🧠:[<think>Okay, let's try to figure out this problem step by step. So, the problem is about an ellipse and a circle. The circle has its diameter as the distance between the two foci of the ellipse. When this circle intersects the ellipse, there are four points. Connecting these four points along with the two foci in sequence forms a regular hexagon. We need to find the eccentricity of the ellipse.First, let me recall some basic properties of an ellipse. The standard equation of an ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) is the semi-major axis, (b) is the semi-minor axis, and the distance from the center to each focus is (c), with the relationship (c^2 = a^2 - b^2). The eccentricity (e) is given by (e = frac{c}{a}).Now, the problem mentions a circle with diameter equal to the distance between the two foci of the ellipse. The distance between the foci is (2c), so the diameter of the circle is (2c), meaning the radius is (c). The center of the circle should be the same as the center of the ellipse, which is the midpoint between the two foci. So, the equation of the circle would be (x^2 + y^2 = c^2) if we assume the ellipse is centered at the origin.This circle intersects the ellipse at four points. Connecting these four points and the two foci in sequence forms a regular hexagon. A regular hexagon has all sides equal and all internal angles equal (120 degrees each). So, each vertex of the hexagon is equally spaced around a circle. Wait, but in this case, the hexagon is formed by connecting four intersection points and the two foci. So, total six points: two foci and four intersection points.Since it's a regular hexagon, all sides should be equal. That means the distance between consecutive points should be the same. Let me visualize this. The two foci are on the major axis of the ellipse, separated by (2c). The other four points are where the circle (x^2 + y^2 = c^2) intersects the ellipse. Let's find the coordinates of these intersection points.To find the intersection points, we can solve the system of equations:1. (frac{x^2}{a^2} + frac{y^2}{b^2} = 1)2. (x^2 + y^2 = c^2)Let me subtract equation 2 multiplied by (frac{1}{a^2}) from equation 1:(frac{x^2}{a^2} + frac{y^2}{b^2} - frac{x^2}{a^2} - frac{y^2}{a^2} = 1 - frac{c^2}{a^2})Simplifying:(left(frac{1}{b^2} - frac{1}{a^2}right)y^2 = 1 - frac{c^2}{a^2})Multiply both sides by (a^2b^2) to eliminate denominators:((a^2 - b^2)y^2 = a^2b^2left(1 - frac{c^2}{a^2}right))But we know that (c^2 = a^2 - b^2), so substituting:((c^2)y^2 = a^2b^2left(1 - frac{a^2 - b^2}{a^2}right))Simplify the right side:(a^2b^2left(1 - 1 + frac{b^2}{a^2}right) = a^2b^2left(frac{b^2}{a^2}right) = b^4)Therefore:(c^2y^2 = b^4 implies y^2 = frac{b^4}{c^2} implies y = pm frac{b^2}{c})So, substituting (y = pm frac{b^2}{c}) into equation 2 to find x:(x^2 + left(frac{b^2}{c}right)^2 = c^2)(x^2 = c^2 - frac{b^4}{c^2} = frac{c^4 - b^4}{c^2})Factor numerator:(c^4 - b^4 = (c^2 - b^2)(c^2 + b^2))But (c^2 = a^2 - b^2), so:(x^2 = frac{(a^2 - b^2 - b^2)(a^2 - b^2 + b^2)}{c^2})Simplify:(x^2 = frac{(a^2 - 2b^2)(a^2)}{c^2})Therefore:(x = pm frac{asqrt{a^2 - 2b^2}}{c})Hmm, so the intersection points are at (left(pm frac{asqrt{a^2 - 2b^2}}{c}, pm frac{b^2}{c}right)). Wait, but for real solutions, we need (a^2 - 2b^2 geq 0), so (a^2 geq 2b^2). That might be an important condition here.Now, the problem states that connecting these four intersection points and the two foci in sequence forms a regular hexagon. Let's think about the coordinates of the six points:1. The two foci are at ((pm c, 0)).2. The four intersection points are at (left(pm frac{asqrt{a^2 - 2b^2}}{c}, pm frac{b^2}{c}right)).So, in total, the six points of the hexagon are:1. ((c, 0))2. (left(frac{asqrt{a^2 - 2b^2}}{c}, frac{b^2}{c}right))3. ((-c, 0))4. (left(-frac{asqrt{a^2 - 2b^2}}{c}, -frac{b^2}{c}right))5. ((c, 0)) again? Wait, but connecting them in sequence. Wait, maybe the order is different. Let me clarify.Wait, a regular hexagon has six vertices. The problem says "connecting these four intersection points and the two foci in sequence forms a regular hexagon." So, the sequence alternates between foci and intersection points? Wait, but there are only two foci. Wait, maybe the two foci are adjacent in the sequence? Let me think.Wait, a regular hexagon has all sides equal and all angles equal. If we have two foci and four intersection points, the hexagon must have two foci and four other points arranged such that all edges are equal. Let's try to figure out the positions.Perhaps the foci are at opposite vertices of the hexagon. Since the foci are on the major axis, separated by 2c, and the other four points are symmetric with respect to both axes. So, the regular hexagon would have vertices at (c, 0), (x, y), (-c, 0), (-x, -y), (c, 0), but that doesn't form a hexagon. Wait, maybe it's arranged differently.Alternatively, the regular hexagon could have points at (c, 0), (x, y), (something), (-c, 0), (-x, -y), (something), and back to (c, 0). But since a regular hexagon is cyclic (all vertices lie on a circle), but in this case, the foci are not on the circle unless the circle is the one passing through all six vertices. Wait, but the circle here is the one with diameter 2c, which is the foci. The four intersection points lie on both the ellipse and the circle. So, the six vertices of the hexagon are the two foci (on the major axis) and four points on the circle (which are intersection points with the ellipse).But in a regular hexagon, all vertices lie on a circle. However, here the two foci are on the major axis, separated by 2c, and the four intersection points lie on the circle of radius c. Wait, if all six vertices lie on a circle, then the circle would need to have radius such that the distance from the center to each vertex is the same.But the foci are at distance c from the center (since they are at ((pm c), 0)), and the four intersection points are at distance sqrt(x^2 + y^2) = sqrt(c^2) = c. So, all six vertices lie on a circle of radius c. Therefore, the regular hexagon is inscribed in a circle of radius c. But in a regular hexagon inscribed in a circle, each vertex is separated by 60 degrees. However, our foci are on the x-axis, so the angle between them is 180 degrees. Wait, but in a regular hexagon, adjacent vertices are 60 degrees apart. So, this seems contradictory.Wait, maybe I need to re-examine the positions. If the regular hexagon is inscribed in a circle of radius c, then all vertices are at a distance c from the center, which matches the foci and the intersection points. So, the two foci are at (c, 0) and (-c, 0), and the other four points are equally spaced around the circle. However, in a regular hexagon, the angle between consecutive points is 60 degrees, but here, between the two foci, there is 180 degrees. So, how can that be?Wait, maybe the sequence is not going around the circle in order. If you connect the two foci and the four intersection points in a certain sequence, it might form a regular hexagon. Let me try to think of the coordinates.Suppose the six vertices of the regular hexagon are:1. (c, 0) [focus]2. (c/2, (sqrt(3)/2)c) [intersection point]3. (-c/2, (sqrt(3)/2)c) [intersection point]4. (-c, 0) [focus]5. (-c/2, -(sqrt(3)/2)c) [intersection point]6. (c/2, -(sqrt(3)/2)c) [intersection point]But in this case, the foci are at (c, 0) and (-c, 0), and the other four points are at 60 degrees, 120 degrees, 240 degrees, 300 degrees on the circle of radius c. Then, connecting them in the order 1-2-3-4-5-6-1 would form a regular hexagon. But in our case, the four intersection points are specific points on the ellipse and the circle. So, are their coordinates as above?Wait, in the regular hexagon inscribed in a circle of radius c, each vertex is at angles 0°, 60°, 120°, 180°, 240°, 300°, right? So, if two of those points are the foci at 0° and 180°, the other four points would be at 60°, 120°, 240°, 300°, which are the intersection points. But the intersection points must lie on both the ellipse and the circle. So, if those four points are at 60°, 120°, 240°, 300°, then their coordinates would be (c*cosθ, c*sinθ). So, for 60°, it's (c/2, (sqrt(3)/2)c), etc.So, according to this, the four intersection points should be at those positions. So, if we can set the coordinates of the intersection points to these, then we can relate them to the ellipse equation.So, let's take one of the intersection points, say, (c/2, (sqrt(3)/2)c). Since this point lies on the ellipse, substituting into the ellipse equation:(frac{(c/2)^2}{a^2} + frac{(sqrt{3}c/2)^2}{b^2} = 1)Simplify:(frac{c^2/4}{a^2} + frac{3c^2/4}{b^2} = 1)Multiply through by 4:(frac{c^2}{a^2} + frac{3c^2}{b^2} = 4)Factor out c^2:(c^2 left( frac{1}{a^2} + frac{3}{b^2} right) = 4)But we know that (c^2 = a^2 - b^2). So substitute that in:((a^2 - b^2) left( frac{1}{a^2} + frac{3}{b^2} right) = 4)Let me compute this expression:First, expand the product:((a^2 - b^2)left(frac{1}{a^2} + frac{3}{b^2}right) = (a^2 - b^2)left(frac{b^2 + 3a^2}{a^2b^2}right))Wait, let me compute each term:First term: (a^2 times frac{1}{a^2} = 1)Second term: (a^2 times frac{3}{b^2} = frac{3a^2}{b^2})Third term: (-b^2 times frac{1}{a^2} = -frac{b^2}{a^2})Fourth term: (-b^2 times frac{3}{b^2} = -3)So, combining all terms:1 + (frac{3a^2}{b^2}) - (frac{b^2}{a^2}) - 3 = 4Simplify:(1 - 3) + (frac{3a^2}{b^2}) - (frac{b^2}{a^2}) = 4-2 + (frac{3a^2}{b^2}) - (frac{b^2}{a^2}) = 4Bring -2 to the right side:(frac{3a^2}{b^2} - frac{b^2}{a^2} = 6)Let me denote (k = frac{a}{b}). Since (a > b), (k > 1). Then, (frac{a^2}{b^2} = k^2) and (frac{b^2}{a^2} = frac{1}{k^2}). Substitute into the equation:(3k^2 - frac{1}{k^2} = 6)Multiply both sides by (k^2) to eliminate the denominator:(3k^4 - 1 = 6k^2)Bring all terms to one side:(3k^4 - 6k^2 - 1 = 0)This is a quadratic equation in terms of (k^2). Let (m = k^2):(3m^2 - 6m - 1 = 0)Use quadratic formula:(m = frac{6 pm sqrt{36 + 12}}{6} = frac{6 pm sqrt{48}}{6} = frac{6 pm 4sqrt{3}}{6} = frac{3 pm 2sqrt{3}}{3})Since (m = k^2) must be positive, both roots are positive, but since (k > 1), (k^2 > 1), so we need to check which root satisfies this.Compute the roots:First root: (frac{3 + 2sqrt{3}}{3} ≈ frac{3 + 3.464}{3} ≈ frac{6.464}{3} ≈ 2.154)Second root: (frac{3 - 2sqrt{3}}{3} ≈ frac{3 - 3.464}{3} ≈ frac{-0.464}{3} ≈ -0.155)Discard the negative root. So, (m = frac{3 + 2sqrt{3}}{3}), so (k^2 = frac{3 + 2sqrt{3}}{3}). Therefore, (k = sqrt{frac{3 + 2sqrt{3}}{3}}).But we need to find the eccentricity (e = frac{c}{a}).Recall that (c^2 = a^2 - b^2), so (e = sqrt{1 - left(frac{b}{a}right)^2} = sqrt{1 - frac{1}{k^2}}).Compute (1 - frac{1}{k^2}):Since (k^2 = frac{3 + 2sqrt{3}}{3}), then (frac{1}{k^2} = frac{3}{3 + 2sqrt{3}}).Rationalize the denominator:Multiply numerator and denominator by (3 - 2sqrt{3}):(frac{3(3 - 2sqrt{3})}{(3 + 2sqrt{3})(3 - 2sqrt{3})} = frac{9 - 6sqrt{3}}{9 - 12} = frac{9 - 6sqrt{3}}{-3} = -3 + 2sqrt{3})Therefore, (frac{1}{k^2} = -3 + 2sqrt{3}), so:(1 - frac{1}{k^2} = 1 - (-3 + 2sqrt{3}) = 1 + 3 - 2sqrt{3} = 4 - 2sqrt{3})Therefore, (e = sqrt{4 - 2sqrt{3}})Simplify (sqrt{4 - 2sqrt{3}}). Let me see if this can be expressed in a simpler radical form.Suppose (sqrt{4 - 2sqrt{3}} = sqrt{a} - sqrt{b}). Squaring both sides:(4 - 2sqrt{3} = a + b - 2sqrt{ab})Comparing the terms:1. Rational part: (a + b = 4)2. Irrational part: (-2sqrt{ab} = -2sqrt{3}) ⇒ (sqrt{ab} = sqrt{3}) ⇒ (ab = 3)So, we have the system:(a + b = 4)(ab = 3)Solving this quadratic: (t^2 - 4t + 3 = 0), which factors as ((t - 1)(t - 3) = 0). Thus, (a = 1), (b = 3) or vice versa.Therefore, (sqrt{4 - 2sqrt{3}} = sqrt{3} - sqrt{1} = sqrt{3} - 1)Check: ((sqrt{3} - 1)^2 = 3 - 2sqrt{3} + 1 = 4 - 2sqrt{3}), which is correct.Thus, the eccentricity (e = sqrt{4 - 2sqrt{3}} = sqrt{3} - 1). Approximately, (sqrt{3} ≈ 1.732, so e ≈ 0.732), which is reasonable since eccentricity of an ellipse is between 0 and 1.So, the eccentricity is (sqrt{3} - 1), which can also be written as (sqrt{3} - 1). Let me confirm this answer.Wait, let's recap. The key steps were:1. Found the intersection points between the ellipse and the circle with radius c.2. Noted that these points must be vertices of a regular hexagon along with the foci.3. Assumed the intersection points are at 60°, 120°, etc., on the circle of radius c.4. Plugged the coordinates into the ellipse equation to find the relationship between a and b.5. Derived the equation leading to eccentricity e = sqrt(3) - 1.Is there another way to check this?Alternatively, consider the regular hexagon. In a regular hexagon, the distance between adjacent vertices is equal to the radius of the circumscribed circle. Wait, in a regular hexagon, the side length is equal to the radius. So, in our case, the circle has radius c, so the distance between any two adjacent vertices should be c. Let's verify this.Take two adjacent vertices: (c, 0) and (c/2, (sqrt(3)/2)c). The distance between them is sqrt[(c - c/2)^2 + (0 - sqrt(3)/2 c)^2] = sqrt[(c/2)^2 + (sqrt(3)c/2)^2] = sqrt[c^2/4 + 3c^2/4] = sqrt[c^2] = c. Similarly, all other adjacent vertices should be separated by distance c, which is consistent.Therefore, in the regular hexagon, all sides are length c. So, the distance between the foci (which is 2c) must correspond to two sides of the hexagon. Wait, but in the regular hexagon, the distance between opposite vertices is 2c (the diameter of the circle). So, in our case, the two foci are at opposite vertices (distance 2c apart), and the other four vertices are spaced at 60°, 120°, 240°, 300°, each separated by 60° from their neighbors.Therefore, this seems consistent. So, the key was realizing that the intersection points must lie at 60° and 120°, etc., on the circle of radius c, leading to the coordinates that we plugged into the ellipse equation.Therefore, the eccentricity is indeed (sqrt{3} - 1), which is approximately 0.732. To check if this makes sense, let's recall that for an ellipse, the eccentricity is between 0 (circle) and 1 (parabola). Since the problem involves a hexagon, which is more "stretched" than a circle, so the eccentricity should be reasonably high, which 0.732 is.Hence, after careful consideration, the eccentricity of the ellipse is (sqrt{3} - 1).Final Answerboxed{sqrt{3} - 1}</think>Given an ellipse with semi-major axis (a), semi-minor axis (b), and distance from the center to each focus (c), we know the relationship (c^2 = a^2 - b^2) and the eccentricity (e = frac{c}{a}). A circle with diameter equal to the distance between the foci (radius (c)) intersects the ellipse at four points. Connecting these points and the foci forms a regular hexagon. The circle's equation is (x^2 + y^2 = c^2). Solving the system of equations for the ellipse (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) and the circle, we find the intersection points. These points, along with the foci, form a regular hexagon inscribed in a circle of radius (c). Considering the regular hexagon's properties, the vertices are spaced at 60° intervals. The coordinates of these vertices are ((c cos theta, c sin theta)) for (theta = 0°, 60°, 120°, 180°, 240°, 300°). Substituting these coordinates into the ellipse equation, we derive the relationship between (a), (b), and (c).For a vertex at ((c/2, (sqrt{3}/2)c)) on the ellipse:[frac{(c/2)^2}{a^2} + frac{(sqrt{3}c/2)^2}{b^2} = 1]Simplifying, we find:[frac{c^2}{4a^2} + frac{3c^2}{4b^2} = 1 implies frac{c^2}{a^2} + frac{3c^2}{b^2} = 4]Using (c^2 = a^2 - b^2) and substituting (k = frac{a}{b}), we solve:[3k^4 - 6k^2 - 1 = 0]Solving this quadratic in (k^2), we find (k^2 = frac{3 + 2sqrt{3}}{3}). The eccentricity (e) is:[e = sqrt{1 - frac{1}{k^2}} = sqrt{4 - 2sqrt{3}} = sqrt{3} - 1][boxed{sqrt{3} - 1}],<think>Okay, let me try to tackle this problem. Hmm, so we have an ellipse with two foci, and there's a circle whose diameter is the line segment connecting these two foci. This circle intersects the ellipse at four different points. When we connect these four points and the two foci in sequence, it forms a regular hexagon. We need to find the eccentricity of the ellipse.First, let me recall some basic properties of ellipses. The standard equation for an ellipse centered at the origin with major axis along the x-axis is (x²/a²) + (y²/b²) = 1, where a is the semi-major axis, b is the semi-minor axis, and the distance from the center to each focus is c, with c² = a² - b². The eccentricity e is c/a.Now, the problem mentions a circle with diameter equal to the distance between the two foci. Since the foci are located at (-c, 0) and (c, 0), the diameter of the circle is 2c, so the radius is c. The circle is also centered at the origin, just like the ellipse. Therefore, the equation of the circle is x² + y² = c².This circle intersects the ellipse at four points. Let me visualize this: the ellipse is longer along the major axis (x-axis), and the circle is centered at the origin with radius c. Since c < a (because in an ellipse, c = ea and e < 1), the circle is entirely inside the ellipse along the major axis but might intersect the ellipse somewhere else. The intersection points are four in total, so they must be symmetric with respect to both axes.Connecting these four points and the two foci in sequence forms a regular hexagon. A regular hexagon has all sides equal and all internal angles equal (120 degrees each). So, the six points (four intersection points and two foci) must be equally spaced around a circle. Wait, but the foci are on the x-axis, and the other four points are the intersections of the circle and the ellipse. Hmm, maybe the regular hexagon is inscribed in the circle with radius c? But the foci are at (-c, 0) and (c, 0), so if the hexagon is regular, the other four points must be equally spaced on the circle.In a regular hexagon, the angle between consecutive vertices is 60 degrees. Starting from one focus, say (c, 0), the next vertex would be at 60 degrees, then 120, 180, 240, 300 degrees, etc. But the foci are at 0 and 180 degrees on the circle. So if we connect (c, 0), then the next point at 60 degrees, then 120, then ( -c, 0), then 240, then 300 degrees, and back to (c, 0). Wait, but the problem says connecting the four intersection points and the two foci in sequence. So the sequence would be: focus, intersection point, intersection point, focus, intersection point, intersection point, and back to the first focus? Hmm, maybe not. If it's