Question

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

• A. 5
• B. 2
• C. 4
• D. 3

Hint:

For inverse trigonometric functions, always start by applying the basic range condition before solving inequalities.

Answer 1

The Correct Option is A

To find the domain of the function \( f(x) = \sin^{-1}\left(\dfrac{1}{x^2-2x-2}\right) \), we need to determine for which values of \( x \) the expression \( \dfrac{1}{x^2-2x-2} \) falls within the principal range of the inverse sine function, that is \([-1, 1]\).

Therefore, we need to solve the inequality:

\[-1 \leq \dfrac{1}{x^2-2x-2} \leq 1\]

This inequality can be broken down into two parts:

\[\dfrac{1}{x^2-2x-2} \leq 1\]

Multiply throughout by \((x^2-2x-2)\) (sign depends on the value of \(x^2-2x-2\)):

Case 1: \(x^2-2x-2 > 0\), then:

\[1 \leq x^2-2x-2\]

Rearranging gives:

\[x^2-2x-3 \geq 0\]

Factoring:

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

Which holds in \(x \leq -1 \) or \( x \geq 3\).

Case 2: \(x^2-2x-2 < 0\), then the inequality is not possible since \(\dfrac{1}{x^2-2x-2}\) is positive while 1 is the upper boundary.

\[-1 \leq \dfrac{1}{x^2-2x-2}\]

Multiply throughout by \((x^2-2x-2)\) (again, the sign of \( x^2-2x-2\) should be considered):

Case 1: \(x^2-2x-2 > 0\), then:

\[-x^2 + 2x + 2 \leq 1\]

Rearranging:

\[-x^2 + 2x + 3 \leq 0\]

Factoring:

\[-(x+1)(x-3) \leq 0\]

This inequality holds true for \(-1 \leq x \leq 3\).

Bringing both cases together, the inequality \( -1 \leq \dfrac{1}{x^2-2x-2} \leq 1 \) holds true in the domains:

• \(x \leq -1\) or \(x \geq 3\) from Case 1.
• \(-1 \leq x \leq 3\) from Case 2.

Therefore, the domain is:

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

From this, we have \(\alpha = -1\), \(\beta = 3\), \(\gamma\) not applicable in this union, and \(\delta = 3\).

Calculating \(\alpha + \beta + \gamma + \delta\):

\[-1 + 3 + 3 = 5\]

Thus, the sum \(\alpha + \beta + \gamma + \delta\) is equal to 5.