The set of all values of k for which the function `f(x) = (k^2 - 3k + 2) (cos^2 "" (x)/4 - sin^2 "" (x)/4) + ( k - 1)x + sin 1` does not possess critical points is

Options

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

Answer

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

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

does not possess critical points. Critical points occur where the first derivative \( f'(x) \) is equal to zero. Therefore, we need to find the derivative and determine the conditions under which it does not equal zero.

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

The function can be rewritten as:

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

Using the identity \( \cos^2(x) - \sin^2(x) = \cos(2x) \), we can express \( f(x) \) as:

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

Now, we differentiate \( f(x) \):

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

This simplifies to:

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

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

To find critical points, we set \( f'(x) = 0 \):

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

Rearranging gives:

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

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

For the function \( f(x) \) to have no critical points, the equation must not have solutions. This occurs if:

1. \( k^2 - 3k + 2 = 0 \) (which would make the coefficient of \( \sin(2x) \) zero)
2. The right-hand side \( (k - 1) \) must also be zero.

First, we solve \( k^2 - 3k + 2 = 0 \):

Factoring gives:

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

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

Next, if \( k = 1 \), then \( k - 1 = 0 \), which means \( f'(x) = 0 \) for all \( x \).

If \( k = 2 \), then \( f'(x) = 0 \) for \( \sin(2x) = 0 \), which has solutions.

### Step 4: Determine the intervals for \( k \)

To ensure \( f'(x) \) does not equal zero, we must have:

- \( k \neq 1 \)
- \( k \neq 2 \)

Now, we analyze the behavior of \( f'(x) \) for values of \( k \):

1. If \( k < 1 \) or \( k > 2 \), \( f'(x) \) can be non-zero.
2. If \( 1 < k < 2 \), \( f'(x) \) can still equal zero.

Thus, the set of values of \( k \) for which \( f(x) \) does not have critical points is:

\[ k \in (-\infty, 1) \cup (1, 2) \cup (2, \infty) \]

### Final Answer

The set of all values of \( k \) for which the function does not possess critical points is:

\[ k \in (-\infty, 1) \cup (1, 2) \cup (2, \infty) \]