For a suitable chosen real constant a, let a function f : R - - a R be defined by f (x)= a -x a +x . Further suppose that for any real number x - a and f (x) - a , (fof) (x)=x . Then f (- 1 5 ) is equal to

Step-by-step solution

Given: f (x)= a -x a +x aligned & f ( f (x))=x & a - f (x) a + f (x) =x & a - ( a -x a +x ) a + ( a -x a +x ) =x & a ^2+ a x- a +x a ^2+ a x+ a -x =x & ( a ^2- a )+( a +1) x= ( a ^2+ a ) x+( a -1) x^2 & ( a -1) x^2+ ( a ^2-1 ) x- a ^2+ a =0 & ( a -1) [x^2+( a +1) x- a ]=0 aligned This is possible when a =1 aligned & f(x)= 1-x 1+x & aligned f ( -1 5 ) & = 1- ( -1 5 ) 1+ ( -1 5 ) & = 1+ 1 5 1- 1 5 & = 6 5 4 5 = 6 4 =1.5 aligned aligned