💡First, I need to determine the unknown percentage that, when applied to the value of ( x = 170 ), equals 20% of 552.50.I'll start by calculating 20% of 552.50:[20% times 552.50 = 0.20 times 552.50 = 110.50]Next, I'll set up the equation where the unknown percentage of 170 equals 110.50:[frac{text{Percentage}}{100} times 170 = 110.50]To solve for the percentage, I'll divide both sides of the equation by 170:[frac{text{Percentage}}{100} = frac{110.50}{170}]Calculating the right side:[frac{110.50}{170} = 0.65]Finally, I'll multiply both sides by 100 to find the percentage:[text{Percentage} = 0.65 times 100 = 65%]

💡First, I need to find an equation involving ( x ) and ( y ) that, when combined with ( 2x + y = 7 ), gives the correct value for ( frac{x + y}{3} ).Given that ( frac{x + y}{3} = 1.888888888888889 ), I can start by isolating ( x + y ) by multiplying both sides of the equation by 3:[x + y = 3 times 1.888888888888889 = 5.666666666666667]Now, I have two equations:1. ( 2x + y = 7 )2. ( x + y = 5.666666666666667 )To find the values of ( x ) and ( y ), I can subtract the second equation from the first:[(2x + y) - (x + y) = 7 - 5.666666666666667][x = 1.3333333333333333]Now that I have ( x ), I can substitute it back into the second equation to find ( y ):[1.3333333333333333 + y = 5.666666666666667][y = 5.666666666666667 - 1.3333333333333333][y = 4.333333333333333]Finally, I can verify the solution by plugging ( x ) and ( y ) back into the original equations to ensure they hold true.

💡First, I need to determine the speed of the boat in still water and the speed of the current. I'll denote the boat's speed in still water as ( b ) miles per hour and the current's speed as ( c ) miles per hour.When the boat is going downstream, the current aids its motion, so the effective speed is ( b + c ). The boat travels 24 miles in 4 hours downstream, which gives the equation:[(b + c) times 4 = 24]Simplifying this, I get:[b + c = 6 quad text{(Equation 1)}]When the boat is going upstream, it works against the current, so the effective speed is ( b - c ). The boat travels the same 24 miles in 6 hours upstream, leading to the equation:[(b - c) times 6 = 24]Simplifying this, I get:[b - c = 4 quad text{(Equation 2)}]Now, I have a system of two equations:[begin{cases}b + c = 6 b - c = 4end{cases}]I can solve this system by adding the two equations together to eliminate ( c ):[(b + c) + (b - c) = 6 + 4][2b = 10][b = 5]With ( b = 5 ), I can substitute back into Equation 1 to find ( c ):[5 + c = 6][c = 1]Therefore, the rate of the current in the river is 1 mile per hour.

💡Alright, so I've got this problem about a ship called the "Wuyang" that sails downstream and upstream, and I need to find the speed of the water flow. Hmm, okay, let me try to break this down step by step.First, I know that when a ship sails downstream, its effective speed increases because the current helps it move faster. Conversely, when it sails upstream, it's going against the current, so its effective speed decreases. The problem gives me two different scenarios with different distances sailed downstream and upstream, but both take the same total time of 12.5 hours. That must mean there's a relationship between the ship's speed in still water and the speed of the current that we can figure out.Let me denote the ship's speed in still water as ( x ) kilometers per hour and the speed of the water flow as ( y ) kilometers per hour. So, when the ship is going downstream, its effective speed would be ( x + y ), and when it's going upstream, it would be ( x - y ).Now, the first scenario says the ship sails downstream for 135 kilometers and then upstream for 70 kilometers, taking a total of 12.5 hours. The time taken to sail a certain distance is equal to the distance divided by the speed. So, the time taken to go downstream