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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

1

This post received KUDOS

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.

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.

_________________

My dad once said to me: Son, nothing succeeds like success.

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 ?
_________________

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?
_________________

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

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.

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.

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

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.

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 \(\frac{x}{|x|}<x\).

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

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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