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

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idreesma

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20 Jun 2012, 06:41

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Hi,
I am having issues with a fundanmental concept. In the below explanation x(x+1)>0 , isnt x>0 and x>-1 (from x+1>0) rather than x<-1 and x>0. clearly by inspection my answer is wrong, but was confused over how you go correct answer (is it possible to show the steps)

If x>0 then x2-x>0 --> x(x-1)>0 --> x<0 and x>1. Since we are considering x>0 range then x>1;
If x<0 then x2-x>0 --> x(x+1)>0 --> x<-1 and x>0. Since we are considering x<0 range then x<-1;
appreciate your input
Thanks

Bunuel

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20 Jun 2012, 06:44

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idreesma

Welcome to GMAT Club. Below links might help you to understand the concept.

x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535

Hope it helps.

Hope it helps.

KarishmaB

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Updated on: 02 Oct 2022, 22:28

Originally posted by KarishmaB on 20 Jun 2012, 22:12.
Last edited by KarishmaB on 02 Oct 2022, 22:28, edited 1 time in total.

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idreesma

Responding to a pm:

The problem you are facing is that you do not know how to handle inequalities.

How do you get the range for which this inequality holds? x(x+1) > 0

Think of it this way: Product of x and (x+1) should be positive. When will that happen? When either both the terms are positive or both are negative.

Case 1: When both are positive
x > 0
x + 1 > 0 i.e. x > -1

For both to be positive, x must be greater than 0. Hence this inequality will hold when x > 0.

Case 2: When both are negative
x < 0
x + 1 < 0 i.e. x < -1

For both to be negative, x must be less than -1. Hence this inequality will hold when x < -1.

So we get two ranges in which this inequality holds: x > 0 or x < -1.

The fastest way to solve it is using the number line.
Check this post for the explanation of this method: https://gmatclub.com/forum/inequalities-trick-91482.html

idreesma

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21 Jun 2012, 19:15

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Hi Krishma,
the link that you posted is probably best for summary. It ends with the comment "Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post ". Can you suggest any such thread where absolute value side is summarized like this as well.
Appreciate your input
Thanks

KarishmaB

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Updated on: 11 Oct 2022, 02:16

Originally posted by KarishmaB on 04 Oct 2012, 21:21.
Last edited by KarishmaB on 11 Oct 2022, 02:16, edited 1 time in total.

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idreesma

Please check my signature for the link to the relevant blog posts.

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28 Oct 2012, 03:31

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Bunuel

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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

what is the fastest way to solbe it ?

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 \(x>1\).

I. x^2>1 --> always true;
II. x>0 --> may or may not be true;
III. x<-1 --> --> may or may not be true.

Answer: A (I only).

Hi bunuel,
sorry , but I didn't undestrand how you reduce |x|<x^2 by |x| to get the solution...x<-1 and x>1
please, could you explain me in more details?
Thanks

Bunuel

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29 Oct 2012, 02:51

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mario1987

Bunuel

mehdiov

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

what is the fastest way to solbe it ?

Hi bunuel,
sorry , but I didn't undestrand how you reduce |x|<x^2 by |x| to get the solution...x<-1 and x>1
please, could you explain me in more details?
Thanks

Express \(x^2\) as \(|x|*|x|\), so we have that: \(|x|<|x|*|x|\) --> reduce by \(|x|\) (notice that \(|x|\) is positive, thus we can safely divide both parts of the inequality by it) --> \(1<|x|\) --> \(x<-1\) or \(x>1\).

Hope it's clear.

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10 Mar 2013, 00:42

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Got the answer from : if-x-x-2-which-of-the-following-must-be-true-99506-20.html#p1098244

But can some one draw the graph

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|X| < x^2 means X is not a decimal. means X < -1 or x > 1. Square of anything will be positive. Hence. A
Hope it helps.

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11 Mar 2013, 01:21

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greatps24

Got the answer from : if-x-x-2-which-of-the-following-must-be-true-99506-20.html#p1098244

But can some one draw the graph

|x| will be less than x^2 in the region where the graph of |x| is below the graph of x^2.
Graph of y = |x| is a V at (0, 0) and graph of y = x^2 is an upward facing parabola at (0, 0)

Attachment:

Ques3.jpg [ 7.72 KiB | Viewed 8273 times ]

Just looking at the graph should tell you that answer must be (A). You don't need to solve. x will lie in one of two ranges x > a or x < -a. II and III will not work in any case. I must work since one of them has to be true according to the options.

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21 Jul 2013, 20:25

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1. |x|<x^2 =>

(i) x < x^2 or
(ii) -x < x^2

2. (1) => x>1 or x<-1 . Only this satisfies both (i) and (ii)

Is x^2>1 always true? Yes follows from (2)
Is x>0 always true, No because x could be <-1
Is x<-1 always true?, No because x could be >1

So answer is choice A.

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Here's the way I look at it:

I think were all in agreement that the conditions for the statement is x<-1 and x>1. From this we can conclude this statement is true if we have values of x<-1 or x>1. But for all values to be true if I plug a value of x<-1 or x>1 into the answer choice inequalities then the inequality must hold up.

So looking at my option choices:

First option: x^2 > 1. Let's pick a number less than -1. Say -2. Is (-2)^2 > 1? Yes. Now let's look for a number x>1. Say 2. Is (2)^2>1. Yes. So this holds true for all of our conditions of x (x>1, x<-1).

Second option: x>0. Let's pick the same numbers we picked before. 2,-2. Is 2>0. Yes. Is -2>0. No this won't work if x<-1 or x is between 0 and 1.

