💡Okay, so I have this problem here: Point A is at (0, 0) and point B is on the line y = 6. The slope of segment AB is 3/4. I need to find the sum of the x- and y-coordinates of point B. Hmm, let me try to figure this out step by step.First, I know that the slope of a line between two points (x1, y1) and (x2, y2) is given by the formula:slope = (y2 - y1) / (x2 - x1)In this case, point A is (0, 0), so x1 is 0 and y1 is 0. Point B is somewhere on the line y = 6, which means its y-coordinate is 6. So, point B has coordinates (x, 6), where x is what I need to find.Given that the slope of AB is 3/4, I can plug these values into the slope formula. Let me write that out:slope = (6 - 0) / (x - 0) = 3/4Simplifying the numerator and denominator:6 / x = 3/4Now, I need to solve for x. To do this, I can cross-multiply:6 * 4 = 3 * xWhich simplifies to:24 = 3xNow, divide both sides by 3:24 / 3 = xSo, x = 8Okay, so point B is at (8, 6). Now, the problem asks for the sum of the x- and y-coordinates of point B. That means I need to add 8 and 6 together.8 + 6 = 14So, the sum is 14.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the slope formula, I substituted the coordinates correctly. The slope is 3/4, and I set up the equation 6/x = 3/4. Cross-multiplying gives 24 = 3x, so x = 8. Adding 8 and 6 gives 14. Yeah, that seems right.I think I got it. The sum of the coordinates of point B is 14.

💡Okay, so I have this problem about a sequence where the sum of the first n terms, denoted as S_n, is given by the formula S_n = 2a_n - 1 for any positive integer n. There are two parts to this problem: first, finding the general formula for the sequence {a_n}, and second, solving an inequality involving combinations and the terms of the sequence.Starting with part (1), I need to find the general formula for a_n. I know that S_n is the sum of the first n terms, so S_n = a_1 + a_2 + ... + a_n. The given formula is S_n = 2a_n - 1. Hmm, that's interesting because it relates the sum of the first n terms directly to the nth term.Maybe I can find a relationship between S_n and S_{n-1}. Since S_n = S_{n-1} + a_n, I can substitute the given formula into this equation. So, S_n = 2a_n - 1 and S_{n-1} = 2a_{n-1} - 1. Therefore, substituting into S_n = S_{n-1} + a_n, we get:2a_n - 1 = (2a_{n-1} - 1) + a_nSimplifying this equation:2a_n - 1 = 2a_{n-1} - 1 + a_nSubtracting 2a_{n-1} - 1 from both sides:2a_n - 1 - 2a_{n-1} + 1 = a_nSimplifying further:2a_n - 2a_{n-1} = a_nSubtracting a_n from both sides:a_n - 2a_{n-1} = 0Which means:a_n = 2a_{n-1}So, this tells me that each term is twice the previous term. That suggests that the sequence is a geometric sequence with a common ratio of 2.Now, I need to find the first term a_1. Using the given formula for S_n when n=1:S_1 = 2a_1 - 1But S_1 is also just a_1, since it's the sum of the first term. So:a_1 = 2a_1 - 1Solving for a_1:a_1 = 2a_1 - 1Subtract 2a_1 from both sides:-a_1 = -1Multiply both sides by -1:a_1 = 1So, the first term is 1, and each subsequent term is twice the previous one. Therefore, the general formula for a_n is:a_n = 2^{n-1}That takes care of part (1). Now, moving on to part (2), which is an inequality involving combinations:a_1 C_n^0 + a_2 C_n^1 + a_3 C_n^2 + ... + a_{n+1} C_n^n < 2023First, let's write out what each term is. From part (1), we know that a_k = 2^{k-1}. So, substituting that into the inequality:1 * C_n^0 + 2 * C_n^1 + 4 * C_n^2 + 8 * C_n^3 + ... + 2^n * C_n^n < 2023Looking at this, it seems like a binomial expansion. Remember that the binomial theorem states that (1 + x)^n = C_n^0 + C_n^1 x + C_n^2 x^2 + ... + C_n^n x^n.In our case, each term is multiplied by a power of 2. So, if we set x = 2, then:(1 + 2)^n = C_n^0 * 2^0 + C_n^1 * 2^1 + C_n^2 * 2^2 + ... + C_n^n * 2^nWhich simplifies to:3^n = 1 + 2C_n^1 + 4C_n^2 + 8C_n^3 + ... + 2^n C_n^nWait a minute, that's exactly the left side of our inequality! So, the sum a_1 C_n^0 + a_2 C_n^1 + ... + a_{n+1} C_n^n is equal to 3^n.Therefore, the inequality simplifies to:3^n < 2023Now, we need to find all positive integers n such that 3^n < 2023.Let's compute powers of 3 to find the largest n satisfying this:3^1 = 33^2 = 93^3 = 273^4 = 813^5 = 2433^6 = 7293^7 = 2187Wait, 3^7 is 2187, which is greater than 2023. So, 3^6 is 729, which is less than 2023.Therefore, the largest integer n for which 3^n < 2023 is n=6.But wait, let me double-check:3^6 = 7293^7 = 2187Yes, 2187 is indeed greater than 2023, so n must be less than 7. Since n is a positive integer, n can be 1, 2, 3, 4, 5, or 6.Therefore, the solution to the inequality is all positive integers n such that n ≤ 6.So, summarizing:1. The general formula for the sequence is a_n = 2^{n-1}.2. The inequality holds for all positive integers n where n is less than or equal to 6.