💡Alright, so I've got this problem about the function ( f(x) = e^{ax} cdot sin x ). It has two parts, and I need to tackle them one by one. Let's start with part (1).**Problem (1):** When ( a = 1 ), find the interval where the function ( f(x) ) is monotonically increasing on the interval ( [0, 2pi] ).Okay, so first, I know that to find where a function is monotonically increasing, I need to look at its derivative. If the derivative is positive on an interval, the function is increasing there.Given ( f(x) = e^{x} cdot sin x ) (since ( a = 1 )), let's compute the derivative ( f'(x) ).Using the product rule: ( f'(x) = frac{d}{dx}(e^x) cdot sin x + e^x cdot frac{d}{dx}(sin x) ).Calculating each part:- ( frac{d}{dx}(e^x) = e^x )- ( frac{d}{dx}(sin x) = cos x )So, putting it together:( f'(x) = e^x cdot sin x + e^x cdot cos x = e^x (sin x + cos x) ).Now, I need to find where ( f'(x) > 0 ). Since ( e^x ) is always positive for all real ( x ), the sign of ( f'(x) ) depends on ( sin x + cos x ).So, I need to solve ( sin x + cos x > 0 ).Hmm, I remember that ( sin x + cos x ) can be rewritten using a trigonometric identity. Let me recall: ( sin x + cos x = sqrt{2} sinleft(x + frac{pi}{4}right) ).Yes, that's right. So, ( sin x + cos x = sqrt{2} sinleft(x + frac{pi}{4}right) ).Therefore, ( f'(x) = e^x cdot sqrt{2} sinleft(x + frac{pi}{4}right) ).Since ( e^x ) is always positive, the sign of ( f'(x) ) depends on ( sinleft(x + frac{pi}{4}right) ).So, ( f'(x) > 0 ) when ( sinleft(x + frac{pi}{4}right) > 0 ).The sine function is positive in the intervals ( (0, pi) ) within each ( 2pi ) period. So, let's solve for ( x ):( 0 < x + frac{pi}{4} < pi ).Subtracting ( frac{pi}{4} ) from all parts:( -frac{pi}{4} < x < frac{3pi}{4} ).But since we're considering ( x ) in ( [0, 2pi] ), the lower bound is adjusted to 0:( 0 leq x < frac{3pi}{4} ).Similarly, sine is positive again in the next period:( 2pi < x + frac{pi}{4} < 3pi ).Subtracting ( frac{pi}{4} ):( frac{7pi}{4} < x < frac{11pi}{4} ).But since our interval is up to ( 2pi ), the upper bound is adjusted to ( 2pi ):( frac{7pi}{4} < x leq 2pi ).So, combining these, the intervals where ( f'(x) > 0 ) are ( [0, frac{3pi}{4}) ) and ( (frac{7pi}{4}, 2pi] ).Therefore, the function ( f(x) ) is monotonically increasing on these intervals.Wait, let me double-check. When ( x + frac{pi}{4} ) is between ( 0 ) and ( pi ), that's ( x ) between ( -frac{pi}{4} ) and ( frac{3pi}{4} ). But since ( x ) starts at 0, the first interval is ( [0, frac{3pi}{4}) ). Then, the next positive interval is when ( x + frac{pi}{4} ) is between ( 2pi ) and ( 3pi ), which translates to ( x ) between ( frac{7pi}{4} ) and ( frac{11pi}{4} ). But since our domain is up to ( 2pi ), the second interval is ( (frac{7pi}{4}, 2pi] ). That seems correct.So, part (1) is done. Now, moving on to part (2).**Problem (2):** When ( a geq 1 ), if ( forall x in [0, frac{pi}{2}] ), ( f(x) leq bx ) holds, find the minimum value of ( b - e^{2}a ).Alright, so we need to find the minimum value of ( b - e^{2}a ) given that ( f(x) = e^{ax} sin x leq bx ) for all ( x ) in ( [0, frac{pi}{2}] ).First, let's understand what this condition implies. For all ( x ) in that interval, ( e^{ax} sin x leq bx ). So, ( b ) must be chosen such that this inequality holds for the entire interval.Our goal is to find the minimum value of ( b - e^{2}a ). So, we need to express ( b ) in terms of ( a ), find the minimum of ( b - e^{2}a ), and then find its minimum value.Let me think about how to approach this.First, let's rearrange the inequality:( e^{ax} sin x leq bx ).We can write this as:( frac{e^{ax} sin x}{x} leq b ) for all ( x in (0, frac{pi}{2}] ).Note that at ( x = 0 ), both sides are 0 (since ( sin 0 = 0 ) and the right side is 0). So, the inequality holds at ( x = 0 ).Therefore, for ( x in (0, frac{pi}{2}] ), ( b ) must be greater than or equal to ( frac{e^{ax} sin x}{x} ).Hence, the minimum value of ( b ) is the maximum of ( frac{e^{ax} sin x}{x} ) over ( x in (0, frac{pi}{2}] ).So, ( b geq max_{x in (0, frac{pi}{2}]} frac{e^{ax} sin x}{x} ).Therefore, to find the minimum ( b ), we need to find the maximum of ( frac{e^{ax} sin x}{x} ) on ( (0, frac{pi}{2}] ).Let me