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How many values can the integer p=|x+3|-|x-3| assume?

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Re: How many values can the integer p=|x+3|-|x-3| assume?  [#permalink]

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New post 06 Jan 2017, 06:14
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Re: How many values can the integer p=|x+3|-|x-3| assume?  [#permalink]

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New post 13 Feb 2017, 05:53
Zarrolou wrote:
How many values can the integer \(p=|x+3|-|x-3|\) assume?

A)6
B)7
C)13
D)12
E)Infinite values

My own question, as always any feedback is appreciated
Click here for the OE.


This is a copy of the following GMAT Club question: https://gmatclub.com/forum/if-y-x-5-x-5 ... 73626.html
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Re: How many values can the integer p=|x+3|-|x-3| assume?  [#permalink]

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New post 06 Apr 2018, 05:37
Zarrolou wrote:
Good job arpanpatnaik, vinaymimani!

Official explanation

The function \(|x+3|-|x-3|\) for values \(\geq{}3\) equals \(6\), and for values \(\leq{}-3\) equals \(-6\)
For the middle values it follows the equation \(2x\) (as the users above correctly say)

However there is a quicker way to get to the answer than counting the possible values.

Its upper limit is \(6\), its lower limit is \(-6\) and the function \(2x\) is monotonic and increasing (and continuous), so will assume all values between 6 and -6 included.
(This is not theory necessary for the GMAT, but if notice the fact that \(2x\) must pass for all values between 6 and -6, you can save time)

So the values that the integer p can assume are \(-6,-5,...,0,...,5,6\) TOT=\(13\)

The correct answer is C

For for clarity, below there is the graph of \(|x+3|-|x-3|\) that will make my explanation more clear.


Can you please explain how you drew the graph with two variables. sort of struggling with that.
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Re: How many values can the integer p=|x+3|-|x-3| assume? &nbs [#permalink] 06 Apr 2018, 05:37

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