Third option: x<-1. Again let's use the same numbers that we have been using to validate all possible solutions. Say x is 2. Is 2<-1? No. Is -2<-1? Yes. However both don't satisfy the equation

The reason why the first one does is because of the x^2. When were squaring both a positive and negative number that is x<-1 or x>1 then the result will always be a positive number that is greater than one.

Bunuel does this interpretation make sense?

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Bunuel

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08 Jul 2014, 09:24

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bankerboy30

Yes, that's correct.

From the stem we know that x is some number either less than -1 or more than 1.

Which of the statement is true about that number?

I says that the square of that number is greater than 1. Well, square of ANY number less than -1 or more than 1 is greater than 1. So, no matter what x actually is, this statement is always true.

II says that that number is greater than 0. That might not be true if it's less than -1.

III says that that number is greater less than -1. That might not be true if it's more than 1.

So, only statement I is true.

Hope it's clear.

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13 Jun 2015, 12:07

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Bunuel

pavanpuneet

Hi Bunuel, I understood your logic that we can safely divide by |x|, however, I tried to solve it this way, just wanted to confirm is this is also right...

x2>|x| => if x is positive then x2-x>0 => x(x-1)>0 => x<0 and x>1................case 1

and if x is negative, then x2+x>0 => x(x+1)>0 =>x<-1 and x>0...case2

From case 1 and case2....x<-1 and x>1..[Same conclusion but lengthier method].. I want to especially confirm because I tried to apply modulus and inequalities theory.. so want to be sure.

x^2>|x|

If x>0 then x2-x>0 --> x(x-1)>0 --> x<0 and x>1. Since we are considering x>0 range then x>1;
If x<0 then x2-x>0 --> x(x+1)>0 --> x<-1 and x>0. Since we are considering x<0 range then x<-1;

So, x^2>|x| holds true for x<-1 and x>1.

Hope it's clear.

Hi Bunuel,

I have a small doubt. Can we solve the obve equation like this:

if x>0,

\(x^2 > x\) (now dividing both the sides by x)

we get x > 1 (this is one range)

the other range is when x <0 and |x| =-x,

we get \(x^2 > -x\) (dividing by x)

x>-1

so the range is x>1 and x>-1.

where am i going wrong?

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13 Jun 2015, 12:49

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arshu27

Bunuel

pavanpuneet

x>1 and x>-1 does not make any sense.

As for the mistake you are making, when dividing x^2 > -x by x, which is negative, you should flip the sign: x < -1.

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27 Jun 2015, 11:36

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Hi,
i wonder can we solve this inequality like this square each side
thus we would have x^2<x^4 or X^2(x-1)(x+1)<0, thus we have following solution x (-1:0) and (0:1).
So based on that NONE is the answer.
Please correct me or am i making some fundamental error.

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Updated on: 21 Dec 2023, 08:21

Originally posted by KarishmaB on 28 Jun 2015, 23:01.
Last edited by KarishmaB on 21 Dec 2023, 08:21, edited 1 time in total.

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mika84

x^2 < x^4
x^2 - x^4 < 0
x^2 (1 - x^2) < 0
x^2(x^2 - 1) > 0 (multiplying by -1 and hence flipping the inequality sign)
x^2(x - 1)(x + 1) > 0

The solution to this is x < -1 or x > 1.
You ignore x^2 because it has even power. The transition points are only -1 and 1.

Check the wavy line method here: https://youtu.be/PWsUOe77__E

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29 Jun 2015, 00:11

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mika84

x^2 < x^4

Since the power is even both sides so both sides must be positive which has two inferences
1) The range of Negative values of x will be valid along with the same range of positive values of x
2) Higher powers of x are greater than lower powers of x only when x is greater than 1 (for positive values of x)

I.e. the Required range must be x >1 or x < -1

ALTERNATIVELY

We can solve the inequation,
x^2 - x^4 < 0
x^2 (1 - x^2) < 0
x^2(x^2 - 1) > 0 (multiplying by -1 and hence flipping the inequality sign)
x^2(x - 1)(x + 1) > 0

Since x^2 is positive for all values of x except zero therefore

(x - 1)(x + 1) > 0
I.e. both the parts must be either positive or both must be Negative

For both of (x-1) and (x+1) to be positive, x must be greater than 1
For both of (x-1) and (x+1) to be negative, x must be less than -1
The solution to this is x < -1 or x > 1.

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03 Oct 2015, 04:55

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Bunuel

mehdiov

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

what is the fastest way to solbe it ?

Hi,

Doubt regarding third statement.

The expression in the question is valid for all x>1 and X<-1
As per this even third option should be right.

Also, please tell me how to identify these options as in some questions, a part range is also among the answers while in others it is not. I always get confused in these questions. Is it because of "must" or "could"? Could you please mention similar questions where I can practice for this?

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mehdiov

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

what is the fastest way to solbe it ?

"Must" Means It will be true for all possible values of x
"Could" Means It may be true for some values of x and may not be true for other values of x

About the third:
III. x<-1
Take some values of x less than -1 and check if the Inequation is satisfied
@x = -2, |x|<x^2 this is satisfied
@x = -3, |x|<x^2 this is satisfied as well
for all values of x less than -1
|x|will have positive and smaller absolute value than absolute values of x^2 which will be positive.
x less than -1 is an acceptable range of values of x but since x may be positive (greater than 1) as well so it is not necessary that x be less than -1

I hope this helps!