💡Okay, so I have this sequence problem here. The sequence is called {a_n}, and they give me the first two terms: a₁ is -1 and a₂ is 2. There's also a recurrence relation given, which is a_{n+1} = a_n + a_{n+2}. Hmm, that seems a bit unusual because usually, recurrence relations express a term in terms of previous terms, but this one is expressing a_{n+1} in terms of a_n and a_{n+2}. That might be a bit tricky because a_{n+2} is actually a term that comes after a_{n+1}, so it's kind of going forward in the sequence rather than backward.Let me write down what I know:- a₁ = -1- a₂ = 2- Recurrence relation: a_{n+1} = a_n + a_{n+2}I need to find a_{2016}. That's a pretty large term, so calculating each term one by one up to 2016 isn't practical. Maybe there's a pattern or cycle in the sequence that repeats after a certain number of terms. If I can find the period of the sequence, I can figure out where 2016 falls within that cycle.First, let me try to compute the next few terms to see if a pattern emerges.Starting with a₁ = -1 and a₂ = 2.Let me try to find a₃. Using the recurrence relation:a_{n+1} = a_n + a_{n+2}If I let n = 1, then:a_{2} = a_{1} + a_{3}We know a₂ is 2 and a₁ is -1, so:2 = (-1) + a₃Solving for a₃:a₃ = 2 + 1 = 3Okay, so a₃ is 3.Now, let's find a₄. Let n = 2:a_{3} = a_{2} + a_{4}We know a₃ is 3 and a₂ is 2, so:3 = 2 + a₄Solving for a₄:a₄ = 3 - 2 = 1So, a₄ is 1.Next, let's find a₅. Let n = 3:a_{4} = a_{3} + a_{5}We know a₄ is 1 and a₃ is 3, so:1 = 3 + a₅Solving for a₅:a₅ = 1 - 3 = -2So, a₅ is -2.Now, let's find a₆. Let n = 4:a_{5} = a_{4} + a_{6}We know a₅ is -2 and a₄ is 1, so:-2 = 1 + a₆Solving for a₆:a₆ = -2 - 1 = -3So, a₆ is -3.Continuing, let's find a₇. Let n = 5:a_{6} = a_{5} + a_{7}We know a₆ is -3 and a₅ is -2, so:-3 = (-2) + a₇Solving for a₇:a₇ = -3 + 2 = -1Wait a minute, a₇ is -1, which is the same as a₁. Let's see if the sequence repeats from here.Let's find a₈. Let n = 6:a_{7} = a_{6} + a_{8}We know a₇ is -1 and a₆ is -3, so:-1 = (-3) + a₈Solving for a₈:a₈ = -1 + 3 = 2Hey, a₈ is 2, which is the same as a₂. So, it seems like the sequence is starting to repeat every 6 terms. Let's confirm this by finding a₉.Let n = 7:a_{8} = a_{7} + a_{9}We know a₈ is 2 and a₇ is -1, so:2 = (-1) + a₉Solving for a₉:a₉ = 2 + 1 = 3Which is the same as a₃. So, yes, the sequence repeats every 6 terms. That means the period of the sequence is 6.So, the sequence goes: -1, 2, 3, 1, -2, -3, and then repeats: -1, 2, 3, 1, -2, -3, etc.Therefore, to find a_{2016}, we can find the remainder when 2016 is divided by 6. If the remainder is r, then a_{2016} will be equal to a_r.Let's compute 2016 divided by 6.2016 ÷ 6 = 336 with a remainder of 0.Wait, so 2016 is exactly divisible by 6, which means the remainder is 0. But in our sequence, the terms are indexed starting from 1, so a remainder of 0 would correspond to the 6th term.So, a_{2016} = a_6 = -3.Therefore, the answer should be -3, which is option A.But let me double-check my calculations to make sure I didn't make any mistakes.Starting from a₁ = -1, a₂ = 2.a₃ = a₂ - a₁ = 2 - (-1) = 3.a₄ = a₃ - a₂ = 3 - 2 = 1.a₅ = a₄ - a₃ = 1 - 3 = -2.a₆ = a₅ - a₄ = -2 - 1 = -3.a₇ = a₆ - a₅ = -3 - (-2) = -3 + 2 = -1.a₈ = a₇ - a₆ = -1 - (-3) = -1 + 3 = 2.Yes, it does repeat every 6 terms. So, 2016 divided by 6 is 336 with no remainder, so it's the 6th term, which is -3.I think that's correct. So, the answer is A: -3.

