answer：Alright, so I have this problem here about sets M and P, and I need to find their intersection. Let me try to understand what each set represents and then figure out how they overlap.First, set M is defined as {x | x > a, a² - 12a + 20 < 0}. Hmm, okay, so M is the set of all x such that x is greater than some value a, but only when a satisfies the inequality a² - 12a + 20 < 0. That means I need to figure out for which values of a this inequality holds true.Let me solve the inequality a² - 12a + 20 < 0. To do that, I'll first find the roots of the quadratic equation a² - 12a + 20 = 0. Using the quadratic formula:a = [12 ± sqrt(144 - 80)] / 2a = [12 ± sqrt(64)] / 2a = [12 ± 8] / 2So, the roots are (12 + 8)/2 = 20/2 = 10 and (12 - 8)/2 = 4/2 = 2. Therefore, the quadratic expression a² - 12a + 20 factors to (a - 2)(a - 10).Since the quadratic opens upwards (the coefficient of a² is positive), the expression (a - 2)(a - 10) will be less than zero between the roots. So, the inequality a² - 12a + 20 < 0 holds true when 2 < a < 10.Okay, so now I know that for set M, the condition a² - 12a + 20 < 0 is true only when a is between 2 and 10. Therefore, set M is the set of all x such that x > a, but since a is between 2 and 10, M is essentially {x | 2 < a < 10 and x > a}.Wait, that might not be the clearest way to put it. Let me think again. Since a is between 2 and 10, the condition x > a means that x is greater than some number between 2 and 10. So, the smallest x can be is just above 2, and the largest x can be is just below 10. But since a can be any value between 2 and 10, x has to be greater than a, which is somewhere in that interval.Now, moving on to set P, which is {x | x ≤ 10}. That's straightforward—it's all real numbers less than or equal to 10.So, to find M ∩ P, I need to find all x that are in both M and P. That means x has to satisfy both conditions: x > a (with a between 2 and 10) and x ≤ 10.Let me visualize this. If a is between 2 and 10, then x has to be greater than a but also less than or equal to 10. So, the lower bound for x is a, which is greater than 2, and the upper bound is 10.Therefore, M ∩ P should be all x such that a < x ≤ 10. But wait, a is between 2 and 10, so the smallest a can be is just above 2, and the largest a can be is just below 10. So, effectively, x has to be greater than 2 and less than or equal to 10.But hold on, is that always the case? If a is, say, 5, then x has to be greater than 5 and less than or equal to 10. If a is 3, x has to be greater than 3 and less than or equal to 10. So, depending on the value of a, the lower bound changes. But since a can be any value between 2 and 10, the intersection M ∩ P would cover all x from just above 2 up to 10.But looking at the answer choices, option C is {x | 2 < x ≤ 10}, which seems to match this description. However, option A is {x | a < x ≤ 10}, which also seems correct but depends on the value of a.Wait, but the problem doesn't specify a particular value for a; it's just given as a parameter. So, in the context of the problem, since a is between 2 and 10, the intersection M ∩ P would be all x such that a < x ≤ 10, but since a can vary, the overall intersection would cover all x from 2 to 10.Hmm, I'm a bit confused now. Let me think again. If a is fixed, then M ∩ P would be {x | a < x ≤ 10}. But since a is variable between 2 and 10, the intersection over all possible a would be {x | 2 < x ≤ 10}.But the problem doesn't specify whether a is fixed or variable. It just defines M and P. So, perhaps I need to consider that a is a parameter, and the intersection is expressed in terms of a.Wait, looking back at the problem, it says "Let set M = {x | x > a, a² - 12a + 20 < 0}" and "P = {x | x ≤ 10}". So, M is defined for x such that x > a and a satisfies the inequality. So, M is actually the set of x where x > a, but a is constrained to be between 2 and 10.Therefore, M is the set of all x such that x > a, where a is between 2 and 10. So, effectively, M is all x greater than some a between 2 and 10. So, the smallest x can be is just above 2, and the largest x can be is unbounded above, but since P is x ≤ 10, the intersection would be x between a and 10.But since a can be any value between 2 and 10, the intersection M ∩ P would be all x such that 2 < x ≤ 10, because for any x in that interval, there exists an a such that x > a and x ≤ 10.Wait, that makes sense. So, regardless of the specific a, as long as a is between 2 and 10, the intersection would cover all x from just above 2 to 10.But looking at the answer choices, option C is {x | 2 < x ≤ 10}, which seems to fit. However, option A is {x | a < x ≤ 10}, which is also correct but expressed in terms of a. But since a is a variable between 2 and 10, the intersection would cover all x from 2 to 10.But I'm not sure if the