🧠:[<think>Alright, let's try to figure out this problem step by step. So, Rachel is on a bus, and at each bus stop, one passenger gets off and x passengers get on, where x is at least 2. After five stops, the number of passengers on the bus is x times the number of passengers that were there before the first stop. We need to find how many passengers were on the bus before the first stop.Hmm, okay. Let me break this down. Let's denote the number of passengers before the first stop as N. That's our starting point. Now, at each stop, one person gets off and x get on. So, after each stop, the number of passengers changes by (x - 1) because you subtract one and add x. Wait, no, actually, let's think carefully. If you start with N passengers, after the first stop, the number becomes N - 1 + x. Then after the second stop, it would be (N - 1 + x) - 1 + x, and so on. So each stop adds (x - 1) passengers. But maybe I need to model each step.Alternatively, maybe I can write a recurrence relation. Let's say after k stops, the number of passengers is P_k. Then, P_{k+1} = P_k - 1 + x. So, P_{k+1} = P_k + (x - 1). But wait, this would be a linear recurrence. Solving that, we can find P_5 in terms of N and then set it equal to x*N, since after five stops, the number is x times the original.Wait, hold on. Let's confirm. The recurrence is P_{k+1} = P_k - 1 + x = P_k + (x - 1). So each term is the previous term plus (x - 1). Therefore, this is an arithmetic sequence where each term increases by (x - 1). Therefore, after each stop, the number of passengers increases by (x - 1). Therefore, after k stops, the number of passengers would be P_k = N + k*(x - 1). But wait, that seems too simple. Let me check with k=1: after first stop, N + 1*(x -1) = N + x -1. Which matches the earlier expression: N -1 + x. So yes, correct. So after 5 stops, P_5 = N + 5*(x -1). But according to the problem, P_5 = x*N. So, setting them equal:N + 5*(x -1) = x*NLet's solve for N. Expanding the left side:N + 5x -5 = x*NBring all terms to one side:N + 5x -5 - x*N = 0Factor N:N*(1 - x) + 5x -5 = 0Then,N*(1 - x) = 5 -5xMultiply both sides by -1:N*(x -1) = 5x -5Factor the right side:N*(x -1) = 5(x -1)Assuming that x ≠ 1, which is given because x ≥ 2, so x -1 is not zero, so we can divide both sides by (x -1):N = 5Wait, so the number of passengers before the first stop was 5? Let me check this.Let's test this. If N = 5 and x is some number ≥ 2. Let's pick x =2, for example.After each stop, number of passengers: 5 -1 +2 =6 after first stop.Second stop: 6 -1 +2=7Third:7-1+2=8Fourth:8-1+2=9Fifth:9-1+2=10Original number was 5, x=2, so 5*2=10. That's correct. If x=3:After each stop, passengers:5-1+3=77-1+3=99-1+3=1111-1+3=1313-1+3=15Original number 5, x=3, 5*3=15. Correct again. If x=4:5-1+4=88-1+4=1111-1+4=1414-1+4=1717-1+4=20Original number 5*4=20. Correct. So seems like N=5 works for any x ≥2. Therefore, the answer is 5.Wait, but let me check with N=5 and x=5.After each stop: 5-1+5=9, 9-1+5=13, 13-1+5=17, 17-1+5=21, 21-1+5=25. Which is 5*5=25. Correct.So regardless of x, N=5. So that must be the answer. Therefore, the number of passengers before the first stop was 5.But let me just think again. The problem says "x ≥2". So in the equation N*(x -1)=5(x -1), then we can divide both sides by (x -1) to get N=5. Since x≠1, which is given, so this is valid.Therefore, N=5. So the answer is 5.Final AnswerThe number of passengers on the bus before the first stop was boxed{5}.</think>Let ( N ) be the number of passengers on the bus before the first stop. At each stop, one passenger gets off and ( x ) passengers get on, where ( x geq 2 ). After each stop, the number of passengers increases by ( x - 1 ). After ( k ) stops, the number of passengers can be modeled by the arithmetic sequence:[ P_k = N + k(x - 1) ]After 5 stops, the number of passengers is given by:[ P_5 = N + 5(x - 1) ]According to the problem, this number is ( x ) times the original number of passengers:[ N + 5(x - 1) = xN ]Solving for ( N ):[begin{align*}N + 5(x - 1) &= xN N + 5x - 5 &= xN N - xN &= 5 - 5x N(1 - x) &= 5(1 - x) N &= 5end{align*}]Thus, the number of passengers on the bus before the first stop was (boxed{5}).,<think>Okay, so I need to solve this problem about Rachel and the bus passengers. Let me read it again to make sure I understand."Rachel travelled to school and noticed that at each bus stop, one passenger got off and x passengers got on, where x is at least 2. After five stops, the number of passengers on the bus was x times the number of passengers before the first stop. How many passengers were on the bus before the first stop?"Hmm. Alright, let's break this down.First, before the first stop, there are some number of passengers on the bus. Let's call that number P. That's what we need to find.Then, at each of the five bus stops, one passenger gets off and x passengers get on. So each stop, the number of passengers changes by (x - 1), because you subtract 1 and add x. But wait, maybe I should model it step by step to be precise.Let me try writing out the number of passengers after each stop.Before the first stop: P passengers.After the first stop:- 1 passenger gets off: P - 1- x passengers get on: (P - 1) + x = P + (x - 1)After the second stop:- 1 passenger gets off: (P + x - 1) - 1 = P + x - 2- x passengers get on: (P + x - 2) + x = P + 2x - 2Wait a second, maybe there's a pattern here. Let me see:After first stop: P + x - 1After second stop: (P + x - 1) - 1 + x = P + x - 1 -1 + x = P + 2x - 2Third stop:-1: P + 2x - 2 -1 = P + 2x - 3+x: P + 2x - 3 + x = P + 3x - 3Fourth stop:-1: P + 3x - 3 -1 = P + 3x - 4+x: P + 3x - 4 + x = P + 4x - 4Fifth stop:-1: P + 4x - 4 -1 = P + 4x -5+x: P + 4x -5 + x = P + 5x -5So after five stops, the number of passengers is P + 5x -5.According to the problem, this number is x times the number of passengers before the first stop. So:P + 5x -5 = x * POkay, so we can set up that equation:P + 5x - 5 = xPLet me write that down:xP = P + 5x -5Let me rearrange terms:xP - P = 5x -5Factor out P on the left:P(x -1) = 5(x -1)Hmm, interesting. So, if I divide both sides by (x -1), assuming x ≠1, which is given because x ≥2. So:P = [5(x -1)] / (x -1) = 5Wait, that simplifies to P =5. So the number of passengers before the first stop is 5?But let me check this again to make sure.Wait, if P =5, then after each stop:After first stop: 5 -1 +x =4 +xAfter second stop: (4 +x) -1 +x =3 +2xThird: (3 +2x) -1 +x =2 +3xFourth: (2 +3x) -1 +x =1 +4xFifth: (1 +4x) -1 +x =0 +5x =5xAnd 5x is indeed x times the original number of passengers (5). So 5x =x*5. Yep, that checks out.So regardless of the value of x, as long as x ≥2, the initial number of passengers is 5. Wait, that seems too straightforward. Let me see if there's a different way this could be interpreted.Wait, does the problem say that after five stops, the number is x times the original? So, starting with P, after five stops, it's x*P, which is 5x, and we see that the equation gives P=5.Yes, seems correct. Maybe I was overcomplicating it, but this seems to hold. Let me test with x=2. If x=2, then each stop, 1 gets off, 2 get on, so net +1 each stop. After five stops, the total increase would be 5*1=5. So starting with P=5, after five stops, 5 +5=10. Which is 2*5=10. Correct.If x=3, each stop, net +2. Five stops, net +10. So starting with 5, ending with 15, which is 3*5=15. Correct.So this seems to work. Therefore, the answer is 5. So the number of passengers before the first stop is 5. Therefore, boxed{5}Final AnswerThe number of passengers on the bus before the first stop was boxed{5}.</think>Let ( P ) be the number of passengers on the bus before the first stop. At each of the five stops, one passenger gets off and ( x ) passengers get on. We need to find ( P ) such that after five stops, the number of passengers is ( x ) times the original number ( P ).We can model the number of passengers after each stop:- After the first stop: ( P - 1 + x = P + x - 1 )- After the second stop: ( (P + x - 1) - 1 + x = P + 2x - 2 )- After the third stop: ( (P + 2x - 2) - 1 + x = P + 3x - 3 )- After the fourth stop: ( (P + 3x - 3) - 1 + x = P + 4x - 4 )- After the fifth stop: ( (P + 4x - 4) - 1 + x = P + 5x - 5 )According to the problem, after five stops, the number of passengers is ( x ) times the original number ( P ):[P + 5x - 5 = xP]Rearranging and solving for ( P ):[xP - P = 5x - 5][P(x - 1) = 5(x - 1)]Since ( x geq 2 ), we can divide both sides by ( x - 1 ):[P = 5]Thus, the number of passengers on the bus before the first stop was (boxed{5}).]

🔑:SolutionLet the number of passengers who were on the bus before the first stop be n. Then, after five stops, the number of passengers on the bus was n-5+5 x. So n-5+5 x=n x, that is, n x-n=5 x-5. Therefore n(x-1)=5(x-1). Hence n=frac{5(x-1)}{x-1}=5, as x neq 1.So there were five passengers on the bus before the first stop.

