Let A = bmatrix 1 & 3 & -1 2 & 1 & 0 & 1 & -1 bmatrix be a singular matrix. Let f(x) = ₀^x (t^2 + 2t + 3) ,dt , x [1, ] . If M and m are respectively the maximum and the minimum values of f in [1, ] , then 3(M - m) is equal to :

Step-by-step solution

Since A is a singular matrix, |A| = 0 . |A| = 1(-1 - ) - 3(-2 - 0) - 1(2 - 0) = 0 -1 - + 6 - 2 = 0 = 3 The function is f(x) = ₀^x (t^2 + 2t + 3) ,dt = x^3 3 + x^2 + 3x Differentiating with respect to x , we get f'(x) = x^2 + 2x + 3 = (x+1)^2 + 2 > 0 for all x . Thus, f(x) is a strictly increasing function on the interval [1, 3] . The maximum value M occurs at x = 3 : M = f(3) = 3^3 3 + 3^2 + 3(3) = 9 + 9 + 9 = 27 The minimum value m occurs at x = 1 : m = f(1) = 1^3 3 + 1^2 + 3(1) = 13 3 We need to find 3(M - m) : 3