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01 Jul 2014, 05:06
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If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?
A. 5
B. 10
C. 11
D. 20
E. 21
M30-21
A. 5
B. 10
C. 11
D. 20
E. 21
M30-21
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01 Jul 2014, 05:07
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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
Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges
When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.
When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.
When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.
Total number of integer values for \(y\) = 1 + 19 + 1 = 21.
If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)
As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.
Try NEW absolute value DS question.
WolframAlpha--yx5-x-5_--2014-07-08_0728.png [ 12.01 KiB | Viewed 93465 times ]
If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?
A. 5
B. 10
C. 11
D. 20
E. 21
Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges
When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.
When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.
When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.
Total number of integer values for \(y\) = 1 + 19 + 1 = 21.
If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)
As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.
Try NEW absolute value DS question.
Attachment:
WolframAlpha--yx5-x-5_--2014-07-08_0728.png [ 12.01 KiB | Viewed 93465 times ]
01 Jul 2014, 20:36
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Bunuel
Use the number line for mod questions:
___________-5 ______________0______________5___________
You want the values of y which is the difference between "distance from -5" and "distance from 5".
Anywhere on the left of -5,
<------------------------------------------------
<-------------
___________-5 ______________0______________5___________
The difference between the two distances will always be -10 (taking difference to mean what it does for GMAT)
In between -5 and 5, can y can the value of -9? Sure. Move 0.5 to the right of -5. At x = -4.5, distance from -5 will be .5 and distance from 5 will be 9.5. So y = 0.5 - 9.5 = -9
Note that x needn't be an integer. Only y needs to be an integer.
Similarly, at various points between -5 and 5, y will take all integer values from -9 to 9.
To the right of 5, the difference between the two distances will always be 10.
So values that y can take: -10, -9, ... 0 ... 9, 10 i.e. a total of 21 values.
