Let a differentiable function f satisfy the equation ₀³⁶ f ( t x 36 ) d t=4 f(x) . If y=f(x) is a standard parabola passing through the points (2,1) and (-4, ) , then ^ is equal to _ _ _ _ .

Step-by-step solution

From ₀³⁶ f ( tx 36 ) dt = 4 f(x) , substitute u = tx 36 to get 36 x ₀^x f(u) du = 4 f(x) . Differentiating: f(x) = 9 [f(x) + xf'(x)] , which simplifies to (9- )f(x) = x f'(x) . This gives f'(x) f(x) = 9- x , integrating to f(x) = Kx^ 9- . For f to be a standard parabola: 9- = 2 = 3 . Thus f(x) = Kx^2 . Using point (2,1) : K = 1 4 , so f(x) = x^2 4 . At (-4, ) : = 16 4 = 4 . Therefore ^ = 4^3 = 64