🧠:[<think>Okay, so I need to find the range of the real number λ such that the inequality λ a_n > b_n + 36(n - 3) + 3λ holds for all n ∈ N*. The given information includes two sequences {a_n} and {b_n} with a_n = 2b_n + 3, and the sum of the first n terms of {b_n} is S_n = (3/2)(3^n - 1). Hmm, let's break this down step by step.First, since S_n is the sum of the first n terms of {b_n}, I can find the general term b_n by using the formula for the nth term of a sequence in terms of its partial sums. The nth term b_n is equal to S_n - S_{n-1} for n ≥ 1, right? Wait, actually for n ≥ 2, and b_1 = S_1. Let me verify that. Yes, because the sum of the first n terms minus the sum of the first n-1 terms gives the nth term. So, for n ≥ 1, S_n = b_1 + b_2 + ... + b_n. Therefore, for n ≥ 2, b_n = S_n - S_{n-1}, and for n = 1, b_1 = S_1.So let's compute b_n. Given S_n = (3/2)(3^n - 1). Then, for n ≥ 2, b_n = S_n - S_{n-1} = (3/2)(3^n - 1) - (3/2)(3^{n-1} - 1). Let's compute that:= (3/2)(3^n - 1 - 3^{n-1} + 1)= (3/2)(3^n - 3^{n-1})= (3/2)(3^{n-1}(3 - 1))= (3/2)(2 * 3^{n-1})= 3 * 3^{n-1}= 3^nWait, so for n ≥ 2, b_n = 3^n. But what about n = 1? Let's check S_1:S_1 = (3/2)(3^1 - 1) = (3/2)(3 - 1) = (3/2)(2) = 3. Therefore, b_1 = 3. But according to the formula for n ≥ 2, if n=1, 3^1 = 3, which matches. Wait, so actually, does the formula b_n = 3^n hold for all n ≥ 1? Let's check:If n=1: 3^1 = 3, which equals S_1 = 3. So yes, b_n = 3^n for all n ≥ 1. Therefore, the general term of {b_n} is 3^n. Therefore, the sequence {b_n} is a geometric sequence with common ratio 3 and first term 3.So, since a_n = 2b_n + 3, then substituting b_n = 3^n, we have:a_n = 2*3^n + 3.Now, the inequality given is λ a_n > b_n + 36(n - 3) + 3λ. Let's substitute a_n and b_n with the expressions in terms of n.So, substituting:λ*(2*3^n + 3) > 3^n + 36(n - 3) + 3λLet me write this out:2λ*3^n + 3λ > 3^n + 36(n - 3) + 3λHmm, interesting. Let's subtract 3λ from both sides to simplify:2λ*3^n > 3^n + 36(n - 3)Then, we can factor out 3^n on the right-hand side:2λ*3^n - 3^n > 36(n - 3)Which is equivalent to:(2λ - 1)3^n > 36(n - 3)So, (2λ - 1)3^n > 36(n - 3)We can write this as:(2λ - 1) > [36(n - 3)] / 3^nTherefore,2λ - 1 > 36(n - 3) / 3^nHence,λ > [36(n - 3) / 3^n + 1] / 2But this inequality has to hold for all n ∈ N*. So, we need to find the range of λ such that λ is greater than [36(n - 3)/3^n + 1]/2 for all n. Therefore, the minimum value of λ that satisfies this for all n is the supremum (least upper bound) of the sequence [36(n - 3)/3^n + 1]/2 over all n ∈ N*.Therefore, we need to compute the maximum value of the expression f(n) = [36(n - 3)/3^n + 1]/2 for n ∈ N* and then λ must be greater than this maximum value.Alternatively, since λ must be greater than f(n) for all n, then λ must be greater than the maximum of f(n). So, the range of λ is (max{f(n)}, ∞). Therefore, we need to find the maximum of f(n) over n ∈ N*.So, let's analyze f(n):f(n) = [36(n - 3)/3^n + 1]/2We can split this into two parts:f(n) = 1/2 + [36(n - 3)/3^n]/2 = 1/2 + 18(n - 3)/3^nTherefore, f(n) = 1/2 + 18(n - 3)/3^nSo, we need to find the maximum of the function f(n) = 1/2 + 18(n - 3)/3^n over positive integers n.To find the maximum, we can consider f(n) as a function of a real variable and find its maximum, then check the integer points around it. Since the function is defined for positive integers, but the maximum might occur at some integer n.First, let's treat n as a real variable and find its critical points. Let’s denote x instead of n for real variable analysis.Define f(x) = 1/2 + 18(x - 3)/3^xWe can compute the derivative of f(x) with respect to x to find critical points.First, note that 3^x = e^{x ln 3}, so the derivative of 3^x is ln 3 * 3^x.Compute f'(x):f'(x) = d/dx [1/2 + 18(x - 3)/3^x]The derivative of 1/2 is 0. For the second term, use the quotient rule or product rule. Let's write it as 18(x - 3) * 3^{-x}Then, f'(x) = 18 [ d/dx (x - 3) * 3^{-x} + (x - 3) * d/dx 3^{-x} ]Wait, actually, better to use the product rule:d/dx [ (x - 3) * 3^{-x} ] = (1) * 3^{-x} + (x - 3) * (-ln 3) * 3^{-x}Therefore,f'(x) = 18 [ 3^{-x} - (x - 3)(ln 3)3^{-x ] = 18 * 3^{-x} [1 - (x - 3) ln 3]Set this derivative equal to zero:18 * 3^{-x} [1 - (x - 3) ln 3] = 0Since 18 * 3^{-x} is never zero, the critical points occur when:1 - (x - 3) ln 3 = 0=> (x - 3) ln 3 = 1=> x - 3 = 1 / ln 3=> x = 3 + 1 / ln 3Compute the numerical value of 1 / ln 3. Since ln 3 ≈ 1.0986, so 1 / ln 3 ≈ 0.9102. Therefore, x ≈ 3 + 0.9102 ≈ 3.9102.Therefore, the critical point is at x ≈ 3.91. Since n must be a positive integer, we need to check n=3 and n=4.But wait, n ∈ N*, which are positive integers starting at 1. So n=1,2,3,4,...But let's compute the values of f(n) for n=1,2,3,4,5,... and see where the maximum occurs.Compute f(n):f(n) = 1/2 + 18(n - 3)/3^nLet's compute for n=1:n=1:18(1 - 3)/3^1 = 18*(-2)/3 = -12f(1) = 1/2 -12 = -11.5n=2:18(2 - 3)/3^2 = 18*(-1)/9 = -2f(2) = 1/2 -2 = -1.5n=3:18(3 -3)/3^3 = 0f(3) = 1/2 + 0 = 0.5n=4:18(4 -3)/3^4 = 18*1 /81 = 18/81 = 2/9 ≈ 0.2222f(4) = 1/2 + 2/9 ≈ 0.5 + 0.2222 ≈ 0.7222n=5:18(5 -3)/3^5 = 18*2 / 243 = 36/243 = 4/27 ≈ 0.1481f(5) ≈ 0.5 + 0.1481 ≈ 0.6481n=6:18(6 -3)/3^6 = 18*3 /729 = 54/729 = 6/81 = 2/27 ≈ 0.07407f(6) ≈ 0.5 + 0.07407 ≈ 0.57407n=7:18*4 /3^7 = 72 / 2187 ≈ 0.03292f(7) ≈ 0.5 + 0.03292 ≈ 0.53292n=8:18*5 /3^8 = 90 / 6561 ≈ 0.0137f(8) ≈ 0.5 + 0.0137 ≈ 0.5137n=9:18*6 /3^9 = 108 / 19683 ≈ 0.00548f(9) ≈ 0.5 + 0.00548 ≈ 0.50548n=10:≈ 0.5 + even smaller, approaching 0.5 from above.So, the function f(n) starts at n=1 with -11.5, increases to -1.5 at n=2, 0.5 at n=3, then peaks at n=4 with approximately 0.7222, then decreases again towards 0.5 as n increases.So the maximum value of f(n) occurs at n=4, with f(4) ≈ 0.7222. Let's compute it exactly.Compute f(4):18*(4 -3)/3^4 = 18*1 /81 = 18/81 = 2/9.So f(4) = 1/2 + 2/9 = (9/18 + 4/18) = 13/18 ≈ 0.7222.Similarly, check n=4 and n=5:At n=4, 13/18 ≈ 0.7222At n=5: 1/2 + 4/27 = (27/54 + 8/54) = 35/54 ≈ 0.6481So yes, the maximum is at n=4 with 13/18. Let's confirm that n=4 is indeed the maximum. Since when we treated n as a real variable, the maximum was around x ≈ 3.91, which is between 3 and 4. Therefore, the maximum on integers would be at n=4. So, indeed, the maximum of f(n) is 13/18 at n=4.Therefore, the supremum of f(n) over all n ∈ N* is 13/18. Therefore, λ must be greater than 13/18. Wait, but let me check for n=4, whether there's any other n with higher f(n). From the computations above, n=4 is the highest. So the maximum is 13/18.But wait, let me check n=4 and n=3:At n=3, f(n) = 1/2 ≈ 0.5, which is less than 13/18 ≈ 0.7222. So yes, n=4 is the maximum.Therefore, since the inequality λ > f(n) must hold for all n, and the maximum of f(n) is 13/18, then λ must be greater than 13/18. Hence, the range of λ is (13/18, ∞). But let's check if this is correct.Wait, but the problem says "the range of the real number λ is ___". So perhaps they need an interval. But let's confirm.But wait, let's make sure that for all n, λ must be greater than [36(n - 3)/3^n + 1]/2. So, since the maximum value of this expression is 13/18, then λ must be greater than 13/18. Therefore, the range is λ > 13/18. Therefore, the range is (13/18, +∞).But let's check if this is correct with some values.For example, take λ = 13/18. Then, we need to check if the inequality holds for all n. Wait, but the problem says λ a_n > b_n + 36(n - 3) + 3λ. If λ = 13/18, then at n=4, we should have equality. Let's check:At n=4,Left-hand side: λ a_4 = (13/18)(2*3^4 + 3) = (13/18)(2*81 + 3) = (13/18)(165) = (13*165)/18 = (2145)/18 = 119.166...Right-hand side: b_4 + 36(4 - 3) + 3λ = 81 + 36(1) + 3*(13/18) = 81 + 36 + 13/6 = 117 + 2.166... = 119.166...So, equality holds at n=4 when λ = 13/18, but the problem states that the inequality must hold for all n ∈ N*. Therefore, λ must be strictly greater than 13/18. Hence, the range is λ > 13/18, so in interval notation, (13/18, ∞).But let's check for another n where f(n) is positive. For example, n=3, f(n)=0.5. So if λ=13/18 ≈ 0.722, which is greater than 0.5, so for n=3, we have:λ a_3 > b_3 + 36(3 - 3) + 3λLeft-hand side: 13/18*(2*27 + 3) = 13/18*(57) = (741)/18 = 41.166...Right-hand side: 27 + 0 + 3*(13/18) = 27 + 13/6 = 27 + 2.166... = 29.166...So, 41.166... > 29.166..., which holds.Similarly, for n=5:f(5) ≈ 0.6481, so λ=13/18 ≈0.722 is greater than 0.6481, so the inequality would hold.But at n=4, when λ=13/18, equality holds. Therefore, the strict inequality requires λ > 13/18.Therefore, the answer should be λ > 13/18, so the range is (13/18, ∞). But let me check if there's any other n where f(n) could be higher. For example, n=0? Wait, n ∈ N*, which is positive integers, starting at 1. So n=1,2,3,...We checked n=1 gives f(n)=-11.5, which is way lower, and the maximum at n=4. Hence, the conclusion is that the range of λ is all real numbers greater than 13/18.Therefore, the final answer is boxed{(dfrac{13}{18}, +infty)}.Final AnswerThe range of the real number (lambda) is boxed{left( dfrac{13}{18}, +infty right)}.</think>Given two sequences ({a_n}) and ({b_n}) satisfying (a_n = 2b_n + 3) for (n in mathbb{N}^*), and the sum of the first (n) terms of ({b_n}) is (S_n = frac{3}{2}(3^n - 1)), we need to find the range of the real number (lambda) such that (lambda a_n > b_n + 36(n - 3) + 3lambda) holds for all (n in mathbb{N}^*).First, we determine the general term (b_n) by using the partial sums (S_n):[b_n = S_n - S_{n-1} = frac{3}{2}(3^n - 1) - frac{3}{2}(3^{n-1} - 1) = 3^n quad text{for all } n geq 1.]Next, we substitute (b_n = 3^n) into the expression for (a_n):[a_n = 2 cdot 3^n + 3.]Substituting (a_n) and (b_n) into the given inequality:[lambda (2 cdot 3^n + 3) > 3^n + 36(n - 3) + 3lambda.]Simplifying this inequality:[2lambda cdot 3^n + 3lambda > 3^n + 36(n - 3) + 3lambda,][2lambda cdot 3^n > 3^n + 36(n - 3),][(2lambda - 1)3^n > 36(n - 3),][2lambda - 1 > frac{36(n - 3)}{3^n},][lambda > frac{1}{2} + frac{18(n - 3)}{3^n}.]We need to find the maximum value of the function (f(n) = frac{1}{2} + frac{18(n - 3)}{3^n}) over all positive integers (n). By analyzing the function, we find that the maximum value occurs at (n = 4):[f(4) = frac{1}{2} + frac{18(4 - 3)}{3^4} = frac{1}{2} + frac{18}{81} = frac{1}{2} + frac{2}{9} = frac{13}{18}.]Thus, the range of (lambda) must be greater than (frac{13}{18}). Therefore, the range of the real number (lambda) is:[boxed{left( dfrac{13}{18}, +infty right)}.,<think>Okay, let's try to tackle this problem step by step. First, let me read through it again to make sure I understand all the parts.We have two sequences, {a_n} and {b_n}, related by the equation a_n = 2b_n + 3 for all natural numbers n. The sum of the first n terms of {b_n} is given by S_n = (3/2)(3^n - 1). We need to find the range of the real number λ such that the inequality λ a_n > b_n + 36(n - 3) + 3λ holds for all natural numbers n.Alright, let's break this down. First, since we know the sum S_n of the first n terms of {b_n}, we can probably find an expression for b_n itself. Then, using the relation a_n = 2b_n + 3, we can express a_n in terms of b_n. Once we have expressions for a_n and b_n, we can substitute them into the inequality λ a_n > b_n + 36(n - 3) + 3λ and solve for λ. The key here is that the inequality must hold for all n in N*, so we need to find the values of λ that satisfy this inequality for every natural number n.Starting with the sum S_n. The sum of the first n terms of {b_n} is S_n = (3/2)(3^n - 1). To find b_n, we can use the fact that b_n = S_n - S_{n-1} for n ≥ 2, and b_1 = S_1. Let me compute that.First, let's compute S_1. When n=1, S_1 = (3/2)(3^1 - 1) = (3/2)(3 - 1) = (3/2)(2) = 3. Therefore, b_1 = 3.For n ≥ 2, b_n = S_n - S_{n-1} = (3/2)(3^n - 1) - (3/2)(3^{n-1} - 1) = (3/2)(3^n - 1 - 3^{n-1} + 1) = (3/2)(3^n - 3^{n-1}) = (3/2)(3^{n-1}(3 - 1)) = (3/2)(2*3^{n-1}) = 3*3^{n-1} = 3^n.Wait, so for n ≥ 2, b_n = 3^n. But for n=1, b_1 = 3, which is 3^1. So actually, maybe b_n = 3^n for all n ≥ 1? Let me check. When n=1, S_1 = 3, which is 3^1. If we take b_n = 3^n, then S_n would be the sum from k=1 to n of 3^k, which is a geometric series. The sum of a geometric series with first term 3 and ratio 3 is 3*(3^n - 1)/(3 - 1) = (3/2)(3^n - 1), which matches S_n. Therefore, yes, b_n = 3^n for all n ≥ 1. So that simplifies things. So {b_n} is a geometric sequence where each term is 3^n.Therefore, a_n = 2b_n + 3 = 2*3^n + 3. Got that.Now, substitute a_n and b_n into the inequality λ a_n > b_n + 36(n - 3) + 3λ.So substituting, we have:λ*(2*3^n + 3) > 3^n + 36(n - 3) + 3λLet me write this out:Left-hand side (LHS): λ*(2*3^n + 3)Right-hand side (RHS): 3^n + 36(n - 3) + 3λSo, moving all terms to one side to solve for λ:λ*(2*3^n + 3) - 3λ > 3^n + 36(n - 3)Factor λ on the left:λ*(2*3^n + 3 - 3) > 3^n + 36(n - 3)Simplify the left-hand side:λ*(2*3^n) > 3^n + 36(n - 3)So, 2*3^n * λ > 3^n + 36(n - 3)Divide both sides by 2*3^n (since 3^n is always positive, so division is allowed without flipping the inequality):λ > [3^n + 36(n - 3)] / (2*3^n)Simplify the right-hand side:[3^n / (2*3^n)] + [36(n - 3) / (2*3^n)] = 1/2 + [18(n - 3)/3^n]So, λ > 1/2 + [18(n - 3)/3^n]Therefore, for the inequality to hold for all n ∈ N*, λ must be greater than the maximum value of the expression 1/2 + 18(n - 3)/3^n. So we need to find the maximum of f(n) = 1/2 + 18(n - 3)/3^n over all natural numbers n ≥ 1, and then λ must be greater than this maximum.Therefore, the range of λ is (max_n f(n), ∞). So our task reduces to finding the maximum of f(n) for n ≥ 1.To find the maximum of f(n), let's analyze the function f(n) = 1/2 + 18(n - 3)/3^n.First, let's compute f(n) for some small n to get an idea.n=1:f(1) = 1/2 + 18*(1 - 3)/3^1 = 1/2 + 18*(-2)/3 = 1/2 - 12 = -11.5n=2:f(2) = 1/2 + 18*(2 - 3)/3^2 = 1/2 + 18*(-1)/9 = 1/2 - 2 = -1.5n=3:f(3) = 1/2 + 18*(3 - 3)/3^3 = 1/2 + 0 = 0.5n=4:f(4) = 1/2 + 18*(4 - 3)/3^4 = 1/2 + 18*(1)/81 = 1/2 + 18/81 = 1/2 + 2/9 = (9/18 + 4/18) = 13/18 ≈ 0.722n=5:f(5) = 1/2 + 18*(5 - 3)/3^5 = 1/2 + 18*2/243 = 1/2 + 36/243 = 1/2 + 12/81 = 1/2 + 4/27 ≈ 0.5 + 0.148 ≈ 0.648n=6:f(6) = 1/2 + 18*(6 - 3)/3^6 = 1/2 + 18*3/729 = 1/2 + 54/729 = 1/2 + 6/81 = 1/2 + 2/27 ≈ 0.5 + 0.074 ≈ 0.574n=7:f(7) = 1/2 + 18*(7 - 3)/3^7 = 1/2 + 18*4/2187 = 1/2 + 72/2187 ≈ 0.5 + 0.0329 ≈ 0.5329n=8:f(8) = 1/2 + 18*(8 - 3)/3^8 = 1/2 + 18*5/6561 ≈ 0.5 + 90/6561 ≈ 0.5 + 0.0137 ≈ 0.5137n=9:Similarly, it's going to get smaller. So, from n=3 onwards, the function f(n) increases to a maximum at n=4 and then decreases. Wait, but at n=3, it's 0.5, n=4 it's ~0.722, n=5 ~0.648, so it peaks at n=4. So the maximum is at n=4 with f(4) ≈ 0.722, which is 13/18. Let me verify that.Compute f(4) exactly:18*(4 - 3)/3^4 = 18*1/81 = 18/81 = 2/9. So 1/2 + 2/9 = 9/18 + 4/18 = 13/18 ≈ 0.7222.Similarly, check n=5:18*(5 - 3)/3^5 = 18*2/243 = 36/243 = 12/81 = 4/27 ≈ 0.1481. So 1/2 + 4/27 = 27/54 + 8/54 = 35/54 ≈ 0.6481.n=6:18*3/729 = 54/729 = 6/81 = 2/27 ≈ 0.074. 