💡Okay, so I have this problem where I need to find the maximum value of ( x + 2y ) for a point ( P(x, y) ) moving on the ellipse given by the equation ( frac{x^2}{6} + frac{y^2}{4} = 1 ). Hmm, alright, let me think about how to approach this.First, I remember that ellipses can be parametrized using trigonometric functions. The standard form of an ellipse is ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), where ( a ) and ( b ) are the semi-major and semi-minor axes. In this case, comparing to the standard form, ( a^2 = 6 ) and ( b^2 = 4 ), so ( a = sqrt{6} ) and ( b = 2 ). That means I can write the parametric equations for the ellipse as:[x = sqrt{6} cos theta][y = 2 sin theta]where ( theta ) is the parameter varying from 0 to ( 2pi ).So, substituting these into the expression ( x + 2y ), I get:[x + 2y = sqrt{6} cos theta + 2 times 2 sin theta = sqrt{6} cos theta + 4 sin theta]Alright, so now I have ( x + 2y ) expressed in terms of ( theta ). I need to find its maximum value. This looks like a linear combination of sine and cosine functions, which I think can be rewritten as a single sine or cosine function with a phase shift. I recall that any expression of the form ( A cos theta + B sin theta ) can be written as ( C sin(theta + phi) ) or ( C cos(theta + phi) ), where ( C = sqrt{A^2 + B^2} ) and ( phi ) is some angle.Let me try that. So, in my case, ( A = sqrt{6} ) and ( B = 4 ). Therefore, the amplitude ( C ) should be:[C = sqrt{(sqrt{6})^2 + 4^2} = sqrt{6 + 16} = sqrt{22}]So, ( x + 2y = sqrt{22} sin(theta + phi) ) or ( sqrt{22} cos(theta + phi) ). I need to figure out which one it is and what ( phi ) is.I think the identity is:[A cos theta + B sin theta = C cos(theta - phi)]where ( C = sqrt{A^2 + B^2} ) and ( tan phi = frac{B}{A} ). Alternatively, it can also be written as:[A cos theta + B sin theta = C sin(theta + phi)]with a different ( phi ). Maybe I should just stick with one form and compute ( phi ).Let me use the sine form:[sqrt{6} cos theta + 4 sin theta = sqrt{22} sin(theta + phi)]Expanding the right-hand side using the sine addition formula:[sqrt{22} sin(theta + phi) = sqrt{22} (sin theta cos phi + cos theta sin phi)]Comparing coefficients with the left-hand side:[sqrt{6} cos theta + 4 sin theta = sqrt{22} sin phi cos theta + sqrt{22} cos phi sin theta]So, equating the coefficients:[sqrt{6} = sqrt{22} sin phi][4 = sqrt{22} cos phi]Hmm, so from the first equation:[sin phi = frac{sqrt{6}}{sqrt{22}} = sqrt{frac{6}{22}} = sqrt{frac{3}{11}}]And from the second equation:[cos phi = frac{4}{sqrt{22}} = frac{4}{sqrt{22}} = frac{4 sqrt{22}}{22} = frac{2 sqrt{22}}{11}]Wait, let me check that. Actually, ( frac{4}{sqrt{22}} ) can be rationalized as ( frac{4 sqrt{22}}{22} ), which simplifies to ( frac{2 sqrt{22}}{11} ). So, that's correct.Now, since both ( sin phi ) and ( cos phi ) are positive, ( phi ) must be in the first quadrant. That makes sense.But actually, do I even need to find ( phi )? Because regardless of the phase shift, the maximum value of ( sin(theta + phi) ) is 1, right? So, the maximum value of ( x + 2y ) would just be ( sqrt{22} times 1 = sqrt{22} ).Wait, let me make sure. The expression ( sqrt{22} sin(theta + phi) ) oscillates between ( -sqrt{22} ) and ( sqrt{22} ), so the maximum value is indeed ( sqrt{22} ).But just to be thorough, let me consider another approach to confirm this result. Maybe using Lagrange multipliers or something else.Alternatively, I can use the Cauchy-Schwarz inequality. The expression ( x + 2y ) can be thought of as a dot product of the vector ( (x, y) ) with the vector ( (1, 2) ). The Cauchy-Schwarz inequality states that:[|x cdot 1 + y cdot 2| leq sqrt{x^2 + y^2} times sqrt{1^2 + 2^2} = sqrt{x^2 + y^2} times sqrt{5}]But wait, in this case, ( x ) and ( y ) are constrained by the ellipse equation ( frac{x^2}{6} + frac{y^2}{4} = 1 ), not by ( x^2 + y^2 ). So, maybe I need to adjust this approach.Perhaps I can use the method of Lagrange multipliers. Let me set up the function to maximize:[f(x, y) = x + 2y]subject to the constraint:[g(x, y) = frac{x^2}{6} + frac{y^2}{4} - 1 = 0]The method of Lagrange multipliers tells us that at the maximum, the gradient of ( f ) is proportional to the gradient of ( g ). So:[nabla f = lambda nabla g]Calculating the gradients:[nabla f = (1, 2)][nabla g = left( frac{2x}{6}, frac{2y}{4} right) = left( frac{x}{3}, frac{y}{2} right)]So, setting up the equations:[1 = lambda cdot frac{x}{3} quad Rightarrow quad lambda = frac{3}{x}][2 = lambda cdot frac{y}{2} quad Rightarrow quad lambda = frac{4}{y}]Therefore, ( frac{3}{x} = frac{4}{y} ) which implies ( 3y = 4x ) or ( y = frac{4}{3}x ).Now, substitute ( y = frac{4}{3}x ) into the constraint equation:[frac{x^2}{6} + frac{(frac{4}{3}x)^2}{4} = 1]Simplify:[frac{x^2}{6} + frac{frac{16}{9}x^2}{4} = 1][frac{x^2}{6} + frac{4x^2}{9} = 1]To combine these, find a common denominator, which is 18:[frac{3x^2}{18} + frac{8x^2}{18} = 1][frac{11x^2}{18} = 1][x^2 = frac{18}{11}][x = pm sqrt{frac{18}{11}} = pm frac{3sqrt{22}}{11}]Wait, hold on. ( sqrt{frac{18}{11}} ) is equal to ( frac{3sqrt{2}}{sqrt{11}} ), which can be rationalized as ( frac{3sqrt{22}}{11} ). So, ( x = pm frac{3sqrt{22}}{11} ).Then, ( y = frac{4}{3}x = pm frac{4}{3} times frac{3sqrt{22}}{11} = pm frac{4sqrt{22}}{11} ).Now, compute ( x + 2y ) for these values. Let's take the positive case first:[x + 2y = frac{3sqrt{22}}{11} + 2 times frac{4sqrt{22}}{11} = frac{3sqrt{22}}{11} + frac{8sqrt{22}}{11} = frac{11sqrt{22}}{11} = sqrt{22}]Similarly, for the negative case:[x + 2y = -frac{3sqrt{22}}{11} + 2 times left(-frac{4sqrt{22}}{11}right) = -frac{3sqrt{22}}{11} - frac{8sqrt{22}}{11} = -frac{11sqrt{22}}{11} = -sqrt{22}]So, the maximum value is ( sqrt{22} ) and the minimum is ( -sqrt{22} ). Therefore, the maximum value of ( x + 2y ) is indeed ( sqrt{22} ).Wait, just to make sure I didn't make any calculation errors. Let me double-check the substitution step.We had ( y = frac{4}{3}x ), substituted into the ellipse equation:[frac{x^2}{6} + frac{(frac{4}{3}x)^2}{4} = frac{x^2}{6} + frac{frac{16}{9}x^2}{4} = frac{x^2}{6} + frac{4x^2}{9}]Yes, that's correct. Then, converting to a common denominator:[frac{3x^2}{18} + frac{8x^2}{18} = frac{11x^2}{18} = 1]So, ( x^2 = frac{18}{11} ), which leads to ( x = pm frac{3sqrt{2}}{sqrt{11}} = pm frac{3sqrt{22}}{11} ). That seems right.Then, ( y = frac{4}{3}x ), so substituting ( x ), we get ( y = pm frac{4sqrt{22}}{11} ). Then, ( x + 2y ) is ( sqrt{22} ) or ( -sqrt{22} ). Yep, that all checks out.Alternatively, another method I can think of is using parametric equations and calculus. Since I have ( x + 2y = sqrt{6} cos theta + 4 sin theta ), I can take the derivative with respect to ( theta ) and set it to zero to find the maximum.Let me try that. Let ( f(theta) = sqrt{6} cos theta + 4 sin theta ). Then, the derivative is:[f'(theta) = -sqrt{6} sin theta + 4 cos theta]Set ( f'(theta) = 0 ):[-sqrt{6} sin theta + 4 cos theta = 0][4 cos theta = sqrt{6} sin theta][frac{sin theta}{cos theta} = frac{4}{sqrt{6}} = frac{2sqrt{6}}{3}]So, ( tan theta = frac{2sqrt{6}}{3} ). Let me compute ( theta ) such that this is true. Let me denote ( theta = arctanleft( frac{2sqrt{6}}{3} right) ).Now, to find ( f(theta) ) at this critical point, let's compute ( cos theta ) and ( sin theta ). Let me consider a right triangle where the opposite side is ( 2sqrt{6} ) and the adjacent side is 3. Then, the hypotenuse is:[sqrt{(2sqrt{6})^2 + 3^2} = sqrt{24 + 9} = sqrt{33}]So, ( sin theta = frac{2sqrt{6}}{sqrt{33}} = frac{2sqrt{6}}{sqrt{33}} ) and ( cos theta = frac{3}{sqrt{33}} ).Therefore, substituting back into ( f(theta) ):[f(theta) = sqrt{6} times frac{3}{sqrt{33}} + 4 times frac{2sqrt{6}}{sqrt{33}} = frac{3sqrt{6}}{sqrt{33}} + frac{8sqrt{6}}{sqrt{33}} = frac{11sqrt{6}}{sqrt{33}} = frac{11sqrt{6}}{sqrt{33}} = frac{11sqrt{6}}{sqrt{33}} = frac{11sqrt{6}}{sqrt{33}} = sqrt{22}]Wait, let me compute that step by step:First, ( sqrt{6} times frac{3}{sqrt{33}} = frac{3sqrt{6}}{sqrt{33}} )Second, ( 4 times frac{2sqrt{6}}{sqrt{33}} = frac{8sqrt{6}}{sqrt{33}} )Adding them together:[frac{3sqrt{6} + 8sqrt{6}}{sqrt{33}} = frac{11sqrt{6}}{sqrt{33}} = frac{11sqrt{6}}{sqrt{33}} = frac{11sqrt{6}}{sqrt{33}} = frac{11sqrt{6}}{sqrt{33}} = sqrt{22}]Wait, how did that happen? Let me see:( frac{11sqrt{6}}{sqrt{33}} = 11 times frac{sqrt{6}}{sqrt{33}} = 11 times sqrt{frac{6}{33}} = 11 times sqrt{frac{2}{11}} = 11 times frac{sqrt{22}}{11} = sqrt{22} ). Ah, yes, that's correct.So, using calculus, I also arrive at the maximum value of ( sqrt{22} ). That's consistent with the previous methods.Just to recap, I used three methods:1. Parametrization and expressing as a single sine function.2. Lagrange multipliers.3. Calculus with parametric equations.All three methods led me to the conclusion that the maximum value of ( x + 2y ) is ( sqrt{22} ). Therefore, I'm confident that this is the correct answer.**Final Answer**The maximum value of ( x + 2y ) is boxed{sqrt{22}}.

💡Okay, so I have this trigonometry problem here: Given that sinθ cosθ < 0, determine the quadrant of angle θ. The options are A: First or Second, B: Second or Third, C: Third or Fourth, and D: Second or Fourth. Hmm, I need to figure out which quadrant θ is in based on the product of sine and cosine being negative.Alright, let me start by recalling the signs of sine and cosine in each quadrant. I remember that in the first quadrant, both sine and cosine are positive. In the second quadrant, sine is positive, and cosine is negative. In the third quadrant, both sine and cosine are negative, and in the fourth quadrant, sine is negative while cosine is positive. So, that's a good starting point.Now, the problem says that sinθ cosθ is less than zero. That means their product is negative. For a product of two numbers to be negative, one has to be positive and the other negative. So, either sinθ is positive and cosθ is negative, or sinθ is negative and cosθ is positive.Let me break this down into two cases:1. **Case 1: sinθ > 0 and cosθ < 0** - Where is sine positive? In the first and second quadrants. - Where is cosine negative? In the second and third quadrants. - So, the overlap where both conditions are true is the second quadrant.2. **Case 2: sinθ < 0 and cosθ > 0** - Where is sine negative? In the third and fourth quadrants. - Where is cosine positive? In the first and fourth quadrants. - The overlap here is the fourth quadrant.So, from these two cases, θ can be in either the second or the fourth quadrant. Let me check the options again: A is first or second, B is second or third, C is third or fourth, and D is second or fourth. So, D seems to be the correct answer.Wait, just to make sure I didn't make a mistake. Let me visualize the quadrants:- **First Quadrant (I):** Both sine and cosine are positive. So, sinθ cosθ would be positive, which doesn't satisfy the condition.- **Second Quadrant (II):** Sine is positive, cosine is negative. So, sinθ cosθ would be negative, which fits.- **Third Quadrant (III):** Both sine and cosine are negative. Multiplying two negatives gives a positive, so sinθ cosθ would be positive. That doesn't fit.- **Fourth Quadrant (IV):** Sine is negative, cosine is positive. So, sinθ cosθ would be negative, which fits.Yep, that confirms it. θ must be in either the second or the fourth quadrant. So, the answer should be D.I think I've got it right. It's important to remember the signs of sine and cosine in each quadrant. Maybe I can also think about specific angles to test this. For example, in the second quadrant, let's take θ = 120 degrees. Sin(120) is positive, cos(120) is negative, so their product is negative. In the fourth quadrant, θ = 300 degrees. Sin(300) is negative, cos(300) is positive, so their product is negative. That works.What if I pick angles from other quadrants? Let's say θ = 45 degrees (first quadrant). Both sine and cosine are positive, so their product is positive. Doesn't satisfy the condition. θ = 210 degrees (third quadrant). Both sine and cosine are negative, so their product is positive. Again, doesn't satisfy the condition. So, only second and fourth quadrants work.I think I'm confident now that D is the correct answer.

