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15 Aug 2017, 00:39
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Thanks a lot for the collection Bunuel, I've been struggling with Inequalities, Probability & Combinatorics from so long. This post is very helpful to strengthen my base in Inequalities.
Do we have any such comprehensive thread for Probability and Combinatorics?



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15 Aug 2017, 00:52



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16 Aug 2017, 00:02
nikhilpoddar wrote: What is the range for a ??
a + a^(1) >2 a+1/a>2 (a1)^2 / a >0 So a > 1 What is the OA?



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16 Aug 2017, 00:38
Bunuel wrote: srikanth9502 wrote: nikhilpoddar wrote: What is the range for a ??
a + a^(1) >2 a+1/a>2 (a1)^2 / a >0 So a > 1 What is the OA? a + 1/a > 2 holds true for 0 < a < 1 and a > 1. Is there any specified approach to solve these kind of questions?



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14 Oct 2017, 14:48
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WilDThiNg wrote: Bunuel wrote: Inequalities with Complications  Part I Above we learned how to handle inequalities with many factors i.e. inequalities of the form \((x – a)(x – b)(x – c)(x – d) > 0\). This week, let’s see what happens in cases where the inequality is not of this form but can be manipulated and converted to this form. We will look at how to handle various complications. Complication No. 1: \((a – x)(x – b)(x – c)(x – d) > 0\) We want our inequality to be of the form \((x – a)\), not \((a – x)\) because according to the logic we discussed last week, when x is greater than a, we want this factor to be positive. The manipulation involved is pretty simple: \((a – x) = (x – a)\) So we get: \(– (x – a)(x – b)(x – c)(x – d) > 0\) But how do we handle the negative sign in the beginning of the expression? We want the values of x for which the negative of this expression should be positive. Therefore, we basically want the value of x for which this expression itself (without the negative sign in the beginning) is negative. We can manipulate the inequality to \((x – a)(x – b)(x – c)(x – d) < 0\) Or simply, multiply \(– (x – a)(x – b)(x – c)(x – d) > 0\) by 1 on both sides. The inequality sign flips and you get \((x – a)(x – b)(x – c)(x – d) < 0\) e.g. Given: \((4 – x)(2 – x)(9 – x) < 0\) We can rewrite this as \(–(x – 4)(2 – x)(9 – x) < 0\) \((x – 4)(x – 2)(9 – x) < 0\) \((x – 4)(x – 2)(x + 9) < 0\) \((x – 4)(x – 2)(x – (9)) > 0\) (multiplying both sides by 1) Now the inequality is in the desired form. Complication No 2: \((mx – a)(x – b)(x – c)(x – d) > 0\) (where m is a positive constant) How do we bring \((mx – a)\) to the form \((x – k)\)? By taking m common! \((mx – a) = m(x – a/m)\) The constant does not affect the sign of the expression so we don’t have to worry about it. e.g. Given: \((2x – 3)(x – 4) < 0\) We can rewrite this as \(2(x – \frac{3}{2})(x – 4) < 0\) When considering the values of x for which the expression is negative, 2 has no role to play since it is just a positive constant. Now let’s look at a question involving both these complications. Question 1: Find the range of x for which the given inequality holds.\(2x^3 + 17x^2 – 30x > 0\) Solution:Given: \(2x^3 + 17x^2 – 30x > 0\) \(x(2x^2 + 17x – 30) > 0\) (taking x common) \(x(2x – 5)(6 – x) > 0\) (factoring the quadratic) \(2x(x – \frac{5}{2})(1)(x – 6) > 0\) (take 2 common) \(2(x – 0)(x – \frac{5}{2})(x – 6) < 0\) (multiply both sides by 1) This inequality is in the required form. Let’s draw it on the number line. We are looking for negative value of the expression. Look at the ranges where we have the negative sign. The ranges where the expression gives us negative values are \(5/2 < x < 6\) and x < 0. Hence, the inequality is satisfied if x lies in the range \(\frac{5}{2} < x < 6\) or in the range \(x < 0\). Plug in some values lying in these ranges to confirm. In the next post, we will look at some more variations which can be brought into this form. Hi: Please could you help me understand how to determine which sign to assign to which range in the above graph? Thanks as per my understanding the extreme right one you can take +ve & then assign an alternate sign to each region. Try yourself with a few equations. You can clarify that yourself. cheers
