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Inequalities Made Easy!
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07 Oct 2015, 06:09
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. Attachment:
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Re: Inequalities Made Easy!
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07 Oct 2015, 07:06
Some Inequalities, Mods and Sets Here, let’s look at a question that involves inequalities and modulus and is best understood using the concept of sets. It is not a difficult question but it is still very tricky. You could easily get it right the first time around but if you get it wrong, it could take someone many trials before he/she is able to convince you of the right answer. Even after I write a whole post on it, I wouldn’t be surprised if I see “but I still don’t get it” in the comments below! Anyway, enough of introduction! Let’s get to the question now. Question: If \(\frac{x}{x} < x\), which of the following must be true about x?(A) x > 1 (B) x > 1 (C) x< 1 (D) x = 1 (E) x^2 > 1 Solution:First thing we do is tackle the mod. We know that x is just the absolute value of x. So, \(\frac{x}{x}\) can take only 2 values: 1 or 1 If x is positive, \(\frac{x}{x} = 1\) e.g. if \(x = 4\), then \(\frac{4}{4} = 1\) If x is negative, \(\frac{x}{x} = 1\) e.g. if \(x = 4\), then \(\frac{4}{4} = \frac{4}{4} = 1\) x cannot be 0 because we cannot have 0 in the denominator of an expression. Now let’s work on the inequality. \(x >\frac{x}{x}\) implies \(x > 1\) if x is positive or \(x > 1\) if x is negative. Hence, for this inequality to hold, either \(x > 1\) (when x is positive) or \(1 < x< 0\) (when x is negative) x can take many values e.g. 1/3, 4/5, 2, 5, 10 etc. Now think – which of the following MUST BE TRUE about every value that x can take? (A) x > 1 or (B) x > 1 I hope that you agree that x > 1 doesn’t hold for every possible value of x whereas x > 1 holds for every possible value of x. Mind you, every value greater than 1 need not be a possible value of x. This concept might need some more work. Let me explain with another example. Forget this question for a minute. Say instead you have this question: Example 1: x > 2 and x < 7. What integral values can x take?I guess most of you will come up with 3, 4, 5, 6. That’s correct. I can represent this on the number line. The top arrow shows x > 2 and the bottom arrow shows x < 7. You see that the overlapping area includes 3, 4, 5 and 6. That is the region that satisfies both the inequalities. Now consider this: Example 2: x > 2 or x > 5. What integral values can x take?Let’s draw that number line again. So is the solution again the overlapping numbers i.e. all integers greater than 5? No. This question is different. x is greater than 2 OR greater than 5. This means that if x satisfies at least one of these conditions, it is included in your answer. Think of sets. AND means it should be in both the sets (i.e. the overlapping part) as was the case in example 1. OR means it should be in at least one of the sets. Hence, which values can x take? All integral values starting from 3 onwards i.e. 3, 4, 5, 6, 7, 8, 9 … Now go back to this question. The solution is a one liner. \(\frac{x}{x}\) is either 1 or 1. So \(x > 1\) or \(x > 1\) So which values can x take? All values included in at least one of the sets. Therefore, x > 1. Note that the confusion lies only between the first two options. The other three options are rejected outright. (C) x< 1 implies 1 < x < 1. Definitely doesn’t hold. (D) x = 1 implies x = 1 or 1. Definitely doesn’t hold. (E) x^2 > 1 implies either x < 1 or x > 1. Definitely doesn’t hold. So what do you say? Are you convinced that the answer is (B)? This question is discussed HERE. Attachment:
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Inequalities Made Easy!
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26 Mar 2017, 03:56
TIPS for inequalities !Case 1a) if (x)(y)>0 Either both x and y are positive or both are negativeb) if (x)/(y)>0 Either both x and y are positive or both are negativeCase 2a) if (x)(y)<0 [color=#fff200]Either x is positive and y is negative or y is positive and x is negative[/color] b) if (x)/(y)<0 Either x is positive and y is negative or y is positive and x is negativeCase 3if (x)/(y)>1 Either both x and y are positive or both are negative. And mod(x) > mod(y)




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Re: Inequalities Made Easy!
