Let a > 0 . The function f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 attains its local maximum and minimum at x₁ and x₂ respectively. If the area bounded by the curve y = f'(x) , the x-axis, and the vertical lines x = x₁ and x = x₂ is 27 sq. units, then the value of x₂^2 - x₁^2 is equal to :

Step-by-step solution

Given f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 . Differentiating with respect to x : f'(x) = 6x^2 - 18ax + 12a^2 = 6(x^2 - 3ax + 2a^2) = 6(x - a)(x - 2a) For local extrema, f'(x) = 0 x = a or x = 2a . Since a > 0 , we have a Between x₁ and x₂ (i.e., a The required area is given by: Area = _ a ^ 2a |f'(x)| dx = - _ a ^ 2a f'(x) dx Using the Fundamental Theorem of Calculus: Area = -[f(x)]_ a ^ 2a = f(a) - f(2a) Evaluating the function at the critical points: f(a) = 2a^3 - 9a(a)^2 + 12a^2(a) + 1 = 2a^3 - 9a^3 + 12a^3 + 1 = 5a