重难点 抽象函数及其应用（解析版）

A.(-2,0(∪(0,2(B.(-1,3(C.(-3,0(∪(0,1(D.(-1,0(∪(0,3(A.函数y=2x+、x-1的值域为[2,+∞(B.函数f(x(=tanx的值域为RC.函数f(x(=ex-1与g(x是同一个函数D.若函数f(x-1(的定义域为[1,4[，则函数f(2x(的定义域为[0,2[又函数y=2x+、x-1在[1,+∞(上单调递增，所以y≥2×1+、1-1=2，1122所以函数y=2x+、x-1的值域为[2,+∞(，故A正确；令t=tanx≠0，则y=t-，所以函数f(x(=tanx的值域为R，故B正确；又g(xex-1，所以函数f(2x(的定义域为(-∞,log23[，选项D错误.有f(f(x(-log2x(=3，则f(22025)-2的值是()【分析】由函数f(x)在定义域(0,+∞)上是单调函数，且f(f(x)-log2x)=3，知f(x)-log2x是一个常-2的值.【详解】由函数f(x)在定义域(0,+∞)上是单调知f(x)-log2x是一个常数，令f(x)-log2x=t，则f(x)=t+log2x，∴f(t)=t+log2t=3，f(x(=x(2-x(，则f(9(=；x∈(-4,-2[时，f(x(=．【答案】16-(x+2((x+4(=f(x+4(，即可得解.33则f(9(=2f(7(=4f(5(=8f(3(=16f(1(=16×1×(2-1(=16；则f(x(=f(x+2(=f(x+4(=(x+4((2-(x+4((=-(x+2((x+4(.故答案为：16；-(x+2((x+4(5.(24-25高二下·广东云浮·期末)已知函数f(x(的定义域为R，满足f(x+y(>f(x(.f(y(，且f(1(=A.f(10(>104B.f(20(>106C.f(10(<104D.f(20(<106【详解】令y=1，则f(x+1(>f(x(.f(1(=2f(x(，f(y(+1=f(x+y(，则()A.f(-2(=-3B.f(2025(=2024C.f(x(有最大值D.f(x(+1是奇函数f(x(=x-1可排除C，设g(x(=f(x(+1，则g(x(+g(y(=g(x+y(，令y=-x即可证明函数g(x(=f(x(+1是奇函数.f(0(+f(0(+1=f(0+0(，解得f(0(=-1，对于A，令x=1,y=-1，则f(1(+f(-1(+1=f(1-1(，解得f(-1(=-2，令x=-1,y=-1，则f(-1(+f(-1(+1=f(-1-1(，解得f(-2(=-3，故A正确；对于B，令y=1，则f(x+1(=f(x(+f(1(+1=f(x(+1，对于C，设函数f(x(=x-1，此时f(1(=0，f(x(+f(y(+1=x-1+y-1+1=(x+y(-1=f(x+y(，符合题意；对于D，设g(x(=f(x(+1，因为∀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，则g(x(+g(-x(=g(0(=f(0(+1=0，所以函数g(x(是奇函数，44f(x(f(y(，且f(1(=2，则++…+=()7.(23-24高一上·广东江门·期末)已知函数f(x(的定义域为R，对于任意实数x，y满足f(x+y(f(x(f(y(，且f(1(=2，则++…+=()f(1(f(3(f(2023(【详解】对于任意实数x，y满足f(x+y(=f(x(f(y(，所以当y=1时，f(x+1)=f(x)f(1)=2f(x)，8.(24-25高一上·广东佛山·期末)已知定义在R上的函数f(x(满足f(x+y且当x>A.奇函数，在(0,+∞(上单调递增B.奇函数，在(0,+∞(上单调递减令y=-x，则f得f(-x)=-f(x)，则函数f(x)为奇函数，设x1,x2<x22-x1>0，则0<f(x2-x1(<1， f(x2-x1)+f(x1)则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(，故函数f(x)在(0,+∞(上单调递增．559.【多选】(23-24高三上·广东汕头·期末)已知定义在(0,+∞(上的函数f(x(满足：∀x,y∈(0,+∞(，f(x(+f(y(=f(xy(，且当0<x<1时，f(x(<0，若f(2(=1，则()A.f(1(=0B.f(x(在(0,+∞(上单调递减C.|f(x(|=|fD.f(2(+f(22(+…+f(220(=55=f(x(+f(xn-1(，从而利用累加法与等差数列的求和公式可判断D.f(x(+f(y(=f(xy(，对于C，令y=，得f(x(+f=f(1(=0，则f(x(=-f，则f(x1(-f(x2(=f(x2.2(=f+f(x2(-f(x2(=f，因为当0<x<1时，f(x(<0，所以f<0，即f(x1(所以f(x(在(0,+∞(上单调递增，故B错误；对于D，令y=xn-1，得f(xn(=f(x(+f(xn-1(，则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错误；10.(23-24高一上·广东珠海·期末)已知定义在R上的函数f(x(满足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(，因为f(x(>0，所以f(-1(>0，所以f(-1(=2.(2)f(x(为偶函数.证明如下：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(的定义域为R，所以f(x(为偶函数.f(ab(=af(b(+bf(a(，若f(3(=2，则f(-1(+f(-)A.BC.