🧠:[<think>Okay, so I need to use Horner's method to evaluate the polynomial ( f(x) = x^6 + 6x^4 + 9x^2 + 208 ) at ( x = -4 ), and find the value of ( v_2 ). The answer choices are A through D: -4, 1, 17, 22. Hmm, let me recall how Horner's method works.Horner's method is a way to evaluate polynomials efficiently by reducing the number of multiplications. It involves rewriting the polynomial in a nested form. For a general polynomial ( f(x) = a_n x^n + a_{n-1} x^{n-1} + dots + a_1 x + a_0 ), Horner's method would express it as ( (((dots(a_n x + a_{n-1})x + a_{n-2})x + dots)x + a_0) ). This way, you start by multiplying the highest coefficient by x, add the next coefficient, multiply by x, and so on until you reach the constant term.But looking at the given polynomial ( f(x) = x^6 + 6x^4 + 9x^2 + 208 ), I notice that it has missing terms. Specifically, the ( x^5 ), ( x^3 ), and ( x ) terms are all missing, which means their coefficients are zero. So, the polynomial can be written as ( f(x) = x^6 + 0x^5 + 6x^4 + 0x^3 + 9x^2 + 0x + 208 ). That's important because when applying Horner's method, we have to include all the coefficients, even the zero ones. Otherwise, the nesting won't work correctly.So, the coefficients in order from the highest power (x^6) to the constant term are: 1 (for x^6), 0 (x^5), 6 (x^4), 0 (x^3), 9 (x^2), 0 (x), and 208 (constant term). Let me list them out:1. ( a_6 = 1 )2. ( a_5 = 0 )3. ( a_4 = 6 )4. ( a_3 = 0 )5. ( a_2 = 9 )6. ( a_1 = 0 )7. ( a_0 = 208 )Now, Horner's method typically requires starting with the highest coefficient and working down. Since the polynomial is degree 6, there should be 6 iterations of multiplication and addition. Wait, but let me confirm: for a polynomial of degree n, you need n steps. So, for degree 6, 6 steps. Let me also recall that in Horner's method, each step involves multiplying the current result by x and adding the next coefficient.But since there are missing terms (coefficients are zero), we have to include them in the process. So, starting with the leading coefficient (1), then multiply by x (-4), add the next coefficient (0), then multiply by x (-4), add next coefficient (6), and so on.But the question is asking for the value of ( v_2 ). Hmm, I need to recall how the steps are labeled. In some explanations, the steps are labeled starting from the first multiplication. So, starting with the coefficient of x^6, which is 1. Then step 1 would be multiplying by x and adding the next coefficient (0). Step 2 would be multiplying the result by x and adding the next coefficient (6), and so on. Wait, but actually, Horner's method is a sequence of operations. Let me think.Let me write down the process step by step. Let me denote ( v_i ) as the value computed at each step. Starting with the leading coefficient:( v_0 = a_6 = 1 )Then, each subsequent ( v_i = v_{i-1} times x + a_{6 - i} )Wait, maybe that's how it's structured. So, for each step i from 1 to 6, compute ( v_i = v_{i-1} times x + a_{6 - i} )So, the coefficients are ordered from ( a_6 ) down to ( a_0 ). So, step 1: ( v_1 = v_0 times x + a_5 )But ( a_5 = 0 ). Then step 2: ( v_2 = v_1 times x + a_4 ), etc.Given that, let's compute each step:First, ( v_0 = a_6 = 1 )Step 1: ( v_1 = v_0 times (-4) + a_5 = 1 times (-4) + 0 = -4 )Step 2: ( v_2 = v_1 times (-4) + a_4 = (-4) times (-4) + 6 = 16 + 6 = 22 )Wait, so ( v_2 = 22 ). That would correspond to answer D. But let me verify this step by step, because the question is specifically asking for ( v_2 ). But let's be careful here. If the steps are labeled starting from 0, but in some sources, the steps might be counted differently. Wait, maybe the user is referring to the index based on the coefficients. Let me check again.Given that the polynomial is degree 6, and written in descending order, Horner's method would proceed as follows:Starting with coefficient 1 (x^6 term). Then, multiply by x and add the next coefficient (0 for x^5). Then multiply by x and add the next coefficient (6 for x^4). Since we're at step 2, which would be after two multiplications and additions. Let me map it out:1. Start with coefficient of x^6: 12. Multiply by x (-4): 1 * (-4) = -4. Add coefficient of x^5 (0): -4 + 0 = -4. This is after first iteration (v1).3. Multiply by x (-4): -4 * (-4) = 16. Add coefficient of x^4 (6): 16 + 6 = 22. This is after second iteration (v2).So, yes, v2 is 22, which is option D. However, let me check if this is correct by actually computing the polynomial value step by step and see if the intermediate steps match.Alternatively, compute the entire polynomial using Horner's method and verify the intermediate steps. Let's do that to be sure.The polynomial is f(x) = x^6 + 6x^4 + 9x^2 + 208. To apply Horner's method, we need to write it in a nested form. However, since there are missing terms, we have to include the zero coefficients. So, the polynomial is:f(x) = x^6 + 0x^5 + 6x^4 + 0x^3 + 9x^2 + 0x + 208To apply Horner's method, we can start nesting from the highest degree. Let me rewrite the polynomial:f(x) = (((((x^2 + 0x) * x^2 + 6) * x^2 + 0) * x^2 + 9) * x^2 + 0) * x^2 + 208Wait, but maybe that's not the standard Horner's method. Alternatively, perhaps factor out x^2 terms.Wait, another approach: since the polynomial has only even powers, maybe we can substitute y = x^2, then the polynomial becomes y^3 + 6y^2 + 9y + 208. Then, evaluate at y = (-4)^2 = 16. Then use Horner's method on the cubic in y. But the problem specifically asks to use Horner's method on the original polynomial, so substituting y = x^2 might not be what they want. Although, it's a valid approach, but since the question is about v2 in Horner's method applied to the original polynomial, maybe the substitution complicates the step counting.Alternatively, let's stick to the original plan. Let's write the polynomial with all coefficients:x^6 + 0x^5 + 6x^4 + 0x^3 + 9x^2 + 0x + 208Therefore, the coefficients are [1, 0, 6, 0, 9, 0, 208]Now, using Horner's method for degree 6:Start with v0 = 1 (coefficient of x^6)v1 = v0 * x + 0 = 1*(-4) + 0 = -4v2 = v1 * x + 6 = (-4)*(-4) + 6 = 16 + 6 = 22v3 = v2 * x + 0 = 22*(-4) + 0 = -88 + 0 = -88v4 = v3 * x + 9 = (-88)*(-4) + 9 = 352 + 9 = 361v5 = v4 * x + 0 = 361*(-4) + 0 = -1444 + 0 = -1444v6 = v5 * x + 208 = (-1444)*(-4) + 208 = 5776 + 208 = 5984So, the final result is 5984. Let me confirm this by plugging x = -4 into the original polynomial:f(-4) = (-4)^6 + 6*(-4)^4 + 9*(-4)^2 + 208Calculate each term:(-4)^6 = 4096 (since even exponent, positive)6*(-4)^4 = 6*(256) = 15369*(-4)^2 = 9*16 = 144208 remains as is.Adding them up: 4096 + 1536 = 5632; 5632 + 144 = 5776; 5776 + 208 = 5984. Correct, matches the Horner's result.Therefore, the intermediate value v2 is 22, which is option D. So, the answer should be D.But just to be thorough, let's check again the steps:v0 = 1v1 = 1*(-4) + 0 = -4v2 = (-4)*(-4) + 6 = 16 + 6 = 22Yes, that's two steps after starting with v0. So, v2 is 22. Therefore, the correct answer is D.Final Answerboxed{D}</think>To evaluate the polynomial ( f(x) = x^6 + 6x^4 + 9x^2 + 208 ) at ( x = -4 ) using Horner's method, we need to include all coefficients, even the zero ones. The polynomial can be written as ( f(x) = x^6 + 0x^5 + 6x^4 + 0x^3 + 9x^2 + 0x + 208 ). The coefficients are [1, 0, 6, 0, 9, 0, 208].Using Horner's method, we start with the highest coefficient and proceed step by step:1. ( v_0 = 1 ) (coefficient of ( x^6 ))2. ( v_1 = v_0 times (-4) + 0 = 1 times (-4) + 0 = -4 )3. ( v_2 = v_1 times (-4) + 6 = (-4) times (-4) + 6 = 16 + 6 = 22 )Thus, the value of ( v_2 ) is (boxed{D}).,<think>Okay, so I need to use Horner's method to evaluate the polynomial f(x) = x^6 + 6x^4 + 9x^2 + 208 at x = -4, and find the value of v₂. The options are A: -4, B: 1, C: 17, D: 22. Hmm, let's start by recalling what Horner's method is. From what I remember, Horner's method is a way to evaluate a polynomial efficiently by reducing the number of multiplications. Instead of computing each term separately and then adding them up, you factor the polynomial in a nested form. For example, a polynomial like 3x³ + 2x² + 5x + 7 can be written as ((3x + 2)x + 5)x + 7. This way, you start from the innermost part and work your way out, which minimizes the number of operations.But the given polynomial here is f(x) = x⁶ + 6x⁴ + 9x² + 208. That seems a bit tricky because the exponents are 6, 4, 2, and 0. They are even and decrease by 2 each time. So maybe I can factor this polynomial in terms of x². Let me try that. Let's set y = x². Then the polynomial becomes y³ + 6y² + 9y + 208. If I do that substitution, then f(x) can be written as (y³ + 6y² + 9y + 208) where y = x². Now, this might be easier to apply Horner's method to because it's a cubic in terms of y. But I need to evaluate this at x = -4, which means y = (-4)² = 16. So I need to compute f(-4) = 16³ + 6*16² + 9*16 + 208. Let me compute that directly first to check later. 16³ is 4096, 6*16² is 6*256=1536, 9*16=144, so adding those up: 4096 + 1536 = 5632, 5632 + 144 = 5776, 5776 + 208 = 5984. So f(-4) should be 5984. But the question isn't asking for the final value, it's asking for v₂. Hmm. Wait, in Horner's method, when you apply the algorithm, you compute intermediate values v₀, v₁, v₂, etc., depending on the coefficients. So maybe I need to set up the polynomial in a way that allows me to use Horner's method step by step and identify what v₂ is.But first, let me make sure I recall how Horner's method works step by step. For a general polynomial aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀, Horner's method rewrites it as (...((aₙx + aₙ₋₁)x + aₙ₋₂)x + ... + a₁)x + a₀. Then, to evaluate at a specific x value, you start with the leading coefficient aₙ, multiply by x, add the next coefficient, multiply by x, add the next coefficient, and so on. Each step gives a value v_i which is used in the next step. The final v_n is the value of the polynomial at x.But in our case, the original polynomial is only in even powers. So if we write it as x⁶ + 6x⁴ + 9x² + 208, which is equivalent to (x²)³ + 6(x²)² + 9(x²) + 208. So substituting y = x² gives us a cubic polynomial in y: y³ + 6y² + 9y + 208. So perhaps we can apply Horner's method to this cubic polynomial first, and then substitute y = x². But I need to evaluate this at x = -4, which makes y = 16. So perhaps first apply Horner's method to the cubic in y, evaluating at y = 16. Let me set that up.The cubic is 1*y³ + 6*y² + 9*y + 208. So in Horner's form, this would be ((1*y + 6)*y + 9)*y + 208. So starting with the coefficient of y³, which is 1. Then multiply by y (16) and add the next coefficient (6): 1*16 + 6 = 22. Then take that result (22), multiply by y (16), add the next coefficient (9): 22*16 + 9. Let's compute that: 22*16 is 352, plus 9 is 361. Then take 361, multiply by y (16), add 208: 361*16 + 208. Let's compute that: 361*16 is 5776, plus 208 is 5984. Which matches the direct calculation. So in this process, the intermediate values are v₀ = 1, v₁ = 1*16 +6 =22, v₂ =22*16 +9= 361, v₃=361*16 +208=5984. So if the question is referring to v₂ in this process, then v₂ would be 361. But 361 isn't one of the options (options are -4, 1, 17, 22). So maybe I'm misunderstanding the problem.Wait, perhaps the question is expecting me to apply Horner's method to the original polynomial in x, not in y. Let's check. The original polynomial is x⁶ +6x⁴ +9x² +208. If we want to apply Horner's method directly to this polynomial without substitution, how would we do it? Because normally Horner's method works for polynomials with consecutive descending exponents. Here, we have exponents 6,4,2,0. So there are missing terms (x⁵, x³, x terms are missing). So if I write the polynomial as x⁶ +0x⁵ +6x⁴ +0x³ +9x² +0x +208. Then, arranging the coefficients in order: 1 (x⁶), 0 (x⁵), 6 (x⁴), 0 (x³), 9 (x²), 0 (x), 208 (constant term). Then, applying Horner's method for x = -4. Let's try this approach.So, the coefficients are [1, 0, 6, 0, 9, 0, 208]. The value x = -4. Let's set up Horner's table. Starting with the leading coefficient 1. Then, multiply by x (-4) and add the next coefficient (0): 1*(-4) +0 = -4. Then take -4, multiply by x (-4): -4*(-4)=16, add next coefficient (6): 16 +6=22. Then take 22, multiply by x (-4):22*(-4)= -88, add next coefficient (0): -88 +0= -88. Then take -88, multiply by x (-4): -88*(-4)=352, add next coefficient (9): 352 +9=361. Then take 361, multiply by x (-4): 361*(-4)= -1444, add next coefficient (0): -1444 +0= -1444. Then take -1444, multiply by x (-4): -1444*(-4)=5776, add next coefficient (208):5776 +208=5984. So the final result is 5984, as before. Now, in this process, the intermediate values are:v₀ =1v₁ =1*(-4) +0 = -4v₂ = -4*(-4) +6 =16 +6=22v₃=22*(-4) +0= -88 +0=-88v₄= -88*(-4) +9=352 +9=361v₅=361*(-4) +0= -1444 +0=-1444v₆= -1444*(-4) +208=5776 +208=5984So the question is asking for the value of v₂. In this case, v₂ is 22. Looking at the options, D is 22. So the answer should be D.But wait, let me make sure. The original problem says "use the Horner's method to calculate the value of the polynomial... What is the value of v₂?" Since Horner's method can be applied in different ways depending on how you structure the polynomial, it's important to structure it as per the standard method. In the standard Horner's method for a polynomial with all exponents present, you would write it in decreasing order and include zero coefficients for missing terms. Therefore, by expanding the polynomial to include all missing terms with zero coefficients, we have to process each coefficient step by step.Therefore, in the expanded coefficient list [1, 0, 6, 0, 9, 0, 208], when applying Horner's method at x = -4, the intermediate values v₀ through v₆ are calculated as above. Here, v₂ is 22, which corresponds to option D. So the answer is D: 22.But let me just cross-verify once again to be thorough. Let's recount the steps:1. Start with coefficient of x⁶: 1. This is v₀ =1.2. Multiply by x (-4): 1*(-4) = -4. Add next coefficient (0): -4 +0 = -4. This is v₁ = -4.3. Multiply v₁ by x: -4*(-4) =16. Add next coefficient (6):16 +6=22. This is v₂=22.Yes, that seems correct. So after the third step (processing the x⁴ coefficient), we get v₂=22. So the correct answer is D.Final Answerboxed{D}</think>To evaluate the polynomial ( f(x) = x^6 + 6x^4 + 9x^2 + 208 ) at ( x = -4 ) using Horner's method, we first note that the polynomial can be written with all coefficients included, including zeros for the missing terms. This gives us the polynomial:[ f(x) = x^6 + 0x^5 + 6x^4 + 0x^3 + 9x^2 + 0x + 208 ]The coefficients are [1, 0, 6, 0, 9, 0, 208]. Applying Horner's method at ( x = -4 ):1. Start with the leading coefficient: ( v_0 = 1 )2. Multiply by ( x = -4 ) and add the next coefficient: ( v_1 = 1 cdot (-4) + 0 = -4 )3. Multiply by ( x = -4 ) and add the next coefficient: ( v_2 = -4 cdot (-4) + 6 = 16 + 6 = 22 )Thus, the value of ( v_2 ) is (boxed{D}).]

