Question

已知函数f(x)=ax-b/x-2lnx,f(1)=0,

Answer 1

1

f(1)=a-b=0,a=b

∴f(X)=ax-a/x-2lnx

f'(X)=a+a/x^2-2/x=(ax^2-2x+a)/x^2

根据定义域,x≠0,

∴x^2≠0,

使(-2)^2-4a^21或a0,为单调递增

f '(1/an+1)=[(1/an)·(an+1)]^2

=[1+1/an]^2

∴a(n+1)=f '(1/an+1)-nan +1

=[1+1/an]^2-nan +1>0

1/an^2+2/an-nan +2>0