1/2 + 2/27 = 27/54 + 4/54 = 31/54 ≈ 0.574.So indeed, the function f(n) increases from n=1 (where it's negative) to n=4, reaching 13/18, then decreases afterwards. Therefore, the maximum value of f(n) for n ∈ N* is 13/18. Therefore, λ must be greater than 13/18. Therefore, the range of λ is (13/18, ∞). But let's check if there could be any higher values for n beyond n=4. Let's check n=0? Wait, n is in N*, so n starts at 1.But even if we considered n=0, which is not part of the problem, but just for completion, f(0) = 1/2 + 18*(-3)/1 = 1/2 - 54 = -53.5. So it's even lower.Therefore, the maximum occurs at n=4 with f(4)=13/18. Thus, the required range for λ is λ > 13/18.But let me confirm if there's another n where f(n) might be higher. For n=4, as we saw, it's 13/18. Let me check n=4.5 just in case, but n must be integer. Since n is discrete, the maximum is indeed at n=4.Alternatively, if we treated n as a continuous variable, we could take the derivative of f(n) with respect to n and find the critical point. Let's try that to see if the maximum occurs around n=4.Let’s define f(n) = 1/2 + 18(n - 3)/3^n. Let’s treat n as a real variable. Then, take the derivative with respect to n:f'(n) = 18 * [ (1)*3^n - (n - 3)*3^n ln3 ] / (3^n)^2Wait, no. Wait, f(n) = 1/2 + 18(n - 3)/3^n. So derivative is:f'(n) = 18 * [ (1)*3^n - (n - 3)*3^n ln3 ] / (3^n)^2 ?Wait, no. Let's compute f(n) as 1/2 + 18(n - 3)/3^n. Then, derivative is:d/dn [18(n - 3)/3^n] = 18 * [ (1)/3^n + (n - 3)(-ln3)/3^n ] = 18/3^n [1 - (n - 3)ln3]Set derivative to zero:18/3^n [1 - (n - 3)ln3] = 0Since 18/3^n is never zero, we set 1 - (n - 3)ln3 = 0So,(n - 3)ln3 = 1n - 3 = 1/ln3n = 3 + 1/ln3 ≈ 3 + 1/1.0986 ≈ 3 + 0.910 ≈ 3.910So the critical point is around n ≈ 3.91. Since n must be an integer, we check n=4 and n=3. But we already saw that at n=4, f(n)=13/18≈0.722, and at n=3, f(n)=0.5. So indeed, the maximum is at n≈3.91, which is between 3 and 4. Since n must be integer, the maximum is at n=4. Therefore, the analysis holds.Therefore, the maximum value of f(n) is 13/18, so λ must be greater than 13/18.But let's check for n=4. Is 13/18 exactly? Let's compute f(4):18*(4 - 3)/3^4 = 18*1/81 = 2/9. Then, 1/2 + 2/9. Convert to eighteenths: 9/18 + 4/18 = 13/18. Yes, that's correct. So 13/18 is exactly the value at n=4.Therefore, since the inequality must hold for all n ∈ N*, λ must be greater than 13/18. Hence, the range of λ is (13/18, ∞).But wait, just to be thorough, let's check n=4 and n=5 with λ=13/18 to ensure that the inequality holds as equality at n=4 and fails for larger λ? Wait, no, if λ is greater than the maximum, then the inequality should hold for all n. Let me verify with λ=13/18.Wait, at λ=13/18, the inequality becomes:(13/18)a_n > b_n + 36(n - 3) + 3*(13/18)But a_n = 2*3^n + 3, so:Left-hand side: (13/18)(2*3^n + 3) = (13/18)(2*3^n) + (13/18)*3 = (13/9)*3^n + 13/6Right-hand side: 3^n + 36(n - 3) + 39/18 = 3^n + 36(n - 3) + 13/6So subtract RHS from LHS:(13/9)*3^n + 13/6 - 3^n - 36(n - 3) - 13/6 = (13/9 - 1)*3^n - 36(n - 3)Compute 13/9 - 1 = 4/9, so:(4/9)*3^n - 36(n - 3)We need this to be greater than 0 for the inequality to hold. But at n=4:(4/9)*3^4 - 36(4 - 3) = (4/9)*81 - 36*1 = 4*9 - 36 = 36 - 36 = 0So equality holds at n=4 when λ=13/18. Therefore, for the inequality to be strict (i.e., >), λ must be greater than 13/18. Therefore, the range is λ > 13/18, so the interval is (13/18, ∞).Therefore, the answer should be boxed{(dfrac{13}{18}, +infty)}.Final AnswerThe range of the real number (lambda) is boxed{left( dfrac{13}{18}, +infty right)}.</think>Given two sequences ({a_n}) and ({b_n}) satisfying (a_n = 2b_n + 3) for all (n in mathbb{N}^*), and the sum of the first (n) terms of ({b_n}) given by (S_n = frac{3}{2}(3^n - 1)), we need to find the range of the real number (lambda) such that the inequality (lambda a_n > b_n + 36(n - 3) + 3lambda) holds for all (n in mathbb{N}^*).First, we determine the general term (b_n). The sum (S_n) of the first (n) terms of ({b_n}) is a geometric series with the first term 3 and common ratio 3. Therefore, the (n)-th term (b_n) is:[b_n = S_n - S_{n-1} = frac{3}{2}(3^n - 1) - frac{3}{2}(3^{n-1} - 1) = 3^n]Thus, (b_n = 3^n) for all (n in mathbb{N}^*). Using the relation (a_n = 2b_n + 3), we find:[a_n = 2 cdot 3^n + 3]Substituting (a_n) and (b_n) into the inequality (lambda a_n > b_n + 36(n - 3) + 3lambda), we get:[lambda(2 cdot 3^n + 3) > 3^n + 36(n - 3) + 3lambda]Simplifying, we rearrange terms to isolate (lambda):[lambda(2 cdot 3^n) > 3^n + 36(n - 3)]Dividing both sides by (2 cdot 3^n), we obtain:[lambda > frac{1}{2} + frac{18(n - 3)}{3^n}]We need to find the maximum value of the expression (frac{1}{2} + frac{18(n - 3)}{3^n}) for (n in mathbb{N}^*). Analyzing the function (f(n) = frac{1}{2} + frac{18(n - 3)}{3^n}) for different values of (n), we find that the maximum value occurs at (n = 4), yielding:[f(4) = frac{1}{2} + frac{18(4 - 3)}{3^4} = frac{1}{2} + frac{18}{81} = frac{1}{2} + frac{2}{9} = frac{13}{18}]Thus, the range of (lambda) must be greater than (frac{13}{18}). Therefore, the range of the real number (lambda) is:[boxed{left( dfrac{13}{18}, +infty right)}]]

🔑:From S_n= frac {3}{2}(3^{n}-1), we get b_1=S_1= frac {3}{2}(3-1)=3.When ngeqslant 2, b_n=S_n-S_{n-1}= frac {3}{2}(3^{n}-1)- frac {3}{2}(3^{n-1}-1)=3^{n}.When n=1, the above equation holds, thus b_n=3^{n}.Substituting into a_n=2b_n+3, we get a_n=2cdot 3^{n}+3.Substituting into lambda a_n > b_n+36(n-3)+3lambda, we get lambda(a_n-3) > b_n+36(n-3),which means 2lambdacdot 3^{n} > 3^{n}+36(n-3),so lambda > frac {1}{2}+ frac {18(n-3)}{3^{n}}.Since frac {18(n-2)}{3^{n+1}}- frac {18(n-3)}{3^{n}}= frac {126-36n}{3^{n+1}} > 0, we have nleqslant 3.Therefore, when n=4, frac {1}{2}+ frac {18(n-3)}{3^{n}} reaches its maximum value of frac {13}{18}.Hence, the answer is: boxed{left( frac {13}{18},+inftyright)}.By deriving b_n from the sum of the first n terms of {b_n} being S_n= frac {3}{2}(3^{n}-1), and further obtaining a_n, and then substituting a_n, b_n into lambda a_n > b_n+36(n-3)+3lambda, isolating lambda, and then finding the maximum value of the function with respect to n, we get the answer.This problem examines the recursive formula of sequences and explores the functional characteristics of sequences, training the method of solving problems that always hold true, making it a medium-level question.

🧠:[<think>Okay, let's see. So the problem is about an infantry column that's 1 km long. Sergeant Kim is on a gyro-scooter and starts from the end of the column, goes to the beginning, and then comes back to the end. During the time he does this, the infantrymen have walked 4/3 km. The question is, how far did Sergeant Kim travel during this time?Hmm. Alright, first, let's parse the problem. The column is 1 km long. So when Sergeant Kim starts, the column is moving, right? Wait, infantrymen are walking, so the column is moving. So the entire column is moving forward, which means that both the front and the end of the column are moving. But when Sergeant Kim starts from the end, he has to catch up to the front, which is moving away from him. Then after reaching the front, he turns around and comes back to the end, which is now moving towards him as well. So this seems like a relative speed problem.So in problems like this, where two objects are moving, you have to consider their speeds relative to each other. Let me recall. If two objects are moving towards each other, their relative speed is the sum of their individual speeds. If they're moving in the same direction, the relative speed is the difference.So here, the infantry column is moving at some speed, let's call it v. Sergeant Kim is on a gyro-scooter, so his speed is different, let's call that u. But we don't know either of these speeds. But we do know that during the time Sergeant Kim goes from the end to the front and back, the infantrymen have moved 4/3 km. So the total time taken for Sergeant Kim's trip is the same as the time it takes the infantry column to walk 4/3 km. So if we can find the time, we can relate it to Sergeant Kim's distance, which is u multiplied by that time.Wait, but maybe we need to set up equations here. Let's denote:Let’s assume the speed of the infantry column is v km/h (or km/min, the units might not matter as they cancel out). Then, the speed of Sergeant Kim is u km/h. But since we don't know either, we might need to express u in terms of v, or vice versa.But here's the thing: when Sergeant Kim is moving from the end to the front, he has to cover the length of the column, which is 1 km, but during that time, the front of the column is moving away from him. So the relative speed when he's moving towards the front would be u - v, since they're moving in the same direction. Wait, no: if the column is moving at speed v, and Sergeant Kim is moving at speed u in the same direction, then his speed relative to the column would be u - v. But actually, when moving towards the front, he needs to cover the 1 km length plus whatever distance the front moves during the time he's traveling. Wait, maybe I need to think differently.Wait, perhaps it's better to model this as moving towards a moving target. So when Sergeant Kim starts at the end of the column, the front is 1 km ahead. But the front is moving forward at speed v, while Sergeant Kim is moving towards it at speed u. So the distance between them is closing at a rate of (u - v). So the time it takes for him to reach the front would be the initial distance divided by the closing speed: time1 = 1 km / (u - v).Then, after reaching the front, he turns around and goes back to the end. Now, the end is moving towards him as well. The end was originally at the starting point, but during the time he was going to the front, the column has moved forward. Wait, actually, the entire column is moving, so when he reaches the front, the end has moved forward by some distance. But now, when he turns around to go back, he's moving against the direction of the column. So now, his speed relative to the column would be u + v, since they're moving towards each other. The distance he needs to cover to get back to the end is again the length of the column, which is still 1 km, because the column maintains its length while moving, right? Wait, no, the column is moving, so when he gets to the front, the entire column has moved forward, so the end is now further ahead. Wait, but actually, the column's length is maintained. So if it's an infantry column, they are marching in formation, so the length remains 1 km regardless of their movement. So even though they are moving forward, the column is always 1 km long. So when Sergeant Kim reaches the front, the end has moved forward, but the front has also moved forward, so the length remains 1 km. Therefore, when he turns around, the distance he needs to cover to get back to the end is still 1 km. But in this case, the end is moving towards him at speed v, and he is moving towards the end at speed u. So their closing speed is u + v. Therefore, the time to return would be 1 km / (u + v).Therefore, the total time for the round trip is time1 + time2 = 1/(u - v) + 1/(u + v). During this total time, the infantry column has moved 4/3 km. Since the column is moving at speed v, the time total_time = (4/3)/v. Therefore, we have the equation:1/(u - v) + 1/(u + v) = (4/3)/v.We need to solve for u in terms of v, and then compute the total distance Sergeant Kim traveled, which is u * total_time.Alternatively, maybe we can find u in terms of v, and then compute the distance. Let's work through this.First, let's compute the left-hand side (LHS) of the equation:1/(u - v) + 1/(u + v) = [ (u + v) + (u - v) ] / [ (u - v)(u + v) ] = [2u]/(u² - v²).So LHS = 2u/(u² - v²).Set this equal to (4/3)/v:2u/(u² - v²) = (4/3)/v.Cross-multiplying:2u * 3v = 4(u² - v²)6uv = 4u² - 4v²Let's bring all terms to one side:4u² - 6uv - 4v² = 0.Hmm, this is a quadratic in terms of u. Let's divide both sides by 2 to simplify:2u² - 3uv - 2v² = 0.Now, let's solve for u. Let's treat this as a quadratic equation:2u² - 3uv - 2v² = 0.Let me use the quadratic formula. Let u be the variable, so:u = [3v ± sqrt( ( -3v )² - 4*2*(-2v²) ) ]/(2*2)Calculate discriminant:(3v)^2 - 4*2*(-2v²) = 9v² + 16v² = 25v².So,u = [3v ± 5v]/4.So two solutions:u = (3v + 5v)/4 = 8v/4 = 2v,oru = (3v - 5v)/4 = (-2v)/4 = -v/2.Since speed can't be negative, we discard the negative solution. So u = 2v.Therefore, Sergeant Kim's speed is twice the speed of the infantry column.Therefore, the total time total_time is (4/3)/v.But we need to find the distance Sergeant Kim traveled, which is u * total_time = 2v * (4/3)/v = 8/3 km.Wait, so the answer would be 8/3 km, which is 2 and 2/3 km. But let me verify this.Wait, but let's check if this makes sense.If the column is moving at speed v, and Sergeant Kim's speed is 2v, then when he goes from the end to the front, his relative speed is 2v - v = v. So time1 = 1 km / v.Then, when he returns, his relative speed is 2v + v = 3v, so time2 = 1 km / 3v.Total time is time1 + time2 = 1/v + 1/(3v) = (4)/(3v).During this time, the infantry column moved v * total_time = v * (4)/(3v) = 4/3 km, which matches the problem statement. Therefore, the total distance Sergeant Kim traveled is 2v * (4)/(3v) = 8/3 km. So that checks out.Therefore, the answer is 8/3 km, or 2 and 2/3 km. So boxed, that would be boxed{dfrac{8}{3}}.But let me think again if there's another way to approach this, maybe by considering reference frames or something else.Alternatively, suppose we take the perspective of the infantry column. In their frame of reference, they are stationary, and the earth is moving beneath them at speed v. But Sergeant Kim is moving towards the front at speed u - v and then back at speed u + v. But this is similar to the previous approach. The distance he has to cover is still 1 km each way, since in their frame, the column is 1 km long. So time1 = 1/(u - v), time2 = 1/(u + v), same as before.But in the earth frame, during the total time, the column has moved 4/3 km. So the time is 4/3 divided by v. So same equation. So we get the same result.Alternatively, maybe think about the distances. The total distance Kim travels is d = u * t, where t is total time. The infantrymen walked 4/3 km, so t = (4/3)/v. Therefore, d = u * (4/3)/v. But we found that u = 2v, so substituting, d = 2v * (4/3)/v = 8/3. Same result.Therefore, the answer should be 8/3 km. Seems consistent.Wait, another sanity check. If the sergeant's speed is twice the column's speed. So when going to the front, his relative speed is v, so takes 1/v hours. Then coming back, relative speed 3v, takes 1/(3v) hours. Total time 4/(3v). In that time, the column moves 4/3 km, which is v * 4/(3v) = 4/3. So correct. Then sergeant's distance is 2v * 4/(3v) = 8/3. Makes sense.Therefore, I think the answer is 8/3 km.Final Answerboxed{dfrac{8}{3}}</think>Given an infantry column stretching 1 km, Sergeant Kim rides a gyro-scooter from the end of the column to the beginning and returns. During this time, the infantrymen walked ( frac{4}{3} ) km. We need to determine the distance Sergeant Kim traveled.1. Define variables: - Let ( v ) be the speed of the infantry column. - Let ( u ) be the speed of Sergeant Kim.2. Relative speeds: - When moving towards the front (same direction), the relative speed is ( u - v ). - When returning to the end (opposite