answer：Okay, so I have this problem about a 15-inch stick with four marks that divide it into five segments of lengths 1, 2, 3, 4, and 5 inches. The goal is to prove that it's impossible to place these four marks in such a way that every integer distance from 1 inch to 15 inches can be measured. Hmm, interesting. Let me try to break this down step by step.First, I need to understand what it means to measure every integer distance from 1 to 15 inches using the segments created by the four marks. Essentially, the stick should act like a ruler where you can measure any length by looking at the difference between two marks. So, if I have marks at positions a, b, c, d on the stick, then the distances between these marks should cover all integers from 1 to 15.The stick is 15 inches long, and the segments between the marks are 1, 2, 3, 4, and 5 inches. That means the total length adds up correctly: 1 + 2 + 3 + 4 + 5 = 15 inches. So, the segments are correctly sized, but the problem is about arranging them in such a way that all distances can be measured.Let me think about how a ruler works. On a standard ruler, you can measure any length by subtracting the position of one mark from another. For example, if you have marks at 0, 1, 2, 3, ..., 15, you can measure any distance by subtracting two marks. But in this case, the marks are not at every inch; instead, they're spaced at intervals of 1, 2, 3, 4, and 5 inches. So, the challenge is to arrange these intervals such that all distances from 1 to 15 can be measured.Maybe I should consider the possible distances that can be measured with these segments. If I have segments of 1, 2, 3, 4, and 5 inches, the distances between the marks would be the differences between their positions. For example, if one mark is at position x and another at position y, the distance between them is |x - y|. To cover all distances from 1 to 15, every integer in that range must be achievable as the difference between some pair of marks.Let me try to visualize this. Suppose I place the first mark at position 0 (the start of the stick). Then, the next mark could be at position 1, making the first segment 1 inch. The next mark could be at position 1 + 2 = 3 inches, making the second segment 2 inches. Then, the next mark at 3 + 3 = 6 inches, the next at 6 + 4 = 10 inches, and finally at 10 + 5 = 15 inches. So, the marks would be at 0, 1, 3, 6, 10, and 15 inches.Now, let's see what distances we can measure with these marks. The differences between consecutive marks are 1, 2, 3, 4, and 5 inches, which are good. But what about the other distances? For example, can we measure 7 inches? That would require a mark at 7 inches, but in this arrangement, the marks are at 0, 1, 3, 6, 10, and 15. The distance between 0 and 7 isn't directly a mark, but can we get 7 inches by subtracting two marks? Let's see: 10 - 3 = 7. Yes, that works. How about 8 inches? 10 - 2 = 8, but we don't have a mark at 2 inches. Wait, our marks are at 0, 1, 3, 6, 10, and 15. So, 10 - 3 = 7, 6 - 1 = 5, 10 - 1 = 9, 6 - 0 = 6, 3 - 0 = 3, 10 - 6 = 4, 15 - 10 = 5, 15 - 6 = 9, 15 - 3 = 12, 15 - 1 = 14, and 15 - 0 = 15. Hmm, so we can get 1, 3, 4, 5, 6, 7, 9, 10, 12, 14, 15 inches. But we're missing 2, 8, 11, and 13 inches. So, this arrangement doesn't cover all distances from 1 to 15.Maybe I need to arrange the segments differently. Let's try another configuration. Suppose I start with the largest segment first. So, the first segment is 5 inches, then 4, 3, 2, and 1. So, the marks would be at 0, 5, 9, 12, 14, and 15 inches. Let's see the distances we can measure here. The differences between consecutive marks are 5, 4, 3, 2, and 1 inches. Now, let's check the other distances: 9 - 5 = 4, 12 - 9 = 3, 14 - 12 = 2, 15 - 14 = 1. But what about 1 inch? We have that. 2 inches? Yes, between 14 and 12. 3 inches? Between 12 and 9. 4 inches? Between 9 and 5. 5 inches? Between 5 and 0. 6 inches? 9 - 3 = 6, but we don't have a mark at 3 inches. Wait, our marks are at 0, 5, 9, 12, 14, 15. So, 9 - 5 = 4, 12 - 5 = 7, 14 - 5 = 9, 15 - 5 = 10. 12 - 9 = 3, 14 - 9 = 5, 15 - 9 = 6. 14 - 12 = 2, 15 - 12 = 3. 15 - 14 = 1. So, the distances we can measure are 1, 2, 3, 4, 5, 6, 7, 9, 10. Missing are 8, 11, 12, 13, 14, 15. Wait, 15 is the total length, so that's covered. But 8, 11, 12, 13 inches are missing. So, this arrangement also doesn't cover all distances.Hmm, maybe I need to try a different approach. Instead of starting with the largest or smallest segment, perhaps interleave them. Let's try placing the segments in the order 1, 3, 5, 2, 4. So, the marks would be at 0, 1, 4, 9, 11, and 15 inches. Let's see the distances: 1, 3, 5, 2, 4 inches between consecutive marks. Now, other distances: 4 - 1 = 3, 9 - 4 = 5, 11 - 9 = 2, 15 - 11 = 4. But what about 6 inches? 9 - 3 = 6, but we don't have a mark at 3 inches. Our marks are at 0, 1, 4, 9, 11, 15. So, 4 - 0 = 4, 9 - 0 = 9, 11 - 0 = 11, 15 - 0 = 15. 9 - 1 = 8, 11 - 1 = 10, 15 - 1 = 14. 9 - 4 = 5, 11 - 4 = 7, 15 - 4 = 11. 11 - 9 = 2, 15 - 9 = 6. 15 - 11 = 4. So, the distances we can measure are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 14, 15. Missing are 12 and 13 inches. So, still missing some distances.This is getting frustrating. Maybe there's a systematic way to approach this. Let me think about the concept of a Golomb ruler. A Golomb ruler is a set of marks at integer positions along a ruler such that no two pairs of marks are the same distance apart. The goal is to have the longest possible ruler with the fewest marks. But in this case, we're not necessarily looking for a Golomb ruler, but rather a ruler that can measure every integer distance up to 15 inches.Wait, actually, this problem is similar to constructing a perfect ruler. A perfect ruler can measure every integer distance up to its total length. For a 15-inch ruler, we need to have marks such that every distance from 1 to 15 can be measured as the difference between two marks. The key here is that the set of differences