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Manager  Joined: 25 Aug 2011
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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

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

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

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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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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.
Manager  Joined: 28 Dec 2010
Posts: 248
Location: India
Re: If |x| < x^2, which of the following must be true?  [#permalink]

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1
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
Manager  Joined: 28 Dec 2010
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Location: India
Re: If |x| < x^2, which of the following must be true?  [#permalink]

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1
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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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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Manager  Joined: 26 Dec 2011
Posts: 89
Re: If |x| < x^2, which of the following must be true?  [#permalink]

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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.
Intern  Joined: 08 May 2012
Posts: 3
Re: If |x| < x^2, which of the following must be true?  [#permalink]

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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
Intern  Joined: 08 May 2012
Posts: 3
Re: If |x| < x^2, which of the following must be true?  [#permalink]

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

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

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Got the answer from : if-x-x-2-which-of-the-following-must-be-true-99506-20.html#p1098244

But can some one draw the graph
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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greatps24 wrote:
Got the answer from : if-x-x-2-which-of-the-following-must-be-true-99506-20.html#p1098244

But can some one draw the graph

|x| will be less than x^2 in the region where the graph of |x| is below the graph of x^2.
Graph of y = |x| is a V at (0, 0) and graph of y = x^2 is an upward facing parabola at (0, 0)

Attachment: Ques3.jpg [ 7.72 KiB | Viewed 2932 times ]

Just looking at the graph should tell you that answer must be (A). You don't need to solve. x will lie in one of two ranges x > a or x < -a. II and III will not work in any case. I must work since one of them has to be true according to the options.
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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bankerboy30 wrote:
Here's the way I look at it:

I think were all in agreement that the conditions for the statement is x<-1 and x>1. From this we can conclude this statement is true if we have values of x<-1 or x>1. But for all values to be true if I plug a value of x<-1 or x>1 into the answer choice inequalities then the inequality must hold up.

So looking at my option choices:

First option: x^2 > 1. Let's pick a number less than -1. Say -2. Is (-2)^2 > 1? Yes. Now let's look for a number x>1. Say 2. Is (2)^2>1. Yes. So this holds true for all of our conditions of x (x>1, x<-1).

Second option: x>0. Let's pick the same numbers we picked before. 2,-2. Is 2>0. Yes. Is -2>0. No this won't work if x<-1 or x is between 0 and 1.

Third option: x<-1. Again let's use the same numbers that we have been using to validate all possible solutions. Say x is 2. Is 2<-1? No. Is -2<-1? Yes. However both don't satisfy the equation

The reason why the first one does is because of the x^2. When were squaring both a positive and negative number that is x<-1 or x>1 then the result will always be a positive number that is greater than one.

Bunuel does this interpretation make sense?

Posted from my mobile device

Yes, that's correct.

From the stem we know that x is some number either less than -1 or more than 1.

Which of the statement is true about that number?

I says that the square of that number is greater than 1. Well, square of ANY number less than -1 or more than 1 is greater than 1. So, no matter what x actually is, this statement is always true.

II says that that number is greater than 0. That might not be true if it's less than -1.

III says that that number is greater less than -1. That might not be true if it's more than 1.

So, only statement I is true.

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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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.

Hi Bunuel,

I have a small doubt. Can we solve the obve equation like this:

if x>0,

$$x^2 > x$$ (now dividing both the sides by x)

we get x > 1 (this is one range)

the other range is when x <0 and |x| =-x,

we get $$x^2 > -x$$ (dividing by x)

x>-1

so the range is x>1 and x>-1.

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

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arshu27 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.

Hi Bunuel,

I have a small doubt. Can we solve the obve equation like this:

if x>0,

$$x^2 > x$$ (now dividing both the sides by x)

we get x > 1 (this is one range)

the other range is when x <0 and |x| =-x,

we get $$x^2 > -x$$ (dividing by x)

x>-1

so the range is x>1 and x>-1.

where am i going wrong?

x>1 and x>-1 does not make any sense.

As for the mistake you are making, when dividing x^2 > -x by x, which is negative, you should flip the sign: x < -1.
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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Hi,
i wonder can we solve this inequality like this square each side
thus we would have x^2<x^4 or X^2(x-1)(x+1)<0, thus we have following solution x (-1:0) and (0:1).
So based on that NONE is the answer.
Please correct me or am i making some fundamental error.
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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mika84 wrote:
Hi,
i wonder can we solve this inequality like this square each side
thus we would have x^2<x^4 or X^2(x-1)(x+1)<0, thus we have following solution x (-1:0) and (0:1).
So based on that NONE is the answer.
Please correct me or am i making some fundamental error.

x^2 < x^4
x^2 - x^4 < 0
x^2 (1 - x^2) < 0
x^2(x^2 - 1) > 0 (multiplying by -1 and hence flipping the inequality sign)
x^2(x - 1)(x + 1) > 0

The solution to this is x < -1 or x > 1.
You ignore x^2 because it has even power. The transition points are only -1 and 1.

Check this post for more: http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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mika84 wrote:
Hi,
i wonder can we solve this inequality like this square each side
thus we would have x^2<x^4 or X^2(x-1)(x+1)<0, thus we have following solution x (-1:0) and (0:1).
So based on that NONE is the answer.
Please correct me or am i making some fundamental error.

x^2 < x^4

Since the power is even both sides so both sides must be positive which has two inferences
1) The range of Negative values of x will be valid along with the same range of positive values of x
2) Higher powers of x are greater than lower powers of x only when x is greater than 1 (for positive values of x)

I.e. the Required range must be x >1 or x < -1

ALTERNATIVELY

We can solve the inequation,
x^2 - x^4 < 0
x^2 (1 - x^2) < 0
x^2(x^2 - 1) > 0 (multiplying by -1 and hence flipping the inequality sign)
x^2(x - 1)(x + 1) > 0

Since x^2 is positive for all values of x except zero therefore

(x - 1)(x + 1) > 0
I.e. both the parts must be either positive or both must be Negative

For both of (x-1) and (x+1) to be positive, x must be greater than 1
For both of (x-1) and (x+1) to be negative, x must be less than -1
The solution to this is x < -1 or x > 1.
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