文科数学答案解析

20232023.03123456789101112ACBDDDACAACC5913.23;14.8+4;15;16120cm.6017ana3a8=4a2ancn=Tn{cn}nana1,18anan=cn==()3133122n124n2.18122022.ABAB10060106[010)[1020)[2030)[3040)[4050)[5060]AB28B[010)2[1020)3[2030)5[3040)15[4050)40[5060]35A100A30B[020)221[010)ABAAABmnBabcmnmambmcnanbncabacbcmambmcnanbncPAA38BBB19PABMNPNM2ANNPANBMAN3BM=1CDABNPCDPBMPABMNMNQCQDQDQDQMPDQBMPMPBMPDQBMPABMNCQMBCQBMPMBBMPCQBMPDQCDQCQCDQDQCQQCDQBMP48CDCDQCDBMPABMNBBEMNANEAEBAE2BE2AB2=AE2+BE2AEBEBEMNANNMANNPNMNPNNM,NPNMPANNMPANANMBANMBNMPPNM2PNMPABMN582012·19.OCF(,0)F3.ACxB,DCBDxCPPl1l2l1l2Cl1l2?.68b1Cx2y24.Bm,nDm,n3m3A2,0ABm2,nADm2,n222m443ABADm2n2m24m41m24m3m3332ABAD0,743.P(s,t)s2t24.t1l1,l20l1l2.P(s,t)lklytk(xs)Cx2+3[kx+(tks)]2=3(3k21)x26k(tks)x3(tks)2307836k2(tks)24(3k21)[3(tks)23]0(3s2)k22stk1t203s20l1,l2k1,k2k1,k2l1l2.CPl1l2.21f(x)=lnx+2x+()f(x)()g(x)xf(x)3x2mn>1,g(m)+g(n)>0()x>0x>f'(x)>000g'(x)=lnx+2x1,g(x)=,x>g(x)>08800g(x)0,.ng(n)>g(1)=0g(m)+g(n)>001n>,g(n)>g(),g(m)+g(n)>g()+g(m)=(m+)(mm+)0g()+g(m)=(m+)(mm+)>0g(m)+g(n)>0,mn>1g(m)+g(n)>0.9822.4-4xOyClx+yxClClABABOx1cosysincosx1sinycos2sin212Cx1y21xcos,ysinl1sin42C2cosl1080,04OA·OB23.4-5fxxxm,(mR)m1f(x)2xfxx3mm1fxxxfx2x1x1x2fx2x4x4fx2x1x13x2fx2x3x1fx2x1x1x2fx2x0x1fx23,4x3,4118x4xmx44m2x4x3,4x32x4m4,10128