Let f: R R be a twice differentiable function satisfying the determinant equation vmatrix f(x+y) & 2x + 2y f(x-y) & 2x - 2y vmatrix = 0 for all x, y R . If f'(0) = 4 , then the value of |f'' ( 12 ) | is:

Step-by-step solution

Expanding the given determinant, we get: f(x+y)( 2x - 2y) - f(x-y)( 2x + 2y) = 0 Using the sum-to-product trigonometric identities: 2x - 2y = 2 (x+y) (x-y) 2x + 2y = 2 (x+y) (x-y) Substituting these into the equation: f(x+y)(2 (x+y) (x-y)) = f(x-y)(2 (x+y) (x-y)) Dividing both sides by 2 (x+y) (x+y) (x-y) (x-y) , we obtain: f(x+y) (x+y) (x+y) = f(x-y) (x-y) (x-y) Multiplying the numerator and denominator by 2: f(x+y) 2(x+y) = f(x-y) 2(x-y) Since this holds for all x, y R , the ratio must be a constant K . Thus, f(x