💡Okay, so I have this problem: Among all triangles ABC, find the maximum value of cos A + cos B cos C. Hmm, interesting. I remember that in a triangle, the sum of the angles is 180 degrees, so A + B + C = π radians. Maybe I can use some trigonometric identities or properties of triangles to solve this.First, let me write down what I need to maximize: cos A + cos B cos C. I wonder if there's a way to express this in terms of a single variable or find some relationship between the angles.I recall that in a triangle, the angles are related, so maybe I can express one angle in terms of the others. For example, since A + B + C = π, I can write C = π - A - B. Then, maybe I can substitute that into the expression.So, let's try that. If C = π - A - B, then cos C = cos(π - A - B). I remember that cos(π - x) = -cos x, so cos(π - A - B) = -cos(A + B). Therefore, cos C = -cos(A + B).Now, let's substitute that into the original expression:cos A + cos B cos C = cos A + cos B (-cos(A + B)) = cos A - cos B cos(A + B).Hmm, that might not be the most straightforward path. Maybe I should consider using some trigonometric identities to simplify cos B cos C. I remember that there's a product-to-sum identity: cos B cos C = [cos(B - C) + cos(B + C)] / 2.Let me apply that here:cos A + cos B cos C = cos A + [cos(B - C) + cos(B + C)] / 2.Now, since in a triangle, B + C = π - A, so cos(B + C) = cos(π - A) = -cos A. So substituting that in:cos A + [cos(B - C) + (-cos A)] / 2 = cos A + [cos(B - C) - cos A] / 2.Let me simplify this expression:= cos A + (cos(B - C) - cos A)/2= (2 cos A)/2 + (cos(B - C) - cos A)/2= [2 cos A + cos(B - C) - cos A] / 2= [cos A + cos(B - C)] / 2.So, the expression simplifies to (cos A + cos(B - C)) / 2. Interesting. Now, I need to maximize this.Since we're dealing with a triangle, all angles are between 0 and π radians, so cos A is between -1 and 1, but actually, since A is an angle in a triangle, it's between 0 and π, so cos A is between -1 and 1, but more specifically, for acute angles, it's positive, and for obtuse, it's negative.Similarly, cos(B - C) is between -1 and 1 because B - C can range from -π to π, but since B and C are both between 0 and π, their difference is between -π and π, so cos(B - C) is between -1 and 1.But wait, in a triangle, B and C are both positive and add up to less than π, so B - C can't be more than π or less than -π. So, cos(B - C) is between -1 and 1.But I need to find the maximum of (cos A + cos(B - C)) / 2. So, to maximize this, I need to maximize both cos A and cos(B - C). Since both are being added, their maximums will contribute to the overall maximum.What's the maximum value of cos A? That's 1, which occurs when A = 0, but in a triangle, angles can't be zero. So, the maximum cos A can approach is 1 as A approaches 0. Similarly, cos(B - C) is maximized when B - C = 0, so when B = C, which would make cos(B - C) = cos(0) = 1.So, if I can have A approaching 0 and B = C approaching π/2, then cos A approaches 1, and cos(B - C) = 1. Therefore, the expression (cos A + cos(B - C)) / 2 approaches (1 + 1)/2 = 1.But wait, in a triangle, if A approaches 0, then B + C approaches π, so if B = C, then each approaches π/2. So, in that case, the triangle becomes very "flat" with A approaching 0 and B and C approaching π/2.But is this the maximum? Let me check another case. Suppose the triangle is equilateral, so A = B = C = π/3. Then, cos A = 1/2, and cos(B - C) = cos(0) = 1. So, the expression becomes (1/2 + 1)/2 = (3/2)/2 = 3/4. That's less than 1, so the maximum must be higher.Another case: suppose A = π/2, so it's a right-angled triangle. Then, B + C = π/2. Let's say B = C = π/4. Then, cos A = 0, cos(B - C) = cos(0) = 1. So, the expression is (0 + 1)/2 = 1/2. That's even less.Wait, so when A approaches 0, and B and C approach π/2, the expression approaches 1. But is that actually achievable? Because in a triangle, all angles must be greater than 0. So, A can't be exactly 0, but it can be made arbitrarily small. So, the supremum of the expression is 1, but is it attainable?In the limit as A approaches 0, the expression approaches 1, but never actually reaches it because A can't be 0. So, does that mean the maximum is 1, or is it slightly less?Wait, but in the problem statement, it says "among all triangles ABC," so we're considering all possible triangles, including degenerate ones where A approaches 0. But technically, a degenerate triangle isn't a valid triangle because one angle is 0, making it a straight line. So, maybe the maximum is