💡Alright, let's tackle this problem step by step. I have the function ( f(x) = cosleft(frac{x}{2} - frac{pi}{2}right) ) and I need to determine which of the statements A, B, C, or D is correct. First, I remember that cosine functions have certain properties, like periodicity, symmetry, and whether they're increasing or decreasing over specific intervals. Let me recall that the general form of a cosine function is ( y = Acos(Bx + C) + D ), where ( A ) affects the amplitude, ( B ) affects the period, ( C ) is the phase shift, and ( D ) is the vertical shift. Looking at my function, ( f(x) = cosleft(frac{x}{2} - frac{pi}{2}right) ), I can rewrite it to make it easier to analyze. Let me factor out the ( frac{1}{2} ) from inside the cosine:( f(x) = cosleft(frac{1}{2}(x - pi)right) )So, this is a cosine function with a horizontal stretch and a phase shift. The coefficient ( frac{1}{2} ) in front of ( x ) affects the period. I remember that the period ( T ) of a cosine function ( cos(Bx + C) ) is ( frac{2pi}{|B|} ). In this case, ( B = frac{1}{2} ), so the period should be:( T = frac{2pi}{frac{1}{2}} = 4pi )Hmm, so the period is ( 4pi ). Let me check statement A: "The smallest positive period of the function ( f(x) ) is ( 2pi )." But I just calculated the period as ( 4pi ), so statement A is incorrect.Next, let's look at statement B: "The function ( f(x) ) is increasing in the interval ( [0, 2pi] )." To determine if the function is increasing or decreasing over an interval, I can find its derivative. The derivative of ( f(x) ) is:( f'(x) = -sinleft(frac{x}{2} - frac{pi}{2}right) times frac{1}{2} = -frac{1}{2}sinleft(frac{x}{2} - frac{pi}{2}right) )Wait, but ( sinleft(theta - frac{pi}{2}right) = -costheta ), so:( f'(x) = -frac{1}{2} times (-cosleft(frac{x}{2}right)) = frac{1}{2}cosleft(frac{x}{2}right) )So, ( f'(x) = frac{1}{2}cosleft(frac{x}{2}right) ). Now, let's analyze the sign of ( f'(x) ) over the interval ( [0, 2pi] ). The cosine function is positive in the first and fourth quadrants, which corresponds to angles between ( -frac{pi}{2} ) and ( frac{pi}{2} ). Let me find the range of ( frac{x}{2} ) when ( x ) is in ( [0, 2pi] ):( frac{x}{2} ) ranges from ( 0 ) to ( pi ).So, ( cosleft(frac{x}{2}right) ) starts at ( cos(0) = 1 ) and decreases to ( cos(pi) = -1 ). Therefore, ( cosleft(frac{x}{2}right) ) is positive from ( 0 ) to ( frac{pi}{2} ) and negative from ( frac{pi}{2} ) to ( pi ).But since ( f'(x) = frac{1}{2}cosleft(frac{x}{2}right) ), the derivative is positive when ( cosleft(frac{x}{2}right) ) is positive and negative when it's negative. Therefore, ( f(x) ) is increasing on ( [0, pi] ) and decreasing on ( [pi, 2pi] ). So, the function isn't increasing throughout the entire interval ( [0, 2pi] ); it first increases and then decreases. Therefore, statement B is incorrect.Moving on to statement C: "The graph of the function ( f(x) ) is symmetric about the line ( x = 0 )." For a function to be symmetric about the y-axis (which is the line ( x = 0 )), it must satisfy ( f(-x) = f(x) ) for all ( x ).Let's test this:( f(-x) = cosleft(frac{-x}{2} - frac{pi}{2}right) = cosleft(-frac{x}{2} - frac{pi}{2}right) )Using the identity ( cos(-theta) = costheta ), we can rewrite this as:( cosleft(frac{x}{2} + frac{pi}{2}right) )But ( cosleft(frac{x}{2} + frac{pi}{2}right) ) is not equal to ( cosleft(frac{x}{2} - frac{pi}{2}right) ) unless specific conditions are met, which aren't generally true for all ( x ). Therefore, ( f(-x) neq f(x) ), so the function isn't symmetric about ( x = 0 ). Hence, statement C is incorrect.Finally, statement D: "The function ( f(x) ) is an odd function." An odd function satisfies ( f(-x) = -f(x) ) for all ( x ).Let's compute ( f(-x) ):( f(-x) = cosleft(frac{-x}{2} - frac{pi}{2}right) = cosleft(-frac{x}{2} - frac{pi}{2}right) )Again, using ( cos(-theta) = costheta ):( cosleft(frac{x}{2} + frac{pi}{2}right) )Now, let's compute ( -f(x) ):( -f(x) = -cosleft(frac{x}{2} - frac{pi}{2}right) )I recall that ( cosleft(theta + frac{pi}{2}right) = -sintheta ) and ( cosleft(theta - frac{pi}{2}right) = sintheta ). Let me verify this:Using the cosine addition formula:( cos(A + B) = cos A cos B - sin A sin B )So, ( cosleft(theta + frac{pi}{2}right) = costheta cosfrac{pi}{2} - sintheta sinfrac{pi}{2} = costheta times 0 - sintheta times 1 = -sintheta )Similarly, ( cosleft(theta - frac{pi}{2}right) = costheta cosfrac{pi}{2} + sintheta sinfrac{pi}{2} = costheta times 0 + sintheta times 1 = sintheta )Therefore, substituting back:( f(-x) = cosleft(frac{x}{2} + frac{pi}{2}right) = -sinleft(frac{x}{2}right) )And:( -f(x) = -cosleft(frac{x}{2} - frac{pi}{2}right) = -sinleft(frac{x}{2}right) )So, ( f(-x) = -f(x) ), which means the function is indeed odd. Therefore, statement D is correct.To recap:- Statement A: Incorrect, period is ( 4pi ).- Statement B: Incorrect, function increases then decreases.- Statement C: Incorrect, not symmetric about ( x = 0 ).- Statement D: Correct, function is odd.**Final Answer**boxed{D}

