💡Okay, so I need to figure out if there are positive integers x, y, and z such that x raised to the 2006th power plus y raised to the 2006th power equals z raised to the 2007th power. Hmm, that's a pretty big exponent equation. Let me break it down.First, I know that exponents can sometimes be tricky, especially with such large numbers. Maybe I can start by testing some small numbers to see if the equation holds. Let's try x = y = z = 2. That seems simple enough.So, if x = 2, y = 2, and z = 2, then the left side of the equation becomes 2^2006 + 2^2006. Since both terms are the same, I can factor that out: 2 * 2^2006. Using exponent rules, that's 2^(1 + 2006) = 2^2007. On the right side, z^2007 is 2^2007. So, both sides are equal to 2^2007. That means the equation holds true for x = 2, y = 2, and z = 2. Wait, does that mean the answer is yes? It seems like it. But maybe I should check another set of numbers to be sure. Let's try x = 1, y = 1, and z = something. If x = 1 and y = 1, then the left side is 1^2006 + 1^2006 = 1 + 1 = 2. So, z^2007 must equal 2. But z has to be a positive integer, and 2^(1/2007) is not an integer. So, that doesn't work. Okay, so x = 1 and y = 1 don't work, but x = 2, y = 2, z = 2 do. Maybe there are other combinations too. Let me think about another example. How about x = 3, y = 3, and z = 3? Then, the left side would be 3^2006 + 3^2006 = 2 * 3^2006. The right side is 3^2007. Is 2 * 3^2006 equal to 3^2007? Let's see: 3^2007 is 3 * 3^2006, which is 3^2006 * 3. So, 2 * 3^2006 is not equal to 3 * 3^2006. Therefore, that doesn't work.Hmm, so x = y = z = 3 doesn't satisfy the equation. Maybe I need to find a different z. Let's see, if x = 2 and y = 2, then z must be 2 because 2^2006 + 2^2006 = 2^2007. If I try x = 2 and y = 4, what happens? Left side: 2^2006 + 4^2006. 4^2006 is (2^2)^2006 = 2^(4012). So, left side is 2^2006 + 2^4012. That's a huge number. The right side would be z^2007. I don't know what z would be here, but it's unlikely to be an integer because 2^4012 is much larger than 2^2006, and adding them together might not result in a perfect 2007th power.Maybe I should think about the structure of the equation. The left side is the sum of two 2006th powers, and the right side is a 2007th power. Since 2006 and 2007 are consecutive integers, maybe there's a relationship there. If I let x = y, then the equation becomes 2x^2006 = z^2007. So, z^2007 must be twice a 2006th power. If x is a power of 2, say x = 2^k, then x^2006 = (2^k)^2006 = 2^(2006k). So, 2x^2006 = 2^(2006k + 1). For this to be a perfect 2007th power, 2006k + 1 must be divisible by 2007. Let's see: 2006k + 1 ≡ 0 mod 2007. That simplifies to 2006k ≡ -1 mod 2007. Since 2006 ≡ -1 mod 2007, this becomes (-1)k ≡ -1 mod 2007, which simplifies to k ≡ 1 mod 2007. So, the smallest positive integer k is 1. Therefore, x = 2^1 = 2, and z = 2^( (2006*1 + 1)/2007 ) = 2^(2007/2007) = 2^1 = 2. That's why x = y = z = 2 works.Is there another solution? Let's try k = 1 + 2007 = 2008. Then x = 2^2008, and z = 2^( (2006*2008 + 1)/2007 ). Let's compute the exponent: 2006*2008 = (2007 -1)(2007 +1) = 2007^2 -1. So, 2006*2008 +1 = 2007^2 -1 +1 = 2007^2. Therefore, z = 2^(2007^2 / 2007) = 2^2007. So, x = 2^2008, y = 2^2008, z = 2^2007. Let's check: x^2006 = (2^2008)^2006 = 2^(2008*2006). Similarly, y^2006 = 2^(2008*2006). So, x^2006 + y^2006 = 2 * 2^(2008*2006) = 2^(2008*2006 +1). On the other hand, z^2007 = (2^2007)^2007 = 2^(2007*2007). Let's see if these exponents are equal: 2008*2006 +1 vs 2007*2007. Compute 2008*2006: (2007 +1)(2007 -1) = 2007^2 -1. So, 2008*2006 +1 = 2007^2 -1 +1 = 2007^2. Therefore, 2^(2007^2) on both sides. So, yes, that works too.So, there are infinitely many solutions where x = y = 2^k, z = 2^(k'), where k and k' satisfy the earlier condition. Therefore, the answer is yes, such positive integers exist.