must cover all integers from 1 to 15.Given that we have five segments of lengths 1, 2, 3, 4, and 5, the marks divide the ruler into these segments. The positions of the marks are cumulative sums of these segments. So, if we denote the segments as s1, s2, s3, s4, s5, then the marks are at positions s1, s1+s2, s1+s2+s3, s1+s2+s3+s4, and s1+s2+s3+s4+s5=15.The distances that can be measured are the differences between these marks. So, the set of distances is the set of all possible differences between any two marks. To cover all distances from 1 to 15, every integer in that range must be present in this set.Now, let's consider the number of possible differences. With five marks (including the start at 0), there are C(5,2) = 10 possible differences. But we need to cover 15 distances. Wait, that's impossible because we only have 10 differences but need to cover 15 distances. So, this seems like a contradiction.Wait, no, actually, the total number of possible differences is more than that because we can have differences between non-consecutive marks. For example, the difference between the first and third mark, first and fourth, etc. So, actually, with five marks (including 0), the number of possible differences is C(5,2) = 10. But we need to cover 15 distances, which is more than 10. Therefore, it's impossible to have all distances from 1 to 15 covered because we don't have enough differences.Wait, but that can't be right because in the standard ruler with marks at every inch, you have 15 marks (including 0), giving C(15,2) = 105 differences, which obviously covers all distances. But in our case, we have only five segments, meaning six marks (including 0). So, C(6,2) = 15 differences. Ah, so we have exactly 15 differences, which is the number of distances we need to cover (1 to 15). So, in theory, it's possible if all differences are unique and cover exactly 1 to 15.But wait, the problem is that the segments are fixed lengths: 1, 2, 3, 4, 5. So, the differences between the marks are determined by these segments. Therefore, the set of differences must be exactly the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}. But with only five segments, the differences are limited.Let me think about the possible sums. The maximum distance is 15, which is the total length. The minimum distance is 1. Now, to get all the distances in between, we need to have combinations of the segments that add up to each integer from 1 to 15.But wait, the segments themselves are 1, 2, 3, 4, 5. So, the possible sums of these segments can give us the distances. For example, 1 can be obtained directly. 2 can be obtained directly. 3 can be obtained directly. 4 can be obtained directly. 5 can be obtained directly. 6 can be obtained as 1+5 or 2+4. 7 can be obtained as 2+5 or 3+4. 8 can be obtained as 3+5 or 4+4, but we only have one 4. 9 can be obtained as 4+5. 10 can be obtained as 5+5, but we only have one 5. 11 would require 5+6, but we don't have a 6 segment. Wait, this is getting complicated.Alternatively, maybe I should consider the concept of additive bases. An additive basis of order 2 is a set of numbers such that every number up to a certain limit can be expressed as the sum of at most two elements from the set. In our case, the set is {1, 2, 3, 4, 5}, and we need to cover up to 15. But since we have only five segments, the maximum sum is 1+2+3+4+5=15. But we need to cover all numbers from 1 to 15, not just the sums.Wait, no, the distances are not sums of segments but differences between marks. So, the marks are at positions that are cumulative sums of the segments. Therefore, the distances are the differences between these cumulative sums. So, if the segments are arranged in a certain order, the cumulative sums will determine the positions of the marks, and the differences between these positions will give the measurable distances.Let me try to formalize this. Let the segments be s1, s2, s3, s4, s5, where each si is one of 1, 2, 3, 4, 5, and all are distinct. The positions of the marks are then:- Mark 0: 0- Mark 1: s1- Mark 2: s1 + s2- Mark 3: s1 + s2 + s3- Mark 4: s1 + s2 + s3 + s4- Mark 5: 15The distances that can be measured are the differences between any two marks. So, the set of distances is:{Mark1 - Mark0, Mark2 - Mark0, Mark3 - Mark0, Mark4 - Mark0, Mark5 - Mark0,Mark2 - Mark1, Mark3 - Mark1, Mark4 - Mark1, Mark5 - Mark1,Mark3 - Mark2, Mark4 - Mark2, Mark5 - Mark2,Mark4 - Mark3, Mark5 - Mark3,Mark5 - Mark4}That's 15 distances, which is exactly what we need. So, in theory, if we can arrange the segments such that all these differences are unique and cover 1 to 15, it's possible. But the problem states that it's impossible. So, why is that?Maybe because the segments are fixed as 1, 2, 3, 4, 5, and no matter how you arrange them, you can't get all the required differences. Let's try to see.Suppose we arrange the segments in the order 1, 2, 3, 4, 5. Then the marks are at 0, 1, 3, 6, 10, 15. The distances are:1, 3, 6, 10, 15,2, 5, 9, 14,3, 6, 11,4, 9,5.So, the distances are: 1, 2, 3, 4, 5, 6, 9, 10, 11, 14, 15. Missing are 7, 8, 12, 13.If we arrange them as 1, 3, 5, 2, 4, the marks are at 0, 1, 4, 9, 11, 15. The distances are:1, 4, 9, 11, 15,3, 8, 10, 14,5, 7, 11,2, 6,4.So, distances: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 14, 15. Missing are 12, 13.If we try another arrangement, say 2, 3, 5, 1, 4. Marks at 0, 2, 5, 10, 11, 15. Distances:2, 5, 10, 11, 15,3, 8, 9, 13,5, 6, 10,1, 6,4.So, distances: 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 13, 15. Missing are 7, 12, 14.Hmm, still missing some. What if I try 3, 1, 4, 2, 5? Marks at 0, 3, 4, 8, 10, 15. Distances:3, 4, 8, 10, 15,1, 5, 7, 12,4, 6, 11,2, 7,5.Distances: 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 15. Missing are 9, 13, 14.Still missing. Maybe I need to try a different approach. Let's consider the concept of a complete ruler. A complete ruler can measure every integer distance up to its length. For a 15-inch ruler, we need