🧠:[<think>Okay, let's see. Marko drew a rectangle with two blue sides of length 24 and two red sides of length 36. He colored each point inside the rectangle based on the nearest side. If a point is equally distant from a blue and a red side, it's colored black. I need to find the area of the red part. Hmm, so first, let me visualize this rectangle. The blue sides are 24 units long, and the red sides are 36 units. Wait, so the rectangle has sides of 24 and 36. But which sides are blue and which are red? The problem says two blue sides of length 24 and two red sides of length 36. So the rectangle must have opposite sides of the same color. So, it's a rectangle with length 36 and width 24, where the two shorter sides (24 units) are blue, and the two longer sides (36 units) are red. Got it.Now, he colored each interior point with the color of the nearest side. Points equidistant to a blue and red side are black. We need to figure out the area of the red region. So, essentially, the rectangle is divided into regions based on proximity to the sides. The regions closest to the red sides will be red, those closest to the blue sides will be blue, and the lines where they are equidistant will be black. The black lines probably form the boundaries between the red and blue regions.First, let's recall that in a rectangle, the set of points equidistant from two opposite sides is a line parallel to those sides, midway between them. So, for example, the set of points equidistant from the top and bottom sides (which are red, since they are length 36) would be a horizontal line in the middle of the rectangle. Similarly, the set of points equidistant from the left and right sides (blue, length 24) would be a vertical line in the middle. Wait, but here, the colors are different for different pairs of sides. The vertical sides (left and right) are blue, and the horizontal sides (top and bottom) are red. So, each point inside the rectangle is closest to either a vertical (blue) side or a horizontal (red) side. But the problem states that points equidistant to a blue and red side are black. So, maybe the black lines are the diagonals? Wait, no. Wait, the distance from a point to a side is the perpendicular distance. So, for example, the distance from a point (x, y) to the left side (x=0) is x, to the right side (x=24) is 24 - x. Similarly, the distance to the top (y=36) is 36 - y, and to the bottom (y=0) is y.But the problem says "the color of the nearest side". So, each point is colored based on the closest side, regardless of color. If it's closest to a blue side, it's blue; closest to a red side, it's red. If it's equally close to a blue and a red side, then it's black. Wait, but the sides of the rectangle are colored blue and red. So there are four sides: left (blue), right (blue), top (red), bottom (red). So, for a point inside the rectangle, we need to compute the distance to each of the four sides and determine which is the smallest. If the smallest distance is to a blue side, color it blue; if to a red side, color it red. If there's a tie between a blue and red side, color it black.Therefore, the regions in the rectangle will be determined by the Voronoi diagram of the four sides. The Voronoi diagram partitions the plane into regions based on proximity to each object. In this case, the objects are the four sides of the rectangle. The regions closest to each side will be adjacent to that side, bounded by the bisectors between pairs of sides.However, since the sides are arranged as a rectangle, the Voronoi regions will be determined by the perpendicular distances. Let me think. For each side, the region closest to it would be the area within the rectangle where the perpendicular distance to that side is less than the distance to any other side.Given that the rectangle has length 36 (red top and bottom) and width 24 (blue left and right), let's set up coordinates. Let me place the rectangle with its bottom-left corner at (0, 0), so the sides are:- Left: x = 0 (blue)- Right: x = 24 (blue)- Bottom: y = 0 (red)- Top: y = 36 (red)So any point inside the rectangle is (x, y) where 0 < x < 24 and 0 < y < 36.The distance from a point (x, y) to the left side (x=0) is x.The distance to the right side (x=24) is 24 - x.The distance to the bottom (y=0) is y.The distance to the top (y=36) is 36 - y.For each point, we need to find the minimum of these four distances. If the minimum is x or 24 - x (distances to blue sides), the point is blue. If the minimum is y or 36 - y (distances to red sides), the point is red. If there's a tie between a blue and red distance, it's black.Therefore, the regions will be as follows:1. Near the left side (blue): points where x is less than the distances to the right, top, and bottom sides. So x < 24 - x, x < y, and x < 36 - y.2. Near the right side (blue): points where 24 - x is less than x, y, and 36 - y.3. Near the bottom side (red): points where y is less than x, 24 - x, and 36 - y.4. Near the top side (red): points where 36 - y is less than x, 24 - x, and y.The black lines occur where a point is equidistant to a blue and a red side. For example, where x = y (distance to left side equals distance to bottom side), or x = 36 - y (distance to left side equals distance to top side), etc. Similarly for other combinations.But since the rectangle's Voronoi regions for the sides will form a sort of "medial axis" where regions are bounded by bisectors between sides.To find the area of the red regions, we need to determine the areas within the rectangle where the minimum distance is to a red side (either top or bottom) and compute their total area.Alternatively, compute the regions closest to the top and closest to the bottom, which are both red, and sum their areas.Similarly, the blue regions will be the regions closest to the left and right sides.The black lines are the boundaries between these regions.So, let's first find the bisectors between the red and blue sides.Starting with the bottom (red) and left (blue) sides. The set of points equidistant to the bottom (y=0) and left (x=0) sides is the line where y = x. Similarly, equidistant to bottom (y=0) and right (x=24): y = 24 - x. Equidistant to top (y=36) and left (x=0): 36 - y = x. Equidistant to top (y=36) and right (x=24): 36 - y = 24 - x. Then, equidistant between top and bottom (red sides): y = 18. Equidistant between left and right (blue sides): x = 12.Wait, but the problem says points equidistant to a blue and a red side are colored black. So the bisectors between a red and blue side are black lines, while the bisectors between two red sides (top and bottom) or two blue sides (left and right) would partition the regions further but not colored black. Wait, but in that case, the Voronoi diagram would have bisectors between all pairs of sides, but only the red-blue bisectors are black. The red-red and blue-blue bisectors would just be internal boundaries within the red and blue regions, respectively.So, for example, the line y = 18 is equidistant from top and bottom (both red), so points on that line are still in the red region but just the boundary between the upper and lower red regions. Similarly, x = 12 is equidistant from left and right (blue), so it's the boundary between left and right blue regions. These lines are not colored black because they are equidistant to two sides of the same color.Therefore, the black lines are only those where a point is equidistant to a red and a blue side. So these are the four bisectors between red and blue sides: bottom-left (y = x), bottom-right (y = 24 - x), top-left (36 - y = x), and top-right (36 - y = 24 - x). These four lines form the boundaries of the red and blue regions.Thus, the red regions are near the top and bottom, bounded by the lines equidistant to the top/bottom and left/right sides.To compute the area of the red regions, let's first consider the regions near the top and bottom.Starting with the bottom red region. The region closest to the bottom (y=0) is where the distance to the bottom (y) is less than the distance to the top (36 - y), and also less than the distance to the left (x) and right (24 - x) sides.So, for a point near the bottom, it must satisfy y < 36 - y (which simplifies to y < 18), and y < x, and y < 24 - x.Similarly, the top red region is where 36 - y < y (which simplifies to y > 18), and 36 - y < x, and 36 - y < 24 - x.Therefore, the red region is the union of two regions:1. Near the bottom: 0 < y < 18, y < x, y < 24 - x.2. Near the top: 18 < y < 36, 36 - y < x, 36 - y < 24 - x.Each of these regions is a polygon whose area we can compute.Let me first compute the bottom red region. The conditions are y < x, y < 24 - x, and y < 18. So, in the lower half of the rectangle (y < 18), the region is bounded by y < x and y < 24 - x. These two lines intersect at x = 12, y = 12. Wait, let me check:The lines y = x and y = 24 - x intersect where x = 24 - x => 2x = 24 => x = 12, so y = 12. So, the two lines meet at (12, 12). But since we're in the region y < 18, which is below y = 18, and the intersection at (12, 12) is below 18, so the region near the bottom bounded by y < x and y < 24 - x is a quadrilateral with vertices at (0, 0), (12, 12), (24, 12), (24, 0), but wait, no. Wait, actually, for the bottom red region, the boundary is where y = x and y = 24 - x. So, between x=0 and x=12, the boundary is y = x; between x=12 and x=24, the boundary is y = 24 - x. However, since we are considering y < x and y < 24 - x, the region under both these lines. So, for each x, the upper limit of y is the minimum of x and 24 - x. So the upper boundary is a V-shape peaking at (12,12). Therefore, the bottom red region is a triangle with vertices at (0, 0), (12, 12), (24, 0). Wait, but actually, since y < x and y < 24 - x, and y < 18. Since 12 < 18, the upper boundary is indeed the V-shape up to (12, 12), and then y cannot exceed 12 in that region. Wait, but if y < x and y < 24 - x, then for x between 0 and 12, the maximum y is x, and for x between 12 and 24, the maximum y is 24 - x. So the shape is a diamond or a triangle?Wait, let's think of it as follows: for each x from 0 to 24, the maximum y is the minimum of x and 24 - x. So for x from 0 to 12, y goes up to x; for x from 12 to 24, y goes up to 24 - x. So if you plot this, it's a triangle with vertices at (0,0), (12,12), (24,0). Thus, the area of the bottom red region is the area of this triangle, which is (base * height)/2 = (24 * 12)/2 = 144.But wait, hold on. The triangle is from (0,0) to (12,12) to (24,0). The base along the bottom is 24, and the height is 12. So area is 24 * 12 / 2 = 144. That's correct.But wait, but we also have the condition that y < 18. However, in this case, the maximum y in the bottom red region is 12, which is less than 18, so the condition y < 18 is automatically satisfied. So the bottom red region is indeed that triangle with area 144.Now, moving to the top red region. Similarly, the region closest to the top is where 36 - y < x and 36 - y < 24 - x, and 36 - y < y (which is y > 18). Let's parse these conditions.First, 36 - y is the distance to the top. So, for a point to be closer to the top than to the left, right, or bottom, we need 36 - y < x (distance to top < distance to left), 36 - y < 24 - x (distance to top < distance to right), and 36 - y < y (distance to top < distance to bottom). Let's solve these inequalities.First, 36 - y < x => x > 36 - y.Second, 36 - y < 24 - x => x < y - 12.Third, 36 - y < y => 36 < 2y => y > 18.So, combining these:y > 18,x > 36 - y,x < y - 12.So, we have y > 18,36 - y < x < y - 12.But we need to check if these inequalities are possible. Let's see. For y > 18, 36 - y is less than 18 (since when y=19, 36 -19=17; when y=36, 36 -36=0). And y -12 when y >18 would be greater than 6 (since 18 -12=6). So for y between 18 and 36, 36 - y ranges from 18 to 0, and y -12 ranges from 6 to 24. Therefore, the interval for x is (36 - y, y -12). However, we need to ensure that 36 - y < y -12. Let's solve 36 - y < y -12:36 +12 < y + y => 48 < 2y => y > 24.Therefore, for y > 24, 36 - y < y -12. So the interval for x is valid (non-empty) only when y > 24. Wait, so between y=18 and y=24, 36 - y is still greater than y -12. For example, at y=20: 36 -20=16, y-12=8. But 16 >8, so the interval x must be >16 and <8, which is impossible. So no solution here. Hence, the top red region exists only when y >24, and then x is between 36 - y and y -12.Therefore, the top red region is a quadrilateral bounded by y >24, x >36 - y, x < y -12, and x between 0 and24. Let's visualize this.When y >24:- 36 - y is less than 12 (since y=25 gives 11, y=30 gives 6, etc.)- y -12 is greater than 12 (since y=25 gives13, y=30 gives18, etc.)Therefore, for each y between24 and36, x must be between36 - y and y -12.So, integrating over y from24 to36, the x range is from36 - y to y -12.Let me compute the area of the top red region. It's the integral from y=24 to y=36 of (upper x - lower x) dy.Upper x is y -12, lower x is36 - y.Thus, the width at each y is (y -12) - (36 - y) = 2y -48.Therefore, area = ∫ from24 to36 (2y -48) dy.Compute the integral:∫(2y -48) dy = y² -48y evaluated from24 to36.At y=36: 36² -48*36 =1296 -1728= -432.At y=24: 24² -48*24=576 -1152= -576.So the difference is (-432) - (-576)=144.Therefore, the area of the top red region is144.Therefore, total red area is bottom red area (144) + top red area (144) =288.But wait, the total area of the rectangle is24*36=864. If red area is288, then blue area would also be288, and the remaining area would be black? But 288 +288=576, which is less than864. The difference is864 -576=288. So that would mean black area is288. But that seems like a lot.Wait, maybe I made a mistake here. Let me check again.First, the bottom red region: triangle with area144. The top red region: another region with area144. Total red area288. Then the blue regions would be similar. Let's check the blue regions.The blue regions are near the left and right sides. Let's compute the left blue region.For the left blue region, the distance to the left side (x) must be less than distance to right (24 -x), top (36 - y), and bottom (y). So:x <24 -x =>x <12,x <36 - y,x < y.So the left blue region is where x <12, x <36 - y, andx < y.Similarly, the right blue region is where24 -x <x =>x >12,24 -x <36 - y,24 -x < y.So let's compute the left blue region.Conditions:x <12,x <36 - y,x < y.So, in the left half of the rectangle (x <12), the region is bounded byx < y andx <36 - y. So for each x from0 to12, y must be greater thanx and less than36 -x.Wait, but x <36 - y implies y <36 -x. So, the region is between y =x and y=36 -x for x from0 to12.Wait, but at x=0, y must be between0 and36. But sincex < y andx <36 - y, whenx=0, y must be greater than0 and less than36. But sincex=0, the left blue region atx=0 is all y except y=0 (distance to left side is0, so it's colored blue, but the bottom side is red. Wait, but atx=0, the distance to left side is0, which is less than distance to bottom (y). So actually, the entire left side (x=0) is blue. Similarly, the entire right side (x=24) is blue. The entire top and bottom sides (y=0 andy=36) are red.Wait, but according to the problem statement, points that are equally distant to a blue and red side are colored black. So the edges themselves: the sides are part of their respective colors? For example, the left side (x=0) is entirely blue, right? Because all points on the left side have distance0 to the left (blue) side, which is less than distance to any other side. Similarly, points on the right side are blue, top and bottom sides are red.But when considering regions near the sides, the regions adjacent to the sides will be colored accordingly. However, the regions computed earlier (the triangles and the top region) are the areas closer to the respective sides than to the others. So perhaps my initial calculation is correct.But if the total red area is288 and blue area is288, then the remaining area is black. But that would be864 -288 -288=288. But why is the black area so large?Wait, maybe not. Because the black regions are only the lines where a point is equidistant to a blue and red side. But lines have zero area. Wait, the problem says points that are equally distant to a blue and red side are colored black. But in a rectangle, the set of points equidistant to a blue and a red side would be the bisectors between them, which are line segments. Since lines have no area, the black region should have area zero. Therefore, the total red and blue areas should add up to the entire area of the rectangle. But according to my previous calculation, red and blue areas sum to576, which is less than864. Therefore, there must be a mistake in my calculation.Wait, that suggests that my reasoning is flawed. Let me re-examine.Wait, perhaps the regions are not as simple as triangles and quadrilaterals. Maybe the Voronoi regions for the sides in the rectangle create more complex shapes. Let's think again.For any point inside the rectangle, its color is determined by the closest side. So, for example, near the bottom-left corner, the point is closest to the bottom (red) or left (blue) side. The bisector between the bottom and left sides is the line y = x. So, below that line, closer to the bottom; above, closer to the left. Similarly, near the bottom-right corner, the bisector is y =24 -x. So below that line, closer to the bottom; above, closer to the right.Similarly, near the top-left corner, the bisector between top (red) and left (blue) is y=36 -x. Above that line, closer to the top; below, closer to the left. Near the top-right corner, the bisector between top (red) and right (blue) is y=x +12. Wait, let's compute.Wait, the distance to the top side is36 - y. The distance to the right side is24 -x. Setting them equal:36 - y=24 -x =>y=12 +x.Similarly, the bisector between top (red) and right (blue) is y =x +12.Similarly, the bisector between top (red) and left (blue) is36 - y =x =>y=36 -x.Similarly, bisector between bottom (red) and left (blue) isy =x.Bisector between bottom (red) and right (blue) isy=24 -x.So the Voronoi diagram consists of these four bisectors:1. y =x (bottom-left bisector)2. y=24 -x (bottom-right bisector)3. y=36 -x (top-left bisector)4. y=x +12 (top-right bisector)These four lines divide the rectangle into four regions:- The region closest to the bottom (red), bounded by y <=x and y <=24 -x.- The region closest to the top (red), bounded by y >=36 -x and y >=x +12.- The region closest to the left (blue), bounded by y >=x and y <=36 -x.- The region closest to the right (blue), bounded by y >=24 -x and y <=x +12.Additionally, the central region might be bounded by these four lines. Wait, but let's see if these four lines intersect each other inside the rectangle.First, check where y =x and y=36 -x intersect. Setx =36 -x =>2x=36 =>x=18. But sincex can only go up to24, and y=18 here. So intersection at(18,18).Similarly, y=24 -x and y=x +12 intersect where24 -x =x +12 =>24 -12=2x =>12=2x =>x=6, y=24 -6=18. So intersection at(6,18).Similarly, y=36 -x and y=x +12 intersect where36 -x =x +12 =>36 -12=2x =>24=2x =>x=12, y=12 +12=24. So intersection at(12,24).And y=24 -x and y=36 -x intersect where24 -x=36 -x =>24=36, which is impossible. So they don't intersect.Wait, maybe I need to check all possible intersections.But given that these four lines form a polygon in the center? Let me plot mentally.The lines:1. y =x from (0,0) to(18,18).2. y=24 -x from(0,24) to(24,0). Wait, but in our case, the rectangle is up toy=36. Hmm, actually, the line y=24 -x would start atx=0, y=24 and go tox=24, y=0. But our rectangle's height is36, so this line goes from(0,24) down to(24,0). Similarly, the line y=36 -x starts atx=0, y=36 and goes tox=36, y=0, but sincex only goes up to24, this line goes from(0,36) to(24,12). The line y=x +12 starts atx=0, y=12 and goes tox=12, y=24, and beyond, but within the rectangle, it goes from(0,12) to(12,24).So now, let's see the intersections:- y=x and y=36 -x intersect at(18,18).- y=24 -x and y=x +12 intersect at(6,18).- y=36 -x and y=x +12 intersect at(12,24).- y=x and y=x +12 don't intersect.- y=24 -x and y=36 -x don't intersect.So, connecting these intersection points: (18,18), (6,18), (12,24). Wait, but there's also intersection points with the edges of the rectangle.But perhaps the central region is a polygon bounded by these lines. Let me try to outline the regions:1. Bottom red region: Bounded by y <=x and y <=24 -x. This is a triangle with vertices at(0,0), (12,12), (24,0). Wait, but in reality, the line y=24 -x intersects the bottom edge (y=0) atx=24, but since our rectangle is up toy=36, the line y=24 -x starts at(0,24) and goes to(24,0). So the bottom red region is indeed the triangle under y= min(x,24 -x), which peaks at(12,12). So area=144 as before.2. Top red region: Bounded by y >=36 -x and y >=x +12. Let's see where these lines are. The line y=36 -x goes from(0,36) to(24,12). The line y=x +12 goes from(0,12) to(12,24). These two lines intersect at(12,24). So the top red region is a quadrilateral with vertices at(12,24), (24,12), (24,36), (0,36). Wait, let's check:Wait, the region where y >=36 -x and y >=x +12. For x from0 to12, y >=36 -x and y >=x +12. But36 -x atx=0 is36, andx +12 atx=0 is12. So forx from0 to12, y must be >=36 -x (which is >=36 -12=24 whenx=12). So in this range, the lower bound is36 -x. Forx from12 to24, y >=x +12 andy >=36 -x. But36 -x whenx=12 is24, andx +12 whenx=12 is24. Then, asx increases,36 -x decreases andx +12 increases. So beyondx=12, the lower bound isx +12. Thus, the top red region is bounded by:- Fromx=0 toy=36 up tox=12, followingy=36 -x.- Fromx=12 tox=24, followingy=x +12.- The top edge of the rectangle aty=36.Wait, no. If we are considering y >=36 -x and y >=x +12, then in the regionx from0 to12, y must be >=36 -x (since36 -x >=x +12 in this range). Let's check atx=0:36 -0=36, x +12=12. So y >=36. But the top of the rectangle isy=36, so atx=0, the only point is(0,36). Similarly, atx=12, y >=24. Then forx from12 to24, y >=x +12, which goes from24 toy=36 asx goes from12 to24.But the top red region is actually a triangle with vertices at(12,24), (24,36), and(0,36). Wait, let's see:Wait, whenx=0, y must be >=36, but the maximum y is36. So the point is(0,36). Whenx=12, y >=24, but the liney=36 -x atx=12 isy=24, andy=x +12 atx=12 is24. So the point is(12,24). Whenx=24, y >=36, but the maximum y is36. So the point is(24,36). But betweenx=12 andx=24, the lower boundary isy=x +12. Atx=24, y=x +12=36. So the top red region is a quadrilateral with vertices at(0,36), (12,24), (24,36). Wait, but connecting these points:From(0,36) to(12,24) along y=36 -x.From(12,24) to(24,36) along y=x +12.And from(24,36) back to(0,36) alongy=36.So this is actually a trapezoid with bases aty=36 (fromx=0 tox=24) and the line from(0,36) to(12,24) to(24,36). Wait, but it's actually two triangles?Wait, let's compute the area.Fromx=0 tox=12, the region is a line from(0,36) to(12,24). So the area under this line is a triangle with vertices at(0,36), (12,24), (12,36). Wait, but y >=36 -x forx from0 to12, so the area is the integral fromx=0 tox=12 of (36 - (36 -x)) dx? Wait, no.Wait, actually, for eachx from0 to12, y ranges from36 -x to36. So the area is ∫ from0 to12 [36 - (36 -x)] dx = ∫ from0 to12x dx= [x²/2] from0 to12=72.Fromx=12 tox=24, y ranges fromx +12 to36. So the area is ∫ from12 to24 [36 - (x +12)] dx= ∫ from12 to24 (24 -x) dx= [24x -x²/2] from12 to24.Atx=24:24*24 -24²/2=576 -288=288.Atx=12:24*12 -12²/2=288 -72=216.Difference:288 -216=72.So total area of top red region is72 +72=144.Therefore, the top red region is also144, same as the bottom red region. Thus total red area is144 +144=288.But as before, this leaves864 -288=576 for blue and black. But the problem states that points equidistant to a blue and red side are black. However, lines have zero area, so the black area should be zero. Therefore, this suggests that my calculation is wrong.Wait, no, perhaps the central region is actually colored blue. Wait, let's check.The central region would be where the closest side is left or right (blue). Wait, but according to the Voronoi regions, the central area is actually the region closest to the left or right sides. Let's verify.Take the center of the rectangle at(12,18). The distance to the left side is12, to the right side is12, to the bottom is18, to the top is18. So equidistant to left/right and top/bottom. But according to the problem statement, points equidistant to a blue and red side are black. Since the center is equidistant to two blue sides (left and right) and two red sides (top and bottom). But the problem says points equidistant to a blue and red side are black. So if a point is equidistant to multiple sides, but at least one blue and one red, it's black. The center is equidistant to all four sides, so it's black. Similarly, the lines where equidistant to blue and red sides are black.However, in our previous calculation, we only accounted for the regions closest to the sides, but maybe the central area is a diamond-shaped black region?Wait, perhaps my mistake was assuming that the regions closest to the top and bottom are only the areas calculated, but in reality, there is a central black region where points are equidistant to blue and red sides, which my calculation didn't account for, leading to the discrepancy in area.Wait, but according to the problem statement, only the points that are equidistant to a blue and a red side are black. So if a point is equidistant to, say, the left (blue) and top (red) sides, it's black. Similarly, equidistant to left and bottom, right and top, etc. But in the Voronoi diagram, these bisectors are the lines y=x, y=24 -x, y=36 -x, and y=x +12. The intersection of these lines forms a central polygon where all points inside are equidistant to two different color sides, hence black.Therefore, the central region bounded by these four lines is entirely black. Therefore, the area of this central black region must be subtracted from the total area to get the red and blue regions.But in my previous calculation, I only considered the top and bottom red regions as 144 each, totaling288, but in reality, the central black region would reduce the red and blue areas. Wait, but how?Wait, let me think again. The Voronoi regions for the four sides are the four regions closest to each side. The areas outside the central black region are the red and blue regions. However, the central black region is where points are equidistant to two sides of different colors. So actually, the red and blue regions are the areas closest to the respective sides, excluding the black lines. But since lines have no area, the total red and blue areas should sum to the entire area of the rectangle. However, my calculation gives red area as288 and blue area must also be288, totaling576, which is less than864. Therefore, my calculation is wrong.Wait, maybe the regions are not just the triangles and the top/bottom areas, but include more areas.Wait, perhaps the regions closest to the left and right sides (blue) are larger. Let's recast the problem.Each point is colored based on the closest side. The four sides are left (x=0, blue), right (x=24, blue), top (y=36, red), and bottom (y=0, red). For any point (x,y), compute the distances to all four sides and determine the minimum.But for points not on the edges, the closest side is either left, right, top, or bottom. The regions where the closest side is left or right are blue; where it's top or bottom, red.Therefore, to find the area of the red part, we need to find all points where the closest side is top or bottom.So let's model this.For a point (x,y), the distance to the left isx, right is24 -x, top is36 -y, bottom isy.The minimum of these four distances determines the color.We need to find the set of points where min(x, 24 -x, 36 -y, y) is either36 -y ory (i.e., top or bottom). The area of this set is the red area.Similarly, the blue area is where the minimum isx or24 -x.Therefore, the total red area is the union of two regions:1. Points wherey < min(x, 24 -x, 36 -y).2. Points where36 -y < min(x, 24 -x, y).Similarly for blue.Alternatively, for a point to be closer to the bottom (y) than to any other side, it must satisfyy <x,y <24 -x, andy <36 -y. The last inequality simplifies toy <18.Similarly, to be closer to the top (36 -y), it must satisfy36 -y <x,36 -y <24 -x, and36 -y <y, which simplifies toy >18.Therefore, the red regions are:- Bottom red:y <18,y <x,y <24 -x.