💡First, I need to determine the length of the train. I'll denote the length of the train as ( L ) meters.When the train crosses a post, it only needs to cover its own length. The time taken to cross the post is 10 seconds. This means the speed of the train is ( frac{L}{10} ) meters per second.Next, when the train crosses a platform that is 150 meters long, it needs to cover both its own length and the length of the platform. The total distance in this case is ( L + 150 ) meters, and the time taken is 20 seconds. Therefore, the speed of the train when crossing the platform is ( frac{L + 150}{20} ) meters per second.Since the speed of the train remains constant, I can set the two expressions for speed equal to each other:[frac{L}{10} = frac{L + 150}{20}]To solve for ( L ), I'll cross-multiply:[20L = 10(L + 150)]Expanding the equation:[20L = 10L + 1500]Subtracting ( 10L ) from both sides:[10L = 1500]Finally, dividing both sides by 10:[L = 150]Therefore, the length of the train is 150 meters.

💡Okay, so I have this trigonometry problem here. It says that tan α equals the square root of 3, and α is between π and 3π/2. I need to find the value of cos(2π - α) minus sin α. The options are given as A, B, C, D with some expressions involving square roots and fractions. Alright, let's break this down step by step.First, I know that tan α is given as √3. I remember that tan is sine over cosine, so maybe I can find sine and cosine values from this. Also, the angle α is between π and 3π/2, which is the third quadrant. In the third quadrant, both sine and cosine are negative. That's something important to remember because it will affect the signs of the trigonometric functions.So, tan α = √3. I recall that tan 60° is √3, which is π/3 radians. But since α is in the third quadrant, the reference angle would be π/3, and the actual angle would be π + π/3, which is 4π/3. So, α is 4π/3 radians. That makes sense because 4π/3 is between π and 3π/2.Now, I need to find cos(2π - α) - sin α. Let's tackle each part separately. First, cos(2π - α). I remember that cosine is an even function, which means cos(-θ) = cos θ. But here, it's 2π - α. I think there's a trigonometric identity for cos(2π - α). Let me recall. Cosine has a period of 2π, so cos(2π - α) is the same as cos(-α), which is equal to cos α because cosine is even. Wait, no, that's not quite right. Let me think again.Actually, cos(2π - α) can be simplified using the cosine of a difference identity. The formula is cos(A - B) = cos A cos B + sin A sin B. So, cos(2π - α) would be cos 2π cos α + sin 2π sin α. I know that cos 2π is 1 and sin 2π is 0. So, cos(2π - α) simplifies to 1 * cos α + 0 * sin α, which is just cos α. Okay, that's helpful. So, cos(2π - α) is equal to cos α.Now, the expression becomes cos α - sin α. So, I need to find the values of cos α and sin α when α is 4π/3. Since α is in the third quadrant, both cosine and sine will be negative. Let's find the exact values.For α = 4π/3, the reference angle is π/3. The cosine of π/3 is 1/2, and the sine of π/3 is √3/2. But since we're in the third quadrant, both cosine and sine will be negative. So, cos(4π/3) is -1/2, and sin(4π/3) is -√3/2.Now, plugging these into the expression cos α - sin α:cos α = -1/2sin α = -√3/2So, cos α - sin α = (-1/2) - (-√3/2) = (-1/2) + (√3/2) = (√3 - 1)/2.Looking at the options, that's option B: (√3 - 1)/2.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from tan α = √3, which gives α = π/3 or 4π/3. Since α is between π and 3π/2, it's 4π/3. Then, using the identity for cos(2π - α), which simplifies to cos α. Then, calculating cos α and sin α for 4π/3, both are negative, so cos α is -1/2 and sin α is -√3/2. Subtracting sin α from cos α gives (-1/2) - (-√3/2) = (-1 + √3)/2, which is the same as (√3 - 1)/2. Yep, that seems correct.I think I've got it right. So, the answer should be option B.

💡First, I need to determine the initial number that, when 17 is added to it, becomes completely divisible by 29.Let's denote the initial number as ( x ). According to the problem, when 17 is added to ( x ), the sum is divisible by 29. This can be expressed as:[x + 17 = 29k]where ( k ) is an integer.To find ( x ), I can rearrange the equation:[x = 29k - 17]Since we are looking for the smallest positive initial number, I'll start with the smallest positive integer value for ( k ), which is 1:[x = 29(1) - 17 = 12]To verify, adding 17 to 12 gives:[12 + 17 = 29]which is indeed divisible by 29.Therefore, the initial number is 12.

