A.(-2,0(∪(0,2(B.(-1,3(C.(-3,0(∪(0,1(D.(-1,0(∪(0,3(A.函數(shù)y=2x+、x-1的值域?yàn)閇2,+∞(B.函數(shù)f(x(=tanx的值域?yàn)镽C.函數(shù)f(x(=ex-1與g(x是同一個(gè)函數(shù)D.若函數(shù)f(x-1(的定義域?yàn)閇1,4[，則函數(shù)f(2x(的定義域?yàn)閇0,2[又函數(shù)y=2x+、x-1在[1,+∞(上單調(diào)遞增，所以y≥2×1+、1-1=2，1122所以函數(shù)y=2x+、x-1的值域?yàn)閇2,+∞(，故A正確；令t=tanx≠0，則y=t-，所以函數(shù)f(x(=tanx的值域?yàn)镽，故B正確；又g(xex-1，所以函數(shù)f(2x(的定義域?yàn)?-∞,log23[，選項(xiàng)D錯(cuò)誤.有f(f(x(-log2x(=3，則f(22025)-2的值是()【分析】由函數(shù)f(x)在定義域(0,+∞)上是單調(diào)函數(shù)，且f(f(x)-log2x)=3，知f(x)-log2x是一個(gè)常-2的值.【詳解】由函數(shù)f(x)在定義域(0,+∞)上是單調(diào)知f(x)-log2x是一個(gè)常數(shù)，令f(x)-log2x=t，則f(x)=t+log2x，∴f(t)=t+log2t=3，f(x(=x(2-x(，則f(9(=；x∈(-4,-2[時(shí)，f(x(=．【答案】16-(x+2((x+4(=f(x+4(，即可得解.33則f(9(=2f(7(=4f(5(=8f(3(=16f(1(=16×1×(2-1(=16；則f(x(=f(x+2(=f(x+4(=(x+4((2-(x+4((=-(x+2((x+4(.故答案為：16；-(x+2((x+4(5.(24-25高二下·廣東云浮·期末)已知函數(shù)f(x(的定義域?yàn)镽，滿足f(x+y(>f(x(.f(y(，且f(1(=A.f(10(>104B.f(20(>106C.f(10(<104D.f(20(<106【詳解】令y=1，則f(x+1(>f(x(.f(1(=2f(x(，f(y(+1=f(x+y(，則()A.f(-2(=-3B.f(2025(=2024C.f(x(有最大值D.f(x(+1是奇函數(shù)f(x(=x-1可排除C，設(shè)g(x(=f(x(+1，則g(x(+g(y(=g(x+y(，令y=-x即可證明函數(shù)g(x(=f(x(+1是奇函數(shù).f(0(+f(0(+1=f(0+0(，解得f(0(=-1，對(duì)于A，令x=1,y=-1，則f(1(+f(-1(+1=f(1-1(，解得f(-1(=-2，令x=-1,y=-1，則f(-1(+f(-1(+1=f(-1-1(，解得f(-2(=-3，故A正確；對(duì)于B，令y=1，則f(x+1(=f(x(+f(1(+1=f(x(+1，對(duì)于C，設(shè)函數(shù)f(x(=x-1，此時(shí)f(1(=0，f(x(+f(y(+1=x-1+y-1+1=(x+y(-1=f(x+y(，符合題意；對(duì)于D，設(shè)g(x(=f(x(+1，因?yàn)?x∈R，有f(x(+f(y(+1=f(x+y(，即?x∈R，有f(x(+1+f(y(+1=f(x+y(+1，令y=-x，則g(x(+g(-x(=g(0(=f(0(+1=0，所以函數(shù)g(x(是奇函數(shù)，44f(x(f(y(，且f(1(=2，則++…+=()7.(23-24高一上·廣東江門·期末)已知函數(shù)f(x(的定義域?yàn)镽，對(duì)于任意實(shí)數(shù)x，y滿足f(x+y(f(x(f(y(，且f(1(=2，則++…+=()f(1(f(3(f(2023(【詳解】對(duì)于任意實(shí)數(shù)x，y滿足f(x+y(=f(x(f(y(，所以當(dāng)y=1時(shí)，f(x+1)=f(x)f(1)=2f(x)，8.(24-25高一上·廣東佛山·期末)已知定義在R上的函數(shù)f(x(滿足f(x+y且當(dāng)x>A.