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Question:

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

  1. $(-\infty,-2] \cup\left[-\frac{3}{2}, \infty\right)$

  2. $(-\infty,-2] \cup[-1, \infty)$

  3. $(-\infty,-2] \cup\left[-\frac{4}{3}, \infty\right)$

  4. $(-\infty,-1] \cup[2, \infty)$


Correct Option: , 3

Solution:

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

$\Rightarrow|g(x)| \leq 1 \quad, \quad g(2)=\frac{3}{7}$

$\left|\frac{x^{2}-x-2}{2 x^{2}-x-6}\right| \leq 1$

$\left|\frac{(x+1)(x-2)}{(2 x+3)(x-2)}\right| \leq 1$

$\frac{x+1}{2 x+3} \leq 1$ and $\frac{x+1}{2 x+3} \geq-1$

$\frac{x+1-2 x-3}{2 x+3} \leq 0$ and $\frac{x+1+2 x+3}{2 x+3} \geq 0$

$\frac{x+2}{2 x+3} \geq 0$ and $\frac{3 x+4}{2 x+3} \geq 0$

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

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