a regular hexagon, all sides and angles are equal, so the six points must be equally spaced around the circle. However, the foci are already at (c, 0) and (-c, 0), which are 180 degrees apart. If we have a regular hexagon, the angle between consecutive vertices should be 60 degrees. Therefore, starting at (c, 0), the next vertex would be at 60 degrees, then 120, 180, 240, 300, and back to 0. But the two foci are at 0 and 180 degrees. Therefore, the four intersection points must be at 60, 120, 240, and 300 degrees. So those four points are located at angles of 60°, 120°, 240°, and 300° on the circle x² + y² = c².Therefore, the coordinates of these four intersection points would be (c cos θ, c sin θ) where θ is 60°, 120°, 240°, and 300°. So let's compute those:For 60°: (c/2, (c√3)/2)For 120°: (-c/2, (c√3)/2)For 240°: (-c/2, -(c√3)/2)For 300°: (c/2, -(c√3)/2)These four points lie on both the circle x² + y² = c² and the ellipse (x²/a²) + (y²/b²) = 1. So plugging these coordinates into the ellipse equation should satisfy it.Let's take one of the points, say (c/2, (c√3)/2). Plugging into the ellipse equation:(c²/4)/a² + ( (3c²)/4 )/b² = 1Simplify:(c²)/(4a²) + (3c²)/(4b²) = 1Multiply both sides by 4:c²/a² + 3c²/b² = 4We know that in an ellipse, c² = a² - b². Let's substitute that into the equation:(a² - b²)/a² + 3(a² - b²)/b² = 4Simplify each term:First term: (a² - b²)/a² = 1 - (b²/a²)Second term: 3(a² - b²)/b² = 3(a²/b² - 1)So combining:1 - (b²/a²) + 3(a²/b² - 1) = 4Expand:1 - b²/a² + 3a²/b² - 3 = 4Combine constants: 1 - 3 = -2So:-2 - (b²/a²) + 3a²/b² = 4Bring the -2 to the right side:- (b²/a²) + 3a²/b² = 6Let me denote k = b²/a². Then, since k is positive and less than 1 (because b < a in an ellipse). Then, we have:- k + 3/k = 6Multiply both sides by k to eliminate the denominator:- k² + 3 = 6kBring all terms to one side:- k² - 6k + 3 = 0Multiply both sides by -1:k² + 6k - 3 = 0This is a quadratic equation in k. Let's solve it:k = [ -6 ± √(36 + 12) ] / 2= [ -6 ± √48 ] / 2= [ -6 ± 4√3 ] / 2= -3 ± 2√3Since k = b²/a² must be positive, we discard the negative solution:k = -3 + 2√3 ≈ -3 + 3.464 ≈ 0.464Wait, but -3 + 2√3 is approximately -3 + 3.464 = 0.464, which is positive. So k = 2√3 - 3Therefore, b²/a² = 2√3 - 3But eccentricity e is c/a, and c² = a² - b², so:c² = a² - b²Divide both sides by a²:(c/a)² = 1 - (b²/a²)Therefore:e² = 1 - kBut k = 2√3 - 3, so:e² = 1 - (2√3 - 3) = 1 - 2√3 + 3 = 4 - 2√3Thus, e = √(4 - 2√3)Let me compute this value:4 - 2√3 ≈ 4 - 3.464 ≈ 0.536So √0.536 ≈ 0.732, but let's see if we can simplify √(4 - 2√3)Hmm, maybe rationalize the expression. Let me check if this can be expressed as √3 - 1 or something similar.Suppose √(4 - 2√3) can be written as √( (√3)^2 - 2√3 + 1 )? Wait, (√3 - 1)^2 = 3 - 2√3 + 1 = 4 - 2√3. Yes! So,√(4 - 2√3) = √( (√3 - 1)^2 ) = √3 - 1Since √3 ≈ 1.732, √3 - 1 ≈ 0.732, which matches the approximate value.Therefore, e = √3 - 1 ≈ 0.732But let's confirm this step again. If (√3 - 1)^2 = 3 - 2√3 + 1 = 4 - 2√3, then yes, √(4 - 2√3) = √3 - 1. Since e is positive, we take the positive root.Therefore, the eccentricity e is √3 - 1.Let me check if this makes sense. The eccentricity of an ellipse is between 0 and 1. √3 is approximately 1.732, so √3 - 1 ≈ 0.732, which is reasonable for an eccentricity. Also, since the regular hexagon requires the points to be equally spaced on the circle of radius c, this seems plausible.Alternatively, let me verify with another intersection point. Take the point at 120°: (-c/2, (c√3)/2). Plugging into the ellipse equation:(c²/4)/a² + (3c²/4)/b² = 1Same equation as before, so it holds. Similarly, the other points will give the same equation, so we only needed to check one.Therefore, the calculations seem consistent.So the eccentricity e is √3 - 1.Expressed as a simplified radical, √3 - 1. So the answer is √3 - 1. In boxed form, that's boxed{sqrt{3} - 1}Final AnswerThe eccentricity of the ellipse is boxed{sqrt{3} - 1}.</think>Given an ellipse with foci at ((-c, 0)) and ((c, 0)), we consider a circle centered at the origin with radius (c) (since the diameter is the distance between the foci). This circle intersects the ellipse at four points, and connecting these points and the foci in sequence forms a regular hexagon. We need to find the eccentricity of the ellipse.1. Equations of the ellipse and circle: - Ellipse: (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) - Circle: (x^2 + y^2 = c^2) - Relationship: (c^2 = a^2 - b^2)2. Intersection points: - The intersection points of the circle and the ellipse are at angles (60^circ, 120^circ, 240^circ, 300^circ) on the circle. - Coordinates of these points: (left(frac{c}{2}, frac{csqrt{3}}{2}right)), (left(-frac{c}{2}, frac{csqrt{3}}{2}right)), (left(-frac{c}{2}, -frac{csqrt{3}}{2}right)), (left(frac{c}{2}, -frac{csqrt{3}}{2}right)).3. Substituting into the ellipse equation: - Using the point (left(frac{c}{2}, frac{csqrt{3}}{2}right)) in the ellipse equation: [ frac{left(frac{c}{2}right)^2}{a^2} + frac{left(frac{csqrt{3}}{2}right)^2}{b^2} = 1 ] - Simplifying: [ frac{c^2}{4a^2} + frac{3c^2}{4b^2} = 1 ] - Multiplying by 4: [ frac{c^2}{a^2} + frac{3c^2}{b^2} = 4 ]4. Using (c^2 = a^2 - b^2): - Substitute (c^2 = a^2 - b^2) into the equation: [ frac{a^2 - b^2}{a^2} + frac{3(a^2 - b^2)}{b^2} = 4 ] - Simplifying: [ 1 - frac{b^2}{a^2} + 3left(frac{a^2}{b^2} - 1right) = 4 ] - Let (k = frac{b^2}{a^2}): [ 1 - k + 3left(frac{1}{k} - 1right) = 4 ] - Solving for (k): [ -k + frac{3}{k} = 6 implies k^2 + 6k - 3 = 0 ] - Solving the quadratic equation: [ k = -3 + 2sqrt{3} ]5. Finding the eccentricity: - Eccentricity (e = frac{c}{a}): [ e^2 = 1 - k = 1 - (-3 + 2sqrt{3}) = 4 - 2sqrt{3} ] - Therefore, (e = sqrt{4 - 2sqrt{3}} = sqrt{3} - 1).The eccentricity of the ellipse is (boxed{sqrt{3} - 1}).]

🔑:2. sqrt{3}-1.Let the equation of the ellipse be frac{x^{2}}{a^{2}}+frac{y^{2}}{b^{2}}=1(a>b>0). From the condition, we know that the point left(frac{c}{2}, frac{sqrt{3} c}{2}right) lies on the ellipse.begin{array}{l}text { Then } frac{c^{2}}{4 a^{2}}+frac{3 c^{2}}{4 b^{2}}=1 Rightarrow frac{e^{2}}{4}+frac{3 e^{2}}{4-4 e^{2}}=1 Rightarrow e^{2}=4-2 sqrt{3} Rightarrow e=sqrt{3}-1 .end{array}

🧠:[<think>Okay, so I need to find the equation of a line that passes through the point (1, 1) and is perpendicular to the line given by x - 2y + 3 = 0. Hmm, let me think step by step how to approach this.First, I remember that to find the equation of a line, I usually need a point on the line and its slope. Since the problem says the line passes through (1, 1), that's my point. But I need the slope. And it has to be perpendicular to the given line x - 2y + 3 = 0. So, I should first find the slope of that given line, and then find the negative reciprocal of that slope because perpendicular lines have slopes that multiply to -1. Right?Let me confirm that. If two lines are perpendicular, their slopes m1 and m2 satisfy m1 * m2 = -1. So, if I can find the slope of the given line, then the slope of the perpendicular line would be -1 divided by that slope. Got it.So, first step: find the slope of the line x - 2y + 3 = 0. To do that, maybe I should rearrange the equation into slope-intercept form, which is y = mx + b, where m is the slope.Starting with x - 2y + 3 = 0. Let me solve for y.Subtract x and 3 from both sides: -2y = -x - 3.Now divide both sides by -2: y = (1/2)x + 3/2.So, the slope of the given line is 1/2. Therefore, the slope of the line perpendicular to it should be the negative reciprocal of 1/2, which is -2. Because reciprocal of 1/2 is 2, and negative is -2. So, the slope of the line we need is -2.Now that I have the slope (-2) and a point (1, 1) that the line passes through, I can use the point-slope form of the equation of a line. The point-slope formula is y - y1 = m(x - x1), where (x1, y1) is the point and m is the slope.Substituting in the values: y - 1 = -2(x - 1). Then, I can simplify this equation to get it into the standard form or slope-intercept form, depending on what's required. The problem didn't specify the form, but since the original line was given in standard form (Ax + By + C = 0), maybe I should present the answer in a similar format.Let me expand the equation first. Starting with y - 1 = -2x + 2. Then, bringing all terms to one side to get it into standard form. Let's add 2x to both sides and subtract 2 from both sides:2x + y - 1 - 2 = 0 → 2x + y - 3 = 0.So, simplifying that, the equation is 2x + y - 3 = 0. Alternatively, if I want to write it in slope-intercept form, it would be y = -2x + 3. But let me check if both forms are correct.Testing the point (1,1) in both equations. For the standard form: 2(1) + 1 - 3 = 2 + 1 - 3 = 0, which works. For the slope-intercept form: plugging x = 1, y = -2(1) + 3 = 1, which also works. So both equations are correct.Wait, but the problem just asks for the equation of the line. Since the original line was given in standard form, maybe the answer expects standard form. But maybe either is acceptable. However, in many cases, unless specified, both forms are okay, but perhaps the standard form is preferred here. Alternatively, if the answer is presented in slope-intercept, that's also correct.Alternatively, maybe the problem expects the answer in the same form as the given line. Let me check the original problem again. The given line was x - 2y + 3 = 0, which is standard form. So maybe the answer should be in standard form as well.Therefore, 2x + y - 3 = 0. Alternatively, sometimes people write the standard form with the x term positive, so that's already the case here. So that should be fine.Alternatively, to confirm, let's check once more.Original line: x - 2y + 3 = 0, slope 1/2.Perpendicular slope: -2.Using point-slope: y - 1 = -2(x - 1).Expanding: y - 1 = -2x + 2.Bringing all terms to left-hand side: 2x + y - 3 = 0. So that's correct.Alternatively, writing in slope-intercept form: y = -2x + 3.Either is correct, but since the problem gave the original line in standard form, perhaps the answer should follow the same structure. So 2x + y - 3 = 0.But let me check if there's another standard form convention. Sometimes they prefer Ax + By = C. So, 2x + y = 3. Either way, the equation is correct.Alternatively, if we want to check for potential errors. Let me verify the slope.Original line: x - 2y + 3 = 0. Solving for y: y = (1/2)x + 3/2. So slope is 1/2. Therefore, perpendicular slope is -2. Correct.Then using point (1,1): y = -2x + b. Plug in x = 1, y =1: 1 = -2(1) + b → 1 = -2 + b → b = 3. So equation is y = -2x + 3. Correct.Therefore, either form is correct, but in standard form, that's 2x + y - 3 = 0. So I think that's the answer.But let me check once again if the arithmetic was correct when converting from point-slope to standard form.Starting from y - 1 = -2(x -1). Distribute the -2: y -1 = -2x + 2. Then add 2x to both sides and add 1 to both sides: 2x + y = 3. Therefore, 2x + y - 3 = 0. Yes, that's correct. So that's the equation in standard form.Alternatively, if the problem wants the answer in a different form, but since it wasn't specified, the standard form is acceptable.Therefore, the equation of the line is 2x + y - 3 = 0.Final AnswerThe equation of the line is boxed{2x + y - 3 = 0}.</think>To find the equation of the line that passes through the point (1, 1) and is perpendicular to the line ( x - 2y + 3 = 0 ):1. Find the slope of the given line: - Convert the given line ( x - 2y + 3 = 0 ) to slope-intercept form: [ x - 2y + 3 = 0 implies -2y = -x - 3 implies y = frac{1}{2}x + frac{3}{2} ] - The slope of the given line is ( frac{1}{2} ).2. Determine the slope of the perpendicular line: - The slope of a line perpendicular to another is the negative reciprocal. Thus, the slope of the perpendicular line is: [ -frac{1}{frac{1}{2}} = -2 ]3. Use the point-slope form to find the equation of the line: - Using the point (1, 1) and the slope -2: [ y - 1 = -2(x - 1) ] - Simplify to convert it to standard form: [ y - 1 = -2(x - 1) implies y - 1 = -2x + 2 implies 2x + y - 3 = 0 ]Final Answer: The equation of the line is boxed{2x + y - 3 = 0}.,<think>Okay, so I need to find the equation of a line that passes through the point (1, 1) and is perpendicular to the line given by the equation x - 2y + 3 = 0. Hmm, let me think step by step how to approach this. First, I remember that to find a line perpendicular to another line, their slopes should be negative reciprocals of each other. So, if I can find the slope of the given line, then the slope of the perpendicular line would be the negative reciprocal of that. But wait, the equation of the given line is in general form, right? x - 2y + 3 = 0. I should probably convert this into slope-intercept form (y = mx + b) to easily identify the slope. Let me try that. Starting with x - 2y + 3 = 0. Let me solve for y. Subtract x and 3 from both sides: -2y = -x - 3. Then divide both sides by -2: y = (1/2)x + 3/2. Okay, so the slope (m) of the given line is 1/2. Therefore, the slope of the line perpendicular to this should be the negative reciprocal, which would be -2. Because if the original slope is 1/2, flipping it gives 2/1, and changing the sign gives -2. Got that. So now I know the slope of the line we need to find is -2, and it passes through the point (1, 1). To find the equation of the line, I can use the point-slope form. The formula for point-slope is y - y1 = m(x - x1), where (x1, y1) is a point the line passes through. Plugging in the values here, m is -2, and the point is (1, 1). So substituting into point-slope:y - 1 = -2(x - 1)Let me simplify that. First, distribute the -2 on the right side:y - 1 = -2x + 2Then, add 1 to both sides to solve for y:y = -2x + 2 + 1y = -2x + 3So, the equation in slope-intercept form is y = -2x + 3. But maybe they want the answer in standard form, like the original equation? Let me check. The original problem didn't specify, but since the given line was in standard form (x - 2y + 3 = 0), maybe they want the same. Let me convert y = -2x + 3 into standard form. Starting with y = -2x + 3. Let's move all terms to the left side. 2x + y - 3 = 0Alternatively, it could be written as 2x + y = 3. But standard form is usually written as Ax + By + C = 0, so 2x + y - 3 = 0. Wait, but just to confirm, the original problem's line was x - 2y + 3 = 0, so coefficients might be positive for x. Similarly, our line has 2x + y - 3 = 0, which is also fine. Alternatively, some people prefer the x coefficient to be positive. So in this case, 2x + y - 3 = 0 already has a positive coefficient for x, so that's acceptable. But just to make sure, let me verify. Let's take both equations and check if they are perpendicular. The original line has a slope of 1/2, and the new line has a slope of -2. Multiplying the two slopes: (1/2)*(-2) = -1, which confirms they are perpendicular. That's a good check. Also, we need to verify that the new line passes through the point (1, 1). Let's substitute x = 1 and y = 1 into the equation 2x + y - 3 = 0. Left side: 2(1) + 1 - 3 = 2 + 1 - 3 = 0. Perfect, that's correct. So, in standard form, the equation is 2x + y - 3 = 0. Alternatively, if written as 2x + y = 3, that's also correct, but depending on the required format. The problem didn't specify, but since the given line was in the form Ax + By + C = 0, maybe the answer should follow the same. But just to be thorough, let me present both forms and see which one is more appropriate. The slope-intercept form is y = -2x + 3, and the standard form is 2x + y - 3 = 0. Both are correct. Since the problem didn't specify, but the original equation was in standard form, perhaps the answer should be in standard form. Alternatively, sometimes problems prefer one form over another. Let me check standard form: usually, in math problems, unless specified, both forms are acceptable, but since the original is in standard form, maybe standard is preferable here. So, I think both forms are correct, but since the original line is presented in standard form, the answer might be expected in standard form as well. So, I'll go with 2x + y - 3 = 0. But just to double-check my calculations again. Let me start from the beginning. Given line: x - 2y + 3 = 0. Solve for y: y = (1/2)x + 3/2. Slope is 1/2. Therefore, the perpendicular slope is -2. Using point (1,1), so point-slope: y - 1 = -2(x - 1). Expand: y - 1 = -2x + 2. Add 1: y = -2x + 3. Convert to standard form: 2x + y = 3. Hence, 2x + y - 3 = 0. Yes, that seems correct. Alternatively, if I wanted to write it as x - 2y +3=0's perpendicular, maybe there's another way? Let me think. Another way to find the equation of a line perpendicular to ax + by + c = 0 is that the direction vector of the given line is (b, -a), so the perpendicular line would have direction vector (a, b). But maybe that's overcomplicating. But in any case, the standard method is using slope. Since we know the slope of the given line, find the negative reciprocal, then use point-slope. Another check: if two lines are perpendicular, the product of their slopes is -1. Original slope 1/2, new slope -2. Multiply them: (1/2)*(-2) = -1. Perfect. Also, plugging the point (1,1) into the equation 2x + y -3 =0: 2(1) + 1 -3 = 0. 2 +1 -3 =0. 0=0. Correct. So everything checks out. Therefore, the equation of the line is 2x + y -3 =0. Alternatively, if written as y = -2x +3, but in the problem statement, the original equation was in standard form, so perhaps standard form is preferred here. So, I think that's the answer. Final AnswerThe equation of the line is boxed{2x + y - 3 = 0}.</think>To find the equation of the line that passes through the point (1, 1) and is perpendicular to the line (x - 2y + 3 = 0):1. Convert the given line (x - 2y + 3 = 0) to slope-intercept form: [ x - 2y + 3 = 0 implies -2y = -x - 3 implies y = frac{1}{2}x + frac{3}{2} ] The slope of this line is (frac{1}{2}).2. The slope of the line perpendicular to this is the negative reciprocal of (frac{1}{2}), which is (-2).3. Use the point-slope form of the equation of a line with slope (-2) passing through the point (1, 1): [ y - 1 = -2(x - 1) ] Simplify this equation: [ y - 1 = -2x + 2 implies y = -2x + 3 ]4. Convert the equation to standard form: [ y = -2x + 3 implies 2x + y - 3 = 0 ]Verify that the line passes through the point (1, 1) by substituting (x = 1) and (y = 1) into the equation (2x + y - 3 = 0): [ 2(1) + 1 - 3 = 0 implies 0 = 0 ]Thus, the equation of the line is (boxed{2x + y - 3 = 0}).]

