If x 0 and x/|x| < x, which of the following must be true?

We can simplify the given inequality:

x/|x| < x

x < (x)|x|

(x)|x| > x

If x is positive, we can divide both sides by x and obtain |x| > 1.

If x is negative, we can also divide both sides by x, but we have to switch the inequality sign, so we have |x| < 1.

We see that if x is positive, |x| > 1, which is choice C, and if x is negative, |x| < 1, which is choice D. However, since we don’t know whether x is positive or negative, both choice C and choice D “can be true,” not “must be true.”

Let’s analyze further. If x is positive, |x| = x. So, |x| > 1 means x > 1, which is choice A. If x is negative, |x| = -x. So, |x| < 1 means -x < 1 or x > -1. However, because x is negative, we have -1 < x < 0, which is choice E. Again, since we don’t know whether x is positive or negative, both choice A and choice E “can be true,” not “must be true.”

This leaves choice B as the correct answer. In fact, it’s the correct choice because the inequality x > -1 includes both x > 1 and -1 < x < 0. So, regardless of whether x is positive or negative, we can say x > -1.

Answer: B
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If in case I multiply |x| both sides then inequality will not change and then if i approach like this then how the inequality should be solved :-

X<X|X|
x|x|-x>0
x(|x|-1)>0
now either x>0 or |x|>1,x<0 |x|<1

how to proceed further to solve it to get the range as per the question.

\(x(|x|-1)>0\)

Case 1: \(x > 0\) and \(|x| > 1\) (\(x < -1\) or \(x > 1\)) --> \(x > 1\).

Case 2: \(x < 0\) and \(|x| < 1\) (\(-1 < x < 1\)) --> \(-1 < x < 0\).

Finally, \(-1 < x < 0\) or \(x > 1\).
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Hello Karishma, I understand everything from this Q. I just have one question.

You can agree that we can say that: x<|x|x since |x|>0 and then reach the same solution by writing x(|x|-1)>0.

I have learned to not divide x/x<x --> x>1 since I lose choices.

In which situation do I deduce that x>1 and in which situation I must write that x^2-x>0 .. and continue from there?

Yes, you can deduce
\(x * (|x|-1)>0\)

For the product of the two factors to be positive, either both are positive or both are negative.

EITHER
x > 0 and |x| - 1 > 0 (which is |x| > 1 ---> x > 1 or x < -1)
So x > 1

OR
x< 0 and |x| - 1 < 0 (which is |x| < 1 --> -1 < x < 1 )
So -1 < x < 0

So any value that x can take will either lie in -1 to 0 to will be greater than 1. So every value will be greater than -1 for sure.

Answer (B)

Bunuel

For the case of x<0, you didn't flip the inequality sign, i.e. x/-x < x. I can follow that the answer works out at the end. Conceptually, when should we not flip the sign when we consider the negative case in inequalities problems?

We get -1 < x from \(\frac{x}{-x} < x\) because x/(-x) = -1, so as you can see we don't reduce the WHOLE inequality by anything.
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x/|x| < x

x < x|x|

0 < x|x| - x

0 < x (|x| - 1)

The CRITICAL POINTS are -1, 0 and 1.
These are the only values where x(|x|-1) = 0.
To determine the ranges where x(|x|-1) > 0, test one value to the left and right of each critical point.

Plug x = -2 into x/|x| < x:
-2/ |-2| < -2
-1 < -2.
Doesn't work.
x < -1 is not a valid range.

Plug x = -1/2 into x/|x| < x:
-1/2/ |-1/2| < -1/2
-1 < -1/2.
This works.
-1<x<0 is a valid range.

Plug x = 1/2 into x/|x| < x:
(1/2)/ |1/2| < 1/2
1 < 1/2
Doesn't work.
0<x<1 is not a valid range.

Plug x = 2 into x/|x| < x:
2/ |2| < 2
1 < 2.
This works
x > 1 is a valid range.

Thus, the valid ranges are -1<x<0 and x>1.

Since it's possible that x=-1/2:
Eliminate A, since it doesn't have to be true that x>1.
Eliminate D, since it doesn't have to be true that |x|>1.

Since it's possible that x=2:
Eliminate C, since it doesn't have to be true that |x| < 1.
Eliminate E, since it doesn't have to be true that -1<x<0.

.

Since both -1<x<0 and x>1 are to the right of -1, it must be true that x > -1.
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Hi I am sure I must be missing something here. But what about values between 0<x<1, they will come inclusive of x> -1

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If we plug \(x=0.5\) into \(\frac{x}{|x|} < x\), we get:
\(\frac{0.5}{|0.5|} < 0.5\)
\(1 < 0.5\)
Doesn't work.
Thus, \(0<x<1\) is not a valid range.

However, any value in the two valid ranges (\(-1<0<x\) and \(x>1\)) will be greater than -1, so it must be true that x>-1.
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But in x> -1 ; 0<x<1 comes under a solution range

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If x/|x| < x and x is nonzero, which of the following must be true?
OA: x > -1

The OA does NOT imply the following:
Every value greater than -1 will satisfy x/|x| < x.
The statement in red is invalid, since values between 0 and 1 do not satisfy x/|x| < x.
The OA implies the following:
Every value that satisfies x/|x| < x is greater than -1.
The statement in green is correct, since any value selected from the two valid ranges (-1<x<0 and x>1) will be greater than -1.
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With the final of range x: (1, infinity) & (-1,0) ...no doubt here...