direction), the relative speed is ( u + v ).3. Time calculations: - Time to reach the front: ( frac{1}{u - v} ) - Time to return to the end: ( frac{1}{u + v} ) - Total time: ( frac{1}{u - v} + frac{1}{u + v} )4. Relate total time to infantry movement: - The total time is also the time the infantrymen walked ( frac{4}{3} ) km: ( frac{4/3}{v} )5. Set up the equation: [ frac{1}{u - v} + frac{1}{u + v} = frac{4}{3v} ] Simplifying the left-hand side: [ frac{2u}{u^2 - v^2} = frac{4}{3v} ] Cross-multiplying and solving the quadratic equation: [ 6uv = 4u^2 - 4v^2 implies 2u^2 - 3uv - 2v^2 = 0 ] Solving for ( u ) using the quadratic formula: [ u = frac{3v pm 5v}{4} ] The positive solution is ( u = 2v ).6. Calculate total distance traveled by Sergeant Kim: - Total time: ( frac{4/3}{v} ) - Distance: ( u times text{total time} = 2v times frac{4}{3v} = frac{8}{3} )Thus, the distance Sergeant Kim traveled is (boxed{dfrac{8}{3}}) km.,<think>Okay, so I need to figure out how far Sergeant Kim traveled on his gyro-scooter. Let's start by understanding the problem step by step.First, there's an infantry column that's 1 km long. So imagine a line of soldiers stretching from point A to point B, where the distance between A and B is 1 km. Sergeant Kim starts at the end of the column, which I assume is point B, and rides his gyro-scooter to the beginning, point A, and then comes back to the end, point B again. During the time he does this, the entire column of infantrymen has moved forward 4/3 km. So the question is, how far did Sergeant Kim travel during this time?Hmm, okay. So both the sergeant and the infantry column are moving. Wait, but when it says the infantrymen walked 4/3 km during this time, does that mean that the entire column moved 4/3 km forward? So their speed is such that in the time it took Sergeant Kim to go from B to A and back to B, they moved 4/3 km. So the column has a certain speed, let's call it v_infantry, and Sergeant Kim has a speed v_sergeant on his gyro-scooter.But here's the thing: when Sergeant Kim is moving from the end (B) to the beginning (A), he's moving against the direction of the infantry column's march. So relative to the ground, his speed would be his speed minus the infantry's speed, right? Because he's going against them. But when he returns from A back to B, he's moving in the same direction as the infantry column, so his relative speed would be his speed plus the infantry's speed. Wait, no, hold on. Wait, actually, if the infantry column is moving forward, then the entire column is moving in a certain direction. Let's say the infantry is moving from point A towards point B? Wait, but the column is stretched from A to B, which is 1 km. If they are marching, they must be moving in a direction. Let me clarify.Assuming the infantry column is moving forward, so from A to B direction. Then Sergeant Kim starts at the end of the column, which would be at the back, point B. To reach the beginning of the column, point A, he needs to go against the direction of the march. So while he's moving towards A, the entire column is moving forward, so point A is moving away from him, and point B is also moving forward. Therefore, his trip from B to A is against the movement of the column, and his trip back from A to B is with the movement of the column.Wait, but actually, when the column is moving forward, the beginning of the column is at the front, so if they're moving from A to B direction, then the front is at A initially, moving towards some point further ahead. But the column is 1 km long. So if the entire column moves forward 4/3 km during the total time, then the front (A) has moved 4/3 km, and the end (B) has also moved 4/3 km. So the column remains 1 km in length but has advanced 4/3 km. So the original positions: let's say at time t=0, the column is from position x=0 (A) to x=1 km (B). At time t=T, when the sergeant has finished his trip, the column is from x=4/3 km (A') to x=4/3 +1 km = 7/3 km (B'). So the entire column has moved forward 4/3 km.Therefore, during the time T, the infantrymen have moved 4/3 km. So their speed is v_infantry = (4/3 km) / T.Now, Sergeant Kim starts at the original end, B (1 km), goes to the original beginning A (0 km), but by the time he reaches A, the entire column has moved forward. Wait, no. Wait, actually, the movement of the column is continuous. So when Sergeant Kim is moving towards A, the column is moving forward, so the distance he needs to cover to reach the beginning is not just 1 km, but actually more, because the beginning is moving away from him. Similarly, when he returns from A to B, the end is moving towards him, so the distance he needs to cover is less.Wait, this is a classic relative speed problem. Let me formalize this.Let’s denote:- Length of the column: L = 1 km- Distance infantrymen walked during the entire time: S = 4/3 km- Therefore, time taken for the entire trip (to A and back to B) is T = S / v_infantry = (4/3) / v_infantryBut we need to find T in terms of the sergeant's speed. Wait, maybe we need to set up equations based on relative speeds.Let’s assume the speed of the infantry column is v (km/h), and the speed of Sergeant Kim is u (km/h). We need to find the total distance Sergeant Kim traveled, which would be u * T, where T is the total time.But we know that during time T, the infantry moved 4/3 km, so:v * T = 4/3 => T = (4/3) / vSo if we can find T in terms of u and v, we can relate them.But how?First, let's analyze Sergeant Kim's trip from the end to the beginning (against the direction of the column's movement).When moving against the column, his effective speed relative to the column is u + v. Wait, no. Wait, if the column is moving at speed v, and he's moving against them, his speed relative to the ground is u - v? Wait, no. Wait, confusion here.Wait, actually, if the column is moving at speed v, then from the ground perspective, the beginning of the column (which is moving away from the starting position) is at position x = v * t. The end of the column is at x = v * t + 1 km (since the column is 1 km long). Wait, no. Wait, if the column is moving forward, then the front is moving at speed v, and the entire column is moving at speed v, so the end is also moving at speed v. Therefore, the distance between the front and the end remains 1 km. So at any time t, the front is at position v * t, and the end is at position v * t - 1 km? Wait, no, that would make the column length negative. Wait, maybe coordinate system.Let me set up a coordinate system where at t=0, the front of the column (A) is at position 0 km, and the end of the column (B) is at position 1 km. Then, as the column moves forward at speed v, at time t, the front is at position v * t, and the end is at position v * t + 1 km. Wait, that can't be. If the entire column is moving forward, the front moves from 0 to v * t, and the end moves from 1 km to 1 km + v * t. So the column is stretching? But the problem says the infantry column stretched out to 1 km. So perhaps the column length remains 1 km. Therefore, if the column is moving forward, both the front and the end move at speed v, maintaining the 1 km distance. So at time t, front is at v * t, end is at v * t + 1 km. Wait, but then the column is moving in the positive x-direction, with the front at v * t and the end trailing 1 km behind. Wait, but in reality, if the column is moving forward, the front leads, and the end follows, maintaining the same distance. So if the front moves from 0 to v * t, then the end moves from 1 km to 1 km + v * t. Wait, that would mean the column is moving forward and stretching, but the problem states that the column is stretched out to 1 km. So perhaps the length remains 1 km. So actually, the column is a rigid body moving forward, maintaining 1 km length. Therefore, if the front moves at speed v, the entire column moves at speed v. Therefore, at time t, the front is at position v * t, and the end is at position v * t - 1 km. Wait, that makes sense. So at t=0, front at 0, end at -1 km. Then moving forward, at t=1, front at v*1, end at v*1 -1. But in the problem, the initial position is that the column is stretched out to 1 km. Maybe the coordinate system is such that at t=0, the end is at 0, and the front is at 1 km. Then, as time progresses, both move forward. So at t=0: end (B) at 0, front (A) at 1 km. At time t, end (B) is at v * t, front (A) is at 1 km + v * t. So the column is moving forward, maintaining its length of 1 km. So Sergeant Kim starts at the end, which at t=0 is at position 0. He needs to reach the front, which at t=0 is at 1 km. But as he moves towards the front, the front is moving away at speed v, and the end is also moving forward at speed v. So his movement relative to the ground is his speed u minus the speed of the column? Wait, no. If he is moving towards the front (which is moving away), his relative speed with respect to the column is u - v? Wait, maybe we need to think in terms of relative velocity.When moving from the end to the front, Sergeant Kim's speed relative to the column (which is moving at v) is u - v. Wait, no. If the column is moving at v, and he is moving towards the front (against the direction of the column's movement), then his speed relative to the ground is u - v. Wait, but if the column is moving in the positive direction at speed v, and he is moving towards the front (which is in the positive direction), then his speed relative to the ground would be u + v? Wait, no, that doesn't make sense. Wait, perhaps I need to clarify directions.Let me define the positive direction as the direction the infantry is moving. So the column is moving in the positive direction. Sergeant Kim starts at the end (B) and wants to reach the front (A). Since the column is moving forward, the front is moving away from him. So to catch up to the front, he needs to move in the positive direction faster than the column's speed. Therefore, his speed relative to the ground is u (positive direction), while the column is moving at v (positive direction). So the relative speed of Sergeant Kim with respect to the column is u - v. But the distance he needs to cover is the length of the column, which is 1 km. Wait, but the column is moving, so the distance isn't fixed? Wait, no. The column is 1 km long, regardless of its movement. So even though the entire column is moving forward, the distance between the front and the end remains 1 km. So Sergeant Kim needs to cover that 1 km distance, but since the front is moving away, his relative speed with respect to the front is u - v. Therefore, the time taken to go from B to A is t1 = 1 / (u - v).Then, when he turns around and goes back from A to B, now he's moving in the negative direction relative to the column. His speed relative to the ground is now u (but in the negative direction), so his speed relative to the column is u + v (since the column is moving in the positive direction). The distance is still 1 km, so the time taken to return is t2 = 1 / (u + v).Therefore, the total time T = t1 + t2 = 1/(u - v) + 1/(u + v).But during this total time T, the infantry column moved 4/3 km. Since the infantry's speed is v, then:v * T = 4/3So,v * [1/(u - v) + 1/(u + v)] = 4/3Simplify the left side:v * [ (u + v + u - v) / ( (u - v)(u + v) ) ] = 4/3That becomes:v * [ (2u) / (u² - v²) ) ] = 4/3So,(2u v) / (u² - v²) = 4/3Multiply both sides by (u² - v²):2u v = (4/3)(u² - v²)Bring all terms to one side:2u v - (4/3)u² + (4/3)v² = 0Multiply both sides by 3 to eliminate denominators:6u v - 4u² + 4v² = 0Divide both sides by 2:3u v - 2u² + 2v² = 0Let’s rearrange:-2u² + 3u v + 2v² = 0Multiply both sides by -1:2u² - 3u v - 2v² = 0Now, this is a quadratic equation in terms of u. Let's solve for u.Let’s treat this as 2u² - 3v u - 2v² = 0Using quadratic formula:u = [3v ± sqrt( (3v)^2 + 16v² ) ] / 4Compute discriminant:(3v)^2 + 16v² = 9v² + 16v² = 25v²Therefore,u = [3v ± 5v] / 4So two solutions:u = (3v + 5v)/4 = 8v/4 = 2voru = (3v - 5v)/4 = (-2v)/4 = -v/2Since speed can't be negative, we discard the negative solution. Therefore, u = 2v.So Sergeant Kim's speed is twice the speed of the infantry column.Now, the total distance he traveled is u * T. We know T is the total time, which is (4/3)/v, as established earlier.So total distance = 2v * (4/3)/v = 8/3 km.Wait, so the answer is 8/3 km? Let me verify this.Wait, let's check with relative speeds.If u = 2v, then going from B to A, his relative speed is u - v = 2v - v = v. So time t1 = 1 km / v.Coming back from A to B, his relative speed is u + v = 2v + v = 3v. So time t2 = 1 km / 3v.Total time T = t1 + t2 = 1/v + 1/(3v) = (3 + 1)/(3v) = 4/(3v). Then, since the infantry moves at speed v, the distance they cover is v * T = v * 4/(3v) = 4/3 km, which matches the problem statement. So that checks out.Therefore, Sergeant Kim's total distance is u * T = 2v * 4/(3v) = 8/3 km. So yes, 8/3 km, which is approximately 2.666... km.So the answer should be 8/3 km, boxed as boxed{dfrac{8}{3}}.Final AnswerThe sergeant traveled a distance of boxed{dfrac{8}{3}} kilometers.</think>Given an infantry column stretching 1 km, Sergeant Kim rides a gyro-scooter from the end of the column to the beginning and returns. During this time, the infantrymen walked ( frac{4}{3} ) km. We need to determine the distance Sergeant Kim traveled.1. Define Variables: - Let ( v ) be the speed of the infantry column. - Let ( u ) be the speed of Sergeant Kim. - The total time taken for Sergeant Kim's trip is ( T ).2. Time Calculation: - The infantry column moves ( frac{4}{3} ) km, so ( vT = frac{4}{3} ). Thus, ( T = frac{4}{3v} ).3. Relative Speeds: - When moving towards the beginning (against the column's direction), Sergeant Kim's relative speed is ( u - v ). Time taken: ( frac{1}{u - v} ). - When returning to the end (with the column's direction), his relative speed is ( u + v ). Time taken: ( frac{1}{u + v} ).4. Total Time Equation: - Total time ( T = frac{1}{u - v} + frac{1}{u + v} ).5. Solve for ( u ): - Substitute ( T = frac{4}{3v} ) into the total time equation: [ frac{4}{3v} = frac{1}{u - v} + frac{1}{u + v} ] - Simplify to find ( u = 2v ).6. Distance Traveled by Sergeant Kim: - Total distance is ( uT ). Substituting ( u = 2v ) and ( T = frac{4}{3v} ): [ text{Distance} = 2v times frac{4}{3v} = frac{8}{3} text{ km} ]The distance Sergeant Kim traveled is (boxed{dfrac{8}{3}}) kilometers.]

🔑:Answer: 8 / 3 kmSolution. Let the speed of the column be x km/h, and the corporal rides k times faster, i.e., at a speed of k x km/h. Kim rode to the end of the column for t_{1}=frac{1}{k x-x} hours (catching up), and in the opposite direction for t_{2}=frac{1}{k x+x} hours (meeting head-on). During this time, the column covered 4 / 3 km, i.e., xleft(t_{1}+t_{2}right)=4 / 3. Substituting the expressions for t_{1} and t_{2}, we getfrac{x}{k x-x}+frac{x}{k x+x}=4 / 3 ; quad frac{1}{k-1}+frac{1}{k+1}=4 / 3 ; quad 2 k=frac{4}{3}left(k^{2}-1right)The resulting quadratic equation has a single positive root k=2. The corporal rides twice as fast as the column. Therefore, the distance he covers will also be twice as much.Evaluation. Full solution: 12 points.

