Marcab
If \(x\neq{0}\) and \(\frac{x}{|x|}<x\), which of the following must be true?
(A) \(x>1\)
(B) \(x>-1\)
(C) \(|x|<1\)
(D) \(|x|>1\)
(E) \(-1<x<0\)
Explanations required for this one.
Not convinced at all with the OA.
My range is -1<x<0 and x>1.
Notice that we are asked to find which of the options MUST be true, not COULD be true.
Let's see what ranges does \(\frac{x}{|x|}< x\) give for \(x\). Two cases:
If \(x<0\) then \(|x|=-x\), hence in this case we would have: \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that we consider the range \(x<0\), so \(-1<x<0\);
If \(x>0\) then \(|x|=x\), hence in this case we would have: \(\frac{x}{x}<x\) --> \(1<x\).
So, \(\frac{x}{|x|}< x\) means that \(-1<x<0\) or \(x>1\).Only option which is ALWAYS true is B. ANY \(x\) from the range \(-1<x<0\) or \(x>1\) will definitely be more the \(-1\).
Answer: B.
As for other options:
A. \(x>1\). Not necessarily true since \(x\) could be -0.5;
C. \(|x|<1\) --> \(-1<x<1\). Not necessarily true since \(x\) could be 2;
D. \(|x|>1\) --> \(x<-1\) or \(x>1\). Not necessarily true since \(x\) could be -0.5;
E. \(-1<x<0\). Not necessarily true since \(x\) could be 2.
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