- Top red:y >18,36 -y <x,36 -y <24 -x.As computed before, each of these has area144, totaling288.But this contradicts the total area. Where is the mistake?Wait, the problem is probably that in the central area, the closest side is not top/bottom or left/right, but actually a combination. Wait, no. Every point must have a closest side. Unless it's equidistant to two sides, which are then colored black. But since those are lines, they don't contribute to the area.Wait, let's take an example point in the center, say(12,18). It is equidistant to all four sides: distance to left=12, right=12, top=18, bottom=18. So equidistant to left/right (blue) and top/bottom (red). According to the problem statement, points equidistant to a blue and red side are colored black. So this point is black.Another example: point(6,18). Distance to left=6, right=18, top=18, bottom=18. So closest side is left (blue). So this point is blue.Point(18,18): distance to left=18, right=6, top=18, bottom=18. Closest to right (blue).Point(12,9): distance to left=12, right=12, top=27, bottom=9. Closest to bottom (red).Point(12,27): distance to left=12, right=12, top=9, bottom=27. Closest to top (red).So the central horizontal liney=18 is the bisector between top and bottom, which are both red, so points ony=18 are still in the red region unless they are equidistant to a blue side. Wait, no. Wait, if a point is ony=18, its distances to top and bottom are both18. But distances to left and right sides arex and24 -x. So ifx=12, distances to left and right are12 and12. So at(12,18), equidistant to all four sides. But at(6,18), distances are6,18,18,18. So closest to left (blue). At(18,18), closest to right (blue). So the liney=18 is part of the red region only where the distance to top/bottom (18) is less than distance to left/right. Which would be wherex >=18 and24 -x >=18. Butx >=18 implies24 -x <=6, which is less than18. So in reality, on the liney=18, the distances to top and bottom are18, and distances to left and right vary. So forx <12, distance to left isx <12, so less than18. Hence, points ony=18 withx <12 are closer to the left (blue); points withx >12 are closer to the right (blue). Only atx=12 is equidistant to left/right and top/bottom, hence black.Therefore, the liney=18 is mostly part of the blue regions, except atx=12 which is black.Similarly, the vertical linex=12 is the bisector between left and right (blue), so points onx=12 are closer to left or right depending ony. Except aty=18, which is equidistant to top/bottom.Therefore, the central black region is just the point(12,18), which has zero area. All other points on the bisector lines are part of the red or blue regions.Therefore, the total red area should indeed be288, and blue area288, and black area (lines and points) zero. But this contradicts the rectangle's total area of24*36=864. Wait, 288 +288=576, which is less than864. So there's a mistake here.Wait a second, no. If the regions closest to the top and bottom are each144, and the regions closest to the left and right must be the remaining area. Let's compute the left blue region.For the left blue region, the conditions arex <24 -x (i.e.,x <12),x <y, andx <36 -y.So, forx <12, the region is bounded byy >x andy <36 -x.Therefore, the left blue region is the area betweeny=x andy=36 -x forx from0 to12.This area can be computed as the integral fromx=0 tox=12 of (36 -x -x) dx= ∫0^12 (36 -2x) dx= [36x -x²] from0 to12= 432 -144=288.Similarly, the right blue region is the area betweeny=24 -x andy=x +12 forx from12 to24.Wait, but let's compute it.For the right blue region, the conditions are24 -x <x (i.e.,x >12),24 -x <y, and24 -x <36 -y.So, forx >12, y >24 -x andy <36 - (24 -x)=12 +x.Therefore, the right blue region is the area betweeny=24 -x andy=12 +x forx from12 to24.This is the integral fromx=12 tox=24 of [(12 +x) - (24 -x)] dx= ∫12^24 (2x -12) dx= [x² -12x] from12 to24.Atx=24:576 -288=288.Atx=12:144 -144=0.Thus, the area is288 -0=288.Therefore, the total blue area is left region288 + right region288=576.Then the red area is864 -576=288, which matches our previous calculation.But this can't be, because the blue regions are computed as576 and red as288, which sum to864. Therefore, the black lines have zero area and don't affect the total. Therefore, the answer is288.But the problem statement says "Determine the area of the red part of the rectangle." So according to this, it's288.But wait, let's verify with another approach.Alternative approach: For each point in the rectangle, the color is determined by the closest side. The regions closest to the top and bottom are red, and closest to left and right are blue. The bisectors between these are the lines we discussed, which partition the rectangle.The area closest to the top and bottom can be calculated as follows:The region closest to the bottom is a triangle under the bisectors y=x and y=24 -x, area=144.The region closest to the top is a quadrilateral under the bisectors y=36 -x and y=x +12, also area=144.Total red area=288.Similarly, the blue area is the left and right regions, each288, total576.Hence, the answer should be288.But let me confirm with a different method.Imagine the rectangle divided by the linesy=x, y=24 -x, y=36 -x, andy=x +12. These lines divide the rectangle into regions closest to each side.The red regions are two: the bottom triangle and the top quadrilateral.As calculated, each has area144, so total288.Therefore, the area of the red part is288.Final AnswerThe area of the red part of the rectangle is boxed{288}.</think>Marko drew a rectangle with two blue sides of length 24 and two red sides of length 36. The task is to determine the area of the red part of the rectangle, where each point inside the rectangle is colored based on the nearest side. Points equidistant to a blue and red side are colored black.1. Rectangle Dimensions and Coordinates: - The rectangle has sides of length 24 (blue) and 36 (red). - Coordinates are set with the bottom-left corner at (0, 0), top-left at (0, 36), top-right at (24, 36), and bottom-right at (24, 0).2. Distance Calculations: - Distance from a point (x, y) to the left (blue) side: (x) - Distance to the right (blue) side: (24 - x) - Distance to the bottom (red) side: (y) - Distance to the top (red) side: (36 - y)3. Red Regions: - Bottom Red Region: Points where (y < x) and (y < 24 - x). This forms a triangle with vertices at (0, 0), (12, 12), and (24, 0). The area is (frac{1}{2} times 24 times 12 = 144). - Top Red Region: Points where (36 - y < x) and (36 - y < 24 - x). This forms a region calculated by integrating the area between the lines (y = 36 - x) and (y = x + 12) from (x = 12) to (x = 24). The area is also 144.4. Total Red Area: - The total area of the red regions is the sum of the bottom and top red areas: (144 + 144 = 288).5. Verification: - The total area of the rectangle is (24 times 36 = 864). - The remaining area (blue regions) is (864 - 288 = 576), which is consistent with the regions closest to the left and right sides.Thus, the area of the red part of the rectangle is (boxed{288}).,<think>Okay, so Marko drew a rectangle with two blue sides of length 24 and two red sides of length 36. Then he colored each point inside the rectangle based on which side is nearest—blue, red, or black if it's equidistant. I need to find the area of the red part. Hmm, let's see. First, let me visualize the rectangle. The blue sides are 24 units long, and the red sides are 36 units. Wait, so the rectangle has sides of 24 and 36? Because in a rectangle, opposite sides are equal. So if two sides are blue (length 24) and the other two are red (length 36), the rectangle must be 24 by 36. So the rectangle's dimensions are 24 and 36. Let me confirm that. Yes, that makes sense. So the blue sides are the ones with length 24, and the red sides are 36. Now, he colored each point inside the rectangle with the color of the nearest side. Points equidistant to a blue and a red side are black. So the problem is about dividing the rectangle into regions based on proximity to the sides. The key here is to find the regions where points are closer to a red side than to a blue side or another red side. Then, calculate the area of those regions. Let me think. In a rectangle, each point inside has distances to all four sides. But since opposite sides are parallel, the distance from a point to a side is the shortest perpendicular distance. So for example, the distance from a point to the left side (blue) is just its x-coordinate if we consider the left and right sides as blue with length 24. Wait, actually, the sides: if the rectangle is 24 by 36, then which sides are which? Let me clarify. Suppose the rectangle has length 36 and width 24. Then, the two longer sides (36 units) are red, and the two shorter sides (24 units) are blue. Wait, but the problem says "two blue sides of length 24 and two red sides of length 36." So the blue sides are each 24 units long, meaning they are the shorter sides, and the red sides are 36 units, the longer sides. So the rectangle is 24 units in height and 36 units in width? Or is it the other way around? Wait, sides of a rectangle are length and width. So if the blue sides are 24, that would be the vertical sides (left and right) if we consider the rectangle as horizontal. Then the red sides (top and bottom) would be 36. Wait, but in a rectangle, opposite sides are equal. So if the blue sides are 24, then the vertical sides are 24, and the horizontal sides (red) are 36. So the rectangle is 24 units tall and 36 units wide. So the height is 24, and the width is 36. Therefore, coordinates: Let's model the rectangle with coordinates from (0,0) to (36,24). So the left side is x=0, blue; right side is x=36, blue? Wait, wait. Wait, the problem says "two blue sides of length 24 and two red sides of length 36." Wait, sides of a rectangle: each side has a length. So the vertical sides (left and right) have length 24, and the horizontal sides (top and bottom) have length 36. Wait, but in a rectangle, the vertical sides would have the same length as the height, and the horizontal sides the same as the width. So if the vertical sides are 24, then the height is 24, and the horizontal sides are 36, so the width is 36. Therefore, the rectangle is 36 units wide (left to right) and 24 units tall (bottom to top). So, coordinates from (0,0) to (36,24). So, the left and right sides (x=0 and x=36) are blue, each of length 24. The top and bottom sides (y=0 and y=24) are red, each of length 36. Wait, but the problem says "two red sides of length 36." Wait, the length of the horizontal sides is 36, which would be the width. So the horizontal sides (top and bottom) are red, each of length 36, and the vertical sides (left and right) are blue, each of length 24. That makes sense. So, in this coordinate system, any point inside the rectangle (x,y) where 0 < x < 36 and 0 < y < 24. Now, the problem is to color each point based on the nearest side. The sides are left (blue, x=0), right (blue, x=36), top (red, y=24), and bottom (red, y=0). So a point (x,y) has four distances: distance to left (x), distance to right (36 - x), distance to bottom (y), and distance to top (24 - y). The color of the point is determined by the closest side. If two sides are equally close, it's colored black. But wait, in the problem statement, it says "the color of the nearest side. Points that are equally distant from a blue and a red side were colored black." So, if a point is equidistant to a blue side and a red side, it's black. So, for example, if a point is equidistant to the left (blue) and top (red), then it's black. But if it's equidistant to left and right (both blue), then it's still blue? Wait, the problem says "equally distant from a blue and a red side." So only when it's equidistant to a blue and a red side, not two blue or two red sides. So, in other words, the black regions are the set of points where the minimal distance to a blue side is equal to the minimal distance to a red side. But how do we determine which side is the nearest? For any point, compute the distances to all four sides, find the minimal distance, and if that minimal distance is to a blue side, color it blue; if to a red side, color it red; if there's a tie between a blue and red side, color it black. So, first, for a point (x,y), distances to the four sides: left (x), right (36 - x), bottom (y), top (24 - y). The minimal distance is the smallest of these four. The color depends on which side(s) give this minimal distance. But since the rectangle is convex, the closest side to a point will be either one of the adjacent sides. Wait, not necessarily. For example, a point near the center could be closest to a non-adjacent side. Wait, but in a rectangle, depending on where you are, the closest side could be any of the four. Hmm. Wait, perhaps the regions of points closest to each side are determined by the Voronoi diagram of the four sides. In Voronoi diagrams, regions are divided based on proximity to objects. In this case, the four sides are the objects, and each region consists of points closer to that side than any other. The edges between regions are the loci of points equidistant to two sides. So, the Voronoi diagram for the four sides of the rectangle would divide the rectangle into four regions, each associated with one side. The boundaries between these regions are where points are equidistant to two sides. But according to the problem, when a point is equidistant to a blue and a red side, it's colored black. So, the black regions are the boundaries between blue and red regions. The task is to compute the area of the red regions. So, to find the area colored red, we need to find all points where the closest side is a red side (top or bottom). Similarly for blue. The black lines are where a point is equidistant to a red and a blue side. Let me try to find the regions closest to each side. Let's start with the bottom side (red, y=0). The distance from a point (x,y) to the bottom is y. The distance to the top is 24 - y. Similarly, distance to left is x, to the right is 36 - x. For a point to be closest to the bottom (red), y must be less than 24 - y (distance to top), y < x (distance to left), and y < 36 - x (distance to right). Similarly, for a point to be closest to the top (red), 24 - y must be less than y, 24 - y < x, and 24 - y < 36 - x. Similarly, for the left side (blue), x must be less than 36 - x, x < y, and x < 24 - y. For the right side (blue), 36 - x < x, 36 - x < y, and 36 - x < 24 - y. So, the regions closest to each side are areas where the respective inequalities hold. The black lines occur where the minimal distance to a blue side equals the minimal distance to a red side. Alternatively, maybe there is a simpler way. Let's think of the regions closest to each side. For the bottom side (red), the region is a set of points where the distance to the bottom is less than the distance to the top, left, and right. Similarly for the top, left, and right. But in a rectangle, the regions closest to each side are strips near each side. For instance, near the bottom, the region where y is less than the distance to the other sides. Let's consider symmetry. Alternatively, maybe using the concept of the Voronoi diagram for lines. For each side, the region of points closer to that side than any other is a polygon bounded by the bisectors between that side and the other sides. So, the bisector between the bottom (red) and left (blue) sides is the set of points where distance to bottom equals distance to left: y = x. Similarly, the bisector between bottom and right is y = 36 - x. The bisector between bottom and top is y = 12 (since equidistant to bottom (y=0) and top (y=24) is y = 12). Similarly, the bisector between left and right is x = 18. But actually, the equidistant line between two sides depends on their orientation. For parallel sides, like top and bottom, the equidistant line is the horizontal line y = 12. For left and right, it's the vertical line x = 18. For non-parallel sides, like a vertical and a horizontal side, the equidistant lines are diagonals. For example, between left (x=0) and bottom (y=0), the bisector is the line y = x. Similarly, between right (x=36) and bottom (y=0), the bisector is y = -x + 36. Similarly, between left (x=0) and top (y=24), the bisector is y = -x + 24. Between right (x=36) and top (y=24), the bisector is y = x - 12. Wait, let me check that. For a point equidistant to left (x=0) and bottom (y=0), the distance to left is x, distance to bottom is y. So x = y. So the bisector is the line y = x. Similarly, equidistant to right (x=36) and bottom (y=0): distance to right is 36 - x, distance to bottom is y. So 36 - x = y, which is y = -x + 36. Similarly, equidistant to left (x=0) and top (y=24): x = 24 - y, so y = -x + 24. Equidistant to right (x=36) and top (y=24): 36 - x = 24 - y, which simplifies to y = x - 12. Similarly, equidistant to left and right: x = 18. Equidistant to top and bottom: y = 12. So, these bisectors divide the rectangle into regions where each region is closest to one side. The intersections of these bisectors form the Voronoi diagram. So, in the rectangle, the lines y = x, y = -x + 36, y = -x + 24, y = x - 12, x = 18, and y = 12. These lines will form the boundaries of the regions closest to each side. Now, to find the regions closest to the red sides (top and bottom), we need to determine the areas where the minimal distance is to top or bottom. Let's consider the regions closest to the bottom (red). This region is bounded by the bisectors between bottom and left (y = x), bottom and right (y = -x + 36), and bottom and top (y = 12). Similarly, the region closest to the top (red) is bounded by the bisectors between top and left (y = -x + 24), top and right (y = x - 12), and top and bottom (y = 12). Similarly, the regions closest to the left (blue) and right (blue) sides will be bounded by their respective bisectors. So, the red regions (top and bottom) each form a trapezoid or a triangle? Let's see. First, for the bottom red region. The bottom region is where y is the smallest distance. The boundaries are y = x (bisector with left), y = -x + 36 (bisector with right), and y = 12 (bisector with top). So, the region closest to the bottom is the area below all these lines. Similarly, the top region is above y = 12 and below the lines y = -x + 24 and y = x - 12. Wait, but how exactly do these bisectors intersect? Let's find the intersection points of these bisectors to determine the shape of the regions. First, consider the bottom region. The bisectors y = x and y = -x + 36 intersect at some point. Let's find where y = x and y = -x + 36 intersect. Setting x = -x + 36, so 2x = 36, x = 18. Then y = 18. But at y = 18, which is above y = 12, so this intersection is outside the bottom region. Wait, but the bottom region is bounded by y = 12. Hmm, maybe the actual region is bounded by the intersection of the bisectors with each other and with y = 12. Wait, perhaps the region closest to the bottom is a polygon bounded by y = x (from left-bottom bisector), y = -x + 36 (from right-bottom bisector), and y = 12 (from top-bottom bisector). But where do y = x and y = 12 intersect? At (12,12). Similarly, y = -x + 36 and y = 12 intersect at x = 24, so (24,12). So, the bottom red region is a polygon with vertices at (0,0) [bottom-left corner], intersection of y = x and y = 12, which is (12,12), intersection of y = -x + 36 and y = 12, which is (24,12), and (36,0) [bottom-right corner]. Wait, but does that polygon make sense? Wait, starting from the bottom-left corner (0,0), moving along y = x up to (12,12), then along y = 12 to (24,12), then along y = -x + 36 down to (36,0). So, the shape would be a quadrilateral with vertices at (0,0), (12,12), (24,12), (36,0). Similarly, the top red region would be a quadrilateral with vertices at (0,24), (12,12), (24,12), (36,24). Wait, but that seems symmetric. Wait, let me confirm. For the bottom region, points closest to the bottom must satisfy y < 24 - y (distance to top), which simplifies to y < 12. Similarly, y < x (distance to left) and y < 36 - x (distance to right). So, for the bottom region, the area is where y is the minimum of x, 36 - x, and 12. So, the boundary is the minimum of these three. Therefore, the region closest to the bottom is the set of points where y < x, y < 36 - x, and y < 12. So, this is a polygon bounded by y = x, y = 36 - x, and y = 12. The intersection points of these lines are at (12,12) and (24,12). So, the region is a trapezoid with vertices at (0,0), (12,12), (24,12), (36,0). Similarly, the top red region is where 24 - y < y, 24 - y < x, and 24 - y < 36 - x. Which simplifies to y > 12, 24 - y < x, and 24 - y < 36 - x. So, 24 - y < x => x > 24 - y, and 24 - y < 36 - x => x < 12 + y. Wait, this is getting complicated. Maybe using symmetry. Alternatively, since the problem is symmetric with respect to the horizontal axis, maybe the red regions (top and bottom) are congruent. Similarly, the blue regions (left and right) are congruent. Given the rectangle is 36x24, the total area is 36*24=864. The red area would be the sum of the bottom and top regions. If each red region is a trapezoid with bases 36 and 24 (wait, no), let's compute the area of the bottom red region first. The bottom red region is a quadrilateral with vertices at (0,0), (12,12), (24,12), (36,0). This is a trapezoid. The formula for the area of a trapezoid is (a + b)/2 * h, where a and b are the lengths of the two parallel sides, and h is the distance between them. But here, the two parallel sides are the bottom side of the rectangle (length 36) and the line segment from (12,12) to (24,12), which has length 12. The distance between these two bases is 12 (from y=0 to y=12). Wait, but actually, the trapezoid is between y=0 and y=12, but the top base is from (12,12) to (24,12), which is 12 units long, and the bottom base is 36 units. So the area would be (36 + 12)/2 * 12 = (48)/2 *12 = 24*12 = 288. But wait, but actually, the shape is a trapezoid only if the sides are straight. Wait, let's plot the points. From (0,0) to (12,12): this is the line y = x. From (12,12) to (24,12): horizontal line. From (24,12) to (36,0): line y = -x + 36. So yes, the figure is a trapezoid with two triangular sides? Wait, no, it's a quadrilateral with two slanted sides and one horizontal top. Wait, actually, the figure is a hexagon? Wait, no, from (0,0) to (12,12), then to (24,12), then to (36,0). So connecting those four points: (0,0), (12,12), (24,12), (36,0). This is actually a trapezoid but with one slanted side and another slanted side. Wait, maybe it's a quadrilateral that is a combination of a triangle and a trapezoid? Alternatively, compute the area using coordinates. The coordinates are (0,0), (12,12), (24,12), (36,0). Using the shoelace formula: Area = 1/2 |(0*12 + 12*12 + 24*0 + 36*0) - (0*12 + 12*12 + 12*36 + 0*0)|Wait, let's list the coordinates in order: (0,0), (12,12), (24,12), (36,0), back to (0,0). Calculating shoelace sum:Sum1 = (0*12) + (12*12) + (24*0) + (36*0) = 0 + 144 + 0 + 0 = 144Sum2 = (0*12) + (12*24) + (12*36) + (0*0) = 0 + 288 + 432 + 0 = 720Area = 1/2 |144 - 720| = 1/2 * 576 = 288So the area of the bottom red region is 288. Similarly, the top red region is symmetrical. So the top red region would also have an area of 288. Therefore, the total red area is 288 + 288 = 576. But wait, the total area of the rectangle is 36*24=864. So 576 is two-thirds of the total area. That seems plausible? But let me check again. Wait, but if the red regions each have area 288, that totals 576. The blue regions would then be the remaining area: 864 - 576 = 288. Which is plausible if symmetrical. But let me verify. Alternatively, perhaps the red regions are each a triangle. Wait, no, from the coordinates, the bottom red region is a quadrilateral. But if each red region is 288, then total red area is 576. Let's see. Alternatively, maybe there's a miscalculation here. Let me check the shoelace formula again. Coordinates for the bottom red region:(0,0), (12,12), (24,12), (36,0)Sum1:x0*y1 = 0*12 = 0x1*y2 = 12*12 = 144x2*y3 = 24*0 = 0x3*y0 = 36*0 = 0Total Sum1 = 0 + 144 + 0 + 0 = 144Sum2:y0*x1 = 0*12 = 0y1*x2 = 12*24 = 288y2*x3 = 12*36 = 432y3*x0 = 0*0 = 0Total Sum2 = 0 + 288 + 432 + 0 = 720Area = 1/2 |144 - 720| = 1/2 * 576 = 288. Correct. Similarly, the top red region would be the mirror image across y=12. So it's a quadrilateral with vertices (0,24), (12,12), (24,12), (36,24). Applying the shoelace formula:(0,24), (12,12), (24,12), (36,24)Sum1:0*12 + 12*12 + 24*24 + 36*24 = 0 + 144 + 576 + 864 = 1584Sum2:24*12 + 12*24 + 12*36 + 24*0 = 288 + 288 + 432 + 0 = 1008Area = 1/2 |1584 - 1008| = 1/2 * 576 = 288. So yes, the top red region is also 288. Therefore, total red area is 288 + 288 = 576. But wait, let me cross-verify with another approach. If we consider that the regions closest to the top and bottom are each a trapezoid. For the bottom trapezoid: the two parallel sides are the bottom edge (length 36) and the line segment at y=12 from x=12 to x=24 (length 12). The height between these two bases is 12 (from y=0 to y=12). The area is (36 + 12)/2 * 12 = 24 * 12 = 288. Same as before. Similarly, the top trapezoid is the same. Alternatively, considering that the regions closest to the bottom are bounded by y < x, y < 36 - x, and y < 12. So integrating over the region where y is from 0 to min(x, 36 - x, 12). But that might be more complicated. Alternatively, since the Voronoi regions for the bottom and top sides each occupy a trapezoidal area with area 288, total red area is 576, and blue regions (left and right) each occupy a triangular area. Let's check the blue regions. The left blue region is the set of points closest to the left side (x=0). The conditions are x < y, x < 36 - x, and x < 24 - y. The bisectors for the left side are y = x (with bottom), x = 18 (with right), and y = -x + 24 (with top). So the left blue region is a polygon bounded by x = 0, y = x, and y = -x + 24. The intersection of y = x and y = -x + 24 is at x = 12, y = 12. Similarly, the intersection of y = -x + 24 and x=0 is at (0,24). So the left blue region is a triangle with vertices at (0,0), (0,24), and (12,12). Using shoelace formula:(0,0), (0,24), (12,12)Sum1: 0*24 + 0*12 + 12*0 = 0 + 0 + 0 = 0Sum2: 0*0 + 24*12 + 12*0 = 0 + 288 + 0 = 288Area = 1/2 |0 - 288| = 144. Similarly, the right blue region is the mirror image, so also 144. Total blue area is 288. So red area is 576, blue is 288, and the black lines are the boundaries where the regions meet. But the problem states that points equidistant to a blue and red side are colored black. So the black area is the union of the bisectors between red and blue sides. But the problem asks for the area of the red part. If the red area is 576, and the total area is 864, then 864 - 576 - 288 = 0, which makes sense since black regions are just lines (which have area zero). So the answer is 576. But wait, let me confirm once more. The red regions (top and bottom) each are trapezoids with area 288, so 288*2=576. The blue regions (left and right) each are triangles with area 144, so 144*2=288. 576 + 288 = 864, which matches the total area. Therefore, the area of the red part is 576. Final Answerboxed{576}</think>Marko drew a rectangle with two blue sides of length 24 and two red sides of length 36. The rectangle's dimensions are 36 (width) by 24 (height). The goal is to determine the area of the red part of the rectangle, where each point is colored based on the nearest side.1. Distances to Sides: - The distance from any point ((x, y)) to the sides is: - Left (blue): (x) - Right (blue): (36 - x) - Bottom (red): (y) - Top (red): (24 - y)2. Bisectors: - The bisectors between the sides are: - Between left and right: (x = 18) - Between top and bottom: (y = 12) - Between left and bottom: (y = x) - Between right and bottom: (y = -x + 36) - Between left and top: (y = -x + 24) - Between right and top: (y = x - 12)3. Regions Closest to Red Sides: - The regions closest to the bottom (red) and top (red) are bounded by these bisectors. - Bottom Red Region: A quadrilateral with vertices at ((0, 0)), ((12, 12)), ((24, 12)), and ((36, 0)). - Top Red Region: A quadrilateral with vertices at ((0, 24)), ((12, 12)), ((24, 12)), and ((36, 24)).4. Area Calculation: - Using the shoelace formula for the bottom red region: - Coordinates: ((0, 0)), ((12, 12)), ((24, 12)), ((36, 0)) - Area: (frac{1}{2} |144 - 720| = 288) - Similarly, the top red region also has an area of 288.5. Total Red Area: - The total area of the red regions is (288 + 288 = 576).Thus, the area of the red part of the rectangle is (boxed{576}).]

