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# If x is an integer, What is the value of |x+5|+ |x-3|?

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If x is an integer, What is the value of |x+5|+ |x-3|?  [#permalink]

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10 Jul 2019, 22:53
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If x is an integer, What is the value of |x+5|+ |x-3|?

$$A. x^2<16$$
$$B. (x+1)^2<25$$

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If x is an integer, What is the value of |x+5|+ |x-3|?  [#permalink]

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Updated on: 07 Aug 2019, 08:03
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The range -5≤x≤3 will give a unique value for |x+5|+ |x-3|

If x is outside this range, we can get multiple values

(1) $$x^2<16$$

|x|<4 or -4<x<4

Since x is an integer, the minimum value x can take is -3 and the maximum is 3

This lies within our favorable region -5≤x≤3

Sufficient

(2) $$(x+1)^2<25$$

|x+1|<5 or -5<x+1<5 or -6<x<4

This means the minimum value x can take is -5 and the maximum is 3

Once again, this is within our favorable region -5≤x≤3

Sufficient

Originally posted by firas92 on 11 Jul 2019, 01:17.
Last edited by firas92 on 07 Aug 2019, 08:03, edited 1 time in total.
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Re: If x is an integer, What is the value of |x+5|+ |x-3|?  [#permalink]

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11 Jul 2019, 23:05
Bunuel I do not really follow the solution above. Some concepts seem to have been used without proper explanation. Could you please provide a solution to this that a beginner would understand? Thanks in advance.
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Re: If x is an integer, What is the value of |x+5|+ |x-3|?  [#permalink]

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13 Jul 2019, 16:45
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Before we dive into this question, let's try and understand the properties of the modulus function

1. |x+a| has a domain (-infinity, infinity) and range [0, infinity).
2. |x+a| can be rewritten as |x-(-a)| and is nothing but the distance of x from (-a).
2. |x+a| divides the x axis into 3 regions:
a. x < -a
b. x = -a
c. x > -a

Therefore, the range of the function |x+5|+ |x-3| can be divided into 3 segments: 1. x<-5 2. -5≤x≤3 3. x>3

When x<-5:

1. One approach would be taking 2 random values of x (x1 and x2) such that x<-5 and checking the values of f(x) = |x+5|+ |x-3|. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range.
2. Second approach would be to check out the graphs of the two functions, |x+5| and |x-3| in this range. We would see ascendent parallel lines with a negative slope. Therefore, f(x) decreases for decreasing x.

When x>3:

1. Here too, approach #1, would be taking 2 random values of x (x1 and x2) such that x>3 and checking the values of f(x) = |x+5|+ |x-3|. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range.
2. Second approach would be to check out the graphs of the two functions, |x+5| and |x-3| in this range. We would see ascendent parallel lines with a positive slope. Therefore, f(x) increases for increasing values of x.

When -5≤x≤3:

The graphs of two functions are straight lines here that converge, intersect and diverge, with increasing x starting at -5 through 3, in this region with equal but opposite slopes. Therefore, for all x in this region, f(x) = 8.
The same can be verified by trying random values of x.

Therefore, let's now have a look at the statements to see whether they can help us determine whether x lies in our favorable range or not.

Statement 1: x^2<16

=> |x|<4 => -4<x<4. Since x is an integer, this translates to -3≤x≤3. This lies in the favorable range. Therefore this is sufficient.

Statement 2: (x+1)^2 <25

=> |x+1|<25 => -5<x+1<5 => -6<x<4. Since x is an integer, this translates to -5≤x≤3. This lies in the favorable range. Therefore this is sufficient.

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If x is an integer, What is the value of |x+5|+ |x-3|?  [#permalink]

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26 Jul 2019, 11:52
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Kinshook wrote:
If x is an integer, What is the value of |x+5|+ |x-3|?

$$A. x^2<16$$
$$B. (x+1)^2<25$$

given: x=integer, |x+5|+|x-3|=?

(1) $$x^2<16$$: then |x|<4,…-4>x>4={-3,-2,-1,0,1,2,3}, when you replace any of these values in the equation, you get the same result, 8; sufficient.

(2) $$(x+1)^2<25$$: then, |x+1|<5,…-6<x<4={-5,-4…}, when you replace any of these values in the equation, you get the same result, 8; sufficient.

Hey Bunuel or chetan2u, instead of testing all the values in the range, isn't there some way better?
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Re: If x is an integer, What is the value of |x+5|+ |x-3|?  [#permalink]

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07 Aug 2019, 08:08
Can you explain why this happened
wizardofoddz
Since x is an integer, this translates to -3≤x≤3
Why suddenly range changed from -4 to 4 to -3 to 3?

wizardofoddz wrote:
Before we dive into this question, let's try and understand the properties of the modulus function

1. |x+a| has a domain (-infinity, infinity) and range [0, infinity).
2. |x+a| can be rewritten as |x-(-a)| and is nothing but the distance of x from (-a).
2. |x+a| divides the x axis into 3 regions:
a. x < -a
b. x = -a
c. x > -a

Therefore, the range of the function |x+5|+ |x-3| can be divided into 3 segments: 1. x<-5 2. -5≤x≤3 3. x>3

When x<-5:

1. One approach would be taking 2 random values of x (x1 and x2) such that x<-5 and checking the values of f(x) = |x+5|+ |x-3|. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range.
2. Second approach would be to check out the graphs of the two functions, |x+5| and |x-3| in this range. We would see ascendent parallel lines with a negative slope. Therefore, f(x) decreases for decreasing x.

When x>3:

1. Here too, approach #1, would be taking 2 random values of x (x1 and x2) such that x>3 and checking the values of f(x) = |x+5|+ |x-3|. f(x1) and f(x2) produce different values. Therefore, value of f(x) can not be determined in this range.
2. Second approach would be to check out the graphs of the two functions, |x+5| and |x-3| in this range. We would see ascendent parallel lines with a positive slope. Therefore, f(x) increases for increasing values of x.

When -5≤x≤3:

The graphs of two functions are straight lines here that converge, intersect and diverge, with increasing x starting at -5 through 3, in this region with equal but opposite slopes. Therefore, for all x in this region, f(x) = 8.
The same can be verified by trying random values of x.

Therefore, let's now have a look at the statements to see whether they can help us determine whether x lies in our favorable range or not.

Statement 1: x^2<16

=> |x|<4 => -4<x<4. Since x is an integer, this translates to -3≤x≤3. This lies in the favorable range. Therefore this is sufficient.

Statement 2: (x+1)^2 <25

=> |x+1|<25 => -5<x+1<5 => -6<x<4. Since x is an integer, this translates to -5≤x≤3. This lies in the favorable range. Therefore this is sufficient.

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If x is an integer, What is the value of |x+5|+ |x-3|?  [#permalink]

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07 Aug 2019, 11:08
1
|x+5|+ |x-3|
= (x+5) - (x-3) = 8
---> if both x+5>0 and x-3<0 are satisfied. That gives us a combined range -5≤x≤3. If x is outside that range, we can get many x value.

(1) x^2<16
-4<x<4 where x is n integer.
x is within the abovementioned range -5≤x≤3. Thus, the equation give a singular result, that is a constant 8.
SUFFICIENT

(2) (x+1)^2<25
-5<x+1<5 or -6<x<4 where x is an integer. In other word, x is within the abovementioned range -5≤x≤3. Thus, the equation gives a singular solution, that is a constant 8.
SUFFICIENT

If x is an integer, What is the value of |x+5|+ |x-3|?   [#permalink] 07 Aug 2019, 11:08
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