to have marks such that every distance from 1 to 15 can be measured.Given that we have five segments, the number of marks is six (including 0). The number of possible distances is C(6,2) = 15, which matches the number of distances we need to cover. So, in theory, it's possible if all distances are unique and cover 1 to 15.But the problem is that the segments are fixed as 1, 2, 3, 4, 5. So, the cumulative sums (mark positions) are determined by the order of these segments. Therefore, the set of distances is determined by the differences between these cumulative sums.Let me denote the segments as s1, s2, s3, s4, s5, which are a permutation of 1, 2, 3, 4, 5. The mark positions are:m0 = 0m1 = s1m2 = s1 + s2m3 = s1 + s2 + s3m4 = s1 + s2 + s3 + s4m5 = 15The distances are:m1 - m0 = s1m2 - m0 = s1 + s2m3 - m0 = s1 + s2 + s3m4 - m0 = s1 + s2 + s3 + s4m5 - m0 = 15m2 - m1 = s2m3 - m1 = s2 + s3m4 - m1 = s2 + s3 + s4m5 - m1 = 15 - s1m3 - m2 = s3m4 - m2 = s3 + s4m5 - m2 = 15 - (s1 + s2)m4 - m3 = s4m5 - m3 = 15 - (s1 + s2 + s3)m5 - m4 = s5So, the set of distances is:{s1, s2, s3, s4, s5, s1+s2, s1+s2+s3, s1+s2+s3+s4, 15,s2, s2+s3, s2+s3+s4, 15 - s1,s3, s3+s4, 15 - (s1+s2),s4, 15 - (s1+s2+s3),s5}But since s1, s2, s3, s4, s5 are 1, 2, 3, 4, 5 in some order, we need to ensure that all distances from 1 to 15 are covered.Notice that 15 is covered as m5 - m0. Now, we need to cover 1 to 14 as well.Looking at the distances, we have s1, s2, s3, s4, s5, which are 1, 2, 3, 4, 5. So, we have 1, 2, 3, 4, 5 covered.Next, we have s1+s2, which is at least 1+2=3 and at most 4+5=9.Similarly, s1+s2+s3 is at least 1+2+3=6 and at most 3+4+5=12.s1+s2+s3+s4 is at least 1+2+3+4=10 and at most 2+3+4+5=14.15 - s1 is at least 15 - 5=10 and at most 15 -1=14.15 - (s1+s2) is at least 15 - (4+5)=6 and at most 15 - (1+2)=12.15 - (s1+s2+s3) is at least 15 - (3+4+5)=3 and at most 15 - (1+2+3)=9.So, the distances from 1 to 15 are covered as follows:1, 2, 3, 4, 5 (direct segments),6 to 14 (from sums and differences),and 15.But wait, we need to ensure that every integer from 1 to 15 is covered without missing any. However, the problem is that some distances might overlap or not cover all numbers.For example, if s1=1, then 15 - s1=14. If s1=2, then 15 - s1=13, and so on. Similarly, s1+s2 can be 3 to 9, but we need to ensure that all numbers in that range are covered.But here's the catch: the sums and differences might not cover all numbers uniquely. For example, if s1=1 and s2=2, then s1+s2=3, which is already covered by s3=3. So, we have a duplicate distance of 3, which means we can't cover another number.Similarly, if s1=1, s2=3, then s1+s2=4, which is already covered by s4=4. Again, a duplicate.This duplication means that some distances are not covered because we're reusing the same distance in both a segment and a sum.Therefore, it's impossible to arrange the segments in such a way that all distances from 1 to 15 are covered without duplication, hence making it impossible to measure every integer distance.

answer：Okay, so I have this problem here about an arithmetic progression. The first three terms are given as y - 3, 3y + 1, and 5y - 7. I need to find the possible value of y. Hmm, arithmetic progression, right? That means the difference between consecutive terms is constant. So, the difference between the second and first term should be the same as the difference between the third and second term. Let me write that down.First, let me find the difference between the second and first term. The second term is 3y + 1, and the first term is y - 3. So, subtracting the first term from the second term gives:(3y + 1) - (y - 3) = 3y + 1 - y + 3 = 2y + 4.Okay, so the common difference is 2y + 4. Now, let me find the difference between the third and second term. The third term is 5y - 7, and the second term is 3y + 1. Subtracting the second term from the third term gives:(5y - 7) - (3y + 1) = 5y - 7 - 3y - 1 = 2y - 8.So, the common difference here is 2y - 8. Since it's an arithmetic progression, these two differences should be equal. Therefore, I can set them equal to each other:2y + 4 = 2y - 8.Wait, let me solve this equation. Subtract 2y from both sides:2y + 4 - 2y = 2y - 8 - 2y,Which simplifies to:4 = -8.Hmm, that doesn't make sense. 4 equals -8? That's not possible. Did I do something wrong? Let me check my calculations again.First difference: (3y + 1) - (y - 3) = 3y + 1 - y + 3 = 2y + 4. That seems correct.Second difference: (5y - 7) - (3y + 1) = 5y - 7 - 3y - 1 = 2y - 8. That also seems correct.So, setting them equal: 2y + 4 = 2y - 8. Subtracting 2y from both sides gives 4 = -8, which is a contradiction. That means there is no solution for y that satisfies this condition. So, the answer must be that there's no solution.But wait, let me think again. Maybe I made a mistake in setting up the equations. Let me try another approach. Maybe I should consider the terms in the order given and see if I can find y such that the sequence is arithmetic.So, the first term is y - 3, the second term is 3y + 1, and the third term is 5y - 7. In an arithmetic progression, the second term is the average of the first and third terms. So, maybe I can use that property.The average of the first and third terms should be equal to the second term. So:( (y - 3) + (5y - 7) ) / 2 = 3y + 1.Let me compute the numerator:(y - 3) + (5y - 7) = y + 5y - 3 - 7 = 6y - 10.So, the average is (6y - 10)/2 = 3y - 5.Set this equal to the second term:3y - 5 = 3y + 1.Subtract 3y from both sides:-5 = 1.Again, that's a contradiction. So, this also shows that there's no solution for y. Therefore, the answer is that there's no solution.But wait, the options given are A) -6, B) 0, C) 1, D) 4, E) No solution. So, E is an option. Therefore, the correct answer must be E) No solution.I think that's it. I tried two different approaches, both leading to contradictions, meaning there's no value of y that satisfies the conditions of an arithmetic progression with the given terms. So, the answer is E.

