The domain of the function `f(x)=sqrt(sinx+cosx)+sqrt(7x-x^2-6)` is

Options

A. `[1,6]`
B. `[1,(3pi)/(4)] cup [(7pi)/(4),6]`
C. `[1, pi] cup [(7pi)/(4),6]`
D. None of these

Explanation

To find the domain of the function \( f(x) = \sqrt{\sin x + \cos x} + \sqrt{7x - x^2 - 6} \), we need to ensure that both square root expressions are defined and non-negative.

### Step 1: Analyze the first term \( \sqrt{\sin x + \cos x} \)

The expression inside the square root, \( \sin x + \cos x \), must be non-negative:

\[ \sin x + \cos x \geq 0 \]

To analyze this, we can rewrite \( \sin x + \cos x \) using the identity:

\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]

Thus, we need:

\[ \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \geq 0 \]

This implies:

\[ \sin\left(x + \frac{\pi}{4}\right) \geq 0 \]

The sine function is non-negative in the intervals:

\[ n\pi \leq x + \frac{\pi}{4} \leq (n+1)\pi \quad \text{for } n \in \mathbb{Z} \]

Solving for \( x \):

\[ n\pi - \frac{\pi}{4} \leq x \leq (n+1)\pi - \frac{\pi}{4} \]

This gives us intervals of \( x \) for each integer \( n \).

### Step 2: Analyze the second term \( \sqrt{7x - x^2 - 6} \)

The expression inside this square root must also be non-negative:

\[ 7x - x^2 - 6 \geq 0 \]

Rearranging gives:

\[ -x^2 + 7x - 6 \geq 0 \]

Factoring the quadratic:

\[ -(x^2 - 7x + 6) \geq 0 \implies (x - 1)(x - 6) \leq 0 \]

The solutions to this inequality are found by determining the intervals where the product is non-positive. The roots are \( x = 1 \) and \( x = 6 \). The intervals to check are:

1. \( (-\infty, 1) \)
2. \( (1, 6) \)
3. \( (6, \infty) \)

Testing points in these intervals, we find that \( (x - 1)(x - 6) \leq 0 \) holds for:

\[ x \in [1, 6] \]

### Step 3: Combine the conditions

Now we need to find the intersection of the intervals derived from both conditions.

- From the first term, we have intervals based on \( \sin\left(x + \frac{\pi}{4}\right) \geq 0 \).
- From the second term, we have \( x \in [1, 6] \).

### Step 4: Finding the intersection

The intervals from the first term will vary based on \( n \). However, we can focus on the specific intervals that intersect with \( [1, 6] \).

1. For \( n = 0 \):
\[ 0 \leq x \leq \frac{3\pi}{4} \quad \text{(approximately 2.36)} \]
This interval intersects with \( [1, 6] \) as \( [1, \frac{3\pi}{4}] \).

2. For \( n = 1 \):
\[ \pi - \frac{\pi}{4} \leq x \leq 2\pi - \frac{\pi}{4} \quad \text{(approximately 3.14 to 5.89)} \]
This interval intersects with \( [1, 6] \) as \( [\pi, 6] \).

### Final Domain

The final domain of \( f(x) \) is:

\[ x \in [1, \frac{3\pi}{4}] \cup [\pi, 6] \]