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.
Attachment:
MSP340910303aagbh2c14gf0000420i437if9d12bhg.gif
Attachment:
MSP2380151735322d2i2d790000280i6giebh17fdf3.gif
Attachment:
MSP8506173e0f30207c4a2700005eg6g59c48d51i5b.gif
Attachment:
MSP111401047g43cgaehag920000640367fgf148e668.gif
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..
Negative x does not mean that you should replace x with -x. x just represents a negative number. You should replace |x| with -x, though because if x < 0, then |x| = -x.