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Re: Inequalities trick
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11 Mar 2011, 05:29
VeritasPrepKarishma wrote: vjsharma25 wrote:
How you have decided on the first sign of the graph?Why it is ve if it has three factors and +ve when four factors?
Check out my post above for explanation. 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?



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07 Aug 2011, 06:24
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=a This way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma However, i am confused about how to solve inequalities such as: (xa)^2(xb) and also ones with root value. could someone please explain.



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08 Aug 2011, 21:22
Quote: 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.
Similarly for (xa)^2(xb) > 0, x > b Thanks Karishma for the explanation. Hope you wouldn't mind clarifying a few more doubts. Firstly, in the above case, since x>b could we say that everything with be positive. would the graph look something like this: positive.. b..postive.. a.. positive On the other hand if (xa)^2(xb) < 0, x < b, (xa)^2 would be positive and for (xb) if x<b the the left side would be negative. would the graph look something like this: negative.. b..postive.. a.. postive Am i right? If it is not too much of a trouble, could you please show the graphical representation. problems with \sqrt{x}.. this is all i could find (googled actually ): 1. √(x+4) ≤ √(x) 2. x^\sqrt{x} =< (\sqrt{x})^x {P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems}



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Re: Inequalities trick
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22 Jul 2012, 02:03
VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation\(\frac {(x+2)(x1)}{(x4)(x7)}\) were \(4\leq x \leq 7\), than the lower part would be equal to zero,thus making it impossible to calculate the whole equation.



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25 Jul 2012, 21:21
pavanpuneet wrote: Hi Karishma,
Just for my reference, say if the equation was (x+2)(x1)/(x4)(x7) and the question was for what values of x is this expression >0, then the roots will be 2,1,4,7 and by placing on the number line and making the extreme right as positive...
(2)(1)(4)(7)then x>7, 1<x<4 and x<2...Please confirm.. However, is say it was >=0 then x>7, 1<=x<4 and x<=2; given that the denominator cannot be zero. Please confirm Yes, you are right in both the cases. Also, if you want to verify that the range you have got is correct, just plug in some values to see. Put x = 0, the expression is ve. Put x = 2, the expression is positive.
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26 Jul 2012, 06:59
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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26 Jul 2012, 08:47
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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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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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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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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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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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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22 Apr 2014, 23:42
VeritasPrepKarishma wrote: Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain. If (xa)(xb)(xc)(xd) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x1)(x7)(x4) < 0, draw the points 2, 1, 7 and 4 on the number line as shown. Attachment: doc.jpg This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive. When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x  7) will be negative since x < 7 in this region. All other factors are still positive. When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x  7) and (x  4) are negative, but negative x negative is positive... and so on till you reach the leftmost section. Since we are looking for values of x where the expression is < 0, here the solution will be 2 < x < 1 or 4< x < 7 It should be obvious that it will also work in cases where factors are divided. e.g. (x  a)(x  b)/(x  c)(x  d) < 0 (x + 2)(x  1)/(x 4)(x  7) < 0 will have exactly the same solution as above. Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion. Hi Karishma, I have a query. I have following question x^3  4x^5 < 0 I can define this as (1+2x).x^3(12x). now I have roots 1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of 1/2 and 0 I will get value as 0. So How I will define them in graph and what range I will consider for this inequality. Thanks




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