🧠:[<think>Okay, so I have this geometry problem here. Let me read it again to make sure I understand all the details. There are two circles with radii √5 and √2. They intersect at point A. The distance between their centers is 3. Then, there's a line passing through point A that intersects the circles again at points B and C, such that AB equals AC, and B isn't the same as C. I need to find the length of AB.Hmm. Let me visualize this. There are two circles intersecting at A. The line through A goes through another point on each circle, B and C, such that AB = AC. So, the line passes through A, and on one circle, it meets B, and on the other circle, it meets C. Since AB = AC, the distances from A to both B and C are the same. Therefore, points B and C are on opposite sides of A on this line, each at distance AB from A. So, the total length BC would be 2AB. But maybe I need to be careful here.Wait, actually, the problem says "A line through point A intersects the circles at points B and C such that AB = AC." So, the line passes through A, and intersects the first circle at B (other than A), and the second circle at C (other than A), and AB equals AC. So, B is on the first circle (radius √5), C is on the second circle (radius √2), and the distances from A to B and from A to C are equal. But the line passes through A, so B and C must be on opposite sides of A, right? Because if they were on the same side, then one of AB or AC would be longer than the other. Wait, but maybe not necessarily. Wait, no. If the line passes through A, then points B and C can be on either side. But if AB = AC, then they must be equidistant from A, but on opposite sides. Because if they were on the same side, then one would be further away. Wait, but maybe the line intersects each circle twice? So, maybe the line passes through A, and intersects the first circle at two points: A and B, and the second circle at two points: A and C. Then, if AB = AC, then B and C are both at the same distance from A along the line. But since the line passes through A, they could be on opposite sides. So, if you go from A in one direction along the line, you reach B, and in the other direction, you reach C, each at the same distance from A. So, the total length BC would be AB + AC = 2AB. But I need to confirm that.So, the line passes through A, and intersects the first circle (radius √5) again at B, and the second circle (radius √2) again at C. The problem states AB = AC, so from A, moving along the line in one direction, you reach B after distance AB, and in the other direction, you reach C after distance AC = AB. Therefore, points B and C are on opposite sides of A, each at distance AB. So, BC is 2AB. But I need to find AB. Okay, so how do I approach this? Let me recall that when two circles intersect, the line connecting their centers is important. Maybe I need to use coordinate geometry here. Let me set up a coordinate system. Let's place the centers of the two circles on the x-axis for simplicity. Let me call the centers O1 and O2. Let me assume O1 is at (0, 0), and O2 is at (3, 0), since the distance between the centers is 3. The radii of the circles are √5 and √2, respectively. So, the first circle has equation x² + y² = 5, and the second circle has equation (x - 3)² + y² = 2. They intersect at point A. Then, there's a line through A that intersects the first circle again at B and the second circle again at C, such that AB = AC. We need to find AB.First, maybe I need to find the coordinates of point A. Since the circles intersect, their intersection points can be found by solving the two equations:1. x² + y² = 52. (x - 3)² + y² = 2Subtracting the first equation from the second gives:(x - 3)² + y² - (x² + y²) = 2 - 5Expanding (x - 3)²:x² - 6x + 9 + y² - x² - y² = -3Simplify:-6x + 9 = -3-6x = -12x = 2So, the x-coordinate of the intersection points is 2. Plugging back into the first equation, x² + y² = 5:2² + y² = 5 => 4 + y² = 5 => y² = 1 => y = 1 or y = -1Therefore, the two intersection points are (2, 1) and (2, -1). The problem mentions point A, so we can take either one. Since the problem doesn't specify, maybe the answer is the same regardless. Let's pick (2, 1) as point A for simplicity.So, point A is (2, 1). Now, we need to find a line passing through A that intersects the first circle (centered at O1 (0,0)) again at B and the second circle (centered at O2 (3,0)) again at C, such that AB = AC. Let me denote the line as passing through A (2,1) with some slope m. Let me parametrize the line. Let's let the direction of the line be such that moving from A in one direction along the line, we reach B on the first circle, and moving in the opposite direction, we reach C on the second circle. Since AB = AC, the distances from A to B and from A to C are equal.Let me denote the parametric equations of the line. Let me set point A as (2, 1). Let me represent the line in parametric form. Let me use a parameter t. So, any point on the line can be written as (2 + t*cosθ, 1 + t*sinθ), where θ is the angle the line makes with the x-axis, and t is the parameter. Here, t=0 corresponds to point A. Then, moving in the direction of θ, t positive gives one direction, and t negative gives the opposite direction.Since AB = AC, the points B and C are located at t = k and t = -k for some k. So, point B would be (2 + k*cosθ, 1 + k*sinθ), and point C would be (2 - k*cosθ, 1 - k*sinθ). Then, since B is on the first circle (radius √5, center (0,0)), plugging into the first circle's equation:(2 + k*cosθ)^2 + (1 + k*sinθ)^2 = 5Similarly, point C is on the second circle (radius √2, center (3,0)):(2 - k*cosθ - 3)^2 + (1 - k*sinθ)^2 = 2Simplify the second equation:(-1 - k*cosθ)^2 + (1 - k*sinθ)^2 = 2Which is:(1 + k*cosθ)^2 + (1 - k*sinθ)^2 = 2So, now we have two equations:1. (2 + k*cosθ)^2 + (1 + k*sinθ)^2 = 52. (1 + k*cosθ)^2 + (1 - k*sinθ)^2 = 2Let me expand both equations.First equation:(4 + 4k*cosθ + k²*cos²θ) + (1 + 2k*sinθ + k²*sin²θ) = 5Combine like terms:4 + 1 + 4k*cosθ + 2k*sinθ + k²*(cos²θ + sin²θ) = 5Simplify:5 + 4k*cosθ + 2k*sinθ + k²*(1) = 5Therefore:k² + 4k*cosθ + 2k*sinθ + 5 = 5Subtract 5:k² + 4k*cosθ + 2k*sinθ = 0Similarly, second equation:(1 + 2k*cosθ + k²*cos²θ) + (1 - 2k*sinθ + k²*sin²θ) = 2Combine like terms:1 + 1 + 2k*cosθ - 2k*sinθ + k²*(cos²θ + sin²θ) = 2Simplify:2 + 2k*cosθ - 2k*sinθ + k²*1 = 2Subtract 2:k² + 2k*cosθ - 2k*sinθ = 0So, now we have two equations:1. k² + 4k*cosθ + 2k*sinθ = 02. k² + 2k*cosθ - 2k*sinθ = 0Let me denote these as Equation (1) and Equation (2).If I subtract Equation (2) from Equation (1):( k² + 4k*cosθ + 2k*sinθ ) - ( k² + 2k*cosθ - 2k*sinθ ) = 0 - 0Simplify:0 + 2k*cosθ + 4k*sinθ = 0Factor out 2k:2k*(cosθ + 2sinθ) = 0Since k is the distance from A to B (and A to C), and points B and C are distinct and not coinciding with A, k cannot be zero. Therefore:cosθ + 2sinθ = 0So:cosθ = -2sinθDivide both sides by cosθ (assuming cosθ ≠ 0):1 = -2tanθTherefore:tanθ = -1/2So, θ is the angle whose tangent is -1/2. Therefore, the slope of the line is tanθ = -1/2. Therefore, the line through A has slope -1/2.Alternatively, if cosθ = 0, then sinθ would have to be ±1, but in that case, from cosθ + 2sinθ = 0, we would have 0 + 2sinθ = 0, so sinθ = 0. But cosθ and sinθ cannot both be zero. So, the only solution is tanθ = -1/2. So, the line has slope -1/2.Therefore, the equation of the line is y - 1 = (-1/2)(x - 2). Let me write that as y = (-1/2)x + 2 + 1 → y = (-1/2)x + 3. Wait, no:Wait, slope is -1/2, passing through (2,1). So, using point-slope form:y - 1 = (-1/2)(x - 2)Thus,y = (-1/2)x + (1) + (1)Wait, (-1/2)(x - 2) = (-1/2)x + 1. Then, adding 1 gives y = (-1/2)x + 2.Wait, let me check:y - 1 = (-1/2)(x - 2)Therefore, y = (-1/2)(x - 2) + 1= (-1/2)x + 1 + 1= (-1/2)x + 2Yes, so the equation is y = (-1/2)x + 2.So, now that we know the slope is -1/2, we can parametrize the line accordingly. Alternatively, since we know θ, we can find sinθ and cosθ. Let's do that. tanθ = -1/2, so θ is in the second or fourth quadrant. Since we have a line passing through (2,1) with slope -1/2, it's going from (2,1) downwards to the right and upwards to the left.But maybe we can find sinθ and cosθ. Since tanθ = -1/2, we can consider a right triangle where the opposite side is -1 (but since tan is negative, the angle is in the fourth quadrant if we take the positive sides as 1 and 2). Alternatively, let's just use tanθ = -1/2. Then, sinθ = -1/√(1 + 4) = -1/√5 and cosθ = 2/√5. But since the slope is negative, the angle is in the fourth quadrant where cosθ is positive and sinθ is negative. Wait, but depending on the direction. Hmm, maybe it's better to compute sinθ and cosθ based on the slope.Alternatively, since the direction vector of the line is (2, -1), because the slope is -1/2, so for every 2 units in x, y decreases by 1. So, the direction vector is (2, -1). Therefore, cosθ = 2/√(2² + (-1)²) = 2/√5, and sinθ = -1/√5.Therefore, cosθ = 2/√5 and sinθ = -1/√5.Therefore, in our earlier equations, we can substitute cosθ and sinθ with these values. Let me recall that in Equations (1) and (2), we had:1. k² + 4k*cosθ + 2k*sinθ = 02. k² + 2k*cosθ - 2k*sinθ = 0Substituting cosθ = 2/√5 and sinθ = -1/√5:Equation (1):k² + 4k*(2/√5) + 2k*(-1/√5) = 0Simplify:k² + (8k/√5 - 2k/√5) = 0k² + (6k/√5) = 0Similarly, Equation (2):k² + 2k*(2/√5) - 2k*(-1/√5) = 0Simplify:k² + (4k/√5 + 2k/√5) = 0k² + (6k/√5) = 0So, both equations reduce to the same equation:k² + (6k/√5) = 0Factor out k:k(k + 6/√5) = 0Since k ≠ 0 (as points B and C are distinct from A), we have:k + 6/√5 = 0 → k = -6/√5But wait, k is the parameter in the parametric equation. Remember, in our parametrization, t is a parameter where t=0 is point A. Then, point B is at t = k, which would be (2 + k*(2/√5), 1 + k*(-1/√5)). Similarly, point C is at t = -k, which is (2 - k*(2/√5), 1 + k*(1/√5)). But wait, here we substituted cosθ = 2/√5 and sinθ = -1/√5. Therefore, the direction vector is (2/√5, -1/√5). So, the parameter t is scaled by the direction vector. Therefore, in terms of distance, each unit of t corresponds to moving 1 unit in the direction of the vector (2, -1). Wait, maybe I need to clarify.Wait, the parametric equations we set were (2 + t*cosθ, 1 + t*sinθ), where t is the parameter. Since we found cosθ = 2/√5 and sinθ = -1/√5, the parametric equations become:x = 2 + t*(2/√5)y = 1 + t*(-1/√5)So, here, t is the actual distance along the line from point A. Because the direction vector is (2/√5, -1/√5), which is a unit vector. Therefore, t is the actual distance from A. Therefore, point B is at distance |k| from A in the direction of the vector (2/√5, -1/√5), and point C is at distance |k| in the opposite direction. But since in our equations, we found that k = -6/√5. Wait, but distance can't be negative. Hmm, so maybe there's a confusion here.Wait, in our parametrization, t is a real number, positive in one direction, negative in the other. But when we solved the equation, we got k = -6/√5, which would mean that point B is in the direction opposite to the direction vector. But in our earlier setup, point B is supposed to be on the first circle (radius √5), and point C on the second circle (radius √2). Let me check if the direction is correct.Wait, the line passes through A (2,1) with direction vector (2/√5, -1/√5). Moving in the direction of the vector (2/√5, -1/√5) from A would take us towards increasing x and decreasing y. But point B is on the first circle (center at (0,0)), so moving in that direction might not reach the first circle. Alternatively, moving in the opposite direction (towards decreasing x and increasing y) might reach the first circle.Wait, let's check with k = -6/√5. Then, point B would be at:x = 2 + (-6/√5)*(2/√5) = 2 - 12/5 = 2 - 2.4 = -0.4y = 1 + (-6/√5)*(-1/√5) = 1 + 6/5 = 1 + 1.2 = 2.2So, point B is at (-0.4, 2.2). Let me check if this is on the first circle x² + y² = 5:(-0.4)^2 + (2.2)^2 = 0.16 + 4.84 = 5.0. Yes, that works. So, indeed, point B is (-0.4, 2.2), which is 5 units from the origin. Then, point C would be at t = -k = 6/√5. So, moving in the opposite direction:x = 2 + (6/√5)*(2/√5) = 2 + 12/5 = 2 + 2.4 = 4.4y = 1 + (6/√5)*(-1/√5) = 1 - 6/5 = 1 - 1.2 = -0.2So, point C is (4.4, -0.2). Let's check if this is on the second circle (x - 3)^2 + y^2 = 2:(4.4 - 3)^2 + (-0.2)^2 = (1.4)^2 + 0.04 = 1.96 + 0.04 = 2.0. Correct. Therefore, points B and C are (-0.4, 2.2) and (4.4, -0.2), respectively. The distance from A (2,1) to B (-0.4, 2.2):AB = sqrt[(2 - (-0.4))^2 + (1 - 2.2)^2] = sqrt[(2.4)^2 + (-1.2)^2] = sqrt[5.76 + 1.44] = sqrt[7.2] = sqrt[36/5] = 6/√5Similarly, AC = sqrt[(2 - 4.4)^2 + (1 - (-0.2))^2] = sqrt[(-2.4)^2 + (1.2)^2] = sqrt[5.76 + 1.44] = sqrt[7.2] = 6/√5So, AB = AC = 6/√5. Therefore, the length AB is 6/√5. Rationalizing the denominator, that's (6√5)/5. So, the answer is 6√5/5, which in boxed form is boxed{dfrac{6sqrt{5}}{5}}.But let me verify if this makes sense. The line passes through A (2,1) with slope -1/2. The direction vector is (2, -1), which we normalized. The distance from A to each point B and C is 6/√5, which is approximately 2.68. Given the radii of the circles (√5 ≈ 2.24 and √2 ≈ 1.41), but wait, point C is on the second circle with radius √2 ≈ 1.41, but AC is 6/√5 ≈ 2.68. Wait, that seems longer than the radius. How is that possible?Wait, the second circle has center at (3,0) and radius √2. The distance from center O2 (3,0) to point C (4.4, -0.2) should be √2. Let's check:Distance from (3,0) to (4.4, -0.2):sqrt[(4.4 - 3)^2 + (-0.2 - 0)^2] = sqrt[(1.4)^2 + (-0.2)^2] = sqrt[1.96 + 0.04] = sqrt[2] ≈ 1.41. Correct. So, the point C is indeed on the second circle. Similarly, point B is on the first circle. Even though the distance from A to C is longer than the radius of the second circle, that's okay because point C is not the center. The distance from the center O2 to point C is √2, but the distance from A to C is different.So, even though the radius of the second circle is √2 ≈ 1.41, the distance from A to C is 6/√5 ≈ 2.68. That's possible because point A is an intersection point of the two circles, so it's located at a certain position relative to both centers. Therefore, the length AC isn't constrained by the radius of the circle but by the position of point A and the line passing through it.Therefore, the answer seems to be 6√5/5.But just to make sure, let's check the calculations again.We found the coordinates of B and C as (-0.4, 2.2) and (4.4, -0.2). Let's compute AB and AC.AB:From (2,1) to (-0.4,2.2):Δx = -0.4 - 2 = -2.4, Δy = 2.2 - 1 = 1.2Distance AB = sqrt[(-2.4)^2 + (1.2)^2] = sqrt[5.76 + 1.44] = sqrt[7.2] = sqrt[36/5] = 6/√5. Correct.AC:From (2,1) to (4.4, -0.2):Δx = 4.4 - 2 = 2.4, Δy = -0.2 - 1 = -1.2Distance AC = sqrt[(2.4)^2 + (-1.2)^2] = same as above, 6/√5. Correct.Therefore, AB = AC = 6/√5, which is 6√5/5. Therefore, the answer is boxed{dfrac{6sqrt{5}}{5}}.Final Answerboxed{dfrac{6sqrt{5}}{5}}</think>Given two circles with radii (sqrt{5}) and (sqrt{2}) intersecting at point (A), with the distance between their centers being 3, we need to find the length (AB) where a line through (A) intersects the circles at points (B) and (C) such that (AB = AC).1. Finding the coordinates of point (A): - The equations of the circles are (x^2 + y^2 = 5) and ((x - 3)^2 + y^2 = 2). - Solving these equations, we find the intersection points ((2, 1)) and ((2, -1)). We choose (A = (2, 1)).2. Determining the line through (A): - The line through (A) with slope (-frac{1}{2}) (since (tan theta = -frac{1}{2})) has the equation (y = -frac{1}{2}x + 2).3. Parametrizing the line: - Using parametric form, the coordinates of points (B) and (C) on the line are ((2 + t cos theta, 1 + t sin theta)) and ((2 - t cos theta, 1 - t sin theta)) where (cos theta = frac{2}{sqrt{5}}) and (sin theta = -frac{1}{sqrt{5}}).4. Solving for (t): - Substituting the coordinates into the circle equations, we find (t = frac{6}{sqrt{5}}).5. Calculating the distance (AB): - The distance (AB) (and (AC)) is found to be (frac{6sqrt{5}}{5}).Thus, the length (AB) is (boxed{dfrac{6sqrt{5}}{5}}).,<think>Okay, so we have two intersecting circles. The radii are sqrt(5) and sqrt(2), and the distance between their centers is 3. A line through their intersection point A intersects the circles again at B and C such that AB = AC. We need to find AB.First, let me visualize the problem. There are two circles intersecting at point A. The line through A passes through another point B on the first circle (radius sqrt(5)) and another point C on the second circle (radius sqrt(2)), with AB equal to AC. So, the line passes through A, and extends to B and C such that the segments from A to B and A to C are equal. So, B and C are on opposite sides of A on the line, right? Because if they were on the same side, then AB = AC would mean B and C coincide, but the problem says they don't. So, A is between B and C, and AB = AC. Therefore, BC is twice AB. So, BC = 2AB. But maybe that's getting ahead of myself.Let me recall some theorems. The problem mentions the Pythagorean Theorem and the Theorem on the sum of the squares of the diagonals. Hmm, the sum of the squares of the diagonals in a parallelogram? Yes, in a parallelogram, the sum of the squares of the diagonals equals twice the sum of the squares of the sides. Maybe that's relevant here.Let me think. Let's denote the centers of the two circles as O1 and O2. The distance between O1 and O2 is 3. The radii are sqrt(5) and sqrt(2). The line through A intersects the first circle again at B and the second circle again at C, such that AB = AC. So, points B and C are on the line such that A is the midpoint of BC? Wait, if AB = AC and B and C are distinct, then A must be between B and C, and AB = AC implies that BA = AC, so BC = 2AB. So, BC is twice AB. So, the length AB is half the length of BC. But BC is the chord passing through A in both circles.Alternatively, perhaps we can model this with coordinates. Let me set up coordinate systems. Let's place O1 at the origin (0,0), and O2 at (3,0), since the distance between them is 3. Then, the two circles have equations:Circle O1: x² + y² = 5Circle O2: (x - 3)² + y² = 2They intersect at point A. Let me find the coordinates of point A. To find the intersection points, we can solve the two equations:x² + y² = 5(x - 3)² + y² = 2Subtracting the first equation from the second:(x - 3)² + y² - x² - y² = 2 - 5Expand (x - 3)²: x² - 6x + 9So, x² - 6x + 9 - x² = -3Simplify: -6x + 9 = -3Then, -6x = -12 => x = 2Substitute x = 2 into the first equation: 2² + y² = 5 => 4 + y² = 5 => y² = 1 => y = 1 or y = -1Therefore, the points of intersection are (2,1) and (2,-1). Let's take point A as (2,1). The other intersection is (2,-1), but since the line through A can be any line, maybe we can consider either of them, but the problem states that the line passes through A, which is one of the points. So, let's fix A as (2,1).Now, the line passing through A (2,1) intersects circle O1 again at B and circle O2 again at C, such that AB = AC. We need to find AB.Let me parametrize the line. Let the line have a direction vector. Let's suppose the line has a slope m. Then, parametric equations can be written as x = 2 + t, y = 1 + mt. Alternatively, since we can have any line through A, perhaps it's better to use a parameter t to represent points along the line.But maybe a better approach is to consider vectors. Let’s denote the direction vector of the line as (a, b). Then, any point on the line can be written as (2 + at, 1 + bt) for some parameter t.Since B is on circle O1 (radius sqrt(5)), so plugging into O1's equation:(2 + at)^2 + (1 + bt)^2 = 5Similarly, point C is on circle O2:(2 + at - 3)^2 + (1 + bt)^2 = 2 => (-1 + at)^2 + (1 + bt)^2 = 2We need AB = AC. Let's express AB and AC in terms of t. Since A is (2,1), and B is (2 + at, 1 + bt), the distance AB is sqrt[(at)^2 + (bt)^2] = |t| sqrt(a² + b²). Similarly, AC would be the distance from A to C. But wait, since C is another intersection with circle O2, but depending on the direction, the parameter t for C might be different.Wait, perhaps the line passes through A, and extends to meet circle O1 again at B and circle O2 again at C. So, starting at A, moving in one direction along the line, you reach B on O1, and moving in the opposite direction, you reach C on O2. But since AB = AC, the distances from A to B and A to C must be equal.But if we parametrize the line as (2 + at, 1 + bt), then B corresponds to some parameter t1 (so that the point is on O1) and C corresponds to some parameter t2 (on O2). Then AB = |t1| sqrt(a² + b²), AC = |t2| sqrt(a² + b²). The condition AB = AC implies |t1| = |t2|. But since B and C are in opposite directions from A, t1 and t2 have opposite signs. So, perhaps t1 = -t2. Therefore, the parameters for B and C are t and -t for some t.Let me formalize this. Let’s set the direction vector as (a, b), then points B and C can be represented as:B: (2 + a t, 1 + b t) on O1C: (2 - a t, 1 - b t) on O2Because if we go t units in the direction (a, b) from A to get B, then going -t units (i.e., t units in the opposite direction) from A would give C. Since AB = AC, the distances are the same.Therefore, substituting B into O1:(2 + a t)^2 + (1 + b t)^2 = 5Similarly, substituting C into O2:(2 - a t - 3)^2 + (1 - b t)^2 = 2Simplify O2's equation:(-1 - a t)^2 + (1 - b t)^2 = 2Expanding both equations:For O1:4 + 4 a t + a² t² + 1 + 2 b t + b² t² = 5Combine terms:(4 + 1) + (4 a t + 2 b t) + (a² + b²) t² = 55 + t(4a + 2b) + t²(a² + b²) = 5Subtract 5:t(4a + 2b) + t²(a² + b²) = 0Similarly, for O2's equation:(1 + 2 a t + a² t²) + (1 - 2 b t + b² t²) = 2Wait, let's expand (-1 - a t)^2:(-1 - a t)^2 = 1 + 2 a t + a² t²(1 - b t)^2 = 1 - 2 b t + b² t²So adding them:1 + 2 a t + a² t² + 1 - 2 b t + b² t² = 2Combine terms:(1 + 1) + (2a t - 2b t) + (a² + b²) t² = 22 + 2t(a - b) + t²(a² + b²) = 2Subtract 2:2t(a - b) + t²(a² + b²) = 0So now we have two equations from O1 and O2:From O1:t(4a + 2b) + t²(a² + b²) = 0From O2:2t(a - b) + t²(a² + b²) = 0Assuming t ≠ 0 (since t = 0 would correspond to point A itself), we can divide both equations by t:From O1:4a + 2b + t(a² + b²) = 0From O2:2(a - b) + t(a² + b²) = 0Let’s denote k = t(a² + b²). Then:From O1: 4a + 2b + k = 0From O2: 2(a - b) + k = 0So, we have:4a + 2b + k = 02a - 2b + k = 0Subtract the second equation from the first:(4a + 2b + k) - (2a - 2b + k) = 0 - 02a + 4b = 0Thus:2a + 4b = 0 => a + 2b = 0 => a = -2bSo, the direction vector (a, b) must satisfy a = -2b. Therefore, the direction vector is (-2b, b). Let's choose b = 1 for simplicity, so the direction vector is (-2, 1). Therefore, the line has direction (-2, 1). But we can choose any scalar multiple, so let's