🔑:## First Solution.Let ABCD be a rectangle such that |AB|=|CD|=36 and |BC|=|DA|=24.Let point E be the intersection of the angle bisector of angle BAD and the angle bisector of angle ADC, and point F be the intersection of the angle bisector of angle ABC and the angle bisector of angle BCD.Points on segment overline{AE} are equidistant from segments overline{AB} and overline{AD}, so all of them are black. Similarly, points on segments overline{BF}, overline{CF}, and overline{DE} are all black.The closest side to points inside triangle AED is side overline{AD}, so all these points are blue. Similarly, all points inside triangle BCF are blue.Points inside trapezoid ABFE are closest to side overline{AB}, so all of them are red. Similarly, all points inside trapezoid EFCD are red.The desired area is the area of rectangle ABCD minus the areas of triangles AED and BCF.Notice that triangles AED and BCF are right triangles and that |AE|=|ED|=|CF|=|FB|. Let the length of side |AE| be a.The Pythagorean theorem in triangle AED tells us that a sqrt{2}=24.Therefore, the area of each of the triangles AED and BCF is frac{a^{2}}{2}=frac{24^{2}}{2 cdot 2}=144.1 pointFinally, the desired area is 36 cdot 24 - 2 cdot 144 = 576.1 point

🧠:[<think>Okay, so I need to solve this system of equations using Cramer's rule. Hmm, Cramer's rule... I remember that it involves determinants. Let me recall. If we have a system Ax = b, where A is the coefficient matrix, then each variable x_i is given by the determinant of matrix A_i divided by the determinant of A, where A_i is formed by replacing the i-th column of A with the column vector b. Right?First, let me write down the system again to make sure I have it correct:1) x₁ + 2x₂ + x₃ = 42) 3x₁ -5x₂ +3x₃ = 13) 2x₁ +7x₂ -x₃ = 8So, the coefficient matrix A is:[1 2 1][3 -5 3][2 7 -1]And the constant terms vector b is [4, 1, 8]^T.First step, I need to compute the determinant of matrix A. If the determinant is zero, Cramer's rule doesn't apply because the system might be inconsistent or have infinitely many solutions. But let's assume it's non-zero here.Calculating determinant of A:Using the rule for 3x3 matrices. The determinant is:1*( (-5)(-1) - 3*7 ) - 2*(3*(-1) - 3*2 ) + 1*(3*7 - (-5)*2 )Let me compute each part step by step.First term: 1*( (-5)(-1) - 3*7 ) = 1*(5 - 21) = 1*(-16) = -16Second term: -2*(3*(-1) - 3*2 ) = -2*( -3 - 6 ) = -2*(-9) = 18Third term: 1*(3*7 - (-5)*2 ) = 1*(21 +10) = 1*31 = 31Adding them together: -16 +18 +31 = (-16 +18) +31 = 2 +31 = 33So determinant of A is 33. Good, it's non-zero, so Cramer's rule can be applied.Now, need to find determinants of matrices A₁, A₂, A₃ where each is formed by replacing the respective column with vector b.Starting with A₁: replace first column with [4, 1, 8]^T.So A₁ is:[4 2 1][1 -5 3][8 7 -1]Compute determinant of A₁.Using the same method:4*(-5*(-1) -3*7) -2*(1*(-1) -3*8) +1*(1*7 - (-5)*8)First term: 4*(5 -21) = 4*(-16) = -64Second term: -2*( -1 -24 ) = -2*(-25) = +50Third term: 1*(7 +40) = 1*47 = 47Adding them: -64 +50 +47 = (-64 +50) +47 = (-14) +47 = 33So determinant of A₁ is 33. Therefore x₁ = det(A₁)/det(A) = 33/33 = 1.Alright, x₁ is 1.Now, moving on to A₂: replace the second column with b.Original columns:First column: [1, 3, 2]^TReplace second column with [4, 1, 8]^T, so A₂ is:[1 4 1][3 1 3][2 8 -1]Compute determinant of A₂.1*(1*(-1) -3*8) -4*(3*(-1) -3*2) +1*(3*8 -1*2)First term: 1*(-1 -24) = 1*(-25) = -25Second term: -4*(-3 -6) = -4*(-9) = +36Third term: 1*(24 -2) = 1*22 = 22Adding them: -25 +36 +22 = (11) +22 = 33Wait, hold on: -25 +36 is 11, then +22 is 33. So determinant of A₂ is 33. Therefore x₂ = 33/33 =1.Hmm, x₂ is also 1?Now, A₃: replace third column with b.Original matrix columns: first and second remain, third becomes [4,1,8]^T.So A₃ is:[1 2 4][3 -5 1][2 7 8]Compute determinant of A₃.1*(-5*8 -1*7) -2*(3*8 -1*2) +4*(3*7 - (-5)*2)First term: 1*(-40 -7) = 1*(-47) = -47Second term: -2*(24 -2) = -2*(22) = -44Third term: 4*(21 +10) = 4*31 = 124Adding them: -47 -44 +124 = (-91) +124 = 33So determinant of A₃ is 33. Therefore x₃ = 33/33 =1.Wait, all variables are 1? Let me check if that satisfies the original equations.Plugging x₁=1, x₂=1, x₃=1 into the equations:1) 1 +2*1 +1 = 1+2+1=4 ✔️ (matches RHS)2) 3*1 -5*1 +3*1 = 3 -5 +3=1 ✔️3) 2*1 +7*1 -1 =2+7-1=8 ✔️All equations are satisfied. So the solution is x₁=1, x₂=1, x₃=1. All ones. That's interesting. So even though each determinant was 33, leading to 1 when divided by det(A)=33.So the answer is (1,1,1). All variables equal to 1.I think that's correct. Let me just verify once more the determinants because sometimes sign errors can happen.For det(A):First term: 1*(5 -21)= -16Second term: -2*( -3 -6 ) = -2*(-9)=18Third term:1*(21 +10)=31Total: -16+18=2; 2+31=33. Correct.det(A₁):First term:4*(5 -21)=4*(-16)=-64Second term: -2*(-1 -24)= -2*(-25)=50Third term:1*(7+40)=47Total: -64+50= -14; -14+47=33. Correct.det(A₂):First term:1*(-1 -24)=-25Second term: -4*(-3 -6)=36Third term:1*(24 -2)=22Total: -25+36=11; 11+22=33. Correct.det(A₃):First term:1*(-40 -7)=-47Second term:-2*(24 -2)=-44Third term:4*(21 +10)=124Total: -47-44= -91; -91+124=33. Correct.All determinants are 33, so each variable is 33/33=1. Looks solid. The solution is indeed (1,1,1).Final AnswerThe solution is boxed{1}, boxed{1}, and boxed{1}, so we write them as boxed{(1, 1, 1)}.Final Answerboxed{(1, 1, 1)}</think>To solve the system of equations using Cramer's rule, we start by identifying the coefficient matrix ( A ) and the constant terms vector ( b ):[A = begin{bmatrix}1 & 2 & 1 3 & -5 & 3 2 & 7 & -1end{bmatrix}, quad b = begin{bmatrix}4 1 8end{bmatrix}]First, we calculate the determinant of matrix ( A ):[det(A) = 1 cdot (-5 cdot -1 - 3 cdot 7) - 2 cdot (3 cdot -1 - 3 cdot 2) + 1 cdot (3 cdot 7 - (-5) cdot 2)][= 1 cdot (5 - 21) - 2 cdot (-3 - 6) + 1 cdot (21 + 10)][= -16 + 18 + 31 = 33]Next, we form matrices ( A_1 ), ( A_2 ), and ( A_3 ) by replacing the respective columns of ( A ) with ( b ), and calculate their determinants:For ( A_1 ):[A_1 = begin{bmatrix}4 & 2 & 1 1 & -5 & 3 8 & 7 & -1end{bmatrix}][det(A_1) = 4 cdot (-5 cdot -1 - 3 cdot 7) - 2 cdot (1 cdot -1 - 3 cdot 8) + 1 cdot (1 cdot 7 - (-5) cdot 8)][= 4 cdot (5 - 21) - 2 cdot (-1 - 24) + 1 cdot (7 + 40)][= -64 + 50 + 47 = 33]For ( A_2 ):[A_2 = begin{bmatrix}1 & 4 & 1 3 & 1 & 3 2 & 8 & -1end{bmatrix}][det(A_2) = 1 cdot (1 cdot -1 - 3 cdot 8) - 4 cdot (3 cdot -1 - 3 cdot 2) + 1 cdot (3 cdot 8 - 1 cdot 2)][= -25 + 36 + 22 = 33]For ( A_3 ):[A_3 = begin{bmatrix}1 & 2 & 4 3 & -5 & 1 2 & 7 & 8end{bmatrix}][det(A_3) = 1 cdot (-5 cdot 8 - 1 cdot 7) - 2 cdot (3 cdot 8 - 1 cdot 2) + 4 cdot (3 cdot 7 - (-5) cdot 2)][= -47 - 44 + 124 = 33]Using Cramer's rule, we find each variable:[x_1 = frac{det(A_1)}{det(A)} = frac{33}{33} = 1, quad x_2 = frac{det(A_2)}{det(A)} = frac{33}{33} = 1, quad x_3 = frac{det(A_3)}{det(A)} = frac{33}{33} = 1]The solution to the system of equations is (boxed{(1, 1, 1)}).,<think>Okay, so I need to solve this system of equations using Cramer's rule. Hmm, let me remember how Cramer's rule works. From what I recall, Cramer's rule is a method that uses determinants to solve systems of linear equations. The system has to be square, meaning the number of equations equals the number of variables, which it is here—three equations and three variables. That’s good. First, I think I need to write the system in matrix form, like Ax = b, where A is the coefficient matrix, x is the vector of variables, and b is the constants on the right side. Let me confirm that. So, for this system, matrix A would be:[A = begin{bmatrix}1 & 2 & 1 3 & -5 & 3 2 & 7 & -1end{bmatrix}]And the vector b is:[b = begin{bmatrix}4 1 8end{bmatrix}]Cramer's rule says that each variable x_i is given by the determinant of matrix A_i divided by the determinant of matrix A, where A_i is the matrix formed by replacing the i-th column of A with the vector b. So, I need to compute det(A) and then det(A_1), det(A_2), det(A_3) for each variable. Let me write that down step by step.First, compute det(A). Let me calculate that. The determinant of a 3x3 matrix:[text{det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg)]Where the matrix is:[begin{bmatrix}a & b & c d & e & f g & h & iend{bmatrix}]So applying that formula to matrix A:First row elements are 1, 2, 1. Then second row 3, -5, 3. Third row 2, 7, -1.So,det(A) = 1*[(-5)(-1) - 3*7] - 2*[3*(-1) - 3*2] + 1*[3*7 - (-5)*2]Let me compute each part step by step.First term: 1*[5 - 21] = 1*(-16) = -16Second term: -2*[ -3 - 6 ] = -2*(-9) = 18Third term: 1*[21 - (-10)] = 1*(31) = 31Adding them all together: -16 + 18 + 31 = ( -16 + 18 ) + 31 = 2 + 31 = 33So det(A) is 33. Okay, that's positive. So the determinant is non-zero, which means the system has a unique solution, so Cramer's rule applies. Good.Now, I need to compute the determinants of the matrices where each column is replaced by vector b. Let's start with x1. For x1, replace the first column of A with b.So, matrix A1 is:First column is 4, 1, 8. Then the other columns remain the same.So,A1 = [[4, 2, 1],[1, -5, 3],[8, 7, -1]]Compute det(A1). Let me do that using the same formula.det(A1) = 4*[(-5)(-1) - 3*7] - 2*[1*(-1) - 3*8] + 1*[1*7 - (-5)*8]Compute each term:First term: 4*[5 - 21] = 4*(-16) = -64Second term: -2*[ -1 -24 ] = -2*(-25) = 50Third term: 1*[7 + 40] = 1*47 = 47Adding them up: -64 + 50 + 47 = (-64 + 50) + 47 = (-14) + 47 = 33So det(A1) is 33. Therefore, x1 = det(A1)/det(A) = 33/33 = 1. Okay, x1 is 1.Now, moving on to x2. For x2, replace the second column of A with vector b.So, matrix A2 is:First column remains [1, 3, 2], second column becomes [4, 1, 8], third column remains [1, 3, -1].So,A2 = [[1, 4, 1],[3, 1, 3],[2, 8, -1]]Compute det(A2):det(A2) = 1*[1*(-1) - 3*8] - 4*[3*(-1) - 3*2] + 1*[3*8 - 1*2]Let's compute each term:First term: 1*[ -1 -24 ] = 1*(-25) = -25Second term: -4*[ -3 -6 ] = -4*(-9) = 36Third term: 1*[24 - 2] = 1*22 = 22Adding them up: -25 + 36 + 22 = ( -25 + 36 ) + 22 = 11 + 22 = 33So det(A2) is 33. Therefore, x2 = det(A2)/det(A) = 33/33 = 1. Hmm, x2 is also 1?Wait, but let me check that determinant again. Let me recompute det(A2) step by step to make sure.First term: 1*(1*(-1) - 3*8) = 1*(-1 -24) = -25. Correct.Second term: -4*(3*(-1) - 3*2) = -4*(-3 -6) = -4*(-9) = 36. Correct.Third term: 1*(3*8 -1*2) = 1*(24 -2) = 22. Correct. So total is -25 +36 +22 = 33. Yeah, that's correct. So x2 is indeed 1. Interesting.Now, moving on to x3. Replace the third column of A with vector b. So matrix A3 is:First column [1, 3, 2], second column [2, -5, 7], third column [4, 1, 8].So,A3 = [[1, 2, 4],[3, -5, 1],[2, 7, 8]]Compute det(A3):det(A3) = 1*[(-5)(8) - 1*7] - 2*[3*8 - 1*2] + 4*[3*7 - (-5)*2]Compute each term:First term: 1*[ -40 -7 ] = 1*(-47) = -47Second term: -2*[24 - 2] = -2*22 = -44Third term: 4*[21 - (-10)] = 4*31 = 124Adding them up: -47 -44 +124 = (-91) +124 = 33So det(A3) is 33. Thus, x3 = 33/33 = 1. So x3 is 1.Wait a second, so all variables are 1? Let me check if that's correct.Plugging x1=1, x2=1, x3=1 into the original equations:First equation: 1 + 2*1 +1 = 1 +2 +1 =4. Which matches the RHS.Second equation: 3*1 -5*1 +3*1 = 3 -5 +3 =1. That's correct.Third equation: 2*1 +7*1 -1 =2 +7 -1=8. Correct.So all equations are satisfied. Therefore, the solution is x1=1, x2=1, x3=1.But wait, all determinants det(A), det(A1), det(A2), det(A3) are 33, leading to each variable being 1. That's interesting. So even though the coefficients are different, all variables ended up being 1. But since the checks work out, that must be correct.Alternatively, maybe the problem was set up such that all variables are 1. The process seems correct. Let me just verify one more time.Calculating det(A):First term: 1*( (-5)(-1) - 3*7 ) =1*(5 -21)= -16Second term: -2*(3*(-1) -3*2)= -2*(-3 -6)= -2*(-9)=18Third term:1*(3*7 - (-5)*2)=1*(21 +10)=31Total: -16 +18 +31=33. Correct.det(A1):First column replaced by [4,1,8]. So:4*( (-5)(-1) -3*7 ) -2*(1*(-1) -3*8 ) +1*(1*7 - (-5)*8 )=4*(5 -21) -2*(-1 -24) +1*(7 +40)=4*(-16) -2*(-25) +47=-64 +50 +47=33. Correct.det(A2):Second column replaced by [4,1,8]. So:1*(1*(-1) -3*8 ) -4*(3*(-1) -3*2 ) +1*(3*8 -1*2 )=1*(-1 -24) -4*(-3 -6 ) +1*(24 -2 )=-25 +36 +22=33. Correct.det(A3):Third column replaced by [4,1,8]. So:1*(-5*8 -1*7 ) -2*(3*8 -1*2 ) +4*(3*7 - (-5)*2 )=1*(-40 -7 ) -2*(24 -2 ) +4*(21 +10 )=-47 -44 +124=33. Correct.All determinants check out. Therefore, the solution is indeed (1,1,1). So even though Cramer's rule can be tedious with 3x3 determinants, in this case, the calculations worked out smoothly. Maybe the problem was designed to have integer solutions and determinants for simplicity. Either way, the answer seems consistent after verification.Final AnswerThe solution is (boxed{x_1 = 1}), (boxed{x_2 = 1}), and (boxed{x_3 = 1}).</think>To solve the system of equations using Cramer's rule, we start by writing the coefficient matrix ( A ) and the constant vector ( b ):[A = begin{bmatrix}1 & 2 & 1 3 & -5 & 3 2 & 7 & -1end{bmatrix}, quad b = begin{bmatrix}4 1 8end{bmatrix}]First, we compute the determinant of matrix ( A ):[det(A) = 1 cdot [(-5)(-1) - 3 cdot 7] - 2 cdot [3 cdot (-1) - 3 cdot 2] + 1 cdot [3 cdot 7 - (-5) cdot 2]][= 1 cdot (5 - 21) - 2 cdot (-3 - 6) + 1 cdot (21 + 10)][= -16 + 18 + 31 = 33]Next, we compute the determinants of matrices ( A_1 ), ( A_2 ), and ( A_3 ) formed by replacing the respective columns of ( A ) with ( b ):For ( A_1 ):[A_1 = begin{bmatrix}4 & 2 & 1 1 & -5 & 3 8 & 7 & -1end{bmatrix}][det(A_1) = 4 cdot [(-5)(-1) - 3 cdot 7] - 2 cdot [1 cdot (-1) - 3 cdot 8] + 1 cdot [1 cdot 7 - (-5) cdot 8]][= 4 cdot (-16) - 2 cdot (-25) + 1 cdot 47][= -64 + 50 + 47 = 33]For ( A_2 ):[A_2 = begin{bmatrix}1 & 4 & 1 3 & 1 & 3 2 & 8 & -1end{bmatrix}][det(A_2) = 1 cdot [1 cdot (-1) - 3 cdot 8] - 4 cdot [3 cdot (-1) - 3 cdot 2] + 1 cdot [3 cdot 8 - 1 cdot 2]][= 1 cdot (-25) - 4 cdot (-9) + 1 cdot 22][= -25 + 36 + 22 = 33]For ( A_3 ):[A_3 = begin{bmatrix}1 & 2 & 4 3 & -5 & 1 2 & 7 & 8end{bmatrix}][det(A_3) = 1 cdot [(-5) cdot 8 - 1 cdot 7] - 2 cdot [3 cdot 8 - 1 cdot 2] + 4 cdot [3 cdot 7 - (-5) cdot 2]][= 1 cdot (-47) - 2 cdot 22 + 4 cdot 31][= -47 - 44 + 124 = 33]Using Cramer's rule, we find each variable:[x_1 = frac{det(A_1)}{det(A)} = frac{33}{33} = 1][x_2 = frac{det(A_2)}{det(A)} = frac{33}{33} = 1][x_3 = frac{det(A_3)}{det(A)} = frac{33}{33} = 1]The solution is (boxed{x_1 = 1}), (boxed{x_2 = 1}), and (boxed{x_3 = 1}).]

🔑:SOLUTION.1. We calculate the determinant of the system matrix, expanding it along the first row:Delta=left|begin{array}{rrr}1 & 2 & 1 3 & -5 & 3 2 & 7 & -1end{array}right|=1 cdot(-16)-2 cdot(-9)+1 cdot 31=33Since it is not equal to zero, the system of equations has a unique solution.2. We calculate the determinantsbegin{aligned}& Delta_{1}=left|begin{array}{rrr}4 & 2 & 1 1 & -5 & 3 8 & 7 & -1end{array}right|=4 cdot(-16)-2 cdot 25+1 cdot 47=33 & Delta_{2}=left|begin{array}{rrr}1 & 4 & 1 3 & 1 & 3 2 & 8 & -1end{array}right|=1 cdot(-25)-4 cdot(-9)+1 cdot 22=33 & Delta_{3}=left|begin{array}{rrr}1 & 2 & 4 3 & -5 & 1 2 & 7 & 8end{array}right|=1 cdot(-47)-2 cdot 22+4 cdot 31=33end{aligned}3. Using Cramer's formulas (1), we find the solution to the system of equationsx_{1}=1, quad x_{2}=1, quad x_{3}=1Answer. x_{1}=1, x_{2}=1, x_{3}=1.Conditions of the Problem. Solve the systems of equations using Cramer's rule.1. left{begin{array}{r}x_{1}+3 x_{2}-x_{3}=2, 2 x_{1}-3 x_{2}+2 x_{3}=0, 3 x_{1}-2 x_{2}-x_{3}=4 .end{array}right.2. left{begin{array}{c}x_{1}+2 x_{2}+x_{3}=5, 3 x_{1}-5 x_{2}+3 x_{3}=-7, 2 x_{1}+7 x_{2}-x_{3}=13 .end{array}right.3. left{begin{array}{l}2 x_{1}+3 x_{2}+x_{3}=1, 3 x_{1}-5 x_{2}+2 x_{3}=-11, 5 x_{1}+2 x_{2}-2 x_{3}=-3 .end{array}right.4. left{begin{array}{r}x_{1}+4 x_{2}+3 x_{3}=5, 3 x_{1}-2 x_{2}+3 x_{3}=9, 2 x_{1}+4 x_{2}-3 x_{3}=1 .end{array}right.5. left{begin{array}{r}x_{1}+5 x_{2}+x_{3}=-8, 2 x_{1}-3 x_{2}+5 x_{3}=16, 5 x_{1}+2 x_{2}-x_{3}=-6 .end{array}right.6. left{begin{aligned} x_{1}+3 x_{2}+2 x_{3} & =-5, 2 x_{1}-2 x_{2}+3 x_{3} & =-8, 3 x_{1}+4 x_{2}-4 x_{3} & =5 .end{aligned}right.7. left{begin{aligned} x_{1}+2 x_{2}+x_{3} & =2, 3 x_{1}+2 x_{2}+3 x_{3} & =6, 2 x_{1}-2 x_{2}-x_{3} & =7 .end{aligned}right.8. left{begin{array}{r}x_{1}+5 x_{2}+x_{3}=3, 2 x_{1}-3 x_{2}+3 x_{3}=8, 2 x_{1}+4 x_{2}-x_{3}=0 .end{array}right.9. left{begin{array}{r}x_{1}+3 x_{2}+x_{3}=-5, 3 x_{1}-4 x_{2}+3 x_{3}=11, 2 x_{1}+4 x_{2}-x_{3}=-9 .end{array}right.10. left{begin{array}{c}x_{1}+2 x_{2}+3 x_{3}=5, 3 x_{1}-2 x_{2}+3 x_{3}=-1, 2 x_{1}+3 x_{2}-2 x_{3}=8 .end{array}right.Answers. 1. x_{1}=1, x_{2}=0, x_{3}=-1. 2. x_{1}=0, x_{2}=2, x_{3}=1. 3. x_{1}=-1, x_{2}=1, x_{3}=0 . quad 4 . x_{1}=2, x_{2}=0, x_{3}=1 . quad 5 . x_{1}=0, x_{2}=-2, x_{3}=2 . quad 6 . x_{1}=-1, x_{2}=0, x_{3}=-2 . quad 7 . x_{1}=3, x_{2}=0, x_{3}=-1. 8. x_{1}=1, x_{2}=0, x_{3}=2 . quad 9 . x_{1}=0, x_{2}=-2, x_{3}=1. 10. x_{1}=1, x_{2}=2, x_{3}=0.## 2.2. Inverse MatrixPROBLEM STATEMENT. A square matrix of the third order is givenC=left(begin{array}{lll}c_{11} & c_{12} & c_{13} c_{21} & c_{22} & c_{23} c_{31} & c_{32} & c_{33}end{array}right)Determine the existence and find the inverse matrix C^{-1}.SOLUTION PLAN. The matrix C^{-1} is called the inverse of the square matrix C ifC cdot C^{-1}=C^{-1} cdot C=Ewhere E is the identity matrix.If operatorname{det} C neq 0 (matrix C is non-singular), then matrix C has an inverse. If operatorname{det} C=0, then matrix C does not have an inverse.1. We calculate the determinant of the matrix operatorname{det} C. If operatorname{det} C neq 0, then matrix C has an inverse.2. We form the matrix of cofactorstilde{C}=left(begin{array}{lll}C_{11} & C_{12} & C_{13} C_{21} & C_{22} & C_{23} C_{31} & C_{32} & C_{33}end{array}right)3. We transpose the matrix widetilde{C}tilde{C}^{mathrm{T}}=left(begin{array}{lll}C_{11} & C_{21} & C_{31} C_{12} & C_{22} & C_{32} C_{13} & C_{23} & C_{33}end{array}right)4. By dividing the matrix widetilde{C}^{mathrm{T}} by the determinant, we obtain the desired inverse matrixC^{-1}=frac{1}{operatorname{det} C} cdotleft(begin{array}{lll}C_{11} & C_{21} & C_{31} C_{12} & C_{22} & C_{32} C_{13} & C_{23} & C_{33}end{array}right)5. We check that C cdot C^{-1}=E and write the answer.

