抽象函數(shù)模型歸納總結(jié)（八大題型）（解析版）

(1)對(duì)于正比例函數(shù)fx=kxk≠0，與其對(duì)應(yīng)的抽象函數(shù)為fx±y=fx±fy.(2)對(duì)于一次函數(shù)fx=kx+bk≠0，與其對(duì)應(yīng)的抽象函數(shù)為fx±y=fx±fy?b.(3)對(duì)于二次函數(shù)fx=ax2+bx+ca≠0，與其對(duì)應(yīng)的抽象函數(shù)為fx+y=fx+fy+2axy-c(4)對(duì)于冪函數(shù)fx=xn，與其對(duì)應(yīng)的抽象函數(shù)為fxy=fxfy.(5)對(duì)于冪函數(shù)fx=xn，其抽象函數(shù)還可以是f=.(6)對(duì)于指數(shù)函數(shù)fx=ax，與其對(duì)應(yīng)的抽象函數(shù)為fx+y=fxfy.11(7)對(duì)于指數(shù)函數(shù)f(x(=ax，其抽象函數(shù)還可以是f(x-y(=.(8)對(duì)于對(duì)數(shù)函數(shù)f(x(=logax，與其對(duì)應(yīng)的抽象函數(shù)為f(xy(=f(x(+f(y(.(9)對(duì)于對(duì)數(shù)函數(shù)f(x(=logax，其抽象函數(shù)還可以是f=f(x(-f(y(.(10)對(duì)于對(duì)數(shù)函數(shù)f(x(=logax，其抽象函數(shù)還可以是f(xn(=nf(x(.(11)對(duì)于正弦函數(shù)f(x(=sinx，與其對(duì)應(yīng)的抽象函數(shù)為f(x+y(f(x-y(=f2(x(-f2(y(注：此抽象函數(shù)對(duì)應(yīng)于正弦平方差公式：sin2α-sin2β=sin(α+β(sin(α-β((12)對(duì)于余弦函數(shù)f(x(=cosx，與其對(duì)應(yīng)的抽象函數(shù)為f(x(+f(y(=2ff(13)對(duì)于余弦函數(shù)f(x(=cosx，其抽象函數(shù)還可以是f(x(f(y(=[f(x+y(+f(x-y([(14)對(duì)于正切函數(shù)f(x(=tanx，與其對(duì)應(yīng)的抽象函數(shù)為f(x±y(=1已知f(x+y(=f(x(+f(y(-1且f(1(=2，則f(1(+f(2(+?+f(n(不等于1A.f(1(+2f(1(+?+nf(1(-B.f+n-1C.D.n(n+1(【解析】∵f(x+y(=f(x(+f(y(-1，∴f(x+y(-1=[f(x(-1[+[f(y(-1[，構(gòu)造函數(shù)g(x(=f(x(-1，則g(x+y(=g(x(+g(y(，且g(1(=f(1(-1=1，令an=g(n(=f(n(-1，則a1=f(1(-1=1，22令x=n，y=1，得g(n+1(=g(n(+g(1(，n+1=an+a1=an+1，即an+1-an=1，=1+(n-1(×1=n，∴f(n(-1=n，則f(n(=n+1.f(1(+f(2(+?+f(n(=2+3+?+(n+1(===，f(1(+2f(1(+?+nf(1(-=f(1(-=n(n+1(-=，合乎f+n-1=+1+n-1=，合乎題意；故選D. 2已知函數(shù)f(x(的定義域?yàn)镽，且f≠0，若f(x+y)+f(x)f(y)=4xy，則下列結(jié)論錯(cuò)誤的是()A.f-=0B.f=-2【解析】對(duì)于A，令x=、y=0，則有f+f×f(0(=f[1+f(0([=0，即f(0(+ff（-=-1，由f(0(=-1，可得ff（-=0，-,則f對(duì)于D，有f（x+1-=-2(x+1(=-2x-2，即f（x+=-2x-2，+2x-3，f(0(=-1，則不等式f(x(>3-2x的解集為()【解析】令x=y=0，得f(-1)=f(0)?f(0)+f(0)-3=-3.令y=0，得f(-1)=f(x)f(0)+f(0)+2x-3，解得f(x)=2x-1，33則不等式f(x)>3-2x轉(zhuǎn)化為2x+2x-4>0，1+21-4=0，所以不等式f(x)>3-2x的解集為(1,+∞).4f2022等于()4令y=y0得，fxfy0=xy0，即fx=y0x，于是?x,y∈R，fxfy=fxy0y=y0xy0y=yxy=xy，則y0=±1，從而fx=±x，有f2022=2022.11且f(-2)=1，則f2nn∈N*=()A.4n+6B.8n-1C.4n2+2n-1D.8n2+2n-5【解析】令x=y=0，則f0=f0+f0+1，所以f0=-1，令x=y=-1，則f-2=f-1+f-1+2+1=2f-1+3=1，所以f-1=-1，令x=1,y=-1，則f0=f1+f-1-2+1=f1-2=-1，所以f1=1，令x=n,y=1,n∈N*，則fn+1=fn+f1+2n+1=fn+2n+2，所以fn+1-fn=2n+2，則當(dāng)n≥2時(shí)，fn-fn-1=2n，則fn=fn-fn-1+fn-1-fn-2+?