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

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

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04 Oct 2015, 04:14
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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!
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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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?
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Posts: 1
Re: If |x|<x^2 , which of the following must be true? [#permalink]

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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]

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19 Jun 2016, 22:21
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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)
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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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?
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Joined: 02 Sep 2009
Posts: 43804
Re: If |x|<x^2 , which of the following must be true? [#permalink]

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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.
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Posts: 642
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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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

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

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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.

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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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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/
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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26 Oct 2017, 01:59
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Re: If |x|<x^2 , which of the following must be true?   [#permalink] 26 Oct 2017, 01:59

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