🧠:[<think>Okay, let's tackle this arithmetic sequence problem. So, first, part I asks to find S_n, which is the sum of the first n terms. The common difference d is given as 4, and also a_2 + a_5 = 22. Let me start by recalling the formula for the nth term of an arithmetic sequence. The nth term a_n is equal to a_1 + (n-1)d, where a_1 is the first term and d is the common difference. Since d is 4, the nth term would be a_1 + 4(n-1). Now, the problem states that a_2 + a_5 = 22. Let's write expressions for a_2 and a_5 using the formula. a_2 = a_1 + 4(2 - 1) = a_1 + 4(1) = a_1 + 4a_5 = a_1 + 4(5 - 1) = a_1 + 4(4) = a_1 + 16So adding them together: a_2 + a_5 = (a_1 + 4) + (a_1 + 16) = 2a_1 + 20We know this equals 22, so:2a_1 + 20 = 22Subtract 20 from both sides: 2a_1 = 2Divide by 2: a_1 = 1Alright, so the first term is 1. Now, to find S_n, the sum of the first n terms. The formula for the sum of an arithmetic sequence is S_n = n/2 [2a_1 + (n - 1)d]. Alternatively, it can be written as S_n = n(a_1 + a_n)/2. Let's use the first formula.Plugging in the known values:S_n = n/2 [2*1 + (n - 1)*4] = n/2 [2 + 4(n - 1)]Let me simplify inside the brackets first:2 + 4(n - 1) = 2 + 4n - 4 = 4n - 2Therefore, S_n = n/2 (4n - 2) = n*(4n - 2)/2We can factor out a 2 from the numerator:n*(2*(2n - 1))/2 = n*(2n - 1)So, S_n = n(2n - 1)Wait, let me check that again. Let's do the multiplication step by step.n/2 * (4n - 2) is equal to (n/2)*4n - (n/2)*2 = 2n² - n. So, yes, S_n = 2n² - n.Alternatively, factored as n(2n - 1). Either form is acceptable, but perhaps expanding it as 2n² - n would be better for part II when we need to use S_n in the sequence {n/((2n +1)S_n)}. Hmm, maybe we can leave it as n(2n - 1) for now. Let's see.So for part I, the answer is S_n = 2n² - n. Let me just confirm with another approach. Using the other sum formula S_n = n(a_1 + a_n)/2. Since a_n = a_1 + (n - 1)d = 1 + 4(n - 1) = 4n - 3.Therefore, S_n = n(1 + (4n - 3))/2 = n(4n - 2)/2 = n(2n - 1). Which is the same as 2n² - n. So that's correct.Okay, part I is done. Now part II is a bit more complicated. We have the sequence {n / [(2n +1)S_n]}, and we need to find T_14, which is the sum of the first 14 terms of this sequence. First, let's write down the general term of this sequence. The nth term is n / [(2n + 1)S_n]. Since we already found S_n = 2n² - n, substitute that in:Term_n = n / [(2n + 1)(2n² - n)]We need to find T_n = sum_{k=1 to n} Term_k, so specifically T_14 = sum_{k=1 to 14} [k / ( (2k +1)(2k² -k) ) ]This looks complicated. Maybe we can simplify the term first. Let's try to decompose the fraction into partial fractions. Let's see.Let me denote Term_n = n / [(2n +1)(2n² -n)]. Let's factor the denominator:First, 2n² - n = n(2n -1). So the denominator is (2n +1)*n*(2n -1). So Term_n = n / [n*(2n -1)(2n +1)] = 1 / [(2n -1)(2n +1)(2n -1)? Wait, wait. Wait, denominator:Wait, 2n² - n is n(2n -1), so the denominator is (2n +1)*n*(2n -1). Then, the numerator is n. So Term_n = n / [n*(2n -1)(2n +1)] = 1 / [(2n -1)(2n +1)]Wait, hold on, that seems like a big simplification! So the n in the numerator cancels with the n in the denominator. So Term_n simplifies to 1 / [(2n -1)(2n +1)]. That's a much simpler expression.So T_n is the sum from k=1 to n of 1 / [(2k -1)(2k +1)]Now, this seems manageable. Let's consider the general term 1/[(2k -1)(2k +1)]. This is a telescoping series. To decompose it into partial fractions, we can write:1/[(2k -1)(2k +1)] = A/(2k -1) + B/(2k +1)Multiply both sides by (2k -1)(2k +1):1 = A(2k +1) + B(2k -1)Let me solve for A and B. Let's expand the right side:1 = A*2k + A + B*2k - B = (2A + 2B)k + (A - B)This must hold for all k, so coefficients of like terms must be equal. Therefore:For the coefficient of k: 2A + 2B = 0For the constant term: A - B = 1So we have the system:2A + 2B = 0A - B = 1Let me solve this. From the first equation, divide both sides by 2: A + B = 0 => B = -ASubstitute into the second equation: A - (-A) = 1 => A + A = 1 => 2A =1 => A = 1/2Then B = -1/2Therefore, the decomposition is:1/[(2k -1)(2k +1)] = (1/2)/(2k -1) - (1/2)/(2k +1)So, Term_k = 1/2 [1/(2k -1) - 1/(2k +1)]Therefore, the sum T_n is:Sum_{k=1 to n} [1/2 (1/(2k -1) - 1/(2k +1))] = (1/2) Sum_{k=1 to n} [1/(2k -1) - 1/(2k +1)]This is a telescoping series. Let's write out the terms:For k=1: 1/2 [1/1 - 1/3]For k=2: 1/2 [1/3 - 1/5]For k=3: 1/2 [1/5 - 1/7]...For k=n: 1/2 [1/(2n -1) - 1/(2n +1)]When we add all these up, most terms cancel:The -1/3 from the first term cancels with the +1/3 from the second term.The -1/5 from the second term cancels with the +1/5 from the third term, and so on.This continues until the last term, which has -1/(2n +1). So all intermediate terms cancel, leaving:T_n = (1/2)[1/1 - 1/(2n +1)]Simplify this:T_n = (1/2)[1 - 1/(2n +1)] = (1/2)[(2n +1 -1)/(2n +1)] = (1/2)[2n / (2n +1)] = n / (2n +1)Wow, that's a nice telescoping sum! So T_n = n / (2n +1)Therefore, for part II, T_14 = 14 / (2*14 +1) = 14 / 29So, let me verify this with an example. Let's take n=1. Then T_1 should be Term_1 = 1 / [(2*1 +1)S_1]. S_1 = 2*1² -1 = 1, so Term_1 = 1 / (3*1) = 1/3. According to the formula T_n = n/(2n +1), T_1 = 1/3. Correct.For n=2, Term_1 + Term_2 = 1/3 + 2 / (5*S_2). S_2 = 2*4 -2 = 8 -2 =6. So Term_2 = 2 / (5*6) = 2/30 =1/15. Then T_2 =1/3 +1/15= 5/15 +1/15=6/15=2/5. According to the formula, T_2=2/(5)=2/5. Correct.Similarly, for n=3: T_3=3/(7). Let's compute manually. Term_1=1/3, Term_2=1/15, Term_3=3/(7*S_3). S_3=2*9 -3=18-3=15. Term_3=3/(7*15)=3/105=1/35. Then T_3=1/3 +1/15 +1/35. Let's compute:Convert to denominators of 105: 35/105 +7/105 +3/105=45/105= 9/21=3/7. Which is 3/7. Correct.Therefore, the formula T_n = n/(2n +1) works. So for part II, T_{14}=14/(2*14 +1)=14/29.So the answers are:I. S_n = 2n² -nII. T_{14}=14/29Final AnswerI. boxed{S_n = 2n^2 - n}II. boxed{dfrac{14}{29}}</think>Given an arithmetic sequence ({a_n}) with a common difference (d = 4) and (a_2 + a_5 = 22), we need to find the sum (S_n) of the first (n) terms and the sum (T_{14}) of the first 14 terms of the sequence (left{ frac{n}{(2n+1)S_n} right}). Part I: Finding (S_n)1. Determine the first term (a_1): - The nth term of an arithmetic sequence is (a_n = a_1 + (n-1)d). - For (a_2) and (a_5): [ a_2 = a_1 + 4 quad text{and} quad a_5 = a_1 + 16 ] - Given (a_2 + a_5 = 22): [ (a_1 + 4) + (a_1 + 16) = 22 implies 2a_1 + 20 = 22 implies 2a_1 = 2 implies a_1 = 1 ]2. Sum of the first (n) terms (S_n): - Using the sum formula (S_n = frac{n}{2} [2a_1 + (n-1)d]): [ S_n = frac{n}{2} [2 cdot 1 + 4(n-1)] = frac{n}{2} [2 + 4n - 4] = frac{n}{2} [4n - 2] = n(2n - 1) ] - Simplifying, we get: [ S_n = 2n^2 - n ] Part II: Finding (T_{14})1. General term of the sequence (left{ frac{n}{(2n+1)S_n} right}): - Substitute (S_n = 2n^2 - n) into the term: [ frac{n}{(2n+1)S_n} = frac{n}{(2n+1)(2n^2 - n)} = frac{n}{n(2n-1)(2n+1)} = frac{1}{(2n-1)(2n+1)} ]2. Partial fraction decomposition: - Decompose (frac{1}{(2n-1)(2n+1)}): [ frac{1}{(2n-1)(2n+1)} = frac{1}{2} left( frac{1}{2n-1} - frac{1}{2n+1} right) ]3. Telescoping series sum (T_n): - The sum (T_n) is: [ T_n = sum_{k=1}^{n} frac{1}{(2k-1)(2k+1)} = frac{1}{2} sum_{k=1}^{n} left( frac{1}{2k-1} - frac{1}{2k+1} right) ] - This telescopes to: [ T_n = frac{1}{2} left( 1 - frac{1}{2n+1} right) = frac{n}{2n+1} ]4. Calculate (T_{14}): - Substitute (n = 14): [ T_{14} = frac{14}{2 cdot 14 + 1} = frac{14}{29} ] Final AnswersI. (boxed{S_n = 2n^2 - n})II. (boxed{dfrac{14}{29}}),<think>Okay, so I need to solve this arithmetic sequence problem. Let's start with part (I). The problem says that in an arithmetic sequence {a_n}, the common difference d is 4, and a_2 + a_5 = 22. Then, we need to find S_n, which is the sum of the first n terms.First, let's recall that in an arithmetic sequence, each term is the previous term plus the common difference. So, the general formula for the nth term is a_n = a_1 + (n-1)d, where a_1 is the first term. Here, d is given as 4, so the formula becomes a_n = a_1 + 4(n-1).Now, the problem states that a_2 + a_5 = 22. Let's write out those terms using the formula. a_2 = a_1 + 4(2-1) = a_1 + 4*1 = a_1 + 4a_5 = a_1 + 4(5-1) = a_1 + 4*4 = a_1 + 16So, adding them together: a_2 + a_5 = (a_1 + 4) + (a_1 + 16) = 2a_1 + 20According to the problem, this equals 22. So:2a_1 + 20 = 22Subtract 20 from both sides:2a_1 = 2Divide by 2:a_1 = 1Okay, so the first term is 1. Now, we need to find S_n, the sum of the first n terms. The formula for the sum of an arithmetic sequence is S_n = n/2 * [2a_1 + (n-1)d]. Alternatively, sometimes written as S_n = n*(a_1 + a_n)/2. Either formula should work. Let me confirm.Since we know a_1 and d, maybe the first formula is better. Let's plug in the values:a_1 = 1, d = 4.So, S_n = n/2 * [2*1 + (n - 1)*4] = n/2 * [2 + 4(n - 1)]Let me simplify inside the brackets first:2 + 4(n - 1) = 2 + 4n - 4 = 4n - 2Therefore, S_n = n/2 * (4n - 2) = n/2 * 2*(2n - 1) = n*(2n - 1)Wait, that's nice. So, S_n simplifies to n*(2n - 1). Let me check that step again to be sure.Original expression: n/2 * (4n - 2). Let's factor out a 2 from the numerator:4n - 2 = 2*(2n - 1)So, n/2 * 2*(2n - 1) = n*(2n - 1). Yes, that cancels out the 2. So S_n = n(2n - 1). That seems correct.Alternatively, let's verify with the other formula S_n = n*(a_1 + a_n)/2. We know a_1 = 1 and a_n = 1 + 4(n - 1) = 1 + 4n - 4 = 4n - 3. So, a_n = 4n - 3.Thus, S_n = n*(1 + (4n - 3))/2 = n*(4n - 2)/2 = n*(2n - 1). Same result. Good, so part (I) is done. S_n = 2n² - n. So, written as S_n = 2n² - n.So that's part (I). Now part (II) says: Let {n / [(2n + 1)S_n]} be a sequence, and T_n be the sum of the first n terms of this sequence. Find T_{14}.First, let me parse this. The sequence is given by the nth term: term_n = n / [ (2n + 1) * S_n ]We have already found S_n = 2n² - n. So, substituting that into the denominator:term_n = n / [ (2n + 1)(2n² - n) ]We need to compute T_n, which is the sum from k = 1 to n of term_k. So, T_n = Σ_{k=1}^n [k / ( (2k + 1)(2k² - k) ) ]But specifically, we need T_{14}. So, we need to compute the sum of the first 14 terms of this sequence. The challenge here is to find a formula for T_n, or to compute it directly. Since the problem is asking for T_{14}, perhaps there's a telescoping series here, which would allow us to compute the sum more easily.Let me try to simplify the general term. Let's write term_k = k / [ (2k + 1)(2k² - k) ]First, note that 2k² - k can be factored: 2k² - k = k(2k - 1). So, the denominator becomes (2k + 1)*k*(2k - 1). Therefore, term_k = k / [k*(2k - 1)(2k + 1)] = 1 / [ (2k - 1)(2k + 1) ) ]So, term_k simplifies to 1 / [ (2k - 1)(2k + 1) ). That's a significant simplification. So term_k = 1 / [ (2k - 1)(2k + 1) )So now, T_n = Σ_{k=1}^n [1 / ( (2k - 1)(2k + 1) ) ]This looks like a telescoping series. To handle this, we can use partial fractions.Let me recall that for expressions like 1/( (2k - 1)(2k + 1) ), we can write them as A/(2k - 1) + B/(2k + 1). Let's find A and B.1 = A(2k + 1) + B(2k - 1)Let me solve for A and B. Let's set up the equation:1 = A(2k + 1) + B(2k - 1)To find A and B, we can choose specific values of k or equate coefficients.First, expand the right side:1 = (2A + 2B)k + (A - B)This must hold for all k, so the coefficients of like terms must be equal.Coefficient of k: 2A + 2B = 0Constant term: A - B = 1So, solving this system:From the first equation: 2A + 2B = 0 ⇒ A + B = 0 ⇒ B = -ASubstitute B = -A into the second equation: A - (-A) = 1 ⇒ 2A = 1 ⇒ A = 1/2Then, B = -A = -1/2Therefore, the partial fraction decomposition is:1 / [ (2k - 1)(2k + 1) ) ] = (1/2)/(2k - 1) - (1/2)/(2k + 1)So, term_k = 1/2 [1/(2k - 1) - 1/(2k + 1)]Therefore, the sum T_n = Σ_{k=1}^n term_k = 1/2 Σ_{k=1}^n [1/(2k - 1) - 1/(2k + 1)]This is a telescoping series. Let's write out the terms to see the cancellation.For k = 1: 1/2 [1/1 - 1/3]For k = 2: 1/2 [1/3 - 1/5]For k = 3: 1/2 [1/5 - 1/7]...For k = n: 1/2 [1/(2n - 1) - 1/(2n + 1)]When we add all these up, most terms will cancel. The sum becomes:1/2 [ (1/1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + ... + (1/(2n -1) - 1/(2n +1)) ]All the intermediate terms cancel out: -1/3 cancels with +1/3, -1/5 cancels with +1/5, etc., leaving:1/2 [ 1 - 1/(2n + 1) ]Therefore, T_n = 1/2 [1 - 1/(2n + 1)] = 1/2 [ (2n +1 -1)/(2n +1) ] = 1/2 [2n/(2n +1)] = n / (2n +1)Wow, that's a neat result. So T_n simplifies to n/(2n +1). Let me verify this with n=1, to check.If n=1, then T_1 = term_1 = 1 / [ (2*1 +1)S_1 ].But S_1 = a_1 =1. So term_1 =1/(3*1)=1/3. According to the formula, T_1 =1/(2*1 +1)=1/3. Correct.Similarly, n=2: T_2 = term_1 + term_2 =1/3 + 1/( (5)(2*4 -2))= wait, let me compute term_2. Wait, S_2 =2*(2*2 -1)=2*3=6. So term_2=2/(5*6)=2/30=1/15. Then T_2=1/3 +1/15=6/15=2/5. According to the formula, T_2=2/(2*2 +1)=2/5. Correct. So the formula works. Great.Therefore, the general formula is T_n = n/(2n +1). Therefore, T_{14}=14/(2*14 +1)=14/29.Wait, so part (II) answer is 14/29. That's straightforward once we derived the telescoping sum. Let me recap:We simplified term_k using partial fractions, observed the telescoping nature, summed up the terms, and found that T_n is n/(2n +1). Therefore, substituting n=14 gives 14/29.So, the answers are:(I) S_n = 2n² -n(II) T_{14}=14/29Final Answer(I) boxed{S_n = 2n^2 - n}(II) boxed{dfrac{14}{29}}</think>Given an arithmetic sequence ({a_n}) with a common difference (d = 4) and (a_2 + a_5 = 22), we need to find the sum (S_n) of the first (n) terms and the sum (T_{14}) of the first 14 terms of the sequence (left{ frac{n}{(2n+1)S_n} right}).(I) Finding (S_n):1. The general term of the arithmetic sequence is (a_n = a_1 + (n-1)d). Given (d = 4), we have (a_n = a_1 + 4(n-1)).2. Using the given condition (a_2 + a_5 = 22): [ a_2 = a_1 + 4 quad text{and} quad a_5 = a_1 + 16 ] [ (a_1 + 4) + (a_1 + 16) = 22 implies 2a_1 + 20 = 22 implies 2a_1 = 2 implies a_1 = 1 ]3. The sum (S_n) of the first (n) terms of an arithmetic sequence is given by: [ S_n = frac{n}{2} [2a_1 + (n-1)d] ] Substituting (a_1 = 1) and (d = 4): [ S_n = frac{n}{2} [2 cdot 1 + 4(n-1)] = frac{n}{2} [2 + 4n - 4] = frac{n}{2} [4n - 2] = n(2n - 1) ] Therefore, (S_n = 2n^2 - n).(II) Finding (T_{14}):1. The sequence given is (left{ frac{n}{(2n+1)S_n} right}). Substituting (S_n = 2n^2 - n): [ frac{n}{(2n + 1)(2n^2 - n)} = frac{n}{(2n + 1)n(2n - 1)} = frac{1}{(2n - 1)(2n + 1)} ]2. Using partial fractions: [ frac{1}{(2n - 1)(2n + 1)} = frac{1}{2} left( frac{1}{2n - 1} - frac{1}{2n + 1} right) ]3. The sum (T_n) is a telescoping series: [ T_n = sum_{k=1}^n frac{1}{(2k - 1)(2k + 1)} = frac{1}{2} sum_{k=1}^n left( frac{1}{2k - 1} - frac{1}{2k + 1} right) ] [ T_n = frac{1}{2} left( 1 - frac{1}{2n + 1} right) = frac{1}{2} left( frac{2n}{2n + 1} right) = frac{n}{2n + 1} ]4. Substituting (n = 14): [ T_{14} = frac{14}{2 cdot 14 + 1} = frac{14}{29} ] Final Answer(I) (boxed{S_n = 2n^2 - n})(II) (boxed{dfrac{14}{29}})]

🔑:(text{I}) Given the common difference d=4, and a_2+a_5=22, we can set up the equation[2a_1 + 5d = 22,]Substituting d = 4, we find[2a_1 + 20 = 22 implies a_1 = 1.]Therefore, the general term of the sequence is[a_n = 1 + 4(n - 1) = 4n - 3.]Using the formula for the sum of the first n terms of an arithmetic sequence, we find[S_n = frac{n}{2}(a_1 + a_n) = frac{n}{2}(1 + 4n - 3) = 2n^2 - n.]So S_n is boxed{2n^2 - n}.(text{II}) The term frac{n}{(2n+1)S_n} simplifies to[frac{n}{(2n+1)S_n} = frac{n}{(2n+1)(2n^2 - n)} = frac{1}{2n^2 - 1} = frac{1}{2} left( frac{1}{2n - 1} - frac{1}{2n + 1} right).]This indicates a telescoping series, so the sum T_{14} is[T_{14} = frac{1}{2} left(1 - frac{1}{3} + frac{1}{3} - frac{1}{5} + ldots + frac{1}{27} - frac{1}{29} right)= frac{1}{2}left(1- frac{1}{29}right) = frac{14}{29}.]Thus, T_{14} is boxed{frac{14}{29}}.

🧠:[<think>Okay, let's tackle this problem step by step. There are two parts here. The first part is about the function g(x) = ln x and determining f(t), which counts the number of real roots for the equation g(x) = t. The second part is more complex, involving a piecewise function and finding the range of a parameter 'a' such that there exists some t where f(t+2) > f(t). Let's start with the first part.First Part: g(x) = ln xSo, we need to find how many real roots the equation ln x = t has for a real number t. Let's recall that the natural logarithm function ln x is only defined for x > 0. The equation ln x = t can be rewritten as x = e^t. Since e^t is always positive for any real t, there's exactly one solution x = e^t for any t. Therefore, regardless of the value of t, there's exactly one real root. Hence, f(t) = 1 for all t in this case. That seems straightforward.Second Part: Piecewise Function g(x)Now, the second part is about the piecewise function:g(x) = { x, when x ≤ 0; -x² + 2a x + a, when x > 0 }We need to find the range of 'a' such that there exists some t where f(t+2) > f(t). Here, f(t) is again the number of real roots of g(x) = t.First, let me understand the function g(x). For x ≤ 0, it's a linear function with slope 1, so it's just the line y = x. For x > 0, it's a quadratic function: -x² + 2a x + a. Let's analyze the quadratic part first.The quadratic part is -x² + 2a x + a. Since the coefficient of x² is negative (-1), this is a downward-opening parabola. The vertex of this parabola will be at x = -b/(2a'), where the quadratic is written as a'x² + b x + c. Here, a' = -1, b = 2a. So the vertex is at x = - (2a)/(2*(-1)) = (2a)/2 = a. Wait, let me check that again. The formula is x = -b/(2a'). Here, a' = -1, so x = - (2a) / (2*(-1)) = -2a / (-2) = a. Yes, the vertex is at x = a. The y-coordinate of the vertex is then g(a) = -(a)^2 + 2a*(a) + a = -a² + 2a² + a = a² + a.So, the maximum value of the quadratic part is a² + a at x = a. Since the parabola opens downward, the quadratic function decreases to the left and right from the vertex. However, since the quadratic part is defined only for x > 0, we need to consider the behavior of this quadratic for x > 0. Depending on the position of the vertex (x = a), the maximum value a² + a occurs at x = a if a > 0. If a ≤ 0, then the vertex is at x = a ≤ 0, which is outside the domain of the quadratic part (x > 0). In that case, the maximum of the quadratic part for x > 0 would be at x = 0, but x=0 is part of the linear piece. Wait, but for x > 0, the quadratic is defined. Let me think again.If a ≤ 0, then the vertex x = a is at or to the left of x = 0, so for x > 0, the quadratic is decreasing because the parabola opens downward and the vertex is at x = a ≤ 0. Therefore, for x > 0, the quadratic part starts at x = 0 with value g(0+) = -(0)^2 + 2a*(0) + a = a. Then, since it's decreasing for x > 0 (as the vertex is at x ≤ 0), the quadratic part decreases from a to negative infinity as x approaches infinity.If a > 0, the vertex is at x = a > 0. So, the quadratic increases from x = 0 (value a) up to x = a (value a² + a) and then decreases towards negative infinity. So, in this case, the quadratic part has a maximum at x = a with value a² + a.Now, let's combine this with the linear part (x ≤ 0). The linear part is y = x, which is a straight line passing through the origin with slope 1. For x ≤ 0, y = x takes all values from negative infinity up to 0.So, the entire function g(x) is:- For x ≤ 0: y = x, which covers (-∞, 0]- For x > 0: y = -x² + 2a x + a, which behaves as described above.Now, we need to analyze the number of real roots for g(x) = t. Let's consider different cases based on the value of a.First, let's sketch or visualize the graph of g(x). The left side (x ≤ 0) is a line increasing from -infty to 0. The right side (x > 0) is a quadratic that either starts at a and decreases (if a ≤ 0) or starts at a, rises to a peak at x = a (value a² + a), then decreases (if a > 0).The number of real roots for g(x) = t depends on how the horizontal line y = t intersects with this graph.Let's first consider the case when a ≤ 0.Case 1: a ≤ 0In this case, the quadratic part (x > 0) starts at (0, a) and decreases. Since a ≤ 0, the starting point is at or below y = 0. The linear part (x ≤ 0) covers all y ≤ 0. Therefore, for t > 0:- The left side (x ≤ 0) can't contribute because y = x ≤ 0.- The right side (quadratic) is decreasing from a (≤ 0) to -infty, so it never reaches t > 0. Therefore, no solutions.For t ≤ 0:- The left side (x ≤ 0) has one solution x = t (since x = t, but x ≤ 0, so t must be ≤ 0).- The right side (quadratic) equation: -x² + 2a x + a = t. Since the quadratic is decreasing from a (≤ 0) to -infty, the equation will have: - If t > a: Since the quadratic starts at a and decreases, t > a would mean there's one solution (since it's continuous and strictly decreasing). - If t = a: Exactly one solution at x = 0 (but x > 0, so x = 0 is part of the linear piece, not the quadratic). Wait, at x = 0, the quadratic part evaluates to a, which is the same as the linear part at x=0 (which is 0). Wait, hold on: At x=0, the linear part is 0 (since x ≤ 0 includes x=0), and the quadratic part at x approaching 0 from the right is approaching a. So unless a = 0, the function has a jump at x=0.Wait, this is important. At x = 0, the function is defined by the linear part (x ≤ 0) as g(0) = 0. The quadratic part, when x approaches 0 from the right, approaches a. So unless a = 0, there's a discontinuity at x = 0. Therefore, if a ≠ 0, there's a jump discontinuity at x=0. If a = 0, then both parts meet at y=0, so the function is continuous.So, for a ≤ 0, but a ≠ 0, there's a jump from y=0 (at x=0-) to y=a (at x=0+). Since a ≤ 0, the jump is downward if a < 0. For example, if a = -1, then at x=0, the left limit is 0, and the right limit is -1. So, there's a drop from 0 to -1 at x=0.This affects the number of solutions. Let's re-examine.For t in (a, 0):- On the left side (x ≤ 0), since t < 0, the equation x = t has a solution x = t (which is ≤ 0).- On the right side (x > 0), the quadratic part at x=0+ is a. Since t is in (a, 0), and the quadratic decreases from a to -infty, there's exactly one solution for the quadratic equation -x² + 2a x + a = t. So total solutions: 2.For t = a:- On the left side: x = a. But a ≤ 0, so x = a is a valid solution (since a ≤ 0, x = a is ≤ 0).- On the right side: The quadratic equation at t = a would have x=0 as a solution, but x=0 is part of the linear piece, so the quadratic part doesn't include x=0. The quadratic equation is -x² + 2a x + a = a => -x² + 2a x = 0 => x(-x + 2a) = 0. Solutions are x=0 and x=2a. But x=0 is not in x > 0, and x=2a: since a ≤ 0, 2a ≤ 0, so x=2a is also not in x > 0. Therefore, on the right side, no solutions. But the left side has x = a, so total solutions: 1.For t < a:- On the left side: x = t, which is ≤ 0. Since t < a ≤ 0, x = t is valid.