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07 Oct 2015, 06:18
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. Attachment:
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Re: Inequalities Made Easy!
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07 Oct 2015, 06:36
Inequalities with Complications  Part II Above we discussed a couple of variations of inequality questions with many factors. Let’s now look at some more complications that we should know how to handle. Complication No 3: Even powers \((x – a)^2(x – b)(x – c)(x – d) > 0\) How would you handle even powers of factors? Say, if the question has a factor (x – a) which has an even power, would you still plot ‘a’ on the number line? No, you will just ignore that factor while making the number line! Why? This is so because this factor will never be negative. It will be either 0 (in case the inequality includes the equality sign e.g. \((x – a)^2(x – b)(x – c)(x – d) = 0\)) or it will be positive. Therefore, a factor with an even power acts just like a positive constant. Then, does it mean factors with even powers have no role to play at all? No, it doesn’t. While writing out the range, they could impact the final answer. We will discuss this in more detail using an example later on. e.g. \((x – 4)(x – 2)(x + 8)^4 < 0\) You don’t need to plot 8 here. Since \((x + 8)^4\) can never be negative, it doesn’t change the sign of the expression. The expression will be negative in the range \(2 < x < 4\). Complication No 4: Odd powers \((x – a)^3(x – b)(x – c)(x – d) > 0\) What will you do in case of odd powers? Notice that in the last two posts, we have handled questions where the power of all the factors is 1. 1 is an odd power. So when you have any other odd power, you will handle it the same way. You can assume that the odd power is equal to 1 and proceed as usual. This is so because the sign of (x – a) will be the same as the sign of \((x – a)^3\). e.g. \((x – 4)(x – 2)(x + 1)^3 < 0\) The expression will be negative in the range \(x < 1\) or \(2 < x < 4\). Now let’s look at a question involving both these complications. Question: Find the range of x for which the given inequality holds.\(\frac{(2x – 5) * (6 – x)^3}{x^2} < 0\) Solution:I hope you agree that it doesn’t matter whether the factors are multiplied or divided. We are only concerned with the sign of the factors. \(\frac{2(x – \frac{5}{2}) * (1)(x – 6)^3}{x^2}< 0\) (take 2 common) \(\frac{2(x – \frac{5}{2}) * (x – 6)^3}{(x – 0)^2} > 0\) (multiply both sides by 1) Now let’s draw the number line. We don’t need to plot 0 since (x – 0) has an even power. We want to find the range where the expression is positive. The required range is \(x < \frac{5}{2}\) or \(x > 6\). But we are missing something here. \(x < \frac{5}{2}\) implies that all values less than 5/2 are acceptable but note that x cannot be 0 since x^2 is in the denominator. Hence the acceptable range is \(x < \frac{5}{2}\) or \(x > 6\) but \(x \neq 0\). When you have the equal to sign, you have to be careful about the way you choose your range. Attachment:
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Re: Inequalities Made Easy!
