Let $f(x) = {\sin ^{ - 1}}x$ and $g(x) = {{{x^2} - x - 2} \over {2{x^2} - x - 6}}$. If $g(2) = \mathop {\lim }\limits_{x \to 2} g(x)$, then the domain of the function fog is :

$$g(2) = \mathop {\lim }\limits_{x \to 2} {{(x - 2)(x + 1)} \over {(2x + 3)(x - 2)}} = {3 \over 7}$$<br><br>Domain of $fog(x) = {\sin ^{ - 1}}(g(x))$<br><br>$\Rightarrow |g(x)|\, \le 1$<br><br>$\left| {{{{x^2} - x - 2} \over {2{x^2} - x - 6}}} \right| \le 1$<br><br>$\left| {{{(x + 1)(x - 2)} \over {(2x + 3)(x - 2)}}} \right| \le 1$<br><br>${{x + 1} \over {2x + 3}} \le 1$ and ${{x + 1} \over {2x + 3}} \ge - 1$<br><br>${{x + 1 - 2x - 3} \over {2x + 3}} \le 0$ and ${{x + 1 + 2x + 3} \over {2x + 3}} \ge 0$<br><br>${{x + 2} \over {2x + 3}} \ge 0$ and ${{3x + 4} \over {2x + 3}} \ge 0$<br><br>$x \in ( - \infty , - 2] \cup \left[ { - {4 \over 3},\infty } \right)$