- On the right side: The quadratic equation. Since the quadratic part goes from a (at x=0+) to -infty, and t < a, then there's exactly one solution on the right side. So total solutions: 2.Wait, but if a ≤ 0, then for t < a (which is even smaller), the right side still has one solution because the quadratic is decreasing. So even for t < a, you have two solutions? Wait, let me test with a specific value.Let’s take a = -1.Then the quadratic part for x > 0 is -x² + 2*(-1)x + (-1) = -x² - 2x -1. The vertex is at x = a = -1, which is outside x > 0. So for x > 0, the quadratic starts at x=0: -0 -0 -1 = -1, and decreases from -1 to -infty. So for t = -2 (which is < a = -1), the quadratic equation is -x² - 2x -1 = -2 → -x² -2x -1 + 2 = 0 → -x² -2x +1 =0 → x² + 2x -1 =0. Solutions x = [-2 ± sqrt(4 +4)]/2 = [-2 ± sqrt(8)]/2 = [-2 ± 2√2]/2 = -1 ± √2. Since x > 0, the positive solution is -1 + √2 ≈ 0.414, which is valid. So one solution on the right, and one solution on the left (x = t = -2). So total solutions: 2. Similarly, for t = -1.5 (between a=-1 and -infty), same thing.But wait, if a = -1, then for t = -1 (which is equal to a), the quadratic equation is -x² -2x -1 = -1 → -x² -2x =0 → x(-x -2) =0. Solutions x=0 and x=-2. But x>0, so no solutions on the right. Left side has x = -1. So total solutions: 1.Similarly, for t = 0, which is greater than a=-1, the left side x=0 (which is valid), and the quadratic equation: -x² -2x -1 =0 → x² +2x +1 =0 → (x+1)^2=0. Solution x=-1, which is not in x>0. So only x=0 on the left side. So total solutions:1.Wait, but earlier I thought for t in (a, 0), which would be t in (-1, 0) when a=-1, the left side has x = t (which is in (-1,0)), and the quadratic equation: does it have a solution?Take t = -0.5. Then quadratic equation: -x² -2x -1 = -0.5 → -x² -2x -0.5 =0 → multiply by -1: x² + 2x +0.5 =0. Discriminant: 4 - 2 = 2. Solutions: (-2 ± sqrt(2))/2. sqrt(2) ≈1.414, so (-2 +1.414)/2 ≈ (-0.586)/2≈ -0.293, which is less than 0. The other solution is negative. So no solutions on x>0. So only left side: x = -0.5. So only one solution. Wait, that contradicts my previous reasoning.Wait, maybe my earlier analysis was incorrect. Let's redo it.If a ≤0, the quadratic part for x>0 is decreasing from a (at x=0+) to -infty. So for t > a: Since the quadratic is continuous (for x>0) and decreasing, starting at a and going to -infty, then for any t between a and +infty, does the quadratic part have a solution?Wait, but if a is negative, say a=-1, then the quadratic starts at -1 and decreases. So for t > a (t > -1), but the quadratic is only going down from -1 to -infty. So if t is between -1 and +infty, but since the quadratic can only take values from -1 down to -infty, then the equation -x² +2a x +a = t can only have solutions when t ≤ a (since a is the starting point). Wait, this seems conflicting.Wait, no. Wait, in the case where a is negative, the quadratic starts at a (negative) and decreases. So for t > a (i.e., t greater than a negative number), but the quadratic is going down. So, for t between a and some upper limit?Wait, but since the quadratic is only defined for x>0, and it's continuous there (but separate from the linear part). Wait, if a ≤0, the quadratic is strictly decreasing from x=0+ (value a) to -infty as x approaches infinity. Therefore, the range of the quadratic part is (-infty, a]. So for the quadratic part, the equation -x² +2a x +a = t has solutions only if t ≤ a. For t > a, the quadratic part has no solutions.But the linear part (x ≤0) is y = x, which can take any t ≤0. So combining both parts:- For t >0: Only possible solutions from quadratic part, but quadratic part can't reach t >a. Since a ≤0, t >0 > a, but quadratic part only goes down to -infty, so no solutions. Therefore, f(t) =0.Wait, but if t >0, the linear part (x ≤0) can't have solutions because x ≤0 implies y =x ≤0. The quadratic part can't reach t >0 because it's decreasing from a ≤0. Therefore, for t >0, no solutions.For t ≤0:- The linear part has x = t, which is valid since t ≤0. So at least one solution.- The quadratic part: equation is -x² +2a x +a = t. Since quadratic part has range (-infty, a], and t ≤0. If a ≤0, then the range is (-infty, a] with a ≤0. So for t ≤a, quadratic equation has one solution (since it's strictly decreasing). For a < t ≤0, quadratic equation has no solution (because t is above the starting point a, but quadratic is decreasing). Wait, but if a is negative, say a=-1, and t= -0.5 (which is >a), then the quadratic starts at a=-1 and decreases, so it never reaches -0.5. So quadratic part has no solution. Therefore:- For t ≤a: quadratic part has one solution, linear part has one solution (x=t). So total f(t)=2.- For a < t ≤0: quadratic part has no solution, linear part has one solution (x=t). So total f(t)=1.- For t >0: no solutions. f(t)=0.But this contradicts my earlier example where a=-1 and t=-0.5, which was supposed to have only one solution. Wait, but according to this analysis, when t=-0.5 which is >a=-1, then quadratic part has no solution, so only one solution from the linear part. That matches the example. Then, for t ≤a, say t=-2 (a=-1), quadratic part has one solution, linear part has one solution, total two.Therefore, for a ≤0:f(t) = 0, t >0f(t) =1, a < t ≤0f(t) =2, t ≤aSo f(t) is 0 for t>0, 1 for t between a and 0, and 2 for t ≤a.Case 2: a >0Now, if a >0, the quadratic part (x>0) has a vertex at x=a. The quadratic starts at x=0 with value a, rises to the vertex at x=a with value a² +a, then decreases to -infty as x approaches infinity.So the range of the quadratic part is (-infty, a² +a]. Since a >0, a² +a > a.The linear part (x ≤0) is y=x, which takes all values ≤0.So combining both parts:For t > a² +a: The quadratic part can't reach t, and the linear part (y=x ≤0) can't reach t >0. So f(t)=0.For a < t ≤ a² +a: The quadratic part has two solutions (since the parabola reaches t at two points: one left of the vertex, one right). But x>0, so both solutions must be in x>0. Wait, but since the parabola is increasing up to x=a and then decreasing. Wait, for t between a and a² +a, the equation -x² +2a x +a = t will have two solutions in x>0. Because when t is between the starting value (a at x=0) and the maximum (a² +a), the parabola crosses the line y=t twice: once between x=0 and x=a, and once between x=a and infinity.Wait, but actually, at x=0, the quadratic has value a. Then it increases to a² +a at x=a, then decreases. So yes, for t between a and a² +a, there are two solutions in x>0. For t = a, the quadratic equation has a solution at x=0 (but x=0 is part of the linear piece). Wait, at x=0, quadratic part is a, but linear part is 0. So there's a discontinuity at x=0 when a>0.So, for t = a:- Quadratic part: -x² +2a x +a = a → -x² +2a x =0 → x(-x + 2a)=0. Solutions x=0 (not in x>0) and x=2a (which is >0 since a>0). So quadratic part has x=2a as a solution. The linear part has x=a (but a>0, so x=a is not in x≤0). Wait, the linear part is y=x, so setting x = t gives x = a. But since x ≤0, and a >0, x = a is not in the domain. Therefore, only the quadratic part contributes, with x=2a. So f(t=a) =1.For t between 0 and a:Wait, if a>0, then 0 < t <a. Let's see. The linear part (x ≤0) can only have solutions for t ≤0. So for t >0, the linear part doesn't contribute. The quadratic part for t between 0 and a: since the quadratic starts at a (at x=0) and goes up to a² +a. If t is between 0 and a, but the quadratic starts at a (which is greater than t). Wait, the quadratic at x=0 is a. If a > t (since t is between 0 and a), then starting from a, going up to a² +a, then down. So if t is between 0 and a, then since the quadratic starts above t (at a) and reaches a maximum higher than a, then comes back down. Wait, but t is less than a. Wait, no. If a >0, and t is between 0 and a, then the quadratic part at x=0 is a, which is greater than t. Then the quadratic first increases to a² +a (which is greater than a, since a>0), so it's even higher. Then decreases to -infty. Therefore, the equation -x² +2a x +a = t would have two solutions: one between x=0 and x=a, and another one after x=a. Wait, but since the quadratic is increasing up to x=a, then decreasing. If t is less than the starting value (a) but greater than the minimum? Wait, but it starts at a, goes up, then down to -infty. So if t is between -infty and a² +a, but in this case, t is between 0 and a. Since a² +a is greater than a (since a>0), then t is between 0 and a < a² +a. Therefore, the equation will have two solutions in x>0. Because from x=0 (value a) to x=a (peak), the quadratic increases, so it doesn't cross t (since t < a). Then, after x=a, it decreases from a² +a down to -infty, crossing t once. Wait, no. Wait, if the quadratic is increasing from a to a² +a, then decreasing to -infty. So if t is between a and a² +a, there are two solutions. If t is less than a, then since the quadratic starts at a and goes up, then down, so once it passes below t on the decreasing part, it will cross t once. Wait, but if t is less than a, the quadratic starts at a (above t), goes up to a² +a (still above t if t < a?), then decreases. So when decreasing from a² +a to -infty, it crosses t once. So total one solution in x>0. But the linear part (x <=0) can't contribute because t >0 (since t is between 0 and a). Wait, but if t is between 0 and a, then t is positive, so the linear part (y=x <=0) can't have solutions. Therefore, for t in (0, a), f(t) =1.Wait, this is conflicting with previous reasoning. Let's take a concrete example. Let a=1. Then the quadratic part for x>0 is -x² +2*1*x +1 = -x² +2x +1. The vertex is at x=1, y=1² +1=2. So the quadratic starts at x=0 with y=1, goes up to y=2 at x=1, then decreases to -infty.Take t=0.5. Then the equation -x² +2x +1 =0.5 → -x² +2x +0.5=0 → x² -2x -0.5=0. Discriminant: 4 +2=6. Solutions: [2 ± sqrt(6)]/2 =1 ± sqrt(6)/2 ≈1 ±1.2247. So two solutions: ≈2.2247 and ≈-0.2247. But x>0, so only x≈2.2247 is valid. Wait, that's only one solution. Wait, but according to the quadratic formula, we have two solutions, but one is negative. So in this case, only one solution on x>0. Therefore, f(t=0.5)=1. But according to earlier logic, since t=0.5 <a=1, and quadratic part starts at a=1, then decreases, crossing t once. So indeed, only one solution.But if t=1.5 (between a=1 and a² +a=2), then equation -x² +2x +1 =1.5 → -x² +2x -0.5=0 → x² -2x +0.5=0. Solutions: [2 ± sqrt(4 -2)]/2 = [2 ± sqrt(2)]/2 ≈ [2 ±1.414]/2 ≈1.707/2≈0.85 and 0.292/2≈0.146. Both positive. So two solutions. Hence, f(t=1.5)=2.So for a=1:- For t >2: f(t)=0- For t=2: equation -x² +2x +1=2 → -x² +2x -1=0 →x² -2x +1=0 →(x-1)^2=0 →x=1. So f(t=2)=1 (from quadratic part). The linear part x=2 is invalid since x≤0.- For 1 < t <2: two solutions on quadratic part, f(t)=2.- For t=1: quadratic part has x=2 (from earlier analysis: x=2a=2*1=2). Linear part x=1 is invalid. So f(t=1)=1.- For 0 < t <1: quadratic part has one solution, linear part none, so f(t)=1.- For t ≤0: quadratic part for x>0 can have solutions if t ≤1 (since quadratic's range is (-infty,2]), but t ≤0. So:Wait, for t ≤0:- Linear part: x=t, which is valid (since t ≤0). So one solution.- Quadratic part: equation -x² +2x +1 =t. Since quadratic part can go down to -infty, for t ≤2. For t ≤0, it will cross the quadratic part twice? Wait, let's test t=0:-x² +2x +1 =0 →x² -2x -1=0. Solutions [2 ± sqrt(4 +4)]/2 = [2 ± sqrt(8)]/2 =1 ± sqrt(2). Both solutions: ≈2.414 and ≈-0.414. Only positive solution ≈2.414. So f(t=0)=1 (linear part x=0 and quadratic part x≈2.414). Wait, but linear part x=0 (t=0) is valid. So actually, two solutions: x=0 and x≈2.414. Wait, but x=0 is part of the linear piece. So f(t=0)=2?Wait, at x=0, the linear part gives y=0. The quadratic part at x=0 is 1. So for t=0, the equation g(x)=0 has x=0 (from linear part) and the quadratic equation solution x≈2.414. Therefore, f(t=0)=2.But according to previous logic, for t ≤0, the linear part gives x=t (valid if t ≤0). So for t=0, x=0 is valid. Quadratic part for t=0 has one solution. Hence, total f(t=0)=2.Wait, so this suggests that for a>0 and t ≤0:- Linear part: one solution (x=t)- Quadratic part: equation -x² +2a x +a =t. Since the quadratic's range is (-infty, a² +a], and t ≤0. Since a² +a >0, t ≤0 is within the quadratic's range. For t ≤0, since the quadratic part is a parabola opening downward with vertex at x=a, for t ≤0, how many solutions?At t=0: quadratic part solutions are x=0 (from linear part) and x≈2.414. But actually, solving -x² +2a x +a =0 when a=1:-x² +2x +1=0 →x² -2x -1=0 → solutions [2 ± sqrt(4 +4)]/2 =1 ± sqrt(2). So x≈2.414 and x≈-0.414. The negative solution is invalid. So only one solution in x>0. Wait, but the linear part already gives x=0. So total two solutions: x=0 (linear) and x≈2.414 (quadratic). So f(t=0)=2.For t <0:Say t=-1. Then the linear part x=-1 (valid). Quadratic part: -x² +2x +1 = -1 → -x² +2x +2=0 →x² -2x -2=0. Solutions [2 ± sqrt(4 +8)]/2 = [2 ± sqrt(12)]/2 = [2 ± 2*sqrt(3)]/2 =1 ± sqrt(3). Approximately 2.732 and -0.732. Only positive solution ≈2.732. So f(t=-1)=2 (x=-1 and x≈2.732). Therefore, for t ≤0, f(t)=2.Wait, so for a>0:- t >a² +a: f(t)=0- a <t ≤a² +a: f(t)=2 (quadratic part two solutions, linear part none)- t=a: f(t)=1 (quadratic part one solution, x=2a)- 0 <t <a: f(t)=1 (quadratic part one solution, linear part none)- t=0: f(t)=2 (linear part x=0 and quadratic part one solution)- t <0: f(t)=2 (linear part one solution, quadratic part one solution)So compiling this:For a >0:f(t) =- 0, t > a² +a- 2, a < t ≤a² +a- 1, t =a-1, 0 < t <a-2, t ≤0Wait, but when t=0, it's included in t ≤0, so f(t)=2 there. And for t <0, also 2.But then, how does this fit together?Wait, summarizing:For a>0:- If t > a² + a: 0 solutions- If t = a² +a:1 solution (vertex)- If a < t <a² +a:2 solutions (quadratic part)- If t =a:1 solution (quadratic part at x=2a)- If 0 < t <a:1 solution (quadratic part)- If t ≤0:2 solutions (linear part and quadratic part)But in our earlier example with a=1, t=0 had two solutions. So the breakdown would be:f(t) =0, t >21, t=22, 1 <t <21, t=11, 0 <t <12, t ≤0Hence, the function f(t) for a>0 has different intervals. Now, the problem states: "there exists t such that f(t +2) > f(t)". We need to find the range of a for which such a t exists.So we need to find a where there exists some t where f(t+2) > f(t). Let's analyze this for both cases (a ≤0 and a >0), but since the problem says "there exists t such that f(t +2) > f(t)", we need to see for which a's this is possible.But first, let's consider the possible f(t) values for different a.Case 1: a ≤0From earlier analysis:f(t) =0, t >01, a <t ≤02, t ≤aWe need to check if there exists t such that f(t +2) > f(t).Let's consider possible t:- If t +2 >0, then f(t +2)=0. So for t +2 >0 (i.e., t >-2), f(t +2)=0. For t > -2: If t >0, then f(t)=0. So f(t+2)=0 and f(t)=0, so 0>0? No. If a <t ≤0 (since a ≤0, this interval is possible if a <0). For example, if a=-1, then t in (-1,0]. For t in (-2,0], with a ≤0: - If t >a (since a ≤0 and t >a, but t can be negative or 0). Then f(t) is 1 if a <t ≤0, and 2 if t ≤a. Wait, no. For a ≤0: - If t >0: f(t)=0 (but t >-2, but t can be between -2 and0 or between0 and something. Wait, let's structure it. Let's take a=-1 (specific example). Then: f(t) = 0, t >0 1, -1 <t ≤0 2, t ≤-1 So, t ranges over real numbers. We need to find t where f(t+2) >f(t). Let's analyze for different intervals of t:1. t +2 >0 → t >-2. - Then f(t +2)=0. - So we need f(t) <0, but f(t) can't be negative. Hence, impossible. - So in this interval, f(t+2)=0, so f(t+2) >f(t) implies 0 >f(t). But f(t) is either 0,1, or2. Never negative. So no solutions here.2. t +2 ≤0 → t ≤-2. - Then f(t +2): - If t +2 >a (since a=-1, t +2 >-1 → t >-3). - If t ≤-2, then t +2 ≤0. So for a=-1: - If t +2 >-1 (i.e., t >-3), then f(t +2)=1 if -1 < t +2 ≤0 → t in (-3, -2]. - If t +2 ≤-1 (i.e., t ≤-3), then f(t +2)=2. - For t ≤-2: - t can be in (-3, -2] or t ≤-3. - Let's split: a. t in (-3, -2]: - f(t +2)=1 (since t +2 is in (-1,0]). - f(t): t ≤-2, so t ≤-2 <a=-1? Wait, a=-1. Wait, t ≤-2, which is less than a=-1 only if t ≤-1, but a=-1. Wait, for a=-1, the interval t ≤a is t ≤-1. So t ≤-2 is a subset of t ≤-1. Therefore, f(t)=2 for t ≤-1. So for t in (-3, -2], t ≤-1 (since -3 <t ≤-2 ≤-1). So f(t)=2. - So f(t +2)=1, f(t)=2. Thus 1 >2? No. b. t ≤-3: - f(t +2)=2 (since t +2 ≤-1). - f(t)=2 (since t ≤-3 ≤-1). - So 2 >2? No. Therefore, in this case (a=-1), there is no t such that f(t+2) >f(t). Let's check another a ≤0. Maybe a=0. If a=0, then: f(t) = 0, t>0 1, 0 <t ≤0 → which is just t=0. Wait, no. Wait, for a=0: The quadratic part is -x² +0x +0 = -x² for x>0. Then: For t >0: f(t)=0 For t=0: f(t)=1 (x=0 from linear part; quadratic part equation -x²=0 gives x=0, but x=0 is part of the linear part. Wait, but g(x)=0: x=0 (linear) and quadratic part -x²=0 →x=0, which is same as linear. So only one solution. Hence, f(t=0)=1. For t <0: - Linear part: x=t <0, valid. - Quadratic part: -x²= t →x²= -t. Since t <0, -t >0, so x=±sqrt(-t). But x>0, so x=sqrt(-t). Hence, one solution. Therefore, for a=0: f(t)=0, t>0 f(t)=1, t=0 f(t)=2, t<0 So for a=0, let's check if there exists t where f(t+2) >f(t): - If t +2 >0, then f(t +2)=0. So we need f(t) <0, which is impossible. - If t +2 ≤0 →t ≤-2. Then f(t +2)=2 (since t +2 ≤0, but actually for a=0, t +2 ≤0 implies t ≤-2. For t ≤-2: - f(t +2): For t +2 ≤0, which it is, but a=0. Wait, when a=0, f(t) is: f(t)=2 for t <0, f(t)=1 for t=0, f(t)=0 for t>0. So for t +2 ≤0 (t ≤-2): f(t +2)=2 (since t +2 <0). f(t)=2 (since t ≤-2 <0). So 2 >2? No. Therefore, for a=0, no such t exists. Therefore, for a ≤0, there is no t such that f(t+2) >f(t). Therefore, the answer must be that a has to be in some range where a >0. Let's check for a >0.Case 2: a >0From earlier analysis, for a >0, f(t) is:- 0, t >a² +a- 1, t =a² +a- 2, a <t <a² +a- 1, t=a- 1, 0 <t <a- 2, t ≤0We need to find if there exists t such that f(t+2) > f(t). Let's check possible t values.First, let's note that t+2 can fall into different intervals depending on t. Let's consider different intervals for t:1. t >a² +a -2: Then t+2 >a² +a. So f(t +2)=0. We need 0 >f(t). But f(t) ≥0, so this is possible only if f(t)=0. But if t >a² +a -2, then t can be in various ranges. For example: - If t >a² +a, then f(t)=0. So f(t +2)=0, so 0>0? No. - If a² +a -2 <t ≤a² +a, then f(t) is: - If t >a² +a:0 - If t=a² +a:1 - If a <t <a² +a:2 So in this range, t can be in (a, a² +a) or t=a² +a. For example, if t is in (a² +a -2, a² +a), then t+2 is in (a² +a, a² +a +2). But f(t +2)=0 for t+2 >a² +a. So f(t +2)=0, and f(t) is either 2 (if a <t <a² +a), 1 (if t=a² +a or t=a), or 1 (if 0 <t <a). So: - If a <t <a² +a: f(t)=2. Then f(t +2)=0. So 0 >2? No. - If t=a² +a: f(t)=1. Then f(t +2)=0. So 0 >1? No. - If 0 <t <a: f(t)=1. Then f(t +2)=0. So 0>1? No. - If t ≤0: Not in this interval since t >a² +a -2. If a² +a -2 <0, then t can be in (a² +a -2,0]. But since a >0, a² +a -2 could be positive or negative. For example, let's suppose a=1. Then a² +a -2=1 +1 -2=0. So t >0. So in this case, t >0. If a=2, a² +a -2=4 +2 -2=4. So t >4. Therefore, in this interval (t >a² +a -2), f(t +2)=0 and f(t) ≥0, so no solution.2. t ≤a² +a -2: Then t +2 ≤a² +a. So f(t +2) can be in different intervals depending on t +2. Let's consider the possible values of t +2: - If t +2 >a² +a: Impossible since t ≤a² +a -2 ⇒ t +2 ≤a² +a. - If t +2 =a² +a: Then f(t +2)=1. For this to happen, t =a² +a -2. Then f(t +2)=1. f(t) is f(a² +a -2). Now, since t =a² +a -2 ≤a² +a -2 (equality), which is the original t. So f(t)=f(a² +a -2). Where does a² +a -2 fall? Depending on a: If a² +a -2 >a ⇒a² +a -2 >a ⇒a² -2 >0 ⇒a² >2 ⇒a >sqrt(2) ≈1.414. If a² +a -2 ≤a ⇒a² -2 ≤0 ⇒a² ≤2 ⇒a ≤sqrt(2). So: - If a >sqrt(2), then a² +a -2 >a. So f(a² +a -2)=0 (since t >a² +a), but wait, t =a² +a -2. If a >sqrt(2), then a² +a -2 >a² +a - (a² +a) )= ? Wait, maybe I need to check whether a² +a -2 is greater than a or not. Let's suppose a >sqrt(2). Then a² +a -2 >a ⇒a² -2 >0 ⇒True. So t =a² +a -2 >a. Then f(t)=0. Therefore, f(t +2)=1 and f(t)=0. So 1>0. Hence, for a >sqrt(2), when t =a² +a -2, f(t +2)=1 >0=f(t). So such t exists. For a ≤sqrt(2), a² +a -2 ≤a. Then t =a² +a -2 ≤a. Then f(t)=: If t >0: since a² +a -2 ≤a, and a >0, t could be positive or negative. For example, if a=1: a² +a -2 =1 +1 -2=0. So t=0. f(t)=2 (since t=0 ≤0). f(t +2)=f(2)=1. So 1 >2? No. If a=1.5: a² +a -2=2.25 +1.5 -2=1.75. So t=1.75. Since a=1.5, a² +a=2.25 +1.5=3.75. So t=1.75 <a² +a=3.75. Now, f(t=1.75). Since a=1.5, a=1.5, so t=1.75 is between a=1.5 and a² +a=3.75. Therefore, f(t=1.75)=2. Then f(t +2)=f(1.75 +2)=f(3.75)=1. So 1 >2? No. If a=sqrt(2): a≈1.414. a²=2. So a² +a -2=2 +1.414 -2=1.414≈a. So t= a≈1.414. Then f(t)=f(a)=1. f(t +2)=f(a +2)=? a +2≈3.414. If a=sqrt(2), a² +a=2 +sqrt(2)≈3.414. So t +2= a +2≈3.414. Which is equal to a² +a. Therefore, f(t +2)=1. So f(t +2)=1, f(t)=1. Not greater. Therefore, for a ≤sqrt(2), t =a² +a -2 doesn't work. But for a >sqrt(2), it works. Hence, for a >sqrt(2), there exists t =a² +a -2 where f(t +2)=1 >0=f(t). - If t +2 <a² +a: Then f(t +2) can be 2,1, or0 depending on t +2. Wait, no. For a >0, f(t +2): - If t +2 >a² +a: 0 - If t +2 =a² +a:1 - If a <t +2 <a² +a:2 - If t +2 =a:1 - If 0 <t +2 <a:1 - If t +2 ≤0:2 So since t +2 ≤a² +a (because t ≤a² +a -2), we have: Depending on t +2's position: - If t +2 ≤0: then f(t +2)=2. - If 0 <t +2 <a: f(t +2)=1. - If a ≤t +2 ≤a² +a: f(t +2)=2 if a <t +2 <a² +a, 1 if t +2=a or t +2=a² +a. Now, we need to see if there exists t ≤a² +a -2 such that f(t +2) >f(t). Let's check different subcases: a. t +2 ≤0: Then f(t +2)=2. To have f(t +2) >f(t), we need 2 >f(t). f(t) can be: - For t ≤a² +a -2: Since t can be anything up to a² +a -2. If t +2 ≤0 ⇒t ≤-2. So t ≤-2. For such t, f(t)=2 (since t ≤0). So f(t +2)=2 and f(t)=2. So 2>2? No. b. 0 <t +2 <a: Then f(t +2)=1. Need 1 >f(t). f(t) can be: For t ≤a² +a -2 and t +2 <a ⇒t <a -2. Since a >0, a -2 could be positive or negative. If a -2 >0 ⇒a >2. Then t <a -2. If a ≤2, then a -2 ≤0, so t <a -2 ≤0. So: - If a >2: t <a -2. Since t ≤a² +a -2 (from original interval), but a >2, and a² +a -2 >a -2 (since a² +a -2 - (a -2)=a² >0). So t can be from (-infty, a² +a -2]. But we also have t <a -2. Therefore, t <a -2. For such t, f(t)=: If t ≤0: f(t)=2. If 0 <t <a: f(t)=1. If a ≤t ≤a² +a -2: f(t)=2 (if a <t <a² +a),1 (if t=a or t=a² +a). But since t <a -2 and a >2, a -2 >0. So t can be: - t ≤0: f(t)=2 - 0 <t <a -2: f(t)=1 (since 0 <t <a -2 <a) Therefore: - For t ≤0: f(t)=2. So need 1 >2? No. - For 0 <t <a -2: f(t)=1. So need 1 >1? No. - If a ≤2: Then a -2 ≤0. So t <a -2 ≤0 ⇒t <0. So t <0. For such t: f(t)=2. Therefore, f(t +2)=1. So 1 >2? No. Therefore, in this subcase, no t satisfies the condition. c. a ≤t +2 <a² +a: Then f(t +2)=2 (if a <t +2 <a² +a) or 1 (if t +2=a or t +2=a² +a). Let's consider: - If t +2 is in (a, a² +a): f(t +2)=2. Need 2 >f(t). - If t +2=a or a² +a: f(t +2)=1. Need 1 >f(t). Let's first consider t +2 in (a, a² +a): Then t ∈(a -2, a² +a -2). Now, t is in this interval, but also t ≤a² +a -2 (from original interval). So t ∈(a -2, a² +a -2). For such t, f(t) can be: If t ≤0: f(t)=2. If 0 <t <a: f(t)=1. If a ≤t <a² +a: f(t)=2. If t=a² +a: f(t)=1. So depending on t: - If t ∈(a -2,0]: f(t)=2. So need 2 >2? No. - If t ∈(0,a): f(t)=1. So need 2 >1? Yes. - If t ∈[a,a² +a -2): f(t)=2. So need 2 >2? No. Therefore, in the interval t ∈(0,a), if t ∈(0,a) and t ∈(a -2,a² +a -2), then we can have f(t +2)=2 >1=f(t). But when is (0,a) overlapping with (a -2,a² +a -2)? Since a >0: The interval (a -2,a² +a -2) overlaps with (0,a) if a -2 <a and a² +a -2 >0. Which is always true for a >0, but need a -2 <a, which is true. But to overlap with (0,a), the lower bound a -2 must be <a and the upper bound a² +a -2 must be >0. The upper bound a² +a -2 >0 ⇒a² +a >2 ⇒a² +a -2 >0. The roots of a² +a -2=0 are a=(-1 ±sqrt(1 +8))/2=(-1 ±3)/2. So a=1 or a=-2. Since a >0, the critical point is a=1. For a >1, a² +a -2 >0. For 0 <a <1, a² +a -2 <0. Therefore: - For a >1, a² +a -2 >0, so t ∈(a -2, a² +a -2). To overlap with (0,a), we need a -2 <a and a² +a -2 >0. For a >1: a -2 can be negative or positive. If a >2: a -2 >0. So interval (a -2, a² +a -2) overlaps with (0,a) only if a -2 <a, which is true, but since a >2, (a -2, a² +a -2) starts at a -2 >0. So overlap with (0,a) is (a -2, a). Therefore, t ∈(a -2, a). For these t, f(t)=1 (since 0 <t <a). Therefore, f(t +2)=2 >1. Hence, there exists t in (a -2, a) where f(t +2)=2 >1=f(t). For example, take a=3: a² +a -2=9+3-2=10. So t ∈(3 -2,10)=(1,10). But t must also be <a=3. So t ∈(1,3). For t ∈(1,3), f(t)=1 (since 0 <t <a=3). Then t +2 ∈(3,5). But a² +a=12, so t +2 ∈(3,12). Therefore, f(t +2)=2 (since 3 <t +2 <12). So 2 >1, which is true. Therefore, for a=3, there exists t in (1,3) where f(t +2) >f(t). - For 1 <a ≤2: a -2 ≤0. So interval (a -2, a² +a -2) includes negative t and positive up to a² +a -2. Overlap with (0,a) is (0,a). So t ∈(0,a). Then f(t)=1, and t +2 ∈(2, a +2). Since a ≤2, a +2 ≤4. But a² +a -2 for a=2 is 4 +2 -2=4. So t +2 ∈(2,4]. For a=2, a² +a=6, so t +2 ∈(2,6). Since t ∈(0,2), t +2 ∈(2,4). For a=2, f(t +2)=2 (since 2 <t +2 <6). Therefore, f(t +2)=2 >1=f(t). Hence, for a=2, exists t in (0,2). - For a ≤1: a² +a -2 <0. So interval (a -2,a² +a -2) is from a -2 to a negative number. So t must be <a² +a -2 <0. But in this case, f(t)=2. Then f(t +2)=2 or1. For example, a=0.5: a² +a -2=0.25+0.5-2=-1.25. So t ∈(a -2,a² +a -2)= (-1.5, -1.25). For t in this interval, f(t)=2. Then t +2 ∈(0.5,0.75). So f(t +2)=1 (since 0 <t +2 <a=0.5? No, t +2=0.5 to0.75, but a=0.5. So t +2 ∈(0.5,0.75) where 0.5 <a=0.5? Wait, a=0.5, so t +2 ∈(0.5,0.75) is greater than a=0.5. Therefore, f(t +2)=2 (since a=0.5 <t +2 <a² +a=0.25 +0.5=0.75). So f(t +2)=2. Hence, f(t +2)=2 and f(t)=2. 