🧠:[<think>Okay, so I need to calculate the sum from n = 0 to 502 of the floor of (305n)/503. Hmm, floor functions can be tricky, but maybe there's a pattern or formula I can use here. Let me recall some math concepts. I remember that sums involving floor functions sometimes relate to arithmetic sequences or can be approached using some number theory. First, let me write down the sum explicitly. The sum S is:S = Σₙ=0^502 floor(305n / 503)So for each n from 0 to 502, I compute 305n divided by 503, take the floor of that value, and sum all those floors up. Since 305 and 503 are constants here, maybe I can find a relationship between them. Let me check if they are coprime. The problem is, 305 factors into 5 * 61, and 503 is a prime number, right? Let me confirm: 503 divided by 5 is 100.6, so not divisible by 5. Divided by 61? 61 * 8 = 488, 503 - 488 = 15, which is not divisible by 61. So 503 is prime, and 305 and 503 share no common factors. Therefore, they are coprime. That might be useful.Since 305 and 503 are coprime, the fraction 305/503 is in its simplest form. I remember there's a formula for sums of floor functions when the coefficients are coprime. Let me recall. Is it something like (a-1)(b-1)/2? Wait, that's for the sum Σₙ=1^{m-1} floor(an/m) where a and m are coprime. Let me check that.Yes, I think the formula is (a - 1)(m - 1)/2 when a and m are coprime. But in this case, our sum is from n = 0 to m-1, where m is 503. Wait, 503 is the denominator here, and n ranges from 0 to 502. So if m = 503, then the sum would be Σₙ=0^{m-1} floor(a n / m) where a = 305. Since a and m are coprime, then the formula should apply. Wait, but when n = 0, floor(0) = 0, so the sum from n=0 to m-1 is the same as from n=1 to m-1, except for the n=0 term which is 0. So maybe the formula (a - 1)(m - 1)/2 applies here. Let me verify this with small numbers. Suppose a = 3 and m = 5, which are coprime. Then sum from n=0 to 4 of floor(3n/5). Let's compute:n=0: 0n=1: floor(3/5) = 0n=2: floor(6/5) = 1n=3: floor(9/5) = 1n=4: floor(12/5) = 2Sum: 0 + 0 + 1 + 1 + 2 = 4Formula gives (3-1)(5-1)/2 = 2*4/2 = 4. Correct. So that seems to work. So in general, for coprime a and m, the sum Σₙ=0^{m-1} floor(a n / m) = (a - 1)(m - 1)/2. Hence, in our problem, a = 305, m = 503, so the sum S should be (305 - 1)(503 - 1)/2 = (304)(502)/2.Wait, but let me check another example to be sure. Take a=2, m=3. Then sum from n=0 to 2 of floor(2n/3):n=0: 0n=1: floor(2/3) = 0n=2: floor(4/3) = 1Sum: 0 + 0 + 1 = 1Formula: (2 -1)(3 -1)/2 = 1*2/2 = 1. Correct. Another one: a=1, m=4. Sum floor(n/4) from n=0 to 3: all terms 0. Formula: (1-1)(4-1)/2 = 0. Correct. Okay, seems like the formula holds. So in our case, S = (305 - 1)(503 - 1)/2 = (304)(502)/2.Let me compute that. 304 divided by 2 is 152, so 152 * 502. Let's compute 150*502 + 2*502. 150*502: 150*500 = 75,000 and 150*2 = 300, so total 75,300. 2*502 = 1,004. So 75,300 + 1,004 = 76,304. Therefore, S = 76,304.Wait, but let me confirm if this is correct. Wait, let me check the formula again. The formula is for Σ floor(a n / m) where n runs from 0 to m-1 (inclusive), and a and m coprime. In our case, m = 503, a = 305. So yes, the formula applies. Therefore, the sum should be (305 -1)(503 -1)/2 = (304)(502)/2 = 304*251 = let's compute that again. 300*251 = 75,300. 4*251 = 1,004. So 75,300 + 1,004 = 76,304. So that's the answer? Hmm. But let me verify once more with another approach.Alternatively, there's another formula for such sums. Let me recall that Σₙ=1^{m-1} floor(a n / m) = (a - 1)(m - 1)/2. Since in our case, the sum from n=0 to 502 is the same as from n=1 to 502 because the n=0 term is 0. So Σₙ=1^{502} floor(305n / 503) = (305 -1)(503 -1)/2 = 304*502/2 = 76,304. So that seems correct.But let me also think about another way. The floor function floor(k) is the integer part of k. So 305n/503 is a fraction less than 305n/503. For each n, floor(305n/503) = (305n - (305n mod 503)) / 503. Wait, no. Wait, floor(x) = x - {x}, where {x} is the fractional part. So floor(305n/503) = 305n/503 - {305n/503}. Therefore, the sum S is Σₙ=0^{502} (305n/503 - {305n/503}) = (305/503) Σₙ=0^{502} n - Σₙ=0^{502} {305n/503}.Compute the first sum: Σₙ=0^{502} n = (502)(503)/2. Then multiply by 305/503:(305/503) * (502*503)/2 = 305 * 502 / 2.That's the same as before, which is 305*251 = let's compute 300*251 = 75,300 and 5*251 = 1,255, so total 75,300 + 1,255 = 76,555. Wait, hold on. Wait, 305*251 is 305*(250 + 1) = 305*250 + 305*1 = 76,250 + 305 = 76,555. So that part is 76,555. Then subtract the sum of fractional parts Σₙ=0^{502} {305n/503}. Therefore, S = 76,555 - Σₙ=0^{502} {305n/503}.But {305n/503} is equal to (305n mod 503)/503. So the sum of fractional parts is (1/503) Σₙ=0^{502} (305n mod 503). Therefore, S = 76,555 - (1/503) Σₙ=0^{502} (305n mod 503).But since 305 and 503 are coprime, the residues 305n mod 503 for n from 0 to 502 will be a permutation of the numbers 0, 1, 2, ..., 502. Wait, when n=0, it's 0. Then for n=1 to 502, since 305 and 503 are coprime, 305n mod 503 cycles through all residues from 1 to 502 exactly once. Therefore, the sum Σₙ=0^{502} (305n mod 503) = 0 + Σₙ=1^{502} n = 0 + (502)(503)/2 = (502*503)/2.Therefore, Σₙ=0^{502} {305n/503} = (1/503) * (502*503)/2 = 502/2 = 251.Therefore, S = 76,555 - 251 = 76,304. Which matches the previous result. So this confirms the answer is 76,304. Therefore, the sum is 76,304. So I think that's the correct answer.But let me just verify once again with a small example. Let's take a=3, m=5 as before. Then Σₙ=0^4 floor(3n/5). We know the answer is 4. Using the formula: (3-1)(5-1)/2 = 2*4/2=4. Correct. Alternatively, compute via the other method: sum = (3/5)*Σn=0^4 n - Σ fractional parts. Σn=0^4 n = 10. 3/5*10 = 6. Σ fractional parts = (1/5)*(sum of residues). The residues for n=0 to 4 are 0,3,1,4,2. Sum = 0+3+1+4+2=10. Therefore, Σ fractional parts = 10/5=2. Then 6 - 2=4. Correct. So the method works. Therefore, in our problem, the answer is 76,304.Alternatively, maybe another way to think about it. Since 305n mod 503 cycles through all residues, except 0 when n=0. So for n=1 to 502, the residues are 1 to 502 in some order, so their sum is (502)(503)/2. Then, the sum of {305n/503} is (sum of residues)/503 = [ (502)(503)/2 ] / 503 = 502/2 = 251. Therefore, the total sum S = (305/503)(sum n) - 251. Sum n from 0 to 502 is (502)(503)/2. Therefore, 305/503 * (502*503)/2 = 305*502/2 = 305*251. 305*251: 300*251 = 75,300; 5*251 = 1,255; total 76,555. Then subtract 251 gives 76,555 - 251 = 76,304. Same result. Therefore, I think the answer is definitely 76,304. Let me check once more with another angle. The original formula: when a and m are coprime, the sum floor(an/m) from n=0 to m-1 is (a-1)(m-1)/2. Here a=305, m=503. So (305-1)(503-1)/2 = 304*502/2 = 304*251. 304*251: 300*251 = 75,300; 4*251 = 1,004; total 75,300 + 1,004 = 76,304. Yep. So all methods converge to the same answer. Therefore, confident that the answer is 76,304.Final Answerboxed{76304}</think>To calculate the sum (sum_{n=0}^{502}leftlfloor frac{305n}{503} rightrfloor), we can use the formula for the sum of floor functions when the coefficients are coprime. Since 305 and 503 are coprime, the formula (sum_{n=0}^{m-1} leftlfloor frac{an}{m} rightrfloor = frac{(a-1)(m-1)}{2}) applies, where (a = 305) and (m = 503).1. Verify coprimality: 305 and 503 are coprime since 305 factors into (5 times 61) and 503 is a prime number.2. Apply the formula: The sum is given by (frac{(305 - 1)(503 - 1)}{2}).3. Calculate the result: [ frac{(304)(502)}{2} = 304 times 251 ]4. Compute the product: [ 304 times 251 = 304 times (250 + 1) = 304 times 250 + 304 times 1 = 76000 + 304 = 76304 ]Thus, the final answer is (boxed{76304}).,<think>Okay, so I need to calculate the sum from n equals 0 to 502 of the floor of (305n divided by 503). Let me write that down more clearly: ∑ₙ=0^502 ⌊305n/503⌋. Hmm, floor functions can be tricky. I remember that there's a formula for sums involving floor functions of linear terms, maybe something related to reciprocals or modular arithmetic? Let me think.First, let me recall that when you have a sum like ∑ₙ=0^{m-1} ⌊(kn)/m⌋ where k and m are coprime integers, there's a formula for that. I think it's related to the number of times the floor function increments. Wait, if k and m are coprime, then the fractional parts {(kn)/m} for n from 0 to m-1 are all distinct and cover the fractions 0/m, 1/m, ..., (m-1)/m in some order. Is that right?Yes, that's a property from number theory. If k and m are coprime, then the numbers kn mod m for n=0,1,...,m-1 are a complete residue system modulo m. So the fractional parts {(kn)/m} are {0/m, 1/m, ..., (m-1)/m}, just in a different order. Therefore, the sum of the floor functions would be the same as summing over all these fractional parts, but since the floor is the integer part, maybe we can express the sum in terms of the sum of (kn)/m minus the sum of the fractional parts.Wait, for each term, ⌊(kn)/m⌋ = (kn)/m - {(kn)/m}. So summing over n, we get ∑ₙ=0^{m-1} ⌊(kn)/m⌋ = (k/m)∑ₙ=0^{m-1} n - ∑ₙ=0^{m-1} {(kn)/m}.Since the fractional parts {(kn)/m} are just a permutation of {0/m, 1/m, ..., (m-1)/m} when k and m are coprime, their sum is ∑_{r=0}^{m-1} r/m = (m-1)/2. Therefore, the sum becomes (k/m)(m(m-1)/2) - (m-1)/2 = (k(m-1)/2) - (m-1)/2 = (k - 1)(m - 1)/2.Wait, so if k and m are coprime, then ∑ₙ=0^{m-1} ⌊(kn)/m⌋ = (k - 1)(m - 1)/2. That seems too straightforward. Let me check with a small example. Let's take m=3 and k=2, which are coprime. Then the sum is ⌊0⌋ + ⌊2/3⌋ + ⌊4/3⌋ = 0 + 0 + 1 = 1. The formula gives (2-1)(3-1)/2 = 1*2/2 = 1. That works. Another example, m=5, k=3. Then the sum is ⌊0⌋ + ⌊3/5⌋ + ⌊6/5⌋ + ⌊9/5⌋ + ⌊12/5⌋ = 0 + 0 + 1 + 1 + 2 = 4. The formula: (3-1)(5-1)/2 = 2*4/2 = 4. Perfect. So the formula holds. But in our problem, m is 503 and k is 305. First, check if 305 and 503 are coprime. Let's compute GCD(305,503). 305 factors into 5*61. 503 divided by 5 is 100.6, not integer. 503 divided by 61: 61*8=488, 503-488=15, which is not a multiple of 61. So 503 is a prime? Wait, 503 is actually a prime number. Let me confirm: 503 is not divisible by 2,3,5,7,11,13,17,19,23. sqrt(503) is about 22.4, so primes up to 23. Since none divide 503, it's prime. Therefore, 503 is prime, and 305 is 5*61, which are both primes as well, and neither 5 nor 61 divide 503, so GCD(305,503)=1. So they are coprime. Therefore, we can apply the formula.But wait, in the problem, the sum is from n=0 to 502, which is exactly m-1 where m=503. So the sum would be (305 - 1)(503 - 1)/2 = 304*502/2. Let me compute that. 304 divided by 2 is 152, so 152*502. Let's compute 150*502 + 2*502 = 75,300 + 1,004 = 76,304. Therefore, the sum should be 76,304. But wait, before I get too confident, let me check with another example. Wait, in the problem, the upper limit is 502, which is m-1=503-1=502, so yes, the formula applies. So according to that, the sum is (k -1)(m -1)/2 = (305 -1)(503 -1)/2 = 304*502/2 = 304*251. Wait, 502 divided by 2 is 251. So 304*251. Let's compute 300*251 + 4*251 = 75,300 + 1,004 = 76,304. So same result. So that should be the answer. But let me verify this approach once more.Alternatively, perhaps I can use another method to confirm. Let's consider that the sum ∑ₙ=0^{m-1} ⌊(kn)/m⌋ where k and m coprime is equal to (k -1)(m -1)/2. So the formula gives the answer 76,304.But wait, let me think again. When k and m are coprime, the number of times the floor function increments is (k-1)(m-1)/2. But perhaps there's another way to compute this. For example, each term ⌊305n/503⌋ can be thought of as the integer division of 305n by 503, which gives how many times 503 fits into 305n. Alternatively, since 305 and 503 are coprime, for each n, 305n mod 503 cycles through all residues. However, maybe another approach is needed here. Wait, actually, the formula we used earlier seems to hold, but let's verify with m=5, k=3 again. The sum is 0 + 0 + 1 + 1 + 2 = 4. The formula gives (3-1)(5-1)/2 = 2*4/2=4. Correct. So the formula works. So in our case, it should apply. Thus, the sum is (305 -1)(503 -1)/2 = 304*502/2 = 304*251 = 76,304. Therefore, the answer should be 76,304. But wait, let's check for another coprime pair. Let's take m=7, k=3. Then the sum is ∑ₙ=0^6 ⌊3n/7⌋. Compute each term:n=0: 0n=1: 0n=2: 0n=3: 1n=4: 1n=5: 2n=6: 2Sum: 0+0+0+1+1+2+2=6. The formula: (3-1)(7-1)/2=2*6/2=6. Correct.Another example: m=4, k=3 (coprime? 