approached but not attained.But in optimization problems, sometimes we consider limits. So, perhaps the maximum is 1, even though it's not achieved by any actual triangle, but approached as A approaches 0.Alternatively, maybe there's another configuration where the expression actually reaches 1. Let me think.Suppose A is 0, but then it's not a triangle. So, no. Alternatively, if A is very small, say ε, approaching 0, then B and C are approaching π/2 each. So, cos A approaches 1, and cos(B - C) approaches cos(0) = 1. Therefore, the expression approaches 1.But is there a way to make the expression equal to 1? Let me see.If I set A = 0, then B + C = π, but then it's not a triangle. So, no. Therefore, the maximum value is 1, but it's not attained by any triangle; it's just the supremum.Wait, but in the initial steps, I transformed the expression into (cos A + cos(B - C))/2. So, to maximize this, I need both cos A and cos(B - C) to be as large as possible.The maximum of cos A is 1, and the maximum of cos(B - C) is 1. So, their sum is 2, divided by 2 is 1. So, the maximum possible value is 1, but it's only achieved in the limit as A approaches 0 and B = C approach π/2.Therefore, the maximum value of cos A + cos B cos C is 1.Wait, but earlier when I considered A approaching 0, B and C approaching π/2, then cos A approaches 1, and cos B cos C approaches 0 * 0 = 0. Wait, hold on, that contradicts my earlier conclusion.Wait, no. Wait, if A approaches 0, then B + C approaches π. If B and C are approaching π/2 each, then cos B and cos C approach 0. So, cos B cos C approaches 0. So, in that case, cos A + cos B cos C approaches 1 + 0 = 1.But earlier, when I transformed the expression, I had (cos A + cos(B - C))/2. So, if A approaches 0, and B = C approach π/2, then cos(B - C) = cos(0) = 1, so (cos A + 1)/2 approaches (1 + 1)/2 = 1. So, both approaches agree.Wait, but in the original expression, cos A + cos B cos C, when A approaches 0, cos A approaches 1, and cos B cos C approaches 0, so the sum approaches 1. So, that's consistent.But earlier, when I thought of the equilateral triangle, I got 3/4, which is less than 1. So, the maximum must be 1.But wait, let me test another case. Suppose A is 60 degrees, B is 60 degrees, and C is 60 degrees. Then, cos A = 0.5, cos B = 0.5, cos C = 0.5. So, cos A + cos B cos C = 0.5 + 0.25 = 0.75, which is 3/4, as I had before.Another case: A = 90 degrees, B = 45 degrees, C = 45 degrees. Then, cos A = 0, cos B = cos C = √2/2 ≈ 0.707. So, cos B cos C ≈ 0.5. So, the expression is 0 + 0.5 = 0.5.Another case: A = 30 degrees, B = 75 degrees, C = 75 degrees. Then, cos A ≈ 0.866, cos B ≈ 0.2588, cos C ≈ 0.2588. So, cos B cos C ≈ 0.06699. So, the expression is ≈ 0.866 + 0.06699 ≈ 0.933.That's higher than 3/4. So, maybe the maximum is higher than 3/4.Wait, so when A is 30 degrees, B and C are 75 degrees each, the expression is approximately 0.933. If I make A smaller, say A = 10 degrees, then B and C are 85 degrees each. Then, cos A ≈ 0.9848, cos B ≈ 0.08716, cos C ≈ 0.08716. So, cos B cos C ≈ 0.0076. So, the expression is ≈ 0.9848 + 0.0076 ≈ 0.9924.That's even closer to 1. So, as A approaches 0, the expression approaches 1. So, it seems that 1 is indeed the supremum.But wait, let me think again. When A approaches 0, B and C approach π/2 each, so cos B and cos C approach 0. So, cos B cos C approaches 0. So, the expression approaches 1 + 0 = 1. So, yes, that's consistent.But in the transformed expression, (cos A + cos(B - C))/2, when A approaches 0, and B = C approach π/2, then cos(B - C) = cos(0) = 1. So, (cos A + 1)/2 approaches (1 + 1)/2 = 1. So, same result.Therefore, the maximum value is 1, achieved in the limit as A approaches 0 and B = C approach π/2.But wait, in the problem statement, it says "among all triangles ABC." So, does that include degenerate triangles? Because a degenerate triangle with A = 0 is not a valid triangle. So, maybe the maximum is not achieved, but approached.But in optimization, sometimes we talk about suprema even if they're not attained. So, perhaps the answer is 1.Alternatively, maybe there's another configuration where the expression actually reaches 1. Let me think.Suppose A is 0, but that's not a triangle. So, no. Alternatively, if A is very small, say ε, then B and C are (π - ε)/2 each. So, cos A = cos ε ≈ 1 - ε²/2, and cos B = cos C ≈ cos((π - ε)/2) = sin(ε/2) ≈ ε/2. So, cos B cos C ≈ (ε/2)^2 = ε²/4. So, the expression is ≈ 1 - ε²/2 + ε²/4 = 1 - ε²/4, which approaches 1 as ε approaches 0.So, yes, the expression approaches 