2>2? No. Therefore, for a ≤1, even if t is in negative, t +2 ends up in positive but less than a² +a=0.75, which is still in the quadratic part's increasing-then-decreasing phase. But a=0.5, a² +a=0.75. So for t +2 in (0.5,0.75), f(t +2)=2. But f(t)=2. So no. Therefore, for 1 <a ≤2 and a >2, there exists t such that f(t +2) >f(t). For a >1, it's possible. Now, the other subcase where t +2=a or t +2=a² +a. - If t +2=a ⇒t=a -2. Then f(t +2)=f(a)=1. Then f(t)=f(a -2). For a -2: If a -2 >a² +a ⇒a -2 >a² +a ⇒0 >a² +2. Impossible. If a -2 ≤0 ⇒a ≤2. Then f(a -2)=2 (if a -2 ≤0) or f(a -2)=1 (if a -2 is in (0,a)). For a ≤2: a -2 ≤0. So f(a -2)=2. So f(t +2)=1 >2? No. For a >2: a -2 >0. Then f(a -2)=1 (if 0 <a -2 <a) which is true since a >2 ⇒a -2 <a. So f(a -2)=1. Then f(t +2)=1 >1? No. - If t +2=a² +a ⇒t=a² +a -2. As before, for a >sqrt(2), this works. For a ≤sqrt(2), doesn't work. d. t +2=a: Covered above. Therefore, the main subcase where f(t +2) >f(t) is when t ∈(max(a -2,0), a) for a >1. In this interval, f(t)=1 and f(t +2)=2, which satisfies 2 >1. Additionally, for a >sqrt(2), t =a² +a -2 works, giving f(t +2)=1 >0=f(t). So combining these, for a >1, there exists t such that f(t +2) >f(t). Wait, but earlier with a=1.5 (which is >1 but ≤sqrt(2)≈1.414), wait, no, sqrt(2)≈1.414. So a=1.5 is greater than sqrt(2). Wait, sqrt(2)≈1.414, so a=1.5 is greater than sqrt(2). So for a >sqrt(2), the t=a² +a -2 works. But also, for a >1, the t in (a -2,a) works. Wait, need to check if a >1 includes a >sqrt(2). Since sqrt(2)≈1.414, so for a >1, we have two cases: - 1 <a ≤sqrt(2): here, t ∈(0,a) since a -2 <0. So t ∈(0,a), f(t)=1, t +2 ∈(2, a +2). For a ≤sqrt(2), a +2 ≤sqrt(2)+2≈3.414. But a² +a: For a=1.5: a² +a=2.25+1.5=3.75. So t +2 ∈(2,3.5). For a=1.5, a² +a=3.75, so t +2 ∈(2,3.5) which is within (a, a² +a)=(1.5,3.75). Hence, f(t +2)=2 >1. So for a=1.5, t=0.5 gives t +2=2.5 ∈(1.5,3.75), so f(t +2)=2 >1. Therefore, combining all the cases where a >1, there exists t such that f(t +2) >f(t). For a >1, the condition is satisfied. For a ≤1, we checked examples like a=1 and a=0.5 and saw no t exists. For a=1: a=1. Then f(t) is: 0, t >2 1, t=2 2,1 <t <2 1, t=1 1,0 <t <1 2,t ≤0 Looking for t where f(t +2) >f(t): For example, t=0.5: f(t +2)=f(2.5)=0. f(t)=f(0.5)=1. 0>1? No. t= -1: f(t +2)=f(1)=1. f(t)=f(-1)=2. 1>2? No. t=1.5: f(t +2)=f(3.5)=0. f(t)=2. 0>2? No. t=2: f(t +2)=f(4)=0. f(t)=1. 0>1? No. So no t exists. For a=1, no solution. Therefore, the range of a is a >1.But wait, let's verify with a=1.5 (which is >1):For a=1.5, the quadratic part is -x² +3x +1.5. Vertex at x=1.5, y= (1.5)^2 +1.5=2.25 +1.5=3.75.For t in (0,1.5):f(t)=1. Let's take t=1:f(t)=1. t +2=3. Since 1.5 <3 <3.75, f(t +2)=2. Therefore, 2>1. So yes, such t exists.Similarly, for a=2:Quadratic part: -x² +4x +2. Vertex at x=2, y=4 +2=6.For t=1 (0 <t <2):f(t)=1. t +2=3, which is in (2,6). So f(t +2)=2 >1.Therefore, for a >1, such t exists.For a=1.0:Quadratic part: -x² +2x +1. Vertex at x=1, y=2.For t=0.5 (0 <t <1):f(t)=1. t +2=2.5. For a=1, a² +a=2. So t +2=2.5 >2. So f(t +2)=0. 0>1? No. Therefore, a=1 doesn't work.Wait, but when a=1.0, which is not greater than sqrt(2)≈1.414. So a=1 is in 1 <a ≤sqrt(2)? No, sqrt(2) is ~1.414, so a=1 is in 0 <a ≤sqrt(2). Wait, no, a=1 is less than sqrt(2). So according to previous analysis, for a >1, which would be a >1. So a=1.5 works, a=2 works, but a=1 doesn't.So the critical point is a >1. Therefore, the range of a is a >1.But wait, let's check a=1.2 (which is >1 but <sqrt(2)≈1.414):a=1.2, a² +a=1.44 +1.2=2.64.Take t=0.5 (0 <t <a=1.2):f(t)=1. t +2=2.5. Now, 2.5 <a² +a=2.64. Therefore, f(t +2)=2. So 2 >1. Hence, works.Therefore, for a >1, even if a <sqrt(2), there exists t such that f(t +2) >f(t). Hence, the critical value is a >1.Therefore, the range of a is a >1, or in interval notation (1, +infty).Final ConclusionFor the first part, when g(x) = ln x, f(t) =1 for all t.For the second part, the range of a is a >1.Final AnswerFor the first blank: boxed{1}For the second blank: boxed{(1, +infty)}</think>For the equation ( g(x) = t ) (( t in mathbb{R} )), the number of real roots is denoted as ( f(t) ).1. When ( g(x) = ln x ): - The natural logarithm function ( ln x ) is defined for ( x > 0 ). - The equation ( ln x = t ) has exactly one solution ( x = e^t ) for any real ( t ). - Therefore, ( f(t) = 1 ) for all ( t ).2. When ( g(x) ) is the piecewise function: - For ( x leq 0 ), ( g(x) = x ). - For ( x > 0 ), ( g(x) = -x^2 + 2ax + a ). - The quadratic part ( -x^2 + 2ax + a ) is a downward-opening parabola with vertex at ( x = a ). Analyzing the number of real roots ( f(t) ): - For ( a leq 0 ), the quadratic part starts at ( a ) (at ( x = 0 )) and decreases. The function ( f(t) ) is: - ( f(t) = 0 ) for ( t > 0 ) - ( f(t) = 1 ) for ( a < t leq 0 ) - ( f(t) = 2 ) for ( t leq a ) - For ( a > 0 ), the quadratic part starts at ( a ), rises to a maximum at ( x = a ), then decreases. The function ( f(t) ) is: - ( f(t) = 0 ) for ( t > a^2 + a ) - ( f(t) = 1 ) for ( t = a ) or ( t = a^2 + a ) - ( f(t) = 2 ) for ( a < t < a^2 + a ) or ( t leq 0 ) - ( f(t) = 1 ) for ( 0 < t < a ) To find the range of ( a ) such that there exists ( t ) with ( f(t+2) > f(t) ): - For ( a > 1 ), there exists ( t ) such that ( f(t+2) = 2 ) and ( f(t) = 1 ) (e.g., ( t ) in ( (a-2, a) )). - For ( a leq 1 ), no such ( t ) exists.Final Answers:- For ( g(x) = ln x ): (boxed{1})- Range of ( a ): (boxed{(1, +infty)}),<think>Okay, let's tackle this problem step by step. So, first, the problem has two parts. The first part is about the function g(x) being the natural logarithm of x, and we need to find f(t), which is the number of real roots for the equation ln(x) = t. The second part is more complex: it involves a piecewise function g(x) and determining the range of a parameter 'a' such that there exists some t where f(t+2) > f(t). Let me start with the first part.Starting with g(x) = ln(x). The equation ln(x) = t. To find the number of real roots, we need to solve for x given a real number t. The natural logarithm function ln(x) is only defined for x > 0. So, solving ln(x) = t gives x = e^t. Since e^t is always positive for any real t, this equation has exactly one real root for any t. Therefore, f(t) = 1 for all t when g(x) is ln(x). So the answer to the first part is 1. That seems straightforward.Now, moving on to the second part. The function g(x) is defined piecewise: it's equal to x when x ≤ 0, and for x > 0, it's a quadratic function: -x² + 2a x + a. We need to find the range of 'a' such that there exists some t where f(t+2) > f(t). Here, f(t) denotes the number of real roots of the equation g(x) = t.First, let me make sure I understand what f(t) represents. For each real number t, f(t) counts how many real numbers x satisfy g(x) = t. So, for each t, we need to solve both parts of the piecewise function and count the total number of solutions.Given that the function is piecewise, let's analyze each piece separately.First, for x ≤ 0: g(x) = x. So solving x = t for x ≤ 0. The solution here is x = t, but only if t ≤ 0. So for t ≤ 0, there's one solution from the left side (x ≤ 0), and for t > 0, there are no solutions from the left side.For x > 0: g(x) = -x² + 2a x + a. So solving -x² + 2a x + a = t. Let's rearrange this equation: -x² + 2a x + a - t = 0. Multiply both sides by -1 to make it a standard quadratic: x² - 2a x - a + t = 0. So, quadratic equation x² - 2a x + (t - a) = 0. Let's write that as x² - 2a x + (t - a) = 0. To find real roots, discriminant D must be non-negative.Discriminant D = (2a)^2 - 4*1*(t - a) = 4a² - 4(t - a) = 4a² - 4t + 4a = 4(a² + a - t). So, D ≥ 0 implies that a² + a - t ≥ 0, so t ≤ a² + a.Therefore, for x > 0, the equation g(x) = t has real roots only if t ≤ a² + a. The number of real roots depends on the number of solutions x > 0. Since it's a quadratic equation, it can have 0, 1, or 2 real roots. However, we must check whether the roots are greater than 0.Wait, but first, since the quadratic is x² - 2a x + (t - a) = 0, the roots can be found using quadratic formula:x = [2a ± sqrt(4a² - 4(t - a))]/2 = [2a ± 2sqrt(a² - (t - a))]/2 = a ± sqrt(a² - t + a) = a ± sqrt(a² + a - t).So, the roots are x = a + sqrt(a² + a - t) and x = a - sqrt(a² + a - t).But since we are looking for x > 0, we need to check whether these roots are positive.But before that, the discriminant must be non-negative, which we established as t ≤ a² + a.So, for each t ≤ a² + a, we have two roots, but we need to check if they are positive.But given that x > 0, so the roots x = a ± sqrt(a² + a - t) must be greater than 0.Let me analyze these roots. Let's denote sqrt(a² + a - t) as S. So, the roots are a + S and a - S.We need to check whether a + S > 0 and a - S > 0.First, a + S > 0: Since S is a square root, it is non-negative. So, a + S > 0 depends on the value of a. If a is positive, then certainly a + S is positive. If a is negative, then even with S, it's possible that a + S might not be positive. Similarly, for a - S > 0: a - S > 0 would require that a > S. Since S = sqrt(a² + a - t), which depends on t.But this seems complicated. Maybe a better approach is to consider that for x > 0, the quadratic function g(x) = -x² + 2a x + a is a downward opening parabola. Its maximum value is at the vertex. The vertex of a parabola given by ax² + bx + c is at x = -b/(2a). In our case, the quadratic is -x² + 2a x + a, so the coefficient of x² is -1, so the vertex is at x = -2a/(2*(-1)) = a. So, the vertex is at x = a, and the maximum value is g(a) = -a² + 2a*a + a = -a² + 2a² + a = a² + a.Therefore, the maximum value of the quadratic part is a² + a, which occurs at x = a. Since it's a downward opening parabola, the function decreases to the left and right of x = a. However, the quadratic is only defined for x > 0.So, the range of the quadratic part (x > 0) is (-∞, a² + a]. Therefore, the equation g(x) = t for x > 0 has:- Two real roots if t < a² + a.- One real root if t = a² + a.- No real roots if t > a² + a.But wait, but the quadratic is only defined for x > 0. So even if t ≤ a² + a, the roots might not lie in x > 0. So we need to check whether the roots are positive.So, let's go back to the roots:x = a ± sqrt(a² + a - t).So, both roots must be greater than 0.First, consider x = a + sqrt(a² + a - t). Since sqrt(...) is non-negative, and a is a real number. If a is positive, then a + sqrt(...) is definitely positive. If a is negative, then a + sqrt(...) could be positive or negative depending on the sqrt term.Similarly, x = a - sqrt(a² + a - t). This could be positive or negative.But since the quadratic is defined for x > 0, we need both roots to be positive. So both x = a + sqrt(a² + a - t) and x = a - sqrt(a² + a - t) must be > 0.Alternatively, since the quadratic is x² - 2a x + (t - a) = 0, the product of the roots is (t - a)/1 = t - a. So, if the product of the roots is positive, both roots have the same sign. If it's negative, they have opposite signs. Since the quadratic is x² - 2a x + (t - a), the product is t - a. So, if t - a > 0, then both roots are positive or both negative. If t - a < 0, then roots have opposite signs. If t - a = 0, then one root is zero (but x=0 is not in the domain x > 0).But since we need x > 0, let's analyze:Case 1: t - a > 0. Then both roots are positive or both negative. However, since the sum of the roots is 2a (from x² - 2a x + (t - a) = 0). So, sum of roots is 2a. If both roots are positive, sum is positive, so 2a > 0 => a > 0. If both roots are negative, sum is negative, so 2a < 0 => a < 0. But given that t - a > 0, and if a < 0, then t - a > t - 0 = t. But t can vary. This seems complicated.Alternatively, maybe we can consider the quadratic function's behavior. The quadratic function for x > 0 is -x² + 2a x + a. Let's consider its graph. It's a downward opening parabola with vertex at x = a. The maximum value is a² + a. So, when x approaches 0 from the right, the value is g(0+) = -0 + 0 + a = a. So, at x = 0, the quadratic part starts at value a. Then it goes up to the vertex at x = a with value a² + a, then decreases to negative infinity as x increases beyond a.Therefore, the range of the quadratic part is (-∞, a² + a]. But the quadratic part starts at x = 0 with value a, so when x approaches 0 from the right, the value is a. Therefore, if a is positive, the quadratic part starts at a positive value a, increases to a² + a, then decreases. If a is negative, the quadratic part starts at a negative value a, then increases to a² + a (but if a is negative, a² + a could be positive or negative depending on a).Wait, for example, if a = -1, then a² + a = 1 -1 = 0. So, the maximum value is 0. So, the quadratic part would start at x=0 with value a = -1, then rise to 0 at x = a = -1, but x = -1 is not in the domain x > 0. Wait, no. The vertex is at x = a. If a is negative, say a = -1, the vertex is at x = -1, which is not in x > 0. Therefore, for a negative a, the quadratic part on x > 0 is a downward parabola whose vertex is at x = a (negative), so for x > 0, the function is decreasing since the vertex is to the left of x = 0. Therefore, starting at x = 0, the value is a (negative), and decreasing from there. So, the quadratic part would be decreasing for all x > 0 if a ≤ 0, because the vertex is at x = a ≤ 0. Therefore, for a ≤ 0, the quadratic part on x > 0 is a decreasing function from a (at x=0) to -infinity as x approaches infinity. Therefore, for a ≤ 0, the quadratic part on x > 0 has range (-∞, a]. But since a is negative or zero, that's just (-∞, a].Wait, but the vertex is at x = a, which is outside x > 0, so the maximum value on x > 0 would be at x = 0, which is a. Then it decreases. Therefore, for a ≤ 0, the quadratic part on x > 0 is decreasing from a to -∞, so its range is (-∞, a]. Therefore, the equation g(x) = t for x > 0 would have:- If t ≤ a, then there is one solution (since the function is decreasing, so each t ≤ a will have exactly one x > 0).But wait, if the quadratic is a decreasing function from a to -∞ on x > 0, then for each t ≤ a, there is exactly one x > 0 such that g(x) = t. But wait, but the quadratic equation might have two roots, but only one in x > 0. However, since it's a decreasing function, it's injective on x > 0, so only one solution. Therefore, for a ≤ 0, the quadratic part has one real root for t ≤ a, and no roots otherwise.Wait, but if the quadratic is part of a downward parabola, but in this case, for a ≤ 0, the quadratic is decreasing on x > 0. Therefore, the equation -x² + 2a x + a = t can be rearranged as x² - 2a x + (t - a) = 0. For a ≤ 0, and x > 0, since the function is decreasing, for each t ≤ a, there is exactly one solution. Let me verify with an example. Let’s take a = -1. Then the quadratic part is -x² + 2*(-1)x + (-1) = -x² -2x -1. So, solving -x² -2x -1 = t. Let's say t = -2. Then equation becomes -x² -2x -1 = -2 => -x² -2x +1 = 0 => x² + 2x -1 = 0. The roots are [-2 ± sqrt(4 +4)]/2 = [-2 ± sqrt(8)]/2 = [-2 ± 2*sqrt(2)]/2 = -1 ± sqrt(2). So, positive root is -1 + sqrt(2) ≈ 0.414, which is positive. The other root is -1 - sqrt(2) ≈ -2.414, which is negative. So, only one positive root. Therefore, even though the quadratic equation has two roots, only one is in x > 0. So, for a ≤ 0, quadratic part gives one root for t ≤ a.For a > 0, the vertex is at x = a > 0. Therefore, the quadratic function on x > 0 first increases from x=0 (value a) to x=a (value a² + a), then decreases to -∞ as x approaches infinity. Therefore, the range is (-∞, a² + a]. So, for t between a and a² + a, there are two solutions (since the function increases then decreases), and for t = a² + a, there is one solution at x = a. For t < a, since the function starts at a and then goes up to a² + a, then back down, but t < a would be below the starting value. Wait, but if a > 0, then the minimum value on x > 0 is negative infinity, but the maximum is a² + a. Wait, no: the function starts at x=0 with value a, goes up to a² + a at x=a, then decreases. So, for t between a and a² + a, there are two solutions. For t = a² + a, one solution. For t > a² + a, no solution. For t < a, but since the function goes down to -infty as x approaches infinity, t < a would have one solution on the decreasing part (x > a). Wait, but the function at x=0 is a, then goes up to a peak at x=a, then decreases. Therefore, for t less than a, but since the function decreases from a² + a to -infty, t can be less than a only if a² + a > a, which is a² > 0, which is true as long as a ≠ 0. Since a > 0 in this case, a² + a > a, so t can be as low as -infty. Wait, but hold on, when a > 0, the quadratic function on x > 0 starts at x=0 with value a, increases to a² + a at x=a, then decreases to -infty. Therefore, the range is (-infty, a² + a]. So, for t in (-infty, a² + a], there is always at least one solution. For t between a and a² + a, two solutions (one on the increasing part, one on the decreasing part). For t = a, one solution at x=0 (but x=0 is not included in x > 0). Wait, but at x=0, the value is a, but x=0 is part of the left piece (x ≤ 0). So, for t = a, the equation g(x) = t would have x=0 as a solution from the left piece (since g(0) = 0). Wait, but if t = a, then for x ≤ 0: x = t = a. So, if a ≤ 0, then x = a is a valid solution (since x ≤ 0). If a > 0, then x = a would not satisfy x ≤ 0, so no solution from the left. But for the right side, when a > 0, the quadratic part at x=0 is a, so t = a would correspond to x approaching 0 from the right. But x=0 is not included in x > 0. Therefore, for a > 0, t = a would have one solution from the left side (x = a if a ≤ 0) but if a > 0, x = a is not in x ≤ 0, so no solution from the left, and on the right side, x approaching 0 from the right gives g(x) approaching a, but x=0 is not included. Therefore, does t = a have a solution? Let me check. For a > 0: left side (x ≤ 0): x = a. Since a > 0, x = a is not ≤ 0, so no solution. Right side: solving -x² + 2a x + a = a. So, -x² + 2a x = 0 => x(-x + 2a) = 0. Solutions x=0 or x=2a. But x > 0, so x=2a is a solution. Therefore, for t = a, even though x=0 is not included, there is a solution at x=2a. Therefore, for a > 0, t = a has one solution on the right side (x=2a). Therefore, when a > 0, t = a has one solution (from the right side). Similarly, for a < 0: left side x = a (which is ≤ 0) is a solution, and on the right side, solving -x² + 2a x + a = a. That gives -x² + 2a x = 0 => x(-x + 2a) = 0. Solutions x=0 or x=2a. But x > 0, so x=0 is excluded. x=2a: if a < 0, then x=2a < 0, which is not in x > 0. So, for a < 0, t = a has only one solution from the left side (x = a). Therefore, regardless of a, t = a has one solution.But maybe this is complicating things. Let's structure the analysis:For g(x) = t:Left side (x ≤ 0): solution x = t, valid if t ≤ 0.Right side (x > 0): solutions as per quadratic equation, with x > 0.So, total number of real roots f(t) is the sum of the roots from left and right sides.Therefore, f(t) = [1 if t ≤ 0] + [number of roots from right side].The number of roots from the right side depends on t and a.Let's analyze the right side:Case 1: a ≤ 0.For a ≤ 0, the quadratic part on x > 0 is decreasing from a (at x=0) to -∞. Therefore, for t ≤ a, there is exactly one solution on the right side. For t > a, no solution on the right side.But since a ≤ 0, t ≤ a would mean t ≤ a ≤ 0. So, for t ≤ a (which is ≤ 0), left side gives x = t (which is ≤ 0), and right side gives one solution (since t ≤ a). So f(t) = 1 (left) + 1 (right) = 2.For a < t ≤ 0 (if a < 0), then t is between a and 0. Since a ≤ 0, t ≤ 0. If a < t ≤ 0, then on the left side x = t is valid (since t ≤ 0), and on the right side, since t > a, but a ≤ 0, the right side equation has no solution. So f(t) = 1 (left) + 0 (right) = 1.For t > 0, left side has no solution, right side also has no solution (since t > a and a ≤ 0, so t > 0 > a, so right side equation has no solution). Therefore, f(t) = 0.For t < a (if a < 0), left side: t < a < 0, so x = t is still ≤ 0, so left side gives 1 solution. Right side: t < a, so right side gives 1 solution. Thus, f(t) = 2.But wait, if a ≤ 0, then the right side equation only has solutions when t ≤ a. If a is negative, t ≤ a would be t ≤ (negative number). For example, a = -1, then t ≤ -1. Then, for t ≤ -1, left side gives x = t (valid since t ≤ -1 ≤ 0), right side gives 1 solution. So f(t) = 2. For -1 < t ≤ 0, left side gives 1 solution, right side gives 0. So f(t) = 1. For t > 0, f(t) = 0.Case 2: a > 0.For a > 0, the quadratic part on x > 0 first increases from a at x=0 to a² + a at x=a, then decreases to -infty. Therefore, the equation -x² + 2a x + a = t has:- Two solutions when a < t < a² + a.- One solution when t = a or t = a² + a.- One solution when t ≤ a.Wait, no. Let's think again. When a > 0, the quadratic part starts at x=0 with value a, increases to a maximum of a² + a at x=a, then decreases. Therefore, the equation g(x) = t for x > 0 has:- Two solutions if t is between a and a² + a (since the function goes above t and then below).- One solution if t = a² + a (at x=a) or if t ≤ a (since the function starts at a and decreases past t as x increases, so there's one solution for t ≤ a).Wait, but when t < a, the function starts at a, goes up to a² + a, then comes back down. So, if t < a, then t is less than the starting value. Since the function then goes up and then down, but if t < a, the function would cross t once on the decreasing part (after the peak). Therefore, even for t < a, there's one solution. Wait, let me verify with an example. Let a = 1. Then the quadratic part is -x² + 2x + 1. At x=0, value is 1. It peaks at x=1 with value 1² +1 = 2, then decreases. So, for t=0.5, which is less than a=1, the equation -x² + 2x + 1 = 0.5 => -x² + 2x + 0.5 = 0 => x² - 2x -0.5 = 0. Solutions x = [2 ± sqrt(4 + 2)]/2 = [2 ± sqrt(6)]/2. sqrt(6) ≈ 2.45, so x ≈ (2 + 2.45)/2 ≈ 2.225 and (2 - 2.45)/2 ≈ -0.225. So, positive solution ≈ 2.225. Therefore, even though t=0.5 < a=1, there's one solution on the right side (x ≈ 2.225). Therefore, for a > 0, the quadratic part on x >0 has one solution for t < a² + a (since the maximum is a² + a). Wait, but the example shows that even for t < a, there's still a solution. Therefore, the correct statement is that for a > 0, the quadratic part on x >0 has:- Two solutions when a < t < a² + a.