💡Okay, so I have this problem where black paint was splashed on a white plane, and I need to prove that there exist two points of the same color such that the distance between them is exactly 1965 meters. Hmm, that seems interesting. Let me try to break it down.First, I need to visualize the scenario. There's a white plane, and someone has splashed black paint on it. So, now the plane has some black regions and the rest remains white. The problem is asking about the existence of two points that are the same color and are exactly 1965 meters apart. I remember something about the pigeonhole principle in mathematics, which might be useful here. The pigeonhole principle states that if you have more pigeons than pigeonholes, at least two pigeons must share a pigeonhole. Maybe I can apply this idea somehow.Let me think about points on the plane. If I can construct a specific shape or figure where the distances between certain points are fixed, then maybe I can use the pigeonhole principle to argue that at least two of those points must be the same color.An equilateral triangle comes to mind because all its sides are equal. If I construct an equilateral triangle with each side measuring 1965 meters, then the distance between any two vertices is exactly 1965 meters. That seems promising.So, let's say I have an equilateral triangle ABC, where each side AB, BC, and AC is 1965 meters. Now, each vertex of this triangle can be either black or white, depending on whether it was covered by the paint or not. Since there are three vertices and only two colors (black and white), by the pigeonhole principle, at least two of these vertices must be the same color. Let's say, for example, points A and B are both black. Then, the distance between A and B is 1965 meters, which is exactly what we need to prove.Wait, but what if the paint doesn't cover any of the vertices? Then all vertices would be white, right? But that still means all three points are the same color, so the distance between any two is 1965 meters. So, regardless of how the paint is splashed, as long as we have this triangle, we can find two points of the same color at the required distance.Is there any flaw in this reasoning? Let me double-check. The key here is that we're constructing a specific triangle with sides of 1965 meters. By doing so, we ensure that the distances between the vertices are fixed. Then, using the pigeonhole principle, we argue that with three points and two colors, at least two must share the same color. I think this makes sense. The pigeonhole principle is a fundamental concept in combinatorics, and it's often used in problems where we need to guarantee the existence of certain properties. In this case, the property is having two points of the same color at a specific distance apart.Another way to think about it is to imagine coloring the plane randomly with black and white. No matter how you color it, if you have an equilateral triangle of side length 1965 meters, you can't avoid having at least two vertices of the same color. This is because there are more vertices (three) than colors (two), so repetition is inevitable.I wonder if there's a more general principle at play here. Maybe something related to graph theory? If I consider each vertex as a node and the edges as the distances between them, then this problem reduces to finding a monochromatic edge in a complete graph with three nodes. But since we're dealing with colors and distances, it's more of a geometric application of the pigeonhole principle.Also, I should consider if the plane being infinite affects this problem. Since the plane is infinite, there are infinitely many points, but the problem only requires the existence of two specific points at a certain distance. The construction of the triangle ensures that regardless of how the paint is splashed, those two points will exist.What if the paint was splashed in a very specific pattern that avoids creating two points of the same color at 1965 meters? Is that possible? I don't think so because the triangle's vertices are fixed in distance, and the pigeonhole principle forces at least two of them to be the same color. So, no matter how you try to splash the paint, you can't avoid having two points of the same color at that exact distance.In summary, by constructing an equilateral triangle with sides of 1965 meters and applying the pigeonhole principle, we can guarantee that there will always be two points of the same color exactly 1965 meters apart. This seems like a solid approach to solving the problem.I think I've covered all the bases here. I've considered the pigeonhole principle, constructed a specific geometric shape, and verified that the reasoning holds under different scenarios. I don't see any holes in this logic, so I'm confident that this is the correct way to approach the problem.

