Question

已知函数f(x)的定义域为R,对任意实数m,n都有f(m+n)=f(m)*f(n),

已知函数f(x)的定义域为R,对任意实数m,n都有f(m+n)=f(m)*f(n),且当x>0时，0<f(x)<1。

（1）证明：f(0)=1,且x<0时，f(x)>1
（2）证明：f(x)在R上单调递减
（3）设A={(x,y)|f(x^2)*f(y^2)>f(1)},B={(x,y)|f(ax-y+2)=1,a∈R}，若A∩B=空集，求a的取值范围

Answer 1

（1）对于任意实数n有f(n)=f(0+n)=f(0)*f(n) ∴f(0)=1
任意实数m有f(m-m)=f(m)*f(-m)=f(0)=1 ∴f(m)=1/f(-m)
当m>0时0<f(m)<1 ∴f(-m)>1 ∴x<0时，f(x)>1

(2)设x1<x2,则x1-x2<0, ∴f(x1-x2)=f(x1)*f(-x2)=f(x1)/f(x2)>1(这步由第（1）小题的结论得来的哦~) ∴x1<x2时有f(x1)>f(x2) ∴f(x)在R上单调递减

（3）f(x^2)*f(y^2)=f(x^2+y^2)>f(1) ∵f(x)在R上单调递减
∴A={(x,y)|x^2+y^2<1}(以原点为圆心1为半径的圆内部),
又f(ax-y+2)=1=f(0) ∴B={(x,y)|ax-y+2=0,a∈R}（过（0，2）的直线）
A∩B=空集 即圆心到直线的距离要大于1
由点到直线的距离公式得 |a*0-0+2|/√(a^2+(-1)^2) >1(√表示开根号哈~)
即a^2+1<4 解得-√3<a<√3 （不能取等号，取等的时候相切，有交点~）

希望能帮到你哈~

Answer 2

1、取m=n=0。有f(0)=f(0)*f(0),所以f(0)=0或1。如果f(0)=0，令n=0时，f(m)=0恒成立，显然是不对的所以f(0)=1。
取令m+n=0,并假定m>0。所以f(0)=f(m)*f(-m)=1,f(-m)=1/f(m),因为x>0时，0<f(x)<1，可以得到x<0时，f(x)>1。
2、设x与x+a,a<0,f(x)-f(x+a)=f(x)[1-f(a)]<0[由上题知f(x)>0,1-f(a)<0],由于x>x+a,所以是减函数。
3、f(x^2)*f(y^2)=f(x^2+y^2)>f(1),由于是减函数，所以x^2+y^2<1[表示圆内的部分]；
f(ax-y+2)=1=f(0),所以ax-y+2=0[表示直线]
由于A∩B=空集，即直线与圆没有交点或只是相切[因为是圆内]。所以圆心到直线的距离>=半径的长度1。
|a*0-0+2|/(a^2+1)^0.5>=1,所以-3^0.5<=a<=3^0.5。
有些地方只是给了大概思路，自己还需要想想！

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Answer 4

（1）令m=n=0，得：f(0)=2f(0)，∴f(0)=0
令m=x，n=-x，则f(0)=f(x)+f(-x)=0
∴f(x)+f(-x)=0，∴f(x)=-f(-x)
而定义域为R，关于原点对称
∴函数f(x)为奇函数
（2）∵f(5)=f(4)+f(1)=2f(2)+f(1)=4f(1)+f(1)=5f(1)
∴5f(1)=5，∴f(1)=1
∴f(2)=2f(1)=2
∴f[log2(x^2-x-2)]<f(2)
∵f(x)是单调函数，而f(1)<f(5)（即函数单增）
∴log2(x^2-x-2)<2
∴0<x^2-x-2<4
∴-2<x<-1或2<x<3
望采纳！有问题请追问！

Answer 5

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