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If |x| < x^2, which of the following must be true?

1 2 3

GMATinsight

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04 Oct 2015, 05:14

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SIMILAR QUESTIONS:

Question 1: What are the values of n that satisfy the condition 1/|n| > n?
(A) 0<n<1 (and) - infinity < n < 0
(B) 0 < n < infinity (or) -infinity < n < -1
(C) 0 < n < 1 (and) -1 < n < 0
(D) - infinity < n < 0 (or) 0 < n < 1
(E) 0<n<1 (or) - infinity < n < 0

Answer - Option A
Link: https://gmatclub.com/forum/what-are-the- ... 74746.html

Question 2: If 0 < x < 1 < y, which of the following must be true?

A. 1 < 1/x < 1/y
B. 1/x < 1 < 1/y
C. 1/x < 1/y < 1
D. 1/y < 1 < 1/x
E. 1/y < 1/x < 1

Answer: Option D
Link: https://gmatclub.com/forum/if-0-x-1-y-wh ... 98235.html

Question 3: If x is positive, which of the following could be correct ordering of 1/x, 2x, and x^2?

(I) X^2 < 2x < 1/x
(II) x^2 < 1/x < 2x
(III) 2x < x^2 < 1/x

(a) none
(b) I only
(c) III only
(d) I and II
(e) I, II and III

Answer: Option D
Link: https://gmatclub.com/forum/if-x-is-posit ... 71070.html

I hope this helps!

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06 Oct 2016, 07:55

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Hi Bunuel/Someone,

Could you please explain what is meant by $|x|<x^2$ --> reduce by $|x|$ and how you arrived at "so $|x|>0$) --> $1<|x|$ --> $x<-1$ or $x>1$." step by step ?

Sorry my knowledge of Absolute Values is very basic.

Bunuel

mehdiov

If |x|<x^2 , which of the following must be true?

I. x^2>1
II. x>0
III. x<-1

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

what is the fastest way to solbe it ?

Given: $|x|<x^2$ --> reduce by $|x|$ (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so $|x|>0$) --> $1<|x|$ --> $x<-1$ or $x>1$.

So we have that $x<-1$ or $x>1$.

I. x^2>1 --> always true;
II. x>0 --> may or may not be true;
III. x<-1 --> --> may or may not be true.

Answer: A (I only).

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06 Oct 2016, 22:04

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Dev1212

Note that $x^2 = |x|^2$

Given:

$|x| < x^2$

$|x| < |x|^2$

Now, if you know that x is not 0, then you know that |x| is certainly positive. So you can divide both sides by |x|.

1 < |x|

When |x| is greater than 1, it translates to x > 1 or x < -1.

For explanation of this, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... edore-did/

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15 Dec 2021, 06:03

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I solved it this way:

|x|<x^2
|x|-x^2<0

a) if x>0 then

x-x^2<0
x(1-x)<0
This implies both have opposite signs. But since x>0 we will discard the case where x<0 & 1-x>0. Therefore, we are left with:

x>0 & 1-x<0
x>0 & x>1
so, x>1

a) if x<0 then

-x-x^2<0
x+x^2>0
x(1+x)>0
This implies both have same signs. But since x<0 we will discard the case where x>0 & 1+x>0. Therefore, we are left with:

x<0 & 1+x<0
x<0 & x<-1
so, x<-1

Thus, either x<-1 or x>1 which means x^2>1 will always be true (A)

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31 Aug 2022, 01:24

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Given that |x| < x²

Let's take two cases to open the absolute value

Case 1: Whatever is inside the Absolute Value >= 0
x >= 0
=> |x| = x (Watch this video to learn more about Basics of Absolute Value)
=> x < $x^2$
=> $x^2$ - x > 0
=> x*(x-1) > 0

Attachment:

x between 0 and 1.JPG [ 17.24 KiB | Viewed 2516 times ]

=> x < 0 or x > 1 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x>=0
=> Intersection will be x > 1

Attachment:

x GT 1.JPG [ 14.98 KiB | Viewed 2507 times ]

Case 2: Whatever is inside the Absolute Value < 0
x < 0
=> |x| = -x (Watch this video to learn more about Basics of Absolute Value)
=> -x < $x^2$
=> $x^2$ + x > 0
=> x*(x+1) > 0

Attachment:

x between -1 and 0.JPG [ 16.8 KiB | Viewed 2515 times ]

=> x < -1 or x > 0 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x < 0
=> Intersection will be x < -1

So, x > 1 or x < -1
=> x^2 > 1

So, Answer will be A
Hope it helps!

Watch the following video to learn the Basics of Absolute Values

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16 Mar 2023, 09:48

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IanStewart please explain why the option is wrong?

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16 Mar 2023, 14:02

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Dipanjan005

IanStewart please explain why the option is wrong?