would be ( frac{135}{x + y} ) hours, and the time taken to go upstream would be ( frac{70}{x - y} ) hours. Adding these two times together gives the total time of 12.5 hours. So, I can write the equation:[frac{135}{x + y} + frac{70}{x - y} = 12.5]Okay, that's the first equation. Now, the second scenario says the ship sails downstream for 75 kilometers and then upstream for 110 kilometers, also taking 12.5 hours. Using the same logic, the time taken downstream is ( frac{75}{x + y} ) hours, and upstream is ( frac{110}{x - y} ) hours. So, the second equation is:[frac{75}{x + y} + frac{110}{x - y} = 12.5]Now I have a system of two equations with two variables, ( x ) and ( y ). I need to solve this system to find the value of ( y ), which is the speed of the water flow.Let me write down the equations again for clarity:1. ( frac{135}{x + y} + frac{70}{x - y} = 12.5 )2. ( frac{75}{x + y} + frac{110}{x - y} = 12.5 )Hmm, both equations equal 12.5, so maybe I can set them equal to each other or subtract one from the other to eliminate some variables. Let me try subtracting the second equation from the first to see what happens:[left( frac{135}{x + y} + frac{70}{x - y} right) - left( frac{75}{x + y} + frac{110}{x - y} right) = 12.5 - 12.5]Simplifying the left side:[frac{135 - 75}{x + y} + frac{70 - 110}{x - y} = 0]Which simplifies to:[frac{60}{x + y} - frac{40}{x - y} = 0]So, moving one term to the other side:[frac{60}{x + y} = frac{40}{x - y}]Now, I can cross-multiply to solve for ( x ) and ( y ):[60(x - y) = 40(x + y)]Expanding both sides:[60x - 60y = 40x + 40y]Bringing like terms to one side:[60x - 40x = 60y + 40y]Simplifying:[20x = 100y]Dividing both sides by 20:[x = 5y]Okay, so now I know that the ship's speed in still water is five times the speed of the water flow. That's a useful relationship. Now, I can substitute ( x = 5y ) back into one of the original equations to solve for ( y ). Let me choose the first equation:[frac{135}{5y + y} + frac{70}{5y - y} = 12.5]Simplifying the denominators:[frac{135}{6y} + frac{70}{4y} = 12.5]Which is:[frac{135}{6y} + frac{70}{4y} = 12.5]Let me simplify each fraction:[frac{135}{6y} = frac{22.5}{y}][frac{70}{4y} = frac{17.5}{y}]So, adding these together:[frac{22.5}{y} + frac{17.5}{y} = 12.5]Combining the fractions:[frac{40}{y} = 12.5]Now, solving for ( y ):[y = frac{40}{12.5}]Calculating that:[y = 3.2]So, the speed of the water flow is 3.2 kilometers per hour. Let me just double-check this with the second equation to make sure I didn't make a mistake.Using ( x = 5y = 5 times 3.2 = 16 ) km/h.Substituting into the second equation:[frac{75}{16 + 3.2} + frac{110}{16 - 3.2} = 12.5]Calculating the denominators:[16 + 3.2 = 19.2][16 - 3.2 = 12.8]So, the equation becomes:[frac{75}{19.2} + frac{110}{12.8} = 12.5]Calculating each term:[frac{75}{19.2} approx 3.90625][frac{110}{12.8} approx 8.59375]Adding them together:[3.90625 + 8.59375 = 12.5]Which matches the total time given. So, my calculations seem correct. Therefore, the speed of the water flow is indeed 3.2 kilometers per hour.

💡First, I'll denote A as the amount that person a has and B as the amount that person b has.From the problem, I know that the total amount they have together is 2530. So, I can write the equation:A + B = 2530Next, the problem states that 3/5 of a's amount is equal to 2/7 of b's amount. This gives me another equation:(3/5)A = (2/7)BTo solve for B, I'll first express A in terms of B using the second equation. By rearranging, I find:A = (10/21)BNow, I'll substitute this expression for A back into the first equation:(10/21)B + B = 2530Combining the terms on the left side, I get:(31/21)B = 2530To isolate B, I'll multiply both sides of the equation by 21:31B = 53130Finally, I'll divide both sides by 31 to find the value of B:B = 53130 / 31B = 1714Therefore, person b has 1714.