denote ( g(x) = frac{e^{ax} sin x}{x} ). We need to find ( max g(x) ) on ( (0, frac{pi}{2}] ).To find the maximum, we can take the derivative of ( g(x) ) and find its critical points.But before that, let's note that as ( x to 0^+ ), ( g(x) ) approaches a limit. Let's compute that limit.Using L'Hospital's Rule since both numerator and denominator approach 0:( lim_{x to 0^+} frac{e^{ax} sin x}{x} = lim_{x to 0^+} frac{a e^{ax} sin x + e^{ax} cos x}{1} ).Wait, actually, that's the derivative of the numerator over the derivative of the denominator. Wait, no, L'Hospital's Rule is for 0/0 or ∞/∞ forms. Here, numerator and denominator both approach 0 as ( x to 0 ).So, applying L'Hospital's Rule once:( lim_{x to 0} frac{e^{ax} sin x}{x} = lim_{x to 0} frac{d}{dx}(e^{ax} sin x) / frac{d}{dx}(x) ).Compute the derivatives:Numerator derivative: ( a e^{ax} sin x + e^{ax} cos x ).Denominator derivative: 1.So, the limit becomes:( lim_{x to 0} (a e^{ax} sin x + e^{ax} cos x) ).Plugging in ( x = 0 ):( a e^{0} cdot 0 + e^{0} cdot 1 = 0 + 1 = 1 ).So, as ( x to 0 ), ( g(x) to 1 ).Now, let's compute ( g(frac{pi}{2}) ):( g(frac{pi}{2}) = frac{e^{a cdot frac{pi}{2}} sin frac{pi}{2}}{frac{pi}{2}} = frac{e^{frac{api}{2}} cdot 1}{frac{pi}{2}} = frac{2}{pi} e^{frac{api}{2}} ).So, at ( x = frac{pi}{2} ), ( g(x) = frac{2}{pi} e^{frac{api}{2}} ).Now, we need to see whether ( g(x) ) has a maximum somewhere in between ( 0 ) and ( frac{pi}{2} ).To find the critical points, let's compute ( g'(x) ) and set it equal to zero.Given ( g(x) = frac{e^{ax} sin x}{x} ).Let me compute ( g'(x) ):Using the quotient rule:( g'(x) = frac{(d/dx)(e^{ax} sin x) cdot x - e^{ax} sin x cdot (d/dx)(x)}{x^2} ).Compute each part:First, ( d/dx (e^{ax} sin x) = a e^{ax} sin x + e^{ax} cos x ).Second, ( d/dx (x) = 1 ).So, putting it all together:( g'(x) = frac{(a e^{ax} sin x + e^{ax} cos x) cdot x - e^{ax} sin x cdot 1}{x^2} ).Factor out ( e^{ax} ):( g'(x) = frac{e^{ax} [ (a sin x + cos x) x - sin x ] }{x^2} ).Simplify the numerator inside the brackets:( (a sin x + cos x) x - sin x = a x sin x + x cos x - sin x ).Factor terms:( = sin x (a x - 1) + x cos x ).So, ( g'(x) = frac{e^{ax} [ sin x (a x - 1) + x cos x ] }{x^2} ).Set ( g'(x) = 0 ):Since ( e^{ax} ) and ( x^2 ) are always positive for ( x > 0 ), the sign of ( g'(x) ) depends on the numerator:( sin x (a x - 1) + x cos x = 0 ).So, we need to solve:( sin x (a x - 1) + x cos x = 0 ).This seems a bit complicated. Maybe we can rearrange terms:( sin x (a x - 1) = -x cos x ).Divide both sides by ( cos x ) (assuming ( cos x neq 0 )):( tan x (a x - 1) = -x ).Hmm, not sure if that helps. Alternatively, let's consider the equation:( sin x (a x - 1) + x cos x = 0 ).Let me write it as:( (a x - 1) sin x + x cos x = 0 ).I wonder if this equation can be simplified or if we can find a substitution.Alternatively, maybe it's better to analyze the behavior of ( g(x) ) to see if it's increasing or decreasing.Given that ( a geq 1 ), let's see:At ( x = 0 ), as we saw, ( g(x) to 1 ).At ( x = frac{pi}{2} ), ( g(x) = frac{2}{pi} e^{frac{api}{2}} ).We need to see if ( g(x) ) has a maximum somewhere in between.Let me test ( g'(x) ) at some points.First, let's consider ( x ) near 0.As ( x to 0^+ ), let's approximate ( sin x approx x ) and ( cos x approx 1 ).So, the numerator in ( g'(x) ):( sin x (a x - 1) + x cos x approx x (a x - 1) + x cdot 1 = a x^2 - x + x = a x^2 ).Since ( a geq 1 ), this is positive. Therefore, near ( x = 0 ), ( g'(x) > 0 ), so ( g(x) ) is increasing.Now, at ( x = frac{pi}{2} ), let's compute the numerator:( sin frac{pi}{2} (a cdot frac{pi}{2} - 1) + frac{pi}{2} cos frac{pi}{2} = 1 ( frac{api}{2} - 1 ) + frac{pi}{2} cdot 0 = frac{api}{2} - 1 ).Since ( a geq 1 ), ( frac{api}{2} - 1 geq frac{pi}{2} - 1 approx 1.57 - 1 = 0.57 > 0 ). Therefore, at ( x = frac{pi}{2} ), the numerator is positive, so ( g'(x) > 0 ) there as well.Wait, so if ( g'(x) > 0 ) near 0 and near ( frac{pi}{2} ), does that mean ( g(x) ) is increasing throughout the interval? But that can't be, because if ( g(x) ) is increasing from 1 to ( frac{2}{pi} e^{frac{api}{2}} ), which is larger than 1, then the maximum would be at ( x = frac{pi}{2} ).But wait, let's check the behavior in between. Maybe ( g'(x) ) is always positive on ( (0, frac{pi}{2}] ), which would mean ( g(x) ) is increasing on that interval. Therefore, the maximum of ( g(x) ) is at ( x = frac{pi}{2} ).But let me verify this by checking the derivative at some intermediate point, say ( x = frac{pi}{4} ).Compute the numerator:( sin frac{pi}{4} (a cdot frac{pi}{4} - 1) + frac{pi}{4} cos frac{pi}{4} ).Compute each term:( sin frac{pi}{4} = frac{sqrt{2}}{2} approx 0.707 ).