Question

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

• A. 36
• B. 24
• C. 40
• D. 32

Answer 1

Answer: (D)

32

Answer 2

Step 1: Conditions for the domain of $f(x)$ The argument of $\sin^{-1}(x)$, $\frac{x-1}{2x+3}$, must satisfy two conditions:

1. $2x + 3 \neq 0$ (denominator cannot be zero), so $x \neq -\frac{3}{2}$.
2. $\left| \frac{x-1}{2x+3} \right| \leq 1$.

Step 2: Solve $\left| \frac{x-1}{2x+3} \right| \leq 1$ Split the inequality into two cases:

1. For $\frac{x-1}{2x+3} \geq -1$:

$x-1 \geq -(2x+3) \implies x-1 \geq -2x-3$.

Simplify:

$3x \geq -2 \implies x \geq -\frac{2}{3}$.

2. For $\frac{x-1}{2x+3} \leq 1$:

$x-1 \leq 2x+3 \implies -x \leq 4$.

Simplify:

$x \geq -4$.

Thus, combining the results:

$x \in [-4, -\frac{2}{3}]$ and exclude $x = -\frac{3}{2}$.

Step 3: Identify the excluded interval To exclude values where $|2x+3| \geq |x-1|$, note the critical points:

1. Solve $|x-1| = |2x+3|$, which gives:

$x = -4, \; x = -\frac{2}{3}$.

Using these results and the behavior of the function, the domain of $f(x)$ is:

$x \in (-\infty, -4] \cup \left(-\frac{2}{3}, \infty\right)$.

Step 4: Determine $\alpha$ and $\beta$ From the excluded interval $\left(-\frac{3}{2}, -\frac{2}{3}\right)$:

$\alpha = -4, \; \beta = -\frac{2}{3}$.

Step 5: Compute $12\alpha\beta$ :

$12\alpha\beta = 12 \times (-4) \times \left(-\frac{2}{3}\right)$.

Simplify:

$12\alpha\beta = 12 \times 8/3 = 32$.

Final Answer is Option (4): 32.