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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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04 Oct 2015, 05:14
2
SIMILAR QUESTIONS:

Question 1: What are the values of n that satisfy the condition 1/|n| > n?
(A) 0<n<1 (and) - infinity < n < 0
(B) 0 < n < infinity (or) -infinity < n < -1
(C) 0 < n < 1 (and) -1 < n < 0
(D) - infinity < n < 0 (or) 0 < n < 1
(E) 0<n<1 (or) - infinity < n < 0

Question 2: If 0 < x < 1 < y, which of the following must be true?

A. 1 < 1/x < 1/y
B. 1/x < 1 < 1/y
C. 1/x < 1/y < 1
D. 1/y < 1 < 1/x
E. 1/y < 1/x < 1

Question 3: If x is positive, which of the following could be correct ordering of 1/x, 2x, and x^2?

(I) X^2 < 2x < 1/x
(II) x^2 < 1/x < 2x
(III) 2x < x^2 < 1/x

(a) none
(b) I only
(c) III only
(d) I and II
(e) I, II and III

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

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19 Jun 2016, 23:21
Expanding Bunuel's explanation

If x>0, x < x2 => x2-x>0 --> x(x-1)>0 --> For this product of x and (x-1) to be +ve (>0) or negative. Currently we are considering x as positive (x>0) condition. This if x is positive, (x-1) also has to be +ve (greater than 0). x is already greather than 0 because we are considering x>0 range. For (x-1)>0 (+ve), x has to be greater than 1 as any no. less than 1 will make (x-1) negative.

Thus root of this eq is x>1 which matches our condition (x>0).

If x<0, -x < x2 => then x2+x>0 --> x(x+1)>0 -->Thus both x and x+1 have to be either +ve or -ve. As we are considering x<0 condition, x is negative. Thus x+1 has to be negative as well. x+1<0 => x<-1 (x+1-1 <0 -1, add -1 on both sides)

Thus root of this eq is x<-1 which matches our condition (x<0)

I. x^2>1 - True for both root#1 and root 2. No matter what value of x we take from root 1 and root 2, its square will be greater than 1.
II. x>0 - not true as root#2 implies that x can be less than -1 as well (we are looking for 'must' option)
III. x<-1 - not true as root#1 implies that x can be greater than 1 as well (we are looking for 'must' option)
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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05 Sep 2016, 04:41
If |x|<x^2 , which of the following must be true?

To start with :- For the ease of calculation look at the LHS = |x|
|x| can be essentially seen as +ve x because mod CANNOT yield a negative value.
Now look at the RHS = x^2
x^2 can also be essentially seen as a positive value because squaring always yield a positive value.
IN SUMMARY the stimulus is telling us a positive number is smaller than its square.
Don't we all already know that.
That a number will always be smaller than its square EXCEPT ... EXCEPT WHEN ???? Except when the number is a positive decimal between 0 to 1, then x will be greater than its square x^2
For example
0.5 is always greater than its square 0.25
NOTICE HOW +0.5 IS A DECIMAL THAT FALLS BETWEEN 0 AND 1
But this is not case here. The stimulus tells us that the number is always less than its square. |x|<x^2
See this
-4 is always less than its square 16
+4 is always less than it's square 16
-0.6 is always less than its square 0.36
BUT +0.6 is always greater than its square 0.36

I. x^2>1 (when the irrefutable condition is |x|<x^2)
OFFCOURSE THIS MUST ALWAYS BE TRUE ... We just proved it earlier that if any number x is less than its square x^2 then it means that x^2 is always be greater than 1 ALWAYS TRUE/ MUST BE TRUE.

