💡Alright, so I've got this problem about a 2017x2017 grid where each cell is colored either black or white. The condition is that every cell has at least one side in common with another cell of the same color. Then, for each color, we form a graph by connecting centers of adjacent cells of the same color. Both of these graphs, G₁ (black) and G₂ (white), are supposed to be connected paths—meaning they're just straight lines without any cycles or splits. The goal is to prove that the center of the grid is one of the endpoints of either G₁ or G₂.Hmm, okay. Let me try to break this down. First, the grid is pretty large—2017x2017. That's an odd number, so there's a single center cell. The coloring condition ensures that no cell is isolated; every cell has at least one neighbor of the same color. So, both black and white regions are connected in some way.Now, G₁ and G₂ are connected paths. So, each color forms a single path that doesn't branch or loop. That means each color's cells form a snake-like path through the grid. Since the grid is finite, these paths must start and end somewhere on the grid.I need to show that the center cell is an endpoint of one of these paths. So, either the black path ends at the center, or the white path does, or both? Wait, no, it's either one or the other. Because if both ended at the center, that would mean the center is part of both paths, which isn't possible since each cell is only one color.So, maybe the idea is that the center can't be in the middle of a path because that would require it to have two neighbors of the same color, but since it's the center, it's surrounded by cells of both colors? Hmm, not necessarily, because the paths can twist and turn.Wait, actually, in a grid, each cell has four neighbors. The center cell has four neighbors, each in a different direction. If the center is part of a path, it must have exactly two neighbors of the same color—since it's part of a path, not a cycle or a branch. So, if the center is part of a path, it must have one predecessor and one successor in the path.But the problem states that both G₁ and G₂ are connected paths. So, if the center is part of, say, the black path, then it must have two black neighbors. But then, those neighbors would have to be connected in a way that doesn't create cycles or branches. Similarly for the white path.But the key is that if the center is part of a path, it's not an endpoint. So, if the center is an endpoint, it must have only one neighbor of the same color. But wait, the problem says every cell has at least one neighbor of the same color. So, if the center is an endpoint, it must have exactly one neighbor of the same color. That seems possible.But how does that help me? Maybe I need to consider the structure of the paths. Since both G₁ and G₂ are paths, they each have two endpoints. The grid has four corners, so maybe the endpoints are at the corners? But the grid is 2017x2017, which is odd, so the center is well-defined.Wait, perhaps I can use induction. Let me think about smaller grids. For example, a 1x1 grid is trivial—just the center, which is both the start and end. For a 3x3 grid, the center must be an endpoint of one of the paths. Let me visualize that.In a 3x3 grid, if I color the center black, then the black path must start or end there. If I color the center white, similarly. But in this case, the paths can't both pass through the center because that would require the center to have two neighbors of each color, which isn't possible since each cell is only one color.Wait, no, actually, the center is only one color, so only one path can pass through it. But if it's part of a path, it's not an endpoint. So, if the center is part of a path, it's in the middle, but the problem wants it to be an endpoint. So, maybe the center can't be part of a path—it has to be an endpoint.But how? Because if the center is an endpoint, it must have only one neighbor of the same color. Let's say the center is black. Then, it must have only one black neighbor. But the center has four neighbors—up, down, left, right. So, one of those is black, and the other three must be white. But then, the white path would have to cover the other three cells adjacent to the center.But wait, the white path is also a single connected path. If the center is black with only one black neighbor, then the white cells adjacent to the center must form a connected path. But the three white cells around the center would need to connect to each other and to the rest of the white cells. But three cells can form a connected path, but would that interfere with the black path?This is getting a bit tangled. Maybe I need to think about the parity or something. Since the grid is odd-sized, the center is unique, and maybe the paths have to spiral around it or something.Alternatively, maybe I can argue that if the center is not an endpoint, then both paths must pass through it, which is impossible because the center can only be one color. Therefore, the center must be an endpoint of one of the paths.Wait, that seems too simplistic. Let me elaborate. Suppose, for contradiction, that the center is not an endpoint of either G₁ or G₂. Then, the center must be an internal node in both G₁ and G₂. But that's impossible because the center is only one color, so it can only be part of one graph, either G₁ or G₂, not both. Therefore, the center cannot be an internal node of both, so it must be an endpoint of at least one.Wait, that seems like a valid argument. If the center is not an endpoint of either, it would have to be an internal node in both, which is impossible. Therefore, the center must be an endpoint of at least one of G₁ or G₂.But let me check if that's rigorous. Suppose the center is part of G₁. Then, as an internal node, it has two black neighbors. Similarly, if it's part of G₂, it has two white neighbors. But if it's not an endpoint, it's internal, meaning it has two neighbors of the same color. But since it's only one color, it can't have neighbors of both colors. So, if it's internal in G₁, it can't be internal in G₂, and vice versa.But the problem is that the center is a single cell, so it can only be part of one graph. Therefore, if it's internal in one graph, it's not part of the other. But the other graph must cover the rest of the grid, including the cells adjacent to the center. So, if the center is internal in G₁, then the three other adjacent cells must be part of G₂. But G₂ is a single path, so those three cells must connect in a path. But three cells can form a path, but then how does that affect the overall structure?Wait, maybe the key is that if the center is internal in one graph, the other graph has to wrap around it, but in doing so, it might create a cycle or a split, which is not allowed because both graphs are supposed to be simple paths.So, perhaps if the center is internal in G₁, then G₂ has to go around it, but that would require G₂ to have a cycle or a split, which contradicts the condition that G₂ is a simple path. Therefore, the center cannot be internal in either G₁ or G₂, so it must be an endpoint of at least one.Yes, that seems to make sense. So, the center can't be an internal node in both graphs because it's only one color, and it can't be internal in one without causing the other graph to have a cycle or split. Therefore, it must be an endpoint of one of the graphs.I think that's the gist of the proof. Maybe I should formalize it a bit more, but the main idea is that the center can't be an internal node in both, and if it's internal in one, it causes problems for the other graph. Therefore, it must be an endpoint of one of them.