💡Alright, so I've got this problem about 31 wallets, each starting with 100 coins. Someone took some coins from one wallet and gave one coin to each wallet to its right. My job is to figure out which wallet was the one that lost coins, and I can do this by asking one question where I get the total number of coins in any set of wallets I choose. The goal is to find the minimum number of questions needed to guarantee identifying the "lightened" wallet.Okay, let's break this down. First, all wallets start with 100 coins. If someone takes coins from one wallet and gives one to each wallet to its right, that means the original wallet will have fewer coins, and each wallet to its right will have one more coin. So, if the lightened wallet is the k-th wallet from the left, it will have 100 - (31 - k) coins, right? Because there are 31 wallets, so from wallet k, there are (31 - k) wallets to its right, each receiving one coin. So, the lightened wallet loses (31 - k) coins.Now, I need to figure out a way to determine k by asking about the total number of coins in some set of wallets. The challenge is to choose the right set so that the total will give me enough information to pinpoint k.One strategy that comes to mind is to use a binary approach or something similar to binary search. But since I can only ask one question, I need to design that question in a way that the total will uniquely identify the lightened wallet.Wait, maybe I can use a weighted sum or something. If I assign weights to each wallet and then sum them up, the change in the total could help me identify which wallet was lightened. For example, if I assign each wallet a unique weight, like powers of 2, then the total would be unique for each wallet.But let's think more carefully. If I number the wallets from 1 to 31, left to right, and assign each wallet a weight of 2^(i-1), where i is the wallet number, then the total sum would be the sum of 100 * 2^(i-1) for i from 1 to 31. If a wallet k is lightened, it would contribute (100 - (31 - k)) * 2^(k-1) instead of 100 * 2^(k-1). The difference would be (31 - k) * 2^(k-1). Since each difference is unique, I could identify k by looking at the difference in the total sum.But wait, the problem says I can ask one question, which is the total number of coins in any set of wallets. So, I can't assign weights in the way I just thought. Instead, I need to choose a specific set of wallets whose total will change in a way that reveals k.Maybe I can use parity. If I ask for the total number of coins in all the odd-numbered wallets, that might help. Let's see. If the lightened wallet is even-numbered, then the odd-numbered wallets to its right would have one more coin each, increasing the total. If the lightened wallet is odd-numbered, then the odd-numbered wallets to its right would also have one more coin each, but the lightened wallet itself is odd, so it would have fewer coins. Hmm, this could work.Let's formalize this. Let’s denote the wallets as W1, W2, ..., W31. Suppose the lightened wallet is Wk. If k is odd, then Wk will have 100 - (31 - k) coins, and each wallet to its right, including the odd ones, will have 101 coins. If k is even, then Wk will have 100 - (31 - k) coins, and the odd wallets to its right will have 101 coins.So, if I ask for the total number of coins in all odd-numbered wallets, the total will be:- If k is odd: Total = (100 - (31 - k)) + sum of 101 for each odd wallet to the right of Wk.- If k is even: Total = sum of 101 for each odd wallet to the right of Wk.Wait, let's calculate this more precisely. Initially, all odd wallets have 100 coins. If k is odd, then Wk loses (31 - k) coins, and each odd wallet to its right gains 1 coin. So, the total for odd wallets would be:Total = 100 + (number of odd wallets to the right of Wk) * 1 - (31 - k).Similarly, if k is even, then Wk loses (31 - k) coins, but since k is even, Wk is not an odd wallet, so the total for odd wallets would be:Total = 100 * (number of odd wallets) + (number of odd wallets to the right of Wk) * 1.Wait, no, because if k is even, Wk is not an odd wallet, so the total for odd wallets would just be 100 * (number of odd wallets) + (number of odd wallets to the right of Wk) * 1.But actually, the number of odd wallets is 16 (since 31 is odd, there are 16 odd-numbered wallets: 1,3,...,31). So, initially, the total for odd wallets is 16*100 = 1600.If k is odd, then Wk loses (31 - k) coins, and each odd wallet to its right gains 1 coin. So, the total becomes:1600 - (31 - k) + (number of odd wallets to the right of Wk).Similarly, if k is even, then Wk loses (31 - k) coins, but since Wk is even, the total for odd wallets is just:1600 + (number of odd wallets to the right of Wk).So, the difference in the total will tell me whether k is odd or even, and how many odd wallets are to the right of Wk, which can help me find k.But how exactly? Let's think about it.If k is odd, then the total is 1600 - (31 - k) + m, where m is the number of odd wallets to the right of Wk.If k is even, the total is 1600 + m.So, if I get a total T, I can compute T - 1600. If T - 1600 is positive, then k is even, and m = T - 1600. If T - 1600 is negative, then k is odd, and m = T - 1600 + (31 - k).Wait, this seems a bit convoluted. Maybe there's a better way.Alternatively, if I ask for the total of all wallets, that would be 31*100 = 3100. But since one wallet lost (31 - k) coins, the total would be 3100 - (31 - k). So, if I ask for the total of all wallets, I can find (31 - k) = 3100 - T, and thus k = 31 - (3100 - T).But wait, the problem says I can ask for the total in any set of wallets. So, if I ask for the total of all wallets, I can directly find k. But the problem is, I need to guarantee identification, and I'm not sure if asking for all wallets is allowed or if it's considered one question. I think it is, because