💡Alright, so I have this problem about two perpendicular planes and four propositions about lines within those planes. I need to figure out how many of these propositions are correct. Let me try to break this down step by step.First, I remember that when two planes are perpendicular, it means that they intersect at a right angle. The line where they intersect is called the line of intersection. I think this is important because it might relate to some of the propositions.Let me look at each proposition one by one.**Proposition (1):** "A known line within one plane must be perpendicular to any line within the other plane."Hmm, okay. So if I have a line in one plane, does it have to be perpendicular to every single line in the other plane? That seems too strong. I mean, if the two planes are perpendicular, it doesn't necessarily mean that every line in one plane is perpendicular to every line in the other plane. I think there must be some specific conditions for that.Wait, I remember something about the line of intersection. If a line in one plane is perpendicular to the line of intersection, then it would be perpendicular to the other plane. But does that mean it's perpendicular to every line in the other plane? I'm not entirely sure, but I think it might only be perpendicular to lines that are also perpendicular to the line of intersection in the other plane. So, maybe proposition (1) is not correct because it's too broad.**Proposition (2):** "A known line within one plane must be perpendicular to countless lines within the other plane."Okay, this seems a bit more reasonable. If a line in one plane is perpendicular to the line of intersection, then it would be perpendicular to all lines in the other plane that are also perpendicular to the line of intersection. Since there are infinitely many such lines in the other plane, this proposition might be correct. I think this is talking about the fact that there are countless lines in the other plane that are perpendicular to the given line. So, proposition (2) might be correct.**Proposition (3):** "Any line within one plane must be perpendicular to the other plane."Wait, so any line in one plane is perpendicular to the entire other plane? That doesn't sound right. For a line to be perpendicular to a plane, it needs to be perpendicular to every line in that plane, which is a very strict condition. I think only specific lines in one plane can be perpendicular to the other plane, specifically those that are perpendicular to the line of intersection. So, proposition (3) is probably incorrect because it's too general.**Proposition (4):** "Drawing a perpendicular from any point within one plane to the line of intersection makes this perpendicular necessarily perpendicular to the other plane."Alright, so if I pick any point in one plane and draw a perpendicular to the line of intersection, will that line be perpendicular to the other plane? Let me visualize this. If I have two perpendicular planes intersecting along a line, and I pick a point in one plane, drawing a perpendicular to the line of intersection would mean that this line is perpendicular to the line of intersection. But does that make it perpendicular to the entire other plane?I think so. Because if a line is perpendicular to the line of intersection and lies in one plane, then it should be perpendicular to the other plane. Wait, but does this hold for any point? What if the point is on the line of intersection? Then, drawing a perpendicular from that point to the line of intersection would just be a point, not a line. Hmm, that might be a problem. Or maybe it's still considered perpendicular? I'm not sure. Maybe the proposition assumes that the point is not on the line of intersection. If that's the case, then the perpendicular would indeed be perpendicular to the other plane. But since the proposition says "any point," including those on the line of intersection, it might not always hold. So, proposition (4) might be incorrect.Let me summarize my thoughts:1. Proposition (1): Incorrect because not every line in one plane is perpendicular to every line in the other plane.2. Proposition (2): Correct because there are countless lines in the other plane that are perpendicular to the given line.3. Proposition (3): Incorrect because only specific lines in one plane are perpendicular to the other plane.4. Proposition (4): Incorrect because if the point is on the line of intersection, the perpendicular might not be well-defined or might not be perpendicular to the entire plane.So, only proposition (2) is correct. That means the number of correct propositions is 1.**Final Answer**boxed{1}