confirm.But since direction vector is (-2, 1), the line through A (2,1) is parametrized as:x = 2 - 2ty = 1 + tSo, points B and C would be:B: moving in the direction of (-2,1) from A, which would be t positive?Wait, but earlier, we had B as (2 + a t, 1 + b t) where a = -2b. But with a = -2 and b = 1, then:B is (2 - 2t, 1 + t)C is (2 + 2t, 1 - t) since going in the opposite direction.Wait, no. Wait, when we set a = -2b, and we set direction vector as (a, b) = (-2,1), then moving in the direction of (a, b) from A gives B: (2 - 2t, 1 + t). But C is supposed to be in the opposite direction, which would be (2 + 2t, 1 - t). Let me verify.But in the parametrization earlier, we had:For B: (2 + a t, 1 + b t) with a = -2, b = 1, so (2 - 2t, 1 + t)For C: (2 - a t, 1 - b t) which would be (2 + 2t, 1 - t)So, yes, that's correct.Now, substituting these into the equations for the circles:First, for point B on circle O1:x² + y² = (2 - 2t)^2 + (1 + t)^2 = 5Expand:(4 - 8t + 4t²) + (1 + 2t + t²) = 5Combine like terms:4 + 1 + (-8t + 2t) + (4t² + t²) = 55 - 6t + 5t² = 5Subtract 5:-6t + 5t² = 0Factor:t(-6 + 5t) = 0Solutions t = 0 (which is point A) and t = 6/5.Similarly, for point C on circle O2:(x - 3)^2 + y^2 = (2 + 2t - 3)^2 + (1 - t)^2 = 2Simplify:(-1 + 2t)^2 + (1 - t)^2 = 2Expand:(1 - 4t + 4t²) + (1 - 2t + t²) = 2Combine:1 + 1 + (-4t - 2t) + (4t² + t²) = 22 - 6t + 5t² = 2Subtract 2:-6t + 5t² = 0Same equation as before: t(-6 + 5t) = 0, solutions t=0 and t=6/5.Therefore, t=6/5 gives the points B and C.Therefore, AB is the distance from A to B. Let's compute that.Point B is (2 - 2*(6/5), 1 + (6/5)) = (2 - 12/5, 1 + 6/5) = (-2/5, 11/5)Point A is (2,1). So, distance AB is sqrt[(-2/5 - 2)^2 + (11/5 - 1)^2]Compute differences:x: (-2/5 - 10/5) = -12/5y: (11/5 - 5/5) = 6/5So, distance AB = sqrt[(-12/5)^2 + (6/5)^2] = sqrt[(144/25) + (36/25)] = sqrt[180/25] = sqrt[36*5/25] = sqrt[36/5] = 6/sqrt(5) = (6 sqrt(5))/5Alternatively, since we parametrized the line with direction vector (-2,1), the distance from A to B is |t| * sqrt[(-2)^2 + 1^2] = t * sqrt(5). Since t = 6/5, AB = (6/5)*sqrt(5) = (6 sqrt(5))/5.Similarly, AC would be the same, since t is 6/5 in the opposite direction. So, AB = AC = 6 sqrt(5)/5.But wait, the problem says "Find AB", so the answer is 6√5⁄5. Let me check if this is correct.Wait, but in the parametrization, we set direction vector as (-2,1). But direction vectors can be scaled. However, in our calculation, we derived t=6/5. So, the parameter t in the parametrization gives the distance multiplied by sqrt(5). Wait, actually, the parameter t is not the distance, but the scaling factor for the direction vector. The actual distance is t times the length of the direction vector. Since the direction vector (-2,1) has length sqrt(4 + 1) = sqrt(5). Therefore, the distance AB is t * sqrt(5), which is (6/5)*sqrt(5) = 6/sqrt(5) as above, but rationalized is 6 sqrt(5)/5. That's the same.Therefore, the answer is 6√5⁄5.But let me verify this with another approach to be sure.Alternatively, using coordinate geometry, since we have points A(2,1), B(-2/5,11/5), and C(2 + 2*(6/5),1 - 6/5) = (2 + 12/5, 1 - 6/5) = (22/5, -1/5). Let's check the distance AC.Point C is (22/5, -1/5). Distance from A(2,1) to C(22/5, -1/5):x difference: 22/5 - 10/5 = 12/5y difference: -1/5 - 5/5 = -6/5Distance AC = sqrt[(12/5)^2 + (-6/5)^2] = sqrt[144/25 + 36/25] = sqrt[180/25] = sqrt(36*5/25) = 6/sqrt(5) = 6√5/5. So, same as AB. So that checks out.Alternatively, perhaps using the theorem on the sum of squares of diagonals. Wait, if we consider quadrilateral O1AO2B or something?Wait, maybe O1, O2, B, C form a quadrilateral? Let me think. The centers O1 and O2, and points B and C. But B is on circle O1, C is on circle O2. Let me consider vectors or distances.Alternatively, using the theorem on the sum of squares of diagonals in a parallelogram. Wait, if we can form a parallelogram with the centers and the points B and C?Alternatively, perhaps using the fact that AB = AC, so the line is such that A is the midpoint between B and C. Wait, if AB = AC and A is between B and C, then yes, A is the midpoint. So, BC = 2AB. Therefore, points B and C are such that A is the midpoint. Then, perhaps using the midpoint theorem.But since B is on circle O1 and C is on circle O2, then coordinates of B and C are related. Let me denote coordinates: Let B = (x1, y1) on O1, so x1² + y1² = 5. C = (x2, y2) on O2, so (x2 - 3)^2 + y2² = 2. Since A is the midpoint of BC, then:( (x1 + x2)/2, (y1 + y2)/2 ) = (2,1)Therefore:x1 + x2 = 4y1 + y2 = 2So, x2 = 4 - x1, y2 = 2 - y1Substitute into O2's equation:(4 - x1 - 3)^2 + (2 - y1)^2 = 2Simplify:(1 - x1)^2 + (2 - y1)^2 = 2Expand:1 - 2x1 + x1² + 4 - 4y1 + y1² = 2Combine:(1 + 4) + (-2x1 -4y1) + (x1² + y1²) = 25 - 2x1 -4y1 + (x1² + y1²) = 2But since B is on O1, x1² + y1² = 5, so substitute:5 - 2x1 -4y1 +5 = 210 - 2x1 -4y1 = 2-2x1 -4y1 = -8Divide by -2:x1 + 2y1 = 4So, we have the equation x1 + 2y1 = 4. Also, since B is on O1: x1² + y1² = 5So, solving the system:x1 + 2y1 = 4x1² + y1² = 5Let me solve for x1 from the first equation: x1 = 4 - 2y1Substitute into the second equation:(4 - 2y1)^2 + y1² = 5Expand:16 - 16y1 + 4y1² + y1² = 516 -16y1 +5y1² =55y1² -16y1 +11 =0Quadratic equation: 5y1² -16y1 +11=0Compute discriminant: 256 - 220 = 36Solutions: y1 = [16 ±6]/10 => (22)/10=11/5 or (10)/10=1So, y1=11/5 or y1=1If y1=1, then x1=4 -2*1=2. So, point B would be (2,1), which is point A. Not allowed, since B and C cannot coincide with A.Therefore, y1=11/5. Then x1=4 -2*(11/5)=4 -22/5= (20/5 -22/5)= -2/5Therefore, B is (-2/5,11/5), which matches the earlier result. Therefore, AB is the distance from (2,1) to (-2/5,11/5), which we already calculated as 6√5⁄5.So, this confirms the answer is 6√5⁄5.Alternatively, another approach using the theorem on sum of squares of diagonals. If we consider quadrilateral O1BO2C. Wait, O1 and O2 are centers, B is on O1's circle, C is on O2's circle. If we can relate the lengths.But since A is the midpoint of BC, maybe consider vectors or coordinate geometry. Wait, but we already solved it.Alternatively, the theorem on the sum of squares of the diagonals in a quadrilateral. For any quadrilateral, the sum of the squares of the diagonals is equal to twice the sum of the squares of the two sides plus twice the sum of the squares of the other two sides. Wait, no, in a parallelogram, the sum of the squares of the diagonals equals twice the sum of the squares of the sides. But perhaps if we can form a parallelogram here.Alternatively, perhaps considering triangles O1AO2, and apply the Pythagorean theorem or the law of cosines.The distance between O1 and O2 is 3. The radii are sqrt(5) and sqrt(2). Let's consider triangle O1AO2. The sides are O1A = sqrt(5), O2A = sqrt(2), and O1O2 = 3.Using the Law of Cosines on triangle O1AO2:(O1O2)^2 = (O1A)^2 + (O2A)^2 - 2*O1A*O2A*cos(theta)Where theta is the angle at A.But wait, actually, the Law of Cosines here would be:3² = (sqrt(5))² + (sqrt(2))² - 2*sqrt(5)*sqrt(2)*cos(angle O1AO2)Compute:9 = 5 + 2 - 2*sqrt(10)*cos(theta)9 = 7 - 2√10 cos(theta)Thus:2√10 cos(theta) = 7 - 9 = -2Therefore:cos(theta) = -2 / (2√10) = -1/√10So, angle theta at A is arccos(-1/√10). But not sure if this helps directly.Alternatively, maybe consider coordinates again. Since we know the coordinates of A, O1, O2, we can compute vectors.Vector AO1 is from A(2,1) to O1(0,0): (-2, -1)Vector AO2 is from A(2,1) to O2(3,0): (1, -1)The angle between AO1 and AO2 can be found using the dot product:cos(theta) = [(-2)(1) + (-1)(-1)] / (|AO1| |AO2|)Compute numerator: -2 +1 = -1Denominator: |AO1| = sqrt(4 +1)=sqrt(5); |AO2|=sqrt(1 +1)=sqrt(2)Thus, cos(theta)= -1/(sqrt(5)*sqrt(2)) = -1/sqrt(10), which matches earlier result.But perhaps this isn't directly helpful. Maybe we need to relate the position of line BC with the centers.Since line BC passes through A, and AB = AC, which we already used to find that A is the midpoint.Alternatively, using power of a point. The power of point B with respect to circle O2 is BO2² - (sqrt(2))² = BO2² - 2. But since B is on line BC which passes through A, and A is the midpoint of BC, we might relate distances.But perhaps this complicates things. Our initial approach using coordinates gave the answer, and cross-checked with another method gave the same result. So, likely the answer is 6√5⁄5.Therefore, the answer is boxed{dfrac{6sqrt{5}}{5}}.Final AnswerThe length of ( AB ) is (boxed{dfrac{6sqrt{5}}{5}}).</think>Given two intersecting circles with radii (sqrt{5}) and (sqrt{2}) and centers (O_1) and (O_2) respectively, where the distance between the centers is 3. A line through the intersection point (A) intersects the circles again at points (B) and (C) such that (AB = AC). We need to find the length (AB).1. Finding Intersection Points: - The equations of the circles are: - Circle (O_1): (x^2 + y^2 = 5) - Circle (O_2): ((x - 3)^2 + y^2 = 2) - Solving these equations, the intersection points are found to be ((2, 1)) and ((2, -1)). We consider point (A) as ((2, 1)).2. Parametrizing the Line: - The line through (A) is parametrized with direction vector ((a, b)). Points (B) and (C) are parameterized as ((2 + at, 1 + bt)) and ((2 - at, 1 - bt)) respectively. - Substituting these points into the circle equations and solving for (t), we find the direction vector ((-2, 1)) and parameter (t = frac{6}{5}).3. Calculating Distance: - Using the parameter (t = frac{6}{5}) and the direction vector ((-2, 1)), the coordinates of (B) and (C) are found. - The distance (AB) is calculated using the distance formula between points (A(2, 1)) and (Bleft(-frac{2}{5}, frac{11}{5}right)): [ AB = sqrt{left(2 - left(-frac{2}{5}right)right)^2 + left(1 - frac{11}{5}right)^2} = sqrt{left(frac{12}{5}right)^2 + left(-frac{6}{5}right)^2} = sqrt{frac{144}{25} + frac{36}{25}} = sqrt{frac{180}{25}} = frac{6sqrt{5}}{5} ]Thus, the length of (AB) is (boxed{dfrac{6sqrt{5}}{5}}).]

🔑:Let O_{1} and O_{2} be the centers of the smaller and larger circles, respectively. Drop perpendiculars O_{1} M and O_{2} N to the line B C and denoteA B=A C=2 x, O_{1} M=a, O_{2} N=b. Then b^{2}-a^{2}=left(5-x^{2}right)-left(2-x^{2}right)=3, O_{1} M^{2}=2-x^{2}, O_{2} N^{2}=5-x^{2}.The midline of the trapezoid O_{1} M N O O_{2} is 1 / 2(b+a). On the other hand, it is the median of the triangle O_{1} A O_{2}, and therefore (b+a)^{2}=2left(O_{1} A^{2}+O_{2} A^{2}right)-O_{1} O_{2}^{2}=5 (see problem underline{55267} ).Thus, 2 x^{2}=left(2-a^{2}right)+left(5-b^{2}right)=7-a^{2}-b^{2}, 9=(b-a)^{2}+4 x^{2}=(b-a)^{2}+14-2 a^{2}-2 b^{2}=14-(b+a)^{2}, from which (b+a)^{2}=5,(b-a)^{2}=left(b^{2}-a^{2}right)^{2} /(b+a)^{2}=9 / 5, a^{2}+b^{2}=1 / 2left((b+a)^{2}+(b-a)^{2}right)=17 / 5, x^{2}=1 / 2left(7-a^{2}-b^{2}right)=9 / 5. Therefore, A B= 2 x=frac{6}{sqrt{5}}## AnswerProblem

🧠:[<think>Okay, so I need to figure out the relationship between profit y and the selling price x for this company's green tea. Let me start by recalling what profit means. Profit is basically total revenue minus total cost, right? First, let's parse the information given. The cost is 50 per kilogram. The sales volume w depends on the selling price x with the equation w = -2x + 240. So as the price x increases, the amount they sell w decreases linearly. That makes sense because if something is more expensive, people buy less of it.So, profit y should be total revenue minus total cost. Let me write that down. Total revenue is the amount they make from selling the tea, which would be the selling price per kilogram times the number of kilograms sold. So that's x * w. Total cost is the cost to produce the tea, which is 50 per kilogram times the number of kilograms sold. So that's 50 * w. Therefore, profit y = revenue - cost = x*w - 50*w. But wait, w is given as a function of x, which is w = -2x + 240. So I can substitute that into the equation for y. Let me do that step by step. First, write revenue and cost:Revenue = x * w = x*(-2x + 240) = -2x² + 240xCost = 50 * w = 50*(-2x + 240) = -100x + 12,000Therefore, profit y = revenue - cost = (-2x² + 240x) - (-100x + 12,000)Let me compute that:y = -2x² + 240x + 100x - 12,000Combine like terms:240x + 100x is 340x, so:y = -2x² + 340x - 12,000Hmm, let me check if that makes sense. Let's make sure units are okay. x is in dollars per kilogram, w is in kilograms. Revenue would be dollars/kg * kg = dollars. Cost is dollars/kg * kg = dollars. Profit is dollars, so that works. The coefficients: -2x² comes from the revenue term, which is quadratic because as price increases, sales decrease, leading to a quadratic relationship. The linear term 340x comes from both revenue and cost. The constant term is -12,000, which is from the cost. So that seems okay. Let me verify with an example. Suppose x is 50 dollars per kg. Then w = -2*50 + 240 = -100 + 240 = 140 kg. Revenue would be 50 * 140 = 7000 dollars. Cost would be 50 * 140 = 7000 dollars. Profit would be 0. Let's plug x=50 into the equation for y:y = -2*(50)^2 + 340*(50) -12,000 = -2*2500 + 17,000 -12,000 = -5,000 +17,000 -12,000 = 0. That checks out.Another example: x=0. If they give the tea away for free, w would be 240 kg. Revenue is 0. Cost is 50*240=12,000. Profit y= -12,000, which matches the equation. Plugging x=0 into y = -0 + 0 -12,000 = -12,000. Perfect.Another example: x=100. Then w = -2*100 +240=40 kg. Revenue=100*40=4,000. Cost=50*40=2,000. Profit=2,000. Plug into equation: y = -2*(100)^2 +340*100 -12,000 = -20,000 +34,000 -12,000=2,000. Correct.Okay, so the equation for y in terms of x is y = -2x² +340x -12,000. That answers part (1).Now part (2) asks for the value of x that maximizes y. Since this is a quadratic function in terms of x, and the coefficient of x² is negative (-2), the parabola opens downward, so the maximum occurs at the vertex.The x-coordinate of the vertex of a parabola given by y = ax² + bx + c is at x = -b/(2a). In this case, a = -2, b = 340. So:x = -340/(2*(-2)) = -340/(-4) = 85.Therefore, the maximum profit occurs when x is 85 dollars per kilogram. Let me verify that. If x=85, then w = -2*85 +240 = -170 +240=70 kg. Revenue=85*70=5,950. Cost=50*70=3,500. Profit=5,950 -3,500=2,450.Plugging into the equation: y = -2*(85)^2 +340*85 -12,000First, calculate 85 squared: 85*85=7,225. Then:-2*7,225 = -14,450340*85: 340*80=27,200; 340*5=1,700; total 28,900Then, y = -14,450 +28,900 -12,000 = (28,900 -14,450) -12,000 =14,450 -12,000=2,450. Correct.So the maximum profit is 2,450 when x is 85 per kilogram. But wait, let me make sure there are no calculation errors here. Let me recheck the vertex formula.Yes, for a quadratic y = ax² +bx +c, vertex at x=-b/(2a). Here a=-2, b=340, so x= -340/(2*(-2))= -340/-4=85. Correct.Alternatively, using calculus, take derivative dy/dx = -4x +340. Set to zero: -4x +340=0 => x=340/4=85. So same result. Therefore, x=85 is indeed the value where profit is maximized. So, to recap:1) The relationship between y and x is y = -2x² +340x -12,0002) The profit is maximized when x is 85 dollars per kilogram.Final Answer(1) The relationship between ( y ) and ( x ) is (boxed{y = -2x^2 + 340x - 12000}).(2) The value of ( x ) that maximizes ( y ) is (boxed{85}).</think>To find the relationship between profit ( y ) and selling price ( x ):1. Revenue Calculation: - Sales volume ( w ) is given by ( w = -2x + 240 ). - Revenue is ( x times w = x(-2x + 240) = -2x^2 + 240x ).2. Cost Calculation: - Cost is ( 50 times w = 50(-2x + 240) = -100x + 12000 ).3. Profit Calculation: - Profit ( y ) is revenue minus cost: [ y = (-2x^2 + 240x) - (-100x + 12000) ] - Simplifying the expression: [ y = -2x^2 + 240x + 100x - 12000 = -2x^2 + 340x - 12000 ]Thus, the relationship between ( y ) and ( x ) is ( y = -2x^2 + 340x - 12000 ).To find the value of ( x ) that maximizes ( y ):1. Quadratic Function Vertex: - The profit function ( y = -2x^2 + 340x - 12000 ) is a quadratic function opening downwards (since the coefficient of ( x^2 ) is negative). - The vertex of a quadratic function ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). - Here, ( a = -2 ) and ( b = 340 ), so: [ x = -frac{340}{2(-2)} = -frac{340}{-4} = 85 ]Therefore, the value of ( x ) that maximizes ( y ) is ( 85 ). Final Answer(1) The relationship between ( y ) and ( x ) is (boxed{y = -2x^2 + 340x - 12000}).(2) The value of ( x ) that maximizes ( y ) is (boxed{85}).,<think>Okay, let's tackle this problem step by step. So, the company sells green tea at a cost of 50 per kilogram. The sales volume, which is the amount they sell, is given by the equation w = -2x + 240, where x is the selling price per kilogram. The profit y depends on both the selling price and the cost, right? First, I need to figure out the relationship between profit y and the selling price x. Profit is usually calculated as total revenue minus total cost. Total revenue would be the selling price per kilogram times the number of kilograms sold, which is x * w. Total cost is the cost per kilogram times the number of kilograms sold, so that's 50 * w. Therefore, profit y should be revenue minus cost, so y = (x * w) - (50 * w). Since w is given as a function of x, which is w = -2x + 240, I can substitute that into the profit equation. Let's write that out. Substituting w into the profit equation gives y = x*(-2x + 240) - 50*(-2x + 240). Let me compute that step by step.First, expand the terms. For the revenue part: x*(-2x + 240) = -2x² + 240x. Then, for the cost part: -50*(-2x + 240) = 100x - 12,000. So combining both parts, the profit y is (-2x² + 240x) + (100x - 12,000). Now, combine like terms. The 240x and 100x add up to 340x. So the equation becomes y = -2x² + 340x - 12,000. Let me check if I did the algebra correctly. Multiplying through: yes, x times -2x is -2x², x times 240 is 240x. Then, -50 times -2x is +100x, and -50 times 240 is -12,000. Then combining 240x + 100x gives 340x. So, y = -2x² + 340x - 12,000. That seems right. So that's part (1) done, the relationship between y and x.Now, part (2) asks for the value of x that maximizes y. Since the profit equation is a quadratic function in terms of x, and the coefficient of x² is negative (-2), the parabola opens downward, meaning the vertex is the maximum point. The x-coordinate of the vertex of a parabola given by y = ax² + bx + c is found by the formula -b/(2a). In this case, a is -2 and b is 340. Plugging into the formula, we get x = -340/(2*(-2)) = -340/(-4) = 85. So, x = 85. Let me verify this calculation. The formula is indeed -b/(2a). Here, b is 340, so -340 divided by (2 times -2) is -340 divided by -4, which is positive 85. That seems correct.Wait, but let me think again. If the selling price is 85 per kilogram, does that make sense? The original cost is 50 per kilogram. So, the profit per kilogram would be 85 - 50 = 35. But according to the sales volume equation, when x is 85, w = -2*85 + 240 = -170 + 240 = 70 kilograms. So the profit would be 70*(85 - 50) = 70*35 = 2,450. But let's check using the profit equation y = -2x² + 340x - 12,000. Plugging x = 85, we get y = -2*(85)^2 + 340*85 - 12,000. Compute 85 squared: 85*85 is 7,225. Multiply by -2: -14,450. Then 340*85: 340*80 = 27,200 and 340*5=1,700, so total 28,900. Then subtract 12,000. So total y = -14,450 + 28,900 - 12,000. Let's compute that: -14,450 + 28,900 is 14,450. Then 14,450 - 12,000 = 2,450. That matches. So the maximum profit is 2,450 when x is 85.But wait, let me check if there's any possible mistake here. For example, did I use the right formula for profit? Profit is indeed (selling price - cost) times quantity sold, which is (x - 50)*w. And since w = -2x + 240, then y = (x - 50)*(-2x + 240). Let's compute that to see if we get the same equation as before.Expanding (x - 50)*(-2x + 240): x*(-2x) + x*240 - 50*(-2x) + (-50)*240 = -2x² + 240x + 100x - 12,000. Combine like terms: -2x² + 340x - 12,000. Yes, same result. So that checks out. So, using vertex formula, x = -b/(2a) = -340/(2*(-2)) = 85, which gives the maximum profit. Therefore, x = 85 is the correct answer. But wait another thought: since the problem is about a real-world scenario, should we check if the value x=85 is within a reasonable range? For example, the sales volume w must be non-negative. Let's check when w = 0: 0 = -2x + 240 => x = 120. So the maximum possible selling price before sales volume becomes zero is 120. Since 85 is less than 120, it's a valid price. Also, the cost is 50, so selling price has to be above 50 to make a profit. If selling price is exactly 50, profit per unit is zero, but total cost would be 50*w, and revenue would be 50*w, so profit is zero. If x is less than 50, they would be selling at a loss. So, the maximum profit should occur somewhere above 50. Since 85 is above 50, that's valid. So, all checks out. Another way to verify is to take the derivative of y with respect to x, set it to zero, and solve for x. Since y = -2x² + 340x - 12,000, the derivative dy/dx = -4x + 340. Setting this equal to zero: -4x + 340 = 0 => 4x = 340 => x = 85. Same result. So calculus also confirms this. Therefore, I'm confident that part (1) the relationship is y = -2x² + 340x - 12,000, and part (2) the maximum profit occurs at x = 85.Final Answer(1) The relationship between ( y ) and ( x ) is ( y = -2x^2 + 340x - 12000 ). (2) The value of ( x ) that maximizes ( y ) is boxed{85}.</think>To find the relationship between profit ( y ) and selling price ( x ):1. Profit Calculation: - Revenue is given by ( x times w ). - Cost is given by ( 50 times w ). - Profit ( y ) is revenue minus cost: ( y = x cdot w - 50 cdot w ).2. Substitute Sales Volume: - Sales volume ( w ) is given by ( w = -2x + 240 ). - Substitute ( w ) into the profit equation: [ y = x(-2x + 240) - 50(-2x + 240) ]3. Expand and Simplify: - Expand the terms: [ y = -2x^2 + 240x + 100x - 12000 ] - Combine like terms: [ y = -2x^2 + 340x - 12000 ]Thus, the relationship between ( y ) and ( x ) is ( y = -2x^2 + 340x - 12000 ).To find the value of ( x ) that maximizes ( y ):1. Vertex Formula: - The quadratic function ( y = -2x^2 + 340x - 12000 ) has a maximum at the vertex. - The x-coordinate of the vertex is given by ( x = -frac{b}{2a} ). - Here, ( a = -2 ) and ( b = 340 ): [ x = -frac{340}{2(-2)} = frac{340}{4} = 85 ]2. Verification: - Check if ( x = 85 ) is within a reasonable range (sales volume ( w ) must be non-negative and selling price must be above cost). - When ( x = 85 ), ( w = -2(85) + 240 = 70 ) (non-negative). - Selling price ( x = 85 ) is above cost 50, ensuring a profit.Thus, the value of ( x ) that maximizes ( y ) is (boxed{85}). Final Answer(1) The relationship between ( y ) and ( x ) is ( y = -2x^2 + 340x - 12000 ). (2) The value of ( x ) that maximizes ( y ) is boxed{85}.]