🔑:Since the line we are looking for is perpendicular to the line x-2y+3=0, let's assume the equation of the line we are looking for is 2x+y+c=0. Given that the line passes through the point (1, 1), we have 2cdot1+1+c=0. Therefore, c=-3. Hence, the equation of the line is 2x+y-3=0. So, the answer is boxed{2x+y-3=0}.

🧠:[<think>Okay, let me try to figure out this problem step by step. So, we have this piecewise function f(x) defined as two different parts: one for x ≤ 2 and another for x > 2. The function is given by:f(x) = { -x² + 2x + 3, for x ≤ 2 6 + log_a(x), for x > 2And we know that the range of the entire function f(x) is (-∞, 4]. We need to find the range of the real number a (where a > 0 and a ≠ 1). The options are given as A, B, C, D. Let's break this down.First, let's analyze each part of the piecewise function separately to understand their ranges. Then, we can combine the information and see how the parameter a affects the overall range.Starting with the first part: f(x) = -x² + 2x + 3 for x ≤ 2. This is a quadratic function. Since the coefficient of x² is negative, it's a downward-opening parabola. The vertex form of a quadratic function f(x) = ax² + bx + c can be found using the formula for the vertex at x = -b/(2a). Here, a = -1, b = 2, so the x-coordinate of the vertex is at x = -2/(2*(-1)) = 1. Then, plugging x = 1 into the function to find the y-coordinate: f(1) = -(1)² + 2*1 + 3 = -1 + 2 + 3 = 4. So the vertex is at (1, 4), which is the maximum point of the parabola.Since the parabola opens downward, the maximum value of this part is 4 at x = 1. As x moves away from 1 towards negative infinity, the function decreases to negative infinity. But since x is limited to x ≤ 2, we need to check the value at x = 2 as well. Let's compute f(2): f(2) = -(2)² + 2*2 + 3 = -4 + 4 + 3 = 3. So at x = 2, the value is 3. Therefore, for the first part of the function (x ≤ 2), the range is (-∞, 4], because it attains a maximum of 4 at x = 1 and decreases towards negative infinity as x approaches negative infinity. However, when x approaches 2 from the left, the value is 3. But since the entire parabola is included for x ≤ 2, the range is indeed all real numbers less than or equal to 4.Now, moving on to the second part: f(x) = 6 + log_a(x) for x > 2. The logarithm function log_a(x) is defined for x > 0, which is satisfied here since x > 2. The range of log_a(x) depends on the base a. If a > 1, the logarithm function increases towards infinity as x increases and approaches negative infinity as x approaches 0 from the right. If 0 < a < 1, the logarithm function decreases towards negative infinity as x increases and approaches positive infinity as x approaches 0 from the right. However, in our case, x is greater than 2, so we need to analyze the behavior as x approaches 2 from the right and as x approaches infinity.But since we need the entire function f(x) to have a range of (-∞, 4], we need to ensure that the second part of the function (6 + log_a(x) for x > 2) does not exceed 4. Because if the second part goes above 4, the overall range of f(x) would include values greater than 4, which contradicts the given range (-∞, 4]. Therefore, the maximum value that the second part can take is 4. Let's write that down:For x > 2, 6 + log_a(x) ≤ 4.Subtracting 6 from both sides:log_a(x) ≤ -2.This inequality must hold for all x > 2. Therefore, log_a(x) must be less than or equal to -2 for all x > 2. The question now is: what values of a will make log_a(x) ≤ -2 for all x > 2?To solve this, let's recall the properties of logarithms. Remember that log_a(x) = ln(x)/ln(a). So, log_a(x) ≤ -2 is equivalent to ln(x)/ln(a) ≤ -2.But since x > 2, ln(x) > ln(2) > 0 (since ln is an increasing function). Therefore, the sign of ln(a) will determine the direction of the inequality when we multiply both sides by ln(a). Let's consider two cases: when a > 1 and when 0 < a < 1.Case 1: a > 1. Then ln(a) > 0. Therefore, multiplying both sides of ln(x)/ln(a) ≤ -2 by ln(a) preserves the inequality:ln(x) ≤ -2 ln(a)But ln(x) > ln(2), so:ln(2) < ln(x) ≤ -2 ln(a)But wait, if a > 1, then -2 ln(a) is negative because ln(a) > 0. However, ln(x) is greater than ln(2) (which is positive since 2 > 1). Therefore, ln(x) must be both greater than ln(2) (positive) and less than or equal to -2 ln(a) (negative). But that's impossible because a positive number can't be less than a negative number. Therefore, in this case, there's no solution when a > 1. So, a cannot be greater than 1.Case 2: 0 < a < 1. Then ln(a) < 0. Therefore, multiplying both sides of ln(x)/ln(a) ≤ -2 by ln(a) (which is negative) reverses the inequality:ln(x) ≥ -2 ln(a)But ln(a) is negative, so -2 ln(a) is positive. Therefore, we have:ln(x) ≥ -2 ln(a)Which can be rewritten as:ln(x) ≥ ln(a^{-2}) (since -2 ln(a) = ln(a^{-2}))Therefore, exponentiating both sides (since the exponential function is increasing):x ≥ a^{-2}But we need this inequality to hold for all x > 2. That is, x ≥ a^{-2} must be true for all x > 2. The only way this is possible is if a^{-2} ≤ 2. Because if a^{-2} were greater than 2, then there would exist some x in (2, a^{-2}) where x < a^{-2}, which would violate the inequality ln(x) ≥ ln(a^{-2}) (since x < a^{-2} implies ln(x) < ln(a^{-2})). Therefore, to ensure that x ≥ a^{-2} for all x > 2, we must have a^{-2} ≤ 2. Solving this:a^{-2} ≤ 2=> 1/a² ≤ 2=> a² ≥ 1/2=> a ≥ 1/√2 or a ≤ -1/√2But since a > 0 and a ≠ 1, we discard the negative solution. Therefore:a ≥ 1/√2 ≈ 0.7071...But in this case, we are considering 0 < a < 1. Therefore, combining these two conditions:1/√2 ≤ a < 1So, the range of a is [1/√2, 1). But let's check if the endpoints are included or not.Wait, when a = 1/√2, then a^{-2} = (1/√2)^{-2} = (√2)^2 = 2. Therefore, the inequality becomes x ≥ 2 for all x > 2. But x > 2, so x ≥ 2 is automatically true. However, we need to check if the function 6 + log_a(x) approaches 4 as x approaches 2 from the right. Let's see:If a = 1/√2, then log_a(2) = log_{1/√2}(2) = ln(2)/ln(1/√2) = ln(2)/(- (1/2) ln 2) = -2. Therefore, 6 + log_{1/√2}(2) = 6 - 2 = 4. But x approaches 2 from the right, so as x approaches 2+, log_{1/√2}(x) approaches log_{1/√2}(2) = -2. Hence, 6 + log_{1/√2}(x) approaches 4 from below? Wait, no. Let's see:If a = 1/√2, then log_a(x) is a decreasing function because 0 < a < 1. So, as x increases, log_a(x) decreases. Therefore, as x approaches 2 from the right (x -> 2+), log_a(x) approaches log_a(2) = -2, so 6 + log_a(x) approaches 4. For x > 2, since log_a(x) is decreasing, as x increases beyond 2, log_a(x) becomes less than -2 (since x increases, log_a(x) decreases), so 6 + log_a(x) becomes less than 4. Therefore, when a = 1/√2, the maximum value of the second part is 4 (approached from below as x approaches 2 from the right), and it decreases towards negative infinity as x approaches infinity. Therefore, including a = 1/√2 is acceptable because the second part never exceeds 4, and the overall range of f(x) remains (-∞, 4].Now, when a approaches 1 from below (a -> 1-), let's see what happens. For a < 1, log_a(x) is decreasing. So, as a approaches 1 from below, log_a(x) approaches log_1(x), but log base 1 is undefined. However, near a = 1, log_a(x) tends to diverge. Wait, actually, when a approaches 1 from below, the logarithm function log_a(x) becomes very flat. Let's compute the limit as a approaches 1 from below. Using the change of base formula, log_a(x) = ln(x)/ln(a). As a approaches 1 from below, ln(a) approaches 0 from below (since a < 1). Therefore, ln(a) is approaching 0 from the negative side, so log_a(x) = ln(x)/ln(a) approaches negative infinity (since ln(x) is positive for x > 1, and ln(a) approaches 0 from below, making the ratio go to negative infinity). Therefore, 6 + log_a(x) approaches negative infinity as a approaches 1 from below. But wait, for a fixed x > 2, as a approaches 1 from below, log_a(x) approaches negative infinity, so 6 + log_a(x) approaches negative infinity. However, we need to ensure that for all x > 2, 6 + log_a(x) ≤ 4. But when a is close to 1 from below, even though log_a(x) is decreasing, for x close to 2, log_a(2) = ln(2)/ln(a). As a approaches 1 from below, ln(a) approaches 0 from below, so log_a(2) approaches ln(2)/0- which is negative infinity. Therefore, 6 + log_a(2) approaches negative infinity. However, as a increases towards 1, log_a(x) for x > 2 will approach log_1(x), which is undefined, but near a = 1, the log function becomes very steep.Wait, maybe I need another approach here. Let's think about it. For 0 < a < 1, log_a(x) is a decreasing function. Therefore, for x > 2, log_a(x) < log_a(2). Since log_a(x) is decreasing, the maximum value of log_a(x) in x > 2 occurs as x approaches 2 from the right. Therefore, log_a(x) approaches log_a(2) from below (since x is approaching 2 from above, and the function is decreasing). Therefore, the maximum value of 6 + log_a(x) for x > 2 is 6 + log_a(2). Therefore, to ensure that 6 + log_a(x) ≤ 4 for all x > 2, we need:6 + log_a(2) ≤ 4=> log_a(2) ≤ -2Now, solving log_a(2) ≤ -2.Again, log_a(2) = ln(2)/ln(a). Let's write the inequality:ln(2)/ln(a) ≤ -2Since 0 < a < 1, ln(a) < 0. Therefore, multiplying both sides by ln(a) (negative) reverses the inequality:ln(2) ≥ -2 ln(a)=> ln(2) ≥ ln(a^{-2})Because -2 ln(a) = ln(a^{-2}).Exponentiating both sides (since exponential function is increasing):2 ≥ a^{-2}=> 2 ≥ 1/a²=> a² ≥ 1/2=> a ≥ 1/√2 (since a > 0)So, combining with the previous condition that 0 < a < 1, we have 1/√2 ≤ a < 1.Therefore, the range of a is [1/√2, 1).But wait, the answer options are:A: (√2/2, 1)B: [√2/2, 1)C: (1, √2]D: (1, √2)But √2/2 is equal to 1/√2, since √2/2 = 1/(√2). So, 1/√2 is approximately 0.707, which is equal to √2 divided by 2.Therefore, option B is [1/√2, 1), which is the same as [√2/2, 1).But let me confirm once again if a = 1/√2 is allowed.When a = 1/√2, log_a(2) = log_{1/√2}(2) = ln(2)/ln(1/√2) = ln(2)/(- (1/2) ln 2) = -2. Therefore, 6 + log_a(2) = 6 - 2 = 4. So, as x approaches 2 from the right, f(x) approaches 4. For x > 2, since log_a(x) is decreasing, the values of f(x) = 6 + log_a(x) will be less than 4. Therefore, the maximum value of the second part is 4, which matches the maximum of the first part. Hence, including a = 1/√2 is acceptable because the function f(x) reaches 4 at x = 1 and approaches 4 as x approaches 2 from the right. Therefore, the overall range is indeed (-∞, 4].If a were less than 1/√2, then a^{-2} would be greater than 2. Therefore, the inequality x ≥ a^{-2} would not hold for all x > 2. For example, take a = 0.7 (which is less than 1/√2 ≈ 0.7071). Then a^{-2} ≈ 1/(0.49) ≈ 2.04. So, x must be ≥ 2.04. But x can be 2.01, which is less than 2.04, so in that case, log_a(2.01) would be greater than -2, leading to 6 + log_a(2.01) > 4, which would exceed the desired range. Therefore, a must be ≥ 1/√2.Similarly, if a = 1, the logarithm is undefined. If a approaches 1 from below, the maximum value of the second part approaches 6 + log_a(2). Let's compute log_a(2) when a approaches 1 from below. As a approaches 1-, ln(a) approaches 0-, so log_a(2) = ln(2)/ln(a) approaches negative infinity. Therefore, 6 + log_a(2) approaches negative infinity. However, when a approaches 1 from below, the behavior near x = 2+ is that log_a(x) approaches negative infinity, so 6 + log_a(x) approaches negative infinity. However, for x slightly larger than 2, log_a(x) is just slightly less than log_a(2) (since the function is decreasing), but as a approaches 1, log_a(2) tends to negative infinity. Therefore, for a approaching 1 from below, the second part's maximum (approached as x approaches 2+) tends to negative infinity, which is still within the overall range (-∞, 4]. However, we need to check if for a approaching 1 from below, the second part doesn't exceed 4. But when a is less than 1 but close to 1, say a = 0.99, then log_a(2) = ln(2)/ln(0.99) ≈ 0.6931 / (-0.0101) ≈ -68.6, so 6 + log_a(2) ≈ 6 - 68.6 ≈ -62.6, which is way below 4. Therefore, as a approaches 1 from below, the second part of the function becomes very negative, which is fine because the overall range is allowed to go to negative infinity. However, the upper limit of the second part when a approaches 1 from below would approach 4? Wait, no. Wait, if a is approaching 1 from below, the maximum value of the second part is 6 + log_a(2). As a approaches 1 from below, log_a(2) approaches negative infinity, so 6 + log_a(2) approaches negative infinity. Therefore, the second part's maximum (at x approaching 2 from the right) is approaching negative infinity. Therefore, even though a approaches 1, the second part doesn't approach 4. Wait, this contradicts the earlier conclusion. Wait, let's clarify.Wait, when a is exactly 1/√2, the maximum value of the second part is 4. If a is greater than 1/√2 but less than 1, then log_a(2) < -2 (since a^{-2} < 2). Wait, no. Wait, if a > 1/√2, then a^{-2} = 1/(a²) < 1/( (1/√2)^2 ) = 1/(1/2) = 2. So, a^{-2} < 2. Therefore, x must be ≥ a^{-2} for all x > 2. But if a^{-2} < 2, then for x in (a^{-2}, 2), but wait x > 2. Wait, x > 2, and a^{-2} < 2, so x > 2 ≥ a^{-2}. Therefore, x ≥ a^{-2} is automatically satisfied since x > 2 and a^{-2} < 2. Therefore, for a > 1/√2, the inequality x ≥ a^{-2} is satisfied for all x > 2, because a^{-2} < 2 and x > 2. Therefore, log_a(x) ≤ -2 for all x > 2. But wait, when a > 1/√2, then a^{-2} < 2. So, log_a(x) ≤ -2 for x ≥ a^{-2}. But since x > 2 and a^{-2} < 2, we have x > 2 > a^{-2}, so x is greater than a^{-2}, hence log_a(x) < log_a(a^{-2}) = log_a(a^{-2}) = -2. Because log_a(a^k) = k. So, log_a(a^{-2}) = -2. Since log_a(x) is decreasing (because 0 < a < 1), x > a^{-2} implies log_a(x) < log_a(a^{-2}) = -2. Therefore, for a > 1/√2, the inequality log_a(x) ≤ -2 holds for all x > 2. Therefore, 6 + log_a(x) ≤ 4 for all x > 2. And when x approaches 2 from the right, log_a(x) approaches log_a(2). Since a > 1/√2, a^{-2} < 2, so log_a(2) < log_a(a^{-2}) = -2. Wait, no: log_a(2) since 2 > a^{-2}, and log_a is decreasing, so log_a(2) < log_a(a^{-2}) = -2. Therefore, as x approaches 2 from the right, log_a(x) approaches log_a(2) < -2, so 6 + log_a(x) approaches 6 + log_a(2) < 4. Therefore, the upper bound of the second part is less than 4. But the first part already attains 4 at x = 1. Therefore, the overall range is (-∞, 4].But wait, if a > 1/√2, then log_a(2) < -2, so 6 + log_a(2) < 4. Therefore, the second part never reaches 4, but the first part does. Therefore, the overall range is still (-∞, 4]. However, if a = 1/√2, then log_a(2) = -2, so 6 + log_a(2) = 4. Therefore, as x approaches 2 from the right, f(x) approaches 4, but doesn't exceed it. Therefore, a = 1/√2 is allowed, because the maximum value of 4 is already achieved by the first part at x = 1. Therefore, including a = 1/√2 is okay because the second part approaches 4 but doesn't exceed it. Hence, the lower bound is inclusive.For the upper bound, when a approaches 1 from below, the second part tends to negative infinity, but since a cannot be 1, the upper limit is open. Therefore, the range of a is [1/√2, 1), which corresponds to option B.But let's check if there's any mistake here. The options are:A: (√2/2, 1)B: [√2/2, 1)C: (1, √2]D: (1, √2)Since √2/2 is equal to 1/√2, which is approximately 0.707, so option B is [1/√2, 1). The answer should be B.Wait, but let me verify with an example. Take a = 1/√2 (~0.707). Then log_a(x) = log_{1/√2}(x). Let's check x = 2: log_{1/√2}(2) = ln(2)/ln(1/√2) = ln(2)/(- (1/2) ln 2) = -2, so 6 + (-2) = 4. For x > 2, say x = 3: log_{1/√2}(3) = ln(3)/(-0.5 ln 2) ≈ (1.0986)/(-0.3466) ≈ -3.17, so 6 + (-3.17) ≈ 2.83 < 4. So, that's good.Another example: a = 0.8 (which is greater than 1/√2 ≈ 0.707). Then a^{-2} = 1/(0.64) ≈ 1.5625 < 2. Therefore, log_a(x) = ln(x)/ln(a). For x > 2, since a^{-2} = 1.5625 < 2, x > 2 > a^{-2}, so log_a(x) < log_a(a^{-2}) = -2. Let's check x = 2: log_a(2) = ln(2)/ln(0.8) ≈ 0.6931 / (-0.2231) ≈ -3.11, so 6 + (-3.11) ≈ 2.89 < 4. For x approaching 2 from the right, log_a(x) approaches log_a(2) ≈ -3.11, which is less than -2. So, 6 + log_a(x) approaches 2.89, which is less than 4. Therefore, all values are below 4. So a = 0.8 is acceptable.If a = 0.6 (which is less than 1/√2 ≈ 0.707), then a^{-2} = 1/0.36 ≈ 2.777 > 2. Therefore, for x in (2, 2.777), log_a(x) > log_a(a^{-2}) = -2, so 6 + log_a(x) > 4. For example, x = 2.5: log_{0.6}(2.5) = ln(2.5)/ln(0.6) ≈ 0.9163 / (-0.5108) ≈ -1.793. Then 6 + (-1.793) ≈ 4.207 > 4. This violates the range requirement, so a cannot be less than 1/√2.Therefore, confirming that a must be ≥ 1/√2. Hence, the correct answer is B: [√2/2, 1).Final Answerboxed{B}</think>Given the piecewise function ( f(x) = begin{cases} -x^2 + 2x + 3 & text{if } x leq 2 6 + log_a x & text{if } x > 2 end{cases} ) with ( a > 0 ) and ( a neq 1 ), we need to determine the range of ( a ) such that the range of ( f(x) ) is ( (-infty, 4] ).1. Analysis of the quadratic part ( ( x leq 2 ) ): - The quadratic function ( -x^2 + 2x + 3 ) has a maximum value at ( x = 1 ), which is 4. The range of this part is ( (-infty, 4] ).2. Analysis of the logarithmic part ( ( x > 2 ) ): - The logarithmic function ( 6 + log_a x ) must not exceed 4 for ( x > 2 ). This requires ( log_a x leq -2 ) for all ( x > 2 ). - Converting the inequality: ( log_a x leq -2 ) implies ( x geq a^{-2} ). For this to hold for all ( x > 2 ), we need ( a^{-2} leq 2 ). - Solving ( a^{-2} leq 2 ): ( a^2 geq frac{1}{2} ) hence ( a geq frac{1}{sqrt{2}} ).3. Range of ( a ): - Since ( 0 < a < 1 ), combining with the above result, we get ( frac{1}{sqrt{2}} leq a < 1 ).4. Verification: - When ( a = frac{1}{sqrt{2}} ), ( log_{1/sqrt{2}} 2 = -2 ), so ( 6 + log_{1/sqrt{2}} 2 = 4 ), which is valid. - For ( a ) approaching 1 from below, ( 6 + log_a x ) approaches negative infinity, which is within the required range.Thus, the range of ( a ) is ( left[ frac{sqrt{2}}{2}, 1 right) ), corresponding to option B.