💡Okay, so I'm trying to figure out this game between players A and B. Let me start by understanding the rules and the initial setup.We have a blackboard with 2020 copies of the number 1. So, initially, there are 2020 ones on the board. In each round, player A erases two numbers, say x and y, and then player B writes either x + y or |x - y| back on the board. The game ends when one of two conditions is met: either one number is larger than the sum of all the others, or all numbers are zeros. When the game ends, player B gives player A as many cookies as there are numbers on the board. Player A wants to maximize the number of cookies, while player B wants to minimize it.So, my goal is to determine how many cookies player A will receive if both play optimally.First, let me think about what happens in each round. Player A removes two numbers, and player B replaces them with either their sum or their absolute difference. Since player A is trying to maximize the number of cookies, they probably want the game to end with as many numbers as possible on the board. Conversely, player B wants to minimize the number of numbers, so they might prefer operations that reduce the count more quickly.Let me consider the possible operations:1. If player B writes x + y, the number of numbers on the board decreases by one each round because two numbers are removed and one is added.2. If player B writes |x - y|, the number of numbers also decreases by one each round for the same reason.Wait, so regardless of what player B chooses, the number of numbers decreases by one each round. That means the total number of numbers on the board is decreasing by one each time. So, starting from 2020, it will take 2019 rounds to get down to one number, assuming the game doesn't end earlier.But the game can end earlier if one of the two conditions is met. So, the game could end either when all numbers are zeros or when one number is larger than the sum of the others.I need to figure out under optimal play, which condition will cause the game to end, and how many numbers will be on the board at that point.Let me think about the possible strategies for both players.Player A wants to maximize the number of cookies, so they want the game to end with as many numbers as possible. Player B wants to minimize the number of cookies, so they want the game to end with as few numbers as possible.Given that, player A might try to avoid situations where a single number becomes too large, which would cause the game to end under condition 1. Instead, player A might try to keep the numbers balanced so that the game can continue longer, possibly leading to more numbers on the board when it ends.On the other hand, player B might try to create a situation where one number becomes too large, forcing the game to end early with fewer numbers on the board.But wait, in each round, player A chooses which two numbers to erase, and player B chooses which operation to perform. So, player A has some control over which numbers are combined, and player B has control over whether to sum or subtract.Let me think about the possible outcomes.If player B always chooses to subtract, then the numbers on the board will tend to decrease, possibly leading to zeros. If player B always chooses to add, the numbers will tend to increase, possibly leading to a single large number.But player A can influence which numbers are combined. For example, if player A always combines equal numbers, then subtracting would result in zero, which might help in creating more zeros, leading to the game ending under condition 2.Alternatively, if player A combines unequal numbers, then subtracting could result in a non-zero number, which might not lead to zeros as quickly.But player A wants to maximize the number of numbers, so they might prefer operations that don't eliminate numbers too quickly.Wait, but regardless of the operations, each round reduces the number of numbers by one. So, the total number of numbers is decreasing by one each round, regardless of whether player B adds or subtracts.Therefore, the game will end when either all numbers are zeros or one number is larger than the sum of the others.I need to figure out which of these two conditions will occur first under optimal play.Let me consider the initial state: 2020 ones.If player A always combines two ones, then player B can choose to either write 2 or 0.If player B writes 2, then the number of ones decreases by two, and a 2 is added. If player B writes 0, then the number of ones decreases by two, and a 0 is added.Player A wants to maximize the number of numbers, so they might prefer that player B writes 0, because that keeps the number of numbers the same (since two ones are removed, and one zero is added). Wait, no, actually, the number of numbers decreases by one each round, regardless of whether a 0 or 2 is added.Wait, no, actually, starting with 2020 ones, if player A combines two ones, and player B writes a 0, then the total number of numbers becomes 2019: 2018 ones and one zero. If player B writes a 2, then the total number of numbers becomes 2019: 2018 ones and one 2.So, in either case, the number of numbers decreases by one.But if player B writes a 0, then we have an additional zero, which might lead to more zeros in the future, potentially leading to the game ending under condition 2.If player B writes a 2, then we have a larger number, which might lead to the game ending under condition 1 if that number becomes too large.So, player B, wanting to minimize the number of cookies, might prefer to write a 2, hoping to create a large number quickly, which would cause the game