answer：Okay, so I have this quadratic expression here: 2x² - 12x + 18. I need to factor it. Hmm, factoring quadratics can be a bit tricky, but I think I remember some steps from my math class. Let me try to work through this step by step.First, I recall that when factoring a quadratic expression of the form ax² + bx + c, one common method is to factor out the greatest common factor (GCF) first, if there is one. Looking at the coefficients 2, -12, and 18, I notice that they all have a common factor of 2. So, maybe I can factor out a 2 from the entire expression.Let me try that. If I factor out a 2 from each term, I get:2(x² - 6x + 9)Okay, so now the expression is 2 times (x² - 6x + 9). That seems simpler, but I still need to factor the quadratic inside the parentheses. I wonder if this quadratic can be factored further.I remember that a quadratic expression can sometimes be factored into two binomials, like (x + m)(x + n), where m and n are numbers that multiply to give the constant term and add up to the coefficient of the middle term. In this case, the quadratic is x² - 6x + 9. So, I need to find two numbers that multiply to 9 and add up to -6.Let me think about the factors of 9. The pairs of factors are (1, 9) and (3, 3). Since the middle term is -6x, the numbers should both be negative because their product is positive and their sum is negative. So, if I take -3 and -3, their product is 9 and their sum is -6. Perfect!So, the quadratic x² - 6x + 9 can be factored as (x - 3)(x - 3), which is the same as (x - 3)². Therefore, the entire expression becomes:2(x - 3)²Let me double-check my work to make sure I didn't make any mistakes. Starting from the original expression:2x² - 12x + 18Factoring out the GCF of 2:2(x² - 6x + 9)Factoring the quadratic inside:2(x - 3)²If I expand this back out, does it match the original expression? Let's see:2(x - 3)² = 2(x - 3)(x - 3) = 2(x² - 6x + 9) = 2x² - 12x + 18Yes, that matches perfectly. So, my factoring seems correct.I also remember that sometimes quadratics can be factored in different ways, but in this case, since the quadratic inside the parentheses is a perfect square, the factoring is straightforward. It's always good to check if the quadratic is a perfect square trinomial because that simplifies the factoring process.Another thing I should consider is whether there are any alternative methods to factor this expression, just to confirm my answer. For example, I could use the quadratic formula to find the roots of the quadratic and then write the factored form based on those roots.The quadratic formula is x = [-b ± √(b² - 4ac)] / (2a). For the quadratic inside the parentheses, x² - 6x + 9, a = 1, b = -6, and c = 9.Plugging these values into the quadratic formula:x = [6 ± √((-6)² - 4*1*9)] / (2*1)x = [6 ± √(36 - 36)] / 2x = [6 ± √0] / 2x = 6/2 = 3So, both roots are x = 3. This means that the quadratic can be written as (x - 3)², which confirms my earlier factoring.Therefore, the factored form of the original expression is indeed 2(x - 3)².I think I've covered all the steps and checked my work thoroughly. It's always reassuring to verify the factoring by expanding it back out and ensuring it matches the original expression. This helps catch any mistakes I might have made during the factoring process.In summary, factoring 2x² - 12x + 18 involves first factoring out the GCF of 2, resulting in 2(x² - 6x + 9). Recognizing that x² - 6x + 9 is a perfect square trinomial allows us to factor it further into (x - 3)². Combining these steps gives the fully factored form of the expression as 2(x - 3)².

answer：First, I need to reduce the fraction ( frac{368}{598} ) to its simplest form. To do this, I'll start by finding the greatest common divisor (GCD) of the numerator and the denominator.I'll list the factors of both numbers:- The factors of 368 are 1, 2, 4, 8, 16, 23, 46, 92, 184, 368.- The factors of 598 are 1, 2, 23, 26, 46, 299, 598.The common factors are 1, 2, 23, and 46. Among these, the greatest common factor is 46.Next, I'll divide both the numerator and the denominator by the GCD:[frac{368 div 46}{598 div 46} = frac{8}{13}]Therefore, the fraction ( frac{368}{598} ) reduced to the lowest terms is ( frac{8}{13} ).

answer：Alright, so I've got this problem about a magical land with 100 cities connected by air routes. The rules are that any two cities can be reached with no more than 11 transfers, and there's only one way to do that. If you can't get from city A to city B in 10 or fewer transfers, both cities are called peripheral. I need to figure out the largest number of peripheral cities that can exist.Okay, let's break this down. First, there are 100 cities. They're connected by air routes, which are two-way, so this is an undirected graph. The condition that any two cities can be reached with no more than 11 transfers means the diameter of the graph is at most 11. Also, there's a unique way to do this, so the graph is a tree? Wait, no, because in a tree, there's exactly one path between any two nodes, but here it says there's a unique way to do it with no more than 11 transfers. Hmm, maybe it's not necessarily a tree, but it's a graph where the shortest path between any two nodes is unique and has length at most 11.Now, peripheral cities are those where you can't get from one to the other in 10 or fewer transfers. So, if two cities are peripheral, the shortest path between them must be exactly 11 transfers. And since the shortest path is unique, that means there's only one way to go from one to the other, and it takes 11 steps.I need to maximize the number of peripheral cities. So, how can I arrange the cities so that as many as possible are peripheral? Let's think about the structure of the graph. If I have a long path, like a straight line of cities, each connected to the next, then the cities at the ends would be peripheral because the path between them is the longest. But in this case, the diameter is 11, so the longest path can't be more than 11 edges.Wait, but if I have a path of 12 cities, the distance between the first and last city would be 11, which fits the diameter condition. But in that case, only the two end cities would be peripheral because any other city can be reached from them in fewer than 11 transfers. So, that's only two peripheral cities, which isn't very many.I need more peripheral cities. Maybe instead of a straight path, I can have multiple paths radiating out from a central hub. If I have a central city connected to many other cities, then those other cities can be peripheral if their paths to each other go through the central city. But wait, if two peripheral cities are connected through the central city, their path length would be 2, which is way less than 11. So that wouldn't make them peripheral.Hmm, maybe I need a different structure. What if I have multiple hubs connected in a chain? Like, a central hub connected to several other hubs, each of which is connected to more cities. But I need to ensure that the shortest path between any two peripheral cities is exactly 11.Wait, maybe I should think about the concept of eccentricity in graphs. The eccentricity of a node is the maximum distance from that node to any other node. A peripheral node is one where its eccentricity is equal to the diameter of the graph. So, in this case, the diameter is 11, so a peripheral city would have eccentricity 11.So, to maximize the number of peripheral cities, I need as many cities as possible to have eccentricity 11. That means each of these cities must be at a distance of 11 from some other city.But how can I arrange the graph so that many cities have eccentricity 