If 2 f (x^2 )+3 f ( 1 x^2 )=x^2-1, x R- 0 , then f (x^8 ) is equal to

Step-by-step solution

Given the equation 2 f(x^2) + 3 f ( 1 x^2 ) = x^2 - 1 . Let t = x^2 . Then the equation becomes 2 f(t) + 3 f ( 1 t ) = t - 1 for all t > 0 . Replacing t with 1 t , we get 2 f ( 1 t ) + 3 f(t) = 1 t - 1 . We have a system of two linear equations in f(t) and f ( 1 t ) : 1) 2 f(t) + 3 f ( 1 t ) = t - 1 2) 3 f(t) + 2 f ( 1 t ) = 1 t - 1 Multiply equation (1) by 2 and equation (2) by 3: 4 f(t) + 6 f ( 1 t ) = 2t - 29 f(t) + 6 f ( 1 t ) = 3 t - 3 Subtracting the first from the second: 5 f(t) = 3 t - 3 - (2t - 2) = 3 t -