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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]

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18 Aug 2008, 13:21
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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 ≥ 1
B. x^3 ≤ 27
C. x^2 ≥ 16
D. 2≤ |x| ≤ 5
E. 2 ≤ 3x+4 ≤ 6

OPEN DISCUSSION OF THIS QUESTION IS HERE: which-of-the-following-inequalities-has-a-solution-set-that-127820.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 20 Sep 2014, 16:53, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: Which of the following inequalities has a solution set that, [#permalink]

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18 Aug 2008, 13:24
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: 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.
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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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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.
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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
???
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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.
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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.
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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, [#permalink]

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20 Sep 2014, 16:55
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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 ≥ 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.

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: which-of-the-following-inequalities-has-a-solution-set-that-127820.html

SIMILAR QUESTION TO PRACTICE: which-of-the-following-inequalities-has-a-solution-set-that-130666.html

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Which of the following inequalities has a solution set that,   [#permalink] 20 Sep 2014, 16:55
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