answer：First, determine the efficiency of both Sakshi and Tanya by calculating the fraction of work each can complete in one day.Sakshi's efficiency is ( frac{1}{5} ) of the work per day, and Tanya's efficiency is ( frac{1}{4} ) of the work per day.Next, find the difference in their efficiencies:[frac{1}{4} - frac{1}{5} = frac{5}{20} - frac{4}{20} = frac{1}{20}]Finally, calculate the percentage increase in efficiency by dividing the increase by Sakshi's efficiency and multiplying by 100:[left( frac{frac{1}{20}}{frac{1}{5}} right) times 100 = left( frac{1}{20} times 5 right) times 100 = frac{5}{20} times 100 = 25%]

answer：Alright, let me try to figure out this problem step by step. So, we have two sets, A and B. Set A is defined as all y such that y is the logarithm of x, where x is greater than 1. Set B is just a simple set containing the numbers -2, -1, 1, and 2. We need to check which of the statements A, B, C, or D is correct.First, let me understand what set A actually is. Since A is defined as {y | y = log x, x > 1}, I need to recall what the logarithm function does. The logarithm of a number x with base 10 (or any base greater than 1) increases as x increases. When x is greater than 1, log x is positive. For example, log 1 is 0, log 10 is 1, log 100 is 2, and so on. So, if x is greater than 1, y = log x will be greater than 0. That means set A is all real numbers greater than 0. So, A = (0, ∞).Set B is straightforward: it's just {-2, -1, 1, 2}. So, it has four elements: two negative numbers and two positive numbers.Now, let's look at each option one by one.Option A: A ∩ B = {-2, -1}. Hmm, okay. So, the intersection of A and B would be the elements that are in both A and B. But wait, A is all positive numbers, right? So, A = (0, ∞). Set B has elements -2, -1, 1, 2. So, the intersection would be the elements in B that are also in A. That would be 1 and 2 because they are positive and greater than 0. So, A ∩ B should be {1, 2}, not {-2, -1}. Therefore, option A is incorrect.Option B: (ℝ A) ∪ B = (-∞, 0). Let me parse this. ℝ A means all real numbers not in A. Since A is (0, ∞), ℝ A would be (-∞, 0]. So, ℝ A is all real numbers less than or equal to 0. Then, we take the union of that with B. B is {-2, -1, 1, 2}. So, the union of (-∞, 0] and B would be (-∞, 0] ∪ {1, 2}. Wait, that's not just (-∞, 0). It's (-∞, 0] plus the points 1 and 2. So, the union is actually (-∞, 0] ∪ {1, 2}, which is not equal to (-∞, 0). Therefore, option B is incorrect.Option C: A ∪ B = (0, ∞). Let's see. A is (0, ∞), and B is {-2, -1, 1, 2}. So, the union of A and B would be all elements in A plus all elements in B. That would be (0, ∞) ∪ {-2, -1}. So, the union is actually (-∞, -2] ∪ (-1, ∞). Wait, no, that's not quite right. Let me think again. A is (0, ∞), which includes all positive numbers. B adds the points -2 and -1. So, the union would be all positive numbers plus -2 and -1. So, it's (-∞, -2] ∪ {-1} ∪ (0, ∞). But that's not the same as (0, ∞). So, option C is incorrect because it includes negative numbers as well.Option D: (ℝ A) ∩ B = {-2, -1}. Let's break this down. ℝ A is (-∞, 0], as we established earlier. Then, we take the intersection of that with B. B is {-2, -1, 1, 2}. So, the intersection would be the elements in B that are also in (-∞, 0]. That would be -2 and -1 because 1 and 2 are positive and not in (-∞, 0]. So, (ℝ A) ∩ B is indeed {-2, -1}. Therefore, option D is correct.Wait, let me double-check to make sure I didn't make a mistake. For option A, I concluded that A ∩ B is {1, 2}, which is correct because A only includes positive numbers. So, A is wrong. For option B, ℝ A is (-∞, 0], and union with B adds 1 and 2, so it's not just (-∞, 0). So, B is wrong. For option C, A ∪ B includes negative numbers, so it's not just (0, ∞). So, C is wrong. For option D, ℝ A is (-∞, 0], and intersecting with B gives {-2, -1}, which is exactly what D says. So, D is correct.I think that's solid. I don't see any mistakes in my reasoning. So, the correct answer is D.

