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If x ≠ 0 and 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
Note: The question and the OA are 100% correct. To fully understand the reasoning, please study the discussion carefully. Read the expert explanations and check the linked similar questions for additional practice.
M09-22
A. x > 1
B. x > -1
C. |x| < 1
D. |x| = 1
E. |x|^2 > 1
Note: The question and the OA are 100% correct. To fully understand the reasoning, please study the discussion carefully. Read the expert explanations and check the linked similar questions for additional practice.
M09-22
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18 Jan 2010, 04:11
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nmohindru
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}xxxxxx
Now, we are asked which of the following must be true about \(x\). Option A cannot 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.
01 Sep 2010, 07:32
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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
We are given that \(\frac{x}{|x|}<x\) (this is a true inequality), so first of all we should find the ranges of \(x\) for which this inequality holds true.
\(\frac{x}{|x|}< x\) 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 \(-1<x<1\) --> \(-1<x<0\).
So we have that: \(-1<x<0\) or \(x>1\). Note \(x\) is ONLY from these ranges.
Option B says: \(x>-1\) --> ANY \(x\) from above two ranges would be more than -1, so B is always true.
Answer: B.
A. x>1
B. x>-1
C. |x|<1
D. |x|=1
E. |x|^2>1
We are given that \(\frac{x}{|x|}<x\) (this is a true inequality), so first of all we should find the ranges of \(x\) for which this inequality holds true.
\(\frac{x}{|x|}< x\) 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 \(-1<x<1\) --> \(-1<x<0\).
So we have that: \(-1<x<0\) or \(x>1\). Note \(x\) is ONLY from these ranges.
Option B says: \(x>-1\) --> ANY \(x\) from above two ranges would be more than -1, so B is always true.
Answer: B.