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Re: Inequalities Made Easy! [#permalink]
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28 Dec 2017, 05:16
Bunuel wrote: Or Just Use Inequalities! If you are wondering about the absurd title of this post, just take a look at the above post's title. It will make much more sense thereafter. This post is a continuation of last week’s post where we discussed number plugging. Today, as per students’ request, we will look at the inequalities approach to the same official question. You will need to go through our inequalities post to understand the method we will use here. Recall that, given \(a < b\), \((x – a)(x – b) < 0\) gives us the range \(a < x < b\) and \((x – a)(x – b) > 0\) gives us the range \(x < a\) or \(x > b\). Question: If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2? (I) x^2 < 2x < 1/x (II) x^2 < 1/x < 2x (III) 2x < x^2 < 1/x(A) none (B) I only (C) III only (D) I and II (E) I, II and III Solution: The question has three complex inequalities. We will take each in turn. Note that each inequality consists of two more inequalities. We will split the complex inequality into two simpler inequalities e.g. x^2 < 2x < 1/x gives us x^2 < 2x and 2x < 1/x. Next we will find the range of values of x which satisfy each of these two inequalities and we will see if the two ranges have an overlap i.e. whether there are any values of x which satisfy both these simpler inequalities. If there are, it means there are values of x which satisfy the entire complex inequality too. Things will become clearer once we start working on it so hold on. Let’s look at each inequality in turn. We start with the first one: (I) x^2 < 2x < 1/x We split it into two inequalities: (i) x^2 < 2x We can rewrite x^2 < 2x as x^2 – 2x < 0 or x(x – 2) < 0. We know the range of x for such inequalities can be easily found using the curve on the number line. This will give us 0 < x < 2. (ii) 2x < 1/x It can be rewritten as x^2 – 1/2 < 0 (Note that since x must be positive, we can easily multiply both sides of the inequality with x) [ Secondly would you explain this step again i did not understand how multiplying by x gives us this equation] \(1/?2< x < 1/?2\) This gives us the range 1 /?2 < x < 1/ ?2 (which is 0 < x < 1/?2 since x must be positive). [ Question mark is square root i think] Is there a region of overlap in these two ranges i.e. can both inequalities hold simultaneously for some values of x? Yes, they can hold for 0 < x < 1/?2. Hence, x^2 < 2x < 1/x will be true for the range 0 < x < 1/?2. So this could be the correct ordering. Let’s go on to the next complex inequality. (II) x^2 < 1/x < 2x Again, let’s break up the inequality into two parts: (i) x^2 < 1/x x^1 < 1/x is rewritten as x^3 – 1 < 0 which gives us x < 1. (ii) 1/x < 2x 1/x < 2x is rewritten as x^2 – 1/2 > 0 which gives us x < 1/?2 (not possible since x must be positive) or x > 1/?2 Can both x < 1 and x > 1/?2 hold simultaneously? Sure! For 1/?2 < x < 1, both inequalities will hold and hence x^2 < 1/x < 2x will be true. So this could be the correct ordering too. (III) 2x < x^2 < 1/x The inequalities here are: (i) 2x < x^2 2x < x^2 can be rewritten as x(x – 2) > 0 which gives us x < 0 (not possible) or x > 2. (ii) x^2 < 1/x x^2 < 1/x gives us x^3 – 1 < 0 i.e. x < 1 Can x be less than 1 and greater than 2 simultaneously? No. Therefore, 2x < x^2 < 1/x cannot be the correct ordering. Answer (D) Is this method simpler? Hi bunuel (Or any other expert) i have highlighted a portion in green and red . Could you clarify those points. Thankyou