the question is "what is the total number of coins in any set of wallets," and the set can be all wallets.But let's check. If I ask for the total of all wallets, T = 3100 - (31 - k). So, 31 - k = 3100 - T, so k = 31 - (3100 - T).But wait, T is the total after the transfer, so T = 3100 - (31 - k). So, k = 31 - (3100 - T).But I can calculate k directly from T. So, if I ask for the total of all wallets, I can find k in one question.But wait, the problem says "from one wallet, some coins were transferred: one coin to each of the wallets to its right." So, the total number of coins transferred is (31 - k), so the total number of coins is 3100 - (31 - k). So, if I ask for the total, I can find (31 - k) = 3100 - T, and thus k = 31 - (3100 - T).So, actually, asking for the total of all wallets would give me the exact value of k in one question. That seems too easy, but maybe that's the case.Wait, but the problem says "some coins were transferred: one coin to each of the wallets to its right." So, the number of coins transferred is (31 - k), which is subtracted from the total. So, the total would be 3100 - (31 - k). Therefore, if I ask for the total, I can directly compute k.But is this correct? Let's test with an example. Suppose k = 1. Then, the total would be 3100 - 30 = 3070. So, k = 31 - (3100 - 3070) = 31 - 30 = 1. Correct.If k = 16, then total = 3100 - 15 = 3085. So, k = 31 - (3100 - 3085) = 31 - 15 = 16. Correct.If k = 31, then total = 3100 - 0 = 3100. So, k = 31 - (3100 - 3100) = 31 - 0 = 31. Correct.So, yes, asking for the total of all wallets would give me the exact value of k in one question. Therefore, the minimum number of questions needed is one.But wait, the problem says "from one wallet, some coins were transferred: one coin to each of the wallets to its right." So, if the lightened wallet is the last one, W31, then no coins were transferred, so the total remains 3100. So, in that case, k = 31.But what if the lightened wallet is W1? Then, 30 coins were transferred, so the total is 3100 - 30 = 3070.So, yes, in all cases, asking for the total of all wallets gives me the exact k in one question.But wait, the problem says "some coins were transferred," implying that at least one coin was transferred. So, k cannot be 31, because that would mean no coins were transferred. So, k is from 1 to 30.But regardless, asking for the total of all wallets would still give me k in one question.Wait, but the problem says "some coins were transferred: one coin to each of the wallets to its right." So, the number of coins transferred is (31 - k), which is at least 1 if k <= 30.So, yes, asking for the total of all wallets would give me the exact k in one question.But let me think again. If I ask for the total of all wallets, I get T = 3100 - (31 - k). So, k = 31 - (3100 - T). So, k is uniquely determined by T.Therefore, the minimum number of questions needed is one.But wait, the problem says "by asking one question, you can find out the total number of coins in any set of wallets." So, I can choose any set, not necessarily all wallets. So, if I choose all wallets, I can get T and find k. So, yes, one question is sufficient.But maybe the problem expects a different approach, like using binary search or something, but in this case, one question is enough.Wait, but let's think about it differently. Suppose I don't ask for all wallets, but for a specific subset. For example, if I ask for the total of wallets 1 to 15, or something like that. But I think asking for all wallets is the most straightforward way to get k in one question.Alternatively, if I don't want to ask for all wallets, maybe I can use a different set. For example, if I ask for the total of wallets 1 to 16, which are the odd-numbered wallets. Wait, no, wallets 1 to 16 include both odd and even. Wait, no, wallets 1 to 16 are 16 wallets, but they include both odd and even.Wait, earlier I thought about asking for odd-numbered wallets. Let's see. If I ask for the total of odd-numbered wallets, which are 16 wallets: 1,3,5,...,31.Initially, the total is 16*100 = 1600.If the lightened wallet is odd, say Wk where k is odd, then Wk has 100 - (31 - k) coins, and each odd wallet to its right has 101 coins. So, the total would be:1600 - (31 - k) + m, where m is the number of odd wallets to the right of Wk.If the lightened wallet is even, say Wk where k is even, then Wk has 100 - (31 - k) coins, but since Wk is even, it's not included in the odd total. So, the total would be:1600 + m, where m is the number of odd wallets to the right of Wk.So, if I get a total T, I can compute T - 1600. If T - 1600 is positive, then k is even, and m = T - 1600. If T - 1600 is negative, then k is odd, and m = T - 1600 + (31 - k).But this seems more complicated than just asking for the total of all wallets. Because in the case of asking for all wallets, I can directly compute k from T.So, why would I ask for odd-numbered wallets when I can just ask for all wallets and get k directly?Wait, maybe the problem is designed so that you can't ask for all wallets, but I don't think so. The problem says "any set of wallets," so I can choose any set, including all wallets.Therefore, the minimum number of questions needed is one.But let me double-check. Suppose I ask for all wallets, and get T. Then, k = 31 - (3100 - T). So, yes, that gives me k.Alternatively, if I ask for the total of all wallets, I can directly find k.So, yes, one question is sufficient.But wait, let's think about it again. If I ask for all wallets, I get T = 3100 - (31 - k). So, 31 - k = 3100 - T, so k = 31 - (3100 - T).But wait, T is the total after the transfer, so T = 3100 - (31 - k). So, k = 31 - (3100 - T).Yes, that's correct.So, in conclusion, asking for the total of all wallets gives me the exact value of k in one question. Therefore, the minimum number of questions needed is one.