[boxed{B}],<think>Okay, so I have this piecewise function f(x) defined as two parts: for x ≤ 2, it's a quadratic function -x² + 2x + 3, and for x > 2, it's 6 plus log base a of x. The problem states that the range of f(x) is (-∞, 4], and we need to find the range of real number a (with a > 0 and a ≠ 1). The answer choices are given as A, B, C, D. Let me try to figure this out step by step.First, let me recall that the range of a piecewise function is the union of the ranges of its individual pieces, but restricted by their domains. So here, the function is split into two parts: the quadratic part for x ≤ 2 and the logarithmic part for x > 2. The total range is given as (-∞, 4], which means that the maximum value of the entire function is 4, and it can go down to negative infinity.Starting with the quadratic part: f(x) = -x² + 2x + 3 for x ≤ 2. To find its range, I need to analyze this quadratic function. Since the coefficient of x² is negative (-1), the parabola opens downward, meaning it has a maximum at its vertex. The vertex form of a quadratic function is useful here. Let me find the vertex.The x-coordinate of the vertex is at -b/(2a) for a quadratic ax² + bx + c. Here, a is -1, b is 2. So x = -2/(2*(-1)) = -2 / (-2) = 1. So the vertex is at x = 1. Plugging this back into the quadratic to find the y-coordinate:f(1) = -(1)² + 2*1 + 3 = -1 + 2 + 3 = 4. So the maximum value of the quadratic is 4, which occurs at x = 1. Since the parabola opens downward, the range of the quadratic part will be from negative infinity up to 4. However, the domain for this part is x ≤ 2. Wait, but the vertex is at x=1, which is within the domain x ≤ 2. So as x approaches negative infinity, the quadratic tends to negative infinity (since the parabola opens downward), and the maximum at x=1 is 4. So the range of the quadratic part is indeed (-∞, 4], right? Because even though the domain is x ≤ 2, the function still attains all values from -∞ up to 4. Let me verify that.For x ≤ 2, the function is a downward parabola with vertex at (1,4). As x decreases from 1 towards negative infinity, f(x) decreases without bound. As x increases from 1 to 2, since the parabola is opening downward, f(x) decreases from 4 to f(2). Let me compute f(2):f(2) = -(2)² + 2*2 + 3 = -4 + 4 + 3 = 3. So at x=2, the quadratic part has a value of 3. So from x=1 to x=2, the function decreases from 4 to 3. Therefore, the range of the quadratic part is (-∞, 4], because as x approaches negative infinity, it goes to -∞, peaks at 4, and then decreases to 3 at x=2. So that's correct.Now, moving on to the logarithmic part: f(x) = 6 + log_a(x) for x > 2. The range of this part depends on the base a of the logarithm. Remember that logarithms have different behaviors depending on whether the base is greater than 1 or between 0 and 1.If a > 1, log_a(x) is increasing, and as x approaches infinity, log_a(x) tends to infinity, and as x approaches 2 from the right, log_a(x) approaches log_a(2). But since the domain here is x > 2, as x approaches 2 from the right, log_a(x) approaches log_a(2), and as x increases, log_a(x) increases without bound if a > 1. However, in this case, f(x) = 6 + log_a(x) would go from 6 + log_a(2) up to infinity, which would exceed the total range of (-∞, 4]. That's a problem because the total range is supposed to be up to 4. Therefore, if a > 1, the logarithmic part would have a minimum value at x=2 (since it's increasing) of 6 + log_a(2), and that must be less than or equal to 4. Otherwise, the logarithmic part would exceed 4, which is not allowed because the total range is (-∞,4]. So for a > 1, we need 6 + log_a(2) ≤ 4. Let's write that down:6 + log_a(2) ≤ 4 ⇒ log_a(2) ≤ -2.Converting this to exponential form: a^{-2} ≥ 2. Since a > 1, a^{-2} is 1/a². So 1/a² ≥ 2 ⇒ a² ≤ 1/2 ⇒ a ≤ 1/√2 ≈ 0.707. But wait, here we are considering a > 1, but this gives a ≤ 1/√2, which is less than 1. So there's no solution in this case. Therefore, if a > 1, the logarithmic function would start at 6 + log_a(2) and go to infinity, which would exceed 4, hence not allowed. So a cannot be greater than 1.Wait, but the answer choices include options where a >1 (C and D). Hmm, maybe I made a mistake here. Let me check again.Wait, when a >1, log_a(x) is increasing. So as x approaches 2 from the right, log_a(2) is a finite number, and as x increases, log_a(x) increases to infinity. Therefore, f(x) = 6 + log_a(x) would go from 6 + log_a(2) to infinity. But the total range of f(x) is (-∞,4], so the logarithmic part must not exceed 4. Therefore, 6 + log_a(x) ≤ 4 for all x >2. But since log_a(x) increases without bound as x increases, unless we restrict the base. Wait, that's impossible. So maybe if a >1, the logarithmic part would exceed 4, which is not allowed. Hence, a cannot be greater than 1.But the answer choices C and D are (1, sqrt(2}] and (1, sqrt(2)). So maybe my reasoning is wrong.Wait, maybe the problem is that when a >1, log_a(x) is increasing, so the minimum value is at x=2, which is log_a(2). Therefore, 6 + log_a(2) must be less than or equal to 4? Because otherwise, even the minimum value of the logarithmic part would be higher than 4. Wait, but 6 + log_a(2) ≤4 ⇒ log_a(2) ≤ -2. Which, as before, would imply a^{-2} ≥2 ⇒ a ≤1/√2. But a>1 in this case, so no solution. Therefore, there's no a>1 that satisfies this. But the answer options include intervals where a>1, so perhaps my initial assumption is wrong. Maybe I need to check when 0 <a <1.So let's consider 0 < a <1. In this case, log_a(x) is a decreasing function. Therefore, as x approaches 2 from the right, log_a(x) approaches log_a(2), and as x increases to infinity, log_a(x) tends to -infinity. Therefore, f(x) =6 + log_a(x) would go from 6 + log_a(2) down to -infinity. Therefore, the range of the logarithmic part is (-∞, 6 + log_a(2)).But the total range of f(x) is (-∞,4]. So we need to make sure that the upper bound of the logarithmic part, which is 6 + log_a(2), is less than or equal to 4. Otherwise, the logarithmic part would have values exceeding 4, but since the quadratic part already attains 4, and the total range is up to 4, maybe that's okay. Wait, actually, the total range is the union of both parts. The quadratic part already covers up to 4, so the logarithmic part can't exceed 4. However, the logarithmic part's maximum is 6 + log_a(2). Therefore, we must have 6 + log_a(2) ≤4. So log_a(2) ≤ -2. Let's solve this inequality.Since 0 < a <1, log_a(2) ≤ -2. Remember that for 0 < a <1, log_a(b) = ln(b)/ln(a), and since ln(a) is negative, log_a(b) is negative if b>1. Here, 2 >1, so log_a(2) is negative. Let's convert the inequality log_a(2) ≤ -2.This is equivalent to 2 ≥ a^{-2} (since for 0 <a <1, log_a(2) ≤ -2 ⇒ 2 ≥ a^{-2} by the property of logarithms with base less than 1). Let's verify:log_a(2) ≤ -2=> 2 ≥ a^{-2} (because when you apply the exponential function with base a, which is less than 1, the inequality reverses)Thus:a^{-2} ≤ 2=> (1/a²) ≤ 2=> 1/a² ≤ 2=> a² ≥ 1/2=> a ≥ 1/√2 ≈0.707But since 0 <a <1, the solution is 1/√2 ≤a <1.Therefore, the range of a is [1/√2, 1). But 1/√2 is equal to sqrt(2)/2, which is approximately 0.707. So written as [sqrt(2)/2, 1). Looking at the answer choices:A: (sqrt(2)/2, 1)B: [sqrt(2)/2, 1)C: (1, sqrt(2)]D: (1, sqrt(2))So the answer should be B. However, I need to confirm whether the endpoint 1/√2 is included or not. Let me check.When a =1/√2, then log_a(2) = log_{1/√2}(2). Let's compute that.log_{1/√2}(2) = ln(2)/ln(1/√2) = ln(2)/( - (1/2) ln(2)) ) = ln(2) / (-0.5 ln(2)) )= -2. So 6 + log_a(2) =6 -2=4. So when a=1/√2, the maximum of the logarithmic part is exactly 4, which matches the upper bound of the total range. Therefore, a=1/√2 is allowed. Hence, the interval should include 1/√2, so the answer is B.But let me check why the quadratic part's upper limit is 4, which is achieved at x=1. The logarithmic part, when a=1/√2, at x=2, it is 6 + log_{1/√2}(2) =6 -2=4, so at x=2, both the quadratic and logarithmic parts meet at y=3 and y=4? Wait, wait. Wait, the quadratic part at x=2 is f(2)=3, as we computed earlier. But the logarithmic part is defined for x>2. So at x approaching 2 from the right, the logarithmic part approaches 6 + log_a(2). So if a=1/√2, then 6 + log_a(2) =4. Therefore, as x approaches 2 from the right, f(x) approaches 4, but since x=2 is included in the quadratic part, which at x=2 is 3. So there's a discontinuity at x=2. But the range includes 4 because the quadratic part reaches 4 at x=1, and the logarithmic part approaches 4 as x approaches 2 from the right (when a=1/√2). However, since x=2 is included in the quadratic part, the value at x=2 is 3, but the limit from the right is 4. So the function approaches 4 but doesn't actually attain it in the logarithmic part. Wait, but the quadratic part already attains 4 at x=1, so the total range does include 4. So even though the logarithmic part approaches 4 as x approaches 2 from the right, the maximum value of 4 is already achieved by the quadratic part. Therefore, including a=1/√2 is acceptable because the upper bound of the logarithmic part is 4 (approached but not attained), but since the quadratic part attains 4, the total range still includes 4. Therefore, a=1/√2 is allowed, hence the interval is [sqrt(2)/2, 1).Therefore, the correct answer is B.But let me just make sure. Suppose a is less than 1/√2, say a=0.5. Then log_a(2) = log_0.5(2) = -1, because 0.5^{-1}=2. Then 6 + log_a(2) =6 -1=5, which is greater than 4, which would make the range of the logarithmic part go from 5 down to -infinity. But the total range would then be (-∞,5], which is not (-∞,4], so a cannot be less than 1/√2. If a is exactly 1/√2, then the upper limit of the logarithmic part is 4, so the total range is (-∞,4]. If a is between 1/√2 and 1, then log_a(2) is between -2 and 0. For example, take a=0.75. Then log_{0.75}(2). Let me compute that.log_{0.75}(2) = ln(2)/ln(0.75) ≈ 0.6931 / (-0.2877) ≈ -2.409. So 6 + (-2.409) ≈3.591, which is less than 4. Therefore, the maximum of the logarithmic part would be less than 4. But wait, if a is between 1/√2 and 1, then log_a(2) is between -2 and something. Wait, when a approaches 1 from below, log_a(2) approaches negative infinity, because log base approaching 1 of 2 is undefined. Wait, no. Wait, if a approaches 1 from below, log_a(2) tends to negative infinity? Wait, let's see:If a approaches 1 from below, then for a =1 - ε where ε is small positive, log_{1 - ε}(2) = ln(2)/ln(1 - ε). As ε approaches 0, ln(1 - ε) ≈ -ε, so log_{1 - ε}(2) ≈ ln(2)/(-ε) → -∞. Therefore, 6 + log_a(2) approaches -∞ as a approaches 1 from below. Wait, but for a in (1/√2,1), log_a(2) is less than -2? Wait, when a=1/√2, log_a(2)= -2. If a increases towards 1, then log_a(2) becomes more negative? Wait, no. Wait, if a increases from 1/√2 to 1, since a is approaching 1, the log base a increases, but since a is still less than 1, log_a(2) is negative and becomes more negative as a approaches 1. So for a=0.9, log_0.9(2) is log(2)/log(0.9) ≈0.6931/(-0.1054)≈-6.58. So 6 + (-6.58)= -0.58. So as a approaches 1 from below, the upper bound of the logarithmic part (which is 6 + log_a(2)) approaches negative infinity. Wait, no. Wait, for a between 0 and 1, log_a(x) is decreasing. So as x increases, log_a(x) decreases. Therefore, for x >2, as x approaches 2 from the right, log_a(x) approaches log_a(2), which is the maximum value of the logarithmic part (since it's decreasing). So 6 + log_a(2) is the maximum value of the logarithmic part. If a is between 1/√2 and 1, then log_a(2) is between -2 and -∞. Wait, but when a is 1/√2, log_a(2)= -2. When a increases towards 1, log_a(2) becomes more negative. So 6 + log_a(2) is between 4 (when a=1/√2) and negative infinity (as a approaches 1). Therefore, the upper bound of the logarithmic part is 4 when a=1/√2 and decreases as a increases. However, the quadratic part already covers up to 4, so even if the logarithmic part's upper bound is less than 4, the total range is still (-∞,4] because the quadratic part reaches 4. Therefore, as long as the logarithmic part doesn't exceed 4, which is already handled by a ≥1/√2, the total range remains (-∞,4].Therefore, the allowed values of a are [1/√2,1), which is option B.Final Answerboxed{B}</think>Given the piecewise function ( f(x) ):[f(x) = begin{cases} -x^2 + 2x + 3 & text{if } x leq 2 6 + log_a x & text{if } x > 2 end{cases}]we need to determine the range of ( a ) such that the range of ( f(x) ) is ((-infty, 4]). 1. Quadratic Part Analysis: - The quadratic function ( -x^2 + 2x + 3 ) has a vertex at ( x = 1 ). - The maximum value at the vertex is ( f(1) = 4 ). - As ( x to -infty ), ( f(x) to -infty ). - At ( x = 2 ), ( f(2) = 3 ). - The range of the quadratic part is ((-infty, 4]).2. Logarithmic Part Analysis: - For ( x > 2 ), the function is ( 6 + log_a x ). - The range of ( log_a x ) depends on the base ( a ): - If ( 0 < a < 1 ), ( log_a x ) is decreasing, so as ( x to infty ), ( log_a x to -infty ). - The maximum value of ( 6 + log_a x ) occurs as ( x to 2^+ ), which is ( 6 + log_a 2 ). - To ensure the range of ( f(x) ) is ((-infty, 4]), we need ( 6 + log_a 2 leq 4 ).3. Solving the Inequality: - ( 6 + log_a 2 leq 4 ) implies ( log_a 2 leq -2 ). - Converting to exponential form: ( 2 leq a^{-2} ) which simplifies to ( a^2 geq frac{1}{2} ). - Therefore, ( a geq frac{1}{sqrt{2}} ) or ( a geq frac{sqrt{2}}{2} ). - Since ( 0 < a < 1 ), the valid range is ( frac{sqrt{2}}{2} leq a < 1 ).4. Verification: - When ( a = frac{sqrt{2}}{2} ), ( log_a 2 = -2 ), so ( 6 + log_a 2 = 4 ), which is within the range. - For ( frac{sqrt{2}}{2} < a < 1 ), the maximum value of the logarithmic part is less than 4, ensuring the total range remains ((-infty, 4]).Thus, the range of ( a ) is (left[ frac{sqrt{2}}{2}, 1 right)), which corresponds to option B.[boxed{B}]]

🔑:To analyze the function f(x) given by f(x)=left{begin{array}{l}-{x}^{2}+2x+3, & xleqslant 2 6+log_{a}x, & x>2end{array}right., we need to consider the two parts of the function separately and then determine the range of a based on the given range of f(x), which is (-infty, 4].For xleqslant 2:The quadratic part of the function can be rewritten as:[f(x) = -x^{2}+2x+3 = -(x^2 - 2x - 3) = -(x-1)^{2} + 4]This form shows that the vertex of the parabola is at (1, 4), indicating that the function reaches its maximum at x=1 with f(1)=4. Since the coefficient of x^2 is negative, the parabola opens downwards, making f(x)leqslant 4 for xleqslant 2. Therefore, for this part of the function, f(x)in (-infty, 4].For x>2:The logarithmic part of the function requires that 6+log_{a}xleqslant 4 to ensure the entire function's range is (-infty, 4]. This inequality simplifies to:[log_{a}x leqslant -2]Depending on the base a of the logarithm, the behavior of the function changes:- When a>1, the function 6+log_{a}x is monotonically increasing for x>2. However, for any x>2, f(x) would be greater than 6, which contradicts the requirement that f(x)leqslant 4. Therefore, a cannot be greater than 1.- When 02. To satisfy 6+log_{a}xleqslant 4, we need log_{a}2leqslant -2. This can be rewritten using the property of logarithms as log_{a}2 leqslant log_{a}a^{-2}, which implies 2 leqslant a^{-2}. Solving for a gives a geq frac{sqrt{2}}{2}, and since we are considering 0<a<1, we conclude that frac{sqrt{2}}{2} leq a < 1.Combining the analysis for both parts of the function, we find that to ensure the range of f(x) is (-infty, 4], the range of a must be [frac{sqrt{2}}{2}, 1). Therefore, the correct answer is boxed{B}.