to end under condition 1 with fewer numbers on the board.Alternatively, player B might prefer to write a 0, hoping to create more zeros, which would cause the game to end under condition 2 with potentially more numbers on the board.Wait, but if player B writes a 0, that could lead to more zeros, which might actually result in more numbers on the board when the game ends, which is bad for player B. So, player B might prefer to write a 2 to try to create a large number quickly, causing the game to end under condition 1 with fewer numbers.But player A is trying to maximize the number of numbers, so they might try to prevent the creation of a large number by combining equal numbers, which would result in zeros if player B chooses to subtract.Wait, but if player A combines two equal numbers, say two ones, and player B subtracts, resulting in zero, then we get a zero. If player A combines unequal numbers, say a one and a two, and player B subtracts, resulting in one, which is still a small number.Alternatively, if player A combines two ones, and player B adds, resulting in two, which is a larger number.So, player A might prefer to combine equal numbers to force player B into a situation where subtracting results in zero, which could lead to more zeros and potentially more numbers on the board when the game ends.But I'm getting a bit confused. Maybe I should think about the invariant or some property that remains constant or changes in a predictable way.Let me think about the sum of all numbers on the board. Initially, the sum is 2020.In each round, player A removes two numbers x and y, and player B adds either x + y or |x - y|.So, the sum after the operation is (sum - x - y) + (x + y or |x - y|).If player B adds x + y, the sum remains the same: (sum - x - y) + (x + y) = sum.If player B adds |x - y|, the sum becomes (sum - x - y) + |x - y|.Which is sum - 2 * min(x, y).So, the sum can either stay the same or decrease by twice the smaller of the two numbers.Therefore, the sum is non-increasing. It either stays the same or decreases.So, the sum can never increase.Now, the game ends when either one number is larger than the sum of the others or all numbers are zeros.If the game ends with all zeros, the sum is zero. If it ends with one number larger than the sum of the others, then that number must be greater than half the total sum.Wait, let me think. If one number is larger than the sum of all others, then that number must be greater than (sum - that number). So, that number > (sum - that number), which implies that 2 * that number > sum, so that number > sum / 2.Therefore, the game ends when either all numbers are zero or there exists a number greater than sum / 2.Given that the sum is non-increasing, the maximum possible sum at any point is 2020.So, if a number exceeds 1010, the game ends.But wait, initially, the sum is 2020. If a number exceeds 1010, the game ends.But player A wants to avoid that, while player B might try to create such a number.Alternatively, player A might try to create as many zeros as possible, leading to the game ending with all zeros.But how does the number of numbers relate to the sum?If the game ends with all zeros, the number of numbers is equal to the number of zeros, which could be any number from 1 to 2020, depending on how many operations were performed.But if the game ends with one number greater than the sum of the others, then the number of numbers is 2: one large number and the rest.Wait, no. If one number is larger than the sum of all others, it doesn't necessarily mean there are only two numbers. There could be more, but one of them is larger than the sum of the rest.For example, if there are three numbers: 1000, 1000, and 1000. Then, none is larger than the sum of the others. But if one number is 1010, and the others sum to 1010, then 1010 is larger than the sum of the others (which is 1010). Wait, no, because 1010 is equal to the sum of the others. So, it needs to be strictly larger.So, if the sum is S, then a number needs to be greater than S - that number, which implies that 2 * that number > S.Therefore, if the sum is S, the threshold is S / 2.So, if a number exceeds S / 2, the game ends.Given that, the sum is non-increasing, so S can only stay the same or decrease.Therefore, the threshold for ending the game under condition 1 can only decrease or stay the same.So, starting with S = 2020, the threshold is 1010.If a number exceeds 1010, the game ends.If the sum decreases, say to 1000, then the threshold becomes 500, so a number needs to exceed 500 to end the game.But player A wants to avoid the game ending under condition 1, so they want to prevent any number from exceeding half the current sum.Player B, on the other hand, might try to create a number that exceeds half the sum, causing the game to end early with fewer numbers on the board.But how does this play out?Let me think about the possible strategies.Player A can choose which two numbers to combine. If they combine two equal numbers, say two ones, then player B can choose to write 2 or 0.If player B writes 2, then we have a 2, which could potentially be built up further. If