🔑:AnalysisThis question tests the knowledge of Horner's method. Mastering the operation rules of Horner's method is key to solving this problem. First, rewrite the polynomial in the following form: (f(x) = (((((x)x + 6)x)x + 9)x)x + 208). Substitute (x = -4) and calculate the values of (v_0), (v_1), and (v_2) in sequence to find the answer.SolutionGiven (f(x) = 208 + 9x^2 + 6x^4 + x^6 = (((((x)x + 6)x)x + 9)x)x + 208),When (x = -4),(v_0 = 1),(v_1 = 1 times (-4) = -4),(v_2 = -4 times (-4) + 6 = 22),Therefore, the correct choice is boxed{D}.

🧠:[<think>Okay, so I need to find the last four digits of 5 raised to the power of 2013. Hmm, the problem also gives some examples: 5^5 is 3125, 5^6 is 15625, 5^7 is 78125, and so on. Let me see if I can spot a pattern here.Looking at those examples:- 5^5 = 3125 → last four digits are 3125- 5^6 = 15625 → last four digits are 5625- 5^7 = 78125 → last four digits are 8125- 5^8 = 390625 → last four digits would be 0625- 5^9 = 1953125 → last four digits are 3125Wait a second, I notice that starting from 5^5, the last four digits cycle every four exponents: 3125, 5625, 8125, 0625, and then back to 3125 for 5^9. Let me check that again.Starting with exponent 5:- 5^5: 3125- 5^6: 15625 → 5625- 5^7: 78125 → 8125- 5^8: 390625 → 0625- 5^9: 1953125 → 3125 (same as 5^5)- 5^10: 9765625 → 5625 (same as 5^6)Yes, the pattern repeats every four exponents. So the cycle length is 4. That is, the last four digits of 5^n repeat every 4 exponents starting from n=5. So for n=5,6,7,8,9,10,… the last four digits cycle through 3125, 5625, 8125, 0625, and repeat.If this pattern holds, then we can find the last four digits of 5^2013 by determining where 2013 falls in this cycle.First, let's confirm the cycle length. The examples given show that the cycle starts at n=5 and repeats every 4. So exponents 5,9,13,… will all end with 3125; 6,10,14,… with 5625; 7,11,15,… with 8125; 8,12,16,… with 0625.Therefore, if we can express 2013 in terms of 5 + 4k + r, where r is 0,1,2,3, then we can find the corresponding last four digits. Wait, but starting from n=5, each subsequent exponent adds 1 to the exponent and cycles the last four digits. So maybe we can subtract 5 from the exponent and then take modulo 4?Let me test that. Let's see:For n=5: 5-5=0, 0 mod 4 =0 → corresponds to 3125 (if we start counting from 0)n=6: 6-5=1, 1 mod 4=1 → 5625n=7:7-5=2,2 mod4=2→8125n=8:8-5=3,3 mod4=3→0625n=9:9-5=4,4 mod4=0→3125, which matches.Therefore, the formula seems to be:If we let m = n -5, then the last four digits cycle every 4 exponents starting from m=0 (n=5). Therefore, the last four digits of 5^n are determined by m mod 4.So for exponent 2013, let's compute m =2013 -5=2008. Then 2008 mod4= 2008/4=502, remainder 0. Therefore, m mod4=0, which corresponds to the first term in the cycle, which is 3125. Therefore, the last four digits of 5^2013 should be 3125?Wait, but let me verify this with a higher exponent. For example, n=13: 13-5=8, 8 mod4=0, so last four digits should be 3125. Let's compute 5^13.5^10 is 9765625 → last four digits 56255^11=48828125 →81255^12=244140625→06255^13=1220703125→3125. Yes, that's correct. So at n=13, which is 5+8, 8 mod4=0, last four digits 3125. So the formula seems to hold.But then, according to this, 5^2013 would have last four digits 3125. But let me check n=9: 9-5=4, 4 mod4=0, which gives 3125. And 5^9 is 1953125, which does end with 3125. So that works.Wait, but let's test n=8. According to this formula, n=8: m=8-5=3, which mod4=3, which corresponds to 0625. 5^8=390625, which ends with 0625. Correct.So if this cycle is indeed every 4 exponents starting at n=5, then 5^2013 should end with 3125. But let me check with another approach to be sure.Another way: The last four digits of a number are equivalent to the number modulo 10000. So we need to compute 5^2013 mod10000.But 5 and 10000 are not coprime, since 5 divides 10000. So Euler's theorem doesn't apply directly here. However, perhaps we can use the Chinese Remainder Theorem by splitting modulus 10000 into 16 and 625, since 10000=16*625 and gcd(16,625)=1.Wait, 16 and 625 are coprime? 16 is 2^4, and 625 is 5^4. Yes, they are coprime. Therefore, by Chinese Remainder Theorem, we can compute 5^2013 mod16 and 5^2013 mod625, then combine the results.But since 5^2013 is a multiple of 5, 5^2013 mod16 can be computed as follows:Note that 5 and 16 are coprime. So Euler's theorem tells us that 5^φ(16)=5^8 ≡1 mod16. Therefore, 5^2013 ≡5^(2013 mod8) mod16.2013 divided by 8: 2013=8*251 +5. So 2013 mod8=5. Therefore, 5^2013 ≡5^5 mod16.5^1=5 mod165^2=25≡9 mod165^3=45≡13 mod165^4=65≡1 mod165^5=5 mod16So 5^5≡5 mod16. Therefore, 5^2013 ≡5 mod16.Now compute 5^2013 mod625. Since 625=5^4, and 5^2013 is 5^(4*503 +1)=5^4*503 *5^1. So 5^2013=5^(4*503 +1)= (5^4)^503 *5. Since 5^4=625, then (5^4)^503=625^503. Therefore, 5^2013 ≡0*5=0 mod625. Wait, but 5^4 ≡0 mod625, so any higher power will also be 0. Therefore, 5^2013 ≡0 mod625.Therefore, we need to find a number x such that:x ≡5 mod16x ≡0 mod625We need to solve this system. Let x=625k for some integer k. Then substitute into the first equation:625k ≡5 mod16But 625 mod16: 625 divided by16 is 39*16=624, so 625≡1 mod16. Therefore, 625k ≡k ≡5 mod16. Therefore, k≡5 mod16.Therefore, k=16m +5 for some integer m. Therefore, x=625*(16m +5)=625*16m +3125. Therefore, the smallest positive solution is x=3125 mod10000. Therefore, 5^2013 ≡3125 mod10000. Hence, the last four digits are 3125.Wait, so both methods give the same result. The first method by observing the cycle and the second method using the Chinese Remainder Theorem both lead to 3125. Therefore, the answer should be 3125.But just to make sure, let's test for a higher exponent where both methods can be verified.For example, take n=9. 5^9=1953125, last four digits 3125. According to the cycle method: n=9, m=9-5=4, 4 mod4=0 →3125. According to the Chinese Remainder Theorem method:Compute 5^9 mod16 and mod625.5^9 mod16: φ(16)=8, so 5^8≡1 mod16. 5^9=5^8*5≡1*5=5 mod16.5^9 mod625: 5^4=625≡0 mod625, so 5^9=5^4*5^5=0*3125=0 mod625. Then solving x≡5 mod16 and x≡0 mod625, gives x=3125 mod10000. Correct.Similarly, n=10: last four digits 5625.Using cycle method: n=10, m=10-5=5, 5 mod4=1 →5625.Using CRT:5^10 mod16: φ(16)=8, 5^10=5^(8+2)=5^8*5^2≡1*25≡9 mod16.5^10 mod625=0. Solve x≡9 mod16 and x≡0 mod625.x=625k. 625k≡9 mod16. 625≡1 mod16, so k≡9 mod16. Thus, k=16m +9, x=625*(16m +9)=10000m +5625. Thus x≡5625 mod10000. Correct.Therefore, the CRT method works. So in our original problem, the answer is 3125. But let me check for n=2013.Wait, but according to the cycle method, since m=2013-5=2008, 2008 mod4=0, so last four digits 3125. And according to CRT, also 3125. So both agree.But just to confirm with another exponent, say n=8, which should be 0625.Using cycle method: n=8, m=3, 3 mod4=3 →0625.Using CRT:5^8 mod16: 5^8≡1 mod16.5^8 mod625=0. Solve x≡1 mod16 and x≡0 mod625. x=625k. 625k≡1 mod16 ⇒k≡1 mod16. Thus k=16m +1, x=625*(16m +1)=10000m +625. So x≡625 mod10000. But 0625 is the last four digits, so 0625. Wait, 625 is three digits, but as the last four digits, it's 0625. So x=0625 mod10000. Therefore, 5^8≡0625 mod10000. Correct.Hence, the methods are consistent.Therefore, confident that the last four digits of 5^2013 are 3125.Final AnswerThe last four digits of (5^{2013}) are boxed{3125}.</think>To find the last four digits of (5^{2013}), we observe the pattern in the last four digits of powers of 5 starting from (5^5):- (5^5 = 3125)- (5^6 = 15625) (last four digits: 5625)- (5^7 = 78125) (last four digits: 8125)- (5^8 = 390625) (last four digits: 0625)- (5^9 = 1953125) (last four digits: 3125)We notice a repeating cycle every 4 exponents starting from (5^5): 3125, 5625, 8125, 0625. To determine where (2013) falls in this cycle, we calculate (2013 - 5 = 2008). Taking (2008 mod 4) gives a remainder of 0, indicating that (5^{2013}) corresponds to the first term in the cycle, which is 3125.To confirm, we use the Chinese Remainder Theorem. We need (5^{2013} mod 10000). We split the modulus into 16 and 625:1. (5^{2013} mod 16): Since (5^8 equiv 1 mod 16), we find (2013 mod 8 = 5). Thus, (5^{2013} equiv 5^5 equiv 5 mod 16).2. (5^{2013} mod 625): Since (5^4 = 625), any higher power of 5 is 0 modulo 625.We solve the system (x equiv 5 mod 16) and (x equiv 0 mod 625). Let (x = 625k). Then (625k equiv 5 mod 16) simplifies to (k equiv 5 mod 16). Thus, (k = 16m + 5) and (x = 625(16m + 5) = 10000m + 3125). Hence, (x equiv 3125 mod 10000).Both methods confirm the last four digits of (5^{2013}) are (boxed{3125}).,<think>Okay, so I need to find the last four digits of 5 raised to the power of 2013. Hmm, let's see. The problem mentions some examples: 5^5 is 3125, 5^6 is 15625, 5^7 is 78125, and so on. I notice that starting from 5^5, the last four digits seem to follow a pattern. Let me check that again.5^5 = 3125 → last four digits are 31255^6 = 15625 → last four digits are 56255^7 = 78125 → last four digits are 81255^8 = 390625 → last four digits are 06255^9 = 1953125 → last four digits are 3125Wait a minute, this looks like a cycle. Let's list them out more clearly:5^5: 31255^6: 56255^7: 81255^8: 06255^9: 31255^10: 5625So the cycle here is 3125, 5625, 8125, 0625, and then repeats every 4 exponents. So starting from exponent 5, the last four digits cycle every 4. Let me confirm this by calculating a few more exponents.5^11: 5^10 is 9765625, multiplied by 5 is 48828125 → last four digits 81255^12: 244140625 → last four digits 06255^13: 1220703125 → 3125Yep, it definitely cycles every 4 exponents starting from 5^5. So the cycle is 3125, 5625, 8125, 0625, repeating every 4. So the exponents where the cycle starts are 5, 6, 7, 8, then 9, 10, 11, 12, etc. So each subsequent exponent increases by 1, and the cycle repeats every 4 exponents.So, if we can model this, starting at exponent n=5, the cycle length is 4. So to find the last four digits of 5^2013, we need to see where 2013 falls in this cycle.But wait, the cycle starts at exponent 5. So exponent 5 corresponds to cycle position 1 (3125), exponent 6 is position 2 (5625), exponent 7 is position 3 (8125), exponent 8 is position 4 (0625), exponent 9 is position 1 again, and so on. So the position in the cycle can be calculated by (exponent - 5) mod 4. If that result is 0, it's position 4; otherwise, it's the result.So let's compute (2013 - 5) mod 4. That's 2008 mod 4. 2008 divided by 4 is 502, with remainder 0. So 2008 mod 4 is 0. Therefore, position 4 in the cycle, which corresponds to 0625. So the last four digits of 5^2013 should be 0625.Wait, but hold on. Let me verify this with an example. For exponent 8, which is 5^8=390625, the last four digits are 0625. Then exponent 12 would be 5^12=244140625, which also ends with 0625. So according to this cycle, every exponent that is 8, 12, 16,... should end with 0625. Similarly, 2013 - 5 = 2008, which is divisible by 4, so exponent 2013 is 2008 above exponent 5, which is 4*502, so that would put it at position 4 in the cycle. So 0625. Therefore, the answer should be 0625.But let me double-check my cycle assumption. Let's verify with exponent 9. 