2009年北京市中学生数学竞赛高一年级初赛试题及参考解答

2009(36,,6,0)1.1[-4,6]y=f(x)y=g(x).f(x)g(x)().(A)[-1,3](B)[-4,3][5,6](C)[-4,2](D)[-4,-1][3,6]1:(D).2.sin12,cos13,cos14().(A)sin12cos13cos14(B)sin12cos14cos13(C)cos13sin12cos14(D)cos14cos13sin12:(B).3124,sin120.413144.5,1314,,cos13cos140sin12.(B).3.ABC,AB=6,AC=8,BAC=90.ADBEBCAC.ADBE().(A)1365(B)32(C)-1365(D)0:(C).2A,2.A(0,0),B(6,0),C(0,8),D(3,4),E(0,4).AD(3,4),BE(-6,4),cos(AD,BE)=3(-6)+4432+42(-6)2+42=-25213=-1365.4.log2sin12+log2sin6+log2sin512().(A)-3(B)-1(C)1(D)3:(A).log2sin12+log2sin6+log2sin512=log2sin1212sin512=log212sin12cos12=log2142sin12cos12=log214sin6=log218=-3.5.f(x)=x2+bx+c(x0),-2(x0),f(4)=f(0),f(1)=-1,f(x)=x().(A)0(B)1(C)2(D)3:(D).f(4)=f(0),f(1)=-1,16+()4b+c=c,1+b+c=-1,b=-4,c=2.f(x)=x2-4x+2(x0),-2(x0),3,3.36.a,f(x)=(x+a)3x-2+a2-(x-a)38-x-3a,a().(A){0,5}(B){-2,5}(C){-5,2}(D){1,2009}:(C).f(x)=(x+a)3x-2+a2-(x-a)38-x-3a,f(-a)=f(a),2a38+a-3a=2a3a-2+a2.,a=0,8+a-3a=a-2+a2.a=0,f(x)=x(3x-2-38-x).a0.8+a-3a=a-2+a2,a=-5,a=2,,f(x)=-(5-x)323+x-(5+x)323-xf(x)=(2+x)32+x(2-x)32-x.(64,8)1.b1,b2,b3,,b59,b60b2b1=b3b2=b4b3==b59b58=b60b59,logb11b50(b1b2b60).:30.b2b1=b3b2=b4b3==b59b58=b60b59=q,b2=b1q,b3=b1q2,b4=b1q3,,b58=b1q57,b59=b1q58,b60=b1q59.logb11b50(b1b2b60)=logb1q10b1q49(b1q0b1q1b1q59)=logb21q10+49(b601q0+1++59)=logb21q59(b21q59)30=30.2.sin(+)=0.8,cos(-)=0.3,(sin-cos)(sin-cos).:-0.5.(sin-cos)(sin-cos)=(coscos+sinsin)(sincos+cossin)=cos(-)-sin(+)=0.3-0.8=-0.5.3.ABCDA(1,2),B(2,5),C(7,3),D(5,1),ABCD.:13.5.4,A,B,C,D,ABCD,M(1,1),N(1,5),P(7,5),T(7,1),ABCDMNPTAMDANBBPCCTD,4SMNPT=24,SAMD=2,SANB=1.5,SBPC=5,SCTD=2,,SABCD=24-(2+1.5+5+2)=13.5.4.[x]x,[log36]+[log37]+[log38]++[log3()(n-1)]+[log3n]=2009,n.:474.[log36]=[log37]=[log38]=1,[log36]+[log37]+[log38]=3,[log39]=[log310]==[log326]=2,[log39]+[log310]++[log326]=36,[log327]=[log328]==[log380]=3,[log327]+[log328]++[log380]=162,[log381]=[log382]==[log3242]=4,[log381]+[log382]++[log3242]=648,3+36+162+648=849=2009-11602009,[log3243]=[log3244]==[log3728]=5,[log3243]+[log3244]++[log3728]=24301160,243243n728.n728.849+5(n-242)=2009,n=474.5.ABC,AB=BCAC,AHAMBC,SAMHSABC=38.cosBAC.:cosBAC=14.ABC,AB=BCAC,AHBC,BM=MC,BDACD.AB=BC=a,SABC=12AHBC=12AHa,SAMH=12AHMH,SAMHSABC=38,12AHMH12AHa=38,MH=38a,HC=MC-MH=12a-38a=18a.BAC=,BCA=.RtBCDDC=acos.ABC,AC=2DC=2acos.RtACHcos=HCAC=18a2acos,cos2=116.,cos=14.6.f(x)=2x2-2x-1g(x)=-5x2+2x+3,y=ax+b.a-b.:-1.f(x)=2x2-2x-1g(x)=-5x2+2x+3y=2x2-2x-1y=-5x2+2x+3,7y=5(2x2-2x-1)+2(-5x2+2x+3)=-6x+1.y=-67x+17,a=-67,b=17,a-b=-1.7.6027abcabcabcabc2009abc91.abc.:1092.91=713,1001=71113,91|1001.abcabc=1001abc,91|abcabc.abcabcabcabc2009abc91,abcabcabcabc2009abc=abcabcabcabc21004abc1001+abc,abcabcabcabc21004abc91,abc91.91182,()910,abc=182+910=1092.58.5,ABCD,ABC=120,BC=63,PBCC,APCDE,BEDPQ.PQBP=15,QBP=35,PQ.:43.6,BD,ABDAPF,BF,DF,FCCQ,DFB=DFP=BFP=120,BFE=ECP=120,B,F,E,C,1=2,DFP=DCP=120,6D,F,C,P,2=3,1=3,B,C,Q,D.QBCD.,PBCC,QBCDCDC.BCD63,633223=6,26=12.QBCD35-15=20,QBCD40,19,,PQQ1Q2129=43.7()(39)(a1+a2++an)(1a1+1a2++1an)(1+1++1)2n1=n2,a1+a2++ann21a1+1a2++1an=n2(1+23)+(1+232)++(1+23n)=n2n+23+232++23n=n2n+1-(13)nn2n+1.,.,,.,.()