If the domain of the function f(x)=sin ^-1 ( 1 x^2-2 x-2 ), is (-∞, α] ∪[β, γ] ∪[δ, ∞), then α+β+γ+δ is equal to

For $\sin^{-1}\left(\frac{1}{x^2-2x-2}\right)$sin−1(x2−2x−21​) to be defined, we need $\left|\frac{1}{x^2-2x-2}\right| \leq 1$​x2−2x−21​​≤1, i.e., $|x^2-2x-2| \geq 1$∣x2−2x−2∣≥1.

Case 1:

$$x^2-2x-2 \geq 1 \Rightarrow x^2-2x-3 \geq 0$$x2−2x−2≥1⇒x2−2x−3≥0

$$\Rightarrow (x-3)(x+1) \geq 0$$⇒(x−3)(x+1)≥0

$\Rightarrow x \leq -1$⇒x≤−1 or $x \geq 3$x≥3.

Case 2:

$$x^2-2x-2 \leq -1 \Rightarrow x^2-2x-1 \leq 0$$x2−2x−2≤−1⇒x2−2x−1≤0

$$\Rightarrow (x-1)^2 \leq 2$$⇒(x−1)2≤2

$\Rightarrow 1-\sqrt{2} \leq x \leq 1+\sqrt{2}$⇒1−2​≤x≤1+2​.

Domain $= (-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty)$=(−∞,−1]∪[1−2​,1+2​]∪[3,∞).

So $\alpha = -1,\ \beta = 1-\sqrt{2},\ \gamma = 1+\sqrt{2},\ \delta = 3$α=−1, β=1−2​, γ=1+2​, δ=3.

$\alpha + \beta + \gamma + \delta = -1 + 1 - \sqrt{2} + 1 + \sqrt{2} + 3 = 4$α+β+γ+δ=−1+1−2​+1+2​+3=4.