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

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Re: If |x|<x^2 , which of the following must be true? [#permalink]  04 Oct 2012, 20:21
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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.
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/
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Re: not that hard [#permalink]  28 Oct 2012, 02:31
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: not that hard [#permalink]  29 Oct 2012, 01:51
Expert's post
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]  09 Nov 2012, 08:34
I just cannot get the answer, Bunuel can you give a more detailed explanation about how you conjugated the problem. When I solve the equation in the question I get this --> |x| < x^2 -->

-|x| < x^2 --> -x < x^2 --> divide by x --> -1 < x
|x| < x^2 --> x < x^2 --> divide by x --> 1 < x

which makes the first equation redundant. How do you get x < -1 ?????

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Re: If |x|<x^2 , which of the following must be true? [#permalink]  05 Dec 2012, 03:04
|x|<x^2

Determine thw value of x first before testing agains I,II and III.
Let us find out x by testing against checkpoints -1, 0 and 1.
Let x < -1: |-2| < 4 Yes!
Let x =-1: 1 < 1 No!
Let -1 < x < 0: x=-1/4 : 1/4 < 1/16 No!
Let 0 < x < 1 : x =1/4 : 1/4 < 1/16 No!
Let x = 1: 1 < 1 No!
Let x > 1: x = 2 : 2 < 4 Yes!

Answer: -1 < x < 1 or |x| > 1

I. x^2>1 Remember, x^2 > 1 is the same as |x| > 1 Yes!
II. x>0 This doesn't cover x < -1.
III. x<-1 This doesn't cover x > 1.

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Re: not that hard [#permalink]  07 Mar 2013, 08:13
Bunuel wrote:
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.

great explanation....
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  07 Mar 2013, 09:50
Here are my two cents.

You cannot divide both the sides by only x since you do not know whether x is positive or not. Granted that the absolute value will always be positive but if x=-1, absolute value would be 1 which would be denoted as -x. Hence, the absolute value, even though positive, could be +/-x.

Now take the negative aspect into the picture.

|x|<x^2
-x<x^2
-1>x
Hence, for values such as x <-1, the equation stands. You can test it with values like x=-2,-3 etc.

Hence, c does not stand as it does not necessarily have to be true. Of course, that was evident from you calculations.

Hope this helps!

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?

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Re: If |x|<x^2 , which of the following must be true? [#permalink]  09 Mar 2013, 22:24
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

|x| - x^2 < 0

If x>0, x ( 1 - x) < 0--> x < 0 (can't be true as condition states x>0or x>1
If x<0, -x - x^2 <0-----> x(1+x) >0 --> x>0 or x> -1.

x>1 & x>-1 only option A is valid. x can't be equal to 0. as 0>0 is not true. Is my solution correct?? Why it is different then Bunuel's solution
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  09 Mar 2013, 23:42
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]  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]  11 Mar 2013, 00:21
Expert's post
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 638 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]  01 Jun 2013, 09:08
The trick is in question, it doesn't ask you to choose possible answers, but it asks for an answer which holds true in all possibilities. Option (i) satisfies this always, but option (iii), not always..
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  09 Jul 2013, 21:52
If |x|<x^2 , which of the following must be true?

|x|<x^2

I. x^2>1
If |x|<x^2 then that means x^2 is greater than positive value x. x cannot be between -1 and 1 inclusive because then |x| would be greater than or equal to x^2, not less than it as the stem stipulates. Therefore, for the stem to be true x^2 must be greater than 1.

II. x>0
If x>0 then x could = 1/2.
|1/2|<(1/2)^2
1/2<1/4 which is invalid

III. x<-1
While any x value less than -1 will provide a correct answer, valid values for x don't have to be less than -1. They can be greater than one as well.

(A)
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  21 Jul 2013, 17:47
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

Having seen such a big red quote here, I was wary of this question before starting.
For questions like these I have created a method which has helped me and would like to share it.

Bunnel, Please correct me if this is wrong or has limited scope.

If |x|<x^2 , which of the following must be true?
this means x>1 and x<-1.

The question stem set must be equal to or a subset of the answer choices i.e. The answer choice must contain at-least all of points of the question

I -
x^2>1 => x>1 and x<-1
this choice consists of exactly the same data points as in the question stem, subset and hence, Yes.

II -
X>0
this choice does not contain all of the points of the question stem, not a subset and hence, No.

III -
x<-1
this choice does not contain all of the points of the question stem, not a subset and hence, No.

Try similar question:
If |x| > 3, which of the following must be true?

1) x > 3
2) x^2 > 9
3) |x - 1| > 2

I only
II only
I and II only
II and III only
I, II, and III

OA -
[Reveal] Spoiler:
D
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  21 Jul 2013, 19:25
1
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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: If |x|<x^2 , which of the following must be true? [#permalink]  21 Jul 2013, 20:51
Expert's post
cumulonimbus 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

Having seen such a big red quote here, I was wary of this question before starting.
For questions like these I have created a method which has helped me and would like to share it.

Bunnel, Please correct me if this is wrong or has limited scope.

If |x|<x^2 , which of the following must be true?
this means x>1 and x<-1.

The question stem set must be equal to or a subset of the answer choices i.e. The answer choice must contain at-least all of points of the question

I -
x^2>1 => x>1 and x<-1
this choice consists of exactly the same data points as in the question stem, subset and hence, Yes.

II -
X>0
this choice does not contain all of the points of the question stem, not a subset and hence, No.

III -
x<-1
this choice does not contain all of the points of the question stem, not a subset and hence, No.

Try similar question:
If |x| > 3, which of the following must be true?

1) x > 3
2) x^2 > 9
3) |x - 1| > 2

I only
II only
I and II only
II and III only
I, II, and III

OA -
[Reveal] Spoiler:
D

Yes, that's correct.
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  01 Jan 2014, 17:44
Bunuel or Karishma, is there a problem with the below:
x^2 > abs(x)
x^2 > sqrt(x^2)
x^4 > x^2
x^2 > 1

x^2 > 1 is only true when abs(x) > 1
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Re: If |x|<x^2 , which of the following must be true? [#permalink]  02 Jan 2014, 04:49
Expert's post
TooLong150 wrote:
Bunuel or Karishma, is there a problem with the below:
x^2 > abs(x)
x^2 > sqrt(x^2)
x^4 > x^2
x^2 > 1

x^2 > 1 is only true when abs(x) > 1

Yes, it is. But you could use easier way by reducing the inequality by |x| as shown in my post here: 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]  21 Jan 2014, 22:49
This is very tricky. If this is a 600-700 level, I am in trouble. I did get it, however, I don't think I would get it if didn't read the explanation for several times. And still a little confused. Thank you.
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Re: not that hard [#permalink]  23 Apr 2014, 02:01
Bunuel wrote:
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.

Hi bunnel,

Can we write this as x<0 or x>0 instead of x<-1 or x>1

Re: not that hard   [#permalink] 23 Apr 2014, 02:01

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