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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If |x| < x^2, which of the following must be true?
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27 Nov 2021, 04:28
27 Nov 2021, 04:28
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Question Stats:
41% (01:09) correct
58% (01:23) wrong based on 24 sessions
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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
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
Re: If |x| < x^2, which of the following must be true?
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01 Dec 2021, 03:20
01 Dec 2021, 03:20
1
Expert Reply
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 \(|x|>0\));
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).
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 \(|x|>0\));
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).
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Re: If |x| < x^2, which of the following must be true?
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01 Dec 2021, 09:00
01 Dec 2021, 09:00
1
To answer this, I would suggest first to consider that |x| and x^2 are both ALWAYS POSITIVE.
Next, notice that this is a MUST question, so finding any good NO case (counterexample) will prove it wrong.
I. Given that |x| is defined as sqrt(x^2), then it follows that the number x must be greater than 1 -- that is, the only cases where the square is smaller than the number itself are fractions between 0 and 1. This MUST BE TRUE.
II.
>NO case make x = -2 and we see |-2| = 2 while (-2)^2 = 4
This one doesn't have to be true.
III.
>NO case x = 2 and we see |2| = 2 while 2^2 = 4
This one doesn't have to be true.
The Answer is A.
Next, notice that this is a MUST question, so finding any good NO case (counterexample) will prove it wrong.
I. Given that |x| is defined as sqrt(x^2), then it follows that the number x must be greater than 1 -- that is, the only cases where the square is smaller than the number itself are fractions between 0 and 1. This MUST BE TRUE.
II.
>NO case make x = -2 and we see |-2| = 2 while (-2)^2 = 4
This one doesn't have to be true.
III.
>NO case x = 2 and we see |2| = 2 while 2^2 = 4
This one doesn't have to be true.
The Answer is A.