🔑:To solve this problem, we will tackle it in two parts as requested: Part 1: Find the Relationship Between y and xGiven:- Cost of green tea per kilogram = 50- Sales volume w (in kilograms) varies with selling price x (in dollars per kilogram) as w = -2x + 240- Profit y (in dollars) from selling the green teaWe start by expressing the profit y in terms of x and w. The profit from selling one kilogram of green tea is the selling price minus the cost, which is (x - 50). Therefore, the total profit y can be expressed as the product of the profit per kilogram and the total sales volume w:[y = (x - 50)w]Substituting the given relationship for w:[y = (x - 50)(-2x + 240)]Expanding this expression:[y = -2x^2 + 240x - 100x + 12000]Simplifying:[y = -2x^2 + 340x - 12000]Thus, the relationship between y and x is:[boxed{y = -2x^2 + 340x - 12000}] Part 2: Maximizing the Value of yGiven the quadratic equation for y:[y = -2x^2 + 340x - 12000]We can complete the square to find the vertex of the parabola, which will give us the value of x that maximizes y. The quadratic term can be rewritten as:[y = -2(x^2 - 170x) - 12000]Completing the square:[y = -2left(x^2 - 170x + (85)^2right) + 2(85)^2 - 12000][y = -2(x - 85)^2 + 14450 - 12000][y = -2(x - 85)^2 + 2450]The vertex of this parabola occurs at x = 85, which is the value of x that maximizes y. Therefore, the selling price per kilogram that maximizes profit is:[boxed{x = 85}]

🧠:[<think>Okay, let's see. The problem says that in triangle ABC, a, b, c are the lengths opposite to angles A, B, C respectively. We know that cos B is 4/5, a is 5, and the area is 12. We need to find the value of (a + c)/(sin A + sin C). Hmm.First, I should recall some trigonometric identities and formulas related to triangles. The area of a triangle can be given by (1/2)*ab*sin C, but here, since we know angle B, maybe we can use another formula. Wait, the area is (1/2)*a*c*sin B, since sides a and c would be adjacent to angle B. Let me confirm that. In triangle ABC, sides a, b, c are opposite angles A, B, C. So, sides a and c are adjacent to angle B if we consider sides a, b, c as opposite. Wait, no, actually, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. So, if we want two sides adjacent to angle B, those would be sides a and c? Because angle B is between sides a and c? Wait, no. In a triangle, each angle is between two sides. For example, angle A is between sides b and c, angle B is between sides a and c, and angle C is between sides a and b. Wait, maybe I need to get that straight.Alternatively, perhaps using the standard formula for area: (1/2)*ab*sin C, but here we have angle B. So the area can also be expressed as (1/2)*a*c*sin B. Wait, let's see: if angle B is between sides a and c? Wait, if angle B is opposite side b, then the sides adjacent to angle B are sides a and c. Therefore, the formula for the area using angle B would be (1/2)*a*c*sin B. Yes, that's right. So the area is (1/2)*a*c*sin B = 12. Given that a = 5 and cos B = 4/5. Since we know cos B, we can find sin B. Because sin^2 B + cos^2 B = 1, so sin B = sqrt(1 - (16/25)) = sqrt(9/25) = 3/5. So sin B is 3/5. Therefore, plugging into the area formula: (1/2)*5*c*(3/5) = 12. Let me compute that.First, (1/2)*5*c*(3/5) = (1/2)*(15/5)*c = (1/2)*3*c = (3/2)*c. So (3/2)*c = 12 => c = 12*(2/3) = 8. So c is 8. Okay, so we found that c = 8. So now we know sides a = 5, c = 8, angle B with cos B = 4/5 and sin B = 3/5. We need to find (a + c)/(sin A + sin C). So that would be (5 + 8)/(sin A + sin C) = 13/(sin A + sin C). So we need to find sin A + sin C.Alternatively, maybe use the Law of Sines. The Law of Sines states that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. So if we can find the value of 2R, then sin A = a/(2R), sin C = c/(2R), so sin A + sin C = (a + c)/(2R). Therefore, (a + c)/(sin A + sin C) would be (a + c)/[(a + c)/(2R)] = 2R. So the expression simplifies to 2R. Therefore, the value we need is 2R, which is equal to a/sin A = b/sin B = c/sin C. So if we can compute 2R, then we can find the answer.Alternatively, since we know sides a and c and angle B, maybe we can find side b using the Law of Cosines. Let's try that. Law of Cosines says that b² = a² + c² - 2ac*cos B. So plugging in a = 5, c = 8, cos B = 4/5. Therefore, b² = 5² + 8² - 2*5*8*(4/5). Let's compute that step by step.First, 5² is 25, 8² is 64, so 25 + 64 = 89. Then, 2*5*8 = 80, and 80*(4/5) = 80*(4)/5 = 16*4 = 64. So then, b² = 89 - 64 = 25. Therefore, b = 5. Wait, so side b is 5? But side a is also 5. So the triangle has sides a = 5, b = 5, c = 8. Interesting. So triangle ABC is isoceles with sides a and b equal. Therefore, angles A and B would be equal? Wait, no. Wait, sides a and b are opposite angles A and B. If sides a and b are equal (both 5), then angles A and B are equal. But wait, the problem states that cos B = 4/5, which would make angle B not equal to angle A unless cos A is also 4/5. But let's verify this.Wait, if sides a and b are both 5, then angles A and B would be equal. But since cos B = 4/5, then angle B is arccos(4/5). So angle A would also be arccos(4/5). But then angle C would be 180 - 2 arccos(4/5). Let me compute that. But before that, let's check if this holds.Alternatively, maybe I made a mistake in computing b. Let me check again. Law of Cosines: b² = a² + c² - 2ac cos B. So a=5, c=8, cos B=4/5. Therefore, b² = 25 + 64 - 2*5*8*(4/5). Let's compute each term:25 + 64 = 89.2*5*8 = 80.80*(4/5) = 80*(0.8) = 64.So 89 - 64 = 25. So b² = 25, so b = 5. So yes, that's correct. So sides a and b are both 5. Therefore, angles A and B are equal. Wait, but angle B is arccos(4/5). So angle A is also arccos(4/5). Hmm, but let's see if that makes sense. Let's check the angles.Sum of angles in a triangle is 180 degrees. So angle A = angle B = arccos(4/5). Then angle C = 180 - 2 arccos(4/5). Let's compute angle C. But maybe we can compute angle C using the Law of Cosines as well. Alternatively, since sides a, b, c are 5, 5, 8, we can use Law of Cosines to find angle C. Wait, angle C is opposite side c, which is 8. So Law of Cosines: c² = a² + b² - 2ab cos C. So 8² = 5² + 5² - 2*5*5*cos C. So 64 = 25 + 25 - 50 cos C. 64 = 50 - 50 cos C. Then 64 - 50 = -50 cos C. 14 = -50 cos C. Therefore, cos C = -14/50 = -7/25. So angle C is arccos(-7/25). Which is an obtuse angle. That seems okay.Alternatively, maybe we can compute angle A using Law of Sines. Since a = 5, sin A = a/(2R), and similarly for sin C. Wait, but since we have sides a = 5, b = 5, c = 8, and angles A, B, C. Since angles A and B are equal (since sides a and b are equal), then sin A = sin B. Wait, but angle B is arccos(4/5), so sin B is 3/5. Therefore, sin A is also 3/5. Wait, but angle A is equal to angle B, so sin A = sin B = 3/5. Then angle C is 180 - 2 angle B. Then sin C = sin (2 angle B), since angle C = 180 - 2 angle B, so sin C = sin (2 angle B). Let me compute that.We know that angle B has cos B = 4/5, sin B = 3/5. Then sin C = sin (2 angle B) = 2 sin B cos B = 2*(3/5)*(4/5) = 24/25. Therefore, sin C = 24/25. Therefore, sin A + sin C = 3/5 + 24/25 = 15/25 + 24/25 = 39/25. Therefore, (a + c)/(sin A + sin C) = (5 + 8)/(39/25) = 13/(39/25) = 13*(25/39) = (13/39)*25 = (1/3)*25 = 25/3. So the value is 25/3. Therefore, the answer is 25/3. Let me check this again.Wait, but we have to be careful here. If angles A and B are equal, then angles A and B are each arccos(4/5), but angle C is 180 - 2 arccos(4/5). But sin(180 - x) = sin x, so sin C = sin(2 angle B). Wait, 180 - 2 angle B is angle C. So sin C = sin(2 angle B). Is that correct? Because angle C = 180 - 2 angle B, so sin C = sin(2 angle B) since sin(180 - x) = sin x. So, yes, sin C = sin(2 angle B). Then sin(2 angle B) = 2 sin angle B cos angle B = 2*(3/5)*(4/5) = 24/25. So sin C = 24/25. Then sin A is sin angle B, which is 3/5. So sin A + sin C = 3/5 + 24/25 = 15/25 + 24/25 = 39/25. Then (a + c) is 5 + 8 = 13. So 13 divided by 39/25 is 13*(25/39) = (13/39)*25. Since 13 and 39 are both divisible by 13: 39 = 13*3. So 13/39 = 1/3. Therefore, it's (1/3)*25 = 25/3. So 25 over 3. That seems correct.Alternatively, using the Law of Sines. If we can compute 2R, which is a/sin A = b/sin B = c/sin C. Since we know a = 5, sin A = 3/5, so 2R = 5/(3/5) = 25/3. Alternatively, using side b = 5, sin B = 3/5, so 2R = 5/(3/5) = 25/3. Similarly, c = 8, sin C = 24/25, so 2R = 8/(24/25) = 8*(25/24) = (25/3). Therefore, 2R is 25/3, which is exactly what we found before. Therefore, the expression (a + c)/(sin A + sin C) simplifies to 2R, which is 25/3. Therefore, the answer is 25/3.But let me just make sure I didn't skip any steps or make any incorrect assumptions. For example, when I assumed that angle C = 180 - 2 angle B, that's only valid if sides a and b are equal. But since we found that sides a and b are both 5, angles A and B should indeed be equal. So that seems okay.Alternatively, let's approach this problem again from another angle to verify. Let's use the Law of Sines directly. Since we have sides a, b, c, and angles A, B, C. We know a = 5, c = 8, angle B with cos B = 4/5. From the Law of Sines: a/sin A = b/sin B = c/sin C. We already found that b = 5. So 5/sin A = 5/sin B = 8/sin C. From this, we can express sin A and sin C in terms of sin B.Since 5/sin A = 5/sin B, so sin A = sin B. Therefore, angle A = angle B or angle A = 180 - angle B. But since the sum of angles in a triangle is 180, angle A cannot be 180 - angle B unless angle C = 0, which is impossible. Therefore, angle A = angle B. Therefore, angles A and B are equal. Thus, as before, angle C = 180 - 2 angle B.Therefore, sin C = sin(2 angle B) = 2 sin B cos B = 24/25. Then, sin A + sin C = sin B + sin(2 angle B) = 3/5 + 24/25 = 39/25. Then (a + c)/(sin A + sin C) = 13/(39/25) = 25/3. Same result.Alternatively, maybe compute all angles numerically. Let's compute angle B. cos B = 4/5, so angle B is arccos(4/5). Let's find that in degrees. 4/5 is 0.8. arccos(0.8) is approximately 36.87 degrees. Therefore, angle B is approximately 36.87 degrees. Then angles A and B are both 36.87 degrees, so angle C is 180 - 2*36.87 = 106.26 degrees. Then sin A = sin 36.87 ≈ 0.6 (which is 3/5), sin C = sin 106.26 ≈ sin(180 - 73.74) = sin 73.74 ≈ 0.96, which is 24/25 = 0.96. Then sin A + sin C ≈ 0.6 + 0.96 = 1.56, which is 39/25 = 1.56. Then (a + c)/(sin A + sin C) ≈ 13 / 1.56 ≈ 8.333..., which is 25/3 ≈ 8.333. So that matches as well.Therefore, all approaches lead to the conclusion that the value is 25/3. So I think that's the correct answer.Final Answerboxed{dfrac{25}{3}}</think>Given that (a), (b), (c) are the lengths of the sides opposite to angles (A), (B), (C) of (triangle ABC) respectively, and (cos B = frac{4}{5}). If (a = 5) and the area of (triangle ABC) is 12, we need to find the value of (frac{a + c}{sin A + sin C}).1. Calculate (sin B): [ sin B = sqrt{1 - cos^2 B} = sqrt{1 - left(frac{4}{5}right)^2} = sqrt{frac{9}{25}} = frac{3}{5} ]2. Use the area formula to find (c): [ text{Area} = frac{1}{2} cdot a cdot c cdot sin B = 12 ] Substituting the known values: [ frac{1}{2} cdot 5 cdot c cdot frac{3}{5} = 12 implies frac{3}{2} cdot c = 12 implies c = 8 ]3. Use the Law of Cosines to find (b): [ b^2 = a^2 + c^2 - 2ac cos B ] Substituting the known values: [ b^2 = 5^2 + 8^2 - 2 cdot 5 cdot 8 cdot frac{4}{5} = 25 + 64 - 64 = 25 implies b = 5 ]4. Using the Law of Sines: [ frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} = 2R ] Since (a = 5) and (sin B = frac{3}{5}): [ 2R = frac{5}{frac{3}{5}} = frac{25}{3} ]5. Calculate (sin A + sin C): - Since (a = b = 5), angles (A) and (B) are equal, so (sin A = sin B = frac{3}{5}). - Angle (C = 180^circ - 2B), so (sin C = sin(2B)): [ sin(2B) = 2 sin B cos B = 2 cdot frac{3}{5} cdot frac{4}{5} = frac{24}{25} ] - Therefore: [ sin A + sin C = frac{3}{5} + frac{24}{25} = frac{15}{25} + frac{24}{25} = frac{39}{25} ]6. Final calculation: [ frac{a + c}{sin A + sin C} = frac{5 + 8}{frac{39}{25}} = frac{13}{frac{39}{25}} = frac{13 cdot 25}{39} = frac{25}{3} ]Thus, the value of (frac{a + c}{sin A + sin C}) is (boxed{dfrac{25}{3}}).,<think>Okay, let me try to figure out this problem step by step. So, we have triangle ABC with sides a, b, c opposite angles A, B, C respectively. We know that cos B is 4/5, a is 5, and the area is 12. We need to find (a + c)/(sin A + sin C). Hmm, that seems like it might involve some trigonometric identities or maybe the Law of Sines or Cosines. Let me start by writing down what I know.First, cos B = 4/5. Since in any triangle, the cosine of an angle can be related to the sides via the Law of Cosines: cos B = (a² + c² - b²)/(2ac). But before jumping into that, maybe we can find sin B because the area of the triangle is given. The area formula is (1/2)*a*c*sin B = 12. Since we know a is 5, maybe we can solve for c*sin B.Wait, let's confirm the area formula. Yes, the area of a triangle is (1/2)*ab*sin C, depending on which sides and angle you use. In this case, since angle B is between sides a and c, right? Wait, no. Wait, in triangle ABC, side a is opposite angle A, side b opposite angle B, side c opposite angle C. So, if we want to use sides a and c with the angle between them, that would be angle B. Because angle B is between sides a and c? Wait, no, actually, in triangle notation, side a is opposite angle A, so the sides adjacent to angle B would be sides a and c. Because in triangle ABC, each angle is between two sides. So angle A is between sides b and c, angle B is between sides a and c, and angle C is between sides a and b. Wait, maybe I need to visualize the triangle.Let me draw triangle ABC with angles A, B, C and sides a, b, c opposite them. So angle A is opposite side a (which is 5), angle B opposite side b, angle C opposite side c. So, sides: a=5, b, c. Angles: A, B, C. Given that cos B = 4/5. Area is 12.So, the area can be calculated using two sides and the sine of the included angle. If we take sides a and c, the angle between them would be angle B. Wait, is that correct? Let me think. If sides a and c are adjacent to angle B, then yes, the included angle between a and c is angle B. So, area = (1/2)*a*c*sin B = 12. We know a=5, so (1/2)*5*c*sin B = 12. Then, 5c sin B = 24 => c sin B = 24/5.But we also know cos B = 4/5. Since cos² B + sin² B = 1, sin B = sqrt(1 - (16/25)) = sqrt(9/25) = 3/5. So sin B is 3/5. Therefore, c*(3/5) = 24/5 => c = (24/5)/(3/5) = 24/3 = 8. So c is 8. Great, so now we know c=8.Now, we need to find (a + c)/(sin A + sin C). Since a=5 and c=8, that would be (5 + 8)/(sin A + sin C) = 13/(sin A + sin C). So, we need to find sin A + sin C.Alternatively, maybe we can use the Law of Sines here. The Law of Sines states that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. So, if we can find the common ratio, then we can express sin A and sin C in terms of a and c. Let's see.Given that a=5, c=8, and sin B=3/5. Let's find the value of 2R. Since a/sin A = 2R, and c/sin C = 2R. Therefore, 2R = 5/sin A = 8/sin C. Therefore, sin A = 5/(2R) and sin C = 8/(2R) = 4/R. Therefore, sin A + sin C = 5/(2R) + 8/(2R) = (5 + 8)/(2R) = 13/(2R). Therefore, (a + c)/(sin A + sin C) = 13/(13/(2R)) = 2R. So, if we can find 2R, that's our answer.Alternatively, since 2R = a/sin A, but we don't know sin A. But maybe we can find 2R another way. Since we know angle B and sides a and c, perhaps we can find side b using the Law of Cosines, and then use the Law of Sines with angle B?Let me try that. Law of Cosines on angle B: cos B = (a² + c² - b²)/(2ac). We have cos B = 4/5, a=5, c=8. So:4/5 = (5² + 8² - b²)/(2*5*8)Compute numerator: 25 + 64 - b² = 89 - b²Denominator: 2*5*8 = 80So, (89 - b²)/80 = 4/5 => Multiply both sides by 80: 89 - b² = (4/5)*80 = 64Therefore, 89 - 64 = b² => 25 = b² => b = 5.Wait, so side b is 5. Hmm, interesting. So sides a and b are both 5? So triangle ABC has sides a=5, b=5, c=8. So it's an isoceles triangle with sides a and b equal? Wait, but in that case, angles A and B would be equal. But wait, angle B has cosine 4/5, so angle B is acute, but angle A... Wait, let's check.If a and b are both 5, then angles A and B would be equal. But we were given that cos B = 4/5, so angle B is arccos(4/5), which is approximately 36.87 degrees. If angles A and B are equal, then angle A would also be 36.87 degrees, and angle C would be 180 - 2*36.87 ≈ 106.26 degrees. But then side c is opposite angle C, which is the largest angle, so side c should be the longest side. Indeed, c=8, which is longer than a and b (both 5). So that seems consistent.But let's confirm with Law of Sines. If a=5, b=5, c=8, then Law of Sines says 5/sin A = 5/sin B = 8/sin C. Since a and b are equal, sin A = sin B. But angle B is arccos(4/5), so sin B is 3/5, so sin A is also 3/5. Therefore, angle A is either equal to angle B or its supplement. But since the sum of angles in a triangle is 180, and angle B is approx 36.87, angle A must be equal to angle B, so angles A and B are both 36.87 degrees, and angle C is 106.26 degrees. Then sin C is sin(106.26). Let's compute that. 106.26 degrees is 180 - 73.74 degrees, so sin(106.26) = sin(73.74). Let's compute sin(73.74). Alternatively, since sin C can be found via Law of Sines: 8/sin C = 5/sin B. Since sin B = 3/5, so 5/(3/5) = 25/3. Therefore, 8/sin C = 25/3 => sin C = 8*3/25 = 24/25. Therefore, sin C = 24/25. Let's verify angle C: arcsin(24/25) is approx 73.74 degrees, but angle C was supposed to be 106.26. Wait, that's a problem. Wait, maybe I made a mistake here.Wait, angle C is 180 - A - B. If A and B are both arccos(4/5) ≈ 36.87 degrees, then angle C is 180 - 73.74 ≈ 106.26 degrees. So sin C should be sin(106.26), which is sin(180 - 73.74) = sin(73.74) ≈ 24/25. Because sin(73.74) ≈ 0.96, and 24/25 is 0.96. So that works. So sin C is 24/25.Therefore, sin A + sin C = 3/5 + 24/25 = 15/25 + 24/25 = 39/25. Then, (a + c)/(sin A + sin C) = 13/(39/25) = 13*(25/39) = (13*25)/39. Simplify: 13 and 39 have a common factor of 13. So 39 = 13*3. Therefore, (13*25)/(13*3) )= 25/3 ≈ 8.333...But wait, earlier we saw that (a + c)/(sin A + sin C) = 2R. Let's check 2R. From Law of Sines, 5/sin A = 2R. Since sin A = 3/5, 5/(3/5) = 25/3 = 2R. So 2R = 25/3. Therefore, the answer is 25/3, which is approximately 8.333... So 25/3 is the exact value.Wait, so that seems to check out. Let me recap the steps to make sure I didn't skip anything.1. Given cos B = 4/5, so sin B = 3/5.2. Area = 12 = (1/2)*a*c*sin B => (1/2)*5*c*(3/5) = 12 => c = 8.3. Law of Cosines on angle B: (25 + 64 - b²)/80 = 4/5 => 89 - b² = 64 => b² = 25 => b=5.4. Now, triangle sides are a=5, b=5, c=8.5. Using Law of Sines: a/sin A = b/sin B = c/sin C. Since a = b =5, sin A = sin B = 3/5, so angles A and B are equal.6. Therefore, angle C = 180 - 2*angle B. Then sin C = sin(2*angle B')? Wait, no. Wait, angle C = 180 - A - B = 180 - 2B. So sin C = sin(2B')? Wait, sin(180 - 2B) = sin(2B). Wait, let me compute angle C.Since angle B = arccos(4/5), so angle B ≈ 36.87 degrees. Then angle C = 180 - 2*36.87 ≈ 106.26 degrees. Then sin C = sin(106.26°). Alternatively, using the identity sin(180 - x) = sin x, so sin(106.26°) = sin(73.74°). Since angle B is 36.87°, then 2*angle B = 73.74°, so sin C = sin(2*angle B). But we can compute sin C using double angle formula: sin(2B) = 2 sin B cos B = 2*(3/5)*(4/5) = 24/25. So sin C = 24/25. Therefore, sin A + sin C = 3/5 + 24/25 = 15/25 + 24/25 = 39/25.Therefore, (a + c)/(sin A + sin C) = 13/(39/25) = 13*(25/39) = (13*25)/39. Since 39 = 13*3, this simplifies to 25/3. So the answer is 25/3.Alternatively, using 2R from the Law of Sines: since a/sin A = 2R, and sin A = 3/5, so 5/(3/5) = 25/3 = 2R. Therefore, 2R = 25/3, which is the same as the answer above.So, I think this is correct. Let me double-check all steps.First, area calculation: (1/2)*5*8*(3/5) = (1/2)*24 = 12. Correct.Law of Cosines: (25 + 64 - 25)/80 = (64)/80 = 4/5. Wait, hold on. Wait, cos B = (a² + c² - b²)/(2ac). Wait, if b=5, then:(5² + 8² -5²)/(2*5*8) = (25 + 64 -25)/80 = 64/80 = 4/5. Correct. So that checks out.Law of Sines: 5/sin A = 5/sin B = 8/sin C. Since sin B = 3/5, then sin A = 3/5, and sin C = 8/(5/sin B) = 8/(5/(3/5)) = 8/(25/3) = 24/25. Correct.Therefore, sin A + sin C = 3/5 + 24/25 = 39/25. (a + c) =13. So 13 divided by 39/25 is 13*(25/39) = 25/3. So yes, 25/3 is correct. I think that's solid.Final Answerboxed{dfrac{25}{3}}</think>Given that (a), (b), (c) are the lengths of the sides opposite to angles (A), (B), (C) of (triangle ABC) respectively, and (cos B = frac{4}{5}). We know (a = 5) and the area of (triangle ABC) is 12. We need to find the value of (frac{a+c}{sin A+sin C}).1. Calculate (sin B): [ cos B = frac{4}{5} implies sin B = sqrt{1 - left(frac{4}{5}right)^2} = sqrt{frac{9}{25}} = frac{3}{5} ]2. Use the area formula to find (c): [ text{Area} = frac{1}{2} cdot a cdot c cdot sin B = 12 implies frac{1}{2} cdot 5 cdot c cdot frac{3}{5} = 12 implies frac{3}{2}c = 12 implies c = 8 ]3. Apply the Law of Cosines to find (b): [ cos B = frac{a^2 + c^2 - b^2}{2ac} implies frac{4}{5} = frac{5^2 + 8^2 - b^2}{2 cdot 5 cdot 8} implies frac{4}{5} = frac{89 - b^2}{80} implies 64 = 89 - b^2 implies b^2 = 25 implies b = 5 ]4. Use the Law of Sines to find (sin A) and (sin C): - Since (a = 5) and (b = 5), (sin A = sin B = frac{3}{5}). - For side (c = 8): [ frac{c}{sin C} = frac{a}{sin A} implies frac{8}{sin C} = frac{5}{frac{3}{5}} implies frac{8}{sin C} = frac{25}{3} implies sin C = frac{24}{25} ]5. Calculate (sin A + sin C): [ sin A + sin C = frac{3}{5} + frac{24}{25} = frac{15}{25} + frac{24}{25} = frac{39}{25} ]6. Compute the final value: [ frac{a + c}{sin A + sin C} = frac{5 + 8}{frac{39}{25}} = frac{13}{frac{39}{25}} = 13 cdot frac{25}{39} = frac{25}{3} ]The value of (frac{a+c}{sin A+sin C}) is (boxed{dfrac{25}{3}}).]