**Final Answer**(1) The general formula is boxed{a_n = 2^{n-1}}.(2) The solution to the inequality is all positive integers n with boxed{n leq 6}.

💡Alright, so I've got this problem here about a depositor who withdrew some money from his savings account in a few steps, and now I need to figure out how much he originally deposited. Let me try to break this down step by step.First, let's try to understand the problem. The depositor withdrew 1/4 of his money, then 4/9 of the remaining money, and then another 640 rubles. After all these withdrawals, he had 3/20 of his original money left. So, I need to find out how much he had initially.Okay, let's denote the original amount of money as ( x ) rubles. That seems like a good starting point.So, the first withdrawal is 1/4 of his money. If he had ( x ) rubles, then he withdrew ( frac{1}{4}x ). That means the remaining amount after the first withdrawal would be ( x - frac{1}{4}x ).Let me calculate that: ( x - frac{1}{4}x = frac{3}{4}x ). So, after the first withdrawal, he has three-fourths of his original amount left.Next, he withdrew 4/9 of the remaining money. The remaining money at this point is ( frac{3}{4}x ). So, he's taking out ( frac{4}{9} ) of that. Let me write that down: ( frac{4}{9} times frac{3}{4}x ).Hmm, multiplying those fractions: ( frac{4}{9} times frac{3}{4} = frac{12}{36} = frac{1}{3} ). So, he withdrew ( frac{1}{3}x ) in the second withdrawal.That means the remaining amount after the second withdrawal is ( frac{3}{4}x - frac{1}{3}x ).To subtract these fractions, I need a common denominator. The denominators are 4 and 3, so the least common denominator is 12. Converting both fractions:( frac{3}{4}x = frac{9}{12}x ) and ( frac{1}{3}x = frac{4}{12}x ).Subtracting them: ( frac{9}{12}x - frac{4}{12}x = frac{5}{12}x ).So, after the second withdrawal, he has ( frac{5}{12}x ) left.Then, he withdrew another 640 rubles. So, the remaining amount after this third withdrawal is ( frac{5}{12}x - 640 ).According to the problem, after all these withdrawals, he had ( frac{3}{20}x ) left. So, we can set up the equation:( frac{5}{12}x - 640 = frac{3}{20}x ).Now, I need to solve for ( x ). Let's get all the terms involving ( x ) on one side and the constants on the other.First, subtract ( frac{3}{20}x ) from both sides:( frac{5}{12}x - frac{3}{20}x - 640 = 0 ).Now, to combine the fractions, I need a common denominator. The denominators are 12 and 20. The least common multiple of 12 and 20 is 60.Convert both fractions:( frac{5}{12}x = frac{25}{60}x ) and ( frac{3}{20}x = frac{9}{60}x ).Subtracting them: ( frac{25}{60}x - frac{9}{60}x = frac{16}{60}x ).So, the equation becomes:( frac{16}{60}x - 640 = 0 ).Simplify ( frac{16}{60} ) to ( frac{4}{15} ):( frac{4}{15}x - 640 = 0 ).Now, add 640 to both sides:( frac{4}{15}x = 640 ).To solve for ( x ), multiply both sides by the reciprocal of ( frac{4}{15} ), which is ( frac{15}{4} ):( x = 640 times frac{15}{4} ).Calculate that:First, divide 640 by 4: ( 640 div 4 = 160 ).Then, multiply by 15: ( 160 times 15 = 2400 ).So, ( x = 2400 ) rubles.Let me double-check my calculations to make sure I didn't make any mistakes.Starting with ( x = 2400 ):First withdrawal: ( frac{1}{4} times 2400 = 600 ). Remaining: ( 2400 - 600 = 1800 ).Second withdrawal: ( frac{4}{9} times 1800 = 800 ). Remaining: ( 1800 - 800 = 1000 ).Third withdrawal: 640 rubles. Remaining: ( 1000 - 640 = 360 ).Now, ( frac{3}{20} times 2400 = 360 ). That matches the remaining amount, so it checks out.Therefore, the original deposit was 2400 rubles.