II. x>0 (when the irrefutable condition is |x|<x^2)
This will falsify our stimulus for values of x between 0 and 1 therefore it CAN BE TRUE FOR VALUES MORE THAN 1 BUT NOT ALWAYS TRUE WHEN X LIES BETWEEN 0 and 1

III. x<-1 (when the irrefutable condition is |x|<x^2)
This does not cover our other possible values when x > 1 So IT IS TRUE. BUT ONLY HALF TRUTH AS IT EXCLUDES THE OTHER POSSIBILITY X>1

I HOPE MY ANSWER WILL HELP THOSE PEOPLE WHO ARE STUCK IN THE -1>X>1 CONFUSION.
YOU HAVE TO KEEP IN MIND THE IRREFUTABLE STIMULS |x|<x^2 IN MIND WHEN EVALUATION THE GIVEN OPTIONS I, II, III

Thanks

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only

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

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

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06 Oct 2016, 07:55
Hi Bunuel/Someone,

Could you please explain what is meant by $$|x|<x^2$$ --> reduce by $$|x|$$ and how you arrived at "so $$|x|>0$$) --> $$1<|x|$$ --> $$x<-1$$ or $$x>1$$." step by step ?

Sorry my knowledge of Absolute Values is very basic.

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.

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

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06 Oct 2016, 22:04
Dev1212 wrote:
Hi Bunuel/Someone,

Could you please explain what is meant by $$|x|<x^2$$ --> reduce by $$|x|$$ and how you arrived at "so $$|x|>0$$) --> $$1<|x|$$ --> $$x<-1$$ or $$x>1$$." step by step ?

Sorry my knowledge of Absolute Values is very basic.

Note that $$x^2 = |x|^2$$

Given:

$$|x| < x^2$$

$$|x| < |x|^2$$

Now, if you know that x is not 0, then you know that |x| is certainly positive. So you can divide both sides by |x|.

1 < |x|

When |x| is greater than 1, it translates to x > 1 or x < -1.

For explanation of this, check: https://www.veritasprep.com/blog/2011/0 ... edore-did/
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If |x| < x^2, which of the following must be true?  [#permalink]

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23 Jun 2018, 04:13
Bunuel 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

Given: $$|x|<x^2$$;

Reduce by $$|x|$$: $$1<|x|$$ (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 $$|x|>0$$);

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.

hello generis, your previous posts were excellent many thanks i am now trying to better develop understanding of absolute values

my reasoning of the above problem is based on the two properties below:

$$|x|$$ = $$\sqrt{x^2}$$ this property means that negative values rurn into positive one ?

$$|x|≥0$$ this property absolutue value is always positive or equals zero, right ?

For example if x is -5

$$|-5|$$ = $$\sqrt{(-5)^2}$$ = $$\sqrt{(-5)*(-5)}$$ = $$\sqrt{5*5}$$ = $$\sqrt{(5)^2}$$ = 5

so $$|x|$$ means distance between x and -x

If $$|x| < x^2$$, which of the following must be true? SO I HAVE THREE OPTIONS COMBINED WITH THREE CONFUSIONS

I. $$x^2 > 1$$

II. $$x > 0$$

III. $$x < -1$$

To tackle answer choices at first from here $$|x| < x^2$$ i need to set condition

Can I square both sides ? $$|x| < x^2$$ ---> like this $$(|x|)^2 < (x^2)^2$$ --- and then i get $$x < x^4$$ what for i am doing this have no idea ... any idea ?

looking at answer choices i see that we need to compare -1 VS +1 ? so we cant plug in any values ? or any numbers we can plug in ?

FIRST OPTION I. $$x^2 > 1$$

let x be 1.

based on this formula |x|≥0

$$1^2 > 1$$ vs $$-1^2 > 1$$ so how can X^2 be aways greater than 1 ? 1*1 = techinally is 1

if x were -5 then $$-5^2 > 1$$ --> $$25 > 1$$

( I LEAVE A BLANK SPACE FOR YOUR FANTASTIC EXPLANATION )

--------------------------------------------------------------------------------------

--------------------------------------------------------------------------------------

--------------------------------------------------------------------------------------

SECOND OPTION II. $$x > 0$$

again based on this formula $$|x|≥0$$ , x can be more than or EQUAL ZERO

( I LEAVE A BLANK SPACE FOR YOUR AWESOME EXPLANATION )

--------------------------------------------------------------------------------------

--------------------------------------------------------------------------------------

--------------------------------------------------------------------------------------