🔑:Since in triangle ABC, cos B = frac{4}{5} and a = 5, we can compute sin B = sqrt{1 - cos^2 B} = sqrt{1 - left(frac{4}{5}right)^2} = sqrt{1 - frac{16}{25}} = sqrt{frac{9}{25}} = frac{3}{5}.The area S_{triangle ABC} can be expressed by the formula: S_{triangle ABC} = frac{1}{2}acsin B = 12.From this equation, we can solve for the length c:c = frac{2S_{triangle ABC}}{asin B} = frac{2 times 12}{5 times frac{3}{5}} = frac{24}{3} = 8.Now, by applying the Law of Cosines, we can find the length b:b = sqrt{a^2 + c^2 - 2accos B} = sqrt{5^2 + 8^2 - 2 times 5 times 8 times frac{4}{5}} = sqrt{25 + 64 - 64} = sqrt{25} = 5.Then, we can relate the unknown ratio to the known one, as sin A and sin C correspond to the angles opposite to sides b and a respectively:frac{a+c}{sin A+sin C} = frac{b}{sin B} = frac{5}{frac{3}{5}} = frac{5 times 5}{3} = frac{25}{3}.The final answer is:boxed{frac{25}{3}}.

🧠:[<think>Alright, so I need to find three prime numbers such that their product is five times their sum. Let me start by writing down what the problem is asking. If we let the three prime numbers be p, q, and r, then according to the problem, p * q * r = 5 * (p + q + r). My goal is to find primes p, q, r that satisfy this equation.First, since the product of the primes is a multiple of 5, that means at least one of the primes must be 5. Because primes are numbers greater than 1 that have no divisors other than 1 and themselves. The only way the product can be a multiple of 5 is if one of the primes is 5 itself. So, without loss of generality, let's assume that one of the primes is 5. Let's say p = 5. Then the equation becomes:5 * q * r = 5 * (5 + q + r)Simplifying both sides by dividing by 5, we get:q * r = 5 + q + rNow, we can rearrange this equation to make it easier to solve. Let's subtract q and r from both sides:q * r - q - r = 5Hmm, maybe factor this equation? Let's see. If I add 1 to both sides:q * r - q - r + 1 = 6Then the left side can be factored as (q - 1)(r - 1) = 6Ah, that's a useful step. So, (q - 1)(r - 1) = 6. Now, since q and r are primes, and we know that primes are either 2 or odd numbers. Let's think about possible pairs (q, r) such that (q - 1)(r - 1) = 6.First, let's list the positive integer factors of 6. The pairs of integers (a, b) such that a * b = 6 are:(1, 6), (2, 3), (3, 2), (6, 1)So, considering these factor pairs:1. If (q - 1, r - 1) = (1, 6), then q = 2, r = 72. If (q - 1, r - 1) = (2, 3), then q = 3, r = 43. If (q - 1, r - 1) = (3, 2), then q = 4, r = 34. If (q - 1, r - 1) = (6, 1), then q = 7, r = 2But we need to check if q and r are primes. Let's check each case:1. q = 2, r = 7: Both are primes.2. q = 3, r = 4: 3 is prime, but 4 is not.3. q = 4, r = 3: 4 is not prime, 3 is prime.4. q = 7, r = 2: Both are primes.So cases 1 and 4 give us valid primes. Let's check case 1 first. If q = 2 and r = 7, then the three primes are 5, 2, and 7. Let's verify the original equation:Product: 5 * 2 * 7 = 70Sum: 5 + 2 + 7 = 145 times the sum: 5 * 14 = 70Yes, that works. Now case 4: q = 7 and r = 2. That's the same set of primes, just reordered. So the primes are 5, 7, 2. The product and sum would be the same as above.Now, let's check if there are any other possibilities. The factors we considered were positive integers. Since primes are positive integers greater than 1, (q - 1) and (r - 1) must be at least 1. But perhaps negative factors? Let me think. If we consider negative factors of 6, for example (-1, -6), (-2, -3), etc. Then (q -1, r -1) would be negative, implying q or r is less than 1, which can't be since primes are at least 2. So negative factors are not possible here. Hence, only the positive factor pairs are valid.Therefore, the only possibilities are when the two primes other than 5 are 2 and 7. Let me check if there could be a case where more than one of the primes is 5. For example, what if two of the primes are 5?Suppose p = 5, q = 5, then the equation becomes:5 * 5 * r = 5 * (5 + 5 + r)Simplify:25r = 5*(10 + r) => 25r = 50 + 5r => 20r = 50 => r = 50 / 20 = 2.5But r must be a prime number, and 2.5 is not an integer, so this is invalid.What if all three primes are 5? Then:5 * 5 * 5 = 5 * (5 + 5 + 5) => 125 = 5*15 = 75. Not equal. So that's invalid.Alternatively, could there be a case where none of the primes is 5? If none of the primes is 5, then the product p*q*r would not contain a factor of 5. However, the right side is 5*(sum). So 5 must divide the product p*q*r, which implies that at least one of the primes must be 5. So, it's necessary that one of the primes is 5. Therefore, all possibilities require one of the primes to be 5, so the other two primes are 2 and 7. Hence, the three primes are 2, 5, 7 in some order.Let me just confirm once again. If I take primes 2, 5, 7:Product: 2*5*7 = 70Sum: 2+5+7 = 145 times sum: 5*14 = 70. Yep, that's equal. So that works.Is there another set of primes? Let's see. Suppose instead of 2, maybe another small prime. Let's test with 3. Suppose the primes are 3, 5, and something else. Let's let p=3, q=5, then solve for r.Equation: 3*5*r = 5*(3+5+r)15r = 5*(8 + r)15r = 40 + 5r10r = 40r = 4. But 4 isn't prime. So that's invalid.How about primes 5, 5, and something else? Wait, we saw earlier that two 5s lead to a non-integer or invalid prime. Similarly, trying 5, 7, and another prime:Let p=5, q=7, solve for r:5*7*r =5*(5+7+r)35r =5*(12 + r)35r =60 +5r30r =60 => r=2. Which is prime. So that gives primes 5,7,2 which we already considered. So it's the same set. So same solution.Alternatively, primes 5, 11, and ? Let's check:Suppose p=5, q=11, solve for r:5*11*r =5*(5+11 +r)55r =5*(16 + r)55r =80 +5r50r =80 => r=80/50=1.6. Not integer. Not prime.What about primes 5, 3, 2:Product: 5*3*2=30Sum: 5+3+2=105*10=50. Not equal. So 30≠50.Wait, but we already saw that if you take 5,2,7 you get 70=70. So that works. Let me see if there's another combination.Suppose primes 5,2,2:Product:5*2*2=20Sum:5+2+2=95*9=45. Not equal.Primes 5,2,3:Product:30, sum:10, 5*10=50≠30.Primes 5,2,11:Product:5*2*11=110Sum:5+2+11=185*18=90≠110.Primes 5,2,13:Product:5*2*13=130Sum:5+2+13=205*20=100≠130.So seems like the only combination that works is 2,5,7.Alternatively, let's think if there's a way to get two primes as 5. But earlier that didn't work. If two primes are 5, then third is non-integer or non-prime. So no.Alternatively, maybe primes 5,7,7:Product:5*7*7=245Sum:5+7+7=195*19=95≠245.Nope.Alternatively, primes 5, 3, and 7:Product:5*3*7=105Sum:5+3+7=155*15=75≠105.Nope.So seems like 2,5,7 is the only possible set. Let me confirm again.Original equation: p*q*r =5*(p+q+r). If one of the primes is 5, then equation reduces to q*r=5+q+r. Then we found that (q-1)(r-1)=6. The solutions to this are (q,r)=(2,7) or (7,2). Therefore, the three primes are 2,5,7.Therefore, the answer is 2,5,7.Final AnswerThe three prime numbers are boxed{2}, boxed{5}, and boxed{7}.</think>Given three prime numbers ( p ), ( q ), and ( r ) such that their product is five times their sum, we need to find these primes. The equation is:[ p cdot q cdot r = 5 cdot (p + q + r) ]Since the product must be a multiple of 5, one of the primes must be 5. Let ( p = 5 ). The equation simplifies to:[ 5 cdot q cdot r = 5 cdot (5 + q + r) ]Dividing both sides by 5, we get:[ q cdot r = 5 + q + r ]Rearranging and factoring:[ q cdot r - q - r = 5 ][ (q - 1)(r - 1) = 6 ]We find the factor pairs of 6: ((1, 6)), ((2, 3)), ((3, 2)), ((6, 1)). Considering these pairs for ((q - 1, r - 1)):1. ((1, 6)) gives ( q = 2 ) and ( r = 7 )2. ((2, 3)) gives ( q = 3 ) and ( r = 4 ) (4 is not prime)3. ((3, 2)) gives ( q = 4 ) and ( r = 3 ) (4 is not prime)4. ((6, 1)) gives ( q = 7 ) and ( r = 2 )The valid prime pairs are ((2, 7)) and ((7, 2)). Thus, the primes are 5, 2, and 7. Verifying:- Product: ( 5 cdot 2 cdot 7 = 70 )- Sum: ( 5 + 2 + 7 = 14 )- Five times the sum: ( 5 cdot 14 = 70 )Both sides are equal, confirming the solution. Checking other possibilities where more than one prime is 5 or different primes leads to non-integer or non-prime results, confirming no other solutions.The three prime numbers are (boxed{2}), (boxed{5}), and (boxed{7}).,<think>Okay, let's see. I need to find three prime numbers such that when I multiply them together, the result is five times their sum. Hmm. Let's break this down step by step.First, let's denote the three prime numbers as p, q, and r. According to the problem, their product is five times their sum. So mathematically, that would be:p * q * r = 5 * (p + q + r)Alright, so we need to find primes p, q, r such that this equation holds. Since primes are numbers greater than 1 that have no divisors other than 1 and themselves, the primes we usually consider are 2, 3, 5, 7, 11, etc.Now, looking at the equation, the right-hand side is 5 times the sum of the primes. The left-hand side is the product of the primes. Since the product is a multiple of 5, that means at least one of the primes must be 5. Because 5 is a prime number, and if none of the primes were 5, then the product wouldn't be divisible by 5, but the right-hand side is a multiple of 5. So, one of p, q, or r has to be 5. Let me note that down.Without loss of generality, let's assume that p = 5. Then the equation becomes:5 * q * r = 5 * (5 + q + r)We can divide both sides by 5 to simplify:q * r = 5 + q + rHmm, so now we need two primes q and r such that their product is equal to their sum plus 5. Let's rearrange the equation:q * r - q - r = 5I can factor this a bit. Let's see, maybe add 1 to both sides:q * r - q - r + 1 = 6Now, factor the left-hand side:(q - 1)(r - 1) = 6Ah, this is useful! Because 6 can be factored into integers in a few ways, and since q and r are primes, (q - 1) and (r - 1) must be positive integers (since primes are at least 2, so q-1 and r-1 are at least 1). So, the possible factor pairs of 6 are:1 and 6,2 and 3,3 and 2,6 and 1.So, each of these pairs would correspond to (q - 1, r - 1). Let's consider each case.Case 1: (q - 1) = 1 and (r - 1) = 6.Then q = 2 and r = 7. Both 2 and 7 are primes. So that's a possible solution.Case 2: (q - 1) = 2 and (r - 1) = 3.Then q = 3 and r = 4. But 4 is not a prime. So this case is invalid.Case 3: (q - 1) = 3 and (r - 1) = 2.Then q = 4 and r = 3. Again, 4 is not a prime. So invalid.Case 4: (q - 1) = 6 and (r - 1) = 1.Then q = 7 and r = 2. Both 7 and 2 are primes. So another possible solution.So, from these cases, the valid solutions for (q, r) are (2,7) and (7,2). Since multiplication is commutative, both pairs will result in the same product. So, the primes are 5, 2, and 7.Let me check this. Let's calculate the product: 5 * 2 * 7 = 70. The sum is 5 + 2 + 7 = 14. Then, 5 times the sum is 5 * 14 = 70. Yep, that matches. So 5, 2, 7 is a valid solution.But wait, the problem says "three prime numbers". It doesn't specify that they have to be distinct. Let me check if there are other possibilities where maybe two of the primes are 5. Let's explore that.Suppose two of the primes are 5. So, p = 5, q = 5, then we need to find r.Then the equation becomes:5 * 5 * r = 5 * (5 + 5 + r)Simplify left-hand side: 25rRight-hand side: 5*(15 + r) = 75 + 5rSet equal: 25r = 75 + 5rSubtract 5r: 20r = 75Divide by 20: r = 75 / 20 = 15/4 = 3.75But 3.75 is not a prime number. So this case is invalid.Alternatively, if all three primes are 5:5*5*5 = 125Sum is 5+5+5=155 times sum is 75. 125 ≠75, so that doesn't work.So, having two or three 5s doesn't yield a solution.Alternatively, maybe the prime 5 is not included? Wait, but the product has to be a multiple of 5. Since primes other than 5 are not multiples of 5, the only way the product can be a multiple of 5 is if one of the primes is 5. So, 5 must be one of the primes. So our initial approach was correct.Therefore, the only solution is 2, 5, 7 in some order.Wait, let me check if there are any other primes possible. For example, let's see if we consider primes larger than 7.Suppose we take q as a larger prime, say 11. Then, in the equation q * r = 5 + q + r, if q is 11:11 * r = 5 + 11 + r => 11r = 16 + r => 10r = 16 => r = 1.6, which is not a prime.Similarly, take q = 3:If q = 3, then equation is 3r = 5 + 3 + r => 3r = 8 + r => 2r = 8 => r =4. Not a prime.q=2: 2r =5+2 + r => 2r =7 + r => r=7. Which is valid, as before.q=7: 7r =5 +7 + r =>7r =12 + r =>6r=12 =>r=2. Which is valid.So, only possible primes for q and r are 2 and 7. So the three primes must be 2,5,7.Wait, but primes can be in any order, so the problem just wants three primes, so the answer is 2,5,7.Let me check one more time. 2*5*7=70. Sum is 2+5+7=14. 5 times the sum is 70. Yep, that works.Is there another set of primes? Let's think. Suppose we have two primes as 2 and 2. Then third prime is 5.Product: 2*2*5=20. Sum:2+2+5=9. 5*9=45. Not equal. So 20≠45. Not valid.What about primes 3,5, something. Let's try. If one prime is 3, another is 5, then third prime?So equation: 3*5*r =5*(3+5+r) =>15r=5*(8 +r)=>15r=40 +5r=>10r=40=>r=4. Not prime.So that's invalid.Alternatively, primes 5,5, something? As before, tried that, didn't work.Primes 5,7, something else. Suppose we take 5,7, and another prime.Wait, if p=5, q=7, then r would be?Original equation: 5*7*r=5*(5+7+r). Let's compute:35r=5*(12 + r)=60 +5r35r -5r=60 =>30r=60 =>r=2. Which is prime. So that's the same solution as before.So regardless of ordering, the third prime is 2. So the triplet is 2,5,7.Is there another triplet where instead of 2,5,7, there's another combination? Let's see.Suppose the primes include 2, but not 5. Wait, but then the product wouldn't be a multiple of 5, but the right-hand side is 5 times the sum. So 5 must divide the product, which requires that one of the primes is 5. Therefore, 5 must be included. So 2 and 7 are the other primes.Alternatively, what if one of the primes is 5, and the other two primes are the same? Let's try.Let primes be 5, p, p.Then equation: 5*p^2 =5*(5 + 2p)Divide both sides by 5: p^2 =5 + 2pBring all terms to left: p^2 -2p -5=0Solve quadratic: p=(2 ± sqrt(4 +20))/2=(2 ±sqrt(24))/2=(2±2*sqrt(6))/2=1±sqrt(6)sqrt(6) is approx 2.45, so p≈1+2.45=3.45 or p≈1-2.45=-1.45. Neither are primes. So no solution here.Therefore, the only possible triplet is 2,5,7.Another check: are there any primes larger than 7 that could work? Let's assume one prime is 5, another is a prime greater than 7, say 11, and then solve for the third.So, 5*11*r =5*(5+11+r). Simplify:55r=5*(16 + r)=80 +5r55r -5r=80 =>50r=80 =>r=1.6. Not a prime.Similarly, trying with prime 13:5*13*r=5*(5+13 +r)65r=5*(18 + r)=90 +5r65r -5r=90 =>60r=90 =>r=1.5. Not prime.So, primes larger than 7 don't seem to work.What if one of the primes is 2, which is the only even prime. Let's see, if we have two 2s and a 5:2*2*5=20. Sum:2+2+5=9. 5*9=45. Not equal. So no.If we have one 2, one 5, and another prime. Wait, we already did that, which gives us 7.Alternatively, having three primes with two 5s and another prime, but as checked earlier, that leads to non-integer.So, after considering all possibilities, the only solution is 2,5,7.Therefore, the three prime numbers are 2, 5, and 7.Final AnswerThe three prime numbers are boxed{2}, boxed{5}, and boxed{7}.</think>Let's denote the three prime numbers as ( p ), ( q ), and ( r ). According to the problem, their product is five times their sum:[p cdot q cdot r = 5 cdot (p + q + r)]Since the product must be a multiple of 5, one of the primes must be 5. Without loss of generality, assume ( p = 5 ). The equation simplifies to:[5 cdot q cdot r = 5 cdot (5 + q + r)]Dividing both sides by 5, we get:[q cdot r = 5 + q + r]Rearranging the equation, we have:[q cdot r - q - r = 5]Adding 1 to both sides allows us to factor the left-hand side:[(q - 1)(r - 1) = 6]The factor pairs of 6 are (1, 6), (2, 3), (3, 2), and (6, 1). Considering each pair:1. ( (q - 1) = 1 ) and ( (r - 1) = 6 ) gives ( q = 2 ) and ( r = 7 ).2. ( (q - 1) = 2 ) and ( (r - 1) = 3 ) gives ( q = 3 ) and ( r = 4 ) (4 is not a prime).3. ( (q - 1) = 3 ) and ( (r - 1) = 2 ) gives ( q = 4 ) and ( r = 3 ) (4 is not a prime).4. ( (q - 1) = 6 ) and ( (r - 1) = 1 ) gives ( q = 7 ) and ( r = 2 ).The valid prime pairs are (2, 7) and (7, 2). Thus, the primes are 5, 2, and 7. Checking the solution:- Product: ( 5 cdot 2 cdot 7 = 70 )- Sum: ( 5 + 2 + 7 = 14 )- Five times the sum: ( 5 cdot 14 = 70 )Both sides match, confirming the solution. Therefore, the three prime numbers are:[boxed{2}, boxed{5}, boxed{7}]]

🔑:9. Let three prime numbers be p, q, r, thenp cdot q cdot r=5(p+q+r) text {, }Thus, one of p, q, r must be 5. Without loss of generality, let r=5, thenbegin{array}{l}5 p q=5(p+q+5), p q=p+q+5, (p-1)(q-1)=6 .end{array}Since p>1, q>1, we haveleft{begin{array} { l } { p - 1 = 2 , } { q - 1 = 3 , }end{array} left{begin{array} { l } { p - 1 = 3 , } { q - 1 = 2 , }end{array} left{begin{array} { l } { p - 1 = 1 , } { q - 1 = 6 , }end{array} left{begin{array}{l}p-1=6, q-1=1 .end{array}right.right.right.right.Since q=4 or p=4 is not a prime number, the only solutions are p=2, q=7, or p=7, q=2. Therefore, the three prime numbers are 2, 5, 7.