+f2-f1+f1=2n+2n-2+?+4+1=+1=n2+n-1，所以fn=n2+n-1n∈N*，所以f2n=4n2+2n-1n∈N*.2(2024·山東濟(jì)南·三模)已知函數(shù)fx的定義域?yàn)镽，且yfx-xfy=xyx-y，則下列結(jié)論一定2A.f1=1B.fx為偶函數(shù)C.fx有最小值D.fx在[0,1[上單調(diào)遞增44【解析】由于函數(shù)fx的定義域?yàn)镽，且yfx-xfy=xyx-y，令y=1，則fx-xf1=xx-1，得fx=x2+[f1-1[x，x=1時(shí)，f1=12+[f1-1[恒成立，無法確定f1=1，A不一定成立；由于f1=1不一定成立，故fx=x2+[f1-1[x不一定為偶函數(shù)，B不確定；由于fx=x2+[f1-1[x的對(duì)稱軸為x=-?[f1-1[與[0,1[的位置關(guān)系不確定，故fx在[0,1[上不一定單調(diào)遞增，D也不確定，由于fx=x2+[f1-1[x表示開口向上的拋物線，故函數(shù)fx必有最小值，C正確，3(2024·陜西西安·模擬預(yù)測(cè))已知函數(shù)f(x)的定義域?yàn)镽，且滿足f(x)+f(y)=f(x+y)-2xy+2,3A.f(4)=12B.方程f(x)=x有解C.f【解析】對(duì)于A，因?yàn)楹瘮?shù)f(x)的定義域?yàn)镽，且滿足f(x)+f(y)=f(x+y)-2xy+2,f(1)=2，取x=y=1，得f(1)+f(1)=f(2)-2+2，則f(2)=4，取x=y=2，得f(2)+f(2)=f(4)-8+2，則f(4)=14，故A錯(cuò)誤；對(duì)于B，取y=1，得f(x)+f(1)=f(x+1)-2x+2，則f(x+1)-f(x)=2x，所以f(x)-f(x-1)=2(x-1),f(x-1)-f(x-2)=2(x-2),?,f(2)-f(1)=2，以上各式相加得f(x)-f(1)==x2-x，所以f(x)=x2-x+2，令f(x)=x2-x+2=x，得x2-2x+2=0，此方程無解，故B錯(cuò)誤.對(duì)于CD，由B知f(x)=x2-x+2，所以fx+=x+2-x++2=x2+是偶函數(shù)，fx-=x-2-x-+2=x2-2x+不是偶函數(shù)，故C正確，D錯(cuò)誤.4(2024·河南·三模)已知函數(shù)fx滿足：f1≥3，且?x,y∈R，fx+y=fx+fy+6xy，則4fi的最小值是()【解析】由fx+y=fx+fy+6xy，得fx+y-3x+y2=fx-3x2+fy-3y2，令gx=fx-，得gx+y=gx+gy，令x=n,y=1，得gn+1-gn=g1，故gn=[gn-gn-1[+[gn-1-gn-2[+???+[g2-g1[+g1=ng1，又gn=fn-所以fn=gn+3n2=3n2+[f1-3[n，所以fi=3i2+[f1-3[i=855+45[f1-3[，因?yàn)閒1≥3，當(dāng)f1=3時(shí)，fi的最小值5 1已知函數(shù)fx的定義域?yàn)?∞,0∪0,+∞,且xfx=y+1fy+1，則()A.fx≥0B.f1=1C.fx是偶函數(shù)D.fx沒有極值點(diǎn)【解析】令gx=xfx，則gy+1=y+1fy+1，所以gx=gy+1，且x,y+1為定義域內(nèi)任意值，故gx為常函數(shù).所以fx≥0不恒成立，f1=1不一定成立，A、B錯(cuò). 2(2024·河北·模擬預(yù)測(cè))已知定義在-∞,0∪0,+∞上的函數(shù)fx滿足fxy=+A.fx是奇函數(shù)且在0,+∞上單調(diào)遞減B.fx是奇函數(shù)且在-∞,0上單調(diào)遞增C.fx是偶函數(shù)且在0,+∞上單調(diào)遞減D.fx是偶函數(shù)且在-∞,0上單調(diào)遞增【解析】令x=y=-1，則f1=-2f1+1，所以f1=，令x=y=1，則f1=2f-1+1，所以f-1=-，令y=-1，則f-x=-f-x+-=-f-x+-=-f-x-，所以f-x=-，令y=1，則fx=f-x++=--+=，所以fx=，因?yàn)閒-x=-=-fx，且定義域關(guān)于原點(diǎn)對(duì)稱，所以函數(shù)fx是由反比例函數(shù)的單調(diào)性可得函數(shù)fx=在0,+∞上單調(diào)遞減. 1(多選題)(2024·山西晉中·三模)已知函數(shù)fx的定義域?yàn)镽，滿足fx+y=fxfy+fx+fy，且f0≠-1，f1>-1，則下列說法正確的是()A.f0=0C.