11? Maybe by having a tree where many leaves are at distance 11 from each other.Wait, but in a tree, the number of leaves is limited by the structure. If I have a central node connected to several branches, each branch can have a certain number of nodes. If I want the leaves to be at distance 11 from each other, the branches need to be long enough.Let me try to visualize this. Suppose I have a central node connected to k branches. Each branch has a path of length 11 from the central node to a leaf. So, each branch has 12 nodes (including the central node). Wait, no, if the central node is connected to a branch, and the branch has length 11, then the branch has 11 edges and 12 nodes. But if I have k branches, each with 12 nodes, the total number of nodes would be 1 + k*11. But we have 100 cities, so 1 + k*11 ≤ 100. That would mean k ≤ (100 - 1)/11 ≈ 9. So, k = 9, giving 1 + 9*11 = 100 nodes. Wait, that's exactly 100 nodes.So, if I have a central node connected to 9 branches, each branch being a path of 11 edges (12 nodes), then the total number of nodes is 1 + 9*11 = 100. In this case, the leaves of each branch are at distance 11 from the central node, and the distance between any two leaves from different branches is 11 + 11 = 22, which is way more than 11. But the problem states that any two cities can be reached with no more than 11 transfers. So, this structure doesn't satisfy the condition because some pairs of cities would require more than 11 transfers.Wait, that's a problem. So, my initial idea of a star-like tree with long branches doesn't work because it violates the diameter condition.I need a different approach. Maybe instead of having a central hub, I need a more balanced tree where the maximum distance between any two nodes is 11. But how can I maximize the number of leaves (peripheral cities) in such a tree?In a tree, the number of leaves is related to the number of internal nodes and the degrees of those internal nodes. For a tree with maximum depth 11, the maximum number of leaves would be achieved when the tree is as "bushy" as possible, i.e., each internal node has as many children as possible.But I'm not sure if that's the right way to think about it. Maybe I should consider that in a tree with diameter 11, the maximum number of leaves is achieved when the tree is a caterpillar tree, where all nodes are within distance 1 of a central path.Wait, but I'm not sure. Maybe I need to think about the properties of trees with a given diameter and maximum number of leaves.Alternatively, maybe I can model this as a graph where each peripheral city is connected in such a way that their shortest paths to each other are exactly 11, and all other paths are shorter.But I'm getting stuck. Maybe I should look for a different approach.Let's think about the problem differently. We need to maximize the number of pairs of cities that are at distance exactly 11. Each such pair contributes two peripheral cities. But we need to ensure that the total number of cities is 100, and that the graph remains connected with diameter 11.Wait, but if I have too many such pairs, the graph might not be connected or might exceed the diameter.Alternatively, maybe I can arrange the cities in a cycle. If I have a cycle of 100 cities, the diameter would be 50, which is way more than 11. So that's not good.What if I arrange the cities in a grid? But grids have larger diameters, so that might not work either.Wait, maybe I can use a hypercube structure. A hypercube of dimension d has 2^d nodes and diameter d. But 2^7 = 128, which is more than 100, so a 7-dimensional hypercube would have diameter 7, which is less than 11. But we need diameter 11, so that's not helpful.Hmm, maybe I need to think about the properties of the graph. Since the diameter is 11, the maximum distance between any two nodes is 11. Also, the shortest path between any two nodes is unique.Wait, unique shortest paths. That's an important condition. So, for any two nodes, there's only one shortest path connecting them. That suggests that the graph is a tree, because in a tree, there's exactly one path between any two nodes. But in a tree, the diameter is the longest path between any two nodes, which would be 11 in this case.But earlier, I thought that a star-like tree with long branches would have a diameter larger than 11, but maybe I was wrong.Wait, let's clarify. If I have a tree with a central node connected to several branches, each branch being a path of length 11, then the diameter of the tree would be the longest path between any two leaves, which would be 11 + 11 = 22, which is way more than 11. So that's not acceptable.But the problem says that the diameter is at most 11. So, the longest path between any two nodes must be 11.Therefore, my initial idea of a star-like tree with branches of length 11 is invalid because it would result in a diameter of 22.So, I need a tree where the longest path between any two nodes is exactly 11. How can I construct such a tree?One way is to have a central path of length 11, and then attach additional nodes to this path in such a way that no new path exceeds length 11.Wait, but if I attach nodes to the central path, I have to ensure that the distance from any attached node to any other node doesn't exceed 11.Let me try to visualize this. Suppose I have a central path with 12 nodes (A1, A2, ..., A12). The distance between A1 and A12 is 11. Now, if I attach a node B1 to A1, the distance from B1 to A12 is 12, which exceeds the diameter. So that's not allowed.Similarly, if I attach a node to A2, the distance from that node to A12 would be 10, which is acceptable. Wait, no, the distance from A2 to A12 is 10, so attaching a node to A2 would make its distance to A12 11, which is acceptable.Wait, let's see:- If I attach a node to A1, its distance to A12 is 12, which is too long.- If I attach a node to A2, its distance to A12 is 11, which is acceptable.Similarly, attaching nodes to A3 would make their distance to A12 10, which is fine.Wait, but if I attach nodes to A2, their distance to A1 is 2, which is fine, but their distance to A12 is 11, which is acceptable.So, maybe I can attach nodes to A2 through A11, but not to A1 or A12.Wait, but if I attach a node to A11, its distance to A1 is 11, which is acceptable.Similarly, attaching a node to A2, its distance to A12 is 11, which is acceptable.So, perhaps I can attach nodes to A2 through A11, ensuring that their distance to the farthest end is 11.But then, how many nodes can I attach to each of these?If I attach k nodes to each of A2 through A11, then the total number of nodes would be 12 + 10k.We have 100 cities, so 12 + 10k = 100 ⇒ 10k = 88 ⇒ k = 8.8, which is not an integer. So, k = 8 would give 12 + 80 = 92 nodes, leaving 8 more nodes to attach.Alternatively, maybe I can attach 9 nodes to some of them.But this is getting complicated. Maybe there's a better way.Alternatively, perhaps the graph is a tree where the longest path is 11, and as many nodes as possible are leaves at distance 11 from some node.Wait, but in a tree, the number of leaves is related to the number of internal nodes. For a tree with maximum degree d, the maximum number of leaves is d times the number of internal nodes.But I'm not sure.Wait, maybe I can think of the tree as having a central path of length 11, and then attaching as many leaves as possible to the nodes along this path, ensuring that the distance from any leaf to any other node doesn't exceed 11.So, if I have a central path A1-A2-...