answer：Okay, so I have this problem about finding the maximum daily sales revenue for a product over 30 days. The problem gives me two functions: one for the selling price ( P ) and another for the daily sales volume ( Q ). Both are functions of time ( t ), which is measured in days. First, let me try to understand the functions given. The selling price ( P ) is defined piecewise. For the first 24 days (since ( t ) is a positive integer and less than 25), the price increases by 1 yuan each day, starting from 21 yuan on day 1 (because when ( t = 1 ), ( P = 1 + 20 = 21 )) up to day 24, where ( P = 24 + 20 = 44 ) yuan. Then, starting from day 25 to day 30, the price decreases by 1 yuan each day, starting from 75 yuan on day 25 (( P = -25 + 100 = 75 )) down to 40 yuan on day 30 (( P = -30 + 100 = 70 )). Wait, hold on, that doesn't add up. If ( t = 30 ), then ( P = -30 + 100 = 70 ). So the price goes from 75 to 70 over days 25 to 30.Next, the daily sales volume ( Q ) is given by ( Q = -t + 40 ). So this is a linear function that decreases by 1 unit each day, starting from 39 units on day 1 (( Q = -1 + 40 = 39 )) down to 10 units on day 30 (( Q = -30 + 40 = 10 )).The goal is to find the maximum daily sales revenue. Revenue is calculated as price multiplied by quantity, so ( R = P times Q ). Since both ( P ) and ( Q ) are functions of ( t ), I need to express the revenue as a function of ( t ) and then find its maximum value over the interval ( t = 1 ) to ( t = 30 ).Given that ( P ) is piecewise, I should handle the two intervals separately: ( 0 < t < 25 ) and ( 25 leq t leq 30 ).Starting with the first interval, ( 0 < t < 25 ). Here, ( P = t + 20 ) and ( Q = -t + 40 ). So the revenue function ( R ) is:[R = (t + 20)(-t + 40)]Let me expand this:[R = -t^2 + 40t - 20t + 800 = -t^2 + 20t + 800]This is a quadratic function in terms of ( t ), and since the coefficient of ( t^2 ) is negative, the parabola opens downward, meaning the vertex is the maximum point. The vertex of a parabola ( at^2 + bt + c ) is at ( t = -frac{b}{2a} ). Here, ( a = -1 ) and ( b = 20 ), so:[t = -frac{20}{2 times -1} = -frac{20}{-2} = 10]So the maximum revenue in this interval occurs at ( t = 10 ). Let me calculate the revenue at ( t = 10 ):[R = -(10)^2 + 20 times 10 + 800 = -100 + 200 + 800 = 900]So, in the first interval, the maximum revenue is 900 yuan on day 10.Now, moving on to the second interval, ( 25 leq t leq 30 ). Here, ( P = -t + 100 ) and ( Q = -t + 40 ). So the revenue function ( R ) is:[R = (-t + 100)(-t + 40)]Let me expand this:[R = t^2 - 40t - 100t + 4000 = t^2 - 140t + 4000]Again, this is a quadratic function, but this time the coefficient of ( t^2 ) is positive, so the parabola opens upward, meaning the vertex is the minimum point. Therefore, the maximum revenue in this interval will occur at one of the endpoints, either at ( t = 25 ) or ( t = 30 ).Let me calculate the revenue at both endpoints.First, at ( t = 25 ):[R = (25)^2 - 140 times 25 + 4000 = 625 - 3500 + 4000 = 625 + 500 = 1125]Wait, that doesn't seem right. Let me recalculate:[R = 25^2 - 140 times 25 + 4000 = 625 - 3500 + 4000][625 - 3500 = -2875][-2875 + 4000 = 1125]Yes, that's correct. So at ( t = 25 ), the revenue is 1125 yuan.Now, at ( t = 30 ):[R = 30^2 - 140 times 30 + 4000 = 900 - 4200 + 4000][900 - 4200 = -3300][-3300 + 4000 = 700]So at ( t = 30 ), the revenue is 700 yuan.Comparing the two endpoints, the revenue is higher at ( t = 25 ) with 1125 yuan.Now, to ensure I haven't missed anything, I should check if there's a possibility of a higher revenue within the interval ( 25 leq t leq 30 ). Since the quadratic opens upwards, the minimum is at the vertex, but the maximum must be at one of the endpoints. So, as calculated, 1125 yuan at ( t = 25 ) is the maximum in this interval.Comparing the maximum revenues from both intervals: 900 yuan at ( t = 10 ) and 1125 yuan at ( t = 25 ). Clearly, 1125 yuan is the higher value.Therefore, the maximum daily sales revenue for this product is 1125 yuan, occurring on day 25.Wait, just to be thorough, I should verify my calculations for the second interval. Let me recompute the revenue at ( t = 25 ):[P = -25 + 100 = 75][Q = -25 + 40 = 15][R = 75 times 15 = 1125]Yes, that's correct. And at ( t = 30 ):[P = -30 + 100 = 70][Q = -30 + 40 = 10][R = 70 times 10 = 700]That's also correct. So, my calculations seem accurate.I also want to make sure that I didn't make a mistake in setting up the revenue functions. For the first interval, ( P = t + 20 ) and ( Q = -t + 40 ), so multiplying them gives ( (t + 20)(-t + 40) ), which expands to ( -t^2 + 20t + 800 ). That seems right.For the second interval, ( P = -t + 100 ) and ( Q = -t + 40 ), so multiplying them gives ( (-t + 100)(-t + 40) ), which expands to ( t^2 - 140t + 4000 ). That also seems correct.Another thing to check is whether the vertex calculation for the first interval was correct. The vertex at ( t = 10 ) gives a maximum revenue of 900 yuan. Let me plug in ( t = 10 ) into the original functions:[P = 10 + 20 = 30][Q = -10 + 40 = 30][R = 30 times 30 = 900]Yes, that's correct. So, all my calculations seem to check out.Therefore, I can confidently say that the maximum daily sales revenue is 1125 yuan, occurring on day 25.