💡Okay, so I have this function f(x) that's defined piecewise. For x greater than 0, it's x times the natural log of x minus a times x squared. And for x less than 0, it's x squared plus a times x. The problem says that this function has exactly three extreme points, and I need to find the range of values for a.Hmm, extreme points are where the derivative is zero, right? So I guess I need to find the derivative of f(x) for both pieces and then figure out when each piece contributes to the total number of extreme points.Let me start by finding the derivative for each piece.For x > 0, f(x) = x ln x - a x². The derivative of that would be f'(x) = ln x + 1 - 2a x. Because the derivative of x ln x is ln x + 1, and the derivative of -a x² is -2a x.For x < 0, f(x) = x² + a x. The derivative here is f'(x) = 2x + a. That's straightforward.Now, I need to check what happens at x = 0. Since the function is defined differently on either side, I should check if the function is continuous there, but actually, the problem doesn't specify anything about x = 0, so maybe it's not necessary. But just in case, let me see.For x approaching 0 from the right, f(x) approaches 0 * ln 0, which is 0 * (-infinity), which is an indeterminate form. But using limits, lim x->0+ x ln x is 0 because x approaches zero faster than ln x approaches negative infinity. So f(x) approaches 0 from the right.From the left, f(x) approaches 0² + a*0, which is also 0. So the function is continuous at x = 0. But since the function isn't defined at x = 0, maybe it's just a point of consideration for the derivative.But since the function is defined for x > 0 and x < 0, maybe x = 0 isn't considered an extreme point. So I can focus on x > 0 and x < 0 separately.So for x < 0, f'(x) = 2x + a. Setting that equal to zero gives 2x + a = 0, so x = -a/2. Now, since x < 0, that means -a/2 < 0, so a must be positive. If a is positive, then x = -a/2 is a valid critical point for x < 0. If a is negative, then x = -a/2 would be positive, which is not in the domain of this piece. So for x < 0, there is a critical point only when a is positive.For x > 0, f'(x) = ln x + 1 - 2a x. I need to find when this derivative is zero. So ln x + 1 - 2a x = 0. Let me denote this as g(x) = ln x + 1 - 2a x. I need to find the number of solutions to g(x) = 0 for x > 0.To find the number of solutions, I can analyze the behavior of g(x). Let's compute its derivative: g'(x) = 1/x - 2a. Setting this equal to zero gives 1/x - 2a = 0, so x = 1/(2a). So g(x) has a critical point at x = 1/(2a). Since the second derivative would be g''(x) = -1/x², which is always negative for x > 0, so the critical point is a maximum.So g(x) increases up to x = 1/(2a) and then decreases after that. Therefore, the maximum value of g(x) is at x = 1/(2a). Let's compute g(1/(2a)):g(1/(2a)) = ln(1/(2a)) + 1 - 2a*(1/(2a)) = ln(1/(2a)) + 1 - 1 = ln(1/(2a)) = -ln(2a).So the maximum value of g(x) is -ln(2a). For g(x) to have two zeros, the maximum must be positive. So we need -ln(2a) > 0, which implies ln(2a) < 0, so 2a < 1, hence a < 1/2.Also, since we're dealing with x > 0, and the function g(x) tends to negative infinity as x approaches 0 from the right, and as x approaches infinity, g(x) tends to negative infinity because the -2a x term dominates. So if the maximum is positive, there will be two zeros for g(x), meaning two critical points for x > 0.So putting it all together:- For x < 0, there is one critical point when a > 0.- For x > 0, there are two critical points when a < 1/2.Therefore, to have exactly three extreme points, we need both conditions to hold: a > 0 and a < 1/2. So the range of a is 0 < a < 1/2.Wait, but what if a = 1/2? Then the maximum of g(x) would be -ln(1) = 0, so g(x) would touch the x-axis at x = 1/(2a) = 1/(2*(1/2)) = 1. So there would be only one critical point for x > 0, meaning total extreme points would be two (one from x < 0 and one from x > 0). So a cannot be equal to 1/2.Similarly, if a = 0, then for x < 0, f'(x) = 2x, which is zero at x = 0, but x = 0 isn't in the domain of x < 0. So for x < 0, there's no critical point. For x > 0, f'(x) = ln x + 1, which is zero when ln x + 1 = 0, so x = e^{-1}. So only one critical point. So total extreme points would be one, which is less than three. So a cannot be zero.Therefore, the range is strictly between 0 and 1/2.I think that's it. So the range of a is 0 < a < 1/2.