3 and 4 are coprime). Sum from n=0 to 3: ⌊0⌋ + ⌊3/4⌋ + ⌊6/4⌋ + ⌊9/4⌋ = 0 + 0 + 1 + 2 = 3. Formula: (3-1)(4-1)/2 = 2*3/2=3. Correct. Hmm, seems solid.Therefore, I think the formula is reliable. So applying it here, the answer should be 76,304.Wait, but wait a second. Let me check with m=503 and k=305. The sum is from n=0 to 502. Let me take a specific n, say n=1: floor(305*1 /503) = floor(305/503)=0. For n=503, but wait n only goes up to 502. So the maximum term is when n=502: floor(305*502 /503). Let's compute that. 305*502 = let's compute 300*502 +5*502=150,600 +2,510=153,110. Then divide by 503: 153,110 ÷503. Let me see how many times 503 goes into 153,110. 503*300=150,900. 153,110 -150,900=2,210. 503*4=2,012. 2,210-2,012=198. So total is 300+4=304, with remainder 198. Therefore, floor(305*502/503)=304. So the terms go from 0 up to 304. Therefore, the maximum term is 304, which is (305*502)/503 - some fractional part. So the sum should be the sum of integers from 0 to 304, but adjusted for how many times each integer appears. Wait, but maybe that's a different approach. Alternatively, maybe the formula accounts for the average over the terms.Wait, but in our formula, we get 76,304. Let's see, if each term is on average (305-1)/2 =152 per term? Wait, 503 terms. 503*152 = 503*(150+2)=503*150 +503*2=75,450 +1,006=76,456. But 76,456 is higher than 76,304, so that doesn't align. Hmm, maybe that's not the way. Wait, maybe another way: since the formula gives (k-1)(m-1)/2, that's (304)(502)/2=304*251=76,304. So if I compute 304*251: 300*251=75,300, 4*251=1,004, so 75,300+1,004=76,304. Correct. Alternatively, perhaps the number of lattice points under the line y=(305/503)x from x=0 to x=503. Wait, by the geometry of numbers, the number of lattice points (n, k) with 1 ≤k ≤ ⌊305n/503⌋ for n from 1 to 502. But I think that connects to the same formula. In any case, since the formula checks out with multiple examples, and the logic is that the fractional parts sum to (m-1)/2 because they are a permutation, then the formula is valid. Therefore, I can trust the formula here.Therefore, the answer is 76,304. But just to be thorough, let me see if there's another way to think about this. For example, using the identity that ∑ₙ=1^{m-1} ⌊(kn)/m⌋ = ((k-1)(m-1))/2 when k and m are coprime. But in our problem, the sum starts at n=0. However, when n=0, the term is ⌊0⌋=0, so including it doesn't change the sum. Therefore, ∑ₙ=0^{m-1} ⌊(kn)/m⌋ = ∑ₙ=1^{m-1} ⌊(kn)/m⌋ = ((k-1)(m-1))/2. Which in our case is ((305-1)(503-1))/2 = 304*502/2 =76,304. So that's consistent.Therefore, I think the answer is 76,304. But just to make sure, let me check with another coprime pair where I can compute the sum manually. Let's take m=5, k=2. Then the sum from n=0 to 4: 0, 0, 0, 1, 1. Sum is 2. Formula: (2-1)(5-1)/2=1*4/2=2. Correct. So formula works. Therefore, in our problem, since 305 and 503 are coprime, the formula applies and the answer is 76,304. But let me check one more thing. If the upper limit of the sum were different, say not m-1, then the formula might not apply. But here, since we're summing up to n=502, which is 503-1, the formula is valid. Therefore, I can be confident the answer is 76,304. Final Answer: boxed{76304}Final Answerboxed{76304}</think>To calculate the sum (sum_{n=0}^{502}leftlfloorfrac{305n}{503}rightrfloor), we can use the formula for sums involving floor functions of linear terms when the coefficients are coprime. Given (k = 305) and (m = 503), we first verify that they are coprime. Since 503 is a prime number and neither 5 nor 61 (the prime factors of 305) divide 503, they are indeed coprime. The formula for the sum (sum_{n=0}^{m-1}leftlfloorfrac{kn}{m}rightrfloor) when (k) and (m) are coprime is (frac{(k-1)(m-1)}{2}). Applying this formula:[sum_{n=0}^{502}leftlfloorfrac{305n}{503}rightrfloor = frac{(305 - 1)(503 - 1)}{2} = frac{304 times 502}{2}]Calculating the product:[304 times 251 = 304 times (250 + 1) = 304 times 250 + 304 times 1 = 76000 + 304 = 76304]Thus, the final answer is:[boxed{76304}]]

🔑:1. Since left{frac{305 n}{503}right}+left{frac{305(503-n)}{503}right}=1, thus left[frac{305 n}{503}right]+left[frac{305(503-n)}{503}right]= 304 , therefore sum_{n=0}^{502}left[frac{305 n}{503}right]=76304.

🧠:[<think>Okay, let's see. I need to find the dot product of vectors a and b. The problem gives me vector a as (2,1) and also tells me that vector a minus vector b is (-1,2). Hmm, so first maybe I should figure out what vector b is. Once I have both vectors, I can compute their dot product. Let me try to break this down step by step.Alright, so vector a is given directly: a = (2,1). Then, the problem says a - b = (-1,2). I need to solve for vector b here. Let's write that equation out in components. If a is (2,1) and a - b is (-1,2), then subtracting vector b from a gives (-1,2). To find vector b, I can rearrange the equation. Let's see:a - b = (-1,2)So, subtracting b from a gives that result. To solve for b, I can subtract (a - b) from both sides, but maybe it's easier to just isolate b. Let me try that.Starting with a - b = (-1,2), if I subtract a from both sides, that would give -b = (-1,2) - a. Then, multiplying both sides by -1 would give b = a - (-1,2). Wait, no. Let's think again. If a - b = (-1,2), then adding b to both sides and subtracting (-1,2) from both sides would give b = a - (-1,2). Wait, maybe not. Let me use components to avoid confusion.Vector a is (2,1). So, a - b = (-1,2). Let me denote vector b as (b1, b2). Then, in components:a_x - b_x = -1a_y - b_y = 2Since a_x is 2 and a_y is 1, substituting these in:2 - b_x = -11 - b_y = 2So solving for b_x and b_y:From the first equation: 2 - b_x = -1 => -b_x = -1 - 2 => -b_x = -3 => b_x = 3From the second equation: 1 - b_y = 2 => -b_y = 2 - 1 => -b_y = 1 => b_y = -1So vector b is (3, -1). Let me check that. If a is (2,1) and b is (3, -1), then a - b is (2 - 3, 1 - (-1)) = (-1, 2), which matches the given a - b. Okay, that seems right.Now that I have vector b as (3, -1), I need to compute the dot product of vectors a and b. The dot product is calculated by multiplying corresponding components and then adding them up. So, for vectors a = (a1, a2) and b = (b1, b2), the dot product is a1*b1 + a2*b2.Applying this to our vectors:a · b = (2)(3) + (1)(-1) = 6 - 1 = 5Wait, so the dot product is 5. Let me double-check the calculations to make sure I didn't make a mistake.First, components of a: (2,1). Components of b: (3, -1). Multiply the first components: 2*3 = 6. Multiply the second components: 1*(-1) = -1. Add them together: 6 + (-1) = 5. Yep, that seems right.Looking at the answer choices: A is -5, B is 0, C is 5, D is 7. So the correct answer must be C: 5.Wait, but let me make sure that I found vector b correctly. Let's verify again. Given that a - b = (-1, 2), then b = a - (a - b). Wait, no, that's not right. Wait, if a - b = (-1,2), then b = a - (-1,2). Wait, no, that's not correct. Let me think.Actually, if a - b = (-1,2), then rearranged, b = a - (a - b). Wait, that might be confusing. Let's use component-wise subtraction. If a - b = (-1,2), then:a_x - b_x = -1 => b_x = a_x - (-1) ? No, wait. Let's write it again. If a - b = (-1,2), then:a_x - b_x = -1So, solving for b_x: b_x = a_x - (-1) ? Wait, no. Let's see. If a_x - b_x = -1, then:b_x = a_x - (-1) ?Wait, no. If a_x - b_x = -1, then rearranged:-b_x = -1 - a_xMultiply both sides by -1:b_x = a_x + 1Wait, but a_x is 2. So, b_x = 2 + 1 = 3. Which matches what I did before. Similarly, for the y-component:a_y - b_y = 2So, -b_y = 2 - a_y => b_y = a_y - 2Since a_y is 1, b_y = 1 - 2 = -1. So that's correct. So vector b is (3, -1). Therefore, the dot product is indeed 2*3 + 1*(-1) = 6 -1 = 5.Therefore, the answer is C:5. Let me check again. Maybe I made a mistake in the sign when solving for b.Wait, the problem says a - b = (-1,2). So, if I have a - b = (-1,2), then b = a - (-1,2)? Wait, no. Wait, if a - b = c, then b = a - c. Yes, that's right. So, if c is (-1,2), then b = a - c. Wait, but hold on:Wait, if a - b = c, then b = a - c. Let me verify this algebraically. Starting from a - b = c, subtract a from both sides: -b = c - a, then multiply both sides by -1: b = a - c. Yes, that's correct.So, in this case, c is (-1,2). Therefore, b = a - c = (2,1) - (-1,2) = (2 - (-1), 1 - 2) = (3, -1). Which matches what I found earlier. So that's correct. Then the dot product is 5. So answer C.Therefore, I think the correct answer is C:5. The options are A:-5, B:0, C:5, D:7. So the answer is C.Final Answerboxed{C}</think>Given vectors (overrightarrow{a}) and (overrightarrow{b}) satisfy (overrightarrow{a} = (2, 1)) and (overrightarrow{a} - overrightarrow{b} = (-1, 2)), we need to find (overrightarrow{a} cdot overrightarrow{b}).First, we determine the components of vector (overrightarrow{b}). Starting from the equation (overrightarrow{a} - overrightarrow{b} = (-1, 2)), we can solve for (overrightarrow{b}):[overrightarrow{b} = overrightarrow{a} - (-1, 2)]Breaking this down into components:[overrightarrow{a} = (2, 1)][overrightarrow{a} - overrightarrow{b} = (-1, 2) implies overrightarrow{b} = overrightarrow{a} - (-1, 2)]Calculating the components of (overrightarrow{b}):[b_x = 2 - (-1) = 2 + 1 = 3][b_y = 1 - 2 = -1]Thus, (overrightarrow{b} = (3, -1)).Next, we compute the dot product (overrightarrow{a} cdot overrightarrow{b}):[overrightarrow{a} cdot overrightarrow{b} = (2)(3) + (1)(-1) = 6 - 1 = 5]Therefore, the final answer is (boxed{C}).,<think>Okay, so I need to find the dot product of vectors a and b. The problem gives me vector a as (2,1) and the vector a minus vector b as (-1,2). Hmm, let me think about how to approach this.First, I know that the dot product of two vectors is calculated by multiplying their corresponding components and then adding those products together. So for vectors a = (a₁, a₂) and b = (b₁, b₂), the dot product a · b is a₁b₁ + a₂b₂. That formula seems straightforward. But right now, I have vector a and the difference between a and b. I need to figure out what vector b is first, right? Because once I have both vectors a and b, I can compute their dot product.Alright, so the problem says a - b = (-1, 2). Let me write that out in components. If a is (2,1) and a - b is (-1,2), then each component of a minus the corresponding component of b equals the components of (-1,2). So for the x-component: 2 - b_x = -1, and for the y-component: 1 - b_y = 2. Let me solve these equations to find b_x and b_y.Starting with the x-component: 2 - b_x = -1. To solve for b_x, I can subtract 2 from both sides: -b_x = -1 - 2, which simplifies to -b_x = -3. Then, multiplying both sides by -1 gives b_x = 3. Okay, that seems right.Now the y-component: 1 - b_y = 2. Similarly, subtract 1 from both sides: -b_y = 2 - 1, so -b_y = 1. Multiplying both sides by -1 gives b_y = -1. Wait, so vector b is (3, -1)? Let me check that. If a is (2,1) and b is (3,-1), then a - b should be (2-3, 1 - (-1)) which is (-1, 2). Yep, that's exactly what was given. So that's correct. Vector b is (3, -1).Now that I have both vectors a and b, I can compute their dot product. Vector a is (2,1) and vector b is (3, -1). Applying the dot product formula: a · b = (2)(3) + (1)(-1). Calculating each term: 2*3 is 6, and 1*(-1) is -1. Adding those together: 6 + (-1) = 5. So the dot product is 5.Looking at the answer choices: A is -5, B is 0, C is 5, D is 7. So the answer should be C: 5. Let me just double-check my steps to make sure I didn't make any mistakes.First, solving for b: a - b = (-1,2), so b = a - (-1,2). Wait, hold on. If a - b = (-1,2), then rearranging that equation gives b = a - (-1,2)? Wait, no, that's not right. Let me re-express the equation.If a - b = (-1, 2), then adding b to both sides gives a = b + (-1, 2). Then subtracting (-1,2) from both sides gives a - (-1,2) = b. But a is (2,1), so (2,1) - (-1,2) would be (2 - (-1), 1 - 2) = (3, -1). Wait, that's the same as before. So my previous calculation was correct. So b is indeed (3, -1). Then the dot product is 2*3 + 1*(-1) = 6 -1 = 5. Yeah, that seems right. So the answer is C:5.Hmm, I don't see any mistakes here. So I think that's the correct answer.Final Answerboxed{C}</think>Given vectors (overrightarrow{a}) and (overrightarrow{b}) satisfy (overrightarrow{a} = (2, 1)) and (overrightarrow{a} - overrightarrow{b} = (-1, 2)). We need to find the dot product (overrightarrow{a} cdot overrightarrow{b}).First, we determine the components of vector (overrightarrow{b}). From the equation (overrightarrow{a} - overrightarrow{b} = (-1, 2)), we can solve for (overrightarrow{b}):[overrightarrow{b} = overrightarrow{a} - (-1, 2)]Substituting (overrightarrow{a} = (2, 1)):[overrightarrow{b} = (2, 1) - (-1, 2) = (2 + 1, 1 - 2) = (3, -1)]Now that we have (overrightarrow{b} = (3, -1)), we calculate the dot product (overrightarrow{a} cdot overrightarrow{b}):[overrightarrow{a} cdot overrightarrow{b} = (2)(3) + (1)(-1) = 6 - 1 = 5]Thus, the final answer is (boxed{C}).]

