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Re: Which of the following inequalities has a solution set that, [#permalink]

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18 Aug 2008, 14:10

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: Which of the following inequalities has a solution set that, [#permalink]

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18 Aug 2008, 14:34

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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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: Which of the following inequalities has a solution set that, [#permalink]

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18 Aug 2008, 14:55

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: Which of the following inequalities has a solution set that, [#permalink]

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18 Aug 2008, 14:57

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: Which of the following inequalities has a solution set that, [#permalink]

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18 Aug 2008, 15:14

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: Which of the following inequalities has a solution set that, [#permalink]

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18 Aug 2008, 15:35

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!!!

Re: Which of the following inequalities has a solution set that, [#permalink]

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20 Sep 2014, 15:26

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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 ≥ 1 B. x^3 ≤ 27 C. x^2 ≥ 16 D. 2≤ |x| ≤ 5 E. 2 ≤ 3x+4 ≤ 6

The key words in the stem are: "a single line segment of finite length"

Now, answer choices A, B, and C can not be correct answers as solutions sets for these exponential functions are not limited at all (>= for even powers and <= for odd power) and thus can not be finite (x can go to + or -infinity for A and C and x can got to -infinity for B). As for D: we have that absolute value of x is between two positive values, thus the solution set for x (because of absolute value) will be two line segments which will be mirror images of each other.

Answer: E.

Just to demonstrate:

A. x^4 >= 1 --> \(x\leq{-1}\) or \(x\geq{1}\): two infinite ranges;

B. x^3 <= 27 --> \(x\leq{3}\): one infinite range;

C. x^2 >= 16 --> \(x\leq{-4}\) or \(x\geq{4}\): two infinite ranges;

D. 2 <= |x| <= 5 --> \(-5\leq{x}\leq{-2}\) or \(2\leq{x}\leq{5}\): two finite ranges;

E. 2 <= 3x+4 <= 6 --> \(-2\leq{3x}\leq{2}\) --> \(-\frac{2}{3}\leq{x}\leq{\frac{2}{3}}\): one finite range.