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07 Oct 2015, 06:53
Questions on Inequalities Now that we have covered some variations that arise in inequalities in GMAT problems, let’s look at some questions to consolidate the learning. We will first take up a relatively easy OG question and then a relatively tougher question which looks harder than it is because of the use of mods in the options (even though, we don’t really need to deal with the mods at all). Question 1: Is n between 0 and 1?Statement 1: n^2 is less than n Statement 2: n^3 is greater than 0 Solution: Let’s take each statement at a time and see what it implies. Statement 1: \(n^2 < n\) \(n^2 – n < 0\) \(n( n – 1) < 0\) This is the required form of the expression. We can now put it on the number line. For the expression to be negative, n should be between 0 and 1. So we can answer the question with a ‘yes’. Statement 1 alone is sufficient. Statement 2: \(n^3 > 0\) This only implies that n > 0 and we do not know whether it is less than 1 or not. Hence this statement alone is not sufficient. Answer: (A) This question is discussed HERE. This question could have been easily solved in a minute if you understand the theory we have been discussing for the past few weeks. Let’s go on to the trickier question now. Question 2: Which of the following represents the complete range of x over which \(x^5 – 4x^7 < 0\)?(A) 0 < x < ½ (B) x > ½ (C) –½ < x < 0 or ½ < x (D) x < –½ or 0 < x < ½ (E) x < –½ or x > 0 Solution: As I said, it looks harder than it is. We can easily do this in a minute too. First, let’s look at the given inequality closely: \(x^5 – 4x^7 < 0\) \(x^5 (1 – 4x^2) < 0\) (taking x^5 common) Just to make things easier right away, take out 4 common and multiply both sides by 1 to get \(4(x^5)(x^2 – \frac{1}{4}) > 0\) (notice that the sign has flipped since we multiplied both sides by 1) \(4(x^5)(x – \frac{1}{2})(x +\frac{1}{2}) > 0\) Think of the points you are going to plot: 0, 1/2 and 1/2. Recall that any positive odd power can be treated as a power of 1. In which region is x positive? \(\frac{1}{2} < x < 0\) or \(x > \frac{1}{2}\). This is our option (C). This question is discussed HERE. A quick word on the other options: What does 0 < x < ½ imply? It implies that distance of x from 0 is less than ½. So x lies between 1/2 and 1/2 (but x cannot be 0). What does \(x > \frac{1}{2}\) imply? It implies that distance of x from 0 is more than 1/2. So x is either greater than 1/2 or less than 1/2. If you are wondering what I am talking about, check out an old QWQW post: http://www.veritasprep.com/blog/2011/01 ... edoredid/We have discussed how to deal with modulus here. We hope this discussion has made such questions easier for you! Attachment:
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Re: Inequalities Made Easy!
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07 Oct 2015, 07:25
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.
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Re: Inequalities Made Easy!
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07 Oct 2015, 07:29
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) This gives us the range 1/?2 < x < 1/?2 (which is 0 < x < 1/?2 since x must be positive). 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?
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Re: Inequalities Made Easy!
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07 Oct 2015, 07:40
Plugging Numbers Without Using Transition Points A few months back, one of our posts talked about knowing which numbers to plugin in case you want to use the number–plugging method. To be more exact, we discussed that you need to find the transition points i.e. the points where the two sides of the inequality become equal. The transition points tend to reverse the relation between the two sides. For a detailed discussion of this concept, revisit this post. A question that arises here is what if the transition points are not apparent? What do we do in that case? First of all, let us say that the use of logic is preferable in every question. There are few questions (but there sure are!) where the number plugging method is the only decent option. But that’s beside the point. Let’s take up a question and see what to do in case the transition point