But, it is worth debating on selecting the right option.

B cannot be correct, for the obvious reasons every other person above is trying to debate. Had this been an IIT JEE question, B would never be correct! It's really strange that experts have been trying to justify option B.

Rather option E holds true and will always be correct.

It seems that you just don't understand the question. Let me assure you that the answer is 100% correct and every mathematician in the world would agree on answer B. Not to repeat myself and others in this thread, I can only suggest you to try other similar questions here: Trickiest Inequality Questions Type: Confusing Ranges. Hope it helps.
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Hi VeritasKarishma,

Thank you very much for explaining this question in detail.
Yes, I indeed had a hard time to understand that why Option B is correct (because I was always referencing the values of x, back to the inequality and discarded Option: B).

I have the below queries:
1) Does such a scenario occurs only on these kinds of questions: MUST OR COULD BE TRUE QUESTIONS?
2) If I understand correctly, we just have to make sure that our answer set (range of values) is completely covered in the correct option choice in such questions? And there will be only 1 such (correct) option in PS.
3) We must understand the fact that one implies the other i.e. the answer holds true (x > -1) for the given inequality. But the reverse isn't true i.e. (every value of x > -1), doesn't satisfy the given inequality. The reverse may or may not be true. In this question, as discussed above, the reverse doesn't hold true.

I am yet to practice these type of questions, will surely come back if I have any other queries.
TIA.

Regards,
Ravish.

Check here:
https://anaprep.com/algebra-must-be-tru ... questions/
https://anaprep.com/algebra-game-must-b ... questions/

You will get clarity on this concept.

This question tests your basic understanding of modulus and how you apply the basic definition of modulus in this expression x/|x|.

Let's start by analyzing how x/|x| works in the different ranges of x (x≠0).


Case 1: When x > 0, |x | = x

Hence , x/|x| = x/x =1
Applying the same in the inequality, x/|x|<x

=> 1 < x

=> x > 1

So we can conclude that in case 1 ( when x > 0) , any value of x > 1 will satisfy the inequality.

Case 2: When x < 0, |x | = -x

Hence , x/|x| = x/-x = -1

Applying the same in the inequality, x/|x|<x

=> -1 < x

=> x > -1

As we are considering only the values of x < 0 in case 2 , we can conclude that any values of x in the range -1 < x < 0 , will satisfy the inequality.

Combining both cases , the range of x are -1 < x < 0 and x > 1.


Since it's a must-be type question, we need to analyze each answer option and figure out which options would satisfy both the range of x.

It is the time to have some fun

(A) x>1

Eliminate Option A as it is not true for the values of x in the range -1 < x < 0

(B) x>−1
Any values of x in the range -1 < x < 0 and x > 1 will satisfy the inequality x > -1 . Hence Option B must be true.

(C) |x|<1 ==> -1 < x < 1
Eliminate as it's only true for the values of x between -1 and 0. But, it is not true for the values of x > 1.

(D) |x|>1 => x < -1 or x > 1
Its not true for the values of x in the range -1 < x < 0 .So, eliminated.


(E) −1<x<0
x >1 is not included here. Hence, eliminated.

Option B is the correct answer.

The above-explained would be the ideal way to handle this question. Alternate approaches can also be used to find the range of x. For example.

Approach 2:

x/|x| < x
x < x|x| since |x| is positive, multiplying on both sides of the inequality will not change the sign of the inequality.

x - x|x| < 0
x (1- |x|) < 0 Note: It's always better to represent inequality as a product of two terms, whenever possible. So that we can have an idea about their signs.


Since the product of two terms is less than 0, we can conclude that one of the product terms should be positive and the other one should be negative.
Let's consider this conditions as 2 cases.

Case 1: x > 0 and 1- |x| < 0
i.e., x >0 and |x| >1 .

When x > 0 , |x| = x
So, we can conclude that x>1

Case 2: x< 0 and 1-|x| >0

i.e. x<0 and |x| < 1
When x < 0 , |x| = -x
=> x<0 and -x < 1 => x >-1
Therefore, -1 < x < 0

So, combining both cases, the range of x will be -1 < x < 0 and x>1


I hope, this explanation helps.

Thanks,
Clifin J Francis
GMAT QUANT SME
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Bunuel why can't we consider mod x as positive x ? and directly solve since mod of any value is always positive ?' pls help me

Hi Bunuel Bunuel,

Referring to the highlighted portion

If \(x<0\) then \(|x|=-x\), hence in this case we would have: \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that we consider the range \(x<0\), so \(-1<x<0\);


if I multiply -x with x, then my inequality sign will reverse and new equation would be
\(\frac{x}{-x}<x\) --> x>-x2 --> x+x2 >0 --> x(1+x) >0 --> x<-1

But your solution is coming as x>-1
can you tell me where I am going wrong? Is it wrong that while multiplying a negative number, inequality sign gets reversed?