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GRE Prep Club Team Member
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If |x| < x^2, which of the following must be true ?
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25 Nov 2021, 04:12
25 Nov 2021, 04:12
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If |x| < x², which of the following must be true ?
A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x² > 1
A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x² > 1
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Re: If |x| < x^2, which of the following must be true ?
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25 Nov 2021, 05:16
25 Nov 2021, 05:16
2
GeminiHeat wrote:
If |x| < x², which of the following must be true ?
A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x² > 1
A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x² > 1
I find that these questions can be solved quickly (and accurately) by testing values.
For example, if |x| < x², then x could equal 2 (since |2| < 2²)
This means we can eliminate choices B and D since they state that x cannot equal 2.
Similarly, if |x| < x², then x could equal -2 (since |-2| < (-2)²)
This means we can eliminate choices A and C since they state that x cannot equal -2.
By the process of elimination, the correct answer is E
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Re: If |x| < x^2, which of the following must be true ?
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29 Dec 2021, 23:00
29 Dec 2021, 23:00
2
Given that |x| < x²
Let's take two cases to open the absolute value
Case 1: Whatever is inside the Absolute Value >= 0
x >= 0
=> |x| = x (Watch this video to learn more about Basics of Absolute Value)
=> x < \(x^2\)
=> \(x^2\) - x > 0
=> x*(x-1) > 0
x between 0 and 1.JPG [ 17.24 KiB | Viewed 1934 times ]
=> x < 0 or x > 1 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x>=0
=> Intersection will be x > 1
x GT 1.JPG [ 14.98 KiB | Viewed 1977 times ]
Case 2: Whatever is inside the Absolute Value < 0
x < 0
=> |x| = -x (Watch this video to learn more about Basics of Absolute Value)
=> -x < \(x^2\)
=> \(x^2\) + x > 0
=> x*(x+1) > 0
x between -1 and 0.JPG [ 16.8 KiB | Viewed 1962 times ]
=> x < -1 or x > 0 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x < 0
=> Intersection will be x < -1
So, x > 1 or x < -1
=> x^2 > 1
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Absolute Values
Let's take two cases to open the absolute value
Case 1: Whatever is inside the Absolute Value >= 0
x >= 0
=> |x| = x (Watch this video to learn more about Basics of Absolute Value)
=> x < \(x^2\)
=> \(x^2\) - x > 0
=> x*(x-1) > 0
Attachment:
x between 0 and 1.JPG [ 17.24 KiB | Viewed 1934 times ]
=> x < 0 or x > 1 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x>=0
=> Intersection will be x > 1
Attachment:
x GT 1.JPG [ 14.98 KiB | Viewed 1977 times ]
Case 2: Whatever is inside the Absolute Value < 0
x < 0
=> |x| = -x (Watch this video to learn more about Basics of Absolute Value)
=> -x < \(x^2\)
=> \(x^2\) + x > 0
=> x*(x+1) > 0
Attachment:
x between -1 and 0.JPG [ 16.8 KiB | Viewed 1962 times ]
=> x < -1 or x > 0 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x < 0
=> Intersection will be x < -1
So, x > 1 or x < -1
=> x^2 > 1
So, Answer will be E
Hope it helps!
Watch the following video to learn the Basics of Absolute Values
