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

• Correct answer
  A

  32
• B

  40
• C

  24
• D

  36

Solution

Domain of $f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$ is $\begin{aligned} & 2 x+3 \neq 0 \& x \neq \frac{-3}{2} \text { and }\left|\frac{(x-1)}{2 x+3}\right| \leq 1 \\ & |x-1| \leq|2 x+3| \end{aligned}$
$\begin{aligned} & \text { For }|2 x+3| \geq|x-1| \\ & x \in(-\infty,-4] \cup\left(-\frac{2}{3}, \infty\right) \\ & \alpha=-4 \& \beta=-\frac{2}{3}: 12 \alpha \beta=32\end{aligned}$

Hence, the correct answer is option: 'A'.