💡Okay, so I have this geometry problem here, and I need to figure out which of the propositions A, B, C, or D is true for any given line ( l ) and plane ( alpha ) in space. Let me try to break this down step by step.First, I need to recall some basic concepts about lines and planes in three-dimensional space. I know that a line can either lie on a plane, intersect the plane at a single point, or be parallel to the plane. These are the three possible relationships between a line and a plane. Let me consider each proposition one by one.**Proposition A**: "There must be a line parallel to line ( l ) within plane ( alpha )."Hmm, so this is saying that no matter how line ( l ) is positioned relative to plane ( alpha ), there will always be some line inside ( alpha ) that's parallel to ( l ). Let me think about the different cases:1. If line ( l ) lies on plane ( alpha ), then obviously there are infinitely many lines within ( alpha ) that are parallel to ( l ). So in this case, proposition A holds true.2. If line ( l ) intersects plane ( alpha ) at a single point, then can there be a line within ( alpha ) that's parallel to ( l )? Well, if two lines are parallel, they must lie on the same plane and never intersect. But since ( l ) intersects ( alpha ) at one point, any line within ( alpha ) that's parallel to ( l ) would have to also pass through that point, which would mean they intersect, contradicting the definition of parallel lines. So in this case, there cannot be a line within ( alpha ) that's parallel to ( l ). Therefore, proposition A is false in this scenario.3. If line ( l ) is parallel to plane ( alpha ), meaning it doesn't intersect ( alpha ) and lies outside of it, then can there be a line within ( alpha ) that's parallel to ( l )? Yes, because if ( l ) is parallel to ( alpha ), there must be lines within ( alpha ) that are parallel to ( l ). In fact, there are infinitely many such lines. So in this case, proposition A holds true.But wait, in the second case where ( l ) intersects ( alpha ), proposition A fails. Since the question says "for any given line ( l ) and plane ( alpha )", proposition A doesn't always hold. So A is not necessarily true.**Proposition B**: "There must be a line perpendicular to line ( l ) within plane ( alpha )."This is saying that no matter how ( l ) is positioned relative to ( alpha ), there will always be a line within ( alpha ) that's perpendicular to ( l ).Let me again consider the different cases:1. If line ( l ) lies on plane ( alpha ), then certainly there are lines within ( alpha ) that are perpendicular to ( l ). For example, any line in ( alpha ) that intersects ( l ) at a right angle would satisfy this. So B holds here.2. If line ( l ) intersects plane ( alpha ) at a single point, can there be a line within ( alpha ) that's perpendicular to ( l )? Yes, because at the point of intersection, we can draw a line in ( alpha ) that's perpendicular to ( l ). This is similar to how, in a plane, you can always draw a perpendicular line at a point. So B holds here as well.3. If line ( l ) is parallel to plane ( alpha ), can there be a line within ( alpha ) that's perpendicular to ( l )? Yes, because even though ( l ) doesn't intersect ( alpha ), we can still find a direction within ( alpha ) that's perpendicular to ( l ). For instance, if ( l ) is parallel to ( alpha ), then the normal vector of ( alpha ) is perpendicular to ( l ), and we can construct a line in ( alpha ) that aligns with this normal vector. So B holds here too.Since in all possible cases, there exists a line within ( alpha ) that's perpendicular to ( l ), proposition B must be true.**Proposition C**: "There cannot be a line parallel to line ( l ) within plane ( alpha )."This is the opposite of proposition A. It claims that under no circumstances can there be a line within ( alpha ) that's parallel to ( l ).But as we saw earlier, if ( l ) lies on ( alpha ) or is parallel to ( alpha ), there can indeed be lines within ( alpha ) that are parallel to ( l ). Therefore, proposition C is false because it's not always the case that such lines don't exist.**Proposition D**: "There might not be a line perpendicular to line ( l ) within plane ( alpha )."This suggests that it's possible that there is no line within ( alpha ) that's perpendicular to ( l ).However, from our analysis of proposition B, we saw that in all cases—whether ( l ) lies on ( alpha ), intersects ( alpha ), or is parallel to ( alpha )—there is always at least one line within ( alpha ) that's perpendicular to ( l ). Therefore, proposition D is false because it's not possible for there to be no such line.So, after evaluating all four propositions, only proposition B holds true in all scenarios. Therefore, the correct answer is B.

