🧠:[<think>Okay, so there's this problem about 10 dwarfs, each of whom is either a truth-teller or a liar. They each love exactly one type of ice cream: butter, chocolate, or fruit. Snow White asked them about their preferences in three steps. First, she asked those who love butter to raise their hands, and everyone raised their hand. Then she asked about chocolate, and half of them (so 5) raised their hands. Finally, for fruit, only one dwarf raised his hand. The question is: how many of the dwarfs are truthful?Hmm. