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

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.

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

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 \(x\geq{1}\)

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 \(x\geq{1}\).

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




≥ ≤ ∞

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?

\(x^3 \leq 27\) 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;



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.



Similar questions:
https://gmatclub.com/forum/which-of-the ... 27820.html (GMAT Prep).
https://gmatclub.com/forum/which-of-the ... 30666.html (Knewton).

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

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

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

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:

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

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.

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: \(x^2 \leq 4\) --> \(-2 \leq x \leq 2\).

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 ???

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.