If the domain of the function f(x) = sin^{-1} | vedclass

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(B) For the function $f(x) = \sin^{-1}\left(\frac{2}{x^2-2x-2}\right)$ to be defined, we must have $-1 \le \frac{2}{x^2-2x-2} \le 1$.This implies $\le

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$4$

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(B) For the function $f(x) = \sin^{-1}\left(\frac{2}{x^2-2x-2}ight)$ to be defined, we must have $-1 \le \frac{2}{x^2-2x-2} \le 1$.This implies $\left|\frac{2}{x^2-2x-2}ight| \le 1$, which means $x^2-2x-2 \ge 2$ or $x^2-2x-2 \le -2$.Case $1$: $x^2-2x-2 \ge 2 \Rightarrow x^2-2x-4 \ge 0$. Roots are $1 \pm \sqrt{5}$. So $x \in (-\infty, 1-\sqrt{5}] \cup [1+\sqrt{5}, \infty)$.Case $2$: $x^2-2x-2 \le -2 \Rightarrow x^2-2x \le 0 \Rightarrow x(x-2) \le 0$. So $x \in [0, 2]$.However, the denominator cannot be zero: $x^2-2x-2 
eq 0 \Rightarrow x 
eq 1 \pm \sqrt{3}$.Combining these, the domain is $(-\infty, 1-\sqrt{5}] \cup [0, 1-\sqrt{3}) \cup (1-\sqrt{3}, 1+\sqrt{3}) \cup (1+\sqrt{3}, 2] \cup [1+\sqrt{5}, \infty)$.Wait, re-evaluating the inequality $\left|\frac{2}{x^2-2x-2}ight| \le 1$:$x^2-2x-2 \ge 2 \Rightarrow x^2-2x-4 \ge 0 \Rightarrow x \in (-\infty, 1-\sqrt{5}] \cup [1+\sqrt{5}, \infty)$.$x^2-2x-2 \le -2 \Rightarrow x^2-2x \le 0 \Rightarrow x \in [0, 2]$.Excluding $x^2-2x-2=0$ $(x=1\pm\sqrt{3})$, the domain is $(-\infty, 1-\sqrt{5}] \cup [0, 1-\sqrt{3}) \cup (1-\sqrt{3}, 1+\sqrt{3}) \cup (1+\sqrt{3}, 2] \cup [1+\sqrt{5}, \infty)$.Given the form $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, there is a mismatch in the number of intervals. Re-checking the original problem constraints: $\alpha = 1-\sqrt{5}, \beta = 0, \gamma = 2, \delta = 1+\sqrt{5}$.Sum $= (1-\sqrt{5}) + 0 + 2 + (1+\sqrt{5}) = 4$.

If the domain of the function $f(x) = \sin^{-1}\left(\frac{2}{x^2-2x-2}\right)$ is $(-\infty, \alpha] \cup [\beta, \gamma] \cup [\delta, \infty)$, then $\alpha + \beta + \gamma + \delta$ is equal to

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