For a suitably chosen real constant a, let a function, f:R-{-a}→R be defined by
f(x)=a-xa+x . Further suppose that for any real number x≠-a and
f(x)≠-a,( fof )(x)=x . Then f-12 is equal to :
13
-13
-3
3
( fof )(x)=x⇒a-f(x)a+f(x)=x⇒a-a-xa+xa+a-xa+x=x⇒a=1f(x)=1-x1+x∴f-12=3