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

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

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New post 07 Jul 2014, 00:32
Bunuel wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for \(\frac{x}{|x|}<x\):

A. \(x<0\) --> \(|x|=-x\) --> \(\frac{x}{-x}<x\) --> \(-1<x\) --> \(-1<x<0\);

B. \(x>0\) --> \(|x|=x\) --> \(\frac{x}{x}<x\) --> \(1<x\).

So given inequality holds true in the ranges: \(-1<x<0\) and \(x>1\). Which means that \(x\) can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about \(x\). Option A can not be ALWAYS true because \(x\) can be from the range \(-1<x<0\), eg \(-\frac{1}{2}\) and \(x=-\frac{1}{2}<1\).

Only option which is ALWAYS true is B. ANY \(x\) from the ranges \(-1<x<0\) and \(x>1\) will definitely be more the \(-1\), all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.


Bunnel, if I were to multiply the original stem with |x| (since |x| is always positive) it would result in x*(|x|-1) > 0.
This would mean x > 0 and |x| > 1
|x| > 1 would lead to x < -1 and x > 1 . This is completely different from the answer you've reached. I see that your method is accurate and the answer justified, but can you please correct my method here.
Thanks in advance!
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 07 Jul 2014, 00:41
Kconfused wrote:
Bunuel wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for \(\frac{x}{|x|}<x\):

A. \(x<0\) --> \(|x|=-x\) --> \(\frac{x}{-x}<x\) --> \(-1<x\) --> \(-1<x<0\);

B. \(x>0\) --> \(|x|=x\) --> \(\frac{x}{x}<x\) --> \(1<x\).

So given inequality holds true in the ranges: \(-1<x<0\) and \(x>1\). Which means that \(x\) can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about \(x\). Option A can not be ALWAYS true because \(x\) can be from the range \(-1<x<0\), eg \(-\frac{1}{2}\) and \(x=-\frac{1}{2}<1\).

Only option which is ALWAYS true is B. ANY \(x\) from the ranges \(-1<x<0\) and \(x>1\) will definitely be more the \(-1\), all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.


Bunnel, if I were to multiply the original stem with |x| (since |x| is always positive) it would result in x*(|x|-1) > 0.
This would mean x > 0 and |x| > 1
|x| > 1 would lead to x < -1 and x > 1 . This is completely different from the answer you've reached. I see that your method is accurate and the answer justified, but can you please correct my method here.
Thanks in advance!


It would give the same answer.

\(x*(|x|-1) > 0\). This implies that both multiples have the same sign.

\(x>0\) and \(|x|>1\) (since we consider positive x, then this transforms to x>1) --> \(x>1\).
\(x<0\) and \(|x|<1\) (since we consider negative x, then this transforms to -x<1 --> -1<x) --> \(-1<x<0\).

The same ranges as in my solution.

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

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New post 07 Jul 2014, 00:43
Bunnel, if I were to multiply the original stem with |x| (since |x| is always positive) it would result in x*(|x|-1) > 0.
This would mean x > 0 and |x| > 1
|x| > 1 would lead to x < -1 and x > 1 . This is completely different from the answer you've reached. I see that your method is accurate and the answer justified, but can you please correct my method here.
Thanks in advance![/quote]

It would give the same answer.

\(x*(|x|-1) > 0\). This implies that both multiples have the same sign.

\(x>0\) and \(|x|>1\) (since we consider positive x, then this transforms to x>1) --> \(x>1\).
\(x<0\) and \(|x|<1\) (since we consider negative x, then this transforms to -x<1 --> -1<x) --> \(-1<x<0\).

The same ranges as in my solution.

Hope it's clear.[/quote]


Thank you! I figured the answer just after posting the question. :)
I ignored considering the part where both the terms are negative.

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

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New post 15 Aug 2014, 06:50
Bunuel,

What if we square the inequality x/ |x| < x. Then we get (x^2 / x ) < x^2 which implies that x^2 < x^3. Is this correct? Please explain. Thanku
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If x/|x|<x which of the following must be true about x? [#permalink]

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New post 15 Aug 2014, 08:28
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sri30kanth wrote:
Bunuel,

What if we square the inequality x/ |x| < x. Then we get (x^2 / x ) < x^2 which implies that x^2 < x^3. Is this correct? Please explain. Thanku


We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality), which is not the case here.

Also, the second step in your solution: never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know the sign of it, we don't know the sign of x, so we cannot multiply x^2/x < x^2 by x here.