💡Okay, so I have this problem here about Cara and her marbles. Let me try to understand it step by step. She has a total of 60 marbles in a bag. Out of these, 20 are yellow. The problem says there are half as many green marbles as yellow ones. So, if there are 20 yellow marbles, then the number of green marbles should be half of that, which is 10. That makes sense so far.Now, the remaining marbles are equally divided between red and blue. Hmm, so first I need to figure out how many marbles are left after accounting for the yellow and green ones. The total is 60 marbles, and we've already accounted for 20 yellow and 10 green marbles. So, 20 plus 10 is 30 marbles. That means there are 60 minus 30, which is 30 marbles left for red and blue.Since these remaining marbles are equally divided between red and blue, I need to split 30 into two equal parts. So, 30 divided by 2 is 15. That means there are 15 red marbles and 15 blue marbles.Now, the question is asking for the odds of picking a blue marble at random, expressed as a percentage. To find this, I need to determine the probability of picking a blue marble and then convert that probability into a percentage.The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. In this case, the favorable outcome is picking a blue marble, and the total number of possible outcomes is the total number of marbles.So, the number of blue marbles is 15, and the total number of marbles is 60. Therefore, the probability of picking a blue marble is 15 divided by 60. Let me compute that: 15 divided by 60 equals 0.25.To express this probability as a percentage, I need to multiply it by 100. So, 0.25 multiplied by 100 equals 25%.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting with 60 marbles, 20 yellow, 10 green, that leaves 30 marbles for red and blue. Dividing 30 equally gives 15 red and 15 blue. So, 15 blue marbles out of 60 total marbles. 15 divided by 60 is indeed 0.25, and multiplying by 100 gives 25%. Yeah, that seems correct.I think I've got it. The odds of picking a blue marble are 25%.

💡First, calculate the weight of one-tenth of the cake, which Nathalie eats:525 grams / 10 = 52.5 gramsNext, determine how much Pierre eats, which is triple what Nathalie ate:3 * 52.5 grams = 157.5 gramsTherefore, Pierre ate 157.5 grams of the cake.

💡First, I need to determine how many female officers were on duty that night. Since half of the 240 officers on duty were female, I divide 240 by 2 to find that there were 120 female officers on duty.Next, I know that the total number of female officers on the police force is 300. To find the percentage of female officers on duty, I divide the number of female officers on duty (120) by the total number of female officers (300) and then multiply by 100 to convert it to a percentage.Calculating this gives me (120 / 300) * 100 = 40%. Therefore, 40% of the female officers were on duty that night.

💡Okay, so I have this problem where I need to find all triplets (x, y, ℓ) in natural numbers such that x³ + y³ - 53 equals 7 raised to the power of ℓ. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I know that natural numbers are positive integers, so x, y, and ℓ are all 1, 2, 3, and so on. The equation is x³ + y³ - 53 = 7^ℓ. So, I need to find all combinations of x, y, and ℓ where this equation holds true.Maybe I should start by considering the possible values of ℓ. Since 7^ℓ grows exponentially, ℓ can't be too large because x³ and y³ also grow quickly, but let's see.Let me think about small values of ℓ first. Let's try ℓ = 0. Then 7^0 is 1, so the equation becomes x³ + y³ - 53 = 1, which simplifies to x³ + y³ = 54. Okay, so I need to find natural numbers x and y such that their cubes add up to 54.Let me list the cubes of small natural numbers:1³ = 12³ = 83³ = 274³ = 645³ = 125Wait, 4³ is already 64, which is larger than 54, so x and y can't be 4 or more because their cubes would exceed 54. So x and y must be 1, 2, or 3.Let me check all possible combinations:x=1: Then y³ = 54 - 1 = 53. But 53 isn't a perfect cube. 3³ is 27, 4³ is 64, so no solution here.x=2: Then y³ = 54 - 8 = 46. Again, 46 isn't a perfect cube. 3³ is 27, 4³ is 64, so no solution here either.x=3: Then y³ = 54 - 27 = 27. Oh, 27 is 3³, so y=3. So, when ℓ=0, we have x=3 and y=3. That's one solution: (3, 3, 0).Okay, so that's one triplet. Now, let's try ℓ=1. Then 7^1 is 7, so the equation becomes x³ + y³ - 53 = 7, which simplifies to x³ + y³ = 60.Again, let's see if there are natural numbers x and y such that their cubes add up to 60.Possible x and y values: 1, 2, 3, 4 because 4³=64, which is close to 60.x=1: y³=60-1=59. Not a cube.x=2: y³=60-8=52. Not a cube.x=3: y³=60-27=33. Not a cube.x=4: y³=60-64=-4. Negative, which isn't a natural number cube.So, no solutions for ℓ=1.Next, ℓ=2. 7²=49. So the equation becomes x³ + y³ -53=49, which simplifies to x³ + y³=102.Looking for x and y such that their cubes add up to 102.Possible x and y: 1, 2, 3, 4, 5 because 5³=125, which is more than 102.x=1: y³=102-1=101. Not a cube.x=2: y³=102-8=94. Not a cube.x=3: y³=102-27=75. Not a cube.x=4: y³=102-64=38. Not a cube.x=5: y³=102-125=-23. Negative, so no solution.So, no solutions for ℓ=2.Now, ℓ=3. 7³=343. So the equation becomes x³ + y³ -53=343, which simplifies to x³ + y³=396.Looking for x and y such that their cubes add up to 396.Possible x and y: Let's see, 7³=343, 8³=512. So x and y can be up to 7.x=1: y³=396-1=395. Not a cube.x=2: y³=396-8=388. Not a cube.x=3: y³=396-27=369. Not a cube.x=4: y³=396-64=332. Not a cube.x=5: y³=396-125=271. Not a cube.x=6: y³=396-216=180. Not a cube.x=7: y³=396-343=53. 53 isn't a cube.So, no solutions for ℓ=3.Hmm, maybe I should check if ℓ can be larger. Let's try ℓ=4. 7⁴=2401. So the equation becomes x³ + y³ -53=2401, which simplifies to x³ + y³=2454.Looking for x and y such that their cubes add up to 2454.Possible x and y: Let's see, 13³=2197, 14³=2744. So x and y can be up to 13.x=13: y³=2454-2197=257. Not a cube.x=12: y³=2454-1728=726. Not a cube.x=11: y³=2454-1331=1123. Not a cube.x=10: y³=2454-1000=1454. Not a cube.x=9: y³=2454-729=1725. Not a cube.x=8: y³=2454-512=1942. Not a cube.x=7: y³=2454-343=2111. Not a cube.x=6: y³=2454-216=2238. Not a cube.x=5: y³=2454-125=2329. Not a cube.x=4: y³=2454-64=2390. Not a cube.x=3: y³=2454-27=2427. Not a cube.x=2: y³=2454-8=2446. Not a cube.x=1: y³=2454-1=2453. Not a cube.So, no solutions for ℓ=4.This is getting tedious, but maybe I should try ℓ=5. 7⁵=16807. So the equation becomes x³ + y³ -53=16807, which simplifies to x³ + y³=16860.Looking for x and y such that their cubes add up to 16860.Possible x and y: Let's see, 25³=15625, 26³=17576. So x and y can be up to 25.x=25: y³=16860-15625=1235. Not a cube.x=24: y³=16860-13824=3036. Not a cube.x=23: y³=16860-12167=4693. Not a cube.x=22: y³=16860-10648=6212. Not a cube.x=21: y³=16860-9261=7599. Not a cube.x=20: y³=16860-8000=8860. Not a cube.x=19: y³=16860-6859=10001. Not a cube.x=18: y³=16860-5832=11028. Not a cube.x=17: y³=16860-4913=11947. Not a cube.x=16: y³=16860-4096=12764. Not a cube.x=15: y³=16860-3375=13485. Not a cube.x=14: y³=16860-2744=14116. Not a cube.x=13: y³=16860-2197=14663. Not a cube.x=12: y³=16860-1728=15132. Not a cube.x=11: y³=16860-1331=15529. Not a cube.x=10: y³=16860-1000=15860. Not a cube.x=9: y³=16860-729=16131. Not a cube.x=8: y³=16860-512=16348. Not a cube.x=7: y³=16860-343=16517. Not a cube.x=6: y³=16860-216=16644. Not a cube.x=5: y³=16860-125=16735. Not a cube.x=4: y³=16860-64=16796. Not a cube.x=3: y³=16860-27=16833. Not a cube.x=2: y³=16860-8=16852. Not a cube.x=1: y³=16860-1=16859. Not a cube.So, no solutions for ℓ=5 either.This is getting too big, and I don't think there are any solutions for larger ℓ because x³ and y³ would have to be enormous, and their sum minus 53 would have to be exactly a power of 7, which is very restrictive.Wait, maybe I should consider negative numbers? But the problem specifies natural numbers, so x, y, and ℓ are all positive integers. So negative numbers aren't allowed.Another approach: Maybe I can consider the equation modulo 7 to see if there are any constraints on x and y.So, x³ + y³ - 53 ≡ 0 mod 7, because 7^ℓ ≡ 0 mod 7 for ℓ ≥ 1.So, x³ + y³ ≡ 53 mod 7.53 divided by 7 is 7*7=49, so 53 ≡ 4 mod 7.So, x³ + y³ ≡ 4 mod 7.Now, let's find all possible cubes modulo 7.Compute n³ mod 7 for n=0 to 6:0³=0 mod7=01³=1 mod7=12³=8 mod7=13³=27 mod7=27-3*7=27-21=64³=64 mod7=64-9*7=64-63=15³=125 mod7=125-17*7=125-119=66³=216 mod7=216-30*7=216-210=6So, the possible cubes modulo7 are 0,1,6.So, x³ and y³ can each be 0,1, or 6 mod7.Now, we need x³ + y³ ≡4 mod7.Let's list all possible combinations:0 + 0 =00 +1=10 +6=61 +1=21 +6=7≡06 +6=12≡5None of these combinations give 4 mod7.Wait, that's strange. So, x³ + y³ can never be ≡4 mod7, which means that for ℓ≥1, there are no solutions because the left-hand side would be ≡4 mod7, but the right-hand side is 0 mod7, which is a contradiction.Therefore, the only possible solution is when ℓ=0, which we already found: (3,3,0).So, that must be the only solution.I think that's it. So, the only triplet is (3,3,0).

💡Okay, so I have this problem here about a power function. The function is given as f(x) = k * x^α, where k and α are real numbers. It says that this function passes through the point (4, 1/2), and I need to find the sum of k and α. The options are A: 1/2, B: 1, C: 3/2, and D: 2.Alright, let's break this down. First, I remember that a power function is generally of the form f(x) = k * x^α, where k is a constant and α is the exponent. Since the function passes through the point (4, 1/2), that means when x is 4, f(x) is 1/2. So I can plug these values into the equation to find a relationship between k and α.Let me write that out:f(4) = k * 4^α = 1/2So, k * 4^α = 1/2Hmm, okay. Now I have this equation with two variables, k and α. I need another equation to solve for both, but it seems like there's only one equation here. Wait, maybe I don't need another equation because the problem is asking for k + α, not each individually. Maybe I can express one variable in terms of the other and then add them.Let me try to solve for k first. If I divide both sides of the equation by 4^α, I get:k = (1/2) / 4^αBut that still leaves me with k in terms of α. Maybe I can express 4^α in terms of something else. I know that 4 is 2 squared, so 4^α is (2^2)^α = 2^(2α). So, 4^α = 2^(2α). Let me substitute that back into the equation:k = (1/2) / 2^(2α)Which simplifies to:k = (1/2) * 2^(-2α) = 2^(-1) * 2^(-2α) = 2^(-1 - 2α)Hmm, that might not be the most straightforward way. Maybe I should take logarithms to solve for α. Let me try that.Taking the natural logarithm of both sides of the original equation:ln(k * 4^α) = ln(1/2)Using logarithm properties, ln(ab) = ln(a) + ln(b), so:ln(k) + ln(4^α) = ln(1/2)Which simplifies to:ln(k) + α * ln(4) = ln(1/2)Now, I have an equation involving ln(k) and α. But I still have two variables here. I need another equation, but I don't have any more information. Wait, maybe I can assume that k is a simple number, like 1, and see if that works.If k = 1, then the equation becomes:1 * 4^α = 1/2So, 4^α = 1/2I know that 4 is 2 squared, so 4^α = (2^2)^α = 2^(2α). And 1/2 is 2^(-1). So:2^(2α) = 2^(-1)Since the bases are the same, the exponents must be equal:2α = -1So, α = -1/2Okay, so if k = 1, then α = -1/2. Then, k + α = 1 + (-1/2) = 1/2, which is option A. That seems to fit.But wait, what if k isn't 1? Maybe k is something else. Let me check if there are other possible values for k and α that satisfy the equation k * 4^α = 1/2.Suppose k = 2. Then, 2 * 4^α = 1/2 => 4^α = 1/4 => α = -1. Then, k + α = 2 + (-1) = 1, which is option B.Hmm, so depending on the value of k, I get different sums. But the problem doesn't specify any additional conditions, so there are infinitely many solutions for k and α that satisfy the equation. However, the question is asking for k + α, and it's giving specific options. So maybe there's a standard assumption here, like k is 1 unless specified otherwise.Wait, no, that doesn't