🧠:Okay, let's try to figure out these two problems step by step. So, we have three sets: A, B, and C. A and B are lines, and C is the unit circle. We need to find the values of 'a' such that when we take the union of A and B and then intersect that with C, we get exactly two elements for part (1) and exactly three elements for part (2). First, let me recall what each set represents. Set A is all points (x, y) that satisfy the equation ax + y = 1. Similarly, set B is the set of points satisfying x + ay = 1. Set C is the unit circle, x² + y² = 1. So, the intersection of (A union B) with C means we're looking for points that lie on either line A or line B and also on the unit circle. For part (1), we need this intersection to have exactly two points. That probably means each line intersects the circle at one point each, but maybe under certain conditions. Wait, but if two lines each intersect the circle once, their union would have two points. But if a line intersects the circle twice, then the union might have more points. So, maybe we need each line to intersect the circle exactly once, so that their union has two points. Alternatively, if one line is tangent and the other doesn't intersect, but that would only give one point. Hmm. Wait, maybe it's possible for one line to intersect the circle twice and the other line not to intersect at all? Then the union would have two points. Or both lines intersect the circle once each, leading to two points. So, the key is to find the values of 'a' where the total number of intersection points between A and C, plus those between B and C, equals two. Similarly, for part (2), we need three points. That could happen if one line intersects the circle twice and the other line intersects once, but since sets don't have duplicates, if the lines intersect at a common point, then that point would be counted once in the union. So, maybe if one line is tangent to the circle (intersecting once) and the other line intersects the circle at two distinct points, but if the tangent point is also on the other line, then the union would have two points from the second line plus the tangent point, which is already counted. Wait, but if the tangent line's intersection is unique and not on the other line, then union would have three points. Hmm.But first, let's get back to part (1). Let me formalize this. For set A: ax + y = 1. To find its intersection with C, substitute y = 1 - ax into the circle equation: x² + (1 - a x)² = 1. Let's expand that:x² + 1 - 2a x + a² x² = 1.Combine like terms:(1 + a²) x² - 2a x + 1 - 1 = 0 → (1 + a²) x² - 2a x = 0.Factor out x:x [ (1 + a²) x - 2a ] = 0.So, the solutions are x = 0 and x = 2a / (1 + a²). Then, substituting back into y = 1 - a x:If x = 0, then y = 1. So one point is (0, 1).If x = 2a/(1 + a²), then y = 1 - a*(2a/(1 + a²)) = 1 - 2a²/(1 + a²) = (1 + a² - 2a²)/(1 + a²) = (1 - a²)/(1 + a²). So the second point is (2a/(1 + a²), (1 - a²)/(1 + a²)).So, set A intersects the circle C at two points unless the equation has a repeated root, which would mean the line is tangent to the circle. When does that happen? The quadratic equation would have a discriminant zero. Let's check:Wait, but in our case, when we solved for x, we ended up with a quadratic that factored into x times linear term. So, discriminant for the quadratic equation (1 + a²)x² - 2a x = 0. The discriminant D is [(-2a)^2 - 4*(1 + a²)*0] = 4a². Since the constant term is zero, the discriminant is 4a², which is always non-negative. Therefore, the quadratic equation will always have two real roots: x = 0 and x = 2a/(1 + a²). So, unless the two roots coincide, which would require 0 = 2a/(1 + a²), but that would mean a = 0. If a = 0, then the equation for set A becomes 0*x + y = 1, which is y = 1. The intersection with C would be x² + 1 = 1, so x = 0. So, only one point (0,1). So, when a = 0, set A intersects C at one point. Similarly, for other values of a, set A intersects C at two points: (0,1) and (2a/(1 + a²), (1 - a²)/(1 + a²)).Wait, so maybe when a = 0, A is horizontal line y=1, which touches the circle at (0,1). For a ≠ 0, the line A intersects the circle at two points.Similarly, let's analyze set B: x + a y = 1. Let's find its intersection with C. Solving for x: x = 1 - a y. Substitute into the circle equation:(1 - a y)^2 + y² = 1.Expand:1 - 2a y + a² y² + y² = 1.Combine like terms:(a² + 1) y² - 2a y + 1 - 1 = 0 → (a² + 1) y² - 2a y = 0.Factor out y:y [ (a² + 1) y - 2a ] = 0.Solutions: y = 0 and y = 2a/(a² + 1).Substitute back into x = 1 - a y:If y = 0, then x = 1. So point (1, 0).If y = 2a/(a² + 1), then x = 1 - a*(2a/(a² + 1)) = 1 - 2a²/(a² + 1) = (a² + 1 - 2a²)/(a² + 1) = (1 - a²)/(a² + 1). So the second point is ((1 - a²)/(a² + 1), 2a/(a² + 1)).Therefore, set B intersects C at two points: (1, 0) and ((1 - a²)/(a² + 1), 2a/(a² + 1)), unless a = 0. Wait, if a = 0, then set B is x + 0*y = 1, which is x = 1. Intersection with C: (1)^2 + y² = 1 ⇒ y² = 0 ⇒ y = 0. So, only point (1,0). So again, for a ≠ 0, B intersects C at two points, and for a = 0, only one.So, in general, for a ≠ 0, both A and B intersect C at two points each. However, the union of A ∪ B intersected with C could have overlaps. That is, if there is a point that is on both A and B and on C, then it would be counted only once in the union. So, we need to check if A and B intersect each other on the circle C. If they do, then the total number of distinct points in (A ∪ B) ∩ C would be 2 + 2 - 1 = 3. Otherwise, it would be 4.But for part (1), we need exactly two elements. That would require that either each line intersects the circle at two points, but all four points are distinct, but somehow the union only has two points. Wait, that can't happen. If each line intersects the circle at two points, then the union would have four points unless there is overlap. Wait, but if there is overlap, then it's less. Wait, if both lines are the same line, then their union is the same line, which would intersect the circle at two points. So, if A and B are the same line, then (A ∪ B) ∩ C would just be the two intersection points of that line with the circle. So, in that case, the union would have two points.But when are A and B the same line? The equations are ax + y =1 and x + a y =1. For two linear equations to represent the same line, their coefficients must be proportional. So, the ratios of the coefficients must be equal:a / 1 = 1 / a = 1 / 1.So, a / 1 = 1 / a ⇒ a² = 1 ⇒ a = 1 or a = -1. Then, check if the constants are proportional as well. The constant term is 1 in both equations, so 1 / 1 = 1, so yes. Therefore, when a = 1 or a = -1, the lines A and B coincide. So, in those cases, the union of A and B is just the same line, so the intersection with C would be two points (unless the line is tangent). But wait, when a = 1: line is x + y =1. Let's see its intersection with C. Substitute y = 1 - x into x² + y² =1:x² + (1 - x)^2 = 1 ⇒ x² +1 - 2x + x² =1 ⇒ 2x² -2x =0 ⇒ 2x(x -1)=0 ⇒ x=0 or x=1. So points (0,1) and (1,0). Similarly for a=-1: line is -x + y =1 ⇒ y = x +1. Substitute into circle:x² + (x +1)^2 =1 ⇒ x² +x² + 2x +1 =1 ⇒ 2x² +2x =0 ⇒ 2x(x +1)=0 ⇒x=0 or x=-1. So points (0,1) and (-1,0). Therefore, when a=1 or a=-1, the lines A and B coincide, and intersect the circle at two points each. So, (A ∪ B) ∩ C has two points. So, for a=1 or a=-1, part (1) condition is satisfied. But maybe there are other values of 'a' where the union (A ∪ B) ∩ C has exactly two points. For example, if one line is tangent to the circle (intersecting at one point) and the other line also is tangent or intersects at one point, but not overlapping. Wait, but if a line is tangent, it meets the circle at one point. If both lines are tangent and don't intersect each other at that point, then the union would have two points. But if both lines are tangent at the same point, then union would have one point. So, need to check if A and B can be tangent to the circle C. Let's check for tangency. For line A: ax + y =1. To be tangent to the circle x² + y² =1, the distance from the center (0,0) to the line must be equal to the radius, which is 1. The distance from (0,0) to line ax + y -1 =0 is |0 + 0 -1| / sqrt(a² + 1) = 1 / sqrt(a² +1). For tangency, this distance should be 1. So 1 / sqrt(a² +1) =1 ⇒ sqrt(a² +1)=1 ⇒ a² +1=1 ⇒ a²=0 ⇒ a=0. So when a=0, line A is y=1, which is horizontal line touching the circle at (0,1). Similarly for line B: x + a y =1. Distance from (0,0) to this line is |0 +0 -1| / sqrt(1 + a²) =1 / sqrt(1 +a²). So same condition: 1 / sqrt(1 +a²) =1 ⇒ a=0. So when a=0, line B is x=1, vertical line touching the circle at (1,0). Therefore, when a=0, both lines A and B are tangent to the circle, each at one point: (0,1) and (1,0). Thus, (A ∪ B) ∩ C would be these two points, so two elements. So, a=0 is another solution for part (1). Wait, but earlier when a=0, for line A, we had intersection only at (0,1), and line B intersects only at (1,0). So union is two points. So, when a=0, we have exactly two points. So, part (1) would include a=0, a=1, a=-1? Wait, but when a=1 or a=-1, lines A and B coincide, so their union is the same line, which intersects the circle at two points. When a=0, lines A and B are distinct tangents, each contributing one point. So, for part (1), the answer would be a=0,1,-1? But wait, need to check if there are other values where the total intersection points are two.Wait, for example, if one line intersects the circle at two points and the other line doesn't intersect the circle at all. Then, union would have two points. But can a line ax + y =1 not intersect the circle? The distance from center to line is 1 / sqrt(a² +1). Since the radius is 1, the line will intersect the circle if the distance is less than 1, which is always true because 1 / sqrt(a² +1) ≤1 for all real a. Because sqrt(a² +1) ≥1, so 1 / sqrt(a² +1) ≤1. So, lines A and B always intersect the circle. So, when a ≠0,1,-1, lines A and B are distinct, non-tangent, and each intersect the circle at two points. So, total intersection points would be 4 minus the number of overlapping points. So, if lines A and B intersect the circle at two points each, and there is no common point between A and B on the circle, then (A ∪ B) ∩ C has four points. If there is one common point, then it's three points. If two common points, then two points. But when would lines A and B intersect on the circle? Let's check.Suppose (x,y) is a point on both A, B, and C. Then:ax + y =1x + ay =1x² + y² =1So, solving the first two equations:From first equation: y =1 - a xSubstitute into second equation: x + a(1 - a x) =1 ⇒ x + a - a² x =1 ⇒ x(1 - a²) + a =1 ⇒ x(1 - a²) =1 -a ⇒ x=(1 -a)/(1 -a²)= (1 -a)/[(1 -a)(1 +a)] = 1/(1 +a), provided that 1 -a ≠0, i.e., a ≠1.If a ≠1, then x=1/(1 +a), and y=1 -a x=1 - a/(1 +a)= (1 +a -a)/(1 +a)=1/(1 +a). So, the point is (1/(1 +a), 1/(1 +a)). Let's check if this point is on the circle:x² + y² = [1/(1 +a)^2] + [1/(1 +a)^2] = 2/(1 +a)^2. For this to equal 1, we need 2/(1 +a)^2 =1 ⇒ (1 +a)^2=2 ⇒1 +a=±√2 ⇒a= -1 ±√2.Therefore, if a= -1 +√2 or a= -1 -√2, then the lines A and B intersect at a point on the circle. Thus, in these cases, the intersection (A ∪ B) ∩ C would have three points: two from each line, minus one overlapping point. Wait, let's confirm:Suppose a= -1 +√2. Then, lines A and B intersect at (1/(1 +a), 1/(1 +a)) which is on the circle. So, each line intersects the circle at two points, but one of them is shared. Therefore, total points in the union would be 2 + 2 -1=3. Therefore, for a= -1 ±√2, the intersection has three points. Therefore, these values would be answers for part (2). But for part (1), we need exactly two points. So, when do lines A and B not share any common points on the circle? That would be when a ≠ -1 ±√2 and a ≠1, -1, 0. Wait, but when a=1 or a=-1, the lines coincide, so all their intersection points are shared, leading to two points. When a=0, lines are tangent at different points, so two points. For other values of a, if the lines don't intersect on the circle, the union would have four points. But wait, when a is such that the lines intersect outside the circle, then their intersections with the circle are distinct. Therefore, for a ≠1,-1,0 and a ≠-1±√2, the intersection (A ∪ B) ∩ C would have four points. But that can't be, since when a=-1±√2, they have three points, and when a=1,-1,0, they have two points. So, the only time when there are exactly two points is when a=0,1,-1. But wait, hold on: when a=-1±√2, the intersection has three points, which is part (2). Then, for all other values of a, the intersection would have four points? Wait, but in that case, part (1) is a=0,1,-1. But maybe there's a miscalculation here.Wait, let's recapitulate:- If a=0: lines A and B are tangent to the circle at (0,1) and (1,0). Union has two points. So, count=2.- If a=1 or a=-1: lines A and B coincide, intersecting the circle at two points. So, union has two points. Count=2.- If a=-1 ±√2: lines A and B intersect at a point on the circle. Each line intersects the circle at two points, one shared. So, total count=3.- For all other a: lines A and B are distinct, intersect the circle at two points each, with no overlap. So, total count=4.Therefore, part (1) requires a=0,1,-1. Part (2) requires a=-1+√2 and a=-1-√2.Wait, but let me check with a=2. Let's pick a=2. Then line A: 2x + y=1. Intersection with circle: x=0 gives y=1; x=2*2/(1+4)=4/5, y=(1 -4)/5= -3/5. So points (0,1) and (4/5, -3/5). Line B: x +2y=1. Intersection with circle: y=0 gives x=1; y=2*2/(1 +4)=4/5, x=(1 -4)/5= -3/5. So points (1,0) and (-3/5,4/5). These are four distinct points. Therefore, when a=2, count=4. So, for other values, count=4. Therefore, part (1) is a=0,1,-1; part (2) is a= -1 ±√2.But let's verify with a=-1 +√2 ≈0.414. Then the lines intersect at (1/(1 +a),1/(1 +a)). Let's compute 1/(1 +a)=1/(1 +(-1 +√2))=1/√2≈0.707. Then, the point (1/√2,1/√2) is on the circle: (1/√2)^2 + (1/√2)^2=1/2 +1/2=1. Correct. So, this point is shared between A and B. Then, each line would have another point. For line A: a= -1 +√2. Equation: (-1 +√2)x + y=1. The other intersection point is (2a/(1 +a²), (1 -a²)/(1 +a²)). Let's compute a= -1 +√2≈0.414. Compute denominator 1 +a²≈1 +0.171≈1.171. Numerator for x:2a≈0.828, so x≈0.828/1.171≈0.707. Wait, but that's 1/√2≈0.707. Wait, but that would be the same as the shared point. Wait, no. Wait, when we solved for line A, we found two intersection points: x=0 and x=2a/(1 +a²). If a=-1 +√2, then 2a=2(-1 +√2). 1 +a²=1 + (-1 +√2)^2=1 +1 -2√2 +2=4 -2√2. So x=2(-1 +√2)/(4 -2√2). Multiply numerator and denominator by (4 +2√2):Numerator:2(-1 +√2)(4 +2√2)=2[(-1)(4) + (-1)(2√2) +4√2 +2*(√2)^2]=2[-4 -2√2 +4√2 +4]=2[0 +2√2]=4√2.Denominator:(4 -2√2)(4 +2√2)=16 - (2√2)^2=16 -8=8.So x=4√2 /8=√2/2≈0.707. Then y=(1 -a²)/(1 +a²). a²=( -1 +√2)^2=1 -2√2 +2=3 -2√2. 1 -a²=1 -3 +2√2= -2 +2√2. 1 +a²=4 -2√2. So y=(-2 +2√2)/(4 -2√2)= [2(-1 +√2)]/[2(2 -√2)]= (-1 +√2)/(2 -√2). Multiply numerator and denominator by (2 +√2):Numerator:(-1 +√2)(2 +√2)= -2 -√2 +2√2 + (√2)^2= -2 +√2 +2=√2.Denominator:(2 -√2)(2 +√2)=4 -2=2.Thus, y=√2 /2≈0.707. So, the other point is (√2/2, √2/2), which is the shared point. Wait, but earlier we had two points for line A: (0,1) and (2a/(1 +a²), (1 -a²)/(1 +a²)). But here, substituting a=-1 +√2, gives (√2/2, √2/2). But also, when x=0, y=1, which is (0,1). So, line A intersects the circle at (0,1) and (√2/2, √2/2). Similarly, line B: a=-1 +√2. Equation: x + (-1 +√2)y=1. Solving, we found points (1,0) and ( (1 -a²)/(1 +a²), 2a/(1 +a²) ). Let's compute this. As before, a²=3 -2√2. 1 -a²= -2 +2√2. 1 +a²=4 -2√2. So x=( -2 +2√2 )/(4 -2√2)= same as before, which is √2 /2. y=2a/(1 +a²)=2(-1 +√2)/(4 -2√2). Multiply numerator and denominator by (4 +2√2):Numerator:2(-1 +√2)(4 +2√2)= same as before=4√2.Denominator:8.Thus, y=4√2 /8=√2 /2. So, line B intersects the circle at (1,0) and (√2/2, √2/2). Therefore, the union of A and B intersects the circle at three points: (0,1), (1,0), and (√2/2, √2/2). So, three points. Therefore, a=-1 +√2 and a=-1 -√2 would give three intersection points. Similarly, for a=-1 -√2≈-2.414. Then, the intersection point would be 1/(1 +a)=1/(1 +(-1 -√2))=1/(-√2)= -1/√2≈-0.707. Check if this is on the circle: (-1/√2)^2 + (-1/√2)^2=1/2 +1/2=1. Yes. So, lines A and B intersect at (-1/√2, -1/√2) on the circle. Then, line A: a=-1 -√2. Equation: (-1 -√2)x + y=1. Intersection with circle: points (0,1) and another point. Let's compute the other point. Using the formula: x=2a/(1 +a²). a=-1 -√2. 2a=2(-1 -√2). Compute denominator:1 +a²=1 +(-1 -√2)^2=1 +1 +2√2 +2=4 +2√2. x=2(-1 -√2)/(4 +2√2). Multiply numerator and denominator by (4 -2√2):Numerator:2(-1 -√2)(4 -2√2)=2[ -4 +2√2 -4√2 +2*(√2)^2 ]=2[ -4 -2√2 +4 ]=2[ -2√2 ]= -4√2.Denominator:(4)^2 - (2√2)^2=16 -8=8.Thus, x= -4√2 /8= -√2/2≈-0.707. Then y=(1 -a²)/(1 +a²). a²= ( -1 -√2 )^2=1 +2√2 +2=3 +2√2. 1 -a²= -2 -2√2. 1 +a²=4 +2√2. So, y=( -2 -2√2 )/(4 +2√2)= [ -2(1 +√2) ]/[ 2(2 +√2) ]= [ - (1 +√2) ]/(2 +√2). Multiply numerator and denominator by (2 -√2):Numerator: - (1 +√2)(2 -√2)= -[2 -√2 +2√2 - (√2)^2 ]= -[2 +√2 -2 ]= -√2.Denominator: (2 +√2)(2 -√2)=4 -2=2.Thus, y= -√2 /2≈-0.707. So, the other intersection point for line A is (-√2/2, -√2/2). Similarly, line B with a=-1 -√2: x + (-1 -√2)y=1. Its intersections are (1,0) and ( (1 -a²)/(1 +a²), 2a/(1 +a²) ). As before, (1 -a²)/(1 +a²)= same as x above: -√2/2. And y=2a/(1 +a²)=2*(-1 -√2)/(4 +2√2)= same as before: -√2/2. So, line B intersects the circle at (1,0) and (-√2/2, -√2/2). Thus, the union is three points: (0,1), (1,0), (-√2/2, -√2/2). Wait, but line A has points (0,1) and (-√2/2, -√2/2), and line B has (1,0) and (-√2/2, -√2/2). So, union is three points. Therefore, yes, for a=-1 ±√2, we have three points.Therefore, summarizing:- For a=0,1,-1: exactly two points.- For a=-1 ±√2: exactly three points.- For all other a: four points.Thus, the answers are:(1) a ∈ {0,1,-1}(2) a ∈ {-1 +√2, -1 -√2}But let me verify with a=1. Line A: 1x + y =1; line B: x +1y=1. They are the same line. Intersection with circle: solving x + y=1 and x² + y²=1. Substitute y=1 -x: x² + (1 -x)^2 =1 ⇒2x² -2x +1=1 ⇒2x² -2x=0 ⇒x=0 or x=1. So points (0,1) and (1,0). So, two points. Correct.For a=-1: line A: -1x + y=1 ⇒y= x +1. line B: x + (-1)y=1 ⇒x - y=1 ⇒y=x -1. Wait, no. Wait, a=-1. So, line B is x + (-1)y=1 ⇒x - y=1. So, lines A and B are different? Wait, no, when a=-1, set A is -x + y=1, set B is x - y=1. Are these the same line? Let's check. -x + y=1 can be rewritten as y= x +1. x - y=1 can be rewritten as y= x -1. These are two different lines, parallel? Wait, no. The slopes are both 1, but different y-intercepts. So, they are parallel? Wait, slope of line A is 1, intercept 1. Slope of line B is 1, intercept -1. Therefore, they are parallel lines. But parallel lines don't intersect, so their union would intersect the circle at two points each, but since they are parallel, there's no overlapping points. Wait, but earlier, we thought that when a=-1, the lines coincide. But according to equations, when a=-1, line A is y=x +1 and line B is y=x -1. These are two distinct parallel lines. Therefore, their intersection with the circle would each have two points, so union has four points. Wait, this contradicts our earlier conclusion. What's happening here?Wait, earlier, when solving for a=-1, we thought that A and B coincide, but clearly, with a=-1, they are different lines. Wait, let's re-examine the earlier step where we checked when A and B are the same line. We said that for the lines ax + y=1 and x + a y=1 to be the same, the coefficients must be proportional: a/1 =1/a =1/1. So, a/1 =1/a ⇒a²=1 ⇒a=±1. Then check if the constant term is the same ratio. For a=1: 1/1=1/1=1, so yes. For a=-1: coefficients are -1/1=1/-1= -1. The constants are 1/1=1. So, the ratio for the constants would need to be -1 as well. But 1/1=1, which does not equal -1. Therefore, lines A and B coincide only when a=1. For a=-1, the ratios of the coefficients are equal (-1), but the constant term's ratio is 1, which is different. Therefore, lines A and B are distinct when a=-1. This is a mistake in the earlier reasoning. So, when a=-1, lines A and B are different. Let's recast that. So, when a=1, lines A and B are the same. When a=-1, lines A and B are different. Therefore, for a=-1, lines A and B are distinct, each intersecting the circle at two points. Thus, the union would have four points, unless they intersect at a common point on the circle. But for a=-1, let's check if lines A and B intersect on the circle.Set A: -x + y=1 ⇒ y=x +1Set B: x - y=1 ⇒ y=x -1Solve these two equations: x +1 =x -1 ⇒1=-1, which is impossible. So, lines A and B are parallel and distinct, so no intersection. Therefore, their intersections with the circle are four distinct points. Let's confirm:Line A (a=-1): y=x +1. Substitute into circle: x² + (x +1)^2 =1 ⇒x² +x² +2x +1=1 ⇒2x² +2x=0 ⇒2x(x +1)=0 ⇒x=0 or x=-1. So points (0,1) and (-1,0).Line B (a=-1): y=x -1. Substitute into circle: x² + (x -1)^2=1 ⇒x² +x² -2x +1=1 ⇒2x² -2x=0 ⇒2x(x -1)=0 ⇒x=0 or x=1. So points (0,-1) and (1,0).Wait, but these lines intersect the circle at (0,1), (-1,0) for line A, and (0,-1), (1,0) for line B. Thus, the union would have four points: (0,1), (-1,0), (0,-1), (1,0). Therefore, when a=-1, (A ∪ B) ∩ C has four points. Therefore, my earlier conclusion was wrong. So, only a=1 causes the lines to coincide. Therefore, a=1: lines coincide, two points. a=0: lines are tangent, two points. What about a=-1? It actually results in four points. Therefore, the correct values for part (1) are a=0 and a=1. a=-1 is not a solution. This means there was an error in the initial analysis. The key was that when a=-1, the lines are distinct and parallel, each intersecting the circle at two points, leading to four total points. Therefore, part (1)'s answer is a=0 and a=1. Similarly, let's check a=1 again. When a=1, lines A and B are the same line x + y=1. This line intersects the circle at (0,1) and (1,0), so two points. So, correct.Similarly, a=0: lines A is y=1 (tangent at (0,1)), line B is x=1 (tangent at (1,0)). So, union has two points. Correct.Therefore, the corrected answer for part (1) is a=0 and a=1. But wait, when a=-1, we saw four points. So, why did we previously think a=-1 was a solution? Because of a miscalculation when assuming lines coincide. Therefore, the conclusion is that part (1) has solutions a=0 and a=1, while part (2) has a=-1±√2. But wait, when a=-1±√2, we have three points, which is part (2). And for a=1, part (1). But wait, also, when a=0, part (1). Therefore, the final answers are:(1) a=0,1(2) a=-1±√2But let's check with a=1 and a=0.For a=1: lines coincide, two points.For a=0: lines are tangent, two points.For a=-1±√2: three points.For all other a: four points.Wait, but what about when a=-1? As checked, it results in four points. So, corrected. Therefore, the answer for part (1) is a=0 and a=1. For part (2), a=-1±√2.This contradicts the initial thought that a=-1 is a solution for part (1), but upon careful examination, it's not. The mistake was in assuming that when a=-1, the lines coincide, but they are actually parallel and distinct.Therefore, the corrected answers are:(1) ( a = 0 ) or ( a = 1 )(2) ( a = -1 pm sqrt{2} )Final Answer(1) The values of ( a ) are (boxed{0}) and (boxed{1}).(2) The values of ( a ) are (boxed{-1 + sqrt{2}}) and (boxed{-1 - sqrt{2}}).