💡Okay, so I have this problem here: Given that ((sqrt{2} - x)^{10} = a_0 + a_1 x + a_2 x^2 + dots + a_{10} x^{10}), I need to determine the value of ((a_0 + a_2 + dots + a_{10})^2 - (a_1 + a_3 + dots + a_9)^2). The options are A: 0, B: 2, C: -1, D: 1.Hmm, let's see. I remember that when dealing with polynomials, evaluating them at specific points can give us information about the coefficients. For example, if I plug in (x = 1), I get the sum of all coefficients, which is (a_0 + a_1 + a_2 + dots + a_{10}). Similarly, plugging in (x = -1) gives (a_0 - a_1 + a_2 - a_3 + dots + a_{10}).Wait, the expression I need to compute is ((a_0 + a_2 + dots + a_{10})^2 - (a_1 + a_3 + dots + a_9)^2). That looks like a difference of squares, which factors into ((a_0 + a_2 + dots + a_{10} + a_1 + a_3 + dots + a_9)(a_0 + a_2 + dots + a_{10} - a_1 - a_3 - dots - a_9)).But (a_0 + a_2 + dots + a_{10} + a_1 + a_3 + dots + a_9) is just the sum of all coefficients, which is (f(1)), where (f(x) = (sqrt{2} - x)^{10}). Similarly, (a_0 + a_2 + dots + a_{10} - a_1 - a_3 - dots - a_9) is (f(-1)).So, the expression simplifies to (f(1) times f(-1)). Let me compute (f(1)) and (f(-1)).First, (f(1) = (sqrt{2} - 1)^{10}). Second, (f(-1) = (sqrt{2} - (-1))^{10} = (sqrt{2} + 1)^{10}).Now, multiplying these together: (f(1) times f(-1) = (sqrt{2} - 1)^{10} times (sqrt{2} + 1)^{10}). I can rewrite this as ([( sqrt{2} - 1)(sqrt{2} + 1)]^{10}).Calculating the product inside the brackets: ((sqrt{2} - 1)(sqrt{2} + 1) = (sqrt{2})^2 - (1)^2 = 2 - 1 = 1). So, the entire expression becomes (1^{10} = 1).Therefore, the value of ((a_0 + a_2 + dots + a_{10})^2 - (a_1 + a_3 + dots + a_9)^2) is 1.Wait, let me double-check that. So, if I factor the difference of squares correctly, it's indeed the product of the sum and the difference of the even and odd coefficients. And evaluating (f(1)) and (f(-1)) gives me the sum and the alternating sum of the coefficients. Multiplying those gives me the desired expression. And since ((sqrt{2} - 1)(sqrt{2} + 1) = 1), raising it to the 10th power is still 1. So, yeah, that seems right.I don't think I made any mistakes here. The key was recognizing the difference of squares and relating it to evaluating the polynomial at 1 and -1. That's a common technique in these kinds of problems. So, I'm confident the answer is 1.

💡Okay, so I have this problem about a rook moving on an n x n chessboard. The rook moves in such a way that it visits every cell exactly once, and each move is to a neighboring cell by a side. Then, Bob numbers each cell from 1 to n² based on the order the rook visited them. The goal is to find the minimal possible value of M, which is the greatest difference between numbers in neighboring cells.First, I need to understand the movement of the rook. Since it's a rook, it can move any number of squares vertically or horizontally, but in this case, it seems like it's moving one square at a time to a neighboring cell. So, it's more like a king's move but restricted to only horizontal and vertical moves, not diagonal.The numbering is based on the order the rook visits the cells. So, the first cell the rook visits is 1, the second is 2, and so on until n². Now, M is the maximum difference between any two neighboring cells. We need to arrange the numbering such that this maximum difference is as small as possible.I think the key here is to find a path for the rook that minimizes the maximum difference between consecutive numbers in neighboring cells. If we can find a path where the numbers increase by as little as possible each time, that would be ideal.One idea is to use a snake-like pattern, where the rook moves back and forth across the rows. For example, it goes right across the first row, then down to the next row, then left across that row, then down to the next row, and so on. This way, the numbers increase sequentially across each row, and when moving to the next row, the difference between the last number of one row and the first number of the next row would be minimal.Let me try to visualize this for a small n, say n=3. The board would look like this:1 2 36 5 47 8 9Here, the rook starts at 1, moves right to 2, then to 3. Then it moves down to 6, left to 5, then to 4. Then down