当a=b=-1时，f(1(=-2f(-1(，可得f(-1(=0，函数f(x(是定义在R上且不恒为零的函数，令a=-1,b=x，可得f(-x(=-f(x(+xf(-1(=-f(x(，则函数f(x(是奇函数，f+f(3(=0,得f=-，所以f(-=，所以f(-1(+f(-=.12.(24-25高一上·广东深圳·期末f(x(=x+1，则f(5(=(A.-2B.0C.2D.6【详解】因为函数f(x(是定义在R上的周期为4所以f(5(=f(-5(=f(-1(，所以f(5(=f(-1(=0，7713.(24-25高二下·广东深圳·期末)定义在R上的奇函数f(x(满足f(2-x(=f(x(，且f(11(+f(10(=-1，则f(1(=．【分析】由已知得出f(x(周期为4，f(0(=f(2(=0，再由f(11(+f(10(=-1即可求解．【详解】因为f(x(=f(2-x(=-f(x-2)=f(x-4)，所以f(x(周期为4，又f(x(是定义在R上的奇函数，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·广东·期末)已知函数f(x(，∀x，y∈R，有f(x+y(=f(x(.f(a-y(+f(y(.f(a-x(，A.4a是f(x(的一个周期B.f(x(是奇函数C.f(x(是偶函数D.f(a(=1项.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，此时f(1(=，f(-x(=f(x(=，故f(x(为偶函数，故B错误，D错误；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，此时f(x(为奇函数，故C错误；令x=y=0，则f(0(=2f(0(.f(a(，令x=a，y=0，则f(a(=f2(a(+f2(0(，若f(0(=0，则f(a(=f2(a(，又f(a(≠0，故f(a(=1，令y=a，则f(x+a(=f(x(.f(0(+f(a(.f(a-x(，所以f(x+a(=f(a-x(，令x=y=a，则f(2a(=2f(a(f(0(=0，令x=2a，则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(，此时令y=-a，则f(a(=f(-a(.f(-a(=1，故f(-a(=1或f(-a(=-1.若f(-a(=1，则f(-y(=f(y(，故f(x(为偶函数，故f(x+a(=f(a-x(=f(x-a(，即f(x(=f(2a+x(，所以f(x(为周期函数且周期为2a.若f(-a(=-1，则f(-y(=-f(y(，故f(x(为奇函数，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(为周期函数且周期为4a.若f(0(≠0，则f(a(=，此时f2(0(=-=，故f(0(=或f(0(=-，若f(0(=，令x=-a，y=a，则f(0(=f(-a(f(0(+f(a(f(2a(，所以f(-a(=，令y=a，则f(x+a(=f(x(f(0(+f(a(f(a-x(=f(x(+f(a-x(，令y=-a，则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(为周期函数且周期为2a.若f(0(=-，令x=y=a，则f(2a(=-×-×=-，令x=-a，y=a，则f(0(=f(-a(f(0(+f(a(f(2a(，所以f(-a(=，令y=a，则f(x+a(=f(x(f(0(+f(a(f(a-x(=-f(x(+f(a-x(，令y=-a，则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(为周期函数且周期为2a．<时，f(x(=(x-2，则f=．f(x(的周期，将f化为f即可.由y=f(x+的图象关于y轴对称，得y=f(x(的图象关于直线x=对称，99:f=f(1+=f=f=．16.(24-25高三上·广东湛江·期末)已知函数f(x(满足f(x-2(+f(-x(=0，且f(x+1(是奇函数，若f(2(=3，则f(9(+f(10(=()A.-6B.-3C.3D.6【详解】因为f(x+1(是奇函数，所以f(1(=0,f(x+1(=-f(-x+1(，所以f(x+2(=-f(-x(．因为f(x-2(+f(-x(=0，所以f(x-2(=f(x+2(，所以f(x(=f(x+4(，即f(x(是周期为4的周期函数，则f(9(+f(10(=f(1(+f(2(=3．log2(1-x)f(x-1)-f(x-2),17.(24-25高二上·广东广州log2(1-x)f(x-1)-f(x-2),A.-1B.0C.1D.2x≤0x>0【详解】当x>0时，f(x)=f(x-1)-f(x-2)，则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·广东深圳·期末)已知定义域为R的函数f(x(满足f(3x+1(是奇函数，f(2x-1(是偶A.f(x(的图象关于直线x=-1对称B.f(x(的图象关于点(1,0)对称C.f(-3(=1D.f(x(的一个周期为8【分析】根据f(3x+1(是奇函数，可得f(x(+f(-x+2(=0，判断B;根据f(2x-1(是偶函数，推出f(-x-2(=f(x(，判断A;继而可得f(x+4(=-f(x(，可判断D；利用赋值法求得f(1)=0，根据对称性【详解】由题意知f(3x+1(是奇函数，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(是偶函数，故f(-2x-1(=f(2x-1(,:f(-x-1(=f(x-1(，即f(-x-2(=f(x(，故f(x(的图象关于直线x=-1对称，A结论正确；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(，则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(的图象关于直线x=-1对称，故f(-3(=0，C结论错误，数.x，其关于原点对称后的图像为yx=-2x(x<0(，易知y=-2x(x<0(与f(x)=-|x+2|(x<0(有两个交点，即f(x)=-|x+2|(x<0(上有两个点，中心对称后在f(xx(x>0(上；20.