🧠:[<think>Okay, so I need to find the number of "good subsets" of the set S = {1, 2, ..., 10}. A good subset is defined as a non-empty subset where the number of even numbers is no less than the number of odd numbers. The answer choices are 631, 633, 635, 637. Let's see how to approach this.First, let's understand the problem. We have a set S with numbers from 1 to 10. Each element is either even or odd. The subset is "good" if the count of even numbers in it is at least the count of odd numbers. Also, the subset must be non-empty. So, we need to count all non-empty subsets where #evens ≥ #odds.First, let's figure out how many even and odd numbers are in S. From 1 to 10, there are 5 even numbers (2,4,6,8,10) and 5 odd numbers (1,3,5,7,9). So, in total, 5 even and 5 odd. Therefore, in any subset, the number of evens (E) and odds (O) must satisfy E ≥ O, with E + O ≥ 1 (since it's non-empty).So, the problem reduces to counting all subsets (excluding the empty set) where E ≥ O. Let's think about how to compute this.One way to approach this is to consider all possible subsets and categorize them based on the number of even and odd elements they contain. Since the even and odd elements are distinct, the choices of even and odd elements are independent. That is, the number of subsets with E even numbers and O odd numbers is C(5, E) * C(5, O), where C(n, k) is the combination of n elements taken k at a time.Therefore, the total number of good subsets would be the sum over all E and O such that E ≥ O and E + O ≥ 1 of C(5, E) * C(5, O).So, we can structure this as:Total good subsets = Σ (from E=0 to 5) Σ (from O=0 to min(E,5)) [C(5, E) * C(5, O)] - 1Wait, hold on. Wait, actually, since E has to be at least O. So for each E from 0 to 5, O can range from 0 to E, but O cannot exceed 5, which is already satisfied since E is at most 5.But also, we need E + O ≥ 1. However, when we compute the sum over E ≥ O, including E=0 and O=0 would give the empty set, which we need to subtract. But let me check:If E ranges from 0 to 5, and for each E, O ranges from 0 to E. But E ≥ O, so O can be from 0 to E. But when E=0, O can only be 0. So the total sum would include E=0, O=0 (empty set) and all other subsets where E ≥ O. Then, subtract 1 to remove the empty set.Alternatively, maybe we can model it as:Total = [Σ_{E=0}^5 Σ_{O=0}^E C(5,E) * C(5,O)] - 1But we need to confirm if the double summation is correctly capturing all subsets where E ≥ O.Yes, because for each E, O can be from 0 to E (but O cannot exceed 5, which it doesn't because E is up to 5). So this should be correct. Then subtract 1 to remove the empty set.Alternatively, since each element is either even or odd, the total number of subsets is 2^10, so 1024. The empty set is one, so non-empty subsets are 1023. But we need to count how many of these 1023 subsets have E ≥ O.Alternatively, another approach is to realize that for each subset, either E ≥ O or O > E. Since the total number of non-empty subsets is 1023, then if we can compute the number of subsets where E ≥ O, it's equal to (Total non-empty subsets + number of subsets where E = O) / 2. Wait, no. Wait, if E and O are not equal, then either E > O or O > E. But if E = O, then it's counted as E ≥ O. So, maybe:Number of good subsets = (Total non-empty subsets + Number of subsets with E = O) / 2But wait, let's check:Total non-empty subsets = subsets with E > O + subsets with O > E + subsets with E = O.But subsets where E ≥ O would be subsets with E > O + subsets with E = O. Similarly, subsets where O ≥ E would be subsets with O > E + subsets with E = O. Since E and O are symmetric except for the counts here. Wait, but in this case, the number of even and odd elements in the original set is equal (5 each). So, perhaps the number of subsets where E > O is equal to the number of subsets where O > E. Therefore, subsets where E > O = subsets where O > E. Let me see.Yes, because if you take a subset with E > O, you can swap the even and odd elements to get a subset with O > E. Since there are 5 evens and 5 odds, this should be a bijection. Therefore, the number of subsets with E > O is equal to the number of subsets with O > E. Therefore, since the total non-empty subsets is 1023, which is equal to subsets with E > O + subsets with O > E + subsets with E = O. Let N be the number of subsets with E > O, so 2N + M = 1023, where M is the number of subsets with E = O. Therefore, the number of good subsets is N + M. Since we need N + M = (1023 - M)/2 + M = (1023 + M)/2.Therefore, Number of good subsets = (1023 + M)/2, where M is the number of subsets with E = O. Therefore, if we can compute M, the number of subsets with equal even and odd numbers, then we can compute the answer.So, M is the number of subsets where E = O. Since the subset is non-empty, E and O can be from 1 to 5 (since E and O must be equal, and at least 1). Wait, no: E and O can be 0 as well, but if E = O = 0, that's the empty set. So, M (including empty set) is Σ_{k=0}^5 C(5,k) * C(5,k). Because for each k, choosing k evens and k odds. So, total with E = O (including empty set) is Σ_{k=0}^5 [C(5,k)]^2. Then, M (excluding empty set) is Σ_{k=1}^5 [C(5,k)]^2. Wait, but the problem states non-empty subsets, so the empty set is excluded. Therefore, if we compute M as Σ_{k=1}^5 [C(5,k)]^2, then Number of good subsets is (1023 + M)/2.But let's check. If M is the number of subsets with E = O (including empty set), then the number of non-empty subsets with E = O is M - 1. Wait, confusion arises here. Let's clarify.Let M_total be the number of subsets with E = O, including the empty set. Then M_total = Σ_{k=0}^5 [C(5,k)]^2. Then the number of non-empty subsets with E = O is M = M_total - 1. Then, according to previous logic, the number of good subsets would be (1023 + M)/2. Wait, no. Wait, in the previous logic, if we have total non-empty subsets as 1023, which is equal to subsets with E > O + subsets with O > E + subsets with E = O (non-empty). So, 1023 = N + N + M, where N is subsets with E > O and N is subsets with O > E, and M is non-empty E = O. Therefore, solving for N + M (good subsets) gives N + M = (1023 - M)/2 + M = (1023 + M)/2. Therefore, yes, the number of good subsets is (1023 + M)/2, where M is the number of non-empty subsets with E = O.Therefore, M is Σ_{k=1}^5 [C(5,k)]^2. Let's compute that.First, compute M_total = Σ_{k=0}^5 [C(5,k)]^2. There is a combinatorial identity that Σ_{k=0}^n [C(n,k)]^2 = C(2n, n). So, when n = 5, Σ_{k=0}^5 [C(5,k)]^2 = C(10,5) = 252. Therefore, M_total = 252. Therefore, the number of non-empty subsets with E = O is 252 - 1 = 251. Wait, but hold on. Wait, the identity Σ_{k=0}^n [C(n,k)]^2 = C(2n, n) is correct. So for n=5, it's 252. Therefore, the total number of subsets (including empty set) with E = O is 252. Therefore, the number of non-empty subsets with E = O is 252 - 1 = 251. Therefore, M = 251.Therefore, the number of good subsets is (1023 + 251)/2 = (1274)/2 = 637. Wait, but 1023 + 251 is 1274? Let's check: 1000 + 200 = 1200, 23 + 51 = 74, so total 1274. Divided by 2 is 637. Hmm, but 637 is option D. However, the answer choices are (A) 631, (B) 633, (C) 635, (D) 637. So, according to this calculation, the answer is D) 637. But let me verify because sometimes these combinatorial problems have nuances.Wait, but let's check again: the total number of non-empty subsets is 2^10 - 1 = 1023. Then, number of subsets with E = O (including empty set) is 252, so non-empty is 251. Then, according to the logic, the number of good subsets is (1023 + 251)/2 = 637. So that's D. But let's check if that's correct.Alternatively, maybe the error is in the assumption that the number of subsets with E > O is equal to the number with O > E. Is that correct? Since there are 5 even and 5 odd numbers, so the set is symmetric in evens and odds. Therefore, swapping even and odd elements should give a bijection between subsets with E > O and O > E. Therefore, their counts should be equal. Therefore, the logic holds. Thus, the calculation should be correct.But wait, let's think differently. Let's compute the number directly.Compute the number of good subsets as Σ_{E=0}^5 Σ_{O=0}^E C(5,E)*C(5,O) - 1 (subtracting the empty set). Let's compute this.For E from 0 to 5:When E=0: O can be 0. So C(5,0)*C(5,0) = 1*1=1. But since E=0 and O=0 is the empty set, which we need to subtract later.When E=1: O can be 0 or 1. So C(5,1)*[C(5,0) + C(5,1)] = 5*(1 + 5) = 5*6=30When E=2: O can be 0,1,2. So C(5,2)*[C(5,0)+C(5,1)+C(5,2)] = 10*(1 +5 +10)=10*16=160When E=3: O can be 0,1,2,3. So C(5,3)*[C(5,0)+C(5,1)+C(5,2)+C(5,3)] = 10*(1 +5 +10 +10)=10*26=260Wait, hold on: C(5,3) is 10, same as C(5,2). Then sum inside is 1+5+10+10=26. So 10*26=260.When E=4: O can be 0,1,2,3,4. So C(5,4)*[C(5,0)+C(5,1)+C(5,2)+C(5,3)+C(5,4)] = 5*(1 +5 +10 +10 +5)=5*31=155When E=5: O can be 0,1,2,3,4,5. So C(5,5)*[C(5,0)+C(5,1)+C(5,2)+C(5,3)+C(5,4)+C(5,5)] =1*(1 +5 +10 +10 +5 +1)=1*32=32Now, summing these up:E=0: 1E=1:30E=2:160E=3:260E=4:155E=5:32Total sum: 1 +30 +160 +260 +155 +32.Let's compute step by step:1 +30 =3131 +160=191191 +260=451451 +155=606606 +32=638Then subtract 1 for the empty set: 638 -1=637. So same answer as before. Therefore, the number of good subsets is 637, which is option D.But wait, but the answer choices are A)631, B)633, C)635, D)637. So, according to both methods, the answer is D)637.But let me check the calculation again because sometimes when adding up, mistakes can happen.Let me recheck the values:For E=0:1E=1:5*(1+5)=30E=2:10*(1+5+10)=10*16=160E=3:10*(1+5+10+10)=10*26=260E=4:5*(1+5+10+10+5)=5*31=155E=5:1*(1+5+10+10+5+1)=1*32=32Total:1+30=31; 31+160=191; 191+260=451; 451+155=606; 606+32=638. Then subtract 1:637. Correct.So, that seems right. Alternatively, if we use the other method:Total non-empty subsets:1023Number of subsets with E=O:252-1=251Thus, number of good subsets=(1023 +251)/2= (1274)/2=637. Same answer.Therefore, the answer is D)637.But why is the answer not among the first three options? Maybe I made a miscalculation. Wait, maybe the problem is that in the original problem, the definition is "the number of even numbers in a non-empty subset is no less than the number of odd numbers". So, E >= O.But when E=0, O must be <= E=0, so O=0. But E=0 and O=0 is the empty set. So when E=0, the only subset is the empty set, which is excluded. Therefore, in the calculation above, when we did the double summation over E and O with E>=O, and then subtracted 1 (the empty set), that's correct. But let's check.In the first approach, when E=0, O can only be 0. So the total sum includes E=0,O=0 which is the empty set. Then we subtract 1, so that's correct.Alternatively, if we think of it as the problem says "non-empty subsets", so we need to exclude the empty set. Therefore, the calculation is correct.Alternatively, let's consider a smaller case to verify.Suppose S={1,2}. Then S has 2 elements: 1 is odd, 2 is even. Compute the number of good subsets. The non-empty subsets:{1}: E=0, O=1. Not good.{2}: E=1, O=0. Good.{1,2}: E=1, O=1. Good.Total good subsets:2. Let's see if our formula works.Total non-empty subsets=3.Number of subsets with E=O (non-empty):1 (the subset {1,2}).Thus, according to formula: (3 +1)/2=2. Correct.Alternatively, using the direct summation:E=0: O=0 (empty set). Not counted.E=1: O can be 0 or 1. So C(1,1)*[C(1,0)+C(1,1)]=1*(1+1)=2. Subtract 1 (empty set). Wait, no. Wait, in the direct summation for S={1,2}, the even numbers are {2}, odds {1}.Wait, for E=0: subsets with 0 evens and O=0 (empty set) and O<=0. So only empty set.E=1: subsets with 1 even and O from 0 to1. So {2} (O=0) and {2,1} (O=1). So two subsets. Therefore, total is 1 (E=0,O=0) +2 (E=1, O=0 and 1). Then subtract 1 (empty set) gives 2, which is correct. So the formula works here.Similarly, if S={1}, then there's 1 non-empty subset: {1}, which has E=0, O=1. Not good. So number of good subsets=0. Let's check with formula:Total non-empty subsets=1.Number of subsets with E=O:0 (since you need E=O, but E can be 0 or 1, but O would have to be same, but {1} has E=0, O=1, so no non-empty subsets with E=O. So (1 +0)/2=0.5, which is not integer. Wait, something's wrong here.Wait, in the case of S={1}, which has only one element (odd), the total non-empty subsets is 1. Since all subsets have E=0 and O=1, which is not E >= O, so good subsets=0. According to our formula, M (number of non-empty subsets with E=O) is 0, so (1 +0)/2=0.5, which is wrong. So the formula must be adjusted for cases where the counts are different. Wait, but in the original problem, we have equal numbers of even and odd elements. In S={1}, we have 1 odd and 0 even. So the formula may not hold. Therefore, the formula (Total + M)/2 works only when the number of even and odd elements are equal?Wait, in the previous problem, the original set S has 5 even and 5 odd, so symmetric. Therefore, the bijection between E > O and O > E holds. But in the case when the numbers are not equal, this symmetry breaks.Therefore, in the original problem, since we have 5 even and 5 odd, the formula (Total non-empty + M)/2 should hold. However, in the case when the numbers are not equal, it may not. So in the original problem, since it's symmetric, it's okay.But let's check another small case where the original set has equal evens and odds.Take S={1,2,3,4}. So 2 even (2,4) and 2 odd (1,3). Compute the number of good subsets.Non-empty subsets where E >= O.Total non-empty subsets:15.Compute manually:All subsets except empty set:Size 1: {1} (O=1, E=0) bad; {2} (E=1, O=0) good; {3} (O=1, E=0) bad; {4} (E=1, O=0) good. So 2 good.Size 2: {1,2} (O=1, E=1) good; {1,3} (O=2, E=0) bad; {1,4} (O=1, E=1) good; {2,3} (E=1, O=1) good; {2,4} (E=2, O=0) good; {3,4} (E=1, O=1) good. So total good size 2: {1,2}, {1,4}, {2,3}, {2,4}, {3,4} => 5.Size 3: {1,2,3} (E=1, O=2) bad; {1,2,4} (E=2, O=1) good; {1,3,4} (E=1, O=2) bad; {2,3,4} (E=2, O=1) good; {1,2,3,4} (E=2, O=2) good. Wait, but size 3 subsets:Wait, in S={1,2,3,4}, subsets of size 3:{1,2,3}: E=1, O=2: bad.{1,2,4}: E=2, O=1: good.{1,3,4}: E=1, O=2: bad.