If |x| < x^2, which of the following must be true?

I solved it this way:

|x|<x^2
|x|-x^2<0

a) if x>0 then

x-x^2<0
x(1-x)<0
This implies both have opposite signs. But since x>0 we will discard the case where x<0 & 1-x>0. Therefore, we are left with:

x>0 & 1-x<0
x>0 & x>1
so, x>1

a) if x<0 then

-x-x^2<0
x+x^2>0
x(1+x)>0
This implies both have same signs. But since x<0 we will discard the case where x>0 & 1+x>0. Therefore, we are left with:

x<0 & 1+x<0
x<0 & x<-1
so, x<-1

Thus, either x<-1 or x>1 which means x^2>1 will always be true (A)

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 < 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


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 < -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 A
Hope it helps!

Watch the following video to learn the Basics of Absolute Values

IanStewart please explain why the option is wrong?

If x is positive, the absolute value isn't doing anything, so the inequality |x| < x^2 just says x < x^2, or, dividing by x (which is fine if x > 0), that 1 < x. The equivalent set of negatives will also work in the inequality, because the absolute value on the left and the square on the right both make negative numbers positive. So the inequality means that either x < -1 or x > 1. So x does not need to be positive (x can be -2, say), and II is not always true. And x is not always less than -1 (x can be 5, say), so III is not always true. Once we've ruled out II and III, looking at the answer choices, we don't need to even check I (there is no answer that says "none of the above"), but if x > 1 or x < -1, then x^2 > 1 is true, so the answer is I only.

Asked: If \(|x| < x^2\), which of the following must be true?

|x |< xˆ2
xˆ2 - |x| > 0
Case 1: x>0
x(x-1) > 0
x > 1
Case 2: x<0
x(x+1) > 0
x < -1

I. \(x^2 > 1\)
If x>1; xˆ2>1
And if x<-1; xˆ2>1
MUST BE TRUE

II. \(x > 0\)
If x>1; x>1>0; YES
But if x<-1<0; NO
COULD BE TRUE

III. \(x < -1\)
If x>1; No
But if x<-1; Yes
COULD BE TRUE


(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

IMO A

Bunuel KarishmaB

Could you please validate my approach?

|x| < x^2

√(x^2) < x^2

Squaring both sides

x^2 < x^4

x^2 - x^4 < 0

x^2 (1-x^2) < 0

x^2 < 0 (invalid as a squared value can only be positive) OR
x^2 > 1 (which means x>1 or x<-1, however I stopped before this step as I got what I was looking for)

Thanks in advance!

x^2 (1-x^2) < 0

does not lead to
x^2 < 0
or
(1 - x^2) < 0


It leads to
x^2 > 0 AND (1-x^2) < 0
or
x^2 < 0 AND (1-x^2) > 0
Since x^2 cannot be negative, we can ignore the second case. That is why it made no difference here. But try (x-1)(x-2) < 0.

Think about it: If xy is negative, it means exactly one of x and y is negative and the other is positive.

­Bunuel KarishmaB MartyMurray

I did it this way by squaring both sides. Does it make sense? Thanks.

­Both sides are positive so squaring is fine. 
You get
\(x^4 - x^2 > 0\)
\(x^2(x^2 - 1) > 0\)

Use wavy line method to get x > 1 or x < -1

I. \(x^2 > 1\)
Again gives  x > 1 or x < -1
So correct.

Other 2 are not necessary.

Answer (A)

Wavy line method discussed here:
https://youtu.be/PWsUOe77__E
 

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