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

Options

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

Explanation

To find 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 conditions for both components of the function.

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

The expression \( \sqrt{(x+3)x} \) must be non-negative for the inverse cotangent function to be defined. Therefore, we need:

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

This inequality holds true when either both factors are non-negative or both are non-positive.

1. **Case 1**: Both factors are non-negative:
- \( x + 3 \geq 0 \) implies \( x \geq -3 \)
- \( x \geq 0 \)

The critical points are \( x = -3 \) and \( x = 0 \). The intervals to consider are:
- \( (-\infty, -3) \)
- \( [-3, 0] \)
- \( (0, \infty) \)

Testing these intervals:
- For \( x < -3 \): Both \( x+3 < 0 \) and \( x < 0 \) (not valid).
- For \( -3 \leq x < 0 \): \( x+3 \geq 0 \) and \( x < 0 \) (valid).
- For \( x \geq 0 \): Both \( x+3 \geq 0 \) and \( x \geq 0 \) (valid).

Thus, the valid interval from this case is \( [-3, 0) \).

2. **Case 2**: Both factors are non-positive:
- \( x + 3 \leq 0 \) implies \( x \leq -3 \)
- \( x \leq 0 \)

The only valid interval here is \( (-\infty, -3] \).

Combining both cases, we have:

\[ S_1 = [-3, 0) \]

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

The expression \( \sqrt{x^2 + 3x + 1} \) must be in the range \( [0, 1] \) for the inverse cosine function to be defined. Thus, we need:

1. **Non-negativity**:
\[ x^2 + 3x + 1 \geq 0 \]

To find the roots, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{9 - 4}}{2} = \frac{-3 \pm \sqrt{5}}{2} \]

Let \( r_1 = \frac{-3 - \sqrt{5}}{2} \) and \( r_2 = \frac{-3 + \sqrt{5}}{2} \).

The quadratic opens upwards (as the coefficient of \( x^2 \) is positive), so it is non-negative outside the roots:
\[ x \leq r_1 \quad \text{or} \quad x \geq r_2 \]

2. **Upper bound**:
\[ x^2 + 3x + 1 \leq 1 \implies x^2 + 3x \leq 0 \]
Factoring gives:
\[ x(x + 3) \leq 0 \]
This is satisfied in the interval \( [-3, 0] \).

### Step 3: Combine Conditions

Now, we need to find the intersection of the intervals obtained from both components:

- From \( \cot^{-1} \): \( S_1 = [-3, 0) \)
- From \( \cos^{-1} \): \( S_2 = (-\infty, r_1] \cup [r_2, \infty) \) and \( [-3, 0] \)

The intersection of \( S_1 \) and \( S_2 \) gives us the final set \( S \):

\[ S = [-3, 0] \]

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