5^9 is 1953125, last four digits 3125. Which is the same as 5^5. Then 5^10 is 9765625, ending with 5625, same as 5^6. So yes, the cycle repeats every 4 exponents. Therefore, the cycle length is 4 starting at exponent 5.So the general formula is: For exponents n ≥ 5, the last four digits of 5^n cycle every 4 exponents. So the last four digits depend on (n - 5) mod 4. If the result is 0, 1, 2, 3, then it corresponds to 0625, 3125, 5625, 8125, respectively? Wait, hold on.Wait, when n=5: (5-5) mod 4 = 0 mod 4 = 0. But n=5 gives 3125, which is the first in the cycle. Wait, maybe my earlier reasoning is off.Wait, perhaps I need to adjust the formula. Let's re-examine:Looking at the exponents:n | 5^n | last four digits5 | 31256 | 15625 → 56257 | 78125 → 81258 | 390625 → 06259 | 1953125 → 312510 | 9765625 → 562511 | 48828125 → 812512 | 244140625 → 0625So starting from n=5, the cycle is 3125, 5625, 8125, 0625, and then repeats every 4 exponents. So the cycle is of length 4. So if we let k = n - 5, then the cycle position is k mod 4.For n=5, k=0, cycle position 0: 3125n=6, k=1: cycle position 1: 5625n=7, k=2: cycle position 2: 8125n=8, k=3: cycle position 3: 0625n=9, k=4: cycle position 0 (since 4 mod 4=0): 3125So the cycle position is k mod 4, where k = n - 5. Then:If cycle position 0: 31251: 56252: 81253: 0625Therefore, to find the last four digits of 5^n for n ≥5, compute k = n -5, then cycle_pos = k mod 4.Given n=2013:k = 2013 -5 = 2008cycle_pos = 2008 mod 42008 divided by 4 is 502 with 0 remainder. So cycle_pos=0Therefore, the last four digits are 3125. Wait, but hold on. Wait, for cycle_pos=0, which corresponds to n=5,9,13,... which indeed give 3125. But wait, when n=8, k=3, cycle_pos=3, last four digits 0625.So if cycle_pos=0 → 3125cycle_pos=1 →5625cycle_pos=2→8125cycle_pos=3→0625But for n=2013, cycle_pos=0, so last four digits 3125.But wait, this contradicts my earlier conclusion. Wait, now I'm confused.Wait, let's recast:k = n -5cycle_pos = k mod 4.So for n=5, k=0, cycle_pos=0: 3125n=6, k=1, cycle_pos=1:5625n=7, k=2, cycle_pos=2:8125n=8, k=3, cycle_pos=3:0625n=9, k=4, cycle_pos=0:3125Therefore, cycle_pos=0 corresponds to 3125But in the earlier calculation, when n=2013, k=2008, cycle_pos=0. So last four digits should be 3125. But in my initial reasoning, I thought exponent 2013 would be 0625. Wait, that was an error.Wait, so where did I go wrong before? Let me recast the problem.If cycle_pos is 0,1,2,3 corresponding to 3125,5625,8125,0625. So when n=5, cycle_pos=0, 3125. For n=2013, cycle_pos=0, so 3125.But earlier, when I thought of (n-5) mod4=0, leading to 0625, that was incorrect. Wait, no. Wait, wait. Wait, let's check for n=8:n=8, k=3, cycle_pos=3:0625So if we have k=2008, cycle_pos=0, which would lead to 3125. So why did I think it was 0625?Because I might have confused cycle_pos as (n-5) mod4, but mapped 0 to the 4th element. That was my mistake. Actually, cycle_pos=0 maps to first element, cycle_pos=1 to the second, etc. So if cycle_pos=0, it's 3125, cycle_pos=1:5625, cycle_pos=2:8125, cycle_pos=3:0625.Therefore, n=2013, k=2008, cycle_pos=0. So last four digits 3125.But wait, but when I thought of exponent 8, which is k=3, cycle_pos=3, which is 0625. So why is that?Wait, so perhaps the answer should be 3125. But let's verify with exponent 13.n=13: 5^13. Let's compute 5^8 is 390625 (0625), then 5^9=1953125 (3125), 5^10=9765625 (5625), 5^11=48828125 (8125), 5^12=244140625 (0625), 5^13=1220703125 (3125). So yes, n=13, which is 5+8=13, k=8, cycle_pos=8 mod4=0, which gives 3125. So correct.Therefore, for n=2013, which is 5 + 2008, so k=2008, cycle_pos=2008 mod4=0, so last four digits 3125.But wait, this contradicts the earlier examples where exponents 8,12,16,... give 0625. But according to the cycle_pos, n=8: k=3, cycle_pos=3:0625. n=12: k=7, cycle_pos=7 mod4=3:0625. Similarly, n=16: k=11, cycle_pos=11 mod4=3:0625. So every time cycle_pos=3, last four digits 0625. So cycle_pos=0 gives 3125, which happens at n=5,9,13,17,...Therefore, for n=2013, which is 2013-5=2008, cycle_pos=2008 mod4=0, which is cycle_pos=0, leading to last four digits 3125.But this contradicts my initial thought. Wait, so maybe I need to check with a larger exponent.Wait, let's take n=9, which should be cycle_pos=0. 5^9=1953125 → last four digits 3125. Correct. n=13:5^13=1220703125 → 3125. Correct.n=17:5^17=762939453125 → last four digits 3125. Correct. So every 4 exponents starting at n=5,9,13,17,... the last four digits are 3125.Similarly, exponents 6,10,14,18,... give 5625.Exponents 7,11,15,19,... give 8125.Exponents 8,12,16,20,... give 0625.Therefore, 2013 is in which category? Let's see: 2013 divided by 4. Wait, no. The cycle is every 4 exponents starting from 5. So the cycle repeats every 4 exponents. So if we subtract 5 from the exponent, then divide by 4, the remainder will tell us the position.But another way: Since the cycle length is 4, starting at exponent 5, then exponent 5 + 4k will have last four digits 3125. 5 + 4k +1:5625, +2:8125, +3:0625.So to find 2013, we can write 2013 =5 +4k +r where r=0,1,2,3.But 2013-5=2008. 2008 divided by 4 is 502, remainder 0. Therefore, r=0. So 2013=5 +4*502 +0. Therefore, it's in the same category as exponent 5 +4*502=2013. So that would correspond to r=0, which is the first in the cycle, 3125.Therefore, the last four digits of 5^2013 should be 3125.But earlier, when I first thought, I thought it was 0625, but that was a miscalculation.Wait, so where was the mistake? Initially, I thought that (2013 -5) mod4=0 corresponds to the fourth position, but actually, according to the cycle_pos=0,1,2,3 mapping to 3125,5625,8125,0625, then cycle_pos=0 is 3125.Therefore, the correct answer should be 3125.Wait, but let's check with an exponent that is 5 +4k. For example, 5+4*2=13. 5^13=1220703125, last four digits 3125. Correct. 5+4*3=17: 5^17=...3125. Correct. So 2013=5+4*502, which is 5+2008=2013. Therefore, same as 5^5, 5^9, etc., so last four digits 3125.Therefore, the answer is 3125.But wait, let me cross-verify with another approach. Maybe using modular arithmetic.We need to compute 5^2013 mod 10000. Since we want the last four digits.But 10000 factors into 16 * 625, and 16 and 625 are coprime (since 16 is 2^4 and 625 is 5^4). Therefore, by the Chinese Remainder Theorem, we can compute 5^2013 mod 16 and 5^2013 mod 625, then combine them.But since 5 and 16 are coprime? Wait, no. 5 and 16 are coprime (gcd(5,16)=1), but 5 and 625 are not coprime (gcd(5,625)=5). Hmm, so maybe this complicates things.Alternatively, since 5^2013 is a multiple of 5^4=625, so 5^2013 mod 10000 can be written as 625 * 5^(2013-4) mod 10000. Wait, but 5^2013 = 5^4 * 5^2009 = 625 * 5^2009. So we can compute 625 * (5^2009 mod 16), because 625 and 16 are coprime? Wait, maybe not straightforward.Alternatively, note that 5^2013 mod 16 and mod 625.First, compute 5^2013 mod 16.Since 5 and 16 are coprime, Euler's theorem tells us that phi(16)=8, so 5^8 ≡1 mod16.So 5^2013=5^(8*251 +5)= (5^8)^251 *5^5 ≡1^251 *5^5 mod16.5^5=3125. 3125 mod16: 3125 divided by16 is 195*16=3120, so 3125-3120=5. So 5^5≡5 mod16. Therefore, 5^2013≡5 mod16.Now compute 5^2013 mod625.But 5^2013 is a multiple of 625, since 5^4=625. So 5^2013=5^4*5^2009=625*5^2009. Therefore, 5^2013≡0 mod625.So we have:5^2013 ≡5 mod165^2013≡0 mod625We need to find x such that x≡0 mod625 and x≡5 mod16.Let x=625k. Then 625k≡5 mod16.625 mod16: 625 divided by16 is 39*16=624, so 625≡1 mod16. Therefore, 625k≡k mod16≡5 mod16.Thus, k≡5 mod16. Therefore, the smallest positive k is 5. Therefore, x=625*5=3125.Check: 3125 mod16=3125-195*16=3125-3120=5. Correct. And 3125 mod625=0. Correct.Therefore, the solution is x≡3125 mod10000. Therefore, the last four digits of 5^2013 are 3125.Therefore, the correct answer is 3125, not 0625. So my initial mistake was due to miscounting the cycle positions. The modulus approach confirms it's 3125.Final AnswerThe last four digits of (5^{2013}) are (boxed{3125}).</think>To find the last four digits of (5^{2013}), we observe the pattern in the last four digits of powers of 5 starting from (5^5):[begin{align*}5^5 & = 3125 5^6 & = 15625 quad (text{last four digits } 5625) 5^7 & = 78125 quad (text{last four digits } 8125) 5^8 & = 390625 quad (text{last four digits } 0625) 5^9 & = 1953125 quad (text{last four digits } 3125) end{align*}]We notice a repeating cycle every 4 exponents starting from (5^5): 3125, 5625, 8125, 0625. To determine the position of (5^{2013}) in this cycle, we calculate the exponent's position relative to the start of the cycle at (5^5):[2013 - 5 = 2008]Finding the position in the cycle:[2008 mod 4 = 0]A result of 0 indicates the fourth position in the cycle (0625, 3125, 5625, 8125), but since the cycle starts at (5^5), we need to confirm using modular arithmetic. Using the Chinese Remainder Theorem, we compute (5^{2013} mod 16) and (5^{2013} mod 625):1. (5^{2013} mod 16): [ 5^8 equiv 1 mod 16 implies 5^{2013} = 5^{8 cdot 251 + 5} equiv 5^5 equiv 5 mod 16 ]2. (5^{2013} mod 625): [ 5^{2013} equiv 0 mod 625 ]Combining these results, we solve (x equiv 0 mod 625) and (x equiv 5 mod 16):[x = 625k implies 625k equiv 5 mod 16 implies k equiv 5 mod 16 implies x = 625 cdot 5 = 3125]Thus, the last four digits of (5^{2013}) are (boxed{3125}).]

🔑:The last four digits of 5^5=3125 are 3125, the last four digits of 5^6=15625 are 5625, the last four digits of 5^7=78125 are 8125, the last four digits of 5^8=390625 are 0625, the last four digits of 5^9=1953125 are 3125..., according to the change in the last four digits, 3125, 5625, 8125, 0625, the pattern of the last four digits changes in a cycle of 4. Therefore, 2013 divided by 4 leaves a remainder of 1, which means the last four digits are 3125. Thus, the last four digits of 5^{2013} are 3125. Hence, the answer is boxed{3125}.

🧠:[<think>Okay, let's see. I need to find the height of an isosceles trapezoid ABCD with bases AB and CD. There's a point P on AB such that AP is 11, BP is 27, CD is 34, and angle CPD is 90 degrees. Hmm, okay. Let me start by drawing this trapezoid to visualize it better.So, an isosceles trapezoid means the legs AD and BC are equal in length, and the base angles are equal. The bases are AB and CD, with AB being the top base and CD the bottom one. Wait, actually, since AP is 11 and BP is 27, AB must be AP + PB = 11 + 27 = 38. So AB is 38 units long, and CD is given as 34. The height is the distance between the two bases, which is what we need to find.Point P is on AB, 11 units from A and 27 units from B. The key piece here is that angle CPD is 90 degrees. So, points C, P, D form a right angle at P. That seems like we can use coordinate geometry or some right triangle properties here. Maybe coordinate geometry would be better because it allows me to set up equations based on coordinates.Let me set up a coordinate system. Let's place the trapezoid such that the bases AB and CD are horizontal. Let me put point A at (0, 0) and point B at (38, 0). Since it's an isosceles trapezoid, the legs AD and BC are symmetric. The base CD is 34 units long, so it should be centered below AB. The center of AB is at (19, 0), so the center of CD will be at (19, h), where h is the height. Wait, actually, since the height is h, CD should be shifted down by h. But coordinates usually go up, so maybe I should place AB at the bottom and CD at the top? Wait, no, the problem doesn't specify orientation, but in standard problems, bases are horizontal, and usually, the lower base is AB and the upper base is CD. But maybe it's the other way around. Wait, AP is on AB, so maybe AB is the lower base? Hmm, maybe not. Wait, perhaps it's better to place AB at the top and CD at the bottom. Wait, but the problem says "bases AB and CD". Maybe the orientation isn't specified, so I can choose either. Let me think.If I place AB as the upper base and CD as the lower base, then the legs AD and BC would slope downwards. The height h is the vertical distance between AB and CD. Let me go with that. So, coordinates: Let's set point A at (0, 0), point B at (38, 0). Then, since the trapezoid is isosceles, the legs AD and BC are symmetric with respect to the vertical line through the midpoint of AB. The midpoint of AB is at (19, 0). Similarly, the midpoint of CD will also be at (19, h), since the trapezoid is isosceles. Therefore, the coordinates of points C and D can be determined. Since CD is 34 units long, which is shorter than AB (38), so CD is centered at (19, h), so point D is at (19 - 17, h) = (2, h) and point C is at (19 + 17, h) = (36, h). Wait, but if CD is the lower base, then h would be negative? Hmm, maybe I should flip it. Maybe AB is the lower base, and CD is the upper base. Wait, but the problem didn't specify. Hmm. Let me clarify.Wait, actually, in standard terminology, the bases are the two parallel sides. The height is the distance between them. So, if AB and CD are the bases, then regardless of their positions, the height is the perpendicular distance between AB and CD. So perhaps it's better to set AB as the lower base on the x-axis and CD as the upper base. But then angle CPD is 90 degrees. Let's see.Alternatively, maybe the orientation doesn't matter as long as we're consistent. Let's proceed with coordinates.Let me place AB on the x-axis, with A at (0, 0) and B at (38, 0). Then point P is on AB, 11 units from A, so P is at (11, 0). Then CD is the other base, length 34. The trapezoid is isosceles, so the legs AD and BC are congruent and symmetric with respect to the vertical line through the midpoint of AB, which is at (19, 0). Therefore, the midpoint of CD should also lie on the vertical line x=19. So CD is 34 units long, so the coordinates of C and D would be (19 - 17, h) and (19 + 17, h), i.e., (2, h) and (36, h). Therefore, CD is from (2, h) to (36, h). So points C(36, h) and D(2, h). Then, point P is at (11, 0).Now, angle CPD is 90 degrees. So, points C(36, h), P(11, 0), D(2, h). The angle at P between points C and D is 90 degrees. So vectors PC and PD are perpendicular. Therefore, the dot product of vectors PC and PD should be zero.Let me compute vectors PC and PD. Vector PC is C - P = (36 - 11, h - 0) = (25, h). Vector PD is D - P = (2 - 11, h - 0) = (-9, h). The dot product of PC and PD should be zero.Dot product = (25)(-9) + (h)(h) = -225 + h² = 0. Therefore, h² = 225, so h = 15. Since height is positive, h = 15. So is that the answer? Wait, let me check my steps.Wait, coordinates: AB is from (0,0) to (38,0). P is at (11,0). CD is from (2,h) to (36,h). Then, vectors PC is (36 - 11, h - 0) = (25, h). PD is (2 - 11, h - 0) = (-9, h). Then the dot product is (25)(-9) + (h)(h) = -225 + h². Setting equal to zero gives h² = 225, so h = 15. That seems straightforward. But let me confirm that this is indeed correct.Wait, but in