💡Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, c respectively. The problem gives me an equation involving sines and cosines of these angles: sin²A + sinA sinC + sin²C + cos²B = 1. Part (1) asks me to find the measure of angle B. Hmm, let's start by looking at the given equation. It has sin²A, sinA sinC, sin²C, and cos²B. I know that in trigonometry, especially in triangles, the Law of Sines and Law of Cosines are often useful. Maybe I can relate these terms somehow.First, I notice that the equation has sin²A + sinA sinC + sin²C. That looks a bit like a quadratic in terms of sinA and sinC. Maybe I can factor it or find a relationship. Let me think, sin²A + sinA sinC + sin²C. Hmm, that reminds me of the formula for (sinA + sinC)², but that would be sin²A + 2 sinA sinC + sin²C. So, it's similar but not quite the same. Maybe I can write it as (sinA + sinC)² - sinA sinC. Let me check:(sinA + sinC)² = sin²A + 2 sinA sinC + sin²C.So, sin²A + sinA sinC + sin²C = (sinA + sinC)² - sinA sinC.But I'm not sure if that helps. Maybe another approach. Let's consider that in a triangle, the sum of angles is π radians (or 180 degrees). So, A + B + C = π. Maybe I can express one angle in terms of the others, but I'm not sure yet.Looking back at the equation: sin²A + sinA sinC + sin²C + cos²B = 1. Let me rearrange it:sin²A + sinA sinC + sin²C = 1 - cos²B.But 1 - cos²B is equal to sin²B, from the Pythagorean identity. So, the equation becomes:sin²A + sinA sinC + sin²C = sin²B.Interesting. So, sin²A + sinA sinC + sin²C equals sin²B. Maybe I can relate this to the sides of the triangle using the Law of Sines.Law of Sines says that a/sinA = b/sinB = c/sinC = 2R, where R is the radius of the circumscribed circle. So, sinA = a/(2R), sinB = b/(2R), sinC = c/(2R). Let me substitute these into the equation.So, sin²A + sinA sinC + sin²C becomes:(a²)/(4R²) + (a c)/(4R²) + (c²)/(4R²) = (b²)/(4R²).Multiplying both sides by 4R²:a² + a c + c² = b².So, we have a² + a c + c² = b².Hmm, that's a relation between the sides. Maybe I can use the Law of Cosines here. Law of Cosines says that b² = a² + c² - 2ac cosB.Wait, but from the equation above, b² = a² + a c + c². So, setting them equal:a² + a c + c² = a² + c² - 2ac cosB.Simplify both sides:Left side: a² + a c + c².Right side: a² + c² - 2ac cosB.Subtract a² + c² from both sides:a c = -2ac cosB.Divide both sides by ac (assuming a ≠ 0 and c ≠ 0, which they are in a triangle):1 = -2 cosB.So, cosB = -1/2.Now, what angle B has cosine equal to -1/2? Well, in the unit circle, cos(120°) = cos(2π/3) = -1/2. Since in a triangle, angles are between 0 and π radians (0 and 180 degrees), so B must be 120 degrees or 2π/3 radians.So, part (1) answer is B = 2π/3 radians.Okay, moving on to part (2). Given a = 5, b = 7, find sinC.We know from part (1) that angle B is 120 degrees or 2π/3 radians. So, we can use the Law of Cosines or Law of Sines.First, let's recall that from part (1), we had the relation a² + a c + c² = b².Given a = 5, b = 7, let's plug into this equation:5² + 5c + c² = 7².So, 25 + 5c + c² = 49.Subtract 49 from both sides:c² + 5c + 25 - 49 = 0.Simplify:c² + 5c - 24 = 0.Now, we have a quadratic equation in c: c² + 5c - 24 = 0.Let's solve for c using quadratic formula:c = [-5 ± sqrt(25 + 96)] / 2.Because discriminant D = 25 + 4*24 = 25 + 96 = 121.So, sqrt(121) = 11.Thus, c = [-5 + 11]/2 = 6/2 = 3, or c = [-5 - 11]/2 = -16/2 = -8.But since side lengths can't be negative, c = 3.So, c = 3.Now, to find sinC, we can use the Law of Sines.Law of Sines: a/sinA = b/sinB = c/sinC.We know a = 5, b = 7, c = 3, and angle B = 120 degrees.So, let's write:c/sinC = b/sinB.So, 3/sinC = 7/sin(120°).We know that sin(120°) = sin(60°) = sqrt(3)/2.Wait, no, sin(120°) is actually sin(60°) because 120° is in the second quadrant, and sin(120°) = sin(60°) = sqrt(3)/2.So, sin(120°) = sqrt(3)/2.Thus, 3/sinC = 7/(sqrt(3)/2).Simplify the right side:7/(sqrt(3)/2) = 7 * 2 / sqrt(3) = 14 / sqrt(3).So, 3/sinC = 14 / sqrt(3).Solving for sinC:sinC = 3 * sqrt(3) / 14.So, sinC = (3 sqrt(3)) / 14.Let me double-check the steps.1. From the given equation, we found that cosB = -1/2, so B = 120°, correct.2. Then, using the relation a² + a c + c² = b², with a=5, b=7, we got c=3, correct.3. Then, using Law of Sines: c/sinC = b/sinB.4. sinB = sin(120°) = sqrt(3)/2.5. So, 3/sinC = 7/(sqrt(3)/2) = 14/sqrt(3).6. Therefore, sinC = 3 * sqrt(3)/14.Yes, that seems correct.Alternatively, I can use Law of Cosines to find angle C.Law of Cosines: cosC = (a² + b² - c²)/(2ab).Wait, but let's see:Wait, Law of Cosines for angle C is:c² = a² + b² - 2ab cosC.So, cosC = (a² + b² - c²)/(2ab).Given a=5, b=7, c=3.So, cosC = (25 + 49 - 9)/(2*5*7) = (65)/(70) = 13/14.Then, sinC = sqrt(1 - cos²C) = sqrt(1 - (169/196)) = sqrt(27/196) = 3 sqrt(3)/14.Same result. So, that confirms it.So, sinC is 3 sqrt(3)/14.