JEE Main 2026 Jan 24 Shift 2, Mathematics Q13: Inverse Trigonometric Functions

For ${\sin }^{-1}$, need $-1\le \frac{1}{x^2-2x-2}\le 1$.

Upper: $\frac{1}{x^2-2x-2}\le 1\Rightarrow \frac{x^2-2x-3}{x^2-2x-2}\ge 0\Rightarrow \frac{(x+1)(x-3)}{(x-1-\sqrt{3})(x-1+\sqrt{3})}\ge 0$.

Solution: $x\in (-\infty ,-1]\cup (1-\sqrt{3},1+\sqrt{3})\cup [3,\infty )$ (excluding zeros of denominator).

Lower: $\frac{1}{x^2-2x-2}\ge -1\Rightarrow \frac{(x-1-\sqrt{2})(x-1+\sqrt{2})}{(x-1-\sqrt{3})(x-1+\sqrt{3})}\ge 0$.

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

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

So $\alpha =-1$, $\beta =1-\sqrt{2}$, $\gamma =1+\sqrt{2}$, $\delta =3$.

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