player B writes 0, then we have a zero, which doesn't contribute to the sum.Player A might prefer that player B writes 0, as that doesn't create a larger number, keeping the sum lower and avoiding the risk of a number exceeding half the sum.But player B, wanting to minimize the number of cookies, might prefer to write 2, increasing the sum or keeping it the same, which could lead to a larger number being created, potentially ending the game under condition 1.Wait, but writing 2 doesn't increase the sum; it keeps it the same. Writing 0 decreases the sum by 2.So, player B, wanting to minimize the number of numbers, might prefer to write 0, decreasing the sum, which could make it easier to create a number exceeding half the sum in the future.Wait, but decreasing the sum would lower the threshold for ending the game under condition 1. So, if the sum decreases, the threshold becomes lower, making it easier for a number to exceed half the sum.Therefore, player B might prefer to write 0, decreasing the sum, which could lead to the game ending under condition 1 with fewer numbers.Alternatively, writing 2 keeps the sum the same, which might delay the possibility of a number exceeding half the sum.But player A wants to avoid the game ending under condition 1, so they might prefer that player B writes 0, as that decreases the sum, but also creates a zero, which could lead to more zeros in the future, potentially ending the game under condition 2 with more numbers.Wait, this is getting complicated. Maybe I should think about the parity or some invariant.Let me consider the binary representation of the numbers. Since we start with all ones, which are 1 in binary.Each time player A combines two numbers, and player B either adds or subtracts.If player B adds, it's like combining two numbers into a larger one. If player B subtracts, it's like removing the smaller number from the larger one.This reminds me of the process of finding the greatest common divisor (GCD) using the Euclidean algorithm, where you subtract the smaller number from the larger one repeatedly.But in this case, player B can choose to either add or subtract, so it's not exactly the same.Wait, but if player B always subtracts, then the numbers would tend to decrease, potentially leading to zeros.If player B always adds, the numbers would tend to increase, potentially leading to a large number.But player A can choose which numbers to combine, so they might try to combine numbers in a way that forces player B into a situation where subtracting leads to zeros, or adding doesn't create a large number.Alternatively, player A might try to create as many zeros as possible, which would lead to the game ending under condition 2 with more numbers on the board.But how does this process work?Let me think about a smaller example to get some intuition.Suppose we start with 4 ones: [1, 1, 1, 1].Player A's first move: combine two ones. Player B can choose to write 2 or 0.Case 1: Player B writes 2. Now the board is [1, 1, 2].Player A's next move: combine two ones. Player B can write 2 or 0.If player B writes 2 again, the board becomes [2, 2]. Then player A combines them, and player B can write 4 or 0.If player B writes 4, the board is [4], which is larger than the sum of the others (which is 0), so the game ends with 1 number, and player A gets 1 cookie.If player B writes 0, the board becomes [0, 2]. Then player A combines 0 and 2. Player B can write 2 or 2 (since |0 - 2| = 2). So, the board becomes [2], and again, the game ends with 1 number.Case 2: Player B writes 0 in the first move. Now the board is [0, 1, 1].Player A's next move: combine two ones. Player B can write 2 or 0.If player B writes 0, the board becomes [0, 0]. Then player A combines them, and player B writes 0, so the board is [0], and the game ends with 1 number.If player B writes 2, the board becomes [0, 2]. Then player A combines 0 and 2, and player B writes 2, so the board is [2], and the game ends with 1 number.Wait, in both cases, the game ends with 1 number. So, in this small example, regardless of player B's choices, the game ends with 1 number.But that's not helpful. Maybe I need a different approach.Let me think about the number of ones and how they can be combined.Each time player A combines two ones, player B can choose to write a 2 or a 0.If player B writes a 2, the number of ones decreases by two, and a 2 is added.If player B writes a 0, the number of ones decreases by two, and a 0 is added.Player A wants to maximize the number of numbers, so they might prefer that player B writes 0, as that keeps the number of numbers the same (since two ones are removed, and one zero is added). Wait, no, the number of numbers decreases by one each round, regardless of whether a 0 or 2 is added.Wait, starting with 4 ones:- After first round: 3 numbers (either [0, 1, 1] or [2, 1, 1])- After second round: 2 numbers- After third round: 1 numberSo, regardless of player B's choices, it takes 3 rounds to get to 1 number.But in the process, if player B writes 0s, we might end up with more zeros, but the number of numbers is still decreasing by one each round.So, in this