If the domain of the function is then is equal to, Inverse trigonometric functions

Question

If the domain of the function $f (x) = \sin^{- 1} (\frac{x - 1}{2 x + 3})$ is $R - (α, β),$ then $12 α β$ is equal to : [2024]

Smart Achievers · Accepted Answer

(2)

We have, $f (x) = \sin^{- 1} (\frac{x - 1}{2 x + 3}),$ since domain of $\sin^{- 1} x is [- 1, 1], \forall x .$

$\Rightarrow - 1 \leq \frac{x - 1}{2 x + 3} \leq 1 \Rightarrow \frac{x - 1}{2 x + 3} + 1 \geq 0 and \frac{x - 1}{2 x + 3} - 1 \leq 0$

$\Rightarrow \frac{x - 1 + 2 x + 3}{2 x + 3} \geq 0 and \frac{x - 1 - 2 x - 3}{2 x + 3} \leq 0$

$\Rightarrow \frac{3 x + 2}{2 x + 3} \geq 0 and \frac{- x - 4}{2 x + 3} \leq 0$

Now, $\frac{3 x + 2}{2 x + 3} \geq 0 \Rightarrow x < - \frac{3}{2} or x \geq - \frac{2}{3}$

$\Rightarrow x \in (- \infty, - \frac{3}{2}) \cup [\frac{- 2}{3}, \infty)$

and $\frac{- x - 4}{2 x + 3} \leq 0 \Rightarrow x \leq - 4 or x > - \frac{3}{2}$

$\Rightarrow x \in (- \infty, - 4] \cup (\frac{- 3}{2}, \infty)$

$Hence, x \in (- \infty, - 4] \cup [\frac{- 2}{3}, \infty)$ $i . e ., x \in R - (- 4, \frac{- 2}{3})$

So, Domain : $R - (- 4, \frac{- 2}{3}) \Rightarrow α = - 4 and β = \frac{- 2}{3}$

$∴$ $12 \times β = 12 \times (- 4) \times (- \frac{2}{3}) = 32$

Want Us to Email you About Special Offers & Updates?