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Re: Inequalities trick
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26 Jul 2012, 20:57
@pavanpuneet and Lalab Each of these scenarios is explained and all of your questions are answered in detail in my last few posts. Check these out: http://www.veritasprep.com/blog/2012/06 ... efactors/http://www.veritasprep.com/blog/2012/07 ... nsparti/http://www.veritasprep.com/blog/2012/07 ... spartii/
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Re: Inequalities trick
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18 Oct 2012, 00:49
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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18 Oct 2012, 04:17
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
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18 Oct 2012, 09:27
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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02 Dec 2012, 04:21
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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02 Dec 2012, 04:45
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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02 Dec 2012, 05:01
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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02 Dec 2012, 05:39
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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20 Dec 2012, 04:18
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
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20 Dec 2012, 20:00
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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20 Dec 2012, 20:04
VeritasPrepKarishma wrote: 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. Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy.
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09 Sep 2013, 13:42
gurpreetsingh wrote: I learnt this trick while I was in school and yesterday while solving one question I recalled. Its good if you guys use it 12 times to get used to it.
Suppose you have the inequality
f(x) = (xa)(xb)(xc)(xd) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "" curves and the ans is : (a < x < b) , (c < x < d) and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  +
If you can not figure out how and why, just remember it. Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively. Hi Gurpreet, Thanks for the wonderful method. I am trying to understand it so that i can apply it in tests. Can you help me in applying this method to the below expression to find range of x. x^3 – 4x^5 < 0? I am getting the roots as 1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.



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09 Sep 2013, 22:35
karannanda wrote: gurpreetsingh wrote: I learnt this trick while I was in school and yesterday while solving one question I recalled. Its good if you guys use it 12 times to get used to it.
Suppose you have the inequality
f(x) = (xa)(xb)(xc)(xd) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "" curves and the ans is : (a < x < b) , (c < x < d) and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  +
If you can not figure out how and why, just remember it. Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively. Hi Gurpreet, Thanks for the wonderful method. I am trying to understand it so that i can apply it in tests. Can you help me in applying this method to the below expression to find range of x. x^3 – 4x^5 < 0? I am getting the roots as 1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help. Before you apply the method, ensure that the factors are of the form (x  a)(x  b) etc \(x^3  4x^5 < 0\) \(x^3 ( 1  4x^2) < 0\) \(x^3(1  2x) (1 + 2x) < 0\) \(4x^3(x  1/2)(x + 1/2) > 0\) (Notice the flipped sign. We multiplied both sides by 1 to convert 1/2  x to x  1/2) Now the transition points are 0, 1/2 and 1/2 so put + in the rightmost region. The solution will be x > 1/2 or 1/2 < x< 0. Check out these posts discussing such complications: http://www.veritasprep.com/blog/2012/06 ... efactors/http://www.veritasprep.com/blog/2012/07 ... nsparti/http://www.veritasprep.com/blog/2012/07 ... spartii/
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Re: Inequalities trick
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15 Sep 2013, 20:12
VeritasPrepKarishma wrote: vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Yaxis for four values and starts from ve Yaxis for three values. What if the equation you mentioned is (x+2)(x1)(x7)<0,will the last two ranges be excluded or the graph will also change?
Ok, look at this expression inequality: (x+2)(x1)(x7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x  1) will be positive and (x7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when 2 , x < 1 and negative when x < 2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < 2. Now let me add another factor: (x+8)(x+2)(x1)(x7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and  signs to the regions. Your starting point is the rightmost region.Note: Make sure that the factors are of the form (ax  b), not (b  ax)... e.g. (x+2)(x1) (7  x)<0 Convert this to: (x+2)(x1) (x7)>0 (Multiply both sides by '1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. Responding to a pm: Quote: i have a doubt in the highlighted region. U have said that it will be always +ve but in a bunel post he has asked to substitute the extreme values and if the f(x) is ve then the right most of the inequality will be ve.
Please clarify me on this.
Different people use different methods of solving problems. Both of us are correct. But you cannot mix up the methods. When you follow one, you have to follow that through and through. When I say that the rightmost region will always be positive, it is after I make appropriate changes. Right below the highlighted portion, notice the note given: Note: Make sure that the factors are of the form (ax  b), not (b  ax)... I convert all factors to (ax  b) form. Now the rightmost region is positive by default. Bunuel prefers to keep the factors as it is and check for the rightmost region. What you would like to follow is your personal choice.
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11 Nov 2013, 05:14
Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3  0 < 0 (2) 1 a^2 > 0
Can you please explain the scenario when (xa)(xB)(xC)(xd)>0? Sorry, but finding it difficult to understand.



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11 Nov 2013, 06:43
Responding to a pm: Is a > O? (1) a^3  a < 0 (2) 1 a^2 > 0 (1) a^3  a < 0 a(a+1)(a  1) < 0 Points of transition are 1, 0 and 1. Make the wave. The expression is negative when 0 < a < 1 or a < 1. a could be positive or negative. Not sufficient. (2) 1 a^2 > 0 a^2  1 < 0 (multiplied the inequality by 1 which flipped the sign) (a1)(a+1) < 0 Points of transition are 1 and 1. Make the wave. The expression is negative when 1 < a < 1. a could be positive or negative. Not sufficient. Using both together, we see that only 0 < a < 1 is possible for both inequalities to hold. In this case a must be positive. Sufficient. Answer (C)
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11 Nov 2013, 06:51
anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3  0 < 0 (2) 1 a^2 > 0
Can you please explain the scenario when (xa)(xB)(xC)(xd)>0? Sorry, but finding it difficult to understand. If you have doubts in any particular steps given above, feel free to ask. Also, check out my posts on this method: http://www.veritasprep.com/blog/2012/06 ... efactors/http://www.veritasprep.com/blog/2012/07 ... nsparti/http://www.veritasprep.com/blog/2012/07 ... spartii/
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