Q.

Let f(x)=αxx+1, x≠-1. Then, for what value of α is f(f(x))=x                     [2001]

1 2  
2 -2  
3 1  
4 -1  

Ans.

(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