🧠:[<think>Alright, so I've got this geometry problem here. Let me try to parse it step by step. The problem says:"Given that triangle ABC is an equilateral triangle, and the ellipse Γ has one focus at A, and the other focus F lies on the line segment BC. If the ellipse Γ passes exactly through points B and C, then its eccentricity is ______."Hmm. Okay, so we need to find the eccentricity of this ellipse. Let's recall that an ellipse has two foci, and the sum of the distances from any point on the ellipse to the two foci is constant and equal to the major axis length, which is 2a, where a is the semi-major axis. The distance between the foci is 2c, and the eccentricity e is c/a. So the key here is to find the values of c and a for this particular ellipse.Given that the ellipse passes through B and C, which are vertices of the equilateral triangle, and the foci are at A and some point F on BC. Let me try to visualize this. Let's sketch an equilateral triangle ABC. Let's say point A is at the top, and BC is the base. The ellipse has one focus at A, and the other focus F is somewhere along BC. The ellipse passes through B and C. So points B and C are on the ellipse, which means that the sum of the distances from B to A and from B to F is equal to 2a. Similarly, the sum of the distances from C to A and from C to F is also equal to 2a.Since ABC is equilateral, all sides are equal. Let's denote the side length as s. So AB = BC = CA = s. Let me assign coordinates to the points to make this easier. Let me place point A at (0, h) and points B and C at (-s/2, 0) and (s/2, 0) respectively. Wait, but in an equilateral triangle, the height h is given by h = (√3/2)s. So maybe coordinate geometry can help here.Alternatively, maybe I can work algebraically without coordinates. Let's see. Let's denote the side length of the triangle as s. The other focus F is somewhere on BC. Let's let BF = x and FC = s - x. Since BC is length s. Then, since B and C lie on the ellipse, we can write the equations based on the definition of an ellipse.For point B: The sum of distances from B to A and from B to F is equal to 2a.Similarly, for point C: The sum of distances from C to A and from C to F is equal to 2a.Since ABC is equilateral, BA = CA = s. The distance from B to F is x, and the distance from C to F is s - x.Wait, but wait. Let me check: If F is on BC, then BF + FC = BC = s. So if BF = x, then FC = s - x. However, the distance from B to F is x, since they are colinear on BC. Similarly, distance from C to F is s - x.But also, the distance from B to A is s, since ABC is equilateral. Similarly, the distance from C to A is s. Therefore, for point B:Distance from B to A (which is s) plus distance from B to F (which is x) equals 2a. So s + x = 2a.For point C:Distance from C to A (which is s) plus distance from C to F (which is s - x) equals 2a. So s + (s - x) = 2a. Therefore, 2s - x = 2a.So now we have two equations:1. s + x = 2a2. 2s - x = 2aIf I add these two equations together: s + x + 2s - x = 4a => 3s = 4a => a = (3/4)s.Then, substituting back into equation 1: s + x = 2*(3/4)s => s + x = (3/2)s => x = (3/2)s - s = (1/2)s.So x = s/2. Therefore, point F is at the midpoint of BC. Interesting. So the other focus is at the midpoint of BC.Wait, but in an equilateral triangle, the centroid, circumcenter, inradius, etc., all coincide. But here, we have an ellipse with foci at A and the midpoint of BC. Let me confirm that.So, F is the midpoint of BC. Therefore, BF = FC = s/2. So from point B, the distance to F is s/2, and from point C, the distance to F is s/2. Then, according to the ellipse definition, the sum from B is s + s/2 = 3s/2 = 2a => a = 3s/4. Similarly, from C: s + s/2 = 3s/2 = 2a, so same result. So a = 3s/4.Now, the distance between the two foci is AF. Since A is one focus, and F is the other focus at the midpoint of BC. Let's compute the distance AF.In an equilateral triangle, the distance from a vertex to the midpoint of the opposite side is equal to the height of the triangle. Wait, the height h is (√3/2)s. So AF = h = (√3/2)s.But wait, in an equilateral triangle, the centroid divides the median in a 2:1 ratio. Wait, but here, is F the centroid? No, the centroid is located at a distance of (2/3)h from the vertex. Wait, but in this case, F is the midpoint of BC, so AF is a median, and its length is h = (√3/2)s, but the centroid is located at 2/3 of that distance from the vertex. However, in this problem, F is the midpoint, not the centroid. Wait, but F is the other focus. So the distance between the two foci is AF = (√3/2)s. Therefore, 2c = AF = (√3/2)s, so c = (√3/4)s.But we already found that a = (3/4)s. Therefore, the eccentricity e = c/a = (√3/4)s / (3/4)s = √3/3 ≈ 0.577. But √3/3 is equal to 1/√3, which is approximately 0.577. So is that the answer?Wait, but let me verify this again step by step to make sure I didn't make a mistake.First, we found that the sum of distances from B to A and B to F is s + x = 2a. Similarly, from C to A and C to F is s + (s - x) = 2a. Then adding gives 3s = 4a => a = 3s/4. Then solving for x gives x = s/2. So F is the midpoint of BC.Therefore, distance between foci is AF. Since in an equilateral triangle, the median, angle bisector, and altitude are the same. So AF is the altitude, which is (√3/2)s. So c = AF / 2 = (√3/4)s. Wait, no. Wait, the distance between the two foci is AF, which is (√3/2)s. So 2c = (√3/2)s, so c = (√3/4)s.Then, since a = 3s/4, then eccentricity e = c/a = (√3/4)s / (3/4)s = √3/3 ≈ 0.577. So e = √3/3.But let me confirm if this is correct. Wait, the formula for eccentricity is e = c/a, where c is the distance from center to focus, but in this case, the distance between foci is 2c. Wait, hold on, maybe I got confused here.Wait, let's recall: In standard terms, for an ellipse, the distance between the two foci is 2c, so each focus is c units away from the center. So the distance between the foci is 2c, and the semi-major axis is a, with eccentricity e = c/a.In this problem, the two foci are at A and F. The distance between A and F is AF, so 2c = AF. Therefore, c = AF / 2.But in an equilateral triangle, AF is the altitude, which is h = (√3/2)s. Therefore, c = h / 2 = (√3/4)s.Then, as we found before, a = 3s/4. Therefore, e = c/a = (√3/4)s / (3/4)s = √3/3.So yes, that seems right. Therefore, the eccentricity is √3/3, which can be written as 1/√3, but rationalized, it's √3/3.But let me check once more to ensure there's no mistake here. Let's go through the steps again.1. ABC is equilateral with side length s.2. Ellipse Γ has foci at A and F (on BC). Passes through B and C.3. For any point on ellipse, sum of distances to A and F is 2a.4. For point B: BA + BF = 2a => s + BF = 2a.5. For point C: CA + CF = 2a => s + CF = 2a.6. Since BF + CF = BC = s, and from 4 and 5: s + BF = s + CF => BF = CF. Therefore, F is the midpoint of BC. So BF = s/2.7. Then, substituting back, 2a = s + s/2 = 3s/2 => a = 3s/4.8. Distance between foci AF: In equilateral triangle, the altitude is h = √3/2 s. Since F is midpoint of BC, AF is the altitude, so AF = h = √3/2 s. Therefore, 2c = √3/2 s => c = √3/4 s.9. Eccentricity e = c/a = (√3/4 s) / (3/4 s) = √3/3.Yes, that seems consistent. Therefore, the eccentricity is √3/3.But just to make sure there's no error in reasoning. For example, sometimes in ellipses, the major axis is along the line connecting the foci, but sometimes not. Wait, in an ellipse, the major axis is the longest diameter, which is along the line connecting the two foci. Therefore, the major axis length is 2a, and the distance between foci is 2c. Therefore, the major axis is longer than the distance between the foci, so a > c.In our case, a = 3s/4, and c = √3/4 s. Since √3 ≈ 1.732, so c ≈ 1.732/4 s ≈ 0.433s, and a = 0.75s. So indeed, a > c, which is correct.Alternatively, we can compute e = c/a ≈ 0.433/0.75 ≈ 0.577, which is approximately √3/3 ≈ 0.577. So that matches.Therefore, the answer should be √3⁄3. So in boxed form, that's boxed{dfrac{sqrt{3}}{3}}.Final Answerboxed{dfrac{sqrt{3}}{3}}</think>Given that (triangle ABC) is an equilateral triangle with side length (s), the ellipse (Gamma) has one focus at (A) and the other focus (F) on the line segment (BC). The ellipse passes through points (B) and (C). We need to find the eccentricity of the ellipse.1. Equilateral Triangle Properties: - All sides are equal: (AB = BC = CA = s). - The height (altitude) from (A) to (BC) is (h = frac{sqrt{3}}{2}s).2. Ellipse Definition: - The sum of the distances from any point on the ellipse to the two foci is constant and equal to (2a). - For points (B) and (C) on the ellipse: - For point (B): (BA + BF = 2a) - For point (C): (CA + CF = 2a)3. Solving for (a): - Since (F) is on (BC), let (BF = x) and (CF = s - x). - For point (B): (s + x = 2a) - For point (C): (s + (s - x) = 2a) - Adding these equations: (s + x + s + (s - x) = 4a implies 3s = 4a implies a = frac{3}{4}s).4. Finding the Position of (F): - Solving (s + x = 2a) gives (x = frac{s}{2}), so (F) is the midpoint of (BC).5. Distance Between Foci: - The distance (AF) is the height of the equilateral triangle: (AF = frac{sqrt{3}}{2}s). - The distance between the foci (2c = AF implies c = frac{sqrt{3}}{4}s).6. Eccentricity Calculation: - Eccentricity (e = frac{c}{a} = frac{frac{sqrt{3}}{4}s}{frac{3}{4}s} = frac{sqrt{3}}{3}).Thus, the eccentricity of the ellipse is (boxed{dfrac{sqrt{3}}{3}}).,<think>Okay, so I need to find the eccentricity of this ellipse Γ. Let me start by recalling some properties of ellipses. An ellipse has two foci, and the sum of the distances from any point on the ellipse to the two foci is constant. The eccentricity e is the ratio of the distance between the foci (2c) to the length of the major axis (2a), so e = c/a. Given that triangle ABC is equilateral, all sides are equal. Let me denote the side length as s. The ellipse has one focus at point A, and the other focus F is on the segment BC. The ellipse passes through points B and C. So, points B and C lie on the ellipse, which means the sum of their distances to the foci A and F is equal to the constant 2a.Let me visualize this. Let's say ABC is an equilateral triangle. Let me place it in a coordinate system to make calculations easier. Maybe place point A at (0, 0), point B at (s, 0), and point C at (s/2, (s√3)/2). Then, segment BC is from (s, 0) to (s/2, (s√3)/2). The other focus F is somewhere on BC. Let me parameterize point F. Since BC is from (s, 0) to (s/2, (s√3)/2), any point F on BC can be written as (s - t*s/2, 0 + t*(s√3)/2) where t is between 0 and 1. Wait, actually, if t=0, we get (s, 0) which is point B, and t=1 gives (s/2, (s√3)/2) which is point C. So t ∈ [0,1]. So, coordinates of F would be (s - (s/2)t, (s√3/2)t). Alternatively, maybe it's easier to set specific coordinates. Let me set the side length s = 2 for simplicity. Then, coordinates would be A(0, 0), B(2, 0), C(1, √3). Then BC is from (2, 0) to (1, √3). So, point F can be parameterized as (2 - t, 0 + t√3) where t ∈ [0,1]. Wait, but actually, if we move along BC from B to C, the x-coordinate decreases by 1 and the y-coordinate increases by √3. So, for parameter t (0 ≤ t ≤ 1), F would be (2 - t, t√3). Let me check when t=0, F is (2,0) which is B, and when t=1, F is (1, √3) which is C. So that's correct.So, ellipse Γ has foci at A(0,0) and F(2 - t, t√3) for some t between 0 and 1. The ellipse passes through points B(2,0) and C(1, √3). Since B and C are on the ellipse, the sum of distances from each to the foci is equal to 2a. Let's write equations for both points.For point B(2,0):Distance from B to A is |AB|. Since AB is a side of the equilateral triangle with side length 2, |AB| = 2.Distance from B to F is the distance between (2,0) and (2 - t, t√3). Let me compute that:sqrt[(2 - (2 - t))² + (0 - t√3)²] = sqrt[t² + 3t²] = sqrt[4t²] = 2t.So, the sum is 2 + 2t = 2a ⇒ a = 1 + t.Similarly, for point C(1, √3):Distance from C to A is |AC| = 2 (since it's an equilateral triangle).Distance from C to F is the distance between (1, √3) and F(2 - t, t√3). Let's compute that:sqrt[(1 - (2 - t))² + (√3 - t√3)²] = sqrt[(t - 1)² + (√3(1 - t))²] = sqrt[(1 - t)² + 3(1 - t)²] = sqrt[4(1 - t)²] = 2|1 - t|. Since t is between 0 and 1, 1 - t is non-negative, so it's 2(1 - t).Therefore, the sum of distances from C to A and F is 2 + 2(1 - t) = 4 - 2t. This must equal 2a as well. So:4 - 2t = 2a ⇒ a = 2 - t.But earlier, we found a = 1 + t from point B. Therefore:1 + t = 2 - t ⇒ 2t = 1 ⇒ t = 1/2.So, t = 1/2. Then, the coordinates of focus F are (2 - 1/2, (1/2)√3) = (3/2, (√3)/2).Now, let's confirm. Then, the major axis length is 2a, and a = 1 + t = 1 + 1/2 = 3/2. So, 2a = 3. The distance between the foci is 2c. The foci are at A(0,0) and F(3/2, √3/2). Let's compute the distance between A and F.Distance AF = sqrt[(3/2 - 0)^2 + (√3/2 - 0)^2] = sqrt[(9/4) + (3/4)] = sqrt[12/4] = sqrt[3] ≈ 1.732.Thus, 2c = sqrt(3), so c = sqrt(3)/2.But then, eccentricity e = c/a = (sqrt(3)/2) / (3/2) = sqrt(3)/3 = 1/√3 ≈ 0.577. But wait, let me check if this is correct.Wait, but hold on. The standard formula is e = c/a, where c is the distance from the center to a focus, and a is the semi-major axis. However, in this case, do we know the orientation of the ellipse? The ellipse has foci at A and F, which are not on the same line as the major axis unless the major axis is along the line AF. Wait, in an ellipse, the major axis is the line segment joining the two farthest points on the ellipse, and it passes through both foci. So the major axis is the line through the two foci. But in this case, the foci are at A(0,0) and F(3/2, √3/2). So the major axis is along the line connecting these two points. The center of the ellipse is the midpoint of the segment AF. Let's compute that.Midpoint M of AF: ((0 + 3/2)/2, (0 + √3/2)/2) = (3/4, √3/4).But maybe we don't need the center. Since we have a = 3/2, c = sqrt(3)/2 (half the distance between the foci), then e = c/a = (sqrt(3)/2)/(3/2) = sqrt(3)/3 ≈ 0.577. So this is the eccentricity.But let me verify if this is correct. Let me check the calculations again.First, t was found to be 1/2 by equating a = 1 + t and a = 2 - t, which gives t = 1/2. Then a = 3/2. The distance between foci is sqrt[(3/2)^2 + (√3/2)^2] = sqrt(9/4 + 3/4) = sqrt(12/4) = sqrt(3). So c = sqrt(3)/2, as the distance from the center to each focus is c, but wait, actually, the distance between the two foci is 2c. Wait, no, the standard definition is that the distance between foci is 2c. So, in that case, if the distance between A and F is sqrt(3), then 2c = sqrt(3) ⇒ c = sqrt(3)/2. Then e = c/a = (sqrt(3)/2)/(3/2) = sqrt(3)/3. So that is correct.But sqrt(3)/3 simplifies to 1/√3, which is approximately 0.577. However, the problem states that the ellipse passes through B and C, which are vertices of the equilateral triangle. The answer sqrt(3)/3 seems plausible, but let me think again.Alternatively, perhaps there is a geometric interpretation here. Since the ellipse passes through B and C, which are two vertices of the triangle, and one focus is at A, the other focus is somewhere on BC. Given that the triangle is equilateral, maybe there is some symmetry here.Alternatively, let's think in terms of reflection properties. In an ellipse, the tangent at any point makes equal angles with the lines from the two foci to that point. But since points B and C are on the ellipse, the tangent at those points can be considered. However, in this case, since B and C are vertices of the triangle, the tangent at B would be perpendicular to the angle bisector at B. Wait, maybe this is complicating things. Alternatively, perhaps using coordinates was the right approach, and the answer is indeed sqrt(3)/3.But let me check once more. Let me recompute the distance between foci. A is at (0,0), F is at (3/2, sqrt(3)/2). The distance between them is sqrt[(3/2)^2 + (sqrt(3)/2)^2] = sqrt[9/4 + 3/4] = sqrt[12/4] = sqrt(3). So 2c = sqrt(3) ⇒ c = sqrt(3)/2. Then a = 3/2, so e = c/a = sqrt(3)/3. Which is approximately 0.577. That's a reasonable eccentricity, less than 1.Alternatively, maybe I made an error in parameterizing point F. Let me check. Earlier, I set the side length as 2 for simplicity. The coordinates of F when t=1/2 are (3/2, sqrt(3)/2). Let me confirm that this point is indeed on BC. In the coordinate system where B is (2,0) and C is (1, sqrt(3)), the line BC can be parametrized as (2 - t, 0 + t*(sqrt(3)/1)) where t ranges from 0 to 1? Wait, actually, moving from B(2,0) to C(1, sqrt(3)), the displacement is (-1, sqrt(3)), so the parametric equations would be x = 2 - t, y = 0 + t*sqrt(3), where t ranges from 0 to 1. So when t=1/2, x = 2 - 1/2 = 3/2, y = (sqrt(3)/2), which is correct. So F is indeed on BC.Therefore, the calculations seem correct. Thus, the eccentricity is sqrt(3)/3. However, sqrt(3)/3 is equal to 1/sqrt(3), which is often rationalized as sqrt(3)/3.But let me check if the problem expects the answer in a specific form. The problem says "its eccentricity is ______". Since it's a mathematical problem, both forms are acceptable, but perhaps the answer should be in the form with the radical in the numerator, so sqrt(3)/3.Alternatively, maybe there's a different approach. Let me consider the general case without coordinates.Given an equilateral triangle ABC. Let the side length be s. The ellipse has foci at A and F (on BC), and passes through B and C. For any point on the ellipse, the sum of distances to A and F is 2a. For point B: distance from B to A is s, distance from B to F is the length BF. Therefore, s + BF = 2a. Similarly, for point C: distance from C to A is s, distance from C to F is CF. So s + CF = 2a. Therefore, BF + s = CF + s ⇒ BF = CF. Wait, but BF + CF = BC = s, since F is on BC. Therefore, if BF = CF, then BF = CF = s/2. Therefore, F must be the midpoint of BC. Therefore, in that case, F is the midpoint of BC. So in an equilateral triangle, the midpoint of BC is also the altitude from C to AB, but wait, no. Wait, in an equilateral triangle, all medians are equal and coincide with altitudes and angle bisectors. So the midpoint of BC is the foot of the altitude from A to BC, but in this case, F is on BC. Wait, but in our problem, F is the other focus, located on BC.But wait, according to this reasoning, if BF = CF, then F must be the midpoint. But in our coordinate example, when t=1/2, F is (3/2, sqrt(3)/2) which is actually not the midpoint of BC. Wait, hold on. Wait, in the coordinate system where B is (2,0) and C is (1, sqrt(3)), the midpoint of BC would be ((2 + 1)/2, (0 + sqrt(3))/2) = (1.5, sqrt(3)/2), which is exactly the point F when t=1/2. Wait, but in our coordinate system, BC is from (2,0) to (1, sqrt(3)). The midpoint is indeed (3/2, sqrt(3)/2). So yes, when t=1/2, F is the midpoint. So in general, for any equilateral triangle, if the ellipse passes through B and C, then F must be the midpoint of BC.Therefore, regardless of coordinate system, F is the midpoint of BC. Therefore, in the general case, F is the midpoint. Then, let's compute the distance between A and F. In an equilateral triangle, the distance from a vertex to the midpoint of the opposite side is equal to the altitude. The altitude h of an equilateral triangle with side s is (s√3)/2. So the distance AF is h = (s√3)/2. Therefore, the distance between the foci is AF = (s√3)/2, so 2c = (s√3)/2 ⇒ c = (s√3)/4. Now, the semi-major axis a can be found from the sum of distances from B to A and F. The distance from B to A is s. The distance from B to F is the distance from B to the midpoint of BC. Since BC is length s, the midpoint is s/2 away from B. Therefore, BF = s/2. So the sum s + s/2 = 3s/2 = 2a ⇒ a = 3s/4. Therefore, the eccentricity e = c/a = (s√3/4) / (3s/4) = √3/3. Therefore, regardless of the side length, the eccentricity is √3/3. So the answer is √3⁄3, which can be written as boxed{dfrac{sqrt{3}}{3}}.Final AnswerThe eccentricity of the ellipse is boxed{dfrac{sqrt{3}}{3}}.</think>Given that triangle ABC is an equilateral triangle with side length s, and the ellipse Gamma has one focus at A and the other focus F on the line segment BC. The ellipse Gamma passes through points B and C. We need to find the eccentricity of the ellipse.1. Properties of the Ellipse: - The sum of the distances from any point on the ellipse to the two foci is constant (equal to 2a). - The eccentricity e is given by e = frac{c}{a}, where 2c is the distance between the foci and 2a is the length of the major axis.2. Coordinates and Parameterization: - Place A at (0, 0), B at (s, 0), and C at (frac{s}{2}, frac{ssqrt{3}}{2}). - Focus F is on segment BC. Parameterize F as (s - frac{st}{2}, frac{stsqrt{3}}{2}) for t in [0, 1].3. Sum of Distances: - For point B: Distance from B to A is s, and distance from B to F is st. Thus, s + st = 2a. - For point C: Distance from C to A is s, and distance from C to F is s(1 - t). Thus, s + s(1 - t) = 2a.4. Solving for t: - Equate the expressions for 2a from points B and C: [ s + st = s + s(1 - t) implies 1 + t = 2 - t implies t = frac{1}{2} ] - Therefore, F is the midpoint of BC.5. Calculating Distances: - The distance between foci A and F (midpoint of BC) is the altitude of the equilateral triangle: [ AF = frac{ssqrt{3}}{2} ] - Semi-major axis a is found from the sum of distances: [ 2a = s + frac{s}{2} = frac{3s}{2} implies a = frac{3s}{4} ] - Distance c (half the distance between foci): [ 2c = frac{ssqrt{3}}{2} implies c = frac{ssqrt{3}}{4} ]6. Eccentricity: - Eccentricity e: [ e = frac{c}{a} = frac{frac{ssqrt{3}}{4}}{frac{3s}{4}} = frac{sqrt{3}}{3} ]The eccentricity of the ellipse is boxed{dfrac{sqrt{3}}{3}}.]