the trapezoid, the legs are AD and BC. So points A(0,0) to D(2,h) and B(38,0) to C(36,h). Let's check the lengths of AD and BC. AD would be sqrt((2 - 0)^2 + (h - 0)^2) = sqrt(4 + h²). BC would be sqrt((36 - 38)^2 + (h - 0)^2) = sqrt(4 + h²). So yes, they are equal, as it's an isosceles trapezoid. So that checks out.Therefore, with h = 15, the height is 15. But let me just visualize angle CPD being 90 degrees. So point P is on AB, 11 from A. So in coordinates, (11,0). Then, points C(36,15) and D(2,15). So PC is from (11,0) to (36,15), which is a distance of sqrt(25² + 15²) = sqrt(625 + 225) = sqrt(850). PD is from (11,0) to (2,15), which is sqrt((-9)^2 + 15^2) = sqrt(81 + 225) = sqrt(306). Then, in triangle CPD, with right angle at P, the Pythagorean theorem should hold: PC² + PD² = CD²? Wait, CD is 34, so CD² is 1156. Let's check PC² + PD² = 850 + 306 = 1156, which equals CD². That's correct! So triangle CPD is right-angled at P, which satisfies the Pythagorean theorem. So this confirms that h = 15 is indeed correct.Therefore, the height of the trapezoid is 15. So the answer should be boxed{15}.Wait, but let me think again. The problem states that it's an isosceles trapezoid with bases AB and CD, point P on AB with AP=11, BP=27, CD=34, angle CPD=90°, find the height. The approach I took was coordinate geometry, setting AB on the x-axis, P at (11,0), and found coordinates of C and D based on the trapezoid being isosceles, leading to h=15. Then verified via vectors and Pythagoras. It all checks out.Alternatively, maybe there's another method without coordinates. Let's try that as a cross-check.Since angle CPD is 90°, the point P lies on the circle with diameter CD. Wait, in the trapezoid, CD is 34. The circle with diameter CD would have radius 17, center at the midpoint of CD. But since the trapezoid is isosceles, the midpoint of CD is vertically below the midpoint of AB. Since AB is 38, its midpoint is at 19 units from A, so midpoint of CD is also 19 units from D and C? Wait, no. Wait, CD is 34, so its midpoint is 17 units from each end. So the midpoint of CD is horizontally aligned with the midpoint of AB. Since it's an isosceles trapezoid, the legs are symmetric. Therefore, midpoint of CD is vertically below midpoint of AB. Wait, in my coordinate system, the midpoint of AB is at (19,0) and midpoint of CD is at (19,h). Wait, but in my previous setup, midpoint of CD is at (19, h), but if CD is the upper base, then h is the height. Wait, no, in my coordinate system, CD was at (2,h) to (36,h), so midpoint at (19, h). But AB is at (0,0) to (38,0), midpoint at (19,0). So the vertical distance between midpoints is h. So the height is h. So in that case, the circle with diameter CD would have center at (19, h) and radius 17. Then, point P lies on this circle. Therefore, the distance from P to the center of the circle should be equal to the radius.So point P is at (11,0). The center of the circle is (19, h). The distance between P and the center is sqrt((19 - 11)^2 + (h - 0)^2) = sqrt(8^2 + h^2) = sqrt(64 + h^2). Since the radius is 17, this distance should equal 17. Therefore, sqrt(64 + h^2) = 17. Squaring both sides, 64 + h² = 289. Therefore, h² = 225, so h=15. That's the same result as before. So this is another way to see it. Since angle CPD is 90°, P lies on the circle with diameter CD. Then, the distance from P to the center is equal to the radius, leading to the same equation. Therefore, h=15. So this confirms the answer again.Therefore, the height is 15. Seems like two different methods both lead to h=15, so I think that's solid.Final Answerboxed{15}</think>Given an isosceles trapezoid (ABCD) with bases (AB) and (CD), where (AP = 11), (BP = 27), (CD = 34), and (angle CPD = 90^circ). We need to find the height of the trapezoid.1. Coordinates Setup: - Place (A) at ((0, 0)) and (B) at ((38, 0)). Point (P) is on (AB) such that (AP = 11), so (P) is at ((11, 0)). - The midpoint of (AB) is ((19, 0)). Since (ABCD) is an isosceles trapezoid, the midpoint of (CD) must be vertically aligned with the midpoint of (AB). Thus, the midpoint of (CD) is ((19, h)), where (h) is the height. - Given (CD = 34), the coordinates of (C) and (D) are ((36, h)) and ((2, h)) respectively.2. Vectors and Dot Product: - Vectors (PC) and (PD) are calculated as follows: - (PC = (36 - 11, h - 0) = (25, h)) - (PD = (2 - 11, h - 0) = (-9, h)) - The dot product of (PC) and (PD) must be zero because (angle CPD = 90^circ): [ (25)(-9) + (h)(h) = 0 implies -225 + h^2 = 0 implies h^2 = 225 implies h = 15 ]3. Verification Using Circle Property: - Since (angle CPD = 90^circ), point (P) lies on the circle with diameter (CD). The center of this circle is the midpoint of (CD), which is ((19, h)), and the radius is (17). - The distance from (P) to the center ((19, h)) must be equal to the radius: [ sqrt{(19 - 11)^2 + (h - 0)^2} = 17 implies sqrt{64 + h^2} = 17 implies 64 + h^2 = 289 implies h^2 = 225 implies h = 15 ]Both methods confirm that the height (h) of the isosceles trapezoid (ABCD) is (boxed{15}).,<think>Alright, so I have this problem about an isosceles trapezoid ABCD with bases AB and CD. There's a point P on AB such that AP is 11 units, BP is 27 units, CD is 34 units, and angle CPD is 90 degrees. I need to find the height of the trapezoid. Hmm, okay, let's start by drawing a diagram in my mind. First, an isosceles trapezoid has the two bases AB and CD parallel, and the legs AD and BC equal in length. The height is the perpendicular distance between these two bases. Point P is somewhere on AB, closer to A since AP is 11 and BP is 27, which makes the total length of AB equal to AP + BP = 11 + 27 = 38 units. CD is given as 34 units. So AB is longer than CD, which makes sense because in an isosceles trapezoid, the legs are equal and the bases are parallel but can be of different lengths.Since it's an isosceles trapezoid, the legs AD and BC are congruent, and the base angles are equal. The height h can be found if we know the legs' length and the difference in the bases, but here we might not have enough information yet. The key here is that angle CPD is 90 degrees. That seems important. So points C, P, D form a right angle at P. I need to recall if there's a theorem related to right angles in trapezoids or if coordinate geometry might help here. Maybe placing the trapezoid on a coordinate system would be useful. Let me try that.Let me place the trapezoid so that base AB is on the x-axis. Let's set point A at (0, 0) and point B at (38, 0) since AB is 38 units long. Since the trapezoid is isosceles, the legs AD and BC are symmetric with respect to the vertical line through the midpoint of AB. The midpoint of AB is at (19, 0). The top base CD is 34 units long, so its midpoint should also be vertically above (19, h), where h is the height. Therefore, points C and D would be located at (19 + 17, h) and (19 - 17, h), which simplifies to (36, h) and (2, h). Wait, because CD is 34, so half of CD is 17, so from the midpoint (19, h), adding and subtracting 17 gives the coordinates for C and D.Wait, but hold on. If AB is 38 and CD is 34, then the difference in the lengths is 4 units. So each side beyond the projection of CD onto AB would be 2 units. So, the coordinates of C and D would be (2, h) and (36, h)? Wait, no. Wait, if AB is 38 and CD is 34, the horizontal distance from each end of CD to the corresponding end of AB is (38 - 34)/2 = 2. So, if AB is from (0,0) to (38,0), then CD would be from (2, h) to (36, h). So yes, points C(36, h) and D(2, h). That seems correct.Point P is on AB with AP = 11, so starting from A(0,0), moving 11 units along AB, which is on the x-axis, so P is at (11, 0). Wait, AP is 11, so the coordinates of P would be (11, 0). Similarly, BP is 27, so from B(38,0) going 27 units to the left, we get to (38 - 27, 0) = (11, 0). So that's consistent. So P is at (11,0).Now, we have points C(36, h), D(2, h), and P(11,0). The angle CPD is 90 degrees. So triangle CPD is a right-angled triangle at P. Therefore, vectors PC and PD should be perpendicular. So the dot product of vectors PC and PD should be zero.Alternatively, the slopes of PC and PD should be negative reciprocals. Let me try both approaches.First, coordinates:C is (36, h), D is (2, h), P is (11,0).Vector PC is from P to C: (36 - 11, h - 0) = (25, h)Vector PD is from P to D: (2 - 11, h - 0) = (-9, h)The dot product of PC and PD is (25)(-9) + (h)(h) = -225 + h²Since angle CPD is 90 degrees, the dot product should be zero. Therefore:-225 + h² = 0Solving for h:h² = 225h = 15So the height is 15 units. Wait, that seems straightforward. Let me verify using slopes.Slope of PC: (h - 0)/(36 - 11) = h/25Slope of PD: (h - 0)/(2 - 11) = h/(-9) = -h/9If they are perpendicular, the product of the slopes should be -1:(h/25) * (-h/9) = -h²/225Set this equal to -1:-h²/225 = -1Multiply both sides by -1:h²/225 = 1h² = 225h = 15Same result. So the height is 15. Wait, that seems too straightforward. Did I miss something?Wait, let me check the coordinates again. If AB is 38 units from (0,0) to (38,0), then CD is 34 units. The midpoint of AB is at (19,0), so the midpoint of CD should be at (19, h). Then, since CD is 34, half of that is 17, so C is (19 + 17, h) = (36, h) and D is (19 - 17, h) = (2, h). Correct.Point P is at (11,0). Then vectors PC and PD are as calculated. The calculations for the dot product and slopes both check out. So h² = 225, h=15. So the height is 15. Hmm, seems right.But let me think again. Is there another way to approach this problem? Maybe coordinate geometry is the straightforward way here, but sometimes these problems have synthetic solutions too. Let me see.Alternatively, in the trapezoid, since it's isosceles, the legs are equal. The height h can be related to the legs. Let's denote the legs as AD and BC. The length of the legs can be found using the Pythagorean theorem. The horizontal distance between the bases is (AB - CD)/2 = (38 - 34)/2 = 2, so each leg forms a right triangle with base 2 and height h. Therefore, the legs have length sqrt(2² + h²) = sqrt(4 + h²). But I don't know if that's necessary here.But in the problem, angle CPD is 90 degrees. So maybe there's a circle involved here. If angle CPD is 90 degrees, then point P lies on the circle with diameter CD. Wait, yes! Because in a circle, the angle subtended by a diameter is a right angle. So if CD is the diameter, then any point on the circle will form a right angle with C and D. But here, point P is on AB, so the circle with diameter CD intersects AB at point P. Therefore, if we can find the radius of the circle and see where it intersects AB, but maybe that's complicating things.Alternatively, since we know coordinates, maybe coordinate geometry is the simplest way. But as above, we already found h=15. Let me just check if that makes sense. If h=15, then the legs AD and BC have length sqrt(2² +15²)=sqrt(4+225)=sqrt(229). Then, the distance from P to C and P to D can be checked.Wait, distance from P(11,0) to C(36,15): sqrt((36-11)^2 + (15-0)^2)=sqrt(25² +15²)=sqrt(625 +225)=sqrt(850)=5*sqrt(34)Similarly, distance from P(11,0) to D(2,15): sqrt((2-11)^2 + (15-0)^2)=sqrt((-9)^2 +15^2)=sqrt(81+225)=sqrt(306)=3*sqrt(34)Then, in triangle CPD, sides are 5√34, 3√34, and CD=34.Check if it's a right triangle: (5√34)^2 + (3√34)^2 = 25*34 + 9*34 = 34*(25+9)=34*34=34². Which is indeed CD squared. So yes, the triangle is right-angled at P. Therefore, the calculations are correct.Hence, the height is 15. So the answer is boxed{15}Final Answerboxed{15}</think>Given an isosceles trapezoid (ABCD) with bases (AB) and (CD), where (AP = 11), (BP = 27), (CD = 34), and (angle CPD = 90^circ). We need to find the height of the trapezoid.1. Coordinates Setup: - Place (A) at ((0, 0)) and (B) at ((38, 0)) since (AB = 38). - The midpoint of (AB) is ((19, 0)). The top base (CD) is 34 units long, so its midpoint is ((19, h)) where (h) is the height. - Coordinates of (C) and (D) are ((36, h)) and ((2, h)) respectively, as they are 17 units each from the midpoint.2. Point (P) on (AB): - (P) is 11 units from (A), so (P) is at ((11, 0)).3. Vectors and Dot Product: - Vectors (PC) and (PD) are calculated as ((25, h)) and ((-9, h)) respectively. - The dot product of (PC) and (PD) is ((25)(-9) + (h)(h) = -225 + h^2). - Since (angle CPD = 90^circ), the dot product must be zero: (-225 + h^2 = 0).4. Solving for (h): - (h^2 = 225) - (h = 15)5. Verification: - Using slopes, the product of the slopes of (PC) and (PD) is (-1), confirming the right angle. - Distances from (P) to (C) and (D) confirm the right triangle properties.Thus, the height of the isosceles trapezoid (ABCD) is (boxed{15}).]

🔑:Answer: 15 squareSolution: Drop projections of A, P, B onto C D to get A^{prime}, P^{prime}, B^{prime}. Since A^{prime} B^{prime}=38 and C D=34, we get that D A^{prime}=C B^{prime}=2. Thus, P^{prime} D=9 and P^{prime} C=25. Hence, the answer is P P^{prime}=sqrt{P^{prime} D cdot P^{prime} C}= 15.