🧠:[<think>Okay, so I need to find the sum of all natural numbers that are multiples of both 20 and 14, and don't exceed 3000. Hmm, let's start by understanding the problem.First, multiples of both 20 and 14? That sounds like common multiples. Oh, right! The least common multiple (LCM) of two numbers is the smallest number that is a multiple of both. So if I find the LCM of 20 and 14, then all multiples of that LCM up to 3000 will be the numbers I need to sum.Let me calculate the LCM of 20 and 14. To do that, I can use the formula: LCM(a, b) = (a * b) / GCD(a, b), where GCD is the greatest common divisor.First, find GCD(20, 14). The factors of 20 are 1, 2, 4, 5, 10, 20. The factors of 14 are 1, 2, 7, 14. The common factors are 1 and 2, so the GCD is 2.So LCM(20, 14) = (20 * 14) / 2 = 280 / 2 = 140. Wait, hold on, 20*14 is 280, divided by 2 is 140. Yes, that seems right. Let me check: 140 divided by 20 is 7, which is an integer, and 140 divided by 14 is 10, also an integer. So 140 is indeed the LCM.Therefore, all multiples of 140 that are less than or equal to 3000 are the numbers we need. So the numbers are 140, 280, 420, ..., up to the largest multiple of 140 not exceeding 3000.Next, I need to find how many terms there are in this sequence and then compute their sum.To find the number of terms, we can divide 3000 by 140 and take the floor of that value. Let me compute 3000 / 140. Let's see, 140 * 20 = 2800. 140 * 21 = 2940. 140 * 22 = 3080. But 3080 is more than 3000, so the largest multiple is 2940. So the number of terms is 21.Wait, but let me check: 140 * 21 = 140*20 + 140*1 = 2800 + 140 = 2940. Yes, that's right. Then 140*22 = 2940 + 140 = 3080, which is over 3000. So the number of terms is 21.But wait, let's compute 3000 divided by 140. 3000 / 140 = 300 / 14 ≈ 21.428... So the integer part is 21. Therefore, there are 21 terms.Now, to find the sum of an arithmetic sequence, the formula is S = n/2 * (a1 + an), where n is the number of terms, a1 is the first term, and an is the last term.Here, a1 is 140, an is 2940, and n is 21.So plugging in the values: S = 21/2 * (140 + 2940). Let me compute that.First, compute 140 + 2940. That's 3080.Then, 21/2 * 3080. Let's compute 21 * 3080 first, then divide by 2.21 * 3080. Let's break that down: 20*3080 = 61,600 and 1*3080 = 3,080. So 61,600 + 3,080 = 64,680.Then divide by 2: 64,680 / 2 = 32,340.So the sum should be 32,340.Wait, but let me check again step by step. Maybe I made a miscalculation.First term: 140, last term: 2940, number of terms: 21.Sum formula: n*(a1 + an)/2 = 21*(140 + 2940)/2 = 21*(3080)/2 = 21*1540.Compute 21*1540. 20*1540 = 30,800 and 1*1540 = 1,540. So 30,800 + 1,540 = 32,340. Yes, same result. So that seems correct.But let me check the number of terms again. If the first term is 140 and the last term is 2940, then the number of terms is ((2940 - 140)/140) + 1 = (2800/140) + 1 = 20 + 1 = 21. Yep, that's 21 terms. So that's correct.Alternatively, using the formula for the nth term of an arithmetic sequence: an = a1 + (n - 1)*d, where d is the common difference, which here is 140.So 2940 = 140 + (n - 1)*140.Subtract 140: 2800 = (n -1)*140.Divide by 140: 20 = n - 1, so n = 21. Correct.So everything checks out. The sum is 32,340.But wait, let me just confirm with another approach. Maybe using the formula for the sum of multiples.The sum of all multiples of k up to N is k * m * (m + 1)/2, where m is the number of terms. Wait, but no. Wait, actually, if you have multiples of k, then the sum is k*(1 + 2 + ... + m) = k*m*(m + 1)/2. But in our case, the multiples are starting from k, but if k is the LCM, which is 140, then the number of terms is m = floor(N / k). So here, N = 3000, k = 140, so m = 21. Therefore, sum = 140*(21)*(22)/2. Let me compute that.140*21*11, since 22/2 is 11. 140*21 = 2940. Then 2940*11. Let's compute 2940*10 = 29,400, plus 2940 = 32,340. So same answer. Therefore, the sum is indeed 32,340.Therefore, the answer should be 32,340. But let me just check once again if I didn't miss any steps or miscalculate.Alternatively, perhaps I can list some terms and check:First term: 140Second term: 280Third term: 420...Last term: 2940Number of terms: 21.Sum using arithmetic series formula: 21*(140 + 2940)/2 = 21*3080/2 = 21*1540 = 32,340.Yes, that's consistent.Alternatively, maybe I can check with a smaller example to see if the approach is correct.Suppose instead of 3000, we have N=140. Then the sum would be 140. If N=280, then sum is 140 + 280 = 420. Let's compute using the formula: floor(280/140)=2 terms. Sum=2/2*(140 + 280)=1*420=420. Correct.Similarly, N=420: floor(420/140)=3 terms. Sum=3/2*(140 + 420)=3/2*560=3*280=840. Which is 140 + 280 + 420 = 840. Correct.So the formula works here. So applying it to N=3000 with floor(3000/140)=21 terms, sum is 21/2*(140 + 2940)=21/2*3080=21*1540=32,340. Correct.Therefore, I think the answer is 32,340.Final Answerboxed{32340}</think>To find the sum of all natural numbers that are multiples of both 20 and 14 and do not exceed 3000, we start by determining the least common multiple (LCM) of 20 and 14. The LCM of 20 and 14 is calculated as follows:- The greatest common divisor (GCD) of 20 and 14 is 2.- Using the formula ( text{LCM}(a, b) = frac{a times b}{text{GCD}(a, b)} ), we get ( text{LCM}(20, 14) = frac{20 times 14}{2} = 140 ).Next, we need to find all multiples of 140 up to 3000. The largest multiple of 140 that does not exceed 3000 is 2940, which is calculated by:- Dividing 3000 by 140: ( leftlfloor frac{3000}{140} rightrfloor = 21 ).This means there are 21 terms in the sequence. The sequence of multiples of 140 is an arithmetic sequence where the first term ( a_1 = 140 ), the last term ( a_{21} = 2940 ), and the number of terms ( n = 21 ).The sum of an arithmetic sequence is given by the formula ( S = frac{n}{2} times (a_1 + a_n) ):- Substituting the values, we get ( S = frac{21}{2} times (140 + 2940) ).- Calculating the sum: ( frac{21}{2} times 3080 = 21 times 1540 = 32340 ).Thus, the sum of all natural numbers that are multiples of both 20 and 14 and do not exceed 3000 is (boxed{32340}).,<think>Okay, so I need to find the sum of all natural numbers that are multiples of both 20 and 14, and these numbers shouldn't exceed 3000. Hmm, let's start by understanding the problem step by step.First, the problem mentions "natural numbers that are multiples of 20" and also "multiples of 14". So, we're looking for numbers that are common multiples of both 20 and 14. That makes me think of the least common multiple (LCM) of these two numbers. Once I find the LCM, I can generate all multiples of that LCM up to 3000 and then sum them up.Alright, let me recall how to find the LCM of two numbers. The LCM of two numbers is the smallest number that is a multiple of both. To compute it, I can use the formula: LCM(a, b) = (a * b) / GCD(a, b), where GCD is the greatest common divisor.So, first, I need to find the GCD of 20 and 14. Let's factor both numbers into primes.20 factors into 2^2 * 5^1.14 factors into 2^1 * 7^1.The GCD is the product of the smallest powers of the common prime factors. The common prime factor here is 2, and the smallest power is 1. So GCD(20, 14) = 2^1 = 2.Now, using the LCM formula: LCM(20, 14) = (20 * 14) / 2 = 280 / 2 = 140. Wait, no, wait: 20*14 is 280, and dividing by GCD 2 gives 140. So LCM is 140. Therefore, the numbers we are interested in are multiples of 140 that do not exceed 3000.Got it. So now, the problem reduces to finding the sum of all multiples of 140 up to 3000.To find this sum, I remember that the sum of an arithmetic series can be calculated by the formula: S = n/2 * (a1 + an), where n is the number of terms, a1 is the first term, and an is the last term.First, let's confirm the first term. Since we're dealing with natural numbers (which are positive integers), the first multiple of 140 is 140 itself. So a1 = 140.Next, we need to find the largest multiple of 140 that is less than or equal to 3000. Let's divide 3000 by 140 to see how many times 140 fits into 3000.Calculating 3000 ÷ 140: Let's do this division. 140*20 = 2800, which is less than 3000. 3000 - 2800 = 200. 140*1 = 140, which is less than 200. 200 - 140 = 60, which is less than 140. So the largest multiple is 140*21 = 2940. Wait, 140*21 = 140*(20 + 1) = 2800 + 140 = 2940. Then 2940 + 140 = 3080, which exceeds 3000. So yes, 2940 is the last term. Therefore, an = 2940.Now, we need to find how many terms there are. Since the multiples start at 140 and go up by 140 each time, the number of terms is equal to the quotient when 3000 is divided by 140, discarding any remainder. But let me check that. Alternatively, since the nth term is given by a_n = 140n. We want a_n ≤ 3000. So 140n ≤ 3000 => n ≤ 3000/140. Let's compute 3000 divided by 140.3000 ÷ 140 = 21.428... So the integer part is 21. Therefore, there are 21 terms. But wait, when n=21, a_n = 140*21 = 2940, which is ≤3000. So n=21 is the last term. Therefore, the number of terms is 21.So, n=21, a1=140, an=2940.Now plug into the sum formula: S = 21/2 * (140 + 2940). Let's compute that.First, compute 140 + 2940. 140 + 2940 = 3080. Then, 21/2 = 10.5. So 10.5 * 3080.Let me compute 10 * 3080 = 30,800. Then 0.5 * 3080 = 1,540. Adding these together: 30,800 + 1,540 = 32,340. So the sum is 32,340.Wait, that seems straightforward, but let me verify the number of terms again. Since the first term is 140 (n=1) and the last term is 2940 (n=21). So numbers of terms is 21. Let me check with a smaller example. For example, multiples of 140 up to 140*3=420. The terms would be 140, 280, 420. So n=3. If we use the formula, 420/140=3, so n=3. So yes, that works. So in this case, 3000/140≈21.428, so n=21.Alternatively, another way to compute the number of terms is floor(3000 / 140). Since 3000 divided by 140 is 21.428..., floor of that is 21. So number of terms is 21. So that's correct.Therefore, the sum should be 21*(140 + 2940)/2 = 21*3080/2 = 21*1540=32,340.But just to make sure, let's compute 21*1540. Let's break it down: 20*1540 = 30,800 and 1*1540=1,540. So 30,800 + 1,540 = 32,340. Correct.So, the answer is 32,340. Let me check once again to make sure I didn't make any miscalculations.Wait, but let me verify with another approach. The sum of an arithmetic series can also be written as S = n * (a1 + an)/2. Which is exactly what we did. So n=21, a1=140, an=2940. So 21*(140 + 2940)/2. So (140 + 2940) is 3080, divided by 2 is 1540. Multiply by 21 gives 32,340. Correct.Alternatively, since all the terms are multiples of 140, the sum can also be calculated as 140*(1 + 2 + 3 + ... + 21). Wait, but 1 + 2 + ... +21 is (21*22)/2 = 231. Then 140*231. Let's compute that. 140*200=28,000. 140*31=4,340. So 28,000 + 4,340=32,340. Same answer. So that's good.Another way, 140*21 = 2940. So the last term. So the sum is 140*(1+2+...+21) = 140*(21*22)/2 = 140*231 = 32,340. Yep, same result. So that's correct.So I think the answer is 32,340.But let me check again if LCM is correct. LCM of 20 and 14. As we did before: 20=2^2*5, 14=2*7. LCM is the product of the highest powers: 2^2 *5 *7=4*5*7=140. Correct. So yes, LCM(20,14)=140. So the common multiples are 140, 280, ..., up to 2940. 2940 is 21*140=2940. 22*140=3080>3000. So 21 terms. So all that checks out.Alternatively, maybe the problem is worded as "multiples of 20" and also "multiples of 14", which would mean numbers that are multiples of both, which is the same as multiples of LCM(20,14). So, that's 140. So sum of multiples of 140 not exceeding 3000. So, yeah, 140, 280,...2940. 21 terms. Sum is 32,340.Therefore, I think the answer is 32,340. But let me just confirm once more.Wait, let's check the sum of the first 21 multiples of 140:Sum = 140 + 280 + 420 + ... + 2940.Alternatively, since each term is 140 times 1, 2, 3,...,21. So sum is 140*(1+2+...+21) = 140*(21*22)/2 = 140*231. Let's compute 140*231.140*200=28,000.140*30=4,200.140*1=140.So 200+30+1=231. So 28,000 + 4,200 + 140 = 32,340. Yep, same answer.Therefore, confident that the answer is 32,340. So the sum is 32,340.Final AnswerThe sum is boxed{32340}.</think>To find the sum of all natural numbers that are multiples of both 20 and 14 and do not exceed 3000, we start by determining the least common multiple (LCM) of 20 and 14. 1. Prime Factorization: - 20 factors into (2^2 times 5) - 14 factors into (2 times 7)2. GCD and LCM Calculation: - The greatest common divisor (GCD) of 20 and 14 is (2). - Using the formula ( text{LCM}(a, b) = frac{a times b}{text{GCD}(a, b)} ), we get: [ text{LCM}(20, 14) = frac{20 times 14}{2} = 140 ]3. Finding Multiples of 140 up to 3000: - The first multiple of 140 is 140. - The largest multiple of 140 not exceeding 3000 is found by dividing 3000 by 140 and taking the floor value: [ leftlfloor frac{3000}{140} rightrfloor = 21 ] - The last term is (140 times 21 = 2940).4. Sum of the Arithmetic Series: - The number of terms (n) is 21. - The first term (a_1) is 140. - The last term (a_n) is 2940. - Using the sum formula for an arithmetic series (S = frac{n}{2} times (a_1 + a_n)): [ S = frac{21}{2} times (140 + 2940) = frac{21}{2} times 3080 = 21 times 1540 = 32340 ]Thus, the sum of all natural numbers that are multiples of both 20 and 14 and do not exceed 3000 is (boxed{32340}).]

🔑:32340 .【Analysis】Among all multiples of 20 that do not exceed 2014 and are also multiples of 14, the smallest is 140, and the largest is 2940, totaling 21. Then, solve using the "Gauss summation" method.【Solution】Solution: 20=2 times 2 times 514=2 times 7The least common multiple of 20 and 14 is: 2 times 2 times 5 times 7=140begin{array}{l}3000 div 140 approx 21.4 140 times 21=2940end{array}So, among all multiples of 20 that do not exceed 3000 and are also multiples of 14, the smallest is 140, and the largest is 2940, totaling 21,(140+2940) times 21 div 2begin{array}{l}=3080 times 21 div 2 =32340 .end{array}Answer: Among all natural numbers that are multiples of 20, those that do not exceed 3000 and are also multiples of 14 have a sum of 32340. Therefore, the answer is: 32340 .

🧠:[<think>Okay, let's try to figure out this problem. The question is about finding the range of m such that the inequality sin²x + m sinx + (m² - 3)/m ≤ 0 holds for all real numbers x. The answer choices are given, so we need to determine which one is correct.First, I need to understand what the problem is asking. It's saying that no matter what x we pick, this quadratic expression in sinx will always be less than or equal to zero. So, the quadratic must be non-positive for all possible values of sinx. Since sinx ranges between -1 and 1, the variable here is essentially sinx, and the quadratic must be ≤ 0 for all t in [-1, 1], where t = sinx.So, maybe I can substitute t = sinx, and the inequality becomes t² + m t + (m² - 3)/m ≤ 0, and this must hold for all t ∈ [-1, 1]. Therefore, the quadratic in t must be non-positive for all t in that interval. That seems like the right approach.But wait, before proceeding, I should note that m cannot be zero because in the term (m² - 3)/m, m is in the denominator. So, m ≠ 0. The answer choices don't include m=0, so that's fine. The options are A: m ≤ -3/2, B: 0 < m ≤1, C: 0 < m ≤3, D: m ≤ -3/2 or 0 < m ≤3. So, D is combining A and C. Let's see.Now, the problem is to find m such that t² + m t + (m² -3)/m ≤0 for all t ∈ [-1,1]. Since this is a quadratic in t, maybe we can analyze its maximum and ensure that the maximum is ≤0 over the interval [-1,1].Alternatively, since the quadratic must be ≤0 for all t in [-1,1], the quadratic must lie entirely below or on the t-axis in that interval. Therefore, the quadratic should not have any points above the t-axis in t ∈ [-1,1]. To ensure this, the quadratic must satisfy two conditions:1. The quadratic evaluated at the endpoints t = -1 and t = 1 must be ≤0.2. The maximum of the quadratic in the interval [-1,1] must also be ≤0.But since it's a quadratic, depending on the leading coefficient, it opens upwards or downwards. The coefficient of t² is 1, which is positive, so the parabola opens upwards. Therefore, the minimum is at the vertex, and the maximum on the interval [-1,1] will occur at one of the endpoints. Wait, if it opens upwards, then the maximum on a closed interval will be at one of the endpoints. Therefore, if both endpoints are ≤0, then the entire interval is ≤0? Hmm, that might not necessarily be true. Because even if the endpoints are ≤0, the vertex (the minimum) could be lower, but since the parabola opens upwards, the maximum is indeed at the endpoints. Wait, no. If the parabola opens upwards, then the vertex is the minimum point, so the maximum on the interval would be at either t = -1 or t =1. So, if both endpoints are ≤0, then the entire parabola between -1 and 1 is ≤0. Because between -1 and 1, the parabola is upward opening, so the highest points are at the edges. Therefore, if both edges are ≤0, the entire interval is ≤0. Therefore, maybe we just need to check the endpoints? Let me verify this with an example.Suppose we have a quadratic f(t) = t² + bt + c, which opens upwards. Then, on the interval [-1,1], the maximum value is at t=-1 or t=1. So, if f(-1) ≤0 and f(1) ≤0, then since the parabola is opening upwards, the entire interval between -1 and 1 must be below zero. Is that correct?Wait, suppose the vertex is below the interval. For example, suppose the vertex is at t=0, with f(0) = c. If f(-1) and f(1) are both ≤0, then even though the vertex is the minimum, which is lower, the entire parabola would be below zero in between. Wait, no. If the parabola opens upwards, the minimum is at the vertex. If the vertex is below some point, but the endpoints are below zero, does that mean all points in between are also below zero? Let me take a specific example.Let f(t) = t² - 2. Then f(-1) = 1 - 2 = -1, f(1) = 1 -2 = -1. The vertex is at t=0, which is -2. So between -1 and1, f(t) ranges from -1 to -2. So yes, it's always ≤0. But if I take another example, f(t)=t² -0.5. Then f(-1)=1 -0.5=0.5, which is positive, but f(1)=0.5 as well. So in that case, the endpoints are positive. If I take f(t)= t² -0.5. If we check at t=0, it's -0.5, which is negative, but at t=1 and -1, it's positive. So in this case, the quadratic is negative in the middle but positive at the ends. Therefore, if the endpoints are positive, then the quadratic is positive there, but if endpoints are negative, then the entire interval is negative. Wait, so if we have a quadratic opening upwards, with f(-1) ≤0 and f(1) ≤0, then between -1 and1, is the quadratic always ≤0?Wait, in the first example, f(t)=t² -2, yes. If we have a quadratic that opens upwards and has its vertex below the x-axis, but the endpoints are also below the x-axis. Wait, but suppose the quadratic is such that the vertex is below the x-axis, but the arms go up. If the arms go up, then even if the endpoints are below zero, there could be points beyond the endpoints where the quadratic is positive. But within the interval [-1,1], if the endpoints are both ≤0, and the quadratic opens upwards, then in between, the quadratic is ≤0. Because the parabola can only go up outside the interval, but between -1 and1, it's opening upwards, with the minimum somewhere in between. If both endpoints are ≤0, then between them, since the parabola is opening upwards, the entire segment must be ≤0. Wait, let's think of a quadratic that opens upwards, with f(-1) ≤0 and f(1) ≤0.Suppose the vertex is between -1 and1. Then the minimum is inside the interval. If the minimum is below zero, and the endpoints are also below zero, then the entire interval is below zero. If the minimum is above zero, then even if the endpoints are below zero, the quadratic would cross the x-axis twice. But if the quadratic opens upwards, has minimum above zero, then it can't be below zero anywhere. So that can't happen. Wait, if the minimum is above zero, then the entire quadratic is above zero. So, if the quadratic is to be ≤0 for all t in [-1,1], then it must have its maximum at the endpoints, which must be ≤0. Therefore, the necessary and sufficient condition is that f(-1) ≤0 and f(1) ≤0. Is that correct?Wait, but suppose the quadratic crosses the t-axis at some points outside the interval [-1,1], but between -1 and1, it is below zero. However, since the quadratic opens upwards, if the vertex is below zero, and the endpoints are above zero, then between the roots, the quadratic is below zero. But if the interval [-1,1] is entirely between the roots, then the quadratic would be below zero there. But in that case, the endpoints would be above zero, so f(-1) and f(1) would be positive, which contradicts the problem's requirement that the quadratic is ≤0 for all t in [-1,1]. So, in order for the quadratic to be ≤0 for all t in [-1,1], given that it opens upwards, the entire interval must lie between the two roots or coincide with the vertex. But since it opens upwards, the only way for the quadratic to be ≤0 on the entire interval is if the interval is between the two roots. However, that would require that the quadratic has two real roots and the interval [-1,1] lies between them. But in that case, the quadratic would be ≤0 between its roots. But since the quadratic opens upwards, it is ≤0 between its two roots. Therefore, if we want the quadratic to be ≤0 for all t in [-1,1], then [-1,1] must lie between the two roots. Therefore, the roots must satisfy root1 ≤ -1 and root2 ≥1. Then, between root1 and root2, which includes [-1,1], the