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x/|x|<x. which of the following must be true about x ?

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x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post Updated on: 17 May 2017, 04:35
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x/|x|<x. Which of the following must be true about x ?

A. x > 1

B. x > -1

C. |x| < 1

D. |x| = 1

E. |x|^2 > 1

THE CORRECT ANSWER IS THERE IS NO DEBATE WHATSOEVER. PLEASE READ THE WHOLE DISCUSSION TO UNDERSTAND THE QUESTION BETTER.

Originally posted by christoph on 06 Feb 2005, 09:16.
Last edited by Bunuel on 17 May 2017, 04:35, edited 5 times in total.
OA added.
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 01 Sep 2010, 07:32
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x/|x|<x, which of the following must be true about x ?

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

We are given that \(\frac{x}{|x|}<x\) (this is a true inequality), so first of all we should find the ranges of \(x\) for which this inequality holds true.

\(\frac{x}{|x|}< x\) multiply both sides of inequality 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) --> \(x<x|x|\) --> \(x(|x|-1)>0\):
Either \(x>0\) and \(|x|-1>0\), so \(x>1\) or \(x<-1\) --> \(x>1\);
Or \(x<0\) and \(|x|-1<0\), so \(-1<x<1\) --> \(-1<x<0\).

So we have that: \(-1<x<0\) or \(x>1\). Note \(x\) is ONLY from these ranges.

Option B says: \(x>-1\) --> ANY \(x\) from above two ranges would be more than -1, so B is always true.

Answer: B.

nonameee wrote:
Quote:
x/|x|<x. which of the following must be true about x ?

a) x>1
b) x>-1
c) |x|<1
d) |x|=1
e) |x|^2>1



Bunuel, I think there's a mistake in the question or in the answer choices:

Here's my solution:
1) x<0:
-x/x < x
-1<x<0

2) x>=0
x/x<x
x>1

The solution of the inequality is then:
-1<x<0 union x>1

The answer can't be B, since let x=1/2. We get:
(1/2)/(1/2) < (1/2)
1< 1/2
Contradiction.

I think the answer should be A since it satisfies all the scenarios.

Can you please clarify?


The options are not supposed to be the solutions of inequality \(\frac{x}{|x|}<x\).

Hope it's clear.
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 06 Feb 2005, 22:32
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x/|x|<x
if x>0, then 1/x<1 ie, x>1
if x<0, then 1/(-x)>1, ie, x>-1

So combine these two, we know that x must be greater than -1.

baner is right, (B).
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 06 Feb 2005, 22:50
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It doesn't say that all x greater than -1 would satisfy the equation. We get two range of x that satisfies the equation (-1,0) and (1,infinite). All that the answer says is that every x in our solution ranges are greater than -1, which is true.
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 02 Sep 2010, 00:20
Bunuel, thanks for your reply. But I have to disagree with you because:

The answer can't be B, since let x=1/2. We get:
(1/2)/(1/2) < (1/2)
1< 1/2
Contradiction.

What do you think about that?
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x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 02 Sep 2010, 03:30
3
nonameee wrote:
Bunuel, thanks for your reply. But I have to disagree with you because:

The answer can't be B, since let x=1/2. We get:
(1/2)/(1/2) < (1/2)
1< 1/2
Contradiction.

What do you think about that?


OA for this question is B and it's not wrong.

Consider following:
If \(x=5\), then which of the following must be true about \(x\):
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

Or:
If \(-1<x<10\), then which of the following must be true about \(x\):
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x<120

Again answer is E, because ANY \(x\) from \(-1<x<10\) will be less than 120 so it's always true about the number from this range to say that it's less than 120.

The same with original question:

If \(-1<x<0\) or \(x>1\), then which of the following must be true about \(x\):
A. x>1
B. x>-1
C. |x|<1
D. |x|=1
E. |x|^2>1

As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.

\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.

On the other hand A says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.

Hope it's clear.
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 26 Oct 2010, 03:12
Hi Bunnuel,
Can it be solved as-

x/|x|<x

Hence 1/|x|<1

1/x<1 OR -1/x<1
when 1/x<1
then 1<x = x>1

when -1/x<1
then -1<x = x>-1

The two possible outcomes are x>1 or x>-1
The more restrictive is x>-1
Hence B

Thanks
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 26 Oct 2010, 09:44
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nades09 wrote:
Hi Bunnuel,
Can it be solved as-

x/|x|<x

Hence 1/|x|<1

1/x<1 OR -1/x<1
when 1/x<1
then 1<x = x>1

when -1/x<1
then -1<x = x>-1

The two possible outcomes are x>1 or x>-1
The more restrictive is x>-1
Hence B

Thanks
Neelam


No, that't not correct.

First of all, when you are writing 1/|x|<1 from x/|x|<x you are reducing (dividing) inequality by x:

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Consider a simple inequality \(4>3\) and some variable \(x\).

Now, you can't multiply (or divide) both parts of this inequality by \(x\) and write: \(4x>3x\), because if \(x=1>0\) then yes \(4*1>3*1\) but if \(x=-1\) then \(4*(-1)=-4<3*(-1)=-3\). Similarly, you can not divide an inequality by \(x\) not knowing its sign.

Next, inequality \(\frac{1}{|x|}<1\) holds true in the following ranges: \(x<-1\) and \(x>1\) (and not: x>1 or x>-1, which by the way simply means x>1).