💡Alright, so I've got this problem here about a function f(x) and its derivative f'(x). The function is non-constant, and both f(x) and f'(x) are defined for all real numbers. The problem states two things: first, that f(2 - x) is an odd function, and second, that f(2x + 4) is an even function. Then, I need to figure out which of the options A, B, C, or D are correct based on these properties.Let me start by recalling what it means for a function to be odd or even. An odd function satisfies the property g(-x) = -g(x) for all x, and an even function satisfies g(-x) = g(x) for all x. So, applying these definitions to the given functions.First, f(2 - x) is odd. Let me denote g(x) = f(2 - x). Since g is odd, g(-x) = -g(x). So, substituting, f(2 - (-x)) = -f(2 - x), which simplifies to f(2 + x) = -f(2 - x). That's an important relationship: f(2 + x) = -f(2 - x). Maybe I can use this to find some symmetry or specific values of f(x).Next, f(2x + 4) is even. Let me denote h(x) = f(2x + 4). Since h is even, h(-x) = h(x). So, f(2*(-x) + 4) = f(2x + 4), which simplifies to f(-2x + 4) = f(2x + 4). Hmm, that's another relationship: f(-2x + 4) = f(2x + 4). Maybe I can manipulate this to find another symmetry or perhaps relate it to the first equation.Let me see if I can combine these two properties. From the first property, f(2 + x) = -f(2 - x). If I let x = t - 2, then f(2 + (t - 2)) = -f(2 - (t - 2)), which simplifies to f(t) = -f(4 - t). So, f(t) = -f(4 - t). That's a nice relationship. It tells me that the function f is symmetric about the point (2, 0), but with a negative sign, so it's like a rotational symmetry.From the second property, f(-2x + 4) = f(2x + 4). Let me make a substitution here as well. Let me set y = 2x. Then, f(-y + 4) = f(y + 4). So, f(4 - y) = f(4 + y). That tells me that f is symmetric about the line x = 4. So, f(4 - y) = f(4 + y). That's another symmetry.Now, combining the two symmetries: from the first, f(t) = -f(4 - t), and from the second, f(4 - t) = f(4 + t). So, substituting the second into the first, f(t) = -f(4 + t). So, f(t) = -f(t + 4). That's interesting. So, shifting t by 4 flips the sign of the function. Let me write that down: f(t) = -f(t + 4).If I apply this again, f(t + 4) = -f(t + 8). But from the previous equation, f(t + 4) = -f(t). So, substituting, f(t + 4) = -f(t) = -f(t + 8). Therefore, -f(t) = -f(t + 8), which implies f(t) = f(t + 8). So, the function f is periodic with period 8. That's a key insight.So, f is periodic with period 8, meaning f(t + 8) = f(t) for all t. Also, from earlier, we have f(t) = -f(t + 4). So, every shift by 4 flips the sign, and every shift by 8 brings it back to the original value.Now, let's look at the options one by one.Option A: f(2) = 1.From the first property, when x = 0, f(2 - 0) = f(2) is equal to -f(2 - 0) = -f(2). So, f(2) = -f(2). The only number that is equal to its own negative is zero. Therefore, f(2) must be zero. So, f(2) = 0, not 1. Therefore, option A is incorrect.Option B: f(2024) = -f(2020).Let's see. Since f is periodic with period 8, f(2024) = f(2024 mod 8). Let's compute 2024 divided by 8. 8*253 = 2024, so 2024 mod 8 is 0. Therefore, f(2024) = f(0). Similarly, f(2020) = f(2020 mod 8). 