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(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.

The X-Y plane of looking at it is the fastest way - I would guess.

Mod(x) looks like a V

x^2 is a parabola which just touches the x-axis (tangent) at the origin.

When x < -1; Mod(x) < x^2 When x = -1; Mod (x) = x^2 When -1<x<0; Mod(x) > x^2 When x = 0; Mod (x) = x^2 When 0<x<1; Mod(x) > x^2 When x = 1; Mod (x) = x^2 When x > 1; Mod(x) < x^2

|x|< x^2, X^2 should always be > 1 Can eliminate II, because if x=1/2, |x|> x^2

Could you give an example Bunuel

Question is: "which of the following statements MUST be true", not COULD be true.

We are GIVEN that |x|<x^2, which means that either x<-1 OR x>1.

So GIVEN that: x<-1ORx>1.

Statement II. is x<-1 always true? NO. As x could be more than 1, eg. 2, 3, 5.7, ... and in this case x<-1 is not true. So statement II which says that x<-1 is not always true.

Bunuel, I understand your reasoning but I don't think you are right. IMO the answer is E. Should we follow your logic:

Quote:

Statement II. is x<-1 always true? NO. As x could be more than 1, eg. 2, 3, 5.7, ... and in this case x<-1 is not true. So statement II which says that x<-1 is not always true.

Then we can't even accept I, since x doesn't always have to be greater than 1 (it can be -10, -50 etc.).

The question is similar to (but I still think that it is different):

Re: If |x|<x^2 , which of the following must be true? [#permalink]
23 May 2012, 05:09

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pavanpuneet wrote:

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;

Re: If |x|<x^2 , which of the following must be true? [#permalink]
15 Jun 2012, 01:59

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pavanpuneet wrote:

Hi Bunuel, I tried to solve it this way, please let me know if this is correct as I am getting confused with the final wording of MUST be true. I am kind of stuck at the final stages to make the decision.

The problem mentions: |x|<x2

Case 1: if x is +ve; then x<x2, given that we do not know if x=0, then x-x2<0 ==> x(1-x)<0 ==> multiplying by negative sign; x(x-1)>0...which means x<0 and X>1

Case 2: if x is -ve; then -x<x2, then -x-x2<0 ===>-(x+x2) <0...==> x(1+x)>0 ===> x<-1 and x>0

From C1 AND C2..on the number x<-1 and x>1; after this I am a little lost, how to make the final call.

Anyway, given that |x|<x^2, which means that x<-1 or x>1. Next we are given three options and are asked which of the following MUST be true.

I. x^2>1. Since x<-1 or x>1 then this option is always true (ANY number less than -1 or more than 1 when squared will be more than 1); II. x>0. This option is NOT always true, for example x could be -2 and in this case x>0 won't not true; III. x<-1. This option is also NOT always true, for example x could be 2 and in this case x<-1 won't not true.

Re: If |x|<x^2 , which of the following must be true? [#permalink]
20 Jun 2012, 05:44

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idreesma wrote:

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

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

Re: If |x|<x^2 , which of the following must be true? [#permalink]
20 Jun 2012, 21:12

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idreesma wrote:

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

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: inequalities-trick-91482.html

Also, your Veritas book discusses this concept too. _________________

Re: If |x|<x^2 , which of the following must be true? [#permalink]
04 Oct 2012, 20:21

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idreesma wrote:

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

Bunuel, Isn't this always true when |x|<x^2 ... Isn't the answer E.

Thanks.

|x|<x^2 is given as fact and then we asked to determine which of the following statements MUST be true.

|x|<x^2 means that either x<-1 or x>1, x can be ANY value from these two ranges, (I think in your own solution you've reached this conclusion: when x<-1 the graph of |x| is below (less than) the graph of x^2 and when x>1 again the graph of |x| is below the graph of x^2).

Now, III says x<-1 this statement is not always true as x can be for example 3 and in this case x<-1 doesn't hold true.

Bunuel, I understand your reasoning but I don't think you are right. IMO the answer is E. Should we follow your logic:

Quote:

Statement II. is x<-1 always true? NO. As x could be more than 1, eg. 2, 3, 5.7, ... and in this case x<-1 is not true. So statement II which says that x<-1 is not always true.

Then we can't even accept I, since x doesn't always have to be greater than 1 (it can be -10, -50 etc.).

The question is similar to (but I still think that it is different):

Hi, If i solve option 1. i get x^2-1>0 or (x+1)(x-1)>0 this implies : x>-1 or x>1.. however the question in the stem can be rephrased as x<-1 or X>1.. How is option 1 true then?am i missing something?

Bunuel wrote:

mehdiov 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

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.