answer：Okay, so I have this math problem here with four options, A through D, and I need to figure out which one is correct. Let me take it step by step because I want to make sure I understand each operation properly.Starting with option A: (ab^{2})^{2}=a^{2}b^{4}. Hmm, I remember that when you have an exponent outside the parentheses, it applies to both the coefficient and the variable inside. So, for (ab^{2})^{2}, I think I need to square both 'a' and 'b squared'. That would mean a^{2} and (b^{2})^{2}. Wait, what's (b^{2})^{2}? I think when you raise a power to another power, you multiply the exponents. So, b^{2*2}, which is b^{4}. Putting it all together, that gives me a^{2}b^{4}, which matches the right side of the equation. Okay, so option A seems correct.Moving on to option B: (m^{2})^{3}=m^{5}. Again, this is a power raised to another power. The rule is to multiply the exponents. So, m^{2*3} should be m^{6}. But the option says m^{5}, which doesn't match. So, option B is incorrect.Next is option C: 2m times 5m^{2}=10m^{2}. Let's break this down. First, multiply the coefficients: 2 times 5 is 10. Then, multiply the variables. 'm' is the same as m^{1}, so m^{1} times m^{2} should be m^{1+2}=m^{3}. So, putting it together, it should be 10m^{3}. But the option says 10m^{2}, which is wrong. So, option C is incorrect.Finally, option D: (m-n)^{2}=m^{2}-n^{2}. I recall that squaring a binomial like (m-n)^{2} should follow the formula (a-b)^{2}=a^{2}-2ab+b^{2}. Applying that here, it should be m^{2}-2mn+n^{2}. But the option only has m^{2}-n^{2}, missing the -2mn term. So, option D is also incorrect.After going through each option, only option A turned out to be correct. I think I did each step carefully, applying the exponent rules I remember. It's easy to mix up the rules, especially with exponents, so I had to double-check each one to make sure I wasn't making a mistake. But yeah, A seems to be the right answer.