is hard to see. Question: If x is an integer, is 4^x < 3^(x+1)?Statement I: x is positive Statement II: x – 1 < 2 Solution:Again, our moral duty is to first give you the logical solution since we would like you to think in those terms as far as possible. (Though in this question, plugging in numbers might seem easier.) We will discuss how to get the answer by plugging in appropriate numbers in this case. Method 1: Logical SolutionThe question stem only tells us that x can take any integral value. We need to find whether \(4^x\) is less than \(3^{(x+1)}\). Consider statement I: x is positive. Given that x is positive, is \(4^x\) less than \(3^{(x+1)}\)? If we put x = 1, it is easy to see that 4^x is less than 3^(x+1). So the inequality holds in this case. The point is how do we prove that this will be true for all positive values of x? It’s tough to prove that something holds for a lot of numbers. It’s easier to show that it doesn’t hold for at least one number since we need only one suitable value in that case. Question Stem: Is \(4^x < 3^{(x+1)}\)? Reframe it as: Is \(4^x < 3^x * 3\)? Or Is \((\frac{4}{3})^x < 3\)? Note that 4/3 (= 1.33) is greater than 1. When you raise it to a very high power, it will take a very large value. There is no reason it should stay less than 3. Hence the inequality will not hold for large values of x. Hence this statement alone is not sufficient. Some properties to note: * When you raise a positive number greater than 1 to a large positive integer power, it takes a large value. * When you raise 1 to a large positive integer power, it stays 1. * When you raise a positive number less than 1 to a large positive integer power, the number becomes even smaller than the original value. These are some number properties you need to work through and be comfortable with. Consider statement 2: \(x – 1 < 2\) Hopefully, you understand modulus well now. We can say that this inequality implies that x is a point at a distance less than 2 from the point 1 on the number line i.e. 1 < x < 3. Is \((\frac{4}{3})^x < 3\)? For small values of x e.g. x = 0, we know the inequality holds. Let’s check for only the largest value x can take i.e. 2 since x must be an integer. Even if x were 2, \((\frac{4}{3})^x = \frac{16}{9}\) i.e. less than 2. \((\frac{4}{3})^x\) would still be less than 3. Hence the inequality will hold in this case. This statement alone is sufficient to answer the question. Note that we did use some basic number plugging here too but that number plugging helps us get a clear picture and makes us ask the right questions. There is nothing wrong with plugging in some numbers here and there to understand the logic. If you know why you are plugging in a particular number, it means you are on the right track. Blindly plugging in is the problem. 2. Putting in NumbersLet’s look at this method too now. Is \(4^x < 3^{(x+1)}\)? Here it’s not easy to find the transition point. We would have to plug in numbers again to find where the two sides of the inequality are equal! So let’s ignore the transition points and directly start plugging in numbers. Statement 1: x is positive. Put x = 1, you get 4 < 9 (Holds) Put x = 2, you get 16 < 27 (Holds) Put x = 3, you get 64 < 81 (Holds) What do we do now? How long are we supposed to keep putting in numbers? We cannot do it for all positive integers. How do we decide when to stop? Note that the relative difference between the left hand side and the right hand side is reducing. 9 is more than twice of 4. 27 is more than 16 but not quite twice. It is more than 1.5 times of 16. 81 is somewhat more than 64 but not more than 1.5 times of 64. What this means is that soon enough, the difference will go to 0 and left hand side will become more than the right hand side. If you want to check and you are comfortable with higher powers of numbers, Put x = 4, you get 256 < 243 (Does not hold) What you need to do in case the transition point is not apparent is focus on the pattern of the numbers. Is the difference between them narrowing or widening? This statement alone is not sufficient to answer the question. Statement II: x – 1 < 2 As above, we get 1 < x < 3 so this is simply a matter of putting in x = 0, 1 and 2 to see that the inequality holds in each case. Sufficient. Answer (B) This question is discussed HERE.