🧠:[<think>Alright, let me try to figure out this problem. So, we have two sets, P and Q. P is {x, y, z} and Q is {1, 2, 3}. We need to find the number of mappings (functions) from P to Q such that f(y) = 2. The options are A: 2, B: 4, C: 6, D: 9. Hmm, okay.First, I remember that a function from set P to set Q assigns each element in P to exactly one element in Q. So, for each element in P, we have to choose an image in Q. Normally, the total number of such functions would be 3^3 = 27, since there are 3 choices for each of the 3 elements in P. But here, we have a restriction: f(y) must equal 2. So, y's image is fixed, but the other two elements, x and z, can still be mapped freely, right?Let me break it down. The function f has to map x, y, z to Q. But f(y) is fixed as 2. So, for element y, there's only 1 choice. For the elements x and z, there are no restrictions, so each can be mapped to any of the 3 elements in Q. Therefore, the number of possible functions should be the product of the number of choices for each element. Since y has 1 choice, x has 3, and z has 3. So, total mappings would be 1 * 3 * 3 = 9? Wait, but 9 is option D. But wait, the answer options are A:2, B:4, C:6, D:9. Hmm.Wait, maybe I'm miscalculating. Let me check again. For each element in the domain P, you choose where it maps to in Q. If f(y) must be 2, then for y, we have only 1 choice. For x and z, since there's no restriction, each can be mapped to 1, 2, or 3. So, for x: 3 choices, for z: 3 choices. Therefore, total number of functions is 3 * 3 = 9. But 9 is one of the options, D. But why are the other options there? Maybe the question is a trick question?Wait, but maybe I'm misunderstanding something. Let me think again. A mapping f: P → Q assigns every element of P to some element in Q. So, each of x, y, z must be assigned. If y must be assigned to 2, then x and z can each be assigned to 1, 2, or 3. So, each of x and z has 3 choices. So, the total number of mappings is 3 * 1 * 3 = 9. So, D is the answer. But why is the answer not 9? The options include D:9, so maybe that's the correct one.But wait, let me check again. Maybe there's a restriction I missed. The problem says "mappings f: P → Q that satisfy f(y) = 2." There's no other condition. So, yes, each mapping is determined by where it sends x, y, and z. Since y must go to 2, but x and z can go anywhere. So, the number of such mappings is the number of choices for x times the number of choices for z. Since y is fixed. So, 3 choices for x, 3 for z. 3*3=9. So, answer D:9.But wait, I recall sometimes in combinatorics, when you fix one element, the total number is 3^(n-1) if you have n elements. Here, n is 3 elements in P. So, 3^(3-1)=9. So, that formula also gives 9. So, that seems correct. Therefore, D:9 is the answer.Wait, but the options are listed as A:2, B:4, C:6, D:9. So, D is the last option. So, perhaps the answer is D. But let me see why someone might think it's 6. Maybe if they thought that once y is fixed to 2, the other elements can't map to 2? But the problem doesn't say that. It just requires f(y)=2. There's no restriction on f(x) or f(z). So, they can still map to 1, 2, or 3. So, 3*3=9. Therefore, D is correct.Alternatively, if the problem had stated that f(y)=2 and f is injective (one-to-one), then the answer would be different. Because if f is injective, then once f(y)=2, f(x) and f(z) can't be 2. So, then f(x) and f(z) would have 2 and 1 choices respectively, leading to 2*1=2, but that's not the case here. The problem doesn't mention injectivity. So, the answer should indeed be 9.Therefore, the correct answer is D:9.Final Answerboxed{D}</think>Given sets ( P = {x, y, z} ) and ( Q = {1, 2, 3} ), we need to find the number of mappings ( f: P rightarrow Q ) that satisfy ( f(y) = 2 ).A function from ( P ) to ( Q ) assigns each element in ( P ) to exactly one element in ( Q ). Since ( f(y) = 2 ) is fixed, the element ( y ) has only 1 choice. The elements ( x ) and ( z ) can be mapped to any of the 3 elements in ( Q ). Therefore, for each of ( x ) and ( z ), there are 3 choices.The total number of such mappings is calculated by multiplying the number of choices for each element:- For ( y ): 1 choice- For ( x ): 3 choices- For ( z ): 3 choicesThus, the total number of mappings is ( 1 times 3 times 3 = 9 ).The correct answer is (boxed{D}).,<think>Okay, so I need to figure out how many mappings there are from set P to set Q where the condition is that f(y) = 2. The sets are given as P = {x, y, z} and Q = {1, 2, 3}. The options are A: 2, B: 4, C: 6, D: 9. Hmm, let me think.First, I remember that a mapping from set P to set Q means that each element in P (which are x, y, z) is assigned to an element in Q (which are 1, 2, 3). Normally, without any restrictions, the number of possible mappings would be 3^3 = 27, since each of the three elements in P can map to any of the three elements in Q. But here, there's a condition: f(y) must equal 2. So that restricts the mapping for y, but not for x and z.Let me break it down. The mapping f: P → Q requires that for each element in P, we specify an element in Q. Since y has to map to 2, that means f(y) is fixed. Therefore, the only choices we have are for f(x) and f(z). For each of these, they can be mapped to any of the elements in Q, which is {1, 2, 3}. So how many choices do we have for x and z?For f(x), since there are 3 elements in Q, there are 3 choices. Similarly, for f(z), there are also 3 choices. Since these choices are independent of each other, the total number of mappings would be the product of the number of choices for each. So that's 3 (for x) multiplied by 3 (for z) which is 3*3 = 9. Wait, but 9 is option D. But the answer options given here are A:2, B:4, C:6, D:9. So does that mean D is the answer?Wait, let me verify. Maybe I made a mistake here. Let me think again. If we fix f(y) = 2, then y is already mapped, so we don't have any choice for y. The other elements x and z can be mapped freely to any of the 3 elements in Q. Therefore, for each of x and z, 3 choices, so total mappings would be 3*3=9. So that would be D:9. But let me check if the problem is perhaps looking for something else, like injective mappings or something else. Wait, the problem just says "mappings", which in standard terminology are functions, not necessarily injective or surjective. So in that case, yes, it's just 3 choices for x, 3 choices for z, so 9 total functions. Therefore, the answer should be D:9.Wait, but the options given are A:2, B:4, C:6, D:9. So D is there. Hmm. Let me think again. Maybe I made a mistake in considering how many elements are in P and Q. P has three elements: x, y, z. Q has three elements: 1, 2, 3. Since y must map to 2, that leaves x and z. Each can map to 1, 2, or 3. So for x: 3 choices, z:3 choices. So 3*3=9. So yes, D is correct. Hmm. Maybe the answer is D. But let me check with an example. If instead, the problem was asking for mappings where f(y)=2 and f(x)=f(z), then the number would be 3, since x and z have to map to the same element in Q, which could be 1, 2, or 3. But here, there's no such restriction. So x and z can be mapped independently. So that's 3*3=9. Therefore, D is the correct answer. So why is the option C:6 present? Maybe if someone thought that after fixing f(y)=2, there are two elements left (x and z), each with 3 choices, but subtract some constraints. Wait, but there's no constraints given. Wait, no. Unless there's a misunderstanding. Wait, maybe the problem was phrased as "the number of mappings f: P → Q that satisfy f(y) = 2 and are injective" or something else. But the problem as given just says "mappings" which are functions. So injective would be different, but here it's not specified. So unless the problem is from a non-English source where "mappings" might imply something else, but in standard math terminology, a mapping is a function. So in that case, 3*3=9. Therefore, answer D.Wait, let me check the original problem again. It says "the number of mappings f: P → Q that satisfy f(y) = 2". So mappings here are functions. Therefore, each element of P must be assigned to exactly one element of Q. Since f(y) is fixed at 2, the other two elements x and z can be assigned to any of the three elements in Q. Therefore, for x: 3 choices, for z: 3 choices, total 9. So answer D. So I think that is correct.Alternatively, maybe the options are mistyped? But given the options, D is 9. So that's the answer. Therefore, the correct choice is D.Final Answerboxed{D}</think>Given sets ( P = {x, y, z} ) and ( Q = {1, 2, 3} ), we need to determine the number of mappings ( f: P rightarrow Q ) that satisfy ( f(y) = 2 ).1. Understanding the problem: A mapping from ( P ) to ( Q ) means each element in ( P ) is assigned to an element in ( Q ). The condition ( f(y) = 2 ) fixes the image of ( y ) to 2, leaving the images of ( x ) and ( z ) to be determined.2. Choices for remaining elements: The elements ( x ) and ( z ) can each be mapped to any of the three elements in ( Q ). Therefore, there are 3 choices for ( f(x) ) and 3 choices for ( f(z) ).3. Calculating the total number of mappings: The total number of mappings is the product of the number of choices for ( x ) and ( z ): [ 3 text{ (choices for } x) times 3 text{ (choices for } z) = 9 ]Thus, the number of mappings ( f: P rightarrow Q ) that satisfy ( f(y) = 2 ) is (boxed{D}).]

🔑:Consider the sets P={x, y, z} and Q={1, 2, 3}. To find mappings f: P rightarrow Q that satisfy f(y) = 2, we need to construct a mapping f: P rightarrow Q where the other two elements x and z in set P each have a unique image in set Q. x can correspond to any one of the three elements in set Q, which gives 3 ways of correspondence, Similarly, z can also correspond to any one of the three elements in set Q, which also gives 3 ways of correspondence, By the principle of multiplication, the total number of mappings f: P rightarrow Q that satisfy f(y) = 2 is 3 times 3 = 9 (mappings). Therefore, the correct answer is: boxed{text{D}}. From the concept of mapping, to construct a mapping f: P rightarrow Q, each element in set P needs to have a unique image in set Q, given that f(y) = 2. The unique images for elements x and z in Q are determined, and then the solution is found by applying the principle of multiplication. This question examines the concept of mapping, focusing on understanding the concept of mapping and utilizing the principle of multiplication to simplify and clarify the solution. It is a basic question.

🧠:[<think>Okay, so I need to figure out how many license plates in the city are between 10000 and 99999, inclusive, that meet two conditions: they have at least one digit that is 9, and the sum of their digits is a multiple of 9. Hmm, let's break this down step by step.First, let's understand the total number of license plates. Since they go from 10000 to 99999, that's 99999 - 10000 + 1 = 90000 plates. Got that part. Now, we need to find how many of these have at least one 9 and have digits summing to a multiple of 9.This feels like a problem where inclusion-exclusion might be useful. Because the two conditions are "at least one 9" and "sum of digits is a multiple of 9," maybe we can compute the total number of plates where the sum is a multiple of 9 and then subtract those that don't have any 9s but still have a sum divisible by 9. That makes sense. So, total plates with sum divisible by 9 minus plates with sum divisible by 9 and no 9s equals the answer.Let me formalize that. Let A be the set of all plates with sum of digits divisible by 9. Let B be the set of all plates with at least one 9. Then, we need |A ∩ B|. By inclusion-exclusion, |A ∩ B| = |A| + |B| - |A ∪ B|, but wait, actually, maybe a different approach. Since we want plates that are in both A and B, which is |A| - |A ∩ not B|. Because if you take all plates in A and subtract those that are in A but not in B (i.e., plates in A with no 9s), you get the ones in A that are also in B. Yeah, that seems right. So:Number of desired plates = |A| - |A ∩ not B|So first, let's compute |A|, the number of plates from 10000 to 99999 where the sum of digits is divisible by 9.Since the plates are 5-digit numbers, each plate can be considered as a number from 10000 to 99999. Alternatively, since leading zeros are not allowed, the first digit is from 1 to 9, and the remaining four digits are from 0 to 9.To compute the number of such numbers with digit sums divisible by 9, we can use the concept that, for numbers in a range, the number of them with digit sums congruent to a particular value modulo 9 is roughly 1/9 of the total, assuming uniform distribution. But since we have a specific range here (10000 to 99999), which is all 5-digit numbers except those starting with 0, we need to check if the count is exactly 1/9 of the total.But wait, the total number is 90000. If 90000 is divisible by 9, then each residue class modulo 9 would have exactly 10000 numbers. Let's check: 90000 / 9 = 10000. So, if the numbers are uniformly distributed modulo 9, then |A| = 10000. But is that actually the case? Hmm.Wait, actually, when dealing with numbers from N to M inclusive, the number of numbers congruent to each residue modulo k is either floor((M - r)/k) - floor((N - 1 - r)/k) for each residue r. But in our case, since we are dealing with digit sums, it's a different story. The digit sum modulo 9 is equivalent to the number itself modulo 9 (because of the divisibility rule for 9). However, here we are considering the sum of the digits, not the number itself. But since a number is congruent to its digit sum modulo 9, the digit sum being divisible by 9 is equivalent to the number itself being divisible by 9. Therefore, the number of 5-digit numbers (from 10000 to 99999) divisible by 9 is the same as the number of numbers in that range with digit sums divisible by 9.So, how many numbers between 10000 and 99999 are divisible by 9?The first number in that range divisible by 9 is 10008 (since 10000 ÷ 9 = 1111.111..., so next integer is 1112*9=10008). The last number is 99999, which is 99999 ÷ 9 = 11111, so divisible by 9. The count is ((99999 - 10008)/9) + 1 = (89991)/9 + 1 = 9999 + 1 = 10000. So exactly 10000 numbers. Therefore, |A| = 10000.Alright, so |A| is 10000. Now, we need |A ∩ not B|, which is the number of plates with digit sum divisible by 9 and no digits equal to 9. Then, the answer is |A| - |A ∩ not B| = 10000 - |A ∩ not B|.So, now we need to compute |A ∩ not B|, the number of 5-digit numbers from 10000 to 99999 with no digits equal to 9 and digit sum divisible by 9.To compute this, we can model it as a problem of counting the number of 5-digit numbers where each digit is from 0-8 (since no 9s allowed), except the first digit which is from 1-8 (since it can't be 0 or 9), such that the sum of the digits is divisible by 9.This is a classic stars and bars problem with constraints, but since we have a modulus condition, generating functions might be useful here.Alternatively, we can use dynamic programming to compute the number of such numbers. Let's consider the digits as positions 1 to 5, where position 1 is the first digit (1-8), and positions 2-5 are the other digits (0-8). We need the sum of all digits to be congruent to 0 modulo 9.But maybe generating functions is the way to go. Let's think about the generating function for each digit.For the first digit (d1), possible digits are 1-8, so the generating function is x^1 + x^2 + ... + x^8.For the other four digits (d2 to d5), each can be 0-8, so their generating functions are each 1 + x + x^2 + ... + x^8.The total generating function is then (x + x^2 + ... + x^8) * (1 + x + ... + x^8)^4.We need the coefficient of x^{9k} in this generating function, summed over all k such that 9k is within the possible range of sums.The minimum possible sum is 1 (from 10000) and the maximum is 8 + 8*4 = 8 + 32 = 40. So possible sums are from 1 to 40. The multiples of 9 in this range are 9, 18, 27, 36.Therefore, we need the sum of the coefficients of x^9, x^{18}, x^{27}, x^{36} in the generating function (x + x^2 + ... + x^8)*(1 + x + ... + x^8)^4.But calculating these coefficients directly might be tedious, but perhaps we can use the roots of unity filter method.Since we need the sum of coefficients where the exponent is congruent to 0 mod 9, we can use the standard formula:Sum_{k} a_{9k} = (1/9) * Σ_{j=0}^8 G(ω^j), where ω is a primitive 9th root of unity.Here, G(x) is the generating function (x + x^2 + ... + x^8)*(1 + x + ... + x^8)^4.But this seems complex, but manageable.First, note that:x + x^2 + ... + x^8 = x*(1 - x^8)/(1 - x)Similarly, 1 + x + ... + x^8 = (1 - x^9)/(1 - x)Therefore, G(x) = [x*(1 - x^8)/(1 - x)] * [(1 - x^9)/(1 - x)]^4 = x*(1 - x^8)*(1 - x^9)^4 / (1 - x)^5Therefore, G(x) = x*(1 - x^8)*(1 - x^9)^4 / (1 - x)^5We need to evaluate G(ω^j) for j = 0,1,...,8 and sum them up, then divide by 9.First, let's compute G(1). However, ω^0 = 1, so let's compute G(1):But (1 - x^9) becomes 0 when x=1, so (1 - x^9)^4 is 0. Similarly, (1 - x^8) is (1 -1) = 0. Therefore, G(1) is 0. So the term for j=0 is 0.Therefore, Sum_{k} a_{9k} = (1/9) * Σ_{j=1}^8 G(ω^j)Now, let's compute G(ω^j) = ω^j * (1 - (ω^j)^8) * (1 - (ω^j)^9)^4 / (1 - ω^j)^5But note that ω^9 = 1, since ω is a primitive 9th root. Therefore, (ω^j)^9 = (ω^9)^j = 1^j = 1. Therefore, (1 - (ω^j)^9) = 0. Therefore, the entire expression becomes 0 for all j=1,...,8.Wait, that can't be right. Wait, no, G(ω^j) would be ω^j * (1 - ω^{8j}) * (1 - ω^{9j})^4 / (1 - ω^j)^5. But ω^{9j} = (ω^9)^j = 1^j = 1. Therefore, (1 - ω^{9j}) = 0. So G(ω^j) is 0 for all j. Therefore, Sum_{k} a_{9k} = 0? That can't be.But clearly, there are numbers with digits summing to multiple of 9 without any 9s, so the count can't be zero. So where did I go wrong?Wait, perhaps because when we expanded G(x), we have (1 - x^9)^4, which when evaluated at x = ω^j, becomes (1 - (ω^j)^9)^4 = (1 - 1)^4 = 0. Therefore, all terms except when (1 - x^9) is canceled out would be zero. But since (1 - x^9) is a factor in G(x), all terms G(ω^j) would be zero. Therefore, the sum is zero? That contradicts the fact that such numbers exist.This suggests that the roots of unity filter might not be directly applicable here because the generating function has factors that vanish at the roots of unity. Hmm. Maybe there's another approach.Alternatively, perhaps we can model this problem using the principle that the number of numbers with digits d1, d2, d3, d4, d5 (d1 from 1-8, others from 0-8) with sum congruent to 0 mod 9 is equal to (total numbers without any 9s) divided by 9, plus some correction term. But since the total number of such numbers is (8 choices for d1) * (9 choices for d2) * ... * (9 choices for d5) = 8 * 9^4 = 8 * 