answer：First, I recognize that the bucket is two-thirds full and contains 9 liters of maple syrup.To find the total capacity of the bucket, I can set up an equation where two-thirds of the capacity equals 9 liters.Let ( C ) represent the capacity of the bucket. The equation becomes:[frac{2}{3}C = 9]To solve for ( C ), I multiply both sides of the equation by the reciprocal of (frac{2}{3}), which is (frac{3}{2}):[C = 9 times frac{3}{2} = 13.5]Therefore, the capacity of the bucket is 13.5 liters.

answer：To determine the minimum number of people wearing both a hat and a glove, I'll use the principle of inclusion-exclusion.Let ( n ) represent the total number of people in the room.- The number of people wearing gloves is ( frac{2}{5}n ).- The number of people wearing hats is ( frac{3}{4}n ).According to the inclusion-exclusion principle:[text{Number of people wearing both} = frac{2}{5}n + frac{3}{4}n - n]Simplifying the equation:[frac{2}{5}n + frac{3}{4}n - n = frac{8}{20}n + frac{15}{20}n - frac{20}{20}n = frac{3}{20}n]To ensure the number of people is an integer, ( frac{3}{20}n ) must be an integer. The smallest such ( n ) is 20.Therefore, the minimum number of people wearing both a hat and a glove is:[frac{3}{20} times 20 = 3]

answer：Okay, so I have this problem where I need to find the number of functions ( f: A rightarrow A ) where ( A = {1, 2, ldots, 10} ). The function has to satisfy two conditions: first, it has to be non-decreasing, meaning if ( x leq y ), then ( f(x) leq f(y) ). Second, it has to satisfy ( f(3) = 4 ). Hmm, let me try to break this down. Since the function is non-decreasing, that means once it takes a certain value, it can't go back down. So, for example, if ( f(1) = 2 ), then ( f(2) ) has to be at least 2, and ( f(3) ) has to be at least whatever ( f(2) ) is, and so on.But we also know that ( f(3) = 4 ). So that gives us a specific point in the function. Let me think about how this affects the rest of the function.First, since ( f ) is non-decreasing, all the values before 3 (i.e., ( f(1) ) and ( f(2) )) have to be less than or equal to 4. Similarly, all the values after 3 (i.e., ( f(4) ) to ( f(10) )) have to be greater than or equal to 4.So, let's split the problem into two parts: the part before 3 and the part after 3.Starting with the first part: ( f(1) ) and ( f(2) ). Since ( f(1) leq f(2) leq f(3) = 4 ), both ( f(1) ) and ( f(2) ) can take values from 1 to 4. But they also have to be non-decreasing. So, how many non-decreasing functions are there from ( {1, 2} ) to ( {1, 2, 3, 4} )?I remember that the number of non-decreasing functions from a set of size ( n ) to a set of size ( m ) is equal to the combination ( C(n + m - 1, m - 1) ). Wait, is that right? Let me think. It's similar to the stars and bars problem, where you want to distribute identical items into distinct bins. So, for each function, it's like choosing values where order matters but repetition is allowed.Yes, so the formula is ( C(n + m - 1, n) ) or ( C(n + m - 1, m - 1) ). Since ( n = 2 ) (two elements: 1 and 2) and ( m = 4 ) (possible values: 1, 2, 3, 4), the number of non-decreasing functions is ( C(2 + 4 - 1, 2) = C(5, 2) = 10 ). So, there are 10 possible ways to define ( f(1) ) and ( f(2) ) such that they are non-decreasing and each is at most 4.Okay, that takes care of the first two elements. Now, moving on to the elements after 3, which are ( f(4) ) to ( f(10) ). Since ( f(3) = 4 ) and the function is non-decreasing, each of these ( f(4) ) to ( f(10) ) must be at least 4. Also, they have to be non-decreasing among themselves.So, how many non-decreasing functions are there from ( {4, 5, ldots, 10} ) to ( {4, 5, ldots, 10} )? Wait, actually, the codomain is still ( A = {1, 2, ldots, 10} ), but since each ( f(x) ) has to be at least 4, the effective codomain is ( {4, 5, ldots, 10} ), which has 7 elements.So, the number of non-decreasing functions from a set of size 7 (elements 4 to 10) to a set of size 7 (values 4 to 10) is ( C(7 + 7 - 1, 7) = C(13, 7) ). Let me compute that. ( C(13, 7) = 1716 ). Hmm, that seems correct.Wait, hold on. Is it 7 elements for the domain? Let me check: from 4 to 10, that's 7 elements (4,5,6,7,8,9,10). So, yes, the domain has 7 elements, and the codomain is also 7 elements (4 to 10). So, the number of non-decreasing functions is ( C(7 + 7 - 1, 7) = C(13, 7) = 1716 ).So, putting it all together, the total number of functions ( f ) is the product of the number of ways to define the first two elements and the number of ways to define the last seven elements. So, that would be ( 10 times 1716 = 17160 ).Wait, but hold on a second. Let me make sure I didn't make a mistake here. The function is defined on the entire set ( A ), so the first two elements (1 and 2) and the last seven elements (4 to 10) are independent in terms of their definitions. So, yes, multiplying the two should give the total number.But let me think again about the formula for non-decreasing functions. The number of non-decreasing functions from a set of size ( n ) to a set of size ( m ) is indeed ( C(n + m - 1, m - 1) ). So, for the first part, ( n = 2 ), ( m = 4 ), so ( C(2 + 4 - 1, 4 - 1) = C(5, 3) = 10 ). Wait, hold on, earlier I thought it was ( C(5, 2) ), but actually, it's ( C(5, 3) ). But both are equal because ( C(n, k) = C(n, n - k) ). So, ( C(5, 2) = C(5, 3) = 10 ). So, that part is correct.Similarly, for the second part, ( n = 7 ), ( m = 7 ), so ( C(7 + 7 - 1, 7 - 1) = C(13, 6) ). Wait, hold on, earlier I said ( C(13, 7) ), but according to the formula, it should be ( C(13, 6) ). But again, ( C(13, 6) = C(13, 7) = 1716 ). So, that part is also correct.Therefore, the total number of functions is indeed ( 10 times 1716 = 17160 ).Wait, but let me think again if there's another way to approach this problem. Maybe using recursion or something else.Alternatively, since the function is non-decreasing, we can model it as a sequence of values where each subsequent value is greater than or equal to the previous one. So, the entire function can be thought of as a sequence ( f(1) leq f(2) leq ldots leq f(10) ) with ( f(3) = 4 ).So, we can break this sequence into two parts: the first three elements and the last seven elements. But actually, since ( f(3) = 4 ), we can think of the first three elements as ( f(1) leq f(2) leq f(3) = 4 ), and the last seven elements as ( f(3) = 4 leq f(4) leq ldots leq f(10) ).Wait, so actually, the first three elements are from 1 to 4, and the last seven elements are from 4 to 10. So, the number of non-decreasing functions for the first three elements is ( C(3 + 4 - 1, 4 - 1) = C(6, 3) = 20 ). But hold on, earlier I thought it was 10 for the first two elements. Hmm, maybe I need to clarify.Wait, no. The function is defined on all 10 elements, but we have a fixed point at 3. So, perhaps it's better to think of the function as two separate non-decreasing sequences: one from 1 to 3, and another from 4 to 10. But since ( f(3) = 4 ), the first sequence must end at 4, and the second sequence must start at 4.So, for the first part, from 1 to 3, we have a non-decreasing function where ( f(3) = 4 ). So, how many non-decreasing functions are there from ( {1, 2, 3} ) to ( {1, 2, 3, 4} ) with ( f(3) = 4 )?Similarly, for the second part, from 4 to 10, we have a non-decreasing function from ( {4, 5, ldots, 10} ) to ( {4, 5, ldots, 10} ).Wait, so maybe I was wrong earlier when I considered only the first two elements. Actually, since ( f(3) = 4 ), the first three elements must form a non-decreasing sequence ending at 4. So, the number of such sequences is the number of non-decreasing functions from ( {1, 2, 3} ) to ( {1, 2, 3, 4} ) with ( f(3) = 4 ).Similarly, the second part is a non-decreasing function from ( {4, 5, ldots, 10} ) to ( {4, 5, ldots, 10} ).So, let's recast the problem.First part: ( f(1) leq f(2) leq f(3) = 4 ). So, how many non-decreasing functions are there from ( {1, 2, 3} ) to ( {1, 2, 3, 4} ) with ( f(3) = 4 )?This is equivalent to the number of non-decreasing sequences of length 3 where the last term is 4. So, the first two terms can be any non-decreasing sequence from 1 to 4, and the third term is fixed at 4.So, the number of such sequences is equal to the number of non-decreasing functions from ( {1, 2} ) to ( {1, 2, 3, 4} ), which is ( C(2 + 4 - 1, 2) = C(5, 2) = 10 ). So, that's consistent with my earlier calculation.So, the first part has 10 possibilities.The second part is from 4 to 10, which is 7 elements, each mapping to 4 to 10, with the function non-decreasing. So, as before, the number of such functions is ( C(7 + 7 - 1, 7) = C(13, 7) = 1716 ).Therefore, the total number of functions is ( 10 times 1716 = 17160 ).Wait, but let me think again. Is the second part starting at 4, so ( f(4) geq 4 ), and each subsequent ( f(x) ) is at least as big as the previous. So, it's a non-decreasing function from 7 elements to 7 elements, starting at 4. So, yes, that would be ( C(13, 7) ).Alternatively, another way to think about it is that for the second part, we can model it as placing 7 indistinct balls into 7 distinct boxes (values 4 to 10), where the order matters but repetition is allowed. So, it's the same as the number of multisets of size 7 from a set of size 7, which is ( C(7 + 7 - 1, 7) = C(13, 7) = 1716 ).So, that seems correct.Therefore, combining both parts, the total number of functions is 10 multiplied by 1716, which is 17160.But wait, let me think if there's an alternative approach. Maybe using recursion or considering the problem as a combination with repetition.Alternatively, since the function is non-decreasing, we can represent it as a sequence of values where each value is greater than or equal to the previous one. So, the entire function can be thought of as a sequence ( f(1) leq f(2) leq ldots leq f(10) ) with ( f(3) = 4 ).So, we can split the sequence into two parts: the first three elements and the last seven elements. The first three must be a non-decreasing sequence ending at 4, and the last seven must be a non-decreasing sequence starting at 4.So, the number of such functions is equal to the number of non-decreasing sequences of length 3 ending at 4 multiplied by the number of non-decreasing sequences of length 7 starting at 4.As calculated earlier, the first part is 10, and the second part is 1716, so the total is 17160.Alternatively, another