- One solution when t = a, t = a² + a, or t ≤ a.Wait, but in the example, t=0.5 < a=1 had one solution. t=1 (a=1) would have x=2a=2 as a solution. Then, t=2 (a² + a = 1 +1=2) has one solution at x=a=1. So, for a >0:- If t < a: one solution (on the right side, decreasing part).- If t = a: one solution (x=2a, since solving -x² +2a x +a =a gives x=0 or x=2a, but x>0 gives x=2a).Wait, but if a >0, x=2a is positive. So, for t=a, right side gives one solution.Wait, perhaps the correct breakdown is:For a >0:- For t < a: right side has one solution (on the decreasing part).- For a ≤ t < a² + a: right side has two solutions (one on increasing, one on decreasing).- For t = a² + a: right side has one solution (x=a).- For t > a² + a: right side has no solution.Therefore, combining with left side (x ≤0):Left side: solution exists if t ≤0.Therefore, total solutions:For t > a² + a: 0 (left: t >0, right: no).For t = a² + a: left side: if a² + a ≤0? No, since a >0, a² +a ≥0 +0=0, and actually a² +a >0 for a >0. So, left side: t = a² +a >0, no solution. Right side: one solution. So f(t) =1.For a < t < a² +a: left side: t >0, no. Right side: two solutions. So f(t)=2.For t =a: left side: if a ≤0, but a>0 here, so t=a>0. Left side: no. Right side: one solution (x=2a). So f(t)=1.For t <a: split into t ≤0 and 0 < t <a.If t ≤0: left side: one solution (x=t). Right side: since t <a (and a>0), right side has one solution. So f(t)=2.If 0 < t <a: left side: no solution. Right side: one solution. So f(t)=1.Therefore, summarizing for a >0:f(t) = - 2, if t ≤0.- 1, if 0 < t <a.- 2, if a ≤ t < a² +a.- 1, if t =a, t =a² +a.- 0, if t > a² +a.Wait, but when a >0, for t ≤0, left side has one solution (x=t), and right side also has one solution (since t <a). So f(t)=2.For 0 < t <a: left side no, right side one. So f(t)=1.For a ≤ t < a² +a: left side no, right side two. So f(t)=2.For t =a² +a: right side one. So f(t)=1.For t >a² +a: no solutions. f(t)=0.Therefore, f(t) is 2 on (-infty,0] and [a, a² +a), 1 on (0,a) and at t=a and t=a² +a, 0 otherwise.For a ≤0:Left side: x =t if t ≤0.Right side: solutions when t ≤a. Since a ≤0, t ≤a ≤0. So for t ≤a, left side x=t (valid) and right side one solution. So f(t)=2.For a <t ≤0: left side x=t (valid), right side no solution. So f(t)=1.For t >0: left side no, right side no. f(t)=0.So f(t) is 2 on (-infty,a], 1 on (a,0], 0 otherwise.Now, the problem states: there exists t such that f(t +2) > f(t). We need to find the range of a for which such a t exists.So, we need to find a such that there exists some t where f(t +2) > f(t).To do this, we need to analyze the possible values of f(t) and f(t+2) depending on a.Let me consider cases based on a.First, case 1: a ≤0.For a ≤0, f(t) is:- 2 if t ≤a,- 1 if a <t ≤0,- 0 if t >0.Similarly, f(t +2):- 2 if t +2 ≤a => t ≤a -2,- 1 if a <t +2 ≤0 => a -2 <t ≤-2,- 0 if t +2 >0 => t >-2.Therefore, for a ≤0, the intervals are:t ≤a -2: f(t +2)=2,a -2 <t ≤-2: f(t +2)=1,t >-2: f(t +2)=0.Now, we need to see if there exists a t where f(t +2) > f(t). Let's compare f(t +2) and f(t) over different intervals.First interval: t ≤a -2.Here, f(t +2)=2, and f(t):Since t ≤a -2 ≤a (since a ≤0), then f(t)=2. So f(t +2)=f(t)=2. So no improvement.Second interval: a -2 <t ≤-2.Here, f(t +2)=1, and f(t):Since t ≤-2 <a (since a ≤0, but a -2 <t, so t >a -2. However, a ≤0, so a -2 ≤ -2. So t is in (a -2, -2]. But a -2 ≤ -2. So t is in (-∞, -2] intersected with (a -2, -2], which is (a -2, -2].But since a ≤0, a -2 ≤-2. So t is in (a -2, -2]. Now, f(t) for t in (a -2, -2]:Since t ≤-2 <a (since a ≤0, but a ≤0 and t ≤-2, so t ≤a only if a ≥-2. Wait, this is getting confusing. Let me take an example.Let a = -1 (so a ≤0). Then a -2 = -3.So, for a =-1:Intervals for f(t +2):t ≤-3: f(t +2)=2,-3 <t ≤-2: f(t +2)=1,t >-2: f(t +2)=0.For f(t):t ≤-1: f(t)=2,-1 <t ≤0: f(t)=1,t >0: f(t)=0.So, for t in (-3, -2], f(t +2)=1, and f(t):Since t is in (-3, -2]. For a=-1, t ≤-1: f(t)=2. So if t is in (-3, -2], then t ≤-1 (since -2 ≤-1), so f(t)=2. Therefore, in this interval, f(t +2)=1 < f(t)=2. Not helpful.For t in (a -2, -2] which is (-3, -2], f(t +2)=1 and f(t)=2. So f(t +2) < f(t).Third interval: t >-2.Here, f(t +2)=0, and f(t):t >-2. Depending on t:If t ≤a, f(t)=2.But since a ≤0 and t >-2:If a >=-2, then t can be in (-2, a] (if a >-2) or not. Wait, let's take a=-1 again. Then t >-2. If t is in (-2, -1], f(t)=2. If t in (-1,0], f(t)=1. If t >0, f(t)=0. So for a=-1:For t in (-2, -1], f(t)=2, f(t +2)=0. So 0 vs 2: f(t +2) < f(t).For t in (-1,0], f(t)=1, f(t +2)=0. Still 0 <1.For t >0, f(t)=0, f(t +2)=0. Equal.Therefore, in all cases for a ≤0, f(t +2) ≤ f(t) for all t. Therefore, there is no t where f(t +2) >f(t). Therefore, for a ≤0, the condition is not satisfied.Now, case 2: a >0.For a >0, f(t) is:- 2 if t ≤0,- 1 if 0 <t <a,- 2 if a ≤t <a² +a,- 1 if t =a or t =a² +a,- 0 if t >a² +a.Similarly, f(t +2) will depend on shifting t by 2.So f(t +2):- 2 if t +2 ≤0 => t ≤-2,- 1 if 0 <t +2 <a => -2 <t <a -2,- 2 if a ≤t +2 <a² +a => a -2 ≤t <a² +a -2,- 1 if t +2 =a or t +2 =a² +a => t =a -2 or t =a² +a -2,- 0 if t +2 >a² +a => t >a² +a -2.But note that the intervals for f(t +2) depend on where a -2 and a² +a -2 fall.We need to compare f(t +2) and f(t) over different intervals.Our goal is to find if there exists a t where f(t +2) > f(t).Let's break down the possible intervals for t:1. t ≤-2:f(t +2)=2 (since t +2 ≤0). f(t):For t ≤-2, since a >0, t ≤-2 <0, so f(t)=2. Therefore, f(t +2)=f(t)=2. No improvement.2. -2 <t ≤0:Here, f(t +2):Since t +2 is in (0,2]. For a >0, we need to check where t +2 falls.If 0 <t +2 <a: which implies -2 <t <a -2.But since t ≤0, this interval becomes -2 <t < min(a -2, 0).But since a >0, a -2 could be less than or greater than 0 depending on a.If a -2 <0 => a <2, then t <a -2 (which is <0).If a -2 ≥0 =>a ≥2, then t <0.Similarly, if a -2 <0, which is a <2, then f(t +2)=1 for -2 <t <a -2, and for a -2 ≤t ≤0 (since t +2 ≥a if t ≥a -2), then f(t +2)=2 if a ≤t +2 <a² +a.Wait, this is getting complex. Let's consider two subcases for a >0:Subcase 1: 0 <a <2.Subcase 2: a ≥2.Let's handle Subcase 1: 0 <a <2.For 0 <a <2:Then a -2 <0.Therefore, f(t +2) is:- For t ≤-2: f(t +2)=2.- For -2 <t <a -2: f(t +2)=1 (since 0 <t +2 <a).But wait, t <a -2 => t +2 <a. But since a <2, a -2 <0, and t is in (-2, a -2). But a -2 is negative. So -2 <t <a -2 (but a -2 <0). For example, a=1: a -2 =-1. So t in (-2,-1). Then t +2 in (0,1). For a=1, f(t +2)=1 since 0 <t +2 <a=1. For a=1, t +2 in (0,1), so f(t +2)=1.For t in [a -2,0], since a -2 <0, t ranges from a -2 to0. Then t +2 ranges from a to2. So:- If a ≤t +2 <a² +a, then f(t +2)=2.But t +2 is between a and2. So, we need to see if a² +a >2 or not.For 0 <a <2, a² +a is a quadratic in a. Let's compute a² +a -2=0. Solutions: a = [-1 ± sqrt(1 +8)]/2 = [-1 ±3]/2. So a=1 or a=-2. Since a >0, critical point at a=1. Therefore:- For 0 <a <1: a² +a <1 +1=2. Therefore, a² +a <2. So, t +2 in [a, a² +a) would be up to a² +a <2. Then for t +2 ≥a² +a, f(t +2)=0. But since t +2 ≤2 (because t ≤0), and a² +a <2, then:For t in [a -2, (a² +a) -2], which is [a -2, a² +a -2], t +2 ∈ [a, a² +a). So f(t +2)=2.For t in (a² +a -2,0], t +2 ∈ (a² +a,2], so f(t +2)=0.For example, take a=0.5:a² +a =0.25 +0.5=0.75 <2.So t +2 in [0.5,0.75): t ∈ [0.5 -2,0.75 -2)=[-1.5, -1.25). So for t ∈[-1.5, -1.25), f(t +2)=2. For t ∈[-1.25,0], f(t +2)=0.For a=0.5, f(t +2)=2 when t ∈[-1.5, -1.25), and 0 otherwise in this range.But in this case, a=0.5, which is between0 and2.Similarly, for a=1:a² +a=1 +1=2. So, t +2 ∈[1,2) (since a=1). So t ∈[-1,0). Then f(t +2)=2 for t ∈[-1,0), but t +2 ∈[1,2), and since a² +a=2, at t +2=2, which is t=0, f(t +2)=0. So for a=1:For t ∈[-1,0), f(t +2)=2; for t=0, f(t +2)=0.But let's get back to general analysis.For 0 <a <2:f(t +2):- For t ≤-2:2- For -2 <t <a -2:1- For a -2 ≤t <a² +a -2:2- For a² +a -2 ≤t ≤0:0And f(t):For t ≤0: f(t)=2 if t ≤0.Wait, no. For a >0, f(t)=2 if t ≤0;1 if 0 <t <a;2 if a ≤t <a² +a; etc.Wait, for a >0:f(t) is:- 2 if t ≤0,- 1 if 0 <t <a,- 2 if a ≤t <a² +a,- 1 if t =a, t=a² +a,- 0 if t >a² +a.Therefore, for t ≤0: f(t)=2.So, in the interval where we are considering t in (-2,0], f(t)=2.Therefore, comparing f(t +2) and f(t):In interval -2 <t <a -2:f(t +2)=1, f(t)=2. So 1 <2. No good.In interval a -2 ≤t <a² +a -2:f(t +2)=2, f(t)=2. Equal.In interval a² +a -2 ≤t ≤0:f(t +2)=0, f(t)=2. So 0 <2. No good.Therefore, in Subcase1 (0 <a <2), in the interval -2 <t ≤0, f(t +2) ≤f(t). No improvement.Next, consider t >0.For t >0:f(t +2) depends on t +2.But for a >0:f(t +2) is:- 2 if t +2 ≤0 => t ≤-2. But t >0, so not applicable.- 1 if 0 <t +2 <a => -2 <t <a -2. But t >0, so this interval is 0 <t <a -2. But since a <2 (Subcase1), a -2 <0. So no overlap. Therefore, next:- 2 if a ≤t +2 <a² +a => a -2 ≤t <a² +a -2.But since t >0, this becomes max(a -2,0) ≤t <a² +a -2.If a² +a -2 >0, then there exists t >0 in this interval.Else, if a² +a -2 ≤0, no such t.So, compute a² +a -2 >0?For 0 <a <2:Solve a² +a -2 >0 => a² +a -2 >0. The roots are at a=(-1 ±√(1 +8))/2=(-1 ±3)/2=1 and -2. So, the inequality holds when a >1 or a < -2. Since 0 <a <2, the inequality holds when 1 <a <2.Therefore, for 1 <a <2, a² +a -2 >0, so there exists t in [a -2, a² +a -2). But since a -2 <0 (because a <2), the interval becomes 0 ≤t <a² +a -2.For example, a=1.5: a² +a -2=2.25 +1.5 -2=1.75>0. So t ∈[0,1.75 -2)= [0,-0.25). Wait, that can't be. Wait, t <a² +a -2. For a=1.5, a² +a -2=1.75, so t +2 <1.75 =>t <-0.25. But we are considering t >0. Contradiction. Wait, there's a mistake here.Wait, the interval is a ≤t +2 <a² +a => a -2 ≤t <a² +a -2.But t must be >0. So, we need a -2 ≤t <a² +a -2 and t >0. Therefore, this interval exists only if a² +a -2 >0 and a -2 <a² +a -2.Which simplifies to a² +a -2 >0 and a -2 <a² +a -2 =>0 <a² +a -2 -a +2 =>0 <a². Which is always true. Therefore, for a >1 (since a² +a -2 >0 when a >1), there exists t in (0, a² +a -2) such that f(t +2)=2.But wait, for t in (0, a² +a -2), we need to check what f(t) is.For t >0:f(t)=1 if 0 <t <a,2 if a ≤t <a² +a,1 if t=a or t=a² +a,0 if t >a² +a.So, for t in (0, a):f(t)=1.For t in [a, a² +a):f(t)=2.So, if we can find t in (0, a² +a -2) such that f(t +2)=2 and f(t)=1 or 2.But in this interval, t +2 is in (2, a² +a). But since a² +a can be greater or less than 2. For example, a=1.5:a² +a=2.25 +1.5=3.75>2.So t +2 ∈(2,3.75). For f(t +2):Since t +2 >a² +a => no, t +2 <a² +a. Wait, no, the interval is a ≤t +2 <a² +a. Since a=1.5, t +2 ∈[1.5,3.75). But t is in (0, a² +a -2)=(0,3.75 -2)=(0,1.75). Therefore, t +2 ∈(2,3.75). Therefore, t +2 ≥a=1.5 and <a² +a=3.75. So f(t +2)=2. But f(t) for t ∈(0,1.75):If t <a=1.5, f(t)=1.If t >=1.5 and <3.75 -2=1.75, t is in [1.5,1.75). For these t, f(t)=2 (since a=1.5 <=t <a² +a=3.75).Therefore, for t ∈[1.5,1.75):f(t +2)=2 (since t +2 ∈[3.5,3.75)), but f(t)=2 (since t ∈[1.5,3.75)). So f(t +2)=f(t)=2. No improvement.For t ∈(0,1.5):f(t +2)=2 (since t +2 ∈(2,3.5)), and f(t)=1 (since t <1.5). Therefore, f(t +2)=2 >1=f(t). Therefore, in this interval, there exists t where f(t +2) >f(t).Specifically, for a=1.5, any t in (0,1.5) would have f(t)=1 and f(t +2)=2. Therefore, the condition is satisfied.Similarly, for 1 <a <2:For t ∈(0,a):f(t)=1,f(t +2)=2 (since t +2 ∈(2, a +2). But a +2 <4, and a² +a >2 since a >1. So t +2 ∈(2, a² +a). Therefore, f(t +2)=2 (since a ≤t +2 <a² +a is true because a >1 and t +2 >2 >a (since a <2). Wait, wait, no. If a=1.5, then a=1.5. So, for t ∈(0,1.5), t +2 ∈(2,3.5). The interval a=1.5 ≤t +2 <a² +a=3.75. So yes, t +2 ∈[1.5,3.75). But t +2 ∈(2,3.5), which is within [1.5,3.75). Therefore, f(t +2)=2.Therefore, for t ∈(0,a), f(t +2)=2 and f(t)=1. Therefore, f(t +2) >f(t). Thus, such t exists.For a=1.5, this works. Therefore, for 1 <a <2, the condition is satisfied.Now, check when a=1:a=1. Then a² +a=2.For t ∈(0,1):f(t)=1,t +2 ∈(2,3). Since a² +a=2, so t +2 ≥2. Therefore, f(t +2)=0 (since t +2 >a² +a=2). Wait, but according to previous breakdown, for t +2 >a² +a, f(t +2)=0.But if a=1, a² +a=2. So, t +2 ∈(2,3). Therefore, f(t +2)=0. Therefore, f(t +2)=0 <1=f(t). So, no good.But wait, a=1 is the boundary case. When a=1, a² +a=2. Therefore, the interval for f(t +2)=2 is when a ≤t +2 <a² +a, i.e.,1 ≤t +2 <2 => t ∈[-1,0). But for t >0, t +2 >2, so f(t +2)=0.Therefore, for a=1, even though 1 <a=1 <2, it doesn't satisfy the condition because a=1 is the root of a² +a -2=0. Therefore, for a=1, a² +a=2, so t +2 <2 =>t <0. But t >0, so no overlap. Therefore, for a=1, there is no t >0 such that f(t +2)=2. Instead, for t >0, f(t +2)=0. Therefore, for a=1, even though a >1, there is no t where f(t +2) >f(t). Wait, but earlier analysis suggested that for 1 <a <2, there exists t. But a=1 is a=1, which is the boundary.Let me check a=1.0.1, say a=1.1:a² +a=1.21 +1.1=2.31>2.So, t ∈(0,1.1):f(t)=1,t +2 ∈(2,3.1). Since a=1.1, a² +a=2.31. So t +2 ∈(2,3.1). For t ∈(0,1.1):t +2 ∈(2,3.1). Now, for t +2 <a² +a=2.31, which requires t <0.31. But t >0, so t ∈(0,0.31). Therefore:For t ∈(0,0.31):t +2 ∈(2,2.31). Since a=1.1, a=1.1 ≤t +2 <2.31. So f(t +2)=2.And f(t)=1. Therefore, f(t +2) >f(t).For t ∈[0.31,1.1):t +2 ∈[2.31,3.1). Since t +2 ≥a² +a=2.31, f(t +2)=0. And f(t)=1. So f(t +2)=0 <1.Therefore, for a=1.1, there exists t ∈(0,0.31) where f(t +2) >f(t). Hence, the condition holds.But for a=1:a² +a=2. For t ∈(0,1):t +2 ∈(2,3). Since a² +a=2, t +2 ≥2. So f(t +2)=0. Therefore, no t in (0,1) where f(t +2)=2. So, for a=1, there is no t where f(t +2) >f(t).Therefore, the critical point is a >1.So, for a >1, there exists t ∈(0, a² +a -2) such that f(t +2)=2 and f(t)=1.But wait, a² +a -2 must be greater than a. Wait, no. For a >1, a² +a -2 >0, but how does it compare to a?Compute a² +a -2 >a =>a² -2 >0 =>a² >2 =>a >√2≈1.414.Therefore, for a >√2, a² +a -2 >a.For a between1 and√2, a² +a -2 <a.But regardless, for a >1:The interval where f(t +2)=2 and t >0 is t ∈(max(a -2,0), a² +a -2). But since a >1, a -2 could be negative (if 1 <a <2) or positive (if a ≥2). Let's check a=1.5 and a=3.For a=1.5 (1 <a <2):max(a -2,0)=max(-0.5,0)=0. So t ∈(0,1.75). For t in (0,1.5), f(t)=1 and f(t +2)=2. For t in [1.5,1.75), f(t)=2 and f(t +2)=2. Therefore, the overlap where f(t +2) >f(t) is t ∈(0,1.5).For a=3 (a ≥2):a -2=1>0. So interval is (1,3² +3 -2)=(1,10). For t ∈(1,10):f(t +2)=2 if a=3 ≤t +2 <3² +3=12. So t ∈[3 -2=1,12 -2=10). For t ∈(1,10):f(t) is:- For t ∈(1,3): f(t)=1 (since 0 <t <a=3).- For t ∈[3,12): f(t)=2.- For t=3 or12:1.- For t ≥12:0.Therefore, for t ∈(1,3):f(t)=1,f(t +2)=2 (since t +2 ∈(3,5) which is within [3,12)).Thus, f(t +2)=2 >1=f(t).For t ∈[3,10):f(t)=2,f(t +2)=2 (since t +2 ∈[5,12)).Equal.Therefore, for a=3, there exists t ∈(1,3) where f(t +2) >f(t).Therefore, for a >1, regardless of whether a is between1 and2 or greater than2, there exists t where f(t +2) >f(t).But for a=1, there is no such t.Hence, the range of a is a >1.Now, check Subcase2: a ≥2.For a ≥2:When considering f(t +2):- For t ≤-2: f(t +2)=2.- For -2 <t <a -2: f(t +2)=1 (since t +2 ∈(0,a)).But a ≥2, so a -2 ≥0.Therefore, intervals:- For t ∈(-2, a -2): f(t +2)=1.- For t ∈[a -2,0]: f(t +2)=2 (since t +2 ∈[a,2 +0]=[a,2]. But since a ≥2, t +2 ∈[a,2] only if a ≤2. If a >2, then t +2 ∈[a,2] is empty, since a >2 and t ≤0. Wait, need to re-examine.Wait, for a ≥2:f(t +2):- 2 if t +2 ≤0 =>t ≤-2.- 1 if 0 <t +2 <a =>-2 <t <a -2.- 2 if a ≤t +2 <a² +a =>a -2 ≤t <a² +a -2.- 1 if t +2=a or t +2=a² +a.- 0 if t +2 >a² +a =>t >a² +a -2.But a ≥2:For t ∈(-2, a -2):t +2 ∈(0,a). So f(t +2)=1.For t ∈[a -2, a² +a -2):t +2 ∈[a, a² +a). So f(t +2)=2.For t ∈[a² +a -2, ∞): f(t +2)=0.Now, compare f(t +2) and f(t) for different t.First, t ≤-2: f(t +2)=2, f(t)=2. Equal.Next, t ∈(-2, a -2):Here, f(t +2)=1.f(t):For t ∈(-2, a -2).But a -2 ≥0 (since a ≥2), so t ∈(-2, a -2). Since a ≥2, a -2 can be from0 upwards.For example, a=2: t ∈(-2,0).For a=3: t ∈(-2,1).But f(t):For t ≤0: f(t)=2.For t >0: f(t)=1 if 0 <t <a;2 if a ≤t <a² +a; etc.So:For t ∈(-2,0]: f(t)=2.For t ∈(0, a -2): f(t)=1.Therefore, in t ∈(-2,0]:f(t +2)=1 (if a -2 >0). Wait, no.Wait, for a ≥2:When considering t ∈(-2, a -2):If a=2, t ∈(-2,0). Then f(t)=2 (since t ≤0).If a=3, t ∈(-2,1). So for t ∈(-2,0], f(t)=2; for t ∈(0,1), f(t)=1.Therefore, for a ≥2:In the interval t ∈(-2, a -2):- For t ∈(-2,0]: f(t)=2, f(t +2)=1. So f(t +2) <f(t).- For t ∈(0, a -2): f(t)=1, f(t +2)=1 (since t +2 ∈(2,a). Wait, t ∈(0, a -2), so t +2 ∈(2, a). For a ≥2, t +2 ∈(2, a). Then, for f(t +2):Since t +2 ∈(2,a), which is between2 and a. For a ≥2, a² +a is greater than or equal to 2² +2=6. So t +2 ∈(2,a) <a² +a. Therefore, f(t +2)=2 (since a ≤t +2 <a² +a). Wait, no:For a ≥2, and t +2 ∈(2,a):But a ≥2, so if a=3, t +2 ∈(2,3). But a=3, so a ≤t +2 <a² +a=12. So, t +2 ∈[3,12) would require t ∈[1,10). But in this case, t +2 ∈(2,3), which is below a=3. Therefore, f(t +2)=1 (since 0 <t +2 <a=3). Wait, no:Wait, the quadratic part for x >0 when a ≥2:The quadratic part is -x² +2a x +a. As before, the maximum is at x=a, which is a² +a. The quadratic part on x >0 increases from a at x=0 to a² +a at x=a, then decreases to -infty. Therefore, for t between a and a² +a, two solutions; for t =a or a² +a, one solution; and for t <a, one solution.But wait, for a ≥2, the quadratic part on x >0 starts at x=0 with value a (which is ≥2). Then it increases to a² +a (which is ≥6), then decreases. Therefore, the equation -x² +2a x +a =t for x >0:- For t <a: one solution (since the function decreases past t).But wait, starting at x=0 with value a (≥2), increasing to a² +a, then decreasing. So, if t <a, but the function's minimum after x=a is -infty, but the function reaches down from a² +a to -infty. So for t <a, there is still one solution on the decreasing part (x >a). For example, take a=2. t=1 <a=2. Solve -x² +4x +2=1 =>-x² +4x +1=0 =>x² -4x -1=0. Solutions x=(4 ±sqrt(16 +4))/2=(4 ±sqrt(20))/2=2 ±sqrt(5). sqrt(5)≈2.24, so positive roots are ≈4.24 and≈-0.24. Therefore, x≈4.24 is the solution. So one solution.Therefore, for a ≥2:For t <a: one solution.For a ≤t <a² +a: two solutions.For t =a or t =a² +a: one solution.For t >a² +a:0.Therefore, for a ≥2:f(t +2):- For t +2 ≤0:2- For 0 <t +2 <a:1- For a ≤t +2 <a² +a:2- For t +2 =a or t +2 =a² +a:1- For t +2 >a² +a:0.But in the interval t ∈(-2, a -2):t +2 ∈(0, a). So f(t +2)=1 if 0 <t +2 <a;2 if a ≤t +2 <a² +a.Wait, for t ∈(-2, a -2):t +2 ∈(0, a). So if a ≥2, then t +2 ∈(0, a).But for a ≥2, the quadratic part on x >0 has f(t +2)=1 if 0 <t +2 <a;2 if a ≤t +2 <a² +a. But since t +2 ∈(0, a), so f(t +2)=1.Therefore, for t ∈(-2, a -2):f(t +2)=1.But f(t):For t ∈(-2, a -2):- If t ≤0: f(t)=2.- If t >0: f(t)=1 (since 0 <t <a).Therefore:For t ∈(-2,0]: f(t)=2, f(t +2)=1. So 1 <2. No good.For t ∈(0, a -2): f(t)=1, f(t +2)=1. Equal.Next interval: t ∈[a -2, a² +a -2).Here, f(t +2)=2 (since a ≤t +2 <a² +a).But t ∈[a -2, a² +a -2).For a ≥2, a -2 ≥0. So t ≥0.Therefore, f(t):For t ∈[a -2, a² +a -2).Break into:- t ∈[a -2, a): f(t)=1.- t ∈[a, a² +a -2): f(t)=2.So:For t ∈[a -2, a):f(t)=1 (since 0 <t <a),f(t +2)=2. Therefore, f(t +2) >f(t).For t ∈[a, a² +a -2):f(t)=2,f(t +2)=2. Equal.Therefore, in t ∈[a -2, a), f(t +2)=2 >f(t)=1.Thus, there exists t in [a -2, a) where f(t +2) >f(t).For example, take a=3:t ∈[1,3).For t ∈[1,3):f(t +2)=2 (since t +2 ∈[3,5), and a=3, a² +a=12. So t +2 ∈[3,5) <12, so f(t +2)=2).f(t):For t ∈[1,3): since a=3, 0 <t <3. So f(t)=1 for t ∈(0,3). But wait, for t ∈[1,3):Wait, a=3, t ∈[1,3):For t ∈[1,3): t is between1 and3. Since a=3, f(t)=1 if 0 <t <3;2 if3 ≤t <12.Therefore, for t ∈[1,3): f(t)=1.Therefore, f(t +2)=2 >1=f(t). So for a=3, t ∈[1,3) satisfies f(t +2) >f(t).Therefore, for a ≥2, there exists t where f(t +2) >f(t).Finally, consider t ∈[a² +a -2, ∞):f(t +2)=0,f(t):For t ∈[a² +a -2, a² +a): f(t)=2.For t ≥a² +a:0.So in this interval, f(t +2)=0 ≤f(t). No improvement.Therefore, summarizing:For a >1, whether 1 <a <2 or a ≥2, there exists t where f(t +2) >f(t). For a ≤1, there is no such t.Therefore, the range of a is a >1.But wait, let's verify with a=1.5 and a=3 as above.For a=1.5 (1 <a <2), there exists t ∈(0,1.5) where f(t +2)=2 >1=f(t).For a=3 (a ≥2), there exists t ∈[1,3) where f(t +2)=2 >1=f(t).For a=1, there is no such t.Therefore, the range of a is (1, ∞).But let's check the boundary case a=1. Is there any t such that f(t +2) >f(t)?For a=1:f(t):- 2 if t ≤0,-1 if0 <t <1,-2 if1 ≤t <2,-0 ift ≥2.f(t +2):-2 ift ≤-2,-1 if-2 <t <-1,-2 if-1 ≤t <0,-0 ift ≥0.For t >0:f(t +2)=0.For t ∈(-1,0):f(t +2)=2,f(t)=2. Equal.For t ∈(-2,-1):f(t +2)=1,f(t)=2. 1 <2.For t ≤-2:f(t +2)=2,f(t)=2. Equal.For t ≥0:f(t +2)=0 ≤f(t).Therefore, no t exists where f(t +2) >f(t). Hence, a=1 is excluded.Thus, the range of a is a >1, or in interval notation, (1, ∞).Final AnswerThe number of real roots for ( g(x) = ln x ) is (boxed{1}). The range of ( a ) is (boxed{(1, +infty)}).</think>To determine the number of real roots for the equation ( g(x) = t ) where ( g(x) = ln x ), we solve ( ln x = t ). This equation has exactly one real root ( x = e^t ) for any real ( t ). Therefore, ( f(t) = 1 ) for all ( t ).Next, we analyze the piecewise function ( g(x) = begin{cases} x, & x leq 0 -x^2 + 2ax + a, & x > 0 end{cases} ) to find the range of ( a ) such that there exists ( t ) where ( f(t+2) > f(t) ).1. Left side ( ( x leq 0 ) ): The equation ( x = t ) has a solution if ( t leq 0 ).2. Right side ( ( x > 0 ) ): The quadratic equation ( -x^2 + 2ax + a = t ) rearranges to ( x^2 - 2ax + (t - a) = 0 ). The discriminant is ( 4(a^2 + a - t) ), which must be non-negative for real roots. The roots are ( x = a pm sqrt{a^2 + a - t} ). The number of positive roots depends on the value of ( t ).For ( a leq 0 ), the quadratic part decreases from ( a ) to ( -infty ), giving one root for ( t leq a ). For ( a > 0 ), the quadratic part increases to a maximum at ( x = a ) then decreases, giving two roots for ( a leq t < a^2 + a ).To find ( a ) such that ( f(t+2) > f(t) ):- For ( a leq 0 ), ( f(t) ) is 2 for ( t leq a ), 1 for ( a < t leq 0 ), and 0 otherwise. No ( t ) satisfies ( f(t+2) > f(t) ).- For ( a > 0 ), ( f(t) ) is 2 for ( t leq 0 ), 1 for ( 0 < t < a ), 2 for ( a leq t < a^2 + a ), and 0 otherwise. For ( 1 < a ), there exists ( t ) where ( f(t+2) = 2 ) and ( f(t) = 1 ).Thus, the range of ( a ) is ( (1, +infty) ). Final AnswerThe number of real roots for ( g(x) = ln x ) is (boxed{1}). The range of ( a ) is (boxed{(1, +infty)}).]

🔑:If g(x) = ln x, then the range of the function is mathbb{R}, and the function is monotonic. Therefore, the equation g(x) = t has exactly one root, Hence, f(t) = 1, For g(x) = begin{cases} x, & xleq 0 -x^{2}+2ax+a, & x>0 end{cases}, When t leq 0, f(t) = 1 always holds. If there exists t such that f(t+2) > f(t), Then for x > 0, the maximum value of the function is greater than 2, and the axis of symmetry is to the right of the y-axis, That is begin{cases} a>0, & frac{-4a-4a^{2}}{-4} > 2 end{cases}, Solving this gives: a > 1, Therefore, the answers are: f(t) = boxed{1}, and the range of a is boxed{a > 1}. If g(x) = ln x, then the range of the function is mathbb{R}, and the function is monotonic, hence the equation g(x) = t has exactly one root, so f(t) = 1. If g(x) = begin{cases} x, & xleq 0 -x^{2}+2ax+a, & x>0 end{cases} (a in mathbb{R}), and there exists t such that f(t+2) > f(t), then for x > 0, the maximum value of the function is greater than 2, and the axis of symmetry is to the right of the y-axis, solving gives the answer. This problem examines the application of piecewise functions, the zeros of functions, and the number of roots, with a medium level of difficulty.

