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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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New post Updated on: 19 Jun 2016, 11:20
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If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

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Originally posted by Skywalker18 on 19 Jun 2016, 11:13.
Last edited by Bunuel on 19 Jun 2016, 11:20, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 19 Jun 2016, 11:20
8
15
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


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 \(x>1\).


A. x > 0. Not always true.
B. x < 0. Not always true.
C. x > 1. Not always true.
D. -1 < x < 1. Not true.
E. x^2 > 1. Always true.

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

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New post 21 Jun 2016, 22:12
29
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Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


|x| < x^2

Both left hand side and Right Hand side are positive, so we can square the inequalities.

\(x^2 < x^4\)

\(x^4 > x^2\)

\(x^4 - x^2 > 0\)
\(x^2(x^2 - 1)> 0\)

x^2 is always positive.

So the condition is

\(x^2 - 1> 0\)

\(x^2 > 1\)

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

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New post 19 Jun 2016, 11:22
Bunuel wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


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 \(x>1\).


A. x > 0. Not always true.
B. x < 0. Not always true.
C. x > 1. Not always true.
D. -1 < x < 1. Not true.
E. x^2 > 1. Always true.

Answer: E.


Similar question to practice: if-x-x-2-which-of-the-following-must-be-true-99506.html
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 21 Jun 2016, 20:39
I used it by plugging in ZONEF.

A. x > 0 : If x is an integer, the inequality holds true. It would be opposite for a fraction
B. x < 0 : Same as A
C. x > 1 : Same as A
D. -1 < x < 1 : Again, we can use Zero to plug in.
E. x^2 > 1 : This must be true.

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

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New post 22 Jun 2016, 08:19
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aditi2013 wrote:
I used it by plugging in ZONEF.

A. x > 0 : If x is an integer, the inequality holds true. It would be opposite for a fraction
B. x < 0 : Same as A
C. x > 1 : Same as A
D. -1 < x < 1 : Again, we can use Zero to plug in.
E. x^2 > 1 : This must be true.

IMO E.


your explanation for C is wrong.if u take fraction of x>1 then it satisfies.
But this choice is just sub part of solutions of |x|<x^2 .the other solution is x<-1 which choice C doesn't say about
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 26 Jun 2016, 01:31
hsbinfy wrote:
aditi2013 wrote:
I used it by plugging in ZONEF.

A. x > 0 : If x is an integer, the inequality holds true. It would be opposite for a fraction
B. x < 0 : Same as A
C. x > 1 : Same as A
D. -1 < x < 1 : Again, we can use Zero to plug in.
E. x^2 > 1 : This must be true.

IMO E.


your explanation for C is wrong.if u take fraction of x>1 then it satisfies.
But this choice is just sub part of solutions of |x|<x^2 .the other solution is x<-1 which choice C doesn't say about


Aah, thank you for catching this error. The improper fraction will satisfy, right?
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 29 Jun 2016, 23:40
aditi2013 wrote:
I used it by plugging in ZONEF.

What is ZONEF? Can you explain this.
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 29 Jun 2016, 23:43
2
I looked for four range of values:

i) x<-1
Is |x| < x^2 in this range? Yes.

ii) -1 < x < 0 N
Is |x| < x^2 in this range? No.

iii) 0 < x < 1 N
Is |x| < x^2 in this range? No.

iv) x > 1 Y
Is |x| < x^2 in this range? Yes.

So, clearly |x| < x^2 only for:

i) x < -1 or
ii) x > 1

In either case, x^2 > 1. So, E.
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 30 Jun 2016, 04:01
2
sauravpaul wrote:
aditi2013 wrote:
I used it by plugging in ZONEF.

What is ZONEF? Can you explain this.


Hi Saurav,

ZONEF is just an abbreviation to ensure no value remains untested to check the sufficiency. So, I always check ZONEF in majority of such data sufficiency questions.

