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|x|<x^2

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by vikram4689 » Thu May 24, 2012 5:10 pm
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


[spoiler]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... [/spoiler]
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by Anurag@Gurome » Thu May 24, 2012 6:05 pm
vikram4689 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


[spoiler]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... [/spoiler]
|x| < x²
Dividing throughout by |x|,
|x| > 1
x < -1 or x > 1

I. x² > 1 is always true.
II. x > 0 may or may not be true.
III. x < -1 may or may not be true.

The correct answer is A.
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by melon » Sat May 26, 2012 8:38 am
can u pls explain the answer
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by Anurag@Gurome » Sat May 26, 2012 8:58 am
x² = |x|*|x| and |x| > 0

--> x² > |x|
--> |x|*|x| > |x|
--> |x| > 1 ............... Dividing both sides by |x|
--> Either x < - 1 or x > 1
--> In both of the above cases x² > 1

Hope it is clear now.
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