2020 divided by 8 is 252*8 = 2016, so 2020 - 2016 = 4. Therefore, f(2020) = f(4). From the earlier relationship, f(t) = -f(t + 4). So, f(0) = -f(4). Therefore, f(2024) = f(0) = -f(4) = -f(2020). So, f(2024) = -f(2020). Therefore, option B is correct.Option C: f'(-1) = f'(7).Since f is periodic with period 8, its derivative f' is also periodic with period 8. Because the derivative of a periodic function is also periodic with the same period. So, f'(x + 8) = f'(x). Therefore, f'(-1) = f'(-1 + 8) = f'(7). So, yes, f'(-1) = f'(7). Therefore, option C is correct.Option D: f'(-2021) = f'(2025).Again, since f' is periodic with period 8, f'(-2021) = f'(-2021 mod 8). Let's compute -2021 mod 8. 2021 divided by 8 is 252*8 = 2016, so 2021 - 2016 = 5. Therefore, -2021 mod 8 is equivalent to -5 mod 8, which is 3 (since -5 + 8 = 3). So, f'(-2021) = f'(3). Similarly, f'(2025) = f'(2025 mod 8). 2025 divided by 8 is 253*8 = 2024, so 2025 - 2024 = 1. Therefore, f'(2025) = f'(1). Now, do we have a relationship between f'(1) and f'(3)?From the earlier relationship, f(t) = -f(t + 4). Let's differentiate both sides with respect to t. f'(t) = -f'(t + 4). So, f'(t) = -f'(t + 4). Let's set t = 1: f'(1) = -f'(5). Similarly, set t = 3: f'(3) = -f'(7). But from option C, we know that f'(7) = f'(-1). Wait, but we also have f'(t) = -f'(t + 4). So, f'(7) = -f'(11). But f'(11) = f'(3) because 11 mod 8 is 3. Therefore, f'(7) = -f'(3). But from option C, f'(-1) = f'(7), so f'(-1) = -f'(3). Therefore, f'(3) = -f'(-1). But f'(-1) = f'(7), so f'(3) = -f'(7). Hmm, this seems a bit convoluted.Wait, maybe I can approach it differently. Since f(t) = -f(t + 4), differentiating both sides gives f'(t) = -f'(t + 4). So, f'(t + 4) = -f'(t). Therefore, f'(t + 8) = -f'(t + 4) = --f'(t) = f'(t). So, the derivative has period 8, which we already knew.But we need to relate f'(1) and f'(3). Let's see. From f'(t) = -f'(t + 4), set t = 1: f'(1) = -f'(5). Similarly, set t = 3: f'(3) = -f'(7). But f'(5) = f'(5 - 8) = f'(-3). And f'(7) = f'(7 - 8) = f'(-1). So, f'(1) = -f'(-3) and f'(3) = -f'(-1). But f'(-3) = f'(5) because of periodicity, and f'(-1) = f'(7). So, it's a bit circular.Wait, maybe I can use the fact that f(t) = -f(4 - t). Let's differentiate both sides: f'(t) = f'(4 - t)*(-1). So, f'(t) = -f'(4 - t). So, f'(t) + f'(4 - t) = 0. So, for any t, the derivative at t plus the derivative at 4 - t equals zero.Let me test this with t = 1: f'(1) + f'(3) = 0. Therefore, f'(1) = -f'(3). Similarly, for t = 3: f'(3) + f'(1) = 0, which is consistent.But how does this help with f'(-2021) and f'(2025)? Earlier, we saw that f'(-2021) = f'(3) and f'(2025) = f'(1). Since f'(1) = -f'(3), then f'(-2021) = f'(3) = -f'(1) = -f'(2025). Wait, that would mean f'(-2021) = -f'(2025), not equal. But that contradicts the earlier conclusion.Wait, maybe I made a mistake in the modular arithmetic. Let me double-check.For f'(-2021): -2021 divided by 8. 8*252 = 2016, so -2021 = -2016 -5 = -2016 -5. So, -2021 mod 8 is equivalent to (-5) mod 8, which is 3, because -5 + 8 = 3. So, f'(-2021) = f'(3).For f'(2025): 2025 divided by 8 is 253*8 = 2024, so 2025 - 2024 = 1. Therefore, f'(2025) = f'(1).From the relationship f'(t) = -f'(4 - t), setting t = 1: f'(1) = -f'(3). Therefore, f'(3) = -f'(1). So, f'(-2021) = f'(3) = -f'(1) = -f'(2025). Therefore, f'(-2021) = -f'(2025). But option D says f'(-2021) = f'(2025). So, unless f'(2025) = 0, which we don't know, this would mean that f'(-2021) = -f'(2025), which is not the same as option D. Therefore, option D seems incorrect.Wait, but earlier I