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

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New post 22 Jul 2016, 22:09
'C' x>1 is a subset of 'E' x^2>1
The question is which of the following must be true so if 'E' must be true then 'C' also must be true?
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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

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New post 05 Nov 2016, 23:24
Cant we do this question by opening the Modulus>>??

My approach is as follows
Given that
|x|<x^2

Now opening the modulus accordingly
CASE 1:x<0
=> -x<x^2
=> x+x^2>0
=> x(x+1)>0

By the Line wave Method:
The range of x is x<-1 U x>0
CASE 2: x>0
=> x<x^2
=> x(x-1)>0

Now the range of x is x<0 U x>1

Combining both operations x <-1 U x>1
So C option directly suits or comes in this range..
As per option E.. if I put a value from the above range say x=-0.25..x^2=0.0625 which is less than 1

So according me C should be the correct answer.....

CAN ANYBODY TELL ME WHERE I AM WRONG???
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If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 17 Oct 2017, 01:09
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


Let's square both the sides, we get,

\(x^4 > x^2\)
->\(x^4 - x^2 > 0\)
-> \(x^2 (x^2 -1) > 0\)
This implies, both the terms have to have same signs.
\(x^2 > 0\) and \(x^2 - 1 > 0\).

Additionally, we can't test \(x^2< 0\) since the square of a real number is never negative.

This gives us \(x^2 > 1\) (Option E.)
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 19 Oct 2017, 10:20
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


If x = 0, then |0| will not be less than 0^2 since they both are equal to 0. Thus, we know x can’t be 0.

Since |x| = x when x is positive and -x when x is negative, let’s rewrite the inequality without the absolute value sign. That is, if x > 0, then we have x < x^2, and if x < 0, then we have -x < x^2.

Case 1: If x > 0,

x < x^2

Dividing both sides by x, we have:

1 < x

Case 2: If x < 0,

-x < x^2

Dividing both sides by x (switching the inequality sign since x is negative), we have:

-1 > x

Thus, we have x > 1 or x < -1; in that case, x^2 must be greater than 1.

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

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New post 24 Jun 2018, 03:52
Bunuel wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


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 \(x>1\).


A. x > 0. Not always true.
B. x < 0. Not always true.
C. x > 1. Not always true.
D. -1 < x < 1. Not true.
E. x^2 > 1. Always true.

Answer: E.



Bunuel hello there :-)

can you please explain in your solution when we \(|x|<x^2\) --> reduce by \(|x|\) you say absolute value is non-negative but doesnt absolute value means either negative or positive value ? :? we always consider two conditions one is negative and another positive....


\(|A| < B\) means \(-B < A < B\), (here \(A\) can be \(-1\), and \(B\) can be \(-2\) ) now getting back to your solution how can you be sure that |x| is positive positive number when reducing by |x| ---> \(\frac{|x|}{|x|}<\frac{x^2}{|x|}\) :? can you explain it somehow :? :) ?

thank you and have a nice day :-)
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 24 Jun 2018, 04:08
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Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


It's possible that x=2, since \(|2| < 2^2\).
Eliminate B and D, since they do not have to be true.
It's possible that x=-2, since \(|-2| < (-2)^2\).
Eliminate A and C, since they do not have to be true.



Algebra:
\(|x| < x^2\)
\(|x| < |x| * |x|\)

Since |x| in the blue inequality must be NONNEGATIVE, we can safely divide each side by |x|:
\(\frac{|x|}{|x|} < |x| * \frac{|x|}{|x|}\)
\(1 < |x|\)

Since each side of the red inequality is NONNEGATIVE, we can safely square each side:
\(1^2 < |x|^2\)
\(1 < x^2\)
\(x^2 > 1\)


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

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New post 24 Jun 2018, 04:23
adiagr wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


|x| < x^2

Both left hand side and Right Hand side are positive, so we can square the inequalities.

\(x^2 < x^4\)

\(x^4 > x^2\)

\(x^4 - x^2 > 0\)
\(x^2(x^2 - 1)> 0\)

x^2 is always positive.