case, the number of numbers is determined by the number of rounds, which is fixed as 2019 in the original problem.Wait, but the game can end earlier if one of the conditions is met.So, the game might end before 2019 rounds if either all numbers are zeros or one number exceeds the sum of the others.Therefore, the number of numbers at the end depends on which condition is met first.Player A wants to maximize the number of numbers, so they want the game to end under condition 2 (all zeros) with as many numbers as possible.Player B wants to minimize the number of numbers, so they want the game to end under condition 1 (one number exceeds the sum) with as few numbers as possible.So, the key is to determine which condition is met first under optimal play.Let me think about the sum again.If player B always writes 0, the sum decreases by 2 each time two ones are combined.Starting with sum 2020, each time two ones are combined and a 0 is written, the sum decreases by 2.So, after k rounds of combining two ones and writing 0, the sum becomes 2020 - 2k.The game ends when all numbers are zeros, which would require that all ones are combined into zeros.But to get all zeros, we need to pair up all the ones, which would take 1010 rounds, resulting in 1010 zeros.But wait, each round reduces the number of numbers by one, so starting from 2020, after 1010 rounds, we would have 1010 numbers left.But if all those numbers are zeros, then the game ends with 1010 zeros, so player A gets 1010 cookies.But player B might try to prevent this by sometimes writing 2 instead of 0, increasing the sum or keeping it the same, which could lead to a number exceeding half the sum.Wait, but if player B writes 2, the sum remains the same, so the threshold for ending under condition 1 remains higher.But player A can choose which numbers to combine. If player A always combines two ones, player B can choose to write 0 or 2.If player B writes 2, then we have a 2, which could be combined with another 2 to make 4, and so on, potentially creating a large number.But player A can choose to combine the 2 with a 1, resulting in either 3 or 1, which might not lead to a large number quickly.Alternatively, player A might prefer to combine equal numbers to create zeros, forcing player B to write 0, which keeps the sum decreasing.But player B might resist by sometimes writing 2 instead of 0, trying to build up a large number.This seems like a balance between player A trying to create zeros and player B trying to create a large number.I think the key lies in the binary representation of the initial number of ones, which is 2020.Let me compute the binary representation of 2020.2020 divided by 2 is 1010, remainder 01010 divided by 2 is 505, remainder 0505 divided by 2 is 252, remainder 1252 divided by 2 is 126, remainder 0126 divided by 2 is 63, remainder 063 divided by 2 is 31, remainder 131 divided by 2 is 15, remainder 115 divided by 2 is 7, remainder 17 divided by 2 is 3, remainder 13 divided by 2 is 1, remainder 11 divided by 2 is 0, remainder 1So, writing the remainders from last to first: 11111100100.So, 2020 in binary is 11111100100, which has 7 ones.This might be relevant because the number of ones in the binary representation often relates to the number of operations needed to reduce a number to zero using certain operations.In this case, each time we combine two ones and write a 0, it's similar to a carry operation in binary addition.Therefore, the number of ones in the binary representation might correspond to the minimum number of numbers that can be left on the board when the game ends under condition 2.So, if player A can force the game to end with 7 numbers (the number of ones in the binary representation), then player A would get 7 cookies.But how does this work?Let me think about it.If we start with 2020 ones, which is 2020 in unary.In binary, 2020 is 11111100100, which has 7 ones.Each time we combine two ones and write a 0, it's like reducing the number of ones by two and adding a zero, which doesn't contribute to the count of ones.But actually, in terms of binary representation, combining two ones and writing a 0 is similar to carrying over in binary addition, which reduces the number of ones by two and adds a one in the next higher bit.Wait, no, in binary addition, when you add two ones, you get 0 and carry over 1. So, it's similar to combining two ones and writing a 0, but also creating a carry.But in our game, when player B writes 0, they are effectively removing two ones and adding a zero, which doesn't create a carry. So, it's slightly different.But the number of ones in the binary representation might still be a relevant invariant.Let me think about the process.If player A always combines two ones, and player B always writes 0, then each such operation reduces the number of ones by two and adds a zero.So, starting with 2020 ones, after 1010 such operations, we would have 1010 zeros.But the game would end when all numbers are zeros, which would be 1010 zeros, so player A gets 1010 cookies.But player B might not cooperate and sometimes write 2 instead of 0, which would create a 2 and keep the sum the