If the function
` f(x) = (2a-3)(x+2 sin3)+(a-1)(sin^4x+cos^4x)+log 2`
does not process critical poits , then

Options

A. `a in (-oo,4//3)cup(2,oo)`
B. `a in (4//3,2)`
C. `a in (4//3,oo)`
D. `a in (2,oo)`

Answer

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

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

does not possess critical points. This means that the derivative \( f'(x) \) should not equal zero for any \( x \).

### Step 1: Differentiate \( f(x) \)

We start by differentiating \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left[ (2a-3)(x + 2 \sin 3) + (a-1)(\sin^4 x + \cos^4 x) + \log 2 \right] \]

Since \( \log 2 \) is a constant, its derivative is zero. Thus, we have:

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

### Step 2: Differentiate \( \sin^4 x + \cos^4 x \)

Using the chain rule, we differentiate \( \sin^4 x \) and \( \cos^4 x \):

\[ \frac{d}{dx}(\sin^4 x) = 4 \sin^3 x \cos x \]
\[ \frac{d}{dx}(\cos^4 x) = -4 \cos^3 x \sin x \]

Thus,

\[ \frac{d}{dx}(\sin^4 x + \cos^4 x) = 4 \sin^3 x \cos x - 4 \cos^3 x \sin x = 4 \sin x \cos x (\sin^2 x - \cos^2 x) \]

### Step 3: Substitute back into \( f'(x) \)

Now substituting this back into our expression for \( f'(x) \):

\[ f'(x) = (2a-3) + (a-1) \cdot 4 \sin x \cos x (\sin^2 x - \cos^2 x) \]

Using the identity \( \sin 2x = 2 \sin x \cos x \), we can rewrite this as:

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

### Step 4: Set \( f'(x) \) not equal to zero

For \( f(x) \) to not have critical points, we need:

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

### Step 5: Analyze the conditions

The term \( \sin^2 x - \cos^2 x \) can take values between -1 and 1. Therefore, we need to analyze the two inequalities:

1. \( 2a - 3 + 2(a-1) \cdot 1 > 0 \)
2. \( 2a - 3 + 2(a-1) \cdot (-1) < 0 \)

#### Solving the first inequality:

\[ 2a - 3 + 2a - 2 > 0 \implies 4a - 5 > 0 \implies a > \frac{5}{4} \]

#### Solving the second inequality:

\[ 2a - 3 - 2a + 2 < 0 \implies -1 < 0 \text{ (always true)} \]

### Step 6: Combine the results

Thus, the only condition we have is:

\[ a > \frac{5}{4} \]

### Final Result

The function \( f(x) \) does not possess critical points for:

\[ a > \frac{5}{4} \]