to 7, right to 8, and finally to 9.Now, let's calculate the differences between neighboring cells:- Between 1 and 2: 1- Between 2 and 3: 1- Between 3 and 6: 3- Between 6 and 5: 1- Between 5 and 4: 1- Between 4 and 7: 3- Between 7 and 8: 1- Between 8 and 9: 1So, the maximum difference here is 3, which is 2n - 1 (since n=3, 2*3 -1=5, wait, that doesn't match). Hmm, maybe my initial assumption is wrong.Wait, in the n=3 case, the maximum difference is 3, which is n. So, maybe the minimal M is n.But let's check another pattern. Suppose the rook moves in a way that alternates between rows and columns more evenly. Maybe a spiral pattern or something else.Alternatively, maybe the minimal M is 2n -1. For n=3, that would be 5, but in my snake pattern, I only got a maximum difference of 3. So, perhaps 2n -1 is not the minimal.Wait, maybe I made a mistake in calculating the difference between 3 and 6. Let's see: 6 - 3 = 3, which is correct. So, in this case, M=3 for n=3.If n=4, let's try the same snake pattern:1 2 3 48 7 6 59 10 11 1216 15 14 13Now, the differences:- Between 1 and 2:1- Between 2 and 3:1- Between 3 and 4:1- Between 4 and 8:4- Between 8 and 7:1- Between 7 and 6:1- Between 6 and 5:1- Between 5 and 9:4- Between 9 and 10:1- Between 10 and 11:1- Between 11 and 12:1- Between 12 and 16:4- Between 16 and 15:1- Between 15 and 14:1- Between 14 and 13:1So, the maximum difference here is 4, which is n=4.Wait, so for n=3, M=3, for n=4, M=4. So, it seems like M=n.But earlier, I thought maybe it's 2n-1, but that doesn't fit.Wait, let's think about how the numbering works. When moving from the end of one row to the start of the next row, the difference is n. For example, from 4 to 8 in n=4: 8-4=4.Similarly, in n=3, from 3 to 6: 6-3=3.So, in general, when moving from the end of a row to the start of the next row, the difference is n.But is there a way to make this difference smaller?If we can find a path where the transition between rows doesn't require jumping n numbers, but less, then M could be smaller.But I don't think that's possible because the rook has to visit each cell exactly once, and the numbering has to be sequential. So, when moving from one row to the next, you have to jump by at least n-1 numbers, but in the snake pattern, it's exactly n.Wait, actually, in the snake pattern, the jump is n. So, maybe n is the minimal possible M.But let's think about another pattern. Suppose instead of going row by row, the rook moves in a way that alternates between rows and columns more evenly.For example, in n=3:1 2 36 5 47 8 9This is the same as before, M=3.Alternatively, maybe a different path:1 4 32 5 87 6 9But let's see the differences:- 1 and 4:3- 4 and 3:1- 3 and 8:5- 8 and 9:1- 9 and 6:3- 6 and 7:1- 7 and 2:5- 2 and 5:3- 5 and 1:4Wait, this seems worse, with differences up to 5.So, maybe the snake pattern is better.Alternatively, maybe a diagonal pattern, but since the rook can't move diagonally, that's not applicable.Another idea: use a pattern where the rook alternates between moving right and down, creating a kind of spiral.But in that case, the differences might still be similar or larger.Wait, let's try for n=3:1 2 38 9 47 6 5Differences:- 1 and 2:1- 2 and 3:1- 3 and 4:1- 4 and 5:1- 5 and 6:1- 6 and 7:1- 7 and 8:1- 8 and 9:1- 9 and 4:5Wait, but 9 and 4 are not neighbors. So, actually, the differences are all 1 except between 9 and 4, which are not adjacent. So, in this case, M=1, which is impossible because the rook has to move through all cells, and the numbering has to be sequential. So, this pattern is invalid because the rook can't move from 3 to 4 directly in this arrangement.Wait, maybe I messed up the path. Let me try to define the path correctly.Starting at 1, moving right to 2, right to 3, down to 4, down to 5, left to 6, left to 7, up to 8, up to 9.So, the numbering would be:1 2 38 9 47 6 5Now, let's check the differences:- 1 and 2:1- 2 and 3:1- 3 and 4:1- 4 and 5:1- 5 and 6:1- 6 and 7:1- 7 and 8:1- 8 and 9:1- 9 and 4:5 (but 9 and 4 are not neighbors)Wait, actually, in this path, the rook moves from 3 to 4, which is down, then from 4 to 5, down, then from 5 to 6, left, etc. So, the differences are all 1 except when moving from one row to another.Wait, no, in this