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07 Oct 2015, 07:45
Biggie's Juicy Secret About GMAT Inequality As Hip Hop Month rolls on in the GMAT Tip of the Week space, we’re reminded that small nuances in the ways that GMAT questions are structured can have big consequences for testtakers. So who would be a more fitting man to teach that lesson – what’s small can have big consequences – than Biggie Smalls? Biggie’s most timeless classic, Juicy, may tell the ragstoriches story you’re hoping to live out once you grab that top tier MBA: “and my whole crew is lounging, celebrating every day no more public housing.” But first you need to get into bschool, and that’s where this lyric can prove helpful: “Damn right I like the life I live, ’cause I went from negative to positive … and if you don’t know, now you know.” What secret is Big Poppa passing along? It’s a critical message in two parts: “ …went from negative to positive” is a word of caution. When you’re dealing with inequalities on the GMAT, you need to remember that when a number goes from negative to positive – when you multiply or divide by a negative number to change the sign from positive to negative or from negative to positive – you must also change the direction of the inequality: If 10 > 5, then 10 is LESS THAN 5 If x > 10, then x < 10 The lesson: Be careful when going from negative to positive – if you’re working with inequalities and need to multiply or divide by a negative, you MUST change the direction of the inequality. Perhaps more useful is the next line, however: “ And if you don’t know, now you know.” If you don’t know whether a variable is positive or negative, here’s what you need to know: The GMAT is baiting you into assuming that it’s positive. If you’re asked to multiply or divide by a variable in an inequality question, it’s almost always a trap, as the testmaker knows that negative numbers are our blind spots – we tend to overlook them until they’re made absolutely explicit. So as Biggie said, if you don’t know (whether a variable is positive or negative)…now you know that there’s a high likelihood that that distinction will be important. Consider this Data Sufficiency example: Question: Is a > 3b? (1) a/b > 3 (2) b > 3Solution:The trap on this question is to select A, thinking that you can simply multiply both sides of that inequality by b: a/b > 3 –> a > 3b But you don’t know whether b is negative or positive. The above technique works if, say, a = 4 and b = 1 –> 4/1 is greater than 3, and 4 (a) is greater than 1 (b). So you get “yes”. But the situation also encompasses a = 4 and b = 1, as 4/1 is 4, which is greater than 3. But in this case 4 (a) is LESS than 1 (b). So you get “no”. The trap is to get you to blindly multiply both sides by b…but as Biggie cautions: If you don’t know, now you know (to be careful). Statement 2 isn’t much value on its own, but as it guarantees that b is positive, when you take both statements together, now you know that you can multiply both sides by b. So the correct answer is C, but the takeaway is most important here: When dealing with inequalities, if you don’t know (the + or – sign of a variable) now you know that the question probably hinges on that point. Heed Biggie’s sage advice and you’ll be on your way to one of the world’s most notorious bschools.
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03 Nov 2015, 09:26
Hi Bunuel,
How do we solve inequalities of the form ax^2 + bx + c >= Y
We have seen examples for > or < but not for >= or <=.
Will there be any change in values or the steps involved in calculation?
Regards, karthik



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03 Nov 2015, 09:29
iikarthik wrote: Hi Bunuel,
How do we solve inequalities of the form ax^2 + bx + c >= Y
We have seen examples for > or < but not for >= or <=.
Will there be any change in values or the steps involved in calculation?
Regards, karthik In this case you should just include the roots in the solution ranges.
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17 Feb 2016, 04:52
