Bunuel wrote:
nmohindru wrote:
If \(\frac{x}{|x|}<x\) which of the following must be true about \(x\)?
(A) \(x>1\)
(B) \(x>-1\)
(C) \(|x|<1\)
(D) \(|x|=1\)
(E) \(|x|^2>1\)
This question was well explained by Durgesh and Ian Stewart, but since there are still some doubts, I'll try to add my 2 cents.
First of all let's solve this inequality step by step and see what is the solution for it, or in other words let's see in which ranges this inequality holds true.
Two cases for \(\frac{x}{|x|}<x\):
A. \(x<0\) --> \(|x|=-x\) --> \(\frac{x}{-x}<x\) --> \(-1<x\) --> \(-1<x<0\);
B. \(x>0\) --> \(|x|=x\) --> \(\frac{x}{x}<x\) --> \(1<x\).
So given inequality holds true in the ranges: \(-1<x<0\) and \(x>1\). Which means that \(x\) can take values only from these ranges.
------{-1}
xxxx{0}----{1}
xxxxxxNow, we are asked which of the following must be true about \(x\). Option A can not be ALWAYS true because \(x\) can be from the range \(-1<x<0\), eg \(-\frac{1}{2}\) and \(x=-\frac{1}{2}<1\).
Only option which is ALWAYS true is B. ANY \(x\) from the ranges \(-1<x<0\) and \(x>1\) will definitely be more the \(-1\), all "red", possible x-es are to the right of -1, which means that all possible x-es are more than -1.
Answer: B.
Bunuel just a small query here
just to stick to a standard, its better not to divide by a variable as we don't know its sign, could be positive or negative
so following that I did this sum , however was having difficulty at the end combining both the x ranges, please can you help
\(\frac{x}{|x|} < x\)
\(x<x|x|\)
\(x-x|x|<0\)
\(x(1-|x|)<0\)
so
first case
when \(x>0\) then \(1-|x|<0\)
so \(1-|x|<0 =|x|>1\)
\(|x| >1 = -1 > x > 1\)
so we have \(x>0\)and \(-1 > x > 1\)
now here is my concern how to combine these two to get the final range of \(x\)
I am having difficulty combining these two to get \(x>1\) , is there any technique ?
second case ( although this was easier to combine )
\(x(1-|x|)<0\)
when \(x<0\) then 1-|x|>0
so \(1-|x|>0 = 1 >|x|\)
\(|x|<1 = -1< x < 1\)
we have \(x<0\) and \(-1< x < 1\) now again for combining these two is there any standard way?
logically i can arrive at \(-1<x<0\)
When you consider x<0, then you automatically have that |x|=-x. Similarly, when you consider x>0, then you automatically have that |x|=x.
x<0 --> 1-(-x)>0 --> x>-1 --> -1<x<0.
x>0 --> 1-x<0 --> x>1.
So, the given inequality holds for -1<x<0. and x>1.