answer expects it in terms of a or as a specific interval. The problem doesn't specify a particular value for a, so I think the answer should be in terms of a, which would be option A. But I'm still a bit confused because option C is also a valid interval.Wait, maybe I need to consider that M is defined for a specific a that satisfies the inequality. So, for each a between 2 and 10, M is {x | x > a}, and P is {x | x ≤ 10}. So, their intersection would be {x | a < x ≤ 10}, which is option A.But if I consider all possible a between 2 and 10, the union of all such intersections would be {x | 2 < x ≤ 10}, which is option C. But the problem is asking for M ∩ P, not the union over all a.So, I think the correct answer is A, because for a specific a between 2 and 10, M ∩ P is {x | a < x ≤ 10}. But I'm still a bit unsure because the problem doesn't specify whether a is fixed or variable.Wait, looking back, the problem says "Let set M = {x | x > a, a² - 12a + 20 < 0}". So, M is defined for x such that x > a and a satisfies the inequality. So, M is actually the set of all x where x > a, with a constrained to be between 2 and 10. So, M is effectively {x | x > 2}, because a can be as low as just above 2.But no, that's not quite right. Because a is between 2 and 10, but for each a, M is {x | x > a}. So, M is not just {x | x > 2}, because a can be up to 10. So, M is the union of all {x | x > a} for a between 2 and 10, which would be {x | x > 2}.Wait, that makes sense. Because if a can be as low as just above 2, then M would include all x greater than 2. But since a can also be up to 10, M would also include x greater than 10, but P is x ≤ 10, so the intersection would be x between 2 and 10.Wait, now I'm really confused. Let me try to approach this differently.Set M is defined as {x | x > a, a² - 12a + 20 < 0}. So, first, solve for a: a² - 12a + 20 < 0 ⇒ 2 < a < 10. So, a is between 2 and 10.Then, M is {x | x > a}, but a is between 2 and 10. So, for each a in (2,10), M is {x | x > a}. So, M is the union of all such sets for a in (2,10). The union of all {x | x > a} for a in (2,10) is {x | x > 2}, because as a approaches 2 from above, x has to be greater than 2.But wait, no. If a is between 2 and 10, then for each a, M is {x | x > a}. So, the union of all these M's would be {x | x > 2}, because for any x > 2, there exists an a (specifically, a just less than x) such that x > a.But in the problem, M is defined as {x | x > a, a² - 12a + 20 < 0}, which is equivalent to {x | x > a} where a is in (2,10). So, M is {x | x > 2}, because for any x > 2, there exists an a in (2,10) such that x > a.Wait, that makes sense. So, M is {x | x > 2}, because a can be as low as just above 2, making x > 2 the lower bound.Then, P is {x | x ≤ 10}. So, the intersection M ∩ P would be {x | x > 2 and x ≤ 10}, which is {x | 2 < x ≤ 10}, which is option C.But earlier, I thought it might be option A, but now I'm leaning towards option C.Wait, but let's double-check. If M is {x | x > a} with a in (2,10), then M is {x | x > 2}, because for any x > 2, you can choose a = (x + 2)/2, which is between 2 and x, hence x > a.Therefore, M is indeed {x | x > 2}, and P is {x | x ≤ 10}, so their intersection is {x | 2 < x ≤ 10}, which is option C.But wait, the problem defines M as {x | x > a, a² - 12a + 20 < 0}, which is equivalent to {x | x > a} where a is in (2,10). So, M is the set of all x such that x > a for some a in (2,10). Therefore, M is {x | x > 2}, because for any x > 2, you can find an a in (2,10) such that x > a.Hence, M ∩ P is {x | 2 < x ≤ 10}, which is option C.I think I confused myself earlier by thinking of M as depending on a specific a, but actually, M is the union over all a in (2,10) of {x | x > a}, which simplifies to {x | x > 2}.So, the correct answer should be option C.

answer：Okay, so I have this function f(x) = x² / (1 + x²). The problem has two parts. Let me tackle them one by one.Starting with part (1): I need to calculate f(2) + f(1/2), f(3) + f(1/3), f(4) + f(1/4), and then make a conjecture about a general conclusion. Hmm, okay. Let's compute each pair step by step.First, let's compute f(2). Plugging in x = 2:f(2) = (2)² / (1 + (2)²) = 4 / (1 + 4) = 4/5.Now, f(1/2). Plugging in x = 1/2:f(1/2) = (1/2)² / (1 + (1/2)²) = (1/4) / (1 + 1/4) = (1/4) / (5/4) = (1/4) * (4/5) = 1/5.Adding them together: f(2) + f(1/2) = 4/5 + 1/5 = 5/5 = 1.Interesting, that sums up to 1. Let me check another pair to see if this pattern holds. Next, f(3) and f(1/3).f(3) = (3)² / (1 + (3)²) = 9 / (1 + 9) = 9/10.f(1/3) = (1/3)² / (1 + (1/3)²) = (1/9) / (1 + 1/9) = (1/9) / (10/9) = (1/9) * (9/10) = 1/10.Adding them: f(3) + f(1/3) = 9/10 + 1/10 = 10/10 = 1.Same result! Let me do one more to be sure. f(4) and f(1/4).f(4) = (4)² / (1 + (4)²) = 16 / (1 + 16) = 16/17.f(1/4) = (1/4)² / (1 + (1/4)²) = (1/16) / (1 + 1/16) = (1/16) / (17/16) = (1/16) * (16/17) = 1/17.Adding them: f(4) + f(1/4) = 16/17 + 1/17 = 17/17 = 1.Wow, so each time, f(x) + f(1/x) equals 1. That seems consistent. So, I can conjecture that for any x ≠ 0, f(x) + f(1/x) = 1. That's a neat property of this function.Moving on to part (2). This looks more complicated. I need to compute the value of:2f(2) + 2f(3) + … + 2f(2017) + f(1/2) + f(1/3) + … + f(1/2017) + (1/2²)f(2) + (1/3²)f(3) + … + (1/2017²)f(2017).Wow, that's a mouthful. Let me try to parse this expression.First, let's break it down into parts:1. The first part is 2f(2) + 2f(3) + … + 2f(2017). So, that's 2 multiplied by f(k) for k from 2 to 2017.2. The second part is f(1/2) + f(1/3) + … + f(1/2017). So, that's f(1/k) for k from 2 to 2017.3. The third part is (1/2²)f(2) + (1/3²)f(3) + … + (1/2017²)f(2017). So, that's (1/k²)f(k) for k from 2 to 2017.So, combining these, the entire expression is:Sum from k=2 to 2017 of [2f(k) + f(1/k) + (1/k²)f(k)].Hmm, maybe I can factor this expression or find a way to simplify it using the property from part (1).From part (1), we know that f(k) + f(1/k) = 1. So, f(1/k) = 1 - f(k). Maybe I can substitute that into the expression.Let me rewrite the expression:Sum from k=2 to 2017 of [2f(k) + (1 - f(k)) + (1/k²)f(k)].Simplify inside the sum:2f(k) + 1 - f(k) + (1/k²)f(k) = (2f(k) - f(k) + (1/k²)f(k)) + 1 = (f(k) + (1/k²)f(k)) + 1.Factor f(k):f(k)(1 + 1/k²) + 1.Hmm, let's compute f(k)(1 + 1/k²):f(k) = k² / (1 + k²), so f(k)(1 + 1/k²) = [k² / (1 + k²)] * [1 + 1/k²] = [k² / (1 + k²)] * [(k² + 1)/k²] = (k²)(k² + 1) / [(1 + k²)(k²)] = (k² + 1) / (k² + 1) = 1.Wait, that's interesting. So, f(k)(1 + 1/k²) = 1.Therefore, the expression inside the sum becomes 1 + 1 = 2.So, each term in the sum from k=2 to 2017 is equal to 2.Therefore, the entire sum is 2 multiplied by the number of terms. How many terms are there?From k=2 to k=2017, that's 2017 - 2 + 1 = 2016 terms.So, the total sum is 2 * 2016 = 4032.Wait, that seems straightforward. Let me double-check my steps to make sure I didn't make a mistake.First, I broke down the expression into three parts and combined them into a single sum. Then, I used the property from part (1) to substitute f(1/k) with 1 - f(k). That led me to simplify the expression inside the sum, which turned out to be 2 for each term. Since there are 2016 terms (from 2 to 2017 inclusive), multiplying 2 by 2016 gives 4032.Yes, that makes sense. I think I did this correctly. So, the value of the entire expression is 4032.**Final Answer**The value is boxed{4032}.

answer：Okay, so I have this problem about two concentric circles. Concentric circles mean they share the same center, right? The radii are in the ratio 2:5. So, if I let the radius of the smaller circle be 2r, then the radius of the larger circle would be 5r. That makes sense because 2:5 is the ratio, so scaling by r keeps the ratio intact.The problem says that AC is a diameter of the larger circle. Since the radius of the larger circle is 5r, the diameter AC would be twice that, so AC = 10r. Got that down.Next, BC is a chord of the larger circle that's tangent to the smaller circle. Hmm, okay. So, BC is a line segment on the larger circle, and it just touches the smaller circle at one point, meaning it's tangent there. That should form a right angle with the radius at the point of tangency. I remember that from geometry.Also, AB is given as 8 units. So, AB = 8. I need to find the radius of the larger circle, which is 5r. So, if I can find r, I can find the radius.Let me visualize this. We have two concentric circles, center O. AC is the diameter of the larger circle, so points A and C are on the larger circle, and O is the midpoint. BC is a chord of the larger circle, tangent to the smaller circle. Let me denote the point of tangency as T. So, OT is the radius of the smaller circle, which is 2r, and OT is perpendicular to BC at point T.So, triangle ABC is formed by points A, B, and C. Since AC is the diameter, and B is a point on the circle, triangle ABC is a right-angled triangle. Wait, is that correct? Yes, because of Thales' theorem, which states that if you have a triangle inscribed in a circle where one side is the diameter, then the triangle is right-angled. So, angle ABC is 90 degrees.So, triangle ABC is right-angled at B. Therefore, AB and BC are the legs, and AC is the hypotenuse. We know AB = 8, AC = 10r, and we need to find BC or something related to it.But BC is a chord tangent to the smaller circle. So, maybe I can relate BC to the radius of the smaller circle. Since BC is tangent to the smaller circle at point T, and OT is the radius, OT is perpendicular to BC. So, OT is 2r, and OT is perpendicular to BC.Let me think about how to relate these. Maybe I can use similar triangles or some properties of circles here.Since