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If |x| < x^2, which of the following must be true?
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20 Aug 2010, 13:18
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24
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
Given: \(|x|<x^2\);
Reduce by \(|x|\): \(1<|x|\) (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 \(|x|>0\));
Re: If |x| < x^2, which of the following must be true?
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20 Aug 2010, 13:22
7
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
Re: If |x| < x^2, which of the following must be true?
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20 Aug 2010, 13:33
2
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.
Re: If |x| < x^2, which of the following must be true?
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21 Aug 2010, 02:24
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2
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.
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07 Sep 2010, 01:40
mehdiov, what's the OA?
Bunuel, I understand your reasoning but I don't think you are right. IMO the answer is E. Should we follow your logic:
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):
Re: If |x| < x^2, which of the following must be true?
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07 Sep 2010, 04:03
1
nonameee wrote:
mehdiov, what's the OA?
Bunuel, I understand your reasoning but I don't think you are right. IMO the answer is E. Should we follow your logic:
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):
Re: If |x| < x^2, which of the following must be true?
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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.
Re: If |x| < x^2, which of the following must 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?
Re: If |x| < x^2, which of the following must be true?
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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.
Answer: A (I only).
Bunuel |x|.|x|=x^2 so when we divide x^2 by |x| we get |x|
Re: If |x| < x^2, which of the following must be true?
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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.
Answer: A (I only).
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.
_________________
Re: If |x| < x^2, which of the following must be true?
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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.
Re: If |x| < x^2, which of the following must be true?
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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;
Re: If |x| < x^2, which of the following must be true?
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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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24 May 2012, 21:16
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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?
Re: If |x| < x^2, which of the following must be true?
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25 May 2012, 02:01
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.
Re: If |x| < x^2, which of the following must be true?
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15 Jun 2012, 00:53
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?
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15 Jun 2012, 02:59
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pavanpuneet wrote:
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.
Anyway, given that \(|x|<x^2\), which means that \(x<-1\) or \(x>1\). Next we are given three options and are asked which of the following MUST be true.
I. x^2>1. Since \(x<-1\) or \(x>1\) then this option is always true (ANY number less than -1 or more than 1 when squared will be more than 1); II. x>0. This option is NOT always true, for example x could be -2 and in this case x>0 won't not true; III. x<-1. This option is also NOT always true, for example x could be 2 and in this case x<-1 won't not true.
close
50% (01:32) correct 
