Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?
(1) x/|x|< x
(2) |x| > x
Will really appreciate if answer is supported by explanation.
\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))
(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 \(3\))
OR: \(\frac{x}{|x|}< x\) multiply both sides of inequality by \(|x|\) (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too) --> \(x<x|x|\) --> \(x(|x|-1)>0\):
Either \(x>0\) and \(|x|-1>0\), so \(x>1\) or \(x<-1\) --> \(x>1\);
Or \(x<0\) and \(|x|-1<0\), so \(-1<x<1\) --> \(-1<x<0\).
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 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 \(-1<x<1\). Sufficient.
Answer: C.
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?