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Absolute values

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by vineet.nitd » Tue Oct 25, 2016 2:16 am
You may multiply each side of the inequality by |x| without flipping the sign and then rearrange.

x(|x|-1)>0

x>0 AND |x|>1 => x>1
OR
x<0 AND |x|<1 => -1<x<0

From the two range of values above, we know for sure that x>-1.

Hence choice B is correct.
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by GMATGuruNY » Tue Oct 25, 2016 2:24 am
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by crackverbal » Wed Oct 26, 2016 10:30 pm
Hi AaronRamsey,

Let just first analyze the given information,

x/|x|<x

Cross-multiply with |x| on both sides; we can do this because the denominator is positive.

x < x|x|,

x(|x|-1) > 0

Once you solve this draw the number line, to understand the range of values,

Just plot the roots (critical points) in the number line, that is,

x =-1, 0, and 1.

Image

So here, x > 1 and -1<x<0

(the value greater than the greatest root is always positive and then just alternate the signs for the ranges from there)

So only answer B covers all the set of values,

So answer is B.

Hope it is clear
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by [email protected] » Thu Oct 27, 2016 11:02 am
Hi AARONRAMSEY,

This question can be solved by TESTing VALUES. With typical prompts, you normally just have to TEST once. However, when a question uses the phrase "must be true", that's often a clue that you'll have to TEST at least two different values. With the addition of absolute values and inequalities, we likely have to consider negatives and fractional values as well.

To start, let's TEST X = 2...

2/|2| = 1 and 1 < 2, so X COULD be 2.
Eliminate Answers C and D

With our next TEST, we should look for something different. It might take a bit of 'playing around' before you find a value that fits though. Positive fractions and negative integers do NOT fit the given inequality.

Let's TEST X = -1/2

(-1/2)/|-1/2| = -1 and -1 < -1/2, so X COULD be -1/2
Eliminate Answers A and E.

There's only one answer left...

Final Answer: B

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by Matt@VeritasPrep » Thu Oct 27, 2016 11:13 pm
Since |x| is positive, you can multiply both sides by |x|:

x < x * |x|

Now we've got two possibilities.

Case 1:

x > 0

In this case, we can divide both sides by x and nothing funny will happen: 1 < |x|. Since x is positive, |x| = x, and we can write this as 1 < x.

Case 2:

x < 0

In this case, we can divide both sides by x, but since x is negative, the sign will flip: 1 > |x|. Since |x| is nonnegative, we can write this as 1 > |x| ≥ 0. Since x itself is negative, this is really -1 < x ≤ 0.

Combining the two cases, we have

-1 < x

with x ≠ 1. (In other words, any x > -1 EXCEPT for x = 1 is fine.)

This means B must be true, and we're done!
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by Matt@VeritasPrep » Thu Oct 27, 2016 11:16 pm
Another approach that doesn't require algebra is exploiting the answer choices and trying a few key numbers.

For instance, say x = 1/2. Then x/|x| < x is true, so x = 1/2 is a solution. Since this works, eliminate A, D, and E.

Now try x = 2. Then x/|x| < x is true, so x = 2 is a solution. This rules out C, so we're done!
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by Matt@VeritasPrep » Thu Oct 27, 2016 11:18 pm
One last trick is subtraction:

x/|x| - x < 0

x * (1/|x| - 1) < 0

If x is positive, we can divide both sides by x and get

(1/|x| - 1) < 0

and

1/|x| < 1

and

1 < |x|

If x is negative, division gives us the opposite:

1 > |x|

and we can arrive at B much as we did in my earlier solution.
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