🧠:[<think>Okay, so I have this problem here: Find the number of pairs of natural numbers a and b such that a is greater than or equal to b, and 1/a + 1/b equals 1/6. Hmm, interesting. Let me try to figure this out step by step.First, natural numbers are positive integers, right? So a and b are both positive integers, and a is at least b. The equation given is 1/a + 1/b = 1/6. I need to find all such pairs (a, b) with a ≥ b. Let me think about how to approach this.Maybe I can start by manipulating the equation algebraically. Let's see:1/a + 1/b = 1/6If I combine the fractions on the left, I get:(b + a) / (ab) = 1/6Cross-multiplying to eliminate the denominators:6(b + a) = abHmm, so ab - 6a - 6b = 0. That's a quadratic in terms of a and b. Maybe I can factor this or rearrange it to make it more manageable. Let me try adding 36 to both sides to complete the square or something. Let's see:ab - 6a - 6b + 36 = 36Now, the left side factors as (a - 6)(b - 6) = 36Oh! That's clever. So by adding 36 to both sides, we get a product of (a - 6)(b - 6) equals 36. That seems like a useful factorization. Let me check that:(a - 6)(b - 6) = ab - 6a - 6b + 36. Yes, that's exactly the left side. So, that's correct.So now, the equation becomes (a - 6)(b - 6) = 36. And since a and b are natural numbers, (a - 6) and (b - 6) must be positive integers as well, right? Wait, not necessarily. If a is 6, then (a - 6) is zero, but since 1/a would be undefined if a is 6? Wait, hold on. If a is 6, then 1/a is 1/6, but then 1/b would have to be 0, which is impossible because b is a natural number. So (a - 6) and (b - 6) must be positive integers. So a must be at least 7, and b must be at least 7 as well. Wait, but the original problem states that a ≥ b. So if both a and b are at least 7, then their difference is such that a is greater or equal to b.Therefore, the problem reduces to finding all pairs of positive integers (x, y) such that x * y = 36, where x = a - 6 and y = b - 6, and since a ≥ b, then x + 6 ≥ y + 6, which simplifies to x ≥ y. So we need to find all pairs (x, y) of positive integers where x * y = 36 and x ≥ y.So now, the problem becomes finding the number of divisors of 36 such that x ≥ y. Let me list all the positive divisors of 36. 36 factors into 2^2 * 3^2, so the number of divisors is (2+1)(2+1) = 9. The divisors are:1, 2, 3, 4, 6, 9, 12, 18, 36.So the factor pairs (x, y) such that x * y = 36 and x ≥ y are:(36, 1), (18, 2), (12, 3), (9, 4), (6, 6).So that's 5 pairs. Therefore, there are 5 such pairs (x, y), which correspond to (a, b) by adding 6 to each component. So:For (36, 1): a = 36 + 6 = 42, b = 1 + 6 = 7. So (42, 7)For (18, 2): a = 18 + 6 = 24, b = 2 + 6 = 8. So (24, 8)For (12, 3): a = 12 + 6 = 18, b = 3 + 6 = 9. So (18, 9)For (9, 4): a = 9 + 6 = 15, b = 4 + 6 = 10. So (15, 10)For (6, 6): a = 6 + 6 = 12, b = 6 + 6 = 12. So (12, 12)Therefore, these are the pairs: (42, 7), (24, 8), (18, 9), (15, 10), (12, 12). So there are 5 pairs where a ≥ b. Therefore, the answer is 5.Wait, but let me verify each of these pairs to make sure they actually satisfy the original equation. Let's check:First pair: 42 and 7.1/42 + 1/7 = 1/42 + 6/42 = 7/42 = 1/6. Correct.Second pair: 24 and 8.1/24 + 1/8 = 1/24 + 3/24 = 4/24 = 1/6. Correct.Third pair: 18 and 9.1/18 + 1/9 = 1/18 + 2/18 = 3/18 = 1/6. Correct.Fourth pair: 15 and 10.1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6. Correct.Fifth pair: 12 and 12.1/12 + 1/12 = 2/12 = 1/6. Correct.All of them check out. So, the answer should be 5.But wait, let me think again. When we considered the factor pairs of 36, we got 5 pairs where x ≥ y. Each of these gives a unique solution for a and b. Since the problem asks for the number of such pairs, and we found 5, that should be the answer.Is there a possibility that we missed any pairs? Let's see. The divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. The factor pairs are indeed (1,36), (2,18), (3,12), (4,9), (6,6). Since we need x ≥ y, we stop there. The next pair would be (9,4) but since x must be greater than or equal to y, we don't include (9,4) again once we have (4,9) reversed. Wait, but we actually ordered them as (36,1), (18,2), (12,3), (9,4), (6,6). So that's all the possible pairs where the first element is greater or equal to the second. So that's 5 pairs, which is correct. So I don't think we missed any.Another way to count the number of factor pairs is to note that the number of positive divisors is 9. Each factor pair is counted once when x ≤ y and once when x ≥ y, except when x = y. Since 36 is a perfect square? Wait, 36 is a perfect square (6^2), so the square root is 6, which is an integer. So the number of unordered factor pairs is (number of divisors + 1)/2. Wait, the formula for the number of unordered pairs is (d(n) + 1)/2 if n is a perfect square, and d(n)/2 otherwise, where d(n) is the number of divisors. Here, since 36 is a perfect square, the number of unordered pairs is (9 + 1)/2 = 5. So that's 5 unordered pairs, which is exactly what we found. So that's consistent.Therefore, the number of ordered pairs (a, b) with a ≥ b is 5. So the answer should be 5.Just to make sure, let me think of another approach. Suppose we didn't do the substitution. Let's solve for b in terms of a.Starting from 1/a + 1/b = 1/6.So 1/b = 1/6 - 1/a = (a - 6)/(6a)Therefore, b = 6a/(a - 6)Since b must be a natural number, 6a must be divisible by (a - 6). So (a - 6) divides 6a. Let's write that as:Let k = a - 6. Then, a = k + 6. Then, b = 6(k + 6)/k = 6 + 36/k.Since b must be a natural number, 36/k must be an integer. Therefore, k must be a positive divisor of 36. So k is in {1, 2, 3, 4, 6, 9, 12, 18, 36}. Therefore, k can be any of these divisors, which would give:For k = 1: a = 7, b = 6 + 36/1 = 42. But wait, here, a = 7, b = 42. But since we require a ≥ b, this would not satisfy unless we swap them. Wait, hold on. Wait, when k = 1, a = 7, b = 42. But a = 7 and b = 42. But since a must be ≥ b, this pair would be (42, 7), which is the same as before, but ordered differently.Wait, maybe I need to be careful here. If we proceed this way, solving for b in terms of a, but then a must be ≥ b, so perhaps this approach requires a bit more attention.Wait, when we set k = a - 6, then since a is a natural number greater than 6 (since k must be positive), and then b = 6 + 36/k. So for k to be a positive divisor of 36, let's list k and compute a and b:k = 1: a = 7, b = 42 → (7, 42). But a = 7 < b = 42, which doesn't satisfy a ≥ b. So we need to reverse them? But a and b are just variables, so maybe if we consider k as (b - 6), but in this case, no. Wait, perhaps this approach is leading to both possibilities, but since a and b are interchangeable in the original equation, except for the condition a ≥ b. So perhaps when we solve for b in terms of a, the solutions where a < b are covered when we consider k as (b - 6). Wait, maybe this is getting complicated.Alternatively, if we consider the substitution (a - 6)(b - 6) = 36, which we did earlier, and since a ≥ b, then x = a - 6 must be greater than or equal to y = b - 6. Therefore, x ≥ y, which led us to the factor pairs where x ≥ y, giving us 5 pairs. So that's a better approach.Alternatively, using this k substitution: since k must be a positive divisor of 36, and a = k + 6. Then, b = 6 + 36/k. So for each k divisor of 36, we have a pair (a, b) = (k + 6, 6 + 36/k). But in order to have a ≥ b, we need:k + 6 ≥ 6 + 36/kSubtract 6 from both sides:k ≥ 36/kMultiply both sides by k (since k is positive):k^2 ≥ 36So k ≥ 6.Therefore, k must be a divisor of 36 and k ≥ 6. The divisors of 36 that are ≥6 are: 6, 9, 12, 18, 36. So that's 5 values. For each of these k:k = 6: a = 6 + 6 = 12, b = 6 + 36/6 = 12. So (12, 12)k = 9: a = 9 + 6 = 15, b = 6 + 36/9 = 10. So (15, 10)k = 12: a = 12 + 6 = 18, b = 6 + 36/12 = 9. So (18, 9)k = 18: a = 18 + 6 = 24, b = 6 + 36/18 = 8. So (24, 8)k = 36: a = 36 + 6 = 42, b = 6 + 36/36 = 7. So (42, 7)So these are the same pairs as before, just listed in reverse order. Therefore, when k ≥ 6, we get the pairs where a ≥ b. Therefore, there are 5 such pairs. Therefore, confirming that the answer is 5.This seems consistent. So both approaches, whether through factor pairs of 36 with x ≥ y or through solving for k and requiring k ≥ 6, lead us to 5 solutions. Thus, I can be confident that the answer is 5.Final AnswerThe number of pairs is boxed{5}.</think>Given the problem of finding the number of pairs of natural numbers (a) and (b) such that (a geq b) and (frac{1}{a} + frac{1}{b} = frac{1}{6}), we start by manipulating the given equation:[frac{1}{a} + frac{1}{b} = frac{1}{6}]Combining the fractions on the left-hand side, we get:[frac{a + b}{ab} = frac{1}{6}]Cross-multiplying to eliminate the denominators, we obtain:[6(a + b) = ab]Rearranging terms, we have:[ab - 6a - 6b = 0]Adding 36 to both sides to complete the square:[ab - 6a - 6b + 36 = 36]This factors to:[(a - 6)(b - 6) = 36]We need to find pairs of natural numbers ((a - 6)) and ((b - 6)) such that their product is 36 and (a geq b). Let (x = a - 6) and (y = b - 6). Then, (x geq y) and (xy = 36). The positive divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. The factor pairs ((x, y)) such that (x geq y) are:[(36, 1), (18, 2), (12, 3), (9, 4), (6, 6)]Each pair ((x, y)) corresponds to the pair ((a, b)) by adding 6 to each component:- ((36, 1)) corresponds to ((42, 7))- ((18, 2)) corresponds to ((24, 8))- ((12, 3)) corresponds to ((18, 9))- ((9, 4)) corresponds to ((15, 10))- ((6, 6)) corresponds to ((12, 12))Verifying each pair, we find that they all satisfy the original equation. Therefore, the number of pairs ((a, b)) such that (a geq b) is:[boxed{5}],<think>Okay, let's see. The problem is asking how many pairs of natural numbers a and b exist such that a is greater than or equal to b, and 1/a plus 1/b equals 1/6. Hmm. Alright, so natural numbers are positive integers, right? So a and b are both positive integers, and a is at least as big as b. And their reciprocals add up to 1/6. I need to find all such pairs. Let me start by trying to manipulate the equation to see if I can express it in a way that's easier to work with. The equation is:1/a + 1/b = 1/6Maybe I can combine the fractions on the left side. Let's try that. The common denominator would be ab, so:(b + a) / ab = 1/6Cross-multiplying both sides gives:6(b + a) = abSo, 6a + 6b = ab. Hmm. Let me rearrange this equation to collect like terms. Maybe bring everything to one side:ab - 6a - 6b = 0Hmm, this looks like a quadratic in terms of a and b. I remember that sometimes factoring equations like this can be done by adding a constant to both sides. Let me see. If I add 36 to both sides:ab - 6a - 6b + 36 = 36Now, can I factor the left side? Let's check. The left side is:ab - 6a - 6b + 36Hmm, this looks like (a - 6)(b - 6) maybe? Let me expand that. (a - 6)(b - 6) = ab - 6a - 6b + 36. Yes! Perfect. So the equation becomes:(a - 6)(b - 6) = 36Ah, okay. So the product of (a - 6) and (b - 6) is 36. Now, since a and b are natural numbers, and a >= b, then (a - 6) and (b - 6) must also be integers. Also, since a and b are positive, (a - 6) and (b - 6) must be non-negative. Wait, but if (a - 6) and (b - 6) multiply to 36, which is positive, then both (a - 6) and (b - 6) must be positive integers. Because if either of them were zero, the product would be zero, which isn't 36. So both (a - 6) and (b - 6) are positive integers. Therefore, a and b are integers greater than 6. Because a - 6 and b - 6 are positive. So a > 6 and b > 6.Therefore, we can think of this as finding pairs of positive integers (x, y) such that x * y = 36, where x = a - 6 and y = b - 6. Then, since a >= b, x + 6 >= y + 6, so x >= y. Therefore, we need to find all pairs of positive integers (x, y) where x * y = 36 and x >= y. Then, the number of such pairs will correspond to the number of (a, b) pairs.So, the problem reduces to finding the number of divisors of 36 where x >= y. To find the number of such pairs, we can list all the positive divisors of 36 and pair them such that x >= y and x * y = 36.First, let's find all the divisors of 36. 36 is 2^2 * 3^2, so its divisors are all numbers of the form 2^a * 3^b where a = 0,1,2 and b = 0,1,2. So, listing them out:1, 2, 3, 4, 6, 9, 12, 18, 36.Wait, let me check:1, 2, 3, 4, 6, 9, 12, 18, 36. That's all the divisors. Let's count: 1, 2, 3, 4, 6, 9, 12, 18, 36. That's 9 divisors. So, each divisor less than or equal to sqrt(36) = 6 will pair with a corresponding divisor greater than or equal to 6. So, the pairs (x, y) where x >= y and x * y = 36 are:(36, 1), (18, 2), (12, 3), (9, 4), (6, 6)Wait, but 36 * 1 = 36, 18 * 2 = 36, 12 * 3 = 36, 9 * 4 = 36, and 6 * 6 = 36. So, these are the pairs where x >= y. Let's confirm that. Since 36 is 6^2, the square root is 6, so divisors above 6 are 9, 12, 18, 36, and divisors below or equal to 6 are 1, 2, 3, 4, 6. So pairing the largest with the smallest, we get these five pairs. So, there are 5 such pairs.Therefore, there are 5 pairs (x, y) such that x * y = 36 and x >= y. Translating back to a and b, since x = a - 6 and y = b - 6, then a = x + 6 and b = y + 6. So, each pair (x, y) gives a pair (a, b) = (x + 6, y + 6).Let me verify each of these pairs to make sure they are valid:1. (36, 1): a = 36 + 6 = 42, b = 1 + 6 = 7. Check if 1/42 + 1/7 = 1/42 + 6/42 = 7/42 = 1/6. Correct.2. (18, 2): a = 18 + 6 = 24, b = 2 + 6 = 8. 1/24 + 1/8 = 1/24 + 3/24 = 4/24 = 1/6. Correct.3. (12, 3): a = 12 + 6 = 18, b = 3 + 6 = 9. 1/18 + 1/9 = 1/18 + 2/18 = 3/18 = 1/6. Correct.4. (9, 4): a = 9 + 6 = 15, b = 4 + 6 = 10. 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6. Correct.5. (6, 6): a = 6 + 6 = 12, b = 6 + 6 = 12. 1/12 + 1/12 = 2/12 = 1/6. Correct.All these pairs check out. So, there are 5 pairs where a >= b. Wait, but the problem is asking for pairs of natural numbers a and b with a >= b. So, each of these 5 pairs satisfies that condition, right?Let me check the a and b values:1. a=42, b=7: 42 >=7, yes.2. a=24, b=8: 24 >=8, yes.3. a=18, b=9: 18 >=9, yes.4. a=15, b=10:15 >=10, yes.5. a=12, b=12:12 >=12, yes.So all pairs satisfy a >= b. Therefore, the total number of such pairs is 5.Wait a minute, but let me think again. When I considered the divisors of 36, and paired them such that x >= y, and found 5 pairs. But is there another way that there might be more pairs? Let me see.Alternatively, we can think of the equation (a - 6)(b - 6) = 36, and since a and b are natural numbers greater than 6, then x and y are positive integers. So, the number of solutions is equal to the number of positive divisors of 36, but considering that a >= b, so x >= y. The number of unordered pairs (x, y) where x * y = 36 is equal to the number of divisors up to sqrt(36). Since 36 is a square, sqrt(36) is 6, which is an integer. So the number of such pairs is equal to the number of divisors less than or equal to 6. The divisors of 36 less than or equal to 6 are 1, 2, 3, 4, 6. So that's 5 divisors. Each of these divisors pairs with 36, 18, 12, 9, 6 respectively, which are the corresponding co-divisors. Hence, 5 pairs. Therefore, the answer is 5.Therefore, the number of pairs (a, b) is 5.But let me just check again to make sure I didn't miss anything. Maybe I should list all possible factor pairs of 36 and see if I missed any:1 * 362 * 183 * 124 * 96 * 6Those are all the pairs where the first number is less than or equal to the second. Since we have to have x >= y, which in terms of factors would mean starting from the largest and going down. Wait, but in the original substitution, x = a -6 and y = b -6. Since a >= b, then x = a -6 >= y = b -6. Therefore, x >= y. Therefore, the pairs (x, y) should be ordered such that x >= y. So the pairs are:(36,1), (18,2), (12,3), (9,4), (6,6). So exactly the 5 pairs I listed before. Therefore, I think that is correct.Therefore, the answer is 5.Wait, but let me check if there's any case where (a -6) or (b -6) could be zero or negative. But since a and b are natural numbers, and we have 1/a and 1/b in the original equation. If a or b were 6, then 1/a or 1/b would be 1/6. Then 1/a +1/b would be 1/6 + something positive, which would be more than 1/6, which contradicts the equation. If a or b were less than 6, then 1/a or 1/b would be greater than 1/6, so their sum would definitely be more than 1/6. So, a and b must be greater than 6. Therefore, x and y must be positive integers, so their product is 36, and the rest follows.Therefore, I think I have covered all possible pairs. So, the answer is 5. Thus, the number of pairs is 5.Final AnswerThe number of such pairs is boxed{5}.</think>Given the equation (frac{1}{a} + frac{1}{b} = frac{1}{6}) where (a) and (b) are natural numbers and (a geqslant b), we start by manipulating the equation:[frac{1}{a} + frac{1}{b} = frac{1}{6}]Combining the fractions on the left side, we get:[frac{a + b}{ab} = frac{1}{6}]Cross-multiplying both sides gives:[6(a + b) = ab]Rearranging the terms, we have:[ab - 6a - 6b = 0]Adding 36 to both sides to factorize:[ab - 6a - 6b + 36 = 36]This can be factored as:[(a - 6)(b - 6) = 36]We need to find pairs of natural numbers ((a - 6)) and ((b - 6)) such that their product is 36 and (a geqslant b). Let (x = a - 6) and (y = b - 6). Then, (x geq y) and (xy = 36). The divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. The pairs ((x, y)) such that (x geq y) and (xy = 36) are:[(36, 1), (18, 2), (12, 3), (9, 4), (6, 6)]Translating back to (a) and (b) by adding 6 to each factor, we get the pairs:[(42, 7), (24, 8), (18, 9), (15, 10), (12, 12)]All these pairs satisfy (a geqslant b) and the original equation. Therefore, the number of such pairs is:[boxed{5}]]

🔑:Answer: 5.Solution. From the condition, it follows that frac{1}{6}=frac{1}{a}+frac{1}{b} leqslant frac{2}{b}, from which b leqslant 12. Also, frac{1}{6}=frac{1}{a}+frac{1}{b}>frac{1}{b}, so b>6. Therefore, b can take values from 7 to 12 inclusive.Using frac{1}{a}=frac{1}{6}-frac{1}{b}=frac{b-6}{6 b}, we geta=frac{6 b}{b-6}=6+frac{36}{b-6}Substitute the possible values of b and check if the number a is natural.If b=7, then a=6+frac{36}{7-6}=42 - fits.If b=8, then a=6+frac{36}{8-6}=24 - fits.If b=9, then a=6+frac{36}{9-6}=18 - fits.If b=10, then a=6+frac{36}{10-6}=15 - fits.If b=11, then a=6+frac{36}{11-6}=6+frac{36}{5} - not an integer, does not fit.If b=12, then a=6+frac{36}{12-6}=12 - fits.It is easy to see that all 5 pairs (42,7),(24,8),(18,9),(15,10),(12,12) are solutions to the original equation.Another solution. Multiplying the equation by the denominators (since a and b are natural, they can be multiplied and divided), we get6 a+6 b=a bThis equation can be easily transformed into the following:(a-6)(b-6)=36From this, it is clear that the numbers a-6 and b-6 must be divisors of 36, and natural, because if they were negative integers, the smaller of them would be no more than -6, which is impossible.The first factor is no less than the second by condition. By considering the possible factorizations(a-6)(b-6)=36 cdot 1=18 cdot 2=12 cdot 3=9 cdot 4=6 cdot 6we get the same 5 solutions.## 10th grade

🧠:[<think>Okay, so I need to find the value of 'a' such that the line x - 2ay - 3 = 0 is a symmetry axis of the given circle. The circle's equation is x² + y² - 2x + 2y - 3 = 0. Hmm, symmetry axis of a circle should be a line that passes through the center of the circle, right? Because the axis of symmetry for a circle must go through its center; otherwise, reflecting over the line wouldn't map the circle onto itself.First, let me recall that the standard form of a circle is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. So, maybe I need to rewrite the given circle equation into this standard form.Starting with the given equation: x² + y² - 2x + 2y - 3 = 0.Let me group the x terms and y terms:(x² - 2x) + (y² + 2y) = 3.Now, completing the square for both x and y.For the x terms: x² - 2x. Take half of the coefficient of x, which is -2/2 = -1. Square it: (-1)² = 1. So add and subtract 1.For the y terms: y² + 2y. Take half of 2, which is 1. Square it: 1² = 1. Add and subtract 1.So, the equation becomes:(x² - 2x + 1 - 1) + (y² + 2y + 1 - 1) = 3.Simplify:(x - 1)² - 1 + (y + 1)² - 1 = 3.Combine constants:(x - 1)² + (y + 1)² - 2 = 3.Then, (x - 1)² + (y + 1)² = 5.So, the center of the circle is (1, -1), and the radius is √5.Okay, so the circle is centered at (1, -1). Then, for the line x - 2ay - 3 = 0 to be a symmetry axis, this line must pass through the center of the circle, right? Because reflecting the circle over this line should leave the circle unchanged, which can only happen if the line goes through the center.Therefore, the point (1, -1) must lie on the line x - 2ay - 3 = 0.So, substituting x = 1 and y = -1 into the line equation:1 - 2a*(-1) - 3 = 0.Let me compute that:1 + 2a - 3 = 0.Simplify:(1 - 3) + 2a = 0.-2 + 2a = 0.Adding 2 to both sides:2a = 2.Divide both sides by 2:a = 1.Wait, so is a equal to 1? Let me verify.If a = 1, then the line becomes x - 2*1*y - 3 = 0 => x - 2y - 3 = 0.Is this line passing through the center (1, -1)?Plugging in x = 1, y = -1:1 - 2*(-1) - 3 = 1 + 2 - 3 = 0. Yes, it does. So that's correct.But wait, another thought: for a line to be a symmetry axis of a circle, it's not only sufficient that it passes through the center, but also, if the line is a symmetry axis, then it must be the case that it's perpendicular to the line joining the center to any point on the circle. Wait, but actually, for a circle, any line through the center is a symmetry axis. So in fact, all lines passing through the center are lines of symmetry. So, in that case, the line given here must pass through the center, so substituting the center into the line equation gives the condition. Hence, the only condition is that the line passes through the center, so the value of 'a' is 1.But wait, perhaps there is something wrong here. Let me check again.Alternatively, another approach: if the line is an axis of symmetry, then reflecting any point on the circle over the line should result in another point on the circle. Since the circle is symmetric about any line through its center, then indeed the line must pass through the center. Therefore, substituting the center into the line equation gives the necessary condition.Therefore, solving for 'a' gives a = 1. So, the answer should be 1. Let me check again.Original line: x - 2ay - 3 = 0.Center of the circle is (1, -1).Plugging in x=1, y=-1:1 - 2a*(-1) -3 = 1 + 2a - 3 = 2a - 2.Set equal to zero: 2a - 2 = 0 => a = 1. Yes, correct.Therefore, the value of a is 1.But wait, just to make sure, maybe there's another condition? For example, maybe the line is also required to be a diameter or something else. But in the case of a circle, any line through the center is a line of symmetry, regardless of its slope. So even if the line is at any angle, as long as it passes through the center, it's a symmetry axis.Therefore, the only condition is that the line passes through the center, so substituting the center into the line equation is sufficient.Therefore, the value of a is 1.Final AnswerThe value of ( a ) is boxed{1}.</think>To find the value of ( a ) such that the line ( x - 2ay - 3 = 0 ) is a symmetry axis of the circle ( x^2 + y^2 - 2x + 2y - 3 = 0 ), we start by determining the center of the circle.First, we rewrite the given circle equation into standard form by completing the square:[x^2 + y^2 - 2x + 2y - 3 = 0]Group and complete the square for ( x ) and ( y ):[(x^2 - 2x) + (y^2 + 2y) = 3][(x - 1)^2 - 1 + (y + 1)^2 - 1 = 3][(x - 1)^2 + (y + 1)^2 - 2 = 3][(x - 1)^2 + (y + 1)^2 = 5]The center of the circle is ( (1, -1) ) and the radius is ( sqrt{5} ).For the line ( x - 2ay - 3 = 0 ) to be a symmetry axis, it must pass through the center of the circle. Substituting the coordinates of the center ( (1, -1) ) into the line equation:[1 - 2a(-1) - 3 = 0]Simplifying this equation:[1 + 2a - 3 = 0][-2 + 2a = 0][2a = 2][a = 1]Thus, the value of ( a ) is boxed{1}.,<think>Okay, so I need to find the value of 'a' such that the line x - 2ay - 3 = 0 is a symmetry axis of the given circle. Hmm, symmetry axis of a circle should be the same as its diameter, right? Which means the line has to pass through the center of the circle. Because any line that is a symmetry axis for a circle must pass through its center. So, if I can find the center of the circle and then ensure that the given line passes through that center, then that line would be a symmetry axis. That seems like a solid approach.First, let me find the center and radius of the circle. The equation of the circle is given by x² + y² - 2x + 2y - 3 = 0. To convert this into the standard form, I need to complete the squares for both x and y terms.Starting with the x terms: x² - 2x. Completing the square here would require taking half of the coefficient of x, which is -2/2 = -1, squaring it to get 1, and adding it inside the equation. Similarly, for the y terms: y² + 2y. Half of 2 is 1, squaring it gives 1, so add that as well.So, rewriting the equation:(x² - 2x + 1) - 1 + (y² + 2y + 1) - 1 - 3 = 0Simplify:(x - 1)² + (y + 1)² - 1 - 1 - 3 = 0Which becomes:(x - 1)² + (y + 1)² - 5 = 0So, (x - 1)² + (y + 1)² = 5Therefore, the center of the circle is at (1, -1) and the radius is √5.Okay, so the center is (1, -1). Now, the line x - 2ay - 3 = 0 must pass through this center for it to be a symmetry axis. So substituting x = 1 and y = -1 into the line equation should satisfy it.Let's do that substitution:1 - 2a*(-1) - 3 = 0Simplify:1 + 2a - 3 = 0Combine constants:(1 - 3) + 2a = 0-2 + 2a = 0Now, solving for 'a':2a = 2Divide both sides by 2:a = 1Wait, that seems straightforward. But let me verify if there's another condition or something else I might be missing. Because symmetry axis for a circle is indeed any line passing through the center, so as long as the line passes through the center, it should be a symmetry axis. So, substituting the center into the line equation gives the value of 'a' directly. Therefore, a = 1. Hmm.Let me check again to make sure I didn't make a mistake. Let's plug a = 1 into the line equation:x - 2*(1)*y - 3 = 0 => x - 2y - 3 = 0Does this line pass through the center (1, -1)?Plug in x = 1, y = -1:1 - 2*(-1) - 3 = 1 + 2 - 3 = 0. Yes, it does. So that works.But wait, another thought: is being a symmetry axis only about passing through the center? Because in the case of circles, any line through the center is a line of symmetry. So yes, that should be sufficient. So I think this answer is correct.Alternatively, maybe there's another method where you reflect a general point over the line and check if it lies on the circle. But that might be more complicated. Let me consider that approach briefly.Suppose we take a point (x, y) on the circle. Then its reflection over the line x - 2ay - 3 = 0 should also lie on the circle. If this holds for all points on the circle, then the line is a symmetry axis. But since the circle is symmetric about all lines through its center, this approach would also lead to the same condition that the line passes through the center. Therefore, either way, the result is the same.Therefore, the value of 'a' is 1.Final AnswerThe value of ( a ) is boxed{1}.</think>To find the value of ( a ) such that the line ( x - 2ay - 3 = 0 ) is a symmetry axis of the circle ( x^2 + y^2 - 2x + 2y - 3 = 0 ), we start by determining the center and radius of the circle.First, we convert the given circle equation into standard form by completing the square:[x^2 + y^2 - 2x + 2y - 3 = 0]Completing the square for the ( x )-terms and ( y )-terms:[(x^2 - 2x + 1) + (y^2 + 2y + 1) - 1 - 1 - 3 = 0]Simplifying this, we get:[(x - 1)^2 + (y + 1)^2 - 5 = 0]Thus, the equation of the circle in standard form is:[(x - 1)^2 + (y + 1)^2 = 5]The center of the circle is ( (1, -1) ) and the radius is ( sqrt{5} ).For the line ( x - 2ay - 3 = 0 ) to be a symmetry axis, it must pass through the center of the circle. Substituting the coordinates of the center ( (1, -1) ) into the line equation:[1 - 2a(-1) - 3 = 0]Simplifying this equation:[1 + 2a - 3 = 0][-2 + 2a = 0]Solving for ( a ):[2a = 2][a = 1]Thus, the value of ( a ) is (boxed{1}).]

🔑:For the line x - 2ay - 3 = 0 to be a symmetry axis of the given circle, it must pass through the center of the circle. To find the center of the circle, we complete the squares for both x and y in the equation of the circle:The equation of the circle x^2 + y^2 - 2x + 2y - 3 = 0 can be rewritten by grouping the x terms and the y terms, and completing the square:(x^2 - 2x) + (y^2 + 2y) - 3 = 0 (x - 1)^2 + (y + 1)^2 - 1^2 - 1^2 - 3 = 0 (x - 1)^2 + (y + 1)^2 - 5 = 0 (x - 1)^2 + (y + 1)^2 = 5The center of the circle is at the point (1, -1).Next, we substitute the coordinates of the center into the equation of the line to find a:1 - 2a(-1) - 3 = 0 1 + 2a - 3 = 0 2a - 2 = 0 2a = 2 a = boxed{1}