The set of all values of a for which the function `f(x)=(a^2-3a+2)(cos^2\ x/4-sin^2\ x/4)+(a-1)x+sin1` does not possess critical points is (A) `[1,oo)` (B) `(0,1) uu (1,4)` (C) `(-2,4)` (D) `(1,3) uu (3,5)`

Options

A. `[1,oo]`
B. `(0,1)cup (1,4)`
C. (-2,4)
D. `(1,3)cup (3,5)`

Explanation

To solve the problem, we need to find the set of all values of \( a \) for which the function

\[ f(x) = (a^2 - 3a + 2)(\cos^2 \frac{x}{4} - \sin^2 \frac{x}{4}) + (a - 1)x + \sin 1 \]

does not possess critical points.

### Step 1: Simplify the function
We can use the identity \( \cos^2 \theta - \sin^2 \theta = \cos 2\theta \). Thus, we can rewrite the function as:

\[ f(x) = (a^2 - 3a + 2) \cos \frac{x}{2} + (a - 1)x + \sin 1 \]

### Step 2: Differentiate the function
Next, we differentiate \( f(x) \) with respect to \( x \):

\[ f'(x) = (a^2 - 3a + 2) \cdot \left(-\frac{1}{2} \sin \frac{x}{2}\right) + (a - 1) \]

This simplifies to:

\[ f'(x) = -\frac{1}{2}(a^2 - 3a + 2) \sin \frac{x}{2} + (a - 1) \]

### Step 3: Set the derivative to zero
For the function to not possess critical points, \( f'(x) \) should never equal zero for any \( x \). Therefore, we set:

\[ -\frac{1}{2}(a^2 - 3a + 2) \sin \frac{x}{2} + (a - 1) = 0 \]

Rearranging gives:

\[ (a - 1) = \frac{1}{2}(a^2 - 3a + 2) \sin \frac{x}{2} \]

### Step 4: Analyze the conditions
For \( f'(x) \) to never be zero, the right-hand side must not be able to equal the left-hand side for any \( x \). This means:

1. \( a - 1 \) must not be zero, hence \( a \neq 1 \).
2. The term \( \frac{1}{2}(a^2 - 3a + 2) \) must also be such that it does not allow the equation to hold true for any value of \( \sin \frac{x}{2} \) (which ranges from -1 to 1).

### Step 5: Solve the quadratic
The quadratic \( a^2 - 3a + 2 = 0 \) can be factored as:

\[ (a - 1)(a - 2) = 0 \]

Thus, the roots are \( a = 1 \) and \( a = 2 \).

### Step 6: Determine intervals
To ensure \( f'(x) \) does not equal zero, we need to analyze the sign of \( a^2 - 3a + 2 \):

- If \( a < 1 \), then \( a^2 - 3a + 2 > 0 \).
- If \( 1 < a < 2 \), then \( a^2 - 3a + 2 < 0 \).
- If \( a > 2 \), then \( a^2 - 3a + 2 > 0 \).

Thus, \( f'(x) \) does not possess critical points when \( a \in (-\infty, 1) \cup (2, \infty) \).

### Step 7: Finalize the answer
The intervals where \( f'(x) \) does not equal zero are:

- \( (-\infty, 1) \)
- \( (2, \infty) \)

However, we need to check the provided options for the correct intervals. The only option that fits this description is:

**Option B: \( (0, 1) \cup (1, 4) \)**