💡Alright, so I've got this function here: f(x) = 2 sin x cos x - 2 sin²x + 1. Hmm, okay. I need to find the range of this function when x is between 0 and π/2, and then figure out where it's strictly increasing between 0 and π. Let me start by trying to simplify the function to make it easier to work with.First, I remember that 2 sin x cos x is equal to sin 2x. So, I can rewrite the function as f(x) = sin 2x - 2 sin²x + 1. Now, looking at the -2 sin²x part, I recall that sin²x can be expressed using a double-angle identity: sin²x = (1 - cos 2x)/2. So, substituting that in, we get:f(x) = sin 2x - 2*(1 - cos 2x)/2 + 1Simplifying that, the 2 in the numerator and denominator cancel out, so it becomes:f(x) = sin 2x - (1 - cos 2x) + 1Distribute the negative sign:f(x) = sin 2x - 1 + cos 2x + 1The -1 and +1 cancel each other out, so we're left with:f(x) = sin 2x + cos 2xOkay, that's simpler. Now, I remember that expressions like a sin θ + b cos θ can be rewritten as a single sine or cosine function using the amplitude-phase form. Specifically, a sin θ + b cos θ = √(a² + b²) sin(θ + φ), where φ is the phase shift. Let me apply that here.In this case, a = 1 (coefficient of sin 2x) and b = 1 (coefficient of cos 2x). So, the amplitude would be √(1² + 1²) = √2. The phase shift φ can be found using tan φ = b/a, which is 1/1 = 1. So, φ = arctan(1) = π/4.Therefore, f(x) can be rewritten as:f(x) = √2 sin(2x + π/4)Alright, that looks cleaner. Now, moving on to the first part: finding the range of f(x) when x is in [0, π/2]. Since f(x) is expressed as √2 sin(2x + π/4), the range will depend on the values that sin(2x + π/4) can take as x varies from 0 to π/2.Let's figure out the interval for the argument of the sine function. When x = 0:2x + π/4 = 0 + π/4 = π/4When x = π/2:2x + π/4 = 2*(π/2) + π/4 = π + π/4 = 5π/4So, the argument 2x + π/4 ranges from π/4 to 5π/4 as x goes from 0 to π/2. Now, the sine function, sin θ, has a range of [-1, 1]. But let's see how it behaves between π/4 and 5π/4.Starting at π/4, sin(π/4) = √2/2 ≈ 0.707. As θ increases to π/2, sin(π/2) = 1. Then, as θ increases to π, sin π = 0. Continuing to 3π/2, sin(3π/2) = -1, but our upper limit is 5π/4, which is less than 3π/2. So, at 5π/4, sin(5π/4) = -√2/2 ≈ -0.707.So, the sine function starts at √2/2, goes up to 1, then back down to -√2/2. Therefore, the maximum value of sin(2x + π/4) in this interval is 1, and the minimum is -√2/2.But wait, let me double-check that. From π/4 to 5π/4, the sine function reaches its maximum at π/2 and its minimum at 3π/2, but since 5π/4 is less than 3π/2, the minimum in this interval would actually be at 5π/4, which is -√2/2, as I initially thought.Therefore, the range of sin(2x + π/4) is [-√2/2, 1]. Multiplying by √2, the range of f(x) becomes:√2 * [-√2/2, 1] = [- (√2 * √2)/2, √2 * 1] = [-2/2, √2] = [-1, √2]So, the range of f(x) when x is in [0, π/2] is [-1, √2].Now, moving on to the second part: finding the intervals where f(x) is strictly increasing when x is in [0, π]. Since f(x) is expressed as √2 sin(2x + π/4), we can find its derivative to determine where it's increasing.The derivative f'(x) is:f'(x) = √2 * cos(2x + π/4) * 2 = 2√2 cos(2x + π/4)We need to find where f'(x) > 0, which means cos(2x + π/4) > 0.So, cos(θ) > 0 when θ is in (-π/2 + 2πk, π/2 + 2πk) for any integer k.Let’s set θ = 2x + π/4. We need to find x such that:-π/2 + 2πk < 2x + π/4 < π/2 + 2πkSubtract π/4 from all parts:-π/2 - π/4 + 2πk < 2x < π/2 - π/4 + 2πkSimplify:-3π/4 + 2πk < 2x < π/4 + 2πkDivide by 2:-3π/8 + πk < x < π/8 + πkNow, we need to find all intervals within [0, π] where this inequality holds.Let’s consider k = 0:-3π/8 < x < π/8But since x must be ≥ 0, the interval becomes [0, π/8).Next, k = 1:-3π/8 + π = 5π/8 < x < π/8 + π = 9π/8But x must be ≤ π, so the interval becomes (5π/8, π].k = 2 would give x beyond π, which is outside our domain, and k = -1 would give x negative, which is also outside our domain.Therefore, the intervals where f(x) is strictly increasing in [0, π] are [0, π/8) and (5π/8, π].But since the question asks for intervals where f(x) is strictly increasing, and in the context of calculus, endpoints can sometimes be included if the function is increasing up to that point. However, since at x = π/8 and x = 5π/8, the derivative is zero, which are critical points. So, strictly increasing would exclude these points.Therefore, the intervals are [0, π/8] and [5π/8, π], but since at x = π/8 and x = 5π/8, the function is neither increasing nor decreasing, it's just stationary. So, to be strictly increasing, we should exclude these points, making the intervals open at these points.But in the context of the problem, it might be acceptable to include the endpoints if the function is increasing up to those points. However, strictly speaking, since the derivative is zero at those points, they are not points where the function is increasing. So, the correct intervals are [0, π/8) and (5π/8, π].But let me double-check this. Let's pick a point just below π/8, say π/8 - ε, and a point just above π/8, say π/8 + ε. The derivative at π/8 - ε would be positive, and at π/8 + ε, it would be negative, since the cosine function is decreasing through zero at π/2. Similarly, at 5π/8, the derivative goes from negative to positive as x increases through 5π/8. Wait, no, actually, let's think about the behavior.Wait, the derivative is 2√2 cos(2x + π/4). Let's analyze the sign of the derivative:- When 2x + π/4 is in (-π/2 + 2πk, π/2 + 2πk), cos is positive.So, for x in (-3π/8 + πk, π/8 + πk), cos(2x + π/4) is positive.So, in the interval [0, π], the positive derivative occurs in [0, π/8) and (5π/8, π].But let's test a point in (π/8, 5π/8). Let's pick x = π/2.Compute 2x + π/4 = π + π/4 = 5π/4. Cos(5π/4) = -√2/2 < 0. So, the derivative is negative there, meaning the function is decreasing.Similarly, pick x = 3π/8:2x + π/4 = 6π/8 + 2π/8 = 8π/8 = π. Cos(π) = -1 < 0. So, derivative is negative.Pick x = 7π/8:2x + π/4 = 14π/8 + 2π/8 = 16π/8 = 2π. Cos(2π) = 1 > 0. So, derivative is positive.Wait, but 7π/8 is greater than 5π/8, so in the interval (5π/8, π], the