Let f(x)=αxx+1, x≠-1. Then, for what value of α is f(f(x))=x [2001]
(4)
f(x)=αxx+1, x≠-1
Now, f(f(x))=x⇒α(αxx+1)αxx+1+1=x
⇒α2x(α+1)x+1=x⇒(α+1)x2+(1-α2)x=0
⇒α+1=0 and 1-α2=0
∴ α=-1