91Ó°ÊÓ

We will first find the first derivative of the given function \(f(x)\): \(\frac{d}{dx}(ax^3 + bx^2 + cx + d) = 3ax^2 + 2bx + c\).

Given that \(f'(\alpha) = f'(\beta) = 0\) with \(\alpha \neq \beta\), we can write the system of equations as follows: (1) \(3a\alpha^2 + 2b\alpha + c = 0\) (2) \(3a\beta^2 + 2b\beta + c = 0\)

To prove or disprove option (C), note that if \(\alpha\) is a point of local maxima and \(\beta\) is a point of local minima, the second derivative of \(f''(x)\) at these points should follow: - \(f''(\alpha) < 0\) (since \(\alpha\) is a local maximum) - \(f''(\beta) > 0\) (since \(\beta\) is a local minimum) Calculate the second derivative \(f''(x)\): \(\frac{d^2}{dx^2}(3ax^2 + 2bx + c) = 6ax+2b\). Now, let's substitute \(\alpha\) and \(\beta\) into the second derivative, obtaining: \(f''(\alpha) = 6a\alpha+2b\) \(f''(\beta) = 6a\beta+2b\) We have two cases: 1) \(a > 0\), it implies \(6a(\alpha - \beta) < 0\), so \(\alpha < \beta\) 2) \(a < 0\), it implies \(6a(\alpha - \beta) > 0\), so \(\alpha > \beta\) In the second case, \(\alpha\) cannot be a local maximum, and \(\beta\) cannot be a local minimum. Thus, option (C) is true.

To address the other options, we will analyze the given conditions and the function \(f(x)\). Using continuous functions, we can conclude the following: - (A): If \(f(\alpha) < 0\) and \(f(\beta) < 0\), and both points \(\alpha\) and \(\beta\) are local maxima or minima, it ensures that \(f(x)\) will change its sign at least once. Therefore, there must be at least one real root between \(\alpha\) and \(\beta\). However, this does not guarantee that all three roots are real. - (B): If \(f(\alpha) > 0\) and \(f(\beta) > 0\), similar to case (A), there is no guarantee that all three roots are real. So, option (B) is false. - (D): If \(a > 0\) and \(\beta > \alpha\), from the analysis in step 3, we have that \(f''(x) = 6ax + 2b\) will be negative for any \(x \in (\alpha, \beta)\) because \(a > 0\). This contradicts with the fact that \(f(x)\) should be decreasing on \((\alpha, \beta)\). Thus, option (D) is false. The answer is: Options (A) and (B) are false, option (C) is true, and option (D) is false.