triangle ABC is right-angled at B, and OT is perpendicular to BC, maybe there's a way to relate the lengths. Let me denote the center as O. So, O is the midpoint of AC, so AO = OC = 5r.Let me consider the distance from O to BC. Since BC is tangent to the smaller circle, the distance from O to BC is equal to the radius of the smaller circle, which is 2r. That's because the tangent is at a distance equal to the radius from the center.So, in triangle ABC, the distance from the center O to the side BC is 2r. Also, in a right-angled triangle, the distance from the circumcenter (which is O here, since AC is the diameter) to the side BC can be related to the sides of the triangle.Wait, maybe I can use the formula for the distance from a point to a line. Let me recall that. The distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a^2 + b^2). But I'm not sure if coordinate geometry is the best approach here.Alternatively, maybe I can use area. In triangle ABC, the area can be expressed in two ways: one using the legs AB and BC, and another using the hypotenuse AC and the height from B to AC.But wait, the distance from O to BC is 2r, which is the height from O to BC. But O is the midpoint of AC, so maybe I can relate this to the area.Wait, let me think again. Triangle ABC is right-angled at B, with AB = 8, BC = let's say x, and AC = 10r. So, by Pythagoras, 8^2 + x^2 = (10r)^2. So, 64 + x^2 = 100r^2. So, x^2 = 100r^2 - 64. So, x = sqrt(100r^2 - 64).But I also know that the distance from O to BC is 2r. In a right-angled triangle, the distance from the circumcenter to a side can be related to the sides. Wait, maybe I can use the formula for the distance from the circumcenter to a side.In a right-angled triangle, the circumradius is half the hypotenuse, which is 5r in this case. The distance from the circumcenter to a side can be found using the formula: distance = (product of the other two sides) / (2 * hypotenuse). Wait, is that correct?Let me recall. In any triangle, the distance from the circumcenter to a side is equal to the circumradius times the cosine of the angle opposite that side. Hmm, maybe that's more complicated.Alternatively, in a right-angled triangle, the distance from the circumcenter (which is the midpoint of the hypotenuse) to the other sides can be calculated using the formula: distance = (length of the other side) / 2. Wait, no, that doesn't sound right.Wait, let me think about coordinate geometry. Let me place the triangle in a coordinate system to make it easier. Let me set point O at (0,0). Since AC is the diameter, let me place point A at (-5r, 0) and point C at (5r, 0). Then, point B is somewhere on the circle such that AB = 8 and BC is tangent to the smaller circle.So, coordinates: A(-5r, 0), C(5r, 0), and B(x, y) on the larger circle. The distance AB is 8, so sqrt[(x + 5r)^2 + y^2] = 8. Also, BC is tangent to the smaller circle, so the distance from O to BC is 2r.The equation of line BC can be found using points B(x, y) and C(5r, 0). The distance from O(0,0) to line BC is 2r. The formula for the distance from a point (x0, y0) to the line through points (x1, y1) and (x2, y2) is |(y2 - y1)x0 - (x2 - x1)y0 + x2 y1 - y2 x1| / sqrt[(y2 - y1)^2 + (x2 - x1)^2].So, applying this formula, the distance from O(0,0) to BC is |(0 - y)*0 - (5r - x)*0 + 5r*y - 0*x| / sqrt[(0 - y)^2 + (5r - x)^2] = |5r y| / sqrt[y^2 + (5r - x)^2] = 2r.So, |5r y| / sqrt[y^2 + (5r - x)^2] = 2r. Since y is positive (assuming B is above the x-axis), we can drop the absolute value: 5r y / sqrt[y^2 + (5r - x)^2] = 2r.Simplify: 5y / sqrt[y^2 + (5r - x)^2] = 2.Square both sides: 25y^2 / [y^2 + (5r - x)^2] = 4.Multiply both sides by denominator: 25y^2 = 4[y^2 + (5r - x)^2].Expand: 25y^2 = 4y^2 + 4(25r^2 - 10r x + x^2).Simplify: 25y^2 = 4y^2 + 100r^2 - 40r x + 4x^2.Bring all terms to left: 25y^2 - 4y^2 - 100r^2 + 40r x - 4x^2 = 0.Simplify: 21y^2 - 4x^2 + 40r x - 100r^2 = 0.Now, we also know that point B lies on the larger circle, so its coordinates satisfy the equation x^2 + y^2 = (5r)^2 = 25r^2.So, x^2 + y^2 = 25r^2.We also have AB = 8, so the distance between A(-5r, 0) and B(x, y) is 8: sqrt[(x + 5r)^2 + y^2] = 8.Square both sides: (x + 5r)^2 + y^2 = 64.Expand: x^2 + 10r x + 25r^2 + y^2 = 64.But from the circle equation, x^2 + y^2 = 25r^2, so substitute that in: 25r^2 + 10r x + 25r^2 = 64.Simplify: 50r^2 + 10r x = 64.Divide both sides by 2: 25r^2 + 5r x = 32.So, 5r x = 32 - 25r^2.Thus, x = (32 - 25r^2)/(5r).Okay, so now we have x in terms of r. Let's plug this back into the earlier equation we had from the distance condition.We had: 21y^2 - 4x^2 + 40r x - 100r^2 = 0.First, let's express y^2 from the circle equation: y^2 = 25r^2 - x^2.So, substitute y^2 into the equation:21(25r^2 - x^2) - 4x^2 + 40r x - 100r^2 = 0.Expand: 525r^2 - 21x^2 - 4x^2 + 40r x - 100r^2 = 0.Combine