🧠:[<think>Okay, let's see. Ana wants to buy apples and pears to make pies for her birthday. The problem is asking for the maximum number of fruits she can buy, but she needs to make pies of both fruits, so she can't just buy all apples or all pears. Also, her bag can hold up to 7 kg, which is 7000 grams. Each apple is 300 grams and each pear is 200 grams. So, we need to figure out how many apples and pears she can buy without exceeding the weight limit, and the total number of fruits should be as large as possible. And she needs at least one of each fruit.Hmm. Let me break this down. Let's denote the number of apples as x and the number of pears as y. Then the total weight would be 300x + 200y grams. This has to be less than or equal to 7000 grams. So the inequality is:300x + 200y ≤ 7000And since she needs at least one of each fruit, x ≥ 1 and y ≥ 1. We want to maximize the total number of fruits, which is x + y.So, the problem is to maximize x + y subject to 300x + 200y ≤ 7000, x ≥ 1, y ≥ 1, and x, y are integers.Alright, so this is a linear programming problem with integer constraints. Since both x and y must be integers, we might need to check some possibilities.Alternatively, maybe we can simplify the equation. Let's divide the entire inequality by 100 to make the numbers smaller:3x + 2y ≤ 70So now, we have 3x + 2y ≤ 70, x ≥ 1, y ≥ 1, integers. Maximize x + y.So, perhaps I can express y in terms of x. From the inequality:2y ≤ 70 - 3xy ≤ (70 - 3x)/2Since y has to be at least 1, then (70 - 3x)/2 ≥ 1 → 70 - 3x ≥ 2 → -3x ≥ -68 → x ≤ 68/3 ≈ 22.666. But since x must be an integer, x ≤ 22. Similarly, since x must be at least 1, x ranges from 1 to 22.But we want to maximize x + y. Let's note that for each x, the maximum possible y is floor[(70 - 3x)/2], and then the total fruits would be x + floor[(70 - 3x)/2]. So, we need to compute this for x from 1 to 22 and see which gives the maximum total.Alternatively, maybe there's a smarter way. Let's consider that each apple is heavier than a pear, so to maximize the number of fruits, Ana should buy more pears than apples. But since she needs at least one apple, perhaps the optimal is to have as many pears as possible with one apple. Let's check that first.If x = 1, then:3(1) + 2y ≤ 70 → 2y ≤ 67 → y ≤ 33.5 → y = 33.Total fruits: 1 + 33 = 34.Is this the maximum? Let's check with x = 2.3(2) + 2y ≤ 70 → 6 + 2y ≤ 70 → 2y ≤ 64 → y ≤ 32. Total fruits: 2 + 32 = 34. Same as before.Hmm, same total. x=3:3*3 + 2y ≤70 →9 +2y ≤70→2y ≤61→y ≤30.5→30. Total fruits 3+30=33. Less than 34.So, decreasing by 1. x=4:3*4=12. 2y ≤58→y ≤29. Total fruits 4+29=33. Still less.Wait, but x=2 and x=1 both give 34. Let me check x=0. But x must be at least 1, so no. What about x=1, y=33. Let's compute total weight: 1*300 +33*200=300 + 6600=6900g. Which is under 7000. So, there's 100g left. Similarly, x=2, y=32: 2*300 +32*200=600 +6400=7000g. Exactly 7000g. So both combinations are valid, but x=2 and y=32 uses the full weight, whereas x=1 and y=33 leaves 100g unused.So, the total number of fruits is the same, 34. So, that's interesting. So, both 1 apple and 33 pears (34 fruits) and 2 apples and 32 pears (34 fruits) work, but the second one uses the entire weight capacity.But the problem says "the maximum number of fruits she can buy to be able to make pies of both fruits". So, since both x=1,y=33 and x=2,y=32 give 34 fruits, and 34 is the maximum possible. Wait, but maybe if we choose different numbers where x and y are higher?Wait, let's check if 34 is indeed the maximum. Let's see if there's a combination with more than 34 fruits. For example, 35 fruits. So x + y =35. Let's see if 3x +2y ≤70.But x + y =35. Let's express y =35 -x. Then substitute into the inequality:3x +2(35 -x) ≤70 → 3x +70 -2x ≤70 →x +70 ≤70 →x ≤0. But x must be at least 1. So impossible. Therefore, 35 fruits is impossible.Therefore, the maximum possible is 34.But let's verify with the two cases:Case 1: 1 apple, 33 pears. Total weight: 1*300 +33*200 =300 +6600=6900g. So, 7000 -6900=100g left. But Ana can't buy any more fruits because 100g is less than the weight of either fruit (200g for pear, 300g for apple). So she can't use that remaining 100g.Case 2: 2 apples, 32 pears. Total weight: 2*300 +32*200=600 +6400=7000g. Exactly full. So, no leftover weight.But both cases give 34 fruits. So, the answer is 34. But wait, let's check if there's another combination with more than 34. For example, maybe more apples and pears but with a different ratio. Suppose we take 3 apples. Then y would be 30.5, which is 30. Total fruits 33. Less than 34.Wait, so even if we take fractions, x=1.5 and y=33.25, but since they have to be integers, we can't. So, no. So, 34 is the maximum.But let's think again. Let's try to find x and y such that 3x +2y ≤70 and x + y is maximized. Since x and y are integers ≥1.Another way: Since apples are heavier, replacing an apple with two pears saves 300 - 2*200 = -100g (i.e., adds 100g). Wait, wait. Let me think. If you have one apple (300g) and replace it with two pears (400g), that's 100g more. So that would exceed the weight. So, instead, replacing a pear with an apple would add 100g. So, maybe exchanging pears for apples would allow more total fruits? Wait, but apples are heavier. So, actually, replacing apples with pears would allow more total fruits. Wait, but apples are 300g, pears are 200g. So, each pear is 100g lighter. So, if you replace an apple with a pear, you save 100g, which allows you to add maybe another pear. So, replacing 1 apple with 1 pear saves 100g, then you can use that 100g to add half a pear? But since we can only have whole fruits, maybe 100g is not enough for another pear. So, perhaps replacing apples with pears allows more total fruits?Wait, in our initial case, when x=1 and y=33, we have 100g left. So, if we replace 1 pear with an apple, we would get x=2 and y=32. The weight becomes 2*300 +32*200=600 +6400=7000. So, we use up the leftover 100g. But the total number of fruits remains the same: 34. So, that's interesting. So, in that case, you can't get more fruits by replacing, but you can use up the leftover weight. But the total number of fruits is the same.Alternatively, if we start with x=2 and y=32, which uses all the weight, and try to see if replacing 2 apples with pears. Each apple is 300g, so replacing 2 apples with pears would save 2*300 -2*200=200g, which allows adding 1 pear (200g). So, replacing 2 apples with 3 pears (since 2 apples removed, 3 pears added). Let's check:Original: 2 apples, 32 pears: total fruits 34.After replacement: 0 apples, 35 pears. But she needs at least one apple. So, invalid.Alternatively, replace 1 apple with pears. Remove 1 apple (300g), add pears. How many pears can be added with 300g? 300g /200g per pear = 1.5 pears. But since we can only add 1 pear, we can replace 1 apple with 1 pear, saving 100g. So, x=1, y=33, which is the original case. So, same total fruits. Again, 34.So, perhaps 34 is indeed the maximum. Let me check another approach.Suppose we model this as an integer linear programming problem. The objective is to maximize x + y, subject to 3x + 2y ≤70, x ≥1, y ≥1.The optimal solution in linear programming (without integer constraints) would be where the line x + y is tangent to the constraint 3x + 2y =70. To find that point, we can solve:Let’s denote the objective function as x + y = k, and we want to maximize k. The point of tangency would occur where the gradient of the objective function is parallel to the gradient of the constraint. The gradient of x + y is (1,1), and the gradient of 3x + 2y is (3,2). These are not scalar multiples, so the maximum occurs at a boundary point. In linear programming, the maximum would be at one of the vertices. The vertices of the feasible region (with x ≥1, y ≥1) are the intersections of 3x +2y=70 with x=1 and y=1.When x=1: 3(1) +2y=70 →2y=67→y=33.5. So the vertex is (1, 33.5). Since y must be integer, the closest integers are (1,33) and (1,34), but (1,34) would give 3*1 +2*34=3 +68=71>70, so invalid. So (1,33).When y=1: 3x +2(1)=70→3x=68→x≈22.666. So, the vertex is (22.666,1). The closest integers are (22,1) and (23,1). Let's check (23,1): 3*23 +2*1=69 +2=71>70, invalid. So (22,1).Then, the other vertex is where 3x +2y=70 intersects with x=1 and y=1. But actually, the feasible region is a polygon bounded by x≥1, y≥1, and 3x +2y ≤70. The vertices are (1,1), (1,33.5), (22.666,1). But since x and y are integers, the maximum occurs near these points.The maximum in the continuous case would be at (1,33.5) with x + y =34.5, but since we need integers, the maximum integer value is 34. Which aligns with our previous calculation.Therefore, the maximum number of fruits is 34, achieved by either (1,33) or (2,32) as we saw earlier.But let's check if there are other combinations where x + y=34. Let's see:Suppose x + y=34. Then y=34 -x.Substitute into the weight equation:3x +2(34 -x) ≤703x +68 -2x ≤70x +68 ≤70x ≤2So x can be 1 or 2. Therefore, the only possible combinations are x=1,y=33 and x=2,y=32. So, those are the only two combinations that give 34 fruits. Any other combinations with x + y=34 would require x>2, but from the inequality, x must be ≤2. So, no.Therefore, the answer is 34. Ana can buy either 1 apple and 33 pears or 2 apples and 32 pears, both totaling 34 fruits. Since the problem asks for the maximum number, and both options give 34, the answer is 34.But wait, let me confirm once more. Let's check if there's a combination with more than 34. Suppose someone thinks of buying 3 apples and 31 pears. Then total weight is 3*300 +31*200=900 +6200=7100>7000. So that's over. So invalid. What about 3 apples and 30 pears? 900 +6000=6900. Then total fruits 33. Less than 34.If we take 4 apples and 29 pears: 1200 +5800=7000. Total fruits 33. Same as before.So, indeed, only 34 is possible. Therefore, the answer is 34.Final AnswerThe maximum number of fruits Ana can buy is boxed{34}.</think>To determine the maximum number of fruits Ana can buy, we need to consider the weights of apples and pears, and the maximum weight her bag can hold. An apple weighs 300 grams and a pear weighs 200 grams. Ana's bag can hold up to 7 kg (7000 grams). We need to maximize the number of fruits (apples + pears) while ensuring the total weight does not exceed 7000 grams and she buys at least one of each fruit.We start with the inequality:[ 300x + 200y leq 7000 ]where ( x ) is the number of apples and ( y ) is the number of pears. Simplifying this inequality by dividing by 100, we get:[ 3x + 2y leq 70 ]We need to maximize ( x + y ) subject to the constraints ( x geq 1 ), ( y geq 1 ), and both ( x ) and ( y ) are integers. By testing possible values, we find:- If ( x = 1 ), then ( 3(1) + 2y leq 70 ) gives ( 2y leq 67 ) or ( y leq 33.5 ). Thus, ( y = 33 ). Total fruits: ( 1 + 33 = 34 ).- If ( x = 2 ), then ( 3(2) + 2y leq 70 ) gives ( 2y leq 64 ) or ( y leq 32 ). Thus, ( y = 32 ). Total fruits: ( 2 + 32 = 34 ).Both combinations (1 apple and 33 pears, or 2 apples and 32 pears) result in a total of 34 fruits. Checking other combinations with ( x geq 1 ) and ( y geq 1 ) confirms that 34 is the maximum number of fruits possible without exceeding the weight limit.Thus, the maximum number of fruits Ana can buy is (boxed{34}).,<think>Okay, let's see. Ana wants to make pear and apple pies, so she needs both pears and apples. The problem is to find the maximum number of fruits she can buy without exceeding her bag's weight limit of 7 kg. Each apple is 300 grams, each pear is 200 grams. Hmm, so we need to maximize the number of fruits (apples + pears) while keeping the total weight under or equal to 7000 grams. Also, she needs at least one of each fruit because she has to make pies of both. First, I should convert the total weight Ana can carry into grams to match the fruit weights. 7 kg is 7000 grams. Each apple is 300g, each pear is 200g. Let me define variables: let a be the number of apples, p be the number of pears. Then, the total weight is 300a + 200p ≤ 7000, and we need to maximize a + p. Also, a ≥ 1 and p ≥ 1 since she needs both.So, the problem is an optimization problem. To maximize a + p under the constraint 300a + 200p ≤ 7000, with a and p being positive integers. This is similar to a linear programming problem, but since we can only have integer values for a and p (since you can't buy a fraction of a fruit), it's an integer linear programming problem. But maybe we can approach it by trying to find the combination where the number of fruits is maximized.Let me think. Since pears are lighter, to maximize the number of fruits, she should buy more pears than apples. Because for the same weight, pears give more quantity. For example, 200g per pear vs 300g per apple. So, if we can substitute an apple with a pear, we can carry more total fruits. Therefore, the strategy would be to buy as many pears as possible, and the remaining weight can be filled with apples. But we need to ensure at least one apple.Alternatively, maybe a balance between apples and pears. Let me formalize this.Let me write the weight constraint: 300a + 200p ≤ 7000. We can divide everything by 100 to simplify: 3a + 2p ≤ 70. And we need to maximize a + p.So, we can express this as 3a + 2p ≤ 70. Let's try to express p in terms of a. 2p ≤ 70 - 3a → p ≤ (70 - 3a)/2. Since p must be an integer ≥1, (70 - 3a)/2 must be at least 1. Therefore, 70 - 3a ≥ 2 → 3a ≤ 68 → a ≤ 22.666... So, a can be at most 22. But since a must be at least 1, a can be from 1 to 22.Similarly, if we express a in terms of p: 3a ≤ 70 - 2p → a ≤ (70 - 2p)/3. So, p can be from 1 to floor((70 - 3)/2) = floor(67/2) = 33. But p must be such that (70 - 2p) is divisible by 3 or floor down. Wait, maybe this is getting complicated.Alternatively, we can model this as: for each possible a from 1 to 22, compute the maximum p, then compute a + p. Then find the maximum among all these.But that might take a lot of time. Maybe there's a smarter way. Since we want to maximize a + p given 3a + 2p ≤ 70. Let me think about the ratio. The weight per fruit for apples is higher, so if we want more fruits, we need to minimize the average weight per fruit. So, the average weight per fruit is (3a + 2p)/(a + p). To minimize this average, we need more pears. So, the maximum number of fruits should be achieved when we have as many pears as possible and as few apples as possible. However, we need at least one apple.Therefore, perhaps the maximum number of fruits is when Ana buys 1 apple and the rest pears. Let's check that. If a = 1, then 3*1 + 2p ≤ 70 → 2p ≤ 67 → p ≤ 33.5. So p = 33. Then total fruits = 1 + 33 = 34. Total weight = 300*1 + 200*33 = 300 + 6600 = 6900 grams. So, 6900g, which is under 7000g. There's 100g left. But since we can't buy a fraction of a fruit, maybe we can use that remaining 100g to buy another fruit? But 100g isn't enough for either an apple or a pear. So, no. So 34 fruits is possible here.But maybe if we buy more apples, we can utilize the remaining weight better. Let's see. For example, if a = 2. Then 3*2 + 2p ≤ 70 → 6 + 2p ≤70 → 2p ≤64 → p=32. Total fruits = 2 + 32 = 34. Total weight = 600 + 6400 = 7000. Wait, that's exactly 7000g. So, 34 fruits here. But same total as before, but this time the weight is exactly 7000g. So, same number of fruits but using all the capacity.So both a=1, p=33 and a=2, p=32 give 34 fruits, with the latter using the entire capacity.Similarly, check a=3. 3*3 + 2p ≤70 →9 + 2p ≤70 →2p ≤61 →p=30.5 →p=30. Total fruits=3+30=33. Less than 34. So, worse.a=4: 12 +2p ≤70→2p ≤58→p=29→Total fruits=4+29=33. Still less.Wait, so a=1 gives p=33 (34 fruits, 6900g), a=2 gives p=32 (34 fruits, 7000g). So a=2, p=32 is better because uses all the weight. So 34 is the maximum? Wait, but wait. Let's check other possibilities.Suppose instead of trying to maximize pears, maybe there's a combination where you can have more apples and pears such that the total number is higher. For instance, maybe replacing some pears with apples in a way that the total number of fruits increases. Wait, but since apples are heavier, replacing a pear with an apple would decrease the total number of fruits. For example, replacing two pears (400g) with one apple (300g) would save 100g but lose one fruit. So perhaps not useful. But maybe if the leftover weight can be used for another fruit.Wait, in the case of a=1, p=33: total weight 6900g. Leftover 100g. If we can somehow use that 100g to get another fruit. But 100g is not enough for either. So no.Alternatively, if we take a=2, p=32: total weight 7000g. No leftover. So 34 fruits. That seems like the maximum. But let's check another case. Suppose we have a=0. But she needs at least one apple. So a=0 is invalid. Similarly p=0 is invalid.Alternatively, maybe buying a different combination where the total number is higher. Let's see.Wait, 3a + 2p ≤70. Let me try to model this as a linear equation. If we ignore the integer constraints for a moment, then the maximum a + p would be when the weight is minimized per fruit. But since we have to have integers, maybe 34 is the maximum. Wait, but maybe there's a combination where the weight adds up to 7000g with more than 34 fruits. Let's see.Suppose that 3a + 2p =70. Then, total fruits a + p. Let me solve for p: p=(70 -3a)/2. So total fruits is a + (70 -3a)/2 = (70 -3a + 2a)/2 = (70 -a)/2. To maximize (70 -a)/2, we need to minimize a. The smallest a is 1. So (70 -1)/2= 69/2=34.5. But since a and p have to be integers, the maximum possible is 34 when a=2, p=32, as above.Alternatively, if a=1, p=33.5, but p must be integer, so 33, leading to total 34.So, in the continuous case, maximum is 34.5, but since we can't have half fruits, 34 is the maximum. So, in integer values, 34 is the maximum. Thus, the answer should be 34.But wait, let's check if there are other combinations where 3a + 2p <70, but a + p =34. For example, if a=3, then p=(70 -9)/2=61/2=30.5→30. So total fruits=33. Less than 34. Similarly, a=4, p=29. So, indeed, the maximum is 34.Alternatively, maybe buying more apples but with some leftover weight. For example, a=3, p=30. Total weight=900 + 6000=6900, same as a=1, p=33. But total fruits=33. So, worse. Similarly, if a=5, p=28. 1500 + 5600=7100. Wait, that's over. So not allowed.Wait, 3*5 + 2*28=15 +56=71>70. So invalid.Wait, maybe another way. Let's think in terms of Diophantine equations. We need 3a + 2p ≤70, a ≥1, p≥1, integers. Find maximum a + p.This is similar to the classic problem: given coin denominations, how to make change for a certain amount with the most number of coins. Except here, coins are different denominations (3 and 2) and we need the maximum number. So, similar logic: use as many of the smaller denomination (2) as possible, then fill with the larger (3).But in this case, we have a budget of 70, and each "coin" is 3 or 2, but we need at least one of each. So, the standard approach would be to use as many 2s as possible, but we need at least one 3. So, use 1*3 and the rest 2s. Then total number is 1 + (70 -3)/2=1 + 67/2=1 +33.5=34.5. So, 34.5, which translates to 34 total (1+33=34). But if we use 2*3=6, then the remaining is 64, which can be 32*2. So total is 2+32=34. So same number.Hence, regardless of the number of apples (as long as minimal), the maximum total fruits is 34.Alternatively, perhaps even if we take more apples, but distribute the weights in such a way that total fruits remain same. For example, a=4, 3*4=12, remaining 70 -12=58, p=29. Total fruits=33. Less.So, the conclusion is 34 is the maximum.Wait, but just to confirm, let's check another possible combination. For example, a=2, p=32: total weight=6 +64=70. Total fruits=34.a=1, p=33: total weight=3 +66=69, total fruits=34.Is there a way to get 35? Let's see. Suppose total fruits=35. So a + p=35. Then, 3a + 2p ≤70.Express p=35 -a. Then, 3a +2(35 -a) ≤70 →3a +70 -2a ≤70 →a +70 ≤70 →a ≤0. But a must be at least 1. So impossible. Therefore, a + p=35 is impossible. Therefore, 34 is indeed the maximum.Therefore, the answer is 34.Final AnswerThe maximum number of fruits Ana can buy is boxed{34}.</think>To determine the maximum number of fruits Ana can buy while staying within her bag's weight limit of 7 kg (7000 grams), we need to consider the weights of apples (300 grams each) and pears (200 grams each). We need to maximize the total number of fruits (apples + pears) while ensuring the total weight does not exceed 7000 grams and she has at least one of each fruit.1. Convert the weight limit to grams: 7 kg = 7000 grams.2. Set up the inequality: The total weight constraint is (300a + 200p leq 7000), where (a) is the number of apples and (p) is the number of pears. Simplifying this, we get (3a + 2p leq 70).3. Maximize the total number of fruits: We need to maximize (a + p) under the constraint (3a + 2p leq 70) with (a geq 1) and (p geq 1).To solve this, we consider the scenarios where we maximize the number of lighter fruits (pears) while ensuring at least one apple. - For (a = 1): [ 3(1) + 2p leq 70 implies 2p leq 67 implies p leq 33.5 ] Since (p) must be an integer, (p = 33). Total fruits = (1 + 33 = 34).- For (a = 2): [ 3(2) + 2p leq 70 implies 6 + 2p leq 70 implies 2p leq 64 implies p leq 32 ] Total fruits = (2 + 32 = 34).Both scenarios give 34 fruits, with the second scenario using the entire weight capacity (7000 grams). Checking higher values of (a) results in fewer total fruits. For example, (a = 3) gives (p = 30) and total fruits = 33.We also verify that (a + p = 35) is impossible because it leads to a contradiction in the weight constraint.Thus, the maximum number of fruits Ana can buy is (boxed{34}).]