thought f'(t) = -f'(t + 4). So, f'(1) = -f'(5). And f'(5) = f'(5 - 8) = f'(-3). Similarly, f'(-3) = f'(-3 + 8) = f'(5). So, f'(1) = -f'(5) = -f'(-3). But f'(-3) = f'(5) = -f'(1). So, it's consistent.But going back to f'(t) = -f'(4 - t). So, f'(1) = -f'(3), and f'(3) = -f'(1). So, f'(1) = -f'(3) = --f'(1) = f'(1). So, this implies that f'(1) = f'(1), which is always true, but doesn't give us new information.Wait, maybe I need to consider the chain rule when differentiating f(2x + 4). Let me go back to the original function h(x) = f(2x + 4), which is even. So, h(-x) = h(x). Therefore, f(2*(-x) + 4) = f(2x + 4), which simplifies to f(-2x + 4) = f(2x + 4). Differentiating both sides with respect to x: h'(-x)*(-1) = h'(x). So, -h'(-x) = h'(x). Therefore, h'(-x) = -h'(x). So, h' is odd.But h(x) = f(2x + 4), so h'(x) = 2f'(2x + 4). Therefore, h'(-x) = 2f'(-2x + 4). And h'(-x) = -h'(x) = -2f'(2x + 4). Therefore, 2f'(-2x + 4) = -2f'(2x + 4). Dividing both sides by 2: f'(-2x + 4) = -f'(2x + 4). Let me set y = 2x. Then, f'(4 - y) = -f'(4 + y). So, f'(4 - y) = -f'(4 + y). That's another relationship for the derivative.So, f'(4 - y) = -f'(4 + y). Let me set y = t - 4. Then, f'(4 - (t - 4)) = -f'(4 + (t - 4)), which simplifies to f'(8 - t) = -f'(t). So, f'(8 - t) = -f'(t). But since f' is periodic with period 8, f'(8 - t) = f'(-t). Therefore, f'(-t) = -f'(t). So, f' is an odd function. Wait, that's interesting.So, f' is odd: f'(-t) = -f'(t). So, for any t, f'(-t) = -f'(t). Therefore, f'(-1) = -f'(1), f'(-3) = -f'(3), etc.But earlier, from the relationship f'(t) = -f'(4 - t), let's see if that aligns with f' being odd.From f'(t) = -f'(4 - t), and f' being odd: f'(4 - t) = -f'(t - 4). So, substituting into the first equation: f'(t) = -(-f'(t - 4)) = f'(t - 4). Therefore, f'(t) = f'(t - 4). But since f' is periodic with period 8, f'(t - 4) = f'(t + 4). Therefore, f'(t) = f'(t + 4). But earlier, we had f'(t) = -f'(t + 4). So, combining these: f'(t) = f'(t + 4) and f'(t) = -f'(t + 4). Therefore, f'(t) = -f'(t), which implies f'(t) = 0 for all t. But the problem states that f is a non-constant function, so its derivative cannot be zero everywhere. Therefore, there must be a mistake in my reasoning.Wait, let's go back. From f'(t) = -f'(4 - t), and f' being odd: f'(4 - t) = -f'(t - 4). So, f'(t) = -(-f'(t - 4)) = f'(t - 4). Therefore, f'(t) = f'(t - 4). But f' is periodic with period 8, so f'(t - 4) = f'(t + 4). Therefore, f'(t) = f'(t + 4). But earlier, from differentiating f(t) = -f(t + 4), we had f'(t) = -f'(t + 4). So, combining these: f'(t) = f'(t + 4) and f'(t) = -f'(t + 4). Therefore, f'(t + 4) = -f'(t + 4), which implies f'(t + 4) = 0. But this would mean f'(t) = 0 for all t, which contradicts the function being non-constant.Hmm, this suggests that there's a contradiction, which probably means I made a mistake in my reasoning somewhere. Let me try to trace back.From the original function h(x) = f(2x + 4) being even, we derived that h'(-x) = -h'(x), which led to f'(-2x + 4) = -f'(2x + 4). Then, setting y = 2x, we got f'(4 - y) = -f'(4 + y). Then, setting y = t - 4, we got f'(8 - t) = -f'(t). Since f' is periodic with period 8, f'(8 - t) = f'(-t). Therefore, f'(-t) = -f'(t), meaning f' is odd.But from the other property, f(t) = -f(t + 4), differentiating gives f'(t) = -f'(t + 4). So, f'(t + 4) = -f'(t). But since f' is odd, f'(t + 4) = -f'(-t - 