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If \(X \ge 1\) , then the inequality turns into \(X - 1 \lt 1\) or \(X \lt 2\) . If \(X \lt 1\) , then the inequality turns into \(1 - X \lt 1\) or \(X \gt 0\) . Combine the intervals to get the answer.

If \(X \ge 1\) , then the inequality turns into \(X - 1 \lt 1\) or \(X \lt 2\) . If \(X \lt 1\) , then the inequality turns into \(1 - X \lt 1\) or \(X \gt 0\) . Combine the intervals to get the answer.

\(|1-x|<{1}\) --> key point is \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus two ranges to check:

\(x<1\) --> \(|1-x|=1-x\) and \(|1-x|<{1}\) becomes: \(1-x<{1}\) --> \(x>0\); \(x\geq{1}\) --> \(|1-x|=-1+x\) and \(|1-x|<{1}\) becomes: \(-1+x<{1}\) --> \(x<2\);

So \(|1-x|<{1}\) holds true for \(0<x<2\).

Answer: D.

As for your question: \(x\) can not equal to 2, because \(0<x<2\) means that \(x\) MUST be less than 2 (and more than zero), for ANY \(x\) from this range given inequality will hold true.

I guess (0,2) should be changed to \(0<x<2\) (as well as all other options) to avoid confusion whether 0 and 2 are inclusive in the range.

If \(X \ge 1\) , then the inequality turns into \(X - 1 \lt 1\) or \(X \lt 2\) . If \(X \lt 1\) , then the inequality turns into \(1 - X \lt 1\) or \(X \gt 0\) . Combine the intervals to get the answer.

\(|1-x|<{1}\) --> key point is \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus two ranges to check:

\(x<1\) --> \(|1-x|=1-x\) and \(|1-x|<{1}\) becomes: \(1-x<{1}\) --> \(x>0\); \(x\geq{1}\) --> \(|1-x|=-1+x\) and \(|1-x|<{1}\) becomes: \(-1+x<{1}\) --> \(x<2\);

So \(|1-x|<{1}\) holds true for \(0<x<2\).

Answer: D.

As for your question: \(x\) can not equal to 2, because \(0<x<2\) means that \(x\) MUST be less than 2 (and more than zero), for ANY \(x\) from this range given inequality will hold true.

Hope it helps.

Thanks for the quick reply! Your explanation with the key-point value is very helpful!

Also, I was not aware, that if they talk about "range" the values given are excluded (hence \(0<x<2\) )

Re: GMAT Club - m11#16 [#permalink]
14 Jun 2011, 23:35

quoting bunuel :

" I guess (0,2) should be changed to 0<x<2 (as well as all other options) to avoid confusion whether 0 and 2 are inclusive in the range. "

This correction is needed IMO. when we say (0,2) we mean than the numbers are included. Also another alternative would be to change the question to |1-x|<=1 _________________

Cheers !!

Quant 47-Striving for 50 Verbal 34-Striving for 40