Let g(x) = f(x^2 - 2x) + f(a - x^2 + 2x) where f''(x) > 0 for all real x . It is given that g(x) has a local minimum at x = 3 . Let M be the sum of the squares of all points of local minima, and N be the sum of the squares of all points of local maxima of g(x) . The value of M + 2N is equal to

Step-by-step solution

Given g(x) = f(x^2 - 2x) + f(a - x^2 + 2x) . Differentiating with respect to x using the chain rule: g'(x) = f'(x^2 - 2x) (2x - 2) + f'(a - x^2 + 2x) (-2x + 2) g'(x) = 2(x - 1) [f'(x^2 - 2x) - f'(a - x^2 + 2x)] Since g(x) has a local minimum at x = 3 , g'(3) = 0 . g'(3) = 2(3 - 1) [f'(3^2 - 2(3)) - f'(a - 3^2 + 2(3))] = 4 [f'(3) - f'(a - 3)] = 0 Since f''(x) > 0 , f'(x) is strictly increasing, which means f' is a one-to-one function. Therefore, f'(3) = f'(a - 3) 3 = a - 3 a = 6 . Now, g'(x) = 2(x - 1) [f'(x^2 - 2x)