If f (x)= ( x) then f (x^2 ) f ( y ^2 )- 1 2 [ f ( x^2 y ^2 )+ f (x^2 y ^2 ) ] has the value

Step-by-step solution

Given f (x) = ( x) , evaluate f (x^2) f (y^2) - 1 2 [ f ( x^2 y^2 ) + f (x^2 y^2) ] . Using logarithmic properties, f (x^2) = (2 x) , f (y^2) = (2 y) , f ( x^2 y^2 ) = (2 x - 2 y) , and f (x^2 y^2) = (2 x + 2 y) . Let A = 2 x and B = 2 y . The expression becomes A B - 1 2 [ (A - B) + (A + B)] . By the product-to-sum identity A B = 1 2 [ (A - B) + (A + B)] , substitution yields 1 2 [ (A - B) + (A + B)] - 1 2 [ (A - B) + (A + B)] = 0 . 0