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Question

Let f(x)=αxx+1,x1. Then write the value of α satisfying f(f(x))=x for all x1.

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Solution

We have,
f(x)=axx+1
Now,
f(f(x))=xf(axx+1)=xfa(axx+1)axx+1+1=xa2xx+1ax+x+1x+1=xa2xax+x+1=xa2ax+x+1=1a2=ax+x+1a2ax(x+1)=0a2a(x+1)+a(x+1)=0a[(a)(x+1)+1[a(x+1)]=0[a(x+1)][a+1]=0a+1=0
[a=x+1 does not satisfies f(f(x)) = x]
a=1


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