💡Okay, so I have this problem here where I'm given real numbers ( x_1, x_2, ldots, x_n ) such that their squares add up to 1. I need to prove that for any integer ( k geq 2 ), there exist integers ( a_1, a_2, ldots, a_n ), not all zero, with each ( |a_i| leq k-1 ), and the absolute value of the linear combination ( |a_1x_1 + a_2x_2 + ldots + a_nx_n| ) is less than or equal to ( frac{(k-1)sqrt{n}}{k^n - 1} ).Hmm, okay. So, first, I should probably think about what this is asking. It seems like it's a problem about approximating a linear combination of these real numbers with integer coefficients bounded in absolute value by ( k-1 ). The bound on the linear combination is given in terms of ( k ) and ( n ).I remember that problems like this often use the pigeonhole principle. Maybe I can apply that here. The pigeonhole principle is all about ensuring that when you have more pigeons than holes, at least one hole has more than one pigeon. In this context, maybe the "pigeons" are the possible sums ( a_1x_1 + ldots + a_nx_n ) and the "holes" are some kind of intervals or regions where these sums can fall.Let me think about how many possible combinations of ( a_i ) there are. Each ( a_i ) can be any integer from ( -(k-1) ) to ( (k-1) ), inclusive. So, for each ( a_i ), there are ( 2k-1 ) choices. Since there are ( n ) variables, the total number of possible vectors ( (a_1, a_2, ldots, a_n) ) is ( (2k-1)^n ).Now, if I consider the sums ( S = a_1x_1 + a_2x_2 + ldots + a_nx_n ), each of these sums is a real number. The question is, how many distinct sums can there be? Well, if the sums are spread out, there could be up to ( (2k-1)^n ) distinct sums. But if they overlap, then some sums must be equal.Wait, but how does that help me? If two different vectors ( (a_1, ldots, a_n) ) and ( (b_1, ldots, b_n) ) give the same sum, then their difference ( (c_1, ldots, c_n) = (a_1 - b_1, ldots, a_n - b_n) ) would give a sum of zero. But I need a non-zero vector ( (c_1, ldots, c_n) ) such that ( c_1x_1 + ldots + c_nx_n = 0 ). But the problem doesn't require the sum to be zero; it just requires it to be small.Maybe I need a different approach. Perhaps instead of looking for two sums that are equal, I should look for a sum that's very close to zero. That way, the absolute value of the sum would be small, which is what I need.To make this precise, maybe I can partition the interval containing all possible sums into smaller intervals and use the pigeonhole principle to ensure that at least two sums fall into the same interval. Then, their difference would be small.But what's the range of possible sums? Since each ( |a_i| leq k-1 ) and ( |x_i| leq 1 ) (because ( x_1^2 + ldots + x_n^2 = 1 )), the maximum possible absolute value of the sum is ( (k-1)sqrt{n} ) by the Cauchy-Schwarz inequality.So, the total range of possible sums is from ( -(k-1)sqrt{n} ) to ( (k-1)sqrt{n} ). If I divide this interval into ( (2k-1)^n - 1 ) subintervals, each of width ( frac{2(k-1)sqrt{n}}{(2k-1)^n - 1} ). Wait, but the number of subintervals should be one less than the number of sums, right? Because if you have ( m ) points, you can divide the interval into ( m-1 ) subintervals.But actually, the number of sums is ( (2k-1)^n ), so the number of subintervals should be ( (2k-1)^n - 1 ). Therefore, the width of each subinterval is ( frac{2(k-1)sqrt{n}}{(2k-1)^n - 1} ).By the pigeonhole principle, if we have more sums than subintervals, at least two sums must fall into the same subinterval. The difference between these two sums would then be less than or equal to the width of the subinterval. Let me denote these two sums as ( S_1 = a_1x_1 + ldots + a_nx_n ) and ( S_2 = b_1x_1 + ldots + b_nx_n ). Then, ( |S_1 - S_2| leq frac{2(k-1)sqrt{n}}{(2k-1)^n - 1} ).But ( S_1 - S_2 = (a_1 - b_1)x_1 + ldots + (a_n - b_n)x_n ). Let me define ( c_i = a_i - b_i ). Then, ( |c_i| leq 2(k-1) ) because ( |a_i| leq k-1 ) and ( |b_i| leq k-1 ). So, ( |c_i| leq 2(k-1) ).But the problem requires ( |a_i| leq k-1 ), not ( 2(k-1) ). Hmm, so maybe I need to adjust my approach. Perhaps instead of considering all possible ( a_i ) from ( -(k-1) ) to ( (k-1) ), I can use a different method.Wait, maybe I can use induction on ( n ). For ( n = 1 ), the problem is trivial because ( x_1^2 = 1 ), so ( x_1 = pm 1 ). Then, choosing ( a_1 = 0 ) gives the sum zero, but we need non-zero coefficients. Hmm, maybe not.Alternatively, perhaps I can use the pigeonhole principle in a different way. Let me consider the fractional parts of the sums ( a_1x_1 + ldots + a_nx_n ). If I can show that one of these sums is very close to an integer, then the difference would be small.But I'm not sure if that's the right direction. Maybe I should think about scaling. Since ( x_1^2 + ldots + x_n^2 = 1 ), the vector ( (x_1, ldots, x_n) ) is a unit vector. The sum ( a_1x_1 + ldots + a_nx_n ) is the dot product of ( (a_1, ldots, a_n) ) and ( (x_1, ldots, x_n) ).So, maybe I can use some result from linear algebra or geometry. The dot product is related to the angle between the vectors. But I'm not sure how that helps with bounding the sum.Wait, another idea: maybe I can use the concept of lattice points. The set of all vectors ( (a_1, ldots, a_n) ) with ( |a_i| leq k-1 ) forms a lattice. The problem is asking for a lattice point such that its projection onto the unit vector ( (x_1, ldots, x_n) ) is small.This seems related to the idea of simultaneous Diophantine approximations. In that context, you try to approximate real numbers with rationals such that the approximations are good in some sense. Maybe I can use a similar technique here.Alternatively, perhaps I can use the pigeonhole principle by considering the projections of the lattice points onto the unit vector. If I can show that two projections are close, then their difference is a small sum.Let me formalize this. Let me define the projection of each lattice point ( (a_1, ldots, a_n) ) onto the unit vector ( (x_1, ldots, x_n) ) as ( S = a_1x_1 + ldots + a_nx_n ). The set of all such ( S ) lies in the interval ( [-(k-1)sqrt{n}, (k-1)sqrt{n}] ).If I divide this interval into ( (2k-1)^n - 1 ) equal parts, each of width ( frac{2(k-1)sqrt{n}}{(2k-1)^n - 1} ). By the pigeonhole principle, since there are ( (2k-1)^n ) lattice points, at least two of them must fall into the same subinterval. Let these two points be ( (a_1, ldots, a_n) ) and ( (b_1, ldots, b_n) ).Then, the difference ( S = (a_1 - b_1)x_1 + ldots + (a_n - b_n)x_n ) must satisfy ( |S| leq frac{2(k-1)sqrt{n}}{(2k-1)^n - 1} ). Let ( c_i = a_i - b_i ). Then, ( |c_i| leq 2(k-1) ), and ( c_i ) are integers, not all zero.But the problem requires ( |a_i| leq k-1 ), not ( 2(k-1) ). So, perhaps I can adjust the coefficients ( c_i ) to fit within ( k-1 ). Wait, but if ( |c_i| leq 2(k-1) ), I can write ( c_i = d_i + e_i ) where ( |d_i| leq k-1 ) and ( |e_i| leq k-1 ). But I'm not sure if that helps.Alternatively, maybe I can scale down the coefficients. If I divide each ( c_i ) by 2, but then they wouldn't be integers anymore. Hmm.Wait, maybe instead of considering all possible ( a_i ) from ( -(k-1) ) to ( (k-1) ), I can use a different modulus. For example, if I consider ( a_i ) modulo ( k ), then each ( a_i ) can be represented as ( a_i = k m_i + r_i ) where ( r_i ) is the remainder when ( a_i ) is divided by ( k ), so ( |r_i| leq k-1 ).But I'm not sure if that directly helps either. Maybe I need to think differently.Let me recall the original problem statement. It says that for every integer ( k geq 2 ), there exist integers ( a_1, ldots, a_n ), not all zero, with ( |a_i| leq k-1 ), such that ( |a_1x_1 + ldots + a_nx_n| leq frac{(k-1)sqrt{n}}{k^n - 1} ).Notice that the denominator is ( k^n - 1 ), which is similar to the number of lattice points ( (2k-1)^n ), but not exactly the same. Maybe there's a connection.Wait, perhaps instead of considering all possible ( a_i ) from ( -(k-1) ) to ( (k-1) ), I can consider them from 0 to ( k-1 ). That would give ( k^n ) possible vectors. Then, the number of possible sums would be ( k^n ), and the interval is still ( [0, (k-1)sqrt{n}] ).If I divide this interval into ( k^n - 1 ) subintervals, each of width ( frac{(k-1)sqrt{n}}{k^n - 1} ). Then, by the pigeonhole principle, two sums must fall into the same subinterval, so their difference is less than or equal to ( frac{(k-1)sqrt{n}}{k^n - 1} ).But wait, the problem allows ( a_i ) to be negative as well, so maybe I need to consider both positive and negative sums. Alternatively, I can adjust the interval accordingly.Let me try this approach. Consider all possible vectors ( (a_1, ldots, a_n) ) where each ( a_i ) is an integer between 0 and ( k-1 ). There are ( k^n ) such vectors. The sum ( S = a_1x_1 + ldots + a_nx_n ) can range from 0 to ( (k-1)sqrt{n} ).Divide this interval into ( k^n - 1 ) equal parts, each of width ( frac{(k-1)sqrt{n}}{k^n - 1} ). By the pigeonhole principle, at least two vectors must produce sums that fall into the same subinterval. Let these vectors be ( (a_1, ldots, a_n) ) and ( (b_1, ldots, b_n) ). Then, the difference ( S = (a_1 - b_1)x_1 + ldots + (a_n - b_n)x_n ) satisfies ( |S| leq frac{(k-1)sqrt{n}}{k^n - 1} ).Now, let ( c_i = a_i - b_i ). Each ( c_i ) is an integer, and since ( 0 leq a_i, b_i leq k-1 ), we have ( - (k-1) leq c_i leq (k-1) ). Therefore, ( |c_i| leq k-1 ). Also, since ( a ) and ( b ) are different vectors, not all ( c_i ) are zero.Thus, we have found integers ( c_1, ldots, c_n ), not all zero, with ( |c_i| leq k-1 ), such that ( |c_1x_1 + ldots + c_nx_n| leq frac{(k-1)sqrt{n}}{k^n - 1} ), which is exactly what we needed to prove.Wait, but in this approach, I considered ( a_i ) from 0 to ( k-1 ), not from ( -(k-1) ) to ( (k-1) ). Does this affect the generality? I think it's okay because if I allow ( a_i ) to be negative, I can still represent any integer within ( -(k-1) ) to ( (k-1) ) by adjusting the coefficients accordingly. But in this case, by considering only non-negative coefficients, I was able to apply the pigeonhole principle more straightforwardly.So, to summarize, by considering all possible non-negative integer vectors ( (a_1, ldots, a_n) ) with each ( a_i ) from 0 to ( k-1 ), there are ( k^n ) such vectors. The sums ( S ) they produce lie in the interval ( [0, (k-1)sqrt{n}] ). Dividing this interval into ( k^n - 1 ) subintervals ensures that at least two sums fall into the same subinterval, making their difference small. This difference corresponds to a non-zero vector ( (c_1, ldots, c_n) ) with coefficients bounded by ( k-1 ), and the sum ( |c_1x_1 + ldots + c_nx_n| ) is bounded by ( frac{(k-1)sqrt{n}}{k^n - 1} ).I think this approach works. It uses the pigeonhole principle effectively by considering the number of possible sums and the interval they can fall into. By carefully dividing the interval, we ensure that at least two sums are close enough, leading to the desired bound.Another thing to check is whether the bound is tight or if it can be improved. The denominator ( k^n - 1 ) seems to come directly from the number of subintervals, which is one less than the number of possible sums. So, it's a natural result of the pigeonhole principle in this context.Also, considering the case when ( k = 2 ), the bound becomes ( frac{sqrt{n}}{2^n - 1} ). For ( n = 1 ), this would be ( frac{1}{1} = 1 ), which makes sense because with ( k = 2 ), the only possible non-zero coefficient is 1, and the sum is just ( x_1 ), which has absolute value 1. But wait, the problem states ( k geq 2 ), so ( k = 2 ) is allowed, but for ( n = 1 ), the bound is 1, which is exactly the value of ( |x_1| ). So, it checks out.For larger ( n ) and ( k ), the bound decreases exponentially with ( k ), which makes sense because as ( k ) increases, the number of possible sums increases, making it more likely to find a small sum.Overall, I think this approach using the pigeonhole principle is solid. It carefully constructs the necessary conditions and applies the principle to ensure the existence of the required integers ( a_i ) with the desired properties.