way to think about it is to consider the entire function as a combination of two independent non-decreasing functions: one on the first three elements with a fixed endpoint, and one on the last seven elements with a fixed starting point.Therefore, I think my calculation is correct.Wait, but let me check with a smaller example to make sure.Suppose instead of A = {1,2,...,10}, we have A = {1,2,3}, and we want to find the number of non-decreasing functions f: A → A with f(2) = 2.So, in this case, the function must satisfy f(1) ≤ f(2) = 2 ≤ f(3). So, f(1) can be 1 or 2, and f(3) can be 2, 3.So, how many such functions are there?For f(1): 2 choices (1 or 2).For f(3): 2 choices (2 or 3).So, total functions: 2 * 2 = 4.Alternatively, using the formula:First part: f(1) ≤ f(2) = 2. So, number of non-decreasing functions from {1,2} to {1,2} with f(2)=2. So, f(1) can be 1 or 2, so 2 choices.Second part: f(3) ≥ 2. Since f(3) can be 2 or 3, so 2 choices.Total: 2 * 2 = 4, which matches.Alternatively, using the combination formula:First part: n=2, m=2. Number of non-decreasing functions is C(2 + 2 -1, 2) = C(3,2)=3. Wait, but we have only 2 choices because f(2)=2 is fixed.Wait, so in this case, the formula doesn't directly apply because we have a fixed value at f(2)=2. So, perhaps in the original problem, the formula applies because we fix f(3)=4, and then the first part is the number of non-decreasing functions from {1,2} to {1,2,3,4} with f(3)=4, which is 10.Similarly, in the smaller example, the first part is the number of non-decreasing functions from {1,2} to {1,2} with f(2)=2, which is 2, not 3. So, the formula C(n + m -1, n) gives 3, but in reality, it's 2 because f(2) is fixed.Wait, so perhaps my initial approach was wrong because I fixed f(3)=4, so the first part is not just the number of non-decreasing functions from {1,2} to {1,2,3,4}, but rather the number of non-decreasing functions from {1,2,3} to {1,2,3,4} with f(3)=4.Wait, but in the smaller example, that approach would give the correct result. Let me see.In the smaller example, A = {1,2,3}, f: A→A, non-decreasing, f(2)=2.So, the first part is f(1) ≤ f(2)=2, so f(1) can be 1 or 2.The second part is f(3) ≥ 2, so f(3) can be 2 or 3.Thus, total functions: 2 * 2 = 4.Alternatively, using the formula for the first part: number of non-decreasing functions from {1,2} to {1,2} with f(2)=2. So, it's equivalent to the number of non-decreasing functions from {1} to {1,2} with f(1) ≤ 2. Which is 2.Similarly, the second part is the number of non-decreasing functions from {3} to {2,3}, which is 2.So, 2 * 2 = 4.So, in the original problem, the first part is the number of non-decreasing functions from {1,2} to {1,2,3,4} with f(2) ≤ 4, but since f(3)=4 is fixed, f(2) can be up to 4.Wait, no, actually, f(3)=4 is fixed, so f(2) can be from 1 to 4, but f(1) ≤ f(2) ≤ 4.So, the number of non-decreasing functions from {1,2} to {1,2,3,4} is C(2 + 4 -1, 2) = C(5,2)=10, which is correct.Similarly, in the smaller example, the number of non-decreasing functions from {1,2} to {1,2} is C(2 + 2 -1, 2)=C(3,2)=3, but since f(2)=2 is fixed, it's actually 2.Wait, so in the original problem, we have f(3)=4 fixed, so the first part is the number of non-decreasing functions from {1,2} to {1,2,3,4} with f(2) ≤ 4, which is 10.Similarly, the second part is the number of non-decreasing functions from {4,5,...,10} to {4,5,...,10}, which is C(7 + 7 -1,7)=1716.So, the total number is 10 * 1716=17160.Alternatively, in the smaller example, if we fix f(2)=2, the first part is the number of non-decreasing functions from {1} to {1,2} with f(1) ≤ 2, which is 2. The second part is the number of non-decreasing functions from {3} to {2,3}, which is 2. So, total is 4, which is correct.Therefore, in the original problem, the approach is correct.So, I think my answer is 17160.But just to be thorough, let me think about another way.Another approach is to model the entire function as a sequence of values where each value is at least as large as the previous one, with f(3)=4.So, the function can be represented as:f(1) ≤ f(2) ≤ 4 ≤ f(4) ≤ f(5) ≤ ... ≤ f(10).So, we can think of this as two separate non-decreasing sequences: one from f(1) to f(3)=4, and another from f(3)=4 to f(10).So, the number of such functions is equal to the number of non-decreasing sequences from f(1) to f(3)=4 multiplied by the number of non-decreasing sequences from f(3)=4 to f(10).As calculated earlier, the first part is 10, and the second part is 1716, so total is 17160.Alternatively, another way to think about it is to consider the entire function as a combination of two independent non-decreasing functions, one on the first three elements and one on the last seven elements, with the constraint that f(3)=4.But since f(3)=4 is fixed, the two parts are independent, so we can multiply their counts.Therefore, I think my answer is correct.**Final Answer**The number of such functions ( f ) is boxed{17160}.