like terms: (525r^2 - 100r^2) + (-21x^2 - 4x^2) + 40r x = 0.Simplify: 425r^2 - 25x^2 + 40r x = 0.Divide the entire equation by 5 to simplify: 85r^2 - 5x^2 + 8r x = 0.Now, substitute x = (32 - 25r^2)/(5r) into this equation.First, compute x^2:x = (32 - 25r^2)/(5r), so x^2 = [(32 - 25r^2)^2]/(25r^2).Similarly, compute 8r x:8r x = 8r * (32 - 25r^2)/(5r) = 8*(32 - 25r^2)/5 = (256 - 200r^2)/5.Now, plug into the equation:85r^2 - 5x^2 + 8r x = 0.Substitute x^2 and 8r x:85r^2 - 5*[(32 - 25r^2)^2]/(25r^2) + (256 - 200r^2)/5 = 0.Simplify term by term:First term: 85r^2.Second term: -5*[(32 - 25r^2)^2]/(25r^2) = -[(32 - 25r^2)^2]/(5r^2).Third term: (256 - 200r^2)/5.So, the equation becomes:85r^2 - [(32 - 25r^2)^2]/(5r^2) + (256 - 200r^2)/5 = 0.This looks complicated, but let's try to simplify step by step.Multiply every term by 5r^2 to eliminate denominators:5r^2*85r^2 - (32 - 25r^2)^2 + 5r^2*(256 - 200r^2)/5 = 0.Simplify each term:First term: 5r^2*85r^2 = 425r^4.Second term: -(32 - 25r^2)^2.Third term: r^2*(256 - 200r^2) = 256r^2 - 200r^4.So, the equation becomes:425r^4 - (32 - 25r^2)^2 + 256r^2 - 200r^4 = 0.Combine like terms:425r^4 - 200r^4 = 225r^4.So, 225r^4 - (32 - 25r^2)^2 + 256r^2 = 0.Now, expand (32 - 25r^2)^2:(32)^2 - 2*32*25r^2 + (25r^2)^2 = 1024 - 1600r^2 + 625r^4.So, substitute back:225r^4 - (1024 - 1600r^2 + 625r^4) + 256r^2 = 0.Distribute the negative sign:225r^4 - 1024 + 1600r^2 - 625r^4 + 256r^2 = 0.Combine like terms:225r^4 - 625r^4 = -400r^4.1600r^2 + 256r^2 = 1856r^2.So, the equation becomes:-400r^4 + 1856r^2 - 1024 = 0.Multiply the entire equation by -1 to make the leading coefficient positive:400r^4 - 1856r^2 + 1024 = 0.Now, let's let u = r^2. Then the equation becomes:400u^2 - 1856u + 1024 = 0.This is a quadratic in u. Let's try to simplify it.First, divide all terms by 16 to reduce the coefficients:400/16 = 25, 1856/16 = 116, 1024/16 = 64.So, 25u^2 - 116u + 64 = 0.Now, let's solve for u using the quadratic formula:u = [116 ± sqrt(116^2 - 4*25*64)] / (2*25).Compute discriminant:116^2 = 13456.4*25*64 = 100*64 = 6400.So, discriminant = 13456 - 6400 = 7056.sqrt(7056) = 84.So, u = [116 ± 84]/50.Compute both possibilities:First solution: (116 + 84)/50 = 200/50 = 4.Second solution: (116 - 84)/50 = 32/50 = 16/25.So, u = 4 or u = 16/25.But u = r^2, so r^2 = 4 or r^2 = 16/25.Thus, r = 2 or r = 4/5.But let's check if both solutions make sense in the context.If r = 2, then the radius of the larger circle is 5r = 10. If r = 4/5, then 5r = 4. But let's check if r = 4/5 works with the earlier equations.From earlier, we had x = (32 - 25r^2)/(5r).If r = 4/5, then r^2 = 16/25.So, x = (32 - 25*(16/25))/(5*(4/5)) = (32 - 16)/4 = 16/4 = 4.Then, from the circle equation, y^2 = 25r^2 - x^2 = 25*(16/25) - 16 = 16 - 16 = 0. So, y = 0. But that would mean point B is on the x-axis, which would make BC a diameter, but BC is supposed to be tangent to the smaller circle. If y = 0, then BC would coincide with the x-axis, which is the diameter AC, but AC is already the diameter, so BC can't be another diameter. Therefore, r = 4/5 is not a valid solution.Thus, the only valid solution is r = 2. Therefore, the radius of the larger circle is 5r = 10.So, the answer is 10, which corresponds to option B.

answer：Okay, so I have this sequence problem here. Let me try to figure it out step by step. The sequence is {a_n}, and they've given me the first two terms: a₁ = 1 and a₂ = 1/2. There's also this recurrence relation: (2)/(a_{n+1}) = (a_n + a_{n+2})/(a_n * a_{n+2}) for all natural numbers n. I need to find a_{20}. The options are A: -17/2, B: 2¹⁹, C: (1/2)¹⁹, D: 1/20.Hmm, okay. Let me write down what I know:a₁ = 1a₂ = 1/2Recurrence relation: 2/a_{n+1} = (a_n + a_{n+2})/(a_n * a_{n+2})I need to find a_{20}. So, maybe I can find a pattern or a formula for a_n.First, let's try to understand the recurrence relation. It looks a bit complicated, but maybe I can manipulate it to find a simpler form.Starting with:2/a_{n+1} = (a_n + a_{n+2})/(a_n * a_{n+2})Let me rewrite the right-hand side:(a_n + a_{n+2})/(a_n * a_{n+2}) = (1/a_{n+2} + 1/a_n)Wait, is that right? Let me check:(a_n + a_{n+2})/(a_n * a_{n+2}) = a_n/(a_n * a_{n+2}) + a_{n+2}/(a_n * a_{n+2}) = 1/a_{n+2} + 1/a_nYes, that's correct. So, the equation becomes:2/a_{n+1} = 1/a_n + 1/a_{n+2}Hmm, that's interesting. So, if I let b_n = 1/a_n, then the equation becomes:2b_{n+1} = b_n + b_{n+2}Which can be rewritten as:b_{n+2} - 2b_{n+1} + b_n = 0That's a linear recurrence relation for the sequence {b_n}. Let me write down the characteristic equation for this recurrence.The characteristic equation is:r² - 2r + 1 = 0Solving this quadratic equation:r = [2 ± sqrt(4 - 4)] / 2 = [2 ± 0]/2 = 1So, the characteristic equation has a repeated root r = 1. That means the general solution for the recurrence is:b_n = (A + Bn)(1)^n = A + BnSo, b_n