6561 = 52488.If 52488 is divisible by 9, then 52488 / 9 = 5832. So if the numbers are evenly distributed modulo 9, then there would be 5832 numbers for each residue. However, 52488 divided by 9 is indeed 5832, so that suggests that |A ∩ not B| is 5832.But wait, this would be the case only if the numbers are uniformly distributed modulo 9. Is that true?But since we're dealing with digits from 0-8 (except the first digit from 1-8), the distribution might not be perfectly uniform. Wait, but for numbers with digits chosen uniformly, the sum modulo 9 is also uniform. But is that the case here?Hmm. Let's think. The first digit is from 1-8, and the rest from 0-8. Each digit contributes to the total sum. Since the digits are in different positions, their contributions are different, but modulo 9, the position doesn't matter. For example, 10 ≡ 1 mod 9, 100 ≡ 1 mod 9, etc. Wait, but in the sum of digits, each digit is just added as is. So the sum modulo 9 is the same as the number modulo 9.But if we fix the digits, then the sum mod 9 is equivalent to the number mod 9. So, for numbers from 10000 to 99999, the sum of digits mod 9 is the same as the number mod 9.But in our restricted case (numbers with no digits 9), does the number mod 9 still cover all residues equally?In other words, if we consider numbers composed of digits 0-8 (with first digit 1-8), does each residue class mod 9 have the same number of numbers?If yes, then |A ∩ not B| would be 52488 / 9 = 5832.But why would that be the case? For the full set of numbers (with digits allowed to be 0-9), the numbers are uniformly distributed modulo 9 because 10 ≡ 1 mod 9, so each digit position contributes a weight of 1, so the sum modulo 9 is equivalent to the number modulo 9, and as numbers are consecutive, they are uniformly distributed modulo 9. However, in our case, we are restricting digits to 0-8. So, does this restriction affect the uniformity?Wait, for numbers with digits 0-8, first digit 1-8, the mapping from the number to its residue mod 9 is still a surjective (onto) function? If yes, then each residue would have the same number of pre-images. But is that true?Suppose we fix the first four digits (d1, d2, d3, d4), which can be from 1-8 followed by three 0-8 digits, then the fifth digit d5 can be chosen such that the total sum is congruent to any desired residue modulo 9. Since d5 can be 0-8, which covers all residues mod 9, for any fixed sum of the first four digits, we can choose d5 to make the total sum congruent to any residue. Therefore, for each combination of d1, d2, d3, d4, there is exactly one d5 that will make the total sum congruent to a specific residue mod 9. Therefore, the number of numbers with sum congruent to 0 mod 9 is equal to the number of possible d1, d2, d3, d4, which is 8 * 9 * 9 * 9 = 5832. Which matches 52488 / 9 = 5832.Therefore, this suggests that the numbers with no 9s and digit sum divisible by 9 is indeed 5832.Therefore, |A ∩ not B| = 5832.Therefore, the number we want is |A| - |A ∩ not B| = 10000 - 5832 = 4168.Wait, so is the answer 4168?But let me verify this with a smaller example to ensure that the logic holds.Suppose we consider 2-digit numbers from 10 to 99, excluding those with 9. So numbers from 10 to 88 where neither digit is 9. Total numbers: 8 (first digit) * 9 (second digit) = 72. Total numbers with digit sum divisible by 9: 72 / 9 = 8. Let's check:Numbers from 10 to 88 with no 9s and digit sum divisible by 9.Possible numbers:18 (1+8=9), 27 (2+7=9), 36 (3+6=9), 45 (4+5=9), 54 (5+4=9), 63 (6+3=9), 72 (7+2=9), 81 (8+1=9). That's 8 numbers, which is 72 / 9 = 8. So the logic works here.If we include numbers with 9s, then total 2-digit numbers from 10 to 99 is 90 numbers. Numbers with digit sum divisible by 9: 90 / 9 = 10. The numbers would be 18, 27, 36, 45, 54, 63, 72, 81, 90, 99. So 10 numbers. Then, numbers without 9s are 8, so numbers with at least one 9 and sum divisible by 9 would be 10 - 8 = 2. Which are 90 and 99. Indeed, 90 has a 9, and 99 has two 9s. So that checks out.Therefore, the same logic seems to hold. Therefore, in the original problem, the answer should be 10000 - 5832 = 4168.But let's verify once again.Total numbers from 10000 to 99999: 90000. Numbers with digit sum divisible by 9: 10000. Numbers with no 9s: 8 * 9^4 = 52488. Numbers with no 9s and digit sum divisible by 9: 52488 / 9 = 5832. Therefore, numbers with at least one 9 and digit sum divisible by 9: 10000 - 5832 = 4168.Yes, that seems consistent.But let me check another angle. Suppose we model the license plates as 5-digit numbers where the first digit is from 1-9 and the rest from 0-9. The total is 9*10^4 = 90000. Numbers with digit sum divisible by 9: 10000 as computed before.Numbers with no 9s: first digit 1-8 (8 choices), others 0-8 (9 choices each). So 8*9^4 = 52488. Numbers among these with digit sum divisible by 9: 52488 /9 = 5832.Thus, the answer is 10000 - 5832 = 4168.Therefore, the answer should be boxed{4168}Final Answerboxed{4168}</think>To determine the number of license plates from 10000 to 99999 that contain at least one digit 9 and have a sum of digits divisible by 9, we use the principle of inclusion-exclusion.1. Total number of license plates: 90000 (from 10000 to 99999 inclusive).2. Number of plates with sum of digits divisible by 9: - The numbers from 10000 to 99999 include 10000 numbers divisible by 9, as each residue class modulo 9 is uniformly distributed. - Therefore, (|A| = 10000).3. Number of plates with no digit 9: - The first digit can be 1-8 (8 choices), and the remaining four digits can be 0-8 (9 choices each). - Total numbers: (8 times 9^4 = 52488).4. Number of plates with no digit 9 and sum of digits divisible by 9: - These numbers are uniformly distributed modulo 9. - Therefore, (|A cap text{not } B| = frac{52488}{9} = 5832).5. Number of plates with at least one digit 9 and sum of digits divisible by 9: - Using inclusion-exclusion: (|A| - |A cap text{not } B| = 10000 - 5832 = 4168).Thus, the final answer is (boxed{4168}).,<think>Okay, so the problem is about license plates in a city that are numbered consecutively from 10000 to 99999. We need to find how many of these plates have at least one digit that is 9 and also have the sum of their digits as a multiple of 9. First, let me make sure I understand the problem correctly. The license plates are 5-digit numbers starting from 10000 up to 99999. That gives a total of 90000 plates because 99999 - 10000 + 1 = 90000. We need to count how many of these numbers satisfy two conditions:1. The number contains at least one digit 9.2. The sum of the digits is a multiple of 9.So, the challenge is to count numbers that meet both conditions. Let me think about how to approach this. One way to tackle such problems is to use combinatorics and perhaps the principle of inclusion-exclusion. Since we need numbers where the sum of digits is divisible by 9 and at least one digit is 9, maybe we can first find all numbers with digit sums divisible by 9, and then subtract those that have digit sums divisible by 9 but no 9s at all. That should give us the count of numbers that have both properties. Let me formalize this. Let S be the set of all license plates (numbers from 10000 to 99999). Let A be the subset of S where the sum of digits is divisible by 9. Let B be the subset of S that contains at least one digit 9. We need the size of the intersection of A and B, which is |A ∩ B|. By the principle of inclusion-exclusion, |A ∩ B| = |A| - |A ∩ B^c|, where B^c is the complement of B, i.e., numbers with no digit 9.So, if I can compute |A|, the total numbers with digit sum divisible by 9, and subtract |A ∩ B^c|, the numbers with digit sum divisible by 9 and no digits 9, then I will have the answer.First, let's compute |A|. How many 5-digit numbers from 10000 to 99999 have a digit sum divisible by 9?This is similar to counting the number of solutions to the equation d1 + d2 + d3 + d4 + d5 ≡ 0 mod 9, where d1 is from 1 to 9 (since the first digit cannot be 0), and d2, d3, d4, d5 are from 0 to 9. In combinatorics, the number of such numbers should be approximately 1/9 of all possible numbers, since the digit sums modulo 9 are equally likely. However, since the first digit cannot be 0, the distribution might not be perfectly uniform. But given that the range is large (from 10000 to 99999), which includes 90000 numbers, and 90000 is divisible by 9 (90000 / 9 = 10000), maybe the number of numbers with digit sum divisible by 9 is exactly 10000. Wait, let me check.Wait, 90000 divided by 9 is 10000. So if the distribution is uniform, then there are 10000 numbers with digit sum divisible by 9. But is that actually true?I recall that for numbers from 00000 to 99999 (all 5-digit numbers with leading zeros allowed), the number of numbers with digit sum congruent to 0 mod 9 is exactly 100000 / 9 ≈ 11111.111. But 100000 isn't divisible by 9. Hmm, so maybe my initial assumption is wrong.Wait, maybe when the numbers are from 10000 to 99999, which is 90000 numbers, and 90000 is divisible by 9, then the number of numbers with digit sum divisible by 9 is exactly 90000 / 9 = 10000. But is that a valid approach?Alternatively, we can model this as a generating function. The generating function for digit sums where the first digit is from 1 to 9 and the rest are from 0 to 9 is:G(x) = (x^1 + x^2 + ... + x^9)(1 + x + x^2 + ... + x^9)^4We need the coefficient of x^{9k} in this generating function. The sum of these coefficients can be found using roots of unity. Specifically, using the discrete Fourier transform at the 9th roots of unity.The formula for the number of solutions is (1/9) * Σ_{j=0}^{8} G(ω^j) where ω is a primitive 9th root of unity.But computing this might be complicated. However, if the distribution is uniform, then the number would be (Total numbers) / 9. Since the total numbers here is 9*10^4 (90000), and 90000 is divisible by 9, if the generating function is symmetric across all residues mod 9, then the number of numbers with digit sum ≡0 mod9 would be 10000.But is there symmetry? Let's think. The first digit is from 1 to 9, which is 9 possibilities, and the rest are 0-9. The first digit contributes a term from 1 to 9, which modulo 9 is 1 to 0 (since 9 ≡0 mod9). So the first digit can be considered as contributing 0 to 8 mod9 (since digits 1-9 correspond to 1-9 mod9, but 9 is 0). The other digits contribute 0-9 mod9. So if the generating function is such that each residue is equally likely, then the total numbers would be equally distributed. However, the first digit's contribution is 1-9 (mod9), which is 1-8 and 0. So maybe the distribution is indeed uniform. If so, then |A| = 10000.Alternatively, perhaps the distribution is not exactly uniform, but since 90000 is divisible by 9, and the number of numbers with each residue is the same. Let me check with a smaller example.Suppose we have numbers from 10 to 99 (two-digit numbers). Total numbers: 90. 90 is divisible by 9. How many have digit sums divisible by 9?The numbers are 10 to 99. The digit sums divisible by 9 are 18, 9. But the maximum digit sum is 9 + 9 = 18. So numbers with digit sum 9: 18, 27, 36, 45, 54, 63, 72, 81, 90, 99. Wait, but numbers from 10 to 99. Let's list them:Digit sum 9: 18 (1+8=9), 27, 36, 45, 54, 63, 72, 81, 90. So that's 9 numbers. Then digit sum 18: 99. So total 10 numbers. Wait, 10 numbers. But 90 /9 =10. So that works. So in this case, the number of numbers with digit sums divisible by 9 is 10, which is 90 /9=10. So that seems correct. Therefore, for two-digit numbers from 10 to 99, it works. So perhaps the same applies here. So for our problem, 90000 numbers, so |A| = 90000 /9=10000.Therefore, assuming uniformity, |A|=10000.Now, we need to compute |A ∩ B^c|, which is the number of numbers with digit sum divisible by 9 and no digits equal to 9. Then subtract this from |A| to get our desired count.So let's compute |A ∩ B^c|. These are numbers from 10000 to 99999 with no digits equal to 9 and digit sum divisible by 9.First, let's model such numbers. Each digit can be from 0-8 except the first digit, which is from 1-8 (since it can't be 0 or 9). Wait, the first digit can be 1-8, and digits 2-5 can be 0-8. So we need to count 5-digit numbers where d1 ∈ {1-8}, d2,d3,d4,d5 ∈ {0-8}, and d1 + d2 + d3 + d4 + d5 ≡0 mod9.So how many such numbers are there?Again, we can model this using generating functions. Let me define the generating function for the first digit: G1(x) = x^1 + x^2 + ... + x^8 = x(1 - x^8)/(1 - x)And for the other digits: G2(x) = 1 + x + x^2 + ... + x^8 = (1 - x^9)/(1 - x)So the total generating function is G1(x) * G2(x)^4We need the coefficient of x^{9k} in this generating function. Again, using roots of unity, the number of such numbers is (1/9) * Σ_{j=0}^{8} G1(ω^j) * G2(ω^j)^4 where ω is a primitive 9th root of unity.But again, perhaps if the distribution is uniform, the number would be (number of such numbers) /9. The total number of numbers with no digits 9 is 8 * 9^4. Let's compute that. First digit: 8 choices (1-8), other digits: 9 choices each (0-8). So total numbers: 8 * 9^4 = 8 * 6561 = 52488. Then 52488 divided by 9 is 5832. So if the distribution is uniform, then |A ∩ B^c| = 5832. So then the desired answer would be 10000 -5832=4168. But is this correct?Wait, similar to the previous example, maybe the distribution is uniform. Let me check with a smaller case. Suppose we have 2-digit numbers with no digits 9. So first digit 1-8, second digit 0-8. Total numbers: 8*9=72. Then how many have digit sums divisible by 9? 72 /9=8. Let's verify. The numbers would be from 10 to 88, excluding any with 9. The digit sums divisible by 9: 18 (but maximum digit sum is 8+8=16), so possible digit sums are 9. Let's list them: 18 (no), 9: numbers where d1 + d2 =9. Since d1 from 1-8 and d2 from 0-8. So possible pairs: (1,8), (2,7), (3,6), (4,5), (5,4), (6,3), (7,2), (8,1). That's 8 numbers. So 8, which is 72 /9=8. So yes, in this case, it's uniform. So perhaps the same applies here. Therefore, |A ∩ B^c|=52488 /9=5832.Therefore, the answer would be |A| - |A ∩ B^c| =10000 -5832=4168.But let me verify this with another approach to make sure.Alternatively, we can model this problem using stars and bars with congruence conditions.For numbers with no 9s, each digit d1, d2, d3, d4, d5 is between 0 and 8 (but d1 is between 1 and 8). Let’s think of the sum d1 + d2 + d3 + d4 + d5 ≡0 mod9.The number of solutions to this equation is equivalent to the number of solutions in the specified ranges. Since each digit is restricted, we can use generating functions or inclusion-exclusion.But this might get complicated, but since we already have the example where the total numbers without 9s is 8*9^4=52488, and if we assume uniform distribution modulo 9, then the number of solutions is 52488 /9=5832. This seems reasonable given the previous example. So perhaps this is correct.Therefore, the answer is 10000 -5832=4168.But let me think again. Let's check with the first approach. For numbers with digits from 1-8 (first digit) and 0-8 (others), the generating function for the first digit is x + x^2 + ... +x^8. For the other digits, it's 1 +x +x^2 + ... +x^8. So the generating function is (x + x^2 + ... +x^8)*(1 +x + ... +x^8)^4.We need the coefficient of x^{9k} in this expansion. Using roots of unity, the number of such numbers is (1/9)*Σ_{j=0}^8 G(ω^j), where ω is a primitive 9th root of unity.Compute G(ω^j):First, note that for each j, G(ω^j) = [ω^j*(1 - ω^{8j}) / (1 - ω^j)] * [(1 - ω^{9j}) / (1 - ω^j)]^4.But ω^9 =1, so ω^{9j}=1^j=1. Therefore, (1 - ω^{9j}) =0 for all j not divisible by 9. But j ranges from 0 to8, so when j=0, ω^{0}=1. For other j, (1 - ω^{9j})=0. Wait, but actually ω^{9j} = (ω^9)^j =1^j=1. So (1 - ω^{9j})=0 for all j. Wait, but this seems contradictory. Wait, no. Wait, G2(x)=1 +x + ... +x^8 = (1 -x^9)/(1 -x). Therefore, G2(ω^j)= (1 - (ω^j)^9)/(1 - ω^j) = (1 - ω^{9j})/(1 - ω^j). But since ω is a 9th root of unity, ω^9=1, so ω^{9j}=1 for any j. Therefore, G2(ω^j)=0 for any j≠0. For j=0, G2(1)= (1 -1^9)/(1 -1)=0/0, which is undefined. But we can compute it as the limit as x approaches 1, which is 9.Similarly, G1(x)=x +x^2 + ... +x^8 =x*(1 -x^8)/(1 -x). So G1(ω^j)= ω^j*(1 - ω^{8j})/(1 - ω^j).Therefore, for the generating function G(x)=G1(x)*G2(x)^4. So when evaluating at ω^j:For j=0:G1(1)= (1 +1 + ... +1) =8 (since x=1, G1(1)=1+1+...+1=8)G2(1)=9So G(1)=8 *9^4=8*6561=52488For j≠0:G2(ω^j)=0, so G(ω^j)=G1(ω^j)*0=0Therefore, the sum over all j=0 to8 of G(ω^j) is 52488. Therefore, the number of numbers is (1/9)*52488=5832. So this confirms our previous result. Therefore, |A ∩ B^c|=5832.Therefore, the answer is 10000 -5832=4168. So the number of license plates that contain at least one digit 9 and have the sum of their digits as a multiple of 9 is 4168.But wait, let me check once more. The total numbers with digit sum divisible by 9 is 10000. From these, 5832 do not have any 9s, so the remaining 10000 -5832=4168 must have at least one 9. Therefore, the answer should be 4168. But let me think if there could be a mistake here. For example, when we subtract |A ∩ B^c| from |A|, we are assuming that all numbers in A ∩ B^c are indeed numbers without any 9s. Which is correct because B^c is the complement of B, which is numbers without any 9s. Therefore, the subtraction is valid.Alternatively, we could approach the problem by considering numbers with at least one 9 and digit sum divisible by 9. But this seems more complex, as we'd have to account for overlapping conditions. The inclusion-exclusion approach seems solid here.Therefore, I think the answer is 4168.Final Answerboxed{4168}</think>To solve the problem of finding the number of license plates that contain at least one digit 9 and have a sum of digits that is a multiple of 9, we use the principle of inclusion-exclusion.1. Total License Plates: The license plates range from 10000 to 99999, giving a total of 90000 numbers.2. Numbers with Digit Sum Divisible by 9 (|A|): Since 90000 is divisible by 9, the number of such plates is ( frac{90000}{9} = 10000 ).3. Numbers with Digit Sum Divisible by 9 and No Digit 9 (|A ∩ B^c|): - First digit (d1) ranges from 1 to 8 (8 choices). - The remaining digits (d2, d3, d4, d5) range from 0 to 8 (9 choices each). - Total numbers without any 9s: ( 8 times 9^4 = 52488 ). - Assuming uniform distribution, the number of these with digit sum divisible by 9 is ( frac{52488}{9} = 5832 ).4. Result: Subtract the count of numbers with no 9s from the total count of numbers with digit sum divisible by 9: [ |A ∩ B| = |A| - |A ∩ B^c| = 10000 - 5832 = 4168 ]Thus, the number of license plates that contain at least one digit 9 and have the sum of their digits as a multiple of 9 is (boxed{4168}).]