若f1=1，則f4=15B.fx為非奇非偶函數(shù)D.fx>-1對(duì)任意x∈N*恒成立66【解析】我們有恒等式：fx+y+1=fxfy+fx+fy+1=fx+1fy+1.f0=0，故A正確；對(duì)于B，由于fx=0滿足條件且是偶函數(shù)，所以fx有可能是偶函數(shù)，故B錯(cuò)誤；對(duì)于C，由恒等式可得fx+1+1=fx+1f1+1，故f4+1=f3+1f1+1=f2+1f1+12=f1+14.若f1=1，則f4=f1+14-1=24-1=15，故C正確；對(duì)于D，由恒等式可得fx+1+1=fx+1f1+1.從而再由f1+1>0可知fx+1>0x∈N*，即fx>-1x∈N*，故D正確.2已知函數(shù)fx滿足，fp+q=fp?fq,f1=3，則++++的值為()2【解析】∵fp+q=fp?fq,∴fn+1=fn?f1,∵f1=3∴fn+1=3fnn-1=3n因此++2+3234+3436+3638+38=+++7=6+6+6+6+6=30+f23+f6f5+10+310+9+f24+f8f7+f25+f10f93 如果fa+b=fafb且f1=2，則++=()3A.B.C.6D.8【解析】∵f1=2，fa+b=fafb，∴f2=f1f1，f4=f3f1，f6=f5f1，∴=f1，=f1，=f1，∴++=3f1=6，4已知函數(shù)fx對(duì)一切實(shí)數(shù)a,b滿足fa+b=fa?fb，且f1=2，若an=4，則數(shù)列{an{的前n項(xiàng)和為()A.nB.2nC.4nD.8n77∴f(n+1(=f(n(?f(1(=2f(n(*所以an===4所以數(shù)列{an{的前n項(xiàng)和為4n.1(多選題)已知函數(shù)f(x(的定義域?yàn)镽，f(xy(=y2f(x(+x2f(y(，則().A.f(0(=0B.f(1(=0C.f(x(是偶函數(shù)D.x=0為f(x(的極小值點(diǎn)因?yàn)閒(xy)=y2f(x)+x2f(y)，對(duì)于A，令x=y=0，f(0)=0f(0)對(duì)于C，令x=y=-1，f(1)=f(-1)+f(-1)=2f(-1)，則f(-1)=0，令y=-1,f(-x)=f(x)+x2f(-1)=f(x)，因?yàn)閒(xy)=y2f(x)+x2f(y)，對(duì)于A，令x=y=0，f(0)=0f(0)對(duì)于C，令x=y=-1，f(1)=f(-1)+f(-1)=2f(-1)，則f(-1)=0，令y=-1,f(-x)=f(x)+x2f(-1)=f(x)，當(dāng)x>0肘，f(x)=x2lnx，則f,(x(=2xlnx+x2?=x(2lnx+1)，1-1令f,(x(<0，得0<x<e-2；令f,(x(>0，得x1-1--上單調(diào)遞減，882.已知定義在0,+∞上的函數(shù)fx，滿足fxy+1=fx+fy，且f=0，則f211=()2A.1B.11C.12D.-1【解析】令x=y=1，則f1+1=f1+f1，解得f1=1，令x=2,y=，則f1+1=f2+f，解得f2=2，令x=y=2，則f22+1=f2+f2，解得f22=3，令x=22,y=2，則f23+1=f22+f2，解得f23=4，??,依次類推可得f211=12。33xfy+yfx成立，則下列說法中正確的有()個(gè).①f0=f1=0；②若當(dāng)x>1時(shí)，fx>0，則函數(shù)gx=在0,+∞單調(diào)遞增；*，fxn=nxn-1fx；④若f=-，則=2n-2.A.1B.2C.3D.4【解析】令x=y=0有f0=0，令x=y=1有f1=0.所以①正確.0<x1<x2，因?yàn)間x=，所以gxy===+=gx+gy，所以gx2=gx1=gx1+g又因?yàn)?gt;1，且當(dāng)x>1時(shí)，fx>0，所以gx2-gx1=g=f>0.所以②正確.n=gx+gxn-1，所以gx=gxn-gxn-1，所以gx=gxn-gxn-1=gxn-1-gxn-2=??=gx2-gx，99fx,所以fxn=nxn-1fx，所以③正確.=nx累加得gxn=gx+n-1gx=ngfx,所以fxn=nxn-1fx，所以③正確.=nx令x= 2=令x= 2=f1=0，又因?yàn)閒=-,所以f2=2，f2+2f由③得fn=2n-1f2=2n，=由③得fn=2n-1f2=2n，i=1f2ini=2i=i=1f2ini+fyx24(2024·山西·一模)已知函數(shù)fx是定義在{xlx士0{上不恒為零的函+fyx24yA.f1=1B.f-1=1C.fx為偶函數(shù)D.fx為奇函數(shù)令x=y=-1，則f1=2f-1，故f-1=0，B選項(xiàng)錯(cuò)誤； =fx，故fx為偶函數(shù)，C選項(xiàng)正確；令y=-1，則f-x=f =fx，故fx為偶函數(shù)，C選項(xiàng)正確；因?yàn)閒x為偶函數(shù)，又函數(shù)fx是定義在{xlx士0{上不恒為零的函數(shù)，D選項(xiàng)錯(cuò)誤.11A.f(x)為偶函數(shù)B.f(3)=-2C.f(-1)=f(5)D.f(k)=-2【解析】令x=y=0，則f20=0才f0=0.