1 as the triangle becomes degenerate.Therefore, the maximum value is 1.Wait, but earlier when I considered the expression (cos A + cos(B - C))/2, I thought that when A approaches 0 and B = C approach π/2, cos(B - C) = 1, so the expression approaches 1. So, that's consistent.But let me check another approach. Maybe using calculus. Let's consider A as a variable, and express B and C in terms of A.Since A + B + C = π, and assuming B = C for maximum symmetry, which might give the maximum.So, let's set B = C = (π - A)/2.Then, cos B = cos((π - A)/2) = sin(A/2).Similarly, cos C = sin(A/2).So, cos B cos C = sin²(A/2).Therefore, the expression becomes cos A + sin²(A/2).Let me write that as:cos A + sin²(A/2).Now, let's express sin²(A/2) in terms of cos A. I know that sin²(x) = (1 - cos(2x))/2. So, sin²(A/2) = (1 - cos A)/2.Therefore, the expression becomes:cos A + (1 - cos A)/2 = (2 cos A + 1 - cos A)/2 = (cos A + 1)/2.So, the expression simplifies to (1 + cos A)/2.Now, to maximize this, we need to maximize cos A. The maximum value of cos A is 1, which occurs when A = 0. So, the expression becomes (1 + 1)/2 = 1.Again, this suggests that the maximum value is 1, achieved when A approaches 0.But wait, in this case, we assumed B = C. So, does this mean that the maximum occurs when B = C? It seems so, because when we set B = C, we were able to express the problem in terms of a single variable and find that the maximum is 1.Therefore, the maximum value of cos A + cos B cos C is 1, achieved in the limit as A approaches 0 and B = C approach π/2.But wait, let me confirm this with another method. Maybe using Lagrange multipliers or some other optimization technique.Let me consider the function f(A, B, C) = cos A + cos B cos C, subject to the constraint A + B + C = π.We can use the method of Lagrange multipliers. Let's set up the Lagrangian:L = cos A + cos B cos C - λ(A + B + C - π).Taking partial derivatives:∂L/∂A = -sin A - λ = 0 → -sin A = λ∂L/∂B = -sin B cos C - λ = 0 → -sin B cos C = λ∂L/∂C = -sin C cos B - λ = 0 → -sin C cos B = λ∂L/∂λ = A + B + C - π = 0From the first equation: λ = -sin AFrom the second equation: λ = -sin B cos CFrom the third equation: λ = -sin C cos BSo, setting the first equal to the second:-sin A = -sin B cos C → sin A = sin B cos CSimilarly, setting the first equal to the third:-sin A = -sin C cos B → sin A = sin C cos BSo, we have:sin A = sin B cos Csin A = sin C cos BTherefore, sin B cos C = sin C cos BWhich implies sin B cos C - sin C cos B = 0 → sin(B - C) = 0.So, sin(B - C) = 0 → B - C = nπ, but since B and C are angles in a triangle, they are between 0 and π, so B - C = 0 or π.But B - C = π would imply one angle is greater than π, which is impossible in a triangle. So, B - C = 0 → B = C.So, from this, we conclude that B = C.Therefore, the maximum occurs when B = C.So, now, with B = C, we can write A + 2B = π → B = (π - A)/2.So, now, the function becomes f(A) = cos A + cos²((π - A)/2).Let me compute this:cos A + cos²((π - A)/2) = cos A + cos²(π/2 - A/2) = cos A + sin²(A/2).As before, sin²(A/2) = (1 - cos A)/2.So, f(A) = cos A + (1 - cos A)/2 = (2 cos A + 1 - cos A)/2 = (cos A + 1)/2.So, f(A) = (1 + cos A)/2.To maximize this, we need to maximize cos A, which is 1 when A = 0.Therefore, the maximum value is (1 + 1)/2 = 1.So, again, we reach the conclusion that the maximum value is 1, achieved when A approaches 0 and B = C approach π/2.But wait, in this case, we used Lagrange multipliers and found that B = C, leading to the same conclusion. So, it seems consistent.Therefore, the maximum value of cos A + cos B cos C is 1.But let me just think about whether this makes sense geometrically. When the triangle becomes very "flat," with one angle approaching 0 and the other two approaching π/2, the expression approaches 1. So, that seems to be the case.Alternatively, if I consider an equilateral triangle, which is a very "balanced" triangle, the expression is 3/4, which is less than 1. So, the maximum must be higher.Another way to think about it: in a triangle, as one angle becomes very small, the other two become almost right angles. So, cos A approaches 1, and cos B and cos C approach 0, but their product approaches 0. So, the sum cos A + cos B cos C approaches 1.Therefore, the maximum value is indeed 1.So, after considering multiple approaches—using trigonometric identities, assuming symmetry by setting B = C, using Lagrange multipliers—it all points to the maximum value being 1.Therefore, the maximum value of cos A + cos B cos C among all triangles ABC is 1.