same.But if player B writes 2, then player A can choose to combine that 2 with another 2, and player B can write 4 or 0.If player B writes 4, then we have a 4, which could be combined further.Alternatively, if player B writes 0, then we have a 0, which doesn't contribute.But player A wants to maximize the number of numbers, so they might prefer that player B writes 0, leading to more zeros.But player B wants to minimize the number of numbers, so they might prefer to write 2, leading to larger numbers.This seems like a stalemate, but perhaps the key is that the number of ones in the binary representation is the minimal number of numbers that can be left when the game ends under condition 2.Therefore, regardless of player B's choices, player A can force the game to end with 7 numbers, corresponding to the 7 ones in the binary representation of 2020.Thus, player A can ensure that the game ends with 7 numbers on the board, so they receive 7 cookies.I think this is the answer, but I'm not entirely sure. Let me try to verify.If we consider that each time we combine two ones and write a 0, it's similar to reducing the number of ones by two and not affecting the higher bits. But in reality, writing a 0 doesn't carry over, so it's not exactly the same as binary addition.However, the number of ones in the binary representation often corresponds to the minimal number of non-zero numbers needed to represent the original number through such operations.Therefore, it's plausible that the minimal number of numbers left when the game ends under condition 2 is equal to the number of ones in the binary representation of the initial count.Hence, the number of cookies player A receives is 7.

💡To evaluate the expression (lfloor (12.1)^2 rfloor - lfloor 12.1 rfloor cdot lfloor 12.1 rfloor), I will follow these steps:1. **Calculate ((12.1)^2):** [ (12.1)^2 = 146.41 ] 2. **Apply the floor function to (146.41):** [ lfloor 146.41 rfloor = 146 ] 3. **Find the floor of (12.1):** [ lfloor 12.1 rfloor = 12 ] 4. **Calculate the square of (lfloor 12.1 rfloor):** [ lfloor 12.1 rfloor cdot lfloor 12.1 rfloor = 12 cdot 12 = 144 ] 5. **Subtract the square of the floor of (12.1) from the floor of ((12.1)^2):** [ lfloor (12.1)^2 rfloor - lfloor 12.1 rfloor cdot lfloor 12.1 rfloor = 146 - 144 = 2 ]

💡Okay, so I have this problem where I'm given two functions, f(x) and g(x), defined on all real numbers. The functional equation provided is:f(x - y) = f(x) * g(y) - g(x) * f(y)And it's also given that f(-2) = f(1) and f(1) is not zero. I need to find the value of g(1) + g(-1). The options are A: -1, B: 1, C: 2, D: -2.Alright, let's start by trying to understand what this functional equation is telling us. It relates the function f at the point x - y to some combination of f and g at points x and y. It seems similar to some trigonometric identities, but I'm not sure yet. Maybe it's related to sine or cosine functions? Or perhaps exponential functions? Hmm.Since f(-2) = f(1), maybe I can plug in specific values for x and y to get some equations involving g(1) and g(-1). Let me try setting x and y to specific values to see if I can get some useful equations.First, let's set y = 1 and x = -2. Then, the equation becomes:f(-2 - 1) = f(-2) * g(1) - g(-2) * f(1)Simplify that:f(-3) = f(-2) * g(1) - g(-2) * f(1)But we know that f(-2) = f(1), so let's substitute that in:f(-3) = f(1) * g(1) - g(-2) * f(1)Factor out f(1):f(-3) = f(1) * (g(1) - g(-2))Since f(1) is not zero, we can divide both sides by f(1):f(-3)/f(1) = g(1) - g(-2) ...(1)Okay, so that's our first equation.Now, let's try another substitution. Let's set y = -1 and x = 1. Then, the equation becomes:f(1 - (-1)) = f(1) * g(-1) - g(1) * f(-1)Simplify that:f(2) = f(1) * g(-1) - g(1) * f(-1)But wait, we know that f(-2) = f(1). Is there a relationship between f(2) and f(-2)? Maybe not directly, but let's see.Wait, actually, if I set x = 2 and y = 1, then:f(2 - 1) = f(2) * g(1) - g(2) * f(1)Simplify:f(1) = f(2) * g(1) - g(2) * f(1)But since f(-2) = f(1), maybe f(2) is related to f(-2)? Hmm, not sure yet.Alternatively, maybe I can express f(2) in terms of f(-2). Let's try setting x = -1 and y = -1:f(-1 - (-1)) = f(-1) * g(-1) - g(-1) * f(-1)Simplify:f(0) = f(-1) * g(-1) - g(-1) * f(-1)Which simplifies to:f(0) = 0So, f(0) = 0. That's a useful piece of information.Now, let's go back to the equation when we set y = -1 and x = 1:f(2) = f(1) * g(-1) - g(1) * f(-1)But we don't know f(2) or f(-1). Maybe we can find another equation involving these terms.Wait, let's set x = 0 and y = 0 in the original equation:f(0 - 0) = f(0) * g(0) - g(0) * f(0)Simplify:f(0) = 0 = 0 * g(0) - g(0) * 0Which is just 0 = 0, so it doesn't give us new information.How about setting x = y? Let's try that. Let x = y, then:f(x - x) = f(x) * g(x) - g(x) * f(x)Simplify:f(0) = 0 = f(x) * g(x) - g(x) * f(x) = 0Again, just 0 = 0, so no new info.Hmm, maybe I can set y = x. Wait, that's the same as above.Alternatively, set y = 0. Let's try that. Let y = 0, then:f(x - 0) = f(x) * g(0) - g(x) * f(0)Simplify:f(x) = f(x) * g(0) - g(x) * 0So:f(x) = f(x) * g(0)Since f(x) is not always zero (given f(1) ≠ 0), we