-A12, and I attach leaves to A2 through A11, then each leaf attached to A2 is at distance 1 from A2, and distance 11 from A12.Similarly, leaves attached to A11 are at distance 11 from A1.But leaves attached to A2 are at distance 12 from A12 if they are attached to A2, which would exceed the diameter. Wait, no, because the distance from A2 to A12 is 10, so a leaf attached to A2 is at distance 11 from A12, which is acceptable.Similarly, a leaf attached to A11 is at distance 11 from A1.So, if I attach leaves to A2 through A11, each leaf is at distance 11 from one end of the central path.Therefore, the number of leaves I can attach is limited by the number of nodes along the central path where I can attach them without exceeding the diameter.So, if I have a central path of 12 nodes, I can attach leaves to A2 through A11, which is 10 nodes. Each of these can have multiple leaves attached.But how many leaves can I attach to each?If I attach k leaves to each of A2 through A11, then the total number of leaves is 10k.The total number of nodes is 12 + 10k.We have 100 nodes, so 12 + 10k = 100 ⇒ 10k = 88 ⇒ k = 8.8, which is not an integer. So, k = 8 gives 88 leaves, total nodes 100.Wait, 12 + 10*8 = 92, so we have 8 more nodes to attach. Maybe we can attach 9 leaves to some nodes.Alternatively, maybe we can have a slightly different structure.Wait, but if we attach 8 leaves to each of A2 through A11, that's 80 leaves, plus the central path of 12, totaling 92 nodes. We have 8 more nodes to attach. Maybe we can attach one more leaf to 8 of the nodes, making it 9 leaves on those 8 nodes, and 8 leaves on the remaining 2 nodes. But that complicates things.Alternatively, maybe we can have a central path of 11 edges (12 nodes), and attach 9 leaves to each of A2 through A11, which would give 9*10 = 90 leaves, plus the central path of 12, totaling 102 nodes, which is more than 100. So that's not possible.Wait, maybe I'm overcomplicating this. Let's think differently.If I have a central path of 12 nodes, and I can attach leaves to A2 through A11, each leaf being at distance 11 from one end.The maximum number of leaves would be when each of A2 through A11 has as many leaves as possible.But since we have 100 nodes, and the central path is 12, we have 88 nodes left to attach as leaves.If we attach 8 leaves to each of A2 through A11, that's 8*10 = 80 leaves, totaling 12 + 80 = 92 nodes. We have 8 more nodes to attach. Maybe we can attach one more leaf to 8 of the nodes, making it 9 leaves on those 8 nodes, and 8 leaves on the remaining 2 nodes. But that would give 8*9 + 2*8 = 72 + 16 = 88 leaves, plus the central path of 12, totaling 100 nodes.So, in this case, we have 10 nodes (A2 through A11) with varying numbers of leaves attached. Some have 9 leaves, some have 8.In this structure, the leaves attached to A2 are at distance 11 from A12, and the leaves attached to A11 are at distance 11 from A1. The leaves attached to A3 through A10 are at distance 11 from both ends? Wait, no. For example, a leaf attached to A3 is at distance 2 from A1 and distance 9 from A12, so its eccentricity is 9, not 11. Therefore, it's not a peripheral city.Wait, so only the leaves attached to A2 and A11 have eccentricity 11, because they are at distance 11 from one end. The leaves attached to A3 through A10 have eccentricity less than 11, so they are not peripheral.Therefore, in this structure, only the leaves attached to A2 and A11 are peripheral cities. So, how many are they?If we have 8 nodes with 9 leaves and 2 nodes with 8 leaves, then the leaves attached to A2 and A11 would be 9 + 9 = 18, assuming A2 and A11 are among the nodes with 9 leaves.Wait, no, because A2 and A11 are specific nodes. If we have 10 nodes (A2 through A11), and we assign 8 of them to have 9 leaves and 2 to have 8 leaves, then A2 and A11 could be among the ones with 9 leaves, giving 9 + 9 = 18 peripheral cities.But wait, if A2 has 9 leaves, each of those leaves is at distance 11 from A12, making them peripheral. Similarly, A11 has 9 leaves, each at distance 11 from A1, making them peripheral. So, total peripheral cities would be 9 + 9 = 18.But we have 100 cities, and in this structure, we have 12 + 88 = 100 nodes. So, 18 peripheral cities.But is this the maximum? Maybe not. Maybe there's a way to have more peripheral cities.Wait, perhaps if we have multiple central paths or a different structure where more nodes have eccentricity 11.Alternatively, maybe instead of a single central path, we can have multiple paths connected in a way that more nodes are at distance 11 from each other.But I'm not sure how to do that without exceeding the diameter.Wait, another idea: if I have a tree where multiple nodes are at distance 11 from each other, not just the leaves.But in a tree, the eccentricity of a node is the maximum distance to any other node. So, if I have a node that is at distance 11 from multiple other nodes, those other nodes would also have to have eccentricity at least 11.But I'm not sure how to arrange that.Alternatively, maybe I can have a tree where there are multiple nodes with eccentricity 11, not just the leaves.But I'm not sure.Wait, maybe I can think of the tree as having a central node connected to several branches, each of length 11. But earlier, I saw that this would result in a diameter of 22, which is too large.But if I limit the branches to length 5, then the diameter would be 10, which is less than 11. So, that's not helpful.Wait, maybe if I have a central node connected to several branches, each of length 5, and then each of those branches has a sub-branch of length 6, making the total length from the central node to the leaf 11.But then, the distance between two leaves on different branches would be 11 + 11 = 22, which is too large.So, that's not acceptable.Hmm, this is tricky.Wait, maybe I need to consider that in a tree with diameter 11, the maximum number of leaves is achieved when the tree is a caterpillar tree, where all nodes are within distance 1 of a central path.In a caterpillar tree, the central path is called the spine, and all other nodes are leaves attached to the spine.So, if I have a spine of length 11 (12 nodes), and attach as many leaves as possible to each spine node, ensuring that the distance from any leaf to any other node doesn't exceed 11.But in this case, the leaves attached to the spine nodes near the ends would be at distance 11 from the opposite end.So, for example, leaves attached to the second node on the spine would be at distance 11 from the last node on the spine.Similarly, leaves attached to the second-to-last node would be at distance 11 from the first node on the spine.Therefore, in this structure, the leaves attached to the second and second-to-last nodes on the spine would be peripheral cities.The number of such leaves would depend on how many we can attach without exceeding the total number of nodes.So, if the spine has 12 nodes, and we attach k leaves to each of the second and second-to-last nodes, then the total number of nodes would be 12 + 2k.We have 100 nodes, so 12 + 2k = 100 ⇒ 2k = 88 ⇒ k = 44.So, we can attach 44 leaves to the second node and 44 leaves to the second-to-last node, making a total of 88 leaves, plus the 12 spine nodes, totaling 100 nodes.In this case, the 44 leaves attached to the second node are at distance 11 from the last node on the spine, and the 44 leaves attached to the second-to-last node are at distance 11 from the first node on the spine.Therefore, all 88 leaves are peripheral cities because their eccentricity is 11.Additionally, the first and last nodes on the spine have eccentricity 11 because their distance to the farthest leaf is 11.Wait, no. The first node on the spine is connected to the second node, which has 44 leaves. The