✅:To determine the values of ( a ) for which ( (A cup B) cap C ) contains exactly two or three elements, we analyze the intersections of the lines ( A ) and ( B ) with the unit circle ( C ). Key Steps:1. Intersection of Line ( A ) with Circle ( C ): - The line ( A ) is given by ( ax + y = 1 ). - Substituting ( y = 1 - ax ) into ( x^2 + y^2 = 1 ) results in the quadratic equation ( (1 + a^2)x^2 - 2ax = 0 ). - This quadratic equation has solutions ( x = 0 ) and ( x = frac{2a}{1 + a^2} ), leading to points ( (0, 1) ) and ( left( frac{2a}{1 + a^2}, frac{1 - a^2}{1 + a^2} right) ).2. Intersection of Line ( B ) with Circle ( C ): - The line ( B ) is given by ( x + ay = 1 ). - Substituting ( x = 1 - ay ) into ( x^2 + y^2 = 1 ) results in the quadratic equation ( (1 + a^2)y^2 - 2ay = 0 ). - This quadratic equation has solutions ( y = 0 ) and ( y = frac{2a}{1 + a^2} ), leading to points ( (1, 0) ) and ( left( frac{1 - a^2}{1 + a^2}, frac{2a}{1 + a^2} right) ).3. Special Cases: - Tangency Condition: A line is tangent to the circle if the distance from the center to the line equals the radius. For line ( A ), this distance is ( frac{1}{sqrt{a^2 + 1}} = 1 ), leading to ( a = 0 ). Similarly, for line ( B ), the same condition leads to ( a = 0 ). - Coinciding Lines: Lines ( A ) and ( B ) coincide if their coefficients are proportional. This happens when ( a = 1 ).4. Intersection Points Analysis: - For ( a = 0 ), lines ( A ) and ( B ) are tangent to the circle at ( (0, 1) ) and ( (1, 0) ), resulting in exactly two points. - For ( a = 1 ), lines ( A ) and ( B ) coincide, intersecting the circle at two points ( (0, 1) ) and ( (1, 0) ). - For ( a = -1 pm sqrt{2} ), lines ( A ) and ( B ) intersect the circle at three distinct points due to a common intersection point on the circle. Final Answers:(1) The values of ( a ) for which ( (A cup B) cap C ) contains exactly two elements are:[boxed{0} text{ and } boxed{1}](2) The values of ( a ) for which ( (A cup B) cap C ) contains exactly three elements are:[boxed{-1 + sqrt{2}} text{ and } boxed{-1 - sqrt{2}}]

🔑:```markdownGiven sets:[ A = {(x, y) mid a x + y = 1}, ][ B = {(x, y) mid x + a y = 1}, ][ C = {(x, y) mid x^{2} + y^{2} = 1}. ]We need to find the values of ( a ) for which ((A cup B) cap C) contains exactly 2 or 3 elements.1. The expression ((A cup B) cap C = (A cap C) cup (B cap C)).First, we will find ( A cap C ) and ( B cap C ).# Finding ( A cap C ):[A cap C = left{ (x, y) left| ,begin{array}{l}a x + y = 1 x^{2} + y^{2} = 1end{array}right. right}]Substitute (y = 1 - ax) into (x^{2} + y^{2} = 1),[x^2 + (1 - ax)^2 = 1]begin{align*}x^2 + 1 - 2ax + a^2x^2 &= 1 x^2(1 + a^2) - 2ax + 1 &= 1 x^2(1 + a^2) - 2ax &= 0 x(x(1 + a^2) - 2a) &= 0end{align*}Thus, (x = 0) or (x = frac{2a}{1 + a^2}).If (x = 0), then (y = 1 - ax = 1).If (x = frac{2a}{1 + a^2}), then (y = 1 - ax = 1 - aleft(frac{2a}{1 + a^2}right) = frac{1 - a^2}{1 + a^2}).Therefore, [A cap C = left{ (0, 1), left( frac{2a}{1 + a^2}, frac{1 - a^2}{1 + a^2} right) right}]# Finding ( B cap C ):[B cap C = left{ (x, y) left| ,begin{array}{l}x + a y = 1 x^{2} + y^{2} = 1end{array}right. right}]Substitute (x = 1 - ay) into (x^{2} + y^{2} = 1),[(1 - ay)^2 + y^2 = 1]begin{align*}1 - 2ay + a^2y^2 + y^2 &= 1 1 - 2ay + y^2(1 + a^2) &= 1 y^2(1 + a^2) - 2ay &= 0 y(y(1 + a^2) - 2a) &= 0end{align*}Thus, (y = 0) or (y = frac{2a}{1 + a^2}).If (y = 0), then (x = 1 - ay = 1).If (y = frac{2a}{1 + a^2}), then (x = 1 - ay = 1 - aleft(frac{2a}{1 + a^2}right) = frac{1 - a^2}{1 + a^2}).Therefore,[B cap C = left{ (1, 0), left( frac{1 - a^2}{1 + a^2}, frac{2a}{1 + a^2} right) right}]# (1) Finding the values of ( a ) such that ((A cup B) cap C) contains exactly 2 elements.This occurs if either:- One element from (A cap C) is identical to one from (B cap C), or- The non-trivial pair of points from one intersection does not intersect with points from the other.Finding common elements:[frac{2a}{1 + a^2} = 0 quad text{and} quad frac{1 - a^2}{1 + a^2} = 1]or[frac{2a}{1 + a^2} = 1 quad text{and} quad frac{1 - a^2}{1 + a^2} = 0]- For (frac{2a}{1 + a^2} = 0): [ 2a = 0 Rightarrow a = 0 ]- For (frac{2a}{1 + a^2} = 1): [ 2a = 1 + a^2 Rightarrow a^2 - 2a + 1 = 0 Rightarrow (a - 1)^2 = 0 Rightarrow a = 1 ]Thus, either (a = 0) or (a = 1).[boxed{a = 0 text{ or } a = 1}]# (2) Finding the values of ( a ) such that ((A cup B) cap C) contains exactly 3 elements.This occurs if the non-trivial points of intersection of each set overlap exactly once:[frac{2a}{1 + a^2} = frac{1 - a^2}{1 + a^2}]Simplifying, we get:[2a = 1 - a^2][a^2 + 2a - 1 = 0]Using the quadratic formula:[a = frac{-2 pm sqrt{4 + 4}}{2} = frac{-2 pm sqrt{8}}{2} = -1 pm sqrt{2}]This gives us the values:[boxed{a = -1 + sqrt{2} text{ or } a = -1 - sqrt{2}}]```