🔑:Let m denote the number of apples and p the number of pears that Ana buys, so the weight she carries in the bag is 300 m + 200 p grams. Since the bag can hold a maximum of 7000 grams, we have 300 m + 200 p leq 7000, which is equivalent to 3 m + 2 p leq 70. As pears weigh less, Ana should take a larger number of pears and, therefore, a smaller number of apples. Since Ana wants to make pies of both fruits, she needs to take at least 1 apple.If she takes one apple, we have 2 p leq 70 - 3 = 67, so p leq 33.5, which means Ana can take up to 33 more pears, for a total of 34 fruits. If she takes two apples, we have 2 p leq 70 - 6 = 64, so p leq 32, which means Ana can take up to 32 more pears, again for a total of 34 fruits. If she takes three apples, we have 2 p leq 70 - 9 = 61, so p leq 30.5, which means Ana can only take up to 30 more pears, for a total of 33 fruits.From the calculations, we see that for each additional apple Ana takes, she needs to take 1.5 fewer pears. Thus, if she takes more than two apples, she will never be able to take more than 30 pears, for a total always less than 34 fruits.Conclusion: the maximum number of fruits Ana can take is 34 fruits (either one apple and 33 pears, or two apples and 32 pears).

🧠:[<think>Okay, let's see. I need to find the trajectory equation of point Q(x, y) given these vector conditions. Hmm. Let me start by understanding the problem step by step.First, we have two non-zero, non-collinear vectors OA and OB. That means points O, A, and B are not on the same straight line, right? So OA and OB can be considered as basis vectors for the plane. Then, we have a point P such that 2OP = xOA + yOB. So OP is expressed as a linear combination of OA and OB with coefficients x/2 and y/2. Wait, so OP is (x/2)OA + (y/2)OB? Yeah, because if 2OP equals that combination, then OP is half of that.Then, there's another condition: PA = λAB, where λ is a real number. So vector PA is some scalar multiple of vector AB. That means points P, A, and B are colinear? Because PA is parallel to AB, so P lies somewhere on the line AB? Wait, not exactly. If PA is a scalar multiple of AB, then P is somewhere on the line extending from A in the direction of AB, but since λ can be any real number, P can be anywhere along that line. But how does this relate to point Q(x, y)?Wait, the question is about the trajectory of point Q(x, y). The coordinates x and y are the coefficients in the expression for 2OP. So as P moves according to the condition PA = λAB, the coefficients x and y in the expression for OP must satisfy some equation, which is the trajectory of Q.So, I need to relate the condition PA = λAB to the coordinates x and y. Let me try to express PA and AB in terms of OA and OB.First, let's recall that vector PA is OA - OP, because PA = A - P. Wait, vector PA is from P to A, so PA = OA - OP. Similarly, vector AB is OB - OA.Given that PA = λ AB, substituting the vectors:OA - OP = λ (OB - OA)Now, substitute OP from the given equation 2OP = x OA + y OB. So OP = (x/2) OA + (y/2) OB.Therefore, substituting OP into the equation:OA - [(x/2) OA + (y/2) OB] = λ (OB - OA)Let me simplify the left side:OA - (x/2 OA) - (y/2 OB) = (1 - x/2) OA - (y/2) OBThe right side is λ OB - λ OA.So, set the left side equal to the right side:(1 - x/2) OA - (y/2) OB = (-λ) OA + λ OBSince OA and OB are non-collinear (and hence linearly independent), their coefficients must be equal on both sides. Therefore, we can equate the coefficients:For OA:1 - x/2 = -λFor OB:-y/2 = λSo now we have two equations:1. 1 - x/2 = -λ2. -y/2 = λWe can solve these equations for λ and relate x and y. Let's substitute λ from the second equation into the first equation.From equation 2: λ = -y/2Substitute into equation 1:1 - x/2 = -(-y/2) => 1 - x/2 = y/2Multiply both sides by 2 to eliminate denominators:2 - x = yRearranging terms:x + y = 2So the equation is x + y - 2 = 0, which is option A.Wait, but let me check my steps again to be sure.Starting from PA = λ AB.PA is OA - OP. AB is OB - OA.Expressed in terms of OA and OB, that becomes OA - OP = λ (OB - OA)OP is (x/2 OA + y/2 OB). So OA - (x/2 OA + y/2 OB) = (1 - x/2) OA - (y/2) OBOn the right side: λ OB - λ OASo equate coefficients:OA: 1 - x/2 = -λOB: -y/2 = λSo substituting λ = -y/2 into the first equation:1 - x/2 = -(-y/2) => 1 - x/2 = y/2Multiply both sides by 2: 2 - x = y => x + y = 2. So equation is x + y - 2 = 0. Therefore, the answer is A.But let me check the options again. The options given are:A: x + y - 2 = 0B: 2x + y -1=0C: x +2y -2=0D:2x + y -2=0So A is correct? Hmm. Alternatively, maybe I made a mistake in the direction of the vectors.Wait, PA is from P to A, so vector PA = OA - OP. That's correct because if you have point P, then PA = A - P. Since OP is the vector from O to P, then PA is OA - OP.Similarly, AB is OB - OA. So that's correct.Then substituting OP = (x/2)OA + (y/2)OB into OA - OP gives OA - (x/2 OA + y/2 OB) = (1 - x/2) OA - (y/2) OB.Which is equal to λ (OB - OA) = -λ OA + λ OB.Therefore, equating coefficients:1 - x/2 = -λand- y/2 = λSo solving these, from the second equation: λ = - y/2Plugging into first equation: 1 - x/2 = -(- y/2) = y/2Multiply both sides by 2: 2 - x = y => x + y = 2 => x + y - 2 = 0, which is answer A.So seems correct. Maybe the answer is A. But let me cross-verify with coordinates.Alternatively, perhaps using coordinate geometry. Let's assume OA and OB are standard basis vectors. Let me set OA as (1, 0) and OB as (0, 1). Then O is the origin.Then OP is (x/2, y/2). So point P has coordinates (x/2, y/2).Then vector PA is OA - OP = (1 - x/2, 0 - y/2) = (1 - x/2, -y/2)Vector AB is OB - OA = (0 - 1, 1 - 0) = (-1, 1)Given that PA = λ AB, so:(1 - x/2, -y/2) = λ (-1, 1)Therefore, components:1 - x/2 = -λand-y/2 = λSame as before. So solving:From second equation: λ = - y/2Substitute into first equation: 1 - x/2 = -(- y/2) = y/2Multiply by 2: 2 - x = y => x + y = 2.So the equation is x + y - 2 = 0, which is A. Therefore, seems correct.But the options include A through D. The problem says "the trajectory equation of point Q(x, y)", so Q is the point with coordinates (x, y). Wait, but in the problem statement, 2OP = x OA + y OB. So in terms of coordinates, if OA and OB are basis vectors, then OP would have coordinates (x/2, y/2). But Q is (x, y). So Q is a scaling of point P? That is, Q is 2OP in terms of OA and OB coordinates.But if OA and OB are the standard basis vectors, then Q is (x, y) where OP is (x/2, y/2). So Q is just twice the coordinates of P.But regardless, the relation between x and y is found to be x + y = 2, so Q's trajectory is x + y - 2 = 0, which is option A. So the answer is A.But let me check again. Suppose OA and OB are not standard basis vectors. But since they are non-zero and non-collinear, they form a basis. So any vector can be expressed as a combination of OA and OB. The coefficients x and y in 2OP = x OA + y OB are scalar coefficients. Therefore, regardless of the coordinate system, the equation derived is x + y - 2 = 0.Therefore, the answer is A.But wait, wait a second. Let me check once more because sometimes these problems can be tricky.We have point Q(x, y). The coordinates x and y are such that 2OP = x OA + y OB. Then, the condition PA = λ AB is given. So in terms of Q, which is (x, y), the relation derived is x + y = 2. Therefore, the trajectory of Q is the line x + y = 2, which is option A.Yes, that seems correct. So I think the answer is A.Final Answerboxed{A}</think>Given non-zero, non-collinear vectors (overrightarrow{OA}) and (overrightarrow{OB}), we need to find the trajectory equation of point (Q(x, y)) given that (2overrightarrow{OP} = xoverrightarrow{OA} + yoverrightarrow{OB}) and (overrightarrow{PA} = lambdaoverrightarrow{AB}) for (lambda in mathbb{R}).1. Express (overrightarrow{OP}) in terms of (overrightarrow{OA}) and (overrightarrow{OB}): [ 2overrightarrow{OP} = xoverrightarrow{OA} + yoverrightarrow{OB} implies overrightarrow{OP} = frac{x}{2}overrightarrow{OA} + frac{y}{2}overrightarrow{OB} ]2. Use the given condition (overrightarrow{PA} = lambdaoverrightarrow{AB}): [ overrightarrow{PA} = overrightarrow{OA} - overrightarrow{OP} ] [ overrightarrow{AB} = overrightarrow{OB} - overrightarrow{OA} ] Substituting (overrightarrow{OP}) into the equation: [ overrightarrow{OA} - left(frac{x}{2}overrightarrow{OA} + frac{y}{2}overrightarrow{OB}right) = lambda (overrightarrow{OB} - overrightarrow{OA}) ] Simplifying the left side: [ left(1 - frac{x}{2}right)overrightarrow{OA} - frac{y}{2}overrightarrow{OB} = -lambdaoverrightarrow{OA} + lambdaoverrightarrow{OB} ]3. Equate coefficients of (overrightarrow{OA}) and (overrightarrow{OB}): [ 1 - frac{x}{2} = -lambda quad text{and} quad -frac{y}{2} = lambda ] Solving for (lambda) from the second equation: [ lambda = -frac{y}{2} ] Substituting (lambda) into the first equation: [ 1 - frac{x}{2} = frac{y}{2} ] Multiplying through by 2: [ 2 - x = y implies x + y = 2 ]Thus, the trajectory equation of point (Q(x, y)) is (x + y - 2 = 0), which corresponds to option (boxed{A}).,<think>Okay, let's see. I need to find the trajectory equation of point Q(x, y) given these vector equations. Alright, let's start by writing down what's given.First, we have non-zero, non-collinear vectors OA and OB. That means OA and OB are vectors from the origin to points A and B, respectively, and they aren't in the same line, so they form a basis for the plane. Then, the vector OP is given by 2OP = xOA + yOB. So, OP is a linear combination of OA and OB with coefficients x/2 and y/2, right? Because if I divide both sides by 2, OP = (x/2)OA + (y/2)OB. That might be useful later.Then, there's another equation: PA = λAB, where λ is a real number. PA is the vector from P to A, and AB is the vector from A to B. So this means that vector PA is some scalar multiple of vector AB. Since vectors can be added and subtracted, maybe I can express this in terms of OA and OB?Let me recall that vectors can be written in terms of position vectors. Let's denote points O, A, B, P as the origin, A, B, and P respectively. Then, the vector PA is A - P, right? Because PA goes from P to A, so it's A minus P. Similarly, vector AB is B - A. So the equation PA = λAB becomes:A - P = λ(B - A)That's an equation involving vectors. Now, since we have expressions for OP in terms of OA and OB, maybe we can substitute P into this equation.Given that 2OP = xOA + yOB, then OP = (x/2)OA + (y/2)OB. Therefore, the position vector of P is (x/2)OA + (y/2)OB. So, in vector terms, P = (x/2)OA + (y/2)OB. Therefore, substituting into A - P = λ(B - A):A - [(x/2)OA + (y/2)OB] = λ(B - A)But wait, OA and OB are vectors from the origin, right? So OA is the vector from O to A, which is just the position vector of A, which is A. Similarly, OB is the position vector of B, which is B. So OA = A and OB = B. Therefore, the expression simplifies to:A - [(x/2)A + (y/2)B] = λ(B - A)Let me write that out:A - (x/2)A - (y/2)B = λB - λANow, let's collect like terms on the left-hand side:(1 - x/2)A - (y/2)B = (-λ)A + λBSo, now we have an equation where the left-hand side is a linear combination of A and B, and the right-hand side is another linear combination. Since A and B are non-collinear (i.e., they form a basis), the coefficients for A and B on both sides must be equal. Therefore, we can set up two equations:For the coefficients of A:1 - x/2 = -λFor the coefficients of B:-y/2 = λSo now we have a system of equations:1 - (x/2) = -λ ...(1)- y/2 = λ ...