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Re: Inequalities trick [#permalink]
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26 Jul 2012, 07:59
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Thankyou Karishma.
Further, say if the same expression was (x+2)(1x)/(x4)(x7) and still the question was for what values of x is the expression positive, then ... make it x1 and with the same roots, have the rightmost as ve. Then we look for the +ve intervals and check for those intervals if the expression is positive. for examples, in this case, 2<x<1 and 4<x<7 both depict positive interval but only first range satisfies the condition. Please confirm
However, if for the same equation as mentioned, say the expression was (x+2)(x1)/(x4)(x7) >0 and then we were asked to give the range where this is valid, then we would also multiply the ve sign and make is <0 and then make the range after extreme right root ve and provide all the intervals where it is negative. Please confirm



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Re: Inequalities trick [#permalink]
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26 Jul 2012, 09:47
VeritasPrepKarishma wrote: When you have (xa)^2(xb) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x  b) should be negative i.e. x  b < 0 or x < b. .
IMHO, it should be x<b and also x is not equal to a . so, one shouldn't totally ignore the squared term. We can ignore it, if the expression is <=0 correct me, if I am wrong my question is  do we always have a sequence of + and  from rightmost to the left side. I mean is it possible to have + and then + again ?
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Re: Inequalities trick [#permalink]
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26 Jul 2012, 21:57



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Re: Inequalities trick [#permalink]
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18 Oct 2012, 01:49
gurpreetsingh wrote: ulm wrote: in addition: if we have smth like (xa)^2(xb) we don't need to change a sign when jump over "a". yes even powers wont contribute to the inequality sign. But be wary of the root value of x=aHi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ? Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.



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Re: Inequalities trick [#permalink]
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18 Oct 2012, 05:17
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GMATBaumgartner wrote: gurpreetsingh wrote: ulm wrote: in addition: if we have smth like (xa)^2(xb) we don't need to change a sign when jump over "a". yes even powers wont contribute to the inequality sign. But be wary of the root value of x=aHi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ? Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. If the powers are even then the inequality won't be affected. eg if u have to find the range of values of x satisfying (xa)^2 *(xb)(xc) >0 just use (xb)*(xc) >0 because xa raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero.
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Re: Inequalities trick [#permalink]
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18 Oct 2012, 10:27
GMATBaumgartner wrote: Hi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ?
Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. In addition, you can check out this post: http://www.veritasprep.com/blog/2012/07 ... spartii/I have discussed how to handle powers in it.
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Re: Inequalities trick [#permalink]
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20 Nov 2012, 08:35
Kudos. Very good post.
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02 Dec 2012, 05:21
This is a great method but using this I am not getting an expected answer! Can Karishma and other please help me find where I am wrong?
The problem is (x  2)(x + 1)/x(x+2) <= 0 The graph will be +++ 2  1 +++++ 0  2 ++++++ So the answer will be 0 < x <= 2 and 2 < x < 1
But in Arun Sharma book, the answer given is 2 < x <= 2
Can you help?



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Re: Inequalities trick [#permalink]
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02 Dec 2012, 05:45
Bunuel wrote: GMATGURU1 wrote: This is a great method but using this I am not getting an expected answer! Can Karishma and other please help me find where I am wrong?
The problem is (x  2)(x + 1)/x(x+2) <= 0 The graph will be +++ 2  1 +++++ 0  2 ++++++ So the answer will be 0 < x <= 2 and 2 < x < 1
But in Arun Sharma book, the answer given is 2 < x <= 2
Can you help? Solution set for \(\frac{(x  2)(x + 1)}{x(x+2)}\leq{0}\) is \(2<x\leq{1}\) and \(0<x\leq{2}\). So, you've done everything right, except < sign for the second range, which should be <=. Yes even I think so that I am correct but in that book (famous Arun Sharma book for CAT), he writes the following... case 1) Numerator positive and denominator negative: This occurs only between 2 < x < 1 case 2) Numerator negative and denominator positive: Numerator is negative when (x  2) and (x + 1) take opposite signs. This can be got for ... case a) x  2 < 0 and x + 1 > 0 i.e. x < 2 and x > 1 case b) x  2 > 0 and x + 1 < 0 i.e. x > 2 and x < 1.  Cannot happen Hence answer is 2 < x <= 2.  The point I want to make here is, using our curve savy method we are getting a different answer. Using our answer x is not coming between 0 and 1. but the answer given in the book take the full range of 2 < x <= 2 So do we need any more tweaks in out curve method? Thanks!