Hope it's clear.
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 24 Jul 2011, 19:12
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A graph helps me visualize these types of problems. If you graph \(\frac{x}{|x|}\) (red line) and \(x\) (blue line), the green areas represent the region in which the inequality \(\frac{x}{|x|}<x\) is satisfied.

The question asks what MUST BE TRUE if the inequality holds. In other words, IF \(\frac{x}{|x|}<x\) IS SATISFIED, then what must be true of \(x\)?

The green areas represent the regions where this inequality holds. What must be true of both green areas?

The \(x\) values in both regions must be greater than \(-1\). B is the correct answer.
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 28 Dec 2012, 00:55
Bunuel wrote:

OA for this question is B and it's not wrong.

Consider following:
If \(x=5\), then which of the following must be true about \(x\):
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

Or:
If \(-1<x<10\), then which of the following must be true about \(x\):
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x<120

Again answer is E, because ANY \(x\) from \(-1<x<10\) will be less than 120 so it's always true about the number from this range to say that it's less than 120.

The same with original question:

If \(-1<x<0\) or \(x>1\), then which of the following must be true about \(x\):
A. x>1
B. x>-1
C. |x|<1
D. |x|=1
E. |x|^2>1

As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.

\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.

On the other hand A says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.

Hope it's clear.


Bunuel,
As you've mentioned that we're to verify the range of x for which the given inequality holds good.
So for x>-1, x can have the value like 1/2. So in that case the inequality doesn't hold good for sure. Aren't we validating the inequality to be true ?Now,'must be true' means it has to satisfy all the possible plug-in values taking one from each of the category i.e. positive fraction and integer and negative fraction and integer as per the given conditions.

Whereas, for x>1 it does satisfy for all the possible values like x=3/2,4 etc. and the inequality holds good.So.how can we ignore the above case where the inequality clearly becomes false ?
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 28 Dec 2012, 05:04
debayan222 wrote:
Bunuel wrote:

OA for this question is B and it's not wrong.

Consider following:
If \(x=5\), then which of the following must be true about \(x\):
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x>-10

Answer is E (x>-10), because as x=5 then it's more than -10.

Or:
If \(-1<x<10\), then which of the following must be true about \(x\):
A. x=3
B. x^2=10
C. x<4
D. |x|=1
E. x<120

Again answer is E, because ANY \(x\) from \(-1<x<10\) will be less than 120 so it's always true about the number from this range to say that it's less than 120.

The same with original question:

If \(-1<x<0\) or \(x>1\), then which of the following must be true about \(x\):
A. x>1
B. x>-1
C. |x|<1
D. |x|=1
E. |x|^2>1

As \(-1<x<0\) or \(x>1\) then ANY \(x\) from these ranges would satisfy \(x>-1\). So B is always true.

\(x\) could be for example -1/2, -3/4, or 10 but no matter what \(x\) actually is it's IN ANY CASE more than -1. So we can say about \(x\) that it's more than -1.

On the other hand A says that \(x>1\), which is not always true as \(x\) could be -1/2 and -1/2 is not more than 1.

Hope it's clear.


Bunuel,
As you've mentioned that we're to verify the range of x for which the given inequality holds good.
So for x>-1, x can have the value like 1/2. So in that case the inequality doesn't hold good for sure. Aren't we validating the inequality to be true ?Now,'must be true' means it has to satisfy all the possible plug-in values taking one from each of the category i.e. positive fraction and integer and negative fraction and integer as per the given conditions.

Whereas, for x>1 it does satisfy for all the possible values like x=3/2,4 etc. and the inequality holds good.So.how can we ignore the above case where the inequality clearly becomes false ?


I think you don't understand what is given and what is asked.

Given: -1<x<0 and x>1 (that's what x/|x|<x means).

Now, the question asks which of the following MUST be true.

You are saying: "so for x>-1, x can have the value like 1/2." That's not correct: if -1<x<0 and x>1, then how x can be 1/2?
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x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 24 Mar 2014, 07:40
Hi Bunuel,
Can you help clear the confusion with X = 1/2.

Thanks.
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x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 24 Mar 2014, 07:51
seabhi wrote:
Hi Bunuel,
Can you help clear the confusion with X = 1/2.

Thanks.


x/|x|<x, which of the following must be true about x ?

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

We are given that \(\frac{x}{|x|}<x\) (this is a true inequality), so first of all we should find the ranges of \(x\) for which this inequality holds true.

\(\frac{x}{|x|}< x\) multiply both sides of inequality 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) --> \(x<x|x|\) --> \(x(|x|-1)>0\):
Either \(x>0\) and \(|x|-1>0\), so \(x>1\) or \(x<-1\) --> \(x>1\);
Or \(x<0\) and \(|x|-1<0\), so \(-1<x<1\) --> \(-1<x<0\).

So we have that: \(-1<x<0\) or \(x>1\). Note \(x\) is ONLY from these ranges.

Option B says: \(x>-1\) --> ANY \(x\) from above two ranges would be more than -1, so B is always true.

Answer: B.
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Re: x/|x|<x. which of the following must be true about x ?  [#permalink]

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New post 02 Sep 2019, 22:36
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Re: x/|x|<x. which of the following must be true about x ?   [#permalink] 02 Sep 2019, 22:36
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