🔑:3. frac{sqrt{3}}{3}.Let the side length of the equilateral triangle A B C be x, the semi-major axis of the ellipse Gamma be a, and the semi-focal distance be c.According to the problem, we have|A C|+|C F|=|A B|+|B F|=2 a text {. }Therefore, F is the midpoint of B C, sox+frac{x}{2}=2 a Rightarrow a=frac{3}{4} x text {. }In the right triangle triangle A C F, by the Pythagorean theorem, we getbegin{array}{l}|A F|=sqrt{|A C|^{2}-|C F|^{2}} =sqrt{x^{2}-left(frac{x}{2}right)^{2}}=frac{sqrt{3}}{2} x Rightarrow c=frac{sqrt{3}}{4} x .end{array}In summary, the eccentricity of the ellipse Gamma is frac{c}{a}=frac{sqrt{3}}{3}.

🧠:[<think>Alright, let's tackle this problem. So, we need to compare the values of a, b, and c and figure out the correct order. The options are given, so maybe I can compute each value numerically and then compare them. Let's start by understanding each variable.First, a is given as 1/2. That's straightforward; a equals 0.5.Next, b is log base 3 of 2. Hmm, logarithms can sometimes be tricky without a calculator, but maybe I can estimate its value. Remember that log base 3 of 2 is the exponent to which we need to raise 3 to get 2. So, 3^x = 2. Let's think about some exponents of 3. 3^0.5 is sqrt(3) which is approximately 1.732. Since 1.732 is less than 2, the exponent must be higher than 0.5. What about 3^0.6? Let's calculate that. 3^0.6 can be written as e^(0.6 ln3). Let's approximate ln3 as 1.0986. So, 0.6*1.0986 ≈ 0.659. Then e^0.659 ≈ e^0.65 is approximately 1.915. That's still less than 2. So 0.6 gives about 1.915, which is close to 2. Let's try 0.63. 0.63*1.0986 ≈ 0.692. e^0.692 is approximately 2.0? Wait, e^0.6931 is approximately 2, since ln2 is approximately 0.6931. Wait, but here we are dealing with log base 3 of 2. Maybe there's another way. Alternatively, using change of base formula: log_3 2 = ln2 / ln3 ≈ 0.6931 / 1.0986 ≈ 0.6309. So, approximately 0.63. So, b is around 0.63.Then c is 2^(1/3). That's the cube root of 2. Let me compute that. 2^(1/3) is approximately 1.26, since 1.26^3 is approximately 2. Let me verify: 1.26*1.26 = 1.5876, then 1.5876*1.26 ≈ 1.5876*1 + 1.5876*0.26 ≈ 1.5876 + 0.4128 ≈ 2.0004. Wow, that's very close. So, 2^(1/3) is approximately 1.26. Therefore, c is about 1.26.Now, let's summarize the approximate values:a = 0.5b ≈ 0.63c ≈ 1.26So, arranging them from largest to smallest: c (1.26) > b (0.63) > a (0.5). Therefore, the correct order is c > b > a, which is option D.Wait, but let me double-check my approximations to make sure I didn't make a mistake.For log base 3 of 2, using the change of base formula: log_3 2 = ln2 / ln3. Let me compute that with more precise values.ln2 is approximately 0.69314718056ln3 is approximately 1.098612288668So, log_3 2 = 0.69314718056 / 1.098612288668 ≈ 0.63092975357. So, approximately 0.6309. So, yes, b ≈ 0.63.Then, cube root of 2: 2^(1/3). Let me check with a calculator. Cube root of 2 is approximately 1.25992105. So, approximately 1.26. Correct.And a is 0.5. So, indeed, c (1.26) is the largest, then b (0.63), then a (0.5). So, the order is c > b > a, which is option D.But wait, let me check the answer options again:A: a > b > cB: b > a > cC: c > a > bD: c > b > aYes, D is c > b > a. So that's the correct answer.But hold on, maybe there's a different approach without approximating? Let me think.Alternatively, we can compare them by exponents or logarithms. Let's see.First, comparing a and b. a is 1/2, which is 0.5. b is log_3 2. Since 3^(1/2) = sqrt(3) ≈ 1.732, and 3^1 = 3. Since 2 is between 1.732 and 3, but log_3 2 is between 0.5 and 1. But we already calculated it as approximately 0.63, which is more than 0.5. So, b > a.Then, comparing c with a and b. c is 2^(1/3). Let's see, 2^(1/3) is equal to approximately 1.26, which is greater than both 0.63 and 0.5, so c is the largest.Therefore, the order is c > b > a. So, D is correct.Alternatively, maybe we can compare b and c without using decimals. Let's see. c = 2^(1/3). Let's take logarithm base 3 of c. log_3 c = log_3 2^(1/3) = (1/3) log_3 2 = (1/3) * b. Since b ≈ 0.63, then log_3 c ≈ 0.21. Therefore, c = 3^(log_3 c) = 3^0.21. Hmm, but 3^0.21 is approximately 3^(0.2) which is about 1.245, which is close to 1.26. Not sure if this helps. Alternatively, to compare c and b: c = 2^(1/3) ≈ 1.26, and b ≈ 0.63. So, c is definitely larger than b. Therefore, c > b > a.Alternatively, can we compare 2^(1/3) and log_3 2 without decimals? Let's see.Wait, 2^(1/3) is the cube root of 2. Let's compare it to 1.25. 1.25 cubed is (5/4)^3 = 125/64 ≈ 1.953, which is less than 2. So, 2^(1/3) is greater than 5/4 (1.25). But 1.25 is 5/4, which is 1.25. So, 2^(1/3) is approximately 1.26, as before.Meanwhile, log_3 2 is approximately 0.63. So, 1.26 > 0.63, so c > b.Therefore, even without exact decimals, we can see that c is greater than 1.25, and b is less than 1, so c > b.So, combining all, c > b > a. So, answer D.Therefore, the correct answer is D.Final Answerboxed{D}</think>Given ( a = frac{1}{2} ), ( b = log_{3} 2 ), and ( c = 2^{frac{1}{3}} ), we need to determine the correct relationship among ( a ), ( b ), and ( c ).1. Value of ( a ): [ a = frac{1}{2} = 0.5 ]2. Value of ( b ): Using the change of base formula: [ b = log_{3} 2 = frac{ln 2}{ln 3} approx frac{0.6931}{1.0986} approx 0.6309 ] So, ( b approx 0.63 ).3. Value of ( c ): [ c = 2^{frac{1}{3}} approx 1.2599 ] So, ( c approx 1.26 ).4. Comparison: - ( a = 0.5 ) - ( b approx 0.63 ) - ( c approx 1.26 ) Arranging these values from largest to smallest, we get ( c > b > a ).Thus, the correct relationship is (boxed{D}).,<think>Okay, let's see. The problem gives me a = 1/2, b = log base 3 of 2, and c = 2^(1/3). I need to figure out the correct relationship among a, b, and c. The options are A through D, each with different orderings. Hmm.First, let me make sure I understand each value. a is straightforward—it's just 0.5. b is log base 3 of 2. That means 3^b = 2. So, I need to figure out what this logarithm is approximately. Maybe convert it to a different base to compute it?c is the cube root of 2, which is 2^(1/3). I know that 2^(1/3) is approximately 1.26 because 1.26 cubed is roughly 2. Let me check that. 1.26 * 1.26 = 1.5876, and then 1.5876 * 1.26 ≈ 2.000. So, yeah, around 1.26.So, c is about 1.26, which is greater than 1. Now, a is 0.5, which is 0.5. Then, b is log base 3 of 2. Let me calculate that. Maybe using natural logarithm? So, log base 3 of 2 is equal to ln(2)/ln(3). Let me compute that. I know ln(2) is approximately 0.6931 and ln(3) is approximately 1.0986. So, 0.6931 divided by 1.0986 is approximately 0.6309. So, b is approximately 0.63.So, putting these together: c is about 1.26, a is 0.5, and b is about 0.63. So, the order from largest to smallest would be c > b > a. Which corresponds to option D. Hmm. Let me check again to be sure.Wait, let me verify the value of b again. Log base 3 of 2. If I remember, 3^0.6 is 3^(3/5). Let me compute that. 3^0.6 is approximately e^(0.6*ln3) ≈ e^(0.6*1.0986) ≈ e^0.659 ≈ 1.933. Which is close to 2. So, 3^0.6 ≈ 1.933, so to get to 2, the exponent should be a bit higher than 0.6. So, maybe 0.63 is correct. Since 3^0.63 ≈ e^(0.63*1.0986) ≈ e^(0.691) ≈ 1.996, which is almost 2. So, yeah, b is approximately 0.63.So, c is 1.26, b is 0.63, and a is 0.5. So, the order is c > b > a. So, the correct answer is D. Let me check the options again. Option D says c > b > a. Yes, that matches.Alternatively, maybe I can compare them without calculating the exact decimal values. Let's see. Compare a and b first. a is 1/2 = 0.5. b is log_3 2. Since 3^(1/2) = sqrt(3) ≈ 1.732, so log_3 1.732 = 0.5. But 2 is greater than 1.732, so log_3 2 is greater than 1/2. Therefore, b > a. Okay, that confirms that b is greater than a without converting to decimals.Now, compare c and a. c is 2^(1/3). 2^(1/3) is definitely greater than 1, since 2 > 1. So, c is greater than 1, and a is 0.5, so c > a.Now, compare c and b. c is 2^(1/3), which is approximately 1.26, and b is approximately 0.63, so numerically, c is larger. But maybe we can do this without converting?Alternatively, let's express all three in exponents with the same base. Let me try base 2.a = 1/2 = 2^(-1)b = log_3 2. Let's see, log_3 2 = 1 / log_2 3. Since log_b a = 1 / log_a b. So, log_3 2 = 1 / log_2 3. log_2 3 is approximately 1.58496, so 1 / 1.58496 ≈ 0.6309, which matches previous result.c = 2^(1/3) = 2^(0.333...)So, in terms of exponents of 2:a is 2^(-1), which is 1/2.b is 1 / log_2 3, which is approximately 0.63, so 2^(log_2 (1 / log_2 3))... Hmm, maybe this isn't the best approach.Alternatively, if we can compare b and c. Let me think. c is 2^(1/3). Let's take the logarithm of both c and b in base 2 to compare them.Wait, c is 2^(1/3), so log_2 c = 1/3 ≈ 0.333.b is log_3 2, which is 1 / log_2 3 ≈ 0.6309. So, log_2 c ≈ 0.333, and log_2 (something) equals 0.6309? Wait, no. Wait, log_2 c is 1/3. But b itself is approximately 0.6309. So, since log_2 c is 0.333, which is less than b. Hmm, perhaps this is confusing.Alternatively, since c = 2^(1/3) ≈ 1.26, and b ≈ 0.63. So, in terms of magnitude, c is larger than b. So, c > b > a.So, the correct order is c > b > a, which is option D. So, the answer should be D.But let me check once more. Maybe if I convert all to exponents with base 3?Wait, a = 1/2. Let's write 1/2 as 3^x. So, 3^x = 1/2. Taking log base 3, x = log_3 (1/2) = - log_3 2 ≈ -0.6309. So, a is 3^(-0.6309). b is log_3 2 ≈ 0.6309, so b is 3^(log_3 2) = 2.Wait, that might not help. Alternatively, perhaps express each in terms of exponents with natural logs. But maybe this complicates.Alternatively, let's raise them all to the power of 6 to eliminate roots and denominators? Maybe.Wait, a = 1/2. a^6 = (1/2)^6 = 1/64 ≈ 0.015625.b = log_3 2. Raising it to the 6th power? Wait, that's not helpful because exponents and logarithms don't mix like that. Wait, maybe another approach.Alternatively, compare b and c. We have c = 2^(1/3) ≈ 1.26. Let's see, is 1.26 greater than log_3 2? Since log_3 2 ≈ 0.63, yes, 1.26 is greater. So, c > b.So, putting it all together: c > b > a. Therefore, the answer is D.Just to make sure, let's see if any of the other options could make sense. Option C is c > a > b. But since a is 0.5 and b is 0.63, b is greater than a. So, C is incorrect. Option B is b > a > c, but c is the largest, so B is wrong. Option A is a > b > c, but that's also wrong because a is the smallest except for... no, a is 0.5, which is bigger than some? Wait, no. Wait, a is 0.5, b is 0.63, c is 1.26. So, c is the largest, then b, then a. So, D is correct. Yes.So, the answer is D.Final Answerboxed{D}</think>Given (a = frac{1}{2}), (b = log_3 2), and (c = 2^{1/3}), we need to determine the correct relationship among (a), (b), and (c).1. Value of (a): - (a = frac{1}{2} = 0.5).2. Value of (b): - (b = log_3 2). Converting this to natural logarithms, we get: [ b = frac{ln 2}{ln 3} approx frac{0.6931}{1.0986} approx 0.6309 ]3. Value of (c): - (c = 2^{1/3}). Approximating this, we get: [ 2^{1/3} approx 1.26 ]4. Comparing the values: - (c approx 1.26) - (b approx 0.6309) - (a = 0.5) Therefore, the order from largest to smallest is (c > b > a).Thus, the correct relationship is (boxed{D}).]

🔑:We have a= dfrac {1}{2} and c=2^{ frac {1}{3}} > 2^{0}=1.Also, dfrac {1}{2}=log _{3} sqrt {3} b > a.Therefore, the correct choice is boxed{D}.By using the properties of logarithmic and exponential functions, we can determine the relationship among a, b, and c. This problem primarily tests the understanding of logarithmic operation rules, using the monotonicity of logarithms and the properties of logarithmic functions to make judgments.