For more check here: inequalities-tips-and-hints-175001.html
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 16 Aug 2014, 21:25
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


I disagree with OA .According to me the Answer should be A.
OA is wrong for insistence take x=1/2 (since 1/2 > -1) but the inequality does not hold true for this value, so how can the answer be correct.
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 17 Aug 2014, 02:58
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dransa wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


I disagree with OA .According to me the Answer should be A.
OA is wrong for insistence take x=1/2 (since 1/2 > -1) but the inequality does not hold true for this value, so how can the answer be correct.


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

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New post 27 Nov 2014, 11:25
Bunuel wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.

First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.

Two cases for \(\frac{x}{|x|}<x\):

A. \(x<0\) --> \(|x|=-x\) --> \(\frac{x}{-x}<x\) --> \(-1<x\) --> \(-1<x<0\);

B. \(x>0\) --> \(|x|=x\) --> \(\frac{x}{x}<x\) --> \(1<x\).

So given inequality holds true in the ranges: \(-1<x<0\) and \(x>1\). Which means that \(x\) can take values only from these ranges.

------{-1}xxxx{0}----{1}xxxxxx

Now, we are asked which of the following must be true about \(x\). Option A can not be ALWAYS true because \(x\) can be from the range \(-1<x<0\), eg \(-\frac{1}{2}\) and \(x=-\frac{1}{2}<1\).

Only option which is ALWAYS true is B. ANY \(x\) from the ranges \(-1<x<0\) and \(x>1\) will definitely be more the \(-1\), all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.

Answer: B.



After forming the inequalities 'x>1' & 0>x>-1 ...it really took some minutes for me to get to the answer. The reason is when we think about answers, we always think ..if x>-1...then 0, 1/2, 1 etc are also greater than -1 but 0, 1/2, 1 can't be the answers..so how will x>-1 be the answer. BUT THE QUESTION IS TRICKY MUST BE TRUE QUESTION...so when all other options fails to accommodate one of the two derived inequalities either x>1 or 0>x>-1 ....we are left only with option B x>-1 which accommodates both x>1 & 0>x>-1.
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 27 Nov 2014, 12:07
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dransa wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


I disagree with OA .According to me the Answer should be A.
OA is wrong for insistence take x=1/2 (since 1/2 > -1) but the inequality does not hold true for this value, so how can the answer be correct.



For the ones who might get confused.

There are two solutions for the inequality.
No1. X>1
No2. 0>x>-1
So on the number line -1 to 0 [and then after the gap 0 to 1 which doesn’t belong to the solution set] 1 to infinity [on the positive side] is the solution.

Choice A is ruled out [doesn’t include the negative side -1 to 0]
Choice C is ruled for 2,3 etc are not less than 1
Choice D is ruled out for 1 & -1 are not at all in the solution.
Choice E is ruled out for [-1/2]^2 gives ¼ which is not greater than 1.
Choice B says x>-1 , though all numbers greater than -1 are not solutions of the inequality, all Xs greater than -1 [i.e. -1/2, -2/3, 2,3,4 etc] are in fact greater than -1 only. Therefore the statement that all Xs [the solution numbers in the number line 0>x>-1 & x>1] are greater than -1 is absolutely true [ Its a must be true question] . Therefore B is the answer.
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If x/|x|<x which of the following must be true about x? [#permalink]

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New post 30 Nov 2014, 07:44
As Bunnel states.
Two cases for \frac{x}{|x|}<x:

A. x<0 --> |x|=-x --> \frac{x}{-x}<x --> -1<x --> -1<x<0;

B. x>0 --> |x|=x --> \frac{x}{x}<x --> 1<x.

Concept is the absolute value of -5 equals 5, or, in mathematical
symbols, I-51 = 5.
From above A.
x<0 mean x is negative Assume x = -1
lxl = l -x l becz x is negative /positive = X is always positive
But not |x|=-x ?????
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 30 Nov 2014, 21:08
kanusha wrote:
x<0 mean x is negative Assume x = -1
lxl = l -x l becz x is negative /positive = X is always positive
But not |x|=-x ?????



