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

Step 1: Apply condition for $\sin^{-1}$.
For $\sin^{-1}(y)$ to be defined, \[ -1 \le y \le 1 \] So, \[ -1 \le \frac{1}{x^2-2x-2} \le 1 \] Step 2: Solve the inequality.
This gives two inequalities: \[ \frac{1}{x^2-2x-2} \le 1 \quad \text{and} \quad \frac{1}{x^2-2x-2} \ge -1 \] Solving, we obtain critical points \[ x = 1-\sqrt{3},\ 1,\ 1+\sqrt{3} \] Step 3: Determine valid intervals.
The domain becomes \[ (-\infty,1-\sqrt{3}) \cup [1,1] \cup [1+\sqrt{3},\infty) \] Thus, \[ \alpha=1-\sqrt{3},\ \beta=1,\ \gamma=1,\ \delta=1+\sqrt{3} \] Step 4: Find the required sum.
\[ \alpha+\beta+\gamma+\delta=(1-\sqrt{3})+1+1+(1+\sqrt{3})=5 \]