If the domain of the function [f(x)= ^-1((x-1)/(2 x+3))] is [R-(, )], then [12 ] is equal …

Question

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 :

Accepted Answer

$$
\[\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}$$
\]

$$\begin{aligned}
& \frac{-x-4}{2 x+3} \leq 0 \
& \frac{x+4}{2 x+3} \geq 0
\end{aligned}$$

$$\begin{aligned}
& x \in(-\infty,-4] \cup\left[\frac{-2}{3}, \infty\right) \
& \text { Domain : } R-\left(-4, \frac{-2}{3}\right) \
& \alpha=-4 \
& \beta=\frac{-2}{3} \
& 12 \times \alpha \beta=12 \times 4 \times \frac{2}{3}=32
\end{aligned}$$

Solutions · 1