make sense. In power functions, k is just a constant multiplier, and α is the exponent. There's no standard assumption that k is 1 unless the function is given as f(x) = x^α. Since it's given as k * x^α, k can be any real number.But since the problem is giving specific options, it's likely that there's a unique solution. Maybe I made a mistake earlier when assuming k = 1. Let me think again.Starting from the original equation:k * 4^α = 1/2I can express this as:k = (1/2) / 4^αBut I need another relation between k and α. Wait, maybe the function is supposed to pass through another point, but it's not given. So, with only one point, I can't uniquely determine both k and α. Therefore, the problem must have some additional information or assumption.Looking back at the problem statement: "Given the power function f(x) = k * x^α (k ∈ R, α ∈ R) passes through the point (4, 1/2), then k + α = ( )"Hmm, it just says it's a power function passing through that point. Maybe in the context of the problem, there's an assumption that the function is of a certain type, like a square root function or something, which would fix α.Wait, if it's a square root function, then α would be 1/2. Let's test that.If α = 1/2, then:k * 4^(1/2) = k * 2 = 1/2 => k = (1/2)/2 = 1/4Then, k + α = 1/4 + 1/2 = 3/4, which isn't one of the options. So that's not it.Alternatively, if it's a reciprocal function, then α = -1.Then, k * 4^(-1) = k * (1/4) = 1/2 => k = (1/2) * 4 = 2So, k + α = 2 + (-1) = 1, which is option B.But earlier, assuming k = 1 gave me α = -1/2 and k + α = 1/2, which is option A.So, depending on the assumption, I get different answers. But the problem doesn't specify any other conditions, so I'm confused.Wait, maybe I should consider that the function is a power function, which typically means that k is non-zero, but that's it. So, without additional information, there are infinitely many solutions for k and α. However, since the problem is giving specific options, perhaps it's expecting a particular solution where k and α are simple fractions or integers.Looking at the options: 1/2, 1, 3/2, 2.If I consider k = 1, then α = -1/2, sum is 1/2.If k = 2, α = -1, sum is 1.If k = 1/2, then 4^α = (1/2)/(1/2) = 1, so α = 0, sum is 1/2 + 0 = 1/2.Wait, that's the same as when k =1.Wait, no, if k = 1/2, then:(1/2) * 4^α = 1/2 => 4^α = 1 => α = 0So, k + α = 1/2 + 0 = 1/2.So, whether k is 1 or 1/2, the sum is 1/2.Wait, but when k =2, α = -1, sum is 1.Hmm, so depending on k, the sum can be 1/2 or 1.But the problem is giving both 1/2 and 1 as options. So, which one is correct?Wait, maybe I need to consider that the function is a power function, and power functions are often defined with α being a real number, but sometimes they are considered with integer exponents. But in the problem, α is given as a real number, so it can be any real number.Wait, but if I take logarithms, I can express α in terms of k or vice versa.From the equation:k * 4^α = 1/2Taking natural logs:ln(k) + α ln(4) = ln(1/2)So,ln(k) = ln(1/2) - α ln(4)But I still have two variables. Unless I can express k in terms of α or vice versa, but without another equation, I can't solve for both.Wait, maybe the problem is assuming that k and α are integers? Let me check.If k and α are integers, then let's see:From k * 4^α = 1/2But 4^α is a positive real number, and 1/2 is also positive. So, k must be positive.If α is an integer, then 4^α is 1, 4, 16, etc., or 1/4, 1/16, etc.So, let's try α = -1:4^(-1) = 1/4Then, k * (1/4) = 1/2 => k = (1/2) / (1/4) = 2So, k = 2, α = -1, sum is 1.If α = -2:4^(-2) = 1/16k * (1/16) = 1/2 => k = (1/2) * 16 = 8Sum is 8 + (-2) = 6, which isn't an option.If α = 0:4^0 = 1k * 1 = 1/2 => k = 1/2Sum is 1/2 + 0 = 1/2.If α = 1:4^1 = 4k * 4 = 1/2 => k = 1/8Sum is 1/8 + 1 = 9/8, which isn't an option.If α = 1/2:4^(1/2) = 2k * 2 = 1/2 => k = 1/4Sum is 1/4 + 1/2 = 3/4, not an option.If α = -1/2:4^(-1/2) = 1/2k * (1/2) = 1/2 => k = 1Sum is 1 + (-1/2) = 1/2.So, when α is -1, k is 2, sum is 1.When α is 0, k