First slide
Functions (XII)
Question

 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 : 

Moderate
Solution

( fof )(x)=xa-f(x)a+f(x)=xa-a-xa+xa+a-xa+x=xa=1f(x)=1-x1+xf-12=3

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