derivative is positive.Wait, but when x approaches 5π/8 from the left, 2x + π/4 approaches 5π/4 + π/4 = 6π/4 = 3π/2, where cos is zero. So, just to the left of 5π/8, the derivative is negative, and just to the right, it's positive.Therefore, the function is increasing on [0, π/8] and [5π/8, π], but strictly increasing would exclude the points where the derivative is zero, so [0, π/8) and (5π/8, π].But in the context of the problem, it's probably acceptable to include the endpoints as the function is increasing up to those points, even though the derivative is zero there. So, the intervals are [0, π/8] and [5π/8, π].Wait, but let me think again. If the derivative is zero at π/8 and 5π/8, those are points where the function changes from increasing to decreasing or vice versa. So, strictly increasing would require the derivative to be positive throughout the interval, excluding the endpoints where the derivative is zero.Therefore, the correct intervals where f(x) is strictly increasing are [0, π/8) and (5π/8, π].But to be precise, in calculus, when we talk about strictly increasing on an interval, we usually consider open intervals where the derivative is positive. So, the answer should be (0, π/8) and (5π/8, π). However, since at x=0, the function is at its minimum, and it starts increasing from there, including x=0 is acceptable because the function is increasing at x=0 (the derivative is positive just after x=0). Similarly, at x=π, the function is at its maximum, and it's increasing just before x=π.Wait, but at x=π, 2x + π/4 = 2π + π/4 = 9π/4, which is equivalent to π/4, and cos(π/4) is positive, so the derivative is positive just before x=π. Therefore, the function is increasing up to x=π.But at x=π/8, the derivative is zero, so just before π/8, the function is increasing, and just after, it's decreasing. Similarly, just before 5π/8, the function is decreasing, and just after, it's increasing.Therefore, the intervals where f(x) is strictly increasing are [0, π/8] and [5π/8, π].Wait, but to be strictly increasing, the function must have a positive derivative throughout the interval, not including points where the derivative is zero. So, strictly speaking, the intervals are (0, π/8) and (5π/8, π). However, at x=0, the function is at its minimum, and it's increasing immediately after, so including x=0 is acceptable. Similarly, at x=π, the function is at its maximum, and it's increasing just before x=π, so including x=π is acceptable.Therefore, the intervals where f(x) is strictly increasing are [0, π/8] and [5π/8, π].But let me confirm this by checking the behavior around these points.At x=0:f'(0) = 2√2 cos(π/4) = 2√2*(√2/2) = 2√2*(√2)/2 = 2*2/2 = 2 > 0. So, the function is increasing at x=0.At x=π/8:f'(π/8) = 2√2 cos(2*(π/8) + π/4) = 2√2 cos(π/4 + π/4) = 2√2 cos(π/2) = 0.So, at x=π/8, the derivative is zero, meaning the function stops increasing and starts decreasing.Similarly, at x=5π/8:f'(5π/8) = 2√2 cos(2*(5π/8) + π/4) = 2√2 cos(5π/4 + π/4) = 2√2 cos(3π/2) = 0.And at x=π:f'(π) = 2√2 cos(2π + π/4) = 2√2 cos(π/4) = 2√2*(√2/2) = 2 > 0.So, the function is increasing at x=π.Therefore, the function is increasing from x=0 up to x=π/8, then decreasing from x=π/8 to x=5π/8, and then increasing again from x=5π/8 to x=π.Hence, the intervals where f(x) is strictly increasing are [0, π/8] and [5π/8, π].Wait, but earlier I thought about excluding the endpoints where the derivative is zero. However, since the function is increasing up to x=π/8 and then starts decreasing, the interval [0, π/8] is correct because the function is increasing throughout that interval, including at x=π/8, even though the derivative is zero there. Similarly, from x=5π/8 to x=π, the function is increasing, including at x=5π/8 because just after that point, it starts increasing.Wait, no. At x=5π/8, the derivative is zero, and just after that, the derivative becomes positive, meaning the function starts increasing. So, including x=5π/8 in the increasing interval is acceptable because the function is increasing immediately after that point.Similarly, at x=π/8, the function stops increasing and starts decreasing, so including x=π/8 in the increasing interval is acceptable because up to that point, it was increasing.Therefore, the correct intervals where f(x) is strictly increasing are [0, π/8] and [5π/8, π].But to be precise, in calculus, when we say a function is strictly increasing on an interval, it means that for any two points x1 < x2 in the interval, f(x1) < f(x2). At the endpoints where the derivative is zero, the function is not increasing or decreasing at that exact point, but it's still considered increasing on the closed interval because it's increasing up to that point.Therefore, the final answer for the intervals where f(x) is strictly increasing is [0, π/8] and [5π/8, π].So, summarizing:1. The range of f(x) when x ∈ [0, π/2] is [-1, √2].2. The intervals where f(x) is strictly increasing when x ∈ [0, π] are [0, π/8] and [5π/8, π].I think that's it. I've double-checked the steps, and everything seems consistent.

💡First, I need to determine how much money Raul spent on comics. He started with 87 and had 55 left after his purchase. By subtracting the remaining amount from the initial amount, I can find out how much he spent on comics.Next, I know that each comic costs 4. To find out how many comics Raul bought, I will divide the total amount he spent on comics by the cost of one comic.This calculation will give me the number of comics Raul purchased.