answer：Okay, so I have this problem here: I need to find the minimum value of (frac{1}{a} + frac{4}{b}) given that the line (ax + by + 1 = 0) passes through the center of the circle defined by (x^2 + y^2 + 8x + 2y + 1 = 0). Both (a) and (b) are positive. Hmm, let me try to break this down step by step.First, I remember that the general equation of a circle is (x^2 + y^2 + Dx + Ey + F = 0), and its center is at ((-D/2, -E/2)). So, for the given circle equation (x^2 + y^2 + 8x + 2y + 1 = 0), the center should be at ((-8/2, -2/2)), which simplifies to ((-4, -1)). Okay, so the center is at ((-4, -1)).Now, the line (ax + by + 1 = 0) passes through this center. That means if I plug the coordinates of the center into the line equation, it should satisfy the equation. Let me do that:Substituting (x = -4) and (y = -1) into (ax + by + 1 = 0):(a(-4) + b(-1) + 1 = 0)Simplifying this:(-4a - b + 1 = 0)Let me rearrange this equation to express it in terms of (a) and (b):(-4a - b = -1)Multiplying both sides by -1 to make it look nicer:(4a + b = 1)So, I have the equation (4a + b = 1). This is a linear constraint on (a) and (b). Since both (a) and (b) are positive, I can think of this as a line in the first quadrant of the (a)-(b) plane.Now, the problem asks for the minimum value of (frac{1}{a} + frac{4}{b}). Hmm, this looks like an optimization problem with a constraint. I remember that for such problems, methods like substitution or using Lagrange multipliers can be applied. But since this is a problem likely intended for algebraic methods, maybe substitution is the way to go.Let me express (b) in terms of (a) from the constraint equation:From (4a + b = 1), we get (b = 1 - 4a).Since (a > 0) and (b > 0), substituting (b = 1 - 4a > 0) gives:(1 - 4a > 0 implies 4a < 1 implies a < frac{1}{4}).So, (a) must be between 0 and (frac{1}{4}).Now, substituting (b = 1 - 4a) into the expression (frac{1}{a} + frac{4}{b}):Let me denote (f(a) = frac{1}{a} + frac{4}{1 - 4a}).So, (f(a) = frac{1}{a} + frac{4}{1 - 4a}), where (0 < a < frac{1}{4}).Now, I need to find the minimum value of (f(a)). To do this, I can take the derivative of (f(a)) with respect to (a), set it equal to zero, and solve for (a). Then, check if that critical point gives a minimum.Let's compute the derivative (f'(a)):First, (f(a) = frac{1}{a} + frac{4}{1 - 4a}).So, the derivative of (frac{1}{a}) is (-frac{1}{a^2}).For the second term, (frac{4}{1 - 4a}), let me rewrite it as (4(1 - 4a)^{-1}). The derivative of this with respect to (a) is (4 times (-1) times (-4)(1 - 4a)^{-2}), which simplifies to (frac{16}{(1 - 4a)^2}).So, putting it together:(f'(a) = -frac{1}{a^2} + frac{16}{(1 - 4a)^2}).To find the critical points, set (f'(a) = 0):(-frac{1}{a^2} + frac{16}{(1 - 4a)^2} = 0)Let me move one term to the other side:(frac{16}{(1 - 4a)^2} = frac{1}{a^2})Cross-multiplying:(16a^2 = (1 - 4a)^2)Let me expand the right-hand side:((1 - 4a)^2 = 1 - 8a + 16a^2)So, substituting back:(16a^2 = 1 - 8a + 16a^2)Subtract (16a^2) from both sides:(0 = 1 - 8a)So, (1 - 8a = 0 implies 8a = 1 implies a = frac{1}{8}).Okay, so (a = frac{1}{8}). Let me check if this is within the interval (0 < a < frac{1}{4}). Since (frac{1}{8} = 0.125) and (frac{1}{4} = 0.25), yes, it is within the interval.Now, let's find the corresponding (b) using (b = 1 - 4a):(b = 1 - 4 times frac{1}{8} = 1 - frac{4}{8} = 1 - frac{1}{2} = frac{1}{2}).So, (a = frac{1}{8}) and (b = frac{1}{2}).Now, let's compute the value of (frac{1}{a} + frac{4}{b}):(frac{1}{frac{1}{8}} + frac{4}{frac{1}{2}} = 8 + 8 = 16).So, the value is 16.Wait, but before I conclude, I should verify whether this critical point indeed gives a minimum. Since the function (f(a)) tends to infinity as (a) approaches 0 or (frac{1}{4}), and we have only one critical point in the interval, this must be the minimum.Alternatively, I can use the second derivative test.Compute the second derivative (f''(a)):We have (f'(a) = -frac{1}{a^2} + frac{16}{(1 - 4a)^2}).So, (f''(a)) is the derivative of (f'(a)):First term: derivative of (-frac{1}{a^2}) is (frac{2}{a^3}).Second term: derivative of (frac{16}{(1 - 4a)^2}) is (16 times (-2) times (-4)(1 - 4a)^{-3}), which is (frac{128}{(1 - 4a)^3}).So, (f''(a) = frac{2}{a^3} + frac{128}{(1 - 4a)^3}).Since (a > 0) and (1 - 4a > 0) (because (a < frac{1}{4})), both terms are positive. Therefore, (f''(a) > 0), which means the function is concave upward at this critical point, confirming it's a minimum.Therefore, the minimum value of (frac{1}{a} + frac{4}{b}) is 16.Wait, hold on a second. Let me make sure I didn't make a mistake in computing the derivatives.First, (f(a) = frac{1}{a} + frac{4}{1 - 4a}).First derivative:- The derivative of (frac{1}{a}) is (-frac{1}{a^2}).