**Final Answer**(1) The measure of angle ( B ) is (boxed{dfrac{2pi}{3}}).(2) (sin C = boxed{dfrac{3sqrt{3}}{14}}).

💡Alright, so I've got this problem about the museum admission fees. Let me try to understand it step by step. The museum charges 30 per adult and 15 per child. Last Friday, they made a total of 2700. They also mention that there were at least two adults and two children. The question is asking for the ratio of adults to children that is closest to 2.Okay, first, I need to set up some variables. Let me call the number of adults 'a' and the number of children 'c'. So, the total money made would be 30 times the number of adults plus 15 times the number of children, right? That gives me the equation:30a + 15c = 2700Hmm, that looks a bit complicated. Maybe I can simplify it. If I divide everything by 15, that should make the numbers smaller and easier to work with. Let's try that:(30a)/15 + (15c)/15 = 2700/15Which simplifies to:2a + c = 180Okay, that's better. Now, I can solve for one variable in terms of the other. Let's solve for c:c = 180 - 2aSo, the number of children is 180 minus twice the number of adults. Got it.Now, the problem also says there were at least two adults and two children. So, a has to be at least 2, and c has to be at least 2. That means:a ≥ 2c ≥ 2But since c = 180 - 2a, I can substitute that in:180 - 2a ≥ 2Let me solve that inequality:180 - 2a ≥ 2-2a ≥ 2 - 180-2a ≥ -178Now, when I divide both sides by -2, I have to remember to flip the inequality sign because I'm dividing by a negative number:a ≤ 89So, a has to be less than or equal to 89. But we also know a has to be at least 2. So, a is between 2 and 89, inclusive.Now, the main goal is to find the ratio of adults to children that is closest to 2. So, we want a/c ≈ 2. Let's write that as an equation:a/c = 2But we know that c = 180 - 2a, so substitute that in:a / (180 - 2a) = 2Now, let's solve for a:a = 2*(180 - 2a)a = 360 - 4aa + 4a = 3605a = 360a = 72Okay, so if a is 72, then c would be:c = 180 - 2*72c = 180 - 144c = 36So, the ratio a/c is 72/36, which is exactly 2. That seems perfect, right? But wait, let me make sure that this is the closest ratio to 2.Is there a possibility that another ratio could be closer to 2? Let's check numbers around 72 for a and see what the ratio becomes.If a = 71, then c = 180 - 2*71 = 180 - 142 = 38Ratio = 71/38 ≈ 1.868If a = 73, then c = 180 - 2*73 = 180 - 146 = 34Ratio = 73/34 ≈ 2.147So, 71/38 is about 1.868, which is less than 2, and 73/34 is about 2.147, which is more than 2. Comparing these to 2, which one is closer?The difference between 1.868 and 2 is 0.132The difference between 2.147 and 2 is 0.147So, 1.868 is closer to 2 than 2.147 is. But wait, our initial calculation gave us exactly 2 when a=72. So, 72/36=2 is exactly 2, which is better than both 1.868 and 2.147.Let me check a=70 and a=74 to see if the trend continues.If a=70, c=180-140=40Ratio=70/40=1.75If a=74, c=180-148=32Ratio=74/32≈2.3125So, 1.75 is further away from 2, and 2.3125 is also further away. So, it seems that as we move away from a=72, the ratio gets further from 2.Therefore, a=72 and c=36 gives us the ratio exactly equal to 2, which is the closest possible.But just to be thorough, let's check a=69 and a=75.a=69, c=180-138=42Ratio=69/42≈1.642a=75, c=180-150=30Ratio=75/30=2.5Again, both are further away from 2. So, it's clear that a=72 and c=36 is the point where the ratio is exactly 2, and moving away from that point only increases the difference from 2.Therefore, the ratio of adults to children that is closest to 2 is 2:1, achieved when there are 72 adults and 36 children.

