If `a lt int_(0)^(2pi) (1)/(10+3 cos x)dx lt b`. Then the ordered pair (a,b) is

Options

A. `((2pi)/(7),(2pi)/(3))`
B. `((2pi)/(13),(2pi)/(7))`
C. `((pi)/(10),(2pi)/(13))`
D. none of these

Answer

To solve the problem, we need to evaluate the integral \( I = \int_{0}^{2\pi} \frac{1}{10 + 3 \cos x} \, dx \) and find the bounds \( a \) and \( b \) such that \( a < I < b \).

### Step 1: Determine the range of the function \( 10 + 3 \cos x \)

The cosine function \( \cos x \) varies between -1 and 1. Therefore, we can find the minimum and maximum values of \( 10 + 3 \cos x \):

- Minimum value:
\[ 10 + 3(-1) = 10 - 3 = 7 \]
- Maximum value:
\[ 10 + 3(1) = 10 + 3 = 13 \]

Thus, we have:
\[ 7 \leq 10 + 3 \cos x \leq 13 \]

### Step 2: Find the bounds for \( \frac{1}{10 + 3 \cos x} \)

Taking the reciprocal of the inequalities (and reversing the inequalities since we are dealing with positive values):
\[ \frac{1}{13} \leq \frac{1}{10 + 3 \cos x} \leq \frac{1}{7} \]

### Step 3: Integrate the bounds

Now, we can integrate the bounds over the interval from \( 0 \) to \( 2\pi \):
\[ \int_{0}^{2\pi} \frac{1}{13} \, dx \leq \int_{0}^{2\pi} \frac{1}{10 + 3 \cos x} \, dx \leq \int_{0}^{2\pi} \frac{1}{7} \, dx \]

Calculating the integrals:
- For the lower bound:
\[ \int_{0}^{2\pi} \frac{1}{13} \, dx = \frac{1}{13} \cdot (2\pi - 0) = \frac{2\pi}{13} \]
- For the upper bound:
\[ \int_{0}^{2\pi} \frac{1}{7} \, dx = \frac{1}{7} \cdot (2\pi - 0) = \frac{2\pi}{7} \]

### Step 4: Combine the results

From the integrations, we have:
\[ \frac{2\pi}{13} \leq I \leq \frac{2\pi}{7} \]

### Conclusion

Thus, the ordered pair \( (a, b) \) is:
\[ \left( \frac{2\pi}{13}, \frac{2\pi}{7} \right) \]