is a linear function of n. Therefore, b_n = A + Bn.Now, let's find the constants A and B using the initial terms.We know that b_n = 1/a_n, so:b₁ = 1/a₁ = 1/1 = 1b₂ = 1/a₂ = 1/(1/2) = 2So, plugging n = 1 into b_n = A + Bn:1 = A + B(1) => A + B = 1Similarly, plugging n = 2:2 = A + B(2) => A + 2B = 2Now, we have a system of two equations:1) A + B = 12) A + 2B = 2Subtracting equation 1 from equation 2:(A + 2B) - (A + B) = 2 - 1 => B = 1Then, plugging back into equation 1:A + 1 = 1 => A = 0So, the general term is:b_n = 0 + 1*n = nTherefore, b_n = n, which means 1/a_n = n, so a_n = 1/n.Wow, okay, so the sequence {a_n} is just the reciprocals of natural numbers. So, a_n = 1/n.Therefore, a_{20} = 1/20.Looking back at the options, D is 1/20, so that should be the answer.Let me just verify this with the recurrence relation to make sure I didn't make a mistake.Given a_n = 1/n, let's check if the recurrence holds.Compute 2/a_{n+1}:2/a_{n+1} = 2/(1/(n+1)) = 2(n+1)Compute (a_n + a_{n+2})/(a_n * a_{n+2}):(a_n + a_{n+2})/(a_n * a_{n+2}) = (1/n + 1/(n+2))/( (1/n)*(1/(n+2)) )Simplify numerator:1/n + 1/(n+2) = (n+2 + n)/(n(n+2)) = (2n + 2)/(n(n+2)) = 2(n + 1)/(n(n+2))Denominator:(1/n)*(1/(n+2)) = 1/(n(n+2))So, the entire expression is:[2(n + 1)/(n(n+2))] / [1/(n(n+2))] = 2(n + 1)Which is equal to 2/a_{n+1} as we computed earlier. So, the recurrence holds.Therefore, my solution seems correct. So, a_{20} is indeed 1/20.**Final Answer**boxed{D}

answer：Okay, so I have this problem here about a continuous function ( f(x) ) that satisfies the inequality ( f(x^2) - (f(x))^2 geq frac{1}{4} ) for all real numbers ( x ). The question is asking whether this function must necessarily have extremum points. Hmm, interesting. I need to figure out if ( f(x) ) has to have a maximum or minimum somewhere.Alright, let's start by understanding the given inequality. It says that for any real number ( x ), the value of the function at ( x^2 ) minus the square of the function at ( x ) is at least ( frac{1}{4} ). So, ( f(x^2) ) is always greater than or equal to ( (f(x))^2 + frac{1}{4} ). First, maybe I should plug in some specific values of ( x ) to see what I can find out about ( f(x) ). Let's try ( x = 0 ). Then the inequality becomes:( f(0^2) - (f(0))^2 geq frac{1}{4} )Which simplifies to:( f(0) - (f(0))^2 geq frac{1}{4} )Let me rearrange this:( (f(0))^2 - f(0) + frac{1}{4} leq 0 )Hmm, this is a quadratic inequality in terms of ( f(0) ). Let me denote ( a = f(0) ) for simplicity. Then the inequality becomes:( a^2 - a + frac{1}{4} leq 0 )I can solve this quadratic equation ( a^2 - a + frac{1}{4} = 0 ) to find the critical points. The discriminant is ( D = (-1)^2 - 4 cdot 1 cdot frac{1}{4} = 1 - 1 = 0 ). So, there's a repeated root. The solution is:( a = frac{1 pm sqrt{0}}{2} = frac{1}{2} )So, the quadratic expression ( a^2 - a + frac{1}{4} ) is a perfect square, specifically ( (a - frac{1}{2})^2 ). Therefore, the inequality ( (a - frac{1}{2})^2 leq 0 ) implies that ( a = frac{1}{2} ). So, ( f(0) = frac{1}{2} ).Okay, so we've found that ( f(0) ) must be ( frac{1}{2} ). Let me try another value, maybe ( x = 1 ). Plugging in ( x = 1 ) into the inequality:( f(1^2) - (f(1))^2 geq frac{1}{4} )Which simplifies to:( f(1) - (f(1))^2 geq frac{1}{4} )Again, let me rearrange:( (f(1))^2 - f(1) + frac{1}{4} leq 0 )This is the same quadratic as before. So, solving it gives ( f(1) = frac{1}{2} ).Interesting, so both ( f(0) ) and ( f(1) ) are ( frac{1}{2} ). Maybe I can try ( x = -1 ) as well, just to see. Plugging in ( x = -1 ):( f((-1)^2) - (f(-1))^2 geq frac{1}{4} )Which is:( f(1) - (f(-1))^2 geq frac{1}{4} )But we already know ( f(1) = frac{1}{2} ), so:( frac{1}{2} - (f(-1))^2 geq frac{1}{4} )Subtracting ( frac{1}{2} ) from both sides:( - (f(-1))^2 geq -frac{1}{4} )Multiplying both sides by -1 (and reversing the inequality):( (f(-1))^2 leq frac{1}{4} )Taking square roots:( |f(-1)| leq frac{1}{2} )So, ( f(-1) ) is between ( -frac{1}{2} ) and ( frac{1}{2} ). Hmm, not as specific as the previous results, but it gives a bound.Let me think about this. We know that ( f(0) = frac{1}{2} ) and ( f(1) = frac{1}{2} ). Since ( f ) is continuous, maybe I can use the Intermediate Value Theorem or something related to that.Wait, the function is continuous everywhere, so between 0 and 1, the function must take all values between ( f(0) ) and ( f(1) ), but since both are ( frac{1}{2} ), does that mean the function is constant on [0,1]? Hmm, that might be the case.Let me suppose that ( f(x) = frac{1}{2} ) for all ( x ) in [0,1]. Then, for any ( x ) in [0,1], ( x^2 ) is also in [0,1], so ( f(x^2) = frac{1}{2} ), and ( (f(x))^2 = (frac{1}{2})^2 = frac{1}{4} ). Therefore, ( f(x^2) - (f(x))^2 = frac{1}{2} - frac{1}{4} = frac{1}{4} ), which satisfies the inequality as equality. So, that works.But what about outside