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

For the inverse-sine function to exist, its argument must lie in the closed interval $$[-1,\,1]$$.
Given $$f(x)=\sin^{-1}\!\left(\dfrac1{x^{2}-2x-2}\right)$$ we therefore require

$$-1 \;\le\; \dfrac1{x^{2}-2x-2}\;\le\; 1$$

and, of course, the denominator must not be zero:

$$x^{2}-2x-2\;\neq\;0.$$

Simplify the double inequality. Taking absolute values is the quickest route.

$$-1\;\le\;\dfrac1{x^{2}-2x-2}\;\le\;1 \;\Longleftrightarrow\; \left|\dfrac1{x^{2}-2x-2}\right|\;\le\;1 \;\Longleftrightarrow\; \dfrac1{|\,x^{2}-2x-2\,|}\;\le\;1$$

Since the reciprocal function is positive, we can invert both sides:

$$|\,x^{2}-2x-2\,|\;\ge\;1.$$

Let $$g(x)=x^{2}-2x-2=(x-1)^{2}-3.$$
The required condition becomes

$$|g(x)|\;\ge\;1.$$

This splits into two cases.

Case 1:

$$g(x)\;\ge\;1 \;\Longrightarrow\; x^{2}-2x-2\;\ge\;1 \;\Longrightarrow\; x^{2}-2x-3\;\ge\;0 \;\Longrightarrow\; (x+1)(x-3)\;\ge\;0.$$

The solution of $$(x+1)(x-3)\ge0$$ is

$$x\;\le\;-1\quad\text{or}\quad x\;\ge\;3.$$

Case 2:

$$g(x)\;\le\;-1 \;\Longrightarrow\; x^{2}-2x-2\;\le\;-1 \;\Longrightarrow\; x^{2}-2x-1\;\le\;0.$$

Solve $$x^{2}-2x-1=0$$:

$$x=\dfrac{2\pm\sqrt{4+4}}{2}=1\pm\sqrt2.$$

A quadratic that opens upward is non-positive between its roots, so

$$1-\sqrt2\;\le\;x\;\le\;1+\sqrt2.$$

Combining both cases, and omitting the points where $$g(x)=0$$ (that is, $$x=1\pm\sqrt3$$), we get the complete domain:

$$(-\infty,\,-1] \;\cup\; [\,1-\sqrt2,\,1+\sqrt2\,] \;\cup\; [\,3,\,\infty).$$

Writing it in the prescribed form $$[-\infty,\,\alpha] \;\cup\; [\beta,\,\gamma] \;\cup\; [\delta,\,\infty),$$ we identify

$$\alpha=-1,\qquad \beta=1-\sqrt2,\qquad \gamma=1+\sqrt2,\qquad \delta=3.$$

The required sum is therefore

$$\alpha+\beta+\gamma+\delta =\;(-1)\;+\;(1-\sqrt2)\;+\;(1+\sqrt2)\;+\;3 =\;4.$$

Hence $$\alpha+\beta+\gamma+\delta=4,$$ which corresponds to Option B.