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)
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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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Thank. Got hold in language.

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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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X / [X] < X


The Output of [X] (since X does NOT equal 0) will Always be (+)Pos.

Thus, we can Cross-Multiply Absolute Value of [X] across the Inequality without Flipping the Inequality Sign

X < [X] * X


Case 1: If X < 0

X < -(X) * X

X < - (X^2)

X^2 + X < 0

----taking X Common-----

(X) * (X + 1) < 0

(X - 0) * (X + 1) < 0


Critical Points: 0 and -1

-------------(-)1----------------- 0 -----------------------

X > 0: We are assuming that this Range does NOT Exist in this Scenario

(-)1 < X < 0:
for All Values of X in this Range ----> (X - 0) * (X + 1) < 0 -----> SATISFIED


X < (-1):
for All Values of X in this Range ----> (X - 0) * (X + 1) > 0 ----> NOT Satisfied


**Summary**
If X < 0:

Then: (-)1 < X < 0



Case 2: X > 0

X < [X] * X

X < (X) * X

X < (X)^2

0 < (X)^2 - X

X * (X -1) > 0

(X - 0) * (X - 1) > 0

Critical Values: 0 and +1

------------ 0 ---------------- +1 ------------

X > 1: For every Value in this Range ----> (X - 0) * (X - 1) > 0 ----> SATISFIED

0 < X < 1: For every Value in this Range -----> (X - 0) * (X - 1) < 0 -----> NOT Satisfied

X < 0: we are Assuming that this Range does NOT Exist in this Scenario


**SUMMARY**
If X > 0

then: X > +1




if X < 0 ------------> (-)1 < X < 0

if X > 0 ------------> +1 < X


Q Asks: W.O.T.F. Answer Choices MUST be a True Statement regarding EVERY Possible Value that X can take?

no matter which Value of X we pick within the 2 Possible Ranges above, we can ALWAYS say that the Value will be:

GREATER THAN > (-)1

because every (+)Pos. Value of X will be Greater Than > (-)1 as well


Answer is -B-

X > (-)1 must be True