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

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07 Sep 2010, 04:31
Oh, yes, you are right, Bunuel. Thanks.
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16 Feb 2012, 22:02
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, 22:06
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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, 07:28
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, 08:00
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, 04:10
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, 11:09
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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, 11:54
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]

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23 May 2012, 23:45
BDSunDevil wrote:
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?

We can divide by |x| not by x ans we'll get 1<|x| --> x<-1 or x>1. Check this: if-x-x-2-which-of-the-following-must-be-true-99506.html#p767256

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

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24 May 2012, 17:09
Bunuel,
Doesn't the question mean - which of the values of x must hold for |X|<X^2
all values of x<-1 satisfy , although there are other solutions(x>1) but E does not say those solution do not exist. In fact if a superset of values are true then subset of values must be true...

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

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24 May 2012, 19:48
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 ?

Hi,

Using number lines:

for x > 0, $$x < x^2$$
x(x+1) > 0, number line (a)

for x < 0,$$-x < x^2$$
x(x-1) > 0, number line (b)

(a)------(-1)--0--------

(b)------------0--1-----

combining both we have, x>1, x<-1

$$x^2 > 1$$, already includes above the above values of x.

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

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24 May 2012, 20:16
vikram 4689 what x <-1 says is that x can ONLY hold values of less than -1. Which actually means that x cannot have values of 2,3.. etc. this we all know is not true as x can also takes values of >1. Hence this statement cannot hold true. makes sense?
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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25 May 2012, 01:01
vikram4689 wrote:
Bunuel,
Doesn't the question mean - which of the values of x must hold for |X|<X^2
all values of x<-1 satisfy , although there are other solutions(x>1) but E does not say those solution do not exist. In fact if a superset of values are true then subset of values must be true...

where am mis-interpreting

No, that's not true.

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

$$|x|<x^2$$ means that $$x<-1$$ or $$x>1$$. Now, evaluate each option:

I. x^2>1. Since $$x<-1$$ or $$x>1$$ then this option is always true;
II. x>0. Since $$x<-1$$ or $$x>1$$ then this option may or may not be true;
III. x<-1. Since $$x<-1$$ or $$x>1$$ then this option 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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25 May 2012, 01:22
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 ?
hope this might help u ..
x^2 >|x|---> (implies)
x^2>x (since x^2 can not be less than ..it will always by +ve)
x^2-x>0
x(x-1)>0
this would give two values x <0 and x>1
but x can not be <0 because values lying between -1 to +1 would nt satisfy the given condition |x|<x^2 ...therefore... x>1
therefore x^2 is also greater than 1
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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25 May 2012, 01:25
deepti1206 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 ?
hope this might help u ..
x^2 >|x|---> (implies)
x^2>x (since x^2 can not be less than ..it will always by +ve)
x^2-x>0
x(x-1)>0
this would give two values x <0 and x>1
but x can not be <0 because values lying between -1 to +1 would nt satisfy the given condition |x|<x^2 ...therefore... x>1
therefore x^2 is also greater than 1

$$|x|<x^2$$ means that $$x<-1$$ or $$x>1$$.
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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25 May 2012, 01:36
Sorry i did a mishtake...i would like to revise it,.....

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 ?
hope this might help u ..
x^2 >|x|---> (implies)
i)x^2>x
x^2-x>0
x(x-1)>0
this would give two values x <0 and x>1

ii) x^2 <-x
x^2-x<0
x(x+1)<0
giving u ..x>0 and x <-1

giving u the common condition that is x >1
therefore x^2 is also greater than 1
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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25 May 2012, 01:40
deepti1206 wrote:
Sorry i did a mishtake...i would like to revise it,.....

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 ?
hope this might help u ..
x^2 >|x|---> (implies)
i)x^2>x
x^2-x>0
x(x-1)>0
this would give two values x <0 and x>1

ii) x^2 <-x
x^2-x<0
x(x+1)<0
giving u ..x>0 and x <-1

giving u the common condition that is x >1
therefore x^2 is also greater than 1

The red part is not correct.

Once more: $$|x|<x^2$$ means that $$x<-1$$ or $$x>1$$. Explained here: if-x-x-2-which-of-the-following-must-be-true-99506.html#p767256 and here: if-x-x-2-which-of-the-following-must-be-true-99506.html#p1088622
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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25 May 2012, 20:41
Bunuel wrote:
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.

Hello Bunuel

I didnot understand this part
x^2>|x| holds true for x<-1 and x>1.

if you have x<-1 then why ans "E" is not true?

i also didnot understand folloiwng part
"If x>0 then x2-x>0 --> x(x-1)>0 -->[highlight]x<0[/highlight] and x>1" how can x<0 ?

I solved this problem and voted for "A"

for x>1 / 1>x>0 / x<0 (I considered these 3 ranged because if 1>x>0 then the square of number is less then number)

if (x>1) --> x^2>|x| holds true

if (1>x>0) --> x^2>|x| doesnot holds true

if (x<0) --> x^2>|x| holds true (-1/2)^2 > (-1/2)
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Re: If |x|<x^2 , which of the following must be true? [#permalink]

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26 May 2012, 01:36
kuttingchai wrote:
Bunuel wrote:
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.

Hello Bunuel

I didnot understand this part
x^2>|x| holds true for x<-1 and x>1.

if you have x<-1 then why ans "E" is not true?

i also didnot understand folloiwng part
"If x>0 then x2-x>0 --> x(x-1)>0 -->[highlight]x<0[/highlight] and x>1" how can x<0 ?

E cannot be correct since

I solved this problem and voted for "A"

for x>1 / 1>x>0 / x<0 (I considered these 3 ranged because if 1>x>0 then the square of number is less then number)

if (x>1) --> x^2>|x| holds true

if (1>x>0) --> x^2>|x| doesnot holds true

if (x<0) --> x^2>|x| holds true (-1/2)^2 > (-1/2)

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

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14 Jun 2012, 23:53
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.
Re: If |x|<x^2 , which of the following must be true?   [#permalink] 14 Jun 2012, 23:53

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