💡Okay, so I have this problem about arranging rooks on a chessboard. It says there are n white rooks and n black rooks, and they shouldn't be able to attack each other if they're different colors. I need to find the maximum possible value of n. Hmm, let me try to figure this out step by step.First, I remember that rooks attack any piece that's in the same row or column. So, if a white rook and a black rook are in the same row or column, they can attack each other. That means I have to place the white and black rooks in such a way that no two rooks of different colors share a row or a column.Let me think about how many rows and columns there are on a chessboard. There are 8 rows and 8 columns, making a total of 16 lines (8 rows + 8 columns). Each rook, whether white or black, occupies one row and one column. So, if I place a white rook on a particular row and column, that row and column are now "claimed" by white, and I can't place a black rook there.Wait, so if I have n white rooks, they will occupy n rows and n columns. Similarly, n black rooks will also occupy n rows and n columns. But since the white and black rooks can't share any rows or columns, the total number of rows and columns used by both colors should be 2n. But we only have 16 rows and columns in total. So, 2n can't exceed 16. That means n can't be more than 8, right? Because 2*8=16.But wait, that doesn't seem right. If n is 8, then white rooks would occupy 8 rows and 8 columns, and black rooks would also need 8 rows and 8 columns, but there are only 8 rows and 8 columns available. So, how can both white and black rooks each occupy 8 rows and 8 columns without overlapping? That seems impossible because the white rooks would already be using all the rows and columns, leaving nothing for the black rooks.Maybe my initial thought was wrong. Let me try a different approach. Instead of thinking about rows and columns separately, maybe I should consider that each rook, regardless of color, occupies one square, which is at the intersection of a row and a column. So, if I have n white rooks and n black rooks, that's a total of 2n rooks on the board.But the chessboard has 64 squares. So, 2n can be up to 64, but of course, we have the restriction that white and black rooks can't attack each other. So, it's not just about the number of squares but also about the arrangement.Wait, maybe I should think about it in terms of non-attacking rooks. For one color, the maximum number of non-attacking rooks is 8, one in each row and column. But here, we have two colors, and they can't attack each other either. So, maybe the total number of rooks is limited by the number of rows and columns.If I have n white rooks, they occupy n rows and n columns. Similarly, n black rooks would need n rows and n columns. But since the white and black rooks can't share any rows or columns, the total number of rows and columns used would be 2n. But there are only 8 rows and 8 columns, so 2n can't exceed 16. Therefore, n can't be more than 8. But earlier, I thought that was impossible because white and black rooks would need separate rows and columns.Wait, maybe I'm misunderstanding the problem. It says rooks of different colors do not attack each other. So, does that mean that a white rook can attack another white rook, and a black rook can attack another black rook? Or does it mean that no rook can attack any other rook, regardless of color?I think it means that rooks of different colors do not attack each other. So, white rooks can attack other white rooks, and black rooks can attack other black rooks, but a white rook cannot attack a black rook and vice versa.So, in that case, the white rooks can be placed in such a way that they don't attack each other, and the black rooks can be placed in such a way that they don't attack each other, but also, no white rook is in the same row or column as a black rook.So, for the white rooks, we can have up to 8 non-attacking rooks, one in each row and column. Similarly, for the black rooks, we can have up to 8 non-attacking rooks, but they have to be placed in rows and columns that are not used by the white rooks.But wait, if white rooks are using 8 rows and 8 columns, there are no rows or columns left for the black rooks. So, that can't be right. Therefore, maybe the maximum n is less than 8.Let me think again. If I have n white rooks, they occupy n rows and n columns. Similarly, n black rooks occupy n rows and n columns. But since the white and black