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Which of the following inequalities has a solution set that,

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Which of the following inequalities has a solution set that, [#permalink] New post 18 Aug 2008, 12:21
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Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

A) x^4 \ge 1
B) x^3 \le 27
C) x^2 \ge 16
D) 2 \le |x| \le 5
E) 2 \le 3x + 4 \le 6
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Re: finite line segment [#permalink] New post 18 Aug 2008, 12:24
brokerbevo wrote:
Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

A) x^4 \ge 1
B) x^3 \le 27
C) x^2 \ge 16
D) 2 \le |x| \le 5
E) 2 \le 3x + 4 \le 6


E. 3x+4 is the equation for a line. If you set boundaries for the equation, you get a line segment.
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Re: finite line segment [#permalink] New post 18 Aug 2008, 13:10
zoinnk wrote:
brokerbevo wrote:
Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

A) x^4 \ge 1
B) x^3 \le 27
C) x^2 \ge 16
D) 2 \le |x| \le 5
E) 2 \le 3x + 4 \le 6


E. 3x+4 is the equation for a line. If you set boundaries for the equation, you get a line segment.


So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further.
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Re: finite line segment [#permalink] New post 18 Aug 2008, 13:34
brokerbevo wrote:
zoinnk wrote:
brokerbevo wrote:
Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

A) x^4 \ge 1
B) x^3 \le 27
C) x^2 \ge 16
D) 2 \le |x| \le 5
E) 2 \le 3x + 4 \le 6


E. 3x+4 is the equation for a line. If you set boundaries for the equation, you get a line segment.


So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further.


D is not the solution because,

2 \le |x| \le 5
Split the inequality,
You have two cases, since we are dealing with a mod function,
2<= -x which means, x<= -2 -----(1)
2<=x which means, x>=2 --------(2)

If you analyze this in a number line, you do not get a finite line. You can do the same with the other half of the inequality (with 5) and get a similar result.

The way, I solved E is,
2<= 3x+4 <= 6
-2 <= 3x <=2
-(2/3) <= x <= (2/3)
Which gives a finite boundary (line) in a number line and hence E.
You can adopt a similar methodology for A, B and C to prove it incorrect.
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Re: finite line segment [#permalink] New post 18 Aug 2008, 13:55
brokerbevo wrote:
zoinnk wrote:
brokerbevo wrote:
Which of the following inequalities has a solution set that, when graphed on the number line, is a single line segment of finite length?

A) x^4 \ge 1
B) x^3 \le 27
C) x^2 \ge 16
D) 2 \le |x| \le 5
E) 2 \le 3x + 4 \le 6


E. 3x+4 is the equation for a line. If you set boundaries for the equation, you get a line segment.


So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further.


The graph for D is disjointed due to the absolute values.
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Re: finite line segment [#permalink] New post 18 Aug 2008, 13:57
leonidas wrote:
brokerbevo wrote:
zoinnk wrote:

E. 3x+4 is the equation for a line. If you set boundaries for the equation, you get a line segment.


So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further.


D is not the solution because,

2 \le |x| \le 5
Split the inequality,
You have two cases, since we are dealing with a mod function,
2<= -x which means, x<= -2 -----(1)
2<=x which means, x>=2 --------(2)

If you analyze this in a number line, you do not get a finite line. You can do the same with the other half of the inequality (with 5) and get a similar result.


But in D, x IS finite because the boundary of x is as follows: -5 < x < 5 --> x can have no value outside this expression.
For example, x can not equal -6, -97, 22, or 9 ---> x is finite in D
???
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Re: finite line segment [#permalink] New post 18 Aug 2008, 14:14
leonidas wrote:
brokerbevo wrote:
zoinnk wrote:

So, you are saying that the only reason it is not D is because D is not an equation for a line b/c its not in the form y = mx + b? Because D does have a set of finite values for x. So, if could explain a little further.


D is not the solution because,

2 \le |x| \le 5
Split the inequality,
You have two cases, since we are dealing with a mod function,
2<= -x which means, x<= -2 -----(1)
2<=x which means, x>=2 --------(2)

If you analyze this in a number line, you do not get a finite line. You can do the same with the other half of the inequality (with 5) and get a similar result.


But in D, x IS finite because the boundary of x is as follows: -5 < x < 5 --> x can have no value outside this expression.
For example, x can not equal -6, -97, 22, or 9 ---> x is finite in D
???



The question stem says that- "single line segment of finite length"
Solving D,
-5 <= x <= 5 and
2 <= x <= -2
If you plot these on the number line, you will get two line segments satisfying the conditions:
-5 <= x <= -2 and 2 <= x<= 5
Hence, not D.
Sorry, I take back what I said earlier about not making a finite line segment for D. It actually gives two finite line segments.
Hope this helps.
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Re: finite line segment [#permalink] New post 18 Aug 2008, 14:35
leonidas wrote:
leonidas wrote:
zoinnk wrote:

The question stem says that- "single line segment of finite length"
Solving D,
-5 <= x <= 5 and
2 <= x <= -2
If you plot these on the number line, you will get two line segments satisfying the conditions:
-5 <= x <= -2 and 2 <= x<= 5
Hence, not D.
Sorry, I take back what I said earlier about not making a finite line segment for D. It actually gives two finite line segments.
Hope this helps.


Ahh yes. That makes sense now. I wasn't taking into account the fact that -5 < x < 5 includes 1 and -1 (for example) which does not satisfy 2 \le |x| \le 5. Its definitely 2 different line segments.
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Re: finite line segment   [#permalink] 18 Aug 2008, 14:35
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