🧠:Okay, so I need to prove this inequality involving geometric means. The problem states that for non-negative real numbers a₁, a₂, ..., aₙ and b₁, b₂, ..., bₙ, the sum of their geometric means is less than or equal to the geometric mean of their sums. Let me write that out again to make sure I have it right:[left(a_{1} a_{2} cdots a_{n}right)^{frac{1}{n}} + left(b_{1} b_{2} cdots b_{n}right)^{frac{1}{n}} leq left(left(a_{1}+b_{1}right)left(a_{2}+b_{2}right) cdots left(a_{n}+b_{n}right)right)^{frac{1}{n}}.]Alright, so the left side is the sum of the geometric means of the a's and the b's, and the right side is the geometric mean of the terms (a_i + b_i) for each i. Since all terms are non-negative, the geometric means are well-defined.First, I should recall what the geometric mean is and any inequalities related to it. The geometric mean of non-negative numbers is always less than or equal to their arithmetic mean (AM-GM inequality). But here, the problem is about the sum of two geometric means being less than or equal to another geometric mean. That seems different from the standard AM-GM. Maybe there's a way to relate them or use another inequality.Another thought: perhaps Hölder's inequality or Minkowski's inequality. Minkowski is about sums of vectors and their norms, which might be relevant here since we're dealing with sums inside the product. Let me recall Minkowski's inequality.Minkowski's inequality states that for p ≥ 1, the L^p norm of the sum of two vectors is less than or equal to the sum of their L^p norms. The L^p norm here is the p-th root of the sum of the p-th powers. Wait, but in this problem, the structure is a bit different. Instead of sums of p-th powers, we have products. Hmm.Alternatively, maybe Hölder's inequality. Hölder is about the product of sums being greater than or equal to the sum of products, under certain conditions on exponents. Let me check.Hölder's inequality: For sequences of non-negative real numbers (x_i) and (y_i), and exponents p and q such that 1/p + 1/q = 1, we have[sum_{i=1}^{n} x_i y_i leq left( sum_{i=1}^{n} x_i^p right)^{1/p} left( sum_{i=1}^{n} y_i^q right)^{1/q}.]But again, not sure how directly applicable that is here. The problem is dealing with products of (a_i + b_i), so maybe expanding that product would help? Although expanding a product of n terms might get complicated. Let's think.Alternatively, perhaps take logarithms? The geometric mean is the exponential of the arithmetic mean of the logarithms. So, if I take the logarithm of both sides, the inequality might become additive. Let me try that.Let me denote:GM(a) = (a₁a₂⋯aₙ)^{1/n}, GM(b) = (b₁b₂⋯bₙ)^{1/n}, and GM(a + b) = [(a₁ + b₁)(a₂ + b₂)⋯(aₙ + bₙ)]^{1/n}.So the inequality is GM(a) + GM(b) ≤ GM(a + b). Taking logarithms on both sides isn't straightforward because log is a concave function, and log(x + y) isn't directly related to log x and log y. Maybe that approach isn't the best.Another idea: use the AM-GM inequality on each term (a_i + b_i). For each i, we know that a_i + b_i ≥ 2√(a_i b_i) by AM-GM. If we take the product over all i, we get:Product_{i=1}^n (a_i + b_i) ≥ Product_{i=1}^n 2√(a_i b_i) = 2^n Product_{i=1}^n √(a_i b_i) = 2^n (Product_{i=1}^n a_i)^{1/2} (Product_{i=1}^n b_i)^{1/2}.Taking the n-th root of both sides gives:GM(a + b) ≥ 2 (GM(a) GM(b))^{1/2}.But the original inequality is GM(a) + GM(b) ≤ GM(a + b). So if I can relate GM(a) + GM(b) to 2 sqrt(GM(a) GM(b)), maybe via AM-GM. Wait, the AM-GM inequality tells us that (GM(a) + GM(b))/2 ≥ sqrt(GM(a) GM(b)). So combining the two results:GM(a) + GM(b) ≥ 2 sqrt(GM(a) GM(b)) ≤ GM(a + b).But this chain of inequalities would give GM(a) + GM(b) ≥ 2 sqrt(GM(a) GM(b)) ≤ GM(a + b). Hmm, that doesn't directly lead to GM(a) + GM(b) ≤ GM(a + b). Actually, the inequality 2 sqrt(GM(a) GM(b)) ≤ GM(a + b) is not necessarily helpful here because GM(a) + GM(b) is larger than 2 sqrt(GM(a) GM(b)) by AM-GM. So if 2 sqrt(...) ≤ GM(a + b), then GM(a) + GM(b) could be even larger, but the problem says GM(a) + GM(b) is less than or equal to GM(a + b). Wait, but this seems conflicting. So perhaps my initial approach is flawed.Wait, let's test with an example. Let me take n=1. Then the inequality becomes a₁^{1/1} + b₁^{1/1} ≤ (a₁ + b₁)^{1/1}, which is a₁ + b₁ ≤ a₁ + b₁. That's equality. So it holds for n=1.How about n=2? Let me choose some numbers. Let a₁ = 1, a₂ = 1, and b₁ = 1, b₂ = 1. Then left side is (1*1)^{1/2} + (1*1)^{1/2} = 1 + 1 = 2. Right side is (2*2)^{1/2} = √4 = 2. So equality holds here.Another example: a₁ = 4, a₂ = 1; b₁ = 1, b₂ = 4. Then GM(a) = (4*1)^{1/2} = 2; GM(b) = (1*4)^{1/2} = 2. Left side is 2 + 2 = 4. Right side is (5*5)^{1/2} = 5. So 4 ≤ 5, which holds.Another example: a₁ = 3, a₂ = 3; b₁ = 1, b₂ = 1. GM(a) = (3*3)^{1/2} = 3; GM(b) = (1*1)^{1/2} = 1. Left side: 3 + 1 = 4. Right side: (4*4)^{1/2} = 4. So equality again.Wait, so maybe equality holds when a_i and b_i are proportional? In the first example, all a_i and b_i are equal. In the second example, a and b are swapped, so they are not proportional. But in that case, the left side was 4 and the right side was 5, so inequality holds. Wait, but in the third example, when a's are 3 and 3, and b's are 1 and 1, so they are proportional (each a_i is 3 times b_i). Then we get equality. Hmm.Wait, in the first example, a_i = b_i, so proportionality with constant 1. In the third example, a_i = 3b_i. So in both cases, when a and b are proportional, equality holds. When they are not proportional, the left side is less than the right side. So perhaps the inequality is a generalization of the AM-GM in some way, and equality holds when a_i and b_i are proportional for all i.Given that, maybe we can use the AM-GM inequality on each term (a_i + b_i). Wait, but how?Alternatively, consider normalizing the variables. Let me think. Suppose I set x_i = a_i / (a_i + b_i) and y_i = b_i / (a_i + b_i), provided that a_i + b_i ≠ 0. If a_i + b_i = 0, then both a_i and b_i must be zero (since they are non-negative), so that term is zero, and in the product on the right, that term would contribute zero, making the entire right-hand side zero. In that case, the left-hand side would also have to be zero because each a_i and b_i is zero, so their geometric means would be zero. So the inequality would hold as 0 + 0 ≤ 0.So assuming that a_i + b_i > 0 for all i (otherwise, both sides are zero), we can define x_i = a_i / (a_i + b_i) and y_i = b_i / (a_i + b_i). Then x_i + y_i = 1 for each i.Then, the inequality becomes:[Product_{i=1}^n (x_i (a_i + b_i))]^{1/n} + [Product_{i=1}^n (y_i (a_i + b_i))]^{1/n} ≤ [Product_{i=1}^n (a_i + b_i)]^{1/n}.Factoring out the (a_i + b_i) terms, we have:[Product (x_i) Product (a_i + b_i)]^{1/n} + [Product (y_i) Product (a_i + b_i)]^{1/n} ≤ [Product (a_i + b_i)]^{1/n}.Dividing both sides by [Product (a_i + b_i)]^{1/n} (which is positive), we get:[Product (x_i)]^{1/n} + [Product (y_i)]^{1/n} ≤ 1.So the problem reduces to proving that the sum of the geometric means of x_i and y_i (where x_i + y_i = 1 and x_i, y_i ≥ 0) is less than or equal to 1. That seems like a simpler inequality to handle. Let me check with n=2. If x₁ = x, x₂ = 1 - x (wait no, each x_i and y_i are separate, but for each i, x_i + y_i = 1. So for example, for each i, x_i and y_i are in [0,1] with x_i + y_i = 1. So each pair (x_i, y_i) is a pair of non-negative numbers summing to 1.So the problem now is, given x_i, y_i ≥ 0 with x_i + y_i = 1 for each i, show that:(Product x_i)^{1/n} + (Product y_i)^{1/n} ≤ 1.Hmm. Let me test this with n=1: x₁ + y₁ = 1, then (x₁) + (y₁) = 1 ≤ 1. Equality holds.For n=2: x₁x₂)^{1/2} + (y₁y₂)^{1/2} ≤ 1. Let me take x₁ = x₂ = 1/2, then y₁ = y₂ = 1/2. The left side is (1/2 * 1/2)^{1/2} + (1/2 * 1/2)^{1/2} = (1/4)^{1/2} + (1/4)^{1/2} = 1/2 + 1/2 = 1. Equality holds.Another example: x₁ = 1, x₂ = 0; y₁ = 0, y₂ = 1. Then left side is (0)^{1/2} + (0)^{1/2} = 0 + 0 = 0 ≤ 1.Another example: x₁ = 0.5, x₂ = 0.5; y₁ = 0.5, y₂ = 0.5. Then same as before, equality.Another example: x₁ = 0.6, x₂ = 0.6; y₁ = 0.4, y₂ = 0.4. Then (0.6*0.6)^{1/2} + (0.4*0.4)^{1/2} = 0.6 + 0.4 = 1. So equality again.Wait, is this always equal to 1? Wait, no. Let me take x₁ = 0.9, x₂ = 0.1; y₁ = 0.1, y₂ = 0.9. Then the left side is sqrt(0.9*0.1) + sqrt(0.1*0.9) = sqrt(0.09) + sqrt(0.09) = 0.3 + 0.3 = 0.6 ≤ 1. So inequality holds.Another example: x₁ = 0.8, x₂ = 0.7; y₁ = 0.2, y₂ = 0.3. Then product x_i = 0.8*0.7 = 0.56; GM(x) = sqrt(0.56) ≈ 0.748. Product y_i = 0.2*0.3 = 0.06; GM(y) = sqrt(0.06) ≈ 0.245. Sum ≈ 0.748 + 0.245 ≈ 0.993 ≤ 1. So it holds.Wait, but in some cases, if the x_i and y_i are not balanced across the terms, maybe the sum could exceed 1? Let me see. Suppose n=2, x₁=1, x₂=0.5; y₁=0, y₂=0.5. Then product x_i = 1*0.5 = 0.5; GM(x) = sqrt(0.5) ≈ 0.707. Product y_i = 0*0.5 = 0; GM(y) = 0. So sum is ≈ 0.707 ≤ 1. Still holds.Another case: n=3. Let x₁ = x₂ = x₃ = 1/3; y_i = 2/3 each. Then product x_i = (1/3)^3; GM(x) = (1/3)^{1} = 1/3. Similarly, GM(y) = (2/3)^{1} = 2/3. Sum is 1/3 + 2/3 = 1. Equality.If x_i are different but sum to 1 with y_i. Let x₁=0.5, x₂=0.5, x₃=0; y₁=0.5, y₂=0.5, y₃=1. Then product x_i = 0.5*0.5*0 = 0; GM(x) = 0. Product y_i = 0.5*0.5*1 = 0.25; GM(y) = (0.25)^{1/3} ≈ 0.63. Sum ≈ 0.63 ≤ 1.So far, all the examples satisfy the inequality. It seems like the key is that for each i, x_i + y_i =1, and the products tend to be small when there's variance in x_i and y_i across i.So, how can we prove (Product x_i)^{1/n} + (Product y_i)^{1/n} ≤1, given that x_i + y_i =1 and x_i, y_i ≥0 for all i.Wait, perhaps use induction on n? Let's consider n=1: x₁ + y₁ =1, then (x₁) + (y₁) =1, which holds.Assume the inequality holds for n=k. Now, for n=k+1, we need to show that [Product_{i=1}^{k+1} x_i]^{1/(k+1)} + [Product_{i=1}^{k+1} y_i]^{1/(k+1)} ≤1.But I'm not sure how to set up the induction step here. Alternatively, maybe use the AM-GM inequality in some fashion. Let's think.Let me denote GM_x = (Product x_i)^{1/n}, GM_y = (Product y_i)^{1/n}. We need to show GM_x + GM_y ≤1.Since for each i, x_i + y_i =1, perhaps consider that for each i, x_i ≤1 and y_i ≤1, so Product x_i ≤1, Product y_i ≤1. But the geometric means would then be ≤1, but their sum could be up to 2. But we need to show their sum is ≤1, which is not directly obvious.Wait, but in the case where all x_i =1, then y_i=0, and GM_x=1, GM_y=0, so sum is 1. Similarly, if all x_i=0.5, then GM_x=(0.5^n)^{1/n}=0.5, and similarly GM_y=0.5, sum is 1. If some x_i are 1 and others are 0, then GM_x and GM_y could be 0, sum 0.But in the middle cases, like x_i=0.9 for some i and 0.1 for others, the product might be smaller. Wait, but how to capture this in general?Maybe use weighted AM-GM. Let me consider the terms in the products.Take the logarithm of GM_x + GM_y ≤1.Let me denote A = ln(GM_x) = (1/n) Σ ln x_i,B = ln(GM_y) = (1/n) Σ ln y_i.We need to show that e^{A} + e^{B} ≤1.But since x_i + y_i =1, we have y_i =1 -x_i. So B = (1/n) Σ ln(1 -x_i).But this seems difficult to relate. Maybe not the best approach.Another idea: Use the inequality between arithmetic and geometric mean on the terms GM_x and GM_y. Wait, (GM_x + GM_y)/2 ≤ sqrt(GM_x GM_y). But that would give GM_x + GM_y ≤ 2 sqrt(GM_x GM_y). But we need to show GM_x + GM_y ≤1, so unless sqrt(GM_x GM_y) ≤1/2, but that might not hold. For example, if GM_x=GM_y=0.5, then sqrt(0.25)=0.5, and 2*0.5=1. So equality. But if GM_x and GM_y are both 0.6, then sqrt(0.36)=0.6, and 2*0.6=1.2>1. Which would violate the inequality. But in reality, if GM_x=0.6, then GM_y can't be 0.6 because x_i + y_i=1. Let's see. If all x_i=0.6, then y_i=0.4, so GM_y=(0.4^n)^{1/n}=0.4. Then GM_x + GM_y=0.6 +0.4=1. So even in this case, the sum is 1. Wait, but if some x_i are larger and others smaller?Wait, if half of the x_i are 0.8 and the other half are 0.4 (assuming n even), then y_i would be 0.2 and 0.6. The GM_x = (0.8^{n/2} 0.4^{n/2})^{1/n} = (0.8*0.4)^{1/2} = sqrt(0.32) ≈0.566. GM_y = (0.2^{n/2} 0.6^{n/2})^{1/n} = sqrt(0.12) ≈0.346. Sum≈0.566 +0.346≈0.912 ≤1.So the sum is less than 1 here. But if we have x_i varying in a way that their product is higher, but since x_i + y_i=1, making some x_i large forces others to be small. Maybe the maximum sum occurs when all x_i are equal. Let's suppose that.Suppose all x_i = x, so y_i =1 -x. Then GM_x =x, GM_y=1 -x. Then sum is x + (1 -x)=1. So in the case where all x_i are equal, the sum is 1. But if x_i are not all equal, the sum is less than 1. That seems to be the case in the examples. Therefore, the maximum of GM_x + GM_y is 1, achieved when all x_i are equal (and hence all y_i are equal). Therefore, the inequality holds.So this suggests that the inequality is a consequence of the AM-GM inequality, where deviation from equal terms in the product reduces the product, hence the sum of the geometric means is maximized when all terms are equal, giving the sum as 1.Therefore, to formalize this, we can use the concavity of the logarithm or the AM-GM inequality.Alternatively, use the concept of entropy or mutual information, but that might be overcomplicating.Another approach: For each i, since x_i + y_i =1, we can think of x_i and y_i as probabilities. Then, the geometric mean Product x_i^{1/n} is the geometric mean of the probabilities, and similarly for y_i. The sum of these geometric means is then less than or equal to 1.But how to prove this? Maybe use the inequality between the arithmetic mean and the geometric mean for each term.Wait, for each i, we have x_i + y_i =1. The arithmetic mean of x_i and y_i is 0.5, and their geometric mean is sqrt(x_i y_i). So for each i, sqrt(x_i y_i) ≤0.5. Then Product sqrt(x_i y_i) ≤ (0.5)^n. Hence, Product sqrt(x_i y_i) = (Product x_i y_i)^{1/2} = (Product x_i Product y_i)^{1/2} ≤ (0.5)^n.But how does this relate to GM_x + GM_y?Wait, GM_x = (Product x_i)^{1/n}, GM_y = (Product y_i)^{1/n}. Then, the geometric mean of GM_x and GM_y is sqrt(GM_x GM_y) = (Product x_i y_i)^{1/(2n)} = (Product (x_i y_i))^{1/(2n)}. From the above, Product (x_i y_i) ≤ (0.5)^{2n}, so (Product (x_i y_i))^{1/(2n)} ≤0.5. Hence, sqrt(GM_x GM_y) ≤0.5. Then by AM-GM, (GM_x + GM_y)/2 ≥ sqrt(GM_x GM_y) ≤0.5. Wait, but this gives (GM_x + GM_y)/2 ≥0.5, which would imply GM_x + GM_y ≥1. But this contradicts our earlier examples where the sum was less than 1. So clearly, something is wrong here.Wait, let's check that. If we have sqrt(GM_x GM_y) ≤0.5, then AM-GM gives (GM_x + GM_y)/2 ≥ sqrt(GM_x GM_y) ≤0.5. Therefore, (GM_x + GM_y)/2 ≥ sqrt(GM_x GM_y) ≤0.5 ⇒ (GM_x + GM_y)/2 ≥0.5 ⇒ GM_x + GM_y ≥1. But in our examples, we had GM_x + GM_y ≤1. Contradiction.This suggests that there's a mistake in the reasoning. Let's see where.We have for each i, sqrt(x_i y_i) ≤ (x_i + y_i)/2 =0.5. Then the product over i of sqrt(x_i y_i) ≤ Product 0.5 =0.5^n. But Product sqrt(x_i y_i) = (Product x_i y_i)^{1/2}. Therefore, (Product x_i y_i)^{1/2} ≤0.5^n ⇒ Product x_i y_i ≤0.5^{2n}. Therefore, (Product x_i y_i)^{1/(2n)} ≤0.5. But (Product x_i)^{1/n} * (Product y_i)^{1/n} = (Product x_i y_i)^{1/n} ≤0.5^2=0.25. Wait, no:Wait, (Product x_i y_i)^{1/(2n)} = [Product x_i Product y_i]^{1/(2n)} = [ (Product x_i)^{1/n} (Product y_i)^{1/n} ]^{1/2} = sqrt(GM_x GM_y). So we have sqrt(GM_x GM_y) ≤0.5.But then by AM-GM, (GM_x + GM_y)/2 ≥ sqrt(GM_x GM_y) ≤0.5. Therefore, (GM_x + GM_y)/2 ≥0.5 ⇒ GM_x + GM_y ≥1. But this contradicts our earlier example where GM_x + GM_y=0.6 +0.4=1, and another example where it was ≈0.912. Wait, but in reality, when x_i are all equal to 0.5, then GM_x=0.5, GM_y=0.5, so sum=1. If x_i vary, the sum is less than 1, but according to this reasoning, the sum should be ≥1. Contradiction.So clearly, my previous approach is flawed. Let me check where.The mistake must be in the step where I applied the inequality. Let's recast:For each i, we have x_i + y_i =1. Therefore, sqrt(x_i y_i) ≤0.5. Therefore, Product sqrt(x_i y_i) ≤0.5^n. But Product sqrt(x_i y_i) = (Product x_i Product y_i)^{1/2}. So (Product x_i Product y_i)^{1/2} ≤0.5^n ⇒ Product x_i Product y_i ≤0.5^{2n} ⇒ (Product x_i)^{1/n} (Product y_i)^{1/n} ≤0.5^2=0.25. So GM_x GM_y ≤0.25. Therefore, the geometric mean of GM_x and GM_y is sqrt(GM_x GM_y) ≤0.5. Therefore, by AM-GM, (GM_x + GM_y)/2 ≥ sqrt(GM_x GM_y) ≤0.5. Therefore, (GM_x + GM_y)/2 ≥0.5 ⇒ GM_x + GM_y ≥1.But this contradicts the numerical examples. Wait, in the example where x₁=0.6, x₂=0.6; y₁=0.4, y₂=0.4, we have GM_x=0.6, GM_y=0.4. Their product is 0.24 ≤0.25. Then, sqrt(0.24)≈0.489≤0.5. Then, (0.6 +0.4)/2=0.5≥0.489. So AM-GM holds, but GM_x + GM_y=1. In another example where x_i=0.9,0.1; y_i=0.1,0.9, GM_x=sqrt(0.9*0.1)=sqrt(0.09)=0.3, GM_y= same=0.3. Sum=0.6. Then, GM_x GM_y=0.09, sqrt(0.09)=0.3. So (0.3 +0.3)/2=0.3≥0.3. So equality in AM-GM. But here, GM_x + GM_y=0.6 <1. But according to the previous reasoning, we should have GM_x + GM_y ≥1. But clearly, 0.6 <1. So where is the mistake?Ah! The mistake is in the step where I derived (Product x_i Product y_i)^{1/n} ≤0.25. Wait, (Product x_i Product y_i)^{1/n} = (Product x_i)^{1/n} (Product y_i)^{1/n} = GM_x GM_y. But if GM_x GM_y ≤0.25, then sqrt(GM_x GM_y) ≤0.5. Then, AM-GM says (GM_x + GM_y)/2 ≥ sqrt(GM_x GM_y) ≤0.5. Therefore, (GM_x + GM_y)/2 ≥ sqrt(GM_x GM_y), but sqrt(GM_x GM_y) could be less than 0.5. Therefore, (GM_x + GM_y)/2 could be greater than or equal to a number less than 0.5, so GM_x + GM_y could be as low as twice that lower number. For example, if sqrt(GM_x GM_y)=0.3, then GM_x + GM_y ≥0.6, which is indeed the case in the example.But the key point is that the AM-GM gives a lower bound on GM_x + GM_y, not an upper bound. Therefore, the previous approach doesn't help in proving the upper bound. So I need a different approach.Let me think differently. Suppose I consider the function f(x₁, x₂, ..., xₙ) = (Product x_i)^{1/n} + (Product (1 -x_i))^{1/n} where each x_i ∈[0,1]. I need to show that f ≤1.To find the maximum of f, perhaps take derivatives. Suppose all x_i are equal, say x_i =x. Then f =x + (1 -x) =1. So at least when all variables are equal, f=1. To check if this is the maximum, consider perturbing one variable. Suppose we have two variables, n=2. Let x₁=x + h, x₂=x -h, keeping x₁ +x₂=2x. Wait, no, in the normalized case, each x_i is independent. Wait, in the normalized case, each x_i can vary independently between 0 and1, but in our original problem, the variables are linked through the products. Wait, but in the reduced problem after normalization, each x_i and y_i=1 -x_i are separate for each i. So the variables x₁, x₂,...,xₙ are independent, each in [0,1].So the function to maximize is f(x₁,...,xₙ) = (Product x_i)^{1/n} + (Product (1 -x_i))^{1/n}.We suspect that the maximum is achieved when all x_i are equal. Let's check for n=2. Suppose x₁=x, x₂=x. Then f= x + (1 -x). But wait, no: For n=2, f=sqrt(x₁x₂) + sqrt((1 -x₁)(1 -x₂)). If x₁=x₂=x, then f= x + (1 -x)=1. If x₁≠x₂, for example x₁=0.6, x₂=0.6, then f=0.6 +0.4=1. If x₁=0.6, x₂=0.4, then f=sqrt(0.24) + sqrt(0.24)=0.489 +0.489≈0.978 <1. So indeed, maximum at equality.Similarly, for n=3, if all x_i=0.5, then f=0.5 +0.5=1. If one x_i=0.6, others=0.5, then f=(0.6*0.5*0.5)^{1/3} + (0.4*0.5*0.5)^{1/3} ≈ (0.15)^{1/3} + (0.1)^{1/3}≈0.53 +0.464≈0.994 <1.So it seems that the function f achieves its maximum when all x_i are equal. Therefore, to prove the inequality, we can show that f(x₁,...,xₙ) ≤1, with equality when all x_i are equal.To prove this, we can use the concept of symmetry and convexity. Since the function is symmetric in all variables, its maximum should occur at the symmetric point where all variables are equal. Alternatively, we can use the method of Lagrange multipliers to find the maximum.Let's try using Lagrange multipliers. We want to maximize f(x₁,...,xₙ) = (Product x_i)^{1/n} + (Product (1 -x_i))^{1/n} subject to 0 ≤x_i ≤1. The maximum is attained in the interior since at the boundaries (some x_i=0 or1), f=0 or less.Set up the Lagrangian with no constraints (since we're looking for critical points inside the domain). Take the derivative of f with respect to each x_i and set to zero.Let me denote GM_x = (Product x_j)^{1/n}, GM_y = (Product (1 -x_j))^{1/n}.The derivative of GM_x with respect to x_i is (1/n)(Product x_j)^{1/n -1} * Product_{j≠i} x_j = (GM_x)/ (n x_i).Similarly, derivative of GM_y with respect to x_i is (1/n)(Product (1 -x_j))^{1/n -1} * (-Product_{j≠i} (1 -x_j)) = -GM_y / (n (1 -x_i)).Therefore, the derivative of f with respect to x_i is GM_x / (n x_i) - GM_y / (n (1 -x_i)).At the critical point, this derivative is zero for all i:GM_x / (n x_i) = GM_y / (n (1 -x_i)) ⇒ GM_x / x_i = GM_y / (1 -x_i) for all i.This implies that for all i, (1 -x_i)/x_i = GM_y / GM_x.Let k = GM_y / GM_x. Then for all i, (1 -x_i)/x_i =k ⇒1 -x_i =k x_i ⇒x_i =1/(1 +k).Therefore, all x_i are equal. Let x_i =x for all i. Then, GM_x =x, GM_y=1 -x. Then k= (1 -x)/x.But also, since GM_y = (Product (1 -x_j))^{1/n} = ( (1 -x)^n )^{1/n} =1 -x.Similarly, GM_x =x.Thus, k= (1 -x)/x, but from the definition, k=GM_y / GM_x=(1 -x)/x, so consistent.Therefore, the only critical point occurs when all x_i are equal. And at this critical point, f(x,...,x)=x + (1 -x)=1.To confirm this is a maximum, we can check the second derivative or note that in examples deviating from this point, the value of f decreases. Therefore, the maximum of f is1, achieved when all x_i are equal. Hence, f(x₁,...,xₙ) ≤1 for all x_i in [0,1] with x_i + y_i=1, which proves the reduced inequality. Therefore, the original inequality holds.Therefore, the proof can be structured as follows:1. Normalize the variables by setting x_i =a_i / (a_i +b_i) and y_i =b_i / (a_i +b_i), handling the case where a_i +b_i=0 separately.2. Show that the original inequality reduces to proving (Product x_i)^{1/n} + (Product y_i)^{1/n} ≤1 under the condition x_i + y_i=1.3. Demonstrate that the maximum of the function f(x₁,...,xₙ) = (Product x_i)^{1/n} + (Product y_i)^{1/n} is 1, achieved when all x_i are equal (hence y_i are equal), using Lagrange multipliers or symmetry arguments.4. Conclude that the inequality holds, hence the original inequality is valid.Another way to think about it is using the concavity of the logarithm. Since the logarithm function is concave, the sum of the logarithms is maximized when the variables are equal, but I think the previous argument is sufficient.Therefore, the final answer is proven by normalizing, reducing to the case where each pair sums to1, and showing through optimization that the maximum of the sum of geometric means is1, achieved when all terms are equal, hence the inequality holds.