💡Okay, so I have this problem here about a function f(x) = x² + 2a x + b. The graph of this function passes through the point (1,3), and it also satisfies the condition f(-1 + x) = f(-1 - x) for all real numbers x. Then, there's another function g(x) which is symmetric to f(x) with respect to the origin. Part (1) asks me to find the explicit formulas for f(x) and g(x). Hmm, okay. Let's start with f(x). Since the graph passes through (1,3), that means when x is 1, f(x) is 3. So I can plug that into the equation:f(1) = (1)² + 2a(1) + b = 3Which simplifies to:1 + 2a + b = 3So, 2a + b = 2. Let me write that down as equation (1).Next, the condition f(-1 + x) = f(-1 - x) for all x. Hmm, this seems like a symmetry condition. If I replace x with 0, then f(-1 + 0) = f(-1 - 0), which is just f(-1) = f(-1). That doesn't tell me much. Maybe I need to expand f(-1 + x) and f(-1 - x) and set them equal.Let me compute f(-1 + x):f(-1 + x) = (-1 + x)² + 2a(-1 + x) + bExpanding that:= (1 - 2x + x²) + (-2a + 2a x) + b= x² - 2x + 1 - 2a + 2a x + bSimilarly, f(-1 - x):f(-1 - x) = (-1 - x)² + 2a(-1 - x) + bExpanding that:= (1 + 2x + x²) + (-2a - 2a x) + b= x² + 2x + 1 - 2a - 2a x + bNow, set f(-1 + x) equal to f(-1 - x):x² - 2x + 1 - 2a + 2a x + b = x² + 2x + 1 - 2a - 2a x + bLet me subtract x² and 1 - 2a + b from both sides to simplify:-2x + 2a x = 2x - 2a xCombine like terms:(-2 + 2a)x = (2 - 2a)xBring all terms to one side:(-2 + 2a - 2 + 2a)x = 0Wait, that would be (-4 + 4a)x = 0Since this must hold for all x, the coefficient must be zero:-4 + 4a = 0So, 4a = 4Therefore, a = 1Okay, so a is 1. Now, going back to equation (1):2a + b = 22(1) + b = 22 + b = 2So, b = 0Wait, that doesn't seem right. If a = 1 and b = 0, then f(x) = x² + 2x. Let me check if this passes through (1,3):f(1) = 1 + 2 = 3. Okay, that works. And the symmetry condition, does f(-1 + x) = f(-1 - x)?Let me test with a specific x, say x = 1:f(-1 + 1) = f(0) = 0² + 2*0 = 0f(-1 - 1) = f(-2) = (-2)² + 2*(-2) = 4 - 4 = 0Okay, that works.Another test, x = 2:f(-1 + 2) = f(1) = 3f(-1 - 2) = f(-3) = 9 - 6 = 3Good, it works.So, f(x) = x² + 2x. Got that.Now, g(x) is symmetric to f(x) with respect to the origin. Symmetric with respect to the origin means that if (x, y) is on f(x), then (-x, -y) is on g(x). So, g(x) = -f(-x).Let me compute that:g(x) = -f(-x) = -[(-x)² + 2*(-x)] = -(x² - 2x) = -x² + 2xSo, g(x) = -x² + 2x.Alright, that seems to make sense. Let me verify with a point. Suppose x = 1, f(1) = 3, so g(-1) should be -3. Let's check:g(-1) = -(-1)² + 2*(-1) = -1 - 2 = -3. Perfect.So, part (1) is done. f(x) = x² + 2x and g(x) = -x² + 2x.Moving on to part (2). We have F(x) = g(x) - λ f(x). We need to find the range of λ such that F(x) is increasing on the interval (-1, 1).First, let's write F(x):F(x) = g(x) - λ f(x) = (-x² + 2x) - λ(x² + 2x) = (-1 - λ)x² + (2 - 2λ)xSo, F(x) is a quadratic function. To determine if it's increasing on (-1, 1), we need to look at its derivative.Compute F'(x):F'(x) = d/dx [(-1 - λ)x² + (2 - 2λ)x] = -2(1 + λ)x + (2 - 2λ)For F(x) to be increasing on (-1, 1), F'(x) must be non-negative for all x in (-1, 1). So, F'(x) ≥ 0 for all x in (-1, 1).Let's analyze F'(x) = -2(1 + λ)x + (2 - 2λ)This is a linear function in x. The behavior of F'(x) depends on the coefficient of x, which is -2(1 + λ). Case 1: If -2(1 + λ) = 0, i.e., λ = -1.Then, F'(x) = 0 + (2 - 2*(-1)) = 2 + 2 = 4, which is positive. So, F'(x) = 4 > 0 for all x. Thus, F(x) is increasing.Case 2: If -2(1 + λ) > 0, i.e., 1 + λ < 0, so λ < -1.In this case, the coefficient of x is positive, meaning F'(x) is increasing as x increases. So, to ensure F'(x) ≥ 0 on (-1, 1), we need to check the minimum value of F'(x) on this interval. The minimum occurs at the left endpoint x = -1.Compute F'(-1):F'(-1) = -2(1 + λ)*(-1) + (2 - 2λ) = 2(1 + λ) + 2 - 2λ = 2 + 2λ + 2 - 2λ = 4Which is positive. So, as long as λ < -1, F'(x) is increasing and its minimum is 4, which is positive. Therefore, F(x) is increasing on (-1, 1).Case 3: If -2(1 + λ) < 0, i.e., 1 + λ > 0, so λ > -1.Here, the coefficient of x is negative, meaning F'(x) is decreasing as x increases. So, the minimum value of F'(x) on (-1, 1) occurs at the right endpoint x = 1.Compute F'(1):F'(1) = -2(1 + λ)*(1) + (2 - 2λ) = -2 - 2λ + 2 - 2λ = (-2 + 2) + (-2λ - 2λ) = 0 - 4λWe need F'(1) ≥ 0:-4λ ≥ 0Which implies λ ≤ 0But in this case, we have λ > -1. So, combining these, we have -1 < λ ≤ 0.But wait, let's check if F'(x) is non-negative throughout (-1,1) in this range.Let me pick λ = -0.5, which is between -1 and 0.Compute F'(x) = -2(1 - 0.5)x + (2 - 2*(-0.5)) = -2(0.5)x + (2 + 1) = -x + 3So, F'(x) = -x + 3. At x = -1: F'(-1) = 1 + 3 = 4 > 0. At x = 1: F'(1) = -1 + 3 = 2 > 0. Since it's linear, it's positive throughout.Another test, λ = 0. Then F'(x) = -2(1 + 0)x + (2 - 0) = -2x + 2. At x = -1: 2 + 2 = 4 > 0. At x = 1: -2 + 2 = 0. So, at x =1, it's zero, which is still non-negative. So, F(x) is non-decreasing on (-1,1).But wait, the problem says "increasing". If F'(x) is non-negative, it's non-decreasing. But sometimes "increasing" is used to mean strictly increasing. Hmm, the problem says "increasing", so maybe non-decreasing is acceptable. If so, then λ can be up to 0.But let's check λ = 0.5, which is greater than 0.F'(x) = -2(1 + 0.5)x + (2 - 1) = -3x +1At x = -1: 3 +1 =4 >0. At x=1: -3 +1 = -2 <0. So, F'(x) becomes negative at x=1, which is bad. So, F(x) is not increasing on the entire interval.Therefore, for λ >0, F'(x) becomes negative somewhere in (-1,1). So, we can't have λ >0.So, combining all cases:- If λ < -1: F'(x) is increasing, minimum at x=-1 is 4, which is positive. So, F(x) is increasing.