Which of the following inequalities has a solution set that, when

Question

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. \(x^4 \geq 1\)

B. \(x^3 \leq 27\)

C. \(x^2\geq 16\)

D. \(2 \leq |x| \leq 5\)

E. \(2 \leq 3x+4 \leq 6\)

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Official Answer

E

Bunuel · Accepted Answer

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. Answer: E. Hope it's clear.

WholeLottaLove · Answer

Which of the following inequalities has a solution set that when graphed on the number line, is a single segment of finite length?

A. x^4 ≥ 1 
4√x^4 ≥ 4√1
x ≥ 1
x is found anywhere from 1 to + infinity. INVALID

B. x^3 ≤ 27 
3√x^3 ≤ 3√27
x ≤ 3
x is found anywhere from 3 to -infinity. INVALID

C. x^2 ≥ 16 
√x^2 ≥ √16
|x| ≥ 4
x≥4 OR x≤-4
x is found anywhere from 4 to +infinity or from -4 to -infinity. INVALID

D. 2≤ |x| ≤ 5 
2≤|x|
2≤x OR -2≥x
|x| ≤ 5
x≤5 OR x≥-5
This option has two segments of finite length, not one as is required by the question. INVALID

E. 2 ≤ 3x+4 ≤ 6  
-2 ≤ 3x ≤ 2
(-2/3) ≤ x ≤ (2/3)
x is found anywhere from (-2/3) to (2/3). VALID

(E)

MacFauz · Answer

The question asks for the option for which the range does not extend to infinity or does not have a break in between.

A) x can be any value greater than 1
B) x can be any value lesser than 3
C) x can be any value greater than 4
D) This option does have a finite range. However, there is a break inbetween for values in the range -2 < x < 2.
E) Same as  -\frac{2}{3} \leq x \leq \frac{2}{3} . Finite straight line.

Bunuel · Answer

Similar question to practice: https://gmatclub.com/forum/which-of-the- ... 30666.html

connectvinoth · Answer

Could someone explain  Option A  in detail? 
I understand upon taking 4th root on both sides it becomes:   x>= +-1  
But, I don't understand how it gets simplified further as its been explained as:  x^4 >= 1 -->  x\leq{-1}  or  x\geq{1}

Bunuel · Answer

x >= +/- 1 does not make any sense. When taking 4th root from both sides we'll get |x| \geq{1} , which is the same as x\leq{-1} or x\geq{1} . Theory on Inequalities: Solving Quadratic Inequalities - Graphic Approach: solving-quadratic-inequalities-graphic-approach-170528.html Inequality tips: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1379270 inequalities-trick-91482.html data-suff-inequalities-109078.html range-for-variable-x-in-a-given-inequality-109468.html everything-is-less-than-zero-108884.html graphic-approach-to-problems-with-inequalities-68037.html All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184 All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189 700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

LogicGuru1 · Answer

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

Lets check one by one 
 x^4 ≥ 1

Nope :- This maps to a two value of x≥1 and x ≤-1 on the number line. This is the not a finite line. This is actually an infinite line broken only from -1 to 1 but extending from  -∞  to  +∞

x^3 ≤ 27

Nope:- This maps to infinite values of ≤3 on the number line. This is the equation of a INFINITE line from 3 to  -∞

x^2 ≥ 16

Nope :- This maps to a two value of x≥4 and x ≤-4 on the number line. This is the not a finite line. This is actually an infinite line broken only from -4 to 4 but otherwise extending from  -∞  to  +∞

2≤ |x| ≤ 5

Looks interesting. Lets check |X| can actually be seen as +x and also =-x
Lets check both cases individually

CASE 1)  2≤ x≤ 5  This gives a finite line.

CASE 2)  2≤ -x ≤ 5  or -5≤ x ≤ -2 This gives a finite line again 
A TOTAL OF 2 FINITE RANGES. We need only one finite range.

