Question

設(shè)函數(shù)y=f(x)定義域為R,當(dāng)x＜0時,f(x)＞1,且對于任意的x,y∈R,有f(x+y)=f(x)·f(y)成立.數(shù)列{a_n}滿足a₁=f(0),且f(a_n+1)=(n∈N^*).

(1)求f(0)的值;求證:函數(shù)y=f(x)在R上是減函數(shù);

(2)求數(shù)列{a_n}的通項公式并證明.

Answer 1

解:(1)令x=-1,y=0,得f(-1)=f(-1)·f(0),即f(0)=1.                          

當(dāng)x＞0時,-x＜0,∴f(0)=f(x)·f(-x)=1.∴0＜f(x)＜1.

設(shè)x₁,x₂∈R,且x₁＜x₂,

f(x₁)-f(x₂)=f(x₁)-f(x₁+x₂-x₁)=f(x₁)［1-f(x₂-x₁)］,

∵x₁＜x₂,∴x₂-x₁＞0.∴f(x₂-x₁)＜1.

∴1-f(x₂-x₁)＞0,而f(x₁)＞0.

∴f(x₁)-f(x₂)＞0,即f(x₁)＞f(x₂).

∴函數(shù)y=f(x)在R上是減函數(shù).                                         

(2)由f(a_n+1)=得f(a_n+1)·f(-2-a_n)=1,

∴f(a_n+1-a_n-2)=f(0).

∴a_n+1-a_n-2=0,

即a_n+1-a_n=2(n∈N^*).

∴{a_n}是等差數(shù)列，其首項為1，公差為d=2.∴a_n=2n-1.