For what values of a does the curve `f(x)=x(a^2-2a-2)+cosx` is always strictly monotonic decreasing `AA x in R`

Options

A. `a in R`
B. `a gt 2`
C. `1-sqrt2 lt a lt 1 + sqrt2`
D. none of these

Answer

To determine the values of \( a \) for which the curve \( f(x) = x(a^2 - 2a - 2) + \cos x \) is always strictly monotonically decreasing for \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Find the derivative of \( f(x) \)
To check if the function is monotonically decreasing, we need to find its derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}[x(a^2 - 2a - 2) + \cos x] \]
Using the product rule and the derivative of cosine:
\[ f'(x) = (a^2 - 2a - 2) + (-\sin x) \]
Thus, we have:
\[ f'(x) = a^2 - 2a - 2 - \sin x \]

### Step 2: Set the condition for strict monotonicity
For \( f(x) \) to be strictly monotonically decreasing, we need:
\[ f'(x) < 0 \quad \forall x \in \mathbb{R} \]
This implies:
\[ a^2 - 2a - 2 - \sin x < 0 \]
Since \( \sin x \) varies between -1 and 1, we can analyze the extremes:
\[ a^2 - 2a - 2 - 1 < 0 \quad \text{and} \quad a^2 - 2a - 2 + 1 < 0 \]

### Step 3: Solve the inequalities
1. **First inequality**:
\[ a^2 - 2a - 3 < 0 \]
Factoring the quadratic:
\[ (a - 3)(a + 1) < 0 \]
The roots are \( a = -1 \) and \( a = 3 \). The intervals where this inequality holds are:
\[ -1 < a < 3 \]

2. **Second inequality**:
\[ a^2 - 2a - 1 < 0 \]
Using the quadratic formula:
\[ a = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]
The roots are \( a = 1 - \sqrt{2} \) and \( a = 1 + \sqrt{2} \). The intervals where this inequality holds are:
\[ 1 - \sqrt{2} < a < 1 + \sqrt{2} \]

### Step 4: Find the intersection of the intervals
Now we need to find the intersection of the two intervals:
1. \( -1 < a < 3 \)
2. \( 1 - \sqrt{2} < a < 1 + \sqrt{2} \)

Calculating \( 1 - \sqrt{2} \) and \( 1 + \sqrt{2} \):
- \( \sqrt{2} \approx 1.414 \)
- Thus, \( 1 - \sqrt{2} \approx -0.414 \) and \( 1 + \sqrt{2} \approx 2.414 \).

The intersection is:
\[ -0.414 < a < 2.414 \]

### Final Result
The values of \( a \) for which the curve \( f(x) \) is always strictly monotonically decreasing are:
\[ 1 - \sqrt{2} < a < 1 + \sqrt{2} \]