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28 Dec 2017, 06:32
mtk10 wrote: Hi bunuel (Or any other expert) i have highlighted a portion in green and red . Could you clarify those points. Thankyou (ii)\(2x < \frac{1}{x}\)... Multiply by x.. \(2x*x < \frac{1}{x}*x.........2x^2<1..........x^2<\frac{1}{2}.....x^21/2<0\)...
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05 Apr 2018, 18:50
Bunuel wrote: Plug Using Transition Points Let’s discuss the concept of transition points. This is especially useful in questions where you are tempted to plug in values. A question often asked is: how do I know which values to plug and how do I know that I have covered the entire range in the 34 values I have tried? What transition points do is that they give you the ranges in which the relationships differ. All you have to do is try one value from each range. If you do, you would have figured out all the different relationships that can hold. We will discuss this concept using a GMAT Prep question. You can solve it using our discussion on inequalities too. But if number plugging is what comes first to your mind in this question, then it will be a good idea to get the transition points. Let’s begin: Question: If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\) and \(x^2\)? (I) \(x^2 < 2x < \frac{1}{x}\) (II) \(x^2 < \frac{1}{x} < 2x\) (III) \(2x < x^2 < \frac{1}{x}\)(A) none (B) I only (C) III only (D) I and II (E) I, II and III Solution:Notice that the question says “could be correct ordering”. This means that for different values of x, different orderings could hold. We need to find the one (or two or three) which will not hold in any case. So what do we do? We cannot try every value that x can take so how do we know for sure that one or more of these relations cannot hold? What if we try 45 values and only one relation holds for all of those values? Can we say for sure that the other two relations will not hold for any value of x? No, we cannot since we haven’t tried all values of x. So there are two options you have in this case: 1. Use logic to figure out which relations can hold and which cannot. This you can do using inequalities (but we will not discuss that today). 2. You can figure out the ranges in which the relationships are different and then try one value from each range. This is our transition points concept which we will discuss today. Let’s discuss the second option in more detail. First of all, we are just dealing with positives so there is less to worry about. That’s good. To picture the relationship between two functions, we first need to figure out the points where they are equal. \(x^2 = 2x\) \(x^2 – 2x = 0\) \(x = 0\) or 2 x cannot be 0 since x must be positive so this equation holds when x = 2 So \(x =2\) is the transition point of their relation. x^2 is less than 2x when x is less than 2 and it will be greater than 2x when x is greater than 2. Let’s try to figure out the relation between \(\frac{1}{x}\) and \(2x\) now. \(\frac{1}{x} = 2x\) \(x = \frac{1}{\sqrt 2}\) Since \(\frac{1}{x}\) is less than \(2x\) when x is greater than \(\frac{1}{\sqrt 2}\), it will be more than 2x when x is less than \(\frac{1}{\sqrt 2}\). Move on to the relation between 1/x and x^2. \(\frac{1}{x} = x^2\) \(x^3 = 1\) (notice that since x must be positive, we can easily multiply/divide by x without any complications) \(x = 1\) So you have got three transition points: \(\frac{1}{\sqrt 2}\), 1 and 2. Now all you need to do is try a number from each of these ranges: (i) \(x < \frac{1}{\sqrt 2}\) (ii) \(\frac{1}{\sqrt 2} < x < 1\) (iii) \(1 < x < 2\) (iv) \(x > 2\) If a relation doesn’t hold in any of these ranges, it will not hold for any value of x. (i) For \(x < \frac{1}{\sqrt 2}\), put x = a little more than 0 (e.g. 0.01) \(\frac{1}{x} = 100\), \(2x = 0.02\), \(x^2 = 0.0001\) We get \(x^2 < 2x < \frac{1}{x}\) is possible. So (I) is possible (ii) For \(\frac{1}{\sqrt 2} < x < 1\), put x = a little less than 1 (e.g. 0.99) 1/x = slightly more than 1, 2x = slightly less than 2, x^2 = slightly less than 1 We get \(x^2 < \frac{1}{x} < 2x\) is possible. So (II) is also possible. (iii) For \(1 < x < 2\), put \(x = \frac{3}{2}\) \(\frac{1}{x}= \frac{2}{3}\), \(2x = 3\), \(x^2 = \frac{9}{4} = 2.25\) We get \(\frac{1}{x} < x^2 < 2x\) is possible. (iv) For \(x > 2\), put \(x = 3\) \(\frac{1}{x} = \frac{1}{3}\), \(2x = 6\), \(x^2 = 9\) We get \(\frac{1}{x} < 2x < x^2\) is possible. We see that for no positive value of x is the third relation possible. We have covered all different ranges of values of x. Answer (D) This question is discussed HERE. Try using inequalities instead of number plugging to see if solving the question becomes easier in that case. Pretty sure i'm missing something very stupid but based on what was this conclusion made after we found the transition point( see below ) So x=2x=2 is the transition point of their relation. x^2 is less than 2x when x is less than 2 and it will be greater than 2x when x is greater than 2.similarly, Since 1x1x is less than 2x2x when x is greater than 12√12, it will be more than 2x when x is less than 12√12.



