If the domain of the function `f(x) = sin^(-1)(frac{x-1}{2x + 3})` is `R - (alpha, beta)`, then `12 alpha beta` is equal to:

Options

A. 32
B. 40
C. 36
D. 24

Answer

To find the value of \(12 \alpha \beta\) where the domain of the function \(f(x) = \sin^{-1}\left(\frac{x-1}{2x + 3}\right)\) is \(R - (\alpha, \beta)\), we need to determine the values of \(\alpha\) and \(\beta\) for which the expression inside the sine inverse function is valid.

### Step 1: Determine the range for the sine inverse function
The domain of the sine inverse function \(y = \sin^{-1}(x)\) is defined for \(x\) in the interval \([-1, 1]\). Therefore, we need to ensure that:
\[ -1 \leq \frac{x-1}{2x + 3} \leq 1 \]

### Step 2: Solve the inequalities
We will break this into two inequalities.

#### Inequality 1:
\[ \frac{x-1}{2x + 3} \geq -1 \]
Multiplying both sides by \(2x + 3\) (noting that we need to consider the sign of \(2x + 3\)):
\[ x - 1 \geq -2x - 3 \]
Rearranging gives:
\[ 3x \geq -2 \quad \Rightarrow \quad x \geq -\frac{2}{3} \]

#### Inequality 2:
\[ \frac{x-1}{2x + 3} \leq 1 \]
Again, multiplying both sides by \(2x + 3\):
\[ x - 1 \leq 2x + 3 \]
Rearranging gives:
\[ -x \leq 4 \quad \Rightarrow \quad x \geq -4 \]

### Step 3: Combine the results
From the two inequalities, we have:
1. \(x \geq -\frac{2}{3}\)
2. \(x \leq -4\)

However, we need to check the points where the denominator \(2x + 3 = 0\) to ensure that the function is defined:
\[ 2x + 3 = 0 \quad \Rightarrow \quad x = -\frac{3}{2} \]

### Step 4: Identify the domain
The function is undefined at \(x = -\frac{3}{2}\). Thus, the domain of \(f(x)\) is:
\[ (-\infty, -4] \cup \left(-\frac{2}{3}, \infty\right) \]
This means that the function is not defined in the interval \((-4, -\frac{3}{2})\) and \((-2/3, \infty)\).

### Step 5: Identify \(\alpha\) and \(\beta\)
From the intervals, we can identify:
- \(\alpha = -4\)
- \(\beta = -\frac{2}{3}\)

### Step 6: Calculate \(12 \alpha \beta\)
Now we can calculate \(12 \alpha \beta\):
\[ 12 \alpha \beta = 12 \cdot (-4) \cdot \left(-\frac{2}{3}\right) \]
Calculating this gives:
\[ = 12 \cdot 4 \cdot \frac{2}{3} = 32 \]

### Final Answer
Thus, the value of \(12 \alpha \beta\) is:
\[ \boxed{32} \]