奇函數(shù)，在(0,+∞(上單調(diào)遞增B.奇函數(shù)，在(0,+∞(上單調(diào)遞減令y=-x，則f得f(-x)=-f(x)，則函數(shù)f(x)為奇函數(shù)，設(shè)x1,x2<x22-x1>0，則0<f(x2-x1(<1， f(x2-x1)+f(x1)則f(x1(-f(x2(=f(x1(-f(x2-x1+x1(=f(x1(-1+ f(x2-x1)+f(x1)得f(x1(-f(x2(<0，得f(x1(<f(x2(，故函數(shù)f(x)在(0,+∞(上單調(diào)遞增．559.【多選】(23-24高三上·廣東汕頭·期末)已知定義在(0,+∞(上的函數(shù)f(x(滿足：?x,y∈(0,+∞(，f(x(+f(y(=f(xy(，且當(dāng)0<x<1時(shí)，f(x(<0，若f(2(=1，則()A.f(1(=0B.f(x(在(0,+∞(上單調(diào)遞減C.|f(x(|=|fD.f(2(+f(22(+…+f(220(=55=f(x(+f(xn-1(，從而利用累加法與等差數(shù)列的求和公式可判斷D.f(x(+f(y(=f(xy(，對(duì)于C，令y=，得f(x(+f=f(1(=0，則f(x(=-f，則f(x1(-f(x2(=f(x2.2(=f+f(x2(-f(x2(=f，因?yàn)楫?dāng)0<x<1時(shí)，f(x(<0，所以f<0，即f(x1(所以f(x(在(0,+∞(上單調(diào)遞增，故B錯(cuò)誤；對(duì)于D，令y=xn-1，得f(xn(=f(x(+f(xn-1(，則f(x2(=f(x(+f(x(，f(x3(=f(x(+f(x2(，…,f(xn(=f(x(+f(xn-1(，上述各式相加，得f(xn(=(n-1(f(x(+f(x(=nf(x(，又f(2(=1，所以f(2(+f(22(+…+f(220(=(1+2…+20(f(2(==210，故D錯(cuò)誤；10.(23-24高一上·廣東珠?！て谀?已知定義在R上的函數(shù)f(x(滿足f(xy(=f(x(f(y(-f(x(-f(y(+f(0(<f(1(，且f(x(>0.(1)求f(-1(的值；【答案】(1)f(-1(=2令x=y=1，得f(1(=[f(1([2-2f(1(+2，66令x=y=-1，得f(1(=[f(-1([2-2f(-1(+2，即[f(-1([2=2f(-1(，因?yàn)閒(x(>0，所以f(-1(>0，所以f(-1(=2.(2)f(x(為偶函數(shù).證明如下：令y=-1，得f(-x(=f(-1(f(x(-f(-1(-f(x(+2，由(1)得f(-x(=2f(x(-2-f(x(+2，即f(-x(=f(x(，又f(x(的定義域?yàn)镽，所以f(x(為偶函數(shù).f(ab(=af(b(+bf(a(，若f(3(=2，則f(-1(+f(-)A.BC.當(dāng)a=b=-1時(shí)，f(1(=-2f(-1(，可得f(-1(=0，函數(shù)f(x(是定義在R上且不恒為零的函數(shù)，令a=-1,b=x，可得f(-x(=-f(x(+xf(-1(=-f(x(，則函數(shù)f(x(是奇函數(shù)，f+f(3(=0,得f=-，所以f(-=，所以f(-1(+f(-=.12.(24-25高一上·廣東深圳·期末f(x(=x+1，則f(5(=(A.-2B.0C.2D.6【詳解】因?yàn)楹瘮?shù)f(x(是定義在R上的周期為4所以f(5(=f(-5(=f(-1(，所以f(5(=f(-1(=0，7713.(24-25高二下·廣東深圳·期末)定義在R上的奇函數(shù)f(x(滿足f(2-x(=f(x(，且f(11(+f(10(=-1，則f(1(=．【分析】由已知得出f(x(周期為4，f(0(=f(2(=0，再由f(11(+f(10(=-1即可求解．【詳解】因?yàn)閒(x(=f(2-x(=-f(x-2)=f(x-4)，所以f(x(周期為4，又f(x(是定義在R上的奇函數(shù)，所以f(0(=f(2(=0，所以f(11(+f(10(=f(-1)+f(2)=-f(1)+f(2)=-1?f(1)=1，14.