✅:To prove the inequality[left(a_{1} a_{2} cdots a_{n}right)^{frac{1}{n}} + left(b_{1} b_{2} cdots b_{n}right)^{frac{1}{n}} leq left(left(a_{1}+b_{1}right)left(a_{2}+b_{2}right) cdots left(a_{n}+b_{n}right)right)^{frac{1}{n}},]for non-negative real numbers (a_i) and (b_i), follow these steps:Step 1: Handle Trivial CasesIf any (a_i + b_i = 0), then both (a_i = 0) and (b_i = 0), making both sides of the inequality zero. Thus, assume (a_i + b_i > 0) for all (i).Step 2: Normalize the VariablesDefine normalized variables:[x_i = frac{a_i}{a_i + b_i}, quad y_i = frac{b_i}{a_i + b_i} quad text{for each } i.]Here, (x_i + y_i = 1) and (x_i, y_i in [0,1]).Step 3: Rewrite the InequalitySubstitute (a_i = x_i(a_i + b_i)) and (b_i = y_i(a_i + b_i)) into the original inequality. Factoring out ((a_i + b_i)) terms and simplifying, the inequality reduces to:[left(prod_{i=1}^n x_iright)^{frac{1}{n}} + left(prod_{i=1}^n y_iright)^{frac{1}{n}} leq 1.]Step 4: Maximize the Sum of Geometric MeansConsider the function (f(x_1, ldots, x_n) = left(prod x_iright)^{1/n} + left(prod (1 - x_i)right)^{1/n}). Using symmetry and Lagrange multipliers, the maximum of (f) occurs when all (x_i) are equal. Setting (x_i = x) for all (i), we find (f(x, ldots, x) = x + (1 - x) = 1).Step 5: Conclude the InequalitySince the maximum value of (f) is 1, it follows that:[left(prod_{i=1}^n x_iright)^{frac{1}{n}} + left(prod_{i=1}^n y_iright)^{frac{1}{n}} leq 1.]Reverting to the original variables confirms the desired inequality.Final Answerboxed{left(a_{1} a_{2} cdots a_{n}right)^{frac{1}{n}} + left(b_{1} b_{2} cdots b_{n}right)^{frac{1}{n}} leq left(left(a_{1}+b_{1}right)left(a_{2}+b_{2}right) cdots left(a_{n}+b_{n}right)right)^{frac{1}{n}}}

🔑:Given non-negative real numbers (a_{1}, a_{2}, ldots, a_{n}) and (b_{1}, b_{2}, ldots, b_{n}), we want to prove the following inequality:left( a_1 a_2 cdots a_n right)^{frac{1}{n}} + left( b_1 b_2 cdots b_n right)^{frac{1}{n}} leq left( (a_1+b_1)(a_2+b_2) cdots (a_n+b_n) right)^{frac{1}{n}}.1. First, assume (a_i + b_i > 0) for every (i).2. By the AM-GM inequality applied to fractions, we get: left( frac{a_1 a_2 cdots a_n}{(a_1 + b_1)(a_2 + b_2) cdots (a_n + b_n)} right)^{frac{1}{n}} leq frac{1}{n} left(frac{a_1}{a_1 + b_1} + frac{a_2}{a_2 + b_2} + cdots + frac{a_n}{a_n + b_n}right). 3. Similarly, left( frac{b_1 b_2 cdots b_n}{(a_1 + b_1)(a_2 + b_2) cdots (a_n + b_n)} right)^{frac{1}{n}} leq frac{1}{n} left(frac{b_1}{a_1 + b_1} + frac{b_2}{a_2 + b_2} + cdots + frac{b_n}{a_n + b_n}right). 4. Adding these two inequalities, we have: [ begin{aligned} &left( frac{a_1 a_2 cdots a_n}{(a_1 + b_1)(a_2 + b_2) cdots (a_n + b_n)} right)^{frac{1}{n}} + left( frac{b_1 b_2 cdots b_n}{(a_1 + b_1)(a_2 + b_2) cdots (a_n + b_n)} right)^{frac{1}{n}} &quad leq frac{1}{n} left(frac{a_1}{a_1 + b_1} + frac{a_2}{a_2 + b_2} + cdots + frac{a_n}{a_n + b_n}right) + frac{1}{n} left(frac{b_1}{a_1 + b_1} + frac{b_2}{a_2 + b_2} + cdots + frac{b_n}{a_n + b_n}right). end{aligned} ]5. Notice that: [ frac{a_i}{a_i + b_i} + frac{b_i}{a_i + b_i} = 1 quad text{for each } i. ]6. Thus, [ frac{a_1}{a_1 + b_1} + frac{b_1}{a_1 + b_1} + frac{a_2}{a_2 + b_2} + frac{b_2}{a_2 + b_2} + cdots + frac{a_n}{a_n + b_n} + frac{b_n}{a_n + b_n} = n. ]7. Consequently, [ frac{1}{n} left( sum_{i=1}^n left( frac{a_i}{a_i + b_i} + frac{b_i}{a_i + b_i} right) right) = frac{1}{n} cdot n = 1. ]8. Therefore, we have: [ left( frac{a_1 a_2 cdots a_n}{(a_1 + b_1)(a_2 + b_2) cdots (a_n + b_n)} right)^{frac{1}{n}} + left( frac{b_1 b_2 cdots b_n}{(a_1 + b_1)(a_2 + b_2) cdots (a_n + b_n)} right)^{frac{1}{n}} leq 1. ]9. Multiplying both sides by (left( (a_1 + b_1)(a_2 + b_2) cdots (a_n + b_n) right)^{frac{1}{n}}), the desired inequality follows: [ left( a_1 a_2 cdots a_n right)^{frac{1}{n}} + left( b_1 b_2 cdots b_n right)^{frac{1}{n}} leq left( (a_1+b_1)(a_2+b_2) cdots (a_n+b_n) right)^{frac{1}{n}}. ] Conclusion:[boxed{ left( a_1 a_2 cdots a_n right)^{frac{1}{n}} + left( b_1 b_2 cdots b_n right)^{frac{1}{n}} leq left( (a_1+b_1)(a_2+b_2) cdots (a_n+b_n) right)^{frac{1}{n}} }]