THIRD OPTION III. $$x < -1$$

i somehow cant wrap my mind around this option even cant ask a right question to understand it

if we compare x = -1 and x = +1 than either of the choises could be correct, same applies to other values (-5 or +5 )

( I LEAVE A BLANK SPACE FOR YOUR MIND BLOWING EXPLANATION )

--------------------------------------------------------------------------------------

--------------------------------------------------------------------------------------

--------------------------------------------------------------------------------------

many thanks and have an absolutely positive weekend
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If |x| < x^2, which of the following must be true?  [#permalink]

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30 Jun 2018, 00:37
1
dave13 wrote:
Bunuel 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

Given: $$|x|<x^2$$;

Reduce by $$|x|$$: $$1<|x|$$ (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 $$|x|>0$$);

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.

hello generis, your previous posts were excellent many thanks i am now trying to better develop understanding of absolute values

my reasoning of the above problem is based on the two properties below:

$$|x|$$ = $$\sqrt{x^2}$$ this property means that negative values rurn into positive one ?

$$|x|≥0$$ this property absolutue value is always positive or equals zero, right?

For example if x is -5

$$|-5|$$ = $$\sqrt{(-5)^2}$$ = $$\sqrt{(-5)*(-5)}$$ = $$\sqrt{5*5}$$ = $$\sqrt{(5)^2}$$ = 5

so $$|x|$$ means distance between x and -x

If $$|x| < x^2$$, which of the following must be true? SO I HAVE THREE OPTIONS COMBINED WITH THREE CONFUSIONS

I. $$x^2 > 1$$

II. $$x > 0$$

III. $$x < -1$$

To tackle answer choices at first from here $$|x| < x^2$$ i need to set condition

Can I square both sides ? $$|x| < x^2$$ ---> like this $$(|x|)^2 < (x^2)^2$$ --- and then i get $$x < x^4$$ what for i am doing this have no idea ... any idea ?

looking at answer choices i see that we need to compare -1 VS +1 ? so we cant plug in any values ? or any numbers we can plug in ?

FIRST OPTION I. $$x^2 > 1$$
let x be 1.
based on this formula |x|≥0
$$1^2 > 1$$ vs $$-1^2 > 1$$ so how can X^2 be aways greater than 1 ? 1*1 = techinally is 1

if x were -5 then $$-5^2 > 1$$ --> $$25 > 1$$

SECOND OPTION II. $$x > 0$$

again based on this formula $$|x|≥0$$ , x can be more than or EQUAL ZERO

THIRD OPTION III. $$x < -1$$

i somehow cant wrap my mind around this option even cant ask a right question to understand it

if we compare x = -1 and x = +1 than either of the choises could be correct, same applies to other values (-5 or +5 )
SECOND OPTION II. $$x > 0$$

again based on this formula $$|x|≥0$$ , x can be more than or EQUAL ZERO
many thanks and have an absolutely positive weekend

dave13 - uhh . . . I am not too sure about "mind-blowing" (you use vernacular well!), but here is your answer.
Sorry for the wait. I'm going to reverse the order of your questions, in a way.

The main concept here keeps getting missed, and as a result, many people are doing way too much work!
Main concept: the given fact and its conditions determine whether the statements in the options MUST be true.
If an option is true for one number but not for another, according to the conditions we find, it is PARTLY FALSE and gets eliminated.

We have a given fact, a posited truth, so we have to accept it. AND... it has conditions.

The statement is true as long as certain conditions are met.
Here is the given fact: $$|x| < x^2$$

The fact and its conditions precede and determine the truth value of all the options.
Quote:
To tackle answer choices at first from here $$|x| < x^2$$ i need to set condition

Excellent. I might add that we need to FIND the conditions (because the conditions follow from the fact).

You started by squaring both sides and trying an absolute value approach.
Then you assessed the options in a few different ways. I want to streamline this process.

Find the conditions.
Assess the answers: given the fact and its conditions, which options MUST be true?

Squaring both sides IS possible, but not ideal, IMO. Squaring requires us to consider the direction of signs
at critical points, and is neither as straightforward nor as efficient as the method Bunuel used.

