The domain of definition of function
`f(x)=log(sqrt(x^(2)-5x-24)-x-2)`, is

Options

A. `(-infty,-3]`
B. `(-infty,-3] cup [8, infty)`
C. `(-infty,(-28)/9)`
D. None of these

Answer

To find the domain of the function \( f(x) = \log(\sqrt{x^2 - 5x - 24} - x - 2) \), we need to ensure that the argument of the logarithm is positive. This means we need to solve the inequality:

\[ \sqrt{x^2 - 5x - 24} - x - 2 > 0 \]

### Step 1: Isolate the square root
First, we isolate the square root:

\[ \sqrt{x^2 - 5x - 24} > x + 2 \]

### Step 2: Square both sides
Next, we square both sides to eliminate the square root (remember to check for extraneous solutions later):

\[ x^2 - 5x - 24 > (x + 2)^2 \]

### Step 3: Expand the right side
Now we expand the right side:

\[ x^2 - 5x - 24 > x^2 + 4x + 4 \]

### Step 4: Simplify the inequality
Subtract \( x^2 \) from both sides:

\[ -5x - 24 > 4x + 4 \]

Now, move all terms involving \( x \) to one side and constant terms to the other:

\[ -5x - 4x > 4 + 24 \]

This simplifies to:

\[ -9x > 28 \]

### Step 5: Solve for \( x \)
Now, divide by -9 (remember to reverse the inequality sign):

\[ x < -\frac{28}{9} \]

### Step 6: Check the second condition
Now we need to ensure that the expression inside the square root is non-negative:

\[ x^2 - 5x - 24 \geq 0 \]

### Step 7: Factor the quadratic
We can factor the quadratic:

\[ (x - 8)(x + 3) \geq 0 \]

### Step 8: Determine the intervals
The critical points are \( x = -3 \) and \( x = 8 \). We analyze the sign of the expression in the intervals:

1. \( (-\infty, -3) \)
2. \( (-3, 8) \)
3. \( (8, \infty) \)

Testing these intervals, we find:

- For \( x < -3 \): the expression is positive.
- For \( -3 < x < 8 \): the expression is negative.
- For \( x > 8 \): the expression is positive.

Thus, the solution for \( x^2 - 5x - 24 \geq 0 \) is:

\[ (-\infty, -3] \cup [8, \infty) \]

### Step 9: Combine the conditions
Now we combine the two conditions:

1. From the logarithm condition: \( x < -\frac{28}{9} \) (approximately -3.11)
2. From the square root condition: \( x \in (-\infty, -3] \cup [8, \infty) \)

Since \( -\frac{28}{9} \) is less than -3, we take the intersection of the intervals:

The valid domain is:

\[ (-\infty, -3] \]

### Final Answer
Thus, the domain of the function \( f(x) \) is:

\[ \boxed{(-\infty, -3]} \]