Let f(x) = ₀^ x^2-4x t^2-4t+3 1+t^2 dt . Let M and m be the x -coordinates of the points where f(x) attains its absolute maximum and absolute minimum respectively, in the interval [1, 4] . The value of M^2 + m^2 is :

Step-by-step solution

Using Leibnitz's rule, we differentiate f(x) with respect to x : f'(x) = (x^2-4x)^2 - 4(x^2-4x) + 3 1+(x^2-4x)^2 (2x - 4) Let y = x^2 - 4x . For x [1, 4] , the minimum value of y occurs at x=2 where y = -4 , and the maximum is at x=4 where y = 0 . Thus, y [-4, 0] . The numerator of the first term is y^2 - 4y + 3 = (y-1)(y-3) . Since y 0 , both (y-1) and (y-3) are strictly negative, making their product (y-1)(y-3) > 0 . The denominator 1+y^2 is always positive. Therefore, the sign of f'(x) is entirely determined by