( cos frac{pi}{4} = frac{sqrt{2}}{2} approx 0.707 ).So,First term: ( 0.707 (a cdot 0.785 - 1) ).Second term: ( 0.785 cdot 0.707 approx 0.555 ).So, overall:( 0.707 (0.785 a - 1) + 0.555 ).Let me compute this for ( a = 1 ):First term: ( 0.707 (0.785 - 1) = 0.707 (-0.215) approx -0.152 ).Second term: ( 0.555 ).Total: ( -0.152 + 0.555 approx 0.403 > 0 ).So, for ( a = 1 ), the numerator is positive at ( x = frac{pi}{4} ).What about for larger ( a )? Let's take ( a = 2 ):First term: ( 0.707 (1.57 - 1) = 0.707 (0.57) approx 0.403 ).Second term: ( 0.555 ).Total: ( 0.403 + 0.555 approx 0.958 > 0 ).So, it's still positive.Wait, so for ( a geq 1 ), at ( x = frac{pi}{4} ), the numerator is positive. Therefore, ( g'(x) > 0 ) at ( x = frac{pi}{4} ).Therefore, it seems that ( g'(x) > 0 ) for all ( x in (0, frac{pi}{2}] ) when ( a geq 1 ). Therefore, ( g(x) ) is increasing on this interval.Therefore, the maximum of ( g(x) ) occurs at ( x = frac{pi}{2} ), which is ( frac{2}{pi} e^{frac{api}{2}} ).Therefore, the minimal ( b ) that satisfies ( f(x) leq bx ) for all ( x in [0, frac{pi}{2}] ) is ( b = frac{2}{pi} e^{frac{api}{2}} ).Now, we need to find the minimum value of ( b - e^{2}a ).So, substituting ( b ):( b - e^{2}a = frac{2}{pi} e^{frac{api}{2}} - e^{2}a ).Let me denote this expression as ( G(a) = frac{2}{pi} e^{frac{api}{2}} - e^{2}a ).We need to find the minimum value of ( G(a) ) for ( a geq 1 ).To find the minimum, we can take the derivative of ( G(a) ) with respect to ( a ) and set it equal to zero.Compute ( G'(a) ):( G'(a) = frac{2}{pi} cdot frac{pi}{2} e^{frac{api}{2}} - e^{2} cdot 1 ).Simplify:( G'(a) = e^{frac{api}{2}} - e^{2} ).Set ( G'(a) = 0 ):( e^{frac{api}{2}} - e^{2} = 0 ).So,( e^{frac{api}{2}} = e^{2} ).Taking natural logarithm on both sides:( frac{api}{2} = 2 ).Solving for ( a ):( a = frac{4}{pi} approx 1.273 ).But wait, our domain is ( a geq 1 ), and ( frac{4}{pi} approx 1.273 ) is greater than 1, so it lies within our domain.Therefore, the critical point is at ( a = frac{4}{pi} ).Now, we need to check whether this critical point is a minimum.Compute the second derivative ( G''(a) ):( G''(a) = frac{d}{da} left( e^{frac{api}{2}} - e^{2} right ) = frac{pi}{2} e^{frac{api}{2}} ).Since ( e^{frac{api}{2}} > 0 ) for all ( a ), ( G''(a) > 0 ). Therefore, the function ( G(a) ) is convex, and the critical point at ( a = frac{4}{pi} ) is a minimum.Therefore, the minimum value of ( G(a) ) occurs at ( a = frac{4}{pi} ).Now, let's compute ( Gleft( frac{4}{pi} right ) ):( Gleft( frac{4}{pi} right ) = frac{2}{pi} e^{frac{pi}{2} cdot frac{4}{pi}} - e^{2} cdot frac{4}{pi} ).Simplify the exponent:( frac{pi}{2} cdot frac{4}{pi} = 2 ).So,( Gleft( frac{4}{pi} right ) = frac{2}{pi} e^{2} - frac{4}{pi} e^{2} = left( frac{2}{pi} - frac{4}{pi} right ) e^{2} = -frac{2}{pi} e^{2} ).Therefore, the minimum value of ( b - e^{2}a ) is ( -frac{2}{pi} e^{2} ).Wait, let me double-check the calculations:( G(a) = frac{2}{pi} e^{frac{api}{2}} - e^{2}a ).At ( a = frac{4}{pi} ):( frac{api}{2} = frac{4}{pi} cdot frac{pi}{2} = 2 ).So, ( e^{frac{api}{2}} = e^{2} ).Thus,( Gleft( frac{4}{pi} right ) = frac{2}{pi} e^{2} - e^{2} cdot frac{4}{pi} = left( frac{2}{pi} - frac{4}{pi} right ) e^{2} = -frac{2}{pi} e^{2} ).Yes, that's correct.Therefore, the minimum value is ( -frac{2}{pi} e^{2} ).So, summarizing:For part (1), the function ( f(x) ) is monotonically increasing on ( [0, frac{3pi}{4}) ) and ( (frac{7pi}{4}, 2pi] ).For part (2), the minimum value of ( b - e^{2}a ) is ( -frac{2}{pi} e^{2} ).**Final Answer**(1) The function is monotonically increasing on the intervals (boxed{[0, frac{3pi}{4})}) and (boxed{(frac{7pi}{4}, 2pi]}).(2) The minimum value of ( b - e^{2}a ) is (boxed{-frac{2e^2}{pi}}).