🔑:Given the vectors overrightarrow{a} and overrightarrow{b}, we have the following information:1. overrightarrow{a}=(2,1)2. overrightarrow{a}-overrightarrow{b}=(-1,2)From these, we want to find the dot product overrightarrow{a}cdot overrightarrow{b}.Step 1: Find overrightarrow{b}Given overrightarrow{a}-overrightarrow{b}=(-1,2), we can solve for overrightarrow{b} as follows:[begin{align*}overrightarrow{b} &= overrightarrow{a} - (overrightarrow{a} - overrightarrow{b}) &= (2,1) - (-1,2) &= (2+1, 1-2) &= (3,-1)end{align*}]Step 2: Calculate overrightarrow{a}cdot overrightarrow{b}Now, we calculate the dot product of overrightarrow{a} and overrightarrow{b}:[begin{align*}overrightarrow{a}cdot overrightarrow{b} &= (2,1) cdot (3,-1) &= 2times 3 + 1times (-1) &= 6 - 1 &= 5end{align*}]Therefore, the dot product overrightarrow{a}cdot overrightarrow{b} is 5. Thus, the correct answer is boxed{C}.

🧠:[<think>Okay, so I need to find the analytic function f(z) such that the real part u(x, y) is given by x³ - 3xy² + 2y, and also f(i) = 2. Hmm, right. Let's recall that if a function is analytic, it satisfies the Cauchy-Riemann equations. So, since we know the real part u(x, y), we can find the imaginary part v(x, y) by integrating the partial derivatives related through the Cauchy-Riemann equations. First, let me write down the Cauchy-Riemann equations. For a function f(z) = u(x, y) + iv(x, y) to be analytic, the partial derivatives of u and v must satisfy:∂u/∂x = ∂v/∂y,∂u/∂y = -∂v/∂x.So, starting with the given u(x, y) = x³ - 3xy² + 2y. Let's compute the partial derivatives of u.First, the partial derivative of u with respect to x, which is ∂u/∂x. Let's compute that:∂u/∂x = 3x² - 3y².Then, the partial derivative of u with respect to y, which is ∂u/∂y:∂u/∂y = -6xy + 2.According to the Cauchy-Riemann equations, ∂u/∂x = ∂v/∂y, so we have:∂v/∂y = 3x² - 3y².To find v, we need to integrate this with respect to y. Let's do that:v = ∫ (3x² - 3y²) dy + C(x),where C(x) is the constant of integration, which may depend on x.Integrating term by term:∫ 3x² dy = 3x² y,∫ -3y² dy = -y³.So, putting it together:v = 3x² y - y³ + C(x).Now, we also know from the other Cauchy-Riemann equation that ∂u/∂y = -∂v/∂x. Let's compute ∂v/∂x using the expression for v we just found:∂v/∂x = ∂/∂x [3x² y - y³ + C(x)] = 6x y + C'(x).But according to the Cauchy-Riemann equation, this should be equal to -∂u/∂y. Let's compute -∂u/∂y:-∂u/∂y = -(-6xy + 2) = 6xy - 2.Therefore, we have:6x y + C'(x) = 6x y - 2.Subtracting 6xy from both sides gives:C'(x) = -2.Integrating C'(x) with respect to x:C(x) = -2x + K,where K is a constant (with respect to x and y, so just a complex constant).Therefore, the imaginary part v(x, y) is:v(x, y) = 3x² y - y³ - 2x + K.Now, putting together the real and imaginary parts, the function f(z) is:f(z) = u(x, y) + i v(x, y) = (x³ - 3x y² + 2y) + i (3x² y - y³ - 2x + K).Now, we need to express this in terms of z. Since z = x + iy, we can try to write f(z) as a function of z. Let's see if we can express the terms in terms of z³, z², etc.Looking at u and v, let's note that x³ - 3x y² is the real part of z³, since:z³ = (x + iy)³ = x³ + 3x²(iy) + 3x(iy)² + (iy)³= x³ + 3i x² y + 3x (i² y²) + i³ y³= x³ + 3i x² y - 3x y² - i y³.So, the real part of z³ is x³ - 3x y², which matches the cubic terms in u. Similarly, the imaginary part of z³ is 3x² y - y³, which are the cubic terms in v (excluding the linear terms). So, the function f(z) seems to be related to z³. Let's check:If we take z³, its real part is x³ - 3x y², and the imaginary part is 3x² y - y³. Comparing to our u and v:u(x, y) = Re(z³) + 2y,v(x, y) = Im(z³) - 2x + K.Therefore, f(z) can be written as:f(z) = z³ + i(-2x + 2y + K).Wait, let's check. Wait, u is Re(z³) + 2y, and v is Im(z³) -2x + K. So, f(z) = z³ + 2y i + (-2x + K)i. Hmm, perhaps not. Wait, let's think again.Wait, let's write f(z) = (x³ - 3x y² + 2y) + i(3x² y - y³ - 2x + K). If z³ is (x + iy)³, which is (x³ - 3x y²) + i(3x² y - y³). So, f(z) = z³ + 2y + i(-2x + K). Hmm, but 2y is a real term, and i(-2x + K) is an imaginary term. So, perhaps f(z) can be written as z³ + i(-2z) + i K. Let's check:If we take z³ - 2i z + i K, then expanding:z³ is (x + iy)³ as above, which gives real part x³ - 3x y², and imaginary part 3x² y - y³.Then, -2i z is -2i(x + iy) = -2i x - 2i² y = -2i x + 2y. So, the real part of -2i z is 2y, and the imaginary part is -2x.Therefore, f(z) = z³ - 2i z + i K. Let's verify:Real part: Re(z³) + Re(-2i z) = (x³ - 3x y²) + 2y, which matches u(x, y).Imaginary part: Im(z³) + Im(-2i z) + K i? Wait, no. Wait, let's compute the imaginary part:Im(f(z)) = Im(z³ - 2i z + i K) = Im(z³) + Im(-2i z) + Im(i K).Compute each term:Im(z³) = 3x² y - y³,Im(-2i z) = Im(-2i x - 2i² y) = Im(-2i x + 2 y) = -2x,Im(i K) = Re(K) (since i K is a purely imaginary constant, its imaginary part is the real part of K? Wait, maybe not. Wait, if K is a real constant, then i K is purely imaginary, so its real part is 0 and imaginary part is K. Wait, but K was introduced as a constant of integration, which in the expression for v(x, y) is a real constant, right? Because when we integrated C'(x) = -2, we got C(x) = -2x + K, where K is a real constant.Therefore, in the expression for v(x, y), we have -2x + K. Therefore, the imaginary part of f(z) is 3x² y - y³ - 2x + K. Therefore, when we write f(z) as z³ - 2i z + i K, let's check:Im(z³ - 2i z + i K) = Im(z³) + Im(-2i z) + Im(i K) = (3x² y - y³) + (-2x) + K.Yes, that's correct. Because Im(-2i z) is Im(-2i x + 2 y) = -2x (since -2i z = -2i x + 2 y, as above), and Im(i K) = K, since i K is a purely imaginary number with imaginary part K (assuming K is real).Therefore, f(z) = z³ - 2i z + i K. Wait, but if K is a real constant, then i K is purely imaginary. However, in our expression for f(z), the constant term is i K. But the problem states that f(i) = 2. Let's use that condition to find K.First, let's write f(z) = z³ - 2i z + i K.Compute f(i): substitute z = i into the function.f(i) = (i)³ - 2i*(i) + i K.Calculate each term:(i)³ = i² * i = (-1)*i = -i,-2i*(i) = -2i² = -2*(-1) = 2,Therefore, f(i) = -i + 2 + i K.Given that f(i) = 2, which is a real number. So, let's equate:- i + 2 + i K = 2 + (-1 + K)i.Since this equals 2, which has no imaginary part, the imaginary part must be zero and the real part must be 2. Therefore:Real part: 2 = 2, which is already satisfied.Imaginary part: -1 + K = 0 ⇒ K = 1.Therefore, K is 1, so the function is f(z) = z³ - 2i z + i*1 = z³ - 2i z + i.Simplify that: f(z) = z³ - 2i z + i.Let me check this again with f(i) = 2:f(i) = (i)^3 - 2i*(i) + i = (-i) - 2i^2 + i = (-i) - 2*(-1) + i = (-i) + 2 + i = 2. Yes, that works.Therefore, the analytic function is f(z) = z³ - 2i z + i.Alternatively, maybe we can write this in a slightly different form. Let's see:f(z) = z³ - 2i z + i. Maybe factor out i from the last two terms?f(z) = z³ + i(-2 z + 1). Not sure if that's necessary, but perhaps it's better to leave it as is.Let me verify once again by expanding f(z) = z³ - 2i z + i into real and imaginary parts.Let z = x + iy,z³ = (x + iy)^3 = x³ + 3x²(iy) + 3x(iy)^2 + (iy)^3= x³ + 3i x² y + 3x (-y²) + i (-y³)= x³ - 3x y² + i(3x² y - y³).Then, -2i z = -2i(x + iy) = -2i x - 2i² y = -2i x + 2 y.Adding these together:z³ - 2i z = (x³ - 3x y² + 2 y) + i(3x² y - y³ - 2x).Then, adding the last term +i:f(z) = z³ - 2i z + i = (x³ - 3x y² + 2 y) + i(3x² y - y³ - 2x + 1).Comparing with the original u and v:u(x, y) = x³ - 3x y² + 2 y,v(x, y) = 3x² y - y³ - 2x + 1.Which matches our earlier result when K = 1. Therefore, this is correct.So, the final answer is f(z) = z³ - 2i z + i.Final AnswerThe analytic function is boxed{f(z) = z^3 - 2iz + i}.</think>To find the analytic function ( f(z) ) such that the real part ( u(x, y) = x^3 - 3xy^2 + 2y ) and ( f(i) = 2 ), we use the Cauchy-Riemann equations. First, we compute the partial derivatives of ( u ):[frac{partial u}{partial x} = 3x^2 - 3y^2 quad text{and} quad frac{partial u}{partial y} = -6xy + 2.]Using the Cauchy-Riemann equations, we know:[frac{partial v}{partial y} = 3x^2 - 3y^2 quad text{and} quad frac{partial v}{partial x} = 6xy - 2.]Integrating ( frac{partial v}{partial y} ) with respect to ( y ):[v = int (3x^2 - 3y^2) , dy + C(x) = 3x^2 y - y^3 + C(x).]Next, we find ( C(x) ) by using the other Cauchy-Riemann equation:[frac{partial v}{partial x} = 6xy + C'(x) = 6xy - 2 implies C'(x) = -2 implies C(x) = -2x + K.]Thus, the imaginary part ( v(x, y) ) is:[v(x, y) = 3x^2 y - y^3 - 2x + K.]Combining ( u ) and ( v ), the function ( f(z) ) is:[f(z) = (x^3 - 3xy^2 + 2y) + i(3x^2 y - y^3 - 2x + K).]Recognizing the terms related to ( z^3 ) and ( -2iz ), we rewrite:[f(z) = z^3 - 2iz + iK.]Using the condition ( f(i) = 2 ):[f(i) = (i)^3 - 2i(i) + iK = -i + 2 + iK = 2 + i(-1 + K).]Since the imaginary part must be zero:[-1 + K = 0 implies K = 1.]Thus, the analytic function is:[f(z) = z^3 - 2iz + i.]The final answer is:[boxed{f(z) = z^3 - 2iz + i}],<think>Okay, so I have this problem where I need to find an analytic function f(z) such that the real part u(x, y) is given by x³ - 3xy² + 2y, and also f(i) = 2. Alright, let me start recalling what I know about analytic functions and the Cauchy-Riemann equations.First, since f is analytic, it must satisfy the Cauchy-Riemann equations. That means if u is the real part and v is the imaginary part, then the partial derivatives of u and v should satisfy u_x = v_y and u_y = -v_x. So my goal is to find v(x, y) such that these equations hold, and then combine u and v into f(z) = u + iv. Also, once I find f(z), I need to check if f(i) = 2, which might help in determining any constants of integration.Let me write down the given u(x, y) = x³ - 3x y² + 2y. Let's compute the partial derivatives of u first.First, the partial derivative of u with respect to x, u_x. That would be 3x² - 3y². Then the partial derivative of u with respect to y, u_y. That would be -6xy + 2.According to the Cauchy-Riemann equations, u_x = v_y and u_y = -v_x. So, from u_x = v_y, we have v_y = 3x² - 3y². Then from u_y = -v_x, we have -v_x = -6xy + 2, so v_x = 6xy - 2.Now, to find v, I can integrate v_x with respect to x and v_y with respect to y, and make sure the results are consistent.Let's first integrate v_x with respect to x. So ∫(6xy - 2) dx. Integrating term by term, the integral of 6xy dx is 3x² y, and the integral of -2 dx is -2x. So that gives v = 3x² y - 2x + C(y), where C(y) is a function of y (since we integrated with