JEE Main 2026 — 24 January, Evening Shift — Question 13

$-1 \leq \frac{2}{x^{2}-2 x-2} \leq 1$

$\frac{1+\mathrm{x}^{2}-2 \mathrm{x}-2}{\mathrm{x}^{2}-2 \mathrm{x}-2} \geq 0$

$\Rightarrow \frac{(\mathrm{x}-1)^{2}-2}{(\mathrm{x}-1)^{2}-3} \geq 0$

$\Rightarrow \frac{(\mathrm{x}-1-\sqrt{2})(\mathrm{x}-1+\sqrt{2})}{(\mathrm{x}-1-\sqrt{3})(\mathrm{x}-1+\sqrt{3})} \geq 0$

$$\begin{gathered} x \in(-\infty, 1-\sqrt{3}) \cup[1-\sqrt{2}, 1+\sqrt{2}] \cup(1+\sqrt{3}, 0) \end{gathered}$$

$1-\frac{1}{\mathrm{x}^{2}-2 \mathrm{x}-2} \geq 0 \Rightarrow \frac{\mathrm{x}^{2}-2 \mathrm{x}-3}{\mathrm{x}^{2}-2 \mathrm{x}-2} \geq 0$

$\Rightarrow \frac{(\mathrm{x}+1)(\mathrm{x}-3)}{(\mathrm{x}-1+\sqrt{3})(\mathrm{x}-1-\sqrt{3})} \geq 0$

$$\begin{gathered} \mathrm{x} \in(-\infty,-1] \cup(1-\sqrt{3}, \sqrt{3}+1) \cup[3, \infty)\cap(2) \end{gathered}$$

$\Rightarrow \mathrm{x} \in(-\infty,-1] \cup[1-\sqrt{2}, 1+\sqrt{2}] \cup[3, \infty)$

$\therefore \alpha+\beta+\gamma+\delta=4$