4). Wait, maybe I can relate these.From f'(t + 4) = -f'(t), and f' being odd: f'(t + 4) = -f'(-t). Therefore, -f'(-t) = -f'(t). So, f'(-t) = f'(t). But this contradicts the earlier conclusion that f' is odd, i.e., f'(-t) = -f'(t). Therefore, f'(t) = f'(-t) and f'(-t) = -f'(t) implies f'(t) = 0, which again is a contradiction.This suggests that my earlier assumption that f' is odd might be incorrect, or perhaps I made a mistake in the differentiation step.Wait, let's go back to h(x) = f(2x + 4). Since h is even, h(-x) = h(x). Therefore, f(2*(-x) + 4) = f(2x + 4), which is f(-2x + 4) = f(2x + 4). Differentiating both sides with respect to x: h'(-x)*(-1) = h'(x). So, -h'(-x) = h'(x). Therefore, h'(-x) = -h'(x). So, h' is odd.But h'(x) = 2f'(2x + 4). Therefore, h'(-x) = 2f'(-2x + 4). And h'(-x) = -h'(x) = -2f'(2x + 4). Therefore, 2f'(-2x + 4) = -2f'(2x + 4). Dividing both sides by 2: f'(-2x + 4) = -f'(2x + 4). Let me set y = 2x. Then, f'(4 - y) = -f'(4 + y). So, f'(4 - y) = -f'(4 + y). Let me set y = t - 4. Then, f'(4 - (t - 4)) = -f'(4 + (t - 4)), which simplifies to f'(8 - t) = -f'(t). Since f' is periodic with period 8, f'(8 - t) = f'(-t). Therefore, f'(-t) = -f'(t). So, f' is odd.But from the other property, f(t) = -f(t + 4), differentiating gives f'(t) = -f'(t + 4). So, f'(t + 4) = -f'(t). But since f' is odd, f'(t + 4) = -f'(-t - 4). Wait, let's see:From f'(t + 4) = -f'(t), and f' being odd: f'(t + 4) = -f'(-t). Therefore, -f'(-t) = -f'(t). So, f'(-t) = f'(t). But this contradicts f' being odd, which requires f'(-t) = -f'(t). Therefore, f'(t) = f'(-t) and f'(-t) = -f'(t) implies f'(t) = 0, which is a contradiction.This suggests that there's an inconsistency in the properties, which probably means that the only way to resolve this is if f'(t) = 0, but since f is non-constant, this is impossible. Therefore, perhaps my initial assumption that f is periodic with period 8 is incorrect, or perhaps I made a mistake in deriving the relationships.Wait, let's go back to the beginning. From f(2 - x) being odd, we have f(2 + x) = -f(2 - x). From f(2x + 4) being even, we have f(-2x + 4) = f(2x + 4). Let me try to relate these without assuming periodicity.From f(2 + x) = -f(2 - x), let me set x = y - 2. Then, f(2 + (y - 2)) = -f(2 - (y - 2)), which simplifies to f(y) = -f(4 - y). So, f(y) = -f(4 - y). That's one relationship.From f(-2x + 4) = f(2x + 4), let me set x = z/2. Then, f(-z + 4) = f(z + 4). So, f(4 - z) = f(4 + z). That's another relationship.Combining these two: from f(4 - z) = f(4 + z), and f(4 - z) = -f(z). Therefore, f(4 + z) = -f(z). So, f(z + 4) = -f(z). Therefore, f(z + 8) = f(z). So, f is periodic with period 8. So, that part seems correct.Therefore, f(z + 4) = -f(z), and f(z + 8) = f(z). So, f is periodic with period 8, and f(z + 4) = -f(z). Therefore, f'(z + 4) = -f'(z). Because differentiating both sides: d/dz [f(z + 4)] = d/dz [-f(z)] => f'(z + 4) = -f'(z).Also, since f is periodic with period 8, f'(z + 8) = f'(z). Therefore, f'(z + 4) = -f'(z), and f'(z + 8) = f'(z). So, combining these, f'(z + 8) = f'(z) = -f'(z + 4) = --f'(z) = f'(z). So, consistent.Now, from the derivative properties: f'(z + 4) = -f'(z), and f'(z + 8) = f'(z). Also, from the earlier relationship, f'(4 - z) = -f'(4 + z). Let me set z = t - 4. Then, f'(4 - (t - 4)) = -f'(4 + (t - 4)), which simplifies to f'(8 - t) = -f'(t). But since f' is periodic with period 