rooks can't share any rows or columns, the total number of rows and columns used is 2n. But there are only 8 rows and 8 columns, so 2n ≤ 16, which means n ≤ 8. But as I thought before, if n=8, white rooks would use all 8 rows and columns, leaving nothing for black rooks.So, maybe n can't be 8. Let's try n=7. If n=7, white rooks use 7 rows and 7 columns, leaving 1 row and 1 column for black rooks. But black rooks need 7 rows and 7 columns, so that's not enough. Hmm, that doesn't work either.Wait, maybe I'm approaching this wrong. Maybe I should consider that each rook, regardless of color, occupies one row and one column, but the constraint is that no two rooks of different colors share a row or column. So, the white rooks and black rooks must be placed on completely separate sets of rows and columns.So, if I have n white rooks, they need n rows and n columns. Similarly, n black rooks need n rows and n columns. But these sets of rows and columns must be disjoint. So, the total number of rows used is n (for white) + n (for black) = 2n. Similarly, the total number of columns used is 2n. But since there are only 8 rows and 8 columns, 2n ≤ 8. Therefore, n ≤ 4.Wait, that seems too restrictive. Because if n=4, then white rooks use 4 rows and 4 columns, and black rooks use another 4 rows and 4 columns, totaling 8 rows and 8 columns. That works. But can we do better?Wait, maybe I'm misunderstanding the problem. It says rooks of different colors do not attack each other. So, does that mean that a white rook can attack another white rook, and a black rook can attack another black rook, but a white rook cannot attack a black rook? Or does it mean that no rook can attack any other rook, regardless of color?I think it's the former: rooks of different colors do not attack each other. So, white rooks can attack other white rooks, and black rooks can attack other black rooks, but a white rook cannot attack a black rook.In that case, the white rooks can be placed in such a way that they don't attack each other, and the black rooks can be placed in such a way that they don't attack each other, but also, no white rook is in the same row or column as a black rook.So, for the white rooks, we can have up to 8 non-attacking rooks, one in each row and column. Similarly, for the black rooks, we can have up to 8 non-attacking rooks, but they have to be placed in rows and columns that are not used by the white rooks.But wait, if white rooks are using 8 rows and 8 columns, there are no rows or columns left for the black rooks. So, that can't be right. Therefore, maybe the maximum n is less than 8.Let me think again. If I have n white rooks, they occupy n rows and n columns. Similarly, n black rooks occupy n rows and n columns. But since the white and black rooks can't share any rows or columns, the total number of rows and columns used is 2n. But there are only 8 rows and 8 columns, so 2n ≤ 16, which means n ≤ 8. But as I thought before, if n=8, white rooks would use all 8 rows and columns, leaving nothing for black rooks.So, maybe n can't be 8. Let's try n=7. If n=7, white rooks use 7 rows and 7 columns, leaving 1 row and 1 column for black rooks. But black rooks need 7 rows and 7 columns, so that's not enough. Hmm, that doesn't work either.Wait, maybe I'm approaching this wrong. Maybe I should consider that each rook, regardless of color, occupies one row and one column, but the constraint is that no two rooks of different colors share a row or column. So, the white rooks and black rooks must be placed on completely separate sets of rows and columns.So, if I have n white rooks, they need n rows and n columns. Similarly, n black rooks need n rows and n columns. But these sets of rows and columns must be disjoint. So, the total number of rows used is n (for white) + n (for black) = 2n. Similarly, the total number of columns used is 2n. But since there are only 8 rows and 8 columns, 2n ≤ 8. Therefore, n ≤ 4.Wait, that seems too restrictive. Because if n=4, then white rooks use 4 rows and 4 columns, and black rooks use another 4 rows and 4 columns, totaling 8 rows and 8 columns. That works. But can we do better?Wait, maybe I'm misunderstanding the problem. It says rooks of different colors do not attack each other. So, does that mean that a white rook can attack another white rook, and a black rook can attack another black rook, but a white rook cannot attack a black rook? Or does it mean that no rook can attack any other rook, regardless of