💡Okay, so I'm trying to figure out if it's possible to place 8 observation stations on an opaque spherical planet with diameter ( D ) such that any celestial body approaching the planet, when it's at a height ( D ) above the planet's surface, is visible from at least two stations. Hmm, that sounds like a geometry problem involving spheres and maybe some optimal placement of points.First, let me visualize the planet as a sphere with diameter ( D ), so its radius is ( frac{D}{2} ). The celestial body is approaching at a height ( D ) above the surface, which means it's at a distance of ( frac{D}{2} + D = frac{3D}{2} ) from the center of the planet. So, the celestial body is moving along a sphere of radius ( frac{3D}{2} ) centered at the planet's center.Now, the observation stations are on the surface of the planet, which is a sphere of radius ( frac{D}{2} ). The goal is to place 8 stations such that any point on the larger sphere (where the celestial body is) is visible from at least two stations. Visibility here probably means that the line of sight from the station to the celestial body doesn't pass through the planet.So, for a station to see the celestial body, the line connecting them shouldn't intersect the planet. That means the angle between the line connecting the station to the center of the planet and the line connecting the center to the celestial body should be greater than 90 degrees. Wait, no, actually, if the angle is greater than 90 degrees, the line would pass through the planet. So, actually, the angle should be less than 90 degrees for visibility.But I need to think about how to ensure that every point on the larger sphere is within the visibility cone of at least two stations. Maybe this relates to covering the sphere with visibility regions from the stations.I recall that placing points on a sphere to maximize coverage often involves regular polyhedrons like the cube, tetrahedron, octahedron, etc. Since we need 8 stations, maybe a cube? A cube has 8 vertices, so placing stations at the vertices of a cube inscribed in the planet's sphere might work.Let me think about that. If I inscribe a cube in the sphere of radius ( frac{D}{2} ), each vertex of the cube will be at a distance ( frac{D}{2} ) from the center. The edge length of the cube can be calculated. For a cube inscribed in a sphere of radius ( r ), the edge length ( a ) is ( a = sqrt{2}r ). Wait, no, that's the space diagonal. The space diagonal of a cube is ( asqrt{3} ), so if the space diagonal equals the diameter of the sphere, which is ( D ), then ( asqrt{3} = D ), so ( a = frac{D}{sqrt{3}} ).So, each edge of the cube is ( frac{D}{sqrt{3}} ). The distance between any two adjacent vertices is ( frac{D}{sqrt{3}} ). But how does this help with visibility?I think I need to consider the angular distance between the stations as seen from the center of the planet. The angular distance between two points on a sphere is the angle between the vectors from the center to each point. For a cube, the angle between adjacent vertices is ( arccosleft(frac{1}{3}right) ) because the dot product of two adjacent vertices is ( frac{1}{3} ) times the product of their magnitudes.Wait, let me verify that. If two adjacent vertices of a cube are at coordinates ( (1,1,1) ) and ( (1,1,-1) ), normalized to the sphere's radius, their dot product is ( 1*1 + 1*1 + 1*(-1) = 1 + 1 - 1 = 1 ). But since the radius is ( frac{D}{2} ), the actual vectors would be scaled by ( frac{D}{2} ). So, the dot product would be ( left(frac{D}{2}right)^2 (1*1 + 1*1 + 1*(-1)) = left(frac{D}{2}right)^2 (1) = frac{D^2}{4} ). The magnitude of each vector is ( frac{D}{2} ), so the cosine of the angle between them is ( frac{frac{D^2}{4}}{left(frac{D}{2}right)^2} = 1 ). Wait, that can't be right because adjacent vertices shouldn't have a zero angle.Maybe I messed up the coordinates. Let's take two adjacent vertices, say ( (1,1,1) ) and ( (1,1,-1) ). The dot product is ( 1*1 + 1*1 + 1*(-1) = 1 + 1 - 1 = 1 ). The magnitude of each vector is ( sqrt{1^2 + 1^2 + 1^2} = sqrt{3} ). So, the cosine of the angle is ( frac{1}{sqrt{3} cdot sqrt{3}} = frac{1}{3} ). So, the angle is ( arccosleft(frac{1}{3}right) ), which is approximately 70.5288 degrees.Okay, so the angle between any two adjacent stations is about 70.5 degrees. Now, how does this relate to the visibility of the celestial body?If a celestial body is at a height ( D ) above the surface, it's at a distance ( frac{3D}{2} ) from the center. The line of sight from a station to the celestial body should not pass through the planet. So, the angle between the station, the center, and the celestial body should be less than 90 degrees.Wait, actually, if the angle is greater than 90 degrees, the line would pass through the planet, so for visibility, the angle should be less than 90 degrees. So, the celestial body must lie within the hemisphere defined by the plane tangent to the planet at the station.But we need to ensure that every point on the larger sphere is within the visibility hemisphere of at least two stations.This seems related to sphere covering problems, where we want to cover the sphere with the minimal number of spherical caps (each cap being the visibility region from a station). The size of each cap is determined by the angular radius, which in this case is 90 degrees because the visibility hemisphere is a cap with angular radius 90 degrees.But wait, actually, the visibility region is a hemisphere, so each station covers a hemisphere. However, to cover the entire sphere with hemispheres, you need at least two stations. But we have 8 stations, so we can certainly cover the sphere multiple times.But the question is not just about covering, but ensuring that every point is covered by at least two stations. So, we need that