2 ≤ 3x+4 ≤ 6

x≥-\frac{2}{3}  and  x≤\frac{2}{3}  or  -\frac{2}{3}≤x≤\frac{2}{3}

YES this is a finite range from -x to +x

E IS THE ANSWER

≥ ≤ ∞

pra1785 · Answer

Hi,

I think I am unable to understand the question here. I was between B and E and chose B because it gave one value which is 3.

I failed to understand what "a single line segment of finite length" means. Can you explain this and what we actually need to find here?

Bunuel · Answer

x^3 \leq 27 gives x\leq{3} , not x = 3. Maybe images could help: A. x^4 >= 1 --> x\leq{-1} or x\geq{1} : two infinite ranges; https://gmatclub.com/forum/download/file.php?id=36852 B. x^3 <= 27 --> x\leq{3} : one infinite range; https://gmatclub.com/forum/download/file.php?id=36848 C. x^2 >= 16 --> x\leq{-4} or x\geq{4} : two infinite ranges; https://gmatclub.com/forum/download/file.php?id=36853 D. 2 <= |x| <= 5 --> -5\leq{x}\leq{-2} or 2\leq{x}\leq{5} : two finite ranges; https://gmatclub.com/forum/download/file.php?id=46258 E. 2 <= 3x+4 <= 6 --> -2\leq{3x}\leq{2} --> -\frac{2}{3}\leq{x}\leq{\frac{2}{3}} : one finite range. https://gmatclub.com/forum/download/file.php?id=36851 Similar questions: https://gmatclub.com/forum/which-of-the ... 27820.html (GMAT Prep). https://gmatclub.com/forum/which-of-the ... 30666.html (Knewton). MSP65914h1996e60ae9h87000025b2397iibih7d09.gif MSP5943231f8gec7gfdfc45000012a38989ihh83g14.gif MSP5943231f8gec7gfdfc45000012a38989ihh83g14.gif MSP461h7hfif5ab0a1bfd0000250gdh577fh80424.gif MSP25821e4cg2fhi8d970g1000028dh8ii72237cfgc.gif MSP67562331eie358a08a31000017b4hh7e5e6ahiih.gif

ScottTargetTestPrep · Answer

We must determine from our answer choices which inequality is a single line segment of finite length when graphed on the number line.
 
A) x^4 ≥ 1
 
∜(x^4) ≥ ∜1
 
|x| ≥ 1
 
x ≥ 1 or x ≤ -1
 
Since the solution set from answer choice A produces two rays of infinite length, answer choice A is not correct.
 
B) x^3 ≤ 27
 
∛(x^3) ≤ ∛27
 
x ≤ 3
 
Since the solution set from answer choice B produces one ray of infinite length, answer choice B is not correct.
 
C) x^4 ≥ 16
 
∜(x^4) ≥ ∜16
 
|x| ≥ 2
 
x ≥ 2 or x ≤ -2
 
Since the solution set from answer choice C produces two rays of infinite length, answer choice C is not correct.
 
D) 2 ≤ |x| ≤ 5
 
We must solve for when x is positive and for when x is negative.
 
When x is positive:
 
2 ≤ x ≤ 5
 
When x is negative:
 
2 ≤ -x ≤ 5
 
-1(2 ≤ -x ≤ 5)
 
-2 ≥ x ≥ -5
 
Since the solution set from answer choice D produces two line segments of finite length, answer choice D is not correct.
 
Since answers A, B, C, and D are not correct, E must be the correct answer. However, for practice, we can solve it anyway.
 
E) 2 ≤ 3x + 4 ≤ 6
 
-2 ≤ 3x ≤ 2
 
-2/3 ≤ x ≤ 2/3
 
Since the solution set from answer choice E produces one line segment of finite length, answer choice E is correct.
 
Answer: E

HolaMaven · Answer

Hi,
Since question is saying single line of finite length,
Graph is going to be a straight line is equation is a linear equation. Linear equation is equation with highest power of variable being one. For higher powers, graph is going to be a curve.
Option A has highest power of x as 4, so it is curve
Option B has highest power of x as 3, so it is also curve
Option C has highest power of x as 2, so it is also curve
Option D has absolute function. An absolute function is combination of two functions, so it is going to give two lines.
Thus option E

BrentGMATPrepNow · Answer

IMPORTANT
This is one of those questions that require us to check/test the answer choices. In these situations,  always check the answer choices from E to A , because the correct answer is typically closer to the bottom than to the top.

