It is currently 19 Feb 2018, 04:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

SVP
Joined: 08 Jul 2010
Posts: 1955
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

04 Oct 2015, 04:14
2
KUDOS
Expert's post
longfellow wrote:
Thanks GMATInsight. I would still like practice on more questions like this one? Could you point out a few?

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

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

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

I hope this helps!
_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Intern
Joined: 23 May 2016
Posts: 11
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

06 Jun 2016, 12:03
Bunuel wrote:
mehdiov wrote:
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.

Bunuel I understand that x < -1 or x > 1. Does that mean when we square an inequality, the sign flips to get x2 > 1 instead of x2 < 1? I thought squares were safe to do?
Intern
Joined: 28 Apr 2016
Posts: 1
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

06 Jun 2016, 12:16
Suppose x>0 then |x|=x. Now solve the equation x-x^2 <0.
x(x-1) >0 for this x should lie either below 0 or above 1
Now suppose x<0 then |x| = -x.
Now solve the equation x^2+x>0
x(x+1) >0 for this x has to be greater than -1. Now combine both possible values of x and ee get x range as -1<x<0 and 1<x

Sent from my LS-5501 using GMAT Club Forum mobile app
Intern
Joined: 01 Feb 2016
Posts: 10
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

19 Jun 2016, 22:21
1
This post was
BOOKMARKED
Expanding Bunuel's explanation

If x>0, x < x2 => x2-x>0 --> x(x-1)>0 --> For this product of x and (x-1) to be +ve (>0) or negative. Currently we are considering x as positive (x>0) condition. This if x is positive, (x-1) also has to be +ve (greater than 0). x is already greather than 0 because we are considering x>0 range. For (x-1)>0 (+ve), x has to be greater than 1 as any no. less than 1 will make (x-1) negative.

Thus root of this eq is x>1 which matches our condition (x>0).

If x<0, -x < x2 => then x2+x>0 --> x(x+1)>0 -->Thus both x and x+1 have to be either +ve or -ve. As we are considering x<0 condition, x is negative. Thus x+1 has to be negative as well. x+1<0 => x<-1 (x+1-1 <0 -1, add -1 on both sides)

Thus root of this eq is x<-1 which matches our condition (x<0)

I. x^2>1 - True for both root#1 and root 2. No matter what value of x we take from root 1 and root 2, its square will be greater than 1.
II. x>0 - not true as root#2 implies that x can be less than -1 as well (we are looking for 'must' option)
III. x<-1 - not true as root#1 implies that x can be greater than 1 as well (we are looking for 'must' option)
Chat Moderator
Joined: 04 Aug 2016
Posts: 596
Location: India
GPA: 4
WE: Engineering (Telecommunications)
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

04 Sep 2016, 22:02
Bunuel wrote:
mehdiov wrote:
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.

Hi Bunuel,

I was unable to follow the reduction part and explanation so forth. Could you please explain in detail?
Math Expert
Joined: 02 Sep 2009
Posts: 43804
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

04 Sep 2016, 22:35
warriorguy wrote:
Bunuel wrote:
mehdiov wrote:
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.

Hi Bunuel,

I was unable to follow the reduction part and explanation so forth. Could you please explain in detail?

You can find detailed explanations on previous 5 pages.
_________________
Director
Joined: 04 Jun 2016
Posts: 642
GMAT 1: 750 Q49 V43
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

05 Sep 2016, 03:41
If |x|<x^2 , which of the following must be true?

To start with :- For the ease of calculation look at the LHS = |x|
|x| can be essentially seen as +ve x because mod CANNOT yield a negative value.
Now look at the RHS = x^2
x^2 can also be essentially seen as a positive value because squaring always yield a positive value.
IN SUMMARY the stimulus is telling us a positive number is smaller than its square.
Don't we all already know that.
That a number will always be smaller than its square EXCEPT ... EXCEPT WHEN ???? Except when the number is a positive decimal between 0 to 1, then x will be greater than its square x^2
For example
0.5 is always greater than its square 0.25
NOTICE HOW +0.5 IS A DECIMAL THAT FALLS BETWEEN 0 AND 1
But this is not case here. The stimulus tells us that the number is always less than its square. |x|<x^2
See this
-4 is always less than its square 16
+4 is always less than it's square 16
-0.6 is always less than its square 0.36
BUT +0.6 is always greater than its square 0.36

I. x^2>1 (when the irrefutable condition is |x|<x^2)
OFFCOURSE THIS MUST ALWAYS BE TRUE ... We just proved it earlier that if any number x is less than its square x^2 then it means that x^2 is always be greater than 1 ALWAYS TRUE/ MUST BE TRUE.

II. x>0 (when the irrefutable condition is |x|<x^2)
This will falsify our stimulus for values of x between 0 and 1 therefore it CAN BE TRUE FOR VALUES MORE THAN 1 BUT NOT ALWAYS TRUE WHEN X LIES BETWEEN 0 and 1

III. x<-1 (when the irrefutable condition is |x|<x^2)
This does not cover our other possible values when x > 1 So IT IS TRUE. BUT ONLY HALF TRUTH AS IT EXCLUDES THE OTHER POSSIBILITY X>1

I HOPE MY ANSWER WILL HELP THOSE PEOPLE WHO ARE STUCK IN THE -1>X>1 CONFUSION.
YOU HAVE TO KEEP IN MIND THE IRREFUTABLE STIMULS |x|<x^2 IN MIND WHEN EVALUATION THE GIVEN OPTIONS I, II, III

Thanks

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

mehdiov wrote:
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

_________________

Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly.
FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.

Intern
Joined: 13 Dec 2015
Posts: 16
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

06 Oct 2016, 06:55
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 wrote:
mehdiov wrote:
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.

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7938
Location: Pune, India
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

06 Oct 2016, 21:04
Dev1212 wrote:
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.

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.veritasprep.com/blog/2011/0 ... edore-did/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Non-Human User
Joined: 09 Sep 2013
Posts: 13832
Re: If |x|<x^2 , which of the following must be true? [#permalink]

### Show Tags

26 Oct 2017, 01:59
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If |x|<x^2 , which of the following must be true?   [#permalink] 26 Oct 2017, 01:59

Go to page   Previous    1   2   3   4   5   [ 90 posts ]

Display posts from previous: Sort by