answer：Alright, so I have this problem here where I need to figure out which of the given inequalities always hold true when ( a > b ). The inequalities are:1. ( frac{a}{c^{2}} > frac{b}{c^{2}} )2. ( frac{1}{a} < frac{1}{b} )3. ( |a| > |b| )4. ( a|c| geqslant b|c| )I need to determine which of these are always true. Let me go through each one step by step.Starting with the first inequality: ( frac{a}{c^{2}} > frac{b}{c^{2}} ). Hmm, since ( a > b ), if I divide both sides by ( c^{2} ), the inequality should still hold as long as ( c^{2} ) is positive. But wait, ( c^{2} ) is always non-negative, right? And since it's in the denominator, ( c ) can't be zero. So, as long as ( c ) isn't zero, ( frac{a}{c^{2}} ) should indeed be greater than ( frac{b}{c^{2}} ). So, I think inequality ① is always true.Moving on to the second inequality: ( frac{1}{a} < frac{1}{b} ). This one is trickier. If both ( a ) and ( b ) are positive, then since ( a > b ), ( frac{1}{a} ) would indeed be less than ( frac{1}{b} ). But what if ( a ) and ( b ) are negative? Let's say ( a = -1 ) and ( b = -2 ). Then ( a > b ) because -1 is greater than -2. But ( frac{1}{a} = -1 ) and ( frac{1}{b} = -0.5 ). So, ( -1 < -0.5 ) is true. Wait, so does that mean it's still true when both are negative?Hold on, let me test another case where ( a ) is positive and ( b ) is negative. Suppose ( a = 2 ) and ( b = -1 ). Then ( a > b ) is true. But ( frac{1}{a} = 0.5 ) and ( frac{1}{b} = -1 ). So, ( 0.5 < -1 ) is false. Therefore, inequality ② doesn't always hold because it fails when ( a ) is positive and ( b ) is negative. So, ② is not always true.Next, the third inequality: ( |a| > |b| ). This is about the absolute values. Let's see. If ( a ) and ( b ) are both positive, then ( a > b ) implies ( |a| > |b| ). Similarly, if both are negative, ( a > b ) (since ( a ) is closer to zero) would mean ( |a| < |b| ), right? For example, ( a = -1 ) and ( b = -2 ). Here, ( a > b ) because -1 is greater than -2, but ( |a| = 1 ) and ( |b| = 2 ), so ( |a| < |b| ). Therefore, inequality ③ doesn't always hold. It fails when both ( a ) and ( b ) are negative.Finally, the fourth inequality: ( a|c| geqslant b|c| ). Let's analyze this. Since ( |c| ) is always non-negative, multiplying both sides of ( a > b ) by ( |c| ) should preserve the inequality. So, ( a|c| geqslant b|c| ) should always be true because even if ( |c| = 0 ), both sides become zero, making the inequality hold as an equality. Therefore, inequality ④ is always true.To summarize:- Inequality ① holds because dividing both sides by a positive number preserves the inequality.- Inequality ② doesn't hold when ( a ) is positive and ( b ) is negative.- Inequality ③ doesn't hold when both ( a ) and ( b ) are negative.- Inequality ④ holds because multiplying both sides by a non-negative number preserves the inequality.So, the inequalities that always hold are ① and ④.

answer：Okay, so I have this differential equation to solve: ( y'' - 2y' - 3y = e^{3t} ), with initial conditions ( y(0) = 0 ) and ( y'(0) = 0 ). Hmm, I remember that differential equations can sometimes be tricky, especially when they're nonhomogeneous like this one. Let me think about how to approach this.First, I recall that for linear differential equations, especially second-order ones, the general solution is the sum of the homogeneous solution and a particular solution. So, I should probably start by solving the homogeneous equation ( y'' - 2y' - 3y = 0 ).To solve the homogeneous equation, I need to find the characteristic equation. The characteristic equation for ( y'' - 2y' - 3y = 0 ) is ( r^2 - 2r - 3 = 0 ). Let me solve this quadratic equation. Using the quadratic formula, ( r = frac{2 pm sqrt{(2)^2 - 4(1)(-3)}}{2(1)} ). That simplifies to ( r = frac{2 pm sqrt{4 + 12}}{2} = frac{2 pm sqrt{16}}{2} = frac{2 pm 4}{2} ). So, the roots are ( r = frac{2 + 4}{2} = 3 ) and ( r = frac{2 - 4}{2} = -1 ).Great, so the roots are real and distinct, which means the homogeneous solution will be ( y_h = C_1 e^{3t} + C_2 e^{-t} ), where ( C_1 ) and ( C_2 ) are constants to be determined later using the initial conditions.Now, I need to find a