🧠:[<think>Okay, so I need to find the probability that when I randomly select six different integers from the set {1,2,3,...,10}, the second smallest number in the selected set is 3. The answer choices are given, and I need to figure out which one is correct. Let's start by understanding the problem.First, the set we're dealing with is {1,2,3,...,10}. We are selecting six different numbers from this set. The question is about the probability that the second smallest number in this selected six-number set is exactly 3. The answer choices are A through E, with E being "none of the above," so I need to compute the probability and see which option matches.Let me recall that probability is generally the number of favorable outcomes divided by the total number of possible outcomes. So, I need to find two things:1. The total number of ways to choose six different integers from the set {1,2,3,...,10}.2. The number of those selections where the second smallest number is exactly 3.Then, divide the second by the first to get the probability.First, let's compute the total number of ways to choose six numbers from ten. This is a combination problem because the order of selection doesn't matter. The formula for combinations is C(n, k) = n! / (k! * (n - k)!), where n is the total number of items, and k is the number we want to choose. So, here, n = 10 and k = 6.Calculating that, C(10, 6) = 10! / (6! * 4!) = (10*9*8*7)/(4*3*2*1) = 210. So, there are 210 possible ways to select six numbers. That's the denominator of the probability.Now, the numerator is the number of favorable outcomes, i.e., the number of six-element subsets where the second smallest number is 3. Let's think about how such a subset must look.If the second smallest number is 3, that means the smallest number must be less than 3. Since we are dealing with integers, the smallest number can only be 1 or 2. So, in the selected subset, one of the numbers is 1 or 2, another number is 3, and the remaining four numbers must be larger than 3. Additionally, since 3 is the second smallest, all the numbers after the second must be greater than 3. So, the structure of such a subset is: {x, 3, y1, y2, y3, y4}, where x ∈ {1, 2}, and y1, y2, y3, y4 ∈ {4, 5, 6, 7, 8, 9, 10}.Therefore, to count the number of such subsets:1. Choose x: there are 2 choices (1 or 2).2. Then, since 3 is already included as the second smallest, we need to choose the remaining four numbers from the numbers greater than 3, which are {4,5,6,7,8,9,10}. There are 7 numbers here, and we need to choose 4. So, the number of ways is C(7, 4).Therefore, the total number of favorable subsets is 2 * C(7, 4). Let's compute that.First, C(7, 4) = 7! / (4! * 3!) = (7*6*5)/(3*2*1) = 35. So, 2 * 35 = 70. Therefore, there are 70 favorable subsets.Wait, but hold on. Let me verify this. If we have x being either 1 or 2, and then choose four numbers from the upper seven, does that cover all possibilities? Let's check.Suppose we fix the two smallest numbers as 1 and 3. Then, the other four numbers must be from 4 to 10. Similarly, if the two smallest numbers are 2 and 3, the other four numbers must be from 4 to 10. So, in both cases (whether the smallest is 1 or 2), the remaining four numbers are chosen from 4-10. Therefore, each case contributes C(7,4) subsets, so total 2*C(7,4) = 70. That seems right.But let me think again. The second smallest number is 3. So, the smallest number must be either 1 or 2. Then, 3 is the next one, and the other four numbers must be greater than 3. So, yes, that logic holds. Therefore, the number of favorable subsets is 70. Therefore, the probability is 70 / 210 = 1/3. So, the answer would be C) 1/3.But wait, the answer choices are A) 1/60, B) 1/6, C) 1/3, D)1/2, E) none. So, 1/3 is option C, which is one of the choices. But let me check again because sometimes when probabilities are calculated, there might be a mistake in the counting.Alternatively, maybe I need to consider the positions or something else. Let me see.Wait, another approach: Fix that the second smallest number is 3. Then, the numbers in the subset must include exactly one number smaller than 3 (i.e., 1 or 2), the number 3, and the remaining four numbers must be from those greater than 3. So, same as before. So, the count is 2 * C(7,4) = 70. The total number of subsets is C(10,6) = 210. So, 70 / 210 = 1/3, which is answer C. So, seems correct.But let me test this with a smaller case to see if the reasoning holds. Suppose instead of 10 numbers, let's say we have numbers {1,2,3,4}, and we want to select 3 numbers. What is the probability that the second smallest is 2?Following the same logic: total subsets C(4,3) = 4. Favorable subsets: those where second smallest is 2. So, smallest number must be 1, second smallest 2, and the remaining number must be greater than 2, i.e., 3 or 4. So, the number of favorable subsets is C(2,1) = 2. Therefore, probability is 2 / 4 = 1/2. Let's list them:Subsets of size 3: {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}. The second smallest numbers:- {1,2,3}: sorted [1,2,3], second smallest is 2.- {1,2,4}: sorted [1,2,4], second smallest is 2.- {1,3,4}: sorted [1,3,4], second smallest is 3.- {2,3,4}: sorted [2,3,4], second smallest is 3.So, indeed, 2 out of 4 have second smallest 2. So, probability is 1/2. Which matches the calculation: number of favorable subsets is C(1,1) [since smallest must be 1] * C(2,1) [numbers greater than 2] = 1 * 2 = 2. Total subsets C(4,3)=4. So, 2/4=1/2. Therefore, the method works.Similarly, if in the original problem, the answer is 1/3, which is C. But let me see another way.Alternatively, let's think in terms of order statistics. The second smallest being 3. In a sample of six numbers from 1 to 10, what is the probability that the second order statistic is 3.In general, the formula for the probability that the k-th order statistic is x is [C(x-1, k-1) * C(n - x, m - k)] / C(n, m), where n is the population size, m is the sample size, k is the order statistic.Wait, let me check. Let's see.The number of ways that the k-th smallest element is x is equal to the number of ways to choose k-1 elements less than x, and m - k elements greater than x.In our problem, the second smallest (k=2) is 3 (x=3). So, we need to choose 1 element (k-1 =1) less than 3 and 4 elements (m -k =6 -2=4) greater than 3. Since x=3, numbers less than 3 are 1 and 2, so 2 numbers. Numbers greater than 3 are 4-10, which are 7 numbers.Therefore, the number of favorable subsets is C(2,1)*C(7,4) = 2*35=70. As before. Total subsets C(10,6)=210. So, 70/210=1/3. So, this formula confirms the answer.Therefore, the answer should be C) 1/3.Wait, but let me think again. Maybe the problem is with the fact that when selecting six numbers, some of the numbers could be duplicates? Wait, no, the problem states six different integers are selected. So, all numbers in the subset are unique, so that's okay.Alternatively, is there a different interpretation where the second number when arranging in order is 3, but perhaps considering the numbers as selected in a sequence rather than a subset? But the problem says "six different integers are randomly selected," so I think it's a combination, not a permutation. So, order doesn't matter. So, the second smallest in the subset. So, yeah, the formula applies.Alternatively, maybe the answer is different? Let me compute another way.Suppose we fix that the second smallest number is 3. Then, the elements in the subset are:- One number from {1, 2}.- The number 3.- Four numbers from {4, 5, 6, 7, 8, 9, 10}.So, the number of subsets is C(2,1)*C(7,4) = 2*35=70. So, same as before.Total subsets: C(10,6)=210.So, 70/210=1/3. So, answer C.Alternatively, let's think of hypergeometric distribution. The probability that in the sample, exactly one element is less than 3, and exactly four elements are greater than 3. Since 3 itself is included.So, the probability is [C(2,1)*C(7,4)] / C(10,6). Which is again 70/210=1/3. So, that's the same.Therefore, I think the answer is C) 1/3.But let me check with another approach. Let's compute the probability step by step.Imagine selecting the numbers one by one. However, since order doesn't matter, perhaps thinking in terms of all combinations is better.Alternatively, perhaps using conditional probability. The probability that the second smallest is 3. Let's see. The second smallest is 3, so that means exactly one number is less than 3 (i.e., 1 or 2), and the remaining four numbers are greater than 3. So, same as before.But maybe someone could think that if the second smallest is 3, then there's 1 number less than 3, 1 number equal to 3, and 4 numbers greater than 3. So, the count is C(2,1)*C(1,1)*C(7,4)=2*1*35=70. Same result.Alternatively, maybe the problem is that we have to include 3 as one of the numbers. So, in the subset, 3 must be present, and exactly one number from {1,2}, and the rest from {4,...,10}. So, same as above.Therefore, all approaches lead to 70/210=1/3. Therefore, the answer should be C. So, why is E an option? Maybe there's a trick here?Wait, let's check once again. The problem says "the second smallest number is 3". So, does that mean that 3 has to be present in the subset? Yes, because the second smallest is 3. So, 3 must be in the subset. Also, there must be exactly one number less than 3, which can be either 1 or 2, and the remaining four numbers must be greater than 3.Therefore, the count is indeed C(2,1)*C(7,4)=70.Alternatively, someone might think that the second smallest is 3, so there is one number smaller than 3 and four numbers larger than 3. But also, since the total numbers selected are six, 1 (smallest) +1 (second smallest) +4 (the rest) =6. So, that matches. Therefore, correct.Alternatively, if we fix the second position in the ordered list as 3, then we need to choose 1 number for the first position (from {1,2}), and the remaining four positions from {4,...,10}. But since order doesn't matter in the subset, this is equivalent to combinations.Alternatively, if someone thought that the position of 3 matters, but in subsets, it's unordered. So, the second smallest is just a label, not an actual position in an ordered list. So, the logic is correct.Another check: total number of subsets with 6 elements: 210.Total number of subsets where 3 is the second smallest: 70.So, 70/210=1/3. So, answer C.Alternatively, maybe the answer is E because the real probability is different? Let's compute the actual value step by step.Number of favorable subsets: To have the second smallest number as 3. So, must include one number from {1,2}, must include 3, and four numbers from {4,5,6,7,8,9,10}.So, the number of such subsets is C(2,1)*C(7,4). C(2,1)=2, C(7,4)=35, so 2*35=70.Total subsets: C(10,6)=210.70 divided by 210 is indeed 1/3. So, unless there's a miscalculation here, the answer should be C.But maybe the original problem was phrased differently? Wait, let me check the problem again."From {1,2,3, ...,10}, six different integers are randomly selected. Among these selections, the probability that the second smallest number is 3 is (A) 1/60; (B)1/6; (C)1/3; (D)1/2; (E) none of the above."No, the problem is as I thought. So, given that, I think the answer is C) 1/3. So, why is E an option? Maybe the test creators consider that the answer is different? Let me think.Wait, perhaps there's a mistake in the reasoning. Let me consider another angle. Suppose we fix the second smallest number as 3. Then, the smallest number is 1 or 2, 3 is included, and the remaining four numbers are greater than 3. So, the number of such subsets is C(2,1)*C(7,4). But another way to calculate this is:First, include 3 in the subset. Then, we need to choose 5 more numbers. However, among these five, exactly one must be from {1,2}, and the remaining four must be from {4,...,10}. Therefore, the number is C(2,1)*C(7,4). So, same as before.Alternatively, suppose someone thought that the second smallest is 3, so the numbers can be arranged such that 3 is in the second position. But since it's a combination, not a permutation, the order doesn't matter. So, the second smallest is just a property of the set, not an actual position. Therefore, the calculation is correct.Alternatively, maybe the problem is that 3 is the second smallest, so we need to make sure that there is exactly one number less than 3 and the rest are higher. But wait, 3 is the second smallest, so exactly one number is less than 3 (i.e., in {1,2}), and the remaining four numbers are greater than 3 (i.e., in {4,...,10}), and 3 is included. Therefore, the count is correct.Wait, another way: Let's compute how many subsets of six numbers include 3 and have exactly one number less than 3.Number of subsets including 3: C(9,5)=126. Because if we include 3, we need to choose 5 more numbers from the remaining 9.But how many of these have exactly one number less than 3?Numbers less than 3 are 1 and 2, so two numbers. So, to have exactly one number less than 3, we choose 1 from these two, and the remaining 4 numbers from those greater than 3, which are 7 numbers. So, C(2,1)*C(7,4)=2*35=70. Therefore, of the 126 subsets that include 3, 70 have exactly one number less than 3. Therefore, the number of subsets where the second smallest is 3 is 70. Therefore, probability is 70/210=1/3. Same result.Alternatively, the total number of subsets that include 3 is C(9,5)=126. So, the probability that a random subset including 3 has exactly one number less than 3 is 70/126=5/9. But that's not the same as the original probability. Wait, but the original probability is over all subsets, not just those that include 3. Wait, but in our problem, we are considering all subsets. So, the total number is C(10,6)=210. The number of subsets that include 3 and have exactly one number less than 3 is 70. Therefore, the probability is 70/210=1/3.Alternatively, if we considered conditional probability given that 3 is included, then it's 70/126=5/9. But since the problem doesn't condition on 3 being included, the probability is 1/3. So, that's correct.Alternatively, someone might think that the second smallest can be 3 in more ways, but no. If 3 is the second smallest, then the structure is fixed: one smaller, 3, and four larger. Therefore, only 2*35=70 subsets.Alternatively, maybe the answer is different because the problem says "the second smallest number is 3", which might mean that there are exactly two numbers less than or equal to 3. Wait, no: if the second smallest is 3, then the smallest number is less than 3, and the second is 3. So, numbers less than 3: exactly one, and numbers equal to 3: exactly one, and numbers greater than 3: four. So, total numbers:1+1+4=6. Therefore, correct.But if someone thought that the second smallest is 3 means there are two numbers less than or equal to 3, that's different. But no, "second smallest" is 3, which is a specific value. So, 3 must be present, and exactly one number less than 3 must be present. Therefore, the count is as above.Alternatively, maybe the answer is different because of overlapping intervals? For example, if 3 is the second smallest, but maybe there's an overlap in selection. But no, the selection is without replacement, different integers, so each number is unique.Alternatively, let's compute another example. Suppose the set is {1,2,3,4,5}, select 3 numbers. What is the probability that the second smallest is 2?Using the same method:Numbers less than 2: only 1. So, need to choose 1 from 1 number (which is 1), and the remaining number from numbers greater than 2 (which are 3,4,5). So, number of favorable subsets: C(1,1)*C(3,1)=1*3=3.Total subsets: C(5,3)=10. So, probability is 3/10. Let's list them:Subsets of size 3:{1,2,3}, {1,2,4}, {1,2,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,3,4}, {2,3,5}, {2,4,5}, {3,4,5}.Second smallest number:- {1,2,3}: 2- {1,2,4}: 2- {1,2,5}: 2- {1,3,4}: 3- {1,3,5}: 3- {1,4,5}: 4- {2,3,4}: 3- {2,3,5}: 3- {2,4,5}: 4- {3,4,5}: 4So, subsets where second smallest is 2: 3. So, 3/10. Which matches the calculation. So, the formula works here as well. So, that gives me more confidence.Therefore, in the original problem, the answer should be C) 1/3. Therefore, unless there is an error in my reasoning that I can't see, the correct answer is C.But the option E is "none of the above," so maybe the answer is E? Wait, but C is listed. Wait, in the problem statement, the options are:(A) 1/60;(B) 1/6;(C) 1/3;(D) 1/2;(E) none of the above.Since 1/3 is option C, which is present. So, unless there is a miscalculation here, C is the answer. But maybe the original answer key says E? But the user hasn't provided any answer key, just the problem.Alternatively, perhaps the original problem was from a source where the answer is different, but according to my calculations, it's 1/3. Therefore, unless I made a mistake, answer C.Wait, just to make 100% sure, let me compute C(7,4). 7 choose 4 is 35? Yes: 7*6*5*4/(4*3*2*1)=7*6*5/(3*2*1)=7*5=35. So, yes. Then, 2*35=70. Total subsets 210. 70/210=1/3.Yes, so answer is C.Final Answerboxed{C}</think>To find the probability that the second smallest number is 3 when six different integers are randomly selected from the set ({1, 2, 3, ldots, 10}), we need to consider the following steps:1. Total number of ways to select six integers from the set: [ binom{10}{6} = frac{10!}{6! cdot 4!} = 210 ]2. Number of favorable outcomes (second smallest number is 3): - The smallest number must be either 1 or 2 (2 choices). - The second smallest number is 3 (1 choice). - The remaining four numbers must be selected from the numbers greater than 3 (i.e., ({4, 5, 6, 7, 8, 9, 10})). [ binom{7}{4} = frac{7!}{4! cdot 3!} = 35 ] - Therefore, the number of favorable outcomes is: [ 2 times binom{7}{4} = 2 times 35 = 70 ]3. Probability calculation: [ text{Probability} = frac{text{Number of favorable outcomes}}{text{Total number of outcomes}} = frac{70}{210} = frac{1}{3} ]Thus, the probability that the second smallest number is 3 is (boxed{C}).,<think>Okay, so I need to figure out the probability that when I select six different integers from the set {1,2,3,...,10}, the second smallest number in my selection is 3. The answer choices are given, and I need to pick the correct one. Let me start by understanding the problem step by step.First, the total number of ways to select six different integers from the set {1,2,3,...,10}. Since the set has 10 elements, the number of possible combinations is C(10,6). Let me compute that. Remember, the combination formula is C(n,k) = n! / (k! (n - k)!).So, C(10,6) = 10! / (6! * 4!) = (10*9*8*7)/(4*3*2*1) = 210. Okay, so there are 210 possible ways to choose six numbers. That's the denominator of my probability.Now, the numerator is the number of favorable outcomes where the second smallest number in the selection is 3. Let me think about what this condition implies.If the second smallest number is 3, then the smallest number must be less than 3, right? Because the second smallest is 3, the smallest has to be either 1 or 2. So, the two smallest numbers in the selection are 1 and 3, or 2 and 3. Wait, but if the second smallest is 3, the smallest must be either 1 or 2, and the remaining four numbers must be greater than 3. Because if there was a number between 1/2 and 3, then that would be the second smallest. So, the numbers selected have to include one number from {1,2}, then 3, and the remaining four numbers must be from {4,5,6,7,8,9,10}.Let me confirm that. If the second smallest is 3, then the smallest number has to be 1 or 2. Because if the smallest number were 3, then the second smallest would have to be at least 4, which contradicts. So, yes, the smallest number is either 1 or 2, and the next one is 3. Then the other four numbers have to be greater than 3. So, how many such combinations are there?First, how many ways to choose the smallest number: it's either 1 or 2. So that's 2 choices. Then, we must include 3. Then, the remaining four numbers have to be selected from the numbers greater than 3, which are {4,5,6,7,8,9,10}. There are 7 numbers there, so the number of ways to choose four numbers from them is C(7,4).Therefore, the number of favorable combinations is 2 * C(7,4). Let me compute that. C(7,4) = 7! / (4! * 3!) = (7*6*5)/(3*2*1) = 35. So, 2 * 35 = 70. Therefore, there are 70 favorable combinations.Therefore, the probability would be 70 / 210 = 1/3. Hmm, so that would be answer choice (C). But wait, let me double-check my reasoning to make sure I didn't miss anything.Wait, perhaps there's a different way to think about this. Let's consider that the second smallest is 3. That means that exactly one of the numbers selected is less than 3 (i.e., 1 or 2), and 3 is included. The other four numbers are greater than 3. So, the number of ways is C(2,1) * C(7,4). Which is exactly what I did before. So that's 2 * 35 = 70. Total combinations C(10,6) = 210. 70/210 simplifies to 1/3. So that's answer (C). But wait, let me check again because sometimes in probability problems with combinatorics, there can be a mistake in the setup.Alternatively, another approach: Let's fix the second smallest number as 3. Then, in the selected six numbers, we must have exactly one number less than 3 (which can be 1 or 2), 3, and four numbers greater than 3. So, again, that leads to C(2,1) * C(7,4) = 2 * 35 = 70. So, that's the same.But another way to think about it is: Suppose we select six numbers. The second smallest is 3. So, the ranks of the numbers are 1st, 2nd, ..., 6th. The second rank is 3. Therefore, the first rank (smallest) is either 1 or 2, the second is 3, and the remaining four are from 4 to 10. So, same conclusion.Alternatively, maybe using hypergeometric distribution or something else? Hmm, not sure. Let's see. The total number of ways is C(10,6). The favorable number is 2 * C(7,4). So, 70/210 is 1/3, so answer (C). The options are (A) 1/60, (B)1/6, (C)1/3, (D)1/2, (E) none.But wait, just to make sure, let me check if maybe there's an alternative interpretation where the answer is different. For example, maybe considering that the second smallest is 3, but perhaps other numbers could interfere? Wait, no. If the second smallest is 3, the smallest must be 1 or 2, and the other four numbers must be greater than 3, because if there was a number between the smallest and 3, then that number would be the second smallest. So, for example, if you have 1,2,3,..., then 2 would be the second smallest, right? So, in our case, since the second smallest is 3, the only numbers allowed below 3 are one number (either 1 or 2). So, that logic holds.Alternatively, maybe someone could think that the two smallest numbers are 1 and 3 or 2 and 3, but that is indeed the case, so the count is correct. Therefore, 2 choices for the smallest number, 1 choice for the second smallest (3), and the rest from the higher numbers.Alternatively, maybe using probability step by step. Let me think. Suppose we pick six numbers. The probability that the second smallest is 3. How else can we compute this?Alternatively, think of all possible positions. The second smallest being 3 implies that exactly one number is less than 3 and 3 is included. So, the number of ways is C(2,1)*C(7,4) as before. So, same count.Alternatively, maybe using the concept of order statistics. The second order statistic is 3. For discrete uniform variables, the probability that the second smallest is 3.But in combinatorics, it's similar. The formula for the probability that the k-th order statistic is x is C(x-1, k-1) * C(n - x, sample_size - k) divided by C(n, sample_size). Wait, let me check that.Wait, in general, if we want the probability that the k-th smallest number is x, then we need to choose k-1 numbers less than x, and sample_size - k numbers greater than x. So, in this case, k=2 (second smallest), x=3.Therefore, the number of favorable combinations is C(3-1, 2-1) * C(10 - 3, 6 - 2) = C(2,1)*C(7,4). Which is exactly what I did earlier. So, this formula gives the same result. Therefore, 2*35=70. Therefore, 70/210=1/3. So, answer (C). Therefore, the answer is (C). So, the answer is C.Wait, but let me check again if this formula is applicable here. The formula is from order statistics in combinatorics. The number of ways for the k-th order statistic to be x is C(x-1, k-1) * C(N - x, n - k), where N is the total number of elements in the population, and n is the sample size. In this case, N=10, n=6, k=2, x=3. So, C(3-1,2-1)*C(10-3,6-2)=C(2,1)*C(7,4)=2*35=70. So, same result. Therefore, 70/210=1/3. So, answer (C).But let me also check an example with smaller numbers to see if the formula holds. Suppose the set is {1,2,3,4}, and we pick 2 numbers. What's the probability the second smallest is 3? Let's compute. Total combinations: C(4,2)=6. The favorable combinations where second smallest is 3. The numbers must include 3 and one number less than 3 (1 or 2). So, there are 2 favorable combinations: {1,3}, {2,3}. So, probability is 2/6=1/3. Applying the formula: C(3-1,2-1)*C(4-3,2-2)=C(2,1)*C(1,0)=2*1=2. 2/6=1/3. Correct. So, formula works here.Similarly, if in this example, the second smallest is 2. Then, formula: C(2-1,2-1)*C(4-2,2-2)=C(1,1)*C(2,0)=1*1=1. So, probability 1/6. Indeed, only {1,2} is the combination where second smallest is 2. So, 1/6. Correct.Therefore, the formula seems to hold. Therefore, in our original problem, the calculation is correct.Therefore, the answer should be (C) 1/3.But wait, just to be thorough, let me compute it in another way. Suppose we fix that the second smallest is 3. Then, the possible selected numbers must include exactly one number from {1,2}, include 3, and four numbers from {4,...,10}. So, number of ways is C(2,1)*C(7,4)=2*35=70. Total number of ways C(10,6)=210. 70/210=1/3. So, same result.Alternatively, think of all possible selections of six numbers. How many of them have 3 as the second smallest? For that, the numbers must include one number from {1,2}, 3, and four numbers from {4-10}. So, same count.Alternatively, imagine arranging the numbers in order. The second one is 3, so the first is 1 or 2, and the rest are above 3. So, same logic.Therefore, I think that the answer is (C) 1/3, which is one of the options. Therefore, the correct answer should be (C). Therefore, the answer is boxed{C}.Final Answerboxed{C}</think>To determine the probability that the second smallest number is 3 when six different integers are randomly selected from the set ({1, 2, 3, ldots, 10}), we start by calculating the total number of ways to select six integers. This is given by the combination formula (C(10, 6)):[C(10, 6) = frac{10!}{6! cdot 4!} = 210]Next, we consider the favorable outcomes where the second smallest number is 3. For this to happen, the smallest number must be either 1 or 2, and the remaining four numbers must be greater than 3. The number of ways to choose the smallest number (1 or 2) is (C(2, 1)), and the number of ways to choose the remaining four numbers from ({4, 5, 6, 7, 8, 9, 10}) is (C(7, 4)):[C(2, 1) times C(7, 4) = 2 times frac{7!}{4! cdot 3!} = 2 times 35 = 70]Thus, the probability is the ratio of favorable outcomes to the total number of outcomes:[frac{70}{210} = frac{1}{3}]Therefore, the correct answer is (boxed{C}).]

🔑:frac{22}{C}22. To choose six different integers from {1,2, cdots, 10}, there are C_{10}^{mathrm{B}}=210 different ways. However, if the second smallest number is 3, then one number must be chosen from {1,2}, and four numbers must be chosen from {4,5, cdots, 10}. This results in C_{2}^{1} C_{7}^{4}=7 C ways. Therefore, the required probability is frac{70}{210}=frac{1}{3}.

