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When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long +1 Kudos if you find this post helpful
Originally posted by Skywalker18 on 19 Jun 2016, 11:13.
Last edited by Bunuel on 19 Jun 2016, 11:20, edited 2 times in total.
Re: If |x| < x^2, which of the following must be true ?
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19 Jun 2016, 11:20
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15
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?
A. x > 0 B. x < 0 C. x > 1 D. -1 < x < 1 E. x^2 > 1
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 \(x>1\).
A. x > 0. Not always true. B. x < 0. Not always true. C. x > 1. Not always true. D. -1 < x < 1. Not true. E. x^2 > 1. Always true.
Re: If |x| < x^2, which of the following must be true ?
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19 Jun 2016, 11:22
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Bunuel wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?
A. x > 0 B. x < 0 C. x > 1 D. -1 < x < 1 E. x^2 > 1
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 \(x>1\).
A. x > 0. Not always true. B. x < 0. Not always true. C. x > 1. Not always true. D. -1 < x < 1. Not true. E. x^2 > 1. Always true.
Re: If |x| < x^2, which of the following must be true ?
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21 Jun 2016, 20:39
I used it by plugging in ZONEF.
A. x > 0 : If x is an integer, the inequality holds true. It would be opposite for a fraction B. x < 0 : Same as A C. x > 1 : Same as A D. -1 < x < 1 : Again, we can use Zero to plug in. E. x^2 > 1 : This must be true.
Re: If |x| < x^2, which of the following must be true ?
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22 Jun 2016, 08:19
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aditi2013 wrote:
I used it by plugging in ZONEF.
A. x > 0 : If x is an integer, the inequality holds true. It would be opposite for a fraction B. x < 0 : Same as A C. x > 1 : Same as A D. -1 < x < 1 : Again, we can use Zero to plug in. E. x^2 > 1 : This must be true.
IMO E.
your explanation for C is wrong.if u take fraction of x>1 then it satisfies. But this choice is just sub part of solutions of |x|<x^2 .the other solution is x<-1 which choice C doesn't say about
Re: If |x| < x^2, which of the following must be true ?
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26 Jun 2016, 01:31
hsbinfy wrote:
aditi2013 wrote:
I used it by plugging in ZONEF.
A. x > 0 : If x is an integer, the inequality holds true. It would be opposite for a fraction B. x < 0 : Same as A C. x > 1 : Same as A D. -1 < x < 1 : Again, we can use Zero to plug in. E. x^2 > 1 : This must be true.
IMO E.
your explanation for C is wrong.if u take fraction of x>1 then it satisfies. But this choice is just sub part of solutions of |x|<x^2 .the other solution is x<-1 which choice C doesn't say about
Aah, thank you for catching this error. The improper fraction will satisfy, right?
Re: If |x| < x^2, which of the following must be true ?
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30 Jun 2016, 04:01
2
sauravpaul wrote:
aditi2013 wrote:
I used it by plugging in ZONEF.
What is ZONEF? Can you explain this.
Hi Saurav,
ZONEF is just an abbreviation to ensure no value remains untested to check the sufficiency. So, I always check ZONEF in majority of such data sufficiency questions.
Zero One Negative Extremes (Lower Limit, High) Fraction
If |x| < x^2, which of the following must be true ?
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05 Nov 2016, 23:24
Cant we do this question by opening the Modulus>>??
My approach is as follows Given that |x|<x^2
Now opening the modulus accordingly CASE 1:x<0 => -x<x^2 => x+x^2>0 => x(x+1)>0
By the Line wave Method: The range of x is x<-1 U x>0 CASE 2: x>0 => x<x^2 => x(x-1)>0
Now the range of x is x<0 U x>1
Combining both operations x <-1 U x>1 So C option directly suits or comes in this range.. As per option E.. if I put a value from the above range say x=-0.25..x^2=0.0625 which is less than 1
So according me C should be the correct answer.....
Re: If |x| < x^2, which of the following must be true ?
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19 Oct 2017, 10:20
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?
A. x > 0 B. x < 0 C. x > 1 D. -1 < x < 1 E. x^2 > 1
If x = 0, then |0| will not be less than 0^2 since they both are equal to 0. Thus, we know x can’t be 0.
