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Re: Inequalities trick
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04 May 2015, 20:29
Hi Guys, I'm in the process of absorbing the fundamentals and neat tricks provided by the experts here. This inequality trick is phenomenal. Would it be safe to conclude that  for odd number of factors or roots, we start with "" (cosine waveform) and for even number of roots we start with "+" (sine wave)?
As explained by Gurpreet & Karishma: If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  +
Similarly: If f(x) has two factors then the graph will have +  + If f(x) has seven factors then the graph will have  +  +  +  + ?
Thanks!



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04 May 2015, 20:55
In addition to my previous post, I'd like to understand:
If the inequality is: f(x) = (x+a)(xb)^2(x+c)(xd)^3 < 0 We are still going to work with the same number line from left to right starting with "+" and factors a,b,c,d placed on it in ascending order? Since the number of factors remain an even number (4 total). Am I on the right track?



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Re: Inequalities trick
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04 May 2015, 21:20
rohitd80 wrote: Hi Guys, I'm in the process of absorbing the fundamentals and neat tricks provided by the experts here. This inequality trick is phenomenal. Would it be safe to conclude that  for odd number of factors or roots, we start with "" (cosine waveform) and for even number of roots we start with "+" (sine wave)?
As explained by Gurpreet & Karishma: If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  +
Similarly: If f(x) has two factors then the graph will have +  + If f(x) has seven factors then the graph will have  +  +  +  + ?
Thanks! Yes, you can look at it like that too. You should always have the positive sign at the extreme right region. Do remember though that the factors must be of the form (x  a), (x  b) etc and not of the form (a  x).
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Re: Inequalities trick
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04 May 2015, 21:25
rohitd80 wrote: In addition to my previous post, I'd like to understand:
If the inequality is: f(x) = (x+a)(xb)^2(x+c)(xd)^3 < 0 We are still going to work with the same number line from left to right starting with "+" and factors a,b,c,d placed on it in ascending order? Since the number of factors remain an even number (4 total). Am I on the right track? No, you will ignore the factor with the even power since that factor will never be negative. You should go through all three of these posts to understand exactly why this happens: 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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04 May 2015, 21:42
rohitd80 wrote: Hi Guys, I'm in the process of absorbing the fundamentals and neat tricks provided by the experts here. This inequality trick is phenomenal. Would it be safe to conclude that  for odd number of factors or roots, we start with "" (cosine waveform) and for even number of roots we start with "+" (sine wave)?
As explained by Gurpreet & Karishma: If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  +
Similarly: If f(x) has two factors then the graph will have +  + If f(x) has seven factors then the graph will have  +  +  +  + ?
Thanks! Dear rohitd80, That's a good observation! However, remember that your inference will hold true when: 1. the factors of f(x) are of the form (xa), (xb)... and so on. (i.e., not of the form (ax)(bx).... If they are of this form, then again you need to make proper adjustments based on the number of factors). AND2. Also, the powers of all the factors are odd. Even if one of the factors has an even power, your patterns (+  +, +++, etc.) will fail to hold true. Below is an example. If \(f(x) = (x  1)^2 * (x  3)\)and we need to find the regions where f(x) > 0, we first draw the wavy line starting from the top right corner. It crosses the "xaxis" at x=3 and enters the negative region. However, at x=1, since there are two roots at the same place, it bounces back into the negative region. As you can see, f(x) > 0 only when x > 3 Notice that if a student doesn't pay attention to the even power of roots, then he/she is prone to draw an incorrect wavy line and therefore arrive at an incorrect answer. You can refer to this post to understand how to draw wavy lines when multiple roots are present. inequalitiestrick9148280.html#p1465609Hope this helps. Regards, Krishna
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Re: Inequalities trick
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27 Aug 2015, 02:15
Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C.



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Re: Inequalities trick
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27 Aug 2015, 03:50
vihavivi wrote: Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C. But it does. Everything you have done is correct (ignoring the formatting issues).
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Re: Inequalities trick
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27 Aug 2015, 08:03
VeritasPrepKarishma wrote: vihavivi wrote: Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C. But it does. Everything you have done is correct (ignoring the formatting issues). I knew the answer makes sense, but if I apply the inequalities trick into this, with 3 roots 1, 0, and 1 and for the f(x) >0, shouldn't it be x <1 and 0 <x<1? Thanks



