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

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If |x|<x^2 , which of the following must be true? [#permalink]  20 Aug 2010, 11:59
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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

[Reveal] Spoiler: OA

Last edited by Bunuel on 20 Jun 2012, 05:46, edited 2 times in total.
Edited the question and added the OA
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Re: not that hard [#permalink]  20 Aug 2010, 12:18
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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: not that hard [#permalink]  20 Aug 2010, 12:22
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Hi,

The X-Y plane of looking at it is the fastest way - I would guess.

Mod(x) looks like a V

x^2 is a parabola which just touches the x-axis (tangent) at the origin.

When x < -1; Mod(x) < x^2
When x = -1; Mod (x) = x^2
When -1<x<0; Mod(x) > x^2
When x = 0; Mod (x) = x^2
When 0<x<1; Mod(x) > x^2
When x = 1; Mod (x) = x^2
When x > 1; Mod(x) < x^2

Hope this helps. Thanks.

Wrong forum ?!?!?
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Re: not that hard [#permalink]  21 Aug 2010, 01:24
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Expert's post
qweert wrote:
I'm confused, because if x= -2 ==> |x|< x^2

|x|< x^2, X^2 should always be > 1
Can eliminate II, because if x=1/2, |x|> x^2

Could you give an example Bunuel

Question is: "which of the following statements MUST be true", not COULD be true.

We are GIVEN that $$|x|<x^2$$, which means that either $$x<-1$$ OR $$x>1$$.

So GIVEN that: $$x<-1$$ OR $$x>1$$.

Statement II. is $$x<-1$$ always true? NO. As $$x$$ could be more than 1, eg. 2, 3, 5.7, ... and in this case $$x<-1$$ is not true. So statement II which says that $$x<-1$$ is not always true.

Hope it's clear.
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Re: not that hard [#permalink]  07 Sep 2010, 03:03
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nonameee wrote:
mehdiov, what's the OA?

Quote:
Statement II. is x<-1 always true? NO. As x could be more than 1, eg. 2, 3, 5.7, ... and in this case x<-1 is not true. So statement II which says that x<-1 is not always true.

Then we can't even accept I, since x doesn't always have to be greater than 1 (it can be -10, -50 etc.).

The question is similar to (but I still think that it is different):

ps-inequality-13943.html

Maybe I don't understand something.

Thanks.

OA is A.

$$|x|<x^2$$ means that either $$x<-1$$ or $$x>1$$. For example $$x$$ could be -5, -3, 2, 3, ...

Basically the question is: If $$x<-1$$ or $$x>1$$ which of the following must be true?

You are saying that answer is: E (I and III only). Is III: $$x<-1$$ ALWAYS true? Is it true for $$x=3$$? NO. So E is not correct.
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  23 May 2012, 05:09
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Expert's post
pavanpuneet wrote:
Hi Bunuel, I understood your logic that we can safely divide by |x|, however, I tried to solve it this way, just wanted to confirm is this is also right...

x2>|x| => if x is positive then x2-x>0 => x(x-1)>0 => x<0 and x>1................case 1

and if x is negative, then x2+x>0 => x(x+1)>0 =>x<-1 and x>0...case2

From case 1 and case2....x<-1 and x>1..[Same conclusion but lengthier method].. I want to especially confirm because I tried to apply modulus and inequalities theory.. so want to be sure.

x^2>|x|

If x>0 then x2-x>0 --> x(x-1)>0 --> x<0 and x>1. Since we are considering x>0 range then x>1;
If x<0 then x2-x>0 --> x(x+1)>0 --> x<-1 and x>0. Since we are considering x<0 range then x<-1;

So, x^2>|x| holds true for x<-1 and x>1.

Hope it's clear.
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  15 Jun 2012, 01:59
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Expert's post
pavanpuneet wrote:
Hi Bunuel, I tried to solve it this way, please let me know if this is correct as I am getting confused with the final wording of MUST be true. I am kind of stuck at the final stages to make the decision.