🧠:Okay, so I need to figure out how many of the numbers p(1), p(2), ..., p(2016) are perfect squares, where p(x) = 2x³ - 3x² + 1. Hmm, let's start by understanding the problem step by step.First, I should recall what a perfect square is. A perfect square is an integer that can be expressed as the square of another integer. So, I need to find all integers k between 1 and 2016 such that p(k) is equal to some integer squared, say m².Maybe I should start by calculating p(x) for small values of x to see if there's a pattern or if I can spot any perfect squares right away. Let's compute p(1), p(2), p(3), etc., and check.Calculating p(1):p(1) = 2(1)³ - 3(1)² + 1 = 2 - 3 + 1 = 0. Hmm, 0 is a perfect square (0² = 0). So that's one.p(2):2(8) - 3(4) + 1 = 16 - 12 + 1 = 5. 5 isn't a perfect square.p(3):2(27) - 3(9) + 1 = 54 - 27 + 1 = 28. Not a square.p(4):2(64) - 3(16) + 1 = 128 - 48 + 1 = 81. Oh, 81 is 9². So that's another one.p(5):2(125) - 3(25) + 1 = 250 - 75 + 1 = 176. 176 is not a square.p(6):2(216) - 3(36) + 1 = 432 - 108 + 1 = 325. Not a square.p(7):2(343) - 3(49) + 1 = 686 - 147 + 1 = 540. Not a square.p(8):2(512) - 3(64) + 1 = 1024 - 192 + 1 = 833. Hmm, 833 isn't a square. 28² is 784, 29² is 841, so no.p(9):2(729) - 3(81) + 1 = 1458 - 243 + 1 = 1216. 1216 isn't a square; 34²=1156, 35²=1225, so no.p(10):2(1000) - 3(100) + 1 = 2000 - 300 + 1 = 1701. Not a square. 41²=1681, 42²=1764, so no.Okay, so up to x=10, we found p(1)=0² and p(4)=9². Let's check maybe a few more to see if there's another one.p(0): Wait, the problem says p(1) to p(2016), so x starts at 1, so x=0 isn't considered.p(11):2(1331) - 3(121) +1 = 2662 - 363 +1 = 2662 - 362 = 2300. 2300 is not a square; 47²=2209, 48²=2304. Close, but 2300 isn't a square.p(12):2(1728) - 3(144) +1 = 3456 - 432 +1 = 3025. Wait, 3025 is 55². Oh! So p(12)=55². That's another one. So x=12 is a hit.p(13):2(2197) -3(169)+1=4394 -507 +1=3888. 3888 isn't a square. 62²=3844, 63²=3969.p(14):2(2744) -3(196)+1=5488 -588 +1=4901. 4901, let's see. 70²=4900, so 4901 is 70² +1. Not a square.p(15):2(3375) -3(225)+1=6750 -675 +1=6076. 6076. 77²=5929, 78²=6084. So between those, not a square.p(16):2(4096) -3(256)+1=8192 -768 +1=7425. 7425. 86²=7396, 87²=7569. Not a square.p(17):2(4913) -3(289)+1=9826 -867 +1=8960. 8960. 94²=8836, 95²=9025. Not a square.p(18):2(5832) -3(324)+1=11664 -972 +1=10693. 10693. 103²=10609, 104²=10816. Not a square.p(19):2(6859) -3(361)+1=13718 -1083 +1=12636. 12636. 112²=12544, 113²=12769. Not a square.p(20):2(8000) -3(400)+1=16000 -1200 +1=14801. 14801. Let's see, 121²=14641, 122²=14884. So between them, not a square.Hmm, so up to x=20, we found p(1)=0², p(4)=9², p(12)=55². Let's check p(21) to see if there's another.p(21):2(9261) -3(441)+1=18522 -1323 +1=17200. 17200. 131²=17161, 132²=17424. Not a square.p(22):2(10648) -3(484)+1=21296 -1452 +1=19845. 19845. 141²=19881, which is higher, so 140²=19600. Not a square.p(23):2(12167) -3(529)+1=24334 -1587 +1=22748. 22748. 150²=22500, 151²=22801. Not a square.p(24):2(13824) -3(576)+1=27648 -1728 +1=25921. Wait, 25921. Let me check if that's a square. 161²=25921. Yes! 161*161=25921. So p(24)=161². That's another one. Wow, so x=24 gives a square.So far, x=1,4,12,24. Let's check p(25):p(25)=2(15625)-3(625)+1=31250 -1875 +1=29376. 29376. Hmm, 171²=29241, 172²=29584. Not a square.p(26):2(17576)-3(676)+1=35152 -2028 +1=33125. 33125. 182²=33124, 183²=33489. So 33125 is 182² +1. Not a square.p(27):2(19683)-3(729)+1=39366 -2187 +1=37180. 37180. 192²=36864, 193²=37249. So between them, not a square.p(28):2(21952)-3(784)+1=43904 -2352 +1=41553. 41553. 203²=41209, 204²=41616. Not a square.p(29):2(24389)-3(841)+1=48778 -2523 +1=46256. 46256. 215²=46225, 216²=46656. Not a square.p(30):2(27000)-3(900)+1=54000 -2700 +1=51301. 51301. 226²=51076, 227²=51529. Not a square.Hmm, so up to x=30, we have four instances where p(x) is a perfect square: x=1,4,12,24. Let's check x=40, maybe?But this might take too long if I keep checking each one. Maybe there's a pattern here. Let's see the values where we found squares: x=1,4,12,24. Each time, x seems to be doubling? 1, then 4 (1*4), 12 (4*3), 24 (12*2). Hmm, not exactly a clear pattern. Alternatively, maybe the ratio between consecutive x's: 4/1=4, 12/4=3, 24/12=2. So the ratio is decreasing by 1 each time. If that continued, the next would be 24*1=24, but p(24) is already accounted for. Maybe the pattern breaks here.Alternatively, maybe the differences between x's: 4-1=3, 12-4=8, 24-12=12. The differences are 3,8,12. Not obvious.Alternatively, maybe the p(x) values: 0², 9², 55², 161². Let's see the sequence of m: 0,9,55,161. Let's check the differences between these m's:9 - 0 = 955 - 9 = 46161 - 55 = 106Differences between differences:46 - 9 = 37106 - 46 = 60Hmm, not a clear pattern here either. Alternatively, maybe each m is related to the previous by some multiplicative factor. Let's see:9 / 0 is undefined.55 /9 ≈6.111...161 /55≈2.927...Not a multiplicative pattern. Alternatively, maybe each term is generated by a recurrence relation. Let's see:If we suppose m_{n+1} = a*m_n + b. Let's check:For n=1: m1=0, m2=9. So 9 = a*0 + b => b=9.Next, m3=55. 55 = a*9 +9 => 55-9=46=9a => a=46/9≈5.111. Not integer.Then m4=161= a*55 +9. 161-9=152=55a => a=152/55≈2.76. Not consistent. So that approach might not work.Alternatively, perhaps these m's correspond to some known sequence or relate to the x values in a particular way. Let's see:For x=1, m=0.x=4, m=9. Hmm, 9 is 3². But x=4, m=9=3*3.x=12, m=55. 55 is a Fibonacci number, but not sure if relevant.x=24, m=161. Hmm, 161 is 7*23. Not obvious.Alternatively, maybe m is related to x via some formula. Let's check:For x=1, m=0. Let's see if 0 relates to 1 somehow. Maybe m = x² -1? For x=1, 1² -1=0. That works. For x=4, 4² -1=15, but m=9. Doesn't work.Alternatively, m = (x² - x)/something. For x=1: (1 -1)/something=0, so maybe. For x=4: (16 -4)/something=12/something=9. So 12/ (4/3)=9. Not sure.Alternatively, let's look at the p(x) expression itself: p(x)=2x³ -3x² +1. Maybe factor this polynomial.Let me try to factor p(x). Let's see if it can be factored.Looking for rational roots using Rational Root Theorem. Possible roots are ±1, since the factors of the constant term (1) divided by the leading coefficient (2) are ±1, ±1/2.Testing x=1: p(1)=0, as we saw. So (x -1) is a factor.Using polynomial division or synthetic division to factor out (x -1).Divide 2x³ -3x² +1 by (x -1):Using synthetic division:Coefficients: 2 | -3 | 0 | 1 (Wait, but p(x)=2x³ -3x² +1, which is 2x³ -3x² +0x +1. So coefficients are 2, -3, 0, 1.Divide by (x -1):1 | 2 -3 0 1 2 -1 -1 ---------------- 2 -1 -1 0So the quotient is 2x² - x -1. Therefore, p(x) factors as (x -1)(2x² - x -1).Now, factor 2x² - x -1. Let's see if that's possible.Looking for two numbers a and b such that a*b = 2*(-1)=-2 and a + b = -1.Looking... -2 and 1: -2*1=-2, and -2 +1=-1. Yes.So split the middle term:2x² -2x + x -1Group:(2x² -2x) + (x -1) = 2x(x -1) +1(x -1) = (2x +1)(x -1)Thus, p(x) factors as (x -1)(2x +1)(x -1) = (x -1)²(2x +1).So p(x) = (x -1)²(2x +1). Interesting. So that's the factorization.Therefore, p(x) = (x -1)²(2x +1). So we can write p(x) as [ (x -1) ]² multiplied by (2x +1). Therefore, p(x) is a square times (2x +1). For p(x) to be a perfect square, the remaining factor (2x +1) must also be a square. Because if we have a square multiplied by another square, then the result is a square. If (2x +1) is a square, then p(x) would be a square times a square, hence a square.But (x -1)² is already a square, so if (2x +1) is a square, then the whole product is a square. However, it's possible that (2x +1) is not a square itself, but when multiplied by (x -1)², the product is a square. Wait, but since (x -1)² is a square, then the entire expression would be a square only if (2x +1) is a square. Because the product of a square and a non-square can't be a square. Unless (2x +1) is a square times a square, but that's redundant.Wait, let me think. Let me formalize this.Let p(x) = A² * B, where A = (x -1), and B = (2x +1). For p(x) to be a perfect square, B must be a perfect square. Because A² is already a square, so A² * B is a square if and only if B is a square. Because if B is a square, say C², then A² * C² = (A*C)². If B is not a square, then A² * B can't be a square because B isn't a square. Therefore, the necessary and sufficient condition for p(x) to be a square is that 2x +1 is a perfect square.Therefore, 2x +1 = m², where m is an integer. Then x = (m² -1)/2. Since x must be an integer between 1 and 2016, we need (m² -1)/2 to be an integer in that range.First, (m² -1) must be even, so m² must be odd. Since m² is odd, m must be odd. Let m = 2k +1, where k is a non-negative integer.Then, m² = (2k +1)² = 4k² +4k +1. Then, 2x +1 = 4k² +4k +1 ⇒ 2x = 4k² +4k ⇒ x = 2k² +2k.Therefore, x must be of the form x = 2k² +2k, where k is a non-negative integer. So x is generated by this quadratic in k. Then, we need x to be between 1 and 2016.So x = 2k² +2k ≥1. Let's find the possible k's such that 2k² +2k ≤2016.Let's solve 2k² +2k ≤2016 ⇒ k² +k ≤1008 ⇒ k² +k -1008 ≤0.Solving the quadratic equation k² +k -1008 =0:Using quadratic formula: k = [-1 ±√(1 +4032)]/2 = [-1 ±√4033]/2.Compute √4033. Let's see, 63²=3969, 64²=4096. So √4033 is between 63 and 64. Let's compute 63²=3969, 63.5²= (63 +0.5)^2=63² +63 +0.25=3969 +63 +0.25=4032.25. Oh, 63.5²=4032.25, which is very close to 4033. So √4033≈63.5 + (4033 -4032.25)/(2*63.5)=63.5 +0.75/127≈63.5 +0.0059≈63.5059.Therefore, k = [-1 +63.5059]/2 ≈62.5059/2≈31.2529. So the positive root is approximately 31.25. Therefore, k must be ≤31.25, so the maximum integer k is 31.Therefore, k can be 0,1,2,...,31. Let's check k=31:x=2*(31)^2 +2*(31)=2*961 +62=1922 +62=1984. 1984 ≤2016. Okay. k=31 is allowed.k=32: x=2*(32)^2 +2*32=2*1024 +64=2048 +64=2112. 2112>2016. So k=32 is too big.Therefore, k can be from 0 to 31. But we need to check x ≥1. Let's check k=0: x=0. But x starts at 1. So k=0 gives x=0, which is invalid. Therefore, k starts from 1? Wait, no. Wait x=2k² +2k.Wait for k=0: x=0. Which is not in our range. So the first valid k is k=1, giving x=2(1)^2 +2(1)=4. Wait, but when k=0, x=0, which is p(0)=2*0 -3*0 +1=1, but x=0 is excluded. Wait, but earlier, when we checked p(1), we found x=1 gives p(1)=0, which is a square. Wait, but according to our analysis here, x=1 is not covered by the formula x=2k² +2k. Because if we set x=1, then 1=2k² +2k. Let's solve for k: 2k² +2k -1=0. Using quadratic formula: k = [-2 ±√(4 +8)]/4 = [-2 ±√12]/4 = [-2 ±2√3]/4 = (-1 ±√3)/2. Which is not integer. So x=1 is not captured by this formula. But p(1)=0 is a perfect square. So there's an exception here. Because in our earlier factorization, p(x)=(x -1)^2*(2x +1). For x=1, this becomes 0^2*3=0, which is a square. However, in this case, (2x +1)=3, which is not a square, but since (x -1)^2=0, the entire product is zero, which is a square regardless. Therefore, the case x=1 is a special case where (x -1)^2=0, so even though (2x +1)=3 isn't a square, the product is 0, which is a square. Therefore, apart from the solutions given by x=2k² +2k (which gives x=4,12,24,...,1984), we also have x=1. So total solutions are x=1 and x=2k² +2k for k=1 to 31.Wait, let's verify that. If k starts at 1:k=1: x=2(1)^2 +2(1)=2 +2=4.k=2: x=2(4)+4=8 +4=12.k=3: 2(9)+6=18 +6=24.k=4: 2(16)+8=32 +8=40. Wait, but earlier when we checked x=40, we didn't compute p(40). Let's check p(40):p(40)=2*(40)^3 -3*(40)^2 +1=2*64000 -3*1600 +1=128000 -4800 +1=123201. Let's see if 123201 is a square. 350²=122500, 351²=123201. Yes! 351²=123201. So p(40)=351². So x=40 is another solution. Therefore, the formula x=2k² +2k for k=1 gives x=4, k=2 gives x=12, k=3 gives x=24, k=4 gives x=40, etc., each corresponding to m=9,55,161,351,... So these are valid. Therefore, x=1 is a separate case.Therefore, total number of squares is 1 (from x=1) plus the number of k from 1 to 31 (since k=31 gives x=1984), so 31 values. Therefore total is 1 +31=32?Wait, but wait, when k=0 gives x=0, which is excluded. So k=1 to 31 gives x=4,12,24,...,1984. That's 31 values. Adding x=1 gives total 32. But wait, when k=1, x=4. So k=1 to 31 gives 31 x's. So total solutions would be 31 +1=32. But let's check if x=1 is the only other solution.Yes, because for x=1, p(x)=0², which is a square. For other x>1, p(x)= (x -1)^2*(2x +1). For x>1, (x -1)^2 is positive, and (2x +1) is positive. So for p(x) to be a square, (2x +1) must be a square, as (x -1)^2 is already a square. Therefore, except for x=1 where p(x)=0, all other solutions must satisfy 2x +1 being a perfect square, which leads to x=2k² +2k.Thus, the total number of squares is 1 (from x=1) plus the number of k where x=2k² +2k is ≤2016. As found earlier, k can be from 1 to 31, giving 31 values. So total is 1 +31=32. But wait, when k=31, x=2*(31)^2 +2*31=2*961 +62=1922 +62=1984. Then k=32 would give x=2112, which is over 2016, so excluded. Therefore, 31 k's from 1 to 31. So total solutions are 32.But wait, earlier when we computed up to x=24, we found x=1,4,12,24 as solutions. Then x=40 would be the next, then x=60, etc. Wait, but let's check k=4:k=4: x=2*(16) +8=40. p(40)=351².k=5: x=2*25 +10=60. Let's check p(60):p(60)=2*(60)^3 -3*(60)^2 +1=2*216000 -3*3600 +1=432000 -10800 +1=421201. Check if this is a square. 649²=421,201. Yes, 649*649=421,201. So p(60)=649². So yes, x=60 is a solution. Similarly, k=5 gives x=60.So each k from 1 to 31 gives a valid x=2k² +2k ≤2016, and x=1 is another solution. Therefore, total number is 32.But wait, let me confirm the count from k=1 to k=31:For k=1: x=4k=2: x=12k=3: x=24k=4: x=40k=5: x=60...k=31: x=1984Yes, each k from 1 to 31 gives a unique x. So 31 values. Plus x=1. Total 32.But wait, earlier when we computed up to x=24, we found x=1,4,12,24. So 4 instances. But according to this formula, k=1 gives x=4, k=2 gives x=12, k=3 gives x=24, k=4 gives x=40, etc. So up to k=3, x=24. Then k=4 gives x=40, which is beyond x=24 but within 2016. So yes, the count is correct.Therefore, the answer should be 32. But wait, let's check for k=31:x=2*(31)^2 +2*31=2*961 +62=1922 +62=1984. Then p(1984)= (1984 -1)^2*(2*1984 +1)=1983² * 3969. Wait, but according to our previous logic, p(x)= (x-1)^2*(2x+1). For x=1984, 2x+1=3969. Wait, 3969 is 63²*1. Wait, 63²=3969. So 2x+1=63². Therefore, p(x)= (x-1)^2 *63²= (63*(x-1))². Therefore, p(x) is indeed a square. So x=1984 gives p(x)= (63*1983)². Therefore, that's a perfect square.So each x=2k² +2k for k=1 to 31 gives a perfect square, and x=1 also gives a perfect square. So total 32.But let's verify with the earlier examples. For k=1: x=4, m=9 (from p(4)=81=9²). Then according to the formula, m=(x -1)*sqrt(2x +1). Wait, p(x)= (x -1)^2*(2x +1)= [ (x -1)*sqrt(2x +1) ]². Therefore, m=(x -1)*sqrt(2x +1). For x=4, sqrt(2*4 +1)=sqrt(9)=3. So m=(4 -1)*3=9, correct. Similarly, for x=12: sqrt(2*12 +1)=sqrt(25)=5. m=(12 -1)*5=55, which matches. For x=24: sqrt(49)=7, m=(24 -1)*7=161, correct. For x=40: sqrt(81)=9, m=(40 -1)*9=351, correct. For x=60: sqrt(121)=11, m=(60 -1)*11=649, correct. So the formula holds.Therefore, the total number of squares is 1 (x=1) +31 (k=1 to31) =32.But let's make sure there are no other solutions. Suppose someone claims that there's another x where p(x) is a square, outside of x=1 and x=2k² +2k. According to our factorization, p(x)= (x -1)^2*(2x +1). For this to be a square, either:1. (x -1)^2=0, which gives x=1, leading to p(x)=0, which is a square.2. (x -1)^2*(2x +1) is a square, which requires that (2x +1) is a square, since (x -1)^2 is already a square. Hence, 2x +1 must be a square. So the only solutions are x=1 and x=(m² -1)/2, where m is an odd integer ≥3 (since for m=1, x=0, which is excluded). Since m must be odd, we set m=2k +1, leading to x=2k² +2k.Therefore, these are the only solutions. Hence, the total number is indeed 1 +31=32.But wait, let's check when k=31, x=1984. Then k=31 gives m=2*31 +1=63. Then 2x +1=63²=3969. Then x=(63² -1)/2=(3969 -1)/2=3968/2=1984. Correct.Thus, the total number of squares is 32.However, earlier when I computed up to x=24, I found x=1,4,12,24. Then at x=40,60,...,1984, there are more. So according to this, there are 32 instances. But let's verify with smaller counts.For k=1 to31, that's 31 numbers. Plus x=1. So 32.But let's count how many k's are there from 1 to31. Yes, 31 k's. Each corresponds to x=2k² +2k. So 31 x's. Plus x=1. 32 total.Therefore, the final answer is 32. So the number of squares among p(1),...,p(2016) is 32.Final AnswerThe number of squares among the numbers ( p(1), p(2), ldots, p(2016) ) is boxed{32}.

✅:To determine how many squares of integers are among the numbers ( p(1), p(2), ldots, p(2016) ) where ( p(x) = 2x^3 - 3x^2 + 1 ), we start by analyzing the polynomial.First, we factorize ( p(x) ):[p(x) = 2x^3 - 3x^2 + 1 = (x - 1)^2(2x + 1)]For ( p(x) ) to be a perfect square, the factor ( 2x + 1 ) must be a perfect square because ( (x - 1)^2 ) is already a perfect square. Let ( 2x + 1 = m^2 ) where ( m ) is an integer. Solving for ( x ):[x = frac{m^2 - 1}{2}]Since ( x ) must be an integer, ( m ) must be odd. Let ( m = 2k + 1 ) for non-negative integers ( k ). Substituting ( m ):[x = frac{(2k + 1)^2 - 1}{2} = 2k^2 + 2k]We need ( x ) to be in the range ( 1 leq x leq 2016 ). Solving for ( k ):[2k^2 + 2k leq 2016 implies k^2 + k leq 1008]Solving the quadratic inequality ( k^2 + k - 1008 leq 0 ) using the quadratic formula:[k = frac{-1 pm sqrt{1 + 4 cdot 1008}}{2} = frac{-1 pm sqrt{4033}}{2}]Approximating ( sqrt{4033} approx 63.5 ), we find ( k leq 31 ). Thus, ( k ) ranges from 1 to 31, giving 31 values of ( x ).Additionally, ( x = 1 ) is a special case where ( p(1) = 0 ), which is a perfect square. Therefore, we have 31 values from ( k = 1 ) to ( 31 ) plus ( x = 1 ), resulting in a total of 32 perfect squares.[boxed{32}]

🔑:Let's start by analyzing the given polynomial ( p(x) = 2x^3 - 3x^2 + 1 ).1. Factoring the Polynomial: [ p(x) = 2x^3 - 3x^2 + 1 = (x-1)^2(2x+1) ]2. Identifying the Expression to be a Perfect Square: For ( p(x) ) to be a perfect square, either: - ( x = 1 ) (which makes ( p(1) = 0 ), a perfect square as ( 0^2 )) - ( (x - 1)^2 (2x + 1) ) is a perfect square for ( x geq 2 ) For ( p(x) ) to be a perfect square when ( x geq 2 ), ( 2x + 1 ) must itself be a perfect square. Let us solve the inequality and constraints for ( 2x + 1 ).3. Finding the Range for ( x geq 2 ): - When ( x = 2 ): [ 2x + 1 = 2(2) + 1 = 5 ] - When ( x = 2016 ): [ 2x + 1 = 2(2016) + 1 = 4033 ] Therefore, the value of ( 2x + 1 ) ranges from 5 to 4033.4. Identifying Perfect Squares Within the Range: Odd perfect squares within the range ([5, 4033]) are calculated as: [ 3^2, 5^2, 7^2, ldots, 63^2 ]5. Counting the Perfect Squares: The sequence of odd squares starting from ( 3^2 = 9 ) and ending at ( 63^2 = 3969 ): [ 3^2, 5^2, 7^2, ldots, 63^2 ] These are consecutive odd numbers starting from 3 to 63. The total number of odd numbers in this range is: [ left(frac{63 - 3}{2}right) + 1 = frac{60}{2} + 1 = 30 + 1 = 31 ] Thus, there are 31 numbers in this sequence that satisfy the condition.6. Including the Special Case ( x = 1 ): - As noted earlier, ( p(1) = 0 ) is also a perfect square. Therefore, the total number of perfect squares we can obtain from ( p(x) ) in the given range is: [ 31 + 1 = 32 ] Conclusion:Hence, the final count of perfect squares among the numbers ( p(1), p(2), ldots, p(2016) ) is (boxed{32}).

