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

The Correct Option is D

Given function:

\( f(x) = \sin^{-1}\left(\frac{x - 1}{2x + 3}\right) \)

For \( f(x) \) to be defined, the following conditions must hold:

\( 2x + 3 \neq 0 \) and \( -1 \leq \frac{x - 1}{2x + 3} \leq 1 \)

⇒ \( \frac{x - 1 - (2x + 3)}{2x + 3} \leq 0 \leq \frac{x - 1 + (2x + 3)}{2x + 3} \)

⇒ \( \frac{-x - 4}{2x + 3} \leq 0 \leq \frac{3x + 2}{2x + 3} \)

Now, solving \( \frac{-x - 4}{2x + 3} \leq 0 \):

Critical points: \( x = -4, -\frac{3}{2} \)

Sign analysis gives: \( x \in (-\infty, -4] \cup \left(-\frac{3}{2}, \infty \right) \)

Similarly, for \( \frac{3x + 2}{2x + 3} \geq 0 \):

Critical points: \( x = -\frac{3}{2}, -\frac{2}{3} \)

Sign analysis gives: \( x \in (-\infty, -\frac{3}{2}) \cup \left[-\frac{2}{3}, \infty \right) \)

Taking the intersection of the two intervals, the domain is:

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

Thus, the range is:

\( R = \left(-4, -\frac{2}{3}\right] \)

Now, let \( \alpha = -4 \) and \( \beta = -\frac{2}{3} \)

Then, \( 12 \times (-4) \times \left(-\frac{2}{3}\right) = 32 \)

Hence, the final answer is:

\( 32 \)

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.