💡Okay, so I have this function g(x) = (x - 2)/(x - 5), and I need to find for what value of x is its inverse function g⁻¹(x) undefined. Hmm, let's see. I remember that for a function to have an inverse, it needs to be bijective, meaning it's both injective (one-to-one) and surjective (onto). I think this function is probably bijective because it's a rational function, and as long as it's not constant, it should have an inverse.First, maybe I should find the inverse function g⁻¹(x). To find the inverse, I need to swap x and y in the equation and then solve for y. So let me write that down.Starting with y = (x - 2)/(x - 5). To find the inverse, I'll swap x and y:x = (y - 2)/(y - 5)Now, I need to solve this equation for y. Let's do that step by step.Multiply both sides by (y - 5) to get rid of the denominator:x(y - 5) = y - 2Now, distribute the x on the left side:xy - 5x = y - 2Now, I need to get all the terms with y on one side and the rest on the other side. Let me subtract y from both sides:xy - 5x - y = -2Hmm, that's a bit messy. Maybe I can factor out y from the terms that have it. Let's see:y(x - 1) - 5x = -2Wait, that doesn't seem right. Let me try again. Starting from:xy - 5x = y - 2Let me bring all the y terms to the left and the rest to the right:xy - y = 5x - 2Now, factor out y from the left side:y(x - 1) = 5x - 2Okay, that looks better. Now, solve for y by dividing both sides by (x - 1):y = (5x - 2)/(x - 1)So, the inverse function g⁻¹(x) is (5x - 2)/(x - 1). Now, I need to find where this inverse function is undefined. A function is undefined where its denominator is zero because division by zero is undefined.So, set the denominator equal to zero and solve for x:x - 1 = 0x = 1Therefore, g⁻¹(x) is undefined at x = 1.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the original function, swapped x and y, solved for y, and ended up with (5x - 2)/(x - 1). The denominator is x - 1, so when x is 1, it's undefined. That seems correct.Just to be thorough, maybe I can check if g⁻¹(1) is indeed undefined. Plugging x = 1 into g⁻¹(x):g⁻¹(1) = (5*1 - 2)/(1 - 1) = (5 - 2)/0 = 3/0Which is undefined, as expected. So, yes, x = 1 is the correct answer.I think that's it. The inverse function is undefined at x = 1.

💡First, I need to compare each of the given numbers with (-frac{3}{7}) to determine which one is smaller.**Option A: (-left(-1right))**[-left(-1right) = 1]Since (1) is a positive number, it is greater than (-frac{3}{7}).**Option B: (-frac{8}{21})**To compare (-frac{8}{21}) with (-frac{3}{7}), I'll find a common denominator:[-frac{3}{7} = -frac{9}{21}]Now, comparing (-frac{8}{21}) and (-frac{9}{21}), since (-frac{8}{21}) is closer to zero, it is greater than (-frac{9}{21}).**Option C: (0)**Zero is greater than any negative number, so (0) is greater than (-frac{3}{7}).**Option D: (-0.43)**Converting (-frac{3}{7}) to a decimal:[-frac{3}{7} approx -0.4286]Comparing (-0.43) and (-0.4286), since (-0.43) is further from zero, it is smaller than (-0.4286).After evaluating all options, only (-0.43) is smaller than (-frac{3}{7}).