color?I think it's the former: rooks of different colors do not attack each other. So, white rooks can attack other white rooks, and black rooks can attack other black rooks, but a white rook cannot attack a black rook.In that case, the white rooks can be placed in such a way that they don't attack each other, and the black rooks can be placed in such a way that they don't attack each other, but also, no white rook is in the same row or column as a black rook.So, for the white rooks, we can have up to 8 non-attacking rooks, one in each row and column. Similarly, for the black rooks, we can have up to 8 non-attacking rooks, but they have to be placed in rows and columns that are not used by the white rooks.But wait, if white rooks are using 8 rows and 8 columns, there are no rows or columns left for the black rooks. So, that can't be right. Therefore, maybe the maximum n is less than 8.Let me try to visualize this. Suppose I place 4 white rooks on the board, each in separate rows and columns. Then, I can place 4 black rooks in the remaining 4 rows and 4 columns. That way, no white and black rook share a row or column, and each color has 4 non-attacking rooks. So, n=4 seems possible.But can I do better? What if I try n=5? If I place 5 white rooks, they will occupy 5 rows and 5 columns. Then, the black rooks would need 5 rows and 5 columns, but there are only 3 rows and 3 columns left. So, that's not enough. Therefore, n=5 is not possible.Wait, but maybe there's a way to arrange them differently. Maybe not all rows and columns need to be used exclusively. For example, if I place some white rooks in such a way that they leave some rows and columns partially available for black rooks.But no, because if a white rook is in a row, that row can't have any black rooks, and similarly for columns. So, each white rook effectively blocks a row and a column for black rooks.Therefore, the maximum n is 4, because 4 white rooks block 4 rows and 4 columns, leaving 4 rows and 4 columns for black rooks. That seems to be the maximum.Wait, but earlier I thought that n=8 was impossible because white rooks would block all rows and columns. But maybe there's a way to interleave them somehow. Let me think.If I place white rooks in such a way that they don't block all rows and columns, but leave some for black rooks. For example, if I place white rooks on the main diagonal, (1,1), (2,2), ..., (8,8), then black rooks could be placed on the anti-diagonal, (1,8), (2,7), ..., (8,1). In this case, each white rook is in a unique row and column, and each black rook is also in a unique row and column, and no white and black rook share a row or column. So, in this case, n=8 is possible.Wait, that seems to contradict my earlier conclusion. So, maybe n=8 is possible after all.Let me check. If I place 8 white rooks on the main diagonal, they occupy rows 1-8 and columns 1-8. Then, placing 8 black rooks on the anti-diagonal, they also occupy rows 1-8 and columns 1-8, but in such a way that no black rook is in the same row or column as a white rook. Wait, but in this case, each row and column has both a white and a black rook. So, they do share rows and columns, which would mean they can attack each other. That's not allowed.So, my mistake. Placing white rooks on the main diagonal and black rooks on the anti-diagonal would result in each row and column having both colors, which violates the condition. Therefore, n=8 is not possible in that arrangement.So, maybe n=8 is impossible because placing 8 white rooks would block all rows and columns, leaving nothing for black rooks. Therefore, the maximum n is 4.Wait, but earlier I thought that n=4 was possible, but maybe there's a better arrangement. Let me think again.Suppose I divide the chessboard into two separate 4x4 boards. Place all white rooks on one 4x4 board and all black rooks on the other 4x4 board. In this case, white rooks are in rows 1-4 and columns 1-4, and black rooks are in rows 5-8 and columns 5-8. This way, no white and black rook share a row or column. So, in this case, n=4 is possible.But can I do better? What if I use a more efficient arrangement where white and black rooks share some rows and columns, but not in a way that they attack each other. For example, if I place white rooks in such a way that they leave some rows and columns partially available for black rooks.Wait, but if a white rook is in a row, that row can't have any black rooks. Similarly for columns. So, any row or column with a