Question

已知定義在R上的函數(shù)f(x)滿(mǎn)足：對(duì)于任意實(shí)數(shù)x，y，恒有f(x-y)=，且當(dāng)x＞0時(shí)，0＜f(x)＜1．

(1)求證：f(0)=1，且當(dāng)x＜0時(shí)，有f(x)＞1；

(2)試判斷函數(shù)f(x)在R上的單調(diào)性，并證明你的結(jié)論；

(3)若實(shí)數(shù)x、y，滿(mǎn)足：f[(x-2)²]·f[(y-2)²]≥f(2)，且f(x+y-4)≤1，求z=x+y的取值范圍．

Answer 1

答案：(1)∵f(x-x)==1，∴f(0)=1

設(shè)x＜0，則-x＞0，0＜f(-x)＜1

f(x)=f[0-(-x)]=＞1．

(2)設(shè)-∞＜x₁＜x₂＜+∞，

則x₂-x₁＞0，0＜f(x₂-x₁)＜1

f(x₁)=f[x₂-(x₂-x₁)]=＞f(x₂)

∵f(x₂)＞0

∴f(x)在R上單調(diào)遞減．

(3)∵f(x-y)·f(y)=f(x)

∴f(x)·f(y)=f(x+Y)

∴f[(x-2)²]·f[(y-2)²]

=f[(x-2)²+(y-2)²]≥f(2)

∴(x-2)²+(y-2)²≤2                                                           ①

又f(x+y-4)≤1=f(0)

∴x+y-4≥0                                                                 ②

∴z∈[4，6](可作圖求解)．