case, moving from 3 to 4 is down, so the difference is 1 (4-3=1). Similarly, moving from 5 to 6 is left, difference 1. Moving from 7 to 8 is up, difference 1. Moving from 9 to 4 would be a jump, but they are not neighbors in the grid.So, in this case, all neighboring cells have a difference of 1, except when moving from one row to another, which also has a difference of 1. Wait, that can't be right because in the numbering, moving from 3 to 4 is a difference of 1, but 3 is in row 1, column 3, and 4 is in row 2, column 3. So, they are neighbors, and the difference is 1.Similarly, moving from 5 to 6 is left, difference 1.Moving from 7 to 8 is up, difference 1.So, in this case, all neighboring cells have a difference of 1. But that's impossible because the rook has to move through all cells, and the numbering has to be sequential. Wait, no, actually, in this case, the numbering is sequential, and the differences are all 1, which would mean M=1. But that's not possible because the rook has to move through all cells, and the numbering has to be sequential, but the differences between some cells would have to be larger.Wait, I'm getting confused. Let me try to list the actual differences between all neighboring cells in this numbering:- 1 is adjacent to 2 and 8. - |2-1|=1 - |8-1|=7- 2 is adjacent to 1, 3, and 9. - |1-2|=1 - |3-2|=1 - |9-2|=7- 3 is adjacent to 2, 4, and 9. - |2-3|=1 - |4-3|=1 - |9-3|=6- 4 is adjacent to 3, 5, and 9. - |3-4|=1 - |5-4|=1 - |9-4|=5- 5 is adjacent to 4, 6, and 8. - |4-5|=1 - |6-5|=1 - |8-5|=3- 6 is adjacent to 5, 7, and 8. - |5-6|=1 - |7-6|=1 - |8-6|=2- 7 is adjacent to 6 and 8. - |6-7|=1 - |8-7|=1- 8 is adjacent to 1, 5, 6, and 7. - |1-8|=7 - |5-8|=3 - |6-8|=2 - |7-8|=1- 9 is adjacent to 2, 3, 4, and 5. - |2-9|=7 - |3-9|=6 - |4-9|=5 - |5-9|=4So, in this case, the maximum difference M is 7, which is much larger than the snake pattern's M=3.Therefore, the snake pattern seems better.Wait, so in the snake pattern, the maximum difference is n, which for n=3 is 3, and for n=4 is 4. So, maybe the minimal M is n.But let's think about larger n. For n=2:1 24 3Differences:- |2-1|=1- |4-2|=2- |3-4|=1- |3-2|=1So, M=2, which is n=2.Similarly, for n=1, trivially M=0.So, it seems like M=n.But wait, in the snake pattern, the maximum difference is n, but is there a way to make it smaller?Suppose we can find a path where the transition between rows doesn't require a jump of n numbers, but less.But how? Since the rook has to visit each cell exactly once, and the numbering is sequential, when moving from one row to another, you have to jump by at least n-1 numbers, but in the snake pattern, it's exactly n.Wait, actually, in the snake pattern, moving from the end of one row to the start of the next row is a difference of n. For example, in n=3, from 3 to 6: 6-3=3.But if we can arrange the numbering such that the transition between rows is less than n, that would be better.But I don't think that's possible because the rook has to visit all cells, and the numbering has to be sequential. So, when moving from one row to another, you have to jump by at least n-1 numbers, but in the snake pattern, it's exactly n.Wait, maybe if we alternate the direction of the rows, but that's what the snake pattern already does.Alternatively, maybe a different path that doesn't require such a large jump.Wait, let's think about a different pattern. Suppose the rook moves in a way that alternates between moving right and down, creating a kind of spiral.For n=3:1 2 38 9 47 6 5But as I saw earlier, this leads to a maximum difference of 7, which is worse.Alternatively, maybe a more efficient path.Wait, perhaps using a knight's move pattern, but the rook can't move like a knight.Alternatively, maybe a pattern where the rook moves in a way that the numbering increases by 1 each time, but that's only possible if the rook moves in a single row or column, which isn't the case.Wait, no, the rook has to move to neighboring cells, so it can't jump over cells.So, the minimal M seems to be n, as achieved by the snake pattern.But let's think about n=4 again.In the snake pattern:1 2 3 48 7 6 59 10 11 1216 15 14 13Differences:- Between 1 and 2:1- Between 2 and 3:1- Between 3 and 4:1- Between 4 and 8:4- Between 8 