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

Consider the function:

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

For \( f(x) \) to be defined, the following inequalities must be satisfied:

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

These inequalities can be rewritten as:

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

Simplifying these expressions yields:

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

Solving the first inequality, \( \frac{-x - 4}{2x + 3} \leq 0 \):

Critical points are \( x = -4 \) and \( x = -\frac{3}{2} \). The solution set is \( x \in (-\infty, -4] \cup \left(-\frac{3}{2}, \infty \right) \).

Solving the second inequality, \( \frac{3x + 2}{2x + 3} \geq 0 \):

Critical points are \( x = -\frac{3}{2} \) and \( x = -\frac{2}{3} \). The solution set is \( x \in (-\infty, -\frac{3}{2}) \cup \left[-\frac{2}{3}, \infty \right) \).

The domain of \( f(x) \) is the intersection of these two solution sets:

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

The range of \( f(x) \) is then:

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

Let \( \alpha = -4 \) and \( \beta = -\frac{2}{3} \).

Calculate the product \( 12 \times \alpha \times \beta \):

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

The final result is \( 32 \).