另令x=0，則fyf-y=-f2y，由f1=2，所以fy=0不成立，所以f-y=-fy，所以函數(shù)fx為奇函數(shù)，故A錯(cuò)誤；令x=2，y=1，則f3f1=f22-f21才f3=-2，故B正確；令x=3，y=2，則f5f1=f23-f22才f5=2，又f-1=-f1=-2，所以f-1=-f5，故C錯(cuò)；令y=2得fx+2fx-2=f2x-f22=f2x.且f1=2，f3=-2，f5=2.所以f7f3=f25才f7=-2；f9f5=f27才f9=2；f11f7=f29才f11=-2…所以f2k+1=-1kx2，又f0=0，f2=2，所以f6f2=f24才f4=0；f8f4=f26才f6=0…所以f2k=0；所以f1+f2+f3+f4=f5+f6+f7+f8=…=f4k+1+f4k+2+f4k+3+f4k+4=0所以fk=fk-f1=0-f1=-2，故D正確. 2(多選題)(2024·全國(guó)·模擬預(yù)測(cè))已知函數(shù)fx的定義域?yàn)镽，且f(x+y)f(x-y)=f2(x)-f2(y),A.f(x)為偶函數(shù)B.f(3)=-1C.f(-1)=-f(5)D.f(k)=12A-sin2B=sin(A+B)sin(A-B).sin2A-sin2B=(sinA+sinB)(sinA-sinB)=sin++sin-?（sin+-sin-===sin(A+B)sin(A-B).因?yàn)閒(3)=-1，故選項(xiàng)B正確.因?yàn)閒(-1)=-1=-f(5)，故選項(xiàng)C正確.因?yàn)門=4,2023÷4=505??3，故f(k)=f(1)+f(2)+f(3)=0，故選項(xiàng)D錯(cuò)誤.令x=0，則f(y)f(-y)=f2(0)-f2(y)，又f(y)不恒為0，故f(-y)=-f(y)，又因?yàn)閒(x)為奇函數(shù)，所以f(-1)=-f(1)=-1，故C正確.對(duì)于選項(xiàng)D，由選項(xiàng)B以及f(x+2)f(x-2)=f2(x)，可得f(7)=-1,f(9)=1,f(11)=-1，所以f(2k+1)=(-1)k，同理可得f(2k)=因?yàn)?023÷4=505??3，故f(k)=f(1)+f(2)+f(3)=0，故D錯(cuò)誤.1(多選題)已知定義域?yàn)镽的函數(shù)fx滿足f(x+y)=f(x)?f(y)-f(2-x)f(2-y)，且f0≠0,f-2=0，則()1A.f(2(=1B.f(x(是偶函數(shù)C.[f(x)]2+[f(2+x)]2=1D.f(1(+f(2(+f(3(+?+f(2024(=1【解析】A.f(x+y(=f(x(?f(y(-f(2-x(?f(2-y(，令y=2，則f(x+2(=f(x(?f(2(-f(2-x(?f(0(=-f(2-x(，所以函數(shù)f(x(關(guān)于(2,0(對(duì)稱，令x=y=2，則f(4(=f(2(?f(2(-f(0(?f(0(=-1，令y=4，且f(-2(=0，則f(x+4(=f(x(?f(4(-f(2-x(?f(-2(=-f(x(=-f(-x(，所以f(x(=f(-x(，又函數(shù)f(x(的定義域R，所以函數(shù)f(x(為偶函數(shù)，故B正確；令y=-x，則f(0(=f(x(?f(-x(-f(2-x(?f(2+x(，又f(0(=1,f(x(=f(-x(,f(x+2(=-f(2-x(，所以[f(x([2+[f(2+x(]2=1，故C正確；因?yàn)閒(x+4(=-f(x(，所以f(x+8(=-f(x+4(=f(x(，所以函數(shù)f(x(的一個(gè)周期為8，令x=2,y=1，則f(3(=f(2(?f(1(-f(0(?f(1(=-f(1(，所以f(3(+f(1(=0，所以f(-3(+f(-1(=0，所以f(5(+f(7(=f(-3(+f(-1(=0，f(6(=f(-2(=f(2(=0,f(8(=f(0(=1，所以f(1(+f(2(+f(3(+f(4(+f(5(+f(6(+f(7(+f(8(=[f(1(+f(3([+[f(5(+f(7([+f(2(+f(4(+f(6(+f(8(=0+0+0-1+0+1=0，所以f(1(+f(2(+f(3(+?+f(2024(=253×[f(1(+f(2(+f(3(+?+f(8([=0，故D錯(cuò)誤. 題目2(多選題)(2024·遼寧·二模)已知定義城為R的函數(shù)f(x(．滿足f(x+y(=f(x(f(y(-f(1-x(f(1-y(，且f(0(≠0，f(-1(=0，則()A.f(1(=0B.f(x(是偶函數(shù)C.