Functions – Domain and Range – JEE Main 24 Jan 2026 Shift 2

Solution:

For domain of $\sin^{-1}(t)$, we need $-1 \le t \le 1$.
$$ -1 \le \frac{1}{x^2-2x-2} \le 1 $$

Case 1: $\frac{1}{x^2-2x-2} \ge -1$
$$\frac{1 + x^2 – 2x – 2}{x^2 – 2x – 2} \ge 0 \Rightarrow \frac{x^2 – 2x – 1}{x^2 – 2x – 2} \ge 0$$
Roots of numerator: $x = \frac{2 \pm \sqrt{4+4}}{2} = 1 \pm \sqrt{2}$.
Roots of denominator: $x = \frac{2 \pm \sqrt{4+8}}{2} = 1 \pm \sqrt{3}$.
Solution $S_1$: $x \in (-\infty, 1-\sqrt{3}) \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup (1+\sqrt{3}, \infty)$.

Case 2: $\frac{1}{x^2-2x-2} \le 1$
$$\frac{1 – (x^2 – 2x – 2)}{x^2 – 2x – 2} \le 0 \Rightarrow \frac{-x^2 + 2x + 3}{x^2 – 2x – 2} \le 0 \Rightarrow \frac{x^2 – 2x – 3}{x^2 – 2x – 2} \ge 0$$
Roots of numerator: $(x-3)(x+1) = 0 \Rightarrow x = 3, -1$.
Solution $S_2$: $x \in (-\infty, -1] \cup (1-\sqrt{3}, 1+\sqrt{3}) \cup [3, \infty)$.

Intersection of $S_1$ and $S_2$:
Comparing values: $1-\sqrt{3} \approx -0.732$, $1-\sqrt{2} \approx -0.414$, $-1$, $1+\sqrt{2} \approx 2.414$, $1+\sqrt{3} \approx 2.732$, $3$.

Domain: $(-\infty, -1] \cup [1-\sqrt{2}, 1+\sqrt{2}] \cup [3, \infty)$.

Here $\alpha = -1, \beta = 1-\sqrt{2}, \gamma = 1+\sqrt{2}, \delta = 3$.
$$\alpha + \beta + \gamma + \delta = -1 + (1-\sqrt{2}) + (1+\sqrt{2}) + 3 = 4$$

Ans. (2)