respect to x, the constant of integration can depend on y).Now, let's also compute v by integrating v_y with respect to y. So ∫(3x² - 3y²) dy. Integrating term by term, the integral of 3x² dy is 3x² y, and the integral of -3y² dy is -y³. So that gives v = 3x² y - y³ + D(x), where D(x) is a function of x (constant of integration after integrating with respect to y).Now we have two expressions for v:1. From integrating v_x: v = 3x² y - 2x + C(y)2. From integrating v_y: v = 3x² y - y³ + D(x)These two expressions must be equal for all x and y. So set them equal:3x² y - 2x + C(y) = 3x² y - y³ + D(x)Subtracting 3x² y from both sides:-2x + C(y) = -y³ + D(x)Now, rearrange terms:C(y) + y³ = D(x) + 2xThe left-hand side is a function of y only, and the right-hand side is a function of x only. The only way this can hold for all x and y is if both sides are equal to the same constant. Let's denote this constant as K. So,C(y) + y³ = K and D(x) + 2x = KTherefore, solving for C(y) and D(x):C(y) = K - y³D(x) = K - 2xBut wait, substituting back into the expressions for v:From the first expression: v = 3x² y - 2x + (K - y³)From the second expression: v = 3x² y - y³ + (K - 2x)So both give the same expression: v = 3x² y - y³ - 2x + KTherefore, the imaginary part v(x, y) is 3x² y - y³ - 2x + K, where K is a constant.But since we need f(z) = u + iv, let's write that:f(z) = (x³ - 3x y² + 2y) + i(3x² y - y³ - 2x + K)Now, since f(z) is analytic, it should be expressible as a function of z alone, not x and y separately. So we need to write this in terms of z = x + iy. Let's see if we can express f(z) in terms of z.Looking at the real and imaginary parts, note that x³ - 3x y² is the real part of z³, because if we expand z³:z³ = (x + iy)³ = x³ + 3x²(iy) + 3x(iy)² + (iy)³ = x³ + 3i x² y - 3x y² - i y³Therefore, Re(z³) = x³ - 3x y², which matches the first part of u(x, y). Similarly, the imaginary part of z³ is 3x² y - y³.Looking at the given u(x, y) = Re(z³) + 2y, and the v(x, y) we found is Im(z³) - 2x + K.Therefore, f(z) = z³ + something. Let's check:If f(z) = z³ + something, then u would be Re(z³) + ... and v would be Im(z³) + ...But our u is Re(z³) + 2y, so the extra term is 2y. Similarly, our v is Im(z³) -2x + K. So perhaps f(z) = z³ + i(-2x) + iK + 2y. Wait, but that needs to be combined into a function of z.Wait, let me think. The terms 2y and -2x. Let's see:If we have 2y - 2i x, that can be written as -2i(x + iy) + ... Hmm, x + iy is z, so maybe:2y - 2i x = -2i x + 2y = -2i(x - (i y)) = ?Wait, maybe factor out terms. Let's see:2y - 2i x = -2i x + 2y = 2i(i y) - 2i x = -2i(x - i y) = -2i zBecause z = x + i y, so x - i y is not exactly z, but if we consider complex conjugates, but maybe that's complicating.Alternatively, note that 2y - 2i x can be written as -2i(x + i y) + 2i(i y + i^2 x). Wait, perhaps that's messy. Let me compute:-2i z = -2i(x + i y) = -2i x - 2i² y = -2i x + 2 yYes! Exactly. So 2y - 2i x is equal to -2i z. Therefore, the term 2y in the real part and -2x in the imaginary part (times i) would combine to -2i z. Then the constant term iK would just be a constant.So putting it all together:f(z) = z³ + (-2i z) + i KWait, let's verify:If f(z) = z³ - 2i z + i K, then:u = Re(z³) + Re(-2i z) + Re(i K) = (x³ - 3x y²) + Re(-2i x - 2i² y) + Re(i K)But Re(-2i z) = Re(-2i x + 2 y) since z = x + i y, so -2i z = -2i x + 2 y. Therefore, Re(-2i z) = 2 y. Then Re(i K) is Re(i K) = Re(K i) = 0 if K is a real constant. Wait, but K was a real constant? Wait, hold on.Wait, in our expression for v, the constant K is a real constant because when we had v = 3x² y - y³ - 2x + K, K is real. Therefore, when we write f(z) = u + i v, the term i K would be i times a real constant, so it's an imaginary constant. Therefore, Re(i K) is 0, and Im(i K) is K. So in the expression f(z) = z³ - 2i z + i K, the real part is indeed Re(z³) + Re(-2i z) which is x³ - 3x y² + 2y, which matches u(x, y). The imaginary part is Im(z³) + Im(-2i z) + K, which is (3x² y - y³) + (-2x) + K, which matches v(x, y) = 3x² y - y³ - 2x + K. So that's correct.Therefore, f(z) = z³ - 2i z + i K. Now, we need to determine K using the condition f(i) = 2.Let's compute f(i). Let's substitute z = i into f(z):f(i) = (i)³ - 2i (i) + i KCompute each term:(i)³ = i² * i = (-1) * i = -i-2i(i) = -2i² = -2(-1) = 2Therefore, f(i) = -i + 2 + i KSo f(i) = 2 + (-i + i K) = 2 + i(-1 + K)But we are given that f(i) = 2. Since f(i) is a complex number, both the real and imaginary parts must match. The real part is 2, and the imaginary part is (-1 + K). Therefore, to have f(i) = 2, the imaginary part must be 0. So:-1 + K = 0 => K = 1Therefore, substituting back into f(z):f(z) = z³ - 2i z + i * 1 = z³ - 2i z + iSo that should be the function. Let me check once again.Let me verify the real and imaginary parts:f(z) = z³ - 2i z + iExpressed in terms of x and y:z³ = (x + iy)³ = x³ + 3x²(iy) + 3x(iy)² + (iy)³ = x³ + 3i x² y - 3x y² - i y³-2i z = -2i(x + iy) = -2i x - 2i² y = -2i x + 2 yAdding i: +iSo f(z) = (x³ - 3x y²) + i(3x² y - y³) + (2 y - 2i x) + iCombine real parts:x³ - 3x y² + 2 yCombine imaginary parts:(3x² y - y³) - 2x + 1So u(x, y) = x³ - 3x y² + 2 y, which matches the given u. The imaginary part is 3x² y - y³ - 2x + 1, which is consistent with our earlier v(x, y) = 3x² y - y³ - 2x + K where K = 1. Then f(i) = 2, as required.So everything checks out. Therefore, the analytic function is f(z) = z³ - 2i z + i.Final AnswerThe analytic function is boxed{z^3 - 2iz + i}.</think>To find the analytic function ( f(z) ) such that the real part ( u(x, y) = x^3 - 3xy^2 + 2y ) and ( f(i) = 2 ), we use the Cauchy-Riemann equations. First, we compute the partial derivatives of ( u ):[u_x = 3x^2 - 3y^2 quad text{and} quad u_y = -6xy + 2]Using the Cauchy-Riemann equations ( u_x = v_y ) and ( u_y = -v_x ), we find:[v_y = 3x^2 - 3y^2 quad text{and} quad v_x = 6xy - 2]Integrate ( v_x ) with respect to ( x ):[v = int (6xy - 2) , dx = 3x^2 y - 2x + C(y)]Integrate ( v_y ) with respect to ( y ):[v = int (3x^2 - 3y^2) , dy = 3x^2 y - y^3 + D(x)]Equating the two expressions for ( v ):[3x^2 y - 2x + C(y) = 3x^2 y - y^3 + D(x)]This gives:[-2x + C(y) = -y^3 + D(x)]Solving for ( C(y) ) and ( D(x) ), we find ( C(y) = -y^3 + K ) and ( D(x) = -2x + K ). Thus, the imaginary part ( v ) is:[v = 3x^2 y - y^3 - 2x + K]The function ( f(z) ) is:[f(z) = u + iv = (x^3 - 3xy^2 + 2y) + i(3x^2 y - y^3 - 2x + K)]Expressing ( f(z) ) in terms of ( z ), we recognize:[f(z) = z^3 - 2iz + iK]Using the condition ( f(i) = 2 ):[f(i) = (-i) + 2 + iK = 2 + i(-1 + K)]Setting the imaginary part to zero gives ( -1 + K = 0 Rightarrow K = 1 ). Thus, the function is:[f(z) = z^3 - 2iz + i]The final answer is:[boxed{z^3 - 2iz + i}]]

🔑:SOLUTION.1. We find the partial derivatives of the function u(x, y)frac{partial u}{partial x}=3 x^{2}-3 y^{2}, quad frac{partial u}{partial y}=-6 x y+22. From the 2nd Cauchy-Riemann condition (1)frac{partial v}{partial x}=-frac{partial u}{partial y}=6 x y-2we findv=int frac{partial v}{partial x} d x+varphi(y)=int(6 x y-2) d x+g(y)=3 x^{2} y-2 x+g(y)Differentiating v with respect to y, we getfrac{partial v}{partial y}=3 x^{2}+g^{prime}(y)To find the function g(y), we use the 1st Cauchy-Riemann condition (1). Equating partial v / partial y=3 x^{2}+g^{prime}(y) to the derivativefrac{partial u}{partial x}=3 x^{2}-3 y^{2}we obtain an ordinary differential equation of the 1st orderg^{prime}(y)=-3 y^{2}from which we determine g(y)g(y)=-int 3 y^{2} d y=-y^{3}+CThus, we obtain the functionv(x, y)=3 x^{2} y-2 x-y^{3}+C3. We write the desired function f(z) in the formf(z)=u+i v=x^{3}-3 x y^{2}+2 y+ileft(3 x^{2} y-y^{3}-2 x+C_{1}right)Transforming the obtained expression into a function of the variable z, using expressions (2) or the equalitiesz=x+i y quad u quad z^{3}=(x+i y)^{3}=x^{3}-3 x y^{2}+ileft(3 x^{2} y-y^{3}right) .We getf(z)=z^{3}-2 i z+Cwhere C is an arbitrary complex constant.4. We find the value of C from the condition f(i)=2 :2=i^{3}-2 i cdot i+CWe get C=i and, consequently,f(z)=z^{3}-2 i z+iAnswer. f(z)=z^{3}-2 i z+i.PROBLEM CONDITIONS. Find the analytic function f(z), if its real or imaginary part and the value of f(z) at some point z_{0} are given.1. operatorname{Re} f(z)=x^{2}-y^{2}+2 x, f(i)=2 i-1. 2 . operatorname{Re} f(z)=x /left(x^{2}+y^{2}right), f(pi)=1 / pi . quad 3 . operatorname{Im} f(z)=operatorname{arctg}(y / x), f(1)=0 . quad 4 . operatorname{Re} f(z)= =3 x^{2}-4 x y-3 y^{2}, f(-i)=-3-2 i . quad 5 . operatorname{Im} f(z)=10 x y-6 y, f(1 / 5)=-1 . quad 6 . operatorname{Im} f(z)=sin y operatorname{ch}(x+1), quad f(-1+pi i / 2)=i. 7. operatorname{Im} f(z)=2 yleft[y^{2}+(x+1)^{2}right], f(i)=i . quad 8 . operatorname{Im} f(z)=x /left(x^{2}+y^{2}right)+x, f(1)=-2 i . quad 9. operatorname{Re} f(z)=e^{-y}(x cos x-y sin x), f(0)=0. 10. operatorname{Re} f(z)=x sin x operatorname{ch} y-y cos x operatorname{sh} y, f(0)=0.Answers.1. f(z)=z^{2}+2 z. 2. f(z)=1 / z .3 . f(z)=ln z .4 . f(z)=z^{2}(3+2 i). 5. f(z)=5 z^{2}-6 z . quad 6 . f(z)=operatorname{sh}(z+1). 7. f(z)=(z-1) /(z+1). 8. f(z)=i / z+i z. 9. f(z)=z e^{i z}. 10. f(z)=z sin z.## 1.5. Integral of a Function of a Complex VariablePROBLEM STATEMENT. Calculate the integralint_{L} f(z) d zwhere L is a piecewise smooth curve lying in a domain D where the function f(z) is continuous.SOLUTION PLAN. The integral of the function f(z)=u(x, y)+i v(x, y), continuous in the domain D, is expressed through line integrals of the second kind by the formulaint_{L} f(z) d z=int_{L} u d x-v d y+i int_{L} u d y+v d x1. We write f(z) in algebraic formf(z)=u(x, y)+i v(x, y)2. Using formula (1), we represent the desired integral as the sum of two line integrals of the second kind of the functions u(x, y) and v(x, y) of two real variables x and y.3. We write the equations of the curve L in explicit form y=y(x) (or x=x(y)) or parametrically y=y(t), x=x(t).4. We compute the line integrals by reducing them to definite integrals and write the answer.Remark. The integral of a function of a complex variable can also be computed using the formulaint_{L} f(z) d z=int_{alpha}^{beta} f(z(t)) z^{prime}(t) d twhere z=z(t) is the parametric equation of the curve L in complex form.