(2)We can substitute equation (2) into equation (1) to find a relation between x and y.From equation (2), λ = -y/2. Plugging that into equation (1):1 - (x/2) = -(-y/2) => 1 - x/2 = y/2Multiply both sides by 2 to eliminate denominators:2 - x = ySo, rearranged:x + y = 2Therefore, the trajectory equation of point Q(x, y) is x + y - 2 = 0, which is option A.Wait, but let me double-check. Let me verify each step again.Starting with 2OP = xOA + yOB, so OP = (x/2)OA + (y/2)OB. Then, vector PA is A - P, which is A - [(x/2)OA + (y/2)OB]. But since OA and OB are position vectors of A and B, OA is A and OB is B, so substituting:A - [(x/2)A + (y/2)B] = (1 - x/2)A - (y/2)B. That's correct.PA = λAB, which is λ(B - A). So:(1 - x/2)A - (y/2)B = λ(B - A) = -λA + λBEquating coefficients:1 - x/2 = -λand- y/2 = λSo substituting λ from the second equation into the first gives:1 - x/2 = -(-y/2) => 1 - x/2 = y/2Multiply both sides by 2: 2 - x = y => x + y = 2. So, x + y - 2 = 0, which is option A. Therefore, the answer should be A.But wait, let me check the options again. The options are:A: x + y - 2 = 0B: 2x + y - 1 = 0C: x + 2y - 2 = 0D: 2x + y - 2 = 0So yes, A is x + y - 2 = 0, which matches our result. Therefore, the correct answer is A.But hold on, the problem mentions "the trajectory equation of point Q(x, y)". Wait, in the problem statement, the point is Q(x, y). But in the given equation, we have 2OP = xOA + yOB. So is OP related to Q(x, y)? Or is Q(x, y) defined as the point with coordinates x and y such that 2OP = xOA + yOB?Wait, maybe the problem is using coordinates with OA and OB as basis vectors. So, in other words, the coordinates of point Q are (x, y) with respect to the basis OA and OB. But since OA and OB are vectors, perhaps the problem is in a coordinate system where OA and OB are the standard basis vectors?Wait, the problem says "the trajectory equation of point Q(x, y)". So Q has coordinates (x, y), but probably in the standard coordinate system. However, the expression 2OP = xOA + yOB suggests that OP is a vector expressed as a linear combination of OA and OB with coefficients x and y. So perhaps in the problem, the coordinates x and y are coefficients in the linear combination of OA and OB that gives 2OP. Therefore, the coordinates of Q are (x, y), and we need to find the equation that x and y satisfy based on the given vector equations.So in that case, my previous reasoning still holds, leading to x + y - 2 = 0, so option A. However, I need to make sure that there's no confusion between the coordinate system and the basis vectors.Wait, in the problem statement, is Q(x, y) a point in the plane with coordinates x and y? Or is Q a point whose coordinates are defined in terms of the coefficients x and y in the linear combination of OA and OB? The problem says "the trajectory equation of point Q(x, y)", which probably means that Q has coordinates (x, y) in the standard coordinate system. But the equation 2OP = xOA + yOB defines OP in terms of OA and OB. So perhaps OP is a vector related to Q? Wait, maybe OP is the position vector of Q? Wait, maybe I need to clarify.Wait, maybe there's a misinterpretation here. Let me read the problem again carefully:"Given non-zero, non-collinear vectors OA and OB, if 2OP = xOA + yOB, and PA = λAB (λ ∈ ℝ), then the trajectory equation of point Q(x, y) is (    )"So Q(x, y) is a point whose coordinates are x and y. The vector OP is given by 2OP = xOA + yOB. So OP is a vector expressed as a linear combination of OA and OB with coefficients x and y divided by 2.But then PA = λAB. PA is the vector from P to A, so as before, PA = A - P. But P is a point such that 2OP = xOA + yOB. So OP is the position vector of point P. Therefore, OP = (x/2)OA + (y/2)OB.Therefore, the position vector of P is (x/2)OA + (y/2)OB, so in terms of coordinates, if OA and OB are the standard basis vectors, then P would have coordinates (x/2, y/2). But OA and OB are arbitrary non-collinear vectors, not necessarily the standard basis.But in the problem, point Q is given as Q(x, y). So maybe the coordinates of Q are x and y, which are the coefficients in the expression 2OP = xOA + yOB. Therefore, Q is defined by the coefficients x and y in the linear combination that gives 2OP. Therefore, the trajectory of Q(x, y) is determined by the condition PA = λAB. So we need to find the equation that x and y satisfy given this condition.In that case, the previous steps hold. We derived x + y - 2 = 0, so the answer is A. But let me check again to be sure.Starting from PA = λAB:PA = A - P = λAB = λ(B - A)Expressed in vectors:A - P = λ(B - A)But P is given by 2OP = xOA + yOB, so OP = (x/2)OA + (y/2)OB. Therefore, P = (x/2)OA + (y/2)OB. So substituting into the equation:A - [ (x/2)OA + (y/2)OB ] = λ(B - A)But OA and OB are vectors OA and OB, so OA = vector from O to A, which is A - O = A, since O is the origin. Similarly, OB is the vector from O to B, which is B. So OA = A and OB = B. Therefore:A - [ (x/2)A + (y/2)B ] = λ(B - A)Left-hand side: A - (x/2)A - (y/2)B = (1 - x/2)A - (y/2)BRight-hand side: λB - λA = -λA + λBEquating coefficients:For A: 1 - x/2 = -λFor B: -y/2 = λFrom the second equation, λ = -y/2. Substitute into the first equation:1 - x/2 = -(-y/2) => 1 - x/2 = y/2Multiply both sides by 2:2 - x = y => x + y = 2 => x + y - 2 = 0So yes, the equation is x + y - 2 = 0, which is option A. Therefore, the correct answer is A.But let me check if perhaps the problem is using affine coordinates or something else. Suppose OA and OB are position vectors, so OA is (a1, a2) and OB is (b1, b2), but since they're non-collinear, they form a basis. However, the point Q(x, y) is defined such that 2OP = xOA + yOB. So in terms of coordinates, if we consider OA and OB as basis vectors, then the coordinates of P would be (x/2, y/2). But the problem states Q(x, y), so maybe Q is the point with coordinates (x, y) in the standard basis, but OP is expressed in terms of OA and OB.But unless there's a coordinate transformation involved, which isn't mentioned, the problem is likely in a general vector space with OA and OB as basis vectors, so the coordinates x and y are coefficients in that basis. Then, the equation derived is x + y - 2 = 0, which would be a line in the plane with coordinates x and y. Therefore, the answer is A.Alternatively, if OA and OB are the standard basis vectors, then OA is (1, 0) and OB is (0, 1), then OP would be (x/2, y/2). Then, PA is A - P = (1, 0) - (x/2, y/2) = (1 - x/2, -y/2). Vector AB is B - A = (0, 1) - (1, 0) = (-1, 1). Then PA = λAB gives:(1 - x/2, -y/2) = λ(-1, 1)So component-wise:1 - x/2 = -λ- y/2 = λSolving:From the second equation, λ = -y/2. Substitute into the first equation:1 - x/2 = -(-y/2) => 1 - x/2 = y/2 => 2 - x = y => x + y = 2 => same as before.Therefore, even if OA and OB are standard basis vectors, we still get x + y - 2 = 0. So the answer is A. Therefore, the correct answer is option A.Final Answerboxed{A}</think>Given non-zero, non-collinear vectors (overrightarrow{OA}) and (overrightarrow{OB}), we need to find the trajectory equation of point (Q(x, y)) given that (2overrightarrow{OP} = xoverrightarrow{OA} + yoverrightarrow{OB}) and (overrightarrow{PA} = lambdaoverrightarrow{AB}) for some (lambda in mathbb{R}).First, we express (overrightarrow{OP}) in terms of (overrightarrow{OA}) and (overrightarrow{OB}):[2overrightarrow{OP} = xoverrightarrow{OA} + yoverrightarrow{OB} implies overrightarrow{OP} = frac{x}{2}overrightarrow{OA} + frac{y}{2}overrightarrow{OB}]Next, we use the given vector equation (overrightarrow{PA} = lambdaoverrightarrow{AB}). The vector (overrightarrow{PA}) can be written as (overrightarrow{A} - overrightarrow{P}), and (overrightarrow{AB}) can be written as (overrightarrow{B} - overrightarrow{A}). Substituting the expression for (overrightarrow{P}) into the equation, we get:[overrightarrow{A} - left( frac{x}{2}overrightarrow{OA} + frac{y}{2}overrightarrow{OB} right) = lambda (overrightarrow{B} - overrightarrow{A})]Since (overrightarrow{OA}) and (overrightarrow{OB}) are position vectors of points (A) and (B), we can rewrite the equation as:[overrightarrow{A} - left( frac{x}{2}overrightarrow{A} + frac{y}{2}overrightarrow{B} right) = lambda (overrightarrow{B} - overrightarrow{A})]Simplifying the left-hand side:[left( 1 - frac{x}{2} right)overrightarrow{A} - frac{y}{2}overrightarrow{B} = -lambda overrightarrow{A} + lambda overrightarrow{B}]Equating the coefficients of (overrightarrow{A}) and (overrightarrow{B}) on both sides, we get:[1 - frac{x}{2} = -lambda quad text{and} quad -frac{y}{2} = lambda]Substituting (lambda = -frac{y}{2}) into the first equation:[1 - frac{x}{2} = frac{y}{2}]Multiplying both sides by 2 to clear the denominators:[2 - x = y implies x + y = 2]Thus, the trajectory equation of point (Q(x, y)) is:[x + y - 2 = 0]The correct answer is (boxed{A}).]

🔑:From overrightarrow{PA}=lambdaoverrightarrow{AB}, we get overrightarrow{OA}-overrightarrow{OP}=lambda(overrightarrow{OB}-overrightarrow{OA}),which means overrightarrow{OP}=(1+lambda)overrightarrow{OA}-lambdaoverrightarrow{OB}.Also, since 2overrightarrow{OP}=xoverrightarrow{OA}+yoverrightarrow{OB},we have begin{cases} x=2+2lambda, y=-2lambda, end{cases} eliminating lambda gives x+y-2=0, hence the correct choice is boxed{text{A}}.

🧠:[<think>Okay, so I need to prove that for x greater than 1, the function f(x) = ln(x) + sqrt(x) - 1 is less than (3/2)(x - 1). Hmm, let's see. First, maybe I should try to understand the behavior of both functions around x = 1. Since the inequality is for x > 1, checking the point x = 1 might give some insight.At x = 1, f(1) = ln(1) + sqrt(1) - 1 = 0 + 1 - 1 = 0. On the right-hand side, (3/2)(1 - 1) = 0. So both sides are equal at x = 1. That means they touch at that point. Maybe the function f(x) is always below the line (3/2)(x - 1) for x > 1? To check this, perhaps I can consider the difference between the two functions and show that it's negative for x > 1.Let me define a new function g(x) = (3/2)(x - 1) - f(x) = (3/2)(x - 1) - ln(x) - sqrt(x) + 1. Simplifying that, let's see: (3/2)x - 3/2 - ln(x) - sqrt(x) + 1. Combine constants: -3/2 + 1 = -1/2. So g(x) = (3/2)x - 1/2 - ln(x) - sqrt(x). I need to show that g(x) > 0 for all x > 1.If I can show that g(x) is positive for all x > 1, then the original inequality holds. To analyze this, maybe compute the derivative of g(x) and see its behavior. If the derivative is increasing or decreasing appropriately, and if at x = 1, g(x) = 0, then check if the derivative ensures that g(x) becomes positive for x > 1.First, let's compute g(1): (3/2)(1) - 1/2 - ln(1) - sqrt(1) = 3/2 - 1/2 - 0 - 1 = (3/2 - 1/2) = 1 - 1 = 0. Correct, as before.Now, compute g'(x). The derivative of (3/2)x is 3/2. The derivative of -1/2 is 0. The derivative of -ln(x) is -1/x. The derivative of -sqrt(x) is -1/(2sqrt(x)). So, g'(x) = 3/2 - 1/x - 1/(2sqrt(x)).We need to check if g'(x) is positive for x > 1. If so, then g(x) is increasing, and since g(1) = 0, then for x > 1, g(x) would be increasing from 0, hence positive. That would prove the result.So, let's check if g'(x) = 3/2 - 1/x - 1/(2sqrt(x)) is positive for x > 1. Let's evaluate this expression for x > 1.First, at x = 1: 3/2 - 1/1 - 1/(2*1) = 3/2 - 1 - 1/2 = 3/2 - 3/2 = 0. So the derivative at x = 1 is 0. Then, we need to see if the derivative increases or decreases beyond x = 1.Let's compute the second derivative g''(x) to check concavity. The derivative of g'(x):g''(x) = derivative of 3/2 is 0. Derivative of -1/x is 1/x^2. Derivative of -1/(2sqrt(x)) is (1)/(4x^(3/2)). So, g''(x) = 1/x^2 + 1/(4x^(3/2)).Since x > 1, both terms are positive. Therefore, g''(x) > 0 for x > 1. This means that g'(x) is concave up (increasing) for x > 1. Since at x = 1, g'(x) = 0, and the derivative g'(x) is increasing beyond x = 1, then g'(x) becomes positive for x > 1. Therefore, g(x) is increasing for x > 1.But wait, let's think again. If g'(x) is increasing (since g''(x) > 0) and at x = 1, g'(x) = 0, then for x > 1, g'(x) is greater than 0. Therefore, g(x) is increasing on x > 1. Since g(1) = 0 and g is increasing, then for x > 1, g(x) > 0. Hence, (3/2)(x - 1) - f(x) > 0 => f(x) < (3/2)(x - 1). That would complete the proof.But maybe I should check with some test value. Let's try x = 2.Compute f(2) = ln(2) + sqrt(2) - 1 ≈ 0.6931 + 1.4142 - 1 ≈ 1.1073.Compute (3/2)(2 - 1) = 1.5. So 1.1073 < 1.5, which is true.Another test: x = 4. f(4) = ln(4) + 2 - 1 ≈ 1.3863 + 1 ≈ 2.3863. (3/2)(4 - 1) = 4.5. 