The set of all values a for which `f(x) = (a^(2) - 3a+2)(cosx/2)+(a-1)x + sin 1` does not posses critical point, is

Options

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

Answer

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

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

does not possess critical points. Critical points occur where the derivative \( f'(x) = 0 \).

### Step 1: Find the derivative \( f'(x) \)

The function can be differentiated term by term. The derivative of \( f(x) \) is:

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

### Step 2: Set the derivative equal to zero

For \( f(x) \) to not have critical points, we need \( f'(x) \neq 0 \) for all \( x \in \mathbb{R} \).

Setting \( f'(x) = 0 \):

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

Rearranging gives:

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

### Step 3: Analyze the conditions for no critical points

For \( f'(x) \) to not equal zero for any \( x \), we need:

1. \( a^2 - 3a + 2 \neq 0 \)
2. \( (a - 1) \neq 0 \)
3. The term \( \sin \frac{x}{2} \) must not allow the equation to hold true for any \( x \).

### Step 4: Solve the quadratic equation

The quadratic \( a^2 - 3a + 2 = 0 \) can be factored as:

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

Thus, \( a = 1 \) or \( a = 2 \).

### Step 5: Determine the conditions

From \( (a - 1) \neq 0 \), we have \( a \neq 1 \).

From \( a^2 - 3a + 2 \neq 0 \), we have \( a \neq 1 \) and \( a \neq 2 \).

### Step 6: Analyze the sine term

The sine function oscillates between -1 and 1. For \( f'(x) \) to not equal zero, we need:

\[ \frac{-(a - 1)}{(a^2 - 3a + 2)(-\frac{1}{2})} \text{ must not be solvable for any } x. \]

This means \( |a - 2| > 2 \) (as derived from the sine function's range).

### Step 7: Solve the inequality

The inequality \( |a - 2| > 2 \) gives us two conditions:

1. \( a - 2 > 2 \) or \( a > 4 \)
2. \( a - 2 < -2 \) or \( a < 0 \)

### Step 8: Combine all conditions

Thus, combining all conditions, we have:

- \( a < 0 \) or \( a > 4 \)
- \( a \neq 1 \)
- \( a \neq 2 \)

### Final Answer

The set of all values \( a \) for which \( f(x) \) does not possess critical points is:

\[ (-\infty, 0) \cup (4, \infty) \]