💡Okay, so I have this problem about an arithmetic sequence. Let me try to figure it out step by step. Hmm, arithmetic sequences have a common difference, right? So each term increases by the same amount. The problem gives me the sum of the first 2 terms, S₂ = 4, and the sum of the first 4 terms, S₄ = 16. I need to find the value of a₅ + a₆. The options are A:11, B:16, C:20, D:28.First, let me recall the formula for the sum of the first n terms of an arithmetic sequence. The sum Sₙ is given by:Sₙ = n/2 [2a₁ + (n - 1)d]where a₁ is the first term and d is the common difference.So, for S₂, which is the sum of the first 2 terms, plugging into the formula:S₂ = 2/2 [2a₁ + (2 - 1)d] = 1 [2a₁ + d] = 2a₁ + dAnd we know S₂ = 4, so:2a₁ + d = 4 ...(1)Similarly, for S₄, the sum of the first 4 terms:S₄ = 4/2 [2a₁ + (4 - 1)d] = 2 [2a₁ + 3d] = 4a₁ + 6dWe know S₄ = 16, so:4a₁ + 6d = 16 ...(2)Now, I have two equations:1) 2a₁ + d = 42) 4a₁ + 6d = 16I can solve these simultaneously to find a₁ and d.Let me solve equation (1) for a₁:2a₁ + d = 4So, 2a₁ = 4 - dTherefore, a₁ = (4 - d)/2Now, substitute this into equation (2):4a₁ + 6d = 16Substituting a₁:4*( (4 - d)/2 ) + 6d = 16Simplify:(16 - 4d)/2 + 6d = 16Which is:8 - 2d + 6d = 16Combine like terms:8 + 4d = 16Subtract 8 from both sides:4d = 8Divide both sides by 4:d = 2Now that I have d, I can find a₁:a₁ = (4 - d)/2 = (4 - 2)/2 = 2/2 = 1So, the first term a₁ is 1, and the common difference d is 2.Now, I need to find a₅ + a₆.In an arithmetic sequence, the nth term is given by:aₙ = a₁ + (n - 1)dSo, let's find a₅ and a₆:a₅ = a₁ + (5 - 1)d = 1 + 4*2 = 1 + 8 = 9a₆ = a₁ + (6 - 1)d = 1 + 5*2 = 1 + 10 = 11Therefore, a₅ + a₆ = 9 + 11 = 20So, the answer should be C:20.Wait, but let me double-check my calculations to make sure I didn't make a mistake.First, checking a₁ and d:From S₂ = 4: 2a₁ + d = 4If a₁ =1, d=2, then 2*1 + 2 = 4, which is correct.From S₄ = 16: 4a₁ + 6d = 16Plugging in a₁=1, d=2: 4*1 + 6*2 = 4 + 12 = 16, which is correct.So, a₁ and d are correct.Calculating a₅ and a₆:a₅ = 1 + 4*2 = 9a₆ = 1 + 5*2 = 11Sum is 20, which is option C.Alternatively, maybe there's another way to approach this problem without finding a₁ and d.I remember that in an arithmetic sequence, the sums of terms also form an arithmetic sequence. So, S₂, S₄ - S₂, S₆ - S₄, etc., should form an arithmetic sequence.Given S₂ = 4 and S₄ = 16, so S₄ - S₂ = 16 - 4 = 12.So, the differences between the sums are 4, 12, and then the next difference should be 20, since the common difference in the sums' differences is 8 (12 - 4 = 8). Wait, is that right?Wait, if S₂ = 4, then S₄ - S₂ = 12, which is the sum of the third and fourth terms. Then, S₆ - S₄ would be the sum of the fifth and sixth terms, which is a₅ + a₆, which is what we're looking for.Since the sums of the terms in arithmetic sequence also form an arithmetic sequence, the differences between consecutive sums should form an arithmetic sequence.So, the differences are S₂ = 4, S₄ - S₂ = 12, and S₆ - S₄ = ?Since the differences themselves form an arithmetic sequence, the difference between 12 and 4 is 8, so the next difference should be 12 + 8 = 20. Therefore, S₆ - S₄ = 20, which is a₅ + a₆.So, that confirms that a₅ + a₆ = 20, which is option C.I think that's solid. Both methods give me the same answer, so I feel confident.**Final Answer**boxed{C}

