Key Concepts and Formulas

• The domain of the inverse sine function, $\sin^{-1}(u)$, is $[-1, 1]$. This means $-1 \leq u \leq 1$.
• When solving inequalities involving rational expressions, it's crucial to consider the sign of the numerator and the denominator. The critical points are where the numerator or denominator equals zero.
• The solution to a compound inequality is the intersection of the solutions to the individual inequalities.

Step-by-Step Solution

1. Identify the argument of the $\sin^{-1}$ function: The given function is $f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$. The argument of the inverse sine function is $u = \frac{x-1}{2 x+3}$.

2. Establish the domain condition for $\sin^{-1}(u)$: For $f(x)$ to be defined, the argument $u$ must lie within the interval $[-1, 1]$. Therefore, we have the inequality: $-1 \leq \frac{x-1}{2 x+3} \leq 1$ We also note that the denominator cannot be zero, so $2x+3 \neq 0$, which means $x \neq -\frac{3}{2}$.

3. Split the compound inequality into two separate inequalities: The compound inequality can be split into two parts that must both be satisfied: (a) $\frac{x-1}{2 x+3} \leq 1$ (b) $\frac{x-1}{2 x+3} \geq -1$

4. Solve inequality (a): $\frac{x-1}{2 x+3} \leq 1$

   • Subtract 1 from both sides to get a zero on one side: $\frac{x-1}{2 x+3} - 1 \leq 0$
   • Combine the terms into a single rational expression: $\frac{(x-1) - (2x+3)}{2 x+3} \leq 0$ $\frac{x-1-2x-3}{2 x+3} \leq 0$ $\frac{-x-4}{2 x+3} \leq 0$
   • Multiply by -1 and reverse the inequality sign to make the leading coefficient of $x$ in the numerator positive: $\frac{x+4}{2 x+3} \geq 0$
   • Find the critical points by setting the numerator and denominator to zero: $x+4 = 0 \implies x = -4$ $2x+3 = 0 \implies x = -\frac{3}{2}$
   • Use the wavy curve method. The critical points are $-4$ and $-\frac{3}{2}$. We test the intervals:
     • For $x < -4$ (e.g., $x=-5$): $\frac{-5+4}{2(-5)+3} = \frac{-1}{-7} = \frac{1}{7} \geq 0$. This interval is valid.
     • For $-4 < x < -\frac{3}{2}$ (e.g., $x=-2$): $\frac{-2+4}{2(-2)+3} = \frac{2}{-1} = -2 < 0$. This interval is not valid.
     • For $x > -\frac{3}{2}$ (e.g., $x=0$): $\frac{0+4}{2(0)+3} = \frac{4}{3} \geq 0$. This interval is valid.
   • Since the inequality is $\geq 0$, the points where the numerator is zero are included. The point where the denominator is zero is excluded.
   • The solution to inequality (a) is $x \in (-\infty, -4] \cup (-\frac{3}{2}, \infty)$.

5. Solve inequality (b): $\frac{x-1}{2 x+3} \geq -1$

   • Add 1 to both sides to get a zero on one side: $\frac{x-1}{2 x+3} + 1 \geq 0$
   • Combine the terms into a single rational expression: $\frac{(x-1) + (2x+3)}{2 x+3} \geq 0$ $\frac{x-1+2x+3}{2 x+3} \geq 0$ $\frac{3x+2}{2 x+3} \geq 0$
   • Find the critical points: $3x+2 = 0 \implies x = -\frac{2}{3}$ $2x+3 = 0 \implies x = -\frac{3}{2}$
   • Use the wavy curve method. The critical points are $-\frac{3}{2}$ and $-\frac{2}{3}$. We test the intervals:
     • For $x < -\frac{3}{2}$ (e.g., $x=-2$): $\frac{3(-2)+2}{2(-2)+3} = \frac{-4}{-1} = 4 \geq 0$. This interval is valid.
     • For $-\frac{3}{2} < x < -\frac{2}{3}$ (e.g., $x=-1$): $\frac{3(-1)+2}{2(-1)+3} = \frac{-1}{1} = -1 < 0$. This interval is not valid.
     • For $x > -\frac{2}{3}$ (e.g., $x=0$): $\frac{3(0)+2}{2(0)+3} = \frac{2}{3} \geq 0$. This interval is valid.
   • Since the inequality is $\geq 0$, the points where the numerator is zero are included. The point where the denominator is zero is excluded.
   • The solution to inequality (b) is $x \in (-\infty, -\frac{3}{2}) \cup [-\frac{2}{3}, \infty)$.

6. Find the intersection of the solutions from (a) and (b): The domain of $f(x)$ is the intersection of the solution sets from step 4 and step 5. Solution (a): $(-\infty, -4] \cup (-\frac{3}{2}, \infty)$ Solution (b): $(-\infty, -\frac{3}{2}) \cup [-\frac{2}{3}, \infty)$

   Let's visualize the intersection on a number line:

   • The interval $(-\infty, -4]$ from (a) intersects with $(-\infty, -\frac{3}{2})$ from (b) to give $(-\infty, -4]$.
   • The interval $(-\frac{3}{2}, \infty)$ from (a) needs to be intersected with $(-\infty, -\frac{3}{2}) \cup [-\frac{2}{3}, \infty)$ from (b). The intersection here is $[-\frac{2}{3}, \infty)$.

   Therefore, the domain of $f(x)$ is $(-\infty, -4] \cup [-\frac{2}{3}, \infty)$.

7. Determine $\alpha$ and $\beta$: The problem states that the domain of the function is $\mathbf{R}-(\alpha, \beta)$. This means the set of all real numbers excluding the open interval $(\alpha, \beta)$. Our derived domain is $(-\infty, -4] \cup [-\frac{2}{3}, \infty)$. The complement of this set is the interval between $-4$ (exclusive) and $-\frac{2}{3}$ (exclusive). So, the excluded interval is $(-4, -\frac{2}{3})$. Comparing this with $(\alpha, \beta)$, we have $\alpha = -4$ and $\beta = -\frac{2}{3}$.

8. Calculate $12 \alpha \beta$: Substitute the values of $\alpha$ and $\beta$: $12 \alpha \beta = 12 \times (-4) \times \left(-\frac{2}{3}\right)$ $12 \alpha \beta = 12 \times \left(\frac{8}{3}\right)$ $12 \alpha \beta = 4 \times 8$ $12 \alpha \beta = 32$

Common Mistakes & Tips

• Sign errors when multiplying by negative numbers: Always remember to reverse the inequality sign when multiplying or dividing by a negative number.
• Forgetting the denominator cannot be zero: Ensure that the values of $x$ that make the denominator zero are always excluded from the domain.
• Confusing union and intersection: The domain of the function requires both inequalities to be true, so we need the intersection of their solution sets. The problem statement defines the domain as $\mathbf{R}$ minus an interval, meaning the complement of the derived domain is the interval $(\alpha, \beta)$.

Summary

To find the domain of the given function, we used the property that the argument of the $\sin^{-1}$ function must be between -1 and 1, inclusive. This led to a compound inequality that was split into two separate inequalities. Solving these inequalities and finding their intersection gave us the domain of the function as $(-\infty, -4] \cup [-\frac{2}{3}, \infty)$. The problem defines the domain as $\mathbf{R}$ excluding an interval $(\alpha, \beta)$, which means the excluded interval is the complement of our derived domain. This excluded interval is $(-4, -\frac{2}{3})$, giving $\alpha = -4$ and $\beta = -\frac{2}{3}$. Finally, we calculated $12 \alpha \beta$.

The final answer is \boxed{32}.