🧠:[<think>Okay, so I need to solve this problem about the function f(x) = 2 sin(ωx + π/6). The problem has several parts, and there are four statements to check for correctness. Let me start by breaking down the problem step by step.First, the problem states that for all real numbers x, f(x - π/4) = -f(x + π/4). This is a functional equation that must hold for every x. Then, after some transformations (shifting right by π/6 and stretching x-coordinates to twice their length), we get the function g(x). The question is which of the statements A, B, C, D are correct.I need to find the correct statements. Let me start with the first part: determining ω from the given functional equation.Given f(x) = 2 sin(ωx + π/6), and the equation f(x - π/4) = -f(x + π/4) holds for all x ∈ R.Let me write out both sides of the equation:Left-hand side (LHS): f(x - π/4) = 2 sin[ω(x - π/4) + π/6] = 2 sin(ωx - ωπ/4 + π/6)Right-hand side (RHS): -f(x + π/4) = -2 sin[ω(x + π/4) + π/6] = -2 sin(ωx + ωπ/4 + π/6)Set LHS equal to RHS:2 sin(ωx - ωπ/4 + π/6) = -2 sin(ωx + ωπ/4 + π/6)Divide both sides by 2:sin(ωx - ωπ/4 + π/6) = -sin(ωx + ωπ/4 + π/6)Recall that sin(θ) = -sin(-θ), so we can write the RHS as sin(-ωx - ωπ/4 - π/6). Wait, but maybe another identity would help here. Let's recall that sin(A) = -sin(B) implies that A = -B + 2πk or A = π + B + 2πk for some integer k.But since this identity must hold for all x, the coefficients of x and the constants must match appropriately. Let me think.Let me denote θ = ωx + π/6. Then, LHS becomes sin(θ - ωπ/4) and RHS becomes -sin(θ + ωπ/4). So:sin(θ - ωπ/4) = -sin(θ + ωπ/4)Using the identity sin(A - B) = sin A cos B - cos A sin B and sin(A + B) = sin A cos B + cos A sin B.So:Left-hand side: sin θ cos(ωπ/4) - cos θ sin(ωπ/4)Right-hand side: -[sin θ cos(ωπ/4) + cos θ sin(ωπ/4)]Set them equal:sin θ cos(ωπ/4) - cos θ sin(ωπ/4) = -sin θ cos(ωπ/4) - cos θ sin(ωπ/4)Bring all terms to the left side:sin θ cos(ωπ/4) - cos θ sin(ωπ/4) + sin θ cos(ωπ/4) + cos θ sin(ωπ/4) = 0Simplify:[sin θ cos(ωπ/4) + sin θ cos(ωπ/4)] + [-cos θ sin(ωπ/4) + cos θ sin(ωπ/4)] = 0Which becomes:2 sin θ cos(ωπ/4) + 0 = 0Therefore:2 sin θ cos(ωπ/4) = 0This equation must hold for all x, which means that θ can be any real number as x varies. But sin θ is not identically zero, so the coefficient must be zero. Therefore:2 cos(ωπ/4) = 0Which implies:cos(ωπ/4) = 0The solutions to cos(φ) = 0 are φ = π/2 + kπ, where k is an integer. Hence:ωπ/4 = π/2 + kπSolve for ω:ω = (π/2 + kπ) * 4 / π = ( (1/2) + k ) * 4 = 2 + 4kSince ω ≥ 0, k can be 0, 1, 2, etc. But we need to check if all these values of ω satisfy the original equation.Wait, but the equation must hold for all x. Let's check with k=0: ω=2. Let's see if that works.If ω=2, then the original equation becomes:f(x - π/4) = 2 sin[2(x - π/4) + π/6] = 2 sin(2x - π/2 + π/6) = 2 sin(2x - π/3)On the other hand, -f(x + π/4) = -2 sin[2(x + π/4) + π/6] = -2 sin(2x + π/2 + π/6) = -2 sin(2x + 2π/3)So, check if sin(2x - π/3) = -sin(2x + 2π/3)Compute the RHS: -sin(2x + 2π/3) = -sin(2x + π - π/3) = -sin(2x + π - π/3) = -[sin(2x + π) cos(π/3) - cos(2x + π) sin(π/3)]Wait, maybe using another identity. Let's recall that sin(A + π) = -sin A. So sin(2x + 2π/3) = sin(2x + π - π/3) = sin(2x + π)cos(π/3) - cos(2x + π)sin(π/3)But sin(2x + π) = -sin(2x), cos(2x + π) = -cos(2x). Therefore:sin(2x + 2π/3) = -sin(2x) * (1/2) - (-cos(2x)) * (√3/2) = - (1/2) sin(2x) + (√3/2) cos(2x)But then -sin(2x + 2π/3) = (1/2) sin(2x) - (√3/2) cos(2x)Compare with sin(2x - π/3):sin(2x - π/3) = sin(2x) cos(π/3) - cos(2x) sin(π/3) = (1/2) sin(2x) - (√3/2) cos(2x)Therefore, yes, sin(2x - π/3) = -sin(2x + 2π/3). So ω=2 works. Similarly, check for k=1, ω=6.Wait, but if ω=6, let's see:f(x - π/4) = 2 sin[6(x - π/4) + π/6] = 2 sin(6x - 6π/4 + π/6) = 2 sin(6x - 3π/2 + π/6) = 2 sin(6x - 4π/3)-f(x + π/4) = -2 sin[6(x + π/4) + π/6] = -2 sin(6x + 6π/4 + π/6) = -2 sin(6x + 3π/2 + π/6) = -2 sin(6x + 10π/6) = -2 sin(6x + 5π/3)Check if sin(6x - 4π/3) = -sin(6x + 5π/3)But sin(6x - 4π/3) = sin(6x + 2π/3 - 2π) = sin(6x + 2π/3 - 2π) = sin(6x + 2π/3) because sine is 2π periodic.Similarly, -sin(6x + 5π/3) = -sin(6x + 5π/3 - 2π) = -sin(6x - π/3). So the equation becomes sin(6x + 2π/3) = -sin(6x - π/3)Is this true for all x? Let's check:Left side: sin(6x + 2π/3)Right side: -sin(6x - π/3) = -[sin6x cosπ/3 - cos6x sinπ/3] = -[ (1/2) sin6x - (√3/2) cos6x ] = -1/2 sin6x + √3/2 cos6xLeft side: sin6x cos2π/3 + cos6x sin2π/3 = sin6x (-1/2) + cos6x (√3/2) = -1/2 sin6x + √3/2 cos6xHence, yes, they are equal. Therefore, ω=6 also satisfies the equation. Similarly, higher k would work.But the problem states ω ≥ 0. But since the equation is for all x, and sine is periodic, the minimal positive ω is 2. However, the problem might require the minimal ω unless specified otherwise. Wait, but the problem doesn't specify uniqueness, just that ω ≥ 0. So there are infinitely many ω satisfying this, but we need to determine ω from the given condition. However, since the problem then proceeds to define transformations leading to g(x), and the answer options refer to properties of g(x). So perhaps multiple ω satisfy the condition, but maybe when transformed, the properties hold regardless? Wait, but that seems unlikely. Therefore, perhaps there is a unique ω. Let me check again.Wait, when we derived ω = 2 + 4k, k integer ≥ 0, but the problem is about a function f(x) which is then shifted and stretched. If ω is not unique, the problem is underdetermined. But given that the problem is from a test question, likely there is a unique ω. Let me check again the functional equation.Wait, in my earlier steps, I concluded that 2 cos(ωπ/4) sinθ = 0 for all θ, which requires cos(ωπ/4) = 0. Hence, ωπ/4 = π/2 + kπ, leading to ω = 2 + 4k. So for k=0, ω=2; k=1, ω=6; k=2, ω=10; etc. So the problem states ω ≥ 0. So unless there is an additional constraint, the answer might not be unique. However, the problem is about a function f(x) which is then transformed. If multiple ω satisfy the original condition, then the transformed function g(x) would depend on ω. However, the answer options might hold for all such ω or only for specific ones. Therefore, perhaps the problem expects the minimal positive ω, which is 2. Alternatively, maybe the answer options are valid regardless of ω. But I need to check.Wait, let's check if the answer options depend on ω. For example, statement B says the distance between adjacent axes of symmetry is π. The axes of symmetry for a sine function are separated by half the period. The period of the original function f(x) is 2π / ω. After the transformations, the period of g(x) would be different. Therefore, the answer options likely depend on ω, so unless ω is uniquely determined, we need to know its value. So perhaps the answer expects ω=2. Let me check again.In the problem statement: "Given the function f(x)=2 sin (ωx + π/6) where ω ≥ 0, and for ∀x∈R, f(x−π/4)=−f(x+π/4) holds." So the problem gives ω as part of the function definition, with the constraint that for all x, the equation holds. So this equation determines ω. But as we saw, there are multiple ω possible. However, in such problems, often the minimal positive ω is considered. But let me check for ω=2 and ω=6 whether the answer options hold.Alternatively, maybe the problem implicitly requires the minimal positive ω. Let me proceed with ω=2 first.Assuming ω=2.Now, the graph of f(x) is shifted to the right by π/6 units. Shifting right by π/6 means replacing x with x - π/6. So the shifted function is f(x - π/6) = 2 sin[2(x - π/6) + π/6] = 2 sin(2x - π/3 + π/6) = 2 sin(2x - π/6).Then, all the x-coordinates are stretched to twice their original length. Stretching x-coordinates by a factor of 2 means replacing x with x/2. So the transformed function g(x) is f((x/2) - π/6) = 2 sin[2((x/2) - π/6) + π/6] = 2 sin(x - π/3 + π/6) = 2 sin(x - π/6).Wait, let me verify that again. The original function is f(x) = 2 sin(2x + π/6). Shift right by π/6: replace x with x - π/6, so f(x - π/6) = 2 sin[2(x - π/6) + π/6] = 2 sin(2x - π/3 + π/6) = 2 sin(2x - π/6). Then, stretching x-coordinates to twice their original length. Stretching in the x-direction by a factor of A replaces x with x/A. So stretching by factor 2 would replace x with x/2. Therefore, the function becomes f((x/2) - π/6) = 2 sin[2((x/2) - π/6) + π/6] = 2 sin(x - π/3 + π/6) = 2 sin(x - π/6). So g(x) = 2 sin(x - π/6).Wait, so g(x) is 2 sin(x - π/6). Let me check this again step by step:Original function: f(x) = 2 sin(2x + π/6).Shift right by π/6: Replace x with x - π/6:f(x - π/6) = 2 sin[2(x - π/6) + π/6] = 2 sin(2x - π/3 + π/6) = 2 sin(2x - π/6).Then, stretching x-coordinates to twice their length: Replace x with x/2:g(x) = f((x/2) - π/6) = 2 sin[2*(x/2 - π/6) + π/6] = 2 sin(x - π/3 + π/6) = 2 sin(x - π/6). Correct.So g(x) = 2 sin(x - π/6). Now, let's check the statements.A: g(π/6 - x) + g(x + π/6) = 0Compute g(π/6 - x):g(π/6 - x) = 2 sin( (π/6 - x) - π/6 ) = 2 sin(-x) = -2 sin xCompute g(x + π/6):g(x + π/6) = 2 sin( (x + π/6) - π/6 ) = 2 sin xThus, g(π/6 - x) + g(x + π/6) = (-2 sin x) + (2 sin x) = 0. So statement A is correct.B: The distance between adjacent axes of symmetry of the function g(x) is π.The function g(x) = 2 sin(x - π/6). The general sine function y = A sin(Bx + C) has period 2π/B. For g(x), B = 1, so the period is 2π. The axes of symmetry (lines of symmetry) for a sine function are located at the midpoints between consecutive peaks and troughs. For the standard sine function y = sin x, the axes of symmetry are at x = π/2 + kπ, which are spaced by π. Since g(x) is a sine function with period 2π, shifted horizontally, its axes of symmetry should also be spaced by π. Therefore, statement B is correct.C: The function g(x + 2π/3) is an odd function.First, let's find g(x + 2π/3):g(x + 2π/3) = 2 sin( (x + 2π/3) - π/6 ) = 2 sin(x + 2π/3 - π/6) = 2 sin(x + π/2)Simplify: sin(x + π/2) = cos x. So g(x + 2π/3) = 2 cos x. Is 2 cos x an odd function? An odd function satisfies h(-x) = -h(x). Let's check:2 cos(-x) = 2 cos x, which is not equal to -2 cos x unless cos x = 0. Since this is not true for all x, 2 cos x is not an odd function. Therefore, statement C is incorrect.Wait, but let me double-check. Maybe there was a miscalculation.Wait, g(x + 2π/3) = 2 sin(x + 2π/3 - π/6) = 2 sin(x + (4π/6 - π/6)) = 2 sin(x + 3π/6) = 2 sin(x + π/2) = 2 cos x. Correct. So 2 cos x is indeed an even function, not odd. Therefore, statement C is incorrect.D: The function g(x) is monotonically increasing on the interval [π/6, π/3].First, find the derivative of g(x):g'(x) = 2 cos(x - π/6)We need to check if g'(x) ≥ 0 on [π/6, π/3].Compute x - π/6 when x ∈ [π/6, π/3]:At x = π/6: π/6 - π/6 = 0At x = π/3: π/3 - π/6 = π/6So x - π/6 ranges from 0 to π/6. Therefore, cos(x - π/6) ranges from cos(0) = 1 to cos(π/6) = √3/2 ≈ 0.866. Both positive. Hence, g'(x) is positive on [π/6, π/3], so g(x) is monotonically increasing there. Therefore, statement D is correct.Wait, but let me check again. If x is in [π/6, π/3], then x - π/6 is in [0, π/6]. Since cosine is positive and decreasing in [0, π/2], so yes, cos(x - π/6) is positive and decreasing, but still positive. Therefore, derivative is positive, so function is increasing. Hence, statement D is correct.But wait, according to the options, A, B, D are correct. But let me verify again.Wait, but earlier, when I assumed ω=2, but the problem might have multiple ω. Wait, but if ω is 6, then the transformations would be different. Wait, let's check if ω could be 6. Let's do the same steps with ω=6.If ω=6, then the original function is f(x) = 2 sin(6x + π/6). Then, shifting right by π/6 gives f(x - π/6) = 2 sin[6(x - π/6) + π/6] = 2 sin(6x - π + π/6) = 2 sin(6x - 5π/6). Then stretching x-coordinates by a factor of 2 gives g(x) = f((x/2) - π/6) = 2 sin[6*(x/2 - π/6) + π/6] = 2 sin(3x - π + π/6) = 2 sin(3x - 5π/6).So g(x) = 2 sin(3x - 5π/6). Now check the statements:A: g(π/6 - x) + g(x + π/6) = 0Compute g(π/6 - x):g(π/6 - x) = 2 sin(3(π/6 - x) - 5π/6) = 2 sin(π/2 - 3x - 5π/6) = 2 sin(-3x - π/3) = -2 sin(3x + π/3)Compute g(x + π/6):g(x + π/6) = 2 sin(3(x + π/6) - 5π/6) = 2 sin(3x + π/2 - 5π/6) = 2 sin(3x - π/3)Now, add them together: -2 sin(3x + π/3) + 2 sin(3x - π/3). Let's compute this:Use the identity sin(A ± B) = sin A cos B ± cos A sin B.sin(3x + π/3) = sin3x cosπ/3 + cos3x sinπ/3 = (1/2) sin3x + (√3/2) cos3xsin(3x - π/3) = sin3x cosπ/3 - cos3x sinπ/3 = (1/2) sin3x - (√3/2) cos3xThen:-2 [ (1/2 sin3x + √3/2 cos3x ) ] + 2 [ (1/2 sin3x - √3/2 cos3x ) ]= -sin3x - √3 cos3x + sin3x - √3 cos3x= (-sin3x + sin3x) + (-√3 cos3x - √3 cos3x )= 0 - 2√3 cos3x= -2√3 cos3xWhich is not identically zero. Therefore, statement A would not hold if ω=6. Therefore, statement A's correctness depends on the value of ω. But earlier, we saw that multiple ω satisfy the original condition. However, the answer options must be correct based on the given problem. Therefore, this suggests that ω must be 2, the minimal positive solution. Otherwise, some statements may not hold. But the problem didn't specify to take the minimal ω. Wait, but maybe the functional equation f(x - π/4) = -f(x + π/4) for all x uniquely determines ω. Wait, but we saw that for ω=2 and ω=6, it holds. Let's check for another value, say ω=10. Let's see:For ω=10, f(x) = 2 sin(10x + π/6). Then f(x - π/4) = 2 sin[10(x - π/4) + π/6] = 2 sin(10x - 10π/4 + π/6) = 2 sin(10x - 5π/2 + π/6) = 2 sin(10x - 15π/6 + π/6) = 2 sin(10x - 14π/6) = 2 sin(10x - 7π/3)-f(x + π/4) = -2 sin[10(x + π/4) + π/6] = -2 sin(10x + 10π/4 + π/6) = -2 sin(10x + 5π/2 + π/6) = -2 sin(10x + 15π/6 + π/6) = -2 sin(10x + 16π/6) = -2 sin(10x + 8π/3)Now, check if sin(10x - 7π/3) = -sin(10x + 8π/3)Compute sin(10x - 7π/3 + 2π) = sin(10x - 7π/3 + 6π/3) = sin(10x - π/3)Similarly, sin(10x + 8π/3 - 2π) = sin(10x + 8π/3 - 6π/3) = sin(10x + 2π/3)So the equation becomes sin(10x - π/3) = -sin(10x + 2π/3)But sin(10x - π/3) = -sin(10x + 2π/3) ?From previous steps, similar to ω=2, let's see:sin(10x - π/3) = -sin(10x + 2π/3)Use the identity sin A = -sin B => A = -B + 2kπ or A = π + B + 2kπ.But this must hold for all x. Therefore, the coefficients of x must match, and the constants must satisfy.So, 10x - π/3 = - (10x + 2π/3) + 2kπ => 10x - π/3 = -10x - 2π/3 + 2kπ => 20x = -π/3 + 2kπ. This can't hold for all x.Alternatively, 10x - π/3 = π + 10x + 2π/3 + 2kπ => -π/3 = π + 2π/3 + 2kπ => -π/3 - π - 2π/3 = 2kπ => - (π + π) = 2kπ => -2π = 2kπ => k = -1. So possible, but it's for all x. So this would require that the equation holds for all x, but the left-hand side and right-hand side differ by a constant. However, when k=-1, we have:10x - π/3 = π + 10x + 2π/3 - 2π => 10x - π/3 = 10x + π + 2π/3 - 2π => 10x - π/3 = 10x - π/3. So this holds for all x. Therefore, yes, the equation is satisfied. Therefore, ω=10 also works.Therefore, the functional equation holds for ω=2,6,10,... So the problem as stated has multiple possible ω. But when we perform the transformations, the resulting g(x) will depend on ω, leading to different functions. However, the answer options might hold for all such ω or only for ω=2. But since the problem didn't specify which ω to use, we need to check whether the answer options hold for all possible ω. However, the problem is presented as a multiple-choice question where typically only one ω is valid. Alternatively, maybe the answer options are independent of ω, which seems unlikely.Alternatively, perhaps the problem has a unique solution for ω. Wait, but when solving the equation cos(ωπ/4)=0, we get ω=2+4k. However, when we stretch the x-coordinates by 2, the resulting function's period is (2π/ω)*2=4π/ω. If ω=2+4k, then the period is 4π/(2+4k)=2π/(1+2k). Then the distance between adjacent axes of symmetry would be half the period, which is π/(1+2k). For example, if k=0, ω=2: period is 2π/2=π, stretching by 2 makes the period 2π, so axes of symmetry every π. If k=1, ω=6: period is 2π/6=π/3, stretching by 2 gives 2*(π/3)=2π/3. Wait, no. Wait, original period is 2π/ω. After stretching x-coordinates by 2, the period becomes 2*(2π)/ω=4π/ω. For ω=2, period is 4π/2=2π. For ω=6, period is 4π/6=2π/3. Then the axes of symmetry would be every half-period, so for ω=2, half-period is π, distance π. For ω=6, half-period is π/3, so distance π/3. Therefore, statement B would only hold if ω=2. If ω=6, the distance is π/3, which is different.But the problem states that the original function satisfies f(x - π/4) = -f(x + π/4) for all x. So ω can be 2,6,10,... However, depending on ω, the answer options might change. Therefore, the problem is ill-posed unless there is a unique ω. But given that the problem is from an exam, likely ω=2 is the expected answer. However, this contradicts the multiple possible solutions. Therefore, perhaps the correct approach is to realize that when the problem says "the graph of the function f(x)=2 sin(ωx + π/6) is shifted...", it implies that the given functional equation f(x - π/4) = -f(x + π/4) must hold for that specific ω. Therefore, the value of ω is uniquely determined, which in our case has multiple solutions, but the problem likely intended ω=2. Alternatively, perhaps the problem assumes the minimal positive ω. Since the answer options A,B,D are correct for ω=2, but if we take another ω, like ω=6, then those statements may not hold. Therefore, in the problem's context, the answer is A,B,D. But in our previous calculation for ω=2, statements A,B,D are correct. For ω=6, A is incorrect, B is incorrect, and D may vary. Thus, if the problem expects a unique answer, ω=2 is the minimal positive solution. Hence, the correct answers are A,B,D. However, in the problem statement, they might have intended ω=2, leading to correct A,B,D. Alternatively, maybe even for general ω, but given the transformations, maybe not. But in our initial steps, assuming ω=2, we derived that A,B,D are correct. Therefore, the answer is A,B,D. But looking at the original problem statement, the options are A,B,C,D, and the user asks which are correct. But in our analysis for ω=2, A,B,D are correct. However, in the original problem statement, maybe the answer is A,B,C,D, but no. Wait, let me check again.Wait, if ω=2, then:C: The function g(x + 2π/3) is an odd function.We found that g(x + 2π/3) = 2 cos x, which is even, not odd. Hence, C is incorrect.But if ω=6, let's check statement C.For ω=6, g(x) = 2 sin(3x - 5π/6). Then, g(x + 2π/3) = 2 sin[3(x + 2π/3) -5π/6] = 2 sin(3x + 2π -5π/6) = 2 sin(3x + 7π/6).Is this an odd function? An odd function satisfies h(-x) = -h(x). Let's check:h(-x) = 2 sin(-3x + 7π/6) = 2 sin[-(3x -7π/6)] = -2 sin(3x -7π/6)Compare to -h(x) = -2 sin(3x +7π/6)So, we need to check if sin(3x -7π/6) = sin(3x +7π/6)But sin(3x -7π/6) = sin(3x +7π/6 - 14π/6) = sin(3x -7π/6). Wait, not helpful. Alternatively, compute:sin(3x -7π/6) = sin(3x - π - π/6) = -sin(3x - π/6)Similarly, sin(3x +7π/6) = sin(3x + π + π/6) = -sin(3x + π/6)Thus, h(-x) = -2 sin(3x -7π/6) = -2*(-sin(3x - π/6)) = 2 sin(3x - π/6)Compare to -h(x) = -2 sin(3x +7π/6) = -2*(-sin(3x + π/6)) = 2 sin(3x + π/6)These are not equal unless sin(3x - π/6) = sin(3x + π/6), which is not true for all x. Therefore, statement C is incorrect for ω=6.Therefore, regardless of ω=2 or 6, statement C is incorrect. Therefore, statement C is incorrect.Hence, the correct answers are A,B,D.Final Answerboxed{ABD}</think>Given the function ( f(x) = 2sin(omega x + frac{pi}{6}) ) where ( omega geq 0 ), and the condition ( f(x - frac{pi}{4}) = -f(x + frac{pi}{4}) ) holds for all ( x in mathbb{R} ). We need to determine the correct statements about the transformed function ( g(x) ).1. Determine ( omega ): - The given functional equation leads to ( sin(omega x - frac{omega pi}{4} + frac{pi}{6}) = -sin(omega x + frac{omega pi}{4} + frac{pi}{6}) ). - This simplifies to ( cos(frac{omega pi}{4}) = 0 ), leading to ( frac{omega pi}{4} = frac{pi}{2} + kpi ). - Solving for ( omega ), we get ( omega = 2 + 4k ) for non-negative integers ( k ). The minimal positive solution is ( omega = 2 ).2. Transformations to obtain ( g(x) ): - Shift ( f(x) ) right by ( frac{pi}{6} ): ( f(x - frac{pi}{6}) = 2sin(2(x - frac{pi}{6}) + frac{pi}{6}) = 2sin(2x - frac{pi}{6}) ). - Stretch x-coordinates by a factor of 2: Replace ( x ) with ( frac{x}{2} ), resulting in ( g(x) = 2sin(x - frac{pi}{6}) ).3. Check statements: - A: ( g(frac{pi}{6} - x) + g(x + frac{pi}{6}) = 0 ) - Compute ( g(frac{pi}{6} - x) = -2sin(x) ) and ( g(x + frac{pi}{6}) = 2sin(x) ), leading to ( -2sin(x) + 2sin(x) = 0 ). Correct. - B: The distance between adjacent axes of symmetry is ( pi ). - The period of ( g(x) ) is ( 2pi ), so axes of symmetry are spaced by ( pi ). Correct. - C: ( g(x + frac{2pi}{3}) ) is an odd function. - ( g(x + frac{2pi}{3}) = 2cos(x) ), which is even, not odd. Incorrect. - D: ( g(x) ) is monotonically increasing on ( [frac{pi}{6}, frac{pi}{3}] ). - Derivative ( g'(x) = 2cos(x - frac{pi}{6}) ), which is positive on ( [frac{pi}{6}, frac{pi}{3}] ). Correct.Thus, the correct statements are A, B, and D.boxed{ABD},<think>Okay, let's try to tackle this problem step by step. First, I need to understand what's being asked here. We have a function f(x) = 2 sin(ωx + π/6), and there's a condition given for all real numbers x: f(x - π/4) = -f(x + π/4). Then, after some transformations—shifting right by π/6 and stretching x-coordinates to twice their length—we get the function g(x). The question is about which of the statements A to D are correct.Let me start by figuring out what ω is. The condition given is f(x - π/4) = -f(x + π/4) for all x. Let's write out both sides using the function definition.First, f(x - π/4) = 2 sin[ω(x - π/4) + π/6] = 2 sin[ωx - ωπ/4 + π/6].On the other hand, -f(x + π/4) = -2 sin[ω(x + π/4) + π/6] = -2 sin[ωx + ωπ/4 + π/6].Setting these equal for all x:2 sin[ωx - ωπ/4 + π/6] = -2 sin[ωx + ωπ/4 + π/6]Divide both sides by 2:sin[ωx - ωπ/4 + π/6] = -sin[ωx + ωπ/4 + π/6]Hmm, sin(A) = -sin(B) implies that either A = -B + 2πk or A = π + B + 2πk for some integer k. Since this must hold for all x, the coefficients of x and the constants must match appropriately. Let's set A = ωx - ωπ/4 + π/6 and B = ωx + ωπ/4 + π/6. Then sin(A) = -sin(B).So, let's consider the two cases:1. A = -B + 2πk2. A = π + B + 2πkLet's check case 1 first:A = -B + 2πkSubstituting A and B:ωx - ωπ/4 + π/6 = - (ωx + ωπ/4 + π/6) + 2πkSimplify the right side:-ωx - ωπ/4 - π/6 + 2πkNow, equate coefficients of x and constants:Coefficient of x: ω = -ω ⇒ ω = 0. But ω ≥ 0, so ω = 0 is possible, but let's check constants:Constants: -ωπ/4 + π/6 = -ωπ/4 - π/6 + 2πkBring similar terms to one side:-ωπ/4 + π/6 + ωπ/4 + π/6 = 2πkSimplify:(π/6 + π/6) = 2πk ⇒ π/3 = 2πk ⇒ k = 1/6But k must be an integer. 