💡Okay, so I need to find the remainder when the 50th term of the Lucas sequence is divided by 5. Hmm, let me start by understanding what the Lucas sequence is. From the problem statement, it says the first term is 1, the second term is 3, and each subsequent term is the sum of the two preceding terms. So, it's similar to the Fibonacci sequence but with different starting values.Let me write down the first few terms to get a sense of the pattern:- L₁ = 1- L₂ = 3- L₃ = L₁ + L₂ = 1 + 3 = 4- L₄ = L₂ + L₃ = 3 + 4 = 7- L₅ = L₃ + L₄ = 4 + 7 = 11- L₆ = L₄ + L₅ = 7 + 11 = 18- L₇ = L₅ + L₆ = 11 + 18 = 29- L₈ = L₆ + L₇ = 18 + 29 = 47- L₉ = L₇ + L₈ = 29 + 47 = 76- L₁₀ = L₈ + L₉ = 47 + 76 = 123Wait, calculating up to the 50th term manually would take forever. There must be a smarter way, especially since we only need the remainder when divided by 5. Maybe I can find a pattern in the remainders of the Lucas sequence modulo 5.Let me compute the terms modulo 5:- L₁ mod 5 = 1 mod 5 = 1- L₂ mod 5 = 3 mod 5 = 3- L₃ = 4 mod 5 = 4- L₄ = 7 mod 5 = 2 (since 7 - 5 = 2)- L₅ = 11 mod 5 = 1 (since 11 - 2*5 = 1)- L₆ = 18 mod 5 = 3 (since 18 - 3*5 = 3)- L₇ = 29 mod 5 = 4 (since 29 - 5*5 = 4)- L₈ = 47 mod 5 = 2 (since 47 - 9*5 = 2)- L₉ = 76 mod 5 = 1 (since 76 - 15*5 = 1)- L₁₀ = 123 mod 5 = 3 (since 123 - 24*5 = 3)Hmm, looking at these remainders: 1, 3, 4, 2, 1, 3, 4, 2, 1, 3. It seems like after L₅, the sequence of remainders starts repeating every 4 terms: 1, 3, 4, 2, and then again 1, 3, 4, 2, etc. So the cycle length here is 4.If the cycle repeats every 4 terms, then to find L₅₀ mod 5, I can figure out where 50 falls in the cycle. Let me see: starting from L₁, the cycle is 1, 3, 4, 2, and then repeats. But wait, the cycle starts repeating from L₅, right? So L₅ is the start of the cycle.So, if I consider the cycle starting at L₅, then L₅ corresponds to the first term of the cycle, L₆ to the second, L₇ to the third, L₈ to the fourth, and then L₉ back to the first, and so on.Therefore, to find L₅₀, I need to see how many terms after L₅ it is. So, 50 - 4 = 46 terms after L₄. Wait, actually, maybe another approach is better. Since the cycle length is 4, starting from L₅, I can calculate the position of L₅₀ in the cycle.Let me think: starting from L₅, the cycle is 1, 3, 4, 2, and repeats every 4 terms. So, the position of L₅₀ in the cycle can be found by calculating (50 - 4) mod 4. Wait, why subtract 4? Because L₅ is the first term of the cycle, so L₅ is position 1, L₆ is position 2, L₇ is position 3, L₈ is position 4, L₉ is position 1 again, and so on.So, to find the position of L₅₀, I can calculate (50 - 4) mod 4, which is 46 mod 4. 46 divided by 4 is 11 with a remainder of 2. So, 46 mod 4 is 2. Therefore, L₅₀ corresponds to the second term in the cycle, which is 3.Wait, let me double-check that. If L₅ is position 1, then L₅₀ is 50 - 4 = 46 terms after L₅. Since the cycle is 4, 46 mod 4 is 2, so it's the second term in the cycle, which is 3. So, L₅₀ mod 5 is 3.Alternatively, I can list the positions:- L₅: position 1 → 1- L₆: position 2 → 3- L₇: position 3 → 4- L₈: position 4 → 2- L₉: position 1 → 1- L₁₀: position 2 → 3- ...- L₅₀: position (50 - 4) mod 4 = 46 mod 4 = 2 → 3Yes, that seems consistent. So, the remainder when the 50th term is divided by 5 is 3.

💡First, I need to simplify the expression ( frac{1}{y} - frac{1}{x} ). To do this, I'll find a common denominator, which is ( xy ).So, ( frac{1}{y} - frac{1}{x} = frac{x}{xy} - frac{y}{xy} = frac{x - y}{xy} ).Next, I'll use the given equation ( xy = x - y ). Substituting this into the simplified expression:( frac{x - y}{xy} = frac{xy}{xy} = 1 ).Therefore, the value of ( frac{1}{y} - frac{1}{x} ) is 1.