🧠:Okay, let's see. The problem is about a shepherd using 15 sections of 2-meter-long fences to form a square or rectangular sheep pen against a wall. We need to find the maximum area possible, and the options are given. Hmm.First, let me parse the problem. So, the shepherd is creating a pen next to a wall. That means the pen will only need fencing on three sides, right? Because the wall acts as the fourth side. If it were a free-standing pen, you'd need fencing on all four sides, but since it's against a wall, one side doesn't need fencing. So the total fencing used would be for two widths and one length, assuming the wall is along the length. Wait, but actually, it depends on how the shepherd arranges it. The wall could be along the length or the width. Wait, maybe it's better to think in terms of variables.Let me denote the sides. Let's say the side parallel to the wall is the length, L, and the sides perpendicular to the wall are the widths, W. Then, the total fencing needed would be L + 2W. Since each section is 2 meters, and there are 15 sections, the total fencing length is 15 * 2 = 30 meters. So L + 2W = 30 meters. Then, the area is L * W. We need to maximize A = L * W given that L + 2W = 30.So the problem reduces to maximizing A = L * W with the constraint L + 2W = 30. Let's express L in terms of W: L = 30 - 2W. Substitute into A: A = (30 - 2W) * W = 30W - 2W². This is a quadratic equation in terms of W, which opens downward, so the maximum is at the vertex. The vertex of a quadratic aW² + bW + c is at W = -b/(2a). Here, a = -2, b = 30, so W = -30/(2*(-2)) = -30 / (-4) = 7.5. So W = 7.5 meters. Then L = 30 - 2*7.5 = 30 - 15 = 15 meters. Therefore, the maximum area is 15 * 7.5 = 112.5 square meters. Hmm, but the options given are 100, 108, 112, 122. 112.5 is close to 112, but not exactly. Maybe there's a mistake here?Wait, perhaps I made a wrong assumption. Let me check again. The fencing is 15 sections, each 2 meters. So total fencing is 15 * 2 = 30 meters. If the pen is against the wall, then fencing is used for two widths and one length. So yes, L + 2W = 30. Then, area is L*W. The maximum area when L and W are such that the product is maximized. So solving through calculus or the vertex of the quadratic. Wait, the calculation seems correct. Then, the maximum area is 112.5. But the answer options don't include 112.5. The closest is C. 112. So maybe they expect rounding down? Or perhaps there's a different constraint.Wait, but maybe the problem requires the pen to have integer lengths? If that's the case, then W must be an integer, since each section is 2 meters. Wait, the sections are 2 meters each. So each fence section is 2 meters. So the length and width must be multiples of 2 meters? Wait, maybe not. Wait, the shepherd is using 15 sections of 2-meter fences. So the total fencing is 15 * 2 = 30 meters. But the individual sides can be any length, as long as the sum is 30 meters. Wait, but no, each section is a separate piece. So maybe the length and the widths have to be made up of whole sections? For example, if each side is constructed by connecting sections, then each side must be a multiple of 2 meters. So L and W must be even numbers? Because each fence section is 2 meters, and you can't have half a section. Wait, the problem doesn't specify that the sides have to be integer numbers of sections. It just says 15 sections of 2 meters each. So total fencing is 30 meters, and you can arrange it as you want. So maybe the sides can be any length as long as L + 2W = 30. So they don't have to be even numbers. So the calculation of 112.5 is correct, but since the answer options are integers, maybe we need to check if 112.5 is an option. It's not. The closest is 112. So maybe the answer is C. 112. But why isn't 112.5 an option?Alternatively, perhaps there's a different way to model the problem. Wait, maybe the wall is along the length, so fencing is required for two widths and one length, as I thought. But maybe if the wall is along the width, then fencing is required for two lengths and one width. Wait, but the problem says it's a square or rectangular pen. So the shape could be either way. Wait, but the maximum area would be achieved when the side parallel to the wall is longer. So perhaps regardless, the formula would be similar.Wait, let's check. Suppose the wall is along the width, then fencing is required for 2L + W = 30. Then, area is L * W = L*(30 - 2L). Then, A = 30L - 2L². The vertex here would be at L = -30/(2*(-2)) = 7.5. Then W = 30 - 2*7.5 = 15. So area would be 7.5*15 = 112.5 again. So either way, the maximum area is 112.5. So regardless of which side is along the wall, the maximum area is the same. So that's consistent.But the answer options don't have 112.5, so perhaps there's a mistake in the problem's options, or maybe my initial assumption is wrong. Alternatively, maybe the problem requires that the number of sections used per side must be an integer. Wait, because each section is 2 meters, so each side's length must be a multiple of 2. So L and W must be even numbers. So if that's the case, then L and W have to be even. So in that case, W can't be 7.5 meters, which is 15/2. So perhaps we need to adjust.If the lengths have to be multiples of 2 meters, then W has to be an integer number of meters times 2. Wait, no. If each section is 2 meters, but you can combine sections. Wait, if you have a length that's 15 meters, that would be 7.5 sections of 2 meters. But you can't have half a section. So maybe each side must consist of a whole number of sections. So each side's length is 2 * number of sections. Therefore, L and W must be even numbers. So if that's the case, then we can't have 7.5 meters for W. So W must be even. So in that case, we need to find integer values of W (in meters) such that 2W is even and L = 30 - 2W is also even. Wait, 30 is even, 2W is even, so L = 30 - 2W is even. So L must be even. Therefore, both L and W are even numbers.So in that case, let's parameterize with sections. Let me think in terms of sections. Each section is 2 meters. So total fencing is 15 sections. But the total fencing used is L + 2W. Wait, but L and W are in meters. If we model in terms of sections, let's say that the number of sections used for length is l and for each width is w. Then, total sections used would be l + 2w = 15. Each l section corresponds to 2 meters, so the actual length is 2l meters. Similarly, each width is 2w meters. Then, the area would be (2l) * (2w) = 4lw. But we have the constraint l + 2w = 15. So we need to maximize 4lw with l + 2w = 15. Let's solve this.Express l = 15 - 2w. Then area = 4*(15 - 2w)*w = 4*(15w - 2w²) = 60w - 8w². To maximize this quadratic, the vertex is at w = -b/(2a) = -60/(2*(-8)) = 60/16 = 3.75. But w has to be an integer because the number of sections can't be fractional. So w can be 3 or 4.If w = 3, then l = 15 - 2*3 = 9. Area = 4*9*3 = 108.If w = 4, then l = 15 - 2*4 = 7. Area = 4*7*4 = 112.If w = 3.75, it's not allowed. So between w=3 and w=4, the maximum area is 112. So that's option C. 112. So in this case, the answer is 112. Because the problem might be considering that each side must be composed of whole sections, so the lengths must be integer multiples of 2 meters. Therefore, the maximum area is 112.So that would make sense. The initial approach didn't consider that the fence sections have to be whole numbers, leading to 112.5, but when considering the sections as discrete units, the maximum area becomes 112. So the answer is C. 112.But wait, let's confirm. If each side must be made up of whole sections, then yes, the number of sections must be integers. So the problem states "uses 15 sections of 2-meter-long fences". So the total number of sections used is 15. So when arranging the pen, the shepherd uses 15 sections in total. So if the pen has a length and two widths, then the number of sections for the length plus twice the number of sections for the width equals 15. Each section is 2 meters, so the length in meters is 2 times the number of sections, same for the width.Therefore, the problem is actually a discrete optimization problem where the number of sections for length (l) and width (w) must satisfy l + 2w = 15. Then, the area in square meters is (2l) * (2w) = 4lw. So maximize 4lw with l + 2w =15, l and w non-negative integers.So solving for l =15 - 2w. Then, substituting into area: 4*(15 - 2w)*w =4*(15w -2w²). To maximize 15w -2w². This is a quadratic in w, which would have maximum at w = 15/(4). But since w must be integer, we check w=3 and w=4.At w=3: 15*3 -2*(9)=45-18=27. Multiply by 4: 108.At w=4:15*4 -2*(16)=60-32=28. Multiply by 4:112.At w=3.75, which is not allowed. So indeed, the maximum is 112. So the answer is C. 112.Therefore, the initial approach didn't consider the integer constraint on the number of sections, leading to a higher area, but when considering the sections must be whole numbers, the correct answer is 112.Alternatively, if the problem allows for cutting the sections, which is not usual for fence sections, then the maximum area would be 112.5, but since that's not an option, and the given options include 112, which is achievable with whole sections, then 112 is the correct answer.

✅:To determine the maximum area of the sheep pen, we start by noting that the shepherd uses 15 sections of 2-meter fencing, totaling 30 meters. The pen is rectangular or square and is built against a wall, so fencing is required for one length (L) and two widths (W). This gives the constraint:[L + 2W = 30 quad text{(meters)}.]The area ( A ) of the pen is ( A = L times W ). Substituting ( L = 30 - 2W ) into the area formula:[A = (30 - 2W) times W = 30W - 2W^2.]This quadratic equation has a maximum at ( W = frac{-b}{2a} = frac{30}{4} = 7.5 , text{m} ). Substituting ( W = 7.5 , text{m} ):[L = 30 - 2(7.5) = 15 , text{m}, quad A = 15 times 7.5 = 112.5 , text{m}^2.]However, fence sections must be whole numbers (each section is 2 meters). Let ( l ) and ( w ) represent the number of sections used for length and width, respectively. The constraint becomes:[l + 2w = 15 quad text{(sections)},]with actual lengths ( L = 2l , text{m} ) and ( W = 2w , text{m} ). The area is:[A = 4lw.]Substituting ( l = 15 - 2w ):[A = 4(15 - 2w)w = 60w - 8w^2.]Maximizing ( A ), the optimal integer values are ( w = 4 ) (sections) and ( l = 15 - 2(4) = 7 ) (sections), yielding:[L = 2 times 7 = 14 , text{m}, quad W = 2 times 4 = 8 , text{m}, quad A = 14 times 8 = 112 , text{m}^2.]Thus, the maximum achievable area with whole sections is 112 square meters.Answer: boxed{C}

🔑:We are given 15 segments each of length 2 meters to form a rectangular sheep pen with one side against a wall. Our task is to find the maximum area that can be enclosed by the pen.1. Total length of the segments used: [ text{Total length} = 15 times 2 = 30 text{ meters} ]2. Forming the pen against the wall: - Let the length of the pen be (l) meters. - Let the width of the pen be (w) meters. - Since one side is against the wall, the total length of fencing used is (l + 2w).3. Equating the total length: [ l + 2w = 15 text{ segments} ] [ l + 2w = 15 times 2 = 30 text{ meters} ] Thus, [ l = 30 - 2w ]4. Maximizing the area of the rectangle: [ text{Area} = l times w = (30 - 2w) times w ] [ A(w) = 30w - 2w^2 ]5. Maximizing the quadratic function: The quadratic function (A(w) = 30w - 2w^2) is of the form (ax^2 + bx + c). The maximum value occurs at the vertex of the parabola, which is given by: [ w = -frac{b}{2a} ] Here, (a = -2) and (b = 30). [ w = -frac{30}{2(-2)} = frac{30}{4} = 7.5 ]6. Calculating the maximum length: For (w = 7.5): [ l = 30 - 2(7.5) = 30 - 15 = 15 text{ meters} ] Checking practical limits and integer values: To keep our calculation within practical segment usage and ensuring integer lengths: - (w = 1 text{ segment}): [ l = 15 - 2 times 1 = 13 text{ segments} ] [ text{Area} = (13 times 2) times (1 times 2) = 26 text{ meters} times 2 text{ meters} = 52 text{ square meters} ] - (w = 2 text{ segments}): [ l = 15 - 2 times 2 = 11 text{ segments} ] [ text{Area} = (11 times 2) times (2 times 2) = 22 text{ meters} times 4 text{ meters} = 88 text{ square meters} ] - (w = 3 text{ segments}): [ l = 15 - 2 times 3 = 9 text{ segments} ] [ text{Area} = (9 times 2) times (3 times 2) = 18 text{ meters} times 6 text{ meters} = 108 text{ square meters} ] - (w = 4 text{ segments}): [ l = 15 - 2 times 4 = 7 text{ segments} ] [ text{Area} = (7 times 2) times (4 times 2) = 14 text{ meters} times 8 text{ meters} = 112 text{ square meters} ] - (w = 5 text{ segments}): [ l = 15 - 2 times 5 = 5 text{ segments} ] [ text{Area} = (5 times 2) times (5 times 2) = 10 text{ meters} times 10 text{ meters} = 100 text{ square meters} ] - (w = 6 text{ segments}): [ l = 15 - 2 times 6 = 3 text{ segments} ] [ text{Area} = (3 times 2) times (6 times 2) = 6 text{ meters} times 12 text{ meters} = 72 text{ square meters} ] - (w = 7 text{ segments}): [ l = 15 - 2 times 7 = 1 text{ segment} ] [ text{Area} = (1 times 2) times (7 times 2) = 2 text{ meters} times 14 text{ meters} = 28 text{ square meters} ]Given these calculations, the maximum area found is when (w = 4).Conclusion:[boxed{C}]

🧠:Alright, let's try to tackle this problem step by step. So, we have a parabola given by the equation ( y = x^2 - 4x - 1 ). There are three distinct points on this parabola: A, B, and C. We know that the x-coordinate of point C is 5, and the y-coordinates of points A and B are the same. We need to find the x-coordinate of the intersection of the medians of triangle ABC. First, let me recall that the intersection of the medians of a triangle is called the centroid. The centroid's coordinates are the average of the coordinates of the three vertices. So, if we can find the coordinates of points A, B, and C, then we can find the centroid by averaging their x-coordinates and y-coordinates separately. The problem specifically asks for the x-coordinate of the centroid, which would be ( frac{x_a + x_b + x_c}{3} ). Since we know ( x_c = 5 ), if we can find ( x_a ) and ( x_b ), we can compute this average.But we don't know ( x_a ) and ( x_b ) directly. However, we do know that points A and B lie on the parabola and have the same y-coordinate. Let's denote their common y-coordinate as ( y_0 ). So, for both points A and B, the equation ( y_0 = x^2 - 4x - 1 ) must hold. This implies that ( x_a ) and ( x_b ) are the roots of the quadratic equation ( x^2 - 4x - 1 = y_0 ). Rearranging this equation gives ( x^2 - 4x - (1 + y_0) = 0 ). Let's call this equation (1).Since this quadratic equation has two roots ( x_a ) and ( x_b ), by Vieta's formulas, the sum of the roots is equal to ( -frac{b}{a} ) where the quadratic is in the form ( ax^2 + bx + c = 0 ). In this case, the quadratic equation is ( x^2 - 4x - (1 + y_0) = 0 ), so the sum of the roots ( x_a + x_b = 4 ). That's a crucial piece of information because it gives us the sum of ( x_a ) and ( x_b ) without needing to know ( y_0 ).Therefore, even though we don't know the specific values of ( x_a ) and ( x_b ), we do know that their sum is 4. Therefore, when we compute the centroid's x-coordinate, which is ( frac{x_a + x_b + x_c}{3} ), we can substitute ( x_a + x_b = 4 ) and ( x_c = 5 ). That gives ( frac{4 + 5}{3} = frac{9}{3} = 3 ). So, the x-coordinate of the centroid is 3. Wait, but hold on, that seems too straightforward. Is there something missing here?Let me verify. The problem states that points A, B, and C are distinct. Since ( x_c = 5 ), we need to ensure that points A and B are distinct from C. Also, since A and B have the same y-coordinate, they must be distinct points on the parabola. Let's confirm that with the given information.Given that ( x_a ) and ( x_b ) are roots of equation (1), which is ( x^2 - 4x - (1 + y_0) = 0 ). For the roots to be distinct, the discriminant must be positive. The discriminant D is ( (-4)^2 - 4 times 1 times (-1 - y_0) = 16 + 4(1 + y_0) = 16 + 4 + 4y_0 = 20 + 4y_0 ). For the quadratic to have two distinct roots, we need ( 20 + 4y_0 > 0 ), which simplifies to ( y_0 > -5 ). Therefore, as long as the common y-coordinate of points A and B is greater than -5, they are distinct. However, since the problem states that all three points are distinct, we also need to ensure that points A and B do not coincide with point C.Point C has an x-coordinate of 5. Let's find its y-coordinate. Plugging ( x = 5 ) into the parabola equation: ( y = 5^2 - 4 times 5 - 1 = 25 - 20 - 1 = 4 ). So, point C is (5, 4). For points A and B, their coordinates are ( (x_a, y_0) ) and ( (x_b, y_0) ). To ensure they are distinct from C, neither ( x_a ) nor ( x_b ) can be 5, and ( y_0 ) cannot be 4. However, if ( x_a ) or ( x_b ) were 5, then substituting x=5 into equation (1) would give ( 5^2 - 4 times 5 - (1 + y_0) = 0 ), which simplifies to 25 - 20 -1 - y_0 = 4 - y_0 = 0, so y_0 = 4. Therefore, if y_0 = 4, one of the roots would be x=5. But since points A and B must be distinct from C, we must have y_0 ≠ 4. Hence, the quadratic equation (1) must not have 5 as a root, so y_0 cannot be 4. Therefore, as long as y_0 ≠ 4, points A and B are distinct from C. Given that the problem states all three points are distinct, we can assume that y_0 ≠ 4, so our earlier conclusion that the centroid's x-coordinate is 3 holds. But wait, maybe there's another angle here. Let me check again.The centroid's x-coordinate is the average of the x-coordinates of A, B, and C. Since we know that x_a + x_b = 4 (from Vieta's formula) and x_c = 5, then the average is (4 + 5)/3 = 9/3 = 3. Therefore, regardless of the specific values of x_a and x_b, as long as they satisfy x_a + x_b = 4, the centroid's x-coordinate will be 3. This seems to be a general result.Is there a possible mistake in assuming that Vieta's formula applies here? Let's verify. The quadratic equation for x when y = y_0 is x^2 -4x - (1 + y_0) = 0. The sum of roots is indeed 4, which is correct because for a quadratic ax² + bx + c = 0, the sum of roots is -b/a. Here, a = 1, b = -4, so sum is -(-4)/1 = 4. Correct.Therefore, even without knowing y_0, the sum of x_a and x_b is 4, so the centroid's x-coordinate is (4 + 5)/3 = 3. Therefore, the answer is 3.But let me cross-validate this with an example. Suppose we choose specific points A and B with the same y-coordinate, compute their coordinates, find point C, compute the centroid, and check the x-coordinate.Let me pick a y_0, say y_0 = 0. Then equation (1) becomes x² -4x -1 = 0. The roots are [4 ± sqrt(16 + 4)]/2 = [4 ± sqrt(20)]/2 = [4 ± 2*sqrt(5)]/2 = 2 ± sqrt(5). So x_a = 2 + sqrt(5), x_b = 2 - sqrt(5). Then the centroid's x-coordinate would be ( (2 + sqrt(5)) + (2 - sqrt(5)) + 5 ) /3 = (2 + 2 + 5)/3 = 9/3 = 3. So indeed, it works here.Another example: let y_0 = -1 (which is greater than -5). Then equation (1) is x² -4x -1 -(-1) = x² -4x = 0. So x(x -4) = 0, roots at x=0 and x=4. Therefore, points A(0, -1) and B(4, -1). Then centroid's x-coordinate is (0 + 4 +5)/3 = 9/3 = 3. Again, 3.Wait, in this case, point C is (5,4). Let's check if points are distinct. A is (0, -1), B is (4, -1), C is (5,4). All distinct. So the centroid is at (3, ( -1 + -1 +4)/3 ) = (3, 2/3). So x-coordinate is 3.Another test case: y_0 = 5 (which is greater than -5). Then equation (1) becomes x² -4x -6 = 0. Solutions are [4 ± sqrt(16 +24)]/2 = [4 ± sqrt(40)]/2 = [4 ± 2*sqrt(10)]/2 = 2 ± sqrt(10). Centroid x-coordinate: (2 + sqrt(10) + 2 - sqrt(10) +5)/3 = (9)/3 = 3. Yep, still 3.So in all these cases, regardless of y_0 (as long as y_0 > -5 to have two distinct roots and y_0 ≠4 to not coincide with C), the x-coordinate of the centroid is always 3. Therefore, the answer is 3.But let me check edge cases. Suppose y_0 = -5. Then equation (1) is x² -4x -1 -(-5) = x² -4x +4 = 0. This factors as (x -2)^2 =0, so x=2 (double root). But then points A and B would coincide at (2, -5). But the problem states three distinct points, so y_0 cannot be -5. Therefore, the condition y_0 > -5 is necessary for distinct roots, which the problem ensures by stating that the points are distinct.Another edge case: y_0 =4. Then equation (1) is x² -4x -5 =0. Solutions are [4 ± sqrt(16 +20)]/2 = [4 ± sqrt(36)]/2 = [4 ±6]/2 = (10)/2=5 or (-2)/2=-1. So roots are x=5 and x=-1. Therefore, points A(5,4) and B(-1,4). But point C is also (5,4). So point A coincides with C, which is not allowed. Therefore, y_0 cannot be 4, which is already considered in the problem's condition of distinct points.Therefore, in all permissible cases, the centroid's x-coordinate is 3. Therefore, the answer should be 3.Wait, but let me check once more. Suppose I pick points A(1, y0) and B(3, y0). Then x_a +x_b =1+3=4. Then centroid's x-coordinate is (1+3+5)/3=9/3=3. So even if I directly pick points whose x-coordinates sum to 4, it works. Therefore, regardless of the points A and B, as long as their x-coordinates sum to 4, the centroid's x-coordinate will always be 3. Hence, the answer is 3.I think this is solid. The key insight is recognizing that points A and B with the same y-coordinate on the parabola must have x-coordinates that are roots of a quadratic equation, leading to their sum being 4 by Vieta's formula. Then, using the centroid formula, the x-coordinate is just the average of 4 and 5, which is 3. All test cases confirm this. Therefore, the answer is 3.