of [0,1]? For ( x > 1 ), ( x^2 ) is greater than ( x ), so we might have a recursive relation here. Let's see.Suppose ( x > 1 ). Then ( x^2 > x ), so ( f(x^2) geq (f(x))^2 + frac{1}{4} ). If ( f(x) ) is greater than ( frac{1}{2} ), then ( (f(x))^2 ) would be greater than ( frac{1}{4} ), so ( f(x^2) ) would have to be greater than ( (f(x))^2 + frac{1}{4} ), which is more than ( frac{1}{2} ). But then, ( f(x^2) ) would be greater than ( frac{1}{2} ), and so on. This might lead to ( f(x) ) increasing without bound, which can't happen if ( f ) is continuous everywhere, unless it's constant.Wait, but if ( f(x) ) is constant ( frac{1}{2} ) everywhere, then it satisfies the inequality as equality, right? Because ( f(x^2) = frac{1}{2} ) and ( (f(x))^2 = frac{1}{4} ), so ( frac{1}{2} - frac{1}{4} = frac{1}{4} ), which is exactly the inequality. So, the constant function ( f(x) = frac{1}{2} ) works.But is this the only possibility? Or can there be other functions that satisfy the inequality?Wait, let's think about ( x ) between 0 and 1. If ( x ) is in (0,1), then ( x^2 ) is also in (0,1), but closer to 0. So, if ( f(x) ) is not constant, but still satisfies the inequality, what would happen?Suppose ( f(x) ) is greater than ( frac{1}{2} ) somewhere in (0,1). Let's say ( f(a) = c > frac{1}{2} ) for some ( a ) in (0,1). Then, ( f(a^2) geq c^2 + frac{1}{4} ). But since ( a^2 < a ), and ( f ) is continuous, if ( f(a) = c ), then ( f(a^2) ) must be at least ( c^2 + frac{1}{4} ). But ( c > frac{1}{2} ), so ( c^2 + frac{1}{4} > (frac{1}{2})^2 + frac{1}{4} = frac{1}{4} + frac{1}{4} = frac{1}{2} ). So, ( f(a^2) > frac{1}{2} ). But if ( f(a^2) > frac{1}{2} ), then applying the same logic, ( f((a^2)^2) geq (f(a^2))^2 + frac{1}{4} > (frac{1}{2})^2 + frac{1}{4} = frac{1}{2} ). So, this seems like it would cause ( f(x) ) to keep increasing as ( x ) approaches 0, which might not be possible if ( f ) is continuous at 0, where ( f(0) = frac{1}{2} ).Similarly, if ( f(x) ) is less than ( frac{1}{2} ) somewhere in (0,1), say ( f(b) = d < frac{1}{2} ), then ( f(b^2) geq d^2 + frac{1}{4} ). But ( d < frac{1}{2} ), so ( d^2 < frac{1}{4} ), hence ( f(b^2) geq d^2 + frac{1}{4} < frac{1}{4} + frac{1}{4} = frac{1}{2} ). So, ( f(b^2) ) is less than ( frac{1}{2} ). But then, applying the same reasoning, ( f((b^2)^2) geq (f(b^2))^2 + frac{1}{4} ). If ( f(b^2) < frac{1}{2} ), then ( (f(b^2))^2 < frac{1}{4} ), so ( f((b^2)^2) geq text{something less than } frac{1}{4} + frac{1}{4} = frac{1}{2} ). Wait, that seems contradictory because ( f((b^2)^2) ) would have to be at least ( frac{1}{2} ), but ( (b^2)^2 ) is closer to 0, and ( f(0) = frac{1}{2} ). So, if ( f ) is continuous, it can't jump above ( frac{1}{2} ) near 0 if it's less than ( frac{1}{2} ) somewhere else.This suggests that ( f(x) ) can't be less than ( frac{1}{2} ) in (0,1) because it would cause a conflict with the value at 0. Similarly, it can't be greater than ( frac{1}{2} ) because it would cause a conflict as ( x ) approaches 0. Therefore, ( f(x) ) must be equal to ( frac{1}{2} ) for all ( x ) in [0,1].Now, what about ( x ) outside [0,1]? Let's consider ( x > 1 ). Suppose ( x = 2 ). Then, ( f(4) - (f(2))^2 geq frac{1}{4} ). If ( f(2) ) is something, then ( f(4) ) has to be at least ( (f(2))^2 + frac{1}{4} ). But if ( f(2) ) is ( frac{1}{2} ), then ( f(4) geq frac{1}{4} + frac{1}{4} = frac{1}{2} ). Similarly, ( f(4) ) would have to be ( frac{1}{2} ), otherwise, it would cause a chain reaction where ( f(x) ) would have to be greater than ( frac{1}{2} ) for larger ( x ), but we already saw that this leads to a problem as ( x ) approaches 0.Wait, but if ( f(x) = frac{1}{2} ) for all ( x ), then it works everywhere. So, maybe the only function that satisfies this inequality is the constant function ( f(x) = frac{1}{2} ).If that's the case, then ( f(x) ) is a constant function, which means it doesn't have any extremum points except everywhere. But in terms of having extremum points, a constant function does have extrema because every point is both a maximum and a minimum.But the question is asking if the function must necessarily have extremum points. So, if ( f(x) ) is constant, it does have extremum points. But is there any other function that satisfies the inequality and doesn't have extremum points?Wait, if ( f(x) ) is not constant, can it satisfy the inequality? From the earlier reasoning, it seems difficult because if ( f(x) ) deviates from ( frac{1}{2} ) at any point, it causes a chain reaction that either forces it to be constant or leads to a contradiction with the continuity at 0.Therefore, it seems that the only continuous function satisfying the inequality is the constant function ( f(x) = frac{1}{2} ), which does have extremum points everywhere.So, the answer is yes, the function must necessarily have extremum points.

