第三届全国大学生数学竞赛决赛试卷及解答(数学类)

2012

(

15

)

A(1,0,1),B(1,1,2),C(1, 1, 2),D(3,1,0),E

(3,1,2).

E

A,B,C

Σ

AD

ABC

→ →

n=AB×AC=(0,1,1)×(0, 1, 3)=( 2,0,0).

l=(a,b,c),

l·n=0,

l· AD →

=0.

l=(0,c,c)(c=0),

l=(0,1,1).

x 3z 2

1

=

A1=A2=···=An=0

f(x)

[0,2π)

2k02k0

A1,A2,···,An

f(x)[0,2π)

F1

1(x)=

k2sinkix

.

i

F1(0)=F1(2π)=0,

F′′1(x)=f(x).

F1(x)[0,2π)

N,

Rolle

F1′

(x)

N

F1′′

(x)

(0,2π)

N 1

N

F0(x)=f(x).F( 1)

s

s(x)=

k2sinkixi

s

,Fs′′+1(x)=Fs(x),s=0,1,2,···,

Fs(x)

[0,2π)

f(x)[0,2π)

N.

g(x)=sink0x+

nAik20

s

i=1

k2s

g(x).

·<kn,

k0<k1<··0

n|Ai|k20

s

2

i=1

k2sinkixi

s

n|Ai|k20s

2

i=1

4

2

k2s>0,

i

mπ πg

k

√0

2

n+

|Ai|k20

s

i=1

4

2

k2s

>0,

i

(0,2π)

F1′′

(x)

[0,2π)

N,

s,

g

mπ+k0

π

√2

+

n |Ai|k2s

i=1

44

n→∞

an=1,

a1a2···an=1.

n→∞

lim

a1+a2+···+an

n→∞

xn=0,

n

n k=1

xk=0.

xn A.

xn 0.

A>0,

0 x ln2

ex 1+2x.

n,

Sn={i∈Z|1 i n,xi ln2},Tn={i∈Z|1 i n,xi>ln2},

1

nk∈T

n

=0.

xk>

|Tn|

n

1

nk∈S

n nk∈S

n

exk+

1

|Tn|2

(1+2xk)+

n

n k=1

(1 eA)+

1

n

e

xk

e

n→∞

lim

1

n

n k=1

exk=1.

z=

xn,xn<0,n=1,2,···.

n0,

xn 0

limzn=0.

n→∞

yn=xn+zn,yn 0,

lim

1

nlim→∞yn<+∞,nlim

→∞

n

neyk=1.

zn 0,xn yn,

k=1

1

e

exk

1

n

nk=1

nn

ak=nlim

1

k=1

→∞

2

2BA.

AB

λ(λ 9)2.

AB

BA

)

BA

9.

BA

C,

CBA=BAC=I2.AB

λ(λ 9),

A(BA 9I2)B=ABAB 9AB=AB(AB 9I3)=0,

BA 9I2=CB·A(BA 9I2)B·AC=0,

BA=9I2=

9009

.

9 6

01

,

(

2

(

20

)

F

{Ai}i∈I,{Bi}i∈Ii∈I

FP

F

P 1AiP=Bi, i∈I.

Q

Q

{Ai}i∈I,{Bi}i∈I,

R

i∈I,

Mn(R)

R

Ui,Q={T∈Mn(Q)|AiT=TBi},

Ui,R={T∈Mn(R)|AiT=TBi}

Q

n

Mn(R)

Mn(Q)

Mn(Q)

UR=

UR=0.

i∈I

Ui,R,UQ=

i∈I

Ui,Q,

n2,

UR

UQ

Ui,R

Ui,Q

Ui,R

Ui,Q

(

AiT=TBi

)

T1,T2,···,Tl.

l

f(t1,t2,···,tl)=dettkTk,

k=1

UR=0f

Q

s1,s2,···,sl

f(s1,s2,···,sl)=0,

f(r1,r2,···,rl)=0,Bi.

P=

l k=1

r1,r2,···,rl

rkTk∈UQ

i∈I,P 1AiP=

(

15

)

F(x),G(x)

[0,+∞)

x→+∞

limx(F(x)+G(x))=0.xF(xt)costdt=0.

(i)(ii)

ε>0,lim

x→+∞

+∞ε

n→∞

lim

+∞

(F(t) G(t))cos

t

2

ε

β

xF(xt)costdt xF(αx)|cost|dt→0,x→+∞,

α

π

α=minε,

x→+∞

lim

2

+∞

π

.

xF(xt)costdt=

2

∞

π

xF(xt)costdt

3π2

=x

k=02+kπ∞ k=0

( 1)k

F(x(t+kπ))costdt.

+∞ π

2π

2

+∞

π

→0,x→+∞,

x→+∞

lim

2

.

lim

ε0

n→∞

n(G(nt) G((n+1)t))costdt=0.

n(G(nt) G((n+1)t))costdt0 nε0 (n+1)ε0

uu

=G(u)cosG(u)cos

n+100

nε0 nε0

uu

G(u)cosG(u)cos

n+10n+10

n

n+1

du

=I1+I2 I3,

ε0

n+1

du

x→+∞

limxG(x)=0,limG(x)=0,

x→+∞

|I1|=

=

1

n

1n

du

ε0

1n

ε0

G(nu)cosudu

G(nu)cosudu+G(nu)cosudu

√G(0)

+G(n

n kε0 n+1

k=1

√

(k 1)ε0

G(u)cos

u

n

n(n+1)

du

)

=

n

du(Lagrange

n k=1

ε0

2(n+1)2

nε0

n

n k=1

G((k 1)ε0)

(k 1)ε0G((k 1)ε0)+

nε20G(0)

ε0

nε0

G((n+1)u)cosudu nG(nε0)

n+1

n+1

→0,n→∞,

ε0

n→∞

limn(G(nt) G((n+1)t))costdt=0.

( )

I1

x→+∞

lim

ε0

(F(xt) G(xt))costdt=0,

ε0

x→+∞

lim[x]

[x]

x

(F(xt) G(xt))costdt=0,

[x]=n,

F(x),G(x)

n x<n+1,

ε0n(F(xt) G(xt))costdt 0ε0 ε0

n(F(nt) G(nt))costdt+n(G(nt) G((n+1)t))costdt,

0ε0n(F(xt) G(xt))costdt 0ε0 ε0

n(F((n+1)t) G((n+1)t))costdt n(G(nt) G((n+1)t))costdt,

00

n→∞

limn

+∞

(F(nt) G(nt))costdt=0,

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