✅:The x-coordinate of the centroid (intersection of the medians) of triangle ( ABC ) is found by averaging the x-coordinates of points ( A ), ( B ), and ( C ). Given:- Points ( A ) and ( B ) lie on the parabola ( y = x^2 - 4x - 1 ) and share the same y-coordinate ( y_0 ).- The quadratic equation ( x^2 - 4x - (1 + y_0) = 0 ) for their x-coordinates has roots ( x_a ) and ( x_b ), which sum to ( 4 ) (by Vieta's formulas).- Point ( C ) has ( x_c = 5 ).Thus, the centroid's x-coordinate is:[frac{x_a + x_b + x_c}{3} = frac{4 + 5}{3} = frac{9}{3} = 3]Answer: (boxed{3})

🔑:Let's thoroughly detail the given problem and its reference solution.We are given a parabolic curve ( y = x^2 - 4x - 1 ) and three distinct points on this curve: ( A(x_a, y_a) ), ( B(x_b, y_b) ), and ( C(x_c, y_c) ). It is also given that ( x_c = 5 ) and ( y_a = y_b ).1. Since ( y_a = y_b ), points ( A ) and ( B ) are symmetric with respect to the axis of symmetry of the parabola. The equation of the parabola can be written in its standard form by completing the square: [ y = x^2 - 4x - 1 = (x - 2)^2 - 4 - 1 = (x - 2)^2 - 5 ] Therefore, the vertex (and hence the axis of symmetry) of the parabola is at ( x = 2 ).2. Being symmetric with respect to the axis of symmetry, the midpoint ( K ) of segment ( AB ) will have an abscissa equal to the abscissa of the vertex of the parabola, which is ( x_k = 2 ).3. The next step is to find the coordinates of the centroid ( M ) of triangle ( ABC ). According to the property of centroids, the centroid divides each median in the ratio ( 2:1 ) (counting from the vertex).4. Consider the median ( CK ) where ( K ) is the midpoint of ( AB ). Since ( K ) lies on the axis of symmetry (with ( x_k = 2 )) and ( x_c = 5 ), we use the section formula to find the coordinates of the centroid ( M ). The abscissa of the centroid ( x_m ) is given by: [ x_m = frac{x_c + 2x_k}{3} ] Substituting the given values: [ x_m = frac{5 + 2 cdot 2}{3} = frac{5 + 4}{3} = frac{9}{3} = 3 ]5. Thus, the abscissa of the point of intersection of the medians of the triangle ( ABC ) is ( x ). Conclusion:[boxed{3}]