and 7:1- Between 7 and 6:1- Between 6 and 5:1- Between 5 and 9:4- Between 9 and 10:1- Between 10 and 11:1- Between 11 and 12:1- Between 12 and 16:4- Between 16 and 15:1- Between 15 and 14:1- Between 14 and 13:1So, the maximum difference is 4, which is n=4.Therefore, it seems that the minimal M is n.But wait, let's think about n=5.In the snake pattern:1 2 3 4 510 9 8 7 611 12 13 14 1520 19 18 17 1621 22 23 24 25Differences:- Between 1 and 2:1- Between 2 and 3:1- Between 3 and 4:1- Between 4 and 5:1- Between 5 and 10:5- Between 10 and 9:1- Between 9 and 8:1- Between 8 and 7:1- Between 7 and 6:1- Between 6 and 11:5- Between 11 and 12:1- Between 12 and 13:1- Between 13 and 14:1- Between 14 and 15:1- Between 15 and 20:5- Between 20 and 19:1- Between 19 and 18:1- Between 18 and 17:1- Between 17 and 16:1- Between 16 and 21:5- Between 21 and 22:1- Between 22 and 23:1- Between 23 and 24:1- Between 24 and 25:1So, the maximum difference is 5, which is n=5.Therefore, it seems consistent that the minimal M is n.But wait, is there a way to make M smaller than n?Suppose we can find a path where the transition between rows is less than n.But how? Since the rook has to visit each cell exactly once, and the numbering is sequential, when moving from one row to another, you have to jump by at least n-1 numbers, but in the snake pattern, it's exactly n.Wait, actually, in the snake pattern, the jump is n, but maybe if we can arrange the numbering such that the jump is n-1.But I don't think that's possible because the rook has to visit all cells, and the numbering has to be sequential. So, when moving from one row to another, you have to jump by at least n-1 numbers, but in the snake pattern, it's exactly n.Wait, no, in the snake pattern, moving from the end of one row to the start of the next row is a difference of n. For example, in n=3, from 3 to 6: 6-3=3.But if we can arrange the numbering such that the transition between rows is less than n, that would be better.But I don't think that's possible because the rook has to visit all cells, and the numbering has to be sequential. So, when moving from one row to another, you have to jump by at least n-1 numbers, but in the snake pattern, it's exactly n.Wait, maybe if we can alternate the direction of the rows in a way that the jump is less than n.But in the snake pattern, the jump is n, which seems to be the minimal possible.Therefore, I think the minimal M is n.But wait, let's think about the initial problem statement again."A rock travelled through an n x n board, stepping at each turn to the cell neighbouring the previous one by a side, so that each cell was visited once. Bob has put the integer numbers from 1 to n^2 into the cells, corresponding to the order in which the rook has passed them. Let M be the greatest difference of the numbers in neighbouring by side cells. What is the minimal possible value of M?"So, the rook's path is a Hamiltonian path on the grid graph, where each move is to a neighboring cell.The numbering is based on the order of the rook's path.M is the maximum difference between numbers in adjacent cells.We need to find the minimal possible M over all possible rook paths.From the examples above, it seems that the minimal M is n.But I recall that in some grid graphs, the minimal maximum difference can be lower.Wait, maybe I'm missing something.Let me think about the problem differently.Suppose we want to arrange the numbers 1 to n² in the grid such that the maximum difference between adjacent cells is minimized, and the arrangement corresponds to a rook's tour (i.e., a Hamiltonian path where each move is to a neighboring cell).This is similar to a graph labeling problem, where we want to label the vertices with numbers 1 to n² such that the maximum difference on edges is minimized, and the labeling corresponds to a Hamiltonian path.In graph theory, this is related to bandwidth minimization. The bandwidth of a graph is the minimum, over all labelings, of the maximum difference between labels of adjacent vertices.So, in this case, we're looking for the bandwidth of the n x n grid graph, considering only Hamiltonian paths.But I'm not sure about the exact bandwidth of the grid graph.Wait, I think the bandwidth of the n x n grid graph is n, achieved by the row-wise snake pattern.Yes, that makes sense.Therefore, the minimal possible value of M is n.So, the answer should be boxed{n}.