So the condition is

\(x^2 - 1> 0\)

\(x^2 > 1\)

E is the answer.



hello there Bunuel GMATGuruNY, niks18, generis chetan2u :) welcome to GMATclub - a life changing learning experience © :lol: (copy right of this slogan belongs to me :grin: )

one question is there a rule when we can square modulus as in the solution above ? i mean in which case can i square both sides and in which case i cant square both sides of mudulus ? how can squaring modulus on both sides facilitate problem solving process under time pressure ? :)

Also modulus means either negative or positive value how can we be sure that number is positive when squaring :?

Also how do we go from here \(x^2(x^2 - 1)> 0\) to this \(x^2 > 1\) :? ? do we divide both sides by (x^2 - 1) --->

---> do we divide by \((x^2 - 1)\)both sides or do we divide both sides by \(x^2\) :-) ?

have a great weekendstart :)
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If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 24 Jun 2018, 06:51
1
dave13 wrote:
adiagr wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


|x| < x^2

Both left hand side and Right Hand side are positive, so we can square the inequalities.

\(x^2 < x^4\)

\(x^4 > x^2\)

\(x^4 - x^2 > 0\)
\(x^2(x^2 - 1)> 0\)

x^2 is always positive.

So the condition is

\(x^2 - 1> 0\)

\(x^2 > 1\)

E is the answer.



hello there Bunuel GMATGuruNY, niks18, generis chetan2u :) welcome to GMATclub - a life changing learning experience © :lol: (copy right of this slogan belongs to me :grin: )

one question is there a rule when we can square modulus as in the solution above ? i mean in which case can i square both sides and in which case i cant square both sides of mudulus ? how can squaring modulus on both sides facilitate problem solving process under time pressure ? :)

Also modulus means either negative or positive value how can we be sure that number is positive when squaring :?

Also how do we go from here \(x^2(x^2 - 1)> 0\) to this \(x^2 > 1\) :? ? do we divide both sides by (x^2 - 1) --->

---> do we divide by \((x^2 - 1)\)both sides or do we divide both sides by \(x^2\) :-) ?

have a great weekendstart :)


Hi dave13

Squaring is a method to remove mod function, because a square is ALWAYS POSITIVE and so is a MOD function. Hence when you square you are not changing any property of the mod function but increasing the degree of the equation.

You can also solve this equation without squaring by taking two scenarios: Scenario 1: if x>0 and scenario 2: if x<0, but this will increase the calculation steps. you can try yourself.

Now once you have squared and arrived at \(x^2(x^2 - 1)> 0\), then this simply implies that product of two number is positive. the two numbers are \(x^2\) & \((x^2-1)\). We know that \(x^2>0\), hence for the product to be positive \(x^2-1>0 =>x^2>1\)
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If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 28 Jun 2018, 09:32
GMATGuruNY wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1


It's possible that x=2, since \(|2| < 2^2\).
Eliminate B and D, since they do not have to be true.
It's possible that x=-2, since \(|-2| < (-2)^2\).
Eliminate A and C, since they do not have to be true.



Algebra:
\(|x| < x^2\)
\(|x| < |x| * |x|\)

Since |x| in the blue inequality must be NONNEGATIVE, we can safely divide each side by |x|:
\(\frac{|x|}{|x|} < |x| * \frac{|x|}{|x|}\)
\(1 < |x|\)

Since each side of the red inequality is NONNEGATIVE, we can safely square each side:
\(1^2 < |x|^2\)
\(1 < x^2\)
\(x^2 > 1\)




hey GMATGuruNY great explanation ! :-) i have just one question

Algebra:
\(|x| < x^2\) (as you see \(x^2\) is without bars / brackets )

\(|x| < |x| * |x|\) Here you rewrite \(x^2\)as |x| * |x| and not as x*x , can you please explain why ? :-)

many thanks ! :)
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