the intersection of the visibility regions (hemispheres) of any two stations covers the entire sphere.Wait, no, actually, it's the union of the visibility regions that needs to cover the sphere, but each point needs to be in at least two visibility regions.So, each point on the larger sphere must lie within the visibility hemisphere of at least two stations.Given that each station's visibility is a hemisphere, the intersection of two hemispheres is a lens-shaped region. So, to ensure that every point is in at least two hemispheres, the arrangement of stations must be such that every point on the larger sphere is within two hemispheres.But how does the placement of the stations affect this? If the stations are too close together, their hemispheres might overlap too much, leaving some regions only covered by one station. If they're too spread out, some regions might not be covered by any.Wait, but with 8 stations, which is the number of vertices of a cube, maybe the cube arrangement ensures that every point is covered by multiple stations.Let me think about the cube again. Each station is at a vertex of the cube. The cube has 8 vertices, and each vertex is connected to three others. The angular distance between any two stations is either ( arccosleft(frac{1}{3}right) ) for adjacent vertices or larger for non-adjacent ones.If I consider a point on the larger sphere, it's at a distance ( frac{3D}{2} ) from the center. The line from the station to this point must make an angle less than 90 degrees with the line from the center to the station.Wait, actually, the angle between the station, the center, and the celestial body must be less than 90 degrees for visibility. So, the celestial body must lie within the hemisphere defined by the plane through the center and the station.But since the celestial body is on a sphere of radius ( frac{3D}{2} ), which is larger than the planet's radius, the visibility condition is that the angle between the station and the celestial body, as seen from the center, is less than 90 degrees.So, for each station, the set of points on the larger sphere that are visible from it form a spherical cap with angular radius ( arccosleft(frac{text{distance from center to station}}{text{distance from center to celestial body}}right) ).Wait, the distance from the center to the station is ( frac{D}{2} ), and the distance from the center to the celestial body is ( frac{3D}{2} ). So, the angular radius ( theta ) of the cap is given by ( costheta = frac{frac{D}{2}}{frac{3D}{2}} = frac{1}{3} ). So, ( theta = arccosleft(frac{1}{3}right) ), which is approximately 70.5288 degrees.So, each station can see a spherical cap of angular radius about 70.5 degrees on the larger sphere. Now, we need to place 8 such caps such that every point on the larger sphere is covered by at least two caps.This is similar to covering the sphere with overlapping caps. The question is whether 8 caps of angular radius ( arccosleft(frac{1}{3}right) ) can cover the sphere such that every point is in at least two caps.I think the cube arrangement might work because the angular distance between stations is also ( arccosleft(frac{1}{3}right) ). So, if two stations are adjacent on the cube, their caps just touch each other at the edges. But wait, if the angular radius of each cap is equal to the angular distance between stations, then the caps would just touch, not overlap. That might not provide double coverage.Wait, let me think again. If two stations are separated by an angle ( alpha ), and each has a cap of angular radius ( theta ), then the overlap region is where both caps cover the same area. The condition for overlap is that ( alpha < 2theta ). In our case, ( alpha = arccosleft(frac{1}{3}right) ) and ( theta = arccosleft(frac{1}{3}right) ). So, ( 2theta = 2arccosleft(frac{1}{3}right) approx 141.0576 ) degrees. Since ( alpha approx 70.5288 ) degrees, which is less than ( 2theta ), the caps do overlap.Therefore, each pair of adjacent stations' caps overlap, meaning that points near the edge of one cap are covered by the adjacent cap. But does this ensure that every point is covered by at least two caps?I think so, because the cube's symmetry ensures that every point on the sphere is within the cap of at least two stations. Each point on the larger sphere is in the vicinity of some station, and due to the cube's arrangement, it's also near another station.Wait, but actually, the larger sphere is three times the radius of the planet. So, the caps on the larger sphere are smaller relative to the planet's sphere. Hmm, maybe I need to think about the geometry differently.Alternatively, perhaps considering the problem in terms of the planet's sphere and the celestial body's sphere. The celestial body is on a sphere of radius ( frac{3D}{2} ), and the stations are on a sphere of radius ( frac{D}{2} ). The visibility condition is that the line from the station to the celestial body doesn't pass through the planet, which is equivalent to the angle between the station and the celestial body, as seen from the center, being less than 90 degrees.So, the set of points visible from a station is a hemisphere on the celestial body's sphere. But since the celestial body's sphere is larger, the hemisphere is "stretched" over a larger area.Wait, maybe it's better to think about the problem in terms of the planet's sphere and the celestial body's position. For any point on the celestial body's sphere, we need at least two stations such that the angle between the station and the point, as seen from the center, is less than 90 degrees.This is equivalent to saying that the point lies in the intersection of two hemispheres defined by the planes through the center and each of the two stations.But how to ensure that every point is in at least two