¿n2N§3˜xn2[0;1]§¦
f(xn)
n
=
Z
1
0
f(x)
n
dx:
2¤y²µlimn!1xn= 1:
yyy²²²:1¤
f(0)
n
Z
1
0
f(x)
n
dxf(1)
n
;
dëY¼ê0Š5Ÿxn35" (3©)
duf´î‚üN¼ê§xn´˜" (5©)
2¤é?¿ >0,dfšK5ÚüN5§
f(xn)
n
Z
1
1
f(1 )
n
=
f(1 )
n
;
f(xn)
n
p
f(1 );
l
lim inf
n!1
f(xn)f(1 ):
12£5¤
6¶:OyÒ:¤3:)Œ Ò:;’:
—µ‚‰KžØ‡‡Ld‚
dfüN5§
lim inf
n!1
xn1 :
d?¿5, limxn= 1. (15 ©)
n!(K15©)V4«m[0;1]þN¢¼ê¤¢•þ˜m§Ù¥
•þ\{†Xþ¦{þÏ~"f1; : : : ; fn2V.y²±eü^dµ
1¤f1; : : : ; fn‚5Ã'¶
2¤9a1; : : : ; an2[0;1]¦det(fi(aj))6= 0,ùpdetL1ª.
yyy²²²2))1).ħ1f1+ +nfn= 0.òa1; : : : ; an©O“\§§
| 8
>
>
<
>
>
:
1f1(a1) + +nfn(a1) = 0
.
.
.
1f1(an) + +nfn(an) = 0
5¿þã§|XêÝ(fi(aj))
T
,Ïdddet(fi(aj))6= 0†1=
=n= 0. (6 ©)
1))2).^8B{.Äk§n= 1ž§(Øw,"
Ùg§n=kž(Øý"Kn=k+1ž§df1; : : : ; fk+1‚5Ã'§f1; : : : ; fk
‚5Ã'.Ïd9a1;
2014年数学竞赛预赛(数学专业)答案