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

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

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06 Feb 2005, 09:16
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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

[Reveal] Spoiler:
B
THERE IS NO DEBATE WHATSOEVER. PLEASE READ THE WHOLE DISCUSSION TO UNDERSTAND THE QUESTION BETTER.
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 May 2017, 04:35, edited 5 times in total.
VP
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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06 Feb 2005, 17:35
"B".

x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1
VP
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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06 Feb 2005, 17:49
banerjeea_98 wrote:
"B".

x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1

lets suppose x be 0.5, which is greater than -1. 0.5/l0.5l=1 which is not less than 0.5.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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06 Feb 2005, 17:53
MA wrote:
banerjeea_98 wrote:
"B".

x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1

lets suppose x be 0.5 which is more than -1. 0.5/l0.5l=1 which is not less than 0.5.

ya but in no case x can be < -1.....it has to be > -1....but not bet 0 and 1.....x > 1 is wrong.....as the eqn also satisfies when x = -1/2
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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06 Feb 2005, 18:09
MA wrote:
banerjeea_98 wrote:
MA wrote:
banerjeea_98 wrote:
"B".

x/|x|<x-------x<x|x|.....for +ve x...|x| > 1...for -ve x .....|x| < 1....that means x can be -1<x<0....or x > 1....in any case it has to be > -1

lets suppose x be 0.5 which is more than -1. 0.5/l0.5l=1 which is not less than 0.5.

ya but in no case x can be < -1.....it has to be > -1....but not bet 0 and 1.....x > 1 is wrong.....as the eqn also satisfies when x = -1/2

baner,
unclear to me.... pls be specific.

Let's imagine a number line with points -1, 0 and 1.....as u can see x can't be <-1...e.g. x = -2......-1 < -2....will not satisfy....

Now let's see -1<x<0......x = -1/2.....-1 < -1/2.....satisfies
Now for 0<x<1.....x = 1/2.....1 < 1/2.....will not satisfy....
For x > 1....x = 2.....1 < 2....satisfies

As u see the soln lies in the following categories:
-1<x<0
x > 1

In all case x > -1 even tho it can't be 0<x<1.....altho I think the ques is lil weird.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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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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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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07 Feb 2005, 00:10
I believe we're debating solution B, which says x>-1 must be always true.

I can pick positive 1 as an integer that is greater than -1, but does not satisfy x/|x|<x.

If x = +1,
x/|x| = 1 which is not < 1. (1=1)

Now, if there's just one value of x > -1 that does not satisfy the inequality, then we can't say x>-1 is always true.
If you look at my previous post, I've not said I'm solving for x, i'm just stating values of x that hold true for the inequality. I came up with x<-1, and x>1.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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07 Feb 2005, 08:26
(A) cannot be correct, people.

Let me give you another example. Lets say we have two sets of integers.
S1=(0), S2=(2,3,4)
Now we know that x belongs to S1 or S2. Ask what must be true for x:
(A) x>1
(B) x>-1

(A) is obviously wrong
Because x could be an element of S1 or S2. If x belongs to S1, then it would be 0, and it is NOT greater than 1.
(B) is, however, correct
No matter which set x belongs, x is greater than -1.
You cannot say that well what about x=1? It satisfies x>-1, but is not in either S1 or S2. The mistake you are having here, is instead of A->B, you are trying to say B->A.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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07 Feb 2005, 09:53
My example is completely parallel to the question, only in the integer and limited form.

Look:
HongHu wrote:
(A) cannot be correct, people.

Let me give you another example. Lets say we have two sets of integers.
S1=(0), S2=(2,3,4)
Now we know that x belongs to S1 or S2. Ask what must be true for x:
(A) x>1
(B) x>-1

Compare it with:
S1=(-1,0), S2=(1,infinity)
We know x belongs to S1 or S2. What must be true for x?

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

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07 Feb 2005, 15:48
greenandwise wrote:
A must be the answer because it works in every case and therefore MUST be true.

One value of x could be x=-1/2
x/|x|=-1<-1/2

However this x is not greater than 1. So it is not always true that x>1.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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07 Feb 2005, 20:14
DLMD wrote:

if X <0, then X < X* (-X) ---------> X< -X^2 ------> 0 < -X^2-X ---------> 0 > X^2 + X ---------> X <0 or X <-1, so X <-1

The direction of an inequality sign would have to be changed when you multiply a negative number on both side.

For example,

-x<1
=>
x>-1
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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08 Feb 2005, 00:11
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Urgh, I just can't talk any sense into you guys. Read my examples!

A leads to B doesn't mean B leads to A!!! And if B doesn't lead to A it doesn't mean A doesn't lead to B! It's basic logic people!!!
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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08 Feb 2005, 10:31
DLMD wrote:

if X <0, then X < X* (-X) ---------> X< -X^2 ------> 0 < -X^2-X ---------> 0 > X^2 + X ---------> X <0 or X <-1, so X <-1

Question is x/|x|<x
if x<0, you multiply (-x) on both sides and change the direction of the inequality:
x > x*(-x)
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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01 Sep 2010, 04:16
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

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

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

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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.

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

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

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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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

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

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02 Sep 2010, 03:30
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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

What do you think about that?

hemanthp wrote:
Yup. IMO - A.

It fails in all other cases.
A. x>1
B. x>-1 => fails for any value between 0 and 1.
C. |x|<1 => fails for any value between 0 and 1.
D. |x|=1 => obviously fails.
E. |x|^2>1 => Fails for negative number less than -1. Take -4. -4/4 < -4 => FALSE.

What is the OA?

Thanks.

hemanthp wrote:
B makes no sense. Either the poster posted the question wrong or the choices wrong (the order probably).

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

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26 Oct 2010, 09:44
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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] 26 Oct 2010, 09:44

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