quadratic is ≤0.Alternatively, if the quadratic does not cross the t-axis at all (discriminant <0), then since it opens upwards, it is always positive, which would not satisfy the inequality. So the quadratic must have two real roots, and the interval [-1,1] must lie between them. Therefore, the conditions are:1. The quadratic has two real roots. So discriminant ≥0.2. The roots are such that root1 ≤ -1 and root2 ≥1.So, combining these two conditions.Let me recall that for a quadratic equation at² + bt + c =0, the roots are given by [-b ± sqrt(b² -4ac)]/(2a). In our case, a=1, b=m, c=(m² -3)/m.Therefore, discriminant D = m² -4*1*( (m² -3)/m ) = m² - 4(m² -3)/m.Simplify D:D = m² - [4(m² -3)/m] = (m³ -4m² +12)/m.Wait, let me compute step by step:First, compute 4ac: 4*1*( (m² -3)/m ) = 4(m² -3)/m.Then discriminant D = b² -4ac = m² - 4(m² -3)/m.So D = m² - (4m² -12)/m = m² -4m + 12/m.Wait, that seems complicated. Let me write it over a common denominator:D = [m³ -4m² +12]/m.So, discriminant D = (m³ -4m² +12)/m.For the quadratic to have real roots, D must be ≥0.But since D is (m³ -4m² +12)/m, the sign of D depends on m and the numerator.But this might be complicated. Alternatively, maybe there's a different approach.Alternatively, as we thought before, if the quadratic t² + m t + (m² -3)/m ≤0 for all t ∈ [-1,1], then since the quadratic opens upwards (coefficient 1 is positive), the maximum in the interval occurs at t=-1 or t=1. Therefore, if both f(-1) and f(1) are ≤0, then the entire interval is ≤0.So maybe just check f(-1) ≤0 and f(1) ≤0.Wait, but let's test this with an example. Suppose m=3, which is in option C and D.Compute f(t)= t² +3 t + (9 -3)/3 = t² +3t + 6/3 = t² +3t +2.So f(t)= t² +3t +2. The roots are t=-1 and t=-2. So this quadratic is ≤0 between t=-2 and t=-1. But we need it to be ≤0 for all t ∈ [-1,1]. However, at t=0, f(0)=2 >0, which violates the inequality. So something is wrong here. Therefore, my previous reasoning must be incorrect.Wait, that's a problem. If m=3, then according to answer options C and D, m=3 is allowed, but the quadratic at t=0 is 0 +0 + (9 -3)/3 = 6/3=2>0, which contradicts the inequality. Therefore, maybe the answer options are wrong, or my reasoning is flawed.But this suggests that m=3 is not a valid solution. But in the options, C and D include m=3. So there's a mistake here. Therefore, my initial approach must be wrong.Wait, maybe I substituted m=3 incorrectly. Let's check again.Original inequality: sin²x + m sinx + (m² -3)/m ≤0. For m=3, this becomes sin²x +3 sinx + (9 -3)/3 = sin²x +3 sinx + 2. So sin²x +3 sinx +2 ≤0. Let's factor this quadratic: (sinx +1)(sinx +2) ≤0. The roots are sinx=-1 and sinx=-2. Since sinx cannot be -2, the quadratic is (sinx +1)(sinx +2). Since sinx +2 is always positive (since sinx ≥-1), then the inequality becomes sinx +1 ≤0, so sinx ≤-1. But sinx=-1 is the only solution. Therefore, the inequality holds only when sinx=-1, but the problem states that the inequality must hold for all x ∈ R. However, for most x, sinx is not -1, so the inequality would not hold. Therefore, m=3 is not a valid solution, which contradicts options C and D. Therefore, there's an error in the problem or the answer choices. Wait, but the options are given, so maybe I made a mistake.Wait, perhaps the problem is stated incorrectly? Let me check again.The inequality is sin²x +m sinx + (m² -3)/m ≤0. For m=3, as above, the quadratic is sin²x +3 sinx +2 ≤0. As we saw, this is only true when sinx=-1. Therefore, it does not hold for all x. Therefore, m=3 is invalid. So why is it included in the options? Maybe my reasoning is wrong.Wait, perhaps I need to approach this differently. The problem says that for any x ∈ R, the inequality holds. Therefore, the inequality must hold for all real numbers x. Since sinx can take any value in [-1,1], the inequality must hold for all t ∈ [-1,1], where t=sinx. So, the quadratic in t must be ≤0 for all t ∈ [-1,1]. Therefore, maybe the correct approach is to ensure that the quadratic is ≤0 for all t in [-1,1], which would require that the maximum of the quadratic on [-1,1] is ≤0.Since the quadratic opens upwards (coefficient 1), the maximum occurs at one of the endpoints, t=-1 or t=1. Therefore, if both f(-1) and f(1) are ≤0, then the quadratic is ≤0 on [-1,1]. But in the example with m=3, f(1)=1 +3 +2=6>0, so it's not ≤0. Therefore, m=3 is invalid. So perhaps the answer options are wrong. Alternatively, maybe I need to consider the minimum of the quadratic?Wait, but if the quadratic has to be ≤0 for all t ∈ [-1,1], then since it opens upwards, the entire interval must lie between the two roots. So the quadratic must have two real roots, and the interval [-1,1] must be between them. Therefore, the roots must satisfy root1 ≤ -1 and root2 ≥1.Therefore, the conditions are:1. The quadratic has two real roots. So discriminant ≥0.2. The smaller root ≤ -1 and the larger root ≥1.Let me write that down.Quadratic equation: t² +mt + (m² -3)/m =0.First, discriminant D = m² -4*(1)*[(m² -3)/m] = m² - 4(m² -3)/m.Let's compute D:D = m² - (4m² -12)/m = (m³ -4m² +12)/m.For real roots, D ≥0. So (m³ -4m² +12)/m ≥0.This is equivalent to (m³ -4m² +12) * m ≥0, considering m ≠0.So, depending on the sign of m, the inequality changes.Case 1: m >0.Then, since m >0, D ≥0 ⇨ m³ -4m² +12 ≥0.So we need m³ -4m² +12 ≥0.Case 2: m <0.Then, m <0, so D ≥0 ⇨ m³ -4m² +12 ≤0.Because multiplying both sides by m (negative) reverses the inequality.So first, let's analyze discriminant conditions.Additionally, the roots must satisfy root1 ≤ -1 and root2 ≥1. Let's express that.For quadratic t² +bt +c =0, the roots are [-b ± sqrt(b² -4ac)]/(2a).In our case, a=1, b=m, c=(m² -3)/m.So the roots are [-m ± sqrt(D)]/2.But D = (m³ -4m² +12)/m.So sqrt(D) is sqrt[(m³ -4m² +12)/m]. Let's denote sqrt(D) as S.Therefore, roots are [-m + S]/2 and [-m - S]/2.We need the smaller root ≤-1 and the larger root ≥1.So:[-m - S]/2 ≤ -1 and [-m + S]/2 ≥1.Let's write these inequalities:First inequality: [-m - S]/2 ≤ -1 ⇒ -m - S ≤ -2 ⇒ -m +2 ≤ S.Second inequality: [-m + S]/2 ≥1 ⇒ -m + S ≥2 ⇒ S ≥ m +2.Therefore, combining both inequalities:From first: S ≥ -m +2From second: S ≥ m +2So S must be ≥ max(-m +2, m +2)But S is sqrt[(m³ -4m² +12)/m], which is non-negative.Therefore, we need both S ≥ -m +2 and S ≥ m +2.But since S ≥0, we have to consider the values of m where -m +2 ≤ S and m +2 ≤ S.But this seems complicated. Maybe there's a better approach.Alternatively, for the quadratic to be ≤0 for all t in [-1,1], the following must hold:1. At t=-1: (-1)^2 + m*(-1) + (m² -3)/m ≤0 ⇒ 1 - m + (m² -3)/m ≤0.2. At t=1: 1^2 + m*1 + (m² -3)/m ≤0 ⇒ 1 + m + (m² -3)/m ≤0.So compute these two inequalities.First, let's compute inequality at t=-1:1 - m + (m² -3)/m ≤0.Combine terms:[ m(1 - m) + m² -3 ] / m ≤0 ⇒ [ m - m² + m² -3 ] /m ≤0 ⇒ (m -3)/m ≤0.Similarly, inequality at t=1:1 + m + (m² -3)/m ≤0.Combine terms:[ m(1 + m) + m² -3 ] /m ≤0 ⇒ [ m + m² + m² -3 ] /m ≤0 ⇒ (2m² + m -3)/m ≤0.So now, we have two inequalities:1. (m -3)/m ≤02. (2m² + m -3)/m ≤0Additionally, since the quadratic must be ≤0 for all t in [-1,1], but as we saw in the example with m=3, even if the endpoints are satisfied (wait, for m=3, t=-1 gives (3-3)/3=0 ≤0, and t=1 gives (2*9 +3 -3)/3=(18)/3=6 >0, which violates the second inequality. Therefore, m=3 does not satisfy the second inequality, which is why it's invalid.Therefore, to solve the problem, we need to solve the two inequalities:1. (m -3)/m ≤02. (2m² + m -3)/m ≤0And also ensure that the quadratic does not have a minimum within [-1,1] that is above zero. Wait, but since the quadratic opens upwards, if the endpoints are ≤0, then the entire interval is ≤0. However, in the case where the vertex is above the interval, but the endpoints are negative, but that can't happen because the quadratic opens upwards.Wait, perhaps if the vertex is within the interval [-1,1], then even if the endpoints are ≤0, the quadratic might dip below zero in between, but since it opens upwards, the minimum is at the vertex, so if the vertex's y-value is ≤0, then combined with endpoints ≤0, the entire interval would be ≤0. Wait, but the problem is to ensure that the quadratic is ≤0 for all t in [-1,1]. So, actually, the necessary and sufficient conditions are:1. The quadratic is ≤0 at both endpoints t=-1 and t=1.2. The quadratic's vertex is ≤0.But since the quadratic opens upwards, if the vertex is ≤0 and the endpoints are ≤0, then the entire interval is ≤0.But maybe just checking the endpoints isn't enough. Let me think.Wait, let's consider the vertex. The vertex occurs at t = -b/(2a) = -m/(2*1) = -m/2.So the t-coordinate of the vertex is -m/2. The y-coordinate is f(-m/2) = (-m/2)^2 + m*(-m/2) + (m² -3)/m.Compute that:= m²/4 - m²/2 + (m² -3)/m= -m²/4 + (m² -3)/m.= (-m³/4 + m² -3)/m.Hmm, complicated expression.Alternatively, maybe instead of evaluating the vertex, since we already have the conditions on the endpoints, and given that the quadratic opens upwards, if both endpoints are ≤0, then the quadratic is ≤0 on the entire interval. Because the parabola opens upwards, so between t=-1 and t=1, the maximum is at the endpoints. If those are ≤0, then everything in between is ≤0. So perhaps checking the endpoints is sufficient.But in the example with m=1:First inequality: (1 -3)/1 = -2 ≤0.Second inequality: (2*1 +1 -3)/1 = (2 +1 -3)/1=0 ≤0.So m=1 satisfies both inequalities.Let's check the quadratic: t² +1*t + (1 -3)/1 = t² + t -2.Factor: (t +2)(t -1). So roots at t=-2 and t=1. Therefore, the quadratic is ≤0 between t=-2 and t=1. Since our interval is t ∈ [-1,1], which is inside [-2,1], the quadratic is ≤0 on [-1,1]. Therefore, m=1 works. So at t=0: 0 +0 -2 = -2 ≤0, which is okay.Another test case: m= -2 (not in the options, but let's check):First inequality: (-2 -3)/(-2) = (-5)/(-2)=2.5 ≤0? No, 2.5 >0. So m=-2 doesn't satisfy the first inequality. Therefore, m=-2 is invalid.Another test case: m= -3/2 (option A and D).First inequality: (-3/2 -3)/(-3/2) = (-9/2)/(-3/2)= (9/2)/(3/2)= 3 ≤0? No, 3 >0. So m=-3/2 doesn't satisfy the first inequality. Wait, that's strange. The answer option A is m ≤ -3/2. But m=-3/2 would not satisfy the first inequality. Therefore, maybe there's a mistake in the options.Wait, let me recast the two inequalities:First inequality: (m -3)/m ≤0.This is equivalent to (m -3)/m ≤0.The critical points are m=0 and m=3.For m >0:If 0 < m <3, then numerator is negative, denominator positive: overall negative, so inequality holds.If m >3, numerator positive, denominator positive: overall positive, inequality doesn't hold.For m <0:Numerator is m -3 (which is negative since m <0), denominator m is negative: overall positive, so inequality doesn't hold.Therefore, the first inequality holds when 0 < m ≤3.Second inequality: (2m² + m -3)/m ≤0.Factor numerator: 2m² +m -3.Let me try to factor it:Looking for two numbers a and b such that a*b = 2*(-3)=-6 and a +b=1.Numbers 3 and -2: 3*(-2)=-6, 3 + (-2)=1. Therefore:2m² +m -3 =2m² +3m -2m -3 =m(2m +3) -1(2m +3)=(2m +3)(m -1)Therefore, 2m² +m -3 factors into (2m +3)(m -1).Therefore, the second inequality becomes:(2m +3)(m -1)/m ≤0.Critical points at m=-3/2, m=0, m=1.Test intervals:1. m < -3/2:(2m +3) negative, (m -1) negative, m negative.Multiply: (-)(-)/(-) = (-)/- = positive. So inequality not satisfied.2. -3/2 <m <0:(2m +3) positive (since m >-3/2), (m -1) negative, m negative.Multiply: (+)(-)/(-) = (-)/(-) = positive. Inequality not satisfied.3. 0 <m <1:(2m +3) positive, (m -1) negative, m positive.Multiply: (+)(-)/(+) = (-)/+ = negative. Inequality satisfied.4. m >1:(2m +3) positive, (m -1) positive, m positive.Multiply: (+)(+)/(+) = positive. Inequality not satisfied.Therefore, the second inequality holds when 0 <m ≤1.Therefore, combining the two inequalities:First inequality holds for 0 <m ≤3.Second inequality holds for 0 <m ≤1.Therefore, the intersection is 0 <m ≤1.But wait, but in the answer options, B is 0 <m ≤1, D includes 0 <m ≤3. But according to this, the correct range is 0 <m ≤1. However, when m=1, we saw that the quadratic is ≤0 for all t ∈ [-1,1]. But when m=2, which is in 0 <m ≤3, let's check:For m=2:First inequality: (2 -3)/2 = (-1)/2 = -0.5 ≤0: holds.Second inequality: (2*(4) +2 -3)/2 = (8 +2 -3)/2 =7/2=3.5 >0: does not hold. Therefore, m=2 does not satisfy the second inequality. Therefore, the valid m must satisfy both inequalities, so 0 <m ≤1. Therefore, answer B.But this conflicts with the answer options given. The options are:(A) m ≤-3/2(B) 0<m ≤1(C) 0<m ≤3(D) m ≤-3/2 or 0<m ≤3But according to the above, the correct answer is B. However, when m=1, let's verify.At m=1, quadratic is t² + t -2. Factor: (t+2)(t-1). Therefore, quadratic ≤0 when t is between -2 and1. Since t ∈ [-1,1], which is within that interval, so yes, quadratic is ≤0. Therefore, m=1 is valid.But what about m=0.5?First inequality: (0.5 -3)/0.5 = (-2.5)/0.5 = -5 ≤0: holds.Second inequality: (2*(0.25) +0.5 -3)/0.5 = (0.5 +0.5 -3)/0.5 = (-2)/0.5 = -4 ≤0: holds.Therefore, m=0.5 is valid.What about m approaching 0 from the right?As m→0+, first inequality: (m -3)/m ≈ (-3)/m → -∞ ≤0: holds.Second inequality: (2m² +m -3)/m ≈ (-3)/m → -∞ ≤0: holds.Therefore, m can approach 0, but m>0.But what about m=1?Wait, but according to our previous analysis, the correct range is 0 <m ≤1, which is option B. However, the answer options do include B, but the given options also have D, which combines A and C. However, our analysis didn't find any valid m ≤-3/2. Let's check m=-2, but m negative:First inequality: (m -3)/m ≤0. For m negative, numerator m-3 is negative, denominator m is negative: negative/negative=positive. So inequality (positive ≤0) not satisfied. Therefore, for m negative, first inequality is not satisfied. Therefore, no solutions for m <0.Therefore, only valid solutions are 0 <m ≤1, which is option B. But the original problem's answer options don't have B as correct? Wait, the problem says:(A) m ≤-3/2(B) 0<m ≤1(C) 0<m ≤3(D) m ≤-3/2 or 0<m ≤3So according to our analysis, the correct answer should be B. However, when I first tested m=1, it worked, but m=3 did not. However, let's check m=1.5, which is in option C and D but not in B.For m=1.5:First inequality: (1.5 -3)/1.5 = (-1.5)/1.5 = -1 ≤0: holds.Second inequality: (2*(2.25) +1.5 -3)/1.5 = (4.5 +1.5 -3)/1.5 =3/1.5=2 >0: does not hold. Therefore, m=1.5 is invalid.Therefore, only m ∈ (0,1] works. Therefore, answer B.But the problem gives answer options, and according to our analysis, B is correct. But in the initial problem statement, maybe there's a mistake.Wait, let's check the problem statement again:"For any x ∈ R, the inequality sin²x +m sinx + (m² -3)/m ≤0 always holds. Then the range of m is (quad)."The options are:(A) m ≤-3/2(B) 0<m ≤1(C) 0<m ≤3(D) m ≤-3/2 or 0<m ≤3But according to the analysis, the correct answer is B. However, in the original problem, answer D is given as an option. So there must be an error in my reasoning.Wait, perhaps I missed something when m is negative. Let's check m negative again.Wait, let's take m=-3/2, which is option A and D.First inequality: (m -3)/m ≤0. For m=-3/2:(-3/2 -3)/(-3/2) = (-9/2)/(-3/2) = 3 ≤0? No, 3 is not ≤0. So first inequality fails.Second inequality: (2m² +m -3)/m ≤0. For m=-3/2:Compute numerator: 2*(9/4) + (-3/2) -3 = 9/2 -3/2 -3 = (9-3)/2 -3 =6/2 -3=3-3=0. So numerator=0. Therefore, inequality becomes 0/m ≤0. Since m=-3/2 ≠0, 0/m=0. So 0 ≤0: holds. Therefore, for m=-3/2, second inequality holds (equality), but first inequality fails. Therefore, m=-3/2 is invalid.But why is D an option? Maybe there is a case for m negative where both inequalities hold.Let's take m=-2:First inequality: (m -3)/m = (-5)/(-2)=2.5 ≤0? No.Second inequality: (2*(-2)^2 + (-2) -3)/(-2) = (8 -2 -3)/(-2)=3/-2=-1.5 ≤0? Yes. But first inequality fails.Take m=-4:First inequality: (-4 -3)/-4= (-7)/-4=1.75 ≤0? No.Second inequality: (2*16 + (-4) -3)/-4=(32 -4 -3)/-4=25/-4=-6.25 ≤0? Yes. But first inequality fails.Therefore, for m negative, the first inequality fails. Therefore, there are no negative m that satisfy both inequalities. Therefore, answer is B.But the options have B and D. Since B is 0<m≤1, and according to our analysis, that's correct. However, the original problem might have a different intended solution.Alternatively, maybe the vertex must also be considered.Wait, in the case of m=1, the quadratic is t² +t -2, which factors to (t+2)(t-1). Therefore, it is ≤0 between t=-2 and t=1. Since t ∈ [-1,1], which is within that interval, it holds.For m approaching 0 from the right, say m=0.1:First inequality: (0.1 -3)/0.1= (-2.9)/0.1=-29 ≤0: holds.Second inequality: (2*(0.01) +0.1 -3)/0.1=(0.02 +0.1 -3)/0.1=(-2.88)/0.1=-28.8 ≤0: holds.The quadratic becomes t² +0.1t + (0.01 -3)/0.1= t² +0.1t -29.9.This quadratic has roots at t = [-0.1 ± sqrt(0.01 +119.6)]/2 = [-0.1 ± sqrt(119.61)]/2.sqrt(119.61)≈10.94, so roots≈(-0.1 +10.94)/2≈5.42 and (-0.1 -10.94)/2≈-5.52. Therefore, the quadratic is ≤0 between t≈-5.52 and t≈5.42. Since our interval is [-1,1], which is inside, so it holds. Therefore, m=0.1 is valid.Similarly, for m approaching 0, the quadratic's roots go to ±infinity, so the interval [-1,1] is always inside.But when m increases beyond 1, say m=2, the quadratic at t=1 gives 1 +2 + (4 -3)/2=1 +2 +0.5=3.5>0, which violates the inequality.Therefore, the maximum valid m is 1. Therefore, the correct range is 0<m≤1, which is option B.But the answer choices given include B and D. The user hasn't provided a correct answer yet, but according to this analysis, B is correct. However, the original problem may have intended different conditions. Alternatively, perhaps there's an error in the problem statement.Wait, let's re-express the quadratic inequality:t² +mt + (m² -3)/m ≤0.Multiply both sides by m (but need to consider the sign of m):If m >0, the inequality direction remains:m t² + m² t + m² -3 ≤0.If m <0, inequality direction reverses:m t² + m² t + m² -3 ≥0.But original inequality is sin²x + m sinx + (m² -3)/m ≤0.Ah, perhaps I should consider the cases when m is positive and negative separately.Case 1: m >0.Then, multiplying the inequality by m (positive, so inequality direction remains):m sin²x + m² sinx + m² -3 ≤0.Which is a quadratic in sinx: m t² + m² t + (m² -3) ≤0, where t=sinx ∈[-1,1].Similarly, for m <0:Multiplying by m (negative) reverses the inequality:m sin²x + m² sinx + m² -3 ≥0.So, for m <0, the inequality becomes:m t² + m² t + (m² -3) ≥0 for all t ∈[-1,1].So, perhaps I didn't consider this before. My previous analysis didn't account for the sign of m when multiplying. Therefore, perhaps the correct approach is to split into cases based on the sign of m.Case 1: m >0.Then, the inequality is quadratic in t: m t² + m² t + (m² -3) ≤0 for all t ∈ [-1,1].Since m >0, the quadratic opens upwards. Therefore, to satisfy the inequality for all t ∈ [-1,1], we need:f(-1) ≤0 and f(1) ≤0.Where f(t)=m t² + m² t + (m² -3).Compute f(-1)=m(-1)^2 +m²*(-1) +m² -3 =m -m² +m² -3 =m -3 ≤0 ⇒ m ≤3.Compute f(1)=m(1)^2 +m²*(1) +m² -3 =m +m² +m² -3=2m² +m -3 ≤0.So for m >0, we need both f(-1)=m -3 ≤0 ⇒m ≤3 and f(1)=2m² +m -3 ≤0.Solve 2m² +m -3 ≤0.Factor: (2m +3)(m -1) ≤0.Critical points m=-3/2 and m=1. Since m >0, consider m ∈ (0,1].So for m >0, solutions are 0 <m ≤1.Case 2: m <0.Then, the inequality becomes m t² + m² t + (m² -3) ≥0 for all t ∈ [-1,1].Since m <0, the quadratic opens downward. Therefore, to satisfy the inequality for all t ∈ [-1,1], the quadratic must be ≥0 everywhere on [-1,1]. For a downward opening parabola, this requires that the minimum on the interval is ≥0. The minimum of a downward opening parabola occurs at the endpoints or the vertex.But since it opens downward, the vertex is the maximum point. Therefore, the minima occur at the endpoints t=-1 and t=1.Therefore, for m <0, we need:f(-1) ≥0 and f(1) ≥0.Compute f(-1)=m(-1)^2 +m²*(-1) +m² -3 =m -m² +m² -3 =m -3 ≥0 ⇒m ≥3. But m <0, so no solution.Compute f(1)=m(1)^2 +m²*1 +m² -3 =m +m² +m² -3=2m² +m -3 ≥0.For m <0, need 2m² +m -3 ≥0.Solve 2m² +m -3 ≥0.Factor: (2m +3)(m -1) ≥0.Critical points m=-3/2, m=1.For m <0, consider intervals:1. m < -3/2: test m=-2:(2*(-2) +3)(-2 -1)=(-4 +3)(-3)=(-1)(-3)=3 ≥0: yes.2. -3/2 <m <0: test m=-1:(2*(-1)+3)(-1 -1)=( -2 +3)(-2)=(1)(-2)=-2 <0: no.Therefore, for m < -3/2, 2m² +m -3 ≥0.But for m < -3/2, we need both f(-1) ≥0 and f(1) ≥0. However, f(-1)=m -3. If m < -3/2, then m -3 < -3/2 -3 = -9/2 <0. Therefore, f(-1)=m -3 <0, which contradicts the requirement f(-1) ≥0.Therefore, even though 2m² +m -3 ≥0 for m ≤-3/2, f(-1)=m -3 <0, so there are no solutions for m <0.Therefore, combining both cases, the only valid solutions are 0 <m ≤1.Therefore, the correct answer is B. But the options include D. However, according to this detailed analysis, B is correct. Therefore, there might be a mistake in the answer options provided, or perhaps a miscalculation in the problem.Alternatively, maybe the original problem had a different inequality. Let me check once again:Original inequality: sin²x +m sinx + (m² -3)/m ≤0.Yes. So when m <0, we have to reverse the inequality when multiplying by m, leading to the quadratic being ≥0. But for m <0, even though the quadratic opens downward, the requirement f(-1) and f(1) ≥0 cannot be satisfied since f(-1)=m -3 <0 for m <0. Therefore, no solutions for m <0.Hence, the correct answer is B, which is option (B) 0<m ≤1.However, the original answer options do not include B as correct in combination with others. The answer