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 of finding the set of all values of \( a \) for which the function

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

does not possess a critical point, we need to follow these steps:

### Step 1: Find the derivative of the function

The first step is to differentiate \( f(x) \) with respect to \( x \):

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

Since \( \sin 1 \) is a constant, its derivative is 0. The derivative of \( (a - 1)x \) is simply \( a - 1 \). For the term involving \( \cos x \):

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

### Step 2: Set the derivative not equal to zero

For the function to not possess a critical point, we need:

\[ f'(x) \neq 0 \]

This leads to:

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

Rearranging gives:

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

### Step 3: Analyze the conditions

This inequality must hold for all \( x \). Since \( \sin x \) varies between -1 and 1, we analyze two cases:

1. **When \( \sin x = 1 \)**:
\[ -\frac{(a^2 - 3a + 2)}{2} + (a - 1) \neq 0 \]

Simplifying gives:
\[ (a - 1) \neq \frac{(a^2 - 3a + 2)}{2} \]

2. **When \( \sin x = -1 \)**:
\[ \frac{(a^2 - 3a + 2)}{2} + (a - 1) \neq 0 \]

Simplifying gives:
\[ (a - 1) \neq -\frac{(a^2 - 3a + 2)}{2} \]

### Step 4: Solve the inequalities

From the first case:
\[ 2(a - 1) \neq a^2 - 3a + 2 \]
\[ 0 \neq a^2 - 5a + 4 \]
Factoring gives:
\[ 0 \neq (a - 1)(a - 4) \]

Thus, \( a \) should not be in the interval \( [1, 4] \).

From the second case:
\[ 2(a - 1) \neq - (a^2 - 3a + 2) \]
\[ 0 \neq a^2 - a \]
Factoring gives:
\[ 0 \neq a(a - 1) \]

Thus, \( a \) should not be equal to 0 or 1.

### Step 5: Combine the conditions

Combining both conditions, we find:
- \( a \) cannot be in the interval \( [1, 4] \)
- \( a \) cannot be equal to 0 or 1

Thus, the set of all values of \( a \) for which \( f(x) \) does not possess a critical point is:

\[ a \in (-\infty, 0) \cup (0, 1) \cup (4, \infty) \]

### Final Answer
The set of all values \( a \) for which \( f(x) \) does not possess a critical point is:

\[ (-\infty, 0) \cup (0, 1) \cup (4, \infty) \]