🔑:It is known that among them, there are 10000 numbers divisible by 9. Now consider the number of these 10000 numbers that do not contain the digit 9. Let the license plate number be A B C D E, and let the remainder of A+B+C+D divided by 9 be T. Then, by taking E=9-T (when T=0, take E=0), the requirement can be met. Thus, A can take values from 1 sim 8, B can take values from 0 sim 8, C can take values from 0 sim 8, D can take values from 0 sim 8. Once A, B, C, D are determined, E is also determined. Therefore, the number of such numbers without the digit 9 is 8 times 9 times 9 times 9=5832. Hence, the total number of license plates that meet the conditions is 10000-5832=4168.

🧠:[<think>Okay, let me try to figure out this problem. So, we have three equations here: I, II, and III. The question is asking which of them represent the same graph. The options are combinations of these equations. Let me take each equation one by one and analyze them.Starting with equation I: y = x - 2. That's a straight line, right? The slope is 1, and the y-intercept is -2. Simple enough. So, this is a linear equation, and its graph is a straight line.Moving on to equation II: y = (x² - 4)/(x + 2). Hmm, this looks like a rational function. The numerator is x squared minus 4, which I recognize as a difference of squares. So, that factors into (x - 2)(x + 2). The denominator is x + 2. So, if I factor the numerator, it becomes (x - 2)(x + 2)/(x + 2). Then, the (x + 2) terms in the numerator and denominator cancel out, leaving y = x - 2. Wait, but hold on! If we cancel those terms, we have to consider the domain, right? Because the original equation II has a denominator of x + 2, so x cannot be -2, since that would make the denominator zero. So, equation II simplifies to y = x - 2, but with the restriction that x ≠ -2. That means the graph of II is the same as the line y = x - 2, except there's a hole at x = -2. Let me calculate the y-coordinate of the hole. If x = -2, then substituting into the simplified equation y = x - 2 gives y = -2 - 2 = -4. So, the graph of II is y = x - 2 with a hole at (-2, -4). Therefore, equation I is a complete line, and equation II is the same line but missing a point. So, their graphs are almost the same but not exactly because of that hole. So, I and II are not exactly the same graph.Now, equation III: (x + 2)y = x² - 4. Let me rearrange this equation to solve for y. Dividing both sides by (x + 2), we get y = (x² - 4)/(x + 2). Wait, that's exactly the same as equation II. So, equation III simplifies to the same expression as equation II. But again, we have to consider the domain. The original equation III is (x + 2)y = x² - 4. If x + 2 ≠ 0, then we can divide both sides by (x + 2) to get y = (x² - 4)/(x + 2) = x - 2 (with x ≠ -2). But what happens if x + 2 = 0? That is, if x = -2. Let's plug x = -2 into equation III. Then, the left side becomes ( -2 + 2 )y = 0 * y = 0. The right side is (-2)² - 4 = 4 - 4 = 0. So, 0 = 0. That's true for any y! So, when x = -2, the equation III becomes 0 = 0, which is always true, meaning that when x = -2, y can be any real number. Therefore, equation III actually represents the line y = x - 2 (same as I and II) but when x = -2, any y is allowed. So, graphically, equation III is the line y = x - 2 combined with the vertical line x = -2? Wait, no. Let me think again. When x = -2, the original equation III is 0 * y = 0, which is true regardless of y. So, all points where x = -2 are solutions. So, the graph of equation III is the line y = x - 2 plus the entire vertical line x = -2. That seems significant.Wait, so equation III is not just the line y = x - 2, but also includes all the points where x = -2. So, that vertical line at x = -2 is part of the graph. Therefore, equation III's graph is different from both I and II. Equation I is just the line y = x - 2. Equation II is that line minus the point (-2, -4). Equation III is the line y = x - 2 plus the entire vertical line x = -2. So, all three have different graphs.Wait, let me check that again. For equation III, when we have (x + 2)y = x² - 4. If x ≠ -2, then y must equal (x² - 4)/(x + 2) = x - 2, so that part is the same as equation II. But if x = -2, then the equation becomes 0 * y = 0, which is true for any y. So, when x = -2, y can be anything. Therefore, the graph of III is the line y = x - 2 (with x ≠ -2) combined with all points where x = -2. So, it's a combination of the line y = x - 2 and the vertical line x = -2. Therefore, equation III's graph is actually two lines: the line y = x - 2 and the line x = -2. That's different from both I and II.Therefore, equation I is just the line y = x - 2. Equation II is the same line but missing a point at x = -2. Equation III is the line y = x - 2 and the vertical line x = -2. So, none of them are exactly the same. Let me verify once more.Equation I: All points (x, y) where y = x - 2. That's a straight line.Equation II: All points (x, y) where y = (x² - 4)/(x + 2). As I said, simplifies to y = x - 2 but x ≠ -2. So, missing one point.Equation III: All points (x, y) where (x + 2)y = x² - 4. If x ≠ -2, then y = x - 2. If x = -2, then any y satisfies the equation. So, the solutions are all points on the line y = x - 2 and all points on the line x = -2. So, two lines. So, equation III's graph is two lines: one slanted and one vertical. So, equation III is different from both I and II.So, equation I is a single line. Equation II is the same line with a hole. Equation III is that line plus a vertical line. Therefore, none of them are the same as each other. Wait, but hold on. Let me check if the vertical line x = -2 is part of equation III. Let me plug x = -2 into equation III: ( -2 + 2 )y = (-2)^2 - 4 ⇒ 0 * y = 4 - 4 ⇒ 0 = 0. So, that's true for any y. So, yes, all points where x = -2 are included in equation III. So, equation III is the combination of the line y = x - 2 and x = -2. Therefore, the graph of III is not the same as I or II.Therefore, equation I is a line, equation II is a line with a hole, and equation III is two lines. Therefore, none of them are the same. So, the answer would be (E) None represent the same graph.Wait, but let me check the answer options again. The options are:(A) I and II.(B) I and III.(C) II and III.(D) I, II, and III.(E) None represent the same graph.So, according to my analysis, all three equations have different graphs, so the answer is E. But let me check again just to be sure.Equation I: y = x - 2. Straight line.Equation II: y = (x² - 4)/(x + 2) = x - 2 with x ≠ -2. So, the line minus a point.Equation III: (x + 2)y = x² - 4. Which simplifies to y = x - 2 when x ≠ -2, but allows any y when x = -2. So, the line y = x - 2 and the line x = -2. Therefore, equation III's graph is two lines: one diagonal and one vertical. So, the graph is different from both I and II, which are a single line (with or without a hole). Therefore, none of the equations have the same graph. So, the answer is E.But wait, wait. Wait, when I first looked at equation III, I thought maybe it's equivalent to equation II. But when you divide both sides by (x + 2), you have to assume x ≠ -2. However, equation III, when x = -2, allows any y. So, equation III is not just y = x - 2 with a restriction, but also includes the vertical line. So, the graph is actually different. So, equation III's graph is two lines, which is different from both I and II. Therefore, equations I, II, and III all have different graphs. Therefore, the answer must be E.But wait, let me check with specific points. Let's take x = -2. For equation I, when x = -2, y = -2 - 2 = -4. So, the point (-2, -4) is on I.For equation II, since x ≠ -2, the point (-2, -4) is not on II; instead, there's a hole there.For equation III, when x = -2, any y is allowed. So, all points (-2, y) are on III. So, for example, (-2, 0) is on III, but not on I or II. So, that's different. So, equation III includes infinitely many points that are not on I or II, so it can't be the same graph as either. Therefore, all three are different.Thus, the correct answer is E.Final Answerboxed{E}</think>Given the equations:I. ( y = x - 2 )II. ( y = frac{x^2 - 4}{x + 2} )III. ( (x + 2)y = x^2 - 4 )We need to determine which represent the same graph.1. Equation I: This is a straight line with slope 1 and y-intercept -2.2. Equation II: Simplifies to ( y = frac{(x - 2)(x + 2)}{x + 2} = x - 2 ) for ( x neq -2 ). The domain excludes ( x = -2 ), creating a hole at ( (-2, -4) ).3. Equation III: Solving for ( y ) when ( x neq -2 ) gives ( y = x - 2 ). When ( x = -2 ), the equation becomes ( 0 cdot y = 0 ), which is true for any ( y ). Thus, the graph includes the line ( y = x - 2 ) and all points where ( x = -2 ).- Equation I is a complete line.- Equation II is the same line as I but with a hole at ( (-2, -4) ).- Equation III includes the line ( y = x - 2 ) and the vertical line ( x = -2 ).Since all three equations have different graphs (a complete line, a line with a hole, and a combination of a line and a vertical line), none represent the same graph.[boxed{E}],<think>Okay, so I need to figure out which of these equations represent the same graph. There are three equations given: I, II, and III. The options are combinations of these, and the answer is one of the choices from A to E. Let me start by understanding each equation individually and then compare them.First, let's look at equation I: y = x - 2. That's a linear equation, so its graph is a straight line with a slope of 1 and a y-intercept at (0, -2). Pretty straightforward.Equation II is y = (x² - 4)/(x + 2). Hmm, this looks like a rational function. The numerator is a quadratic, and the denominator is linear. I remember that when dealing with rational functions, we should check for any common factors between the numerator and the denominator because that could indicate a hole in the graph. Let's factor the numerator: x² - 4 is a difference of squares, so it factors into (x - 2)(x + 2). The denominator is x + 2, so there's a common factor of (x + 2). That means we can simplify the expression as long as x ≠ -2 because at x = -2, the original function is undefined (since the denominator would be zero). So simplifying equation II gives y = (x - 2)(x + 2)/(x + 2) = x - 2, but with the restriction that x ≠ -2. Therefore, equation II is equivalent to y = x - 2 except at x = -2, where there's a hole. So graph II is almost the same as graph I, except it's missing the point where x = -2. If I plug x = -2 into equation I, y would be -2 - 2 = -4, so the hole is at (-2, -4).Now equation III: (x + 2)y = x² - 4. Let me rearrange this equation to solve for y. Dividing both sides by (x + 2) gives y = (x² - 4)/(x + 2). Wait a second, that's the same as equation II! So equation III simplifies to equation II, right? But hold on, let me check if there are any differences. If I start with equation III: (x + 2)y = x² - 4. If I move everything to one side, it's (x + 2)y - (x² - 4) = 0. Let me see if this can be factored or simplified further. If I distribute y into (x + 2), it becomes xy + 2y - x² + 4 = 0. That seems more complicated. But when we solve for y, it's y = (x² - 4)/(x + 2), same as equation II. So equations II and III are algebraically the same. However, similar to equation II, equation III would have the same restriction: x ≠ -2 because when solving for y, we divided by (x + 2), which can't be zero. Therefore, equation III also results in y = x - 2 with a hole at x = -2. Wait, but hold on. Let me check equation III again. If (x + 2)y = x² - 4, what happens when x = -2? Let's substitute x = -2 into equation III. Then we have (-2 + 2)y = (-2)² - 4 => 0 * y = 4 - 4 => 0 = 0. So when x = -2, the equation becomes 0 = 0, which is always true. That means when x = -2, y can be any real number. So for equation III, when x = -2, there are infinitely many solutions for y, meaning the graph includes the vertical line x = -2 with all possible y-values. Wait, but that's conflicting with my previous thought. Let me verify this.If I consider equation III: (x + 2)y = x² - 4. If x ≠ -2, then we can divide both sides by (x + 2) and get y = (x² - 4)/(x + 2) = x - 2, as before. However, if x = -2, then the left side becomes 0 * y = 0, and the right side is (-2)^2 - 4 = 4 - 4 = 0. So 0 = 0 is true for any y. Therefore, when x = -2, any y satisfies the equation. So the graph of equation III is the line y = x - 2 except when x = -2, where instead of being undefined (like in equation II), it allows any y. So in the graph, there's the line y = x - 2 and also the vertical line x = -2. But wait, that doesn't seem right. Let me think again.No, actually, when x = -2, the equation becomes 0 * y = 0, which is true for all y. So in the graph, all points where x = -2 are included, which is the entire vertical line x = -2. But also, when x ≠ -2, we have the line y = x - 2. Therefore, equation III's graph is actually the combination of the line y = x - 2 and the vertical line x = -2. But this is different from equation II, which is just the line y = x - 2 with a hole at x = -2. So equation III includes more points than equation II. Therefore, equations II and III are not the same. Equation III includes the entire vertical line x = -2 in addition to the line y = x - 2, whereas equation II is just y = x - 2 with a hole at (-2, -4). Wait, this is a critical point. So equation III isn't just y = x - 2 with a hole, but it's actually the union of the line y = x - 2 and the vertical line x = -2. So that makes equation III's graph different from both I and II. Let me confirm this again.Original equation III: (x + 2)y = x² - 4. Let's rearrange this equation. Expand the right side: x² - 4 can be written as (x - 2)(x + 2). So equation III becomes (x + 2)y = (x - 2)(x + 2). If we subtract the right side from both sides, we get (x + 2)y - (x - 2)(x + 2) = 0. Factoring out (x + 2), we have (x + 2)(y - (x - 2)) = 0. So this equation is equivalent to (x + 2)(y - x + 2) = 0. Therefore, the product is zero when either factor is zero. That means either x + 2 = 0 (i.e., x = -2) or y - x + 2 = 0 (i.e., y = x - 2). Therefore, the graph of equation III is the combination of the vertical line x = -2 and the line y = x - 2. That's two different lines. So equation III is actually two lines: one vertical at x = -2 and the other with slope 1 and y-intercept -2. So now, comparing all three equations:- Equation I is just the line y = x - 2.- Equation II is the line y = x - 2 but missing the point where x = -2 (a hole at (-2, -4)).- Equation III is both the line y = x - 2 and the vertical line x = -2. Therefore, the graphs are different. Equation I is a single line, equation II is the same line with a hole, and equation III is the line plus another vertical line. Therefore, none of them are exactly the same. Wait, but let me check again.Wait, equation III is two lines: x = -2 and y = x - 2. So when plotted, it's not just the line y = x - 2 but also the vertical line at x = -2. So equation III's graph is different from both I and II. Equation II is almost like equation I but missing a point. Therefore, equations I and II are not the same because II has a hole, equations I and III are not the same because III includes an extra vertical line, and equations II and III are not the same because II is missing a point while III has an extra line. Therefore, none of them represent the same graph. So the answer should be E: None represent the same graph.But let me check again to be thorough. Maybe there's a mistake in my reasoning.Equation I: y = x - 2. Straight line.Equation II: y = (x² - 4)/(x + 2) simplifies to y = x - 2 with x ≠ -2. So it's the line y = x - 2 with a hole at (-2, -4).Equation III: (x + 2)y = x² - 4. As we saw, simplifies to (x + 2)(y - x + 2) = 0, so x = -2 or y = x - 2. Therefore, the graph is the union of the line x = -2 and the line y = x - 2. So two lines crossing each other at (-2, -4). Wait, but if x = -2, then for y in the original equation III, (x + 2)y = x² - 4. If x = -2, then 0 * y = (-2)^2 - 4 = 4 - 4 = 0. So 0 = 0, which is true for any y. Therefore, when x = -2, any y satisfies the equation. So the graph is the vertical line x = -2 (all points where x = -2, y can be anything) and the line y = x - 2. Therefore, equation III is two lines: x = -2 and y = x - 2. So the graph is not the same as equation I, which is a single line, nor equation II, which is the same line as I but missing a point. Therefore, all three equations have different graphs. Hence, the answer is E: None represent the same graph.But wait, the original answer options don't include E? Wait, no: options are A to E, with E being "None represent the same graph." Yes, that's option E. So according to this reasoning, the answer is E.But let me cross-check with the original problem statement. The problem is from the 43rd American High School Mathematics Examination, 1992. Let me confirm if my reasoning is correct.Another way to check is to test specific points. For example, take x = -2.In equation I: y = -2 - 2 = -4. So the point (-2, -4) is on graph I.In equation II: When x = -2, the function is undefined, so there's a hole there. So (-2, -4) is not on graph II.In equation III: When x = -2, any y is allowed. So the entire vertical line x = -2 is on the graph, which includes infinitely many points, including (-2, -4), but also (-2, 0), (-2, 1), etc. So equation III includes all points where x = -2, not just (-2, -4). Therefore, the point (-2, -4) is on both I and III, but III has more points. Therefore, I and III are different. So equation I and III share the point (-2, -4), but III has more points. Equations I and II differ at x = -2. Equations II and III also differ because II is missing the point at x = -2, and III includes the entire vertical line. Therefore, none of the equations have identical graphs. Therefore, the correct answer is E.Final Answerboxed{E}</think>Given the equations:I. ( y = x - 2 )II. ( y = frac{x^2 - 4}{x + 2} )III. ( (x + 2)y = x^2 - 4 )We need to determine which represent the same graph.1. Equation I: This is a linear equation representing the line ( y = x - 2 ).2. Equation II: Simplifying ( frac{x^2 - 4}{x + 2} ) gives ( x - 2 ) with a restriction ( x neq -2 ). Thus, it represents the line ( y = x - 2 ) with a hole at ( (-2, -4) ).3. Equation III: Rearranging ( (x + 2)y = x^2 - 4 ) gives ( (x + 2)(y - x + 2) = 0 ). This represents the union of the line ( y = x - 2 ) and the vertical line ( x = -2 ). When ( x = -2 ), any ( y ) satisfies the equation, resulting in the vertical line ( x = -2 ).Key differences:- Equation I is the line ( y = x - 2 ).- Equation II is the line ( y = x - 2 ) with a hole at ( (-2, -4) ).- Equation III includes both the line ( y = x - 2 ) and the vertical line ( x = -2 ).None of the equations represent the same graph. Therefore, the answer is (boxed{E}).]

