2005年安徽省高中数学竞赛初赛

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2005(6,36)1.m1,a=m+1-m,b=m-m-1.,().(A)ab(B)ab(C)a=b(D)abm2.(x2+3x+2)5,x().(A)160(B)240(C)360(D)80013.1,ABCD-A1B1C1D1,EFAB1BC1(),AE=BF.,4:AA1EF;A1C1EF;EFA1B1C1D1;EFA1C1.().(A)(B)(C)(D)4.:11,21,12,31,22,13,41,32,23,14,,10,2005a2005().(A)1a20052(B)12a20051(C)0a200512(D)a200525.,,.,().(A)1013(B)1011(C)1010(D)10016.A={-2,0,1},B={1,2,3,4,5},f:ABxAx+f(x)+xf(x).f().(A)45(B)27(C)15(D)11(9,54)7.y=f(x)y=f-1(x),y=f(x-1)(3,3).y=f-1(x+2).8.x2a2+y2b2=1(ab0)A,B,F,ABF=90..9.S-ABC,MNSBSC.AMNSBC,.10.M={x|x=limn2n+1-2n+2n,,+20}.M.11.f(x)=4sinxsin24+x2+cos2x.|f(x)-m|26x23,m.12.6m+2n+2(mnN),(m,n)=.13.(20)f(x)=ax2+bx+c(abcR),x[-1,1],|f(x)|1.(1):|b|1;(2)f(0)=-1,f(1)=1,a.14.(20)a0,p=(1,62©1994-2007ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.),q=(0,a),M(0,-a)p+qN(0,a)p+2qR,R.(1)R;(2)a=22,F(0,1)lRAB,FAFB.15.(20){an},a1=t,a2=t2,t0,x=tf(x)=an-1x3-3[(t+1)an-an+1]x+1(n2).(1){an};(2)12t2,bn=2an1+a2n(nN+),:1b1+1b2++1bn2n-2-n2.1.B.a=1m+1+m,b=1m+m-1,m+1mm-10,ab.2.B.(x2+3x+2)5,x,42,3x,xC45324=240.23.D.2,EPABP,FQBCQ,PQ.,PEQF.PEAC,PEPQ,PEFQ.,AA1EF,EFAC,EFA1C1..4.C.11,21,12,31,22,13,41,32,23,14,.a2005n.n(n-1)22005n(n+1)2,n=63.,a2005632005-63622=52.a2005=1252=313.5.A.k(2k10)Ck10.10k=2Ck10=10k=0Ck10-C110-C010=1013.6.A.x=-2,x+f(x)+xf(x)=-2-f(-2),f(-2)1,3,5,3;x=0,x+f(x)+xf(x)=f(0),f(0)1,3,5,3;x=1,x+f(x)+xf(x)=1+2f(1),f(1)1,2,3,4,5,5.,335=45.7.(1,2).f-1(3)=2,f-1(1+2)=2.(1,2).8.5-12.BF2+AB2=AF2,c2+b2+b2+a2=(a+c)2.ca2+ca-1=0.ca=5-12.9.arccos66.AMNSBC,MNG,AG,AGMN.BCD,SDAD,SDA.a,SA=AD=32a.,SD=22a,GD=24a.cosSDA=24a32a=66.10.3.||2,x=limn22n-21n1+2n=0;7220063©1994-2007ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.=2,x=limn1-12n=1;||2,x=limn2-212n2n+1=2.11.(1,4).f(x)=4sinx1-cos2+x2+cos2x=2sinx(1+sinx)+1-2sin2x=1+2sinx.6x23,|f(x)-m|2,f(x)-2mf(x)+2.(f(x)-2)maxm(f(x)+2)min.f(x)max=3,f(x)min=2.1m4.12.(0,0),(1,0),(1,3).m2,n2,6m+2n+2=2(36m-1+2n-1+1),,.m1n1.m=0,6m+2n+2=2n+3,n2,2n+33(mod4),2n+3,n=01,,(m,n)=(0,0);m=1,6m+2n+2=2n+8,n4,2n+8=8(2n-3+1),n=0,1,23.(m,n)=(1,0),(1,3).n=0,1,(m,n)=(0,0),(1,0).13.(1)f(1)=a+b+c,f(-1)=a-b+c,b=12(f(1)-f(-1)).|f(1)|1,|f(-1)|1.|b|=12|f(1)-f(-1)|12(|f(1)|+|f(-1)|)1.(2)f(0)=-1,f(1)=1,c=-1,b=2-a.f(x)=ax2+(2-a)x-1=ax-a-22a2-(a-2)24a-1.|f(-1)|1,|2a-3|1,1a2.,a-22a=12-1a[-1,1].,fa-22a1.(a-2)24a+11.(a-2)24a0,(a-2)24a+11,(a-2)24a+1=1,a=2.14.(1)R(x,y),MR=(x,y+a),NR=(x,y-a).p+q=(,a),p+2q=(1,2a),MR(p+q),NR(p+2q),(y+a)=ax,y-a=2ax.,R(y+a)(y-a)=2a2x2,y2-2a2x2=a2((0,-a)).(2)a=22,R2y2-2x2=1(0,-22).l,x=0,lA0,22.l,y=kx+1.2y2-2x2=1,2(k2-1)x2+4kx+1=0.k2-10,=16k2-8(k2-1)0,k1.A(x1,y1)B(x2,y2),x1x2=12(k2-1).FAFB=(x1,y1-1)(x2,y2-1)=(x1,kx1)(x2,kx2)=x1x2+k2x1x2=k2+12(k2-1)=121+2k2-1.-1k1,k2-10,k=0,FAFB-12;k1k-1,k2-10,FAFB12.FAFB(-,-12](12,+).15.(1)f(t)=0,3an-1t-3[(t+1)an-an+1]=0,an+1-an=t(an-an-1)(n2).t1,{an+1-an}t2-tt.,82©1994-2007ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.=(t2-t)tn-1,an+1-tn+1=an-tn,an-tn=a1-t=0.an=tn(nN+),t=1.(2)1bn=12an+1an=12(tn+t-n).12t2,,(2t)n1,tn2n.(2n+2-n)-(tn+t-n)=1(2t)n(2n-tn)[(2t)n-1]0.1bn12(2n+2-n).1b1+1b2++1bn122+12+22+122++2n+12n=122(1-2n)1-2+121-12n1-12=2n-121+12n2n-12212n=2n-2-n2.:y=tn+t-nt12,2.()2005()(5,30)1.p:|x+1|2,q:13-x1.?p?q().(A)(B)(C)(D)2.1(12).5().(A)68(B)132(C)133(D)260112345673.ax2+bx+40{x|-2x1},y=bx2+4x+a[0,3]().(A)0,-8(B)0,-4(C)4,0(D)8,014.1,f(x)[-4,4],g(x)=af(x)+b.g(x),().(A)a0,g(x)(B)a=-1,-2b0,g(x)=02(C)a0,b=2,g(x)=0(D)a1,b0,g(x)=05.ABC,x=cosA+cosB+cosC,y=sinA2+sinB2+sinC2.xy().(A)x=y(B)xy(C)xy(D)9220063©1994-2007ChinaAcademicJournalElectronicPublishingHouse.Allrightsreserved.

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