Basic Operations for GMAT Inequalities We know that we can perform all basic operations of addition, subtraction, multiplication and division on two equations: \(a = b\) \(c = d\) When these numbers are equal, we know that: \(a + c = b + d\) (Valid) \(a – c = b – d\) (Valid) \(a * c = b * d\) (Valid) \(\frac{a}{c} =\frac{b}{d}\) (Valid assuming c and d are not 0) When can we add, subtract, multiply or divide two inequalities? There are rules that we need to follow for those. Today let’s discuss those rules and the concepts behind them. Addition:We can add two inequalities when they have the same inequality sign. \(a < b\) \(c < d\) \(a + c < b + d\) (Valid) Conceptually, it makes sense, right? If a is less than b and c is less than d, then the sum of a and c will be less than the sum of b and d.On the same lines: \(a > b\) \(c > d\) \(a + c > b + d\) (Valid) Case 2: What happens when the inequalities have opposite signs? \(a > b\) \(c < d\) We need to multiply one inequality by 1 to get the two to have the same inequality sign. \(c > d\) Now we can add them. \(a – c > b – d\) Subtraction:We can subtract two inequalities when they have opposite signs: \(a > b\) \(c < d\) \(a – c > b – d\) (The result will take the sign of the first inequality) Conceptually, think about it like this: from a greater number (a is greater than b), if we subtract a smaller number (c is smaller than d), the result (a – c) will be greater than the result obtained when we subtract the greater number from the smaller number (b – d).Note that this result is the same as that obtained when we added the two inequalities after changing the sign (see Case 2 above). We cannot subtract inequalities if they have the same sign, so it is better to always stick to addition. If the inequalities have the same sign, we simply add them. If the inequalities have opposite signs, we multiply one of them by 1 (to get the same sign) and then add them (in effect, we subtract them). Why can we not subtract two inequalities when they have the same inequality sign, such as when a > b and c > d? Say, we have 3 > 1 and 5 > 1. If we subtract these two, we get 3 – 5 > 1 – 1, or 2 > 0 which is not valid. If instead it were 3 > 1 and 2 > 1, we would get 1 > 0 which is valid. We don’t know how much greater one term is from the other and hence we cannot subtract inequalities when their inequality signs are the same. Multiplication:Here, the constraint is the same as that in addition (the inequality signs should be the same) with an extra constraint: both sides of both inequalities should be nonnegative. If we do not know whether both sides are nonnegative or not, we cannot multiply the inequalities. If a, b, c and d are all non negative, \(a < b\) \(c < d\) \(a*c < b*d\) (Valid) When two greater numbers are multiplied together, the result will be greater. Take some examples to see what happens in Case 1, or more numbers are negative: 2 < 1 10 < 30 Multiply to get: 20 < 30 (Not valid) 2 < 7 8 < 1 Multiply to get: 16 < 7 (Not valid) Division:Here, the constraint is the same as that in subtraction (the inequality signs should be opposite) with an extra constraint: both sides of both inequalities should be nonnegative (obviously, 0 should not be in the denominator). If we do not know whether both sides are positive or not, we cannot divide the inequalities. \(a < b\) \(c > d\) \(\frac{a}{c} < \frac{b}{d}\) (given all a, b, c and d are positive) The final inequality takes the sign of the numerator. Think of it conceptually: a smaller number is divided by a greater number, so the result will be a smaller number. Take some examples to see what happens in Case 1, or more numbers are negative. 1 < 2 10 > 30 Divide to get 1/10 < 2/30 (Not valid) Takeaways: Addition: We can add two inequalities when they have the same inequality signs.
Subtraction: We can subtract two inequalities when they have opposite inequality signs.
Multiplication: We can multiply two inequalities when they have the same inequality signs and both sides of both inequalities are nonnegative.
Division: We can divide two inequalities when they have opposite inequality signs and both sides of both inequalities are nonnegative (0 should not be in the denominator).
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31 May 2016, 07:46
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



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Updated on: 05 Jun 2016, 05:57
Is it possible that the last graph in your third post is wrong? In my opinion the graph should be positive between 2,5 und 6 and negative for x>6 and x<0? Thanks!
Originally posted by 22gmat on 05 Jun 2016, 05:49.
Last edited by 22gmat on 05 Jun 2016, 05:57, edited 2 times in total.



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05 Jun 2016, 05:52
22gmat wrote: Is it possible that the last graph in your third post is wrong? In my opinion the graph should be positive between 2,5 und 6 and negative for x>6 and x<0? Thanks! Which graph exactly are you talking about? Please quote the post itself. Thank you.