🔑:[Solution] For II, x neq-2, which means its graph is the line y=x-2 excluding the point (-2,-4).For III, its graph consists of two linesx+2=0 text { and } y=x-2 text {. }Clearly, the graphs of I, II, and III are all different.Therefore, the answer is (E).

🧠:[<think>Okay, so I need to plot the set of points (x, y) that satisfy the equation |5x| + |12y| + |60 - 5x - 12y| = 60, and then find the area of that figure. Hmm, this looks like it involves absolute values, which often result in piecewise functions and linear equations. Let me try to break this down step by step.First, let me recall that absolute value equations can be tackled by considering different cases based on the signs of the expressions inside the absolute values. In this case, we have three absolute value terms: |5x|, |12y|, and |60 - 5x - 12y|. Each of these can be positive or negative, so maybe I need to consider different regions where each of these expressions is positive or negative, and then solve the equation in each region. However, since there are three absolute value terms, that could lead to 2^3 = 8 different cases. That seems like a lot, but maybe some of these cases are redundant or impossible. Let me see.Alternatively, maybe there's a smarter way to analyze this equation without considering all eight cases. Let me think about the structure of the equation. The left-hand side is the sum of three absolute values, and the right-hand side is 60. Let's denote the expressions inside the absolute values:A = 5x,B = 12y,C = 60 - 5x - 12y.So the equation is |A| + |B| + |C| = 60. Now, notice that A + B + C = 5x + 12y + (60 - 5x - 12y) = 60. So, A + B + C = 60. Interesting. So the sum of A, B, and C is a constant 60. So the equation becomes |A| + |B| + |C| = A + B + C = 60. Therefore, the sum of the absolute values of A, B, C equals their regular sum, which is 60. When does that happen?In general, |a| + |b| + |c| = a + b + c if and only if each of a, b, c is non-negative. Because if any of them is negative, their absolute value would be greater than themselves, so the sum of absolute values would exceed the sum of the original terms. Therefore, the equation |A| + |B| + |C| = 60 holds if and only if A, B, C are all non-negative. Because A + B + C = 60, so if all of them are non-negative, then |A| + |B| + |C| = A + B + C = 60. Conversely, if any of A, B, or C is negative, then |A| + |B| + |C| would be greater than 60, which can't happen because the right-hand side is 60. Therefore, the equation is equivalent to the system:A ≥ 0,B ≥ 0,C ≥ 0,and A + B + C = 60.So substituting back:5x ≥ 0,12y ≥ 0,60 - 5x - 12y ≥ 0,and 5x + 12y + (60 - 5x - 12y) = 60 (which is always true).Therefore, the set of points (x, y) satisfying the original equation is exactly the set of points where:5x ≥ 0 ⇒ x ≥ 0,12y ≥ 0 ⇒ y ≥ 0,and 60 - 5x - 12y ≥ 0 ⇒ 5x + 12y ≤ 60.So combining these inequalities, we have x ≥ 0, y ≥ 0, and 5x + 12y ≤ 60.Wait, that seems too straightforward. So the original equation simplifies to these three inequalities? Let me check that again.Given that A + B + C = 60, and |A| + |B| + |C| = 60. Since A + B + C = |A| + |B| + |C|, this equality holds if and only if each of A, B, C is non-negative, because otherwise, the absolute values would make the left-hand side larger. So yes, the equation is equivalent to A ≥ 0, B ≥ 0, C ≥ 0.Therefore, the solution set is the intersection of the regions defined by 5x ≥ 0, 12y ≥ 0, and 60 - 5x - 12y ≥ 0.Simplifying these inequalities:5x ≥ 0 ⇒ x ≥ 0,12y ≥ 0 ⇒ y ≥ 0,and 5x + 12y ≤ 60.So this defines a region in the first quadrant (since x ≥ 0 and y ≥ 0) bounded by the line 5x + 12y = 60.So, plotting this region would be a triangle with vertices at the intercepts of the line 5x + 12y = 60 with the x-axis and y-axis.Let me find the intercepts.When y = 0, 5x = 60 ⇒ x = 12.When x = 0, 12y = 60 ⇒ y = 5.So the line intersects the x-axis at (12, 0) and the y-axis at (0, 5). Since we are in the first quadrant, the region is a triangle with vertices at (0, 0), (12, 0), and (0, 5). Wait, hold on, when x and y are zero, 5x + 12y = 0 ≤ 60, so (0,0) is part of the region, but the intercepts are (12,0) and (0,5). Therefore, the region is the triangle connecting (0,0), (12,0), and (0,5). Hmm, but wait, let's see.Wait, if x and y are greater than or equal to zero, and 5x + 12y ≤ 60, then yes, the feasible region is a triangle with vertices at (0,0), (12,0), and (0,5). Because when x=0, y can be up to 5; when y=0, x can be up to 12; and when both x and y are positive, they are bounded by the line connecting (12,0) and (0,5).Therefore, the set of points is a right triangle in the first quadrant with legs along the x-axis and y-axis, and hypotenuse from (12,0) to (0,5). The area of this triangle would be (base * height)/2 = (12 * 5)/2 = 30. Therefore, the area is 30.Wait, but hold on a second. Let me verify that this is indeed the case.But wait, the original equation was |5x| + |12y| + |60 - 5x - 12y| = 60. If we follow the logic, then the solution set is the region where all three terms inside the absolute values are non-negative. However, perhaps I should check with some test points to see if that's the case.Take the point (0,0): |0| + |0| + |60 - 0 - 0| = 0 + 0 + 60 = 60, which satisfies the equation. Good.Take the point (12,0): |5*12| + |12*0| + |60 - 5*12 - 12*0| = |60| + 0 + |60 - 60| = 60 + 0 + 0 = 60. That works.Take the point (0,5): |0| + |12*5| + |60 - 0 - 12*5| = 0 + 60 + |60 - 60| = 60 + 0 = 60. That works too.Take a point inside the triangle, say (6, 2.5). Compute |5*6| + |12*2.5| + |60 -5*6 -12*2.5| = |30| + |30| + |60 -30 -30| = 30 + 30 + 0 = 60. So that works.Now take a point outside the triangle but in the first quadrant, say (12,5). Then compute |5*12| + |12*5| + |60 -5*12 -12*5| = 60 + 60 + |60 -60 -60| = 60 +60 + | -60 | = 60 +60 +60 = 180 ≠60. So that's outside.What about a point in the first quadrant but below the line 5x +12y =60? Let's say (6,1). Then |30| + |12| + |60 -30 -12| =30 +12 +18=60. Wait, 30 +12 +18=60? 30+12=42, 42+18=60. So yes, that's equal to 60. But (6,1) is inside the triangle, as 5*6 +12*1=30+12=42 ≤60. So that works.Wait, but if I take a point in the first quadrant but where 5x +12y >60, like (12,1). Then |60| + |12| + |60 -60 -12| =60 +12 +12=84≠60. So that's outside. So seems like the equation only holds when 5x +12y ≤60, x ≥0, y≥0.Therefore, the solution set is exactly the triangle with vertices at (0,0), (12,0), and (0,5). Therefore, the area is ½ *12*5=30.But wait, the original problem was presented as an absolute value equation, which I translated into linear inequalities by recognizing that the sum of absolute values equals the sum of the terms only when each term is non-negative. But let me make sure there isn't another interpretation or another set of points that might satisfy the equation even if some terms inside the absolute values are negative. Because sometimes, absolute value equations can have solutions where some terms are negative if their contributions cancel out, but in this case, since the sum of absolute values equals a fixed number, maybe not. Let me test a point outside the first quadrant.Take a point where x is negative, say (-1,0). Compute |5*(-1)| + |12*0| + |60 -5*(-1) -12*0| =5 +0 +|60 +5|=5 +0 +65=70≠60. So that doesn't work.Take a point where y is negative, say (0,-1). |0| + |12*(-1)| + |60 -0 -12*(-1)|=0 +12 +|60 +12|=0 +12 +72=84≠60.Take a point where 5x +12y >60, say (12,5). As before, 60 +60 + |-60|=60+60+60=180≠60.What if two terms are positive and one is negative? Let's see. Suppose A=5x is negative, B=12y positive, C=60 -5x -12y=60 - (negative) -12y. So if A is negative, then 5x <0, so x <0. Then C=60 -5x -12y. Since x is negative, -5x is positive, so C=60 + positive -12y. Suppose y is positive. Then C=60 + positive -12y. Depending on y, C could be positive or negative. Let's try a point where x is negative, y positive, such that C is negative.Let's say x=-1, y=6. Then A=5*(-1)=-5, B=12*6=72, C=60 -(-5) -72=60 +5 -72= -7. So compute |A| + |B| + |C|=5 +72 +7=84≠60. Doesn't satisfy.Alternatively, suppose A and B are negative, but C is positive. Let x=-1, y=-1. Then A=-5, B=-12, C=60 - (-5) - (-12)=60 +5 +12=77. Then |A| + |B| + |C|=5 +12 +77=94≠60. Not equal.Alternatively, maybe in some regions, two terms are negative and one is positive, but how would that sum up? For example, let's suppose that A and C are negative, B is positive. Then |A| + |B| + |C| = (-A) + B + (-C) = -A + B - C. But since A + B + C =60, substituting C=60 -A -B, then -A + B - (60 - A - B)= -A + B -60 + A + B= 2B -60. So set this equal to 60: 2B -60=60 ⇒ 2B=120 ⇒ B=60. But B=12y, so 12y=60 ⇒ y=5. But if A and C are negative, then A=5x <0 ⇒ x<0, and C=60 -A -B <0 ⇒60 -5x -12y <0. But if B=60, then 60 -5x -60 <0 ⇒ -5x <0 ⇒ x>0. But x<0 from A<0. Contradiction. Therefore, this case is impossible.Similarly, other combinations of signs might lead to contradictions. Therefore, it seems that the only solution is when all three terms A, B, C are non-negative, which reduces to the triangle in the first quadrant.Therefore, the resulting figure is a triangle with vertices at (0,0), (12,0), and (0,5), and the area is 30.But just to be thorough, let me check another point where maybe two coordinates are on the boundary. For example, take (6, 2.5). Wait, we already checked that. But let's check another one, say (4, 3). Then 5x +12y=20 +36=56 ≤60. So |20| + |36| + |60 -20 -36|=20 +36 +4=60. Yes, that works. So that's inside the triangle.Another point on the line 5x +12y=60, say (6, (60 -30)/12=2.5). Then 5*6=30, 12*2.5=30, so 30+30=60. Then |30| + |30| + |60 -30 -30|=30 +30 +0=60. Correct.But what if I take a point on the line 5x +12y=60 but outside the first quadrant? For example, x=24, y= -5. Then 5*24 +12*(-5)=120 -60=60. But compute |5*24| + |12*(-5)| + |60 -120 +60|=120 +60 +0=180≠60. So even though 5x +12y=60, since x and y are not both non-negative, the equation doesn't hold. Therefore, only the part of the line in the first quadrant, combined with x≥0 and y≥0, gives the solution.Therefore, the conclusion is that the set of points is the triangle with vertices at (0,0), (12,0), and (0,5), and its area is 30.Final AnswerThe area of the resulting figure is boxed{30}.</think>To solve the equation ( |5x| + |12y| + |60 - 5x - 12y| = 60 ) and find the area of the resulting figure, we start by analyzing the absolute value terms. We denote the expressions inside the absolute values as:- ( A = 5x )- ( B = 12y )- ( C = 60 - 5x - 12y )The equation becomes ( |A| + |B| + |C| = 60 ). Since ( A + B + C = 60 ), the equation holds if and only if each of ( A ), ( B ), and ( C ) is non-negative. This is because the sum of absolute values equals the sum of the original terms only when all terms are non-negative.Thus, we need:1. ( 5x geq 0 ) (implies ( x geq 0 ))2. ( 12y geq 0 ) (implies ( y geq 0 ))3. ( 60 - 5x - 12y geq 0 ) (implies ( 5x + 12y leq 60 ))These inequalities define a region in the first quadrant bounded by the line ( 5x + 12y = 60 ). The intercepts of this line are:- When ( y = 0 ), ( x = 12 )- When ( x = 0 ), ( y = 5 )Thus, the vertices of the resulting triangle are ( (0, 0) ), ( (12, 0) ), and ( (0, 5) ). The area of this triangle is calculated as:[text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 12 times 5 = 30]Therefore, the area of the resulting figure is (boxed{30}).,<think>Okay, so I have this equation to plot on the (x; y) plane: |5x| + |12y| + |60 - 5x - 12y| = 60. And then I need to find the area of the resulting figure. Hmm, let's break this down step by step. First, dealing with absolute values can be tricky because they can split equations into different cases depending on the signs of the expressions inside. So, maybe I should start by understanding each term in the equation.The equation is |5x| + |12y| + |60 - 5x - 12y| = 60. Let me see. Each of these absolute value terms can be positive or negative, so depending on the signs of 5x, 12y, and (60 - 5x - 12y), the equation will change. But handling all possible cases might be too cumbersome. Maybe there's a smarter way to approach this.Wait a second, the coefficients 5 and 12 remind me of the Pythagorean theorem, 5-12-13 triangle. Maybe that's a hint? Also, 5x and 12y – if I think of 5x + 12y, that might relate to a line. The third term is |60 - 5x -12y|. So, if I let z = 5x + 12y, then the equation becomes |5x| + |12y| + |60 - z| = 60, where z = 5x + 12y.Hmm, not sure if that substitution helps immediately. Let me think. Maybe consider the three terms: |5x|, |12y|, and |60 - 5x -12y|. Let's denote A = 5x, B = 12y, so the equation becomes |A| + |B| + |60 - A - B| = 60.That's interesting. So, |A| + |B| + |60 - A - B| = 60. Maybe this equation is simpler in terms of A and B. Let's consider variables A and B. Then the equation is |A| + |B| + |60 - A - B| = 60. Let me analyze this equation.First, note that 60 - A - B = 60 - (A + B). So, we have |A| + |B| + |60 - (A + B)| = 60. Maybe we can set C = A + B, so the equation becomes |A| + |B| + |60 - C| = 60. But C = A + B, so maybe not directly helpful. Alternatively, perhaps consider the three variables A, B, and 60 - A - B.Wait, if I denote three terms: |A|, |B|, and |60 - A - B|, then the sum of these three absolute values equals 60. Let me think about when each of these terms is positive or negative. For the absolute value expressions to drop, we need to know the sign inside. So, let's consider different cases based on the signs of A, B, and (60 - A - B).But this could be quite a few cases. Let's see. Each absolute value term can be either positive or negative, but since they are absolute values, they are always non-negative. Wait, but the expressions inside can be positive or negative, so the absolute value will change accordingly.Alternatively, maybe there's a geometric interpretation. If I think of