[f(x([2+[f(1+x([2=1D.f(i(=-1【解析】對(duì)于A項(xiàng)，由f(x+y(=f(x(f(y(-f(1-x(f(1-y(，令x=y=，則f(1(=f2-f2=0，故A項(xiàng)正確；對(duì)于B項(xiàng)，令x=y=0，則f(0(=[f(0([2-[f(1([2=[f(0([2，f(0(=1，令y=1，則f(x+1(=f(x(f(1(-f(1-x(f(0(=-f(1-x(①,所以函數(shù)f(x(關(guān)于點(diǎn)(1,0(成中心對(duì)稱，令x=y=1，則f(2(=[f(1([2-[f(0([2=-1，令y=2，則f(x+2(=f(x(f(2(-f(1-x(f(-1(=-f(x(②,由①可得：f(x+2(=-f(-x(③,由②③可知：f(-x(=f(x(，且函數(shù)f(x(的定義域?yàn)镽，則函數(shù)f(x(是偶函數(shù)，故B項(xiàng)正確；對(duì)于C項(xiàng)，令y=-x，則f(0(=f(x(f(-x(-f(1-x(f(1+x(，因?yàn)閒(0(=1，f(-x(=f(x(，f(x+1(=-f(1-x(，代入上式中得，故得：[f(x([2+[f(1+x([2=1，故C項(xiàng)正確；對(duì)于D項(xiàng)，由上可知：fx+2=-fx，則fx+4=-fx+2=fx，故函數(shù)fx的一個(gè)周期為4，故f4=f0=1，令x=2,y=1，則f3=f2f1-f-1f0=0，所以f1+f2+f3+f4=0+-1+0+1=0， f(3x+1)=-f(-3x+1)，則f(k)=()A.-2B.-1C.0D.1【解析】由題意知函數(shù)f(x)的定義域?yàn)镽，且f(x+y)+f(x-y)=2f(x)f(y)，f(0)=1，令x=0，則f(y)+f(-y)=2f(y)，即f(-y)=f(y)，故f(x)為偶函數(shù)；又f(3x+1)=-f(-3x+1)，令x=0，則f(1)=-f(1),∴f(1)=0，又由f(3x+1)=-f(-3x+1)，得f(x+1)+f(-x+1)=0，f(x+1)+f(-x+1)=0，即f(x+2)=-f(-x)，又結(jié)合f(x)為偶函數(shù)，則f(x+2)=-f(x)，故f(x+4)=-f(x+2)=f(x)，即4為f(x)的周期，故f(3)=f(-1)=f(1)=0，f(4)=f(0)=1故f(k)=f(0)+[f(1)+f(2)+?+f(2024)]=1+506[f(1)+f(2)+f(3)+f(4)]=1+506[0-1+0+1]=1，4(2024·安徽·模擬預(yù)測(cè))若定義在R上的函數(shù)fx，滿足2fx+yfx-y=f2x+f2y，且f1=-1，則f0+f1+f2+???+f2024=()A.0B.-1C.2D.1【解析】令x=y=，則有2f1f0=f1+f1，又f1=-1，∴f0=1.令x=，y=0.則有2ff=f1+f0=-1+1=0，∴f=0.令y=x-，則有2f（2x-f=f2x+f2x-1.∵f=0，∴f2x+f2x-1=0，∴fx+fx-1=0，∴f0+f1+f2+???+f2024=f0+[f1+f2[+???+[f2023+f2024[=1+1012×0=1.f-=1．設(shè)m=f+f+?+f,n≥2,n∈N?,則實(shí)數(shù)m與-1的大小關(guān)系為()A.m<-1B.m=-1C.m>-1D.不確定【解析】∵函數(shù)f(x(滿足f(x(-f(y(=f（1x--令x=y=0得f(0(=0；令x=0得-f(y(=f(-y(,∴f(x(在(-1,1(為奇函數(shù)，f(x(<0，設(shè)-1<x1<x2<1，所以x1x2<1，x2-x1>0，所以>0，則f(x2(-f(x1(=f<0，所以f(x2(-f(x1(<0，即f(x2(<f(x1(，又f=-1,∴f=f=f-f,∴m=f+f+...+f==f-f=-1-f>-1，即m>-1，=f-fn=f，n∈N*1+a2+a3+?+a2024=()A.fB.fC.fD.f（(∴令x=n+2,y=n+3，則1x--==，n=f=f-f1+a2+a3+?+a2024=f-f+f-f+?+f-f1已知函數(shù)f(x)對(duì)于一切實(shí)數(shù)x,y均有f(x+y)-f(y)=x(x+2y+1)成立，且f(1)=0，則當(dāng)x∈【解析】∵函數(shù)f(x)對(duì)于一切實(shí)數(shù)x,y均有f(x+y)-f(y)=x(x+2y+1)成立，∴f(0)=-2，∴令y=0得，f(x)-f(0)=x(x+1)，即f(x)=x2+x-2，f(x)+2<logax恒成立，2+x<logax恒成立，2+x<logax恒成立，當(dāng)0<a<1時(shí)，則有l(wèi)oga≥，所以≤a<1. f(2017(=()【解析】∵f(xy+1(=f(x(f(y(-f(y(-x+2，令x=y=0，得f(1(=1-1-0+2，令y=0,f(1(=f(x(f(0(-f(0(-x+2，又f(0(=1，∴f(x(=x+1，3f(x(滿足對(duì)任意的實(shí)數(shù)a,b都有f(a+b(=f(a(?f(b(，且f(1(=2，則+++?