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# If k is a positive constant and y = |x - k| - |x + k|, what

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If k is a positive constant and y = |x - k| - |x + k|, what [#permalink]  23 May 2006, 02:34
If k is a positive constant and y = |x - k| - |x + k|, what is the maximum value of y?

(1) x < 0

(2) k = 3
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[#permalink]  23 May 2006, 02:48
I go with B.
The maximum value we get is 6 whatever the value of x
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[#permalink]  23 May 2006, 04:26
remgeo wrote:
I go with B.
The maximum value we get is 6 whatever the value of x

Answer should be E,

Take X =1
=> |1-3| - |1+3|
=> 2 - 4
=> -2

Take x=0
=> |0-3| - |0+3|
=> 3 - 3
=> 0

Hence E
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[#permalink]  23 May 2006, 04:56
Agree with B

There are four cases:
1) y = -2k
2) y = -2x
3) y = 2x
4) y = 2k

Since we are looking for the maximum value (1) is out immediately since
it's less then (4). Now , considering (2), -k < x < k , so (4) is still the winner here. In (3) , x > k and x < -k is impossible since k is positive by definition. So (4) is the biggest possible value.

Since B provides the value of k , we can find y. y = 2 * 3 = 6
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[#permalink]  23 May 2006, 04:57
jaynayak wrote:
remgeo wrote:
I go with B.
The maximum value we get is 6 whatever the value of x

Answer should be E,

Take X =1
=> |1-3| - |1+3|
=> 2 - 4
=> -2

Take x=0
=> |0-3| - |0+3|
=> 3 - 3
=> 0

Hence E

jaynayax,
(1) says that x<0

Agree with b. The max value of y is 2k

I choose C.
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[#permalink]  23 May 2006, 06:41
deowl wrote:
Agree with B

There are four cases:
1) y = -2k
2) y = -2x
3) y = 2x
4) y = 2k

Since we are looking for the maximum value (1) is out immediately since
it's less then (4). Now , considering (2), -k < x < k , so (4) is still the winner here. In (3) , x > k and x < -k is impossible since k is positive by definition. So (4) is the biggest possible value.

Since B provides the value of k , we can find y. y = 2 * 3 = 6

Deowl, can you go back a step and explain how you came up with the 4 cases? I got B as well, but I did the problem in a more lengthy, plug in numbers type of way.
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[#permalink]  23 May 2006, 06:55
How do you get 6 as a max? I'm getting 0
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[#permalink]  23 May 2006, 07:06
tl372 wrote:

Deowl, can you go back a step and explain how you came up with the 4 cases? I got B as well, but I did the problem in a more lengthy, plug in numbers type of way.

1) x-k > 0, x+k > 0 => y=-2k, x > k, x > -k
2) x-k < 0, x+k > 0 => y=-2x, x < k , x > -k
3) x-k > 0, x+k < 0 => y=2x, x > k, x < -k
4) x-k < 0, x+k < 0 => y=2k, x < k, x < -k
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Re: MHTN Challenge [#permalink]  23 May 2006, 07:09
BG wrote:
If k is a positive constant and y = |x - k| - |x + k|, what is the maximum value of y?

(1) x < 0

(2) k = 3

C it is...

If x = -1 Y = 2
If X= -2 Y= 4
If X= -3 Y = 6
If X= -4 Y = 6
....

Y can not be over 6

hence, C
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[#permalink]  23 May 2006, 07:11
oops....so B...not C
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[#permalink] 23 May 2006, 07:11
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