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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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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.

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.

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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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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15 Aug 2014, 08:28
1
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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If x/|x|<x which of the following must be true about x?  [#permalink]

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

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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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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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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]

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14 Apr 2018, 01:04
Hi

The main point in this question is to understand what the question is asking.
which of the following must be true about x?

It means the question is asking about a Set of Values, which contains all the value of x which satisfy the given inequality.
But the Big Idea is that the set may contain other values also which does not satisfy the inequality.

It simply means the set must contains all the value of x satisfying the inequality but vice versa is not required.

only after solving the inequality as explained above,
-1<x<0 , when x is negative , or
x>1 when x is positive.

So option B x>-1 is the only the set of values which contains all the above required values of x.

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

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26 May 2018, 04:50
Its always a good idea to simply or manipulate the Question Stem

X divided by |X| will either give a "1" or "-1", depending on "sign of X"

Incase "1" then X > 1 and Incase "-1" X > "-1",

If you think on this X >1 might be true but not always true, but X > -1 will always be true.

Even if "X" is a fraction, X>-1 is true and this also matches our answer.
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Re: If x/|x|<x which of the following must be true about x?  [#permalink]

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08 Sep 2018, 15:50
If x=0, then

0/0 < 0

undefined < 0

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

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09 Sep 2018, 09:24
Quote:
Think again.
Every value greater than -1 need not satisfy the inequality but every value satisfying the inequality must be greater than -1.

Earlier I marked A as the answer, but this line made it crystal clear that the correct choice should be B
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Re: If x/|x|<x which of the following must be true about x?  [#permalink]

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06 Oct 2018, 05:00
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.

in the 1 case when x<0, why arent changing the sign of the inequality. i thought when we open a mod with a negative sign we change the sign of the inequality.
p.s- this question has jolted all my concepts of mod!
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Re: If x/|x|<x which of the following must be true about x?  [#permalink]

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12 Oct 2018, 06:53
Hi

We change the sign of inequality only when we divide or multiply by a negative number.

Jasveensingh wrote:

in the 1 case when x<0, why arent changing the sign of the inequality. i thought when we open a mod with a negative sign we change the sign of the inequality.
p.s- this question has jolted all my concepts of mod![/quote]
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Re: If x/|x|<x which of the following must be true about x?  [#permalink]

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08 Jan 2019, 19:16
x/lxl<x
=x/x<lxl
=1<lXl
or
lxl>1

which gives -1>x>1

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

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09 Jan 2019, 05:53
Did you read the above posts?

Rupesh1Nonly wrote:
x/lxl<x
=x/x<lxl
=1<lXl
or
lxl>1

which gives -1>x>1

how could it be "B"?

Posted from my mobile device
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Re: If x/|x|<x which of the following must be true about x? &nbs [#permalink] 09 Jan 2019, 05:53

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