- The derivative of (frac{4}{1 - 4a}) is (4 times frac{d}{da} (1 - 4a)^{-1} = 4 times (-1) times (-4) times (1 - 4a)^{-2} = frac{16}{(1 - 4a)^2}).So, that seems correct.Setting derivative to zero:(-frac{1}{a^2} + frac{16}{(1 - 4a)^2} = 0)Which leads to:(frac{16}{(1 - 4a)^2} = frac{1}{a^2})Cross-multiplying:(16a^2 = (1 - 4a)^2)Which expands to:(16a^2 = 1 - 8a + 16a^2)Subtracting (16a^2):(0 = 1 - 8a)So, (a = frac{1}{8}). That seems correct.Then, (b = 1 - 4a = 1 - frac{4}{8} = frac{1}{2}). Correct.Then, plugging back into (frac{1}{a} + frac{4}{b}):(frac{1}{1/8} + frac{4}{1/2} = 8 + 8 = 16). Correct.So, all the steps seem to check out. Therefore, the minimum value is indeed 16.Alternatively, another approach could be using the Cauchy-Schwarz inequality or the AM-GM inequality. Let me see if that's possible.Given the constraint (4a + b = 1), and we need to minimize (frac{1}{a} + frac{4}{b}).Let me consider using the Cauchy-Schwarz inequality in the form:((frac{1}{a} + frac{4}{b})(4a + b) geq (sqrt{frac{1}{a} times 4a} + sqrt{frac{4}{b} times b})^2)Wait, actually, the Cauchy-Schwarz inequality in the form of ((u_1v_1 + u_2v_2)^2 leq (u_1^2 + u_2^2)(v_1^2 + v_2^2)). Maybe not directly applicable here.Alternatively, let me try the method of Lagrange multipliers, just to confirm.Let me define the function to minimize as (f(a, b) = frac{1}{a} + frac{4}{b}), subject to the constraint (g(a, b) = 4a + b - 1 = 0).The method of Lagrange multipliers tells us that at the minimum, the gradient of (f) is proportional to the gradient of (g). So,(nabla f = lambda nabla g)Compute the gradients:(nabla f = left( -frac{1}{a^2}, -frac{4}{b^2} right))(nabla g = (4, 1))So, setting up the equations:(-frac{1}{a^2} = 4lambda)(-frac{4}{b^2} = lambda)So, from the first equation: (lambda = -frac{1}{4a^2})From the second equation: (lambda = -frac{4}{b^2})Setting them equal:(-frac{1}{4a^2} = -frac{4}{b^2})Multiply both sides by -1:(frac{1}{4a^2} = frac{4}{b^2})Cross-multiplying:(b^2 = 16a^2)Taking square roots (since (a, b > 0)):(b = 4a)So, (b = 4a). Now, plugging this into the constraint (4a + b = 1):(4a + 4a = 1 implies 8a = 1 implies a = frac{1}{8})Then, (b = 4a = 4 times frac{1}{8} = frac{1}{2}). So, same result as before.Therefore, the minimum value is indeed 16.Alternatively, another approach is to use substitution and then apply AM-GM inequality.Given (4a + b = 1), let me express (b = 1 - 4a), as before.We need to minimize (f(a) = frac{1}{a} + frac{4}{1 - 4a}).Let me set (t = a), so (0 < t < frac{1}{4}), and (f(t) = frac{1}{t} + frac{4}{1 - 4t}).Let me consider substituting (u = 4t), so (u = 4t), which implies (t = frac{u}{4}), and since (0 < t < frac{1}{4}), (0 < u < 1).Then, (f(t)) becomes:(frac{1}{frac{u}{4}} + frac{4}{1 - u} = frac{4}{u} + frac{4}{1 - u}).So, (f(u) = frac{4}{u} + frac{4}{1 - u}), where (0 < u < 1).Now, this looks symmetric in a way. Let me factor out the 4:(f(u) = 4left( frac{1}{u} + frac{1}{1 - u} right)).So, I need to minimize (frac{1}{u} + frac{1}{1 - u}).Let me denote (g(u) = frac{1}{u} + frac{1}{1 - u}).To find the minimum of (g(u)), take derivative:(g'(u) = -frac{1}{u^2} + frac{1}{(1 - u)^2}).Set derivative to zero:(-frac{1}{u^2} + frac{1}{(1 - u)^2} = 0)Move one term to the other side:(frac{1}{(1 - u)^2} = frac{1}{u^2})Taking reciprocals:((1 - u)^2 = u^2)Expanding:(1 - 2u + u^2 = u^2)Simplify:(1 - 2u = 0 implies u = frac{1}{2})So, (u = frac{1}{2}). Then, (t = frac{u}{4} = frac{1}{8}), which is consistent with our earlier result.Then, (g(u) = frac{1}{frac{1}{2}} + frac{1}{1 - frac{1}{2}} = 2 + 2 = 4).Therefore, (f(u) = 4 times 4 = 16). So, same result again.Alternatively, using AM-GM inequality on (g(u) = frac{1}{u} + frac{1}{1 - u}).Let me consider that for positive numbers (u) and (1 - u), since (0 < u < 1), we can apply AM-GM.But, AM-GM states that for positive real numbers, the arithmetic mean is greater than or equal to the geometric mean. However, in this case, we have reciprocals, so it's a bit tricky.Alternatively, let me set (v = u) and (w = 1 - u), so (v + w = 1). Then, we have (g(u) = frac{1}{v} + frac{1}{w}).We can use the Cauchy-Schwarz inequality here:((frac{1}{v} + frac{1}{w})(v + w) geq (1 + 1)^2 = 4)Since (v + w = 1), this becomes:(frac{1}{v} + frac{1}{w} geq 4)So, the minimum of (frac{1}{v} + frac{1}{w}) is 4, achieved when (v = w = frac{1}{2}). Therefore, (g(u) geq 4), with equality when (u = frac{1}{2}).Thus, (f(u) = 4 times g(u) geq 4 times 4 = 16), which is consistent with our previous results.So, regardless of the method—whether calculus, substitution, or inequalities—we arrive at the same conclusion that the minimum value is 16.Therefore, I'm confident that the minimum value of (frac{1}{a} + frac{4}{b}) is 16.**Final Answer**The minimum value is boxed{16}.

answer：Okay, so I have this problem here: I need to find the largest real number ( k ) such that the inequality ( frac{1 + sin x}{2 + cos x} geqslant k ) has a solution. The options are given as (A) ( -frac{1}{3} ), (B) ( -frac{3}{4} ), (C) ( frac{3}{4} ), and (D) ( frac{4}{3} ). Hmm, let me think. So, I need to find the maximum value of ( k ) for which there exists some ( x ) such that ( frac{1 + sin x}{2 + cos x} ) is greater than or equal to ( k ). That means ( k ) can't be larger than the maximum value that ( frac{1 + sin x}{2 + cos x} ) can take. So essentially, I need to find the maximum value of the function ( f(x) = frac{1 + sin x}{2 + cos x} ).Alright, so my task is to find the maximum of ( f(x) = frac{1 + sin x}{2 + cos x} ). Once I find that maximum value, that will be the largest ( k ) such that the inequality holds for some ( x ). How do I find the maximum of such a function? It's a trigonometric function, so maybe I can use calculus. I can take the derivative of ( f(x) ) with respect to ( x ), set it equal to zero, and solve for ( x ). That should give me the critical points, and then I can