Question
If the domain of the function $f(x) = \sin^{- 1}\left( \frac{1}{x^{2} - 2x - 2} \right)$, is $( - \infty,\alpha\rbrack \cup \lbrack\beta,\gamma\rbrack \cup \lbrack\delta,\infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to
Options
Solution
$- 1 \leq \frac{2}{x^{2} - 2x - 2} \leq 1$
$${\frac{1 + x^{2} - 2x - 2}{x^{2} - 2x - 2} \geq 0 \Rightarrow \frac{(x - 1)^{2} - 2}{(x - 1)^{2} - 3} \geq 0 }{\Rightarrow \frac{(x - 1 - \sqrt{2})(x - 1 + \sqrt{2})}{(x - 1 - \sqrt{3})(x - 1 + \sqrt{3})} \geq 0}$$
$$\begin{array}{r} x \in ( - \infty,1 - \sqrt{3}) \cup \lbrack 1 - \sqrt{2},1 + \sqrt{2}\rbrack \cup (1 + \sqrt{3},0)\#(1) \end{array}$$
$${1 - \frac{1}{x^{2} - 2x - 2} \geq 0 \Rightarrow \frac{x^{2} - 2x - 3}{x^{2} - 2x - 2} \geq 0 }{\Rightarrow \frac{(x + 1)(x - 3)}{(x - 1 + \sqrt{3})(x - 1 - \sqrt{3})} \geq 0}$$
$$\begin{array}{r} x \in ( - \infty, - 1\rbrack \cup (1 - \sqrt{3},\sqrt{3} + 1) \cup \lbrack 3,\infty)\#(2) \end{array}$$
(1) $\cap (2)$
$${\Rightarrow x \in ( - \infty, - 1\rbrack \cup \lbrack 1 - \sqrt{2},1 + \sqrt{2}\rbrack \cup \lbrack 3,\infty) }{\therefore\alpha + \beta + \gamma + \delta = 4}$$
Create a free account to view solution
View Solution Free