is 1/2, sum is 1/2.When α is -1/2, k is 1, sum is 1/2.So, depending on α, the sum can be 1/2 or 1.But the problem is giving both 1/2 and 1 as options, so which one is correct?Wait, maybe the problem is assuming that the function is in its simplest form, meaning that k is 1. So, if k =1, then α = -1/2, sum is 1/2.Alternatively, if the function is supposed to be a simple reciprocal function, then α = -1, k = 2, sum is 1.But without additional information, it's ambiguous.Wait, looking back at the problem statement: "Given the power function f(x)=k⋅x^α (k∈R,α∈R) passes through the point (4,1/2), then k+α=( )"It just says it's a power function, no additional constraints. So, technically, there are infinitely many solutions for k and α, leading to different sums. However, the problem is giving specific options, so perhaps it's expecting the simplest solution where k =1.In many math problems, unless specified otherwise, they often assume the simplest case, which would be k =1. So, if k =1, then α = -1/2, and sum is 1/2, which is option A.Alternatively, if they consider the function to be a reciprocal function, which is a common power function, then α = -1, k =2, sum is 1, which is option B.But since the problem didn't specify any particular form, I think the safest assumption is that k =1, leading to sum 1/2.Wait, but earlier when I assumed k =1, I got α = -1/2, sum 1/2.But when I assumed k =2, I got α = -1, sum 1.So, both are possible, but the problem is giving both options. Hmm.Wait, maybe I should consider that the function is a power function, which is f(x) = kx^α, and in many cases, especially in problems like this, they often set k =1 for simplicity unless told otherwise. So, I think the intended answer is 1/2.But I'm not entirely sure. Maybe I should check both possibilities.If k =1, α = -1/2, sum is 1/2.If k =2, α = -1, sum is 1.Both are valid, but since the problem is giving both options, I need to see which one is more likely.Wait, let me think about the function f(x) = kx^α passing through (4, 1/2). If k =1, then f(x) = x^(-1/2), which is 1/sqrt(x). So, f(4) = 1/sqrt(4) = 1/2, which matches.If k =2, then f(x) = 2x^(-1), which is 2/x. So, f(4) = 2/4 = 1/2, which also matches.So, both functions pass through (4, 1/2). Therefore, both solutions are valid.But the problem is asking for k + α, and both sums are possible: 1/2 and 1.But the options include both, so how do I choose?Wait, maybe the problem is expecting the answer where k and α are both simple fractions, like 1 and -1/2, leading to sum 1/2.Alternatively, if k is 2 and α is -1, sum is 1.Hmm.Wait, let me think about the exponents. If α is -1/2, then the function is f(x) = x^(-1/2) = 1/sqrt(x). That's a common function.If α is -1, then f(x) = 2/x, which is also a common function.But without more information, I can't determine which one is intended.Wait, maybe the problem is from a textbook or a standard test, and in such cases, they often assume k =1 unless told otherwise. So, I think the intended answer is 1/2.But I'm not 100% sure. Maybe I should look for another way.Wait, another approach: since the function is a power function, it's determined uniquely by two points. But here, we only have one point. So, unless there's another condition, like the function being increasing or decreasing, or something else, we can't uniquely determine k and α.But the problem doesn't provide that, so I think the answer is not uniquely determined, but since the options are given, I have to choose between them.Wait, maybe the problem is assuming that the function is of the form f(x) = x^α, meaning k =1. So, in that case, α = -1/2, sum is 1/2.Yes, that makes sense. If the function is given as f(x) = kx^α, but in many cases, especially in problems like this, they might assume k =1 unless specified otherwise. So, I think the answer is 1/2.But I'm still a bit confused because both solutions are valid. Maybe I should go with the answer that is more commonly expected, which is 1/2.So, I think the answer is A: 1/2.