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Let $\mathrm{f}(\mathrm{x})=\sqrt{\frac{x+1}{x+3}}$ and $\mathrm{g}(\mathrm{x})=\sqrt{\frac{2-x}{x+3}}$ be two real valued functions. Then the domain of $f / g$ is
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The correct answer is:
$[-1,2)$
$f(x)=\sqrt{\frac{x+1}{x-3}}$ and $g(x)=\sqrt{\frac{2-x}{x+3}}$
Domain for $\mathrm{f} / \mathrm{g}$
$\frac{x+1}{x+3} \geq 0 \frac{+-+}{-\infty-3-1}$
$x \in\left(-{ }^{\circ},-3\right) \cup\left[-1,{ }^{\circ}\right)$ ........(i)
$\frac{2-x}{x+3}>\Rightarrow \frac{x-2}{x+3} < 0 \frac{+-+}{-\infty-3-2 \infty}$
$x \in(-3,2)$ ........(ii)
from (i) and (ii)
$x \in[-1,2)$
Domain for $\mathrm{f} / \mathrm{g}$
$\frac{x+1}{x+3} \geq 0 \frac{+-+}{-\infty-3-1}$
$x \in\left(-{ }^{\circ},-3\right) \cup\left[-1,{ }^{\circ}\right)$ ........(i)
$\frac{2-x}{x+3}>\Rightarrow \frac{x-2}{x+3} < 0 \frac{+-+}{-\infty-3-2 \infty}$
$x \in(-3,2)$ ........(ii)
from (i) and (ii)
$x \in[-1,2)$
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