such hemispheres.I think the cube arrangement might work because each point on the celestial body's sphere is in the vicinity of some station, and due to the cube's symmetry, it's also near another station.But I'm not entirely sure. Maybe I should look for a more rigorous approach.Perhaps considering the concept of spherical codes or sphere coverings. The problem is similar to placing points on a sphere such that every point on another sphere is within a certain distance from at least two points.Wait, but in our case, the distance is angular, and the condition is visibility, which translates to being within a certain angular distance.I recall that the cube vertices provide a good covering of the sphere, but I'm not sure about the exact covering properties.Alternatively, maybe considering that the cube's dual is the octahedron, which has 6 vertices. But we have 8 stations, so cube seems more appropriate.Wait, another approach: if we place 8 stations at the vertices of a cube, then any point on the celestial body's sphere will be in the vicinity of at least two stations because the cube's vertices are spread out enough.But I need to verify this.Let me consider a point on the celestial body's sphere. The angle between this point and the nearest station is less than 90 degrees, so it's visible from that station. But is it also visible from another station?Given the cube's symmetry, for any point, there should be multiple stations within less than 90 degrees. But how many?Actually, for a cube, each point on the sphere is in the vicinity of three stations, corresponding to the three adjacent vertices. But wait, no, because the cube's vertices are only 8, and the sphere is continuous.Wait, maybe considering that each face of the cube is a square, and the center of each face is equidistant from four stations. So, a point near the center of a face would be visible from four stations.But what about points near the edges or vertices? A point near an edge would be visible from two stations, and a point near a vertex would be visible from three stations.Wait, but the celestial body is on a larger sphere, so the angular distances might be different.I think I need to calculate the maximum angular distance from any point on the celestial body's sphere to the nearest station. If this maximum distance is less than 90 degrees, then every point is visible from at least one station. But we need it to be visible from at least two stations.So, perhaps the maximum angular distance to the second nearest station should also be less than 90 degrees.Given that the stations are at the cube's vertices, the angular distances between stations are known. The minimal angular distance is ( arccosleft(frac{1}{3}right) approx 70.5288^circ ), and the maximal is ( 180^circ ) for antipodal points.But since we have 8 stations, antipodal points would be covered by two stations.Wait, but the celestial body is on a sphere of radius ( frac{3D}{2} ), so antipodal points on the planet's sphere correspond to points on the celestial body's sphere that are diametrically opposite.But I'm not sure if that's directly relevant.Alternatively, maybe considering that for any point on the celestial body's sphere, the two closest stations are within a certain angular distance, ensuring that both can see the point.Given that the minimal angular distance between stations is about 70.5 degrees, and the maximal is 180 degrees, but with 8 stations, the average angular distance is probably less.Wait, perhaps the key is that the cube's vertices are such that any point on the sphere is within a certain angular distance from at least two vertices.I think this might be the case, but I'm not entirely sure. Maybe I should look for a theorem or result related to sphere coverings with multiple points.Alternatively, maybe considering that the cube's dual, the octahedron, has 6 vertices, and that might be related, but we have 8 stations.Wait, another idea: if we place 8 stations at the cube's vertices, then any great circle on the sphere will pass through or near at least two stations. Since the celestial body's path is along a great circle (assuming it's approaching along a straight line), it would pass near two stations, making them visible.But I'm not sure if that's necessarily true.Alternatively, maybe considering that the cube's vertices are such that any point on the sphere is within a certain angular distance from at least two vertices, ensuring visibility.I think I need to calculate the maximum angular distance from any point on the sphere to the nearest two stations.Given that the stations are at the cube's vertices, the maximum angular distance to the nearest station is the maximum of the minimal angular distances from any point to the 8 stations.I recall that for a cube, the covering radius is the maximum angular distance from any point on the sphere to the nearest vertex. For a cube, the covering radius is ( arccosleft(frac{1}{sqrt{3}}right) approx 54.7356^circ ). So, every point on the sphere is within about 54.7 degrees of at least one station.But we need to ensure that every point is within 90 degrees of at least two stations. Since 54.7 degrees is less than 90 degrees, and given the cube's symmetry, it's likely that every point is within 90 degrees of multiple stations.In fact, for any point on the sphere, it's within 90 degrees of at least three stations, I believe. Because the cube's vertices are arranged such that each point is near a vertex, and the adjacent vertices are within 70.5 degrees, which is less than 90 degrees.Therefore, placing 8 stations at the vertices of a cube inscribed in the planet's sphere should ensure that any celestial body approaching at height ( D ) is visible from at least two stations.So, the answer is yes, it is possible.

