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

Question

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

Accepted Answer

$\begin{gathered} f(x)={\sin }^{-1}(\frac{x-1}{2x+3}) \ -1\le \frac{x-1}{2x+3}\le 1 \frac{x-1}{2x+3}+1\ge 0 \ \frac{x-1}{2x+3}-1\le 0 \frac{x-1+2x+3}{2x+3}\ge 0 \ \frac{x-1-2x-3}{2x+3}\le 0 \frac{3x+2}{2x+3}\ge 0 \end{gathered}$

$\begin{gathered} \frac{-x-4}{2x+3}\le 0 \ \frac{x+4}{2x+3}\ge 0 \end{gathered}$

$\begin{gathered} x\in (-\infty,-4]\cup [\frac{-2}{3},\infty) \ R-(-4,\frac{-2}{3}) \ \alpha =-4 \ \beta =\frac{-2}{3} \ 12	imes \alpha \beta =12	imes 4	imes \frac{2}{3}=32 \end{gathered}$

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