Question

已知定義在R上的函數(shù)f(x)滿足:f(1)=,且對(duì)于任意實(shí)數(shù)x,y,總有f(x) f(y)=f(x+y) +f(x-y)成立.

(1)求f(0)的值,并證明函數(shù)f(x)為偶函數(shù);

(2)定義數(shù)列{a_n}:a_n=2f(n+1)-f(n)(n=1,2,3,…),求證:{a_n}為等比數(shù)列;

(3)若對(duì)于任意的非零實(shí)數(shù)y,總有f(y)＞2.設(shè)有理數(shù)x₁,x₂滿足:|x₁|＜|x₂|,判斷f(x₁)和f(x₂)的大小關(guān)系,并證明你的結(jié)論.

Answer 1

(1)解:令x=1,y=0,∴f(1)f(0)=f(1)+f(1).∵f(1)=,∴f(0)=2.

令x=0,∴f(0)f(y)=f(y)+f(-y),即2f(y)=f(y)+f(-y).

∴f(y)=f(-y),對(duì)任意實(shí)數(shù)y總成立.

∴f(x)為偶函數(shù).

(2)證明:令x=y=1,得f(1)f(1)=f(2)+f(0).

∴=f(2)+2.∴f(2)=.

∴a₁=2f(2)-f(1)==6.

令x=n+1,y=1,得f(n+1)f(1)=f(n+2)+f(n).

∴f(n+2)=f(n+1)-f(n).

∴a_n+1=2f(n+2)-f(n+1)

=2［f(n+1)-f(n)］-f(n+1)

=4f(n+1)-2f(n)

=2［2f(n+1)-f(n)］

=2a_n(n≥1).

∴{a_n}是以6為首項(xiàng),以2為公比的等比數(shù)列.

(3)解:結(jié)論:f(x₁)＜f(x₂).

證明:設(shè)y≠0,

∵y≠0時(shí),f(y)＞2,

∴f(x+y)+f(x-y)=f(x)f(y)＞2f(x),即f(x+y)-f(x)＞f(x)-f(x-y).

∴對(duì)于k∈N,總有f［(k+1)y］-f(ky)＞f(ky)-f［(k-1)y］成立.

∴f［(k+1)y］-f(ky)＞f(ky)-f［(k-1)y］＞f［(k-1)y］-f［(k-2)y］

＞…＞f(y)-f(0)＞0.

∴對(duì)于k∈N總有f［(k+1)y］＞f(ky)成立.

∴對(duì)于m,n∈N,若n＜m,則有f(ny)＜…＜f(my)成立.

∵x₁,x₂∈Q,∴可設(shè)|x₁|=,

|x₂|=,其中q₁,q₂是非負(fù)整數(shù),p₁,p₂都是正整數(shù),

則|x₁|=,|x₂|=.

令y=,t=q₁p₂,s=p₁q₂,則t,s∈N.

∵|x₁|＜|x₂|,∴t＜s.

∴f(ty)＜f(sy),即f(|x₁|)＜f(|x₂|).

∵函數(shù)f(x)為偶函數(shù),

∴f(|x₁|)=f(x₁),f(|x₂|)=f(x₂).

∴f(x₁)＜f(x₂).