The function used is f(f(x)).

First, x =1>; 0, with f (x) = lgx (x >; 0), you can get f( 1)=lg 1=0.

Bring f( 1)=0 into f(f(x)) and get f(f( 1))=f(0)= 1.

At this time, for f(x), x takes 0 and is calculated by another expression. After expansion, f (x) = x+a 3 (x

Bring in x=0, f (0) = a 3 = 1,

So a= 1