Q39 / 75 · JEE Main 2026 (24 January Shift 2) · Mathematics
If the domain of the function $f(x)=\sin ^{-1}\left(\frac{1}{x^{2}-2 x-2}\right)$, is $(-\infty, \alpha] \cup[\beta, \gamma] \cup[\delta, \infty)$, then $\alpha+\beta+\gamma+\delta$ is equal to
- A2
- B3
- Correct answerC4
- D5
Solution
For $\sin^{-1}\left(\frac{1}{x^2-2x-2}\right)$ to be defined, we need $\left|\frac{1}{x^2-2x-2}\right| \leq 1$, i.e., $|x^2-2x-2| \geq 1$.
Case 1:
$x^2-2x-2 \geq 1 \Rightarrow x^2-2x-3 \geq 0$
$\Rightarrow (x-3)(x+1) \geq 0$
$\Rightarrow x \leq -1$ or $x \geq 3$.
Case 2:
$x^2-2x-2 \leq -1 \Rightarrow x^2-2x-1 \leq 0$
$\Rightarrow (x-1)^2 \leq 2$
$\Rightarrow 1-\sqrt{2} \leq x \leq 1+\sqrt{2}$.
Domain $= (-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty)$.
So $\alpha = -1,\ \beta = 1-\sqrt{2},\ \gamma = 1+\sqrt{2},\ \delta = 3$.
$\alpha + \beta + \gamma + \delta = -1 + 1 - \sqrt{2} + 1 + \sqrt{2} + 3 = 4$.
Case 1:
$x^2-2x-2 \geq 1 \Rightarrow x^2-2x-3 \geq 0$
$\Rightarrow (x-3)(x+1) \geq 0$
$\Rightarrow x \leq -1$ or $x \geq 3$.
Case 2:
$x^2-2x-2 \leq -1 \Rightarrow x^2-2x-1 \leq 0$
$\Rightarrow (x-1)^2 \leq 2$
$\Rightarrow 1-\sqrt{2} \leq x \leq 1+\sqrt{2}$.
Domain $= (-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty)$.
So $\alpha = -1,\ \beta = 1-\sqrt{2},\ \gamma = 1+\sqrt{2},\ \delta = 3$.
$\alpha + \beta + \gamma + \delta = -1 + 1 - \sqrt{2} + 1 + \sqrt{2} + 3 = 4$.
Hence, the correct answer is option: 'C'.