💡Alright, so I've got these two calculus problems here, and I need to figure them out step by step. Let's start with the first one.**Problem I: Finding the value of b**We have the function f(x) = -x³ + x² + b. We need to find the value of b such that the maximum value of f(x) on the interval x ∈ [-1/2, 1) is 3/8.Okay, so to find the maximum value of a function on a closed interval, I remember that I need to check both the critical points inside the interval and the endpoints. Critical points occur where the derivative is zero or undefined. Since f(x) is a polynomial, its derivative will be defined everywhere, so I just need to find where the derivative is zero.First, let's find the derivative of f(x):f'(x) = d/dx [-x³ + x² + b] = -3x² + 2x.Now, set the derivative equal to zero to find critical points:-3x² + 2x = 0.Factor out an x:x(-3x + 2) = 0.So, x = 0 or -3x + 2 = 0 → x = 2/3.Therefore, the critical points are at x = 0 and x = 2/3.Now, we need to evaluate f(x) at these critical points and at the endpoints of the interval, which are x = -1/2 and x = 1 (though since the interval is up to 1, not including 1, we might need to consider the limit as x approaches 1 from the left, but let's see).Let's compute f(x) at each of these points:1. At x = -1/2:f(-1/2) = -(-1/2)³ + (-1/2)² + b = -(-1/8) + 1/4 + b = 1/8 + 1/4 + b = 3/8 + b.2. At x = 0:f(0) = -(0)³ + (0)² + b = 0 + 0 + b = b.3. At x = 2/3:f(2/3) = -(2/3)³ + (2/3)² + b = -(8/27) + (4/9) + b. Let's compute this:-8/27 + 4/9 = -8/27 + 12/27 = 4/27. So, f(2/3) = 4/27 + b.4. At x approaching 1 from the left:f(1) = -(1)³ + (1)² + b = -1 + 1 + b = 0 + b = b. But since the interval is up to 1, not including 1, we might need to check the limit as x approaches 1 from the left, but since f(x) is continuous, the limit will just be b.Now, we have the function values at these points:- f(-1/2) = 3/8 + b- f(0) = b- f(2/3) = 4/27 + b- f(1^-) = bWe are told that the maximum value on this interval is 3/8. So, we need to see which of these function values is the maximum.Looking at the values:- 3/8 + b is at x = -1/2- b is at x = 0 and x approaching 1- 4/27 + b is at x = 2/3We need to find which of these is the largest. Let's compare them:First, 3/8 is approximately 0.375, and 4/27 is approximately 0.148. So, 3/8 is larger than 4/27.Therefore, the maximum value is either 3/8 + b or b, depending on the value of b.But wait, we are told that the maximum value is 3/8. So, 3/8 + b must be equal to 3/8, which would mean that b = 0. Alternatively, if b were greater than 3/8, then the maximum would be b, but since the maximum is given as 3/8, b must be 0.Wait, let me think again. If b were positive, then f(-1/2) would be 3/8 + b, which would be greater than 3/8. But since the maximum is given as 3/8, that implies that 3/8 + b = 3/8, so b = 0.Alternatively, if b were negative, then f(-1/2) would be less than 3/8, and the maximum would be either b or 4/27 + b, but since 4/27 is about 0.148, if b were negative, 4/27 + b could be less than b if b is negative enough. Wait, no, because 4/27 is positive, so 4/27 + b would be greater than b if b is negative. Hmm, this is getting a bit confusing.Wait, let's approach it differently. The maximum value is 3/8, so the highest value among f(-1/2), f(0), f(2/3), and f(1^-) must be 3/8.So, we have:f(-1/2) = 3/8 + bf(0) = bf(2/3) = 4/27 + bf(1^-) = bWe need the maximum of these to be 3/8. So, the maximum of 3/8 + b, b, 4/27 + b, and b is 3/8.So, 3/8 + b must be equal to 3/8, because if b were greater than 0, then 3/8 + b would be greater than 3/8, which contradicts the given maximum. If b were less than 0, then 3/8 + b would be less than 3/8, and the maximum would be either b or 4/27 + b, but since 4/27 is positive, 4/27 + b would be greater than b if b is negative. Wait, no, if b is negative, 4/27 + b could be less than 3/8.Wait, let's set up the equation. The maximum is 3/8, so:max{3/8 + b, b, 4/27 + b, b} = 3/8This implies that 3/8 + b ≤ 3/8, because if 3/8 + b were greater than 3/8, then the maximum would be greater than 3/8, which contradicts the given. Therefore, 