evaluate ( f(x) ) at those points to find the maximum.Let me write down the function again: ( f(x) = frac{1 + sin x}{2 + cos x} ). To find its derivative, I'll use the quotient rule. The quotient rule says that if I have a function ( frac{u}{v} ), its derivative is ( frac{u'v - uv'}{v^2} ).So, let me set ( u = 1 + sin x ) and ( v = 2 + cos x ). Then, ( u' = cos x ) and ( v' = -sin x ). Applying the quotient rule, the derivative ( f'(x) ) is:( f'(x) = frac{ cos x cdot (2 + cos x) - (1 + sin x) cdot (-sin x) }{(2 + cos x)^2} ).Let me simplify the numerator step by step. First, expand the terms:Numerator = ( cos x (2 + cos x) + sin x (1 + sin x) ).Let me compute each part:1. ( cos x cdot 2 = 2 cos x )2. ( cos x cdot cos x = cos^2 x )3. ( sin x cdot 1 = sin x )4. ( sin x cdot sin x = sin^2 x )So putting it all together, the numerator becomes:( 2 cos x + cos^2 x + sin x + sin^2 x ).Hmm, I notice that ( cos^2 x + sin^2 x = 1 ), so that simplifies to:( 2 cos x + 1 + sin x ).So, the numerator simplifies to ( 2 cos x + sin x + 1 ). Therefore, the derivative ( f'(x) ) is:( f'(x) = frac{2 cos x + sin x + 1}{(2 + cos x)^2} ).Now, to find the critical points, I need to set ( f'(x) = 0 ). Since the denominator ( (2 + cos x)^2 ) is always positive (because ( 2 + cos x ) is always positive, as ( cos x ) ranges between -1 and 1, so ( 2 + cos x ) ranges from 1 to 3), the sign of the derivative is determined by the numerator.So, setting the numerator equal to zero:( 2 cos x + sin x + 1 = 0 ).So, the equation to solve is:( 2 cos x + sin x = -1 ).Hmm, this is a linear combination of sine and cosine. Maybe I can write this as a single sine or cosine function using the amplitude-phase form. Remember, ( a cos x + b sin x = R cos(x - phi) ) where ( R = sqrt{a^2 + b^2} ) and ( phi = arctanleft(frac{b}{a}right) ).Let me compute ( R ):( R = sqrt{2^2 + 1^2} = sqrt{4 + 1} = sqrt{5} ).So, ( 2 cos x + sin x = sqrt{5} cos(x - phi) ), where ( phi = arctanleft(frac{1}{2}right) ).Therefore, the equation becomes:( sqrt{5} cos(x - phi) = -1 ).Divide both sides by ( sqrt{5} ):( cos(x - phi) = -frac{1}{sqrt{5}} ).So, ( x - phi = arccosleft(-frac{1}{sqrt{5}}right) ) or ( x - phi = -arccosleft(-frac{1}{sqrt{5}}right) ).Therefore, the solutions are:( x = phi pm arccosleft(-frac{1}{sqrt{5}}right) + 2pi n ), where ( n ) is any integer.Hmm, okay. So, these are the critical points. Now, I need to evaluate ( f(x) ) at these points to find the maximum value.But before I proceed, let me think if there's another way. Maybe instead of using calculus, I can use some trigonometric identities or inequalities to find the maximum of ( f(x) ).Alternatively, I can consider using the method of Lagrange multipliers or some substitution to simplify the expression. Wait, another idea: since both numerator and denominator are linear in sine and cosine, maybe I can write this as a function and find its maximum using the Cauchy-Schwarz inequality or something similar.Alternatively, I can set ( t = tan(x/2) ), which is the Weierstrass substitution, and express ( sin x ) and ( cos x ) in terms of ( t ). Let me try that.So, using the substitution ( t = tan(x/2) ), we have:( sin x = frac{2t}{1 + t^2} ) and ( cos x = frac{1 - t^2}{1 + t^2} ).Substituting these into ( f(x) ):( f(x) = frac{1 + frac{2t}{1 + t^2}}{2 + frac{1 - t^2}{1 + t^2}} ).Let me simplify numerator and denominator:Numerator: ( 1 + frac{2t}{1 + t^2} = frac{(1 + t^2) + 2t}{1 + t^2} = frac{1 + 2t + t^2}{1 + t^2} = frac{(1 + t)^2}{1 + t^2} ).Denominator: ( 2 + frac{1 - t^2}{1 + t^2} = frac{2(1 + t^2) + 1 - t^2}{1 + t^2} = frac{2 + 2t^2 + 1 - t^2}{1 + t^2} = frac{3 + t^2}{1 + t^2} ).Therefore, ( f(x) = frac{(1 + t)^2 / (1 + t^2)}{(3 + t^2)/(1 + t^2)} = frac{(1 + t)^2}{3 + t^2} ).So, now, ( f(x) ) is expressed in terms of ( t ) as ( frac{(1 + t)^2}{3 + t^2} ). Now, this is a rational function in ( t ), and I can find its maximum by taking the derivative with respect to ( t ) and setting it to zero.Let me compute the derivative of ( f(t) = frac{(1 + t)^2}{3 + t^2} ).Using the quotient rule again, with ( u = (1 + t)^2 ) and ( v = 3 + t^2 ).Then, ( u' = 2(1 + t) ) and ( v' = 2t ).So, the derivative ( f'(t) ) is:( f'(t) = frac{2(1 + t)(3 + t^2) - (1 + t)^2 (2t)}{(3 + t^2)^2} ).Let me expand the numerator:First, compute ( 2(1 + t)(3 + t^2) ):= ( 2 times [1 times 3 + 1 times t^2 + t times 3 + t times t^2] )= ( 2 times [3 + t^2 + 3t + t^3] )= ( 6 + 2t^2 + 6t + 2t^3 ).Next, compute ( (1 + t)^2 times 2t ):= ( (1 + 2t + t^2) times 2t )= ( 2t + 4t^2 + 2t^3 ).Now, subtract the second expression from the first:Numerator = ( (6 + 2t^2 + 6t + 2t^3) - (2t + 4t^2 + 2t^3) )= ( 6 + 2t^2 + 6t + 2t^3 - 2t - 4t^2 - 2t^3 )= ( 6 + (2t^2 - 4t^2) + (6t - 2t) + (2t^3 - 2t^3) )= ( 6 - 2t^2 + 4t + 0 )= ( 6 - 2t^2 + 4t ).So, the derivative is:( f'(t) = frac{6 - 2t^2 + 4t}{(3 + t^2)^2} ).To find critical points, set numerator equal to zero:( 6 - 2t^2 + 4t = 0 ).Let me rewrite this:( -2t^2 + 4t + 6 = 0 ).Multiply both sides by -1 to make it easier:( 2t^2 - 4t - 6 = 0 ).Divide both sides by 2:( t^2 - 2t - 3 = 0 ).Now, solve for ( t ):Using quadratic formula: ( t = frac{2 pm sqrt{4 + 12}}{2} = frac{2 pm sqrt{16}}{2} = frac{2 pm 4}{2} ).So, ( t = frac{2 + 4}{2} = 3 ) or ( t = frac{2 - 4}{2} = -1 ).So, critical points at ( t = 3 ) and ( t = -1 ).Now, I need to evaluate ( f(t) ) at these points to find the maximum.First, compute ( f(3) ):( f(3) = frac{(1 + 3)^2}{3 + 3^2} = frac{16}{3 + 9} = frac{16}{12} = frac{4}{3} ).Next, compute ( f(-1) ):( f(-1) = frac{(1 + (-1))^2}{3 + (-1)^2} = frac{0}{3 + 1} = 0 ).So, the function ( f(t) ) has a maximum value of ( frac{4}{3} ) at ( t = 3 ) and a minimum value of 0 at ( t = -1 ).Therefore, the maximum value of ( f(x) = frac{1 + sin x}{2 + cos x} ) is ( frac{4}{3} ). So, the largest value of ( k ) for which the inequality ( frac{1 + sin x}{2 + cos x} geqslant k ) has a solution is ( frac{4}{3} ).Wait a second, but looking back at the options, (D) is ( frac{4}{3} ), which seems to be the answer. But let me verify if this is indeed attainable.So, when ( t = 3 ), which is ( tan(x/2) = 3 ), so ( x/2 = arctan(3) ), so ( x = 2 arctan(3) ). Let me compute ( sin x ) and ( cos x ) at this ( x ).Alternatively, since ( t = 3 ), we can compute ( sin x = frac{2t}{1 + t^2} = frac{6}{1 + 9} = frac{6}{10} = frac{3}{5} ).Similarly, ( cos x = frac{1 - t^2}{1 + t^2} = frac{1 - 9}{10} = frac{-8}{10} = -frac{4}{5} ).So, plugging back into ( f(x) ):( f(x) = frac{1 + frac{3}{5}}{2 + (-frac{4}{5})} = frac{frac{8}{5}}{frac{6}{5}} = frac{8}{5} times frac{5}{6} = frac{8}{6} = frac{4}{3} ).Yes, that checks out. So, the maximum value is indeed ( frac{4}{3} ).But wait, hold on. Let me think again. The question is asking for the largest ( k ) such that the inequality ( frac{1 + sin x}{2 + cos x} geqslant k ) has a solution. So, the maximum value of the function is ( frac{4}{3} ), so the largest ( k ) is ( frac{4}{3} ). So, the answer should be (D).But wait, looking back at the options, (D) is ( frac{4}{3} ), which is greater than 1, but let me check if ( frac{1 + sin x}{2 + cos x} ) can actually reach ( frac{4}{3} ).Wait, when ( sin x = frac{3}{5} ) and ( cos x = -frac{4}{5} ), then ( 1 + sin x = 1 + frac{3}{5} = frac{8}{5} ), and ( 2 + cos x = 2 - frac{4}{5} = frac{6}{5} ). So, ( frac{8/5}{6/5} = frac{8}{6} = frac{4}{3} ). So yes, it does reach ( frac{4}{3} ).Therefore, the maximum value is indeed ( frac{4}{3} ), so the largest ( k ) is ( frac{4}{3} ), which is option (D).But wait, let me check the options again. The options are (A) ( -frac{1}{3} ), (B) ( -frac{3}{4} ), (C) ( frac{3}{4} ), (D) ( frac{4}{3} ). So, (D) is the correct answer.But just to make sure, let me think if there's another way to approach this problem without calculus. Maybe using the method of expressing the function in terms of sine or cosine only.Let me consider ( f(x) = frac{1 + sin x}{2 + cos x} ). Let me denote ( y = sin x ) and ( z = cos x ). Then, ( y^2 + z^2 = 1 ). So, the function becomes ( frac{1 + y}{2 + z} ), with the constraint ( y^2 + z^2 = 1 ).So, I can think of this as maximizing ( frac{1 + y}{2 + z} ) subject to ( y^2 + z^2 = 1 ). This is a constrained optimization problem.Using the method of Lagrange multipliers, let me set up the Lagrangian:( mathcal{L}(y, z, lambda) = frac{1 + y}{2 + z} - lambda(y^2 + z^2 - 1) ).Wait, but actually, Lagrange multipliers are for differentiable functions, and the function ( frac{1 + y}{2 + z} ) is differentiable except when ( 2 + z = 0 ), which is not possible since ( z = cos x ) and ( 2 + cos x geq 1 ). So, we can proceed.Taking partial derivatives:1. ( frac{partial mathcal{L}}{partial y} = frac{1}{2 + z} - 2 lambda y = 0 ).2. ( frac{partial mathcal{L}}{partial z} = frac{-(1 + y)}{(2 + z)^2} - 2 lambda z = 0 ).3. ( frac{partial mathcal{L}}{partial lambda} = -(y^2 + z^2 - 1) = 0 ).So, from the first equation:( frac{1}{2 + z} = 2 lambda y ) --> ( lambda = frac{1}{2 y (2 + z)} ).From the second equation:( frac{-(1 + y)}{(2 + z)^2} = 2 lambda z ).Substituting ( lambda ) from the first equation into the second:( frac{-(1 + y)}{(2 + z)^2} = 2 times frac{1}{2 y (2 + z)} times z ).Simplify:Left side: ( frac{-(1 + y)}{(2 + z)^2} ).Right side: ( frac{z}{y (2 + z)} ).Multiply both sides by ( (2 + z)^2 ):( -(1 + y) = frac{z (2 + z)}{y} ).Multiply both sides by ( y ):( -y(1 + y) = z(2 + z) ).So,( -y - y^2 = 2 z + z^2 ).Bring all terms to one side:( -y - y^2 - 2 z - z^2 = 0 ).But since ( y^2 + z^2 = 1 ), we can substitute ( y^2 = 1 - z^2 ).So,( -y - (1 - z^2) - 2 z - z^2 = 0 ).Simplify:( -y - 1 + z^2 - 2 z - z^2 = 0 ).Simplify further:( -y - 1 - 2 z = 0 ).So,( -y - 2 z = 1 ).Multiply both sides by -1:( y + 2 z = -1 ).So, we have the equation ( y + 2 z = -1 ) and the constraint ( y^2 + z^2 = 1 ).Now, let's solve this system of equations.From ( y + 2 z = -1 ), we can express ( y = -1 - 2 z ).Substitute into the constraint:( (-1 - 2 z)^2 + z^2 = 1 ).Expand ( (-1 - 2 z)^2 ):= ( 1 + 4 z + 4 z^2 ).So, the equation becomes:( 1 + 4 z + 4 z^2 + z^2 = 1 ).Simplify:( 1 + 4 z + 5 z^2 = 1 ).Subtract 1 from both sides:( 4 z + 5 z^2 = 0 ).Factor:( z (4 + 5 z) = 0 ).So, solutions are ( z = 0 ) or ( 4 + 5 z = 0 ) --> ( z = -frac{4}{5} ).Case 1: ( z = 0 ).Then, from ( y = -1 - 2 z ), ( y = -1 - 0 = -1 ).Check if ( y^2 + z^2 = 1 ): ( (-1)^2 + 0^2 = 1 + 0 = 1 ). Yes, it satisfies.Case 2: ( z = -frac{4}{5} ).Then, ( y = -1 - 2 (-frac{4}{5}) = -1 + frac{8}{5} = frac{3}{5} ).Check ( y^2 + z^2 = (frac{3}{5})^2 + (-frac{4}{5})^2 = frac{9}{25} + frac{16}{25} = frac{25}{25} = 1 ). Yes, it satisfies.So, we have two critical points:1. ( y = -1 ), ( z = 0 ).2. ( y = frac{3}{5} ), ( z = -frac{4}{5} ).Now, compute ( f(x) = frac{1 + y}{2 + z} ) for each case.Case 1: ( y = -1 ), ( z = 0 ).