+=()【解析】∵fx滿足對(duì)任意的實(shí)數(shù)a,b都有fa+b=fa?fb,∴令b=1得fa+1=fa?f1,∴f1=f1=2，∴===...===2，∴f2+f4+f6++f2016+f2018=1009×2=2018，故選Bf1f3f5...f2015f2017. +數(shù)f(x)對(duì)任意a,b滿足f(a+b)=f(a)f(b)，且f(1)=2，則+++?+++?+=2+2+2+?+2=2×1008=2016，故選B. A.一定是奇函數(shù)B.一定是偶函數(shù)C.既是奇函數(shù)又是偶函數(shù)D.既不是奇函數(shù)【解析】令y=x,則fx2==，∵fx2=f(-x)2==-,∴=-,即f(-x)=-f(x),其中x≠0,∵f(0)=f(0×1)==f(0)+f(1),∴f(1)=0,f(-1)=-f(1)=0.6(2024·全國(guó)·模擬預(yù)測(cè))已知函數(shù)fx的定義域?yàn)镽，fxfy-fx=xy-y，則()A.f0=0B.f-1=1C.fx+1為偶函數(shù)D.fx+1為奇函數(shù)【解析】當(dāng)f(x(=0時(shí)，f(x(f(y(-f(x(=xy-y不恒成立，故f(0(=1，A錯(cuò)誤.B：解法一令x=0，得f(0(f(y(-f(0(=-y，又f(0(=1，所以f(y(=1-y，故f(-1(=1+1=2，B錯(cuò)誤.解法二令x=0,y=-1，得f(0(f(-1(-f(0(=1，又f(0(=1，所以f(-1(=2，B錯(cuò)誤.C：解法一由B選項(xiàng)的解法一可知f(x(=1-x，則f(x+1(=-x，所以f(x+1(為奇函數(shù)，C錯(cuò)誤，D正確.解法二令x=0,y=2，得f(0(f(2(-f(0(=-2，又f(0(=1，所以f(2(=-1， 7設(shè)函數(shù)y=f(x(的定義域?yàn)?0,+∞)，f(xy(=f(x(+f(y(，若f(9(=6，則f(33(等于()A.B.2C.D.【解析】因?yàn)閒(xy(=f(x(+f(y(，令x=y=3，則f(3(+f(3(=f(9(，即2f(3(=6，可得f(3(=3；令x=y=3，則f(3(+f(3(=f(3(，即2f(3(=3，可得f(3(=；令x=3,y=3，可得f(33(=f(3(+f(3(=3+=. f(x+y(-f(x-y(=2f-xA.f(x(為偶函數(shù)B.f(x(=2ffC.f(1(+2f(2(+3f(3(+???+2023f(2023(=1D.[f(x([2+f-x（2=1【解析】因?yàn)?x,y∈R,f(x+y(-f(x+y(=2f-x（f(y(，令x=0，得f(y(-f(-y(=2ff(y(=2f(y(，即f(-y(=-f(y(，所以函數(shù)f(x(為奇函數(shù)，故選項(xiàng)A不用替換x，令y=，得f+-f-=2f-?f，即f(x(-f(0(=2ff,又函數(shù)f(x(為奇函數(shù)，所以f(0(=0，所以f(x(=2ff，故選項(xiàng)B正確；令x=，得f+y（-f-y（=2f(0(?f(y(=0，即f+y（=f-y（=-f（y-，即f(y+1(=-f(y(，所以f(y+2(=-f(y+1(=f(y(，所以函數(shù)f(x(的周期為2，再由f(x(=2ff，令x=1，可得f(1(=2f(0(?f=0，由函數(shù)的周期性可知，f(1(=f(3(=f(5(=???=f(2023(=0，f(2(=f(4(=f(6(=???=f(2022(=0，由f(x+y(-f(x-y(=2f-x（f(y(，即1-f由f(x+y(-f(x-y(=2f-x（f(y(，令x=-y，得f-y+y（-f-y-y（=2[f(y([2，由①+②整理后可得2=2f-x（2+2[f(x([2，即[f(x([2+f-x（2=1，故選項(xiàng)D正確.99()A.f=0B.f(0)=-2C.f(x)的一個(gè)周期為3D.f=2【解析】令x=y=0，則f(0)-f(0)=f2，所以f=0，A選項(xiàng)正確；令x=0，則f(y)-f(-y)=ff（y+=0，即f(y)=f(-y)，所以f(3)=f(-3)，令x=y=，則f(3)-f(0)=f2(3)，令x=y=-，則f(-3)-f(0)=f2(0)=f(3)-f(0)，所以f2(0)=f2(3)，因?yàn)閒2(0)+f(0)=f(3)，所以[f2(0)+f(0)[2=f2(3)，)+f(0)(2=f2(0)，所以f所以fx-=f（x+，所以f(x(=f(x+6(，因?yàn)閒x-+f（x+=0，且f(y)=f(-y)，令x=1，f（-+f=f+f=0， 2 2,f(-1)+f(2)=f(1)+f(2)=0，且f(3)=2，f=0，所以f+f(1)+f+f(2)+f+f(3)=2，由ff+f(1)+f+f(2)+f+f(3)+f+f(4)+f+f(5)+f+f(6)=0，所以f=f(3)=2，D選項(xiàng)正確.f(xy(+xy=xf(y(+A.f(x(為奇函數(shù)C.