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 | = xHence , x/|x| = x/x =1
Applying the same in the inequality, x/|x|<x
=> 1 < x
=>
x > 1So 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 | = -xHence , 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