💡Alright, so I've got this problem here: "A 9% stock yields 25%. What is the market value of the stock?" Hmm, okay, let's break this down. First, I need to understand what each part of this statement means.Starting with "9% stock." I think this refers to the dividend rate. So, if a stock is a 9% stock, that means it pays a dividend of 9% of its face value or par value. But wait, what's the face value? I'm not entirely sure, but I think it's usually 100 unless stated otherwise. So, if the face value is 100, then the annual dividend would be 9% of 100, which is 9. Okay, that makes sense.Now, the stock "yields 25%." Yield, in this context, I believe refers to the dividend yield, which is the percentage return on the investment based on the market price of the stock. So, if the yield is 25%, that means the annual dividend (9) is 25% of the market value of the stock.Wait, so if the yield is 25%, and the annual dividend is 9, then I can set up an equation to find the market value. Let me write that down:Yield = (Annual Dividend / Market Value) × 100Plugging in the numbers I have:25% = (9 / Market Value) × 100Now, I need to solve for Market Value. Let's rearrange the equation:Market Value = (9 / 25%) × 100But wait, 25% is the same as 0.25 in decimal form. So,Market Value = (9 / 0.25) × 100Calculating that:9 divided by 0.25 is 36.Then, 36 multiplied by 100 is 3600.Hmm, that seems really high. I mean, if the face value is 100 and the market value is 3600, that's a huge premium. Is that realistic? Maybe I made a mistake somewhere.Let me double-check. The yield is 25%, which is pretty high. If the market value is 3600, then the dividend of 9 would indeed give a 25% yield because 9 is 25% of 3600. But in reality, stocks don't usually trade at such high multiples of their face value unless they're in a speculative bubble or something. Maybe the face value isn't 100? Or perhaps I misinterpreted the yield.Wait, another thought: maybe the yield is 25%, and the dividend is 9% of the market value, not the face value. Let me think about that. If the stock is a 9% stock, does that mean the dividend is 9% of the market value or the face value?I think it's typically 9% of the face value, which is usually 100. So, the annual dividend is 9. But if the yield is 25%, that's based on the market value. So, the formula should be:Yield = (Dividend / Market Value) × 100So, 25 = (9 / Market Value) × 100Then, solving for Market Value:Market Value = (9 / 25) × 100Which is:Market Value = 0.36 × 100 = 36Oh, wait, that's different. So, the market value is 36.But earlier, I thought it was 3600 because I multiplied by 100 again. I see where I went wrong. I shouldn't multiply by 100 again because the yield is already a percentage. So, the correct market value is 36.That makes more sense because 36 is closer to the face value of 100, although it's still a discount. But a 25% yield on a 36 stock with a 9 dividend does check out because 9 divided by 36 is 0.25, which is 25%.So, I think the correct market value is 36.

💡Okay, so I have this inequality to solve: |2x + 5| ≥ 7. Hmm, absolute value inequalities can sometimes be tricky, but I remember that they involve two separate cases. Let me try to recall how to approach this.First, I know that if |A| ≥ B, then either A ≥ B or A ≤ -B. So applying that to my inequality, I can split it into two separate inequalities:1. 2x + 5 ≥ 72. 2x + 5 ≤ -7Let me solve each of these separately.Starting with the first one: 2x + 5 ≥ 7.Subtract 5 from both sides: 2x ≥ 2.Then divide both sides by 2: x ≥ 1.Okay, so that gives me one part of the solution: x is greater than or equal to 1.Now, the second inequality: 2x + 5 ≤ -7.Subtract 5 from both sides: 2x ≤ -12.Divide both sides by 2: x ≤ -6.So, putting both solutions together, the inequality |2x + 5| ≥ 7 holds when x is either greater than or equal to 1 or less than or equal to -6. So, in interval notation, that would be (-∞, -6] ∪ [1, ∞).Now, the question is asking for a necessary but not sufficient condition for this inequality to hold. Let me make sure I understand what that means.A necessary condition is something that must be true for the inequality to hold, but it's not enough on its own to guarantee the inequality. In other words, if the inequality is true, then the necessary condition must also be true, but the necessary condition could be true even if the inequality isn't.Looking at the options:A: x ≥ 1B: x ≤ -6C: x ≥ 1 or x ≤ -6D: x ≠ 0Hmm. Let me analyze each option.Option A: x ≥ 1. Well, from my solution, x needs to be ≥1 or ≤-6. So if x is ≥1, then the inequality holds. But is x ≥1 a necessary condition? That is, does the inequality imply that x must be ≥1? No, because x could also be ≤-6. So x ≥1 is not a necessary condition because the inequality can hold without x being ≥1 (it could be ≤-6 instead). So A is not a necessary condition.Option B: x ≤ -6. Similar reasoning as above. If x is ≤-6, then the inequality holds. But is it necessary? No, because x could be ≥1 instead. So B is also not a necessary condition.Option C: x ≥1 or x ≤-6. Wait, that's exactly the solution set I found earlier. So this is both necessary and sufficient. If x is in this set, the inequality holds, and if the inequality holds, x must be in this set. So this is a necessary and sufficient condition, but the question is asking for a necessary but not sufficient condition. So C is not the answer.Option D: x ≠ 0. Let's see. If x is not equal to 0, does that mean the inequality holds? Well, let's test x = 0. Plugging into |2(0) + 5| = |5| = 5, which is less than 7. So at x=0, the inequality does not hold. Therefore, if the inequality holds, x cannot be 0. So x ≠ 0 is a necessary condition because whenever the inequality is true, x must not be 0. But is it sufficient? No, because x could be, say, 0.5, which is not 0, but |2(0.5) +5| = |6| = 6, which is still less than 7. So x ≠ 0 doesn't guarantee the inequality holds. Therefore, D is a necessary but not sufficient condition.Let me double-check my reasoning. If the inequality holds, x must be either ≥1 or ≤-6, so x cannot be between -6 and 1, including 0. So x ≠ 0 is indeed a necessary condition because whenever the inequality is true, x is not 0. But it's not sufficient because x could be something else, like 0.5, which doesn't satisfy the inequality. So D seems correct.I think I might have confused necessary and sufficient earlier, but now I'm pretty sure D is the right answer.

