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

• A. 0
• B. 4
• C. 3
• D. 1

Hint:

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

Answer 1

The Correct Option is B

Step 1: Use condition for inverse sine.
For $\sin^{-1}(x)$ to be defined, \[ -1\le x\le 1 \] So, \[ -1\le \frac{1}{x^2-2x-2}\le 1 \] Step 2: Solve the inequalities.
\[ |x^2-2x-2|\ge 1 \] Step 3: Solve $x^2-2x-2=1$ and $x^2-2x-2=-1$.
\[ x^2-2x-3=0 \Rightarrow x=3,-1 \] \[ x^2-2x-1=0 \Rightarrow x=1\pm\sqrt{2} \] Step 4: Write the domain.
\[ (-\infty,-1]\cup[1-\sqrt2,\,1+\sqrt2]\cup[3,\infty) \] Step 5: Identify values.
\[ \alpha=-1,\quad \beta=1-\sqrt2,\quad \gamma=1+\sqrt2,\quad \delta=3 \] Step 6: Final calculation.
\[ \alpha+\beta+\gamma+\delta=-1+(1-\sqrt2)+(1+\sqrt2)+3=4 \] Final conclusion.
The required value is 4.