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Re: If |x|<x^2 , which of the following must be true? [#permalink]
23 May 2012, 23:45
Expert's post
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?
Re: If |x|<x^2 , which of the following must be true? [#permalink]
24 May 2012, 17:09
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...
Re: If |x|<x^2 , which of the following must be true? [#permalink]
24 May 2012, 20:16
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? [#permalink]
25 May 2012, 01:01
Expert's post
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? [#permalink]
25 May 2012, 01:22
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 ? hope this might help u .. x^2 >|x|---> (implies) x^2>x (since x^2 can not be less than ..it will always by +ve) x^2-x>0 x(x-1)>0 this would give two values x <0 and x>1 but x can not be <0 because values lying between -1 to +1 would nt satisfy the given condition |x|<x^2 ...therefore... x>1 therefore x^2 is also greater than 1
Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 01:25
Expert's post
deepti1206 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 ? hope this might help u .. x^2 >|x|---> (implies) x^2>x (since x^2 can not be less than ..it will always by +ve) x^2-x>0 x(x-1)>0 this would give two values x <0 and x>1 but x can not be <0 because values lying between -1 to +1 would nt satisfy the given condition |x|<x^2 ...therefore... x>1 therefore x^2 is also greater than 1
Please read the thread.
\(|x|<x^2\) means that \(x<-1\) or \(x>1\). _________________
Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 01:36
Sorry i did a mishtake...i would like to revise it,.....
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 ? hope this might help u .. x^2 >|x|---> (implies) i)x^2>x x^2-x>0 x(x-1)>0 this would give two values x <0 and x>1
ii) x^2 <-x x^2-x<0 x(x+1)<0 giving u ..x>0 and x <-1
giving u the common condition that is x >1 therefore x^2 is also greater than 1
Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 01:40
Expert's post
deepti1206 wrote:
Sorry i did a mishtake...i would like to revise it,.....
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 ? hope this might help u .. x^2 >|x|---> (implies) i)x^2>x x^2-x>0 x(x-1)>0 this would give two values x <0 and x>1
ii) x^2 <-x x^2-x<0 x(x+1)<0 giving u ..x>0 and x <-1
giving u the common condition that is x >1 therefore x^2 is also greater than 1
Re: If |x|<x^2 , which of the following must be true? [#permalink]
25 May 2012, 20:41
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.
Hello Bunuel
I didnot understand this part x^2>|x| holds true for x<-1 and x>1.
if you have x<-1 then why ans "E" is not true?
i also didnot understand folloiwng part "If x>0 then x2-x>0 --> x(x-1)>0 -->[highlight]x<0[/highlight] and x>1" how can x<0 ?
I solved this problem and voted for "A"
for x>1 / 1>x>0 / x<0 (I considered these 3 ranged because if 1>x>0 then the square of number is less then number)
Re: If |x|<x^2 , which of the following must be true? [#permalink]
26 May 2012, 01:36
Expert's post
kuttingchai 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.
Hello Bunuel
I didnot understand this part x^2>|x| holds true for x<-1 and x>1.
if you have x<-1 then why ans "E" is not true?
i also didnot understand folloiwng part "If x>0 then x2-x>0 --> x(x-1)>0 -->[highlight]x<0[/highlight] and x>1" how can x<0 ?
E cannot be correct since
I solved this problem and voted for "A"
for x>1 / 1>x>0 / x<0 (I considered these 3 ranged because if 1>x>0 then the square of number is less then number)
Re: If |x|<x^2 , which of the following must be true? [#permalink]
14 Jun 2012, 23:53
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? [#permalink]
15 Jun 2012, 01:59
1
This post received KUDOS
Expert's post
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.
Re: If |x|<x^2 , which of the following must be true? [#permalink]
20 Jun 2012, 05:41
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; appreciate your input Thanks
Re: If |x|<x^2 , which of the following must be true? [#permalink]
20 Jun 2012, 05:44
1
This post received KUDOS
Expert's post
idreesma wrote:
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; appreciate your input Thanks
Welcome to GMAT Club. Below links might help you to understand the concept.
Re: If |x|<x^2 , which of the following must be true? [#permalink]
20 Jun 2012, 21:12
1
This post received KUDOS
Expert's post
idreesma wrote:
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; appreciate your input Thanks
Responding to a pm:
The problem you are facing is that you do not know how to handle inequalities.
How do you get the range for which this inequality holds? x(x+1) > 0
Think of it this way: Product of x and (x+1) should be positive. When will that happen? When either both the terms are positive or both are negative.
Case 1: When both are positive x > 0 x + 1 > 0 i.e. x > -1
For both to be positive, x must be greater than 0. Hence this inequality will hold when x > 0.
Case 2: When both are negative x < 0 x + 1 < 0 i.e. x < -1
For both to be negative, x must be less than -1. Hence this inequality will hold when x < -1.
So we get two ranges in which this inequality holds: x > 0 or x < -1.
The fastest way to solve it is using the number line. Check this post for the explanation of this method: inequalities-trick-91482.html
Also, your Veritas book discusses this concept too. _________________
Re: If |x|<x^2 , which of the following must be true? [#permalink]
21 Jun 2012, 18:15
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. Appreciate your input Thanks
Re: If |x|<x^2 , which of the following must be true? [#permalink]
04 Oct 2012, 10:09
A quicker way of solving take the square of both sides x^2 < x^4 since both sides ought to be positive we can simplify by x^2 w/t changing the direction of the inequality 1 < x^2 or x^2 >1 Brother Karamazov
gmatclubot
Re: If |x|<x^2 , which of the following must be true?
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04 Oct 2012, 10:09
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