2023年北京高考数学仿真卷九(含详解)

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5ᜓᡊ2023ÈᓅÊËὃᦪÍÎÏᔁ
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6/
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13ᡠt|SA"SB=(a2)2=/,ᡠt](x+஻)2+y2][(x-a)2+y2]=/,ᡠt](x+a)(x-a)]2+y2(2x2+2a2)+y4=a4,ᡠt1/)2+2k3+2022+4=/,ᡠtx*+y4+2x2y2+2a2y2=2a2x2,ᡠt+)2=2/12),ᦑঝ(x2+y2)2=2a2(x2-y2)„2a2x2,¢(x?+y?>..x4,x4„2a2x2,ᡠt¤,2/,ᡠt-¥¦ky/2a,ᦑজ§YZ[ḄWXz{x¨yḄᢣᦪª\ᏔᦪᡠtYZ[CᐵTᙶ᪗VᙠPḄWX¤8(®¯),ᓽᨵ(°2+/2)=2/(±-”2),ᑣP(m,n)ᐵTG[y=x.VḄ(n,m),ᓽ(w2+m2)-2a2(n2-m2),ᡠt>=஻2_ᶭBp2=n2,mᡠtT·¸[¤8Ḅ(¹஻)ºᡠtYZ[ºᐵTG[y=xVᦑঞº⌕¼|t|=|PB|,ᑣ|PA=|PBᡠt(x+a)2+y2=*_஻)2+y2,ᓽx=0,

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1839Q(X=2)=C^()2(_)°=Z,-5525ᡠXḄᑖGᑡ^X0124129p252525ᦪopqẚ=0x§+1XU+2X2=9,2525255ᡈEX=2x?=9.55(3)ᵫ⚪ª«:X]~5(2,3,x_B(2,|),2ᡠEXLEX?.19.(14ᑖ)¯°ᦪ/(*)=஺-1)[0^+(4+1)*+1],ᐸaeR.(1)Q¶·y=/(x)ᙠ(1,f(1))ᜐḄᑗ·/Ḅ»᳛4,X¼ᦪ஺Ḅr(2)½஻>0ÀQ°ᦪf(x)ᙠx=X]ᜐ
«᩽ᜧXÃ^f{x)..0^x(3)Q°ᦪ/(x)ាᨵkÅÆÇḄÈᑏÉÊᩩÌḄᡠᨵaḄ.ூᫀ௃(1)a=1(2)Í᪆(3)஺=ᡈ஺=1ᡈa=—1ூ௃^/(%)=(x-l)[av2+(a+l)x+1]=ar3+x2-ar-1,:.f(1)=a+l-a-l=O,0/(x)XÔ«f'M=3ax2+2x-a,:[(1)=3a+2—a=2a+2,.•¶·y=/(x)ᙠ(1,f(1))ᜐḄᑗ·/Ḅ»᳛4,2tz+2=4,a=1.

19(2)ÃØ^ff(x)=3ax2+2x-a,,//\=b2-4ac=4+12a2>0,f(x)°ᦪᨵkÅÆÙÇḄÚg(x)k᪷X],x,ÜÝ0,½%£(%,%2)À/(1)<0½XW(%2+࠷>)À/'(X)>O,.,.½X=MÀ/(%)
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21o,=pqr^^srᢣu.6224ᡠMᱏ==᳝ᡠMx2yz={={o|…2+|~ᡠMQ=PM,B[j\PQM*Ὺ,-..iiWM%,y?ᐭ2*±Xx⌲ᡝ0.PQ,PM,QM᳛ᑖT°,G%kQM%ᡠM%=3=9”=3X,x-x,x+᳝22EPQ_LPM,ᡠM2-&M=-1.+᱄o3U-->->x+%x-᳝2225^-+y
=1因6£-112,ᡠM毛ὃ-X
6-+£=i6ᡠM%/,kQM=~~OᡠM9=6,ᓽPQ1QMḄ2᳛346,kQMEP/0ᙊCḄ⚔,ᡠMPQ,PM,QMḄ2᳛ᙠ¡/0,WPMḄFGy=kx+m(kmw0),

22y=kx+mᵫ2p1+622/+“ᑗ¦+6m2-6=0,x—+y=116ᵫ¨>(),ᡠM6ª+1-«>0.WP(%,)¬(sy),PMḄ7(%)%).272kmE¯/=1+6°ᡠM/=±=²%=[+,"=&EOT//QM,ᡠM¯=%=£=■oEPQJ_PM%ᡠMlip஺=—.Pk121.(15ᑖ)·¸¹Ḅº»ᔣ½1¾¿ᦪᑡA:q,a,Äᙠx,,...»x,Æ|%s_||=4z=1,2,...,%ᡂÉᐸ᳝/],i=0,kk1....k,ᑣÍᦪᑡAΓÏᐭ”º»[m,t].(I)ᑖᢣÐÑᑡᦪᑡ0ᔲΓÏᐭ”º»[0,2];জA'2,3
ঝ4'1,0,1.

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