A cubic polynomial curve y = f(x) = ax^3 + bx^2 + cx + d has a point of inflection at x = 1 and passes through the point (1, 1) . If the tangent to the curve at the point (0, 2) is given by the line y = 3x + 2 , then the value of a^2 + b^2 + c^2 + d^2 is equal to:

Step-by-step solution

Let f(x) = ax^3 + bx^2 + cx + d . Since the tangent to the curve at (0, 2) is y = 3x + 2 , the curve passes through (0, 2) . Thus, f(0) = 2 d = 2 . The slope of the tangent at x = 0 is 3 . f'(x) = 3ax^2 + 2bx + c f'(0) = 3 c = 3 . The curve has a point of inflection at x = 1 , so f''(1) = 0 . f''(x) = 6ax + 2b f''(1) = 6a + 2b = 0 3a + b = 0 b = -3a . The curve passes through (1, 1) , so f(1) = 1 . a + b + c + d = 1 a + (-3a) + 3 + 2 = 1 -2a + 5 = 1 -2a = -4 a = 2 . Then b = -3(2) = -6 . We have a = 2 , b = -6 , c