💡Okay, so I've got this question about two football teams and their defensive performance last year. It's asking me to figure out how many of the four given statements are correct based on the average goals conceded and the standard deviation provided for each team. Let me try to break this down step by step.First, let's list out the data:- **Team 1**: - Average goals conceded per match: 1.5 - Standard deviation of total goals conceded throughout the year: 1.1- **Team 2**: - Average goals conceded per match: 2.1 - Standard deviation of total goals conceded throughout the year: 0.4Now, the four statements to evaluate are:1. On average, Team 1 has better defensive skills than Team 2.2. Team 2 has more stable defensive skills than Team 1.3. Team 1's defense sometimes performs very poorly, and sometimes very well.4. Team 2 rarely concedes no goals.Alright, let's tackle each statement one by one.**Statement (1): On average, Team 1 has better defensive skills than Team 2.**Hmm, okay. So, defensive skills in football can be measured by how few goals a team concedes. The lower the average goals conceded, the better the defense, right? So, Team 1's average is 1.5 goals per match, while Team 2's average is 2.1 goals per match. Since 1.5 is less than 2.1, it does seem like Team 1 concedes fewer goals on average. That would imply that Team 1 has better defensive skills. So, I think statement (1) is correct.**Statement (2): Team 2 has more stable defensive skills than Team 1.**Alright, stability in performance is often measured by the standard deviation. A lower standard deviation means the data points are closer to the mean, indicating more consistency. Team 1 has a standard deviation of 1.1, while Team 2 has a standard deviation of 0.4. Since 0.4 is less than 1.1, Team 2's goals conceded are more tightly clustered around their average. This suggests that Team 2's defensive performance is more stable compared to Team 1. So, statement (2) seems correct as well.**Statement (3): Team 1's defense sometimes performs very poorly, and sometimes very well.**This statement is talking about variability in performance. If a team has a higher standard deviation, it means their performance is more spread out from the average. So, with Team 1 having a standard deviation of 1.1, which is higher than Team 2's 0.4, it indicates that Team 1's defensive performance is more variable. That could mean they have matches where they concede a lot of goals (poor performance) and matches where they concede very few (good performance). So, statement (3) appears to be correct.**Statement (4): Team 2 rarely concedes no goals.**This one is a bit trickier. It's saying that Team 2 doesn't often have matches where they don't concede any goals. To evaluate this, I need to think about what the average and standard deviation tell us about the likelihood of conceding zero goals.Team 2 has an average of 2.1 goals conceded per match. That's relatively high. A standard deviation of 0.4 suggests that their goals conceded don't vary much from the average. So, if they average 2.1 goals per match, and their performance is quite consistent (low standard deviation), it's less likely that they have matches where they concede zero goals. Most of their matches are probably clustered around 2 goals, give or take a few.But wait, is it possible for them to have matches with zero goals conceded? Well, in football, it's not impossible, but if their average is 2.1, and they're consistent, it's less likely. So, I think statement (4) is correct too.Wait, hold on. Let me double-check statement (4). If Team 2 concedes an average of 2.1 goals per match, does that mean they rarely concede zero? It depends on the distribution of their goals conceded. If their goals conceded follow a Poisson distribution, which is common in football, the probability of conceding zero goals can be calculated.The Poisson probability formula is:[ P(k) = frac{lambda^k e^{-lambda}}{k!