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If y = |x + 5|-|x - 5|, then y can take how many integer

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Joined: 11 Jun 2019
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 11 Sep 2019, 20:15
For Y = | X+5 | - | X - 5 |, we can identify critical points as -5 and 5.
So X must be in following ranges. X < -5 or -5 <= X < 5 or X > 5

1. Considering X < -5, we can modify modulus as
Y = -(X + 5) + (X - 5)
= -X -5 + X -5
Y = - 10...….(1 integer solution of Y for X < -5)

2. Considering X >= 5, we can modify modulus as
Y = (X +5 ) - (X - 5)
= X + 5 - X + 5
Y = 10...……(1 integer solution of Y for X >= 5)

3. Considering 3rd and last possible range of X. -5 <= X < 5,
Y = (X + 5) + (X -5)
Y = 2X...….Thus X can take any value between -5 till 4.9. But need integer values of Y.
Y= 2X can give integer values of Y only when X is an integer or X is a fraction with .5 at the end. So the value X can take are ( -5, -4.5, -4, -3.5, -3, -2.5, -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5) so total 20 values. But X = -5 will give Y to be -10. We have already considered Y = -10 in 1st range. hence in this range we get 19 possible integer values of Y.

Thus total possible integer values of Y are 1 + 1 + 19 = 21

hence (E)
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer  [#permalink]

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New post 22 Sep 2019, 08:27
Bunuel wrote:


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

A. 5
B. 10
C. 11
D. 20
E. 21

Kudos for a correct solution.



Given: \(y=|x+5|-|x-5|\),

Asked: \(y\) can take how many integer values?

When x = 5; y = 10
when x=-5; y = -10
When x <-5; y= -10
When x>5; y =10
-5<=x<=5; -10<=y<=10; 21 integer values

IMO E
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Re: If y = |x + 5|-|x - 5|, then y can take how many integer   [#permalink] 22 Sep 2019, 08:27

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