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05 Jun 2016, 06:01
Bunuel wrote: Inequalities with Complications  Part II Now let’s look at a question involving both these complications. Question: Find the range of x for which the given inequality holds.\(\frac{(2x – 5) * (6 – x)^3}{x^2} < 0\) Solution:I hope you agree that it doesn’t matter whether the factors are multiplied or divided. We are only concerned with the sign of the factors. \(\frac{2(x – \frac{5}{2}) * (1)(x – 6)^3}{x^2}< 0\) (take 2 common) \(\frac{2(x – \frac{5}{2}) * (x – 6)^3}{(x – 0)^2} > 0\) (multiply both sides by 1) Now let’s draw the number line. We don’t need to plot 0 since (x – 0) has an even power. We want to find the range where the expression is positive. The required range is \(x < \frac{5}{2}\) or \(x > 6\). But we are missing something here. \(x < \frac{5}{2}\) implies that all values less than 5/2 are acceptable but note that x cannot be 0 since x^2 is in the denominator. Hence the acceptable range is \(x < \frac{5}{2}\) or \(x > 6\) but \(x \neq 0\). When you have the equal to sign, you have to be careful about the way you choose your range. Attachment: Ques5.jpg Attachment: Ques6.jpg Attachment: Ques7 (1).jpg I am sorry. I just have seen the initial equation and the graph and thought it would be wrong. However you have multiplied it with (1), so everything is correct. Thanks!



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15 Jun 2017, 07:13
What is the range for a ??
a + a^(1) >2



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30 Jul 2017, 19:41
Bunuel wrote: Questions on Inequalities Now that we have covered some variations that arise in inequalities in GMAT problems, let’s look at some questions to consolidate the learning. We will first take up a relatively easy OG question and then a relatively tougher question which looks harder than it is because of the use of mods in the options (even though, we don’t really need to deal with the mods at all). Question 1: Is n between 0 and 1?Statement 1: n^2 is less than n Statement 2: n^3 is greater than 0 Solution: Let’s take each statement at a time and see what it implies. Statement 1: \(n^2 < n\) \(n^2 – n < 0\) \(n( n – 1) < 0\) This is the required form of the expression. We can now put it on the number line. For the expression to be negative, n should be between 0 and 1. So we can answer the question with a ‘yes’. Statement 1 alone is sufficient. Statement 2: \(n^3 > 0\) This only implies that n > 0 and we do not know whether it is less than 1 or not. Hence this statement alone is not sufficient. Answer: (A) This question is discussed HERE. This question could have been easily solved in a minute if you understand the theory we have been discussing for the past few weeks. Let’s go on to the trickier question now. Question 2: Which of the following represents the complete range of x over which \(x^5 – 4x^7 < 0\)?(A) 0 < x < ½ (B) x > ½ (C) –½ < x < 0 or ½ < x (D) x < –½ or 0 < x < ½ (E) x < –½ or x > 0 Solution: As I said, it looks harder than it is. We can easily do this in a minute too. First, let’s look at the given inequality closely: \(x^5 – 4x^7 < 0\) \(x^5 (1 – 4x^2) < 0\) (taking x^5 common) Just to make things easier right away, take out 4 common and multiply both sides by 1 to get \(4(x^5)(x^2 – \frac{1}{4}) > 0\) (notice that the sign has flipped since we multiplied both sides by 1) \(4(x^5)(x – \frac{1}{2})(x +\frac{1}{2}) > 0\) Think of the points you are going to plot: 0, 1/2 and 1/2. Recall that any positive odd power can be treated as a power of 1. In which region is x positive? \(\frac{1}{2} < x < 0\) or \(x > \frac{1}{2}\). This is our option (C). This question is discussed HERE. A quick word on the other options: What does 0 < x < ½ imply? It implies that distance of x from 0 is less than ½. So x lies between 1/2 and 1/2 (but x cannot be 0). What does \(x > \frac{1}{2}\) imply? It implies that distance of x from 0 is more than 1/2. So x is either greater than 1/2 or less than 1/2. If you are wondering what I am talking about, check out an old QWQW post: http://www.veritasprep.com/blog/2011/01 ... edoredid/We have discussed how to deal with modulus here. We hope this discussion has made such questions easier for you! Attachment: Ques8.jpg Attachment: Ques9.jpg Sorry Bunuel, I have a silly confusion For the Question 1:[/color] Is n between 0 and 1?[/b] And from the statement, we get when n between 0 and 1, it is negative. How does that relate to n between 0 and 1? Thank you so much. You really such a big help, God bless you




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