2023年北京高考数学仿真卷3(含详解)

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20(2)Xa=g,Ycdxe(0,1)e>/(x)<00(3)Xfg(Ḅhᦪxe(0,+=o),/(x)..Oឤᡂj>Y”Ḅᨬᜧ3.ூoᫀ௃(1)y=x(2)rs᪆(3)--12ூus௃s(1)d஺=0e>fM=ex-x2-1,ᑣr(x)=e*-2x,/(0)=1,x/(0)=0,.••ᑗ[`az=/0(2)c{(c|1)d“=~e>f(x)=ex-x2-x-i,2dxw(0,1)e>f(x)<0oe*<*2+x+1=1<'+'+1ex•n.,,x?+x+l஻ஹtஹmil>/ஹ(2x+l)e*—(x~+x+l)e*x-g(x)=——0-,xe(0,l).ᑣg\x)=-------------1X---------=—e(e')z.d0g'(x)>0,g(x)ᓫ⌴>(ᑮg(0)=l,.•.dxe(0,l)e>g(x)>g(0)=l,ᡂj..,.dX€(0,1)e>/(X)<0:(c|2)da=1e>/(x)=er-x2-x-1,xe(0,l).2ᑣf'()=e*2x-1,f"(x)=ex-2Zx/"(x)=0,s=2.d00.ᦑx=/஻2f(x)Ḅ᩽3]>/(Ml)=e"12-2ln2-\=\-2ln2<0.(IJf'(0)=0,/(1)=e—3<0.

21,dxw(O,l),r(x)<0,e/(O)=o0/.dX£(0/)>/(x)<00(3)ᵫ(2))*>da.~e>>—X?—2ar—L,e"—X?—x—1vOᙠ0vxvlឤᡂj;2.,de>/(X)..O)ឤᡂj>.•.4V0,VXG(O,-Ko)9ex-x2-2ar-l..Oe*../+2ax+1L'+2—+1exஹ7>ஹx2+2ar+l/7ஹg(x)=-------0——(x>0),eᑣg>஺+2◥+2.)J*+Sy+(2…g^o,sᵫ⚪஺Ḅᨬᜧ3>ᡃ¡ᐜᩭẆ¥0¯◤gপ>,1ᓽ);´•µ1)=+"·na--1»••91>0,Ḅᨬᜧ3)ᙠ0ᑮ>e22.,fg(ḄhᦪX£(0,”)>/(X)..Oឤᡂj>4Ḅᨬᜧ31-1.r221v20.(14ᑖ)O*¸ᙊC:)+4=13>"0)Ḅºᯖ]6(1,0),¼᳛.A¸ᙊ஺Ḅ½⚔]>P,a-b-2஺¸ᙊC¿AḄºÀ]>Á[AP,AQ«Á[/:x=4ᑖ·ÂM,N].

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