choices are A: m ≤-3/2, B: 0<m ≤1, C:0<m ≤3, D: m ≤-3/2 or 0<m ≤3. According to our analysis, B is correct, but the presence of D suggests that maybe the intended answer was different. However, based on the detailed steps, I think the correct answer is B.But let's re-examine the problem once again:The inequality must hold for all real x. So for all x, sin²x +m sinx + (m² -3)/m ≤0. Let's take m=1:The expression becomes sin²x + sinx + (1 -3)/1= sin²x + sinx -2.We need sin²x + sinx -2 ≤0 for all x.Let t=sinx. Then t² + t -2 ≤0 for all t ∈ [-1,1].The roots of t² +t -2=0 are t=(-1±√(1+8))/2=(-1±3)/2=1 or -2. Therefore, the quadratic is ≤0 between t=-2 and t=1. Since t ∈ [-1,1] is within that interval, the inequality holds. Therefore, m=1 works.For m=2:Expression is sin²x +2 sinx + (4 -3)/2= sin²x +2 sinx +0.5.Which is t² +2t +0.5. The discriminant is 4 -2=2, roots at t=(-2±√2)/2=-1±√2/2≈-1.707 and -0.293. Therefore, the quadratic is ≤0 between these roots. But the interval [-1,1] includes t=-0.5, where f(-0.5)=0.25 -1 +0.5=-0.25 ≤0. Wait, but the quadratic is t² +2t +0.5. At t=1:1 +2 +0.5=3.5>0, which violates the inequality. Therefore, m=2 does not work.But according to the quadratic in t for m=2, it's ≤0 between -1 -√0.5 and -1 +√0.5≈-1.707 and -0.293. Therefore, for t ∈ [-1,1], the inequality is only satisfied between -1 and -0.293. But for t ∈ (-0.293,1], the quadratic is positive. Therefore, m=2 does not satisfy the inequality for all x. Therefore, m=2 is invalid.Similarly, for m=3:Quadratic is t² +3t +2. Which factors as (t+1)(t+2), ≤0 between -2 and -1. But since t ∈ [-1,1], the inequality holds only at t=-1. For other t, it's positive. Therefore, m=3 is invalid.Therefore, the only valid m are 0<m ≤1, which is option B.Final Answerboxed{B}</think>To determine the range of ( m ) such that the inequality [sin^2 x + m sin x + frac{m^2 - 3}{m} leq 0]holds for all ( x in mathbb{R} ), we substitute ( t = sin x ) where ( t in [-1, 1] ). The inequality becomes:[t^2 + mt + frac{m^2 - 3}{m} leq 0]for all ( t in [-1, 1] ). Key Steps:1. Quadratic Analysis: The quadratic ( t^2 + mt + frac{m^2 - 3}{m} ) must be non-positive for all ( t in [-1, 1] ). Since the quadratic opens upwards (coefficient of ( t^2 ) is 1), the maximum values occur at the endpoints ( t = -1 ) and ( t = 1 ).2. Evaluate at Endpoints: - At ( t = -1 ): [ 1 - m + frac{m^2 - 3}{m} leq 0 implies frac{m - 3}{m} leq 0 ] - At ( t = 1 ): [ 1 + m + frac{m^2 - 3}{m} leq 0 implies frac{2m^2 + m - 3}{m} leq 0 ]3. Solve Inequalities: - For ( frac{m - 3}{m} leq 0 ): - This holds when ( 0 < m leq 3 ). - For ( frac{2m^2 + m - 3}{m} leq 0 ): - Factor the numerator: ( (2m + 3)(m - 1) ). - This holds when ( 0 < m leq 1 ).4. Intersection of Solutions: - The intersection of ( 0 < m leq 3 ) and ( 0 < m leq 1 ) is ( 0 < m leq 1 ).5. Negative ( m ) Analysis: - For ( m < 0 ), the quadratic becomes non-negative, which contradicts the requirement. Thus, no solutions exist for ( m < 0 ). Conclusion:The valid range of ( m ) is ( 0 < m leq 1 ), corresponding to option (boxed{B}).,<think>Okay, so I need to find the range of m such that the inequality sin²x + m sinx + (m² -3)/m ≤ 0 holds for all real numbers x. Hmm, let's start by understanding what this inequality means. Since it's a quadratic in terms of sinx, maybe I can treat sinx as a variable and analyze the quadratic inequality. First, let's denote t = sinx. Then, the inequality becomes t² + m t + (m² - 3)/m ≤ 0. But we have to remember that t is bounded between -1 and 1 because it's the sine function. So, the quadratic in t must be non-positive for all t in [-1, 1]. Wait, actually, the original inequality must hold for all real x, which translates to the quadratic in t being non-positive for all t in the range of sinx, which is [-1, 1]. So, the problem reduces to finding m such that the quadratic function f(t) = t² + m t + (m² - 3)/m ≤ 0 for all t ∈ [-1, 1]. But before that, I need to check if m is zero or not. Wait, the term (m² - 3)/m would be undefined if m is zero. So, m cannot be zero. Therefore, m ≠ 0. The answer choices have options with m ≤ -3/2, 0 < m ≤1, 0 < m ≤3, etc., so m can be positive or negative but not zero. So, moving forward, assuming m ≠0. Let's write the quadratic as f(t) = t² + m t + (m² -3)/m. To have this quadratic ≤0 for all t ∈ [-1,1], the quadratic must lie entirely below or on the t-axis in that interval. But quadratics can open upwards or downwards. The coefficient of t² is 1, which is positive. So the parabola opens upwards. Therefore, the quadratic will have a minimum at its vertex. Since it opens upwards, to ensure that f(t) ≤0 for all t in [-1,1], the entire interval [-1,1] must lie between the roots of the quadratic. Wait, but if it opens upwards, the quadratic is below zero between its two roots. Therefore, for the quadratic to be ≤0 for all t ∈ [-1,1], the interval [-1,1] must be contained between the two roots. That is, the quadratic must cross the t-axis at two points, and the interval [-1,1] must be entirely between them. Alternatively, if the quadratic never crosses the t-axis, then it's always positive because the coefficient of t² is positive, which would not satisfy the inequality. Therefore, the quadratic must have two real roots, and the interval [-1,1] must lie between them. Therefore, the necessary conditions are:1. The quadratic has two real roots. So discriminant D ≥0.2. The interval [-1,1] lies between the roots. For a quadratic opening upwards, this means that f(-1) ≤0 and f(1) ≤0. Because if the quadratic is below zero at both ends of the interval, and since it opens upwards, the entire interval is between the roots. Wait, but actually, if the interval is between the roots, then the quadratic is negative between the roots, so the maximum values at the endpoints should be negative. Wait, no. Let me think again.If the quadratic opens upwards and the interval [-1,1] is between the two roots, then the quadratic is negative between the roots. So, in order for [-1,1] to be between the roots, the quadratic must be negative on [-1,1], which would require that both f(-1) and f(1) are ≤0. However, since the quadratic is positive outside the roots, but if the roots are outside the interval, then at t=-1 and t=1, which are outside the roots, the quadratic would be positive. Wait, this seems conflicting. Let me recall the properties.Wait, for a quadratic that opens upwards (positive leading coefficient), it is negative between its two roots. So, if the interval [-1,1] is entirely within the interval between the roots, then the quadratic is negative on [-1,1]. But for that to happen, the roots must be outside the interval [-1,1]. Therefore, both roots should be less than -1 or greater than 1? Wait, no. Wait, the roots are two points. If the quadratic is negative between the roots, then to have the entire interval [-1,1] between the roots, the roots must be such that one root is less than or equal to -1 and the other root is greater than or equal to 1. Because then between those roots, which includes [-1,1], the quadratic is negative. Yes, that makes sense. So, the quadratic f(t) must have two real roots, say t1 and t2, such that t1 ≤ -1 and t2 ≥1. Therefore, the quadratic is negative between t1 and t2, which includes [-1,1]. Hence, to ensure that, we need f(-1) ≤0 and f(1) ≤0. Because if the quadratic at t=-1 is ≤0 and at t=1 is ≤0, given that it opens upwards, the roots must lie outside the interval [-1,1]. Wait, but actually, if the quadratic is opening upwards and we have f(-1) ≤0 and f(1) ≤0, then since the parabola tends to +infinity as t approaches ±infinity, there must be a root less than -1 and a root greater than 1. Therefore, the entire interval [-1,1] is between the roots, so the quadratic is negative there. Therefore, the key conditions are f(-1) ≤0 and f(1) ≤0. Additionally, the quadratic must have real roots, so discriminant D ≥0. So, let's compute these conditions step by step.First, compute f(-1):f(-1) = (-1)² + m*(-1) + (m² -3)/m = 1 - m + (m² -3)/m.Simplify this expression:1 - m + (m² -3)/m = (m(1 - m) + m² -3)/m = (m - m² + m² -3)/m = (m - 3)/m.Similarly, compute f(1):f(1) = (1)² + m*(1) + (m² -3)/m = 1 + m + (m² -3)/m.Simplify:1 + m + (m² -3)/m = (m(1 + m) + m² -3)/m = (m + m² + m² -3)/m = (2m² + m -3)/m.Wait, let me check that again. Wait, expanding the numerator:1 + m + (m² -3)/m. Let's combine terms:First, write 1 as m/m, so:m/m + m + (m² -3)/m = [m + m² + m² -3]/m = [2m² + m -3]/m. Wait, that seems correct.Wait, hold on. Let's do it step by step:1 + m + (m² -3)/m.Express 1 as m/m:m/m + m + (m² -3)/m = [m + m² + m² -3]/m.Wait, no. Wait, m/m is 1, then adding m: m/m + m = (m + m²)/m. Then add (m² -3)/m: so total is (m + m² + m² -3)/m = (2m² + m -3)/m. Yes, that's correct.So, f(-1) = (m -3)/m ≤0 and f(1) = (2m² + m -3)/m ≤0.Additionally, discriminant D of the quadratic f(t) must be ≥0.Compute discriminant D:For quadratic t² + m t + (m² -3)/m, discriminant D is m² - 4*1*(m² -3)/m.So, D = m² - 4*(m² -3)/m = m² - (4m² -12)/m = m² -4m +12/m.Wait, let's compute step by step:D = b² -4ac, where a=1, b=m, c=(m² -3)/m.Therefore, D = m² -4*1*(m² -3)/m = m² -4*(m² -3)/m.To combine terms, let's write m² as m³/m:So, D = (m³ -4(m² -3))/m = (m³ -4m² +12)/m.So, D = (m³ -4m² +12)/m ≥0.So, the discriminant condition is (m³ -4m² +12)/m ≥0.Therefore, the three conditions are:1. (m -3)/m ≤0,2. (2m² + m -3)/m ≤0,3. (m³ -4m² +12)/m ≥0.We need to solve these inequalities for m ≠0 and find the intersection of their solutions.Let me tackle each inequality one by one.First, condition 1: (m -3)/m ≤0.This inequality is equivalent to (m -3)/m ≤0. To solve this, find when the numerator and denominator have opposite signs or zero.Case 1: m >0. Then, denominator is positive, so numerator must be ≤0: m -3 ≤0 → m ≤3. So, 0 < m ≤3.Case 2: m <0. Then, denominator is negative, so numerator must be ≥0: m -3 ≥0 → m ≥3. But m <0 and m ≥3 can't happen. So, no solution in this case.Therefore, solution to condition 1 is 0 < m ≤3.Second, condition 2: (2m² + m -3)/m ≤0.Again, solve (2m² + m -3)/m ≤0. Let's factor numerator if possible.Quadratic in numerator: 2m² + m -3. Let's try factoring:Looking for two numbers a and b such that a*b = 2*(-3) = -6 and a + b =1.Looking for factors of -6 that add to 1. 3 and -2: 3*(-2) = -6; 3 + (-2) =1. Therefore:2m² + m -3 = 2m² +3m -2m -3 = m(2m +3) -1(2m +3) = (m -1)(2m +3). Wait, let's check:Wait, (m -1)(2m +3) = 2m² +3m -2m -3 = 2m² +m -3. Yes, correct.Therefore, numerator factors as (m -1)(2m +3).Therefore, the inequality becomes [(m -1)(2m +3)]/m ≤0.So, [(m -1)(2m +3)]/m ≤0. Let's analyze this.Find when the expression is negative or zero. Critical points are m = -3/2, m=0, m=1.Set up a sign chart:Intervals:1. m < -3/2,2. -3/2 <m <0,3. 0 <m <1,4. m >1.Test each interval:1. m < -3/2: Let m = -2. Then, (m -1) = -3, (2m +3)= -4 +3 = -1, m = -2. So, numerator: (-3)*(-1) =3, denominator: -2. So, 3/-2 = -1.5 <0. So, inequality holds here.2. -3/2 <m <0: Let m = -1. (m -1)= -2, (2m +3)= -2 +3 =1, denominator m = -1. Numerator: (-2)*1 = -2. Denominator: -1. So, (-2)/-1 =2 >0. So, inequality does not hold here.3. 0 <m <1: Let m=0.5. (m -1)= -0.5, (2m +3)=1 +3=4, denominator m=0.5. Numerator: (-0.5)*4 = -2. Denominator:0.5. So, -2 /0.5 =-4 <0. So, inequality holds here.4. m >1: Let m=2. (m -1)=1, (2m +3)=7, denominator m=2. Numerator:1*7=7. Denominator:2. So, 7/2=3.5 >0. So, inequality does not hold here.At critical points:m=-3/2: numerator becomes (-3/2 -1)(2*(-3/2)+3)= (-5/2)(-3 +3)= (-5/2)(0)=0. So, expression is 0. So, included.m=0: denominator zero, undefined. Excluded.m=1: numerator becomes (0)(5)=0. So, expression is 0. Included.Therefore, the solution to condition 2 is m ≤-3/2 or 0 <m ≤1.Third, condition 3: (m³ -4m² +12)/m ≥0.We need to analyze (m³ -4m² +12)/m ≥0. Let me factor the numerator if possible.Numerator is m³ -4m² +12. Let's try rational roots. Possible rational roots are ±1, ±2, ±3, ±4, ±6, ±12.Test m=1: 1 -4 +12=9 ≠0.m=2: 8 -16 +12=4 ≠0.m=3:27 -36 +12=3≠0.m=4:64 -64 +12=12≠0.m=-1: -1 -4 +12=7≠0.So, no rational roots. Therefore, the cubic may not factor nicely. Let's check its behavior.Let’s denote the numerator as g(m) = m³ -4m² +12. Let's analyze its sign.Compute g(m) for different m:For m >0:As m approaches +infinity, g(m) approaches +infinity.At m=0: g(0)=0 -0 +12=12 >0.At m=1:1 -4 +12=9>0.At m=2:8 -16 +12=4>0.At m=3:27 -36 +12=3>0.At m=4:64 -64 +12=12>0.So, for m >0, g(m) is always positive.For m <0:Let m = -a, a>0.g(-a) = (-a)^3 -4*(-a)^2 +12 = -a³ -4a² +12.So, for m <0, the numerator is -a³ -4a² +12. Let's see when this is positive.Find when -a³ -4a² +12 ≥0 → -a³ -4a² ≥-12 → a³ +4a² ≤12.Let’s check for a>0:a=1:1 +4=5 ≤12 → yes, so g(-1)= -1 -4 +12=7>0.a=2:8 +16=24 ≤12? No. So, g(-2)= -8 -16 +12= -12 <0.a=1.5: (3.375) + 4*(2.25)=3.375 +9=12.375 ≤12? No. So, a=1.5 gives 12.375>12. So, at a=1.5, the left side is 12.375 >12, so inequality not satisfied. Therefore, between a=1 and a=2, there is a root.So, the cubic crosses zero somewhere between m=-2 and m=-1.5. Let's check m=-1.5:g(-1.5)= (-1.5)^3 -4*(-1.5)^2 +12 = (-3.375) -4*(2.25) +12 = -3.375 -9 +12= -0.375 <0.m=-1: g(-1)= -1 -4 +12=7>0.Therefore, there is a root between m=-1 and m=-1.5 where g(m)=0.But exact root is hard to find. So, for m <0, g(m) is positive when m is between the root and 0. Since at m=-1, it's positive, and at m=-1.5, it's negative. So, there's a root between m=-1.5 and m=-1. So, for m <0, g(m) ≥0 when m ≥ the root (but m is negative). Therefore, m is between the root and 0. But since we can't find the exact root, let's think about how this affects the inequality.The original inequality for condition 3 is (g(m))/m ≥0. Since for m >0, g(m) >0 and m>0, so (g(m))/m >0. So, for m>0, condition 3 is satisfied.For m <0, g(m) can be positive or negative. Let's check for m <0:Case 1: m <0. Let’s denote m = -k, k>0.Then, g(m) = (-k)^3 -4*(-k)^2 +12 = -k³ -4k² +12.So, (g(m))/m = (-k³ -4k² +12)/(-k) = (k³ +4k² -12)/k.We need this to be ≥0. So, (k³ +4k² -12)/k ≥0.Since k>0, we can multiply both sides by k without changing inequality direction:k³ +4k² -12 ≥0.So, solving k³ +4k² -12 ≥0.Let’s find when k³ +4k² -12 =0. Let’s try k=2: 8 +16 -12=12>0. k=1:1 +4 -12= -7 <0. k=1.5: 3.375 +9 -12=0.375>0. So, the equation crosses zero between k=1 and k=1.5. Let's approximate.Let’s try k=1.2:k=1.2: (1.2)^3 +4*(1.2)^2 -12 =1.728 +4*1.44 -12=1.728 +5.76 -12= (7.488 -12)= -4.512 <0.k=1.3: 2.197 +4*1.69=2.197 +6.76=8.957 -12≈-3.043 <0.k=1.4: 2.744 +4*1.96=2.744 +7.84=10.584 -12≈-1.416 <0.k=1.45: ~1.45^3≈3.05, 4*(1.45)^2≈4*2.1025≈8.41, so total 3.05 +8.41=11.46 -12≈-0.54 <0.k=1.47: 1.47^3≈3.17, 4*(1.47)^2≈4*2.16≈8.64, so total≈3.17 +8.64=11.81 -12≈-0.19 <0.k=1.48: 1.48^3≈3.24, 4*(1.48)^2≈4*2.19≈8.76, total≈3.24 +8.76=12 -12=0.Therefore, k≈1.48. Therefore, k³ +4k² -12 ≥0 when k≥1.48, so m =-k ≤-1.48. Approximately, m ≤-1.48. Since -1.48 is approximately -3/2=-1.5. So, roughly, m ≤-1.5 would satisfy the inequality. But we need to confirm. Since the exact root is around k≈1.48, which is m≈-1.48. So, the inequality (g(m))/m ≥0 for m <0 is satisfied when m ≤-1.48 (approximately). However, the answer choices have m ≤-3/2, which is -1.5. So, since the exact root is near -1.48, which is greater than -1.5, then m ≤-1.5 (which is more restrictive) would satisfy the inequality. Wait, but if the root is at m≈-1.48, then m ≤-1.48 would satisfy the inequality. So, m ≤-1.5 is included in m ≤-1.48, but -1.5 is less than -1.48. Wait, no. Wait, m is negative. So, m ≤-1.48 is equivalent to k ≥1.48. But in the problem, answer options have m ≤-3/2 (which is -1.5). Since -1.5 is less than -1.48, so m ≤-1.5 would be more restrictive. But if the inequality holds for m ≤-1.48, then m ≤-1.5 would also hold. But need to see if the exact value crosses at m≈-1.48. Therefore, perhaps the exact root is m=-1.5? Let's check m=-1.5.Compute g(-1.5)= (-1.5)^3 -4*(-1.5)^2 +12= -3.375 -4*(2.25) +12= -3.375 -9 +12= -0.375. So, g(-1.5)= -0.375. Therefore, (g(m))/m = (-0.375)/(-1.5)=0.25>0. So, at m=-1.5, the expression is 0.25>0. So, the inequality holds at m=-1.5. But earlier, we saw that for k=1.5, which is m=-1.5, the expression k³ +4k² -12= (3.375)+ (4*2.25)=3.375 +9=12.375 -12=0.375≥0. Wait, wait. Wait, when m=-1.5, then k=1.5, and the expression k³ +4k² -12=1.5³ +4*(1.5)^2 -12=3.375 +4*2.25 -12=3.375 +9 -12=0.375≥0. So, yes, so for k=1.5, the inequality is satisfied, which corresponds to m=-1.5. So, the boundary is m=-1.5 is included. Wait, but when m=-1.5, the expression (g(m))/m= (-0.375)/(-1.5)=0.25>0. So, the inequality holds here.But when m=-1.48, we had approximately g(m)=0. So, (g(m))/m=0/m=0. So, the inequality holds as equality. So, the exact solution for condition3 is m≥ the root (in negative) where g(m)=0, but since m is negative, this corresponds to m between the root and 0. But since for m <0, the inequality (g(m))/m≥0 requires that (k³ +4k² -12)/k≥0, so when k³ +4k² -12≥0, which is when k≥≈1.48, so m≤-1.48. But since answer options have m ≤-3/2 (which is -1.5), which is less than -1.48, so the values m ≤-1.5 would satisfy the inequality (since the inequality holds for m ≤-1.48). But actually, m=-1.5 is less than -1.48, so m=-1.5 is in the solution set. Wait, but when m=-1.5, the expression (g(m))/m=0.25>0, so it is included. Therefore, the exact solution for condition3 is m≤-1.48 or m>0. But since answer options have m≤-3/2 or m>0 up to 3, maybe the question considers m≤-3/2 and 0 <m ≤3. But need to check.Wait, let me recap:Condition3 is (m³ -4m² +12)/m ≥0.For m>0:Numerator is m³ -4m² +12. As tested, it's positive for all m>0. So, (positive)/positive= positive. So, m>0 satisfies the inequality.For m<0:Numerator is negative when m < some value between -1.5 and -1.4, but let's re-examine.Wait, when m=-1.5:g(m)= (-1.5)^3 -4*(-1.5)^2 +12= -3.375 -4*(2.25) +12= -3.375 -9 +12= -0.375.Thus, (g(m))/m= (-0.375)/(-1.5)=0.25>0.When m=-1:g(-1)= -1 -4 +12=7>0. Therefore, (g(m))/m=7/-1= -7<0. So, for m=-1, the expression is negative, which does not satisfy the inequality. So, between m=-1.5 and m=0, there is a point where (g(m))/m transitions from positive to negative. Wait, but earlier analysis with k=1.5 gives that (k³ +4k² -12)/k= (3.375 +9 -12)/1.5=0.375/1.5=0.25>0. Wait, this is conflicting. Wait, perhaps my earlier approach was incorrect.Wait, m is negative, so k=-m>0. Then, the expression (m³ -4m² +12)/m = [(-k)^3 -4*(-k)^2 +12]/(-k)= (-k³ -4k² +12)/(-k)= (k³ +4k² -12)/k.Therefore, (k³ +4k² -12)/k ≥0. Since k>0, this is equivalent to k³ +4k² -12 ≥0.So, solving k³ +4k² -12 ≥0.Let’s find k such that k³ +4k² -12 =0. Let’s check k=2: 8+16-12=12>0. So, for k≥2, this is true. Wait, no. Wait, when k=1.5:1.5³ +4*(1.5)^2 -12=3.375 +9 -12=0.375>0.k=1.4: 2.744 +7.84 -12= -1.416<0.k=1.45: approx 3.05 +8.41 -12≈-0.54<0.k=1.475: Let's compute:1.475³ ≈1.475*1.475=2.176*1.475≈3.213.4*(1.475)^2≈4*2.176≈8.704.Total≈3.213 +8.704 -12≈11.917 -12≈-0.083<0.k=1.48:1.48³≈1.48*1.48=2.1904*1.48≈3.242.4*(1.48)^2≈4*2.1904≈8.7616.Total≈3.242 +8.7616 -12≈11.0036 -12≈-0.9964<0. Wait, no, that's not right. Wait, 3.242 +8.7616=12.0036. So, 12.0036 -12=0.0036≈0. So, k≈1.48 is the approximate root. So, k≈1.48. Therefore, when k≥1.48, which corresponds to m≤-1.48, the inequality holds.Therefore, the solution for condition3 is m≤-1.48 or m>0. But since 1.48 is approximately 1.5, the answer options use m≤-3/2 (which is -1.5). Therefore, considering the options given, perhaps the exact value is m≤-3/2. But need to check.But in reality, the exact solution is m≤-1.48 or m>0. However, the problem is given with answer choices, so D is "m ≤-3/2 or 0 <m ≤3". So, given that -3/2 is -1.5, which is less than -1.48, so the interval m≤-1.5 is included in m≤-1.48, but since at m=-1.5, the discriminant is positive (as we saw, D=0.25>0), so m=-1.5 is allowed. But according to the cubic solution, m≤-1.48 is needed. So, in the options, m≤-3/2 is a subset of m≤-1.48. Therefore, perhaps the correct answer is D, which combines m≤-3/2 and 0 <m ≤3.But let's cross-validate with the answer choices. The options are:(A) m ≤−3/2(B) 0<m ≤1(C) 0<m ≤3(D) m ≤−3/2 or 0<m ≤3So, need to find the intersection of the solutions of the three conditions.Recall:Condition1: 0 <m ≤3.Condition2: m ≤-3/2 or 0 <m ≤1.Condition3: m≤-1.48 (approx) or m >0.Therefore, combining all three conditions:For m >0:Condition1: 0 <m ≤3.Condition2: 0 <m ≤1.Condition3: m >0.So intersection is 0 <m ≤1.For m <0:Condition1: No solution.Condition2: m ≤-3/2.Condition3: m≤-1.48 (approx). But since -3/2 is -1.5 which is less than -1.48, so m ≤-3/2 would satisfy condition3 as well.But condition1 does not allow m <0. Wait, but hold on:Wait, the three conditions must all be satisfied. So:The overall solution is the intersection of conditions1,2,3.But for m >0:Need to satisfy condition1 (0 <m ≤3), condition2 (0 <m ≤1), and condition3 (m >0). So intersection is 0 <m ≤1.For m <0:Need to satisfy condition1 (invalid, since condition1 requires 0 <m ≤3). Therefore, m <0 is excluded by condition1. Wait, but condition1 was (m -3)/m ≤0, which for m <0 had no solution. Wait, no:Wait, condition1 is (m -3)/m ≤0. We found that the solution was 0 <m ≤3. Therefore, m <0 is not in the solution of condition1. Therefore, for the overall problem, m must satisfy all three conditions, so m must be in the intersection of conditions1,2,3. Since conditions1 excludes m <0, even if conditions2 and3 allow m ≤-3/2, the intersection is only 0 <m ≤1. But wait, but in the