Question

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

Hint:

Always apply the range condition of inverse trigonometric functions before solving inequalities.

Answer 1

Correct Answer: 4

Step 1: Apply the domain condition of inverse sine

For sin⁻¹(u) to be defined,

−1 ≤ u ≤ 1

Here,

u = 1 / (x² − 2x − 2)

So the condition becomes:

−1 ≤ 1 / (x² − 2x − 2) ≤ 1

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Step 2: Convert the double inequality

The above inequality is equivalent to:

|1 / (x² − 2x − 2)| ≤ 1

⇒ |x² − 2x − 2| ≥ 1

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Step 3: Solve the boundary equations

Case 1:

x² − 2x − 2 = 1

x² − 2x − 3 = 0

(x − 3)(x + 1) = 0

x = 3, −1

Case 2:

x² − 2x − 2 = −1

x² − 2x − 1 = 0

x = 1 ± √2

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Step 4: Write the domain of x

From |x² − 2x − 2| ≥ 1, the solution set is:

(−∞, −1] ∪ [1 − √2, 1 + √2] ∪ [3, ∞)

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Step 5: Identify the required values

α = −1

β = 1 − √2

γ = 1 + √2

δ = 3

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Step 6: Required sum

α + β + γ + δ

= −1 + (1 − √2) + (1 + √2) + 3

= 4

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Final Answer:

The required value is
4