💡Okay, so I have this problem where I need to find ( a^3 + b^3 + c^3 ) given some information about ( a ), ( b ), and ( c ). The given information is:1. ( a + b + c = 8 )2. ( ab + ac + bc = 10 )3. ( abc = -15 )Hmm, I remember there's a formula that relates ( a^3 + b^3 + c^3 ) to these sums. Let me try to recall it. I think it's something like:[ a^3 + b^3 + c^3 = (a + b + c)^3 - 3(a + b + c)(ab + ac + bc) + 3abc ]Wait, is that right? Or is it another formula? Maybe I should check. Alternatively, I think there's another identity:[ a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) + 3abc ]Yeah, that sounds more familiar. So, I need to find ( a^2 + b^2 + c^2 ) first because I don't have that value directly. How can I find ( a^2 + b^2 + c^2 )?Oh, right! There's another identity that relates ( (a + b + c)^2 ) to ( a^2 + b^2 + c^2 ) and ( ab + ac + bc ). Specifically:[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) ]So, plugging in the known values:[ 8^2 = a^2 + b^2 + c^2 + 2 times 10 ][ 64 = a^2 + b^2 + c^2 + 20 ]Subtracting 20 from both sides:[ a^2 + b^2 + c^2 = 64 - 20 = 44 ]Alright, so ( a^2 + b^2 + c^2 = 44 ). Now, going back to the formula for ( a^3 + b^3 + c^3 ):[ a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) + 3abc ]Let me plug in the known values step by step. First, ( a + b + c = 8 ), ( a^2 + b^2 + c^2 = 44 ), and ( ab + ac + bc = 10 ). So, the term inside the parentheses becomes:[ 44 - 10 = 34 ]So now, the formula simplifies to:[ a^3 + b^3 + c^3 = 8 times 34 + 3abc ]We know ( abc = -15 ), so:[ a^3 + b^3 + c^3 = 8 times 34 + 3 times (-15) ]Calculating each part:- ( 8 times 34 = 272 )- ( 3 times (-15) = -45 )Adding these together:[ 272 - 45 = 227 ]So, ( a^3 + b^3 + c^3 = 227 ).Wait, let me double-check my steps to make sure I didn't make any mistakes. First, I used the identity for ( (a + b + c)^2 ) correctly to find ( a^2 + b^2 + c^2 ). Then, I substituted into the sum of cubes formula. The calculations seem straightforward, and I don't see any arithmetic errors. So, I think 227 is the correct answer.**Final Answer**The value of ( a^3 + b^3 + c^3 ) is boxed{227}.

💡Alright, so I've got this problem here about CD thickness and quality standards. Let me try to wrap my head around it step by step. Okay, the problem says that a CD is qualified if its thickness is (1.2 ± 0.1) mm. That means the thickness can vary a little around 1.2 mm, right? Specifically, it can be 0.1 mm more or less than 1.2 mm.So, first things first, I need to figure out what the acceptable range of thickness is for a CD to be considered qualified. If it's 1.2 ± 0.1 mm, that should mean the thickness can be as low as 1.2 - 0.1 mm and as high as 1.2 + 0.1 mm. Let me write that down:Lower limit = 1.2 mm - 0.1 mm = 1.1 mmUpper limit = 1.2 mm + 0.1 mm = 1.3 mmSo, any CD that's between 1.1 mm and 1.3 mm thick is good. Now, the question is asking which of the given measurements is not qualified. The options are:A: 1.12 mmB: 1.22 mmC: 1.28 mmD: 1.32 mmAlright, let's go through each option one by one and see if they fall within the acceptable range.Starting with option A: 1.12 mm. Hmm, 1.12 mm is just slightly above the lower limit of 1.1 mm. So, 1.12 mm is within the range of 1.1 mm to 1.3 mm. That means option A is qualified.Moving on to option B: 1.22 mm. Okay, 1.22 mm is a bit higher than 1.2 mm but still below the upper limit of 1.3 mm. So, 1.22 mm is also within the acceptable range. Option B is qualified.Next is option C: 1.28 mm. Let's see, 1.28 mm is getting closer to the upper limit. Is 1.28 mm less than or equal to 1.3 mm? Yes, it is. So, 1.28 mm is still within the qualified range. Option C is okay.Now, option D: 1.32 mm. Hmm, this one is a bit tricky. The upper limit is 1.3 mm, and 1.32 mm is just a little over that. So, 1.32 mm exceeds the maximum allowable thickness of 1.3 mm. That means option D is outside the acceptable range and therefore not qualified.Wait, let me double-check to make sure I didn't make a mistake. The lower limit is 1.1 mm, and the upper limit is 1.3 mm. So, any measurement below 1.1 mm or above 1.3 mm is not qualified. Let's verify each option again:- A: 1.12 mm is above 1.1 mm and below 1.3 mm. Qualified.- B: 1.22 mm is also within the range. Qualified.- C: 1.28 mm is still within the range. Qualified.- D: 1.32 mm is above 1.3 mm. Not qualified.Yep, that seems correct. So, option D is the one that doesn't meet the quality standard.I wonder if there's any chance I misread the tolerance or the limits. The problem says 1.2 ± 0.1 mm, which clearly means from 1.1 mm to 1.3 mm. So, I think my calculations are right. There's no ambiguity there.Another thing to consider is whether the measurements are precise enough. For example, 1.12 mm is given to two decimal places, but the limits are given to one decimal place. However, in quality control, measurements are usually precise, so 1.12 mm is still within the 1.1 mm to 1.3 mm range.Also, thinking about the manufacturing process, having a tolerance of ±0.1 mm is pretty standard. It allows for slight variations without affecting the functionality of the CD. So, any CD that's too thin or too thick might not work properly or could be more prone to damage.In conclusion, after carefully analyzing each option and ensuring my calculations are correct, I'm confident that option D: 1.32 mm is the measurement that doesn't meet the quality standard.