Let f: R - 0, 1 R be a function satisfying a f(x) + b f ( x-1 x ) = x for all x R - 0, 1 , where a and b are real constants. If f(2) = 2 3 and f ( 1 2 ) = 2 3 , then the value of a^2 + 4b^2 is equal to :

Step-by-step solution

Given functional equation: a f(x) + b f ( x-1 x ) = x Substitute x = 2 : a f(2) + b f ( 1 2 ) = 2 Since f(2) = 2 3 and f ( 1 2 ) = 2 3 , we have: a ( 2 3 ) + b ( 2 3 ) = 2 a + b = 3 (1) Now, substitute x = 1 2 in the given equation: a f ( 1 2 ) + b f(-1) = 1 2 2a 3 + b f(-1) = 1 2 b f(-1) = 3 - 4a 6 (2) Substitute x = -1 in the given equation: a f(-1) + b f(2) = -1 a f(-1) + 2b 3 = -1 a f(-1) = -3 - 2b 3 (3) To eliminate f(-1) , multiply equation (2) by a and equation (3) by b : ab f(-1) = 3a - 4a^2 6 ab f(-1) = -3