} ]Where ( lambda ) is the average rate (in this case, 2.1 goals per match), and ( k ) is the number of occurrences (0 goals conceded).So, plugging in the numbers:[ P(0) = frac{2.1^0 e^{-2.1}}{0!} = e^{-2.1} approx 0.122 ]So, approximately 12.2% chance of conceding zero goals in a match. That's not extremely rare, but it's also not very common. However, considering that they have a low standard deviation, their goals conceded are tightly clustered around the mean, which is 2.1. So, the probability of them conceding zero goals is lower than if they had a higher standard deviation.But 12.2% is still a noticeable probability. It's not like they never concede zero goals. So, does "rarely" mean less than 10%, or is 12% considered rare? This is a bit subjective. If we consider "rarely" as something less than 10%, then 12.2% is not rare. But if "rarely" is just less frequent than other outcomes, then maybe it's still considered rare.Given that their average is 2.1, and the standard deviation is 0.4, the distribution is quite peaked around 2 goals. So, the probability of them conceding zero goals is lower than, say, 1 or 2 goals, but it's still not negligible. Therefore, I might lean towards statement (4) being correct, but I'm a bit uncertain here.Wait, let's think about it differently. If Team 2 concedes an average of 2.1 goals per match, and they have a low standard deviation, it means their performance is quite consistent. So, they don't have many matches where they concede a lot or very few. Therefore, the number of matches where they concede zero goals would be relatively low. So, even though 12% is not extremely rare, in the context of their consistent performance, it might still be considered rare compared to other teams or compared to their own average.Hmm, I'm still a bit torn on this one. But considering the low standard deviation, I think it's safe to say that Team 2 doesn't have many matches where they concede zero goals, so statement (4) is correct.Wait, but let's not forget that in football, even teams that concede a lot can have clean sheets (matches where they concede zero goals). It's just that it's less likely. So, if Team 2 concedes an average of 2.1 goals per match, they probably have some matches where they don't concede any goals, but it's not their typical performance. So, "rarely" might be a fair description.Alright, I think I'll go with statement (4) being correct.So, summarizing:1. Correct2. Correct3. Correct4. CorrectWait, that would mean all four statements are correct. But looking back at the options, the choices are A:1, B:2, C:3, D:4. So, is it possible that all four are correct? Let me double-check each statement.**Re-evaluating Statement (4): Team 2 rarely concedes no goals.**Given that their average is 2.1 and standard deviation is 0.4, the probability of conceding zero goals is about 12.2%, as calculated earlier. Is 12.2% considered "rarely"? It's about 1 in 8 matches. In football terms, that's not extremely rare, but it's also not common. It's somewhat infrequent. So, depending on the interpretation of "rarely," this could be subjective. If "rarely" means less than 10%, then it's not rare. If it's just less frequent than other outcomes, then it is rare.Given that the question is likely expecting a more objective answer, and considering that 12.2% is not extremely low, maybe statement (4) is incorrect. So, perhaps only the first three statements are correct.But I'm not entirely sure. It's a bit ambiguous. However, since the standard deviation is low, it does suggest that their performance is tightly around the mean, so the number of matches where they concede zero goals is lower than if the standard deviation were higher. Therefore, even though 12.2% is not extremely low, in the context of their consistent performance, it might still be considered rare.I think I'll stick with statement (4) being correct, making all four statements correct. But I'm still a bit uncertain about statement (4). Maybe the answer expects only three statements to be correct.Wait, let's think about it differently. If Team 2 has a low standard deviation, their goals conceded are tightly clustered around 2.1. So, the probability of them conceding zero goals is lower than if they had a higher standard deviation. Therefore, even though 12.2% is not extremely rare, it's still less likely than if they had a higher standard deviation. So, in that sense, it's relatively rare for them to concede zero goals. So, statement (4) is correct.Alright, I think I'll go with all four statements being correct. So, the answer is D: 4 statements.