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09 Apr 2018, 07:45
Bunuel wrote: Inequalities with Multiple Factors Students often wonder why ‘\(x(x3) < 0\)’ doesn’t imply ‘\(x < 0\) or \((x – 3) < 0\)’. In this post, we will discuss why and we will see what it actually implies. Also, we will look at how we can handle such questions quickly. When you see ‘< 0’ or ‘> 0’, read it as ‘negative’ or ‘positive’ respectively. It will help you think clearly. So the question we are considering today is: Question: For what values of x will x(x3) be negative?Solution: Before we try to answer this question, think – when will the product of 2 numbers be negative? When one and only one of the factors is negative. Therefore, either x should be negative or (x – 3) should be negative, but not both. Let’s consider each case. Case 1: x is negative and (x – 3) is positive \(x < 0\) and \((x – 3) > 0\) which implies \(x > 3\) This is not possible. x cannot be less than 0 and greater than 3 at the same time. Hence this case gives us no appropriate values for x. Case 2: x is positive and (x – 3) is negative \(x > 0\) and \((x – 3) < 0\) which implies \(x < 3\) x must be greater than 0 but less than 3. Therefore, the range of values for which this inequality will be satisfied is \(0 < x < 3\). Will we have to do this every time we have multiple factors in an inequality? What will happen in case there are lots of factors? There is an easier way of handling such situations. I will first discuss the method and later explain the logic behind it. Method: Say we have an inequality of the form \((x – a)(x – b)(x – c) < 0\). (For clarity, we will work with the example \(x(x – 3) < 0\) discussed above.) This is how we solve for x: Step 1: Make a number line and plot the points a, b and c on it. In our example, a = 0 and b = 3. The number line is divided into sections by these points. In our example, it is divided into 3 sections – greater than 3, between 0 and 3 and less than 0. Step 2: Starting from the rightmost section, mark the sections with alternate positive and negative signs. The inequality will be positive in the sections where you have the positive signs and it will be negative in the sections where you have the negative signs. Therefore, \(x(x – 3)\) will be negative in the section 0 < x < 3 and positive in the other two sections. Hence, the values of x for which \(x(x – 3) < 0\) is satisfied is \(0 < x < 3\). Explanation: When we plot the points on the line, the number line is divided into various sections. Values of x in the right most section will always give you positive value of the expression. The reason for this is that if \(x > 3\), all factors will be positive i.e. x and (x – 3), both will be positive. When you jump to the next region i.e. between x = 0 and x = 3, the values of x will give you negative values for the entire expression because now, only one factor, (x – 3), will be negative. All other factors will be positive. When you jump to the next region on the left where x < 0, expression will be positive again because now both factors x and (x – 3) are negative. The product of two negatives is positive so the expression will be positive again and so on… Similarly, you can solve a question with any number of factors. In the next post, we will look at how to easily handle numerous complications that can arise. Just a small query; how is a=0 in the highlighted part above



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10 Apr 2018, 08:38
Bunuel wrote: Inequalities with Multiple Factors Students often wonder why ‘\(x(x3) < 0\)’ doesn’t imply ‘\(x < 0\) or \((x – 3) < 0\)’. In this post, we will discuss why and we will see what it actually implies. Also, we will look at how we can handle such questions quickly. When you see ‘< 0’ or ‘> 0’, read it as ‘negative’ or ‘positive’ respectively. It will help you think clearly. So the question we are considering today is: Question: For what values of x will x(x3) be negative?Solution: Before we try to answer this question, think – when will the product of 2 numbers be negative? When one and only one of the factors is negative. Therefore, either x should be negative or (x – 3) should be negative, but not both. Let’s consider each case. Case 1: x is negative and (x – 3) is positive \(x < 0\) and \((x – 3) > 0\) which implies \(x > 3\) This is not possible. x cannot be less than 0 and greater than 3 at the same time. Hence this case gives us no appropriate values for x. Case 2: x is positive and (x – 3) is negative \(x > 0\) and \((x – 3) < 0\) which implies \(x < 3\) x must be greater than 0 but less than 3. Therefore, the range of values for which this inequality will be satisfied is \(0 < x < 3\). Will we have to do this every time we have multiple factors in an inequality? What will happen in case there are lots of factors? There is an easier way of handling such situations. I will first discuss the method and later explain the logic behind it. Method: Say we have an inequality of the form \((x – a)(x – b)(x – c) < 0\). (For clarity, we will work with the example \(x(x – 3) < 0\) discussed above.) This is how we solve for x: Step 1: Make a number line and plot the points a, b and c on it. In our example, a = 0 and b = 3. The number line is divided into sections by these points. In our example, it is divided into 3 sections – greater than 3, between 0 and 3 and less than 0. Step 2: Starting from the rightmost section, mark the sections with alternate positive and negative signs. The inequality will be positive in the sections where you have the positive signs and it will be negative in the sections where you have the negative signs. Therefore, \(x(x – 3)\) will be negative in the section 0 < x < 3 and positive in the other two sections. Hence, the values of x for which \(x(x – 3) < 0\) is satisfied is \(0 < x < 3\). Explanation: When we plot the points on the line, the number line is divided into various sections. Values of x in the right most section will always give you positive value of the expression. The reason for this is that if \(x > 3\), all factors will be positive i.e. x and (x – 3), both will be positive. When you jump to the next region i.e. between x = 0 and x = 3, the values of x will give you negative values for the entire expression because now, only one factor, (x – 3), will be negative. All other factors will be positive. When you jump to the next region on the left where x < 0, expression will be positive again because now both factors x and (x – 3) are negative. The product of two negatives is positive so the expression will be positive again and so on… Similarly, you can solve a question with any number of factors. In the next post, we will look at how to easily handle numerous complications that can arise. Hi BunuelPlz help me understand the logic of the highlighted part above as why we taking a = 0 and b = 3. Thanks