The problem mentions: |x|<x2

Case 1: if x is +ve; then x<x2, given that we do not know if x=0, then x-x2<0 ==> x(1-x)<0 ==> multiplying by negative sign; x(x-1)>0...which means x<0 and X>1

Case 2: if x is -ve; then -x<x2, then -x-x2<0 ===>-(x+x2) <0...==> x(1+x)>0 ===> x<-1 and x>0

From C1 AND C2..on the number x<-1 and x>1; after this I am a little lost, how to make the final call.

First of all please study these posts:
if-x-x-2-which-of-the-following-must-be-true-99506.html#p767256
if-x-x-2-which-of-the-following-must-be-true-99506.html#p767264
if-x-x-2-which-of-the-following-must-be-true-99506.html#p767437
if-x-x-2-which-of-the-following-must-be-true-99506.html#p776341
if-x-x-2-which-of-the-following-must-be-true-99506.html#p1088622
if-x-x-2-which-of-the-following-must-be-true-99506-20.html#p1089273

Anyway, given that $$|x|<x^2$$, which means that $$x<-1$$ or $$x>1$$. Next we are given three options and are asked which of the following MUST be true.

I. x^2>1. Since $$x<-1$$ or $$x>1$$ then this option is always true (ANY number less than -1 or more than 1 when squared will be more than 1);
II. x>0. This option is NOT always true, for example x could be -2 and in this case x>0 won't not true;
III. x<-1. This option is also NOT always true, for example x could be 2 and in this case x<-1 won't not true.

So, we have that only option I is always true.

Check our Must or Could be True Questions to practice more: search.php?search_id=tag&tag_id=193

Hope it helps.
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  20 Jun 2012, 05:44
1
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Expert's post
idreesma wrote:
Hi,
I am having issues with a fundanmental concept. In the below explanation x(x+1)>0 , isnt x>0 and x>-1 (from x+1>0) rather than x<-1 and x>0. clearly by inspection my answer is wrong, but was confused over how you go correct answer (is it possible to show the steps)

If x>0 then x2-x>0 --> x(x-1)>0 --> x<0 and x>1. Since we are considering x>0 range then x>1;
If x<0 then x2-x>0 --> x(x+1)>0 --> x<-1 and x>0. Since we are considering x<0 range then x<-1;
Thanks

x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535

Hope it helps.

Hope it helps.
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  20 Jun 2012, 21:12
1
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Expert's post
idreesma wrote:
Hi,
I am having issues with a fundanmental concept. In the below explanation x(x+1)>0 , isnt x>0 and x>-1 (from x+1>0) rather than x<-1 and x>0. clearly by inspection my answer is wrong, but was confused over how you go correct answer (is it possible to show the steps)

If x>0 then x2-x>0 --> x(x-1)>0 --> x<0 and x>1. Since we are considering x>0 range then x>1;
If x<0 then x2-x>0 --> x(x+1)>0 --> x<-1 and x>0. Since we are considering x<0 range then x<-1;
Thanks

Responding to a pm:

The problem you are facing is that you do not know how to handle inequalities.

How do you get the range for which this inequality holds? x(x+1) > 0

Think of it this way: Product of x and (x+1) should be positive. When will that happen? When either both the terms are positive or both are negative.

Case 1: When both are positive
x > 0
x + 1 > 0 i.e. x > -1

For both to be positive, x must be greater than 0. Hence this inequality will hold when x > 0.

Case 2: When both are negative
x < 0
x + 1 < 0 i.e. x < -1

For both to be negative, x must be less than -1. Hence this inequality will hold when x < -1.

So we get two ranges in which this inequality holds: x > 0 or x < -1.

