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.

Answer: E.

To understand the underline concept better practice other Trickiest Inequality Questions Type: Confusing Ranges.

Hope it helps.
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|x| < x^2

Both left hand side and Right Hand side are positive, so we can square the inequalities.

\(x^2 < x^4\)

\(x^4 > x^2\)

\(x^4 - x^2 > 0\)
\(x^2(x^2 - 1)> 0\)

x^2 is always positive.

So the condition is

\(x^2 - 1> 0\)

\(x^2 > 1\)

E is the answer.

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

What is ZONEF? Can you explain this.

I looked for four range of values:

i) x<-1
Is |x| < x^2 in this range? Yes.

ii) -1 < x < 0 N
Is |x| < x^2 in this range? No.

iii) 0 < x < 1 N
Is |x| < x^2 in this range? No.

iv) x > 1 Y
Is |x| < x^2 in this range? Yes.

So, clearly |x| < x^2 only for:

i) x < -1 or
ii) x > 1

In either case, x^2 > 1. So, E.

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

'C' x>1 is a subset of 'E' x^2>1
The question is which of the following must be true so if 'E' must be true then 'C' also must be true?

No. We are given that x < -1 or x > 1. Now, if x < -1, then C is not true.
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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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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.



Algebra:
\(|x| < x^2\)
\(|x| < |x| * |x|\)

Since |x| in the blue inequality must be NONNEGATIVE, we can safely divide each side by |x|:
\(\frac{|x|}{|x|} < |x| * \frac{|x|}{|x|}\)
\(1 < |x|\)

Since each side of the red inequality is NONNEGATIVE, we can safely square each side:
\(1^2 < |x|^2\)
\(1 < x^2\)
\(x^2 > 1\)


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hey GMATGuruNY great explanation ! i have just one question

Algebra:
\(|x| < x^2\) (as you see \(x^2\) is without bars / brackets )

\(|x| < |x| * |x|\) Here you rewrite \(x^2\)as |x| * |x| and not as x*x , can you please explain why ?

many thanks !

\(x^2 = x * x = |x| * |x|\)

For example:
\(3^2 = 3 * 3 = |3| * |3| = 9\)
\((-3)^2 = -3 * -3 = |-3| * |-3| = 9\)

Given inequality:
\(|x| < x^2\)

Here, I chose to replace x^2 with |x| * |x| so that |x| would appear on both sides, enabling me to divide both sides by |x|:
\(|x| < x^2\)
\(|x| < |x| * |x|\)
\(\frac{|x|}{|x|} < |x| * \frac{|x|}{|x|}\)
\(1 < |x|\)
\(|x| > 1\)
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Whenever you are stuck with an equation with an absolute sign, you know you can get rid of the sign by taking positive and negative values of x.

If x >= 0
x < x^2
x^2 - x > 0
x(x - 1) > 0
So x > 1 or x < 0. But x must be positive so x > 1 only.

If x < 0
-x < x^2
x^2 + x > 0
x(x + 1) > 0
So x > 0 or x < -1. But x must be negative so x < -1 only.

We see that x is either > 1 or < -1. This is the same range as option (E).
x^2 - 1 > 0
(x + 1)(x - 1) > 0
x > 1 or x < -1

Answer (E)
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@banuel Sir, Can you please quote an example to prove C is not always true?

Thanks in advance

If x < -1, then C is not true. For example, x = -2 satisfy |x| < x^2 but in this case x > 1 (option C) is not true.
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Bunuel VeritasKarishma Can you please explain whats wrong with this approach. Even I fell for this one!

x(x - 1) > 0
does not imply x > 0 or x > 1

Similarly, x(x+1) > 0
does not imply x > 0 or x > -1

Check these posts to know why and to know how to solve these inequalities:
https://www.veritasprep.com/blog/2012/0 ... e-factors/
https://www.veritasprep.com/blog/2012/0 ... ns-part-i/
https://www.veritasprep.com/blog/2012/0 ... s-part-ii/
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When x < 0, how can -x < x^2? When you are taking the absolute value of a negative number, isn't the outcome always positive? In this case, when x < 0, shouldn't |-x| = x ? and not -x?

Bunuel VeritasKarishma
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