can divide both sides by f(x):1 = g(0)So, g(0) = 1. That's another useful piece of information.Now, let's see if we can find f(-1). Maybe set x = 1 and y = 2:f(1 - 2) = f(1) * g(2) - g(1) * f(2)Simplify:f(-1) = f(1) * g(2) - g(1) * f(2)But we don't know f(2) or f(-1). Maybe we can relate f(2) to f(-2). Let's try setting x = 2 and y = 1:f(2 - 1) = f(2) * g(1) - g(2) * f(1)Simplify:f(1) = f(2) * g(1) - g(2) * f(1)We know f(1) = f(-2), so:f(-2) = f(2) * g(1) - g(2) * f(-2)Let's rearrange this:f(2) * g(1) = f(-2) + g(2) * f(-2)Factor out f(-2):f(2) * g(1) = f(-2) * (1 + g(2))But f(-2) = f(1), so:f(2) * g(1) = f(1) * (1 + g(2))Hmm, not sure if that helps yet.Wait, earlier we had:From equation (1):f(-3)/f(1) = g(1) - g(-2)And from the substitution with y = -1 and x = 1:f(2) = f(1) * g(-1) - g(1) * f(-1)But we also have from x = 1 and y = 2:f(-1) = f(1) * g(2) - g(1) * f(2)So, let's substitute f(2) from the second equation into the first equation.From the second equation:f(-1) = f(1) * g(2) - g(1) * f(2)But f(2) = f(1) * g(-1) - g(1) * f(-1)So substitute:f(-1) = f(1) * g(2) - g(1) * [f(1) * g(-1) - g(1) * f(-1)]Expand:f(-1) = f(1) * g(2) - g(1) * f(1) * g(-1) + [g(1)]^2 * f(-1)Bring all terms involving f(-1) to the left:f(-1) - [g(1)]^2 * f(-1) = f(1) * g(2) - g(1) * f(1) * g(-1)Factor f(-1):f(-1) * [1 - [g(1)]^2] = f(1) * [g(2) - g(1) * g(-1)]Hmm, this is getting complicated. Maybe there's a simpler approach.Wait, let's think about the functional equation again:f(x - y) = f(x) * g(y) - g(x) * f(y)This resembles the sine subtraction formula:sin(a - b) = sin(a)cos(b) - cos(a)sin(b)So, if we think of f(x) as sin(kx) and g(x) as cos(kx), then this functional equation would hold. Is that possible?Let me test this idea. Suppose f(x) = sin(kx) and g(x) = cos(kx). Then:f(x - y) = sin(k(x - y)) = sin(kx - ky) = sin(kx)cos(ky) - cos(kx)sin(ky) = f(x)g(y) - g(x)f(y)Yes, that works! So, f and g could be sine and cosine functions with the same frequency k.Given that, let's see what the condition f(-2) = f(1) implies.f(-2) = sin(-2k) = -sin(2k)f(1) = sin(k)Given that f(-2) = f(1), so:-sin(2k) = sin(k)Which implies:sin(2k) + sin(k) = 0Using the identity sin(2k) = 2 sin(k) cos(k), we get:2 sin(k) cos(k) + sin(k) = 0Factor out sin(k):sin(k)(2 cos(k) + 1) = 0So, either sin(k) = 0 or 2 cos(k) + 1 = 0.If sin(k) = 0, then k = nπ for integer n. But then f(x) = sin(nπ x) would be zero for all x, which contradicts f(1) ≠ 0. So, we discard sin(k) = 0.Thus, 2 cos(k) + 1 = 0 => cos(k) = -1/2So, k = 2π/3 + 2πn or k = 4π/3 + 2πn for integer n.But since k is just a constant, we can take k = 2π/3 for simplicity.Therefore, f(x) = sin(2πx/3) and g(x) = cos(2πx/3)Now, we need to find g(1) + g(-1).Compute g(1):g(1) = cos(2π/3) = -1/2Compute g(-1):g(-1) = cos(-2π/3) = cos(2π/3) = -1/2So, g(1) + g(-1) = (-1/2) + (-1/2) = -1Therefore, the answer is A: -1.Wait, but let me double-check. If f(x) = sin(2πx/3), then f(-2) = sin(-4π/3) = sin(4π/3) = -√3/2, and f(1) = sin(2π/3) = √3/2. But the problem states f(-2) = f(1). However, sin(-4π/3) = -sin(4π/3) = -(-√3/2) = √3/2, which is equal to sin(2π/3). So, yes, f(-2) = f(1) holds.Thus, my conclusion is correct. The value of g(1) + g(-1) is -1.

💡Okay, so I have this problem about an ellipse, and I need to find the distance from point P to the x-axis. Let me try to understand the problem step by step.First, the equation of the ellipse is given as (frac{x^2}{16} + frac{y^2}{9} = 1). I remember that the standard form of an ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) is the semi-major axis and (b) is the semi-minor axis. So in this case, (a^2 = 16) which means (a = 4), and (b^2 = 9) so (b = 3). Next, the foci of the ellipse. I recall that for an ellipse, the distance from the center to each focus is (c), where (c^2 = a^2 - b^2). Plugging in the values, (c^2 = 16 - 9 = 7), so (c = sqrt{7}). Since the ellipse is centered at the origin (because the equation is (frac{x^2}{16} + frac{y^2}{9} = 1)), the foci are located at ((-c, 0)) and ((c, 0)), which means (F_1 = (-sqrt{7}, 0)) and (F_2 = (sqrt{7}, 0)).Now, the problem states that point P is on the ellipse, and the three points P, (F_1), and (F_2) form a right-angled triangle. I need to figure out where this right angle is and then find the distance from P to the x-axis, which is essentially the absolute value of the y-coordinate of P.So, let's think about the possible cases where the right angle can be. In a triangle with three vertices, the right angle can be at any one of the three vertices. So, the right angle could be at P, at (F_1), or at (F_2).Case 1: Right angle at P. That means the lines PF1 and PF2 are perpendicular. Case 2: Right angle at (F_1). That means the lines (F_1P) and (F_1F_2) are perpendicular.Case 3: Right angle at (F_2). That means the lines (F_2P) and (F_2F_1) are perpendicular.I need to analyze each case to see which one gives a valid point P on the ellipse.Let me start with Case 1: Right angle at P.If the right angle is at P, then the vectors PF1 and PF2 are perpendicular. Let me denote point P as ((x, y)). Then, vector PF1 would be ((x + sqrt{7}, y)) and vector PF2 would be ((x - sqrt{7}, y)). For these vectors to be perpendicular, their dot product should be zero.So, the dot product is:[(x + sqrt{7})(x - sqrt{7}) + y cdot y = 0]Simplifying this:[x^2 - 7 + y^2 = 0][x^2 + y^2 = 7]But we also know that P lies on the ellipse, so it must satisfy (frac{x^2}{16} + frac{y^2}{9} = 1). Let me write both equations:1. (x^2 + y^2 = 7)2. (frac{x^2}{16} + frac{y^2}{9} = 1)I can solve these two equations simultaneously. Let me express equation 2 in terms of (x^2) and (y^2):Multiply equation 2 by 144 (the least common multiple of 16 and 9) to eliminate denominators:[9x^2 + 16y^2 = 144]From equation 1, (x^2 = 7 - y^2). Substitute this into the above equation:[9(7 - y^2) + 16y^2 = 144][63 - 9y^2 + 16y^2 = 144][63 + 7y^2 = 144][7y^2 = 81][y^2 = frac{81}{7}][y = pm frac{9sqrt{7}}{7}]So, if the right angle is at P, the y-coordinate of P is (pm frac{9sqrt{7}}{7}), which is approximately (pm 3.4). But wait, the semi-minor axis is 3, so the maximum y-coordinate on the ellipse is 3. So, this result suggests that (y = pm frac{9sqrt{7}}{7}) is approximately (pm 3.4), which is outside the ellipse's bounds because (b = 3). Therefore, this case is not possible because such a point P would lie outside the ellipse.Hmm, that's a problem. Maybe I made a mistake. Let me double-check my calculations.Wait, no, the semi-minor axis is indeed 3, so the maximum y is 3. So, (y = pm frac{9sqrt{7}}{7}) is approximately (pm 3.4), which is beyond 3, so that's not possible. Therefore, Case 1 is invalid because such a point P does not lie on the ellipse.So, moving on to Case 2: Right angle at (F_1).If the right angle is at (F_1), then the vectors (F_1P) and (F_1F_2) are perpendicular. Let me denote point P as ((x, y)). Vector (F_1P) is ((x + sqrt{7}, y)), and vector (F_1F_2) is ((2sqrt{7}, 0)). For these vectors to be perpendicular, their dot product should be zero.So, the dot product is:[(x + sqrt{7})(2sqrt{7}) + y cdot 0 = 0][2sqrt{7}(x + sqrt{7}) = 0]Since (2sqrt{7}) is not zero, we have:[x + sqrt{7} = 0][x = -sqrt{7}]So, point P must have an x-coordinate of (-sqrt{7}). Now, let's find the corresponding y-coordinate. Since P lies on the ellipse, substitute (x = -sqrt{7}) into the ellipse equation:[frac{(-sqrt{7})^2}{16} + frac{y^2}{9} = 1][frac{7}{16} + frac{y^2}{9} = 1][frac{y^2}{9} = 1 - frac{7}{16} = frac{9}{16}][y^2 = frac{81}{16}][y = pm frac{9}{4}]So, the y-coordinate is (pm frac{9}{4}), which is 2.25. That's within the ellipse's bounds since (b = 3). Therefore, in this case, point P is ((- sqrt{7}, pm frac{9}{4})), and the distance from P to the x-axis is (|y| = frac{9}{4}).Now, let's check Case 3: Right angle at (F_2).Similarly, if the right angle is at (F_2), then vectors (F_2P) and (F_2F_1) are perpendicular. Point P is ((x, y)), so vector (F_2P) is ((x - sqrt{7}, y)), and vector (F_2F_1) is ((-2sqrt{7}, 0)). The dot product should be zero:[(x - sqrt{7})(-2sqrt{7}) + y cdot 0 = 0][-2sqrt{7}(x - sqrt{7}) = 0]Again, since (-2sqrt{7}) is not zero, we have:[x - sqrt{7} = 0][x = sqrt{7}]So, point P must have an x-coordinate of (sqrt{7}). Substitute (x = sqrt{7}) into the ellipse equation:[frac{(sqrt{7})^2}{16} + frac{y^2}{9} = 1][frac{7}{16} + frac{y^2}{9} = 1][frac{y^2}{9} = frac{9}{16}][y^2 = frac{81}{16}][y = pm frac{9}{4}]So, similar to Case 2, point P is ((sqrt{7}, pm frac{9}{4})), and the distance from P to the x-axis is (|y| = frac{9}{4}).Wait, so both Case 2 and Case 3 give the same distance from P to the x-axis, which is (frac{9}{4}). But in Case 1, we saw that the y-coordinate was beyond the ellipse's bounds, so that case is invalid.Therefore, the only valid cases are when the right angle is at (F_1) or (F_2), both leading to the same distance from P to the x-axis, which is (frac{9}{4}).But hold on, let me make sure I didn't miss anything. The problem says "the distance from point P to the x-axis," which is the absolute value of the y-coordinate. So, regardless of whether P is above or below the x-axis, the distance is the same, (frac{9}{4}).Looking back at the options given:A: (frac{9}{5}) B: 3 C: (frac{9sqrt{7}}{7}) D: (frac{9}{4})So, (frac{9}{4}) is option D. Therefore, the answer should be D.But just to ensure, let me visualize the ellipse. The major axis is along the x-axis because (a > b). The foci are at ((pm sqrt{7}, 0)). So, when the right angle is at (F_1) or (F_2), the point P is vertically above or below these foci at a distance of (frac{9}{4}). That seems reasonable because (frac{9}{4}) is less than the semi-minor axis length of 3, so it's within the ellipse.I think I covered all the cases, and the calculations seem consistent. So, I feel confident that the answer is D: (frac{9}{4}).**Final Answer**boxed{D}

💡First, I identify the given sets ( M = {1, 2, 5} ) and ( N = {1, 3, 5, 7} ).The union of two sets ( M ) and ( N ), denoted as ( M cup N ), consists of all elements that are in ( M ), in ( N ), or in both.I list the elements of ( M ): 1, 2, 5.I list the elements of ( N ): 1, 3, 5, 7.Combining these elements and removing duplicates, the union ( M cup N ) is ( {1, 2, 3, 5, 7} ).Comparing this result with the provided options, the correct answer is option D.