- If λ = -1: F'(x)=4, which is positive. So, F(x) is increasing.- If -1 < λ ≤0: F'(x) is decreasing, but F'(1) = -4λ ≥0, so F'(x) remains non-negative on (-1,1). So, F(x) is non-decreasing.But the problem says "increasing". If "increasing" allows for non-decreasing, then λ can be up to 0. But if "increasing" means strictly increasing, then we need F'(x) >0 for all x in (-1,1). In that case, when λ=0, F'(1)=0, which would not be strictly increasing.But the problem doesn't specify strictly, so I think non-decreasing is acceptable.Wait, but in the case when λ=0, F(x) = g(x) -0 = g(x). So, F(x)= -x² +2x. Its derivative is F'(x)= -2x +2. At x=1, F'(1)=0. So, it's not strictly increasing on (-1,1). It's increasing on (-1,1) except at x=1 where it's flat.But the problem says "increasing on the interval (-1,1)". If it's increasing on the open interval (-1,1), then at x=1, it's not included. Hmm, but the interval is open, so maybe it's okay.Wait, actually, the interval is open, so x=1 is not included. So, at x approaching 1 from the left, F'(x) approaches 0 from the positive side. So, F'(x) is positive on (-1,1). So, F(x) is increasing on (-1,1).Similarly, when λ=0, F'(x) = -2x +2. For x in (-1,1), F'(x) is positive because at x=1, it's 0, but approaching from the left, it's positive. So, F(x) is increasing on (-1,1).Similarly, when λ approaches 0 from below, say λ= -ε, ε>0 small, then F'(x)= -2(1 - ε)x + (2 - 2*(-ε))= -2(1 - ε)x + 2 + 2ε. At x=1, F'(1)= -2(1 - ε) + 2 + 2ε= -2 + 2ε +2 + 2ε=4ε>0. So, still positive.So, overall, for λ ≤0, F'(x) is non-negative on (-1,1). But wait, earlier when λ < -1, we had F'(x) is increasing, with minimum at x=-1 being 4, which is positive. So, for λ ≤0, F(x) is increasing on (-1,1).But wait, earlier when I considered λ > -1, I concluded that for -1 < λ ≤0, F'(x) is decreasing but remains non-negative on (-1,1). So, combining both cases, λ can be any real number less than or equal to 0.Wait, but hold on. When λ < -1, F'(x) is increasing, so the minimum is at x=-1, which is 4, which is positive. So, for λ < -1, F'(x) is positive on (-1,1). For λ = -1, F'(x)=4, positive. For -1 < λ ≤0, F'(x) is decreasing, but still positive on (-1,1). So, overall, for λ ≤0, F(x) is increasing on (-1,1).But wait, earlier when I thought λ can be up to 0, but when I considered λ=0, F'(1)=0, but since the interval is open, it's okay. So, the range is λ ≤0.But wait, let me check λ=0.5 again. If λ=0.5, F'(x)= -2(1 +0.5)x + (2 -1)= -3x +1. At x=1, F'(1)= -3 +1= -2 <0. So, F(x) is decreasing at x=1, which is not allowed. So, λ cannot be greater than 0.Similarly, for λ=0, F'(x)= -2x +2. At x=1, F'(1)=0, but since the interval is open, it's not included. So, on (-1,1), F'(x) is positive except approaching 1, which is still okay because it's open.Therefore, the range of λ is all real numbers less than or equal to 0.But wait, in the initial analysis, I thought λ can be up to 0, but when I considered λ < -1 and -1 < λ ≤0, I concluded λ ≤0. So, the range is λ ≤0.But let me double-check with λ= -2, which is less than -1.F'(x)= -2(1 + (-2))x + (2 - 2*(-2))= -2(-1)x + (2 +4)= 2x +6So, F'(x)=2x +6. At x=-1: 2*(-1)+6=4>0. At x=1: 2*1 +6=8>0. Since it's linear and increasing, F'(x) is positive throughout (-1,1). So, F(x) is increasing.Another test, λ= -0.5:F'(x)= -2(1 -0.5)x + (2 -2*(-0.5))= -2(0.5)x + (2 +1)= -x +3At x=-1: 1 +3=4>0. At x=1: -1 +3=2>0. So, positive throughout.And λ=0:F'(x)= -2x +2. At x=-1: 2 +2=4>0. At x=1: -2 +2=0. So, positive on (-1,1).Thus, the range of λ is all real numbers less than or equal to 0.Wait, but in the initial problem statement, part (2) says "determine the range of the real number λ". So, I think the answer is λ ≤0.But let me think again. If λ=0, F(x)=g(x). Is g(x) increasing on (-1,1)?g(x)= -x² +2x. Its derivative is g'(x)= -2x +2. So, at x=-1: 2 +2=4>0. At x=1: -2 +2=0. So, on (-1,1), g'(x) is positive except at x=1, which is not included. So, yes, g(x) is increasing on (-1,1).Similarly, for λ= -1, F(x)=g(x) - (-1)f(x)=g(x)+f(x)= (-x² +2x) + (x² +2x)=4x. So, F(x)=4x, which is linear with slope 4>0, so increasing.For λ= -2, F(x)=g(x) - (-2)f(x)=g(x)+2f(x)= (-x² +2x) +2(x² +2x)= (-x² +2x) +2x² +4x= x² +6x. Wait, that's a quadratic opening upwards. Its derivative is 2x +6. At x=-1: 2*(-1)+6=4>0. At x=1: 2*1 +6=8>0. So, it's increasing on (-1,1).Wait, but F(x)=x² +6x. Its vertex is at x=-3, which is outside the interval (-1,1). So, on (-1,1), it's increasing because the derivative is positive.So, yeah, for λ ≤0, F(x) is increasing on (-1,1).But wait, when λ=0, F(x)=g(x)= -x² +2x, which is a downward opening parabola. Its vertex is at x=1, which is the maximum point. So, on (-1,1), it's increasing because the derivative is positive.So, all in all, the range of λ is (-∞, 0].But wait, in the initial problem, part (2) says "determine the range of the real number λ". So, I think the answer is λ ≤0.But let me check if λ can be greater than 0. As I saw earlier, for λ=0.5, F'(x)= -3x +1. At x=1, it's -2, which is negative. So, F(x) is decreasing at x=1, which is not allowed. So, λ cannot be greater than 0.Therefore, the range is λ ≤0.**Final Answer**(1) The explicit formulas are boxed{f(x) = x^2 + 2x} and boxed{g(x) = -x^2 + 2x}.(2) The range of lambda is boxed{(-infty, 0]}.