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Re: Inequalities trick [#permalink]
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02 Dec 2012, 06:01
GMATGURU1 wrote: Bunuel wrote: GMATGURU1 wrote: This is a great method but using this I am not getting an expected answer! Can Karishma and other please help me find where I am wrong?
The problem is (x  2)(x + 1)/x(x+2) <= 0 The graph will be +++ 2  1 +++++ 0  2 ++++++ So the answer will be 0 < x <= 2 and 2 < x < 1
But in Arun Sharma book, the answer given is 2 < x <= 2
Can you help? Solution set for \(\frac{(x  2)(x + 1)}{x(x+2)}\leq{0}\) is \(2<x\leq{1}\) and \(0<x\leq{2}\). So, you've done everything right, except < sign for the second range, which should be <=. Yes even I think so that I am correct but in that book (famous Arun Sharma book for CAT), he writes the following... case 1) Numerator positive and denominator negative: This occurs only between 2 < x < 1 case 2) Numerator negative and denominator positive: Numerator is negative when (x  2) and (x + 1) take opposite signs. This can be got for ... case a) x  2 < 0 and x + 1 > 0 i.e. x < 2 and x > 1 case b) x  2 > 0 and x + 1 < 0 i.e. x > 2 and x < 1.  Cannot happen Hence answer is 2 < x <= 2.  The point I want to make here is, using our curve savy method we are getting a different answer. Using our answer x is not coming between 0 and 1. but the answer given in the book take the full range of 2 < x <= 2 So do we need any more tweaks in out curve method? Thanks! I'm not familiar with that source but with this particular question it's wrong. The numerator is positive and the denominator is negative: \(2<x\leq{1}\); The numerator is negative and the denominator positive: \(0<x\leq{2}\) (the source didn't consider the case when the denominator is positive). Jut to check, try x=1/2 to see that for this value the inequality does not hold true. Hope it's clear.
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Re: Inequalities trick [#permalink]
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02 Dec 2012, 06:36
@ Bunuel
Thanks! Yes I put 1/2 and it gives a positive value, so its wrong.



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Re: Inequalities trick [#permalink]
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02 Dec 2012, 06:39
Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny



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Re: Inequalities trick [#permalink]
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02 Dec 2012, 07:20
GMATGURU1 wrote: Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny I got the list, here it is: Problem solving OG 12: 12th Edition: 161, 173 QR 2nd Edition: 156 DS: 12th Edition: 97, 153, 162, D38 Quantitative Review: 66, 67, 85, 114 OR 2nd Edition: 68, 69, 89, 120 Thanks, Danny



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02 Dec 2012, 07:25
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02 Dec 2012, 07:37
Bunuel wrote: GMATGURU1 wrote: Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny Search for hundreds of question with solutions by tags: viewforumtags.phpDS questions on inequalities: search.php?search_id=tag&tag_id=184PS questions on inequalities: search.php?search_id=tag&tag_id=189Hardest DS inequality questions with detailed solutions: inequalityandabsolutevaluequestionsfrommycollection86939.htmlHope it helps. Wow!!! Thanks a lot!!! Regards, Danny



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Re: Inequalities trick [#permalink]
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05 Dec 2012, 10:24
Wow, GClub continues to amaze me. Some great tips and tricks indeed. Appreciate if Someone can give a REAL GMAT problem where this concept is used.
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Re: Inequalities trick [#permalink]
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20 Dec 2012, 05:18
Excellent post. Till now we have all seen problems in the format f(x) < 0 where f(x) is written in its factors for (xa)(xb)...
what if we have something like f(x) < k "k is a constant" (xa)(xb)(xc) < k How do we solve these kind of questions?



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Re: Inequalities trick [#permalink]
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20 Dec 2012, 07:53
maddyboiler wrote: Excellent post. Till now we have all seen problems in the format f(x) < 0 where f(x) is written in its factors for (xa)(xb)...
what if we have something like f(x) < k "k is a constant" (xa)(xb)(xc) < k How do we solve these kind of questions? Then u can not use this formula. You have to manipulate with values and as per the question.
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Re: Inequalities trick [#permalink]
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20 Dec 2012, 21:00
maddyboiler wrote: Excellent post. Till now we have all seen problems in the format f(x) < 0 where f(x) is written in its factors for (xa)(xb)...
what if we have something like f(x) < k "k is a constant" (xa)(xb)(xc) < k How do we solve these kind of questions? The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side. e.g. (x + 2)(x + 3) < 2 x^2 + 5x + 6  2 < 0 x^2 + 5x + 4 < 0 (x+4)(x+1) < 0 Now use the concept.
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