The domain of  sin⁻¹( 1 / √(x² − 2x − 1) ) is (−∞, α] ∪ [β, δ] ∪ [λ, ∞) Find the value of α + β + δ + λ.

The correct option is 4.

To find the domain of the function $f\left(x\right)={sin}^{\mathrm{-1}}\left(\frac{1}{\sqrt{x^2-2x-1}}\right)$, we must satisfy two conditions:

Condition 1: The term inside the square root in the denominator must be strictly positive.

$x^2-2x-1>0$

To find the roots of $x^2-2x-1=0$, we use the quadratic formula:

$x=\frac{2\pm \sqrt{{(-2)}^2-4\left(1\right)(-1)}}{2}=1\pm \sqrt{2}$

Thus, Condition 1 requires:

$x\in (-\infty ,1-\sqrt{2})\cup (1+\sqrt{2},\infty )$

Condition 2: The argument of the arcsine function must lie in the interval [-1, 1].

$-1\le \frac{1}{\sqrt{x^2-2x-1}}\le 1$

Since the square root is always positive, the term $\frac{1}{\sqrt{x^2-2x-1}}$ is strictly greater than 0. Thus, we only need to solve:

$\frac{1}{\sqrt{x^2-2x-1}}\le 1$

Taking the reciprocal and squaring both sides gives:

$x^2-2x-1\ge 1$

$x^2-2x-2\ge 0$

Finding the roots of $x^2-2x-2=0$:

$x=\frac{2\pm \sqrt{{(-2)}^2-4\left(1\right)(-2)}}{2}=1\pm \sqrt{3}$

Thus, Condition 2 requires:

$x\in (-\infty ,1-\sqrt{3}]\cup [1+\sqrt{3},\infty )$

Comparing the critical points, we identify the boundary values from the inequalities:

$\alpha =1-\sqrt{3}$

$\beta =1-\sqrt{2}$

$\delta =1+\sqrt{2}$

$\lambda =1+\sqrt{3}$

Now, we compute the required sum:

$\alpha +\beta +\delta +\lambda =(1-\sqrt{3})+(1-\sqrt{2})+(1+\sqrt{2})+(1+\sqrt{3})$

$\alpha +\beta +\delta +\lambda =4$