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Which of the following represents the range for all x which [#permalink]
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02 Sep 2008, 23:11
Which of the following represents the range for all x which satisfy 1  x < 1 ?
A. (1, 1) B. (1, 2) C. (0, 1) D. (0, 2) E. (1, 2)



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Re: modulus [#permalink]
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02 Sep 2008, 23:58
arjtryarjtry wrote: Which of the following represents the range for all X which satisfy \(1  X \lt 1\) ?
* (1, 1) * (1, 2) * (0, 1) * (0, 2) * (1, 2)
guys, show the working in detail 1< 1x < 1 x<2, x> 0 D.



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Re: modulus [#permalink]
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03 Sep 2008, 06:46
arjtryarjtry wrote: alpha_plus_gamma wrote: arjtryarjtry wrote: Which of the following represents the range for all X which satisfy \(1  X \lt 1\) ?
* (1, 1) * (1, 2) * (0, 1) * (0, 2) * (1, 2)
guys, show the working in detail 1< 1x < 1........how????? i didnt get it x<2, x> 0 D. say 1x =y y<1 means > y values must be between 1 and 1 when y>0 > y<1 > 1x<1 > x>0 when y<0 > y<1 > y>1 > 1x>1 > x<2 (0,2) is it clear
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Re: modulus [#permalink]
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03 Sep 2008, 06:55
hmmm.... i went to the basics.. and it struck me suddenly thanks suresh... cos ur explanation reinforced what i thought. in maths when u discover something u feel like.... ..



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Re: modulus [#permalink]
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03 Sep 2008, 09:51
When ever you see modulo you have two possibilities
(1x) < 1 & (1x) < 1
If you solve both of them, 0<x<2 is what you will get.



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Re: modulus [#permalink]
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03 Sep 2008, 11:10
Yep,
x> 2 and x < 0.
so it is in between 0 and 2. so, (0,2).
So IMO is D



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Re: modulus [#permalink]
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03 Sep 2008, 12:00
Mathematicians usually think of absolute value as a measure of distance (more abstractly, a metric)
abs(ab) is the distance from a to b (or vice versa)
abs(b) = abs(b0) is the distance from b to zero
To find the values that satisfy the equation
abs(1x) <1
we want to find all values whose distance from 1 is less than 1.
Think visually using the number line and visualize all the values that are less than 1 unit from the number 1
(0,2) is obviously those values



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GMAT Club  m11#16 [#permalink]
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07 Sep 2010, 12:16
Which of the following represents the range for all \(X\) which satisfy \(1  X \lt 1\) ? (C) 2008 GMAT Club  m11#16 * (1, 1) * (1, 2) * (0, 1) * (0, 2) * (1, 2) Answer: Explanation: 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. My problem: 12=1 not smaller than 1 ... ???



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Re: GMAT Club  m11#16 [#permalink]
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07 Sep 2010, 12:30
AndreG wrote: Which of the following represents the range for all \(X\) which satisfy \(1  X \lt 1\) ? (C) 2008 GMAT Club  m11#16 * (1, 1) * (1, 2) * (0, 1) * (0, 2) * (1, 2) Answer: Explanation: 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. My problem: 12=1 not smaller than 1 ... ??? \(1x<{1}\) > key point is \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus two ranges to check: \(x<1\) > \(1x=1x\) and \(1x<{1}\) becomes: \(1x<{1}\) > \(x>0\); \(x\geq{1}\) > \(1x=1+x\) and \(1x<{1}\) becomes: \(1+x<{1}\) > \(x<2\); So \(1x<{1}\) 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. Hope it helps.
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Re: GMAT Club  m11#16 [#permalink]
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07 Sep 2010, 12:35
Bunuel wrote: AndreG wrote: Which of the following represents the range for all \(X\) which satisfy \(1  X \lt 1\) ? (C) 2008 GMAT Club  m11#16 * (1, 1) * (1, 2) * (0, 1) * (0, 2) * (1, 2) Answer: Explanation: 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. My problem: 12=1 not smaller than 1 ... ??? \(1x<{1}\) > key point is \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus two ranges to check: \(x<1\) > \(1x=1x\) and \(1x<{1}\) becomes: \(1x<{1}\) > \(x>0\); \(x\geq{1}\) > \(1x=1+x\) and \(1x<{1}\) becomes: \(1+x<{1}\) > \(x<2\); So \(1x<{1}\) 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 keypoint value is very helpful! Also, I was not aware, that if they talk about "range" the values given are excluded (hence \(0<x<2\) ) Thanks André



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Re: GMAT Club  m11#16 [#permalink]
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15 Jun 2011, 00:35
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 1x<=1
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Re: modulus [#permalink]
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26 May 2012, 12:32
I was able to get to the equation 2> x > 0 but it will be wrong to say (0,2) as this will include 0 and 2 as well where as x not equal to 0 or 2. Please comment
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