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1.Relation and Function
hard
If the domain of the function $f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$ is $R-(\alpha, \beta)$ then $12 \alpha \beta$ is equal to :
A
$36$
B
$24$
C
$40$
D
$32$
(JEE MAIN-2024)
Solution

Domain of $f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right)$ is
$ 2 x+3 \neq 0 \& x \neq \frac{-3}{2} \text { and }\left|\frac{(x-1)}{2 x+3}\right| \leq 1 $
$|x-1| \leq|2 x+3|$
$Image$
$ \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$
Standard 12
Mathematics