{2,3,4}: E=2, O=1: good.Size 3: 2 good.Size 4: {1,2,3,4}: E=2, O=2: good.So total good subsets:Size1:2Size2:5Size3:2Size4:1Total:2+5+2+1=10.Now, according to formula:Total non-empty subsets=15.Number of non-empty subsets with E=O: Let's compute M.Subsets with E=O (including empty set): Σ_{k=0}^2 [C(2,k)]^2 = C(4,2)=6. So including empty set, 6. Therefore, non-empty subsets with E=O:6-1=5.Therefore, formula gives (15 +5)/2=10. Which matches. So the formula works here.Therefore, returning to the original problem, the formula (1023 +251)/2=637 is correct.Alternatively, another way to compute M is Σ_{k=1}^5 [C(5,k)]^2. As we know that Σ_{k=0}^5 [C(5,k)]^2=252, so Σ_{k=1}^5 [C(5,k)]^2=252 -1=251. Therefore, M=251.Therefore, the answer is D)637. Therefore, I think the answer is D.But since the options are given, and my calculation leads to 637, which is option D, and the step-by-step seems correct, probably D is the answer. But let me check once more to be absolutely sure.Alternative approach:Let’s compute the number of good subsets directly by considering all possible pairs where E ≥ O.For each possible number of odds O from 0 to5, and for each O, number of evens E can be from O to5. So for each O, compute C(5,O)*Σ_{E=O}^5 C(5,E). Then sum over O=0 to5, and subtract 1 for the empty set.But let's try:For O=0: E can be 1 to5. So C(5,0)*Σ_{E=1}^5 C(5,E)=1*(2^5 -1)=31Wait, Σ_{E=0}^5 C(5,E)=2^5=32. So Σ_{E=1}^5=31.For O=1: E can be1 to5. C(5,1)*Σ_{E=1}^5 C(5,E)=5*31=155For O=2: E can be2 to5. C(5,2)*Σ_{E=2}^5 C(5,E)=10*(C(5,2)+C(5,3)+C(5,4)+C(5,5))=10*(10+10+5+1)=10*26=260For O=3: E=3 to5. C(5,3)*Σ_{E=3}^5 C(5,E)=10*(10+5+1)=10*16=160For O=4: E=4 to5. C(5,4)*Σ_{E=4}^5 C(5,E)=5*(5+1)=5*6=30For O=5: E=5. C(5,5)*C(5,5)=1*1=1Summing these:O=0:31O=1:155O=2:260O=3:160O=4:30O=5:1Total:31+155=186; 186+260=446; 446+160=606; 606+30=636; 636+1=637. Then subtract 1? Wait, no, because in this approach, O=0, E>=1. So the empty set is only when O=0 and E=0, which is excluded here. So, in this calculation, O=0 includes E=1 to5, which are non-empty. O=1 to5 with E>=O are also non-empty. Therefore, the total is 637, without subtracting anything. Wait, why?Wait, in this approach, for each O from0 to5, and E >= O, but with E >=1 when O=0 (since O=0, E >=0, but since subsets are non-empty, E must be >=1). So in this case, we are already counting all subsets where E >= O and non-empty. Therefore, the total is 637. So same answer.Therefore, regardless of the method, the answer is 637, which is D.Thus, the correct answer is D)637.Final Answerboxed{D}</think>To determine the number of "good subsets" of the set ( S = {1, 2, ldots, 10} ), we need to count all non-empty subsets where the number of even numbers is at least the number of odd numbers. First, we note that the set ( S ) contains 5 even numbers and 5 odd numbers. A subset is "good" if the number of even numbers ( E ) is at least the number of odd numbers ( O ). We use the combinatorial identity that the total number of subsets where ( E = O ) (including the empty set) is given by ( sum_{k=0}^{5} binom{5}{k}^2 = binom{10}{5} = 252 ). Subtracting the empty set, we get 251 non-empty subsets where ( E = O ).The total number of non-empty subsets of ( S ) is ( 2^{10} - 1 = 1023 ). Since the number of subsets where ( E > O ) is equal to the number of subsets where ( O > E ), we can use the formula:[text{Number of good subsets} = frac{text{Total non-empty subsets} + text{Number of subsets with } E = O}{2}]Substituting the values, we get:[text{Number of good subsets} = frac{1023 + 251}{2} = frac{1274}{2} = 637]We verified this result by directly summing the number of valid subsets for each possible value of ( E ) and ( O ) and confirmed the result using different methods. Both approaches led to the same conclusion.Thus, the number of "good subsets" is (boxed{D}).,<think>Alright, let's tackle this problem step by step. So, we have the set S = {1, 2, ..., 10}, and we need to find the number of non-empty subsets where the number of even numbers is no less than the number of odd numbers. These are called "good subsets". The answer choices are given, so we need to compute this number and see which one it matches. Let me break this down.First, let's recall that a subset of S can have any number of elements from 1 to 10. Each element is either included or excluded. The total number of subsets is 2^10 = 1024, including the empty set. But we need non-empty subsets, so that would be 1023. However, we have this additional condition regarding the number of even and odd numbers in each subset.Let me note down some key points:1. The set S has 10 elements: 1 through 10.2. In S, there are 5 even numbers (2, 4, 6, 8, 10) and 5 odd numbers (1, 3, 5, 7, 9).3. A "good subset" is a non-empty subset where the number of even numbers (let's denote this as E) is at least the number of odd numbers (denoted as O). So, E ≥ O.Therefore, we need to count all non-empty subsets where E ≥ O. Since E and O are non-negative integers, and E + O is the size of the subset (which is at least 1). Let's think about how to approach this.One way to approach subset counting problems with conditions is to use combinatorial methods, possibly considering generating functions or symmetry arguments. Let me consider the possible approaches.First, let's denote the number of even numbers in the set as 5 and odd numbers as 5. For any subset, we can choose k even numbers and l odd numbers, where k ranges from 0 to 5 and l ranges from 0 to 5, but k + l ≥ 1 (since the subset is non-empty). The condition for a good subset is k ≥ l. So, we need to compute the sum over all k and l where k ≥ l and k + l ≥ 1 of (5 choose k)*(5 choose l). Then, subtract 1 if necessary (for the empty set), but since we are already considering non-empty subsets, maybe not.Wait, actually, the problem specifies "non-empty subsets", so we don't have to subtract the empty set. So, the total number of good subsets would be the sum over k from 0 to 5 and l from 0 to 5, with k ≥ l and k + l ≥ 1, of (5 choose k)*(5 choose l).But that seems a bit complicated. Alternatively, we can note that the number of subsets with k evens and l odds is C(5, k)*C(5, l), and we need to sum over all k ≥ l where k + l ≥ 1.Alternatively, maybe there's a generating function approach. Let's think about generating functions for even and odd elements.The generating function for the even numbers is (1 + x)^5, since each even number can be either included or not. Similarly, the generating function for the odd numbers is (1 + x)^5. To find the number of subsets where the number of evens is at least the number of odds, we need to consider the coefficient of x^m in the product of these generating functions where m ≥ 2l (since if we have l odds, then we need evens to be at least l). Wait, no, more accurately, for each term in the expansion, the exponent is the total number of elements, but we need to relate the exponents contributed by the even and odd parts.Wait, maybe another way. Let's separate the generating functions for evens and odds. Let’s denote E(x) = (1 + x)^5 for the evens and O(x) = (1 + x)^5 for the odds. The total generating function is E(x)*O(x) = (1 + x)^10. But we need to find the sum over all subsets where the number of evens (let's say 'k') is at least the number of odds ('l'). So, we need to compute the sum over k and l with k ≥ l and k + l ≥ 1 of C(5, k)*C(5, l).Alternatively, since the set is split evenly into 5 evens and 5 odds, maybe there's a symmetry here. Let's consider that for each subset, we can pair it with its complement. But not sure if that helps.Alternatively, think about the problem as having two independent sets: the evens and the odds. For each subset, we can choose some number of evens and some number of odds. The condition is that the number of evens is at least the number of odds. So, the problem reduces to counting the number of pairs (k, l) where k is from 0 to 5, l is from 0 to 5, k ≥ l, and k + l ≥ 1. Each such pair contributes C(5, k)*C(5, l) subsets.Therefore, we can write the total number of good subsets as:Sum_{k=0 to 5} Sum_{l=0 to min(k,5)} C(5, k)*C(5, l) - 1 (if we include the case k=0 and l=0). Wait, but since k + l ≥ 1, we don't need to subtract anything. Wait, no. The term k=0 and l=0 is the empty set, which is excluded. So, actually, the total number of good subsets is:Sum_{k=0 to 5} Sum_{l=0 to k} C(5, k)*C(5, l) - 1 (to subtract the empty set). Because in the double sum from l=0 to k and k=0 to 5, we include the case k=0 and l=0, which is the empty set. So, to get non-empty subsets, we subtract 1.Wait, but is that correct? Let's check.If we do Sum_{k=0 to 5} Sum_{l=0 to k} C(5, k)*C(5, l), this includes all subsets where the number of evens is at least the number of odds, including the empty set (when k=0 and l=0). Then, subtracting 1 gives the number of non-empty good subsets. So yes, that formula would work.But let's verify. For example, when k=0, l can only be 0 (since l ≤ k=0). So that's just 1 subset: the empty set. Then, when k=1, l can be 0 or 1. But wait, l is the number of odds. Wait, no, hold on. Wait, in this setup, k is the number of evens, and l is the number of odds. Wait, but in the problem statement, the "good subset" requires that the number of evens is at least the number of odds, i.e., k ≥ l. Therefore, for each k (number of evens), l can range from 0 to k (but since there are only 5 odds, l is from 0 to min(k,5)). However, since k ranges up to 5, and l can be up to 5, but l ≤ k, and k can be up to 5. So actually, l ranges from 0 to k for each k from 0 to 5.But when k=5, l can be from 0 to 5. Wait, but k is the number of evens, which is up to 5, and l is the number of odds, which is up to 5. So, for example, when k=5, l can be from 0 to 5, as 5 ≥ l. However, the problem is that when k=5, we need l ≤ 5, which it already is. So, actually, Sum_{k=0 to 5} Sum_{l=0 to k} C(5, k)*C(5, l) includes all possible subsets where the number of evens is at least the number of odds, including the empty set. Then, subtract 1 to remove the empty set. Therefore, the total number of good subsets is:Total = [Sum_{k=0 to 5} Sum_{l=0 to k} C(5, k)*C(5, l)] - 1So, our job is to compute this double sum and subtract 1. Let's compute this step by step.First, let's compute the inner sum for each k: Sum_{l=0 to k} C(5, l). Because for each k (number of evens), we sum over l from 0 to k the number of ways to choose l odds. However, note that l cannot exceed 5, but since k is up to 5, and l ≤ k, it's okay.But hold on: C(5, l) is the number of ways to choose l odds from 5. So, if l exceeds 5, it would be zero, but since l ≤ k ≤5, it's okay.Therefore, for each k from 0 to 5, compute C(5, k) * [Sum_{l=0 to k} C(5, l)]. Then sum over k from 0 to 5, and subtract 1.Let me compute each term for k=0 to 5:For k=0:C(5, 0) * [Sum_{l=0 to 0} C(5, l)] = 1 * C(5, 0) = 1*1 = 1For k=1:C(5, 1) * [Sum_{l=0 to 1} C(5, l)] = 5 * (1 + 5) = 5*6 = 30For k=2:C(5, 2) * [Sum_{l=0 to 2} C(5, l)] = 10 * (1 + 5 + 10) = 10*16 = 160Wait, hold on. Wait, Sum_{l=0 to 2} C(5, l) is C(5,0) + C(5,1) + C(5,2) = 1 + 5 + 10 = 16. So 10 * 16 = 160. That seems right.For k=3:C(5, 3) * [Sum_{l=0 to 3} C(5, l)] = 10 * (1 + 5 + 10 + 10) = 10*26 = 260Wait, Sum_{l=0 to 3} C(5, l) = 1 + 5 + 10 + 10 = 26? Wait, C(5,3)=10, yes. So 10*26=260.For k=4:C(5,4) * [Sum_{l=0 to 4} C(5, l)] = 5 * (1 + 5 + 10 + 10 + 5) = 5*31 = 155Sum_{l=0 to 4} C(5, l) = 1 + 5 +10 +10 +5 =31. So, 5*31=155.For k=5:C(5,5) * [Sum_{l=0 to 5} C(5, l)] = 1 * (1 + 5 +10 +10 +5 +1) =1*32=32Sum_{l=0 to5} C(5,l)=32. So, 1*32=32.Now, adding all these up:k=0:1k=1:30k=2:160k=3:260k=4:155k=5:32Total sum =1 +30=31; 31 +160=191; 191 +260=451; 451 +155=606; 606 +32=638.Then subtract 1 for the empty set: 638 -1=637.Wait, but 637 is one of the options, option D. But hold on, let me check my calculations again because the answer is among the choices, but let me verify each step for possible errors.First, verifying the individual terms:For k=0: C(5,0)=1, Sum_{l=0 to 0}=C(5,0)=1. So 1*1=1. Correct.k=1: C(5,1)=5, Sum_{l=0 to1}=1+5=6. 5*6=30. Correct.k=2: C(5,2)=10, Sum_{l=0 to2}=1+5+10=16. 10*16=160. Correct.k=3: C(5,3)=10, Sum_{l=0 to3}=1+5+10+10=26. 10*26=260. Correct.k=4: C(5,4)=5, Sum_{l=0 to4}=1+5+10+10+5=31. 5*31=155. Correct.k=5: C(5,5)=1, Sum_{l=0 to5}=32. 1*32=32. Correct.Adding them up:1+30=31, +160=191, +260=451, +155=606, +32=638. Then subtract 1:637. So, 637. That's option D. Hmm. But wait, the answer options are (A)631 (B)633 (C)635 (D)637. So, 637 is D. But let me think again—maybe my approach is incorrect?Wait, but another way to approach this problem is to note that for each subset, the number of evens can be from 0 to5 and the number of odds from 0 to5. The condition is E ≥ O. The total number of subsets where E ≥ O is equal to half of the total subsets plus half the difference between subsets with E=O. Wait, that might be a standard combinatorial identity.In other words, for symmetric problems where you have two groups with equal size, the number of subsets where one group has at least as many as the other can be expressed as (Total + Number of subsets with E=O)/2. Because for every subset where E > O, there's a complementary subset where O > E, except when E=O.Wait, actually, the number of subsets where E > O is equal to the number of subsets where O > E, due to the symmetry between evens and odds here (since there are 5 evens and 5 odds). So, the total number of subsets (including empty set) is 2^10 =1024. The number of subsets where E > O is equal to the number where O > E, so each is (1024 - N)/2, where N is the number of subsets with E=O. Therefore, the number of subsets with E ≥ O is equal to (1024 - N)/2 + N = (1024 + N)/2.Therefore, another way to compute the number of good subsets (including empty set) is (1024 + N)/2, where N is the number of subsets with E=O.Then, subtract 1 if the empty set is included in that count.Wait, let's see. Let's compute N, the number of subsets with E=O.For subsets with E=O, we have k evens and k odds, where k can be from 0 to5 (since there are 5 evens and 5 odds). Therefore, N = Sum_{k=0 to5} C(5,k)*C(5,k). Let's compute that.C(5,0)*C(5,0)=1*1=1C(5,1)*C(5,1)=5*5=25C(5,2)*C(5,2)=10*10=100C(5,3)*C(5,3)=10*10=100C(5,4)*C(5,4)=5*5=25C(5,5)*C(5,5)=1*1=1Sum these up:1+25=26, +100=126, +100=226, +25=251, +1=252. So N=252.Therefore, the number of subsets with E ≥ O (including empty set) is (1024 +252)/2 = 1276/2 = 638. Then, subtract 1 to exclude the empty set:638 -1=637. So, same result. Therefore, the answer is D)637.Wait, so both approaches give 637, which is option D. That seems solid.But let me just check if the empty set was considered in N. When k=0, E=O=0, so the empty set is included in N. Then, when we compute (1024 +252)/2=638, which includes the empty set, so subtract 1 gives 637 non-empty good subsets. Therefore, correct.Alternatively, maybe my first approach was correct. So, either way, answer is D)637.But let me cross-verify with another method. Let's compute the number of good subsets by considering the possible numbers of elements.For a subset to be good, the number of evens E must be ≥ number of odds O. Let's note that E - O ≥0. Let's consider all possible E and O such that E + O ≥1 and E ≥ O.Alternatively, think of E and O as variables where E ranges from 0 to5, O ranges from0 to5, E >= O, E + O >=1.Alternatively, fix O and compute E from O to min(5, 10 - O). Wait, maybe not necessary.Alternatively, for each possible number of odds l (from0 to5), compute the number of evens k >= l, with k from l to5, and sum over l.Wait, that's similar to my first approach, but switching the order of summation.So, total good subsets = Sum_{l=0 to5} Sum_{k=l to5} C(5,k)*C(5,l) -1 (for empty set). Let's compute it this way.For l=0:k can be 0 to5. But since E >= O=0, so k >=0. But since we need E + O >=1, when l=0, k can be1 to5. Wait, actually, no. Because when l=0, k can be0 to5, but E >= O=0 implies k >=0. However, E + O >=1 implies that either k >=1 or l >=1. So, for l=0, the possible k is1 to5. Similarly, for l >=1, k can be from l to5. Wait, perhaps my original approach included l=0 and k=0, which is the empty set. So perhaps the correct way is:Total good subsets = [Sum_{l=0 to5} Sum_{k=max(l,1) to5} C(5,k)*C(5,l)}] when l=0, k starts at1, else k starts at l.But this seems complicated. Alternatively, just use the earlier approach where we computed the entire sum and subtracted 1. Since we know that the total with E >= O including empty set is638, subtract1 gives637. So, same answer.Therefore, regardless of the approach, the answer is637, option D.Alternatively, maybe there's a mistake in assuming that the number of subsets with E >= O is (1024 +252)/2. Let me check this identity.In general, for any symmetric situation where you have two groups with the same size, the number of subsets where the first group has more elements than the second is equal to (Total - N)/2, where N is the number of subsets with equal elements. Then, the number of subsets where the first is >= the second is equal to (Total - N)/2 + N = (Total + N)/2. Yes, that's a standard identity. Therefore, in our case, with groups of size5 and5, the number of subsets with E >= O is(1024 +252)/2=638, including the empty set. So subtract1, get637. So, correct.Alternatively, the number of subsets with E=O is252. So, the number of non-empty subsets with E=O is252 -1=251 (subtracting the empty set). Then, the remaining subsets (1024 -252=772) are split equally between E > O and O > E. So772 /2=386 subsets with E > O. Then, the number of good subsets is386 +251=637. Yes, that's another way to compute it. Therefore, 386 +251=637. Correct.Therefore, regardless of the method, the answer is637, option D. So I think that is the correct answer.Final Answerboxed{D}</think>To determine the number of "good subsets" of the set ( S = {1, 2, ldots, 10} ), we need to count all non-empty subsets where the number of even numbers is at least the number of odd numbers. First, we note that the set ( S ) contains 5 even numbers (2, 4, 6, 8, 10) and 5 odd numbers (1, 3, 5, 7, 9). A "good subset" is defined as a non-empty subset where the number of even numbers ( E ) is at least the number of odd numbers ( O ).We use the combinatorial identity that the number of subsets where ( E geq O ) is half of the total subsets plus half the number of subsets where ( E = O ). The total number of subsets of ( S ) is ( 2^{10} = 1024 ). The number of subsets where ( E = O ) is calculated as follows:[N = sum_{k=0}^{5} binom{5}{k} binom{5}{k} = 1 + 25 + 100 + 100 + 25 + 1 = 252]Using the identity, the number of subsets where ( E geq O ) (including the empty set) is:[frac{1024 + 252}{2} = 638]Subtracting 1 to exclude the empty set, we get:[638 - 1 = 637]Thus, the number of "good subsets" is (boxed{D}).]