8, f'(8 - t) = f'(-t). Therefore, f'(-t) = -f'(t). So, f' is odd.But from f'(z + 4) = -f'(z), and f' being odd: f'(z + 4) = -f'(-z). Therefore, -f'(-z) = -f'(z). So, f'(-z) = f'(z). But this contradicts f' being odd, which requires f'(-z) = -f'(z). Therefore, f'(z) = f'(-z) and f'(-z) = -f'(z) implies f'(z) = 0, which is a contradiction.This suggests that the only way to resolve this is if f'(z) = 0, but since f is non-constant, this is impossible. Therefore, there must be a mistake in my reasoning.Wait, perhaps the mistake is in assuming that f'(4 - z) = -f'(4 + z). Let me re-examine that step.From f(4 - z) = f(4 + z), differentiating both sides: f'(-z) = f'(z). Wait, no. Let me differentiate f(4 - z) = f(4 + z) with respect to z.Left side: d/dz [f(4 - z)] = -f'(4 - z).Right side: d/dz [f(4 + z)] = f'(4 + z).Therefore, -f'(4 - z) = f'(4 + z). So, f'(4 + z) = -f'(4 - z). That's the correct relationship.So, f'(4 + z) = -f'(4 - z). Let me set z = t - 4. Then, f'(4 + (t - 4)) = -f'(4 - (t - 4)), which simplifies to f'(t) = -f'(8 - t). But since f' is periodic with period 8, f'(8 - t) = f'(-t). Therefore, f'(t) = -f'(-t). So, f' is odd: f'(-t) = -f'(t).Therefore, f' is odd, and from f'(t + 4) = -f'(t), we have f'(t + 4) = -f'(t). But since f' is odd, f'(t + 4) = -f'(-t - 4). Wait, let's see:From f'(t + 4) = -f'(t), and f' being odd: f'(t + 4) = -f'(-t). Therefore, -f'(-t) = -f'(t). So, f'(-t) = f'(t). But this contradicts f' being odd, which requires f'(-t) = -f'(t). Therefore, f'(t) = f'(-t) and f'(-t) = -f'(t) implies f'(t) = 0, which is a contradiction.This suggests that the only way to resolve this is if f'(t) = 0, but since f is non-constant, this is impossible. Therefore, perhaps the initial conditions are contradictory, but the problem states that f is non-constant, so there must be a way to resolve this.Wait, maybe I'm overcomplicating it. Let's try to use the periodicity and the oddness of f' to evaluate the options.From f' being odd: f'(-t) = -f'(t).From f'(t + 4) = -f'(t).So, f'(-1) = -f'(1).But from f'(t + 4) = -f'(t), f'(1) = -f'(5). And f'(5) = f'(5 - 8) = f'(-3). So, f'(1) = -f'(-3). But f'(-3) = -f'(3). Therefore, f'(1) = -(-f'(3)) = f'(3). So, f'(1) = f'(3).But from f'(t + 4) = -f'(t), f'(3) = -f'(7). And f'(7) = f'(7 - 8) = f'(-1). So, f'(3) = -f'(-1). But f'(-1) = -f'(1). Therefore, f'(3) = -(-f'(1)) = f'(1). So, consistent.Therefore, f'(1) = f'(3) = -f'(7) = -f'(-1). So, f'(-1) = -f'(1).Now, let's look at option D: f'(-2021) = f'(2025).From earlier, f'(-2021) = f'(3) and f'(2025) = f'(1). From above, f'(3) = f'(1). Therefore, f'(-2021) = f'(3) = f'(1) = f'(2025). Therefore, f'(-2021) = f'(2025). So, option D is correct.Wait, but earlier I thought f'(3) = -f'(7) and f'(7) = f'(-1), so f'(3) = -f'(-1). But since f'(-1) = -f'(1), then f'(3) = -(-f'(1)) = f'(1). So, f'(3) = f'(1). Therefore, f'(-2021) = f'(3) = f'(1) = f'(2025). Therefore, option D is correct.So, despite the earlier confusion, option D is correct.To summarize:- Option A: f(2) = 1 is incorrect because f(2) = 0.- Option B: f(2024) = -f(2020) is correct because f(2024) = f(0) and f(2020) = f(4), and f(0) = -f(4).- Option C: f'(-1) = f'(7) is correct because f' is periodic with period 8, so f'(-1) = f'(7).- Option D: f'(-2021) = f'(2025) is correct because f'(-2021) = f'(3) and f'(2025) = f'(1), and f'(3) = f'(1).Therefore, the correct options are B, C, and D.