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

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

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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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24 Sep 2011, 11:48
praveenvino wrote:
I am bit confused here. i agree the range of x, -1<x<0 or x>1. But doesn't x>-1 covers the range 0<x<1 also?

Yes it does. But question says, if we pick any number in the range -1<x<0(ex: -0.5,-0.25) or x>1(ex: 2,3,4) will that number be greater than -1 (x>-1). We can see that yes that number will be greater than -1.

praveenvino wrote:
So we have that: -1<x<0 or x>1. Note x is ONLY from these ranges.

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

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24 Sep 2011, 18:05
x/mod(x) < x

when x>=0 , mod(x) = x

=> given expression becomes x/x < x => 1<x => x>1 -------expression 1

when x<0, mod(x) = -x

=> given expression becomes x/(-x) < x => -1<x => x>-1--------expression 2

a. x>1 , need not be true

x <0 can be possible too which is not greater than 1.

b. x>-1 , must be true.

both the expressions 1 and 2 , x>1 and x>-1 means x is greater than -1 for sure.

c. mod(x) <1, need not be true

when x>=0, mod(x)<1 => x<1 ( this is not true from expression 1)

d. mod(x)= 1 , need not be true.

lets say x = -1/2 => mod(x) = 1/2 and this is not equal to 1.
e. mod(x)^2 > 1, need not be true.

lets say x= -1/2 => mod(x)^2 = 1/4 and this is not greater than 1.

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

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13 Aug 2012, 10:07
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Expert's post
stne wrote:
I also think answer should be A, if expert can confirm solution , it would be wonderful

OA for this question is B.

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

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24 Aug 2012, 15:04
+1 B

Solving the inequality, we have two ranges:
x > 1 or x > -1

Whatever the real value of x is, it is always higher than -1.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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04 Oct 2012, 07:38
HERE! the key is "must be true" and not could be or should be true. so, answer is B by above explanation of bunuel.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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

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12 May 2013, 09:11
The question ( must be true about x) asks about x not the solution of the inequality , thus we solve inequality and we see from answer choices if x always is inside our solution of the inequality

x-/x/*x < 0 , i.e x (1-/x/) <0 holds true in 2 cases

a) x+ve and /x/>1 , i.e. x+ve in the range x<-1 ( this is equivalent to x>1) or x>1 thus in this case x is always >1

b) x-ve and /x/<1 , i.e. x-ve and -1<x<1 ( from /x/<1) but since x is always -ve in this assumption therefore the range becomes -1<x<0

now we have 2 ranges that is x>1 and -1<x<0 now we check each answer choice vs. those ranges , x>-1 is always true ( must be true about x) in those ranges

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

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24 Mar 2014, 07:40
Bunuel wrote:
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.

Hi Bunuel,
Even I have the same questions, can you help clear the confusion with X = 1/2.

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

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24 Mar 2014, 07:51
seabhi wrote:
Hi Bunuel,
Even I have the same questions, 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.

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 Jun 2014, 21:11
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Buneul, you are saying that x>-1, suppose we substitute 0 in the inequality we get 0<0. How is this true????
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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03 Jun 2014, 01:18
Buneul, you are saying that x>-1, suppose we substitute 0 in the inequality we get 0<0. How is this true????

x=0 does not satisfy x/|x| < x, so x cannot be 0.

x/|x| < x, means that -1<x<0 or x>1. ANY x from these ranges will be greater than -1.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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15 Sep 2015, 01:18
I hope this helps.

From x/|x|<x, we know that :

1. x cannot be 0, since we can't divide a number by 0
2. x cannot be greater than 1, since we are dividing the number by the same number
3. x cannot be less than -1, since we are dividing the number by then same number

So we get, -1<x<1
From here, if we want x/|x|<x to be true, pick some numbers and you will see that only negative fraction works.
So it means,
-1<x<0

a) x>1 --> false
b) x>-1 --> correct
c) |x|<1 -->false, since x can be positive fraction
d) |x|=1 --> false, since x cannot be greater than 0
e) |x|^2>1 --> false, since squaring fraction gives you less than 0
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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27 Dec 2015, 12:48
Condition 1:

x > 0:

x/|x| = 1

x/|x| < x

1 < x

× -> (1, infinity]

Condition 2:

x < 0:

x/|x| = -1

-1 < x and x < 0

x -> (-1,0)

Conclusion: x -> (-1, 0) or (1, infinity]

A) x > 1 excludes values of x (-1, 0)
B) x > -1 includes all possible values of x. CORRECT
C) |x| < 1 covers (-1, 0) but not (1, infinity]
D) |x| = 1 incorrect as |x| != 1 in any circumstance.
E) |x|^2 >1 fails to cover values of x -> (-1, 0)
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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10 Mar 2016, 11:07
Excellent Question Here we just need to dilute the condition with remarks that x>0 => |x|=x and if x<0 => |x|=-x
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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22 Sep 2016, 01:15
Hi friends! Saw that mainly there is a dispute between answer as A) or B).

As per me, answer is A). Here is the reasoning why B) cannot be an answer ( x > -1).

Take x=0.5 for an example. Then, You'll get the equation not being satisfied... So, x> -1 isn't a possibility... Hence, A is the answer.
Thanks

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

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22 Sep 2016, 02:20
Uttunnu123 wrote:
Hi friends! Saw that mainly there is a dispute between answer as A) or B).

As per me, answer is A). Here is the reasoning why B) cannot be an answer ( x > -1).

Take x=0.5 for an example. Then, You'll get the equation not being satisfied... So, x> -1 isn't a possibility... Hence, A is the answer.
Thanks

Posted from my mobile device

There is no dispute. The correct answer is B. You can find several solutions on previous pages.
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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10 Oct 2016, 05:58
Ans. B
Reason:
x/|x|<x => x/|x| - x <0 => x(1-|x|)/|x| <0

For+ve x,
1-|x| must be less than 0 i.e. 1-|x| <0 => x>1

For-ve x,
1-|x|>0 => -1<x <0

From this we can say that in both cases, -1<x
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Re: x/|x|<x. which of the following must be true about x ? [#permalink]

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13 Oct 2016, 00:08
christoph wrote:
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 did it this way, please tell me if there is anything wrong in my approach?

On the LHS we have: x/|x|. This value will either be +1 or -1 depending on the value of x.

Therefore, 1<x or -1<x. The larger range of the two is 1-<x, hence I chose Option B as it covers both ranges.
Re: x/|x|<x. which of the following must be true about x ?   [#permalink] 13 Oct 2016, 00:08

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