The fastest way to solve it is using the number line.
Check this post for the explanation of this method: inequalities-trick-91482.html

Also, your Veritas book discusses this concept too.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5853 Location: Pune, India Followers: 1480 Kudos [?]: 7955 [1] , given: 190 Re: If |x|<x^2 , which of the following must be true? [#permalink] 04 Oct 2012, 20:21 1 This post received KUDOS Expert's post idreesma wrote: Hi Krishma, the link that you posted is probably best for summary. It ends with the comment "Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post ". Can you suggest any such thread where absolute value side is summarized like this as well. Appreciate your input Thanks I have discussed mods here: http://www.veritasprep.com/blog/2011/01 ... edore-did/ http://www.veritasprep.com/blog/2011/01 ... h-to-mods/ http://www.veritasprep.com/blog/2011/01 ... s-part-ii/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: not that hard [#permalink]  29 Oct 2012, 01:51
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mario1987 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,
sorry , but I didn't undestrand how you reduce |x|<x^2 by |x| to get the solution...x<-1 and x>1
please, could you explain me in more details?
Thanks

Express $$x^2$$ as $$|x|*|x|$$, so we have that: $$|x|<|x|*|x|$$ --> reduce by $$|x|$$ (notice that $$|x|$$ is positive, thus we can safely divide both parts of the inequality by it) --> $$1<|x|$$ --> $$x<-1$$ or $$x>1$$.

Hope it's clear.
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  10 Mar 2013, 01:01
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|X| < x^2 means X is not a decimal. means X < -1 or x > 1. Square of anything will be positive. Hence. A
Hope it helps.
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  21 Jul 2013, 19:25
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1. |x|<x^2 =>

(i) x < x^2 or
(ii) -x < x^2

2. (1) => x>1 or x<-1 . Only this satisfies both (i) and (ii)

Is x^2>1 always true? Yes follows from (2)
Is x>0 always true, No because x could be <-1
Is x<-1 always true?, No because x could be >1

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Re: not that hard [#permalink]  20 Aug 2010, 12:25
Bunuel wrote:
III. x<-1 --> --> may or may not be true.

Bunuel, Isn't this always true when |x|<x^2 ... Isn't the answer E.

Thanks.
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Re: not that hard [#permalink]  20 Aug 2010, 12:33
Expert's post
4gmatmumbai wrote:
Bunuel wrote:
III. x<-1 --> --> may or may not be true.

Bunuel, Isn't this always true when |x|<x^2 ... Isn't the answer E.

Thanks.

$$|x|<x^2$$ is given as fact and then we asked to determine which of the following statements MUST be true.

$$|x|<x^2$$ means that either $$x<-1$$ or $$x>1$$, $$x$$ can be ANY value from these two ranges, (I think in your own solution you've reached this conclusion: when $$x<-1$$ the graph of $$|x|$$ is below (less than) the graph of $$x^2$$ and when $$x>$$1 again the graph of $$|x|$$ is below the graph of $$x^2$$).

Now, III says $$x<-1$$ this statement is not always true as $$x$$ can be for example 3 and in this case $$x<-1$$ doesn't hold true.

Hope it's clear.
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Re: not that hard [#permalink]  20 Aug 2010, 23:10
thanks bunuel. great explanation.
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Re: not that hard [#permalink]  21 Aug 2010, 00:01
I'm confused, because if x= -2 ==> |x|< x^2

|x|< x^2, X^2 should always be > 1
Can eliminate II, because if x=1/2, |x|> x^2

Could you give an example Bunuel
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Re: not that hard [#permalink]  21 Aug 2010, 07:09
Thank you.

I get it now...
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Re: not that hard [#permalink]  07 Sep 2010, 00:40
mehdiov, what's the OA?

Quote:
Statement II. is x<-1 always true? NO. As x could be more than 1, eg. 2, 3, 5.7, ... and in this case x<-1 is not true. So statement II which says that x<-1 is not always true.

Then we can't even accept I, since x doesn't always have to be greater than 1 (it can be -10, -50 etc.).

The question is similar to (but I still think that it is different):

ps-inequality-13943.html

Maybe I don't understand something.

Thanks.
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Re: not that hard [#permalink]  07 Sep 2010, 04:31
Oh, yes, you are right, Bunuel. Thanks.
Re: not that hard   [#permalink] 07 Sep 2010, 04:31

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