distance from the first node to any of those leaves is 2, which is much less than 11. So, the first node's eccentricity is the distance to the last node on the spine, which is 11. Similarly, the last node's eccentricity is 11.So, the first and last nodes on the spine are also peripheral cities.Therefore, in this structure, we have 88 leaves (44 on each side) plus 2 spine nodes (first and last), totaling 90 peripheral cities.Wait, that seems like a lot. Let me check.Total nodes: 12 spine nodes + 44 leaves on second node + 44 leaves on second-to-last node = 12 + 44 + 44 = 100.Peripheral cities: 44 (from second node) + 44 (from second-to-last node) + 2 (first and last spine nodes) = 90.Yes, that seems to fit.But wait, are the first and last spine nodes considered peripheral? Their eccentricity is 11 because their distance to the farthest leaf is 11. So, yes, they are peripheral.Therefore, in this structure, we have 90 peripheral cities.Is this the maximum possible?Well, in this structure, we've maximized the number of leaves attached to the second and second-to-last nodes, which are the only nodes that can have leaves at distance 11 from the opposite end.By attaching 44 leaves to each, we've used up all the remaining nodes (100 - 12 = 88) as leaves, plus the first and last spine nodes, making 90 peripheral cities.I don't think we can have more than 90 because we've used up all the available nodes as leaves or spine ends.Therefore, the maximum number of peripheral cities is 90.Wait, but earlier I thought it was 89. Did I make a mistake?Wait, in the previous calculation, I had 12 spine nodes, 44 leaves on each of two nodes, totaling 12 + 88 = 100. Then, the peripheral cities are the 88 leaves plus the first and last spine nodes, totaling 90.But the problem says that if two cities cannot be reached in 10 or fewer transfers, both are called peripheral. So, in this structure, the first and last spine nodes are peripheral because their distance to the farthest leaf is 11.But are the leaves attached to the second node peripheral? Yes, because their distance to the last spine node is 11.Similarly, the leaves attached to the second-to-last node are peripheral because their distance to the first spine node is 11.So, all 88 leaves plus the first and last spine nodes are peripheral, totaling 90.Therefore, the maximum number of peripheral cities is 90.But wait, the problem says "if it is not possible to reach city A from city B with 10 or fewer transfers, we call both cities peripheral."So, for two cities to be peripheral, they must be at distance exactly 11.In our structure, the first spine node is at distance 11 from the last spine node. Also, the first spine node is at distance 11 from the leaves attached to the second-to-last node.Similarly, the last spine node is at distance 11 from the leaves attached to the second node.The leaves attached to the second node are at distance 11 from the last spine node and the leaves attached to the second-to-last node.Wait, no, the leaves attached to the second node are at distance 11 from the last spine node, but their distance to the leaves attached to the second-to-last node is 11 + 11 = 22, which is way more than 11. But the problem states that any two cities can be reached with no more than 11 transfers, so this would violate the condition.Wait, that's a problem. In this structure, some pairs of leaves are at distance 22, which exceeds the diameter of 11.Therefore, this structure is invalid because it violates the condition that any two cities can be reached with no more than 11 transfers.So, my previous idea is incorrect because it results in some pairs of cities being too far apart.Therefore, I need to find a structure where the maximum distance between any two cities is 11, and as many cities as possible are at distance 11 from some other city.Wait, maybe instead of attaching all the leaves to just two nodes, I can distribute them more evenly along the spine so that the distance between any two leaves is at most 11.But how?If I have a spine of 12 nodes, and I attach leaves to multiple spine nodes, ensuring that the distance between any two leaves is at most 11.For example, if I attach leaves to spine nodes A2, A3, ..., A11, then the distance between a leaf attached to A2 and a leaf attached to A11 is 10 + 1 + 10 = 21, which is too large.Wait, no, the distance between a leaf attached to A2 and a leaf attached to A11 would be the distance from A2 to A11 plus 2, which is 9 + 2 = 11.Wait, let me clarify.If I have a spine A1-A2-...-A12.A leaf attached to A2 is at distance 1 from A2, and a leaf attached to A11 is at distance 1 from A11.The distance between these two leaves is the distance from A2 to A11 plus 2, which is 9 + 2 = 11.Similarly, the distance between a leaf attached to A3 and a leaf attached to A10 is 7 + 2 = 9, which is less than 11.Wait, so if I attach leaves to multiple spine nodes, the distance between any two leaves is at most 11.Therefore, in this case, all leaves attached to the spine nodes from A2 to A11 are at distance 11 from some other leaf.Therefore, all these leaves are peripheral cities.Additionally, the first and last spine nodes (A1 and A12) are also peripheral because their distance to the farthest leaf is 11.So, in this structure, the number of peripheral cities would be:- All leaves attached to A2 through A11: let's say k leaves per spine node.- Plus A1 and A12.Total nodes: 12 spine nodes + 10k leaves = 100 ⇒ 10k = 88 ⇒ k = 8.8, which is not an integer.So, k = 8 gives 80 leaves, plus 12 spine nodes, totaling 92. We have 8 more nodes to attach. Maybe we can attach one more leaf to 8 of the spine nodes, making it 9 leaves on those 8 nodes, and 8 leaves on the remaining 2 nodes.But in this case, the leaves attached to A2 through A11 are all peripheral because their distance to some other leaf is 11.Additionally, A1 and A12 are peripheral.So, total peripheral cities would be:- 8 leaves on 2 spine nodes: 8*2 = 16- 9 leaves on 8 spine nodes: 9*8 = 72- Plus A1 and A12: 2Total: 16 + 72 + 2 = 90Wait, that's the same as before, but now the distance between any two leaves is at most 11, satisfying the diameter condition.Therefore, in this structure, we have 90 peripheral cities.But earlier, I thought that attaching leaves to multiple spine nodes would result in some pairs of leaves being too far apart, but actually, the distance between any two leaves is at most 11, so it's acceptable.Therefore, the maximum number of peripheral cities is 90.Wait, but earlier I thought it was 89. Did I make a mistake?Wait, in the previous calculation, I had 12 spine nodes, 44 leaves on each of two nodes, totaling 100 nodes, but that resulted in some pairs of leaves being too far apart.But in this new structure, attaching leaves to multiple spine nodes ensures that the distance between any two leaves is at most 11, satisfying the diameter condition, and allowing us to have 90 peripheral cities.Therefore, the maximum number of peripheral cities is 90.But wait, let me double-check.If I have a spine of 12 nodes, and attach 8 leaves to each of A2 through A11, that's 8*10 = 80 leaves, plus 12 spine nodes, totaling 92. Then, attach 8 more leaves to 8 of the spine nodes, making it 9 leaves on those 8 nodes, and 8 leaves on the remaining 2 nodes.So, total leaves: 8*2 + 9*8 = 16 + 72 = 88Plus spine nodes: 12Total: 100Peripheral cities:- All 88 leaves: because each leaf is at distance 11 from some other leaf.- Plus A1 and A12: because their distance to the farthest leaf is 11.Total: 88 + 2 = 90Yes, that seems correct.Therefore, the largest number of peripheral cities that can exist in the land is 90.