2.3863 < 4.5, yes.x approaching 1 from the right: Let's take x = 1.1.f(1.1) ≈ 0.0953 + 1.0488 - 1 ≈ 0.1441. (3/2)(0.1) = 0.15. So 0.1441 < 0.15, which holds.So these test points support the inequality. The initial analysis using derivatives seems correct.Alternatively, another approach: Consider the function h(x) = (3/2)(x - 1) - ln(x) - sqrt(x) + 1. Wait, that's the same as g(x). So as before, analyzing h(x) and its derivatives.Alternatively, perhaps using Taylor series expansion around x = 1. Let's see. Expand f(x) around x = 1.First, f(1) = 0.f'(x) = 1/x + 1/(2sqrt(x)). So f'(1) = 1 + 1/2 = 3/2.f''(x) = -1/x^2 - 1/(4x^(3/2)). f''(1) = -1 - 1/4 = -5/4.f'''(x) = 2/x^3 + 3/(8x^(5/2)).So the Taylor expansion of f(x) around x = 1 up to the second order is:f(x) ≈ f(1) + f'(1)(x - 1) + (f''(1)/2)(x - 1)^2 = 0 + (3/2)(x - 1) + (-5/8)(x - 1)^2.So f(x) ≈ (3/2)(x - 1) - (5/8)(x - 1)^2 + higher order terms.Therefore, near x = 1, f(x) is approximately (3/2)(x - 1) minus a positive term (since (x - 1)^2 is positive and coefficient is negative). Therefore, near x = 1, f(x) is less than (3/2)(x - 1). Then, if the subsequent terms in the Taylor series don't reverse this trend, the inequality might hold. However, this is only a local analysis. But combining with the previous derivative analysis, which showed that the difference g(x) is increasing for x > 1, this gives more confidence.Alternatively, since f(x) and (3/2)(x - 1) share the same value and first derivative at x = 1, but f(x) has a negative second derivative, which implies that f(x) curves downward, while (3/2)(x - 1) is a straight line. So near x = 1, f(x) is below the tangent line. If we can show that f(x) stays below this tangent line for all x > 1, that would establish the inequality.To show that f(x) is always below its tangent line at x = 1 for x > 1, we can use the concept of concavity. If f(x) is concave down (i.e., f''(x) < 0) on the interval x > 1, then it lies below its tangent lines. Let's check the second derivative of f(x):f''(x) = -1/x^2 - 1/(4x^(3/2)). For x > 1, both terms are negative, so f''(x) < 0. Therefore, f(x) is concave down for all x > 1. Therefore, the graph of f(x) lies below its tangent line at x = 1 for all x > 1. Since the tangent line at x = 1 is y = (3/2)(x - 1), this would imply that f(x) < (3/2)(x - 1) for all x > 1.That's another way to see it using concavity. So that's a more straightforward argument perhaps. Since f is concave down and the tangent line at x =1 is y = (3/2)(x -1), then for x >1, f(x) lies below this tangent line. Therefore, the inequality holds.So, to recap: If a function is concave down on an interval, then any tangent line lies above the function on that interval. Since we showed f''(x) < 0 for x > 1, f is concave down there, and the tangent at x =1 is y = (3/2)(x -1). Therefore, the inequality f(x) < (3/2)(x -1) holds for all x >1.That's a concise argument. Maybe the problem expects this approach. But let me check if f''(x) is indeed negative for all x >1.Compute f''(x) again:f'(x) = 1/x + 1/(2sqrt(x))f''(x) = -1/x^2 - 1/(4x^(3/2))Yes, both terms are negative for x >0, so definitely for x >1. Therefore, f''(x) <0, concave down. Therefore, tangent line at x=1 is above the function for x >1.Therefore, since the tangent line at x=1 is y = f(1) + f'(1)(x -1) = 0 + (3/2)(x -1), which is exactly the right-hand side of the inequality. Thus, the inequality holds.Therefore, the key idea is using the concavity of f(x). Since f is concave down, it lies below its tangent lines, and the given line is the tangent line at x=1. Hence, the inequality is proven.Alternatively, if not using concavity, we can use the function g(x) as before and show it's increasing and positive for x >1, which also works.Either way, the conclusion is the same. So, both approaches are valid, but the concavity argument is more straightforward once you recognize it.Final Answerboxed{f(x) < frac{3}{2}(x - 1) text{ for } x > 1}</think>To prove that ( f(x) = ln{x} + sqrt{x} - 1 ) is less than ( frac{3}{2}(x - 1) ) for ( x > 1 ), we analyze the function ( g(x) = frac{3}{2}(x - 1) - f(x) ).First, we check the value at ( x = 1 ):[g(1) = frac{3}{2}(1 - 1) - (ln{1} + sqrt{1} - 1) = 0 - (0 + 1 - 1) = 0.]Next, we compute the first derivative ( g'(x) ):[g'(x) = frac{3}{2} - left( frac{1}{x} + frac{1}{2sqrt{x}} right) = frac{3}{2} - frac{1}{x} - frac{1}{2sqrt{x}}.]At ( x = 1 ), the derivative is:[g'(1) = frac{3}{2} - 1 - frac{1}{2} = 0.]To determine the behavior of ( g'(x) ) for ( x > 1 ), we compute the second derivative ( g''(x) ):[g''(x) = frac{1}{x^2} + frac{1}{4x^{3/2}}.]Since ( g''(x) > 0 ) for all ( x > 1 ), ( g'(x) ) is increasing. Given that ( g'(1) = 0 ), it follows that ( g'(x) > 0 ) for ( x > 1 ). This implies ( g(x) ) is increasing for ( x > 1 ).Since ( g(1) = 0 ) and ( g(x) ) is increasing, ( g(x) > 0 ) for ( x > 1 ). Therefore, ( frac{3}{2}(x - 1) - f(x) > 0 ), which simplifies to ( f(x) < frac{3}{2}(x - 1) ).Additionally, considering the concavity of ( f(x) ), the second derivative ( f''(x) = -frac{1}{x^2} - frac{1}{4x^{3/2}} ) is negative for ( x > 1 ), indicating ( f(x) ) is concave down. Thus, the tangent line at ( x = 1 ), ( y = frac{3}{2}(x - 1) ), lies above ( f(x) ) for ( x > 1 ).Hence, we conclude:[boxed{f(x) < frac{3}{2}(x - 1) text{ for } x > 1}],<think>Okay, so I need to prove that for x > 1, the function f(x) = ln x + sqrt(x) - 1 is less than (3/2)(x - 1). Hmm, let's see. Maybe I can start by analyzing the difference between the two sides. If I subtract f(x) from (3/2)(x - 1), the inequality should hold true. Let me write that down:(3/2)(x - 1) - [ln x + sqrt(x) - 1] > 0 for x > 1.Simplifying this expression, I get:(3/2)(x - 1) - ln x - sqrt(x) + 1 > 0.Combine the constants:(3/2)x - 3/2 - ln x - sqrt(x) + 1 > 0.Which simplifies to:(3/2)x - 1/2 - ln x - sqrt(x) > 0.So, if I define a new function g(x) = (3/2)x - 1/2 - ln x - sqrt(x), then I need to show that g(x) > 0 for all x > 1.Maybe I can use calculus here. If I find the derivative of g(x) and analyze its behavior, that might help. Let's compute g'(x):g'(x) = d/dx [ (3/2)x - 1/2 - ln x - sqrt(x) ]= 3/2 - (1/x) - (1/(2sqrt(x))).So, g'(x) = 3/2 - 1/x - 1/(2sqrt(x)).I need to check if g(x) is increasing or decreasing for x > 1. Let's evaluate g'(x) at x = 1 first, to get a sense:g'(1) = 3/2 - 1/1 - 1/(2*1) = 3/2 - 1 - 1/2 = (3/2 - 1) - 1/2 = 1/2 - 1/2 = 0.So at x = 1, the derivative is zero. Hmm. Then maybe check the behavior as x increases beyond 1. Let's pick x = 2:g'(2) = 3/2 - 1/2 - 1/(2*sqrt(2)) = (3/2 - 1/2) - 1/(2*1.414) ≈ 1 - 0.353 ≈ 0.647 > 0.So at x = 2, the derivative is positive. How about x approaching infinity? Let's see:As x approaches infinity, the terms 3/2 dominates, while 1/x and 1/(2sqrt(x)) approach zero. Therefore, g'(x) approaches 3/2, which is positive. So as x becomes very large, the derivative is positive and increasing. Wait, actually, 3/2 is a constant, but as x increases, the negative terms vanish, so the derivative tends to 3/2. Therefore, for large x, the function g(x) is increasing.But at x =1, the derivative is zero. So maybe the derivative starts at zero and then becomes positive? Let's check another point between 1 and 2, say x = 1.5:g'(1.5) = 3/2 - 1/1.5 - 1/(2*sqrt(1.5)).Compute each term:1/1.5 ≈ 0.6667sqrt(1.5) ≈ 1.2247, so 1/(2*1.2247) ≈ 0.4082Thus, g'(1.5) ≈ 1.5 - 0.6667 - 0.4082 ≈ 1.5 - 1.0749 ≈ 0.4251 > 0.So even at x =1.5, the derivative is positive. So between x=1 and x approaching infinity, the derivative is positive. But wait, at x=1, the derivative is zero. So maybe the function g(x) has a minimum at x=1? Let's check the second derivative.Compute g''(x):g''(x) = derivative of g'(x) = 0 - (-1/x²) - (-1/(4x^(3/2))) = 1/x² + 1/(4x^(3/2)).Since x >1, both terms are positive, so g''(x) is positive. Therefore, the function g(x) is concave up at x=1, and the derivative is increasing. So since the derivative at x=1 is zero, and the derivative is increasing, then for x >1, the derivative g'(x) becomes positive. Therefore, g(x) is increasing for x >1.Now, check the value of g(x) at x=1:g(1) = (3/2)(1) -1/2 - ln1 - sqrt(1) = 3/2 -1/2 -0 -1 = (3/2 -1/2) -1 = 1 -1 =0.So at x=1, g(x)=0. Since the derivative is zero at x=1, and then becomes positive for x>1, that means g(x) is increasing for x>1, starting from zero. Therefore, for x>1, g(x) >0, which is exactly what we needed to prove. Hence, f(x) < (3/2)(x -1) for x>1.Wait, but let me double-check. If the function g(x) is increasing for x>1, starting at zero, then yes, for any x>1, g(x) is positive. That seems correct. But let me test with an example. Take x=2:g(2) = (3/2)(2) -1/2 - ln2 - sqrt(2) ≈ 3 -0.5 -0.6931 -1.4142 ≈ 3 -0.5 =2.5; 2.5 -0.6931≈1.8069; 1.8069 -1.4142≈0.3927>0.So positive, which matches the conclusion. How about x approaching 1 from the right? Let's take x=1.1:g(1.1) = (3/2)(1.1) -0.5 - ln(1.1) - sqrt(1.1).Calculate each term:(3/2)(1.1)=1.65ln(1.1)≈0.0953sqrt(1.1)≈1.0488Thus:1.65 -0.5 =1.151.15 -0.0953≈1.05471.0547 -1.0488≈0.0059>0.So even very close to 1, x=1.1 gives a positive value, albeit small. Since the function is increasing, that makes sense. So the argument holds.Alternatively, maybe use the Taylor series expansion around x=1 to approximate f(x) and compare it with (3/2)(x-1). Let's see. For f(x) = lnx + sqrt(x) -1.Compute the expansion at x=1. Let me denote h = x -1, so as h approaches 0.ln(1+h) ≈ h - h²/2 + h³/3 - ...sqrt(1+h) ≈1 + h/2 - h²/8 + h³/16 - ...Therefore, f(x) = ln(1+h) + sqrt(1+h) -1 ≈ [h - h²/2 + h³/3] + [1 + h/2 - h²/8 + h³/16] -1.Simplify:= h - h²/2 + h³/3 +1 + h/2 - h²/8 + h³/16 -1= (h + h/2) + (-h²/2 - h²/8) + (h³/3 + h³/16)= (3h/2) + (-5h²/8) + (19h³/48)So f(x) ≈ (3/2)h - (5/8)h² + (19/48)h³ + ... when h is small.But the right-hand side is (3/2)(x -1) = (3/2)h.Therefore, the difference (3/2)(x -1) - f(x) ≈ (3/2)h - [ (3/2)h - (5/8)h² + ... ] = (5/8)h² - ... So for small h >0 (i.e., x slightly larger than 1), the difference is positive, which aligns with our previous result. Thus, near x=1, the inequality holds, and since the function g(x) is increasing, it continues to hold for all x >1.Another approach could be using the Mean Value Theorem. Maybe consider the function h(x) = (3/2)(x -1) - f(x). We can show that h(x) >0 for x>1. We know h(1)=0. Then if we can show that h'(x) >0 for x>1, then h(x) is increasing, thus h(x) >0 for x>1.Compute h'(x) = (3/2) - f'(x).But f'(x) = 1/x + 1/(2sqrt(x)).Therefore, h'(x) = 3/2 - [1/x + 1/(2sqrt(x))].Which is the same as g'(x) earlier, which we showed is positive for x>1. Therefore, h'(x) >0, so h(x) is increasing, starting from h(1)=0, hence h(x) >0 for x>1. Therefore, the inequality holds.Therefore, multiple approaches confirm that the inequality is true. The key was analyzing the derivative and showing that the difference function is increasing after x=1.Final Answerboxed{frac{3}{2}(x - 1)}</think>To prove that ( f(x) = ln{x} + sqrt{x} - 1 ) is less than ( frac{3}{2}(x - 1) ) when ( x > 1 ), we start by defining the function ( g(x) = frac{3}{2}(x - 1) - (ln{x} + sqrt{x} - 1) ). We need to show that ( g(x) > 0 ) for ( x > 1 ).First, we simplify ( g(x) ):[g(x) = frac{3}{2}x - frac{3}{2} - ln{x} - sqrt{x} + 1 = frac{3}{2}x - frac{1}{2} - ln{x} - sqrt{x}]Next, we compute the derivative ( g'(x) ):[g'(x) = frac{3}{2} - frac{1}{x} - frac{1}{2sqrt{x}}]We evaluate ( g'(x) ) at ( x = 1 ):[g'(1) = frac{3}{2} - 1 - frac{1}{2} = 0]To understand the behavior of ( g'(x) ) for ( x > 1 ), we check values such as ( x = 2 ) and ( x = 1.5 ), finding that ( g'(x) ) is positive. As ( x ) approaches infinity, ( g'(x) ) approaches ( frac{3}{2} ), confirming it remains positive. We also compute the second derivative ( g''(x) ):[g''(x) = frac{1}{x^2} + frac{1}{4x^{3/2}}]Since ( g''(x) > 0 ) for ( x > 1 ), ( g(x) ) is concave up and ( g'(x) ) is increasing. Thus, ( g(x) ) is increasing for ( x > 1 ).Evaluating ( g(x) ) at ( x = 1 ):[g(1) = frac{3}{2}(1) - frac{1}{2} - ln{1} - sqrt{1} = 0]Since ( g(x) ) is increasing for ( x > 1 ) and starts at 0, ( g(x) > 0 ) for ( x > 1 ). This confirms that ( f(x) < frac{3}{2}(x - 1) ) for ( x > 1 ).Thus, the final answer is:[boxed{frac{3}{2}(x - 1)}]]

🔑:Proof: Define g(x) = frac{3}{2}(x - 1) - f(x) = frac{3}{2}(x - 1) - ln{x} - sqrt{x} + 1 (x > 1).Then, g'(x) = frac{3}{2} - frac{1}{x} - frac{1}{2sqrt{x}} = frac{3x - 2 - sqrt{x}}{2x}.Solving g'(x) = 0 gives 3x - sqrt{x} - 2 = 0, which leads to sqrt{x} = 1 or sqrt{x} = -frac{2}{3} (rejected as x>1).Thus, g'(x) = frac{(sqrt{x} + frac{2}{3})(sqrt{x} - 1)}{2x}.Given that x > 1, it follows that g'(x) > 0 always holds true.Hence, g(x) is increasing.So, g(x) > g(1) = 0.Therefore, when x > 1, f(x) < boxed{frac{3}{2}(x - 1)}.