E) 2 ≤ 3x + 4 ≤ 6
Subtract 4 from all sides to get: -2 ≤ 3x ≤ 2
Divide all sides by 3 to get: -2/3 ≤ x ≤ 2/3
So, x can have any value from -2/3 to 2/3
So, if we were to graph the possible values of x, the line segment would have a FINITE length.

Answer: E

Cheers,
Brent

EMPOWERgmatRichC · Answer

Hi All,

The equations in this question can be physically drawn or handled as a Number Property concept:

Note that the question asks for a SINGLE SEGMENT of FINITE LENGTH. This means that the group of numbers that make up the solution set would be consecutive and limited (as opposed to non-consecutive and/or infinite).

Here are the Number Properties:

A: X^4 >= 1

X can be 1, 2, 3, etc. or -1, -2, -3, etc. This group of answers is infinite and non-consecutive. Eliminate A.

B: X^3 <= 27

X can be 3, 2, 1, 0, -1, etc. This group is infinite. Eliminate B.

C: X^2 >= 16

X can be 4, 5, 6, etc. or -4, -5, -6, etc. This group is infinite and non-consecutive. Eliminate C.

D: 2 <= |X| <= 5

X can be 2, 3, 4, 5 or -2, -3, -4, -5, This group is finite BUT non-consecutive. Eliminate D.

E: 2 <= 3X + 4 <= 6

Let's simplify with algebra…

-2 <= 3X <= 2

-2/3 <= X <= 2/3

X is a finite set of answers and the answers are consecutive. This is a MATCH to what we're looking for.

Final Answer:

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E

GMAT assassins aren't born, they're made,
Rich

Nikhil30 · Answer

hey , i have a small doubt with option B.
cant we write x^3<=27 as X<=3 as it should be between the roots , than it is X<=3 and X>=-3 that is between roots i have learn this from quant Wizako.
got confused here.

Bunuel · Answer

We can always take odd positive root from both sides of an inequality. So,  x^3 \leq 27  is the same as  x\leq{3} . I think you are thinking of a square root:  x^2 \leq 4  -->  -2 \leq x \leq 2 .

Apeksha2101 · Answer

How can you solve x^4 >= 1
I am stuck at this step

x^4>=1=>(x^4-1)>=0
=> (x^2-1)(x^2+1)>=0
=> (x+ 1)(x-1) (x^2+1)>=0
for (x+ 1)(x-1) => x<-1 or x>1
But how are we gonna solve (X^2 + 1) >=0 ???

Bunuel · Answer

x^2 + 1 = (nonnegative) + 1, so its is always positive and thus can be safely reduced to get x^2 - 1 ≥ 0, which gives x^2 ≥ 1 and finally we get x ≥ 1 or x ≤ -1. However, we can simply take the fourth root from x^4 ≥ 1 to get |x| ≥ 1, which gives x ≥ 1 or x ≤ -1.

Hope it helps.

ManishaPrepMinds · Answer

Solution:

Identify the inequality that results in a single, finite segment on the number line.

Option A: x^4 ≥ 1

Solutions: x ≤ -1 or x ≥ 1 (two infinite segments).
Option B: x^3 ≤ 27

Solutions: x ≤ 3 (half-infinite segment).
Option C: x^2 ≥ 16

Solutions: x ≤ -4 or x ≥ 4 (two infinite segments).
Option D: 2 ≤ |x| ≤ 5

Solutions: Two segments, -5 ≤ x ≤ -2 and 2 ≤ x ≤ 5.
Option E: 2 ≤ 3x + 4 ≤ 6

Solutions: -2/3 ≤ x ≤ 2/3 (single finite segment).
Conclusion: Only Option E represents a single, finite segment.

Final Answer:
(E) 2 ≤ 3x + 4 ≤ 6

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