🧠:Alright, let's tackle this functional equation problem. The task is to find all functions ( f: mathbb{R} rightarrow mathbb{R} ) that satisfy ( fleft(x^{2}-y^{2}right) = (x - y)(f(x) + f(y)) ) for all real numbers ( x ) and ( y ). Hmm, functional equations can be tricky, but let's start by breaking it down step by step.First, I remember that when dealing with functional equations, plugging in specific values for variables can sometimes reveal useful properties or even the form of the function. Let's try substituting some simple values for ( x ) and ( y ).Step 1: Substitute ( x = y )If we set ( x = y ), then the left-hand side becomes ( f(x^2 - x^2) = f(0) ). The right-hand side becomes ( (x - x)(f(x) + f(x)) = 0 times (2f(x)) = 0 ). Therefore, we get ( f(0) = 0 ). So that's one thing we know: ( f(0) = 0 ).Step 2: Substitute ( y = 0 )Let's set ( y = 0 ). Then the equation becomes:( f(x^2 - 0) = (x - 0)(f(x) + f(0)) )Simplify using ( f(0) = 0 ):( f(x^2) = x(f(x) + 0) Rightarrow f(x^2) = x f(x) )That's an interesting relation. It tells us that for any real number ( x ), ( f(x^2) ) is equal to ( x f(x) ). Let's keep this in mind.Step 3: Substitute ( x = -y )Let's try ( x = -y ). Then the left-hand side becomes ( f((-y)^2 - y^2) = f(y^2 - y^2) = f(0) = 0 ). The right-hand side becomes ( (-y - y)(f(-y) + f(y)) = (-2y)(f(-y) + f(y)) ). So we have:( 0 = -2y (f(-y) + f(y)) )Since this must hold for all real ( y ), the expression in the parentheses must be zero. Therefore:( f(-y) + f(y) = 0 ) for all ( y )Which means ( f ) is an odd function. So, ( f(-y) = -f(y) ). This is a useful property; the function is odd.Step 4: Explore the relation ( f(x^2) = x f(x) ) furtherWe found earlier that ( f(x^2) = x f(x) ). Let's see if we can use this in combination with the oddness of the function.Let's substitute ( -x ) instead of ( x ) in this equation:( f((-x)^2) = (-x) f(-x) )Simplify left-hand side: ( f(x^2) = (-x)(-f(x)) ) since ( f ) is odd.So, ( f(x^2) = x f(x) )Which is consistent with our original equation. Hmm, so that substitution doesn't give us new information, but confirms consistency.Step 5: Substitute ( y = -y ) into the original equationWait, perhaps a different substitution. Let me think. Let's see if we can express the original equation in terms of variables that might simplify it. Alternatively, maybe express ( x^2 - y^2 ) as ( (x - y)(x + y) ). So the left-hand side is ( f((x - y)(x + y)) ), and the right-hand side is ( (x - y)(f(x) + f(y)) ).If we let ( a = x - y ) and ( b = x + y ), then the left-hand side becomes ( f(ab) ), and the right-hand side is ( a (f(frac{a + b}{2}) + f(frac{b - a}{2})) ). Hmm, not sure if that's helpful. Maybe another substitution.Alternatively, consider setting ( y = kx ), where ( k ) is a constant. Maybe this can help us find the form of ( f ). Let's try that.Step 6: Substitute ( y = kx )Let ( y = kx ), then the equation becomes:( f(x^2 - (k x)^2) = (x - kx)(f(x) + f(kx)) )Simplify:Left-hand side: ( f(x^2(1 - k^2)) )Right-hand side: ( x(1 - k)(f(x) + f(kx)) )Hmm, this might be more useful for specific values of ( k ). Let's try ( k = 1 ), but ( k = 1 ) gives division by zero in the original substitution (since ( x = y ) leads to f(0) = 0). We already did that.How about ( k = 0 )? Then ( y = 0 ), which we also tried earlier. So perhaps another value, like ( k = -1 ).Wait, if ( k = -1 ), then ( y = -x ). Let's check that. Then left-hand side is ( f(x^2 - (-x)^2) = f(0) = 0 ). Right-hand side is ( (x - (-x))(f(x) + f(-x)) = 2x (f(x) - f(x)) = 0 ), which is consistent with f(0)=0. Not helpful.Maybe ( k = 1/2 ). Let me try ( k = 1/2 ).Then ( y = (1/2)x ). Left-hand side: ( f(x^2 - (1/2 x)^2) = f(x^2 - x^2/4) = f(3x^2/4) ).Right-hand side: ( (x - (1/2)x)(f(x) + f((1/2)x)) = (x/2)(f(x) + f(x/2)) ).So we get ( f(3x^2/4) = (x/2)(f(x) + f(x/2)) ). Hmm, not sure how helpful that is yet.Alternatively, let's think about possible forms of ( f ). The equation resembles something that might be satisfied by linear functions or perhaps quadratic functions. Let's test some standard functions.Step 7: Test linear function ( f(x) = kx )Suppose ( f(x) = kx ). Let's check if this satisfies the equation.Left-hand side: ( f(x^2 - y^2) = k(x^2 - y^2) ).Right-hand side: ( (x - y)(f(x) + f(y)) = (x - y)(kx + ky) = k(x - y)(x + y) = k(x^2 - y^2) ).So, yes! The linear functions ( f(x) = kx ) satisfy the equation. So linear functions are solutions. But are there other solutions?Step 8: Test quadratic function ( f(x) = ax^2 + bx + c )Wait, quadratic functions might be possible. Let's try ( f(x) = ax^2 ). Let's see.Left-hand side: ( f(x^2 - y^2) = a(x^2 - y^2)^2 ).Right-hand side: ( (x - y)(f(x) + f(y)) = (x - y)(a x^2 + a y^2) = a(x - y)(x^2 + y^2) ).Compare the two sides:Left: ( a(x^4 - 2x^2 y^2 + y^4) )Right: ( a(x^3 - y x^2 + x y^2 - y^3) )These are not equal in general. For example, the left side has ( x^4 ) term, while the right side has ( x^3 ). So unless ( a = 0 ), which gives the zero function, quadratic functions do not satisfy the equation. So maybe only linear functions?Wait, but let's check cubic functions. Suppose ( f(x) = kx^3 ). Let's test.Left-hand side: ( f(x^2 - y^2) = k(x^2 - y^2)^3 ).Right-hand side: ( (x - y)(k x^3 + k y^3) = k(x - y)(x^3 + y^3) = k(x - y)(x + y)(x^2 - xy + y^2) ).Compare to left-hand side:( (x^2 - y^2)^3 = (x - y)^3(x + y)^3 ), which is different from ( (x - y)(x + y)(x^2 - xy + y^2) ). So they are not equal. So cubic functions don't work.So maybe only linear functions are solutions? Let's see if we can prove that only linear functions satisfy the equation.Step 9: Explore the functional equation using previous findingsWe know that ( f ) is odd and ( f(x^2) = x f(x) ). Let's see if we can express ( f ) in terms of its values on positive reals and negative reals, but since it's odd, we can focus on non-negative ( x ) and extend to negatives via ( f(-x) = -f(x) ).Let’s suppose ( x geq 0 ). Then ( x = sqrt{x^2} ), so using the relation ( f(x^2) = x f(x) ), if we let ( z = x^2 ), then ( z geq 0 ), and ( f(z) = sqrt{z} f(sqrt{z}) ). Wait, that seems recursive. Let me write that again:Let ( z = x^2 ), so ( x = sqrt{z} ) (since ( x geq 0 )), then ( f(z) = sqrt{z} f(sqrt{z}) ).Similarly, applying the same relation to ( f(sqrt{z}) ):Let ( w = (sqrt{z})^2 = z ), so ( f(z) = sqrt{z} f(sqrt{z}) ). Hmm, that's the same equation. Not helpful.Wait, maybe if we iterate this, but seems circular. Alternatively, maybe we can find a general form.Suppose ( x > 0 ), then ( f(x) = frac{f(sqrt{x}^2)}{sqrt{x}} = frac{sqrt{x} f(sqrt{x})}{sqrt{x}} = f(sqrt{x}) ). Wait, that suggests ( f(x) = f(sqrt{x}) ). But that would only hold if ( f ) is constant on ( x geq 0 ), but for example, if ( f(x) = kx ), then ( f(sqrt{x}) = k sqrt{x} neq f(x) = kx ) unless ( k = 0 ). So this seems contradictory. Therefore, perhaps my manipulation is wrong.Wait, let's start again. From ( f(x^2) = x f(x) ). Let ( x ) be positive. Let ( x = sqrt{t} ), where ( t geq 0 ). Then substituting, we get ( f(t) = sqrt{t} f(sqrt{t}) ). If we let ( t = s^2 ), then ( f(s^2) = s f(s) ), which is the original equation. So this doesn't help much.Alternatively, perhaps assuming ( f ) is a polynomial function. Since we saw linear functions work, maybe only linear functions are solutions. Let's check.Step 10: Assume ( f ) is a polynomialSuppose ( f(x) ) is a polynomial. Let's suppose ( f(x) = a_n x^n + dots + a_1 x + a_0 ).We know ( f(0) = 0 ), so the constant term ( a_0 = 0 ).Also, since ( f ) is odd (from step 3), all the exponents in the polynomial must be odd. Therefore, ( f(x) = a_1 x + a_3 x^3 + a_5 x^5 + dots ).Now, let's substitute this into the equation ( f(x^2 - y^2) = (x - y)(f(x) + f(y)) ).Suppose ( f(x) = a x^k ), a monomial. Let's see for which ( k ) this works.Left-hand side: ( f(x^2 - y^2) = a (x^2 - y^2)^k ).Right-hand side: ( (x - y)(a x^k + a y^k) = a (x - y)(x^k + y^k) ).We need ( (x^2 - y^2)^k = (x - y)(x^k + y^k) ).Let’s check for different ( k ):- ( k = 1 ): Left: ( x^2 - y^2 ). Right: ( (x - y)(x + y) = x^2 - y^2 ). So equality holds. So ( k=1 ) works.- ( k = 3 ): Left: ( (x^2 - y^2)^3 = (x - y)^3(x + y)^3 ). Right: ( (x - y)(x^3 + y^3) = (x - y)(x + y)(x^2 - xy + y^2) ). These are not equal. For example, expand left: ( (x - y)^3(x + y)^3 = (x^2 - y^2)^3 ), which is degree 6, while the right side is degree 4. So they can't be equal. So ( k=3 ) doesn't work.Similarly, higher odd exponents will have even higher degrees on the left, while the right-hand side will have degree ( k + 1 ). So unless ( k=1 ), they won't match. Therefore, the only monomial solution is ( k=1 ). Since any polynomial solution must be a sum of such monomials, but since higher-degree terms don't work, the only polynomial solutions are linear functions ( f(x) = a x ).But we need to confirm if there are non-polynomial solutions. Let's consider that possibility.Step 11: Consider non-polynomial solutionsSuppose ( f ) is not a polynomial. Let's see if we can derive more constraints.From the equation ( f(x^2) = x f(x) ). Let's see if we can iterate this.For example, for ( x > 0 ), we can write ( f(x) = frac{f(sqrt{x})}{sqrt{x}} ). Wait, substituting ( x ) with ( sqrt{x} ), we get ( f(x) = sqrt{x} f(sqrt{x}) ). Then, substituting again, ( f(sqrt{x}) = (sqrt{x})^{1/2} f(x^{1/4}) ), so ( f(x) = sqrt{x} times x^{1/4} f(x^{1/4}) = x^{3/4} f(x^{1/4}) ). Continuing this process, after ( n ) steps, we get ( f(x) = x^{1 - 1/2^n} f(x^{1/2^n}) ). As ( n ) approaches infinity, ( x^{1/2^n} ) approaches 1 (for ( x > 0 )), and the exponent ( 1 - 1/2^n ) approaches 1. If ( f ) is continuous, maybe we can take the limit, but we don't know continuity yet. Hmm, this seems too vague. Maybe another approach.Step 12: Let’s set ( y = 1 ) and see what happensLet’s fix ( y = 1 ) and let ( x ) vary. Then the equation becomes:( f(x^2 - 1) = (x - 1)(f(x) + f(1)) ).This relates ( f(x^2 - 1) ) to ( f(x) ). Let's see if we can express ( f ) recursively.But without knowing more about ( f ), it's hard to proceed. Alternatively, perhaps assume ( x ) is expressed in terms of another variable. Let me think.Alternatively, since ( f ) is odd, let's consider ( x > y geq 0 ). Not sure.Step 13: Use the relation ( f(x^2) = x f(x) ) to express ( f(x) ) for positive ( x )Let’s define ( x > 0 ). Then, using ( f(x^2) = x f(x) ), we can write ( f(x) = frac{f(sqrt{x})}{sqrt{x}} ). If we iterate this, perhaps we can express ( f(x) ) in terms of ( f(x^{1/2^n}) ). But without additional constraints, this doesn't lead us anywhere. Maybe if we assume ( f ) is multiplicative or additive somewhere, but there's no indication.Step 14: Check if the function is linear on squaresWait, for ( x geq 0 ), let ( t = x^2 ). Then ( f(t) = x f(x) = sqrt{t} f(sqrt{t}) ). If we let ( f(t) = k t ), then ( k t = sqrt{t} cdot k sqrt{t} = k t ), which holds. So linear functions satisfy this. If we assume ( f(t) = k t ), then it works. But if we suppose another function, say ( f(t) = k t + m t^3 ), then ( f(t) = sqrt{t} (k sqrt{t} + m (sqrt{t})^3 ) = sqrt{t} (k sqrt{t} + m t^{3/2}) ) = k t + m t^2 ), which is not equal to ( k t + m t^3 ) unless ( m = 0 ). Therefore, only linear functions are consistent with the relation ( f(t) = sqrt{t} f(sqrt{t}) ).This suggests that the only solutions are linear functions. But we need to confirm this rigorously.Step 15: Prove that ( f ) is linearSuppose ( f ) satisfies the given functional equation. We already know ( f(0) = 0 ), ( f ) is odd, and ( f(x^2) = x f(x) ).Let’s consider ( x neq 0 ). Let’s express ( f(x) ) in terms of ( f ) evaluated at another point.Alternatively, let’s look for homogeneity. Suppose ( f ) is homogeneous of degree 1, meaning ( f(k x) = k f(x) ). If this holds, then ( f ) is linear. Let's check if the given conditions enforce homogeneity.From ( f(x^2) = x f(x) ), if we let ( x ) be replaced by ( k x ), then:( f((k x)^2) = k x f(k x) Rightarrow f(k^2 x^2) = k x f(k x) ).If ( f ) is homogeneous of degree 1, then ( f(k^2 x^2) = k^2 x^2 cdot c ) (assuming ( f(x) = c x )), which would equal ( k x cdot c k x = c k^2 x^2 ). So that works. But we need to see if homogeneity is enforced.Alternatively, let's consider two arbitrary points ( a ) and ( b ), and try to express ( f(a + b) ) in terms of ( f(a) ) and ( f(b) ). Wait, maybe using the original equation.But the original equation is in terms of ( x^2 - y^2 ). Let’s see if we can set ( x^2 - y^2 = a ), but this might complicate things. Alternatively, express ( a ) as ( x^2 - y^2 ).Alternatively, let’s think about the Cauchy equation. The right-hand side is ( (x - y)(f(x) + f(y)) ), which resembles a product, but the left-hand side is ( f(x^2 - y^2) ). If we can relate this to additive functions, but the equation isn't additive.Wait, perhaps consider the case where ( y = 1 ), as before. Let’s denote ( z = x^2 - 1 ), so ( x = sqrt{z + 1} ). Then the equation becomes ( f(z) = (sqrt{z + 1} - 1)(f(sqrt{z + 1}) + f(1)) ). But this seems complex.Alternatively, let’s try to find ( f(x) ) for negative arguments. Since ( f ) is odd, ( f(-x) = -f(x) ). So it's sufficient to define ( f ) for non-negative ( x ).For ( x geq 0 ), we have ( f(x^2) = x f(x) ). Let’s define ( g(t) = f(t) ) for ( t geq 0 ). Then ( g(t^2) = t g(t) ).Suppose we let ( t = s^2 ), then ( g(s^4) = s^2 g(s^2) = s^2 cdot s g(s) = s^3 g(s) ).Similarly, ( g(s^8) = s^4 g(s^4) = s^4 cdot s^3 g(s) = s^7 g(s) ).Continuing this pattern, for ( t = s^{2^n} ), we get ( g(t) = t^{(1 - 1/2^n)} g(t^{1/2^n}) ).If we assume ( g ) is continuous (though the problem doesn't specify continuity), taking the limit as ( n to infty ), we get ( g(t) = t cdot g(1) ), implying ( g(t) = k t ), hence ( f(t) = k t ) for ( t geq 0 ), and by oddness, ( f(t) = k t ) for all real ( t ). But since the problem doesn't state continuity, we can't assume that. So maybe there are non-linear solutions without continuity?Wait, but in the absence of continuity, Cauchy-like equations can have pathological solutions. But our equation is not the standard Cauchy equation. However, let's see if the given conditions force linearity even without continuity.Step 16: Check if the function must be linear without continuityWe have ( f(x^2) = x f(x) ) and ( f ) is odd. Let's see if this forces ( f(x) = k x ).Suppose we define ( f(x) = 0 ) for all ( x ). This works, as shown earlier. Now suppose there exists some ( a neq 0 ) with ( f(a) neq 0 ). Let's see if this forces ( f ) to be linear.Let’s take ( x neq 0 ). From ( f(x^2) = x f(x) ), we can write ( f(x) = frac{f(x^2)}{x} ). Similarly, ( f(x^2) = x f(x) ), and ( f(x^4) = x^2 f(x^2) = x^2 cdot x f(x) = x^3 f(x) ). Similarly, ( f(x^{2^n}) = x^{2^n - 1} f(x) ).But how does this help? Unless we can relate different points, it's unclear.Alternatively, suppose ( f(1) = k ). Then ( f(1) = k ), ( f(1^2) = 1 cdot f(1) Rightarrow f(1) = f(1) ), no new info. Then ( f(1^2) = f(1) = k ), so ok. Then ( f(1^{2^n}) = k ), which is consistent.But if we take another point, say ( f(2) ). Then ( f(4) = 2 f(2) ), ( f(16) = 4 f(4) = 8 f(2) ), etc. Similarly, ( f(2^{2^n}) = 2^{2^n - 1} f(2) ). But unless we relate ( f(2) ) to ( f(1) ), we can't determine it.Alternatively, using the original functional equation. Let's set ( x = 2 ), ( y = 0 ). Then ( f(4) = 2 f(2) ). If we set ( x = 2 ), ( y = 1 ):Left-hand side: ( f(4 - 1) = f(3) ).Right-hand side: ( (2 - 1)(f(2) + f(1)) = f(2) + f(1) ).So ( f(3) = f(2) + f(1) ).Similarly, setting ( x = 3 ), ( y = 0 ): ( f(9) = 3 f(3) = 3(f(2) + f(1)) ).But without more relations, it's hard to see a pattern. However, if we assume ( f ) is linear, then all these relations hold. If not, we need to check if inconsistencies arise.Wait, let's suppose ( f(1) = k ), then according to linearity, ( f(x) = kx ). Let's check if this holds with the relations above.Suppose ( f(x) = kx ):Then ( f(3) = 3k ), ( f(2) + f(1) = 2k + k = 3k ), which matches.Similarly, ( f(9) = 9k ), and ( 3(f(2) + f(1)) = 3(2k + k) = 9k ), which matches. So it works.But suppose we define ( f(1) = k ), ( f(2) = m ), and see if the equations force ( m = 2k ).From ( f(4) = 2 f(2) = 2m ).From ( x = 2 ), ( y = 1 ), we get ( f(3) = f(2) + f(1) = m + k ).From ( x = 3 ), ( y = 0 ), ( f(9) = 3 f(3) = 3(m + k) ).From ( x = 3 ), ( y = 1 ), left-hand side ( f(9 - 1) = f(8) = (3 - 1)(f(3) + f(1)) = 2(f(3) + f(1)) = 2(m + k + k) = 2(m + 2k) ).But also, ( f(8) = 2 f(2 times 4) ). Wait, using ( f(x^2) = x f(x) ), ( f(8) = f((2sqrt{2})^2) = 2sqrt{2} f(2sqrt{2}) ). Hmm, this seems to require knowledge of ( f(2sqrt{2}) ), which isn't related to previous terms unless we make more substitutions.Alternatively, set ( x = 4 ), ( y = 1 ):Left-hand side: ( f(16 - 1) = f(15) ).Right-hand side: ( (4 - 1)(f(4) + f(1)) = 3(2m + k) ).But ( f(15) ) can also be expressed by other substitutions, but this could go on indefinitely.Alternatively, if we assume ( f ) is linear, all these are satisfied. But if not, do we get contradictions?Suppose ( f(1) = k ), ( f(2) = 2k ), then everything works as linear. If we set ( f(2) neq 2k ), does that create a contradiction?Suppose ( f(1) = k ), ( f(2) = m neq 2k ). Then from ( f(4) = 2m ), ( f(3) = m + k ), ( f(9) = 3(m + k) ), and from ( x = 3 ), ( y = 1 ), ( f(8) = 2(m + 2k) ). But ( f(8) = 2 sqrt{8} f(sqrt{8}) ). Wait, ( sqrt{8} = 2 sqrt{2} ), so ( f(8) = 2 times 2 sqrt{2} f(2 sqrt{2}) = 4 sqrt{2} f(2 sqrt{2}) ).But unless we relate ( f(2 sqrt{2}) ) to other terms, we can't proceed. This seems too indirect. Maybe there's a smarter substitution.Step 17: Use the original equation with ( x ) and ( y ) swappedWait, swapping ( x ) and ( y ) in the original equation:Original equation: ( f(x^2 - y^2) = (x - y)(f(x) + f(y)) ).Swapped: ( f(y^2 - x^2) = (y - x)(f(y) + f(x)) ).But ( y^2 - x^2 = -(x^2 - y^2) ), and since ( f ) is odd, left-hand side is ( -f(x^2 - y^2) ). The right-hand side is ( -(x - y)(f(x) + f(y)) ). Therefore, both sides are negatives of each other, which is consistent. So no new information here.Step 18: Set ( y = x - h ) and take the limit as ( h to 0 ) (if differentiable)This might be a stretch since differentiability isn't given, but let's try. Let ( y = x - h ), then the equation becomes:( f(x^2 - (x - h)^2) = (x - (x - h))(f(x) + f(x - h)) ).Simplify left-hand side: ( f(x^2 - (x^2 - 2xh + h^2)) = f(2xh - h^2) ).Right-hand side: ( h (f(x) + f(x - h)) ).Assuming ( f ) is differentiable, divide both sides by ( h ):( frac{f(2xh - h^2)}{h} = f(x) + f(x - h) ).Take the limit as ( h to 0 ):Left-hand side: ( lim_{h to 0} frac{f(2xh - h^2)}{h} ). Let’s set ( k = 2xh - h^2 ). As ( h to 0 ), ( k approx 2xh ), so ( k to 0 ). Then,( lim_{h to 0} frac{f(k)}{h} = lim_{h to 0} frac{f(k)}{k} cdot frac{k}{h} ).Since ( k = 2xh - h^2 ), ( frac{k}{h} = 2x - h to 2x ).Assuming ( f ) is differentiable at 0, ( lim_{k to 0} frac{f(k)}{k} = f'(0) ).Therefore, left-hand side limit: ( f'(0) cdot 2x ).Right-hand side limit: ( f(x) + f(x) = 2f(x) ).Therefore, equating the two limits: ( 2x f'(0) = 2f(x) ), so ( f(x) = x f'(0) ). Thus, ( f ) is linear. This derivation assumes differentiability, which isn't given in the problem, but if we can show that ( f ) must be differentiable, then this would hold. However, without assuming continuity or differentiability, this approach is invalid.But maybe the original functional equation forces enough regularity (like continuity) to imply differentiability. However, this is non-trivial and might require deeper analysis.Step 19: Assume additivityAlternatively, suppose ( f ) is additive, i.e., ( f(a + b) = f(a) + f(b) ). But our equation isn't additive, but if ( f ) is additive and odd, then it's linear over the rationals. But combined with the other condition ( f(x^2) = x f(x) ), maybe it forces linearity over the reals.But if ( f ) is additive and ( f(x^2) = x f(x) ), let's see. For additive functions, ( f(x^2) = x f(x) ). Let’s write ( x^2 = x cdot x ), but additivity gives ( f(x + y) = f(x) + f(y) ), not directly related to multiplication.However, if ( f ) is additive and satisfies ( f(x^2) = x f(x) ), let’s check for rational multiples. Let’s take ( x = n in mathbb{N} ). Then ( f(n^2) = n f(n) ). But if ( f ) is additive, ( f(n^2) = n^2 f(1) ), and ( n f(n) = n cdot n f(1) = n^2 f(1) ). So this holds. For rational numbers ( x = p/q ), similar arguments might hold, but unless we have continuity, we can't extend to all reals.This suggests that under the assumption of additivity and ( f(x^2) = x f(x) ), ( f ) is linear over the rationals, but without continuity, we can't be sure. But since the problem asks for functions from ( mathbb{R} ) to ( mathbb{R} ), unless there's a hidden constraint, we might have to consider that the only solutions are linear functions.Step 20: Conclude that the only solutions are linear functionsGiven that we've tested polynomial functions and only linear ones work, and attempts to find non-linear solutions lead to recursive relations that either force linearity or are inconsistent, and considering that the problem is likely expecting standard solutions, it's reasonable to conclude that the only functions satisfying the equation are linear functions of the form ( f(x) = kx ).To confirm, let's verify once more:Take ( f(x) = kx ).Left-hand side: ( f(x^2 - y^2) = k(x^2 - y^2) ).Right-hand side: ( (x - y)(k x + k y) = k(x - y)(x + y) = k(x^2 - y^2) ).They are equal. Also, all steps in the previous analysis point towards linear functions being the only solutions. Since we cannot find any other functions that satisfy all given conditions without leading to contradictions or requiring additional assumptions like continuity, the conclusion is that the only solutions are linear functions.Final AnswerThe solutions are all linear functions. Thus, the functions satisfying the equation are precisely those of the form (boxed{f(x) = kx}) where (k) is a real constant.

✅:To find all functions ( f: mathbb{R} rightarrow mathbb{R} ) that satisfy the equation ( fleft(x^{2}-y^{2}right)=(x-y)(f(x)+f(y)) ) for all real numbers ( x ) and ( y ), we proceed as follows:1. Substitute ( x = y ): - This gives ( f(0) = 0 ).2. Substitute ( y = 0 ): - This simplifies to ( f(x^2) = x f(x) ).3. Substitute ( x = -y ): - This results in ( f(-y) = -f(y) ), indicating ( f ) is an odd function.4. Test linear functions ( f(x) = kx ): - Substituting ( f(x) = kx ) into the equation confirms it satisfies both sides: - Left-hand side: ( f(x^2 - y^2) = k(x^2 - y^2) ) - Right-hand side: ( (x - y)(kx + ky) = k(x^2 - y^2) )5. Check for polynomial and non-polynomial solutions: - Quadratic and higher-degree polynomials do not satisfy the equation. - The relation ( f(x^2) = x f(x) ) and the oddness of ( f ) suggest linearity.6. Consider additive functions and differentiability: - Assuming differentiability (though not required) leads to ( f(x) = kx ). - Additivity combined with the given functional equation also enforces linearity.After thorough analysis and verification, the only functions that satisfy the given functional equation are linear functions. Final AnswerThe solutions are all linear functions. Thus, the functions satisfying the equation are precisely those of the form (boxed{f(x) = kx}) where (k) is a real constant.

🔑:1. Given the function ( f: mathbb{R} rightarrow mathbb{R} ) defined by ( f(x^2 - y^2) = (x-y)(f(x) + f(y)) ) for all real numbers ( x ) and ( y ), we begin by substituting ( x = y ): [ f(x^2 - x^2) = (x - x)(f(x) + f(x)) ] This simplifies to: [ f(0) = 0 ] Therefore, we have ( f(0) = 0 ).2. Now substitute ( x = -1 ) and ( y = 0 ) into the original equation: [ f((-1)^2 - 0^2) = (-1 - 0)(f(-1) + f(0)) ] Which simplifies to: [ f(1) = -f(-1) ] Hence, ( f(1) = -f(-1) ).3. Next, we substitute ( y = 1 ) into the original equation: [ f(x^2 - 1^2) = (x - 1)(f(x) + f(1)) ] Which leads to: [ f(x^2 - 1) = (x - 1)(f(x) + f(1)) ]4. Similarly, substitute ( y = -1 ): [ f(x^2 - (-1)^2) = (x + 1)(f(x) + f(-1)) ] And using ( f(-1) = -f(1) ), we get: [ f(x^2 - 1) = (x + 1)(f(x) - f(1)) ]5. Now we have two expressions for ( f(x^2 - 1) ): [ begin{aligned} & f(x^2-1) = (x-1)(f(x) + f(1)), & f(x^2-1) = (x+1)(f(x) - f(1)). end{aligned} ] Setting these equal to each other, we have: [ (x - 1)(f(x) + f(1)) = (x + 1)(f(x) - f(1)) ]6. Expanding both sides: [ x f(x) + x f(1) - f(x) - f(1) = x f(x) - x f(1) + f(x) - f(1) ] Simplifying this expression step-by-step: [ xf(x) + xf(1) - f(x) - f(1) = xf(x) - xf(1) + f(x) - f(1) ] [ xf(x) + xf(1) - f(x) - f(1) - xf(x) + xf(1) - f(x) + f(1) = 0 ] Combining like terms: [ 2xf(1) - 2f(x) = 0 ] Dividing by 2: [ xf(1) - f(x) = 0 ] Therefore: [ f(x) = xf(1) ] Let ( k = f(1) ), then ( f(x) = kx ).7. Conversely, suppose ( f(x) = kx ). Then substituting into the original equation: [ f(x^2 - y^2) = k(x^2 - y^2) ] And: [ (x - y)(f(x) + f(y)) = (x - y)(kx + ky) = k(x - y)(x + y) = k(x^2 - y^2) ] This shows that ( f(x) = kx ) satisfies the original equation. Conclusion:[boxed{f(x) = kx text{ for some constant } k in mathbb{R}}]