FACT AND CONDITIONS
$$|x| < x^2$$
Typically with absolute value questions and/or inequalities, there are conditions.

We can use logic, algebra, or numbers to try simplifying the given fact.
Bunuel used algebra immediately.

(1) What are the conditions under which this fact is true?
If we test values in an absolute value inequality, we might well find the conditions.
Almost always, we can simplify with algebra, too.

Test numbers: under what conditions is this statement NOT true? TRUE?
Test a set of "weird" values (sometimes they do not behave like most positive and negative numbers).

Test: 0, 1, -1, $$\frac{1}{2}$$, and $$-\frac{1}{2}$$
Is the statement true if:
$$x=0$$? NO.
The statement is not true. $$|0|=0$$ and $$0^2=0$$.
So $$|0|$$ is not LESS THAN $$0^2$$ - they are equal

$$x=1$$? NO. $$|1|=1^2$$. LHS is NOT less than RHS (they are equal).
(-1) is exactly the same as 1.

$$x=\frac{1}{2}$$? No. $$|\frac{1}{2}|=\frac{1}{2}$$ and
$$\frac{1}{2})^2=\frac{1}{4}$$. LHS is GREATER than RHS. The same thing will happen with $$-\frac{1}{2}$$

Now test 2 and -2. See what happens. Is the statement true?
From those tests, you should be able to sketch a number line, darken it to reflect conditions,
and extrapolate mathematical statements from it to indicate those conditions. (Hint: we know -1, -1/2, 0, 1/2, and 1 do not work.)

Algebra: more difficult, but faster. We need values (and their limits) to assess options

Background material
• absolute value is always non-negative, i.e. 0 or positive
• for this |x|, $$x\neq0$$ because |0| is equal to $$0^2$$, not less than $$0^2$$
• Because $$x\neq0$$, we are guaranteed that $$|x|$$ is positive. Hence we can divide both sides of the inequality by |x|
• $$x^2=|x|^2=(|x|*|x|)$$
Use numbers to see it. Positive $$x=5$$ is easy.
$$x = -5$$: Now $$x^2=(|x|*|x|)=(5*5)=25=(-5)^2$$

$$|x| < x^2$$
$$|x| < |x|*|x|$$
Divide both sides by |x|
$$\frac{|x|}{|x|}<\frac{|x|*|x|}{|x|}$$
$$1<|x|$$, which is equivalent to
$$|x|>1$$

(2) Once simplified, find values for $$|x|>1$$

$$|x|>1$$
If $$|x| > 1$$ , then

Case 1: $$x > 1$$

Case 2: $$-x > 1$$
Divide by (-1), flip the sign
$$x < -1$$

$$x>1$$ OR $$x<-1$$

If we draw the range of solutions on the number line, we have
Attachment:

number line 2018 06 29.jpg [ 28.32 KiB | Viewed 761 times ]

$$x$$ can be any value in the two ranges on the number line

(3) Assess options. $$|x| < x^2$$. And $$x>1$$ or $$x<-1$$.

Which options MUST BE TRUE?

I. $$x^2 > 1$$ MUST be true
We can look at the number line: ALWAYS true. Bunuel wrote:
Quote:
Option 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.

Pick any point on the blue part of the line as a value for $$x$$. Square it. It will always be greater than 1.

Or (algebra that involves critical points): $$x^2>1$$
$$x<-1$$ or $$x>1$$

If that solution set does not make sense, please see this post and read all the linked material. Looking at the number line is easier

II. $$x>0$$ - NOT ALWAYS TRUE.
Our conditions are $$x>1$$ or $$x<-1$$
This statement is not true if $$x=-7$$ (-7 is not greater than 0)

III. $$x<-1$$ - NOT ALWAYS TRUE. If $$x=8$$, it is not true. (8 is NOT < -1)

I hope that helps!

Quote:
$$|x|$$ = $$\sqrt{x^2}$$ this property means that negative values rurn into positive one ?

Not quite. That equation means: this is the positive root of $$x^2$$
$$-\sqrt{x^2} = -|x|$$
That equation means: this is the negative root of $$\sqrt{x^2}$$
Quote:
$$|x|≥0$$ this property absolutue value is always positive or equals zero, right?

