Let [f(x) = ^ - 1x] and [g(x) = x^2 - x - 2 2x^2 - x - 6].

Question

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 :

Accepted Answer

$$g(2) = \mathop {\lim }\limits_{x \to 2} {{(x - 2)(x + 1)} \over {(2x + 3)(x - 2)}} = {3 \over 7}$$

Domain of $$fog(x) = {\sin ^{ - 1}}(g(x))$$

$$ \Rightarrow |g(x)|\, \le 1$$

$$\left| {{{{x^2} - x - 2} \over {2{x^2} - x - 6}}} \right| \le 1$$

$$\left| {{{(x + 1)(x - 2)} \over {(2x + 3)(x - 2)}}} \right| \le 1$$

$${{x + 1} \over {2x + 3}} \le 1$$ and $${{x + 1} \over {2x + 3}} \ge  - 1$$

$${{x + 1 - 2x - 3} \over {2x + 3}} \le 0$$ and $${{x + 1 + 2x + 3} \over {2x + 3}} \ge 0$$

$${{x + 2} \over {2x + 3}} \ge 0$$ and $${{3x + 4} \over {2x + 3}} \ge 0$$

$$x \in ( - \infty , - 2] \cup \left[ { - {4 \over 3},\infty } \right)$$

Solutions · 1