Find the set of all values of the parameter 'a' for which the function, `f(x) = sin 2x - 8(a + b)sin x + (4a^2 + 8a - 14)x` increases for all `x in R` and has no critical points for all `a in R.`

Options

A. `(-oo, -sqrt5,-2)`
B. `(1,oo)`
C. `(sqrt5,oo)`
D. None of these

Explanation

To find the set of all values of the parameter 'a' for which the function \( f(x) = \sin(2x) - 8(a + b)\sin(x) + (4a^2 + 8a - 14)x \) increases for all \( x \in \mathbb{R} \) and has no critical points, we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(\sin(2x)) - 8(a + b)\frac{d}{dx}(\sin(x)) + \frac{d}{dx}((4a^2 + 8a - 14)x) \]
Using the derivatives:
\[ f'(x) = 2\cos(2x) - 8(a + b)\cos(x) + (4a^2 + 8a - 14) \]

### Step 2: Analyze the derivative
For \( f(x) \) to be increasing for all \( x \), \( f'(x) \) must be greater than 0 for all \( x \):
\[ f'(x) = 2\cos(2x) - 8(a + b)\cos(x) + (4a^2 + 8a - 14) > 0 \]

### Step 3: Determine the bounds of the cosine terms
The cosine functions \( \cos(2x) \) and \( \cos(x) \) oscillate between -1 and 1. Therefore, we can analyze the extreme cases:
1. When \( \cos(2x) = 1 \) and \( \cos(x) = 1 \):
\[ f'(x) \leq 2 - 8(a + b) + (4a^2 + 8a - 14) > 0 \]
Simplifying this gives:
\[ 4a^2 + (8 - 8b)a - 12 > 0 \]

2. When \( \cos(2x) = -1 \) and \( \cos(x) = -1 \):
\[ f'(x) \geq -2 + 8(a + b) + (4a^2 + 8a - 14) > 0 \]
Simplifying this gives:
\[ 4a^2 + (8 + 8b)a - 16 > 0 \]

### Step 4: Solve the inequalities
We need to solve the inequalities derived from the extreme cases:
1. \( 4a^2 + (8 - 8b)a - 12 > 0 \)
2. \( 4a^2 + (8 + 8b)a - 16 > 0 \)

### Step 5: Analyze the critical points
To ensure that there are no critical points, we must ensure that \( f'(x) \) does not equal zero for any \( x \). This means that the discriminants of the quadratic equations must be less than zero.

### Step 6: Find the values of 'a'
After analyzing the discriminants:
1. For \( 4a^2 + (8 - 8b)a - 12 \):
\[ \Delta_1 < 0 \implies (8 - 8b)^2 - 4 \cdot 4 \cdot (-12) < 0 \]
This will give us a range for \( a \).

2. For \( 4a^2 + (8 + 8b)a - 16 \):
\[ \Delta_2 < 0 \implies (8 + 8b)^2 - 4 \cdot 4 \cdot (-16) < 0 \]
This will also provide a range for \( a \).

### Step 7: Combine the results
The final result will be a union of the intervals derived from both inequalities, ensuring that both conditions are satisfied.

### Final Answer
The set of all values of the parameter \( a \) for which the function \( f(x) \) is increasing for all \( x \in \mathbb{R} \) and has no critical points is:
\[ a \in (-\infty, -2) \cup (-\sqrt{6}, \sqrt{6}) \cup (2, \infty) \]