💡Okay, so I'm trying to figure out this problem: "In the numbers between a certain number and 50, exactly one number has two of the digits as 5. What is the starting number?" Hmm, okay, let's break this down step by step.First, I need to understand what the question is asking. It says there's a range of numbers starting from some number up to 50, and within this range, there's exactly one number that has two digits which are both 5. My task is to find what that starting number is.Alright, so the range is from some number (let's call it 'X') up to 50. Within this range, there's only one number that has two 5s in its digits. I need to find what 'X' is.Let me think about numbers that have two 5s in their digits. Well, numbers like 55, 155, 255, etc., come to mind. But since we're only going up to 50, the only number that fits is 55. Wait, but 55 is greater than 50, so it's not within the range up to 50. Hmm, that's confusing.Wait, maybe I'm misunderstanding. The problem says "between a certain number and 50." So does that mean the range is from 'X' to 50, inclusive? If so, then 55 isn't in that range because 55 is greater than 50. So, there must be another number between 'X' and 50 that has two 5s in its digits.But wait, in the range up to 50, the only number with two 5s would be 55, which is outside the range. So, maybe the starting number is just below 55, so that when we include 55, it's within the range. But the problem says "between a certain number and 50," which suggests the upper limit is 50, not 55.This is a bit confusing. Let me try to visualize it. If I consider numbers from, say, 50 down to some lower number, I need to find a range where only one number has two 5s in its digits. But since 55 is the only number with two 5s below 100, and it's greater than 50, it's not in the range up to 50.Wait, maybe the starting number is 55, and the range is from 55 to 50, but that doesn't make sense because 55 is greater than 50. Maybe the problem is phrased differently. Perhaps it's saying that in the numbers from 'X' up to 50, there's exactly one number with two 5s. But since 55 isn't in that range, maybe there's no such number, which contradicts the problem statement.Alternatively, maybe I'm missing something. Perhaps the number with two 5s is 50 itself? But 50 only has one 5. Hmm. Maybe the number is 55, but it's outside the range, so the starting number must be just below 55 to include it. But the upper limit is 50, so that doesn't work.Wait, maybe the problem is considering numbers from a lower starting point up to 50, and within that range, there's exactly one number with two 5s. But since 55 is the only such number below 100, and it's above 50, there's no such number in the range up to 50. So, perhaps the starting number is 55, but that doesn't fit because 55 is greater than 50.I'm getting stuck here. Maybe I need to reconsider the problem. Perhaps the number with two 5s is not necessarily a two-digit number. For example, 55 is a two-digit number with two 5s, but maybe there are other numbers with two 5s in their digits, like 155, 255, etc., but those are all above 50. So, in the range up to 50, there are no numbers with two 5s.But the problem says there is exactly one such number in the range. So, maybe the starting number is just below 55, so that when we include 55, it's within the range up to 50. But that doesn't make sense because 55 is greater than 50.Wait, perhaps the problem is phrased differently. Maybe it's saying that in the numbers from 'X' to 50, there's exactly one number with two 5s. But since 55 isn't in that range, maybe 'X' is 55, but that's above 50. This is confusing.Alternatively, maybe the number with two 5s is 50 itself, but 50 only has one 5. So that doesn't work either.I'm going in circles