💡Okay, so I have this problem here asking me which of the given events is a certain event. There are four options: A, B, C, and D. Let me try to understand each one step by step.Starting with Event A: Tossing a coin and getting heads up. Hmm, I know that when you toss a coin, there are two possible outcomes: heads or tails. Since it's a fair coin, the chance of getting heads is 50%, and the same for tails. So, getting heads isn't guaranteed; it's just one of two equally likely possibilities. Therefore, Event A isn't a certain event because it doesn't always happen.Moving on to Event B: Drawing a card from a set of cards numbered 1, 3, 5, 7, 9 and getting an even number. Let me think about this. The cards are all odd numbers: 1, 3, 5, 7, 9. There are no even numbers in this set. So, if I draw a card, it's impossible to get an even number because there are none to draw. This means Event B is actually an impossible event, not a certain one.Next is Event C: Rolling a standard six-sided die and getting a 6. A standard die has numbers 1 through 6. Each number has an equal chance of appearing when you roll it, so the probability of getting a 6 is 1 out of 6. That means it's not guaranteed; it's just one possible outcome among six. So, Event C isn't certain either.Finally, Event D: Drawing two balls from a bag containing one yellow ball and three red balls, and getting at least one red ball. Let me visualize this. There's a total of four balls: one yellow and three red. If I draw two balls, what are the possible combinations? Well, I could draw two red balls, one red and one yellow, or two yellow balls. Wait, but there's only one yellow ball. So, it's impossible to draw two yellow balls because there's only one. Therefore, the only possible outcomes are either two red balls or one red and one yellow. In both cases, I always end up with at least one red ball. There's no way to draw two balls without getting at least one red ball because there's only one yellow ball. So, this event is certain to happen.To summarize:- Event A: 50% chance of heads, not certain.- Event B: Impossible to draw an even number, so not certain.- Event C: 1/6 chance of rolling a 6, not certain.- Event D: Always results in at least one red ball, so it's certain.Therefore, the correct answer must be Event D.

💡Okay, so I have this problem where I need to find the range of the real number ( a ) such that the intersection of sets ( A ) and ( B ) is not empty. The sets are defined as:- ( A = { y mid y > a^2 + 1 text{ or } y < a } )- ( B = { y mid 2 leq y leq 4 } )Alright, let's break this down step by step.First, I need to understand what each set represents. Set ( A ) consists of all real numbers ( y ) that are either greater than ( a^2 + 1 ) or less than ( a ). So, ( A ) is essentially two intervals on the real number line: ( (-infty, a) ) and ( (a^2 + 1, infty) ).Set ( B ) is a closed interval from 2 to 4, inclusive. So, ( B = [2, 4] ).Now, the problem asks for the range of ( a ) such that ( A cap B neq emptyset ). That means there should be at least one element common to both sets ( A ) and ( B ).To find this, it might help to first consider when ( A cap B ) is empty. If I can find the values of ( a ) for which ( A ) and ( B ) don't overlap, then the complement of that set will give me the values where they do overlap.So, ( A cap B = emptyset ) implies that there are no elements in both ( A ) and ( B ). That would mean that the entire interval ( B ) is not in ( A ). Since ( A ) is the union of ( (-infty, a) ) and ( (a^2 + 1, infty) ), for ( B ) to be entirely outside ( A ), ( B ) must lie entirely within the complement of ( A ), which is ( [a, a^2 + 1] ).Therefore, for ( A cap B = emptyset ), the interval ( [2, 4] ) must be contained within ( [a, a^2 + 1] ). So, we have:1. ( a leq 2 ) (since the lower bound of ( B ) is 2, which must be greater than or equal to ( a ))2. ( a^2 + 1 geq 4 ) (since the upper bound of ( B ) is 4, which must be less than or equal to ( a^2 + 1 ))Let's solve these inequalities.Starting with the second inequality:( a^2 + 1 geq 4 )Subtract 1 from both sides:( a^2 geq 3 )Taking square roots:( |a| geq sqrt{3} )Which means:( a geq sqrt{3} ) or ( a leq -sqrt{3} )Now, combining this with the first inequality ( a leq 2 ), we have two cases:1. ( a leq -sqrt{3} )2. ( sqrt{3} leq a leq 2 )So, when ( a ) is in these ranges, ( A cap B = emptyset ).Therefore, to find when ( A cap B neq emptyset ), we take the complement of these ranges. That is, ( a ) must not be in ( (-infty, -sqrt{3}] cup [sqrt{3}, 2] ).So, the range of ( a ) where ( A cap B neq emptyset ) is:( -sqrt{3} < a < sqrt{3} ) or ( a > 2 )Let me double-check this.If ( a ) is between ( -sqrt{3} ) and ( sqrt{3} ), then ( a^2 + 1 ) is less than or equal to 4 because ( (sqrt{3})^2 + 1 = 3 + 1 = 4 ). So, the upper bound of ( A ) is 4 or less, meaning the interval ( (a^2 + 1, infty) ) will overlap with ( B ) since ( B ) goes up to 4.Similarly, if ( a ) is greater than 2, then the lower bound of ( A ) is above 2, but since ( a^2 + 1 ) will be greater than ( 2^2 + 1 = 5 ), the upper part of ( A ) will overlap with ( B ) because ( B ) goes up to 4, which is less than 5. Wait, actually, if ( a > 2 ), then ( a^2 + 1 ) is