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LeoN88
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What values of x will satisfy the inequality \(\frac{|x|-2}{|x|+3}≤0?\)
A) x∈ (-3,2)
B) x∈ (-∞, -3) ∪ (2, ∞)
C) x∈ (-3,3)
D) x∈ (-2,2)
E) x∈ (-∞, -3) ∪ (-2, ∞)


\(\frac{|x|-2}{|x|+3}≤0\) will always have |x|+3>0, |x|-2≤0......|x|≤2.....
Thus x lies between -2 and 2.

D
Shouldn't this be [-2,2]; i.e. inclusive of -2 & 2?


Any thoughts on this VeritasKarishma / GMATNinja
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gaurav2m
LeoN88
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Asad
What values of x will satisfy the inequality \(\frac{|x|-2}{|x|+3}≤0?\)
A) x∈ (-3,2)
B) x∈ (-∞, -3) ∪ (2, ∞)
C) x∈ (-3,3)
D) x∈ (-2,2)
E) x∈ (-∞, -3) ∪ (-2, ∞)


\(\frac{|x|-2}{|x|+3}≤0\) will always have |x|+3>0, |x|-2≤0......|x|≤2.....
Thus x lies between -2 and 2.

D
Shouldn't this be [-2,2]; i.e. inclusive of -2 & 2?


Any thoughts on this VeritasKarishma / GMATNinja

Yes, it should be [-2,2]. None of the options satisfy.
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Here's how I solved it, please confirm if this method is correct.

\(\frac{|x|-2}{|x|+3}\) \(\leq{0}\).

Now, we know that the denominator will be positive no matter what. So, we can go ahead and multiply the whole inequality with |x|+3 to get |x|-2 \(\leq{0.\)

This implies that \(-2\leq{x\leq{2\), and hence the range [-2,2] is valid and the answer is (D).
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I feel there is a mistake here. That answer ought to be [-2 2]
Thank you.
Proof
Since |x|+3> 0 (a positive number), we can multiply both sides of the inequality by |x|+3 to get
|x|-2≤0
Implies
|x|≤2 implies -2≤x≤2
Hence [-2 2]

Posted from my mobile device
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The Ans is D.
Reason:
When we have such type of question where what we get as an answer is not present in the solution then we have to choose the option which is closest to the original solution.

Assuming value of x=3 and putting it in the equation the ans is wrong so B and E is out of scope.

So closest ans is option D.
Assuming value of x=-2.8 and putting it in the equation the ans is wrong so A and C is out of scope.
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