91Ó°ÊÓ

To find the critical points of the function, we first need to find the derivative of \(f(x)\). We will find the derivative by calculating: \(f'(x) = \frac{d}{dx} (3+2(a+1) x+\left(a^{2}+1\right) x^{2}-x^{3})\) Applying the power rule for differentiation, we have: \(f'(x) = 2(a+1) + 2(a^{2}+1)(2x) - 3x^{2}\)

Critical points occur where the derivative of the function is equal to zero or is undefined. In this case, the derivative is always defined. So, we solve for \(x\) when: \(2(a+1) + 2(a^{2}+1)(2x) - 3x^{2} = 0\) Rearrange the equation: \(3x^{2} - 4(a^{2}+1)x - 2(a+1) = 0\) This is a quadratic equation in x.

We will find the roots of the quadratic equation using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) For our quadratic equation, we have a = 3, b = -4(a^{2}+1), and c = -2(a+1). Plugging these values into the formula, we get: \(x = \frac{4(a^{2}+1) \pm \sqrt{(-4(a^{2}+1))^{2} - 4(3)(-2(a+1))}}{2(3)}\) \(x = \frac{4(a^{2}+1) \pm \sqrt{16(a^{2}+1)^{2} + 24(a+1)}}{6}\) We now have two solutions for \(x\), \(x = x_{1}\) and \(x = x_{2}\).

Given that there is a local minimum at \(x=x_{1}\) and a local maximum at \(x=x_{2}\), this means the second derivative of \(f(x)\) will be positive for the local minimum and negative for the local maximum. We also know \(x_1 < 2 < x_2\). Let's find the second derivative and use these conditions: The second derivative, \(f''(x)\), is given by: \(f''(x) = \frac{d}{dx} (2(a+1) + 2(a^{2}+1)(2x) - 3x^{2})\) \(f''(x) = 4(a^{2}+1) - 6x\) Now, we apply the conditions: \(f''(x_1) > 0 \Rightarrow 4(a^{2}+1) - 6x_1 > 0\) \(f''(x_2) < 0 \Rightarrow 4(a^{2}+1) - 6x_2 < 0\) By using the fact that \(x_1 < 2 < x_2\), we can deduce that: \(4(a^{2}+1) - 6x_1 > 4(a^{2}+1) - 12 > 4(a^{2}+1) - 6x_2\) This inequality gives the required range for \(a\):

From the inequality we derived in step 4: \(4(a^{2}+1) - 12 > 0\) \(a^{2} > \frac{3}{2}\) Now, we can see that \(a\) belongs to the interval: \(a \in \left(-\infty, -\sqrt{\frac{3}{2}}\right) \cup \left(\sqrt{\frac{3}{2}}, \infty\right)\) Comparing this with the given options (A) \(\left(-\infty,-\frac{3}{2}\right)\) (B) \(\left(-\frac{3}{2}, 1\right)\) (C) \((0, \infty)\) (D) \((1, \infty)\), we can see that the final answer is a combination of options (A) and (D): The variable \(a\) belongs to the interval(s) \(\left(-\infty, -\sqrt{\frac{3}{2}}\right) \cup \left(\sqrt{\frac{3}{2}}, \infty\right)\), which corresponds to options (A) and (D).