Yes. In this case, $$x$$ does NOT equal 0, so $$x$$ is positive
Quote:
For example if x is -5
$$|-5|$$ = $$\sqrt{(-5)^2}$$ = $$\sqrt{(-5)*(-5)}$$ = $$\sqrt{5*5}$$ $$= 25 = (-5)^2$$
$$\sqrt{(5)^2}$$ = 5 OMIT THIS PART

Mostly true. I have amended slightly. Also note that if $$x = -5, |x| = -x = -(-5)$$
Quote:
so $$|x|$$ means distance between x and -x

No. |x| means the distance of $$x$$ OR $$-x$$ from $$0$$, but not the distance between them
Quote:
Can I square both sides ? $$|x| < x^2$$ ---> like this $$(|x|)^2 < (x^2)^2$$ --- and then i get $$x < x^4$$ what for i am doing this have no idea ... any idea ?

At this point I am not sure I have any idea!
If you are asking, "is it possible"?
Yes. You wouldn't want to, IMO.
We found the conditions without squaring. We just divided by $$|x|$$, got to $$|x| > 1$$, and found our conditions.
Why are you doing it?
Best guess: Because you have seen others do it in different contexts.

Squaring both sides will work. VeritasPrepKarishma discusses such squaring in this case
in this post.
GMATinsight also discusses squaring both sides here.
I square both sides only when absolutely necessary.
Quote:
looking at answer choices i see that we need to compare -1 VS +1 ?

I don't understand here . . . We need to see whether the options
fulfill the conditions laid out in the diagram of the number line: $$x>1$$ or $$x<-1$$
Maybe that is what you mean.
Quote:
so we cant plug in any values ? or any numbers we can plug in ?

Sure you can! While searching for conditions or some simplification, we tested some unique numbers (-1, 1, 0 . . .).
From there, with some logic, we could have drawn the number line.

Or, if we have the conditions and we are assessing options, plug in numbers that are given or possible from the option.
See whether the numbers in the context of that option satisfy our conditions.
In fact, with numbers you can prove that Option I must be true, and that Options 2 and 3 might not be true. If needed, sketch a number line.
Quote:
FIRST OPTION I. $$x^2 > 1$$
let x be 1.
based on this formula |x|≥0

You are analyzing based on a rule that is important but . . .that rule is not the same as the conditions.

See above where we tested numbers. We KNOW we cannot use 1 or -1
x is greater than 1 and less than -1
You are testing and violating explicit conditions that say you may not use those numbers.
Quote:
$$1^2 > 1$$ vs $$-1^2 > 1$$ so how can X^2 be aways greater than 1 ? 1*1 = techinally is 1

if x were -5 then $$-5^2 > 1$$ --> $$25 > 1$$

AHA! So . . . -1 and 1 do not work. -5 DOES work.
We know that 0, numbers between 0 and 1, and numbers between 0 and -1 do not work with the statement.
I think if you had carried this line of thought a little further, you would have arrived at
what we can conclude with logic in just a few seconds:
if |x| < x^2, $$x>1$$ or $$x<-1$$
Quote:
$$|x|≥0$$

This expression got you mixed up, I think.

For this problem, that equation works best to remind us that we can divide the terms of the given inequality by |x|.
Normally, we do not divide sides of inequalities by variables,
because we do not know whether the variables are positive or negative.

In this case, that equation tells us: the absolute value of $$x$$ is either 0 or positive.
And in this case, $$x \neq 0.$$ (See where we tested numbers.)
So if x cannot be 0, it MUST be positive.
In that case, we CAN divide by sides by $$|x|$$
When we did, we got
$$|x| < x^2$$
$$|x| < |x|*|x|$$
Divide both sides by |x|
$$\frac{|x|}{|x|}<\frac{|x|*|x|}{|x|}$$
$$1<|x|$$, which is equivalent to
$$|x|>1$$
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If |x| < x^2, which of the following must be true?  [#permalink]

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03 Dec 2018, 17:24
The general rule is that if $$x^2$$ $$= |x|$$then x can be $$1, 0, or -1$$

Thus if |x| < x^2, x is not equal to 1,0 or -1

Testing values, we also find that x cannot be a fraction for example
$$x= 1/2 (1/2)^2 = 1/4 |1/2|=1/4$$ ? No.

