Let `f(theta)= sin theta - cos^(2) theta-1`, where `theta in R` and `m le f(theta) le M` .
The value of `(4m+13)` is equal to

Options

A. 0
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to analyze the function \( f(\theta) = \sin \theta - \cos^2 \theta - 1 \) and find its minimum and maximum values, denoted as \( m \) and \( M \), respectively. We will then compute \( 4m + 13 \).

### Step-by-Step Solution:

1. **Rewrite the Function**:
\[ f(\theta) = \sin \theta - \cos^2 \theta - 1 \]
We can also express \( \cos^2 \theta \) in terms of \( \sin \theta \):
\[ \cos^2 \theta = 1 - \sin^2 \theta \]
Thus,
\[ f(\theta) = \sin \theta - (1 - \sin^2 \theta) - 1 = \sin \theta + \sin^2 \theta - 2 \]

2. **Differentiate the Function**:
To find the critical points, we differentiate \( f(\theta) \):
\[ f'(\theta) = \cos \theta + 2 \sin \theta \cos \theta = \cos \theta(1 + 2 \sin \theta) \]
Setting \( f'(\theta) = 0 \):
\[ \cos \theta = 0 \quad \text{or} \quad 1 + 2 \sin \theta = 0 \]

3. **Solve for Critical Points**:
- From \( \cos \theta = 0 \), we get:
\[ \theta = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
- From \( 1 + 2 \sin \theta = 0 \):
\[ \sin \theta = -\frac{1}{2} \]
This gives:
\[ \theta = \frac{7\pi}{6}, \frac{11\pi}{6} \quad \text{(in the third and fourth quadrants)} \]

4. **Evaluate \( f(\theta) \) at Critical Points**:
- For \( \theta = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) - \cos^2\left(\frac{\pi}{2}\right) - 1 = 1 - 0 - 1 = 0 \]
- For \( \theta = \frac{7\pi}{6} \):
\[ f\left(\frac{7\pi}{6}\right) = \sin\left(\frac{7\pi}{6}\right) - \cos^2\left(\frac{7\pi}{6}\right) - 1 \]
\[ = -\frac{1}{2} - \left(-\frac{\sqrt{3}}{2}\right)^2 - 1 = -\frac{1}{2} - \frac{3}{4} - 1 = -\frac{1}{2} - \frac{3}{4} - \frac{4}{4} = -\frac{9}{4} \]
- For \( \theta = \frac{11\pi}{6} \):
\[ f\left(\frac{11\pi}{6}\right) = \sin\left(\frac{11\pi}{6}\right) - \cos^2\left(\frac{11\pi}{6}\right) - 1 \]
\[ = -\frac{1}{2} - \left(\frac{\sqrt{3}}{2}\right)^2 - 1 = -\frac{1}{2} - \frac{3}{4} - 1 = -\frac{1}{2} - \frac{3}{4} - \frac{4}{4} = -\frac{9}{4} \]

5. **Determine Minimum and Maximum Values**:
- The maximum value \( M = 0 \) (from \( \theta = \frac{\pi}{2} \)).
- The minimum value \( m = -\frac{9}{4} \) (from \( \theta = \frac{7\pi}{6} \) and \( \theta = \frac{11\pi}{6} \)).

6. **Calculate \( 4m + 13 \)**:
\[ 4m + 13 = 4\left(-\frac{9}{4}\right) + 13 = -9 + 13 = 4 \]

### Final Answer:
\[ \boxed{4} \]