🧠:[<think>Okay, let's see. The problem is to find the equation of the tangent line to the curve y = e^x at a point P, and this tangent line passes through the origin. Hmm, tangent lines... right. So first, I need to recall how to find the equation of a tangent line to a curve at a particular point. Alright, so generally, the equation of a tangent line at a point (a, f(a)) is given by y = f'(a)(x - a) + f(a). That's the formula I remember. Here, the function is f(x) = e^x, so the derivative f'(x) is also e^x, right? Because the derivative of e^x is e^x. So that part is straightforward.So, let's say point P has coordinates (a, e^a). Then, the slope of the tangent line at that point is f'(a) = e^a. Therefore, the equation of the tangent line at P would be y = e^a(x - a) + e^a. Simplifying that, it becomes y = e^a x - a e^a + e^a, which can also be written as y = e^a x + e^a (1 - a). But the problem states that this tangent line passes through the origin, which is (0, 0). So, if we plug in x = 0 and y = 0 into the equation of the tangent line, it should satisfy the equation. Let me do that. Substitute x = 0, y = 0 into y = e^a x + e^a (1 - a):0 = e^a * 0 + e^a (1 - a)Simplify the right-hand side: 0 = 0 + e^a (1 - a), so 0 = e^a (1 - a). Now, e^a is never zero for any real number a, right? Because the exponential function is always positive. So the only way this product can be zero is if (1 - a) is zero. Therefore, 1 - a = 0, which implies a = 1. So, the point P is (1, e^1) or (1, e). Now, we can write the equation of the tangent line using this value of a. Let's substitute a = 1 into the tangent line equation:y = e^1 x + e^1 (1 - 1) = e x + e * 0 = e x. So, the equation of the tangent line is y = e x. Let me just double-check that. If the tangent line at (1, e) has slope e, then the line should pass through (0, 0). Let's verify:Using the point-slope form: y - e = e(x - 1). If we expand this, we get y = e x - e + e = e x. Yes, that simplifies to y = e x. And if we check whether (0, 0) is on this line, plug in x = 0: y = e*0 = 0. So, yes, it passes through the origin. Wait, but just to be thorough, maybe I should check my steps again. Let's start from the beginning.1. The curve is y = e^x. The derivative is y' = e^x. So at any point (a, e^a), the slope of the tangent line is e^a.2. The equation of the tangent line at (a, e^a) is y = e^a (x - a) + e^a.3. This line must pass through (0, 0). Plugging in x = 0, y = 0:0 = e^a (0 - a) + e^a => 0 = -a e^a + e^a => 0 = e^a (1 - a).4. Since e^a ≠ 0, then 1 - a = 0 => a = 1.5. Therefore, the tangent line is at (1, e), equation y = e x - e*1 + e = e x. Yep, that checks out.Is there another way to approach this problem? Maybe by parametrizing the tangent line and setting the intercept to zero? Let's see. Suppose the tangent line has the equation y = m x + b. Since it's tangent to y = e^x at some point (a, e^a), the slope m must equal e^a. Also, the line must pass through (a, e^a), so e^a = m a + b. But since m = e^a, substituting gives e^a = e^a * a + b => b = e^a - e^a * a = e^a (1 - a). But we also know that the line passes through the origin (0,0), so plugging x=0, y=0 into the equation gives 0 = 0 + b => b = 0. Therefore, e^a (1 - a) = 0. Again, since e^a ≠ 0, 1 - a = 0 => a = 1. Therefore, the equation is y = e x + 0 => y = e x. Same result.So, both methods lead to the same conclusion. Therefore, I can be confident that the equation of the tangent line is y = e x.Just to visualize, the curve y = e^x is exponential, increasing rapidly. The tangent at (1, e) would have a slope equal to e, which is approximately 2.718. The line y = e x is a straight line passing through the origin with that slope. It makes sense that this line just touches the exponential curve at (1, e), as the exponential function is convex and its slope increases faster than any linear function. Hence, there's only one such tangent line that passes through the origin.I wonder if there's a geometric intuition here. The origin is below the curve y = e^x for all x, because e^x is always positive. The tangent line at (1, e) has to rise steeply enough to just touch the curve there and still pass through the origin. Since the slope is e, which is the same as the y-coordinate at x=1, that seems to balance out the position. If the slope were less, the line might not reach the origin; if it were more, maybe it would cross the curve somewhere else. But since the exponential function's derivative is equal to its value, the condition works out perfectly at x=1.Another check: let's confirm that the line y = e x only intersects the curve y = e^x at (1, e). Suppose e x = e^x. At x=1, both sides are e. What about x=0? Left side is 0, right side is 1. So they don't intersect there. For x >1, e^x grows much faster than e x, so no intersection. For x <1, let's see: take x=0.5. e*0.5 ≈1.359, e^0.5≈1.648. So e^x is above e x here. At x=2, e x ≈5.436, e^2≈7.389, still e^x above. So the only intersection is at x=1, which confirms that the line is tangent there, since they only touch at that single point.Therefore, all steps and checks seem consistent. I think the answer is correct.Final AnswerThe equation of the tangent line is boxed{y = ex}.</think>To find the equation of the tangent line to the curve ( y = e^x ) at point ( P ) that passes through the origin, we start by noting that the derivative of ( y = e^x ) is ( y' = e^x ). This means the slope of the tangent line at any point ( (a, e^a) ) is ( e^a ).The equation of the tangent line at point ( (a, e^a) ) is given by:[y = e^a (x - a) + e^a]Simplifying this, we get:[y = e^a x - a e^a + e^a = e^a x + e^a (1 - a)]Since the tangent line passes through the origin ( (0, 0) ), substituting ( x = 0 ) and ( y = 0 ) into the equation gives:[0 = e^a (0 - a) + e^a][0 = -a e^a + e^a][0 = e^a (1 - a)]Since ( e^a neq 0 ), we must have ( 1 - a = 0 ), which implies ( a = 1 ). Therefore, the point ( P ) is ( (1, e) ).Substituting ( a = 1 ) into the tangent line equation:[y = e^1 x + e^1 (1 - 1) = e x + e cdot 0 = e x]Thus, the equation of the tangent line is (boxed{y = ex}).,<think>Okay, so I need to find the equation of the tangent line to the curve y = e^x at some point P, and this tangent line has to pass through the origin. Hmm, let's see. I remember that the equation of a tangent line at a point (a, f(a)) is given by y = f(a) + f’(a)(x - a). Right? Because the derivative at that point gives the slope of the tangent line. So first, let's figure out what f(a) and f’(a) are here. The function is y = e^x, so f(a) would be e^a. The derivative of e^x is still e^x, so f’(a) is also e^a. That simplifies things a bit. So the equation of the tangent line at point P, which is (a, e^a), is y = e^a + e^a(x - a). Let me write that out again: y = e^a + e^a(x - a). Maybe I can simplify this equation. Let's distribute the e^a: y = e^a + e^a * x - e^a * a. Combine like terms: y = e^a * x + e^a - e^a * a. Hmm, or factor out e^a from the first term? Wait, maybe not necessary. But we know that this tangent line passes through the origin (0, 0). So if we plug x = 0 and y = 0 into the equation, it should satisfy. Let's do that. Substitute x = 0, y = 0:0 = e^a + e^a*(0 - a)Simplify the right-hand side: 0 = e^a + e^a*(-a)So 0 = e^a - a e^aFactor out e^a: 0 = e^a (1 - a)Since e^a is never zero for any real number a, we can divide both sides by e^a, which gives 0 = 1 - a. Therefore, solving for a: a = 1. So the point P is (1, e^1) which is (1, e). Now, let's find the equation of the tangent line at a = 1. We already have the general equation, so plugging a = 1 into y = e^a + e^a(x - a):y = e^1 + e^1(x - 1)Simplify: y = e + e x - eCombine like terms: e x - e + e = e x. So y = e x. Wait, that seems too simple. Let me check my steps again. When I substituted a = 1 into the equation, it becomes y = e + e(x - 1). Expanding that, it's y = e + e x - e, which indeed simplifies to y = e x. But let's confirm this. The tangent line at (1, e) has slope e^1 = e. So the equation is y - e = e(x - 1). Distribute the e: y = e x - e + e, which is y = e x. Yeah, that's correct. So the tangent line is y = e x, which passes through the origin since when x = 0, y = 0. That makes sense. Just to make sure I didn't make any mistakes, let me visualize this. The curve y = e^x is exponential, increasing rapidly. At the point (1, e), the slope is e, so the tangent line would be steeper there. The line y = e x is a straight line passing through the origin with slope e. Since e is approximately 2.718, that's a pretty steep line. Let me check if (1, e) is on the line y = e x. Plugging x = 1 into the line: y = e*1 = e, which matches the point. So yes, that's correct. Alternatively, maybe I can approach this problem using the general equation of a line passing through the origin. Since the tangent line passes through (0,0) and has slope m, its equation is y = m x. But this line must also be tangent to y = e^x at some point (a, e^a). So the slope m must be equal to the derivative at that point, which is e^a. Therefore, the equation is y = e^a x. But this line must also pass through the point (a, e^a). So substituting x = a, y = e^a into the line equation: e^a = e^a * a. So e^a = a e^a. Divide both sides by e^a (since e^a ≠ 0), we get 1 = a. So again, a = 1. Therefore, the slope is e^1 = e, so the equation is y = e x. Same result. So that confirms it. Wait, so both methods lead to the same answer. First method was using the point-slope form of the tangent line, then substituting the origin into it. Second method was assuming the line passes through the origin with slope m, equating that to the derivative, and solving. Both give a = 1, leading to the equation y = e x. Therefore, the equation of the tangent line is y = e x. But just to be thorough, let's check if there's another possible solution. The equation 0 = e^a (1 - a) gives only a = 1 as a solution because e^a is never zero. So no other solutions. Therefore, the answer is y = e x. Final AnswerThe equation of the tangent line is boxed{y = ex}.</think>To find the equation of the tangent line to the curve ( y = e^x ) at point ( P ) that passes through the origin, we start by noting that the equation of the tangent line at point ( (a, e^a) ) is given by:[ y = e^a + e^a(x - a) ]Since the tangent line passes through the origin ((0, 0)), we substitute ( x = 0 ) and ( y = 0 ) into the equation:[ 0 = e^a + e^a(0 - a) ]Simplifying the right-hand side:[ 0 = e^a - ae^a ][ 0 = e^a(1 - a) ]Since ( e^a neq 0 ), we have:[ 1 - a = 0 ][ a = 1 ]Thus, the point ( P ) is ( (1, e) ). The equation of the tangent line at this point is:[ y = e + e(x - 1) ]Simplifying this equation:[ y = e + ex - e ][ y = ex ]Therefore, the equation of the tangent line is (boxed{y = ex}).]