1/6 is not an integer, so this case is invalid. So case 1 doesn't work.Now check case 2:A = π + B + 2πkSubstituting A and B:ωx - ωπ/4 + π/6 = π + ωx + ωπ/4 + π/6 + 2πkSubtract ωx and π/6 from both sides:-ωπ/4 = π + ωπ/4 + 2πkBring terms with ω to one side:-ωπ/4 - ωπ/4 = π + 2πkCombine like terms:-ωπ/2 = π(1 + 2k)Multiply both sides by -1:ωπ/2 = -π(1 + 2k)Divide both sides by π:ω/2 = -(1 + 2k)Multiply both sides by 2:ω = -2(1 + 2k)But ω ≥ 0, so -2(1 + 2k) ≥ 0 ⇒ 1 + 2k ≤ 0 ⇒ 2k ≤ -1 ⇒ k ≤ -0.5Since k is an integer, possible k values are k = -1, -2, ...Let's take k = -1:ω = -2(1 + 2*(-1)) = -2(1 - 2) = -2(-1) = 2If k = -2:ω = -2(1 + 2*(-2)) = -2(1 - 4) = -2(-3) = 6But we need to check which of these satisfy the original equation for all x. Since the equation must hold for all x, the value of ω should be such that the equality holds regardless of x. Let's check ω = 2.Let’s verify with ω = 2:Compute A and B:A = 2x - 2*(π/4) + π/6 = 2x - π/2 + π/6 = 2x - π/3B = 2x + 2*(π/4) + π/6 = 2x + π/2 + π/6 = 2x + 2π/3Check sin(A) = -sin(B):sin(2x - π/3) = -sin(2x + 2π/3)Let’s compute the right side:-sin(2x + 2π/3) = -sin(2x + π - π/3) = -sin(2x + π - π/3)Wait, maybe use angle addition formula:sin(2x + 2π/3) = sin(2x) cos(2π/3) + cos(2x) sin(2π/3) = sin(2x)*(-1/2) + cos(2x)*(√3/2)So -sin(2x + 2π/3) = (1/2) sin(2x) - (√3/2) cos(2x)On the left side, sin(2x - π/3) = sin(2x) cos(π/3) - cos(2x) sin(π/3) = sin(2x)*(1/2) - cos(2x)*(√3/2)Which is equal to the right side. So yes, sin(2x - π/3) = -sin(2x + 2π/3). Therefore, ω = 2 works.If we take ω = 6, let's check:A = 6x - 6*(π/4) + π/6 = 6x - 3π/2 + π/6 = 6x - 9π/6 + π/6 = 6x - 8π/6 = 6x - 4π/3B = 6x + 6*(π/4) + π/6 = 6x + 3π/2 + π/6 = 6x + 9π/6 + π/6 = 6x + 10π/6 = 6x + 5π/3Then sin(A) = sin(6x - 4π/3), -sin(B) = -sin(6x + 5π/3)Check if sin(6x - 4π/3) = -sin(6x + 5π/3)Note that -sin(6x + 5π/3) = sin(-6x -5π/3 + π) = sin(-6x -5π/3 + 3π/3) = sin(-6x -2π/3) = sin(-6x -2π/3)But sin(-θ) = -sinθ, so sin(-6x -2π/3) = -sin(6x + 2π/3)Wait, maybe another approach. Let's compute sin(6x -4π/3) and -sin(6x +5π/3)sin(6x -4π/3) = sin(6x - π - π/3) = sin(6x - π - π/3) = sin(6x - 4π/3) which is equal to -sin(6x -4π/3 + π) = -sin(6x - π/3)Wait, maybe not. Let's use angle addition:sin(6x -4π/3) = sin(6x) cos(4π/3) - cos(6x) sin(4π/3) = sin(6x)*(-1/2) - cos(6x)*(-√3/2) = -1/2 sin(6x) + (√3/2) cos(6x)-sin(6x +5π/3) = - [sin(6x) cos(5π/3) + cos(6x) sin(5π/3)] = - [sin(6x)*(1/2) + cos(6x)*(-√3/2)] = -1/2 sin(6x) + (√3/2) cos(6x)So yes, they are equal. Therefore, ω = 6 also works.But wait, does this hold for all x? For ω = 6, does the original condition hold?Wait, but if ω = 6, then the equation sin(6x -4π/3) = -sin(6x +5π/3) holds as we saw. So both ω = 2 and ω = 6 satisfy the condition. But we need to check which is the minimal ω. However, since the problem states ω ≥ 0, but doesn't specify uniqueness, but given that the problem is about the function being shifted and stretched, I think we need to find all possible ω. However, since when k = -1, we get ω = 2, and k = -2 gives ω = 6, k = -3 gives ω = 10, etc. So there are infinitely many ω. But in the problem statement, it's given that the function f(x) = 2 sin(ωx + π/6). Maybe the problem expects the minimal positive ω? Because otherwise, there are multiple possible ω, which would make the subsequent transformations ambiguous. Hmm, but maybe the condition holds for all ω of the form ω = 2 + 4k, but need to check.Wait, let's check the original equation again. If the equation f(x - π/4) = -f(x + π/4) holds for all x, then the function f must satisfy this relation. Let's recall that for a sinusoidal function, such a condition would imply a certain relationship between the frequency and the phase shift.Alternatively, perhaps we can approach this problem by considering that f is a sinusoidal function, and such an identity must hold for all x. Let's think about the period of f(x). The period is 2π/ω. The condition given is similar to a shift leading to a negation, which might imply that shifting by π/4 in x leads to a negation. So the shift of π/2 (since shifting left and right by π/4 and the total shift is π/2) would correspond to a half-period shift. Because shifting a sine function by half its period negates it. So if T is the period, then shifting by T/2 would give sin(x + T/2) = -sin(x). Therefore, if shifting by π/2 (the total shift here is π/2, from x - π/4 to x + π/4, the difference is π/2), then T/2 = π/2 ⇒ T = π. Therefore, 2π/ω = π ⇒ ω = 2.Ah, that's another way to think about it. The total shift is π/2, which should be half the period. Therefore, T = π, so ω = 2π / T = 2π / π = 2. So ω must be 2. Therefore, the minimal ω is 2. Maybe the problem expects ω = 2. The other solutions (like ω = 6) would correspond to higher frequencies where shifting by π/2 is equivalent to shifting by multiple half-periods. However, in that case, for ω = 6, the period is 2π/6 = π/3, so half-period is π/6. Shifting by π/2 is equivalent to shifting by 3 half-periods (since π/2 = 3*(π/6)), which would also result in negation. So actually, ω can be 2, 6, 10, etc., but since the problem states "for all x ∈ R", maybe all such ω are possible. However, typically in such problems, unless specified, the minimal positive solution is considered. But maybe we need to confirm.Wait, but let's test with ω = 6. If ω = 6, then shifting x by π/2 (i.e., from x - π/4 to x + π/4) would correspond to a phase shift of 6*(π/2) = 3π. So sin(θ + 3π) = -sinθ, so f(x + π/2) = 2 sin(6x + 3π + π/6) = 2 sin(6x + 19π/6) = 2 sin(6x + 19π/6 - 2π) = 2 sin(6x + 7π/6) = 2 sin(6x + π + π/6) = -2 sin(6x + π/6) = -f(x). So indeed, f(x + π/2) = -f(x). Therefore, shifting by π/2 gives the negative. Therefore, f(x - π/4) = -f(x + π/4) is equivalent to shifting left by π/4 gives the negative of shifting right by π/4. Which would hold if shifting by π/2 in total gives a negation, which as we saw, is true for ω such that ω*(π/2) is an odd multiple of π, i.e., ω*(π/2) = (2k + 1)π ⇒ ω = 2*(2k + 1), where k is a non-negative integer. So ω = 2, 6, 10, etc.But in the problem statement, it says "ω ≥ 0". So possible ω are 2, 6, 10, etc. However, without additional constraints, the problem is underdetermined. But given that the problem is likely expecting a single answer, maybe ω is 2. Alternatively, perhaps all such ω are acceptable, but the subsequent transformations might depend on ω. However, since the problem mentions shifting and stretching, which are transformations that are independent of ω, but actually, the stretching is applied after shifting, so maybe the final function g(x) will depend on ω, so we need to know ω to check the options.Wait, but maybe there is a unique solution. Let me check again. When we set up the equation for the identity to hold for all x, we found that ω = -2(1 + 2k). Since ω must be non-negative, k must be ≤ -1. But k is an integer, so k = -1, -2, ..., leading to ω = 2, 6, 10, etc. But perhaps in the problem, they want the minimal positive ω, which is 2. Let me check the answer options. The options are about properties of g(x), which is obtained by shifting f(x) right by π/6 and then stretching x-coordinates to twice their length. Let's figure out what g(x) is.First, the original function f(x) = 2 sin(ωx + π/6). Shift right by π/6: replacing x with x - π/6, so we get f(x - π/6) = 2 sin[ω(x - π/6) + π/6] = 2 sin(ωx - ωπ/6 + π/6). Then, stretching x-coordinates to twice their length means replacing x with x/2. So g(x) = 2 sin(ω(x/2) - ωπ/6 + π/6) = 2 sin[(ω/2)x - ωπ/6 + π/6].But since ω can be 2, 6, 10, etc., let's see. Suppose ω = 2, then g(x) = 2 sin[(2/2)x - 2π/6 + π/6] = 2 sin(x - π/3 + π/6) = 2 sin(x - π/6).If ω = 6, then g(x) = 2 sin[(6/2)x - 6π/6 + π/6] = 2 sin(3x - π + π/6) = 2 sin(3x - 5π/6).But the problem is that without knowing ω, we can't determine g(x). However, the problem statement probably expects a unique answer, so maybe we need to take ω = 2, the minimal positive solution. Alternatively, perhaps the given condition only holds for ω = 2. Let me check again.If we assume ω = 2, then f(x) = 2 sin(2x + π/6). Then shifting right by π/6 gives 2 sin[2(x - π/6) + π/6] = 2 sin(2x - π/3 + π/6) = 2 sin(2x - π/6). Then stretching x-coordinates to twice their length replaces x with x/2, so g(x) = 2 sin(2*(x/2) - π/6) = 2 sin(x - π/6). So g(x) = 2 sin(x - π/6).But if ω = 6, then f(x) = 2 sin(6x + π/6). Shifting right by π/6: 2 sin[6(x - π/6) + π/6] = 2 sin(6x - π + π/6) = 2 sin(6x - 5π/6). Then stretching x-coordinates by 2: replace x with x/2, so g(x) = 2 sin(6*(x/2) - 5π/6) = 2 sin(3x - 5π/6).So depending on ω, g(x) is different. Therefore, we need to know the correct ω. The problem states that "the graph of the function f(x)=2sin(ωx+π/6) is shifted..." but the value of ω is determined by the given condition. Since multiple ω satisfy the condition, but the problem probably expects the minimal ω, which is 2.Alternatively, maybe the problem is designed such that all possible ω lead to the same conclusions for options A to D. Let's check.Wait, but if ω can be multiple values, then the period of g(x) would vary, affecting options B, C, D. So it's necessary to determine ω. Therefore, the problem must have a unique ω. Let me verify again.Earlier, when solving for ω, we obtained ω = -2(1 + 2k). To have ω ≥ 0, 1 + 2k ≤ 0 ⇒ k ≤ -0.5. Since k is an integer, k = -1, -2, etc. So ω = 2, 6, 10, etc. However, if we take ω = 2, then the function f(x) = 2 sin(2x + π/6). Let's check if this satisfies the given condition f(x - π/4) = -f(x + π/4).Compute f(x - π/4) = 2 sin[2(x - π/4) + π/6] = 2 sin(2x - π/2 + π/6) = 2 sin(2x - π/3).Compute -f(x + π/4) = -2 sin[2(x + π/4) + π/6] = -2 sin(2x + π/2 + π/6) = -2 sin(2x + 2π/3).Now, sin(2x - π/3) vs -sin(2x + 2π/3). Let's compute -sin(2x + 2π/3):-sin(2x + 2π/3) = sin(2x + 2π/3 + π) = sin(2x + 5π/3). Wait, no, sin(θ + π) = -sinθ, so actually:-sin(θ) = sin(θ + π). So -sin(2x + 2π/3) = sin(2x + 2π/3 + π) = sin(2x + 5π/3). But sin(2x - π/3) is equal to sin(2x + 5π/3 - 2π) = sin(2x - π/3). So sin(2x - π/3) = sin(2x + 5π/3). Are these equal?Wait, sin(A) = sin(B) implies A = B + 2πk or A = π - B + 2πk. But for this to hold for all x, we need 2x - π/3 = 2x + 5π/3 + 2πk ⇒ -π/3 = 5π/3 + 2πk ⇒ -2π = 2πk ⇒ k = -1. Then for k = -1, it's valid. So sin(2x - π/3) = sin(2x + 5π/3 - 2π) = sin(2x - π/3). So yes, it's true. Alternatively, using angle identities:sin(2x - π/3) = sin(2x) cos(π/3) - cos(2x) sin(π/3) = (1/2) sin(2x) - (√3/2) cos(2x)-sin(2x + 2π/3) = -[sin(2x) cos(2π/3) + cos(2x) sin(2π/3)] = -[sin(2x)(-1/2) + cos(2x)(√3/2)] = (1/2) sin(2x) - (√3/2) cos(2x)Which matches. Therefore, ω = 2 is valid. Similarly, ω = 6 would also satisfy, but with higher frequency. However, since the problem asks for correctness of statements A to D, which may or may not hold depending on ω. Therefore, I need to verify if the answer options hold regardless of ω or only for specific ω.But given that the problem likely expects a unique solution, and given that shifting by π/2 corresponds to half the period, which leads to ω = 2, I think we should proceed with ω = 2. So let's take ω = 2.So, with ω = 2, the original function is f(x) = 2 sin(2x + π/6). Then shifting to the right by π/6 units: replacing x with x - π/6, we get:f(x - π/6) = 2 sin[2(x - π/6) + π/6] = 2 sin(2x - π/3 + π/6) = 2 sin(2x - π/6).Then stretching the x-coordinates to twice their length means replacing x with x/2, so:g(x) = 2 sin(2*(x/2) - π/6) = 2 sin(x - π/6).So g(x) = 2 sin(x - π/6). Now, let's check the options A to D.A: g(π/6 - x) + g(x + π/6) = 0.Compute g(π/6 - x) = 2 sin(π/6 - x - π/6) = 2 sin(-x) = -2 sinx.g(x + π/6) = 2 sin((x + π/6) - π/6) = 2 sinx.So g(π/6 - x) + g(x + π/6) = -2 sinx + 2 sinx = 0. Therefore, A is correct.B: The distance between adjacent axes of symmetry of the function g(x) is π.Axes of symmetry for a sine function are the vertical lines where it reaches its maxima/minima or midpoints. Wait, for a sine function, the "axes of symmetry" might refer to the lines of symmetry which are the vertical lines passing through the midpoints between extrema. Alternatively, for a standard sine function y = sinx, the lines x = π/2 + kπ are axes of symmetry (they are lines of symmetry for the peaks and troughs). Wait, actually, for the standard sine function, it's an odd function, so it's symmetric about the origin, but individual peaks and troughs have vertical lines of symmetry. For example, the maximum at x = π/2 is symmetric about x = π/2, and similarly for minima at x = 3π/2, etc.For the function g(x) = 2 sin(x - π/6), the standard sine function shifted right by π/6. Its maxima are at x - π/6 = π/2 + 2πk ⇒ x = 2π/3 + 2πk, and minima at x - π/6 = 3π/2 + 2πk ⇒ x = 10π/6 + 2πk = 5π/3 + 2πk. The axes of symmetry would be the midpoints between these extrema. The distance between adjacent maxima and minima is π, so the midpoints between a maximum and the next minimum would be at (2π/3 + 5π/3)/2 = (7π/3)/2 = 7π/6, but that seems complicated. Wait, actually, between a maximum at x = 2π/3 and the next maximum at x = 2π/3 + 2π, the midpoint is at x = 2π/3 + π. Similarly, between a minimum at x = 5π/3 and the next minimum at x = 5π/3 + 2π, the midpoint is 5π/3 + π. But axes of symmetry are spaced every π units. For a standard sine function with period 2π, the axes of symmetry (the lines x = π/2 + kπ) are spaced π apart. Similarly, for g(x) = 2 sin(x - π/6), which is a phase-shifted sine function with period 2π, the axes of symmetry should be at x = π/2 + π/6 + kπ = 2π/3 + kπ, and the distance between these is π. Therefore, statement B is correct.C: The function g(x + 2π/3) is an odd function.Compute g(x + 2π/3) = 2 sin((x + 2π/3) - π/6) = 2 sin(x + 2π/3 - π/6) = 2 sin(x + π/2).So g(x + 2π/3) = 2 sin(x + π/2) = 2 cosx, because sin(x + π/2) = cosx. The function 2 cosx is an even function, not odd. Therefore, statement C is incorrect.Wait, but let me check again. An odd function satisfies h(-x) = -h(x). Let h(x) = g(x + 2π/3) = 2 cosx. Then h(-x) = 2 cos(-x) = 2 cosx = h(x), which is even. So C is incorrect.D: The function g(x) is monotonically increasing on the interval [π/6, π/3].First, find the derivative of g(x): g'(x) = 2 cos(x - π/6). To determine where g is increasing, we need g'(x) > 0. So cos(x - π/6) > 0. Let's analyze x in [π/6, π/3]. Then x - π/6 is in [0, π/6]. The cosine of angles between 0 and π/2 is positive. Since π/6 is 30 degrees, which is less than π/2 (90 degrees), cos(x - π/6) is positive in this interval. Therefore, g'(x) > 0, so g(x) is monotonically increasing on [π/6, π/3]. Therefore, statement D is correct.Wait, but let's confirm. For x in [π/6, π/3], x - π/6 ranges from 0 to π/6. The cosine of these values is positive (since they're in the first quadrant), so yes, the derivative is positive, so the function is increasing. Therefore, D is correct.So, the correct statements are A, B, D. But let me double-check statement B.For g(x) = 2 sin(x - π/6), the axes of symmetry. The standard sine function has symmetry axes at x = π/2 + kπ. For g(x), shifted right by π/6, those axes become x = π/2 + π/6 + kπ = 2π/3 + kπ. The distance between adjacent axes is π, since 2π/3 + (k+1)π - (2π/3 + kπ) = π. Therefore, B is correct.So, A, B, D are correct. But let me check the options again.The options are A, B, C, D. According to my analysis:A: CorrectB: CorrectC: IncorrectD: CorrectTherefore, the correct answers are A, B, D. But let me check once more to avoid mistakes.A: We saw that g(π/6 - x) + g(x + π/6) = -2 sinx + 2 sinx = 0. Correct.B: The distance between adjacent symmetry axes is π. Correct.C: g(x + 2π/3) = 2 cosx, which is even, not odd. Incorrect.D: g(x) is increasing on [π/6, π/3]. Derivative is positive there. Correct.Hence, the correct options are A, B, D.Final Answerboxed{ABD}</think>Given the function ( f(x) = 2sin(omega x + frac{pi}{6}) ) where ( omega geq 0 ), and the condition ( f(x - frac{pi}{4}) = -f(x + frac{pi}{4}) ) for all ( x in mathbb{R} ), we need to determine the correct statements about the function ( g(x) ) obtained by shifting ( f(x) ) right by ( frac{pi}{6} ) and stretching the x-coordinates to twice their length.1. Determine ( omega ): - The given condition implies ( sin(omega x - omega frac{pi}{4} + frac{pi}{6}) = -sin(omega x + omega frac{pi}{4} + frac{pi}{6}) ). - This leads to ( omega = 2 ) (the minimal positive solution).2. Transformations to get ( g(x) ): - Shift right by ( frac{pi}{6} ): ( f(x - frac{pi}{6}) = 2sin(2(x - frac{pi}{6}) + frac{pi}{6}) = 2sin(2x - frac{pi}{3} + frac{pi}{6}) = 2sin(2x - frac{pi}{6}) ). - Stretch x-coordinates by 2: Replace ( x ) with ( frac{x}{2} ): ( g(x) = 2sin(x - frac{pi}{6}) ).3. Check the statements: - A: ( g(frac{pi}{6} - x) + g(x + frac{pi}{6}) = 0 ) - ( g(frac{pi}{6} - x) = 2sin(-x) = -2sin(x) ) - ( g(x + frac{pi}{6}) = 2sin(x) ) - Sum: ( -2sin(x) + 2sin(x) = 0 ). Correct. - B: Distance between adjacent axes of symmetry is ( pi ) - The function ( g(x) = 2sin(x - frac{pi}{6}) ) has period ( 2pi ), axes of symmetry spaced by ( pi ). Correct. - C: ( g(x + frac{2pi}{3}) ) is odd - ( g(x + frac{2pi}{3}) = 2cos(x) ), which is even. Incorrect. - D: ( g(x) ) is monotonically increasing on ( [frac{pi}{6}, frac{pi}{3}] ) - Derivative ( g'(x) = 2cos(x - frac{pi}{6}) ), positive in this interval. Correct.Final Answer: (boxed{ABD})]

🔑:Given the function f(x)=2sin (omega x+frac{π}{6}) where omega geq 0, and for forall xin R, f(x-frac{π}{4})=-f(x+frac{π}{4}) holds, we can derive the following:1. From the given condition f(x-frac{π}{4})=-f(x+frac{π}{4}), we infer that f(x)=-f(x+frac{π}{2}). This implies that f(x)=f(x+pi), indicating the function repeats itself every pi units. Therefore, the minimum positive period of f(x) is pi, leading to the equation frac{2π}{ω}=pi. Solving for omega, we find omega =2. Thus, f(x)=2sin (2x+frac{π}{6}).2. Shifting the graph of f(x)=2sin (2x+frac{π}{6}) to the right by frac{π}{6} units results in y=2sin (2x-frac{π}{6}).3. Stretching all x-coordinates to twice their original length while keeping y-coordinates unchanged transforms the function to g(x)=2sin (x-frac{π}{6}).4. For statement A, we evaluate g(frac{π}{6}-x)+g(x+frac{π}{6}): begin{align*} g(frac{π}{6}-x)+g(x+frac{π}{6}) &= 2sin left(-xright)+2sin x &= 0. end{align*} Therefore, statement A is correct.5. The distance between adjacent axes of symmetry of g(x) is half a period, which is frac{1}{2}times frac{2π}{1}=pi. Hence, statement B is correct.6. The function g(x+frac{2π}{3})=2sin (x+frac{π}{2})=2cos x is an even function, not an odd function, making statement C incorrect.7. On the interval [frac{π}{6}, frac{π}{3}], x-frac{π}{6} falls within [0, frac{π}{6}]. Within this interval, g(x) is monotonically increasing. Thus, statement D is correct.Therefore, the correct choices are boxed{ABD}.

