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

JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Inverse Trigonometric Functions

If the domain of the function

f (x) = sin^{- 1} (\frac{1}{x ^{2} - 2 x - 2})

, is

(- \infty, α] \cup [β, γ] \cup [δ, \infty)

, then

α + β + γ + δ

is equal to

Answer: (B)

The expression

sin^{- 1} (\frac{1}{x ^{2} - 2 x - 2})

is defined when

\frac{1}{x ^{2} - 2 x - 2} \leq 1

, i.e.

∣ x^{2} - 2 x - 2∣ \geq 1

. From

x^{2} - 2 x - 2 \geq 1

(x + 1) (x - 3) \geq 0

, giving

x \leq - 1

x \geq 3

. From

x^{2} - 2 x - 2 \leq - 1

x^{2} - 2 x - 1 \leq 0

, giving

1 - 2 \leq x \leq 1 + 2

. Hence the domain is

(- \infty, - 1] \cup [1 - 2, 1 + 2] \cup [3, \infty)

, so

α = - 1, β = 1 - 2, γ = 1 + 2, δ = 3

. Therefore

α + β + γ + δ = 4

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