Domain of the function `f(x)=log(sin^(-1)sqrt(x^(2)+3x+2))` is :

Options

A. `(-oo, -2) cup (-1, oo)`
B. `((-3-sqrt(5))/(2), (-3+sqrt(5))/(2))`
C. `[(-3-sqrt(5))/(2), -2) cup (-1, (-3+sqrt(5))/(2)]`
D. None of these

Answer

To find the domain of the function \( f(x) = \log(\sin^{-1}(\sqrt{x^2 + 3x + 2})) \), we need to ensure that the argument of the logarithm is defined and positive. Let's break this down step by step.

### Step 1: Determine when the argument of the logarithm is defined
The function \( \sin^{-1}(y) \) is defined for \( y \) in the range \([-1, 1]\). Therefore, we need to ensure that:
\[ \sqrt{x^2 + 3x + 2} \in [0, 1] \]
This means:
\[ 0 \leq \sqrt{x^2 + 3x + 2} \leq 1 \]

### Step 2: Square the inequality
Since the square root is non-negative, we can square the inequality to eliminate the square root:
\[ 0 \leq x^2 + 3x + 2 \leq 1 \]

### Step 3: Solve the inequalities
1. **First Inequality**: \( x^2 + 3x + 2 \geq 0 \)

Factor the quadratic:
\[ (x + 1)(x + 2) \geq 0 \]
The roots of this equation are \( x = -1 \) and \( x = -2 \). To find the intervals where this product is non-negative, we can test the intervals:
- For \( x < -2 \): both factors are negative, product is positive.
- For \( -2 < x < -1 \): one factor is negative, product is negative.
- For \( x > -1 \): both factors are positive, product is positive.

Thus, the solution for this inequality is:
\[ x \in (-\infty, -2] \cup [-1, \infty) \]

2. **Second Inequality**: \( x^2 + 3x + 2 \leq 1 \)

Rearranging gives:
\[ x^2 + 3x + 1 \leq 0 \]
We can find the roots using 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:
\[ x_1 = \frac{-3 - \sqrt{5}}{2}, \quad x_2 = \frac{-3 + \sqrt{5}}{2} \]
To find the intervals where this product is non-positive, we can test the intervals:
- For \( x < x_1 \): the quadratic is positive.
- For \( x_1 < x < x_2 \): the quadratic is negative.
- For \( x > x_2 \): the quadratic is positive.

Thus, the solution for this inequality is:
\[ x \in \left[ \frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2} \right] \]

### Step 4: Find the intersection of the two solutions
Now, we need to find the intersection of the two sets:
1. From \( x^2 + 3x + 2 \geq 0 \): \( (-\infty, -2] \cup [-1, \infty) \)
2. From \( x^2 + 3x + 1 \leq 0 \): \( \left[ \frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2} \right] \)

The intersection will be:
- For the interval \( (-\infty, -2] \): it intersects with \( \left[ \frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2} \right] \) at \( (-\infty, -2] \).
- For the interval \( [-1, \infty) \): it does not intersect with \( \left[ \frac{-3 - \sqrt{5}}{2}, \frac{-3 + \sqrt{5}}{2} \right] \).

### Final Domain
Thus, the domain of \( f(x) \) is:
\[ \left[ \frac{-3 - \sqrt{5}}{2}, -2 \right] \cup [-1, \frac{-3 + \sqrt{5}}{2}] \]