Since |x| = x when x is positive and -x when x is negative, let’s rewrite the inequality without the absolute value sign. That is, if x > 0, then we have x < x^2, and if x < 0, then we have -x < x^2.
Case 1: If x > 0,
x < x^2
Dividing both sides by x, we have:
1 < x
Case 2: If x < 0,
-x < x^2
Dividing both sides by x (switching the inequality sign since x is negative), we have:
-1 > x
Thus, we have x > 1 or x < -1; in that case, x^2 must be greater than 1.
Answer: E
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Re: If |x| < x^2, which of the following must be true ?
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24 Jun 2018, 03:52
Bunuel wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?
A. x > 0 B. x < 0 C. x > 1 D. -1 < x < 1 E. x^2 > 1
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 \(x>1\).
A. x > 0. Not always true. B. x < 0. Not always true. C. x > 1. Not always true. D. -1 < x < 1. Not true. E. x^2 > 1. Always true.
can you please explain in your solution when we \(|x|<x^2\) --> reduce by \(|x|\) you say absolute value is non-negative but doesnt absolute value means either negative or positive value ? we always consider two conditions one is negative and another positive....
\(|A| < B\) means \(-B < A < B\), (here \(A\) can be \(-1\), and \(B\) can be \(-2\) ) now getting back to your solution how can you be sure that |x| is positive positive number when reducing by |x| ---> \(\frac{|x|}{|x|}<\frac{x^2}{|x|}\) can you explain it somehow ?
Re: If |x| < x^2, which of the following must be true ?
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24 Jun 2018, 04:08
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Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?
A. x > 0 B. x < 0 C. x > 1 D. -1 < x < 1 E. x^2 > 1
It's possible that x=2, since \(|2| < 2^2\). Eliminate B and D, since they do not have to be true. It's possible that x=-2, since \(|-2| < (-2)^2\). Eliminate A and C, since they do not have to be true.
GMAT and GRE Tutor Over 1800 followers Click here to learn more GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com.
one question is there a rule when we can square modulus as in the solution above ? i mean in which case can i square both sides and in which case i cant square both sides of mudulus ? how can squaring modulus on both sides facilitate problem solving process under time pressure ?
Also modulus means either negative or positive value how can we be sure that number is positive when squaring
Also how do we go from here \(x^2(x^2 - 1)> 0\) to this \(x^2 > 1\) ? do we divide both sides by (x^2 - 1) --->
---> do we divide by \((x^2 - 1)\)both sides or do we divide both sides by \(x^2\) ?
one question is there a rule when we can square modulus as in the solution above ? i mean in which case can i square both sides and in which case i cant square both sides of mudulus ? how can squaring modulus on both sides facilitate problem solving process under time pressure ?
Also modulus means either negative or positive value how can we be sure that number is positive when squaring
Also how do we go from here \(x^2(x^2 - 1)> 0\) to this \(x^2 > 1\) ? do we divide both sides by (x^2 - 1) --->
---> do we divide by \((x^2 - 1)\)both sides or do we divide both sides by \(x^2\) ?
Squaring is a method to remove mod function, because a square is ALWAYS POSITIVE and so is a MOD function. Hence when you square you are not changing any property of the mod function but increasing the degree of the equation.
You can also solve this equation without squaring by taking two scenarios: Scenario 1: if x>0 and scenario 2: if x<0, but this will increase the calculation steps. you can try yourself.
Now once you have squared and arrived at \(x^2(x^2 - 1)> 0\), then this simply implies that product of two number is positive. the two numbers are \(x^2\) & \((x^2-1)\). We know that \(x^2>0\), hence for the product to be positive \(x^2-1>0 =>x^2>1\)
If |x| < x^2, which of the following must be true ?
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28 Jun 2018, 09:32
GMATGuruNY wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?
A. x > 0 B. x < 0 C. x > 1 D. -1 < x < 1 E. x^2 > 1
It's possible that x=2, since \(|2| < 2^2\). Eliminate B and D, since they do not have to be true. It's possible that x=-2, since \(|-2| < (-2)^2\). Eliminate A and C, since they do not have to be true.
close
59% (01:21) correct 
we always consider two conditions one is negative and another positive....
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