A and B as variables, perhaps this equation represents some geometric figure. Let's see. Let's consider A and B as coordinates. Then, |A| + |B| + |60 - A - B| = 60. Hmm, this seems similar to taxicab geometry but with an extra term. Maybe we can interpret this in three dimensions, but we need to stay in two dimensions since the original problem is in the (x; y) plane.Wait, but A = 5x and B = 12y. So, if we transform the coordinates, scaling x by 5 and y by 12, the equation becomes |A| + |B| + |60 - A - B| = 60. So, maybe in the (A, B) plane, this equation defines a polygon, and we can find its vertices and then transform back to (x, y) coordinates.Yes, that sounds promising. Let's work in the (A, B) coordinate system first. So the equation is |A| + |B| + |60 - A - B| = 60. Let's analyze this equation.First, note that 60 - A - B is the same as 60 - (A + B). So, depending on the sign of 60 - (A + B), that absolute value term will split into two cases: either 60 - A - B or A + B - 60.Similarly, |A| splits into A or -A, depending on whether A is positive or negative. Similarly for |B|. So, there are multiple cases to consider here. However, let's try to approach this systematically.Case 1: All expressions inside the absolute values are non-negative. That is:A ≥ 0, B ≥ 0, and 60 - A - B ≥ 0.In this case, the equation becomes:A + B + 60 - A - B = 60 ⇒ 60 = 60.So, in this case, the equation is always true. Therefore, the region where A ≥ 0, B ≥ 0, and A + B ≤ 60 is part of the solution set.So, in the first quadrant (since A and B are non-negative), the region under the line A + B = 60 is part of the solution.Case 2: Suppose A ≥ 0, B ≥ 0, but 60 - A - B < 0. Then, the equation becomes:A + B + (A + B - 60) = 60 ⇒ 2A + 2B - 60 = 60 ⇒ 2A + 2B = 120 ⇒ A + B = 60.But in this case, we're in the region where A + B > 60. However, substituting into the equation gives A + B = 60, which contradicts A + B > 60. Therefore, there are no solutions in this subcase.Case 3: Let's consider A ≥ 0, B < 0. Then, |B| = -B, and we need to check the sign of 60 - A - B.So, 60 - A - B = 60 - A - B. Since B is negative, -B is positive, so 60 - A - B = 60 - A + |B|. Hmm, but I need to determine the sign of 60 - A - B.Alternatively, since B is negative, let's let B = -b where b > 0. Then, 60 - A - B = 60 - A + b. Since A ≥ 0 and b > 0, 60 - A + b could be positive or negative. So, perhaps we need to split this further.But this is getting complicated. Maybe there's a better approach.Alternatively, maybe we can note that |a| + |b| + |c| is a symmetric expression. But in this case, the third term is |60 - a - b|, which complicates things. Wait, but perhaps if we set variables such that we can relate them to each other.Alternatively, consider the equation |A| + |B| + |60 - A - B| = 60. Let's define C = A + B. Then the equation becomes |A| + |B| + |60 - C| = 60. But since C = A + B, perhaps we can find relationships between A, B, and C.Alternatively, since A and B can be positive or negative, let's consider different quadrants.First, Quadrant I: A ≥ 0, B ≥ 0. As before, in this quadrant, if A + B ≤ 60, the equation holds. If A + B > 60, then |60 - A - B| = A + B - 60, so the equation becomes A + B + (A + B - 60) = 60 ⇒ 2A + 2B = 120 ⇒ A + B = 60. But this line is the boundary of the region where A + B ≤ 60. Therefore, in Quadrant I, the solution is the area under A + B ≤ 60.Now, Quadrant II: A < 0, B ≥ 0. Then |A| = -A, |B| = B. Let's see:-A + B + |60 - A - B| = 60.Let me compute 60 - A - B. Since A is negative (-A positive), 60 - A - B = 60 + |A| - B. Depending on the sign of this expression, we split into two cases.Subcase 1: 60 - A - B ≥ 0 ⇒ 60 + |A| - B ≥ 0. Then:-A + B + 60 - A - B = 60 ⇒ -2A + 60 = 60 ⇒ -2A = 0 ⇒ A = 0. But A < 0 in Quadrant II, so no solution here.Subcase 2: 60 - A - B < 0 ⇒ 60 + |A| - B < 0. Then:-A + B + (A + B - 60) = 60 ⇒ (-A + A) + (B + B) - 60 = 60 ⇒ 0 + 2B - 60 = 60 ⇒ 2B = 120 ⇒ B = 60.But in Quadrant II, B ≥ 0. So, B = 60. However, since A < 0, let's check if 60 - A - B < 0.60 - A - B = 60 - (-|A|) - 60 = |A|. But |A| is positive, so 60 - A - B = |A| > 0, which contradicts the subcase condition 60 - A - B < 0. Therefore, there's no solution here as well.Thus, in Quadrant II, there are no solutions.Quadrant III: A < 0, B < 0. Then |A| = -A, |B| = -B. Let's compute:-A - B + |60 - A - B| = 60.Again, 60 - A - B = 60 - A - B. Since A and B are negative, -A and -B are positive, so 60 - A - B = 60 + |A| + |B|. Since |A| and |B| are positive, 60 + |A| + |B| is always greater than 60, so 60 - A - B > 60. Therefore, |60 - A - B| = 60 - A - B - 60 = -A - B - 60. Wait, no: |60 - A - B| when 60 - A - B is negative would be -(60 - A - B) = A + B - 60. Wait, wait:If 60 - A - B is positive, |60 - A - B| = 60 - A - B. If it's negative, |60 - A - B| = -(60 - A - B) = A + B - 60.But in Quadrant III, since A and B are negative, 60 - A - B = 60 + |A| + |B|, which is definitely positive. Therefore, |60 - A - B| = 60 - A - B. Therefore, equation becomes:-A - B + 60 - A - B = 60 ⇒ -2A - 2B + 60 = 60 ⇒ -2A - 2B = 0 ⇒ A + B = 0.But in Quadrant III, A < 0 and B < 0, so A + B = 0 implies A = -B. So, the line A = -B in Quadrant III. However, in Quadrant III, both A and B are negative, so A = -B implies that one is the negative of the other. For example, if A = -1, then B = 1, but B is supposed to be negative. Therefore, the only point where A + B = 0 in Quadrant III is A = B = 0, which is the origin. But the origin is in Quadrant I. Therefore, there are no solutions in Quadrant III.Quadrant IV: A ≥ 0, B < 0. Then |A| = A, |B| = -B. The equation becomes:A - B + |60 - A - B| = 60.Again, 60 - A - B. Let's analyze its sign. Since B is negative, -B is positive, so 60 - A - B = 60 - A + |B|. So, it's 60 minus A plus |B|. Since A ≥ 0 and |B| > 0, this expression is 60 + (|B| - A). Depending on whether |B| - A is positive or negative, 60 - A - B can be greater or less than 60.But maybe split into cases based on the sign of 60 - A - B.Subcase 1: 60 - A - B ≥ 0. Then:A - B + 60 - A - B = 60 ⇒ (A - A) + (-B - B) + 60 = 60 ⇒ -2B + 60 = 60 ⇒ -2B = 0 ⇒ B = 0. But B < 0 in Quadrant IV, so no solution here.Subcase 2: 60 - A - B < 0. Then |60 - A - B| = A + B - 60. So, the equation becomes:A - B + A + B - 60 = 60 ⇒ 2A - 60 = 60 ⇒ 2A = 120 ⇒ A = 60. So, A = 60. But in Quadrant IV, A ≥ 0, B < 0. Let's check if 60 - A - B < 0. If A = 60, then 60 - 60 - B = -B. Since B < 0, -B > 0, so 60 - A - B = -B > 0. But we assumed 60 - A - B < 0, which contradicts. Therefore, no solution here.Therefore, in Quadrant IV, there are no solutions.So, summarizing the cases:- In Quadrant I: All points where A + B ≤ 60.- In Quadrants II, III, IV: No solutions.Therefore, in the (A, B) plane, the solution set is the triangle in the first quadrant bounded by A ≥ 0, B ≥ 0, and A + B ≤ 60.But wait, that's only if we assume A and B are non-negative, which is Quadrant I. So, the equation reduces to the triangle with vertices at (0,0), (60,0), and (0,60) in the (A, B) plane.But remember that A = 5x and B = 12y. So, transforming back to the (x, y) plane, we can substitute A = 5x and B = 12y.Therefore, the region in the (x, y) plane is defined by 5x ≥ 0, 12y ≥ 0, and 5x + 12y ≤ 60.Which is equivalent to x ≥ 0, y ≥ 0, and 5x + 12y ≤ 60.So, this is a triangle in the first quadrant of the (x, y) plane with vertices at:- When x = 0, 12y ≤ 60 ⇒ y ≤ 5. So, vertex at (0, 5).- When y = 0, 5x ≤ 60 ⇒ x ≤ 12. So, vertex at (12, 0).- The third vertex is the intersection of the axes at (0,0). Wait, but if we have 5x + 12y = 60, then when x=0, y=5; when y=0, x=12. So, actually, the three vertices of the triangle in (x, y) plane are (0,0), (12,0), and (0,5). Wait, but hold on, in the (A, B) plane, the triangle had vertices at (0,0), (60,0), and (0,60). Then, transforming back, A = 5x and B = 12y, so x = A/5, y = B/12. Therefore, the vertices in the (x, y) plane would be:- (A=0, B=0) → (x=0, y=0)- (A=60, B=0) → (x=60/5=12, y=0)- (A=0, B=60) → (x=0, y=60/12=5)So yes, the triangle in (x, y) plane is with vertices at (0,0), (12,0), and (0,5). Therefore, this is a right triangle with legs of length 12 and 5. The area of a triangle is (base * height)/2, so (12 * 5)/2 = 30. Therefore, the area is 30.Wait, but hold on. The original equation was |5x| + |12y| + |60 -5x -12y| = 60. When we analyzed the (A, B) plane, we considered A = 5x and B = 12y, leading to the conclusion that the solution set is the triangle in Quadrant I with vertices at (0,0), (12,0), and (0,5). But the original problem didn't restrict x and y to be positive. However, in our case analysis, we saw that only Quadrant I (i.e., x ≥ 0 and y ≥ 0) had solutions, and other quadrants did not. Is this accurate?Wait, let me check again. When we considered Quadrants II, III, IV, we found no solutions. So, in the original (x, y) plane, does that mean the solution is only the triangle in the first quadrant? But what if x or y is negative? Let me verify with an example.Suppose x is negative. Let's take x = -1, y = 0. Then compute |5*(-1)| + |12*0| + |60 -5*(-1) -12*0| = 5 + 0 + |60 +5| = 5 + 0 + 65 = 70 ≠ 60. So, that doesn't satisfy the equation. Similarly, take x = 0, y = -1. |0| + |12*(-1)| + |60 -0 -12*(-1)| = 0 + 12 + |60 +12| = 12 + 72 = 84 ≠ 60. So, points with negative x or y don't satisfy the equation. Therefore, indeed, the solution set is entirely in the first quadrant.Therefore, the figure is a triangle with vertices at (0,0), (12,0), and (0,5). The area is (12 * 5)/2 = 30.But wait, let me make sure that this is indeed the case. Let's take a point on the line 5x + 12y = 60. For example, the midpoint between (12,0) and (0,5). Let's say x=6, y=2.5. Then, 5*6 +12*2.5 = 30 +30=60. So, compute |5*6| + |12*2.5| + |60 -5*6 -12*2.5| = |30| + |30| + |60 -30 -30| =30 +30 +0=60, which satisfies the equation.Another point inside the triangle: x=1, y=1. Then, |5*1| + |12*1| + |60 -5*1 -12*1| =5 +12 + |60 -5 -12|=17 +43=60. Wait, 5+12+43=60. So, 5 +12=17, 60-17=43. So, yes, 17 +43=60. So, indeed, points inside the triangle satisfy the equation.Wait, but that seems confusing. Because in the (A, B) plane, when we had the triangle A + B ≤60, with A and B positive, then all points inside and on the triangle would satisfy A + B + (60 - A - B) =60, which simplifies to 60=60. But here, in the (x,y) plane, when we plug in a point inside the triangle, we get |5x| + |12y| + |60 -5x -12y| = (5x +12y) + (60 -5x -12y) =60. Wait, but only if 5x and12y are positive. But x and y are positive in the triangle, so |5x|=5x, |12y|=12y, and |60 -5x -12y|=60 -5x -12y because 5x +12y ≤60.Therefore, in that case, the equation becomes 5x +12y +60 -5x -12y=60, which is 60=60. So, all points in the triangle satisfy the equation. However, when we take a point on the edge, like (12,0), |5*12| + |12*0| + |60 -60 -0|=60 +0 +0=60. Similarly, (0,5): |0| +60 +0=60. So, all the edges and interior points satisfy the equation. So, the entire triangle is the solution set.Therefore, the area is indeed 30.But wait, let me confirm once again. The original equation is |5x| + |12y| + |60 -5x -12y|=60. If I consider a point outside the triangle in the first quadrant, say x=12, y=5. Then 5x=60,12y=60, and 60 -5x -12y=60 -60 -60= -60. So, |5x| + |12y| + |60 -5x -12y|=60 +60 +60=180≠60. So, points outside the triangle do not satisfy the equation.Therefore, the solution set is exactly the triangle with vertices at (0,0), (12,0), and (0,5). Thus, the area is 30.Wait, but just to make sure, maybe there's a mistake in the case analysis. Let me check another point outside the first quadrant. Let's say x=1, y=1. Wait, x and y are positive, so in the first quadrant, but if I take x=6, y=5, which is outside the triangle. Wait, 5x +12y=30 +60=90, so |5x| + |12y| + |60 -5x -12y|=30 +60 +| -30|=30 +60 +30=120≠60. So, indeed, outside the triangle, the equation isn't satisfied.Therefore, all checks confirm that the solution is the triangle with area 30. Therefore, the answer should be 30.Final AnswerThe area of the resulting figure is boxed{30}.</think>To solve the equation ( |5x| + |12y| + |60 - 5x - 12y| = 60 ) and find the area of the resulting figure, we start by analyzing the equation in terms of transformed coordinates ( A = 5x ) and ( B = 12y ). This transforms the equation into ( |A| + |B| + |60 - A - B| = 60 ).We consider different cases based on the signs of ( A ), ( B ), and ( 60 - A - B ):1. Case 1: ( A geq 0 ), ( B geq 0 ), and ( 60 - A - B geq 0 ): - The equation simplifies to ( A + B + (60 - A - B) = 60 ), which is always true. This region is the triangle in the first quadrant bounded by ( A geq 0 ), ( B geq 0 ), and ( A + B leq 60 ).2. Other Quadrants (II, III, IV): - For other quadrants, we analyze the signs of ( A ), ( B ), and ( 60 - A - B ). However, we find no solutions in these regions because the resulting equations lead to contradictions or require values outside the valid ranges.Thus, the solution set in the ( (A, B) ) plane is the triangle with vertices at ( (0, 0) ), ( (60, 0) ), and ( (0, 60) ).Transforming back to the ( (x, y) ) plane using ( A = 5x ) and ( B = 12y ), the vertices become:- ( (0, 0) )- ( (12, 0) ) (since ( 60/5 = 12 ))- ( (0, 5) ) (since ( 60/12 = 5 ))This forms a right triangle with legs 12 and 5. The area of this triangle is calculated as:[text{Area} = frac{1}{2} times 12 times 5 = 30]Thus, the area of the resulting figure is (boxed{30}).]

🔑:Answer: 30.Solution. Note that the equality |a|+|b|+|c|=a+b+c holds if and only if the numbers a, b, and c are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalitiesleft{begin{array} { l } { 5 x geq 0 } { 1 2 y geq 0 , } { 6 0 - 5 x - 1 2 y geq 0 }end{array} Leftrightarrow left{begin{array}{l}x geq 0 y geq 0 5 x+12 y leq 60end{array}right.right.This system defines a triangle on the plane with vertices E(12 ; 0), G(0 ; 5), N(0 ; 0), the area of which is 30.