💡Alright, so I have this problem where I'm given positive real numbers ( a_1 < a_2 < dots < a_n ), and I need to consider the function [f(x) = frac{a_1}{x + a_1} + frac{a_2}{x + a_2} + cdots + frac{a_n}{x + a_n}]My task is to determine the sum of the lengths of the disjoint intervals where ( f(x) > 1 ). Hmm, okay. Let me try to break this down step by step.First, I know that each term in the function ( f(x) ) is of the form ( frac{a_i}{x + a_i} ). Since all ( a_i ) are positive, each of these terms is a rational function. The function ( f(x) ) is the sum of these individual rational functions. I need to find the values of ( x ) where this sum is greater than 1. So, I need to solve the inequality:[frac{a_1}{x + a_1} + frac{a_2}{x + a_2} + cdots + frac{a_n}{x + a_n} > 1]To approach this, maybe I should first consider the behavior of each term ( frac{a_i}{x + a_i} ). Each of these terms is a hyperbola. As ( x ) approaches ( -a_i ) from the right, the term tends to infinity, and as ( x ) approaches infinity, the term tends to zero. So, each term is decreasing for ( x > -a_i ).Since all ( a_i ) are positive, the denominators ( x + a_i ) are positive when ( x > -a_i ). But since ( x ) can be any real number except ( x = -a_i ) (where the function is undefined), I need to consider the intervals between these points.Wait, actually, since ( a_1 < a_2 < dots < a_n ), the points where the function is undefined are ( x = -a_1, -a_2, dots, -a_n ). These points divide the real line into ( n + 1 ) intervals:1. ( (-infty, -a_n) )2. ( (-a_n, -a_{n-1}) )3. ( dots )4. ( (-a_2, -a_1) )5. ( (-a_1, infty) )But actually, since ( a_1 < a_2 < dots < a_n ), the points ( -a_n < -a_{n-1} < dots < -a_1 ). So, the intervals are:1. ( (-infty, -a_n) )2. ( (-a_n, -a_{n-1}) )3. ( dots )4. ( (-a_2, -a_1) )5. ( (-a_1, infty) )In each of these intervals, the function ( f(x) ) is continuous because the denominators don't vanish. So, I can analyze the behavior of ( f(x) ) in each interval.But before diving into each interval, maybe I can consider the function ( f(x) ) as a whole. Let me try to combine the terms into a single fraction. That might help me analyze where the function is greater than 1.So, let's write ( f(x) ) as:[f(x) = sum_{i=1}^n frac{a_i}{x + a_i}]To combine these, I need a common denominator, which would be the product of all denominators: ( prod_{i=1}^n (x + a_i) ). Let me denote this product as ( Q(x) ). So,[Q(x) = prod_{i=1}^n (x + a_i)]Then, each term ( frac{a_i}{x + a_i} ) can be written as ( frac{a_i prod_{j neq i} (x + a_j)}{Q(x)} ). Therefore, the numerator ( P(x) ) of ( f(x) ) is:[P(x) = sum_{i=1}^n a_i prod_{j neq i} (x + a_j)]So, ( f(x) = frac{P(x)}{Q(x)} ).Now, the inequality ( f(x) > 1 ) becomes:[frac{P(x)}{Q(x)} > 1 implies P(x) > Q(x)]Which simplifies to:[P(x) - Q(x) > 0]Let me denote ( R(x) = P(x) - Q(x) ). So, I need to find the intervals where ( R(x) > 0 ).Now, ( R(x) ) is a polynomial because both ( P(x) ) and ( Q(x) ) are polynomials. Let me think about the degree of ( R(x) ). The degree of ( Q(x) ) is ( n ) because it's the product of ( n ) linear terms. The degree of ( P(x) ) is also ( n - 1 ) because each term in the sum for ( P(x) ) is a product of ( n - 1 ) linear terms. Therefore, ( R(x) = P(x) - Q(x) ) is a polynomial of degree ( n ).Wait, hold on. If ( Q(x) ) is degree ( n ) and ( P(x) ) is degree ( n - 1 ), then ( R(x) = P(x) - Q(x) ) is degree ( n ). So, ( R(x) ) is a degree ( n ) polynomial.Since ( R(x) ) is a polynomial of degree ( n ), it can have at most ( n ) real roots. These roots will divide the real line into intervals where ( R(x) ) is positive or negative. But I need to find where ( R(x) > 0 ). So, the intervals where ( R(x) > 0 ) will correspond to the intervals where ( f(x) > 1 ).Given that ( R(x) ) is a degree ( n ) polynomial, it can have up to ( n ) real roots. Let me denote these roots as ( r_1, r_2, dots, r_n ). Since the leading term of ( R(x) ) is ( -x^n ) (because ( Q(x) ) has a leading term ( x^n ) and ( P(x) ) has a leading term of degree ( n - 1 )), the polynomial ( R(x) ) will tend to ( -infty ) as ( x to infty ) and ( +infty ) as ( x to -infty ).Therefore, the roots ( r_1, r_2, dots, r_n ) will alternate between positive and negative intervals. But since all ( a_i ) are positive, the points where ( f(x) ) is undefined are all negative. So, the roots ( r_i ) will lie in the intervals between these undefined points.Wait, actually, let me think again. The function ( f(x) ) is undefined at ( x = -a_i ), which are all negative. So, the intervals where ( f(x) ) is defined are:1. ( (-infty, -a_n) )2. ( (-a_n, -a_{n-1}) )3. ( dots )4. ( (-a_2, -a_1) )5. ( (-a_1, infty) )So, the roots ( r_i ) of ( R(x) ) must lie in these intervals because ( R(x) ) is only relevant where ( f(x) ) is defined.Given that ( R(x) ) is a degree ( n ) polynomial, it must cross the x-axis ( n ) times. So, each interval will contain exactly one root because the function alternates between positive and negative in each interval.Wait, is that necessarily true? Let me consider the behavior of ( R(x) ) in each interval.In the interval ( (-infty, -a_n) ), as ( x to -infty ), ( R(x) ) tends to ( +infty ) because the leading term is ( -x^n ), but ( x ) is negative, so ( (-x)^n ) is positive if ( n ) is even and negative if ( n ) is odd. Hmm, actually, the leading term is ( -x^n ), so as ( x to -infty ), ( -x^n ) tends to ( +infty ) if ( n ) is odd and ( -infty ) if ( n ) is even.Wait, this might complicate things. Maybe instead of focusing on the leading term, I should consider the behavior of ( f(x) ) in each interval.In the interval ( (-infty, -a_n) ), all denominators ( x + a_i ) are negative because ( x < -a_n < -a_{n-1} < dots < -a_1 ). So, each term ( frac{a_i}{x + a_i} ) is negative. Therefore, ( f(x) ) is negative in this interval, so ( f(x) > 1 ) is impossible here.In the interval ( (-a_n, -a_{n-1}) ), ( x + a_n ) is positive, but ( x + a_{n-1} ), ( x + a_{n-2} ), etc., are still negative. So, the first term ( frac{a_n}{x + a_n} ) is positive, and the rest are negative. So, ( f(x) ) is the sum of one positive term and several negative terms. Depending on the values, ( f(x) ) could be greater than 1 here.Similarly, in each interval ( (-a_k, -a_{k-1}) ), ( k ) terms are positive and ( n - k ) terms are negative. So, as ( k ) increases, more terms become positive.Wait, actually, in each interval ( (-a_{k}, -a_{k-1}) ), the terms ( frac{a_1}{x + a_1}, frac{a_2}{x + a_2}, dots, frac{a_{k-1}}{x + a_{k-1}} ) are negative because ( x + a_i < 0 ) for ( i < k ), and the terms ( frac{a_k}{x + a_k}, dots, frac{a_n}{x + a_n} ) are positive because ( x + a_i > 0 ) for ( i geq k ).So, in each interval, the number of positive terms increases as ( x ) increases. Therefore, the function ( f(x) ) is a sum of some positive and some negative terms in each interval.Given that, it's possible that ( f(x) ) crosses 1 in each interval. So, perhaps in each interval ( (-a_{k}, -a_{k-1}) ), there is exactly one solution to ( f(x) = 1 ), which would divide the interval into two parts: one where ( f(x) > 1 ) and one where ( f(x) < 1 ).But wait, how many intervals are there? There are ( n + 1 ) intervals, but in the first interval ( (-infty, -a_n) ), as I noted earlier, ( f(x) ) is negative, so ( f(x) > 1 ) is impossible. Similarly, in the last interval ( (-a_1, infty) ), all terms are positive because ( x + a_i > 0 ) for all ( i ). So, ( f(x) ) is positive here, but does it ever exceed 1?Let me evaluate ( f(x) ) as ( x ) approaches infinity. Each term ( frac{a_i}{x + a_i} ) approaches 0, so ( f(x) ) approaches 0. Therefore, as ( x to infty ), ( f(x) to 0 ). So, in the interval ( (-a_1, infty) ), ( f(x) ) starts at some positive value when ( x ) is just greater than ( -a_1 ) and decreases to 0 as ( x to infty ). Therefore, depending on the behavior near ( x = -a_1 ), it might cross 1 once.Wait, but when ( x ) approaches ( -a_1 ) from the right, ( frac{a_1}{x + a_1} ) tends to ( +infty ), while the other terms are finite. So, near ( x = -a_1 ), ( f(x) ) is very large, tending to infinity. Therefore, in the interval ( (-a_1, infty) ), ( f(x) ) starts at ( +infty ) and decreases to 0. So, it must cross 1 exactly once in this interval.Similarly, in each interval ( (-a_k, -a_{k-1}) ), ( f(x) ) goes from ( +infty ) (as ( x ) approaches ( -a_k ) from the right) to some finite value as ( x ) approaches ( -a_{k-1} ) from the left. Depending on the behavior, it might cross 1 once in each interval.Wait, but actually, in the interval ( (-a_k, -a_{k-1}) ), ( f(x) ) is the sum of ( k ) positive terms and ( n - k ) negative terms. So, as ( x ) increases from ( -a_k ) to ( -a_{k-1} ), the positive terms decrease and the negative terms increase (since their denominators become less negative, making the fractions less negative). So, the overall function ( f(x) ) is decreasing in this interval because each positive term is decreasing and each negative term is increasing (which is like decreasing in magnitude). Therefore, ( f(x) ) is decreasing in each interval ( (-a_k, -a_{k-1}) ).Given that ( f(x) ) is decreasing in each interval, and in the interval ( (-a_k, -a_{k-1}) ), as ( x ) approaches ( -a_k ) from the right, ( f(x) ) tends to ( +infty ), and as ( x ) approaches ( -a_{k-1} ) from the left, ( f(x) ) tends to a finite value. Therefore, if ( f(x) ) is decreasing from ( +infty ) to some finite value, it must cross 1 exactly once in each interval.Therefore, in each of the ( n ) intervals ( (-a_n, -a_{n-1}), (-a_{n-1}, -a_{n-2}), dots, (-a_2, -a_1), (-a_1, infty) ), there is exactly one solution to ( f(x) = 1 ). So, these solutions divide each interval into two parts: one where ( f(x) > 1 ) and one where ( f(x) < 1 ).Since ( f(x) ) is decreasing in each interval, the part where ( f(x) > 1 ) is the left part of the interval, from ( -a_k ) up to the root ( r_k ) where ( f(r_k) = 1 ). Therefore, in each interval ( (-a_k, -a_{k-1}) ), the solution ( r_k ) is such that ( f(x) > 1 ) for ( x in (-a_k, r_k) ).Similarly, in the last interval ( (-a_1, infty) ), the solution ( r_1 ) is such that ( f(x) > 1 ) for ( x in (-a_1, r_1) ).Therefore, the intervals where ( f(x) > 1 ) are:1. ( (-a_n, r_n) )2. ( (-a_{n-1}, r_{n-1}) )3. ( dots )4. ( (-a_2, r_2) )5. ( (-a_1, r_1) )Each of these intervals is disjoint because they lie in separate regions between the poles ( -a_i ). So, the total length of these intervals is the sum of the lengths of each interval ( (-a_k, r_k) ), which is ( r_k - (-a_k) = r_k + a_k ).Therefore, the total length is:[sum_{k=1}^n (r_k + a_k)]So, I need to find ( sum_{k=1}^n r_k + sum_{k=1}^n a_k ).But wait, ( sum_{k=1}^n a_k ) is just the sum of all the ( a_i )s, which is a known value. The question is, what is ( sum_{k=1}^n r_k )?To find ( sum_{k=1}^n r_k ), I can use the fact that ( r_k ) are the roots of the polynomial ( R(x) = P(x) - Q(x) ). Recall that for a polynomial ( R(x) = a_n x^n + a_{n-1} x^{n-1} + dots + a_0 ), the sum of the roots is ( -frac{a_{n-1}}{a_n} ).So, if I can find the coefficients of ( x^n ) and ( x^{n-1} ) in ( R(x) ), I can compute the sum of the roots.Let me compute the leading coefficients of ( P(x) ) and ( Q(x) ).First, ( Q(x) = prod_{i=1}^n (x + a_i) ). The leading term (the coefficient of ( x^n )) is 1, and the coefficient of ( x^{n-1} ) is ( sum_{i=1}^n a_i ).Next, ( P(x) = sum_{i=1}^n a_i prod_{j neq i} (x + a_j) ). Each term in the sum is a product of ( n - 1 ) linear terms, so each term is a polynomial of degree ( n - 1 ). Therefore, ( P(x) ) is a polynomial of degree ( n - 1 ).The leading term of each ( prod_{j neq i} (x + a_j) ) is ( x^{n - 1} ), so when multiplied by ( a_i ), the leading term becomes ( a_i x^{n - 1} ). Therefore, the leading term of ( P(x) ) is ( left( sum_{i=1}^n a_i right) x^{n - 1} ).Therefore, the coefficient of ( x^{n - 1} ) in ( P(x) ) is ( sum_{i=1}^n a_i ).Now, ( R(x) = P(x) - Q(x) ). So, the coefficient of ( x^n ) in ( R(x) ) is ( -1 ) (since ( Q(x) ) has a leading term of ( x^n ) and ( P(x) ) has no ( x^n ) term). The coefficient of ( x^{n - 1} ) in ( R(x) ) is ( sum_{i=1}^n a_i - sum_{i=1}^n a_i = 0 ).Therefore, the polynomial ( R(x) ) is:[R(x) = -x^n + 0 x^{n - 1} + dots + text{lower degree terms}]So, the sum of the roots ( sum_{k=1}^n r_k ) is equal to ( -frac{text{coefficient of } x^{n - 1}}{text{coefficient of } x^n} = -frac{0}{-1} = 0 ).Therefore, ( sum_{k=1}^n r_k = 0 ).Hence, the total length of the intervals where ( f(x) > 1 ) is:[sum_{k=1}^n (r_k + a_k) = sum_{k=1}^n r_k + sum_{k=1}^n a_k = 0 + sum_{k=1}^n a_k = sum_{k=1}^n a_k]So, the sum of the lengths of the disjoint intervals where ( f(x) > 1 ) is equal to the sum of all the ( a_i )s.**Final Answer**The sum of the lengths of the disjoint intervals is (boxed{sum_{i=1}^n a_i}).