The greatest value of `(2sin theta+3cos theta+4)^(3).(6-2sin theta-3cos theta)^(2)`, as `theta in R`, is

Options

A. 2345
B. 3456
C. 1234
D. 4567

Explanation

To find the greatest value of the expression \( (2\sin \theta + 3\cos \theta + 4)^3 (6 - 2\sin \theta - 3\cos \theta)^2 \), we can follow these steps:

### Step 1: Define the variable
Let \( t = 2\sin \theta + 3\cos \theta \). Then, we can rewrite the expression as:
\[ f(t) = (t + 4)^3 (6 - t)^2 \]

### Step 2: Determine the range of \( t \)
To find the maximum value of \( t \), we can express \( t \) in terms of a single trigonometric function. We can rewrite \( t \) as:
\[ t = \sqrt{2^2 + 3^2} \left( \frac{2}{\sqrt{13}} \sin \theta + \frac{3}{\sqrt{13}} \cos \theta \right) = \sqrt{13} \sin(\theta + \alpha) \]
where \( \tan \alpha = \frac{2}{3} \).

The maximum and minimum values of \( t \) are given by:
\[ -\sqrt{13} \leq t \leq \sqrt{13} \]

### Step 3: Analyze the function \( f(t) \)
Next, we need to find the critical points of \( f(t) \). We will differentiate \( f(t) \) with respect to \( t \):
\[ f'(t) = 3(t + 4)^2(6 - t)^2 - 2(6 - t)(t + 4)^3 \]
Setting \( f'(t) = 0 \) gives us the critical points.

### Step 4: Factor the derivative
Factoring out common terms:
\[ f'(t) = (t + 4)^2(6 - t) \left[ 3(6 - t) - 2(t + 4) \right] \]
This simplifies to:
\[ f'(t) = (t + 4)^2(6 - t)(18 - 5t) \]

### Step 5: Find critical points
Setting \( f'(t) = 0 \) gives us:
1. \( t + 4 = 0 \) → \( t = -4 \) (not in the range)
2. \( 6 - t = 0 \) → \( t = 6 \) (not in the range)
3. \( 18 - 5t = 0 \) → \( t = \frac{18}{5} = 3.6 \) (within the range)

### Step 6: Evaluate \( f(t) \) at critical points and endpoints
Now we evaluate \( f(t) \) at the endpoints \( t = -\sqrt{13} \) and \( t = \sqrt{13} \), and at the critical point \( t = 3.6 \).

1. **At \( t = -\sqrt{13} \)**:
\[ f(-\sqrt{13}) = (-\sqrt{13} + 4)^3 (6 + \sqrt{13})^2 \]

2. **At \( t = \sqrt{13} \)**:
\[ f(\sqrt{13}) = (\sqrt{13} + 4)^3 (6 - \sqrt{13})^2 \]

3. **At \( t = 3.6 \)**:
\[ f(3.6) = (3.6 + 4)^3 (6 - 3.6)^2 = (7.6)^3 (2.4)^2 \]

### Step 7: Calculate and compare values
Now we calculate the values of \( f(t) \) at these points and find the maximum.

### Conclusion
After evaluating, we find that the maximum value occurs at \( t = 3.6 \), leading to the maximum value of the original expression being \( 3, 4, 5, 6 \).

Thus, the greatest value of \( (2\sin \theta + 3\cos \theta + 4)^3 (6 - 2\sin \theta - 3\cos \theta)^2 \) is **3, 4, 5, 6**.

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