🔑:Method 1: Let a "good subset" contain i (i=1,2,3,4,5) even numbers, then the number of odd numbers can be j=0,1, cdots, i. Therefore, the number of "good subsets" isbegin{aligned}sum_{i=1}^{5}left(mathrm{C}_{5}^{i} sum_{j=0}^{i} mathrm{C}_{5}^{j}right)= & mathrm{C}_{5}^{1}left(mathrm{C}_{5}^{0}+mathrm{C}_{5}^{1}right)+mathrm{C}_{5}^{2}left(mathrm{C}_{5}^{0}+mathrm{C}_{5}^{1}+mathrm{C}_{5}^{2}right) & +mathrm{C}_{5}^{3}left(mathrm{C}_{5}^{0}+mathrm{C}_{5}^{1}+mathrm{C}_{5}^{2}+mathrm{C}_{5}^{3}right)+mathrm{C}_{5}^{4}left(mathrm{C}_{5}^{0}+mathrm{C}_{5}^{1}+mathrm{C}_{5}^{2}right. & left.+mathrm{C}_{5}^{3}+mathrm{C}_{5}^{4}right)+mathrm{C}_{5}^{5}left(mathrm{C}_{5}^{0}+mathrm{C}_{5}^{1}+mathrm{C}_{5}^{2}+mathrm{C}_{5}^{3}+mathrm{C}_{5}^{4}+mathrm{C}_{5}^{5}right) = & 637 .end{aligned}Therefore, the answer is D.Method 2: The non-empty subsets of S are 2^{10}-1=1023 (sets). According to the number of even and odd numbers in the subsets, they can be divided into three categories: (1) even numbers outnumber odd numbers; (2) odd numbers outnumber even numbers; (3) the number of odd and even numbers is equal. Since the 10 elements in S have an equal number of even and odd numbers, the number of subsets in (1) and (2) are equal. Now consider the third category, and consider the number of subsets containing 2, 4, 6, 8, 10 elements, then the total number of subsets ismathrm{C}_{5}^{1} cdot mathrm{C}_{5}^{1}+mathrm{C}_{5}^{2} cdot mathrm{C}_{5}^{2}+mathrm{C}_{5}^{3} cdot mathrm{C}_{5}^{3}+mathrm{C}_{5}^{4} cdot mathrm{C}_{5}^{4}+1=251 .Therefore, the number of subsets in the first category is frac{1}{2}(1023-251)=386.Thus, the number of "good subsets" is 386+251=637 (sets). Therefore, the answer is D.

🧠:[<think>Alright, so I need to figure out how many individuals should be randomly removed from the population of 252 students so that a systematic sampling method can be used to select a sample of size 50. The options are A: 2, B: 3, C: 4, D: 5. Hmm, okay, let me think step by step.First, I remember that systematic sampling involves selecting individuals from a population at regular intervals. To do this, you typically need to determine a sampling interval, often denoted as 'k'. The formula for the sampling interval is usually the population size divided by the desired sample size. So, if we have a population of N individuals and we want a sample of size n, then k = N / n. Then, you randomly select a starting point between 1 and k and select every k-th individual thereafter.But in this problem, the population size is 252, and the desired sample size is 50. If I calculate k directly, that would be 252 divided by 50. Let me compute that: 252 ÷ 50 equals 5.04. Wait, that's not an integer. But for systematic sampling, the interval k needs to be an integer, right? Because you can't select a fraction of a person. So, maybe the population size needs to be adjusted so that when you divide by the sample size, you get an integer. Therefore, they might need to remove some individuals to make N divisible by n.So, the problem is asking how many people to remove so that 252 minus that number is divisible by 50. Then, the sampling interval k would be an integer, allowing systematic sampling. Let me check that. Let's let N' = 252 - x, where x is the number of individuals to remove. We need N' to be divisible by 50. So, 252 - x ≡ 0 mod 50. Therefore, x ≡ 252 mod 50. Let's compute 252 divided by 50. 50*5=250, so 252=50*5+2. Therefore, 252 ≡ 2 mod 50. So, x should be congruent to 2 mod 50. Since x is the number of people to remove, and x has to be between 0 and 252, and we have options A:2, B:3, C:4, D:5, so the answer is A:2. Because 252 -2 =250, which is divisible by 50. Then, k=250/50=5, so the interval would be 5. Then, you can start at a random number between 1 and 5 and pick every 5th person.Wait, but let me verify. If we remove 2 people, the population becomes 250, which is 50*5. So, the interval is 5, which works. If we remove 3 people, population becomes 249. 249 divided by 50 is 4.98, so that's not an integer. Similarly, 252 -4=248, which divided by 50 is 4.96, still not integer. 252-5=247, which divided by 50 is 4.94. So, only when removing 2, we get 250, which is divisible by 50. Therefore, answer should be A:2.But wait, the options given are A:2, B:3, C:4, D:5. So according to this reasoning, it's A:2. But let me think again. Maybe there's another approach. Because sometimes in systematic sampling, even if the population size isn't perfectly divisible by the sample size, they might use a different method, like circular systematic sampling. But the problem mentions a systematic sampling method, not specifying which type, so maybe the standard one. The standard systematic sampling requires the interval to be N/n, but if N isn't divisible by n, then you have to adjust either N or n.Alternatively, maybe the question is about making the population size a multiple of the sample size. If we need a sample size of 50, then population should be a multiple of 50. The closest multiple of 50 less than 252 is 250, so you need to remove 2. Therefore, answer is A:2. So yeah, that seems consistent.But let me check again. If we remove 2 people, then N=250, so k=250/50=5. So, starting at a random number between 1 and 5, then every 5th person. That works. If we didn't remove anyone, then k=252/50=5.04, which is not an integer. So you can't really have an interval of 5.04. So you need to adjust the population. Therefore, answer is A:2.But hold on, maybe there's a different way. Another thought: in systematic sampling, sometimes people round the interval. For example, if k=5.04, you could round it to 5 and proceed, but then you might end up with a sample slightly larger or smaller than 50. However, the problem states that the sample size must be exactly 50. Therefore, in order to guarantee that exactly 50 individuals are selected, you need the population size to be exactly a multiple of 50. Otherwise, depending on the starting point, you might end up with 50 or 51 individuals. For example, if k=5, then 252/5=50.4, so starting at 1, you get 1,6,...,251 which is 51 elements. Wait, but 5*50=250, so the 50th element would be 1 + 5*49=1 +245=246, and the next would be 251. So 251 is the 51st element. But 251 is within 252. So if you use k=5, you might end up with 51 elements if the population is 252. So to ensure exactly 50, you need to have N = k*n. Therefore, N needs to be 250, so remove 2.Alternatively, if you use k=5, and just stop at 50 elements, but then you might not cover the entire population. But systematic sampling usually cycles through the entire population. Hmm. So perhaps in order to make sure that when you apply the interval, you end up with exactly 50 elements, you need N to be a multiple of n. Therefore, the answer is A:2.Alternatively, maybe the question is not about the interval but about something else. Wait, the problem says "a systematic sampling method is used to select a sample of size 50". So perhaps the original population is 252, and they need to adjust it to a number where 50 can be systematically sampled. So removing 2 makes it 250, which allows systematic sampling with interval 5. So that seems correct. Therefore, answer is A:2.But wait, the options include A:2, which is what we got, but I want to check again if perhaps there's a miscalculation. 252 divided by 50 is 5.04. So if we take the floor of that, which is 5, then 5*50=250. Therefore, to get exactly 50 samples with interval 5, population should be 250. Therefore, remove 2. Yes, that's consistent.Alternatively, if you use a different interval. Wait, let's check. If you take k=5, starting at random between 1-5, then the sample would be 5, 10, ..., 250, which is 50 elements. Wait, starting at 1: 1,6,...,246 (total 50 elements: 1 + 49*5=246). Wait, 1 + (50-1)*5=1 +245=246. So if you start at 1, you end at 246. Then 251 and 256 would be beyond. But 246 is less than 252. So you actually only get 50 elements. Wait, maybe my earlier thought was wrong. Let me calculate the number of elements.If the interval is 5, starting at 1, the elements are 1,6,11,..., up to 1 + 5*(49)=1 +245=246. So that's 50 elements. Starting at 2: 2,7,...,247 (50 elements). Similarly, starting at 5: 5,10,...,250. So if you start at 5, the last element is 250. Then, in the original population of 252, starting at 5, you go up to 250, which is 50 elements. The remaining 2 elements (251,252) would not be included. Wait, but that's only if you start at 5. If you start at 1, you end at 246, so 251 and 252 aren't included either. So in that case, the last few elements are excluded. But does that mean that the sampling isn't covering the entire population? So systematic sampling with interval 5 in a population of 252 would result in some elements not being included. Therefore, perhaps in order to have an equal chance for all elements to be included, the population should be a multiple of the interval. Therefore, removing 2 elements would make the population 250, which is a multiple of 5, so that when you do the systematic sampling, you can cover the entire population. Otherwise, the last few elements have a lower chance of being selected. Therefore, to make it fair, you need to have N = k*n, hence remove 2.Therefore, the correct answer is A:2. So the answer is A.Final Answerboxed{A}</think>To determine how many individuals should be randomly removed from the population of 252 students so that a systematic sampling method can be used to select a sample of size 50, we need to ensure the population size is a multiple of the sample size. 1. Calculate the sampling interval: The sampling interval ( k ) is typically the population size ( N ) divided by the sample size ( n ). However, ( k ) must be an integer for systematic sampling.2. Check divisibility: The population size 252 divided by the sample size 50 gives ( frac{252}{50} = 5.04 ), which is not an integer.3. Adjust the population size: We need to find the nearest multiple of 50 less than 252. The closest multiple is 250, which is ( 50 times 5 ).4. Calculate the number to remove: Subtract 250 from 252 to find the number of individuals to remove: ( 252 - 250 = 2 ).By removing 2 individuals, the population size becomes 250, which is divisible by 50. This allows the use of a systematic sampling interval of 5, ensuring a sample size of exactly 50.[boxed{A}],<think>Okay, so I need to figure out how many students should be randomly removed from the population of 252 students so that a systematic sample of size 50 can be taken. The options are 2, 3, 4, or 5. Let me think step by step.First, I remember that systematic sampling involves selecting individuals from a population at regular intervals. The key idea is to determine the sampling interval, usually denoted as 'k', which is calculated by dividing the population size by the desired sample size. The formula is k = N / n, where N is the population size and n is the sample size. But wait, in this case, the population size is 252 and the sample size needed is 50. Let me calculate the interval first. So k = 252 / 50. Let me compute that: 252 divided by 50. Hmm, 50 times 5 is 250, so 5 times 50 is 250, which is 2 less than 252. So 252 / 50 is 5.04. Hmm, that's not a whole number. But in systematic sampling, the interval k should ideally be an integer to make the selection process straightforward. If it's not an integer, you might have to adjust the population size by removing some individuals so that the new population size divided by the sample size gives an integer. That way, you can have a consistent interval.So the problem is asking how many individuals should be removed so that after removal, the remaining population size is divisible by 50, giving an integer for k. Let me verify that. So if we remove 'r' individuals, the new population size would be 252 - r. We want 252 - r to be divisible by 50. Therefore, (252 - r) should be a multiple of 50.Looking at the options given, the possible numbers to remove are 2, 3, 4, or 5. Let's check each one:- If we remove 2: 252 - 2 = 250. Then 250 / 50 = 5. That's an integer. So k would be 5. So this works. So removing 2 individuals would allow a systematic sample of 50 with interval 5.But wait, let's check the other options just to be sure.- Remove 3: 252 - 3 = 249. 249 / 50 is 4.98, not an integer.- Remove 4: 252 - 4 = 248. 248 / 50 = 4.96, not an integer.- Remove 5: 252 - 5 = 247. 247 / 50 = 4.94, not an integer.So only when we remove 2 individuals does the population size become 250, which is exactly divisible by 50, giving an interval of 5. Therefore, the correct answer should be A: 2.Wait, but let me confirm my understanding. In systematic sampling, if the population size isn't perfectly divisible by the sample size, some methods suggest using a fractional interval or using a circular method, but maybe in this problem, they want the traditional approach where the interval must be an integer. So adjusting the population to the nearest lower multiple of the sample size. So 50 times 5 is 250, which is 2 less than 252. So remove 2. That seems to make sense. Alternatively, sometimes systematic sampling is done by calculating the interval as N/n and then rounding to the nearest whole number, but if N/n isn't an integer, there's a possibility of having a sample size slightly different than desired. However, the problem here explicitly states that a systematic sampling method is used to select a sample of size 50. So to ensure that exactly 50 samples are taken, the interval must be an integer, and hence the population size must be adjusted to a multiple of 50. So yes, removing 2 students would give 250, which divided by 50 is exactly 5. Then the sampling interval is 5. Starting at a random number between 1 and 5, then selecting every 5th individual. For example, starting at 3, then 3, 8, 13,... up to 250. That would give exactly 50 individuals. So that works. Therefore, the answer should be A: 2. Let me check the options again. The options are A:2, B:3, C:4, D:5. So yes, A is the correct one. But wait, I just want to make sure there isn't a different approach here. For example, some sources might mention that if N is not a multiple of n, you can still perform systematic sampling by using a fractional sampling interval. For instance, if N = 252 and n = 50, then k = 252 / 50 = 5.04. Then, you could use a fractional interval, but that complicates the selection process because you have to deal with decimals. However, systematic sampling typically uses integer intervals. So if the problem is assuming that we need an integer interval, then adjusting the population by removing 2 makes sense. Alternatively, maybe the question is about making the population size such that when you apply the systematic sampling, you end up with exactly 50 samples. So if you take k as the integer part of N/n, which is 5, then the sample size would be N / k. Wait, that's 252 / 5 = 50.4, which would be 50 or 51. But since we need exactly 50, perhaps we need to reduce the population so that N - r = 50 * k, where k is an integer. So yes, N - r = 50 * k. So solving for N - r = 50 * k. So 252 - r = 50 * k. We need k to be an integer. The closest multiple of 50 less than 252 is 250 (50*5), which is 2 less. Then next is 200 (50*4), which would require removing 52, which is not among the options. So the closest is 250. Therefore, remove 2. Therefore, the answer is A:2. So I think that's the correct approach here.Final Answerboxed{A}</think>To determine how many individuals should be randomly removed from the population of 252 students to use a systematic sampling method for a sample size of 50, we need to ensure the population size after removal is a multiple of 50. 1. Calculate the sampling interval ( k ) using the formula ( k = frac{N}{n} ), where ( N ) is the population size and ( n ) is the sample size.2. The original population size is 252, and the desired sample size is 50. 3. The interval ( k ) should be an integer. Therefore, we need to adjust the population size ( N ) such that ( frac{N - r}{50} ) is an integer, where ( r ) is the number of individuals removed.4. Check the options given (2, 3, 4, 5) to see which value of ( r ) makes ( 252 - r ) a multiple of 50: - Removing 2 individuals: ( 252 - 2 = 250 ), which is a multiple of 50 (50 * 5 = 250). - Removing 3, 4, or 5 individuals results in 249, 248, or 247, none of which are multiples of 50.Thus, removing 2 individuals from the population makes the new population size 250, which is divisible by 50, allowing a systematic sampling interval of 5. [boxed{A}]]

🔑:The total number of students is not divisible by the sample size. According to the systematic sampling method, some individuals should be randomly removed from the population to ensure divisibility.We have 1252 = 50 times 25 + 2. Therefore, 2 individuals should be randomly removed from the population to make it divisible by the sample size.The answer is: boxed{A}.Systematic sampling steps for selecting a sample of size n from a population of size N are as follows:1. Number all N individuals in the population.2. Divide the entire set of numbers into k segments, where k = frac{N}{n} if frac{N}{n} is an integer. If frac{N}{n} is not an integer, remove some individuals from the population so that the remaining number of individuals N' is divisible by n.In this problem, the total number of students is not divisible by the sample size. Hence, some individuals should be randomly removed from the population to ensure divisibility.This problem focuses on systematic sampling and its steps. The key to solving this problem is to understand that when the total number is not divisible by the sample size, some individuals should be randomly removed from the population to ensure divisibility. This is a basic-level problem.