greater than 5, so the upper part of ( A ) is ( (a^2 + 1, infty) ), which doesn't overlap with ( B ) since ( B ) only goes up to 4. Hmm, that seems contradictory.Wait, maybe I made a mistake here. Let me think again.If ( a > 2 ), then ( a ) is greater than 2, so the lower part of ( A ) is ( (-infty, a) ). Since ( a > 2 ), this interval includes all numbers less than ( a ), which includes the entire interval ( B = [2, 4] ) only if ( a > 4 ). But if ( a ) is between 2 and 4, then ( (-infty, a) ) would include ( [2, a) ), which overlaps with ( B ). If ( a > 4 ), then ( (-infty, a) ) includes all of ( B ).Wait, so actually, if ( a > 2 ), regardless of how big ( a ) is, the lower part of ( A ) will always include some part of ( B ). Because ( B ) starts at 2, and if ( a > 2 ), then ( (-infty, a) ) includes ( [2, a) ), which overlaps with ( B ) as long as ( a > 2 ).But earlier, I thought that ( a > 2 ) would mean ( a^2 + 1 > 5 ), so the upper part of ( A ) doesn't overlap with ( B ). But the lower part does overlap. So, actually, as long as ( a > 2 ), ( A ) will overlap with ( B ) because of the lower interval.Similarly, if ( a ) is between ( -sqrt{3} ) and ( sqrt{3} ), then ( a^2 + 1 leq 4 ), so the upper interval of ( A ) will overlap with ( B ).But wait, if ( a ) is negative, say ( a = -2 ), then ( a^2 + 1 = 4 + 1 = 5 ), so the upper interval is ( (5, infty) ), which doesn't overlap with ( B ). But the lower interval is ( (-infty, -2) ), which doesn't overlap with ( B ) either because ( B ) starts at 2. So, in that case, ( A cap B = emptyset ).Wait, so my earlier conclusion that ( A cap B neq emptyset ) when ( -sqrt{3} < a < sqrt{3} ) or ( a > 2 ) might not be entirely correct because when ( a ) is negative, even if ( a > -sqrt{3} ), the lower interval of ( A ) doesn't overlap with ( B ), but the upper interval might.Wait, let's re-examine.If ( a ) is between ( -sqrt{3} ) and ( sqrt{3} ), then ( a^2 + 1 leq 4 ). So, the upper interval of ( A ) is ( (a^2 + 1, infty) ). Since ( a^2 + 1 leq 4 ), this interval starts at or before 4, so it will overlap with ( B ) which is up to 4. Therefore, ( A cap B ) will include ( (a^2 + 1, 4] ), which is non-empty.Similarly, if ( a > 2 ), then the lower interval of ( A ) is ( (-infty, a) ). Since ( a > 2 ), this interval includes ( [2, a) ), which overlaps with ( B ) as long as ( a > 2 ). So, even if ( a ) is greater than 4, the lower interval still includes ( [2, 4] ), so ( A cap B ) is non-empty.However, if ( a leq -sqrt{3} ), then ( a^2 + 1 geq 4 ), so the upper interval of ( A ) starts at or above 4, which doesn't overlap with ( B ). Also, the lower interval ( (-infty, a) ) is entirely below ( a leq -sqrt{3} ), which is less than 2, so it doesn't overlap with ( B ) either. Therefore, ( A cap B = emptyset ) in this case.Similarly, if ( sqrt{3} leq a leq 2 ), then ( a^2 + 1 geq 4 ) (since ( a geq sqrt{3} )), so the upper interval of ( A ) starts at or above 4, which doesn't overlap with ( B ). The lower interval ( (-infty, a) ) includes ( (-infty, 2] ) since ( a leq 2 ), but ( B ) starts at 2, so the overlap would be at 2. But ( A ) is defined as ( y < a ), so if ( a = 2 ), then ( y < 2 ), which doesn't include 2. Therefore, ( A cap B ) would be empty in this case as well.So, putting it all together, ( A cap B ) is empty when ( a leq -sqrt{3} ) or ( sqrt{3} leq a leq 2 ). Therefore, ( A cap B ) is non-empty when ( a ) is in the complement of these intervals, which is ( -sqrt{3} < a < sqrt{3} ) or ( a > 2 ).But wait, earlier I thought that when ( a > 2 ), the lower interval of ( A ) overlaps with ( B ). However, if ( a > 4 ), the lower interval ( (-infty, a) ) includes all of ( B ) because ( B ) is up to 4, and ( a > 4 ). So, in that case, ( A cap B = B ), which is non-empty.Similarly, if ( 2 < a leq 4 ), then ( (-infty, a) ) includes ( [2, a) ), which overlaps with ( B ) as ( B ) is ( [2,4] ). So, yes, even if ( a ) is between 2 and 4, ( A cap B ) is non-empty.Therefore, the correct range for ( a ) is ( -sqrt{3} < a < sqrt{3} ) or ( a > 2 ).But wait, let me check if ( a = sqrt{3} ) or ( a = 2 ) are included. If ( a = sqrt{3} ), then ( a^2 + 1 = 3 + 1 = 4 ), so the upper interval of ( A ) is ( (4, infty) ), which doesn't overlap with ( B ). The lower interval is ( (-infty, sqrt{3}) ), which includes ( [2, sqrt{3}) ), but since ( sqrt{3} approx 1.732 ), which is less than 2, the lower interval doesn't overlap with ( B ). Therefore, ( A cap B = emptyset ) when ( a = sqrt{3} ).Similarly, if ( a = 2 ), the lower interval is ( (-infty, 2) ), which doesn't include 2, and the upper interval is ( (2^2 + 1, infty) = (5, infty) ), which doesn't overlap with ( B ). So, ( A cap B = emptyset ) when ( a = 2 ).Therefore, the endpoints ( a = sqrt{3} ) and ( a = 2 ) are excluded from the range where ( A cap B neq emptyset ).So, the final range is ( -sqrt{3} < a < sqrt{3} ) or ( a > 2 ).I think that's correct. Let me summarize:- When ( a leq -sqrt{3} ) or ( sqrt{3} leq a leq 2 ), ( A cap B = emptyset ).- Therefore, ( A cap B neq emptyset ) when ( a ) is in the complement of these intervals, which is ( -sqrt{3} < a < sqrt{3} ) or ( a > 2 ).Yes, that makes sense. I think I've covered all the cases and checked the boundary points.