(22-23高三上·廣東·期末)已知函數(shù)f(x(，?x，y∈R，有f(x+y(=f(x(.f(a-y(+f(y(.f(a-x(，A.4a是f(x(的一個(gè)周期B.f(x(是奇函數(shù)C.f(x(是偶函數(shù)D.f(a(=1項(xiàng).f(a(=≠0，f(x+y(=，f(x(.f(a-y(+f(y(.f(a-x(=×2=因此f(x+y(=f(x(.f(a-y(+f(y(.f(a-x(成立，此時(shí)f(1(=，f(-x(=f(x(=，故f(x(為偶函數(shù)，故B錯(cuò)誤，D錯(cuò)誤；f(x+y(=sin(x+y(，f(x(.f(a-y(+f(y(.f(a-x(=sinxcosy+cosxsiny=sin(x+y)，因此f(x+y(=f(x(.f(a-y(+f(y(.f(a-x(成立，此時(shí)f(x(為奇函數(shù)，故C錯(cuò)誤；令x=y=0，則f(0(=2f(0(.f(a(，令x=a，y=0，則f(a(=f2(a(+f2(0(，若f(0(=0，則f(a(=f2(a(，又f(a(≠0，故f(a(=1，令y=a，則f(x+a(=f(x(.f(0(+f(a(.f(a-x(，所以f(x+a(=f(a-x(，令x=y=a，則f(2a(=2f(a(f(0(=0，令x=2a，則f(2a+y(=f(2a(.f(a-y(+f(y(.f(-a(=f(y(.f(-a(，又f(2a+y(=f(-y(，故f(-y(=f(y(.f(-a(，此時(shí)令y=-a，則f(a(=f(-a(.f(-a(=1，故f(-a(=1或f(-a(=-1.若f(-a(=1，則f(-y(=f(y(，故f(x(為偶函數(shù)，故f(x+a(=f(a-x(=f(x-a(，即f(x(=f(2a+x(，所以f(x(為周期函數(shù)且周期為2a.若f(-a(=-1，則f(-y(=-f(y(，故f(x(為奇函數(shù)，故f(x+a(=f(a-x(=-f(x-a(，即f(2a+x(=-f(x(，88故f(4a+x(=-f(x+2a(=f(x(，所以f(x(為周期函數(shù)且周期為4a.若f(0(≠0，則f(a(=，此時(shí)f2(0(=-=，故f(0(=或f(0(=-，若f(0(=，令x=-a，y=a，則f(0(=f(-a(f(0(+f(a(f(2a(，所以f(-a(=，令y=a，則f(x+a(=f(x(f(0(+f(a(f(a-x(=f(x(+f(a-x(，令y=-a，則f(x-a(=f(x(f(2a(+f(-a(f(a-x(=f(x(+f(a-x(，故f(x+a(=f(x-a)即f(x+2a(=f(x)，故f(x(為周期函數(shù)且周期為2a.若f(0(=-，令x=y=a，則f(2a(=-×-×=-，令x=-a，y=a，則f(0(=f(-a(f(0(+f(a(f(2a(，所以f(-a(=，令y=a，則f(x+a(=f(x(f(0(+f(a(f(a-x(=-f(x(+f(a-x(，令y=-a，則f(x-a(=f(x(f(2a(+f(-a(f(a-x(=-f(x(+f(a-x(，故f(x+a(=f(x-a)即f(x+2a(=f(x)，故f(x(為周期函數(shù)且周期為2a．<時(shí)，f(x(=(x-2，則f=．f(x(的周期，將f化為f即可.由y=f(x+的圖象關(guān)于y軸對(duì)稱，得y=f(x(的圖象關(guān)于直線x=對(duì)稱，99:f=f(1+=f=f=．16.(24-25高三上·廣東湛江·期末)已知函數(shù)f(x(滿足f(x-2(+f(-x(=0，且f(x+1(是奇函數(shù)，若f(2(=3，則f(9(+f(10(=()A.-6B.-3C.3D.6【詳解】因?yàn)閒(x+1(是奇函數(shù)，所以f(1(=0,f(x+1(=-f(-x+1(，所以f(x+2(=-f(-x(．因?yàn)閒(x-2(+f(-x(=0，所以f(x-2(=f(x+2(，所以f(x(=f(x+4(，即f(x(是周期為4的周期函數(shù)，則f(9(+f(10(=f(1(+f(2(=3．log2(1-x)f(x-1)-f(x-2),17.(24-25高二上·廣東廣州log2(1-x)f(x-1)-f(x-2),A.