Question

已知函数f(x)x∈R满足f(x)=2bx/ax-1,a≠0,f(1)=1,且使f(x)=2x成立的实数只有一个,若数列

Answer 1

1=f(1)=2b/(a-1),a=2b+1.

2x=f(x)=2bx/(ax-1),x不等于1/a.

2ax^2 - 2x = 2bx,

0=ax^2-(1+b)x=ax[x-(1+b)/a],(1+b)/a=0,b=-1,a=2b+1=-1.

f(x)=-2x/(-x-1)=2x/(x+1).

a(n+1)=f[a(n)]=2a(n)/[a(n)+1],

若a(n+1)=0,则a(n)=0,...,a(1)=0,与a(1)=2/3矛盾.

因此,a(n)不等于0.

1/a(n+1)=[a(n)+1]/[2a(n)]=(1/2)[1/a(n)] + 1/2,

1/a(n+1) - 1 = (1/2)[1/a(n) - 1]

{1/a(n)-1}是首项为1/a(1)-1=1/2,公比为(1/2)的等比数列.

1/a(n) - 1 = (1/2)^n = b(n),

1/a(n) = 1 + 1/2^n = [2^n+1]/2^n

a(n)=2^n/[1+2^n],

a(n)b(n)=1/[1+2^n]