original problem, the inequality must hold for all x ∈ R, which requires that the quadratic in t is ≤0 for all t ∈ [-1,1]. So, perhaps I made a mistake in combining the conditions.Wait, let me re-examine. The three conditions must all hold for the inequality to hold for all x.So, if m >0, then:Condition1: 0 <m ≤3,Condition2: 0 <m ≤1,Condition3: m >0.Therefore, intersection is 0 <m ≤1.But also, need to check if discriminant D≥0 for m in 0 <m ≤1.Wait, condition3 is (m³ -4m² +12)/m ≥0. For m >0, this is equivalent to m³ -4m² +12 ≥0.Let’s check for m in (0,1]:At m=1: 1 -4 +12=9≥0.For m approaching 0: m³ -4m² +12 approaches 0 -0 +12=12≥0.Therefore, m³ -4m² +12 is always ≥12 -4*1 +0=8 for m in (0,1]. Wait, actually, for m in (0,1], m³ -4m² +12 = m²(m -4) +12. Since m -4 is negative, m²(m -4) is negative, but 12 dominates. For m=0.5:0.125 -1 +12=11.125>0. So, yes, for m in (0,1], m³ -4m² +12 is positive, so condition3 is satisfied. Therefore, discriminant D is positive. So, the quadratic has real roots. So, the combination of conditions1,2,3 for m>0 is 0 <m ≤1.For m <0:Condition1 excludes m <0. Therefore, no solution here.Wait, but according to answer choices, there is an option D which includes m ≤-3/2 or 0 <m ≤3. But according to this analysis, the correct range is 0 <m ≤1. But this conflicts with the answer choices. So, perhaps there's an error in the analysis.Wait, maybe I missed something. Let's re-examine the problem.The original problem says the inequality sin²x + m sinx + (m² -3)/m ≤0 holds for all real x. So, the quadratic in t=sinx must be ≤0 for all t ∈ [-1,1].We considered that the quadratic must be ≤0 on [-1,1], which requires that the quadratic is non-positive on the entire interval. Since the quadratic opens upwards (coefficient 1), the maximum values on the interval occur at the endpoints. Therefore, if both endpoints are ≤0, then the entire interval is ≤0. Therefore, the key conditions are f(-1) ≤0 and f(1) ≤0. Additionally, the quadratic must have real roots to ensure that it actually dips below the t-axis; otherwise, if discriminant is negative, the quadratic is always positive, which would not satisfy the inequality.But wait, if the discriminant is negative, the quadratic is always positive (since coefficient of t² is positive), which would make the inequality sin²x + ... ≤0 never true. Therefore, discriminant must be non-negative. So, we need the quadratic to have real roots and the interval [-1,1] to be between the roots, which requires f(-1) ≤0 and f(1) ≤0. But in our earlier analysis, the intersection of all three conditions for m >0 was 0 <m ≤1, but answer choice D includes 0 <m ≤3. So, where is the mistake?Let me check with m=2, which is in option C and D.Take m=2. Check if the inequality holds for all x.First, compute the quadratic: t² +2t + (4 -3)/2 = t² +2t +0.5. We need t² +2t +0.5 ≤0 for all t ∈ [-1,1].Find the minimum of this quadratic. Since it opens upwards, the minimum is at t=-b/(2a)=-2/2=-1.Compute f(-1)= (-1)^2 +2*(-1)+0.5=1 -2 +0.5=-0.5 ≤0.Compute f(1)=1 +2 +0.5=3.5 >0. So, the quadratic is positive at t=1, which is within the interval [-1,1]. Therefore, the inequality does not hold for m=2. Therefore, m=2 is invalid. So, option C and D are incorrect if they include m up to 3.But according to our previous analysis, the valid range is 0 <m ≤1. Which is option B. But option D includes m ≤-3/2 or 0 <m ≤3. But when m=2, the inequality does not hold. Therefore, the correct answer cannot include m up to 3. So, where is the mistake in our analysis?Wait, going back to the discriminant. When m=1, discriminant D=(1³ -4*1² +12)/1=1 -4 +12=9>0. So, quadratic has real roots. For m=1, f(-1)=(1 -3)/1=-2 ≤0, f(1)=(2 +1 -3)/1=0 ≤0. So, endpoints are ≤0, and quadratic has real roots, so interval is between the roots. Therefore, the inequality holds for m=1.But for m=2, f(-1)=(2-3)/2=(-1)/2=-0.5 ≤0, f(1)=(2*4 +2 -3)/2=(8 +2 -3)/2=7/2=3.5>0. Therefore, f(1)=3.5>0. So, the inequality fails at t=1. Therefore, m=2 is invalid.Therefore, our previous conclusion that 0 <m ≤1 is correct, but answer options don't have this. Wait, the options are:(A) m ≤−3/2(B) 0<m ≤1(C) 0<m ≤3(D) m ≤−3/2 or 0<m ≤3So, according to our analysis, the answer should be B. But the answer is given as D? Or perhaps not. Wait, let's check for m negative.Earlier, we considered that condition1 (0 <m ≤3) excludes negative m. But maybe the analysis is wrong. Wait, let's check m negative.Suppose m is negative. Let's take m=-2. Check if the inequality holds for all x.So, quadratic becomes t² + (-2)t + (4 -3)/(-2)=t² -2t + (1)/(-2)=t² -2t -0.5. We need this quadratic ≤0 for all t ∈ [-1,1].Since the coefficient of t² is positive, parabola opens upwards. So, minimum at vertex. Vertex at t=-b/(2a)=2/2=1. So, the minimum is at t=1. Compute f(1)=1 -2 -0.5=-1.5 ≤0. The maximum occurs at t=-1. Compute f(-1)=1 +2 -0.5=2.5 >0. Therefore, the quadratic is positive at t=-1, which is in the interval [-1,1]. Therefore, the inequality does not hold. Therefore, m=-2 is invalid. But according to answer option A and D, m ≤-3/2 is allowed. Let's check m=-2 (which is ≤-3/2). It doesn't work. Then why option D includes m ≤-3/2? Maybe take m=-3/2.Take m=-1.5. Compute the quadratic:t² + (-1.5)t + ((2.25)-3)/(-1.5)=t² -1.5t + (-0.75)/(-1.5)=t² -1.5t +0.5.So, quadratic is t² -1.5t +0.5.Check if this is ≤0 for all t ∈ [-1,1].The quadratic opens upwards. Let's compute its discriminant: D= (-1.5)^2 -4*1*0.5=2.25 -2=0.25>0. So, two real roots. Compute roots:t=(1.5 ±√0.25)/2=(1.5 ±0.5)/2.So, t=(1.5 +0.5)/2=2/2=1, and t=(1.5 -0.5)/2=1/2=0.5.So, the quadratic is t² -1.5t +0.5 factors as (t-1)(t-0.5). Since it opens upwards, it is ≤0 between t=0.5 and t=1. Therefore, the quadratic is ≤0 for t ∈ [0.5,1], but we need it to be ≤0 for all t ∈ [-1,1]. However, for t <0.5, say t=0, the value is 0 -0 +0.5=0.5>0. Therefore, the inequality does not hold for m=-1.5. Thus, even m=-3/2 does not satisfy the condition.But according to our earlier analysis, for m <0, the solutions were m ≤-1.48 (approx). But when m=-1.5, the quadratic is positive in parts of the interval, so it's invalid. So, what's wrong here?Wait, the problem is that when m is negative, we have to re-express the conditions. Maybe my initial approach was incorrect.Alternatively, perhaps the correct approach is to ensure that the maximum of the quadratic on [-1,1] is ≤0. Since the quadratic opens upwards, its maximum on [-1,1] occurs at one of the endpoints. Therefore, if both f(-1) and f(1) are ≤0, then the quadratic is ≤0 on the entire interval. But when m is negative, let's check:For m <0, compute f(-1) and f(1):f(-1)= (m -3)/m. Since m <0, let's compute sign:(m -3)/m. If m <0, then numerator m-3 is negative (since m is negative and 3 is positive). Therefore, (negative)/negative= positive. Therefore, f(-1)= positive >0. Which violates the inequality. Therefore, for m <0, f(-1) >0, so the quadratic is positive at t=-1, which is in [-1,1]. Therefore, the inequality cannot hold for any m <0. Therefore, m must be positive.Therefore, the only possible solutions are for m >0.Therefore, the earlier analysis for m >0 is correct: intersection is 0 <m ≤1. Therefore, answer is B.But the answer choices don't include B as correct in the options. Wait, no, the answer choices are:(A) m ≤−3/2(B) 0<m ≤1(C) 0<m ≤3(D) m ≤−3/2 or 0<m ≤3So, option B is 0<m ≤1. But according to the problem, the answer should be B. But when I look at the options, the original problem might have a different condition. Wait, but according to our analysis, m=1 is valid, m=2 is invalid. m=0.5 is valid. And m <0 are invalid. Therefore, the correct answer is B. But according to the options given, option D includes 0<m ≤3 and m ≤−3/2, which are incorrect. But why is there a discrepancy?Wait, maybe I made a mistake in the discriminant condition. Let's check for m=1:Quadratic is t² + t + (1 -3)/1= t² +t -2. Which factors as (t+2)(t-1). So roots at t=-2 and t=1. Therefore, the quadratic is ≤0 between t=-2 and t=1. Since our interval is [-1,1], which is inside [-2,1], the quadratic is ≤0 on [-1,1]. Therefore, m=1 works.For m=0.5:Quadratic is t² +0.5t + (0.25 -3)/0.5= t² +0.5t + (-2.75)/0.5= t² +0.5t -5.5. This quadratic has roots at t= [-0.5 ±√(0.25 +22)]/2= [-0.5 ±√22.25]/2= [-0.5 ±4.7]/2. So, approximately, roots at t=(4.2)/2≈2.1 and t=(-5.2)/2≈-2.6. So, the quadratic is ≤0 between t≈-2.6 and t≈2.1. Since our interval is [-1,1], which is inside, the quadratic is ≤0 there. Therefore, m=0.5 works.For m approaching 0:As m approaches 0+, (m² -3)/m approaches -infinity. So, quadratic becomes t² +mt + (m² -3)/m ≈ t² +0*t -infty, which goes to -infty. So, the quadratic would be negative for all t, but since it's a quadratic in t, the term (m² -3)/m dominates. Wait, but as m approaches 0 from the right, (m² -3)/m ~ -3/m which approaches -infty. Therefore, the quadratic would have a very large negative constant term, so it would be negative everywhere? Wait, but a quadratic with a positive t² coefficient and a very large negative constant term would open upwards and have two real roots, crossing the t-axis at very large |t|. Therefore, between the roots, which are very far apart, the quadratic is negative. Therefore, in the interval [-1,1], the quadratic is negative. Therefore, as m approaches 0+, the quadratic is negative on [-1,1]. So, m approaching 0+ is valid. Therefore, m can be as small as approaching 0 from the right. Therefore, the interval is 0 <m ≤1.But when m=1.5, which is in option C and D:Quadratic is t² +1.5t + (2.25 -3)/1.5= t² +1.5t -0.75/1.5= t² +1.5t -0.5. Compute f(1)=1 +1.5 -0.5=2>0. Therefore, inequality does not hold. So, m=1.5 is invalid.Therefore, the correct answer should be B: 0<m ≤1.But why is option D given? Maybe I made a mistake in the conditions.Wait, maybe I need to re-express the original inequality differently.The inequality is sin²x + m sinx + (m² -3)/m ≤0.Multiply both sides by m. But since m can be positive or negative, the inequality direction may change.But this is risky because m can be positive or negative. However, perhaps consider two cases: m >0 and m <0.Case1: m >0.Multiply both sides by m (inequality direction remains):m sin²x + m² sinx + m² -3 ≤0.But this complicates things. Alternatively, since m >0, the original inequality can be written as sin²x + m sinx + (m² -3)/m ≤0.We have already analyzed this case.Case2: m <0.Multiply both sides by m (inequality direction reverses):m sin²x + m² sinx + m² -3 ≥0.So, the inequality becomes m sin²x + m² sinx + m² -3 ≥0 for all x ∈ R.Now, since m <0, this is equivalent to a quadratic in sinx:Let t=sinx, then m t² + m² t + m² -3 ≥0 for all t ∈ [-1,1].Since m <0, the coefficient of t² is negative. So, the quadratic opens downward. Therefore, to have the quadratic ≥0 for all t ∈ [-1,1], the quadratic must be ≥0 at the minimum points in the interval. Since it opens downward, the minimum occurs at the endpoints or at the vertex if it's inside the interval.But the quadratic opens downward, so it has a maximum at the vertex. To ensure the quadratic is ≥0 for all t ∈ [-1,1], the minimum values on the interval must be ≥0. For a downward opening quadratic, the minima are at the endpoints. Therefore, we need f(-1) ≥0 and f(1) ≥0.Additionally, if the quadratic has real roots, the interval [-1,1] must be outside the roots. But since it opens downward, the quadratic is ≥0 outside the roots. Therefore, to have the quadratic ≥0 on [-1,1], the interval must be entirely outside the roots. That is, both roots are ≤-1 or both roots are ≥1.Alternatively, since the quadratic is ≥0 for t ≤ t1 or t ≥t2, where t1 ≤ t2 are the roots. So, to have [-1,1] ⊂ (-∞, t1] ∪ [t2, ∞), we need that t1 ≥1 or t2 ≤-1.But since the quadratic opens downward, the roots are real if discriminant ≥0.Let’s proceed step by step for m <0:Quadratic in t: m t² + m² t + m² -3 ≥0 for all t ∈ [-1,1].Conditions:1. Leading coefficient m <0.2. Quadratic is ≥0 at both endpoints: f(-1) ≥0 and f(1) ≥0.3. If discriminant ≥0, then the interval [-1,1] must lie outside the roots. Since quadratic is ≥0 outside the roots.Let’s compute f(-1) and f(1):f(-1)= m*(-1)^2 + m²*(-1) + m² -3 = m -m² +m² -3 = m -3 ≥0 → m ≥3. But m <0, so no solution here.f(1)= m*(1)^2 + m²*(1) + m² -3 = m + m² +m² -3=2m² +m -3 ≥0.So, 2m² +m -3 ≥0. Let’s solve this inequality.The quadratic 2m² +m -3 ≥0. Let’s find roots:m=(-1 ±√(1 +24))/4=(-1 ±5)/4.Roots: m=(-1 +5)/4=1, and m=(-1 -5)/4=-1.5.Since the coefficient of m² is positive, the quadratic is ≥0 for m ≤-1.5 or m ≥1. But m <0, so the solution is m ≤-1.5.Additionally, discriminant condition for quadratic in t: discriminant D=(m²)^2 -4*m*(m² -3)=m⁴ -4m(m² -3)=m⁴ -4m³ +12m.Need D ≥0 for quadratic to have real roots.But even if D <0, the quadratic is always ≥0 (since coefficient of t² is negative and no real roots). Wait, if discriminant D <0, the quadratic m t² +m² t +m² -3 has no real roots and opens downward, so it's always negative. But we need it to be ≥0 for all t ∈ [-1,1]. Therefore, if D <0, the quadratic is always negative, which violates the inequality. Therefore, we need D ≥0, so that the quadratic has real roots and is ≥0 outside the roots.Thus, for m <0, the conditions are:1. 2m² +m -3 ≥0 → m ≤-1.5.2. D ≥0: m⁴ -4m³ +12m ≥0.Let’s compute D= m⁴ -4m³ +12m.For m <0, let’s substitute m=-k, k>0.D= (-k)^4 -4*(-k)^3 +12*(-k)=k⁴ +4k³ -12k.Need D ≥0 → k⁴ +4k³ -12k ≥0 for k>0.Factor out k: k(k³ +4k² -12) ≥0.Since k>0, the inequality reduces to k³ +4k² -12 ≥0.This is the same cubic in k as before. As before, the cubic k³ +4k² -12=0 has a real root near k=1.48. So, for k≥1.48, the cubic is ≥0, which means for m=-k ≤-1.48, D ≥0.But m ≤-3/2 is m ≤-1.5, which is less than -1.48, so for m ≤-1.5, k≥1.5, and since the cubic is increasing for k>0 (derivative 3k² +8k >0), so for k≥1.5, cubic k³ +4k² -12≥1.5³ +4*(1.5)^2 -12=3.375 +9 -12=0.375≥0. So, yes, for k≥1.5, cubic ≥0.375≥0. Therefore, D ≥0 for m ≤-1.5.Therefore, for m <0, the conditions are:1. m ≤-1.5.2. D ≥0, which is satisfied for m ≤-1.5.Additionally, in this case, the quadratic m t² +m² t +m² -3 ≥0 for all t ∈ [-1,1]. But since quadratic opens downward, and the roots are t1 and t2 (where t1 ≤t2), the quadratic is ≥0 for t ≤t1 or t ≥t2. For the quadratic to be ≥0 on [-1,1], we need [-1,1] ⊂ (-∞, t1] ∪ [t2, ∞). Since the quadratic is ≥0 outside the roots. Therefore, either t1 ≥1 or t2 ≤-1.But given the quadratic m t² +m² t +m² -3 with m <0 and roots t1, t2. Let’s find expressions for t1 and t2.Roots are t=[-m² ±√D]/(2m). Since m <0, denominator is negative.But this is getting complicated. Alternatively, since we already forced the quadratic to be ≥0 at t=-1 and t=1 (for m <0), which in this case for m ≤-1.5, we have f(-1)= m -3 ≥0 → m ≥3, which contradicts m <0. Wait, earlier calculation for m <0, f(-1)= m -3. Wait, in the case when m <0, we are considering the inequality m sin²x + m² sinx +m² -3 ≥0. Then f(-1)= m*(-1)^2 +m²*(-1) +m² -3= m -m² +m² -3= m -3. So, for m <0, f(-1)= m -3. Since we need f(-1) ≥0, which implies m -3 ≥0 → m ≥3. But m <0, so this is impossible. Therefore, contradiction. Therefore, there is no solution for m <0.But this contradicts earlier analysis. Where is the mistake?Wait, in the case when m <0, after multiplying both sides by m (negative), the inequality flips:Original inequality: sin²x +m sinx + (m² -3)/m ≤0.Multiply by m (negative), inequality becomes:m sin²x + m² sinx +m² -3 ≥0.We have to ensure this holds for all x. Therefore, for m <0, the quadratic in t=sinx is m t² +m² t +m² -3 ≥0 for all t ∈ [-1,1].To ensure this, the quadratic must be ≥0 for all t ∈ [-1,1]. Since it opens downward (m <0), the quadratic has a maximum at its vertex. To be ≥0 everywhere on [-1,1], the minimum on the interval must be ≥0. But since it opens downward, the minima occur at the endpoints. Therefore, need f(-1) ≥0 and f(1) ≥0.But f(-1)= m -3 ≥0 → m ≥3. But m <0, impossible.f(1)= 2m² +m -3 ≥0. For m <0, this is the same as earlier. So, f(1)=2m² +m -3 ≥0. Roots at m=1 and m=-1.5. For m <0, solution is m ≤-1.5. But even if m ≤-1.5, f(-1)=m -3 <0 because m is negative. Therefore, it is impossible for the quadratic to be ≥0 at both endpoints when m <0. Therefore, no solutions for m <0.Therefore, the only valid solutions are for m >0, and intersection of conditions1,2,3 gives 0 <m ≤1.But the answer choices don't include this as an option. Wait, no, option B is 0<m ≤1.But the options are (A) m ≤−3/2; (B) 0<m ≤1; (C)0<m ≤3; (D)m ≤−3/2 or 0<m ≤3.So, according to analysis, the correct answer is B. But the given options might have D as the answer. However, according to rigorous checking, for m >1, like m=2, the inequality does not hold. Therefore, the correct answer is B.But wait, let me check with m=1.5:Wait, for m=1.5, which is not in option B but is in options C and D.The quadratic is t² +1.5 t + (2.25 -3)/1.5= t² +1.5 t -0.75/1.5= t² +1.5 t -0.5.We need this ≤0 for all t ∈ [-1,1].Compute f(1)=1 +1.5 -0.5=2>0. So, inequality fails. Therefore, m=1.5 is invalid.Therefore, answer B is correct.But the problem states that the answer is D. But according to the detailed analysis, D is incorrect. So, this suggests that there might be a mistake in the problem's answer choices, or in the analysis.Alternatively, perhaps there is an error in the discriminant condition. Let me recheck.For m >0:Conditions:1. (m -3)/m ≤0 → 0 <m ≤3.2. (2m² +m -3)/m ≤0 → 0 <m ≤1.3. (m³ -4m² +12)/m ≥0 → for m >0, numerator is m³ -4m² +12. Check for m=1:1 -4 +12=9>0; m=2:8 -16 +12=4>0; m=3:27 -36 +12=3>0. So, for all m >0, numerator is positive. Therefore, condition3 is automatically satisfied for m >0. Therefore, discriminant D ≥0 is satisfied for all m >0, because D=(m³ -4m² +12)/m ≥0 since numerator and denominator are positive.Wait, this is a key mistake in previous analysis! For m >0, discriminant D=(m³ -4m² +12)/m ≥0. Since m >0, we have D ≥0 ⇨ m³ -4m² +12 ≥0.But for m >0, m³ -4m² +12 ≥0. Let's check:For m=0: 0 -0 +12=12>0.For m=1:1 -4 +12=9>0.For m=2:8 -16 +12=4>0.For m=3:27 -36 +12=3>0.For m=4:64 -64 +12=12>0.It seems that for m >0, m³ -4m² +12 is always ≥3>0. Therefore, discriminant is always positive for m >0. Therefore, condition3 is automatically satisfied for m >0. Therefore, the only conditions are conditions1 and conditions2.Condition1: 0 <m ≤3.Condition2:0 <m ≤1.Therefore, intersection is 0 <m ≤1. Therefore, answer is B.But where did the previous thought about condition3 come from? The discriminant is always positive for m >0, so no additional restriction.Therefore, the correct answer is B.But the given options include D. But according to our analysis, D is incorrect. Therefore, possibly a mistake in the problem's options, or in the analysis.Wait, but let's verify with m=1.5.For m=1.5:The quadratic is t² +1.5t + (2.25 -3)/1.5 = t² +1.5t -0.5.The quadratic equation t² +1.5t -0.5 ≤0 for all t ∈ [-1,1].But compute f(1)=1 +1.5 -0.5=2>0. Therefore, inequality does not hold. Therefore, m=1.5 is invalid.Therefore, the correct answer is B: 0 <m ≤1.But the original answer options have D: m ≤−3/2 or 0<m ≤3. This suggests that the problem may have a different solution, but according to our analysis, only 0 <m ≤1 is valid. Therefore, the correct answer should be B. But in the options provided, B is an option, so likely B is the answer. But the problem's original options may have been different. However, the user provided the options, so according to the analysis, the answer is B.But the user might have intended the answer to be D. However, given the step-by-step analysis, the correct answer is B.Final Answerboxed{B}</think>To solve the inequality (sin^2 x + m sin x + frac{m^2 - 3}{m} leq 0) for all (x in mathbf{R}), we start by treating it as a quadratic in (sin x). Let (t = sin x), then the inequality becomes (t^2 + mt + frac{m^2 - 3}{m} leq 0) for all (t in [-1, 1]). Key Steps:1. Quadratic Analysis: - The quadratic (t^2 + mt + frac{m^2 - 3}{m}) must be non-positive for all (t in [-1, 1]). - The quadratic opens upwards (coefficient of (t^2) is 1), so it must be non-positive between its roots. Therefore, the interval ([-1, 1]) must lie between the roots.2. Endpoints Evaluation: - Evaluate the quadratic at (t = 1) and (t = -1): - (f(1) = 1 + m + frac{m^2 - 3}{m} leq 0) - (f(-1) = 1 - m + frac{m^2 - 3}{m} leq 0)3. Solving Inequalities: - For (f(-1) leq 0): [ frac{m - 3}{m} leq 0 implies 0 < m leq 3 ] - For (f(1) leq 0): [ frac{2m^2 + m - 3}{m} leq 0 implies m leq -frac{3}{2} text{ or } 0 < m leq 1 ]4. Discriminant Condition: - The quadratic must have real roots: [ frac{m^3 - 4m^2 + 12}{m} geq 0 ] - For (m > 0), this is always true as the numerator (m^3 - 4m^2 + 12) is positive.5. Intersection of Conditions: - For (m > 0), the valid range is the intersection of (0 < m leq 3) and (0 < m leq 1), which is (0 < m leq 1). - For (m < 0), the analysis shows no valid solutions as the quadratic cannot be non-positive for all (t in [-1, 1]). Conclusion:The range of (m) that satisfies the inequality for all (x in mathbf{R}) is (0 < m leq 1).[boxed{B}]]

🔑:5. B.Let t=sin x, and denote g(t)=t^{2}+m t+frac{m^{2}-3}{m}.Then the given condition is transformed into, when t in[-1,1], g(t) leqslant 0 always holds. This is equivalent toleft{begin{array}{l}g(1)=1+m+frac{m^{2}-3}{m}=frac{2 m^{2}+m-3}{m} leqslant 0, g(-1)=1-m+frac{m^{2}-3}{m}=frac{m-3}{m} leqslant 0 .end{array}right.Solving this, we get left{begin{array}{l}m leqslant-frac{3}{2} text { or } 0<m leqslant 1, 0<m leqslant 3 .end{array}right.Therefore, 0<m leqslant 1.