💡Okay, so I have this problem with four conclusions, and I need to figure out how many of them are correct. Let me take it step by step.First, conclusion 1: "If x > 0, then x > sin x always holds." Hmm, I remember that for small angles, sin x is approximately equal to x, but as x increases, sin x oscillates between -1 and 1. So, for x > 0, especially as x gets larger, sin x will always be less than or equal to 1, while x keeps increasing. So, intuitively, x should be greater than sin x for all x > 0. Maybe I can think about the function f(x) = x - sin x. If I take the derivative, f'(x) = 1 - cos x. Since cos x is always between -1 and 1, 1 - cos x is always non-negative. That means f(x) is increasing. At x = 0, f(0) = 0. So, for x > 0, f(x) > 0, which means x > sin x. So, conclusion 1 seems correct.Moving on to conclusion 2: "The contrapositive of the proposition 'If x - sin x = 0, then x = 0' is 'If x ≠ 0, then x - sin x ≠ 0'." I recall that the contrapositive of "If P, then Q" is "If not Q, then not P." So, in this case, P is "x - sin x = 0" and Q is "x = 0." Therefore, the contrapositive should be "If x ≠ 0, then x - sin x ≠ 0." That's exactly what conclusion 2 states. So, conclusion 2 is correct.Conclusion 3: "'The proposition p ∧ q is true' is a sufficient but not necessary condition for 'The proposition p ∨ q is true'." Let me think about this. If p ∧ q is true, that means both p and q are true. If both are true, then certainly p ∨ q is true because at least one of them is true. So, p ∧ q being true is a sufficient condition for p ∨ q being true. However, it's not necessary because p ∨ q can be true even if only one of p or q is true. For example, if p is true and q is false, p ∨ q is still true, but p ∧ q is false. So, conclusion 3 is correct.Finally, conclusion 4: "The negation of the proposition 'For all x ∈ ℝ, x - ln x > 0' is 'There exists an x₀ ∈ ℝ, such that x₀ - ln x₀ < 0'." I know that the negation of a universal statement ("for all") is an existential statement ("there exists") with the negation inside. So, the negation of "For all x, P(x)" is "There exists x such that not P(x)." In this case, P(x) is "x - ln x > 0," so the negation should be "There exists x₀ such that x₀ - ln x₀ ≤ 0." However, conclusion 4 says "< 0" instead of "≤ 0." Is this a problem? Well, the original statement is strict inequality, so the negation should include the possibility of equality. Therefore, conclusion 4 is incorrect because it doesn't account for the case where x₀ - ln x₀ could be equal to zero.So, out of the four conclusions, the first three are correct, and the fourth is incorrect. That means the number of correct conclusions is 3.**Final Answer**boxed{C}

