If the domain of the function f ( x ) = sin - 1 ( 1 x 2 - 2 x - 2 ) , is , then is equal to [2026]

Question

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 [2026]

Smart Achievers · Accepted Answer

(1)

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

$\frac{1 + x^{2} - 2 x - 2}{x^{2} - 2 x - 2} \geq 0 \Rightarrow \frac{{(x - 1)}^{2} - 2}{{(x - 1)}^{2} - 3} \geq 0$

$\Rightarrow \frac{(x - 1 - \sqrt{2}) (x - 1 + \sqrt{2})}{(x - 1 - \sqrt{3}) (x - 1 + \sqrt{3})} \geq 0$

$x \in (- \infty, 1 - \sqrt{3}) \cup [1 - \sqrt{2}, 1 + \sqrt{2}] \cup (1 + \sqrt{3}, 0) . . . (1)$

$1 - \frac{1}{x^{2} - 2 x - 2} \geq 0 \Rightarrow \frac{x^{2} - 2 x - 3}{x^{2} - 2 x - 2} \geq 0$

$\Rightarrow \frac{(x + 1) (x - 3)}{(x - 1 - \sqrt{3}) (x - 1 + \sqrt{3})} \geq 0$

$x \in (- \infty, - 1] \cup (1 - \sqrt{3}, 1 + \sqrt{3}) \cup [3, \infty) . . . (2)$

$(1) \cap (2)$

$\Rightarrow x \in (- \infty, - 1] \cup [1 - \sqrt{2}, 1 + \sqrt{2}] \cup [3, \infty)$

$∴$ $α + β + γ + δ = 4$

Solution

Q.
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 [2026]

1 4

2 2

3 3

4 5

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Solution

Q. If the domain of the function f(x)=sin-1(1x2-2x-2), is (-∞,α]∪[β,γ]∪[δ,∞), then α+β+γ+δ is equal to [2026]

1 4

2 2

3 3

4 5

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Q.
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 [2026]