white rook can't have a black rook, and vice versa. Therefore, the rows and columns used by white and black rooks must be completely separate.Therefore, the total number of rows and columns used by both colors is 2n. Since there are only 8 rows and 8 columns, 2n ≤ 16, so n ≤ 8. But as we saw earlier, n=8 is impossible because white rooks would block all rows and columns, leaving nothing for black rooks.Wait, but if I arrange the white and black rooks in such a way that they don't block all rows and columns, but still don't share any. For example, if I place white rooks in rows 1-4 and columns 1-4, and black rooks in rows 5-8 and columns 5-8, then n=4 is possible. But what if I interleave them differently?Suppose I place white rooks in rows 1,3,5,7 and columns 1,3,5,7, and black rooks in rows 2,4,6,8 and columns 2,4,6,8. In this case, each color has 4 rooks, and they don't share any rows or columns. So, n=4 is possible.But can I have more than 4? Let's try n=5. If I place 5 white rooks, they will occupy 5 rows and 5 columns. Then, black rooks would need 5 rows and 5 columns, but there are only 3 rows and 3 columns left. So, that's not enough. Therefore, n=5 is impossible.Wait, but maybe there's a way to arrange them so that some rows and columns are shared in a way that doesn't allow attacks. For example, if I place white rooks in rows 1-5 and columns 1-5, and black rooks in rows 4-8 and columns 4-8. But then, rows 4-5 and columns 4-5 would have both white and black rooks, which is not allowed.So, it seems that n=4 is the maximum possible. Therefore, the maximum value of n is 4.Wait, but earlier I thought that n=8 was impossible because white rooks would block all rows and columns, but maybe there's a way to arrange them so that they don't block all. For example, if I place white rooks in such a way that they leave some rows and columns available for black rooks.But no, because each white rook blocks a row and a column, and if I have n white rooks, they block n rows and n columns. Similarly, black rooks block n rows and n columns. So, the total blocked rows and columns are 2n. Since there are only 8 rows and 8 columns, 2n ≤ 16, so n ≤ 8. But as we saw, n=8 is impossible because white rooks would block all rows and columns, leaving nothing for black rooks.Therefore, the maximum n is 4.Wait, but I'm confused because earlier I thought that n=8 was possible by placing white and black rooks on separate 4x4 boards, but that only allows n=4. So, maybe n=4 is the correct answer.But let me double-check. If I place 4 white rooks on the top-left 4x4 board and 4 black rooks on the bottom-right 4x4 board, then no white and black rook share a row or column. So, n=4 is possible.But can I place more? Let's try n=5. If I place 5 white rooks, they will occupy 5 rows and 5 columns. Then, black rooks would need 5 rows and 5 columns, but there are only 3 rows and 3 columns left. So, that's not enough. Therefore, n=5 is impossible.Therefore, the maximum possible value of n is 4.

💡Okay, so I have this function f(x) = -x² + 2x, and I need to do two things with it. First, I have to prove that it's a decreasing function on the interval [1, +∞). Second, when x is between 2 and 5, I need to find the maximum and minimum values of f(x). Hmm, let's take this step by step.Starting with the first part: proving that f(x) is decreasing on [1, +∞). I remember that a function is decreasing on an interval if, as x increases, the function's value decreases. In other words, for any two points x₁ and x₂ in the interval where x₁ < x₂, f(x₁) should be greater than f(x₂). So, maybe I can use the definition of a decreasing function here.Alternatively, I recall that the derivative of a function can tell us about its increasing or decreasing behavior. If the derivative is negative on an interval, the function is decreasing there. Let me try that approach. So, first, find the derivative of f(x).f(x) = -x² + 2xThe derivative f’(x) would be the rate at which f(x) changes with respect to x. Applying basic differentiation rules:f’(x) = d/dx (-x² + 2x) = -2x + 2.Okay, so f’(x) = -2x + 2. Now, to determine where the function is decreasing, I need to find where f’(x) is negative.Set f’(x) < 0:-2x + 2 < 0Let's solve for x:-2x + 2 < 0 Subtract 2 from both sides: -2x < -2 Divide both sides by -2. Remember, when you divide or multiply both sides of an inequality by a negative number, the inequality sign flips. So:x > 1.So, f’(x) is negative when x > 1, which means the function