If x is positive, the absolute value isn't doing anything, so the inequality |x| < x^2 just says x < x^2, or, dividing by x (which is fine if x > 0), that 1 < x. The equivalent set of negatives will also work in the inequality, because the absolute value on the left and the square on the right both make negative numbers positive. So the inequality means that either x < -1 or x > 1. So x does not need to be positive (x can be -2, say), and II is not always true. And x is not always less than -1 (x can be 5, say), so III is not always true. Once we've ruled out II and III, looking at the answer choices, we don't need to even check I (there is no answer that says "none of the above"), but if x > 1 or x < -1, then x^2 > 1 is true, so the answer is I only.

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17 Mar 2023, 01:42

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Asked: If $|x| < x^2$, which of the following must be true?

|x |< xˆ2
xˆ2 - |x| > 0
Case 1: x>0
x(x-1) > 0
x > 1
Case 2: x<0
x(x+1) > 0
x < -1

I. $x^2 > 1$
If x>1; xˆ2>1
And if x<-1; xˆ2>1
MUST BE TRUE

II. $x > 0$
If x>1; x>1>0; YES
But if x<-1<0; NO
COULD BE TRUE

III. $x < -1$
If x>1; No
But if x<-1; Yes
COULD BE TRUE

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

IMO A

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30 May 2023, 02:56

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Bunuel KarishmaB

Could you please validate my approach?

|x| < x^2

√(x^2) < x^2

Squaring both sides

x^2 < x^4

x^2 - x^4 < 0

x^2 (1-x^2) < 0

x^2 < 0 (invalid as a squared value can only be positive) OR
x^2 > 1 (which means x>1 or x<-1, however I stopped before this step as I got what I was looking for)

Thanks in advance!

KarishmaB

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30 May 2023, 03:29

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achloes

x^2 (1-x^2) < 0

does not lead to
x^2 < 0
or
(1 - x^2) < 0

It leads to
x^2 > 0 AND (1-x^2) < 0
or
x^2 < 0 AND (1-x^2) > 0
Since x^2 cannot be negative, we can ignore the second case. That is why it made no difference here. But try (x-1)(x-2) < 0.

Think about it: If xy is negative, it means exactly one of x and y is negative and the other is positive.

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26 Apr 2024, 07:28

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mehdiov

If $|x| < x^2$, which of the following must be true?

I. $x^2 > 1$

II. $x > 0$

III. $x < -1$

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

PLEASE READ THE WHOLE THREAD AND FOLLOW THE LINKS PROVIDED IN ORDER TO UNDERSTAND THE QUESTION/SOLUTIONS CORRECTLY.

Bunuel KarishmaB MartyMurray

I did it this way by squaring both sides. Does it make sense? Thanks.

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image0 (11).jpeg [ 2.23 MiB | Viewed 1484 times ]

KarishmaB

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27 Apr 2024, 23:26

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Engineer1

mehdiov

Bunuel KarishmaB MartyMurray

I did it this way by squaring both sides. Does it make sense? Thanks.

Both sides are positive so squaring is fine.
You get
$x^4 - x^2 > 0$
$x^2(x^2 - 1) > 0$

Use wavy line method to get x > 1 or x < -1

I. $x^2 > 1$
Again gives x > 1 or x < -1
So correct.

Other 2 are not necessary.

Answer (A)

Wavy line method discussed here:
https://youtu.be/PWsUOe77__E

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03 Jun 2025, 00:28

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Do we have to assume in absolute value questions that the value of a variable like x here is an integer?
if X is not an integer and a fraction say "1/2" which also fulfills the condition 0<x<1 here then even statement 1 would fail, isn't it?

Bunuel

If $|x| < x^2$, which of the following must be true?

I. $x^2 > 1$

II. $x > 0$

III. $x < -1$

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

Given: $|x|<x^2$;

Reduce by $|x|$: $1<|x|$ (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so $|x|>0$);

So we have that $x<-1$ or $x>1$.

I. $x^2>1$ --> always true;

II. $x>0$ --> may or may not be true;

III. $x<-1$ --> may or may not be true.

Answer: A (I only).

Bunuel

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03 Jun 2025, 00:31

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Abhishek_Relan

Bunuel

We are not assuming that x is an integer. Your doubt is addressed many times. Please review the topic carefully.

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18 Jul 2025, 10:51

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Dear Bunuel,

Hope you are doing well. I just wanted to check that in option II x>0, can I say it is actually an invalid option? My logic is that because the range of x is greater than 1 or less than -1, so, x>0 will include values like 0.5, 0.1 and for those |x|>x^2. Hence, we cannot say may or maybe true.

Kindly let me know if my logic is correct.
Thank you!

Best,
Komal

Bunuel

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18 Jul 2025, 13:26

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Komal324

Bunuel

Your doubt is addressed many times. Please review the topic carefully.

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20 Jul 2025, 03:18

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Hello, Bunuel.

Thank you for your response. As suggested by you, I will surely review the topic thoroughly. However, this is actually the first time I have asked a query related to this particular question. I am not sure where the impression came from that it had been addressed earlier, but I appreciate your guidance and will make sure to revisit the concept carefully.

Best,
Komal

Bunuel

Komal324

Bunuel