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2014年华约数学自主招生真题

发布时间:2014-04-14 13:02:35  

2014cu??gì?)ê?áK9??)‰

2014/03/01

1!x1,x2,x3,x4,x5′????ê§???o?ùú|¤??8ü?{44,45,46,47}§|ùê?ê")‰μê?ê???o??±????ê?ú?§??7kü?ú??ó"??ùê?ú??ú?4(x1+x2+x3+x4+x5)§?4????ꧤ±ù??ó??ú???U′46§l??

x1+x2+x3+x4+x5=44+45+46+47+46=574

ùê?ê?O?57?44=13,57?45=12,57?46=11,57?47=10,57?46=11§=10,11,11,12,13"

1)§`I??'m??V?′q§2!???¥'m§ê?n‘?"???`‘???V?′p(p>|p?????§q?p???????"

)‰μe??'m??3?§K`I??'m??V??p3?e??'m??4?§K?????`‘§

23`I??'m??V??C3p(1?p)?e??'m??5?§K?????`‘§`I??'m??V?

23?C4p(1?p)2§?d

2323p(1?p)2p(1?p)+C4q=p3+C3

2323p(1?p)2?p=6p5?15p4+10p3?pp(1?p)+C4q?p=p3+C3

1??f(p)=6p5?15p4+10p3?p(p>)§Kf??(p)=30p4?60p3+30p2?1=30(p2?p+????111111112????)(p?p?)§??p∈(,+1?)§f(p)>0?p∈[+?,1]§f(p)≤0"1¤±f(p)3(,1]kO??~§??f(p)????????

p=3!?êf(x)=|a,b"√1+2??11?4>0)??????1§?????-4§π(cosx?sinx)sin(x+)?2asinx+b(a

11(cos2x?sin2x)?2asinx+b=?sin2x?2asinx+b+§ˉK??dug(t)=)‰μf(x)=1?t2?2at+b+3[?1,1]t?????ú?????O?1ú-4"e???é??t0=?a?[?1,1]???é ??1μ

e?a≤?1=a≥1§Kg(t)3[?1,1]tü~§

g(?1)=1,g(1)=?4

5)??a=,b=?1"

1

e?1<?a<0=0<a<1§K

g(?a)=1,g(1)=?4

)??a=?1±√????¤"

5,b=?1"nt§a=4!£1¤y2y=f(g(x))????ê?y=g?1(f?1(x))?£2¤F(x)=f(?x),G(x)=f?1(?x)§eG(x)????ê′F(x)§y2f(x)???ê"

y2μ£1¤y=f(g(x))????ê?x=f(g(y))§??

f?1(x)=f?1(f(g(y)))=g(y)

g?1(f?1(x))=g?1(g(y))=y

¤±y=g?1(f?1(x))?y=f(g(x))????ê"

£2¤duG(x)????ê′F(x)§??

G(F(x))=G(G?1(x))=x

f(x)=f(G(F(x)))=f(G(F(x)))=f(f?1(?F(x)))=?F(x)=?f(?x)

¤±f(x)???ê"

5!??y??x+2y2=1???x2+y2=b2§Ly??t?:M?????ü^??§?:?O

?P,Q§??PQ?x?§y??O??u:E,F§|S?EOF??????"

)‰μ??M(acosθ,bsinθ)(θ∈[0,2π))§??PQ?:M'u??x2+y2=b2??4?§ù?§?

acosθ·x+bsinθ·y=b2

l??

bb2

,yF=xE=acosθsinθ

S?EOF

√b3b31≥=|xE||yF|=2a|sin2θ|a√??…=??M?I?(±a,±b)t???ò¤á"

6!??ê??{an}÷vμa1=0,an+1=npn+qan"£1¤eq=1§|an?£2¤e|p|<1,|q|<1§|yμê??{an}k."

)‰μ£1¤an+1?an=npn§K

an=a1+

??p=1?§an=

??p=1?§

pan=p2+2p3+···+(n?1)pn

an?pan=p+p2+p3+···+pn?1?(n?1)pn

an=p(1?pn?1)n?1??k=1n(n?1)?(ak+1?ak)=p+2p2+···+(n?1)pn?1?(n?1)pn

1?p

2(n?1)pn+1?npn+p=(1?p)2

£2¤|an+1|=|npn+qan|≤|npn|+|qan|≤n|p|n+|an|¤±§|an+1|?|an|≤n|p|n§u′

|an|≤|p|+2|p|+···+(n?1)|p|2n?1(n?1)|p|n+1?n|p|n+|p|=(1?|p|)2

??(n?1)|p|n+1?n|p|n≤(n?1)|p|n?n|p|n=?|p|n<0§¤±

|an|<

=yank."

xnx)e≤x2"7!??n∈N+,x≤n§|yμn?n(1?x)·e)n"??x2≥n?§t????a?>???§)‰μ?????a??dun?x2≤n((1?x|p|(1?|p|)???a¤á???x2<n?§duey≥1+y(y≥0)9???????a(1+y)n≥1+ny£ù¥n≥1§y>?1¤§l??

xxxnx2nx2xnn((1?)·e)≥n((1?)·(1+))=n(1?2)≥n(1?n·2)=n?x2

nnnnn

=y"

3

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