3/8 + b ≤ 3/8 ⇒ b ≤ 0.But also, we need to ensure that none of the other function values exceed 3/8. So, f(0) = b ≤ 3/8, which is already satisfied if b ≤ 0. Similarly, f(2/3) = 4/27 + b ≤ 3/8.So, 4/27 + b ≤ 3/8 ⇒ b ≤ 3/8 - 4/27.Let's compute 3/8 - 4/27:Convert to common denominator, which is 216.3/8 = 81/2164/27 = 32/216So, 81/216 - 32/216 = 49/216 ≈ 0.22685.So, b ≤ 49/216.But from earlier, we have b ≤ 0.Therefore, the most restrictive condition is b ≤ 0.But we also need to ensure that the maximum is exactly 3/8. So, we need that at least one of the function values equals 3/8.Looking at f(-1/2) = 3/8 + b. If b = 0, then f(-1/2) = 3/8, which is the maximum. If b < 0, then f(-1/2) < 3/8, and the maximum would have to be achieved at another point.Wait, but if b < 0, then f(-1/2) = 3/8 + b < 3/8, and f(2/3) = 4/27 + b. Since 4/27 ≈ 0.148, if b is negative, 4/27 + b could be less than 3/8, but we need to ensure that the maximum is 3/8.Wait, perhaps I made a mistake earlier. Let's consider that the maximum is achieved at x = -1/2, which is 3/8 + b, and we are told that this maximum is 3/8. Therefore, 3/8 + b = 3/8 ⇒ b = 0.Alternatively, if the maximum were achieved at another point, say x = 2/3, then f(2/3) = 4/27 + b = 3/8. Solving for b:4/27 + b = 3/8 ⇒ b = 3/8 - 4/27 = (81 - 32)/216 = 49/216 ≈ 0.22685.But then, we need to check if with b = 49/216, the function f(x) doesn't exceed 3/8 elsewhere in the interval.So, let's compute f(-1/2) with b = 49/216:f(-1/2) = 3/8 + 49/216 = (81 + 49)/216 = 130/216 ≈ 0.60185, which is greater than 3/8 ≈ 0.375. That's a problem because we were told the maximum is 3/8.Therefore, b cannot be 49/216 because then f(-1/2) would exceed 3/8. So, the only way for the maximum to be exactly 3/8 is if f(-1/2) = 3/8, which implies b = 0.Therefore, b = 0.Wait, let me double-check. If b = 0, then f(-1/2) = 3/8, f(0) = 0, f(2/3) = 4/27 ≈ 0.148, and f(1^-) = 0. So, the maximum is indeed 3/8 at x = -1/2. That makes sense.So, the value of b is 0.**Problem II: Finding the range of a**We have g(x) = a ln x. For any x ∈ [1, e], g(x) ≥ -x² + (a + 2)x.We need to find the range of a such that this inequality holds for all x in [1, e].So, the inequality is:a ln x ≥ -x² + (a + 2)x.Let's rearrange this inequality to solve for a.First, bring all terms to one side:a ln x + x² - (a + 2)x ≥ 0.Let's factor out a:a (ln x - x) + x² - 2x ≥ 0.Now, let's solve for a:a (ln x - x) ≥ -x² + 2x.So, a ≤ (-x² + 2x)/(ln x - x).Wait, because ln x - x is negative for x > 1, right? Because ln x grows slower than x, so ln x - x is negative.Therefore, when we divide both sides by (ln x - x), which is negative, the inequality sign flips.So, a ≤ [ -x² + 2x ] / [ ln x - x ].But let's write it as:a ≤ [ -x² + 2x ] / [ ln x - x ].Alternatively, we can write it as:a ≤ [ 2x - x² ] / [ ln x - x ].Let me denote the right-hand side as T(x):T(x) = (2x - x²)/(ln x - x).We need to find the minimum value of T(x) over x ∈ [1, e], because a must be less than or equal to T(x) for all x in [1, e]. Therefore, the maximum lower bound for a is the minimum of T(x) over [1, e].Wait, no. Let me think again.We have a ≤ T(x) for all x ∈ [1, e]. Therefore, a must be less than or equal to the minimum value of T(x) on [1, e]. Because if a is less than or equal to the smallest T(x), then it will be less than or equal to all T(x).Wait, no, actually, if a ≤ T(x) for all x, then a must be less than or equal to the infimum of T(x) over x ∈ [1, e]. But since T(x) is continuous on [1, e], it will attain its minimum, so a must be ≤ the minimum of T(x).Wait, but let's check the behavior of T(x).First, let's compute T(x) at the endpoints:At x = 1:T(1) = (2*1 - 1²)/(ln 1 - 1) = (2 - 1)/(0 - 1) = 1/(-1) = -1.At x = e:T(e) = (2e - e²)/(ln e - e) = (2e - e²)/(1 - e).Compute numerator: 2e - e² = e(2 - e) ≈ e*(-0.718) ≈ negative.Denominator: 1 - e ≈ -1.718.So, T(e) = [negative]/[negative] = positive.Let me compute it exactly:T(e) = (2e - e²)/(1 - e) = e(2 - e)/(1 - e) = e(2 - e)/(- (e - 1)) = -e(2 - e)/(e - 1).But 2 - e is negative, so -e(2 - e) is positive, and e - 1 is positive, so overall T(e) is positive.So, T(e) is positive, and T(1) is -1.Now, we need to see if T(x) has any critical points in (1, e) where it might attain a