is decreasing on the interval (1, +∞). But the problem asks for [1, +∞). At x = 1, what's the derivative?f’(1) = -2(1) + 2 = -2 + 2 = 0.Hmm, so at x = 1, the derivative is zero. That means the function has a horizontal tangent there. But does that affect the decreasing nature? Well, a single point with a derivative of zero doesn't necessarily mean the function isn't decreasing on the interval. It just means that at that exact point, the function isn't increasing or decreasing—it's stationary.But to be thorough, maybe I should check the behavior around x = 1. For x just less than 1, say x = 0.5:f’(0.5) = -2(0.5) + 2 = -1 + 2 = 1 > 0.So, the function is increasing before x = 1. At x = 1, the derivative is zero, and for x > 1, the derivative is negative, so the function is decreasing.Therefore, on the interval [1, +∞), the function is decreasing because after x = 1, it starts to decrease, and the point at x = 1 is just the peak where it changes from increasing to decreasing.Alternatively, maybe I can use the definition of a decreasing function without calculus. Let's try that as well to reinforce my understanding.Take any two points x₁ and x₂ in [1, +∞) with x₁ < x₂. I need to show that f(x₁) > f(x₂).Compute f(x₁) - f(x₂):f(x₁) - f(x₂) = (-x₁² + 2x₁) - (-x₂² + 2x₂) = -x₁² + 2x₁ + x₂² - 2x₂ = (x₂² - x₁²) + 2(x₁ - x₂) Factor the difference of squares: = (x₂ - x₁)(x₂ + x₁) + 2(x₁ - x₂) Factor out (x₂ - x₁): = (x₂ - x₁)(x₂ + x₁ - 2)Now, since x₁ < x₂, (x₂ - x₁) is positive. What about (x₂ + x₁ - 2)?Given that x₁ and x₂ are both in [1, +∞), the smallest they can be is 1. So, x₁ + x₂ is at least 1 + 1 = 2. Therefore, x₂ + x₁ - 2 is at least 0. But since x₁ and x₂ are greater than or equal to 1, and x₁ < x₂, unless both are exactly 1, x₂ + x₁ - 2 will be positive.Wait, if x₁ = x₂ = 1, then x₂ + x₁ - 2 = 0. But in that case, x₁ = x₂, which isn't allowed since x₁ < x₂. So, for x₁ < x₂ in [1, +∞), x₂ + x₁ - 2 is positive because x₁ and x₂ are both at least 1, so their sum is at least 2, and since x₁ < x₂, their sum is strictly greater than 2 when x₁ > 1. Wait, no—if x₁ = 1 and x₂ > 1, then x₂ + x₁ - 2 = (1 + x₂) - 2 = x₂ - 1 > 0 because x₂ > 1. If x₁ > 1, then x₂ > x₁ > 1, so x₂ + x₁ - 2 > 2 - 2 = 0. So, in all cases, x₂ + x₁ - 2 is positive.Therefore, both factors (x₂ - x₁) and (x₂ + x₁ - 2) are positive, so their product is positive. Hence, f(x₁) - f(x₂) > 0, which implies f(x₁) > f(x₂). Therefore, f(x) is decreasing on [1, +∞). That checks out.Okay, so part (1) is done. Now, moving on to part (2): When x is in [2, 5], find the maximum and minimum values of f(x).Since we already know from part (1) that f(x) is decreasing on [1, +∞), and [2, 5] is a subset of [1, +∞), f(x) is decreasing on [2, 5] as well. That means the function starts at its highest point at x = 2 and decreases as x increases towards 5.Therefore, the maximum value of f(x) on [2, 5] should be at x = 2, and the minimum value should be at x = 5.Let me compute f(2) and f(5) to confirm.First, f(2):f(2) = -(2)² + 2*(2) = -4 + 4 = 0.Next, f(5):f(5) = -(5)² + 2*(5) = -25 + 10 = -15.So, f(2) = 0 and f(5) = -15.Since the function is decreasing, these should indeed be the maximum and minimum values on the interval [2, 5].Just to be thorough, maybe I should check another point in the interval to ensure the function is indeed decreasing. Let's pick x = 3.f(3) = -(3)² + 2*(3) = -9 + 6 = -3.Compare that to f(2) = 0 and f(5) = -15. So, 0 > -3 > -15, which is consistent with the function decreasing as x increases.Alternatively, I can also compute the derivative on [2, 5] to confirm it's negative, ensuring the function is decreasing.We already found f’(x) = -2x + 2.At x = 2: f’(2) = -4 + 2 = -2 < 0.At x = 5: f’(5) = -10 + 2 = -8 < 0.Since the derivative is negative throughout [2, 5], the function is decreasing on that interval, so the maximum is at the left endpoint (x=2) and the minimum at the right endpoint (x=5).Therefore, the maximum value is 0 and the minimum value is -15.I think that covers both parts of the problem. I used calculus to find the derivative and determine the increasing/decreasing nature, then applied that to find the extrema on the given interval. I also double-checked using the definition of a decreasing function without calculus, which was a good exercise to ensure my understanding. All the computations seem consistent, so I feel confident about the results.**Final Answer**The maximum value of ( f(x) ) is (boxed{0}) and the minimum value is (boxed{-15}).