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Re M30-21 [#permalink ]

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13 Oct 2014, 06:57
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Official Solution: If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values? A. 5

B. 10

C. 11

D. 20

E. 21

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\).

Hence in this case \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\).

1 integer value of \(y\) for this range.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\).

Hence in this case \(y=|x+5|-|x-5|=x+5-(5-x)=2x\).

Therefore for this range \(-10 < (y=2x) < 10\).

19 integer values of \(y\) for this range (from -9 to 9, inclusive).

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\).

Hence in this case \(y=|x+5|-|x-5|=x+5-(x-5)=10\).

1 integer value of \(y\) for this range.

Total = 1 + 19 + 1 = 21.

If anyone interested here is a graph of \(y=|x+5|−|x−5|:\)

As you can see y is a continuous function from -10 to 10, inclusive.

Answer: E

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Re: M30-21 [#permalink ]

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02 Dec 2014, 21:34

when -5<x< 5 Why do you multiply I X-5 I by -1 ? ? I am trying to understand the logic behind this calculation When -5<x<5, then |x+5|=x+5 and |x-5|=-(x-5)=5-x.

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Re: M30-21 [#permalink ]

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03 Dec 2014, 04:16
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rsamant wrote:

when -5<x< 5 Why do you multiply I X-5 I by -1 ? ? I am trying to understand the logic behind this calculation When -5<x<5, then |x+5|=x+5 and |x-5|=-(x-5)=5-x.

Absolute value properties: When \(x \le 0\) then \(|x|=-x\), or more generally when \(\text{some expression} \le 0\) then \(|\text{some expression}| = -(\text{some expression})\). For example: \(|-5|=5=-(-5)\); When \(x \ge 0\) then \(|x|=x\), or more generally when \(\text{some expression} \ge 0\) then \(|\text{some expression}| = \text{some expression}\). For example: \(|5|=5\).

Therefore, when -5 < x < 5, then x + 5 > 0 and x - 5 < 0, thus |x + 5| = x + 5 and |x - 5| = -(x - 5) = 5 - x. Hope it helps.

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Re: M30-21 [#permalink ]

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17 Dec 2014, 22:00

Can you please explain that in condition -5<x<5 , how we calulated that which side should have +ve and which side should have -ve.

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Re: M30-21 [#permalink ]

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20 Nov 2015, 08:52

Hello Bunuel, Could you please explain the solution in some other way?? Thanks

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Re: M30-21 [#permalink ]

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21 Nov 2015, 02:47
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Re: M30-21 [#permalink ]

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02 Oct 2016, 14:14

Bunuel, I fell for the trap and assumed x was an integer as well. If you can add, "x does not have to be an integer" in the explanation, I think it will help. Thanks for a great question!

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Re: M30-21 [#permalink ]

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23 Oct 2016, 06:27

Bunuel wrote:

Official Solution: If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values? A. 5

B. 10

C. 11

D. 20

E. 21

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\).

Hence in this case \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\).

1 integer value of \(y\) for this range.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\).

Hence in this case \(y=|x+5|-|x-5|=x+5-(5-x)=2x\).

Therefore for this range \(-10 < (y=2x) < 10\).

19 integer values of \(y\) for this range (from -9 to 9, inclusive).

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\).

Hence in this case \(y=|x+5|-|x-5|=x+5-(x-5)=10\).

1 integer value of \(y\) for this range.

Total = 1 + 19 + 1 = 21.

If anyone interested here is a graph of \(y=|x+5|−|x−5|:\)

As you can see y is a continuous function from -10 to 10, inclusive.

Answer: E

Dear Bunuel,

Are the ranges correctly defined , shouldn't the ranges be

For critical points [-5,5]

the solution range would be

1) x< -5

2) -5 <= x < 5

3) x>=5

Thanks

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Re: M30-21 [#permalink ]

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23 Oct 2016, 06:41
rt1601 wrote:

Bunuel wrote:

Official Solution: If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values? A. 5

B. 10

C. 11

D. 20

E. 21

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\).

Hence in this case \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\).

1 integer value of \(y\) for this range.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\).

Hence in this case \(y=|x+5|-|x-5|=x+5-(5-x)=2x\).

Therefore for this range \(-10 < (y=2x) < 10\).

19 integer values of \(y\) for this range (from -9 to 9, inclusive).

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\).

Hence in this case \(y=|x+5|-|x-5|=x+5-(x-5)=10\).

1 integer value of \(y\) for this range.

Total = 1 + 19 + 1 = 21.

If anyone interested here is a graph of \(y=|x+5|−|x−5|:\)

As you can see y is a continuous function from -10 to 10, inclusive.

Answer: E

Dear Bunuel,

Are the ranges correctly defined , shouldn't the ranges be

For critical points [-5,5]

the solution range would be

1) x< -5

2) -5 <= x < 5

3) x>=5

Thanks

You have the same ranges - it does not matter in which range you include = sign.

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Re: M30-21 [#permalink ]

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13 Jan 2017, 08:50

I always get a bit lost with regards to >= versus >. Could someone please clarify the following: if I have |X-5| which of the following is correct: X<=-5 --> -X + 5 X<5 --> -X + 5

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Re: M30-21 [#permalink ]

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19 Jul 2017, 21:42

hi brunel when x>5, y = 10 when x<5, y = -10 when -5<x<5, y= 2x...These 3 steps are clear to me.But my concern is as per your solution you have taken the following range: Therefore for this range −10<(y=2x)<10.- still not clear with this part as why we are taking value from -10 to 10.What is the logic behind this

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Re: M30-21 [#permalink ]

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19 Jul 2017, 21:52
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Re M30-21 [#permalink ]

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26 Aug 2017, 07:55

I think this is a poor-quality question and I don't agree with the explanation. possible mistake? in the range of -5<x<5, according to the explanation, y can get 19 values. however, if we assign x= 4,3,2,1,0,-1,-2,-3,-4 it results return exactly 9 options. respectively y=2x as mentioned.

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Re: M30-21 [#permalink ]

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26 Aug 2017, 07:59
gleshem wrote:

I think this is a poor-quality question and I don't agree with the explanation. possible mistake? in the range of -5<x<5, according to the explanation, y can get 19 values. however, if we assign x= 4,3,2,1,0,-1,-2,-3,-4 it results return exactly 9 options. respectively y=2x as mentioned.

You should read the question more carefully. We are not told that x is an integer so there are more integer values of y possible when x is not integer.

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This is a min/max and count of set problem in disguise. For convenience, let |x+5| = a and |x-5| = b, so y = a-b, where a≥0 and b≥0 (modulus' result can't be less than 0) y is minimized when a=0, and maximized when b=0, so set a and b to these numbers: a = |x+5| = 0 --> x1=-5 b = |x-5| = 0 --> x2=5 Sub x1 & x2 into the original question for the min/max of y Min(y) = |-5+5| - |-5-5| = -10 Max(y) = |5+5| - |5-5| = 10 Thus the subset of y containing only integers has 10 - (-10) + 1 = 21 members in it

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Re M30-21 [#permalink ]

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18 Nov 2017, 01:29

I think this is a high-quality question and I agree with explanation. How do we get to know about the range which we have to check? Like x<-5 -5<x<5 and x>5 Please explain @bunnel

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