JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 1)
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 :
40
36
24
32
Explanation
$$ \begin{array}{ll} f(x)=\sin ^{-1}\left(\frac{x-1}{2 x+3}\right) & \\ -1 \leq \frac{x-1}{2 x+3} \leq 1 & \frac{x-1}{2 x+3}+1 \geq 0 \\ \frac{x-1}{2 x+3}-1 \leq 0 & \frac{x-1+2 x+3}{2 x+3} \geq 0 \\ \frac{x-1-2 x-3}{2 x+3} \leq 0 & \frac{3 x+2}{2 x+3} \geq 0 \end{array} $$
_9th_April_Morning_Shift_en_1_1.png)
$$\begin{aligned} & \frac{-x-4}{2 x+3} \leq 0 \\ & \frac{x+4}{2 x+3} \geq 0 \end{aligned}$$
_9th_April_Morning_Shift_en_1_2.png)
$$\begin{aligned} & x \in(-\infty,-4] \cup\left[\frac{-2}{3}, \infty\right) \\ & \text { Domain : } R-\left(-4, \frac{-2}{3}\right) \\ & \alpha=-4 \\ & \beta=\frac{-2}{3} \\ & 12 \times \alpha \beta=12 \times 4 \times \frac{2}{3}=32 \end{aligned}$$
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