-1B.0C.1D.2x≤0x>0【詳解】當(dāng)x>0時(shí)，f(x)=f(x-1)-f(x-2)，則f(x+1)=f(x)-f(x-1)=-f(x-2)，即f(x+3)=-f(x)，于是f(x+6)=-f(x+3)=f(x)，所以f(2024)=f(337×6+2)=f(2)=-f(-1)=-log22=-1.18.(22-23高二上·廣東深圳·期末)已知定義域?yàn)镽的函數(shù)f(x(滿足f(3x+1(是奇函數(shù)，f(2x-1(是偶A.f(x(的圖象關(guān)于直線x=-1對(duì)稱B.f(x(的圖象關(guān)于點(diǎn)(1,0)對(duì)稱C.f(-3(=1D.f(x(的一個(gè)周期為8【分析】根據(jù)f(3x+1(是奇函數(shù)，可得f(x(+f(-x+2(=0，判斷B;根據(jù)f(2x-1(是偶函數(shù)，推出f(-x-2(=f(x(，判斷A;繼而可得f(x+4(=-f(x(，可判斷D；利用賦值法求得f(1)=0，根據(jù)對(duì)稱性【詳解】由題意知f(3x+1(是奇函數(shù)，即f(-3x+1(=-f(3x+1(,:f(-x+1(=-f(x+1(，即f(-x+2(=-f(x(，即f(x(+f(-x+2(=0，又f(2x-1(是偶函數(shù)，故f(-2x-1(=f(2x-1(,:f(-x-1(=f(x-1(，即f(-x-2(=f(x(，故f(x(的圖象關(guān)于直線x=-1對(duì)稱，A結(jié)論正確；由以上可知f(x(=f(-x-2(=-f(-x+2(，即f(x-2(=-f(x+2(，所以f(x+4(=-f(x(，則f(x+8(=-f(x+4(=f(x)，由于f(-3x+1(=-f(3x+1(，令x=0，可得f(1)=-f(1),:f(1)=0，而f(x(的圖象關(guān)于直線x=-1對(duì)稱，故f(-3(=0，C結(jié)論錯(cuò)誤，數(shù).x，其關(guān)于原點(diǎn)對(duì)稱后的圖像為yx=-2x(x<0(，易知y=-2x(x<0(與f(x)=-|x+2|(x<0(有兩個(gè)交點(diǎn)，即f(x)=-|x+2|(x<0(上有兩個(gè)點(diǎn)，中心對(duì)稱后在f(xx(x>0(上；20.(24-25高一下·廣東廣州·期末A.f(-x(=-f(x(B.g(-x(=-g(x(C.=6066Df(x)+g(1-x)=3，可得f(-x)+g(1+x)=3，故f(x)+f(-x)=6，故A錯(cuò)誤；對(duì)于B，在f(x)+f(-x)=6中，取x=0，可得f(0)=3因f(x)+g(1-x)=3，則兩式相加，可得f(x)+f(x-2)=6，則f(x-2)+f(x-4)=6，故可得f(x)=f(x-4)，即4為函數(shù)f(x(的一個(gè)周期.期.由g(x(+f(x-3)=3，可得g(x-1(+f(x-4)=3，與f(x)+g(1-x)=3聯(lián)立，可得g(x-1(=g(1-x)，故得g(x(=g(-x(，故B錯(cuò)誤；對(duì)于C，由f(x)+f(x-2)=6可得f(3)+f(1)=6，但f(3),f(1)的值不能確定，又f(0)=3，f(2)=3，則f(0)+f(1)+f(2)+f(3)=12，A.4是f(x)的一個(gè)周期B.f(x)的圖象關(guān)于直線x=2對(duì)稱CD.方程f(x)=ln|x|恰有8不同的實(shí)數(shù)根【詳解】對(duì)于A，因?yàn)間(x)=f(x+1)是偶函數(shù)，所以g(-x)=g(x)，即f(1-f(2+x)=f(1-(1+x))=f(-x)=-f(x),f(x+4)=-f(x+2)=-(-f(x))=f(x)，對(duì)于B，由A得f(4-x)=f(-x)=-f(x)，函數(shù)f(x)的圖象關(guān)于點(diǎn)(2,0)對(duì)稱，故B錯(cuò)誤；f(1(=2，由f(-x(=f(x+2(，令x=0則f(2(=0，令x=1則f(3(=f(-1(=-f(1(=-2，所以曲線y=f(x)與y=ln|x|有8個(gè)交點(diǎn)，故D正確．故選:ACD.22.