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

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If |x| < x^2, which of the following must be true ?  [#permalink]

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Updated on: 19 Jun 2016, 10:20
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If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

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Originally posted by Skywalker18 on 19 Jun 2016, 10:13.
Last edited by Bunuel on 19 Jun 2016, 10:20, edited 2 times in total.
Renamed the topic and edited the question.
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Posts: 65014
Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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19 Jun 2016, 10:20
10
26
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

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.

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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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21 Jun 2016, 21:12
56
24
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

|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

$$x^2 - 1> 0$$

$$x^2 > 1$$

##### General Discussion
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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19 Jun 2016, 10:22
Bunuel wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

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.

Similar question to practice: if-x-x-2-which-of-the-following-must-be-true-99506.html
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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21 Jun 2016, 19:39
1
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.
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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22 Jun 2016, 07:19
1
1
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
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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29 Jun 2016, 22:40
I used it by plugging in ZONEF.

What is ZONEF? Can you explain this.
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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29 Jun 2016, 22:43
2
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.
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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30 Jun 2016, 03:01
3
sauravpaul wrote:
I used it by plugging in ZONEF.

What is ZONEF? Can you explain this.

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
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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22 Jul 2016, 21:09
'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?
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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22 Jul 2016, 23:17
1
gaurav0480 wrote:
'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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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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19 Oct 2017, 09:20
6
1
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

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.

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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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24 Jun 2018, 03:08
3
1
2
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

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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If |x| < x^2, which of the following must be true ?  [#permalink]

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28 Jun 2018, 08:32
GMATGuruNY wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

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:
$$1^2 < |x|^2$$
$$1 < x^2$$
$$x^2 > 1$$

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 !
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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28 Jun 2018, 09:21
1
1
dave13 wrote:
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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If |x| < x^2, which of the following must be true ?  [#permalink]

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23 Aug 2019, 23:35
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

Given: |x| < x^2

Asked: Which of the following must be true ?

|x|< x^2 => x is not in the region -1<x<1 & $$x \neq$$ 0=> x^2 >1

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

IMO E
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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24 Aug 2019, 02:50
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

the given relation |x| < x^2 would be true whenever x is an integer
so out of given options only E; x^2 > 1 would be valid
IMO E
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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22 Sep 2019, 10:48
1
Bunuel wrote:
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

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.

Hi Bunuel, what am I doing wrong in the below pic?

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If |x| < x^2, which of the following must be true ?  [#permalink]

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02 Nov 2019, 10:20
1
Skywalker18 wrote:
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

√[(x)^2] < x^2, square both sides
x^2 < x^4
Since x cannot be 0 (initial inequality shows this), we can divide by x^2
1 < x^2

Another, slightly longer way:
√[(x)^2] < x^2
x^2 < x^4
0 < (x^4)(x^2)
0 < x^2*(x^2 - 1)
0 < x^2*(x-1)(x+1)
The trap here is C x > 1, which is true, but the opposite is also true (x < -1)
When we square both sides in either case, we get x^2 > 1
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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11 Nov 2019, 11:06
Re: If |x| < x^2, which of the following must be true ?   [#permalink] 11 Nov 2019, 11:06

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