🧠:[<think>Okay, let's try to solve these two equations step by step. I remember that A and C in combinatorics stand for permutations and combinations, right? So A(n, k) is the number of permutations of n things taken k at a time, and C(n, k) is the number of combinations. The formulas are A(n, k) = n! / (n - k)! and C(n, k) = n! / (k!(n - k)!).Starting with equation (1): 3A_x^3 = 2A_{x+1}^2 + 6A_x^2. Let me make sure I understand the notation here. The problem uses subscripts, so A_x^3 would be permutations of x things taken 3 at a time, which is x! / (x - 3)!. Similarly, A_{x+1}^2 would be (x+1)! / (x+1 - 2)! = (x+1)! / (x - 1)! and A_x^2 is x! / (x - 2)!.So substituting the formulas into the equation:3 * [x! / (x - 3)!] = 2 * [(x + 1)! / (x - 1)!] + 6 * [x! / (x - 2)!]Hmm, let's simplify each term. First, note that (x + 1)! = (x + 1) * x!, so maybe we can factor x! out of each term.Left side: 3 * x! / (x - 3)!Right side: 2 * (x + 1)! / (x - 1)! + 6 * x! / (x - 2)! = 2 * (x + 1) * x! / (x - 1)! + 6 * x! / (x - 2)! Let me write all terms in terms of x! divided by some factorial. Let's see:First term on the right: 2*(x + 1)*x! / (x - 1)! = 2*(x + 1)*x! / (x - 1)!.Second term on the right: 6*x! / (x - 2)!.So maybe we can factor out 2*x! from the right side?Wait, let's see. Let me write each denominator:For the first term on the right: (x - 1)! = (x - 1)(x - 2)! So 2*(x + 1)*x! / [(x - 1)(x - 2)!] = 2*(x + 1)/ (x - 1) * x! / (x - 2)!.Similarly, the second term is 6*x! / (x - 2)!.Therefore, the right side can be written as [2*(x + 1)/(x - 1) + 6] * x! / (x - 2)!.Hmm, so combining these terms. Let's factor x! / (x - 2)! out of both terms on the right.So the right side becomes:x! / (x - 2)! * [2*(x + 1)/(x - 1) + 6]Similarly, the left side is 3*x! / (x - 3)!.Note that (x - 3)! is related to (x - 2)!: (x - 2)! = (x - 2)*(x - 3)!.So left side: 3*x! / (x - 3)! = 3*(x!)*(x - 2)! / (x - 2)!*(x - 3)!) = Wait, maybe another approach. Let's express both sides in terms of x! divided by (x - 3)!.Since (x - 2)! = (x - 2)(x - 3)!.Therefore, x! / (x - 2)! = x! / [(x - 2)(x - 3)!] = [x! / (x - 3)!] / (x - 2) = [A(x, 3)] / (x - 2).But perhaps it's better to convert all terms to have (x - 3)! in the denominator. Let's see:Left side is 3 * x! / (x - 3)!.Right side: 2*(x + 1)! / (x - 1)! + 6*x! / (x - 2)!.Let's express both terms on the right in terms of (x - 3)! denominator.First term: 2*(x + 1)! / (x - 1)! = 2*(x + 1)*x! / (x - 1)!.But (x - 1)! = (x - 1)(x - 2)(x - 3)!.So 2*(x + 1)*x! / [(x - 1)(x - 2)(x - 3)!] = 2*(x + 1)/[(x - 1)(x - 2)] * x! / (x - 3)!.Second term: 6*x! / (x - 2)! = 6*x! / [(x - 2)(x - 3)!] = 6/(x - 2) * x! / (x - 3)!.Therefore, the right side is:[2*(x + 1)/[(x - 1)(x - 2)] + 6/(x - 2)] * x! / (x - 3)!.So now, the equation becomes:3 * x! / (x - 3)! = [2*(x + 1)/[(x - 1)(x - 2)] + 6/(x - 2)] * x! / (x - 3)!.Since x! / (x - 3)! is non-zero (assuming x ≥ 3, which it has to be for the permutation to be defined), we can divide both sides by x! / (x - 3)! to get:3 = 2*(x + 1)/[(x - 1)(x - 2)] + 6/(x - 2)Now, let's simplify the right side:Let me write 6/(x - 2) as 6*(x - 1)/[(x - 1)(x - 2)] to get a common denominator with the first term.So:2*(x + 1)/[(x - 1)(x - 2)] + 6*(x - 1)/[(x - 1)(x - 2)] = [2(x + 1) + 6(x - 1)] / [(x - 1)(x - 2)]Compute the numerator:2(x + 1) + 6(x - 1) = 2x + 2 + 6x - 6 = (2x + 6x) + (2 - 6) = 8x - 4So the equation becomes:3 = (8x - 4)/[(x - 1)(x - 2)]Multiply both sides by (x - 1)(x - 2):3(x - 1)(x - 2) = 8x - 4Expand the left side:3[(x - 1)(x - 2)] = 3[x^2 - 3x + 2] = 3x^2 - 9x + 6So:3x^2 - 9x + 6 = 8x - 4Bring all terms to the left side:3x^2 - 9x + 6 - 8x + 4 = 0Combine like terms:3x^2 -17x +10 = 0Now, solve the quadratic equation 3x^2 -17x +10 = 0.Using the quadratic formula: x = [17 ± sqrt(17^2 - 4*3*10)] / (2*3)Compute discriminant:17^2 = 289, 4*3*10 = 120, so sqrt(289 - 120) = sqrt(169) = 13Therefore, x = [17 ±13]/6So two solutions:x = (17 +13)/6 =30/6=5x=(17-13)/6=4/6=2/3≈0.666...But x must be an integer greater than or equal to 3 for A_x^3 to be defined (since permutations of x taken 3 at a time requires x ≥3). So x=5 is valid, but x=2/3 is not an integer and less than 3, so we discard it.Thus, equation (1) solution is x=5.Now equation (2): C_8^x = C_8^{5x -4}Recall that C(n, k) = C(n, n - k). So, either the two exponents are equal, or they add up to n.So in this case, either:1) x = 5x -4, or2) x + (5x -4) = 8Let's solve both possibilities.Case 1: x = 5x -4Subtract x from both sides: 0 =4x -4Add 4:4=4xDivide by4:x=1Case 2: x +5x -4=8Combine terms:6x -4=8Add4:6x=12Divide by6:x=2So possible solutions x=1 and x=2. Now check if these are valid.For combinations, C(8, x) requires that x is an integer between 0 and 8, inclusive.x=1: yes, since 1 is between 0 and8. Similarly, x=2 is valid.Check if both satisfy the original equation.First, x=1:C(8,1)=8, and C(8,5*1-4)=C(8,1)=8. So 8=8, true.x=2:C(8,2)=28, and C(8,5*2 -4)=C(8,6)=28. So 28=28, which is true.Thus, both x=1 and x=2 are solutions.Therefore, the solutions are x=5 for equation (1), and x=1 and x=2 for equation (2).Wait, but the problem says "Solve the equations: (1)... (2)...". So I need to present solutions for both equations. Let me double-check my work for equation (1) to make sure.Equation (1) was 3A_x^3 = 2A_{x+1}^2 +6A_x^2. We found x=5. Let's verify:Left side:3*A_5^3=3*(5*4*3)=3*60=180Right side:2*A_6^2 +6*A_5^2=2*(6*5)+6*(5*4)=2*30 +6*20=60 +120=180. Yes, that's correct.So equation (1) x=5 is correct.Equation (2) solutions x=1 and x=2.So all solutions are x=5 for the first equation, and x=1, x=2 for the second.Final AnswerThe solutions are (boxed{5}) for equation (1) and (boxed{1}), (boxed{2}) for equation (2).</think>To solve the given equations:1. (3A_x^3 = 2A_{x+1}^2 + 6A_x^2) - Convert the permutation formulas: [ 3 cdot frac{x!}{(x-3)!} = 2 cdot frac{(x+1)!}{(x-1)!} + 6 cdot frac{x!}{(x-2)!} ] - Simplify each term: [ 3 cdot frac{x!}{(x-3)!} = 2 cdot frac{(x+1)x!}{(x-1)(x-2)(x-3)!} + 6 cdot frac{x!}{(x-2)(x-3)!} ] - Factor out common terms and solve: [ 3 = frac{2(x+1)}{(x-1)(x-2)} + frac{6}{x-2} ] - Combine and simplify: [ 3 = frac{8x - 4}{(x-1)(x-2)} ] - Solve the quadratic equation: [ 3x^2 - 17x + 10 = 0 ] - Solutions: (x = 5) (valid) and (x = frac{2}{3}) (invalid).2. (C_8^x = C_8^{5x-4}) - Use the property (C(n, k) = C(n, n-k)): - Case 1: (x = 5x - 4) leads to (x = 1) - Case 2: (x + (5x - 4) = 8) leads to (x = 2) - Verify solutions: Both (x = 1) and (x = 2) are valid. Final AnswerThe solutions are (boxed{5}) for equation (1) and (boxed{1}), (boxed{2}) for equation (2).,<think>Okay, let's tackle these two equations step by step. Hmm, they both involve combinations or permutations, I think. Let me start with the first one: (1) 3A_x^3 = 2A_{x+1}^2 + 6A_x^2. First off, I need to remember what A_n^k stands for. I think it's the number of permutations of n things taken k at a time, so A_n^k = n! / (n - k)! Right? Yeah, that's the permutation formula. So, A_x^3 would be x! / (x - 3)!, and similarly for the others. Let me write that down.So equation (1) becomes:3 * [x! / (x - 3)!] = 2 * [(x + 1)! / (x + 1 - 2)!] + 6 * [x! / (x - 2)!]Simplify each term step by step. Let's start with the left side: 3 * A_x^3. That's 3 * [x! / (x - 3)!]. On the right side, the first term is 2 * A_{x+1}^2. A_{x+1}^2 is (x + 1)! / (x + 1 - 2)! = (x + 1)! / (x - 1)! Similarly, the second term on the right is 6 * A_x^2, which is 6 * [x! / (x - 2)!]Let me rewrite the equation with these substitutions:3 * [x! / (x - 3)!] = 2 * [(x + 1)! / (x - 1)!] + 6 * [x! / (x - 2)!]Hmm, maybe we can factor out some terms or simplify factorials. Let me recall that (x + 1)! = (x + 1) * x!, so that might help. Let's apply that to the first term on the right side:2 * [(x + 1)! / (x - 1)!] = 2 * (x + 1) * x! / (x - 1)! So, the equation becomes:3 * [x! / (x - 3)!] = 2 * (x + 1) * x! / (x - 1)! + 6 * [x! / (x - 2)!]Now, notice that x! is a common factor in all terms. Let's factor that out:x! [ 3 / (x - 3)! ] = x! [ 2(x + 1) / (x - 1)! + 6 / (x - 2)! ]Since x! is non-zero (assuming x is a positive integer, which it should be in permutations), we can divide both sides by x!:3 / (x - 3)! = 2(x + 1) / (x - 1)! + 6 / (x - 2)! Now, let's express all denominators in terms of (x - 3)! to make the equation easier. Let's note that:(x - 1)! = (x - 1)(x - 2)(x - 3)!(x - 2)! = (x - 2)(x - 3)!So, substituting these into the equation:Left side: 3 / (x - 3)! Right side: 2(x + 1) / [ (x - 1)(x - 2)(x - 3)! ] + 6 / [ (x - 2)(x - 3)! ]So, factoring out 1 / (x - 3)! from all terms, we can multiply both sides by (x - 3)! to eliminate the denominators:3 = 2(x + 1) / [ (x - 1)(x - 2) ] + 6 / (x - 2)Now, let's write this as:3 = [2(x + 1)] / [(x - 1)(x - 2)] + 6 / (x - 2)To combine these terms, we can get a common denominator. The denominators are (x - 1)(x - 2) and (x - 2). So, the common denominator is (x - 1)(x - 2). Let's adjust the second term accordingly:6 / (x - 2) = 6(x - 1) / [ (x - 1)(x - 2) ]So, substituting back in:3 = [2(x + 1) + 6(x - 1)] / [ (x - 1)(x - 2) ]Multiply numerator terms:2(x + 1) + 6(x - 1) = 2x + 2 + 6x - 6 = 8x - 4So, the equation becomes:3 = (8x - 4) / [ (x - 1)(x - 2) ]Multiply both sides by (x - 1)(x - 2):3(x - 1)(x - 2) = 8x - 4Expand the left side:3[(x - 1)(x - 2)] = 3[x^2 - 3x + 2] = 3x^2 - 9x + 6So:3x^2 - 9x + 6 = 8x - 4Bring all terms to the left side:3x^2 - 9x + 6 - 8x + 4 = 0Combine like terms:3x^2 - 17x + 10 = 0Now, solve this quadratic equation. Let's check the discriminant:D = (-17)^2 - 4*3*10 = 289 - 120 = 169Square root of 169 is 13. So, solutions:x = [17 ± 13] / (2*3)First solution: (17 + 13)/6 = 30/6 = 5Second solution: (17 - 13)/6 = 4/6 = 2/3But x must be a positive integer since it's the upper number in permutations. Also, in permutations like A_x^3, x must be at least 3. Similarly, A_{x+1}^2 requires x + 1 >= 2 => x >=1. So, x must be an integer >=3. Therefore, x = 5 is the only valid solution here. x = 2/3 is not an integer and too small, so we discard it. So equation (1) solution is x = 5.Now, moving to equation (2): C_8^x = C_8^{5x - 4}C here stands for combinations, right? So C(n, k) = n! / [k!(n - k)!]A property of combinations is that C(n, k) = C(n, n - k). So, either the two exponents are equal, or they add up to n. So in this case, n = 8.Therefore, either:x = 5x - 4orx + (5x - 4) = 8Let's solve both possibilities.First equation: x = 5x - 4Subtract x from both sides:0 = 4x - 4Add 4:4 = 4xDivide by 4:x = 1Second equation: x + 5x -4 = 8Combine like terms:6x - 4 = 8Add 4:6x = 12Divide by 6:x = 2Now, we need to check if these solutions are valid. For combinations, C(8, x) requires that x is a non-negative integer not exceeding 8. Let's check x =1 and x=2.For x=1: 5x -4 = 5*1 -4 =1. So C(8,1) = C(8,1), which is valid.For x=2: 5x -4 =10 -4 =6. So C(8,2) = C(8,6). Since 8 -2 =6, this is also valid by the combination symmetry property.So both x=1 and x=2 are valid solutions. Therefore, equation (2) has solutions x=1 and x=2.Let me just recap to make sure I didn't make any mistakes. For the first equation, I converted permutations into factorial expressions, simplified, solved the quadratic, and checked the validity. For the second, used the combination symmetry property, solved the two cases, and verified the solutions. Seems solid. So final answers: equation (1) x=5; equation (2) x=1 and x=2.Final AnswerThe solutions are (boxed{5}) for equation (1) and (boxed{1}) and (boxed{2}) for equation (2).</think>To solve the given equations, we start with the first equation:1) (3A_x^3 = 2A_{x+1}^2 + 6A_x^2)We use the permutation formula (A_n^k = frac{n!}{(n-k)!}):[3 cdot frac{x!}{(x-3)!} = 2 cdot frac{(x+1)!}{(x-1)!} + 6 cdot frac{x!}{(x-2)!}]Factoring out (x!) and simplifying each term:[3 cdot frac{1}{(x-3)!} = 2 cdot frac{(x+1)}{(x-1)(x-2)(x-3)!} + 6 cdot frac{1}{(x-2)(x-3)!}]Multiplying through by ((x-3)!):[3 = frac{2(x+1)}{(x-1)(x-2)} + frac{6}{(x-2)}]Combining terms over a common denominator:[3 = frac{2(x+1) + 6(x-1)}{(x-1)(x-2)}]Simplifying the numerator:[3 = frac{8x - 4}{(x-1)(x-2)}]Multiplying both sides by ((x-1)(x-2)) and expanding:[3(x^2 - 3x + 2) = 8x - 4]Solving the quadratic equation:[3x^2 - 17x + 10 = 0]The solutions are (x = 5) and (x = frac{2}{3}). Since (x) must be a positive integer, the valid solution is (x = 5).2) (C_8^x = C_8^{5x-4})Using the combination property (C(n, k) = C(n, n-k)):Either (x = 5x - 4) or (x + (5x - 4) = 8).Solving both equations:For (x = 5x - 4):[4x - 4 = 0 implies x = 1]For (x + 5x - 4 = 8):[6x - 4 = 8 implies x = 2]Both solutions (x = 1) and (x = 2) are valid as they are non-negative integers not exceeding 8. Final AnswerThe solutions are (boxed{5}) for equation (1) and (boxed{1}) and (boxed{2}) for equation (2).]

🔑:Step-by-Step Solution:Solving Equation (1):Given: 3A_x^3 = 2A_{x+1}^2 + 6A_x^2From the properties of combinations, we have:3A_x^3 = 3left[frac{x!}{3!(x-3)!}right] = 3x(x-1)(x-2)2A_{x+1}^2 = 2left[frac{(x+1)!}{2!(x-1)!}right] = 2x(x+1)6A_x^2 = 6left[frac{x!}{2!(x-2)!}right] = 6x(x-1)Substituting these back into the equation, we get:3x(x-1)(x-2) = 2x(x+1) + 6x(x-1)Simplifying, we obtain:3x^3 - 9x^2 + 6x = 2x^2 + 2x + 6x^2 - 6xBringing terms to one side and simplifying further:3x^3 - 17x^2 + 10x = 0Factoring out x, we get:x(3x^2 - 17x + 10) = 0Solving the quadratic equation 3x^2 - 17x + 10 = 0, we find:x = 5 or x = frac{2}{3}Given that x in N_{+}, we discard x = frac{2}{3}.Thus, the solution to equation (1) is boxed{x = 5}.Solving Equation (2):Given: C_8^x = C_8^{5x-4}This implies two conditions:1. x = 5x - 4 2. 8 - x = 5x - 4Solving the first condition:x = 5x - 4Rearranging and solving for x, we get:4x = 4Thus, x = 1.Solving the second condition:8 - x = 5x - 4Rearranging and solving for x, we get:6x = 12Thus, x = 2.Both x = 1 and x = 2 satisfy the conditions of the equation. Therefore, the solutions to equation (2) are boxed{x = 1 text{ or } x = 2}.