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The set of all real $x$ satisfying the inequality
$\frac{3-|x|}{4-|x|} \geq 0,$ is
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$\frac{3-|x|}{4-|x|} \geq 0,$ is
Solution:
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Verified Answer
The correct answer is:
[-3,3]$\cup(-\infty,-4) \cup(4, \infty)$
Given, $\frac{3-|x|}{4-|x|} \geq 0$
$\Rightarrow 3-|x| \leq 0$ and $4-|x|<0$
or $3-|x| \geq 0$ and $4-|x|<0$
$\Rightarrow|x| \geq 3$ and $|x|>4$
or $|x| \leq 3$ and $|x|<4$
$\Rightarrow|x|>4$ or $|x| \leq 3$
$\Rightarrow x \in(-\infty,-4) \cup[-3,3] \cup(4, \infty)$
$\Rightarrow 3-|x| \leq 0$ and $4-|x|<0$
or $3-|x| \geq 0$ and $4-|x|<0$
$\Rightarrow|x| \geq 3$ and $|x|>4$
or $|x| \leq 3$ and $|x|<4$
$\Rightarrow|x|>4$ or $|x| \leq 3$
$\Rightarrow x \in(-\infty,-4) \cup[-3,3] \cup(4, \infty)$
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