(24-25高一下·广东广州·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错误；对于B，在f(x)+f(-x)=6中，取x=0，可得f(0)=3因f(x)+g(1-x)=3，则两式相加，可得f(x)+f(x-2)=6，则f(x-2)+f(x-4)=6，故可得f(x)=f(x-4)，即4为函数f(x(的一个周期.期.由g(x(+f(x-3)=3，可得g(x-1(+f(x-4)=3，与f(x)+g(1-x)=3联立，可得g(x-1(=g(1-x)，故得g(x(=g(-x(，故B错误；对于C，由f(x)+f(x-2)=6可得f(3)+f(1)=6，但f(3),f(1)的值不能确定，又f(0)=3，f(2)=3，则f(0)+f(1)+f(2)+f(3)=12，A.4是f(x)的一个周期B.f(x)的图象关于直线x=2对称CD.方程f(x)=ln|x|恰有8不同的实数根【详解】对于A，因为g(x)=f(x+1)是偶函数，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)，对于B，由A得f(4-x)=f(-x)=-f(x)，函数f(x)的图象关于点(2,0)对称，故B错误；f(1(=2，由f(-x(=f(x+2(，令x=0则f(2(=0，令x=1则f(3(=f(-1(=-f(1(=-2，所以曲线y=f(x)与y=ln|x|有8个交点，故D正确．故选:ACD.22.(24-25高一上·广东汕尾·期末)若函数f(x)=(x2-2x)(x2+mx+n)的图象关于直线x=-1对称，则m+n=，f(x)的最小值为.由f(x)的图象关于直线x=-1对称，得-2,-4必为方程x2+mx+n=0的二根，此时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)的最小值为-16.A.充要条件B.充分不必要条件C.必要不充分条件D.既不充分也不必要条件性和必要性.因为函数f(x)的图象关于(0,1(对称艹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，化简得(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(的解集为()A.(-1,5(B.(-∞,-1(U(2,5(C.(-∞,-1(U(5,+∞(D.(-1,2(U(5,+∞(【详解】因为f(x(为R上的偶函数，且在(-∞,0[上单调递增，所以f(x(在[0,+∞(上单调递减.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为：(-∞,-1(U(5,+∞(.26.(24-25高一上·广东深圳·期末)已知函数f(x)是定义域为R的奇函数，且f(x)在[0,+∞)上单调递减．若f(3+m)+f(3m-7)>0，则m的取值范围为()A.(-∞,0)B.(0,+∞)C.(-∞,1)D.(1,+∞)则f(0)=0,f(x(在(-∞,0(上单调递减，即f(x)在R上单调递减.又f(3+m)+f(3m-7)>0，则f(3+m)>-f(3m-7)=f(7-3m(，27.(22-23高一上·广东深圳·期末)若f(x(是偶函数且在[0,+∞(上单调递增，又f(-2(=1，则不等式f(x(>1的解集为()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】根据偶函数的性质有f(x(在(-∞,0)上单调递减，在[0,+∞(上单调递增，且f(-2(=f(2)=【详解】由题设，偶函数f(x(在(-∞,0)上单调递减，在[0,+∞(上单调递增，且f(-2(=f(2)=1，所以f(x(>1=f(|±2|)，故x<-2或x>2，解集为{x|x<-2或x>2}.28.(22-23高一上·广东肇庆·期末)已知函数f(x(的定义域是R，函数f(x+1(的图象的对称中心是f(x(-x>0的解集为()A.(-∞,-1(∪(1,+∞(B.(-1,1(C.(-∞,-1(∪(0,1(D.(-1,0(∪(1,+∞(【分析】利用函数f(x+1(的图象的对称中心是(-1,0(可得f(x(是R上的奇函数，由【详解】因为f(x+1(是f(x(向左平移1个单位长度得到，且函数f(x+1(的图象的对称中心是(-1,0(，所以f(x(的图象的对称中心是(0,0(，故f(x(是R上的奇函数，所以f(-1(=-f(1(=-1，≠x2令g(x，所以根据单调性的定义可得g(x(在(0,+∞(上单调递增，由f(x(是R上的奇函数可得g(x(是(-∞,0(∪(0,+∞(上的偶函数所以g(x(在(-∞,0(上单调递减，综上所述，不等式f(x(-x>0的解集为(-1,0(∪(1,+∞(29.(25-26高一上·广东·期末)设偶-7)，f(π(，f(-3)的大小关系是()A.f(π(>f(-3(>f(-