(24-25高一上·廣東汕尾·期末)若函數(shù)f(x)=(x2-2x)(x2+mx+n)的圖象關(guān)于直線x=-1對(duì)稱，則m+n=，f(x)的最小值為.由f(x)的圖象關(guān)于直線x=-1對(duì)稱，得-2,-4必為方程x2+mx+n=0的二根，此時(shí)f(x)=(x2-2x)(x2+6x+8)=x(x-2)(x+2f(-2-x)=(-2-x)(-4-x)(-x)(2-x)=x(x-2)(x+2)(x+4)=f(x)，所以f(x)的最小值為-16.A.充要條件B.充分不必要條件C.必要不充分條件D.既不充分也不必要條件性和必要性.因?yàn)楹瘮?shù)f(x)的圖象關(guān)于(0,1(對(duì)稱艸f(x(+(-x(=2恒成立，即x3-(a-1(x2+x+b+(-x(3-(a-1((-x(2+(-x(+b=2恒成立，化簡(jiǎn)得(a-1(x2-b+1=0恒成立，A.(-∞,1)B.(-1,1)C.(-3,1)D.(-∞,1]不等式f(x-2(<f(3(的解集為()A.(-1,5(B.(-∞,-1(U(2,5(C.(-∞,-1(U(5,+∞(D.(-1,2(U(5,+∞(【詳解】因?yàn)閒(x(為R上的偶函數(shù)，且在(-∞,0[上單調(diào)遞增，所以f(x(在[0,+∞(上單調(diào)遞減.所以f(x-2(<f(3(→|x-2|>3→x-2<-3或x-2>3，即x<-1或x>5.所以所求不等式的解集為：(-∞,-1(U(5,+∞(.26.(24-25高一上·廣東深圳·期末)已知函數(shù)f(x)是定義域?yàn)镽的奇函數(shù)，且f(x)在[0,+∞)上單調(diào)遞減．若f(3+m)+f(3m-7)>0，則m的取值范圍為()A.(-∞,0)B.(0,+∞)C.(-∞,1)D.(1,+∞)則f(0)=0,f(x(在(-∞,0(上單調(diào)遞減，即f(x)在R上單調(diào)遞減.又f(3+m)+f(3m-7)>0，則f(3+m)>-f(3m-7)=f(7-3m(，27.(22-23高一上·廣東深圳·期末)若f(x(是偶函數(shù)且在[0,+∞(上單調(diào)遞增，又f(-2(=1，則不等式f(x(>1的解集為()A.{x|-2<x<2{B.{x|x<-2或x>2}C.{x|x<-2或0<x<2}D.{x|x>2或-2<x<0}【分析】根據(jù)偶函數(shù)的性質(zhì)有f(x(在(-∞,0)上單調(diào)遞減，在[0,+∞(上單調(diào)遞增，且f(-2(=f(2)=【詳解】由題設(shè)，偶函數(shù)f(x(在(-∞,0)上單調(diào)遞減，在[0,+∞(上單調(diào)遞增，且f(-2(=f(2)=1，所以f(x(>1=f(|±2|)，故x<-2或x>2，解集為{x|x<-2或x>2}.28.(22-23高一上·廣東肇慶·期末)已知函數(shù)f(x(的定義域是R，函數(shù)f(x+1(的圖象的對(duì)稱中心是f(x(-x>0的解集為()A.(-∞,-1(∪(1,+∞(B.(-1,1(C.(-∞,-1(∪(0,1(D.(-1,0(∪(1,+∞(【分析】利用函數(shù)f(x+1(的圖象的對(duì)稱中心是(-1,0(可得f(x(是R上的奇函數(shù)，由【詳解】因?yàn)閒(x+1(是f(x(向左平移1個(gè)單位長(zhǎng)度得到，且函數(shù)f(x+1(的圖象的對(duì)稱中心是(-1,0(，所以f(x(的圖象的對(duì)稱中心是(0,0(，故f(x(是R上的奇函數(shù)，所以f(-1(=-f(1(=-1，≠x2令g(x，所以根據(jù)單調(diào)性的定義可得g(x(在(0,+∞(上單調(diào)遞增，由f(x(是R上的奇函數(shù)可得g(x(是(-∞,0(∪(0,+∞(上的偶函數(shù)所以g(x(在(-∞,0(上單調(diào)遞減，綜上所述，不等式f(x(-x>0的解集為(-1,0(∪(1,+∞(29.(25-26高一上·廣東·期末)設(shè)偶-7)，f(π(，f(-3)的大小關(guān)系是()A.f(π(>f(-3(>f(-