🧠:[<think>Okay, let's tackle part (1) first. We have triangle ABC with sides a, b, c opposite angles A, B, C respectively. Given that c = 2√3 and sin B = 2 sin A. Also, angle C is π/3 radians, which is 60 degrees. We need to find the values of a and b.Hmm, so let's recall the Law of Sines here. The Law of Sines states that in any triangle, a/sin A = b/sin B = c/sin C. Since we know c and angle C, maybe we can use that to find the other sides?Given that sin B = 2 sin A, maybe we can express sin B in terms of sin A and then use the Law of Sines to relate sides a and b. Let's write down the Law of Sines ratios:a / sin A = b / sin B = c / sin CWe know c = 2√3 and C = π/3, so sin C = sin(π/3) = √3/2. Therefore, c / sin C = (2√3) / (√3/2) = (2√3) * (2/√3) = 4. So, the common ratio is 4. That means:a = 4 sin Ab = 4 sin BBut we also know that sin B = 2 sin A. So, substituting that into the expression for b:b = 4 * 2 sin A = 8 sin ABut from a = 4 sin A, we can write sin A = a / 4. Then, substituting into the expression for b:b = 8 * (a / 4) = 2aSo, b = 2a. So, we have a relationship between a and b. Now, maybe we can use the Law of Cosines to relate the sides. Since we know angle C and sides a, b, c.Law of Cosines states that c² = a² + b² - 2ab cos C.We know c = 2√3, angle C = π/3, so cos C = cos(π/3) = 0.5.Substituting into the formula:(2√3)² = a² + b² - 2ab*(0.5)Calculating left side: (2√3)² = 4*3 = 12Right side: a² + b² - abSo, 12 = a² + b² - abBut since we know b = 2a, substitute that into the equation:12 = a² + (2a)² - a*(2a)12 = a² + 4a² - 2a²12 = (1 + 4 - 2)a²12 = 3a²So, a² = 4Thus, a = 2 (since side lengths are positive)Then, b = 2a = 4So, the values are a = 2 and b = 4. Let me check if this makes sense.Check Law of Sines:a/sin A = 2 / sin Ab/sin B = 4 / sin BGiven sin B = 2 sin A, then 4 / (2 sin A) = 2 / sin A, which matches a/sin A. So that's consistent.Also, check Law of Cosines:a² + b² - ab = 4 + 16 - 8 = 12, which equals c². So that's correct.Okay, part (1) seems solved. a = 2, b = 4.Now, part (2): Given cos C = 1/4, find the area of triangle ABC.Again, we have sides a, b, c with c = 2√3, sin B = 2 sin A, and now cos C = 1/4. Need to find the area.First, let's recall that the area of a triangle can be calculated by (1/2)ab sin C. But we need to find a and b first, maybe?Alternatively, if we can find two sides and the included angle, but in this case, angle C is given with cos C = 1/4, so sin C can be found since sin² C + cos² C = 1. So sin C = sqrt(1 - (1/4)^2) = sqrt(1 - 1/16) = sqrt(15/16) = √15 / 4.So sin C = √15 / 4. Therefore, if we can find sides a and b, then area is (1/2)ab*(√15 /4).But to find a and b, we need to relate them using given conditions.Again, we have sin B = 2 sin A. Let's use the Law of Sines again. From Law of Sines:a / sin A = b / sin B = c / sin CWe know c = 2√3 and sin C = √15 /4. So, c / sin C = (2√3) / (√15 /4) = 2√3 * 4 / √15 = 8√3 / √15 = 8√(3/15) = 8√(1/5) = 8 / √5 = (8√5)/5. So the common ratio is (8√5)/5.Therefore, a = (8√5)/5 * sin ASimilarly, b = (8√5)/5 * sin BBut sin B = 2 sin A, so:b = (8√5)/5 * 2 sin A = (16√5)/5 sin ABut since a = (8√5)/5 sin A, we can write sin A = (5a)/(8√5)Substituting into expression for b:b = (16√5)/5 * (5a)/(8√5) = (16√5 * 5a) / (5 * 8√5) ) = (16a)/8 = 2aSo again, we have b = 2a, similar to part (1). So, similar relationship here.Now, to relate a and b with angle C. Let's use the Law of Cosines again. Since we know angle C and side c.Law of Cosines: c² = a² + b² - 2ab cos CAgain, c = 2√3, so c² = 12cos C = 1/4So,12 = a² + (2a)^2 - 2*a*(2a)*(1/4)Calculate each term:a² + 4a² - 2*a*2a*(1/4) = 5a² - (4a²)*(1/4) = 5a² - a² = 4a²Therefore, 12 = 4a² => a² = 3 => a = √3Then, b = 2a = 2√3Wait, but hold on. If a = √3 and b = 2√3, then sides a, b, c would be √3, 2√3, and 2√3. Wait, but c is given as 2√3. So sides would be a = √3, b = 2√3, c = 2√3. Hmm, but in a triangle, the sum of any two sides should be greater than the third side. Let's check:√3 + 2√3 = 3√3 ≈ 5.196, which is greater than 2√3 ≈ 3.464. Similarly, other combinations are okay. So that's fine.But wait, let's verify with the Law of Cosines.Compute a² + b² - 2ab cos C:( (√3)^2 + (2√3)^2 ) - 2*(√3)*(2√3)*(1/4)= (3 + 12) - 2*(√3*2√3)*(1/4)= 15 - 2*(6)*(1/4)= 15 - 12*(1/4)= 15 - 3 = 12Which matches c² = 12. So that's correct.Now, the area would be (1/2)*a*b*sin CWe have a = √3, b = 2√3, sin C = √15 /4So:Area = (1/2)*(√3)*(2√3)*(√15 /4 )Simplify:(1/2)*(2√3*√3)*(√15 /4 )= (1/2)*(2*3)*(√15 /4 )= (1/2)*(6)*(√15 /4 )= 3*(√15 /4 ) = (3√15)/4So the area is 3√15 /4.Wait, let's check that again. Let's compute step by step:First, a*b = √3 * 2√3 = 2*(√3)^2 = 2*3 = 6Then, (1/2)*a*b*sin C = (1/2)*6*(√15 /4 ) = 3*(√15 /4 ) = 3√15 /4. Yes, that's correct.Alternatively, could we have used Heron's formula? Let's check:With sides a = √3, b = 2√3, c = 2√3. The semi-perimeter s = (a + b + c)/2 = (√3 + 2√3 + 2√3)/2 = (5√3)/2Area = √[s(s - a)(s - b)(s - c)]Compute each term:s - a = (5√3/2) - √3 = (5√3/2 - 2√3/2) = 3√3/2s - b = (5√3/2) - 2√3 = (5√3/2 - 4√3/2) = √3/2s - c = same as s - b = √3/2So, Area = √[ (5√3/2)*(3√3/2)*(√3/2)*(√3/2) ]Multiply the terms inside the root:First, multiply constants: 5 * 3 * 1 * 1 = 15Multiply √3 terms: √3 * √3 * √3 * √3 = (3)*(3) = 9Denominator: 2*2*2*2 = 16So inside the root: (15 * 9) /16 = 135 /16Therefore, Area = √(135/16) = (√135)/4 = (3√15)/4, which matches our previous result. So that's correct.Hence, the area is 3√15 /4.So, summarizing:1. For part (1), a = 2, b = 4.2. For part (2), area is 3√15 /4.Final Answer(1) (a = boxed{2}) and (b = boxed{4}); (2) The area of (triangle ABC) is (boxed{dfrac{3sqrt{15}}{4}}).</think>Given triangle ( triangle ABC ) with sides ( a ), ( b ), and ( c ) opposite angles ( A ), ( B ), and ( C ) respectively, we are provided with the following information: ( c = 2sqrt{3} ) and ( sin B = 2 sin A ). Part (1): ( C = frac{pi}{3} )1. Using the Law of Sines: [ frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} ] Given ( c = 2sqrt{3} ) and ( C = frac{pi}{3} ), we find: [ sin C = sin frac{pi}{3} = frac{sqrt{3}}{2} ] Therefore, [ frac{c}{sin C} = frac{2sqrt{3}}{frac{sqrt{3}}{2}} = 4 ] This gives: [ a = 4 sin A quad text{and} quad b = 4 sin B ] Since ( sin B = 2 sin A ), we substitute to get: [ b = 8 sin A quad text{and} quad a = 4 sin A implies b = 2a ]2. Using the Law of Cosines: [ c^2 = a^2 + b^2 - 2ab cos C ] Substituting ( c = 2sqrt{3} ), ( cos C = frac{1}{2} ), and ( b = 2a ): [ (2sqrt{3})^2 = a^2 + (2a)^2 - 2a(2a) cdot frac{1}{2} ] Simplifying: [ 12 = a^2 + 4a^2 - 2a^2 implies 12 = 3a^2 implies a^2 = 4 implies a = 2 ] Therefore, ( b = 2a = 4 ). Part (2): ( cos C = frac{1}{4} )1. Using the Law of Sines: [ frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} ] Given ( cos C = frac{1}{4} ), we find: [ sin C = sqrt{1 - left(frac{1}{4}right)^2} = frac{sqrt{15}}{4} ] Therefore, [ frac{c}{sin C} = frac{2sqrt{3}}{frac{sqrt{15}}{4}} = frac{8sqrt{5}}{5} ] This gives: [ a = frac{8sqrt{5}}{5} sin A quad text{and} quad b = frac{16sqrt{5}}{5} sin A implies b = 2a ]2. Using the Law of Cosines: [ c^2 = a^2 + b^2 - 2ab cos C ] Substituting ( c = 2sqrt{3} ), ( cos C = frac{1}{4} ), and ( b = 2a ): [ 12 = a^2 + 4a^2 - 2a(2a) cdot frac{1}{4} ] Simplifying: [ 12 = 5a^2 - a^2 implies 12 = 4a^2 implies a^2 = 3 implies a = sqrt{3} ] Therefore, ( b = 2sqrt{3} ).3. Calculating the area: [ text{Area} = frac{1}{2} ab sin C = frac{1}{2} (sqrt{3})(2sqrt{3}) cdot frac{sqrt{15}}{4} = frac{3sqrt{15}}{4} ] Final Answers(1) ( a = boxed{2} ) and ( b = boxed{4} ); (2) The area of ( triangle ABC ) is ( boxed{dfrac{3sqrt{15}}{4}} ).,<think>Okay, so I have this triangle ABC, with sides a, b, c opposite angles A, B, C respectively. The problem has two parts. Let's start with part (1): given that c = 2√3 and angle C is π/3 radians (which is 60 degrees), and also that sin B = 2 sin A. I need to find the values of a and b.First, I remember the Law of Sines, which states that in any triangle, a/sin A = b/sin B = c/sin C. That might be useful here since we have relationships between the sines of angles and the sides.Given that sin B = 2 sin A, maybe I can express sin B in terms of sin A and use the Law of Sines to relate sides a and b. Let's see.From the Law of Sines, we have:a / sin A = b / sin B = c / sin CWe know c = 2√3 and angle C = π/3, so sin C = sin(π/3) = √3/2. Therefore, the common ratio for the Law of Sines is c / sin C = (2√3) / (√3/2) = (2√3) * (2/√3) = 4. So, each side divided by the sine of its opposite angle is 4. So:a = 4 sin Ab = 4 sin BBut we also know that sin B = 2 sin A. Therefore, substituting into the equation for b:b = 4*(2 sin A) = 8 sin ABut from the equation for a, a = 4 sin A, so sin A = a/4. Therefore, substituting back into the equation for b:b = 8*(a/4) = 2aSo, b = 2a. So we have a relationship between a and b: b is twice as long as a.Now, we can use the Law of Cosines to relate the sides. Since we know angle C and sides a, b, c, the Law of Cosines states:c² = a² + b² - 2ab cos CWe know c = 2√3, angle C = π/3, so cos C = cos(π/3) = 1/2. Substituting into the equation:(2√3)² = a² + (2a)² - 2*a*(2a)*(1/2)Calculating each term:Left side: (2√3)² = 4*3 = 12Right side: a² + 4a² - 2*a*2a*(1/2) = 5a² - (4a²*(1/2)) = 5a² - 2a² = 3a²So, 12 = 3a² => a² = 4 => a = 2 (since side lengths are positive)Then, since b = 2a, b = 4.So, part (1) answer is a = 2 and b = 4. Let me check if this makes sense.Using Law of Sines: a/sin A = 2/sin A = 4, so sin A = 2/4 = 1/2. Therefore, angle A is π/6 (30 degrees) or 5π/6 (150 degrees). But since angle C is already π/3 (60 degrees), the sum of angles in a triangle is π. If angle A were 150 degrees, then angle B would have to be negative, which is impossible. So angle A is π/6, angle B is π - π/3 - π/6 = π/2 (90 degrees). Then sin B = sin(π/2) = 1, and 2 sin A = 2*(1/2) = 1. So that checks out. Then sides: a = 2, b = 4, c = 2√3. Let's check Law of Cosines again: c² = 12, a² + b² - 2ab cos C = 4 + 16 - 2*2*4*(1/2) = 20 - 8 = 12. Correct. So part (1) seems solved.Now part (2): If cos C = 1/4, find the area of triangle ABC. The given data is still c = 2√3 and sin B = 2 sin A. So similar to part (1), but now angle C is such that cos C = 1/4, so we need to compute the area.Again, using Law of Sines. Let's recall from part (1) that sin B = 2 sin A, and using Law of Sines: a = 4 sin A, b = 4 sin B = 8 sin A. So again, b = 2a.But now, angle C is different. So we can't assume the same angle C as before. So angle C is arccos(1/4). So maybe we need to use the Law of Cosines here as well. Let's think.First, let's note that in any triangle, angles add up to π. So A + B + C = π. Given that sin B = 2 sin A, perhaps we can express angle B in terms of angle A, then use angle C = π - A - B. But since we know cos C = 1/4, maybe we can use some trigonometric identities here.Alternatively, use the Law of Sines and Law of Cosines together.Let me try to express angles in terms of A. Let angle A = α, then sin B = 2 sin α. But angle B = π - α - C. So maybe we can write sin B = sin(π - α - C) = sin(α + C). Wait, sin(π - x) = sin x, so sin(π - α - C) = sin(α + C). Therefore, we have:sin(α + C) = 2 sin αExpanding sin(α + C) = sin α cos C + cos α sin C = 2 sin αSo:sin α cos C + cos α sin C = 2 sin αLet's rearrange:sin α cos C - 2 sin α + cos α sin C = 0Factor sin α:sin α (cos C - 2) + cos α sin C = 0Divide both sides by cos α (assuming cos α ≠ 0):sin α (cos C - 2)/cos α + sin C = 0Which can be written as:tan α (cos C - 2) + sin C = 0Solve for tan α:tan α = sin C / (2 - cos C)Given that cos C = 1/4, so sin C = sqrt(1 - (1/4)^2) = sqrt(15/16) = sqrt(15)/4Therefore:tan α = (sqrt(15)/4) / (2 - 1/4) = (sqrt(15)/4) / (7/4) = sqrt(15)/7So tan α = sqrt(15)/7, which gives α = arctan(sqrt(15)/7)But maybe we can find sin α and cos α from this.If tan α = sqrt(15)/7, then we can think of a right triangle with opposite side sqrt(15), adjacent side 7, hypotenuse sqrt( (sqrt(15))² + 7² ) = sqrt(15 + 49) = sqrt(64) = 8. So sin α = sqrt(15)/8, cos α = 7/8Wait, wait: tan α = opposite/adjacent = sqrt(15)/7, so if the sides are sqrt(15) and 7, hypotenuse is sqrt(15 + 49) = sqrt(64) = 8. Therefore, sin α = sqrt(15)/8, cos α = 7/8Therefore, sin α = sqrt(15)/8, cos α = 7/8So angle α is known now.From here, let's recall that in the Law of Sines, we have a = 4 sin A = 4 sin α = 4*(sqrt(15)/8) = sqrt(15)/2Similarly, b = 2a = 2*(sqrt(15)/2) = sqrt(15)So sides a = sqrt(15)/2, b = sqrt(15), c = 2√3Now, to find the area of triangle ABC. There are several ways to compute the area:1. Using Heron's formula: area = sqrt(s(s - a)(s - b)(s - c)), where s is the semi-perimeter.2. Using formula (1/2)ab sin C3. Using formula (1/2)bc sin A, etc.Given that we know