here. Maybe I need to approach this differently. Let's list out numbers from, say, 40 to 50 and see if any have two 5s.40: 4 and 0 – no41: 4 and 1 – no42: 4 and 2 – no43: 4 and 3 – no44: 4 and 4 – no45: 4 and 5 – one 546: 4 and 6 – no47: 4 and 7 – no48: 4 and 8 – no49: 4 and 9 – no50: 5 and 0 – one 5So, between 40 and 50, there are no numbers with two 5s. Therefore, if the starting number is 40, there are zero numbers with two 5s in the range up to 50. But the problem says there's exactly one such number.Wait, maybe the starting number is lower. Let's try starting from 30.30: 3 and 0 – no31: 3 and 1 – no...39: 3 and 9 – no40: 4 and 0 – no...50: 5 and 0 – one 5Still, no numbers with two 5s. So, starting from 30, there are zero numbers with two 5s up to 50.Wait, maybe the starting number is 55, but that's above 50. So, perhaps the problem is considering numbers above 50, but the upper limit is 50, which doesn't make sense.I'm really stuck here. Maybe the problem is misphrased or there's a trick to it. Alternatively, perhaps the number with two 5s is 55, and the starting number is 55, but that's above 50, so it's not included in the range up to 50. Therefore, there are zero numbers with two 5s in the range up to 50, which contradicts the problem statement.Wait, maybe the problem is considering numbers from a higher starting point, like 55, down to 50, making the range from 55 to 50, which includes 55. But that would mean the starting number is 55, and the range is from 55 to 50, which is a bit unconventional since we usually go from lower to higher numbers.But if we consider that, then 55 is included, and it's the only number with two 5s in that range. So, the starting number would be 55.But the problem says "between a certain number and 50," which implies the starting number is below 50. So, this is conflicting.Alternatively, maybe the problem is considering numbers from 50 down to some lower number, and within that range, there's exactly one number with two 5s. But as we've seen, below 50, there are no numbers with two 5s.Wait, unless we consider three-digit numbers, but the problem doesn't specify the number of digits. If we consider three-digit numbers, like 155, 255, etc., but those are all above 50, so they wouldn't be in the range up to 50.I'm really confused. Maybe the problem is designed to have the starting number as 55, even though it's above 50, to include 55 in the range. So, the range would be from 55 to 50, which is a bit unusual, but perhaps that's the intended answer.Alternatively, maybe the problem is misworded, and it should say "between 50 and a certain number," which would make more sense. In that case, the starting number would be 55, and the range would be from 50 to 55, including 55, which has two 5s.But as it stands, the problem says "between a certain number and 50," so the starting number must be below 50. But in that case, there are no numbers with two 5s in the range up to 50, which contradicts the problem statement.Maybe I'm overcomplicating this. Perhaps the starting number is 55, and the range is from 55 to 50, even though it's unconventional. So, the starting number is 55.But I'm not sure. I think the problem might have a typo or is misphrased. Alternatively, maybe I'm missing something obvious.Wait, let's think differently. Maybe the number with two 5s is 50 itself, but as I thought earlier, 50 only has one 5. So that doesn't work.Alternatively, maybe the number is 55, but it's outside the range, so the starting number must be just below 55 to include it. But since the upper limit is 50, that doesn't work.I'm really stuck. Maybe I need to conclude that the starting number is 55, even though it's above 50, to include the only number with two 5s in the range.So, I think the starting number is 55.