minimum.To find the minimum of T(x), we can take its derivative and find where it's zero.Let me compute T'(x):T(x) = (2x - x²)/(ln x - x).Let me denote numerator as N = 2x - x², denominator as D = ln x - x.Then, T'(x) = (N' D - N D') / D².Compute N' = 2 - 2x.Compute D' = (1/x) - 1.So, T'(x) = [ (2 - 2x)(ln x - x) - (2x - x²)(1/x - 1) ] / (ln x - x)².Let's simplify the numerator:First term: (2 - 2x)(ln x - x)Second term: - (2x - x²)(1/x - 1)Let's expand both terms.First term:(2 - 2x)(ln x - x) = 2(ln x - x) - 2x(ln x - x) = 2 ln x - 2x - 2x ln x + 2x².Second term:- (2x - x²)(1/x - 1) = - [ (2x)(1/x) - 2x*1 - x²*(1/x) + x²*1 ] = - [ 2 - 2x - x + x² ] = - [ 2 - 3x + x² ] = -2 + 3x - x².Now, combine the two terms:First term: 2 ln x - 2x - 2x ln x + 2x²Second term: -2 + 3x - x²So, total numerator:2 ln x - 2x - 2x ln x + 2x² - 2 + 3x - x²Combine like terms:- For ln x terms: 2 ln x - 2x ln x- For x² terms: 2x² - x² = x²- For x terms: -2x + 3x = x- Constants: -2So, numerator becomes:(2 - 2x) ln x + x² + x - 2.Wait, let me factor out the ln x:= (2 - 2x) ln x + x² + x - 2.Hmm, this seems a bit complicated. Maybe we can factor further or see if it can be simplified.Alternatively, perhaps we can factor the numerator differently.Wait, let's see:Numerator: (2 - 2x) ln x + x² + x - 2.Let me factor out 2(1 - x) from the first term:= 2(1 - x) ln x + x² + x - 2.Hmm, not sure if that helps.Alternatively, perhaps we can factor the entire expression.Wait, let me try to see if x = 1 is a root of the numerator.Plug x = 1:2(1 - 1) ln 1 + 1 + 1 - 2 = 0 + 2 - 2 = 0.So, x = 1 is a root. Therefore, (x - 1) is a factor.Let's perform polynomial division or factorization.Let me write the numerator as:(2 - 2x) ln x + x² + x - 2.But since x = 1 is a root, let's see if we can factor (x - 1) out.Wait, perhaps it's better to consider that the numerator is a function of x, say, N(x) = (2 - 2x) ln x + x² + x - 2.We can factor (x - 1) out of N(x).Let me try to factor N(x):N(x) = (2 - 2x) ln x + x² + x - 2.Let me rearrange:= (2(1 - x)) ln x + (x² + x - 2).Now, x² + x - 2 can be factored as (x + 2)(x - 1).So, N(x) = 2(1 - x) ln x + (x + 2)(x - 1).Factor out (x - 1):= (x - 1)[ -2 ln x + (x + 2) ].Wait, let me check:(x - 1)[ -2 ln x + (x + 2) ] = (x - 1)(x + 2 - 2 ln x).Yes, that's correct.So, N(x) = (x - 1)(x + 2 - 2 ln x).Therefore, the numerator of T'(x) is (x - 1)(x + 2 - 2 ln x).So, T'(x) = [ (x - 1)(x + 2 - 2 ln x) ] / (ln x - x)².Now, since (ln x - x)² is always positive for x ≠ 1, the sign of T'(x) depends on the numerator: (x - 1)(x + 2 - 2 ln x).We need to analyze the sign of this expression for x ∈ (1, e).First, x - 1 is positive for x > 1.So, the sign of T'(x) depends on the sign of (x + 2 - 2 ln x).Let me define S(x) = x + 2 - 2 ln x.We need to determine where S(x) is positive or negative in (1, e).Compute S(1):S(1) = 1 + 2 - 2 ln 1 = 3 - 0 = 3 > 0.Compute S(e):S(e) = e + 2 - 2 ln e = e + 2 - 2*1 = e + 0 ≈ 2.718 > 0.Now, let's check if S(x) ever becomes zero or negative in (1, e).Compute the derivative of S(x):S'(x) = 1 - 2*(1/x).Set S'(x) = 0:1 - 2/x = 0 ⇒ x = 2.So, S(x) has a critical point at x = 2.Compute S(2):S(2) = 2 + 2 - 2 ln 2 ≈ 4 - 2*0.693 ≈ 4 - 1.386 ≈ 2.614 > 0.So, S(x) is always positive in (1, e), because at x=1, S=3; at x=2, S≈2.614; at x=e, S≈2.718. Since S'(x) = 1 - 2/x, which is negative for x < 2 and positive for x > 2. So, S(x) decreases from x=1 to x=2, reaching a minimum at x=2, then increases from x=2 to x=e. But since S(2) is still positive, S(x) remains positive throughout (1, e).Therefore, S(x) > 0 for all x ∈ (1, e).Thus, the numerator of T'(x) is (x - 1)*positive, which is positive for x > 1.Therefore, T'(x) > 0 for x ∈ (1, e).This means that T(x) is increasing on [1, e].Therefore, the minimum value of T(x) on [1, e] occurs at x = 1, which is T(1) = -1.Therefore, to satisfy a ≤ T(x) for all x ∈ [1, e], a must be ≤ the minimum of T(x), which is -1.Hence, the range of a is (-∞, -1].Wait, let me confirm.Since T(x) is increasing on [1, e], its minimum is at x=1, which is -1, and its maximum is at x=e, which is positive. Therefore, for a to be ≤ T(x) for all x ∈ [1, e], a must be ≤ the smallest T(x), which is -1.Therefore, a ≤ -1.So, the range of a is (-∞, -1].I think that's the answer.