You assumed x = -1
You got |x| = 1

Is |x| = x? No. |x| is 1 but x is -1
Then what is |x| in terms of x?

|x| = -x
1 = -(-1) = 1

That is why you say that |x| = -x when x is negative because then -x becomes positive.
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 23 Feb 2015, 00:53
So, since we know that the absolute value is positive, we can multiply both sides by abs(x) without having to change the sign.

x<x * |x| This means that x has to be greater than 1 or in between -1 and 0. You can figure this out from intuition or by testing number. 0 and -1 don't work because that would make both sides equal.

We are looking for something that must be true, so if we can find a scenario for x that works outside the given parameters, we can eliminate it right away.
A) doesn't have to be true, because x could because -1/2 works for x
B) does have to be true there is no value for x that works and is below -1
C) doesn't have to be true, because -1/2 works
D) doesn't have to be true, because x=1 doesn't even work
E) doesn't have to be true because -1/2 works
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If x/|x|<x which of the following must be true about x? [#permalink]

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New post 02 Mar 2015, 12:52
First I solved the inequation \(\frac{{x}}{{|x|}}<x\)
If x>0; \(\frac{{x}}{{|x|}}=1<x\); so x>1 (when x>0, which is redundant)
If x>0; \(\frac{{x}}{{|x|}}=-1<x\); so -1<x<0
The inequation has its solution in (-1;0) and (1;∞).

After solving the inequation the problem is to understand which of the options is a mandatory condition that is met by all the possible solutions of the inequation.

a) \(X>1\)
Doesn't have to be true because there are solutions for X<1 in the interval (-1;0): Incorrect

b) \(X>-1\)
Both intervals meet this condition, so it is required to solve the inequation: correct

c) \(|X| < 1\)
Doesn't have to be true because there are solutions for |X| > 1 in the interval (-1;): Incorrect

d) \(|X| = 1\)
No solution of the inequation meets this condition. Incorrect

e) \(|X|^2 > 1\)
This equals to say that |X|>1, and the solutions for the interval (-1;0) do not meet this requirement. Incorrect
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 06 Apr 2015, 16:16
Hi Bunuel, I have a doubt about a concept I do need some clarification.
it is about this question if-x-x-x-which-of-the-following-must-be-true-about-83605.html

the particular concern i have is about this statement :
Quote:
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 --> but as we are told that x is an integer then this range is out as there is no integer in this range."


I do understand how we factor out the given inequality but after that you say x>0 and /x/-1>0 and x<0 and /x/-1<0 , so my question is why we have fliped the sign or what is the logic and when we have to do that. I have learned that when there is absolute inequality we have for example /x+1/>2 and to solve this we need to get rid of the absolute value by x+1>2 and -(x+1)>2 . So if you will can you briefly explain what Im missing here.

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

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New post 06 Apr 2015, 17:53
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)

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

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New post 11 Oct 2015, 12:49
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


I have a little doubt here. The question asks which among the options are true? Consider option B, if x assumes 0, then the whole expression takes an undefined form. However, for option A , all the values holds true. So according to me A must be the answer. Please explain why B is correct
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 11 Oct 2015, 13:26
shreekantkhaitan wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?

(A) \(x>1\)

(B) \(x>-1\)

(C) \(|x|<1\)

(D) \(|x|=1\)

(E) \(|x|^2>1\)


I have a little doubt here. The question asks which among the options are true? Consider option B, if x assumes 0, then the whole expression takes an undefined form. However, for option A , all the values holds true. So according to me A must be the answer. Please explain why B is correct


To clear your doubts please read previous 4 pages of discussion.
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 01 Feb 2016, 11:08
IMO, the question can be interpreted as, "All possible values of x satisfy which of the below conditions". Only B gives the answer. For the answer to be A, the question would be worded a bit differently.
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 08 Feb 2016, 23:29
CAN I SOLVE THE QUESTION IN THIS WAY..
x/|x|<x---->|x|/X>1/X----->|x|>1

Two Cases---> X>1 or X<-1

since,

|ax+b|>s--->ax+b>1 or ax+b<-1
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Re: If x/|x|<x which of the following must be true about x? [#permalink]

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New post 09 Feb 2016, 02:52
akshay4gmat wrote:
CAN I SOLVE THE QUESTION IN THIS WAY..
x/|x|<x---->|x|/X>1/X----->|x|>1

Two Cases---> X>1 or X<-1

since,

|ax+b|>s--->ax+b>1 or ax+b<-1



You cannot cancel off x's from the denominator without knowing the sign of x.
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Re: If x/|x|<x which of the following must be true about x?   [#permalink] 09 Feb 2016, 02:52

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