💡Okay, so I have this problem here about a parabola intersecting the x-axis at two points, A_n and B_n, for each natural number n. The equation of the parabola is given as y = (n² + n)x² - (2n + 1)x + 1. I need to find the sum of the distances between these two points for n from 1 to 1992. The answer choices are given, so that might help if I get stuck.First, I remember that to find where a parabola intersects the x-axis, I need to set y = 0 and solve for x. So, for each n, I can write the equation as:(n² + n)x² - (2n + 1)x + 1 = 0.This is a quadratic equation in terms of x. I can use the quadratic formula to find the roots. The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a). In this case, a = n² + n, b = -(2n + 1), and c = 1.Plugging these into the quadratic formula:x = [2n + 1 ± sqrt((2n + 1)² - 4*(n² + n)*1)] / [2*(n² + n)].Let me compute the discriminant first, which is the part under the square root:(2n + 1)² - 4*(n² + n)*1.Expanding (2n + 1)² gives 4n² + 4n + 1. Then subtract 4*(n² + n), which is 4n² + 4n. So, 4n² + 4n + 1 - 4n² - 4n = 1. That's nice, the discriminant is 1.So, the roots simplify to:x = [2n + 1 ± 1] / [2*(n² + n)].Calculating both roots:First root: [2n + 1 + 1] / [2*(n² + n)] = (2n + 2) / [2*(n² + n)] = (2(n + 1)) / [2n(n + 1)] = 1/n.Second root: [2n + 1 - 1] / [2*(n² + n)] = (2n) / [2*(n² + n)] = (2n) / [2n(n + 1)] = 1/(n + 1).So, the two points where the parabola intersects the x-axis are at x = 1/n and x = 1/(n + 1). Therefore, the distance between A_n and B_n is |1/n - 1/(n + 1)|.Calculating this distance:|1/n - 1/(n + 1)| = |(n + 1 - n)/(n(n + 1))| = |1/(n(n + 1))| = 1/(n(n + 1)).Wait, hold on, I think I made a mistake here. Because 1/n - 1/(n + 1) is equal to (n + 1 - n)/(n(n + 1)) = 1/(n(n + 1)). So, the distance is 1/(n(n + 1)).But wait, let me double-check. If the roots are at 1/n and 1/(n + 1), then the distance between them is |1/n - 1/(n + 1)|. Let me compute that:1/n - 1/(n + 1) = (n + 1 - n)/(n(n + 1)) = 1/(n(n + 1)). So, yes, that's correct.So, each distance |A_n B_n| is equal to 1/(n(n + 1)). Therefore, the sum we need is the sum from n = 1 to n = 1992 of 1/(n(n + 1)).Hmm, I remember that 1/(n(n + 1)) can be written as 1/n - 1/(n + 1). Let me verify that:1/n - 1/(n + 1) = (n + 1 - n)/(n(n + 1)) = 1/(n(n + 1)). Yes, that's correct.So, the sum becomes:Sum from n = 1 to 1992 of [1/n - 1/(n + 1)].This is a telescoping series, which means that most terms will cancel out. Let me write out the first few terms to see the pattern:For n = 1: 1/1 - 1/2For n = 2: 1/2 - 1/3For n = 3: 1/3 - 1/4...For n = 1992: 1/1992 - 1/1993When we add all these up, the -1/2 from the first term cancels with the +1/2 from the second term, the -1/3 from the second term cancels with the +1/3 from the third term, and so on. This cancellation continues all the way up to the last term.So, after all the cancellations, we are left with the first term of the first expression, which is 1/1, and the last term of the last expression, which is -1/1993.Therefore, the sum simplifies to:1 - 1/1993.Calculating this:1 - 1/1993 = (1993/1993) - (1/1993) = (1993 - 1)/1993 = 1992/1993.So, the total sum of the distances from n = 1 to n = 1992 is 1992/1993.Looking back at the answer choices, option B is 1992/1993, which matches my result.I think that's the correct answer. Let me just recap to make sure I didn't skip any steps or make any mistakes.1. Found the roots of the quadratic equation by setting y = 0.2. Used the quadratic formula and simplified the discriminant to 1.3. Calculated the two roots, which turned out to be 1/n and 1/(n + 1).4. Found the distance between these two points as 1/(n(n + 1)).5. Recognized that this can be expressed as a telescoping series: 1/n - 1/(n + 1).6. Summed the series from n = 1 to 1992, resulting in 1 - 1/1993 = 1992/1993.Everything seems to check out, so I'm confident that the answer is 1992/1993.**Final Answer**boxed{B}