若x≠0，則xf+f(x(為定值B.f(x(為R上減函數(shù)令x=y=1，可得f(1(+1=f(1(+f(1(，則f(1(=1，令x=y=-1，可得f(1(+1=-f(-1(-f(-1(，則f(-1(=-1，令x=y=0，可得f(0(=0，令y=-1，可得f(-x(-x=xf(-1(-f(x(，所以f(-x(=-f(x(，所以f(x(為奇函數(shù)，故A正確；因?yàn)閒(-1(=-1、f(1(=1，所以f(x(不可能為R上減函數(shù)，故B錯(cuò)誤；令y=(x≠0(可得f(1(+1=xf+f(x(，所以xf+f(x(=2，故C正確；令y=2可得f(2x(+2x=xf(2(+2f(x(，因?yàn)閒(2(=2，所以f(2x(=2f(x(，所以f(4(=2f(2(=22，f(8(=2f(4(=23，??,所以f(2n(=2n，所以f(2k(=21+22+?+210==2046，故D正確.A.f(0)=1B.f(1)=-1C.f(x)是偶函數(shù)D.f(x)是奇函數(shù)令x=y=0，則[f(0([2=f(0(，解得f(0(=0或f(0(=1，函數(shù)f(x(=1+|x|，定義域?yàn)镽，f(-x(=1+|-x|=1+|x|=f(x(，f(x(為偶函數(shù)，C正確，D錯(cuò)誤.(多選題)(2024·廣西·二模)已知函數(shù)y=f(x(的定義域與值域均為Q+，且f(y(f=f(x(+f2(y(+txf(y((t∈N*(，則()A.f(1(=1B.函數(shù)f(x(的周期為4C.f(x(=x2(x∈Q+(D.t=2【解析】令x=y得f(x(f(x+1(=f(x(+f2(x(+txf(x((t∈Z*(，即f(x+1(=1+f(x(+tx(t∈Z*(①,令y=1，得f(1(f(x+1(=f(x(+f2(1(+txf(1((t∈Z*(②,令x=y2=4，得f(2(f(4(=f(4(+f2(2(+4tf(2((t∈Z*(③,由①,f(2(=1+f(1(+t=t+2(t∈Z*(，f(4(=1+f(3(+3t=1+[1+f(2(+2t[+3t=6t+4(t∈Z*(，由f(x+1(=1+f(x(+2x可知f(x(為一元二次函數(shù)，設(shè)f(x(=ax2+bx+c，則有a(x+1(2+b(x+1(+c=ax2+bx+c+2x+1，整理得2ax+a+b=2x+1?a=1,b=0，又由f(1(=a+b+c=1?c=0，(多選題)已知非常數(shù)函數(shù)f(x(的定義域?yàn)镽，且f(x(f(y(=f(xy(+xy(x+y(，則()A.f(0(=0B.f(1(=-2或f(1(=1D.f(x(是R上的增函數(shù)【解析】在f(x(f(y(=f(xy(+xy(x+y(中，因?yàn)楹瘮?shù)f(x(為非常數(shù)函數(shù)，所以f(0(=0，A正確.令x=y=-1，則[g(-1([2=g(1(-2，①(多選題)已知f(x(是定義在R上的函數(shù)，?x∈R,f(x(>0，且f(xy(=f(x(?f(y(-x2-()A.f(1(=1B.f(x(是偶函數(shù)C.f(x(的最小值是1D.不等式f(x-2(<10的解集是(-1,5(對(duì)于BC，令y=1，得fx=fxf1-x2-1=2fx-x2-1，則fx=x2+1，從而fx是偶函數(shù)，且fx≥1，故B，C正確.對(duì)于D，因?yàn)閒x=x2+1,fx是偶函數(shù)，在(0,+∞)上單調(diào)遞增，且f3=10，所以不等式fx-2<10等價(jià)于f-3<fx-2<f3，所以-3<x-2<3，解得-1<x<5，則D正確.15(多選題)已知函數(shù)f(x)滿足f(x+y)=f(x)+f(y),x,y∈R，則()A.f(0)=0B.f(k)=kC.f(x)=kf,(k≠0)D.f(-x)f(x)<0對(duì)于B，f(k)=f(k-1)+f(1)=f(k-2)+f(1)+f(1)=?=f(1)+f(1)+?+f(1)=kf(1)，故B正確；對(duì)于C，f(x)=f?x+=f?x（+f=f?x（+f+f=?=f+f+?+f=kf,k≠0，故C正確；對(duì)于D，f(x-x)=f(x+(-x))=f(x)+f(-x)=f(0)=0，f(x)=-f(-x),f(x)f(-x)=-(f(x))2≤0，故D錯(cuò)誤. fy-2xy，則()A.f(0)=0B.f(2)=4C.y=f(x)-2x是奇函數(shù)D.y=fx-2x2是偶函數(shù)【解析】令x=y=0，則f0=2f0，即f0=0.A正確.