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

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.

Hope it's clear.

I'm confused, because if x= -2 ==> |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: \(x<-1\) OR \(x>1\).

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.

Hope it's clear.

mehdiov, what's the OA?

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



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):

ps-inequality-13943.html

Maybe I don't understand something.

Thanks.

OA is A.

\(|x|<x^2\) means that either \(x<-1\) or \(x>1\). For example \(x\) could be -5, -3, 2, 3, ...

Basically the question is: If \(x<-1\) or \(x>1\) which of the following must be true?

You are saying that answer is: E (I and III only). Is III: \(x<-1\) ALWAYS true? Is it true for \(x=3\)? NO. So E is not correct.

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?

x^2-1>0 --> x<-1 or x>1.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

Bunuel |x|.|x|=x^2 so when we divide x^2 by |x| we get |x|

so |x|.|x|= x^2 is this a formula

Yes, |x|*|x|=x^2. Consider x=-2 then |x|*|x|=2*2=4=(-2)^2.

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.

picked A in 46 secs. for |X| to be less than x^2 just means that it is x is not a decimal. it can be less than -1 and greater than 1. you can only say this in totality with certainty. And square of nay no less than -1 and greater than 1 will be greater than 1. so A

vikram 4689 what x <-1 says is that x can ONLY hold values of less than -1. Which actually means that x cannot have values of 2,3.. etc. this we all know is not true as x can also takes values of >1. Hence this statement cannot hold true. makes sense?

No, that's not true.

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

\(|x|<x^2\) means that \(x<-1\) or \(x>1\). Now, evaluate each option:

I. x^2>1. Since \(x<-1\) or \(x>1\) then this option is always true;
II. x>0. Since \(x<-1\) or \(x>1\) then this option may or may not be true;
III. x<-1. Since \(x<-1\) or \(x>1\) then this option may or may not be true.

Answer: A (I only).

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.

First of all please study these posts:
if-x-x-2-which-of-the-following-must-be-true-99506.html#p767256
if-x-x-2-which-of-the-following-must-be-true-99506.html#p767264
if-x-x-2-which-of-the-following-must-be-true-99506.html#p767437
if-x-x-2-which-of-the-following-must-be-true-99506.html#p776341
if-x-x-2-which-of-the-following-must-be-true-99506.html#p1088622
if-x-x-2-which-of-the-following-must-be-true-99506-20.html#p1089273

Because the question you ask was answered before.

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.

So, we have that only option I is always true.

Answer: A (I only).

Check our Must or Could be True Questions to practice more: search.php?search_id=tag&tag_id=193

Hope it helps.