Test integers $$<-1$$ and greater than $$>1$$
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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06 Dec 2018, 06:23
Bunuel wrote:
qweert wrote:
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.

We are GIVEN that |x|<x2|x|<x2, which means that either x<−1x<−1 OR x>1x>1.
how is this possible? please explain more.
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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06 Dec 2018, 07:01
rashedBhai wrote:
Bunuel wrote:
qweert wrote:
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.

We are GIVEN that |x|<x2|x|<x2, which means that either x<−1x<−1 OR x>1x>1.
how is this possible? please explain more.

I tried explaining this on previous three pages. Please re-read. You'll find many useful posts there.
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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14 Dec 2018, 09:21
With absolute value, inequalities, and number properties problems, I always consider

(x=-2,x=-1, x=neg fraction, 0, pos fraction, 1, 2) let x = -2 |-2| <(-2)^2 ? Yes let x = -1 |-1|<(-1)^2 No Let x =-1/2
|-1/2| < (-1/2)^2 No . No Similarly with x =-1/2, 0,1. let x =2 |2|<2^2 Yes. So x = (...-4,-3,-2) U (2,3,4...)
I. x^2>1 Always
II. X>0 Not always
III. X<-1 Not always
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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23 Sep 2019, 03:48
Bunuel wrote:

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.

How did you obtain the bold? If x<0 then x^2 > -x therefore -x^2<x which means in turn that x+x^2 > 0. This means that x(1+x)>0 so x>0 and x>-1?

What have I done wrong?
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If |x| < x^2, which of the following must be true?  [#permalink]

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24 Nov 2019, 06:29
jamalabdullah100 wrote:
Bunuel wrote:

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.

How did you obtain the bold? If x<0 then x^2 > -x therefore -x^2<x which means in turn that x+x^2 > 0. This means that x(1+x)>0 so x>0 and x>-1?

What have I done wrong?

Hey, a quick suggestion-

For questions like this there are multiple ways to find the range of x. I'm sure you have seen people using number line, graph, modulus properties etc.

lxl < x^2

By seeing this equation you must develop the ability to deduce
#1 x can't be zero or 1 or -1
#2 Both sides of the equation is +ve
#3 Yes x can be +ve or -ve as we are dealing with modulus
#4 2^2 is always more than 2 and (-3)^2 is always more than l-3l

Hence, x must be R - [-1,1], where R is a real number. Hope this approach helps.
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Re: If |x| < x^2, which of the following must be true?  [#permalink]

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01 Dec 2019, 05:15
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

I always struggle with the choice of answer for this type of questions, eventhough the formula is easy to digest.

When I was in high school or higher education, this type of questions would take III as the correct answer. For example, if we have answer choice IV that says, x=-2, we will take it as correct as well. Because x=-2 will always make the equation correct. And it is not about if the range of x that cover every x that can be plug into the formula. It makes less sense for these questions to have combination choices, if there's only one answer that is correct.

Need to change my thought process a bit....

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

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01 Dec 2019, 23:25
X^2>|X|
We can say that X^2= |X|^2
Therefore above equation can also be written as:
|X|^2-|X|>0
Or |X|*(|X|-1)>0
Therefore either both|X|>0 or (|X|-1)>0 , or both are smaller than 0.
But modulus of a number cannot be smaller than 0.
Therefore:
|X|>0 or |X|-1>0
|X|>0 => X>0 or X<0, option 2 only answers one part, so wrong.

|X|-1>0 => X>1 or X<-1 option 3 only answers one part, so wrong.

Now |X|-1>0 or
|X|>1 or X^2>1 so only option 1 always hold true.

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Re: If |x| < x^2, which of the following must be true?   [#permalink] 01 Dec 2019, 23:25

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