令y=0，則fx=f-x.令y=x，則f-x+fx-2x2=f(0)=0，則fx=x2.故f2=4.B正確.y=fx-2x=x2-2x是非奇非偶函數(shù).C不正確.y=fx-2x2=-x2是偶函數(shù).D正確.A.f0=0B.f2=8C.f,1=4D.f(k)=(n-1)?2n+1+2fx1x2所以f(k)=f(1)+f(2)+?+f(n)=1×2+2×22+3×23+n?2n，2+32+23+34+?+n?2n+1，所以Sn-2Sn=2+22+23+?+2n-n?2n+1，所以-Sn=-n?2n+1，所以Sn=(n-1)?2n+1+2，故D正確.(多選題)(2024·全國(guó)·模擬預(yù)測(cè))已知函數(shù)fx的定義域?yàn)镽，且fx+y=fxfy+fx+fy，x>0時(shí)，fx>0，f2=3，則()A.f1=1B.函數(shù)fx在區(qū)間0,+∞單調(diào)遞增C.函數(shù)fx是奇函數(shù)D.函數(shù)fx的一個(gè)解析式為fx=2x-1【解析】A項(xiàng)：因?yàn)閒x+y=fxfy+fx+fy，當(dāng)x>0時(shí)，fx>0,f2=3，令x=y=1，則f2=[f1[2+2f1=3，解得f1=1，A正確；則fx2=f[x1+x2-x1[=fx1fx2-x1+fx1+fx2-x1，因?yàn)楫?dāng)x>0時(shí)，fx>0，所以fx2-x1>0，fx1>0，所以fx1fx2-x1+fx1+fx2-x1>fx1，即fx2>fx1，所以函數(shù)fx在區(qū)間0,+∞單調(diào)遞增，B正確；C項(xiàng)：令x=y=0，則f0=[f0[2+2f0，解得f0=0或f0=-1，當(dāng)f0=0，且x>0時(shí)，令y=-x，則0=fxf-x+fx+f-x，若fx為奇函數(shù)，則f-x=-fx，即0=-f2x+fx-fx，解得fx=0，與題意矛盾；當(dāng)f0=-1時(shí)fx不為奇函數(shù).D項(xiàng)：當(dāng)fx=2x-1，則fx+y=2x+y-1，fxfy+fx+fy=2x-12y-1+2x-1+2y-1=2x+y-2x-2y+1+2x-1+2y-1=2x+y-1，所以f(x+y(=f(x(f(y(+f(x(+f(y(，易得f(x(=2x-1在R上單調(diào)遞增，所以x>0時(shí)，f(x(=2x-1>20-1=0，f(2(=22-1=3，故函數(shù)f(x(的一個(gè)解析式為f(x(=2x-1，D正確.(多選題)已知函數(shù)y=f(x(，對(duì)于任意x,y∈R，=f(x-y(，則()A.f(0(=1B.f(x2(=2f(x(C.f(x(>0D.≥f由已知=f(x-y(?f(x(=f(y(f(x-y(?f(x+y(=f(x(f(y(，①由①可知，令x=y=，則f(x(=ff=f2，又因?yàn)?f(x-y(，則f≠0，所以f(x(=f2>0，故C正確；因?yàn)閒(x(>0，所以f(x(+f(y(≥2f(x(f(y(=2f(x+y(，又由①,令x=y=，則f(x+y(=ff=f2，所以≥f，故D正確.A.f(0(=0B.f(x(是偶函數(shù)C.當(dāng)A，B是銳角△ABC的內(nèi)角時(shí)，f(sinA(<f(cosB(D.當(dāng)xn>0，且=，x1=時(shí)，f(xn(=2n-1令x=0，則-f(y(=f(-y(，所以f(x(為奇函數(shù)，故B錯(cuò)誤.<x2，則f(x2(-f(x1(=f.因?yàn)?1+x1((1-x2(=1-x2+x1-x1x2>0，所以x2-x1<1-x1x2，所以0<<1.f(x(>0，所以f>0，f(x1(<f(x2(，即f(x(在(-1,1(上單調(diào)遞增.因?yàn)锳，B是銳角△ABC的內(nèi)角，所以A+B>，所以A>-B，所以sinA>sin-B（=cosB.f(sinA(>f(cosB(，故C錯(cuò)誤.令y=-x，則2f(x)=f,令x=xn，則2f(xn(=f=f(xn+1(，所以f(因?yàn)閒(x1(=1，所以f(xn(=2n-1，故D正確.(多選題)函數(shù)f(x(的定義域?yàn)镽，f≠0，若f(x+y(+f(x(f(y(=4xy，則下列選項(xiàng)正確的有()A.f-=0B.f=-2【解析】令x=，y=0，得f+f×f(0(=f×[1+f(0([=0，令x=，y=-得f(0(+f×f（-=-1，令f(x(=kx+b，則所以f(x(=-2x-1，因?yàn)閒=-2，所以選項(xiàng)B正確；因?yàn)閒x+=-2x-2是減函數(shù)，所以選項(xiàng)C錯(cuò)誤；因?yàn)閒(多選題)定義在(0