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(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.
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
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.
|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.
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):
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):
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.
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?
(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|
(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? [#permalink]
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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.
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? [#permalink]
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23 May 2012, 12:09
1
This post was BOOKMARKED
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
Re: If |x|<x^2 , which of the following must be true? [#permalink]
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23 May 2012, 12:54
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?
gmatclubot
Re: If |x|<x^2 , which of the following must be true?
[#permalink]
23 May 2012, 12:54
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46% (02:02) correct
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