if-x-3-1-which-of-the-following-must-be-true-41133.html#p1193086
_________________

I have deleted my earlier posts as there is point to agree with the OA -- ALL the values satisfying the given inequality also satisfies the range in option B. All other options are wrong as there are some x's which will satisfy x / |x| < x but will not be in that range. Examples are:
A) x>1 is not true for x = -0.5
C) |x|<1 is not true for x = 2
D) |x|>1 is not true for x = -0.5
E) −1<x<0 is not true for x = 2
I admit that I read the question in incorrect way and from that I went to a different direction.

For \(x>0\) you get \(x>1\) --> \(x>1\) not 0<x<1.
_________________

The two ranges which you have got are absolutely correct :

-1<x<0 OR 1<x

Now, which trait for any value of x, which subscribes to either of the range is ALWAYS true?

That's x>-1. Having said that, it is NOT necessary that any value of x>-1 MUST satisfy the given inequality.

However,any value of x satisfying the given inequality WILL be greater than -1.
_________________

Thank you for that link. Unfortunately I am still a bit lost.

When we take the negative case (x<0) we have -1<x therefore we join x<0 and -1<x thus: -1<x<0

When we take the positive case (x>0) we have 1<x I would assume we join the two cases thus: x>0, x>1

In your link you made reference to the AND/OR concepts and I am vaguely familiar with them but I am still a bit confused as to why we count the range of the negative cases (which makes sense because we are given that x falls within -1 and 0) but we don't count any # between zero and one but we do count numbers greater than one (is it because only values that fall within both x>0 and X>1 count? And if so, why?)

Thanks!

If x doesn't equal 0 and (x)/|x|<x, which of the following must be true?

(I made the mistake of assuming that all values of x would make the statement true [i.e. what is the value of (x)/|x|<x] as opposed to looking what values of x are possible given the constraint of (x)/|x|<x)

So: (x)/|x|<x

x/x<x 1<x
OR
x/-x<x -1<x

(A) x>1 x is greater than 1. This lies entirely outside of the desired range. INVALID

(B) x>-1 As long as x>-1 it will fall within EITHER desired range of x>-1 or x>1 VALID

(C) |x|<1 x<1 OR x<-1 X may fall in the range (less than one) but it may also fall out of it (x<-1). INVALID

(D) |x|>1 x>1 OR x<-1 Both values of x fall entirely out of the desired range. INVALID

(E) -1<x<0

I'm still a little unsure as to why (E) isn't valid...I think it's because (E) states that x MUST be between -1 and 0 when in fact x simply has to be greater than -1.

Is that correct?

Question: if \(-1<x<0\) or \(x>1\), then which of the following must be true? Notice that \(-1<x<0\) or \(x>1\) is given to be true: x is either from {-1, 0} or from {1, +infinity}

Only option which is ALWAYS true is B. ANY \(x\) from the range \(-1<x<0\) or \(x>1\) will definitely be more the \(-1\). (Complete solution is here: if-x-0-and-x-x-x-which-of-the-following-must-be-true-143572.html#p1150594)

The following posts might help:
if-x-0-and-x-x-x-which-of-the-following-must-be-true-143572.html#p1150609
if-x-0-and-x-x-x-which-of-the-following-must-be-true-143572.html#p1181573

All Must or Could be True Questions to practice: search.php?search_id=tag&tag_id=193

Hope it helps.
_________________

I think it is though it is a little hard to understand. Let me try to write it clearly:

x((1/|x|)-1)<0

Two cases:

Case 1: x < 0 AND (1/|x|-1) > 0
From (1/|x|-1) > 0 we get that |x| < 1 i.e. -1 < x < 1
So combining, we get -1 < x < 0

Case 2: x > 0 AND (1/|x|-1) < 0
From (1/|x|-1) < 0 we get that |x| > 1 i.e. x > 1 or x < -1
So combining, we get x > 1

x could lie between -1 and 0 or it could be more than 1. Hence in any case, it will be more than -1.
_________________

Hi, I tried solving the problem by multiplying both sides by |x| since |x| will always be positive. This makes the equation as x<x|x|.
Case 1: x>0 => x<sq(x)=>x>1
Case 2:x<0=>x<-sq(x)=>x<-1

I believe I am going wrong in multiplying both sides by |x|, however, I fail to understand how since |x| will always be positive and hence I could multiply it.

Please assist.

Thanks

Here is the problem:
Case 2:x<0=>x<-sq(x)=>x<-1
If x is negative, |x| = -x

x < x|x| becomes x < -x^2
Now if you divide both sides by x, remember that x is negative so inequality sign will flip.

1 > -x

Multiply both sides by -1 and that will flip the sign again.

-1 < x

So you get that x should be greater than -1 in case x is negative.

So in any case, x will always be greater than -1.
_________________