Question

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

• A. $(-\infty ,-2]\cup \left[-\frac{3}{2},\infty \right)$
• B. $(-\infty ,-2]\cup [-1,\infty )$
• C. $(-\infty ,-2]\cup \left[-\frac{4}{3},\infty \right)$
• D. $(-\infty ,-1]\cup [2,\infty )$

Answer 1

Answer: (C)

Domain of $\mathrm{fog}(x)={\sin }^{-1}(g(x))$
$\Rightarrow |g(x)|\le 1g(2)=\frac{3}{7}$
$\left|\frac{x^2-x-2}{2x^2-x-6}\right|\le 1$
$\left|\frac{(x+1)(x-2)}{(2x+3)(x-2)}\right|\le 1$
$\frac{x+1}{2x+3}\le 1$ and $\frac{x+1}{2x+3}\ge -1$
$\frac{x+1-2x-3}{2x+3}\le 0$ and $\frac{x+1+2x+3}{2x+3}\ge 0$
$\frac{x+2}{2x+3}\ge 0$ and $\frac{3x+4}{2x+3}\ge 0$
$x\in (-\infty ,-2]\cup \left[-\frac{4}{3},\infty \right)$