answer：Alright, so I'm trying to figure out for which values of ( n ) it's possible to have a cube with edge length ( n ) made up of white and black unit cubes. The condition is that each white cube shares exactly one face with three black cubes, and each black cube shares exactly one face with three white cubes. Hmm, that sounds a bit tricky, but let's break it down step by step.First, let's understand the problem better. We have a cube made up of smaller cubes, each of which is either white or black. The key condition here is about how these white and black cubes are arranged relative to each other. Specifically, every white cube must have exactly three black neighbors, and every black cube must have exactly three white neighbors. I think this means that each cube, whether white or black, has exactly three neighbors of the opposite color. Now, in a cube, each unit cube can have up to six neighbors (one on each face). So, if each white cube has exactly three black neighbors, that means the other three neighbors must be white. Similarly, each black cube has exactly three white neighbors and three black neighbors. Wait, but if that's the case, then the number of white and black cubes must be equal because each white cube has three black neighbors and each black cube has three white neighbors. So, the total number of white cubes must equal the total number of black cubes. That implies that the total number of unit cubes, which is ( n^3 ), must be even. Therefore, ( n^3 ) must be even, which means ( n ) itself must be even because the cube of an odd number is odd, and the cube of an even number is even.So, that gives me a starting point: ( n ) must be even. But is that the only condition? Or are there more constraints?Let me think about smaller values of ( n ) to see if this holds. Let's start with ( n = 2 ). A ( 2 times 2 times 2 ) cube has 8 unit cubes. If we alternate colors in a checkerboard pattern, each cube will have three neighbors of the opposite color. Let me visualize this:Imagine the bottom layer of the cube as a ( 2 times 2 ) grid. If I color it like a checkerboard, alternating black and white, then the top layer should be the inverse of the bottom layer to maintain the three opposite color neighbors for each cube. So, the bottom layer would be:[begin{matrix}B & W W & B end{matrix}]And the top layer would be:[begin{matrix}W & B B & W end{matrix}]This way, each white cube in the bottom layer is adjacent to three black cubes (two in the bottom layer and one in the top layer), and each black cube is adjacent to three white cubes. Similarly, the cubes in the top layer follow the same pattern. So, for ( n = 2 ), this works.What about ( n = 4 )? If ( n = 4 ), the cube is ( 4 times 4 times 4 ), which has 64 unit cubes. Following the same logic, we can divide this cube into smaller ( 2 times 2 times 2 ) cubes, each arranged in the checkerboard pattern we discussed earlier. Since each ( 2 times 2 times 2 ) block satisfies the condition, the entire ( 4 times 4 times 4 ) cube should also satisfy the condition. But wait, is that always the case? Let me think about how these smaller blocks interact with each other. If I have multiple ( 2 times 2 times 2 ) blocks, the way they are arranged might affect the neighbors of the cubes on the edges between the blocks. If the coloring is consistent across the entire cube, then the neighbors should still be correctly alternating. For example, if I have a ( 4 times 4 times 4 ) cube, I can think of it as a ( 2 times 2 times 2 ) arrangement of ( 2 times 2 times 2 ) blocks. Each of these blocks is colored in a checkerboard pattern, and if I ensure that adjacent blocks are colored consistently, then the entire cube should maintain the required adjacency conditions. So, it seems like as long as ( n ) is even, we can construct such a cube by dividing it into smaller ( 2 times 2 times 2 ) blocks, each of which satisfies the condition. Therefore, ( n ) must be even.But let me test this with ( n = 1 ) just to be thorough. A ( 1 times 1 times 1 ) cube has only one unit cube. But since there's only one cube, it can't have any neighbors. The condition requires that each white cube has three black neighbors and vice versa, but with only one cube, this is impossible. So, ( n = 1 ) doesn't work, which aligns with our earlier conclusion that ( n ) must be even.What about ( n = 3 )? A ( 3 times 3 times 3 ) cube has 27 unit cubes. Since 27 is odd, it's impossible to have an equal number of white and black cubes. Therefore, the condition that each white cube has three black neighbors and each black cube has three white neighbors can't be satisfied because we can't have an equal number of white and black cubes. So, ( n = 3 ) doesn't work either.This pattern seems consistent. For any odd ( n ), ( n^3 ) is odd, making it impossible to have an equal number of white and black cubes. Therefore, the condition can't be satisfied. On the other hand, for even ( n ), ( n^3 ) is even, allowing for an equal number of white and black cubes, and we can arrange them in a checkerboard pattern to satisfy the adjacency conditions.I think I've covered the possible cases and the reasoning seems solid. The key takeaway is that the cube must have an even number of unit cubes to allow for equal numbers of white and black cubes, and that an even ( n ) allows for such a configuration through a checkerboard pattern.