sides a, b, c and angle C, maybe the easiest is (1/2)ab sin C.But we need to confirm that.Wait, we know sides a, b, c and angle C. So yes, if we can use sides a and b and angle C between them? Wait, angle C is opposite side c, so sides a and b are adjacent to angle C?Wait, in triangle ABC, angle C is opposite side c. So sides a and b are adjacent to angle C. Wait, no. Wait, side a is opposite angle A, side b opposite angle B, side c opposite angle C. So angle C is between sides a and b. Because in a triangle, each angle is between the two other sides. For example, angle A is between sides b and c, angle B is between sides a and c, angle C is between sides a and b.Therefore, yes, angle C is between sides a and b. Therefore, the area can be calculated as (1/2)*a*b*sin C.Since we know a, b, and sin C (since cos C = 1/4, so sin C = sqrt(15)/4 as calculated earlier), then:Area = (1/2)*a*b*sin C = (1/2)*(sqrt(15)/2)*sqrt(15)*(sqrt(15)/4)Let me compute this step by step:First, a = sqrt(15)/2, b = sqrt(15), sin C = sqrt(15)/4Multiply them together:a*b = (sqrt(15)/2)*sqrt(15) = (15)/2Then multiply by sin C: (15/2)*(sqrt(15)/4) = (15 sqrt(15))/8Multiply by 1/2: (1/2)*(15 sqrt(15)/8) = (15 sqrt(15))/16Wait, but let me check again:Wait, Area = (1/2)*a*b*sin Ca = sqrt(15)/2b = sqrt(15)sin C = sqrt(15)/4Therefore:(1/2)*(sqrt(15)/2)*(sqrt(15))*(sqrt(15)/4)First multiply sqrt(15)/2 * sqrt(15) = (15)/2Then multiply by sqrt(15)/4: (15)/2 * sqrt(15)/4 = (15 sqrt(15))/8Multiply by 1/2: (15 sqrt(15))/16So area is 15√15 / 16But let me verify this with another method to make sure.Alternatively, using Heron's formula. We have sides a = sqrt(15)/2, b = sqrt(15), c = 2√3First compute the semi-perimeter, s = (a + b + c)/2Compute a + b + c:sqrt(15)/2 + sqrt(15) + 2√3 = sqrt(15)/2 + 2 sqrt(15)/2 + 2√3 = (3 sqrt(15))/2 + 2√3Therefore, s = (3 sqrt(15)/2 + 2√3)/2 = (3 sqrt(15) + 4√3)/4This seems complicated, but let's try to compute Heron's formula:Area = sqrt( s(s - a)(s - b)(s - c) )But this might get very messy. Let's see:s - a = (3 sqrt(15) + 4√3)/4 - sqrt(15)/2 = (3 sqrt(15) + 4√3 - 2 sqrt(15))/4 = (sqrt(15) + 4√3)/4s - b = (3 sqrt(15) + 4√3)/4 - sqrt(15) = (3 sqrt(15) + 4√3 - 4 sqrt(15))/4 = (-sqrt(15) + 4√3)/4s - c = (3 sqrt(15) + 4√3)/4 - 2√3 = (3 sqrt(15) + 4√3 - 8√3)/4 = (3 sqrt(15) - 4√3)/4Therefore, the product s(s - a)(s - b)(s - c) is:[ (3 sqrt(15) + 4√3)/4 ] * [ (sqrt(15) + 4√3)/4 ] * [ (-sqrt(15) + 4√3)/4 ] * [ (3 sqrt(15) - 4√3)/4 ]This looks complex, but notice that terms can be paired to use the difference of squares.First pair: [ (sqrt(15) + 4√3)/4 ] * [ (-sqrt(15) + 4√3)/4 ] = [ (4√3)^2 - (sqrt(15))^2 ] / 16 = (16*3 - 15)/16 = (48 - 15)/16 = 33/16Second pair: [ (3 sqrt(15) + 4√3)/4 ] * [ (3 sqrt(15) - 4√3)/4 ] = [ (3 sqrt(15))^2 - (4√3)^2 ] / 16 = (9*15 - 16*3)/16 = (135 - 48)/16 = 87/16So the product becomes (33/16)*(87/16) = (33*87)/(16*16)Compute 33*87: 33*80 = 2640, 33*7=231, total 2640+231=2871So 2871/(256)Then, the area is sqrt(2871/256) = sqrt(2871)/16Wait, but 2871 is equal to 3*957 = 3*3*319, 319 is prime? Let's check:319 divided by 11: 11*29=319. So 319=11*29. Therefore, 2871=3*3*11*29. Therefore, sqrt(2871)=3*sqrt(319). But 319 is 11*29, which are primes, so sqrt(2871)=3*sqrt(319). Hmm, but this is different from 15√15 /16. Wait, so there must be a mistake.Wait, this inconsistency suggests that either my calculation using Heron's formula is wrong, or the previous calculation was wrong. Let me check.First, let's compute 15√15 /16 numerically and sqrt(2871)/16 numerically.15√15 ≈15*3.87298 ≈58.0947, then divided by 16 ≈3.6309sqrt(2871) ≈ sqrt(2871). Let's see, 53²=2809, 54²=2916, so sqrt(2871)≈53.58, divided by 16≈3.34875These are different, which implies a mistake.But where?Alternatively, maybe my Heron's formula approach messed up.Wait, let's check the sides again.In part (2), the sides are:From the Law of Sines:a = 4 sin A = 4*(sqrt(15)/8) = sqrt(15)/2b = 8 sin A = sqrt(15)c = 2√3Is that correct? Let's recap.Given sin B = 2 sin A, and from Law of Sines, a/sin A = b/sin B = c/sin C. Given that c = 2√3 and angle C has cos C =1/4. So sin C = sqrt(15)/4, so c/sin C = 2√3 / (sqrt(15)/4) = 2√3 * 4 / sqrt(15) = 8√3 / sqrt(15) = 8√(3/15) = 8√(1/5) = 8/(√5) = (8√5)/5. Wait, hold on, this is different from part (1). Wait, in part (1), c/sin C was 4. But in part (2), since angle C is different, c/sin C is different.Wait, wait, hold on. Maybe my mistake is assuming that a = 4 sin A here. But in part (1), we had c/sin C =4 because angle C was π/3. In part (2), angle C is different, so c/sin C is different. So the ratio in the Law of Sines is different here. Therefore, my previous approach is wrong because I assumed the same ratio as part (1), which is not the case.Oh no! I made a mistake here. I assumed the same ratio a/sin A = c/sin C as in part (1), but in part (2), angle C is different, so the ratio changes. So this is incorrect.So let's start over for part (2).Given triangle ABC with sides a, b, c opposite angles A, B, C respectively. Given that c = 2√3, sin B = 2 sin A, and cos C =1/4. Find the area.First, let's use the Law of Sines:a/sin A = b/sin B = c/sin C = 2R (where R is the circumradius)But we need to relate these given the conditions.Given sin B = 2 sin A. So, from Law of Sines:b = 2R sin B = 2R * 2 sin A = 4R sin ABut a = 2R sin A. Therefore, b = 2a. So again, we have b = 2a, similar to part (1), but here the ratio a/sin A is different.But now, since we have c = 2√3 and angle C, which has cos C = 1/4, sin C = sqrt(1 - (1/4)^2) = sqrt(15)/4.Therefore, c/sin C = 2√3 / (sqrt(15)/4) = 8√3 / sqrt(15) = 8√(3/15) = 8√(1/5) = 8/√5 = (8√5)/5. Therefore, the common ratio in Law of Sines is (8√5)/5.Therefore:a = (8√5)/5 * sin Ab = (8√5)/5 * sin B = (8√5)/5 * 2 sin A = (16√5)/5 sin AThus, b = 2a, as before.But now, in terms of a, we can express:a = (8√5)/5 sin Aand b = (16√5)/5 sin A = 2aBut we need to find a and b first to compute the area. Alternatively, use Law of Cosines with angle C.We know sides a, b, c with c = 2√3, angle C has cos C = 1/4. Let's apply the Law of Cosines:c² = a² + b² - 2ab cos CSubstitute c = 2√3, cos C = 1/4, and b = 2a:(2√3)^2 = a² + (2a)^2 - 2*a*(2a)*(1/4)Compute left side: 4*3 = 12Right side: a² + 4a² - 2*a*2a*(1/4) = 5a² - (4a²*(1/4)) = 5a² - a² = 4a²Therefore, 12 = 4a² => a² = 3 => a = √3Therefore, a = √3, b = 2√3Wait, but c is also 2√3. So sides are a = √3, b = 2√3, c = 2√3. So triangle has sides √3, 2√3, 2√3. That seems possible.But let's verify with Law of Sines.From Law of Sines, a/sin A = √3 / sin A = c/sin C = 2√3 / (sqrt(15)/4) = 8√3 / sqrt(15) = 8√(3/15) = 8/√5 as before.Therefore, √3 / sin A = 8/√5 => sin A = √3 * √5 /8 = sqrt(15)/8, which matches our previous result. So angle A is arcsin(sqrt(15)/8), and angle B is arcsin(2*sqrt(15)/8)=sqrt(15)/4, but angle B = π - A - C. Anyway, seems consistent.But the sides are a = √3, b = 2√3, c = 2√3. So two sides are equal: b and c, both 2√3. So it's an isoceles triangle with sides b and c equal. Wait, but in a triangle, if two sides are equal, the opposite angles are equal. So if b = c, then angles B and C are equal. But in our case, angle C has cos C =1/4, so angle C is not equal to angle B unless angle B also has cos B =1/4. But from sin B = 2 sin A, angles B and A are related. Wait, but if sides b and c are equal, then angles B and C should be equal, but angle C is arccos(1/4), so angle B should also be arccos(1/4), but sin B = 2 sin A. Is this possible?Wait, if b = c, then angle B = angle C. But angle C = arccos(1/4), so angle B = arccos(1/4). Then sin B = sin C = sqrt(15)/4, so from sin B = 2 sin A, sin A = sqrt(15)/8. Then angle A = arcsin(sqrt(15)/8). Then angles add up to A + B + C = arcsin(sqrt(15)/8) + 2 arccos(1/4). Let's check numerically.Compute angle C: arccos(1/4) ≈ 1.3181 radians ≈75.522 degreesAngle B is same as angle C, so ≈75.522 degreesThen angle A ≈ 180 - 2*75.522 ≈28.956 degreesCompute sin A: sin(28.956°) ≈0.487, sqrt(15)/8 ≈0.487. So that works. So angles add up. Therefore, the triangle is isoceles with sides b = c =2√3, and a =√3.Wait, but the problem didn't state that b = c, but according to Law of Cosines calculation, we arrived at a =√3, b =2√3, c=2√3. Therefore, sides b and c are equal. Therefore, angles B and C are equal. But from the given data, sin B = 2 sin A. If angles B and C are equal, then sin B = sin C = sqrt(15)/4, so sin A = sqrt(15)/8, which matches.Therefore, seems okay.But then, for the area, we can use (1/2)*a*b*sin C.We have a =√3, b=2√3, sin C =sqrt(15)/4.So area = (1/2)*√3*2√3*(sqrt(15)/4) = (1/2)*2*3*(sqrt(15)/4) = 3*(sqrt(15)/4) = (3 sqrt(15))/4Alternatively, using Heron's formula:s = (a + b + c)/2 = (√3 + 2√3 + 2√3)/2 = (5√3)/2s =5√3/2Area = sqrt(s(s -a)(s -b)(s -c)) = sqrt(5√3/2 * (5√3/2 - √3) * (5√3/2 - 2√3) * (5√3/2 - 2√3))Compute each term:s -a =5√3/2 - √3 =5√3/2 - 2√3/2=3√3/2s -b =5√3/2 -2√3=5√3/2 -4√3/2=√3/2s -c = same as s -b =√3/2Therefore:Area = sqrt(5√3/2 *3√3/2 *√3/2 *√3/2 )Multiply inside:First, multiply constants: 5*3*1*1=15Multiply √3 terms: √3 *√3 *√3 *√3 = (√3)^4 =9Multiply denominators: 2*2*2*2=16So inside sqrt:15*9 /16=135/16Therefore, area =sqrt(135/16)= (sqrt(135))/4= (3*sqrt(15))/4Which matches the previous result. Therefore, the correct area is 3√15 /4But earlier, when I did it the wrong way (assuming the same ratio as part (1)), I got 15√15 /16, which is different. Therefore, my initial approach was wrong because I incorrectly assumed the ratio from part (1), but in part (2), angle C is different, so the ratio a/sin A is different. After correcting that with Law of Cosines, we find a=√3, b=2√3, c=2√3, and area=3√15 /4.So final answer for part (2) is 3√15 /4.Let me check with another method. For example, using formula (1/2)ab sin C:a =√3, b=2√3, sin C =sqrt(15)/4Thus, area = (1/2)*√3*2√3*sqrt(15)/4 = (1/2)*2*3*sqrt(15)/4 = 3*sqrt(15)/4. Yep, same result.Therefore, part (2) answer is 3√15 /4.Final Answer(1) ( a = boxed{2} ) and ( b = boxed{4} ); (2) The area of ( triangle ABC ) is ( boxed{dfrac{3sqrt{15}}{4}} ).</think>Given triangle ( triangle ABC ) with sides ( a ), ( b ), and ( c ) opposite angles ( A ), ( B ), and ( C ) respectively, we need to solve two parts:(1) If ( C = frac{pi}{3} ), find the values of ( a ) and ( b ):Using the Law of Sines, we know:[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}]Given ( c = 2sqrt{3} ) and ( sin C = sin frac{pi}{3} = frac{sqrt{3}}{2} ), the common ratio is:[frac{c}{sin C} = frac{2sqrt{3}}{frac{sqrt{3}}{2}} = 4]Thus, ( a = 4 sin A ) and ( b = 4 sin B ). Given ( sin B = 2 sin A ), we substitute to get:[b = 4 cdot 2 sin A = 8 sin A = 2a]Using the Law of Cosines for ( c ):[c^2 = a^2 + b^2 - 2ab cos C]Substituting ( c = 2sqrt{3} ), ( cos C = frac{1}{2} ), and ( b = 2a ):[(2sqrt{3})^2 = a^2 + (2a)^2 - 2a(2a) cdot frac{1}{2}][12 = 5a^2 - 2a^2 implies 12 = 3a^2 implies a^2 = 4 implies a = 2]Thus, ( b = 2a = 4 ).(2) If ( cos C = frac{1}{4} ), find the area of ( triangle ABC ):Using the Law of Sines again with ( cos C = frac{1}{4} ), we find ( sin C = sqrt{1 - left(frac{1}{4}right)^2} = frac{sqrt{15}}{4} ). The common ratio is:[frac{c}{sin C} = frac{2sqrt{3}}{frac{sqrt{15}}{4}} = frac{8sqrt{3}}{sqrt{15}} = frac{8sqrt{5}}{5}]Given ( sin B = 2 sin A ), we use the relationship ( b = 2a ). Using the Law of Cosines:[c^2 = a^2 + b^2 - 2ab cos C]Substituting ( c = 2sqrt{3} ), ( cos C = frac{1}{4} ), and ( b = 2a ):[(2sqrt{3})^2 = a^2 + (2a)^2 - 2a(2a) cdot frac{1}{4}][12 = 5a^2 - a^2 implies 12 = 4a^2 implies a^2 = 3 implies a = sqrt{3}]Thus, ( b = 2sqrt{3} ). The area is calculated using:[text{Area} = frac{1}{2}ab sin C = frac{1}{2} cdot sqrt{3} cdot 2sqrt{3} cdot frac{sqrt{15}}{4} = frac{3sqrt{15}}{4}] Final Answer(1) ( a = boxed{2} ) and ( b = boxed{4} ); (2) The area of ( triangle ABC ) is ( boxed{dfrac{3sqrt{15}}{4}} ).]

🔑:(The full score for this question is 12 points.)Solution: (1) Since C= frac{pi}{3} and sin B=2sin A, By the Law of Sines, we have: b=2a,...(2 points) Since c=2sqrt{3}, By the Law of Cosines, we have: c^{2}=a^{2}+b^{2}-2abcos C, which simplifies to: 12=a^{2}+4a^{2}-2a^{2}, Solving this gives: a=2, b=4...(6 points) (2) Since cos C= frac{1}{4}, Then sin C= sqrt{1-cos^{2}C}= frac{sqrt{15}}{4}, Also, since b=2a, By the Law of Cosines, we have: c^{2}=a^{2}+b^{2}-2abcos C=a^{2}+4a^{2}-a^{2}=4a^{2}, solving this gives: c=2a,...(9 points) Since c=2sqrt{3}, we find: a=sqrt{3}, b=2sqrt{3}, Therefore, the area of triangle ABC is S_{triangle ABC}= frac{1}{2}absin C= frac{1}{2}times sqrt{3}times2sqrt{3}times frac{sqrt{15}}{4}= frac{3sqrt{15}}{4}...(12 points)Thus, the final answers are: (1) a=2, b=4, so a=2, b=4 are boxed{a=2, b=4}. (2) The area of triangle ABC is boxed{frac{3sqrt{15}}{4}}.

