gmat6087 wrote:If 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

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.
Eliminate E, since it doesn't have to be true that |x|² > 1.

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

The correct answer is B.

Since both -1<x<0 and x>1 are to the right of -1, it must be true that x > -1.

gmat6087 wrote:If 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

Since the mod is always positive we multiply both sides by positive x and get x<x^2 for x distinct from zero. The viable solution lies within intervals on -1<x<0 and x>1, excluding segment [0;1] and interval x=<-1

a) X>1, limits to x>1 and excludes -1<x<0

b) X>-1, includes ostensibly zero too, i.e. x=0 but the division by zero isn't possible (x/|x|) hence we count this as possible choice.

c) |X| < 1, excludes the whole ranges of values x>1

d) |X| = 1, wrong as falls into x<x ?

e) |X|^2 > 1, limits only to segment (-1;1)

yea, this question is ambitious as the choice b) suggests impossible value x=0. I doubt such question would appear on actual GMAT.

Question stem says "what must be true about X"
and if -1<X is true then it includes 1/2 or 3/4(0<x<1) then inequality X/|X|<X does not hold, then how a thing which is not true can become "must be true".
for must be true, -1<x<0 and x>1 are separately correct.
If question stem would have been like for what values of X inequality is ALWAYS true then answer choice has to mention full range i.e. all values of X on order to be correct.

I overlooked in the previous post, the value of 0 and segment [0;1] are excluded from the solution on the number line -1<x. It's like the original statement X/|X| < X is true, and given it's true we find the range of values including all solutions of this inequality. On the number line we put restriction zones, as the first condition of original statement is always true, and then we decide about choice b) as correct.

See Img