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14 Apr 2018, 01:59
Bunuel wrote: Inequalities with Multiple Factors Students often wonder why ‘\(x(x3) < 0\)’ doesn’t imply ‘\(x < 0\) or \((x – 3) < 0\)’. In this post, we will discuss why and we will see what it actually implies. Also, we will look at how we can handle such questions quickly. When you see ‘< 0’ or ‘> 0’, read it as ‘negative’ or ‘positive’ respectively. It will help you think clearly. So the question we are considering today is: Question: For what values of x will x(x3) be negative?Solution: Before we try to answer this question, think – when will the product of 2 numbers be negative? When one and only one of the factors is negative. Therefore, either x should be negative or (x – 3) should be negative, but not both. Let’s consider each case. Case 1: x is negative and (x – 3) is positive \(x < 0\) and \((x – 3) > 0\) which implies \(x > 3\) This is not possible. x cannot be less than 0 and greater than 3 at the same time. Hence this case gives us no appropriate values for x. Case 2: x is positive and (x – 3) is negative \(x > 0\) and \((x – 3) < 0\) which implies \(x < 3\) x must be greater than 0 but less than 3. Therefore, the range of values for which this inequality will be satisfied is \(0 < x < 3\). Will we have to do this every time we have multiple factors in an inequality? What will happen in case there are lots of factors? There is an easier way of handling such situations. I will first discuss the method and later explain the logic behind it. Method: Say we have an inequality of the form \((x – a)(x – b)(x – c) < 0\). (For clarity, we will work with the example \(x(x – 3) < 0\) discussed above.) This is how we solve for x: Step 1: Make a number line and plot the points a, b and c on it. In our example, a = 0 and b = 3. The number line is divided into sections by these points. In our example, it is divided into 3 sections – greater than 3, between 0 and 3 and less than 0. Step 2: Starting from the rightmost section, mark the sections with alternate positive and negative signs. The inequality will be positive in the sections where you have the positive signs and it will be negative in the sections where you have the negative signs. Therefore, \(x(x – 3)\) will be negative in the section 0 < x < 3 and positive in the other two sections. Hence, the values of x for which \(x(x – 3) < 0\) is satisfied is \(0 < x < 3\). Explanation: When we plot the points on the line, the number line is divided into various sections. Values of x in the right most section will always give you positive value of the expression. The reason for this is that if \(x > 3\), all factors will be positive i.e. x and (x – 3), both will be positive. When you jump to the next region i.e. between x = 0 and x = 3, the values of x will give you negative values for the entire expression because now, only one factor, (x – 3), will be negative. All other factors will be positive. When you jump to the next region on the left where x < 0, expression will be positive again because now both factors x and (x – 3) are negative. The product of two negatives is positive so the expression will be positive again and so on… Similarly, you can solve a question with any number of factors. In the next post, we will look at how to easily handle numerous complications that can arise. Hi Bunuel Thanks soo much for this wonderful collection.. helped me understand the concepts well. Is there any direct link to the PDF version for the topics numbered 110. Thanks again.




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