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Re: finite line segment [#permalink]
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. _________________

Factorials were someone's attempt to make math look exciting!!!

Re: finite line segment [#permalink]
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. _________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

Re: finite line segment [#permalink]
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.

Re: finite line segment [#permalink]
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 ??? _________________

Factorials were someone's attempt to make math look exciting!!!

Re: finite line segment [#permalink]
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. _________________

To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton

Re: finite line segment [#permalink]
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. _________________

Factorials were someone's attempt to make math look exciting!!!

gmatclubot

Re: finite line segment
[#permalink]
18 Aug 2008, 14:35