answer：Okay, so I have this function f(x) = a^x + 1/(a^x), where a is greater than 1. I need to figure out two things: first, whether the function is even or odd, and whether it's increasing or decreasing. Second, I need to solve the inequality f(x - 1) > f(2x + 1). Hmm, let's take it step by step.Starting with part (1): determining the parity and monotonicity of f(x). Parity refers to whether the function is even or odd. I remember that an even function satisfies f(-x) = f(x), and an odd function satisfies f(-x) = -f(x). So, let me compute f(-x) and see what happens.f(-x) = a^{-x} + 1/(a^{-x}). Hmm, a^{-x} is the same as 1/(a^x), right? So, f(-x) = 1/(a^x) + a^x. Wait, that's exactly the same as f(x) = a^x + 1/(a^x). So, f(-x) = f(x), which means the function is even. Got it, so f(x) is an even function.Now, for monotonicity. Monotonicity refers to whether the function is increasing or decreasing over its domain. Since a > 1, a^x is an exponential function that increases as x increases. Similarly, 1/(a^x) is the same as a^{-x}, which decreases as x increases because the exponent is negative. So, f(x) is the sum of an increasing function and a decreasing function. Hmm, that's a bit tricky. I need to figure out whether the overall function is increasing or decreasing.Maybe I can take the derivative of f(x) to determine its monotonicity. Let's try that. The derivative f’(x) would be the derivative of a^x plus the derivative of 1/(a^x). The derivative of a^x is ln(a) * a^x, and the derivative of 1/(a^x) is -ln(a) * a^{-x}. So, f’(x) = ln(a) * a^x - ln(a) * a^{-x}.Factor out ln(a): f’(x) = ln(a) * (a^x - a^{-x}). Since a > 1, ln(a) is positive. So, the sign of f’(x) depends on (a^x - a^{-x}).Let's analyze (a^x - a^{-x}). If x > 0, then a^x > 1 and a^{-x} < 1, so a^x - a^{-x} is positive. Therefore, f’(x) is positive when x > 0, meaning f(x) is increasing for x > 0.If x < 0, then a^x < 1 and a^{-x} > 1, so a^x - a^{-x} is negative. Therefore, f’(x) is negative when x < 0, meaning f(x) is decreasing for x < 0.At x = 0, f’(0) = ln(a) * (1 - 1) = 0. So, the function has a critical point at x = 0.Putting it all together, f(x) is decreasing on (-∞, 0) and increasing on (0, ∞). So, it's an even function with a minimum at x = 0.Alright, that was part (1). Now, moving on to part (2): solving the inequality f(x - 1) > f(2x + 1). Since f(x) is even, I know that f(x) = f(-x). Also, from the monotonicity, f(x) is decreasing for x < 0 and increasing for x > 0.So, to solve f(x - 1) > f(2x + 1), I need to consider the behavior of f(x) based on the arguments x - 1 and 2x + 1.First, let's note that since f(x) is even, f(x - 1) = f(|x - 1|) and f(2x + 1) = f(|2x + 1|). Therefore, the inequality f(x - 1) > f(2x + 1) is equivalent to f(|x - 1|) > f(|2x + 1|).Since f(x) is decreasing on (-∞, 0) and increasing on (0, ∞), and it's even, the function f(|x|) is increasing for x ≥ 0. So, if I have f(|x - 1|) > f(|2x + 1|), this implies that |x - 1| > |2x + 1| because f is increasing for positive arguments.So, the inequality reduces to |x - 1| > |2x + 1|. Now, I need to solve this inequality.To solve |x - 1| > |2x + 1|, I can square both sides to eliminate the absolute values, since both sides are non-negative. Squaring both sides gives:(x - 1)^2 > (2x + 1)^2.Let's expand both sides:Left side: (x - 1)^2 = x^2 - 2x + 1.Right side: (2x + 1)^2 = 4x^2 + 4x + 1.So, the inequality becomes:x^2 - 2x + 1 > 4x^2 + 4x + 1.Subtracting the left side from both sides:0 > 4x^2 + 4x + 1 - x^2 + 2x - 1.Simplify:0 > 3x^2 + 6x.Which is the same as:3x^2 + 6x < 0.Factor out 3x:3x(x + 2) < 0.So, the inequality is 3x(x + 2) < 0. Since 3 is positive, we can divide both sides by 3 without changing the inequality:x(x + 2) < 0.Now, we need to find the values of x for which x(x + 2) < 0. Let's find the critical points where the expression equals zero: x = 0 and x = -2.These points divide the real line into three intervals:1. x < -22. -2 < x < 03. x > 0We can test each interval to see where the product is negative.1. For x < -2, let's pick x = -3: (-3)(-3 + 2) = (-3)(-1) = 3 > 0.2. For -2 < x < 0, let's pick x = -1: (-1)(-1 + 2) = (-1)(1) = -1 < 0.3. For x > 0, let's pick x = 1: (1)(1 + 2) = 3 > 0.So, the inequality x(x + 2) < 0 holds when -2 < x < 0.Therefore, the solution to the inequality |x - 1| > |2x + 1| is -2 < x < 0.But wait, I should check if squaring both sides introduced any extraneous solutions. Let me verify with a test value in the interval (-2, 0), say x = -1.Compute |x - 1| = |-1 - 1| = | -2 | = 2.Compute |2x + 1| = |2*(-1) + 1| = |-2 + 1| = | -1 | = 1.So, 2 > 1, which is true.Now, let's check a value outside the interval, say x = 1.|x - 1| = |1 - 1| = 0.|2x + 1| = |2*1 + 1| = 3.So, 0 > 3 is false, which is consistent.Another test at x = -3:|x - 1| = |-3 -1| = 4.|2x + 1| = |-6 +1| = 5.So, 4 > 5 is false, which is consistent.Therefore, the solution is indeed -2 < x < 0.So, putting it all together, the solution to the inequality f(x - 1) > f(2x + 1) is all real numbers x such that -2 < x < 0.**Final Answer**The solution to the inequality is boxed{(-2, 0)}.