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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]

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20 Aug 2010, 12:59
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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, 06:46, edited 2 times in total.
Edited the question and added the OA

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20 Aug 2010, 13:18
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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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20 Aug 2010, 13:22
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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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20 Aug 2010, 13:25
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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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20 Aug 2010, 13:33
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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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21 Aug 2010, 00:10
thanks bunuel. great explanation.

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21 Aug 2010, 01:01
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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21 Aug 2010, 02:24
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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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21 Aug 2010, 08:09
Thank you.

I get it now...

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07 Sep 2010, 01:40
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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07 Sep 2010, 04:03
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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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07 Sep 2010, 05:31
Oh, yes, you are right, Bunuel. Thanks.

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16 Feb 2012, 23:02
Hi, If i solve option 1. i get x^2-1>0 or (x+1)(x-1)>0 this implies : x>-1 or x>1.. however the question in the stem can be rephrased as x<-1 or X>1.. How is option 1 true then?am i missing something?

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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16 Feb 2012, 23:06
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devinawilliam83 wrote:
Hi, If i solve option 1. i get x^2-1>0 or (x+1)(x-1)>0 this implies : x>-1 or x>1.. however the question in the stem can be rephrased as x<-1 or X>1.. How is option 1 true then?am i missing something?

x^2-1>0 --> x<-1 or x>1.

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

Hope it helps.
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21 May 2012, 08:28
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 |x|.|x|=x^2 so when we divide x^2 by |x| we get |x|

so |x|.|x|= x^2 is this a formula

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21 May 2012, 09:00
Joy111 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.

Bunuel |x|.|x|=x^2 so when we divide x^2 by |x| we get |x|

so |x|.|x|= x^2 is this a formula

Yes, |x|*|x|=x^2. Consider x=-2 then |x|*|x|=2*2=4=(-2)^2.
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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23 May 2012, 05:10
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.

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

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23 May 2012, 06:09
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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]

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23 May 2012, 12:09
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picked A in 46 secs. for |X| to be less than x^2 just means that it is x is not a decimal. it can be less than -1 and greater than 1. you can only say this in totality with certainty. And square of nay no less than -1 and greater than 1 will be greater than 1. so A

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

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23 May 2012, 12:54
Bunuel:
I have solved the problem in the following manner:
Absolute value of x is positive. So, we can divide both side by x. that leaves us, 1<x or x>1.
or x^2>1
Though i have the OA, am i making any fundamental error?

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Re: If |x|<x^2 , which of the following must be true?   [#permalink] 23 May 2012, 12:54

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