💡Okay, so John wants to fence a rectangular garden, and he has some specific requirements. The area needs to be at least 150 square feet, and the length should be 10 feet longer than the width. I need to figure out the minimum width required for the garden. Hmm, let me break this down step by step.First, let's denote the width of the garden as ( w ) feet. Since the length is 10 feet longer than the width, the length would be ( w + 10 ) feet. That makes sense so far.Now, the area of a rectangle is calculated by multiplying its length by its width. So, the area ( A ) of John's garden would be:[ A = w times (w + 10) ]John wants this area to be at least 150 square feet. So, we can set up the inequality:[ w(w + 10) geq 150 ]Let me expand this equation to make it easier to solve:[ w^2 + 10w geq 150 ]To solve this inequality, I should bring all terms to one side to form a standard quadratic inequality:[ w^2 + 10w - 150 geq 0 ]Now, I need to solve the quadratic equation ( w^2 + 10w - 150 = 0 ) to find the critical points. I can use the quadratic formula for this:[ w = frac{-b pm sqrt{b^2 - 4ac}}{2a} ]Here, ( a = 1 ), ( b = 10 ), and ( c = -150 ). Plugging these values into the formula:[ w = frac{-10 pm sqrt{10^2 - 4 times 1 times (-150)}}{2 times 1} ][ w = frac{-10 pm sqrt{100 + 600}}{2} ][ w = frac{-10 pm sqrt{700}}{2} ][ w = frac{-10 pm 26.4575}{2} ]So, we have two solutions:1. ( w = frac{-10 + 26.4575}{2} = frac{16.4575}{2} = 8.22875 )2. ( w = frac{-10 - 26.4575}{2} = frac{-36.4575}{2} = -18.22875 )Since the width cannot be negative, we discard the negative solution. So, the critical point is approximately ( w = 8.22875 ) feet.Now, to determine the intervals where the quadratic expression ( w^2 + 10w - 150 ) is greater than or equal to zero, we consider the critical point ( w = 8.22875 ). The quadratic opens upwards (since the coefficient of ( w^2 ) is positive), so the expression will be greater than or equal to zero when ( w leq -18.22875 ) or ( w geq 8.22875 ). Again, since width cannot be negative, we only consider ( w geq 8.22875 ).Therefore, the minimum width required is approximately 8.23 feet. However, since we're dealing with measurements, it's practical to round this to a reasonable decimal place. Rounding to two decimal places, it's 8.23 feet. But if we consider standard fencing measurements, which often use whole numbers, we might round up to the next whole number, which is 9 feet.Wait, but let me double-check. If I take ( w = 8.23 ) feet, then the length would be ( 8.23 + 10 = 18.23 ) feet. Calculating the area:[ 8.23 times 18.23 approx 150 text{ square feet} ]That's exactly the minimum area required. So, 8.23 feet is the precise minimum width. However, in practical terms, John might prefer to use a whole number for simplicity. If he uses 8 feet, let's see what the area would be:[ 8 times (8 + 10) = 8 times 18 = 144 text{ square feet} ]That's less than 150, so 8 feet isn't enough. If he uses 9 feet:[ 9 times (9 + 10) = 9 times 19 = 171 text{ square feet} ]That's more than 150, so 9 feet would work. Therefore, depending on whether John wants the exact minimum or a practical whole number, the answer could be either 8.23 feet or 9 feet.But since the question asks for the minimum width, and not necessarily a whole number, 8.23 feet is the precise answer. However, in the context of fencing, it's common to use whole numbers, so 9 feet might be the more practical answer.Wait, let me check the exact value without rounding:[ sqrt{700} = 26.4575 ]So,[ w = frac{-10 + 26.4575}{2} = frac{16.4575}{2} = 8.22875 ]So, it's approximately 8.23 feet. If John uses 8.23 feet, the area is exactly 150 square feet. If he uses a slightly larger width, the area increases. Therefore, the minimum width is 8.23 feet.But to express this as a whole number, since 8 feet gives an area of 144, which is less than 150, and 9 feet gives 171, which is more than 150, the minimum whole number width that satisfies the area requirement is 9 feet.So, depending on whether we need an exact decimal or a whole number, the answer could be 8.23 feet or 9 feet. However, since the problem doesn't specify, and in mathematical terms, the exact minimum width is 8.23 feet, but in practical terms, it's 9 feet.But wait, let me think again. If we solve the quadratic inequality ( w^2 + 10w - 150 geq 0 ), the solution is ( w geq 8.22875 ) or ( w leq -18.22875 ). Since width can't be negative, we only consider ( w geq 8.22875 ). Therefore, the minimum width is 8.22875 feet, which is approximately 8.23 feet.So, in conclusion, the minimum width required is approximately 8.23 feet. If we need to express this as a whole number, it would be 9 feet. But since the problem doesn't specify, I think the exact value is 8.23 feet.Wait, but let me check the calculation again to ensure I didn't make a mistake:[ w = frac{-10 + sqrt{100 + 600}}{2} = frac{-10 + sqrt{700}}{2} ][ sqrt{700} approx 26.4575 ][ w approx frac{-10 + 26.4575}{2} = frac{16.4575}{2} = 8.22875 ]Yes, that's correct. So, 8.22875 feet is the exact minimum width.Therefore, the minimum width required is approximately 8.23 feet. If we need to round to the nearest whole number, it would be 8 feet, but as we saw earlier, 8 feet gives an area of 144, which is less than 150. Therefore, to meet the requirement of at least 150 square feet, the width must be at least 8.23 feet.So, in conclusion, the minimum width is approximately 8.23 feet. However, since fencing materials are often sold in whole numbers, John might need to purchase 9 feet of fencing for the width to ensure the area meets the requirement.But wait, let me think about this again. If John uses 8.23 feet for the width, the length would be 18.23 feet, and the area would be exactly 150 square feet. Therefore, 8.23 feet is the precise minimum width needed. If he uses a width less than that, the area would be less than 150, which doesn't meet the requirement.So, to answer the question precisely, the minimum width is 8.23 feet. However, if we need to provide a whole number, it would be 9 feet. But since the problem doesn't specify, I think the exact value is 8.23 feet.Wait, but let me check the quadratic equation again to ensure I didn't make a calculation error. The quadratic equation is ( w^2 + 10w - 150 = 0 ). Using the quadratic formula:[ w = frac{-10 pm sqrt{10^2 - 4(1)(-150)}}{2(1)} ][ w = frac{-10 pm sqrt{100 + 600}}{2} ][ w = frac{-10 pm sqrt{700}}{2} ][ sqrt{700} approx 26.4575 ][ w = frac{-10 + 26.4575}{2} = frac{16.4575}{2} = 8.22875 ]Yes, that's correct. So, the minimum width is 8.22875 feet, which is approximately 8.23 feet.Therefore, the minimum width required is approximately 8.23 feet. If we need to express this as a whole number, it would be 9 feet, but the exact minimum is 8.23 feet.In conclusion, the minimum width of the garden should be approximately 8.23 feet to ensure the area is at least 150 square feet.