Let the function f(x) = 2x^3 - 3(p-3)x^2 - 6(2p+1)x + 5 . The function f(x) has a local maximum at x₁ and a local minimum at x₂ such that x₁ 1 . Then, the set of all possible values of p is

Step-by-step solution

Given f(x) = 2x^3 - 3(p-3)x^2 - 6(2p+1)x + 5 . Differentiating with respect to x , we get: f'(x) = 6x^2 - 6(p-3)x - 6(2p+1) = 6(x^2 - (p-3)x - (2p+1)) For f(x) to have a local maximum at x₁ and a local minimum at x₂ , x₁ and x₂ must be the roots of f'(x) = 0 . Since the leading coefficient of the cubic is positive, the local maximum occurs at the smaller root and the local minimum at the larger root. Thus, x₁ We are given that x₁ 1 . This implies that both x = -1 and x = 1 lie strictly between the roots of the quad