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Inequalities trick
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Updated on: 26 Apr 2018, 03:34
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. Attachment:
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Originally posted by gurpreetsingh on 16 Mar 2010, 10:11.
Last edited by Bunuel on 26 Apr 2018, 03:34, edited 1 time in total.
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Re: Inequalities trick
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17 Oct 2010, 16:14
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.
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)
I don't understand this part alone. Can you please explain?
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Re: Inequalities trick
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17 Oct 2010, 17:19
Dreamy 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.
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)
I don't understand this part alone. Can you please explain? Suppose you have the inequality f(x) = (xa)(xb)(xc)(xd) < 0 you will consider the curve with ve inside it.. check the attached image. f(x) = (xa)(xb)(xc)(xd) > 0 consider the +ve of the curve
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Re: Inequalities trick
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22 Oct 2010, 06:33
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:
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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.
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Re: Inequalities trick
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22 Oct 2010, 11:45
if = sign is included with < then <= will be there in solution like for (x+2)(x1)(x7)(x4) <=0 the solution will be 2 <= x <= 1 or 4<= x <= 7
in case when factors are divided then the numerator will contain = sign
like for (x + 2)(x  1)/(x 4)(x  7) < =0 the solution will be 2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite
correct me if i am wrong



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Re: Inequalities trick
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11 Mar 2011, 06:29
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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Re: Inequalities trick
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11 Mar 2011, 06:49
vjsharma25 wrote: 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? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x1)(x7) < 0 Here the roots are; 2,1,7 Arrange them in ascending order; 2,1,7; These are three points where the wave will alternate. The ranges are; x<2 2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(10001)(10007) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7> ve between 2 and 1> +ve Before 2 > ve Since the inequality has the less than sign; consider only the ve side of the graph; 1<x<7 or x<2 is the complete range of x that satisfies the inequality.
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Re: Inequalities trick
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11 Mar 2011, 19:57
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.
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Re: Inequalities trick
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07 Aug 2011, 07:24
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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Re: Inequalities trick
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08 Aug 2011, 11:59
Asher 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=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. 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 As for roots, you have to keep in mind that given \(\sqrt{x}\), x cannot be negative. \(\sqrt{x}\) < 10 implies 0 < \(\sqrt{x}\) < 10 Squaring, 0 < x < 100 Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
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Re: Inequalities trick
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08 Aug 2011, 22:22
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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09 Aug 2011, 03:28
Asher wrote: 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 So when you have \((x  a)^2(x  b) < 0\), you ignore x = a and just plot x = b. It is positive in the rightmost region and negative on the left. So the graph looks like this: negative ... b ... positive 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} Squared terms are ignored. You do not put them in the graph. They are always positive so they do not change the sign of the expression. e.g. \((x4)^2(x  9)(x+11) < 0\) We do not plot x = 4 here, only x = 11 and x = 9. We start with the rightmost section as positive. So it looks something like this: positive... 11 ... negative ... 9 ... positive Since we need the region where x is negative, we get 11 < x < 9. Basically, the squared term is like a positive number in that it doesn't affect the sign of the expression. I would be happy to solve inequalities questions related to roots but please put them in a separate post and pm the link to me. That way, everybody can try them.
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Re: Inequalities trick
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10 Aug 2011, 07:00
hey , can u please tel me the solution for this ques
a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?
ans 11



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Re: Inequalities trick
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10 Aug 2011, 17:01
WoW  This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help. 1) CORE CONCEPT@gurpreetsingh  Suppose you have the inequality f(x) = (xa)(xb)(xc)(xd) < 0
Arrange the NUMBERS in ascending order from left to right. a<b<c<d 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. 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. Note: Make sure that the factors are of the form (ax  b), not (b  ax)...example  (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. 2) Variation  ODD/EVEN POWER@ulm/Karishma  if we have even powers like (xa)^2(xb) we don't need to change a sign when jump over "a". This will be same as (xb)We can ignore squares BUT SHOULD consider ODD powersexample  2.a (xa)^3(xb)<0 is the same as (xa)(xb) <0 2.b (x  a)(x  b)/(x  c)(x  d) < 0 ==> (x  a)(x  b)(xc)^1(xd)^1 <0 is the same as (x  a)(x  b)(x  c)(x  d) < 0 3) Variation <= in FRACTION@mrinal2100  if = sign is included with < then <= will be there in solutionlike for (x+2)(x1)(x7)(x4) <=0 the solution will be 2 <= x <= 1 or 4<= x <= 7 BUT if it is a fraction the denominator in the solution will not have = SIGNexample  3.a (x + 2)(x  1)/(x 4)(x  7) < =0 the solution will be 2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite 4) Variation  ROOTS @Karishma  As for roots, you have to keep in mind that given \(\sqrt{x}\), x cannot be negative. \(\sqrt{x}\) < 10 implies 0 < \(\sqrt{x}\) < 10 Squaring, 0 < x < 100 Root questions are specific. You have to be careful. If you have a particular question in mind, send it. Refer  inequalitiesandroots118619.html#p959939 Some more useful tips for ROOTS....I am too lazy to consolidate<5> THESIS  @gmat1220  Once algebra teacher told me  signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane. I will save this future references.... Please add anything that you feel will help. Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post
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Re: Inequalities trick
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11 Aug 2011, 22:57
sushantarora wrote: hey , can u please tel me the solution for this ques
a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?
ans 11 Please put questions in new posts. Put it in the same post only if it is totally related or a variation of the question that we are discussing. Now for the solution: There are 42 cars on the lot. 1/7 of sports cars and 1/2 of luxury cars have sunroofs. This means that 1/7 of number of sports cars and 1/2 of number of luxury cars should be integers (You cannot have 1.5 cars with sunroofs, right?) We want to minimize the sunroofs. Since 1/2 of luxury cars have sunroofs and only 1/7 of sports cars have them, it will be good to have fewer luxury cars and more sports cars. Best would be to have all sports cars. But, the question says there are some of each kind at any time. So let's say there are 2 luxury cars (since 1/2 of them should be an integer value). But 1/7 of 40 (the rest of the cars are sports cars) is not an integer number. Let's instead look for the multiple of 7 that is less than 42. The multiple of 7 that is less than 42 is 35. So we could have 35 sports cars. But then, 1/2 of 7 (since 42  35 = 7 are luxury cars) is not an integer. The next smaller multiple of 7 is 28. This works. 1/2 of 14 (since 42  28 = 14 are luxury cars) is 7. So we can have 14 luxury cars and 28 sports cars. That is the maximum number of sports cars that we can have. 1/7 of 28 sports cars = 4 cars have sunroofs 1/2 of 14 luxury cars = 7 cars have sunroofs So at least 11 cars will have sunroofs.
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Re: Inequalities trick
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22 Jul 2012, 03:03
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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Re: Inequalities trick
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23 Jul 2012, 03:13
Stiv wrote: 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. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined.
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Re: Inequalities trick
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25 Jul 2012, 22:21
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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Re: Inequalities trick
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26 Jul 2012, 07:59
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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Re: Inequalities trick
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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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