💡Okay, so I have this problem about a circle in polar coordinates and a line intersecting it. Let me try to figure it out step by step.First, part (1) asks for the polar equation of the circle C. The center of the circle is given in polar coordinates as C(√2, π/4), and the radius is √3. Hmm, I remember that converting between polar and Cartesian coordinates can be helpful here. The general polar equation of a circle with center at (ρ₀, θ₀) and radius a is ρ² - 2ρρ₀ cos(θ - θ₀) + ρ₀² = a². Let me write that down:ρ² - 2ρρ₀ cos(θ - θ₀) + ρ₀² = a².So, plugging in the given values, ρ₀ is √2, θ₀ is π/4, and a is √3. Let me substitute these:ρ² - 2ρ(√2) cos(θ - π/4) + (√2)² = (√3)².Simplifying the constants:(√2)² is 2, and (√3)² is 3. So,ρ² - 2√2 ρ cos(θ - π/4) + 2 = 3.Subtracting 3 from both sides:ρ² - 2√2 ρ cos(θ - π/4) - 1 = 0.Hmm, that seems right. But wait, I remember another way to write the polar equation of a circle is by converting the Cartesian equation to polar. Maybe I should double-check that.The center in Cartesian coordinates would be (ρ₀ cos θ₀, ρ₀ sin θ₀). So, plugging in √2 and π/4:x = √2 cos(π/4) = √2*(√2/2) = 1,y = √2 sin(π/4) = √2*(√2/2) = 1.So, the center is (1,1) in Cartesian coordinates. The radius is √3, so the Cartesian equation is (x - 1)² + (y - 1)² = 3.Now, converting this to polar coordinates. Remember that x = ρ cos θ and y = ρ sin θ. So,(ρ cos θ - 1)² + (ρ sin θ - 1)² = 3.Expanding both squares:(ρ² cos² θ - 2ρ cos θ + 1) + (ρ² sin² θ - 2ρ sin θ + 1) = 3.Combine like terms:ρ² (cos² θ + sin² θ) - 2ρ (cos θ + sin θ) + 2 = 3.Since cos² θ + sin² θ = 1,ρ² - 2ρ (cos θ + sin θ) + 2 = 3.Subtract 3:ρ² - 2ρ (cos θ + sin θ) - 1 = 0.Wait, that's a different equation than what I got earlier. Which one is correct?Let me see. The first method gave me ρ² - 2√2 ρ cos(θ - π/4) - 1 = 0.The second method gave me ρ² - 2ρ (cos θ + sin θ) - 1 = 0.Hmm, are these equivalent? Let me check.I know that cos(θ - π/4) can be expanded as cos θ cos π/4 + sin θ sin π/4.Since cos π/4 = sin π/4 = √2/2,cos(θ - π/4) = (√2/2)(cos θ + sin θ).So, 2√2 ρ cos(θ - π/4) = 2√2 ρ*(√2/2)(cos θ + sin θ) = 2√2*(√2/2) ρ (cos θ + sin θ) = 2*1*ρ (cos θ + sin θ) = 2ρ (cos θ + sin θ).So, the first equation becomes:ρ² - 2ρ (cos θ + sin θ) - 1 = 0.Which matches the second equation. So, both methods give the same result. Therefore, the polar equation is ρ² - 2ρ (cos θ + sin θ) - 1 = 0.Alright, so part (1) is done.Now, part (2). We have a line l with parametric equations:x = 2 + t cos α,y = 2 + t sin α,where t is the parameter, and α is in [0, π/4). The line intersects the circle C at points A and B. We need to find the range of the chord length |AB|.First, let me visualize this. The circle is centered at (1,1) with radius √3. The line starts at (2,2) and goes in the direction determined by α. Since α is between 0 and π/4, the line is going from (2,2) towards the first quadrant, but not beyond 45 degrees.To find the points of intersection, I need to substitute the parametric equations into the circle's Cartesian equation.The circle's equation is (x - 1)² + (y - 1)² = 3.Substituting x and y:(2 + t cos α - 1)² + (2 + t sin α - 1)² = 3.Simplify:(1 + t cos α)² + (1 + t sin α)² = 3.Expanding both squares:(1 + 2t cos α + t² cos² α) + (1 + 2t sin α + t² sin² α) = 3.Combine like terms:1 + 1 + 2t cos α + 2t sin α + t² (cos² α + sin² α) = 3.Simplify:2 + 2t (cos α + sin α) + t² (1) = 3.So,t² + 2t (cos α + sin α) + 2 - 3 = 0,which simplifies to:t² + 2t (cos α + sin α) - 1 = 0.That's a quadratic equation in t. Let me denote this as:t² + 2t (cos α + sin α) - 1 = 0.Let me write it as:t² + 2t (cos α + sin α) - 1 = 0.Let me denote this quadratic as t² + bt + c = 0, where b = 2 (cos α + sin α) and c = -1.The solutions are t = [-b ± sqrt(b² - 4ac)] / 2a.Here, a = 1, so:t = [-2 (cos α + sin α) ± sqrt{(2 (cos α + sin α))² - 4*1*(-1)}]/2.Simplify the discriminant:(2 (cos α + sin α))² - 4*1*(-1) = 4 (cos α + sin α)² + 4.Factor out 4:4[(cos α + sin α)² + 1].So, sqrt of discriminant is 2 sqrt[(cos α + sin α)² + 1].Therefore, t = [-2 (cos α + sin α) ± 2 sqrt{(cos α + sin α)² + 1}]/2.Simplify:t = - (cos α + sin α) ± sqrt{(cos α + sin α)² + 1}.So, the two solutions are:t₁ = - (cos α + sin α) + sqrt{(cos α + sin α)² + 1},t₂ = - (cos α + sin α) - sqrt{(cos α + sin α)² + 1}.Therefore, the two points A and B correspond to parameters t₁ and t₂.The distance between A and B is |AB|, which is |t₁ - t₂| times the direction vector's magnitude. Since the direction vector is (cos α, sin α), which has magnitude 1, the chord length |AB| is just |t₁ - t₂|.Compute |t₁ - t₂|:t₁ - t₂ = [ - (cos α + sin α) + sqrt{(cos α + sin α)² + 1} ] - [ - (cos α + sin α) - sqrt{(cos α + sin α)² + 1} ]Simplify:t₁ - t₂ = [ - (cos α + sin α) + sqrt{(cos α + sin α)² + 1} + (cos α + sin α) + sqrt{(cos α + sin α)² + 1} ]The - (cos α + sin α) and + (cos α + sin α) cancel out.So, t₁ - t₂ = 2 sqrt{(cos α + sin α)² + 1}.Therefore, |AB| = 2 sqrt{(cos α + sin α)² + 1}.Let me simplify the expression inside the square root:(cos α + sin α)² + 1 = cos² α + 2 cos α sin α + sin² α + 1.Since cos² α + sin² α = 1,= 1 + 2 cos α sin α + 1 = 2 + 2 cos α sin α.We know that 2 cos α sin α = sin 2α, so:= 2 + sin 2α.Therefore, |AB| = 2 sqrt{2 + sin 2α}.So, |AB| = 2 sqrt{2 + sin 2α}.Now, we need to find the range of |AB| as α varies in [0, π/4).First, let's analyze the expression inside the square root: 2 + sin 2α.Since α ∈ [0, π/4), 2α ∈ [0, π/2). So, sin 2α ∈ [0, 1).Therefore, 2 + sin 2α ∈ [2, 3).Thus, sqrt{2 + sin 2α} ∈ [sqrt{2}, sqrt{3}).Multiplying by 2:|AB| ∈ [2 sqrt{2}, 2 sqrt{3}).Therefore, the range of the chord length |AB| is [2√2, 2√3).Wait, let me make sure I didn't make a mistake. When α = 0, sin 2α = 0, so |AB| = 2 sqrt{2 + 0} = 2√2.When α approaches π/4, 2α approaches π/2, so sin 2α approaches 1, so |AB| approaches 2 sqrt{3}.Therefore, the chord length varies from 2√2 up to, but not including, 2√3.Hence, the range is [2√2, 2√3).I think that makes sense. Let me just recap:1. Converted the circle's center from polar to Cartesian coordinates.2. Wrote the Cartesian equation of the circle.3. Substituted the parametric line equations into the circle's equation.4. Solved the quadratic for t, found the difference between the roots to get the chord length.5. Simplified the expression and found the range based on α's interval.Everything seems to check out. So, I think the final answer is [2√2, 2√3).**Final Answer**The range of the chord length (|AB|) is boxed{[2sqrt{2}, 2sqrt{3})}).