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Re: Inequalities trick
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27 Aug 2015, 16:33
vihavivi wrote: VeritasPrepKarishma wrote: vihavivi wrote: Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C. But it does. Everything you have done is correct (ignoring the formatting issues). I knew the answer makes sense, but if I apply the inequalities trick into this, with 3 roots 1, 0, and 1 and for the f(x) >0, shouldn't it be x <1 and 0 <x<1? Thanks Hi Vihavivi, I think you have your inequalities curve flipped upside down. When you have the roots plotted on your line, it should always open up positive on the right and everything alternates from there. There was a previous post on the first page by fluke where he suggest plugging in large numbers to intuitively highlight what's going on. In your case being \(x^9(x+1)(x1)\) just choose say x=1000... In this case all three components will be positiive...\((1000^x)(10001)(1000+1)\) will surely yield a positive answer. Use this as a reminder that in these cases the inequalities line will open up positive to the right. Have a look at my freehand below. If you find this useful maybe throw me my first Kudos???
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Re: Inequalities trick
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27 Aug 2015, 19:56
vihavivi wrote: VeritasPrepKarishma wrote: vihavivi wrote: Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C. But it does. Everything you have done is correct (ignoring the formatting issues). I knew the answer makes sense, but if I apply the inequalities trick into this, with 3 roots 1, 0, and 1 and for the f(x) >0, shouldn't it be x <1 and 0 <x<1? Thanks Remember, the rightmost region is always the positive region. That is how you start assigning  from the right. Of course, keep in mind that the inequality must be of the form (xa)*(xb)... and not of the form (a  x)*(xb)... (x+1)x^9(x−1)>0 > roots are 1, 0 and 1 So to the right of 1 will be positive and between 1 and 0, it will be positive. So x > 1 or 1 < x < 0
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Re: Inequalities trick
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24 Sep 2015, 05:59
Dear All, Is x a negative number? (1) 9x > 10x (2) is positive. GMAT OG claims A is sufficient. But, what if x is a fraction number? Such as 2/90. In this case 2/90 also satisfies ST 1 .. What is the exact solution for this?



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Re: Inequalities trick
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24 Sep 2015, 06:18
mrslee wrote: Dear All, Is x a negative number? (1) 9x > 10x (2) is positive. GMAT OG claims A is sufficient. But, what if x is a fraction number? Such as 2/90. In this case 2/90 also satisfies ST 1 .. What is the exact solution for this? This question is discussed here: isxanegativenumber136882.htmlHope it helps.
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Re: Inequalities trick
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24 Sep 2015, 22:07
mrslee wrote: Dear All, Is x a negative number? (1) 9x > 10x (2) is positive. GMAT OG claims A is sufficient. But, what if x is a fraction number? Such as 2/90. In this case 2/90 also satisfies ST 1 .. What is the exact solution for this? 2/90 does not satisfy statement 1. No positive value satisfies statement 1. 9*2/90 > 10*2/90 18/90 > 20/90  this is false
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Re: Inequalities trick
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07 Nov 2015, 00:12
if (k – 5)(k – 1)(k – 6) < 0 or (k – 5)(k – 1)(k – 6) > 0 what will be range for both? I have tried method suggested by gurpreetsingh, but got confused when to take ">" and "<". can anyone explain? It will be more helpful if anyone can upload an image of the solution. Thanks.



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Re: Inequalities trick
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07 Nov 2015, 22:34
Manali888 wrote: if (k – 5)(k – 1)(k – 6) < 0 or (k – 5)(k – 1)(k – 6) > 0 what will be range for both? I have tried method suggested by gurpreetsingh, but got confused when to take ">" and "<". can anyone explain? It will be more helpful if anyone can upload an image of the solution. Thanks. See the attached solution. Hope its clear!
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Re: Inequalities trick
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06 Mar 2016, 06:35
lets assume we have a,b,c,d and e in ascending order as follows:
(xa)(xb)(xc)(xd)<0
As number of factors are EVEN so for any x<a (i.e. the starting sign) we have (ve)*(ve)*(ive)*(ive) = +ve Then alternate signs will follow.
(xa)(xb)(xc)(xd)(xe)<0
As number of factors are ODD so for any x<a (i.e. the starting sign) we have (ve)*(ve)*(ive)*(ive)*(ve) = ve Then alternate signs will follow.
The same is true for for any number of factors in any form i.e. Numerator, Denominator, Combination of both.



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Re: Inequalities trick
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13 Mar 2016, 00:28
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. Please explain how will it be different when you have <= or >=



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Re: Inequalities trick
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14 Mar 2016, 00:32
RohitPrakash88 wrote: 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. Please explain how will it be different when you have <= or >= \((x+2)(x1)(x7)(x4) < 0\) gives \(2 < x < 1\) or \(4 < x < 7\) \((x+2)(x1)(x7)(x4) <= 0\) gives \(2 <= x <= 1\) or \(4 <= x <= 7\) x can be equal to each transition point such that the expression will take the value 0 in that case. The range is pretty much the same as above. With division, things are a bit different. \(\frac{(x+2)(x1)}{(x7)(x4)} < 0\) gives \(2 < x < 1\) or \(4< x < 7\) \(\frac{(x+2)(x1)}{(x7)(x4)} <= 0\) gives \(2 <= x <= 1\) or \(4< x < 7\) Note that here x cannot be 4 or 7 because then you will get a 0 in the denominator which is not acceptable.
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Re: Inequalities trick
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28 Jul 2016, 16:48
VeritasPrepKarishma wrote: 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. Dear Ma'am! Please try to respond to my pm... really needed that help, if possible then revert on forum so that others can benefit for it too. thanks



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Re: Inequalities trick
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28 Jul 2016, 21:01
Celestial09 wrote: Dear Ma'am! Please try to respond to my pm... really needed that help, if possible then revert on forum so that others can benefit for it too. thanks Celestial09, When you send me a pm, I automatically get a notification. You don't need to separately request me to check the pm on the forum. Check your messages  I have responded to your query.
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