The function `f(x)=cot^(-1)sqrt((x+3)x)+cos^(-1)sqrt(x^(2)+3x+1)` is defined on the set `S`, where `S` is equal to (A) `{0,3}` (B) `(0,3)` (C) `{0,-3}` (D) `[-3,0]`

Options

A. `{0,3}`
B. `(0,3)`
C. `{0,-3}`
D. `[-3,0]`

Explanation

To determine the set \( S \) on which the function \( f(x) = \cot^{-1} \sqrt{(x+3)x} + \cos^{-1} \sqrt{x^2 + 3x + 1} \) is defined, we need to analyze the domains of both components of the function.

### Step 1: Analyze \( \cot^{-1} \sqrt{(x+3)x} \)

1. The expression under the square root, \( (x+3)x \), must be non-negative:
\[ (x+3)x \geq 0 \]
This can be factored as:
\[ x(x + 3) \geq 0 \]

2. The roots of the equation \( x(x + 3) = 0 \) are \( x = 0 \) and \( x = -3 \).

3. We can analyze the sign of \( x(x + 3) \) using a number line:
- For \( x < -3 \): both factors are negative, so the product is positive.
- For \( -3 < x < 0 \): \( x \) is negative and \( x + 3 \) is positive, so the product is negative.
- For \( x > 0 \): both factors are positive, so the product is positive.

4. Therefore, the intervals where \( x(x + 3) \geq 0 \) are:
\[ (-\infty, -3] \cup [0, \infty) \]

### Step 2: Analyze \( \cos^{-1} \sqrt{x^2 + 3x + 1} \)

1. The expression \( \sqrt{x^2 + 3x + 1} \) must be between 0 and 1 (inclusive):
\[ 0 \leq \sqrt{x^2 + 3x + 1} \leq 1 \]

2. Squaring the inequality gives:
\[ 0 \leq x^2 + 3x + 1 \leq 1 \]

3. This leads to two inequalities:
- \( x^2 + 3x + 1 \geq 0 \)
- \( x^2 + 3x + 1 \leq 1 \)

4. For \( x^2 + 3x + 1 \geq 0 \):
- The roots are found using the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-3 \pm \sqrt{5}}{2} \]
- The quadratic opens upwards, so it is non-negative outside the roots.

5. For \( x^2 + 3x + 1 \leq 1 \):
- Rearranging gives:
\[ x^2 + 3x \leq 0 \]
- Factoring gives:
\[ x(x + 3) \leq 0 \]
- The roots are again \( x = 0 \) and \( x = -3 \).

6. The intervals where \( x(x + 3) \leq 0 \) are:
\[ [-3, 0] \]

### Step 3: Find the intersection of both conditions

1. The first condition gives us:
\[ (-\infty, -3] \cup [0, \infty) \]

2. The second condition gives us:
\[ [-3, 0] \]

3. The intersection of these two sets is:
\[ \{-3, 0\} \]

### Conclusion

Thus, the function \( f(x) \) is defined on the set \( S = \{-3, 0\} \).

### Final Answer
The correct option is (C) \(\{0, -3\}\).