🧠:[<think>Okay, so I have this problem: I need to find how many rational terms are in the expansion of the binomial (√2013 + ∛2013)^2012. Hmm, binomial expansions, right? Let me recall how that works.First, the binomial theorem says that (a + b)^n = Σ_{k=0}^n C(n, k) * a^{n-k} * b^k. So each term in the expansion is C(n, k) multiplied by a^(n - k) multiplied by b^k. In this case, a is √2013, which is 2013^(1/2), and b is ∛2013, which is 2013^(1/3). The exponent is 2012.So each term in the expansion will look like C(2012, k) * (2013^(1/2))^{2012 - k} * (2013^(1/3))^k. Simplifying the exponents, that would be 2013^{(2012 - k)/2} * 2013^{k/3} = 2013^{( (2012 - k)/2 + k/3 )}. Combining the exponents: let's compute (2012 - k)/2 + k/3.To combine those fractions, find a common denominator. 6 is the least common multiple of 2 and 3, so:[(2012 - k)/2] = [3(2012 - k)]/6[k/3] = [2k]/6Adding them together: [3(2012 - k) + 2k]/6 = [6036 - 3k + 2k]/6 = [6036 - k]/6.So the exponent on 2013 for each term is (6036 - k)/6. Therefore, the term is rational if and only if (6036 - k)/6 is an integer. Because 2013 is an integer, and if the exponent is an integer, then 2013 to that exponent is also an integer, hence rational. If the exponent is not an integer, then 2013^(something non-integer) is irrational. So we need (6036 - k)/6 to be integer. So that means that 6036 - k must be divisible by 6. Which is equivalent to k ≡ 6036 mod 6. Since 6036 divided by 6 is 1006, which is an integer. So 6036 mod 6 is 0. So that means k ≡ 0 mod 6. Therefore, k must be a multiple of 6. So k = 0, 6, 12, ..., up to 2012.But wait, let's check. So k needs to be between 0 and 2012 inclusive. And k must be a multiple of 6. So how many multiples of 6 are there between 0 and 2012 inclusive?To find the number of terms, we can compute floor(2012 / 6) + 1. Wait, floor(2012 / 6) is 335 because 6*335 = 2010, and 2012 - 2010 = 2. So floor(2012/6) = 335, then +1 gives 336. But wait, but 6*335 = 2010, which is less than 2012, so the next multiple would be 2016, which is beyond 2012. So the number of multiples of 6 from 0 to 2012 is 335 + 1 = 336? Wait, but hold on. Let's check 2012 divided by 6: 2012 ÷ 6 = 335.333... So the integer division gives 335 with a remainder of 2. So yes, the last multiple of 6 less than or equal to 2012 is 6*335 = 2010. Therefore, the number of terms is 335 + 1 = 336.But hold on, let me confirm this. Let me compute 6*335: 6*300=1800, 6*35=210, so 1800+210=2010. Then 2010 +6=2016, which is more than 2012, so 2010 is the last multiple of 6 less than or equal to 2012. Therefore, the number of multiples of 6 from 0 to 2012 inclusive is 335 +1=336. Therefore, there are 336 rational terms. Wait, but is that correct?Wait, but hold on. Let me check an example with smaller numbers to see if my reasoning is correct. Suppose instead of 2012, n was 6. Then the number of multiples of 6 from 0 to 6 would be 0,6: two terms. But 6/6=1, so floor(6/6)=1, 1+1=2. So that works. For n=7, multiples of 6 up to 7: 0,6. So two terms. 7/6=1.166..., floor is 1, 1+1=2. Correct. For n=5, multiples of 6 up to 5: just 0. So 0. 5/6=0.833..., floor 0, 0+1=1. Correct. So yes, formula is floor(n /6) +1. Therefore, in our case, floor(2012 /6)=335, so 335 +1=336.Therefore, the answer is 336. Hmm. Wait, but let me think again. Is that the correct approach?Wait, but the exponent on 2013 is (6036 -k)/6. For this exponent to be integer, (6036 -k) must be divisible by 6. So 6036 -k ≡0 mod6. Therefore, k ≡6036 mod6. But 6036 divided by6 is 1006, so 6036 ≡0 mod6. Therefore, k≡0 mod6. So k must be congruent to 0 mod6. So k=6t, where t is integer. Then k can be 0,6,12,..., up to the maximum k<=2012. So the maximum t is such that 6t<=2012. Then t_max= floor(2012 /6)=335, as above. So k=0,6,12,...,6*335=2010. Therefore, the number of terms is t from 0 to 335: 336 terms. So that seems correct.But wait, hold on a second. Let me check with a small n. Let's say n=6 instead of 2012, and same bases. So (sqrt(a) +cbrt(a))^6. Then the exponents would be ( (6 -k)/2 +k/3 ). Let's compute when that is integer.So (6 -k)/2 +k/3 = (18 -3k +2k)/6 = (18 -k)/6. So (18 -k)/6 must be integer. So 18 -k must be divisible by6, so k ≡18 mod6. But 18 mod6=0, so k≡0 mod6. So k=0,6. Therefore, two terms. Let's check:Term k=0: C(6,0)*(sqrt(a))^6*(cbrt(a))^0 = 1*a^(3)*1=a^3, which is rational.Term k=6: C(6,6)*(sqrt(a))^0*(cbrt(a))^6=1*1*a^2=a^2, which is rational.Other terms, say k=1:Exponent: (18 -1)/6=17/6. So a^(17/6), which is irrational. So yes, only two terms. So the formula gives floor(6/6)+1=1+1=2, which is correct. So seems like the formula works. So in the original problem, with n=2012, the answer is 336. Hmm. But wait, let me check n=12. Then floor(12/6)=2, so 2+1=3. Let's see:In (sqrt(a)+cbrt(a))^12, how many rational terms?Exponent is (12*3 -k)/6=(36 -k)/6. Wait, wait, the general formula is ( (3n -k)/6 )? Wait, wait, for n=6, we had (18 -k)/6. For n=2012, the exponent was (6036 -k)/6. Wait, 2012*3=6036. So in general, exponent is (3n -k)/6. Therefore, for general n, the exponent is (3n -k)/6, so to have integer exponent, (3n -k) must be divisible by6. So 3n -k ≡0 mod6 => k≡3n mod6. Therefore, k≡3n mod6. So depending on n, the congruence changes. Wait, in the original problem, n=2012. So 3n=6036. 6036 mod6=0, so k≡0 mod6. Hence, k must be multiples of6. If n was different, say n=5, then 3n=15≡3 mod6. Then k≡3 mod6. So k=3,9,... etc. So in that case, the number of terms would be different.So in our problem, since n=2012, which is even. 3n=6036, which is divisible by6, so 6036≡0 mod6, so k≡0 mod6. Therefore, k must be multiples of6. Then the number of such k is floor(n /6) +1, but wait, n=2012. Wait, but the maximum k is n=2012. So if k must be multiple of6 up to 2012, then number of terms is floor(2012/6) +1=335 +1=336. So that seems correct.But to be thorough, let me check for another n. Suppose n=7. Then 3n=21≡3 mod6. So k≡21 mod6≡3 mod6. Therefore, k≡3 mod6. So possible k=3,9,15,... but n=7, so maximum k=7. So k=3. Then only one term. Let's check:In (sqrt(a)+cbrt(a))^7, the exponents would be (21 -k)/6. So for k=3: (21 -3)/6=18/6=3, which is integer. For k=9, but n=7, so k can't be 9. Therefore, only one term when k=3. So number of terms would be floor((7 -3)/6) +1? Wait, maybe not. Wait, how do we compute the number of solutions to k≡3 mod6, where 0<=k<=7.So k=3,9,... but 9>7, so only k=3. So one term. So in this case, the number of terms is floor((7 -3)/6)+1= floor(4/6)+1=0+1=1. Hmm, but how to generalize this.Alternatively, for k≡c mod m, in the range 0<=k<=n, the number of solutions is floor((n -c)/m) +1. If c <=n mod m, then it's floor(n/m) +1. Wait, perhaps this is more complicated. Wait, let's see.In our original problem, we have k≡0 mod6, 0<=k<=2012. So the number of solutions is floor((2012 -0)/6) +1 if 0<=2012 mod6 <0? Wait, maybe not. Actually, the formula for the number of integers ≡a mod m in [0, n] is floor((n -a)/m) +1. Here, a=0, m=6, n=2012. So floor((2012 -0)/6) +1= floor(2012/6)+1=335 +1=336. So that works. So that formula gives the correct number. Therefore, in our case, the answer is 336.But wait, let me check with n=6 again. Then floor(6/6)+1=1+1=2. Correct. For n=5, k≡0 mod6, so floor(5/6)+1=0 +1=1. Which is correct (k=0). So yes, the formula works. So in our problem, the answer is 336. Therefore, I think that is the correct answer.But just to be absolutely sure, let me verify another example. Let's take n=12. Then k≡0 mod6. So k=0,6,12. Number of terms= floor(12/6)+1=2+1=3. Let's check exponent:For k=0: (36 -0)/6=6. 2013^6, rational.For k=6: (36 -6)/6=30/6=5. 2013^5, rational.For k=12: (36 -12)/6=24/6=4. 2013^4, rational.Yes, three terms. So the formula works here. So, all checks out.Therefore, in the given problem, the number of rational terms is 336. So I think that's the answer.Final Answerboxed{336}</think>To determine the number of rational terms in the expansion of the binomial ((sqrt{2013} + sqrt[3]{2013})^{2012}), we use the binomial theorem. Each term in the expansion is given by:[binom{2012}{k} cdot (sqrt{2013})^{2012 - k} cdot (sqrt[3]{2013})^k]Simplifying the exponents, we get:[(sqrt{2013})^{2012 - k} = 2013^{(2012 - k)/2} quad text{and} quad (sqrt[3]{2013})^k = 2013^{k/3}]Combining these exponents, we have:[2013^{(2012 - k)/2 + k/3}]To combine the fractions, we find a common denominator (6):[frac{2012 - k}{2} + frac{k}{3} = frac{3(2012 - k) + 2k}{6} = frac{6036 - 3k + 2k}{6} = frac{6036 - k}{6}]For the term to be rational, the exponent (frac{6036 - k}{6}) must be an integer. This requires (6036 - k) to be divisible by 6, which simplifies to (k equiv 0 pmod{6}).Next, we need to count the number of values (k) such that (0 leq k leq 2012) and (k equiv 0 pmod{6}). These values are (0, 6, 12, ldots, 2010). The number of such terms is given by:[leftlfloor frac{2012}{6} rightrfloor + 1 = leftlfloor 335.333 ldots rightrfloor + 1 = 335 + 1 = 336]Thus, the number of rational terms in the expansion is:[boxed{336}],<think>Okay, so I need to find how many rational terms are in the expansion of ((sqrt{2013} + sqrt[3]{2013})^{2012}). Hmm, binomial expansions, right? Let me recall how that works. When you expand ((a + b)^n), each term is (binom{n}{k} a^{n - k} b^k) for (k) from 0 to (n). In this case, (a) is (sqrt{2013}) and (b) is (sqrt[3]{2013}). The exponents on these terms need to result in rational numbers for the entire term to be rational. Since 2013 is just a constant here, the key is figuring out when the exponents on the radicals result in integer exponents, thereby making the term rational.First, let me rewrite the radicals as exponents. (sqrt{2013}) is (2013^{1/2}) and (sqrt[3]{2013}) is (2013^{1/3}). So each term in the expansion will be (binom{2012}{k} times (2013^{1/2})^{2012 - k} times (2013^{1/3})^k). Simplifying the exponents, that becomes (binom{2012}{k} times 2013^{(2012 - k)/2 + k/3}). So the exponent on 2013 in each term is (frac{2012 - k}{2} + frac{k}{3}). For the term to be rational, this exponent needs to be an integer because 2013 is an integer and raising it to an integer power will keep it rational (assuming the result is an integer, but even if not, it's still a rational number if the exponent is integer). If the exponent is fractional, then it would be irrational. So the critical thing is that (frac{2012 - k}{2} + frac{k}{3}) must be an integer.Let me combine those fractions. Let's see: the common denominator for 2 and 3 is 6. So converting both fractions:[frac{2012 - k}{2} = frac{3(2012 - k)}{6}, quad frac{k}{3} = frac{2k}{6}]Adding them together gives:[frac{3(2012 - k) + 2k}{6} = frac{6036 - 3k + 2k}{6} = frac{6036 - k}{6}]So the exponent on 2013 is (frac{6036 - k}{6}). For this to be an integer, (frac{6036 - k}{6}) must be an integer. That means (6036 - k) must be divisible by 6. Which implies that (k equiv 6036 mod 6). Since 6036 divided by 6 is 1006 exactly, because 6*1006=6036. So (6036 equiv 0 mod 6), therefore (k equiv 0 mod 6). So k must be a multiple of 6.But wait, k can range from 0 to 2012. So how many multiples of 6 are there between 0 and 2012 inclusive?To find the number of integers k such that k = 6m where m is an integer and 0 ≤ 6m ≤ 2012.So the maximum m is floor(2012 / 6). Let's compute 2012 divided by 6. 6*335=2010, so 2010 is the largest multiple of 6 less than or equal to 2012. Therefore m ranges from 0 to 335. So there are 335 - 0 + 1 = 336 values. But wait, 6*335=2010, which is less than 2012. The next multiple is 2016, which is beyond 2012. So indeed, the number of multiples of 6 from 0 to 2012 is 336.But hold on, let me verify this. The formula for the number of multiples of n between 0 and N inclusive is floor(N/n) + 1. So here, floor(2012/6) = floor(335.333...) = 335. Then 335 + 1 = 336. So yes, 336 terms. Therefore, there are 336 rational terms in the expansion.Wait, but let me check again. Let's confirm if this is correct because sometimes when dealing with inclusive counts, it's easy to make a mistake. Let's take a smaller example. Suppose N=6. Then floor(6/6)=1, 1+1=2. The multiples are 0,6. Yes, that's two terms. If N=5, floor(5/6)=0, 0+1=1. Which is just 0. Correct. So in our case, floor(2012/6)=335, 335+1=336. So the answer should be 336.But wait, hold on. Let me check if k must be a multiple of 6. So we derived that 6036 - k must be divisible by 6, so k ≡ 6036 mod 6. Since 6036 is divisible by 6, then k ≡ 0 mod 6. So yes, k must be congruent to 0 mod 6. Therefore, k=0,6,12,...,2010. So how many terms is that?The first term is 0, the last term is 2010. The step is 6. The number of terms is ((2010 - 0)/6) +1 = 335 +1 = 336. So that seems correct.But wait, is there a possibility that the exponent (frac{6036 - k}{6}) could be an integer even if k is not a multiple of 6? For example, suppose k ≡ something else mod 6. Let's check. Let's suppose k ≡ 1 mod 6. Then 6036 -1=6035. 6035 divided by 6 is 1005.833..., which is not an integer. Similarly, if k ≡ 2 mod 6, 6036 -2=6034, 6034/6≈1005.666..., not integer. Similarly, k≡3: 6036-3=6033, which is 6033/6=1005.5, still not integer. k≡4: 6036-4=6032, 6032/6≈1005.333... nope. k≡5: 6036-5=6031, which is 6031/6≈1005.166... So only when k≡0 mod 6 does 6036 -k become divisible by 6. Therefore, yes, only multiples of 6. So the number of such k is 336.Therefore, the answer should be 336. But let me verify once again with an example. Let's take a smaller exponent. Suppose instead of 2012, we have n=6. Then the exponent on 2013 in each term would be (6 -k)/2 +k/3. Let's compute this for k=0: 6/2 +0=3, integer. k=6: (6-6)/2 +6/3=0 +2=2, integer. So terms at k=0 and k=6. But how many terms? From k=0 to 6, step 6. So 0 and 6: 2 terms. Which is (6/6)+1=1+1=2. Correct. So that formula works.Another example: n=12. Then the exponent would be (12 -k)/2 +k/3. Let's see for k=0: 6 +0=6, k=6: (12-6)/2 +6/3=3 +2=5, k=12: 0 +12/3=4. Wait, but here, maybe n=12 would have more terms. Wait, but according to our formula, k must be multiples of 6. So k=0,6,12. Let's check each:k=0: exponent 6, which is integer.k=6: exponent (12 -6)/2 +6/3=3 + 2=5, integer.k=12: exponent 0 +4=4, integer. So 3 terms, which is (12/6) +1=2 +1=3. So that works.But if n=5, then (5 -k)/2 +k/3. For k=0: 5/2=2.5, not integer. k=6 is beyond n=5. Wait, n=5, the maximum k is 5. So in this case, are there any k such that the exponent is integer? Let's check k=0: 2.5, k=1: (5-1)/2 +1/3=2 + 0.333...=2.333..., k=2: (5-2)/2 +2/3=1.5 + 0.666...=2.166..., k=3: (5-3)/2 +3/3=1 +1=2, which is integer. So here, k=3 gives an exponent of 2. So even though 3 is not a multiple of 6, in this case, the exponent is integer. Wait, so in this case, our previous logic might not hold?Wait, let's apply our formula to n=5. The exponent would be (5 -k)/2 +k/3. Let me compute that as:(5 -k)/2 +k/3 = (15 -3k + 2k)/6 = (15 -k)/6. So to have this be integer, (15 -k)/6 must be integer. Therefore, 15 -k must be divisible by 6. So k ≡15 mod 6, which is 15 mod 6=3. So k≡3 mod 6. So k=3,9,... But in n=5, k can only be up to 5, so k=3 is the only solution. Which matches our previous calculation. So even though 3 is not a multiple of 6, in this case, the congruence is k≡3 mod6.But in our original problem, the congruence was k≡0 mod6. So the difference comes from the value of n. In the original problem, we had n=2012. So substituting back into the formula:Exponent is (2012 -k)/2 +k/3 = (6036 -3k +2k)/6=(6036 -k)/6. Therefore, 6036 -k must be divisible by 6, so k≡6036 mod6. Since 6036 is divisible by 6 (6036 ÷6=1006), then k≡0 mod6. Therefore, k must be 0 mod6. But in the n=5 case, we had (15 -k)/6, so 15 -k≡0 mod6 ⇒k≡15 mod6≡3 mod6. Therefore, depending on the original n, the congruence for k changes.Therefore, in our original problem, since n=2012, which leads to 6036 -k ≡0 mod6 ⇒k≡0 mod6, which gives the number of terms as 336.But in the n=5 case, even though k must be≡3 mod6, which in that case gives only one term (k=3). So the formula depends on the value of n. Therefore, in our problem, with n=2012, we must have k≡0 mod6, leading to 336 terms.Wait, so to generalize, in the expansion of ((sqrt{a} + sqrt[3]{a})^n), the exponent on a in each term is ((n -k)/2 +k/3). To make this an integer, the expression must be integral. Let me denote this as:[frac{n -k}{2} + frac{k}{3} = frac{3(n -k) + 2k}{6} = frac{3n -k}{6}]Therefore, for the exponent to be integer, (3n -k) must be divisible by 6. Therefore, (3n -k equiv0 mod6), which implies that (k ≡3n mod6). Therefore, k must be congruent to 3n modulo6.In our case, n=2012. So 3n=6036. 6036 mod6 is 0, since 6036 is divisible by6. Therefore, k≡0 mod6. So k must be multiples of6. Therefore, number of such k is floor((n)/6) +1? Wait, no. Wait, n=2012. But k must be between0 and2012, and k≡0 mod6. So the number of such k is floor(2012/6) +1=335 +1=336. So that's the formula.But in the previous example with n=5, 3n=15, so k≡15 mod6≡3 mod6. So k must be≡3 mod6. So k=3,9,15,... But since n=5, maximum k=5. So only k=3. Hence, 1 term. So the formula would be floor((n - (3n mod6))/6) +1? Hmm, maybe not. Alternatively, it's the number of integers k in [0,n] such that k≡3n mod6.Therefore, in general, the number of solutions is equal to the number of integers k in [0,n] congruent to (3n mod6). The number of such terms is floor((n - c)/6) +1, where c is the residue (3n mod6). Wait, perhaps. Let me check.Alternatively, since k ≡3n mod6. Let me let r =3n mod6. Then k must be≡r mod6. Then the number of k in [0,n] with k≡r mod6 is equal to floor((n - r)/6) +1. If r <=n mod6, then floor(n/6) +1. Hmm, perhaps this is getting too complicated.Alternatively, in our problem, since n=2012, and 3n=6036≡0 mod6, so r=0. So k≡0 mod6. Then the number of k is floor(2012/6) +1=335 +1=336. So this works. If r≠0, then we need to adjust.For example, in n=5, 3n=15≡3 mod6, so r=3. Then k≡3 mod6. The numbers are 3,9,15,... but since n=5, only k=3. So floor((5 -3)/6) +1= floor(2/6)+1=0 +1=1. Which is correct. If n=9, 3n=27≡3 mod6. Then k≡3 mod6. So k=3,9. So floor((9 -3)/6) +1= floor(6/6)+1=1 +1=2. Which is correct. So the formula would be floor((n - r)/6) +1 where r=3n mod6.Therefore, in general, for ((sqrt{a} + sqrt[3]{a})^n), the number of rational terms is floor((n - (3n mod6))/6) +1. Wait, but 3n mod6 can be 0,3. So if r=3n mod6, then the number of terms is floor((n - r)/6) +1. Since r is either 0 or 3. Let me check for n=6. 3n=18≡0 mod6. Then floor((6 -0)/6) +1=1 +1=2. Which is correct (k=0,6). For n=7: 3*7=21≡3 mod6. Then floor((7 -3)/6) +1= floor(4/6) +1=0 +1=1. Which would be k=3,9,... but n=7, so only k=3. So 1 term. Correct.Therefore, in our problem, n=2012, r=3*2012=6036≡0 mod6. So floor((2012 -0)/6) +1= floor(2012/6)+1=335 +1=336. Therefore, the answer is 336.But let me just check if there's any possible mistake here. For example, in the original exponent expression:[frac{2012 -k}{2} + frac{k}{3} = frac{3(2012) -k}{6} = frac{6036 -k}{6}]Yes, correct. So for this to be integer, 6036 -k must be divisible by6. So 6036 is divisible by6, so k must be congruent to0 mod6. Therefore, k=0,6,...,2010. So total terms: 2010 /6 =335, so 335+1=336. That's correct.Another sanity check: Let's take n=6. Then the exponent would be (6 -k)/2 +k/3. For k=0: 3, which is integer. k=6: 0 +2=2, integer. So two terms. Which is 6/6 +1=1+1=2. Correct. So the formula works here.Therefore, I think the answer is 336. So I should put that in a box.Final AnswerThe number of rational terms is boxed{336}.</think>To determine the number of rational terms in the expansion of ((sqrt{2013} + sqrt[3]{2013})^{2012}), we start by expressing each term in the binomial expansion. Each term is given by:[binom{2012}{k} left(2013^{1/2}right)^{2012 - k} left(2013^{1/3}right)^k]Simplifying the exponents, we get:[binom{2012}{k} times 2013^{frac{2012 - k}{2} + frac{k}{3}}]For the term to be rational, the exponent on 2013 must be an integer. Combining the fractions, the exponent is:[frac{2012 - k}{2} + frac{k}{3} = frac{3(2012 - k) + 2k}{6} = frac{6036 - k}{6}]For this to be an integer, (6036 - k) must be divisible by 6. This implies (k equiv 0 pmod{6}). Next, we count the number of integers (k) from 0 to 2012 that are multiples of 6. The largest multiple of 6 less than or equal to 2012 is 2010. The number of such terms is given by:[leftlfloor frac{2012}{6} rightrfloor + 1 = leftlfloor 335.333ldots rightrfloor + 1 = 335 + 1 = 336]Thus, the number of rational terms in the expansion is (boxed{336}).]

🔑:## Solution.The general term is given by the formulabegin{aligned}& binom{2012}{k} 2013^{frac{2012-k}{2}} cdot 2013^{frac{k}{3}}= = & binom{2012}{k} 2013^{1006-frac{k}{6}}end{aligned}The exponent 1006-frac{k}{6} must be a natural number or 0.Therefore, k must be of the form k=6n, 0 leq k leq 2012, n in mathbb{N}_{0}.These are the numbers 0, 6, 12, 18, ldots, 2010. They form an arithmetic sequence with a difference of 6. There are 336 of them.