Let f(x)=αxx+1,x≠−1. Then write the value of α satisfying f(f(x))=x for all x≠−1.
We have,
f(x)=axx+1
Now,
f(f(x))=xf(axx+1)=x⇒fa(axx+1)axx+1+1=x⇒a2xx+1ax+x+1x+1=x⇒a2xax+x+1=x⇒a2ax+x+1=1⇒a2=ax+x+1⇒a2−ax−(x+1)=0⇒a2−a(x+1)+a−(x+1)=0⇒a[(a)−(x+1)+1[a−(x+1)]=0⇒[a−(x+1)][a+1]=0⇒a+1=0
[∵a=x+1 does not satisfies f(f(x)) = x]
⇒a=−1