第1个回答 2012-01-15
(1)
由题意f(1)=f(1+0)=f(1)f(0),因为f(1)≠0,所以f(0)=1
(2)
对任意x<0,有 f(0)=f(-x + x)= f(-x)f(x) = 1,
所以 f(x) = 1/f(-x)
因为此时 -x>0,所以 0<f(-x)<1
所以 f(x) = 1/f(-x) > 1
(3)
设x1<x2,则x1-x2<0,f(x1-x2)>1
f(x1) = f[x2+x1-x2] = f[x2] f[x1-x2]
f[x1]/f[x2]=f[x1-x2]>1
所以f(x1)>f(x2)
所以函数f(x)在R上是减函数。
(4)
f(x1)+f(x2)- 2f((x1+x2)/2)
= [ f(x1/2) ]^2 + [ f(x2/2) ]^2 - 2f(x1/2) f(x1/2)
=[ f(x1/2) - f(x1/2) ]^2 ≥0
即 (f(x1)+f(x2))/2 ≥ f((x1+x2)/2)
作差比较大小
f(x1)=f(x1/2 + x1/2)= [ f(x1/2) ]^2
f(x2)=f(x2/2 + x2/2)= [ f(x2/2) ]^2
2f((x1+x2)/2)= 2f(x1/2) f(x1/2)
所以
f(x1)+f(x2)- 2f((x1+x2)/2)
= [ f(x1/2) ]^2 + [ f(x2/2) ]^2 - 2f(x1/2) f(x1/2)
=[ f(x1/2) - f(x1/2) ]^2 ≥0
即 (f(x1)+f(x2))/2 ≥ f((x1+x2)/2)
第2个回答 2012-01-15
设函数f(X)是定义域在R上的函数,且对于任意实数x y都有f(x+y)=f(x)f(y),且当x>0时,0<f(x)<1。
(1)证明x<0时,f(x)>1
(2)证明f(x)是R上的减函数
(3)设集合M=P{(x,y)\f(x^2)·f(y^2)>f(1)},P={(x,y)\f(ax-y+2)=1,a属于R},且M与P交集为空集,求a的取值范围。