particular solution ( y_p ) to the nonhomogeneous equation ( y'' - 2y' - 3y = e^{3t} ). The right-hand side is ( e^{3t} ), which is an exponential function. I remember that when the nonhomogeneous term is of the form ( e^{kt} ), we can try a particular solution of the same form, provided that ( k ) is not a root of the characteristic equation. Wait, in this case, ( k = 3 ), and looking back at the characteristic equation, one of the roots was indeed 3. That means ( e^{3t} ) is already part of the homogeneous solution. So, I can't just use ( y_p = A e^{3t} ) because it would be a solution to the homogeneous equation. Instead, I need to multiply by ( t ) to make it linearly independent. So, I'll try ( y_p = A t e^{3t} ).Let me compute the first and second derivatives of ( y_p ). First derivative: ( y_p' = A e^{3t} + 3A t e^{3t} ).Second derivative: ( y_p'' = 3A e^{3t} + 3A e^{3t} + 9A t e^{3t} = 6A e^{3t} + 9A t e^{3t} ).Now, substitute ( y_p ), ( y_p' ), and ( y_p'' ) into the differential equation:( y_p'' - 2y_p' - 3y_p = (6A e^{3t} + 9A t e^{3t}) - 2(A e^{3t} + 3A t e^{3t}) - 3(A t e^{3t}) ).Let me simplify this step by step.First, expand each term:1. ( y_p'' = 6A e^{3t} + 9A t e^{3t} )2. ( -2y_p' = -2A e^{3t} - 6A t e^{3t} )3. ( -3y_p = -3A t e^{3t} )Now, combine all these terms:( (6A e^{3t} + 9A t e^{3t}) + (-2A e^{3t} - 6A t e^{3t}) + (-3A t e^{3t}) ).Combine like terms:For ( e^{3t} ) terms: ( 6A e^{3t} - 2A e^{3t} = 4A e^{3t} ).For ( t e^{3t} ) terms: ( 9A t e^{3t} - 6A t e^{3t} - 3A t e^{3t} = 0 ).So, the entire expression simplifies to ( 4A e^{3t} ).But according to the differential equation, this should equal ( e^{3t} ). Therefore:( 4A e^{3t} = e^{3t} ).Divide both sides by ( e^{3t} ) (which is never zero), so we get ( 4A = 1 ), which implies ( A = frac{1}{4} ).So, the particular solution is ( y_p = frac{1}{4} t e^{3t} ).Now, the general solution to the differential equation is the sum of the homogeneous and particular solutions:( y(t) = y_h + y_p = C_1 e^{3t} + C_2 e^{-t} + frac{1}{4} t e^{3t} ).Next, I need to apply the initial conditions to find the constants ( C_1 ) and ( C_2 ).First, apply ( y(0) = 0 ):( y(0) = C_1 e^{0} + C_2 e^{0} + frac{1}{4} cdot 0 cdot e^{0} = C_1 + C_2 = 0 ).So, equation (1): ( C_1 + C_2 = 0 ).Now, compute the first derivative of ( y(t) ):( y'(t) = 3C_1 e^{3t} - C_2 e^{-t} + frac{1}{4} e^{3t} + frac{3}{4} t e^{3t} ).Simplify:( y'(t) = (3C_1 + frac{1}{4}) e^{3t} - C_2 e^{-t} + frac{3}{4} t e^{3t} ).Now, apply the initial condition ( y'(0) = 0 ):( y'(0) = (3C_1 + frac{1}{4}) e^{0} - C_2 e^{0} + frac{3}{4} cdot 0 cdot e^{0} = 3C_1 + frac{1}{4} - C_2 = 0 ).So, equation (2): ( 3C_1 - C_2 + frac{1}{4} = 0 ).Now, we have a system of two equations:1. ( C_1 + C_2 = 0 )2. ( 3C_1 - C_2 = -frac{1}{4} )Let me solve this system.From equation (1): ( C_2 = -C_1 ).Substitute ( C_2 = -C_1 ) into equation (2):( 3C_1 - (-C_1) = -frac{1}{4} Rightarrow 3C_1 + C_1 = -frac{1}{4} Rightarrow 4C_1 = -frac{1}{4} Rightarrow C_1 = -frac{1}{16} ).Then, ( C_2 = -C_1 = frac{1}{16} ).So, now we can write the final solution:( y(t) = -frac{1}{16} e^{3t} + frac{1}{16} e^{-t} + frac{1}{4} t e^{3t} ).Let me just double-check my calculations to make sure I didn't make any mistakes.First, the homogeneous solution: correct, roots at 3 and -1.Particular solution: since 3 is a root, multiplied by t, got ( y_p = A t e^{3t} ). Then found A = 1/4, which seems correct.General solution: correct, sum of homogeneous and particular.Applying initial conditions:At t=0, y(0) = C1 + C2 = 0.Computed y'(t): careful with the derivative of the particular solution, which has two terms, so the derivative is 1/4 e^{3t} + 3/4 t e^{3t}. Then, plus the derivatives of the homogeneous solutions.At t=0, y'(0) = 3C1 + 1/4 - C2 = 0.Solving the system: C1 = -1/16, C2 = 1/16.So, plugging back into y(t):( y(t) = -frac{1}{16} e^{3t} + frac{1}{16} e^{-t} + frac{1}{4} t e^{3t} ).Yes, that seems consistent.I think that's the final answer. Let me write it neatly:( y(t) = frac{1}{4} t e^{3t} - frac{1}{16} e^{3t} + frac{1}{16} e^{-t} ).Yeah, that looks good. I don't see any mistakes in the steps. So, I think this is the correct solution.