( f(x) = frac{1 + (-1)}{2 + 0} = frac{0}{2} = 0 ).Case 2: ( y = frac{3}{5} ), ( z = -frac{4}{5} ).( f(x) = frac{1 + frac{3}{5}}{2 + (-frac{4}{5})} = frac{frac{8}{5}}{frac{6}{5}} = frac{8}{6} = frac{4}{3} ).So, the maximum value is ( frac{4}{3} ) and the minimum is 0. Therefore, the largest ( k ) is ( frac{4}{3} ), which is option (D).Wait, but just to make sure, let me think if there's any other critical point or if I missed something. When I used the substitution ( t = tan(x/2) ), I found critical points at ( t = 3 ) and ( t = -1 ), which correspond to ( x = 2 arctan(3) ) and ( x = 2 arctan(-1) = -pi/2 ). At ( x = -pi/2 ), ( sin x = -1 ), ( cos x = 0 ), so ( f(x) = frac{1 - 1}{2 + 0} = 0 ), which is consistent with the Lagrange multiplier result.Therefore, both methods give the same result, so I can be confident that ( frac{4}{3} ) is indeed the maximum value.So, the answer should be (D) ( frac{4}{3} ).**Final Answer**boxed{D}

answer：Okay, so I have this problem here: If the complex number ( z = m^{2} - 1 + (m - 1)i ) where ( m ) is a real number, is a real number, then what is the value of ( m )? Hmm, let me think about how to approach this.First, I remember that a complex number is generally written as ( a + bi ), where ( a ) is the real part and ( b ) is the imaginary part. For a complex number to be purely real, the imaginary part must be zero. So, in this case, the imaginary part of ( z ) is ( (m - 1) ). That means I need to set ( (m - 1) ) equal to zero and solve for ( m ).Let me write that down: ( m - 1 = 0 ). Solving for ( m ), I just add 1 to both sides, so ( m = 1 ). Hmm, that seems straightforward. But wait, let me double-check to make sure I didn't miss anything.The complex number is given by ( z = m^{2} - 1 + (m - 1)i ). If ( m = 1 ), then substituting back into ( z ), we get ( z = (1)^2 - 1 + (1 - 1)i ). Calculating that, ( 1 - 1 = 0 ) for the real part, and ( (1 - 1)i = 0i ) for the imaginary part. So, ( z = 0 + 0i ), which is indeed a real number (specifically, zero). Wait, but the problem says ( z ) is a real number, but doesn't specify it has to be non-zero. So, zero is a real number, so that's okay. So, ( m = 1 ) is a valid solution.Is there any other solution? Let me think. The imaginary part is ( m - 1 ), so setting that equal to zero only gives ( m = 1 ). So, that's the only solution.But just to make sure, let me consider if there's any other way the imaginary part could be zero. For example, sometimes in equations, especially quadratic ones, you can have multiple solutions. But here, the imaginary part is linear in ( m ), so it's a straight line equation, which only crosses zero once. So, no, only one solution.Alternatively, if I think about the definition of a complex number being real, it's when the coefficient of ( i ) is zero. So, in ( z = m^{2} - 1 + (m - 1)i ), the coefficient of ( i ) is ( m - 1 ). So, setting that equal to zero gives ( m = 1 ). Just to recap, the steps are:1. Identify the imaginary part of the complex number ( z ).2. Set the imaginary part equal to zero because ( z ) is real.3. Solve the resulting equation for ( m ).4. Verify the solution by plugging it back into the original expression for ( z ).So, step 1: Imaginary part is ( m - 1 ).Step 2: ( m - 1 = 0 ).Step 3: ( m = 1 ).Step 4: Substitute ( m = 1 ) into ( z ): ( z = 1 - 1 + (1 - 1)i = 0 + 0i ), which is real.Therefore, the value of ( m ) is 1.I don't think I made any mistakes here. It seems pretty straightforward. The key was recognizing that for a complex number to be real, its imaginary component must be zero. So, focusing on that part of the expression was the right move.Another way to think about it is that if ( z ) is real, then ( z = overline{z} ), where ( overline{z} ) is the complex conjugate. The complex conjugate of ( z = a + bi ) is ( a - bi ). So, setting ( z = overline{z} ), we have:( m^{2} - 1 + (m - 1)i = m^{2} - 1 - (m - 1)i )Subtracting ( m^{2} - 1 ) from both sides:( (m - 1)i = - (m - 1)i )Adding ( (m - 1)i ) to both sides:( 2(m - 1)i = 0 )Since ( i ) is not zero, this implies ( 2(m - 1) = 0 ), so ( m - 1 = 0 ), leading again to ( m = 1 ). This method also confirms the solution.I think that's thorough enough. I considered two different approaches: directly setting the imaginary part to zero and using the property that a complex number equals its conjugate if and only if it's real. Both methods led me to the same answer, so I feel confident that ( m = 1 ) is correct.Just to make sure, let me consider if ( m ) could be something else. Suppose ( m = 0 ). Then, ( z = 0 - 1 + (-1)i = -1 - i ), which is not real. If ( m = 2 ), then ( z = 4 - 1 + (2 - 1)i = 3 + i ), which is also not real. So, only when ( m = 1 ) do we get a real number. That makes sense.Therefore, the value of ( m ) is indeed 1.**Final Answer**The value of ( m ) is boxed{1}.