🔑:To find the equation of the tangent line, we begin by differentiating the given curve. For the curve y = e^x, the derivative with respect to x is:y' = frac{dy}{dx} = e^x.Let's denote the point of tangency as (x_0, e^{x_0}). The slope k of the tangent line at this point is equal to the derivative at this point, so:k = e^{x_0}.Using the point-slope form of the equation of a line, the equation of the tangent line is:y - e^{x_0} = e^{x_0} (x - x_0).Since the tangent line passes through the origin (0, 0), we can substitute these values into the equation to find x_0:0 - e^{x_0} = e^{x_0} (0 - x_0),- e^{x_0} = -x_0 e^{x_0},x_0 = 1.Now that we have x_0 = 1, the coordinates of the point of tangency are (1, e), and the slope of the tangent line is e.The equation of the tangent line can be rewritten as:y - e = e(x - 1),which simplifies to:y = ex.Thus, the equation of the tangent line is boxed{y = ex}.

🧠:[<think>Okay, so I need to solve this problem about the function f(x) and g(x). Let me start with part (1) where a=1. First, let's find set A, which is the domain of f(x). The function f(x) is a square root, so the expression inside the square root must be non-negative. That means I need to solve the inequality:- x² + (a + 2)x - a - 1 ≥ 0When a=1, substitute that into the equation:- x² + (1 + 2)x - 1 - 1 ≥ 0 → -x² + 3x - 2 ≥ 0Hmm, quadratic inequality. Let me rearrange it to standard form. Multiply both sides by -1 (remember to reverse the inequality sign):x² - 3x + 2 ≤ 0Factor the quadratic:(x - 1)(x - 2) ≤ 0So the solutions are x between 1 and 2, inclusive. So the domain A is [1, 2]. Let me double-check. If x=1, then inside the sqrt: -1 + 3 -2 = 0, which is okay. x=2: -4 + 6 -2 = 0, also okay. And between 1 and 2, let's pick x=1.5: - (2.25) + 4.5 -2 = 0.25, which is positive. So yes, domain is [1,2].Now set B, the range of g(x) = 2^x -1 for x ≤ 2. Let's think about the function 2^x. Since x ≤ 2, the maximum value of 2^x is 2^2 =4. As x approaches negative infinity, 2^x approaches 0. So the range of 2^x is (0,4], then subtract 1, so the range becomes (-1,3]. But wait, when x approaches negative infinity, 2^x approaches 0, so 0 -1 = -1. But does x actually reach negative infinity? The problem says x ≤ 2, so x can be any real number less than or equal to 2. So the lower limit as x approaches -infty is approaching -1, but never actually reaching it. So the range is (-1,3]. But wait, when x=2, 2^2 -1 =4 -1=3. So 3 is included. So set B is (-1,3].Wait, but maybe I should check if there's a minimum value. Let's see, 2^x is always positive, so 2^x -1 is greater than -1. So yeah, the range is (-1,3]. So set B is (-1, 3].But let me confirm. For example, when x=0: 2^0 -1 =1 -1=0. So 0 is in the range. x=1: 2 -1=1. x=2:4 -1=3. x approaching -infty: approaches -1. So yes, set B is (-1,3]. So for part (1), A is [1,2] and B is (-1,3]. Moving on to part (2): If sets A and B satisfy A ∪ B = B, find the range of real number a (with a >0). So A ∪ B = B means that A is a subset of B. Therefore, all elements in set A must be in set B. Since set A is the domain of f(x), which is the interval where the quadratic inside the sqrt is non-negative. Let's first express set A in terms of a. Original inequality: -x² + (a +2)x -a -1 ≥0. Let's rearrange that:Let me rewrite the quadratic: -x² + (a+2)x -a -1 ≥0. Multiply by -1 (reversing inequality):x² - (a +2)x + a +1 ≤0So we need to find the roots of the quadratic x² - (a +2)x + a +1 =0.Let me compute the discriminant D:D = [-(a +2)]² -4*1*(a +1) = (a +2)^2 -4(a +1) = a² +4a +4 -4a -4 = a²So discriminant is a², which is always non-negative since a is real. Therefore, roots are:x = [ (a +2) ± sqrt(a²) ] / 2Since a >0, sqrt(a²)=a. Therefore,x = [ (a +2) + a ] /2 = (2a +2)/2 = a +1andx = [ (a +2) -a ] /2 = 2/2 =1So the quadratic factors as (x -1)(x - (a +1)) ≤0.Therefore, the solution is the interval between the roots, [1, a +1]. But since the quadratic coefficient is positive, the inequality x² - (a +2)x +a +1 ≤0 is satisfied between the roots. Therefore, set A is [1, a +1].So set A is [1, a +1], and set B is (-1,3] as before.The condition A ∪ B = B is equivalent to A ⊆ B. So we need [1, a +1] ⊆ (-1,3].Therefore, all elements of A must be in B. Since A starts at 1, which is in B (since B is up to 3 and includes 3, and starts from -1 not including -1). Since 1 is greater than -1, 1 is in B. The upper limit of A is a +1. So to have A ⊆ B, we need a +1 ≤3. Because the upper bound of A, which is a +1, must be less than or equal to the upper bound of B, which is 3. But also, since a >0, so a +1 >1. So the interval A is [1, a +1], and B is (-1,3]. So as long as a +1 ≤3, then A is within B. Therefore:a +1 ≤3 → a ≤2.But since a >0, the range of a is (0,2].But wait, let me confirm. Since B is (-1,3], which includes all numbers from just above -1 up to 3, including 3. Since A is [1, a+1], which is entirely within B as long as a +1 ≤3. Because 1 is already in B, and the upper limit a +1 just needs to not exceed 3. So yes, a ≤2. Since a >0, the range is 0 < a ≤2. So in interval notation, (0, 2].But let me check if there's any other conditions. For example, the set A is [1, a+1]. All elements of A are between 1 and a+1. Since B is from -1 to 3, the only critical point is the upper limit. Since 1 is already within B, as long as a+1 ≤3, all of A is in B. So yes, a must be ≤2. Since a>0, then the range is (0,2].Wait, but let me check a=2. Then set A is [1,3]. But set B is (-1,3]. So A is [1,3], which is a subset of (-1,3] because 3 is included in B. So 3 is in B, so [1,3] is entirely in B. If a=2, then okay.If a is 0.5, then A is [1,1.5], which is in B. If a approaches 0, then a +1 approaches 1, so A becomes [1,1], which is just {1}, which is still in B. But since a>0, a can't be 0. So the lower limit is a approaching 0 from the right. Therefore, the answer is 0 < a ≤2.So the range of a is (0,2].But let me double-check. Suppose a=2, then A is [1,3]. B is (-1,3]. Since 3 is in B, then A is [1,3] which is a subset of (-1,3]. Correct. If a=2.5, then A is [1,3.5], which would go beyond 3, so 3.5 is not in B. Hence, a must be ≤2. Therefore, the answer is a ∈ (0,2].Final Answer(1) When (a=1), set (A) is (boxed{[1, 2]}) and set (B) is (boxed{(-1, 3]}).(2) The range of (a) is (boxed{(0, 2]}).</think>To solve the given problem, we start by analyzing the domain of the function ( f(x) ) and the range of the function ( g(x) ). Part (1) when ( a = 1 ):1. Finding set A (Domain of ( f(x) )): - The function ( f(x) = sqrt{-x^2 + (a+2)x - a - 1} ) requires the expression inside the square root to be non-negative. - Substituting ( a = 1 ), we get the inequality: [ -x^2 + 3x - 2 geq 0 ] - Rearranging and factoring: [ x^2 - 3x + 2 leq 0 implies (x - 1)(x - 2) leq 0 ] - The solution is the interval ([1, 2]). Thus, set ( A ) is ([1, 2]).2. Finding set B (Range of ( g(x) )): - The function ( g(x) = 2^x - 1 ) for ( x leq 2 ). - The range of ( 2^x ) for ( x leq 2 ) is ((0, 4]). Subtracting 1, the range becomes ((-1, 3]). Thus, set ( B ) is ((-1, 3]). Part (2) when ( A cup B = B ):1. Expressing set A in terms of ( a ): - The quadratic inequality for the domain of ( f(x) ) is: [ -x^2 + (a + 2)x - a - 1 geq 0 ] - Rearranging and factoring: [ x^2 - (a + 2)x + a + 1 leq 0 implies (x - 1)(x - (a + 1)) leq 0 ] - The solution is the interval ([1, a + 1]). Thus, set ( A ) is ([1, a + 1]).2. Condition for ( A subseteq B ): - Set ( B ) is ((-1, 3]). For ( A subseteq B ), the upper bound of ( A ) must be less than or equal to 3: [ a + 1 leq 3 implies a leq 2 ] - Since ( a > 0 ), the range of ( a ) is ((0, 2]). Final Answers(1) When ( a = 1 ), set ( A ) is (boxed{[1, 2]}) and set ( B ) is (boxed{(-1, 3]}).(2) The range of ( a ) is (boxed{(0, 2]}).,<think>Okay, let me try to work through this problem step by step. First, let's read the problem again carefully to make sure I understand what's being asked.We have two functions here: f(x) and g(x). The domain of f(x) is set A, and the range of g(x) (for x ≤ 2) is set B. There are two parts to the problem. Part (1) asks us to find sets A and B when a=1. Part (2) is about finding the range of the real number a (given that a > 0) such that A ∪ B = B. Let me tackle part (1) first. Starting with set A, which is the domain of f(x). The function f(x) is a square root, so the expression inside the square root must be non-negative. Therefore, we need to find all x such that -x² + (a+2)x - a -1 ≥ 0. When a=1, this inequality becomes -x² + (1+2)x -1 -1 ≥ 0, which simplifies to -x² + 3x -2 ≥ 0.To solve this quadratic inequality, let me first multiply both sides by -1 to make it a standard quadratic. However, multiplying by a negative number reverses the inequality sign. So, multiplying both sides by -1 gives x² -3x +2 ≤ 0. Now, let's factor the quadratic: x² -3x +2 factors into (x -1)(x -2). So, (x -1)(x -2) ≤ 0. The solutions to this inequality are the x values where the product is less than or equal to zero. The critical points are x=1 and x=2. Testing intervals around these points:- For x < 1, say x=0: (0-1)(0-2) = (-1)(-2)=2 >0- For 1 < x < 2, say x=1.5: (0.5)(-0.5)=-0.25 <0- For x >2, say x=3: (3-1)(3-2)=2*1=2 >0Since the inequality is ≤0, the solution is the interval where the product is negative or zero, which is [1,2]. Therefore, the domain of f(x) when a=1 is the closed interval from 1 to 2. So, set A is [1, 2].Now, moving on to set B, which is the range of the function g(x) = 2^x -1 for x ≤2. Let's analyze the function g(x). The function 2^x is an exponential function, which is always increasing. Since x ≤2, as x approaches 2, 2^x approaches 4, so 4 -1 =3. As x decreases, 2^x approaches 0 (as x approaches negative infinity), so 0 -1 = -1. However, since x can be any real number less than or equal to 2, the range of g(x) is (-1, 3]. Wait, but when x approaches negative infinity, 2^x approaches 0, so 2^x -1 approaches -1. But does x actually reach negative infinity? No, x can be any real number less than or equal to 2, so the lower limit is approaching -1, but not including -1. However, when x=2, g(2)=2^2 -1=4-1=3. So the maximum value is 3, which is included. Therefore, the range B is (-1, 3]. Wait, hold on. Let me check that again. The function g(x) = 2^x -1. When x is approaching negative infinity, 2^x approaches 0, so g(x) approaches -1. But since x can be any real number ≤2, the lower bound is approaching -1 but never actually reaching it. Therefore, the range should be (-1, 3]. But is there a value of x where g(x) equals -1? Only when 2^x -1 = -1, which implies 2^x =0. But 2^x is never zero, so there is no x such that g(x) = -1. Hence, the range is (-1, 3].Wait, but sometimes when dealing with intervals, we have to check endpoints. Since x is allowed to approach negative infinity, but can't reach it, so the lower bound is an open interval. The upper bound at x=2 is included, so the upper limit is 3. Therefore, set B is (-1, 3].So, for part (1), set A is [1,2] and set B is (-1,3].But let me confirm once more. For set A: When a=1, the quadratic inside the square root is -x² +3x -2. We solved the inequality and found x between 1 and 2 inclusive. So that's correct.For set B: As x approaches negative infinity, 2^x approaches 0, so g(x) approaches -1, but never reaches it. At x=2, it's 3. So, the range is all real numbers greater than -1 and up to 3. So yes, (-1,3]. Therefore, part (1) answer is A = [1,2] and B = (-1,3].Now, moving on to part (2). We need to find the range of real number a >0 such that A ∪ B = B. First, let's recall that A is the domain of f(x), which is the set of x where -x² + (a+2)x -a -1 ≥0. B is the range of g(x)=2^x -1 for x ≤2, which we found in part (1) as (-1,3]. But wait, actually, in part (2), is set B still the same as in part (1)? Wait, the problem says "the range of the function g(x) = 2^x -1 (for x ≤2) is set B". So regardless of a, set B is always (-1,3] because a is a parameter for function f, not for g. So set B is fixed as (-1,3]. Therefore, in part (2), we need to find a >0 such that when we take the union of set A (domain of f(x)) and set B (range of g(x)), the result is equal to set B. Since B is fixed, A ∪ B = B implies that A is a subset of B. Because if you take the union of A and B and get B, then all elements of A must already be in B. So, A must be a subset of B. Therefore, the problem reduces to finding all a >0 such that set A (the domain of f(x)) is a subset of set B (which is (-1,3]). Therefore, we need to ensure that every x in A is also in B. Since B is (-1,3], this means that for all x in A, -1 < x ≤3. But actually, since A is the domain of f(x), which is a set of real numbers where the expression under the square root is non-negative. Let's first characterize set A for general a.Set A is the solution to the inequality -x² + (a+2)x -a -1 ≥0.Let me write this inequality again:- x² + (a + 2) x - a -1 ≥0.Multiplying both sides by -1 (and reversing the inequality):x² - (a + 2)x + a +1 ≤0.So, the quadratic equation x² - (a + 2)x + a +1 =0.Let me find its discriminant to check the roots.Discriminant D = [-(a + 2)]² -4*1*(a +1) = (a + 2)^2 -4(a +1).Expanding (a +2)^2: a² +4a +4Subtracting 4(a +1): a² +4a +4 -4a -4 = a².Therefore, discriminant D =a². Since a >0, D is positive, so the quadratic has two real roots.The roots are given by:x = [ (a +2) ± sqrt(D) ] / 2 = [ (a +2) ± a ] /2So, two roots:First root: [ (a +2) + a ] /2 = (2a +2)/2 = a +1Second root: [ (a +2) -a ] /2 = 2 /2 =1Therefore, the quadratic x² - (a +2)x +a +1 factors as (x -1)(x - (a +1)).Therefore, the inequality x² - (a +2)x +a +1 ≤0 is equivalent to (x -1)(x - (a +1)) ≤0.So, solving this inequality, we need to find the intervals where the product is ≤0.The critical points are x=1 and x=a+1.Since a >0, a+1 >1. Therefore, the critical points are ordered as 1 < a +1.Therefore, the solution to the inequality (x -1)(x - (a +1)) ≤0 is x between 1 and a +1, inclusive.Therefore, set A is [1, a +1].Therefore, for any a >0, set A is the interval [1, a +1].So, set A is [1, a +1], and set B is (-1,3].We need A ∪ B = B, which implies that A is a subset of B. Since B is (-1,3], all elements of A must lie within (-1,3].But set A is [1, a +1]. So, [1, a +1] must be a subset of (-1,3]. Therefore, all elements of [1, a +1] must be less than or equal to 3 and greater than -1. Since the lower bound of A is 1, which is already greater than -1, we only need to ensure that the upper bound of A, which is a +1, is less than or equal to 3. Therefore, a +1 ≤3 => a ≤2.But since a >0, the range of a is (0, 2].Wait, but let me confirm. The interval A is [1, a +1]. To be a subset of (-1,3], the entire interval [1, a +1] must lie within (-1,3]. Since 1 is already greater than -1, the only constraint is that a +1 ≤3. Because if a +1 is greater than 3, then A would include numbers greater than 3, which are not in B. Therefore, indeed, a +1 ≤3 => a ≤2. Since a >0, the range of a is (0,2].But let me check if there's any other constraints. For example, the interval [1, a +1] must not extend beyond 3. But if a +1 ≤3, then A is [1, a +1] which is within (-1,3], since 1 > -1 and a +1 ≤3. So, yes, the only constraint is a ≤2. Therefore, the range of a is (0,2].But wait, hold on. Let me check with an example. If a=2, then set A is [1,3]. Then A ∪ B would be [1,3] ∪ (-1,3] = (-1,3] which is equal to B. So that works. If a=3, which is beyond 2, then set A would be [1,4], and A ∪ B would be (-1,4], which is not equal to B. Therefore, a must be ≤2.Therefore, the answer for part (2) is that a must satisfy 0 < a ≤2.Wait, but let me check another edge case. If a approaches 0, say a=0.1, then set A is [1,1.1], which is within (-1,3]. So that's okay. If a is exactly 2, then A is [1,3], which is still within B since B is up to 3, inclusive. So 3 is included in B. Therefore, [1,3] is a subset of (-1,3], because 3 is in B.But wait, set B is (-1,3]. So 3 is included in B. Therefore, even if a=2, A=[1,3], which includes 3, which is in B. Therefore, the union A ∪ B is (-1,3] which is equal to B. So, correct.Therefore, the answer for part (2) is a ∈ (0,2].But let me make sure there are no mistakes here. Let me rephrase:Set A is [1, a+1]. Set B is (-1, 3].For A to be a subset of B, every element of A must be in B. Since A starts at 1, which is greater than -1, the lower bound is fine. The upper bound of A is a+1, which must be ≤3. Therefore, a+1 ≤3 => a ≤2. Since a>0, the range is 0 < a ≤2. Therefore, the answer is (0,2]. Yes, that seems correct.Therefore, summarizing:1) When a=1, set A is [1,2] and set B is (-1,3].2) The range of a is 0 < a ≤2.Final Answer(1) When (a=1), set (A) is (boxed{[1, 2]}) and set (B) is (boxed{(-1, 3]}).(2) The range of (a) is (boxed{(0, 2]}).</think>To solve the given problem, we need to analyze the domain of the function ( f(x) = sqrt{-x^2 + (a+2)x - a - 1} ) and the range of the function ( g(x) = 2^x - 1 ) for ( x leq 2 ). Part (1)When ( a = 1 ):1. Domain of ( f(x) ): - The expression inside the square root must be non-negative: [ -x^2 + (1+2)x - 1 - 1 geq 0 implies -x^2 + 3x - 2 geq 0 ] - Multiplying by -1 and solving the quadratic inequality: [ x^2 - 3x + 2 leq 0 implies (x - 1)(x - 2) leq 0 ] - The solution is ( 1 leq x leq 2 ). Thus, set ( A ) is ([1, 2]).2. Range of ( g(x) ): - The function ( g(x) = 2^x - 1 ) is an exponential function. For ( x leq 2 ): - As ( x to -infty ), ( 2^x to 0 ) and ( g(x) to -1 ). - At ( x = 2 ), ( g(2) = 2^2 - 1 = 3 ). - The range is all real numbers greater than -1 and up to 3, which is ((-1, 3]). Thus, set ( B ) is ((-1, 3]). Part (2)To find the range of ( a ) such that ( A cup B = B ):- Domain of ( f(x) ) for general ( a ): - Solve the inequality: [ -x^2 + (a + 2)x - a - 1 geq 0 implies x^2 - (a + 2)x + a + 1 leq 0 ] - The quadratic factors as ((x - 1)(x - (a + 1)) leq 0). The solution is ( 1 leq x leq a + 1 ). Thus, set ( A ) is ([1, a + 1]).- Condition for ( A cup B = B ): - Set ( A ) must be a subset of ( B ). Since ( B = (-1, 3] ), the upper bound of ( A ) must be ( leq 3 ): [ a + 1 leq 3 implies a leq 2 ] - Given ( a > 0 ), the range of ( a ) is ( (0, 2] ). Final Answers(1) When ( a = 1 ), set ( A ) is (boxed{[1, 2]}) and set ( B ) is (boxed{(-1, 3]}).(2) The range of ( a ) is (boxed{(0, 2]}).]

🔑:(1) When a=1, according to the problem we have -x^{2} + 3x - 2 geq 0, which can be rewritten as x^{2} - 3x + 2 leq 0.Solving the inequality, we get: (x - 1)(x - 2) leq 0 which yields the solution 1 leq x leq 2.Hence, the set A is [1, 2].For the function g(x) = 2^x - 1 (with x leq 2), it is monotonically increasing on the interval (-infty, 2].Therefore, we can calculate the range of g(x) as: -1 leq 2^x - 1 leq 2^2 - 1 = 3 which means the set B is (-1, 3].(2) Since we have A cup B = B, it follows that A subseteq B.From the condition -x^{2} + (a+2)x - a - 1 geq 0, we get: x^{2} - (a+2)x + a + 1 leq 0 factoring the quadratic expression gives us: (x - 1)(x - (a + 1)) leq 0 Solving the equation (x - 1)(x - (a + 1)) = 0, we find x_1 = 1 and x_2 = a + 1.Given that a > 0, the solution set for the inequality is [1, a+1], which means the set A is [1, a+1].In order for A subseteq B to hold, we must have: a + 1 leq 3 This leads us to: a leq 2 thus, the range of the real number a is boxed{0 < a leq 2}.