Bunuel wrote:

If x is not equal to 0, is |x| less than 1?Is \(|x|<1\)?

Is \(-1<x<1\)? (\(x\neq{0}\))

So, the question asks whether x is in the range shown below:

(1) \(\frac{x}{|x|}< x\)

Two cases:

A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values:

1 for \(x>0\) --> so we would have: \(1<x\);

Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\):

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or consider two cases again:\(x<0\)--> \(-x>x\)--> \(x<0\).

\(x>0\) --> \(x>x\): never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\)

(\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)):

Every \(x\) from this range is definitely in the range \(-1<x<1\). So, we have a definite YES answer to the question. Sufficient.

Answer: C.

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Hi

BunuelSurely i am missing something here..my basic doubt is in scenario when we take X<0

If we assume, x<0, then shouldn't the statement x/|x|<x = -x/-x <-x (my thinking here is if we considering x<0,

then x should be negative throughout )

So, the equation becomes 1<-x = -1>x, we can keep this option as we have initially assumed X<0

Then if we take any values of x less then -1, |x| will always be greater than 1

if we assume X>0, then x/|x|<x = x/x<x.....which is 1<x, we can keep this option as initially we have assumed X>0

Then we take any values of x greater than 1 then |x| will always be greater than 1.

So statement 1 in either case is sufficient..

Thanks in advance for your help..