💡Okay, so I have this inequality to prove:[frac{a^{4}+b^{4}}{a^{6}+b^{6}}+frac{b^{4}+c^{4}}{b^{6}+c^{6}}+frac{c^{4}+a^{4}}{c^{6}+a^{6}} leqslant frac{1}{abc}]given that (a), (b), and (c) are positive numbers and their sum is 1, i.e., (a + b + c = 1).Hmm, let's break this down. I need to show that the sum of these three fractions is less than or equal to (frac{1}{abc}). Each fraction has a similar structure: the numerator is the sum of fourth powers, and the denominator is the sum of sixth powers of two variables. First, maybe I can look at one of these fractions individually and see if I can find an upper bound for it. Let's take the first term:[frac{a^{4}+b^{4}}{a^{6}+b^{6}}]I wonder if I can relate this to something simpler. Maybe I can manipulate it to find a relationship with (ab), since the right-hand side of the inequality involves (abc). Let me try to compare (frac{a^{4}+b^{4}}{a^{6}+b^{6}}) with (frac{1}{ab}). If I can show that:[frac{a^{4}+b^{4}}{a^{6}+b^{6}} leq frac{1}{ab}]then adding similar inequalities for the other terms might lead me to the desired result.So, let's cross-multiply to see if this holds:[(a^{4} + b^{4}) cdot ab leq a^{6} + b^{6}]Simplifying the left side:[a^{5}b + ab^{5} leq a^{6} + b^{6}]Hmm, is this true? Let's rearrange the terms:[a^{6} + b^{6} - a^{5}b - ab^{5} geq 0]Factor this expression. Maybe factor by grouping:[a^{6} - a^{5}b + b^{6} - ab^{5}]Factor (a^{5}) from the first two terms and (b^{5}) from the last two:[a^{5}(a - b) + b^{5}(b - a)]Which simplifies to:[(a^{5} - b^{5})(a - b)]Wait, that's interesting. So, we have:[(a^{5} - b^{5})(a - b) geq 0]Now, since (a) and (b) are positive, if (a > b), then (a^{5} > b^{5}) and (a - b > 0), so the product is positive. Similarly, if (a < b), then (a^{5} < b^{5}) and (a - b < 0), so the product is still positive. If (a = b), the expression is zero. So, in all cases, this expression is non-negative. Therefore, the inequality:[frac{a^{4}+b^{4}}{a^{6}+b^{6}} leq frac{1}{ab}]holds true.Great, so I can apply this to each of the three fractions in the original inequality. That means:[frac{a^{4}+b^{4}}{a^{6}+b^{6}} leq frac{1}{ab}][frac{b^{4}+c^{4}}{b^{6}+c^{6}} leq frac{1}{bc}][frac{c^{4}+a^{4}}{c^{6}+a^{6}} leq frac{1}{ca}]Now, if I add these three inequalities together, I get:[frac{a^{4}+b^{4}}{a^{6}+b^{6}} + frac{b^{4}+c^{4}}{b^{6}+c^{6}} + frac{c^{4}+a^{4}}{c^{6}+a^{6}} leq frac{1}{ab} + frac{1}{bc} + frac{1}{ca}]So, I need to show that:[frac{1}{ab} + frac{1}{bc} + frac{1}{ca} leq frac{1}{abc}]Wait, is that true? Let me check. Given that (a + b + c = 1), can I relate (frac{1}{ab} + frac{1}{bc} + frac{1}{ca}) to (frac{1}{abc})?Let's see:[frac{1}{ab} + frac{1}{bc} + frac{1}{ca} = frac{c + a + b}{abc}]Since (a + b + c = 1), this becomes:[frac{1}{abc}]So, indeed:[frac{1}{ab} + frac{1}{bc} + frac{1}{ca} = frac{1}{abc}]Therefore, combining everything, we have:[frac{a^{4}+b^{4}}{a^{6}+b^{6}} + frac{b^{4}+c^{4}}{b^{6}+c^{6}} + frac{c^{4}+a^{4}}{c^{6}+a^{6}} leq frac{1}{ab} + frac{1}{bc} + frac{1}{ca} = frac{1}{abc}]Which is exactly what we needed to prove. So, the inequality holds.I think that makes sense. I started by looking at one term, found an upper bound for it, then applied the same logic to the other terms, summed them up, and used the given condition (a + b + c = 1) to simplify the expression. It all ties together nicely. I don't see any mistakes in the steps, so I believe this is a valid proof.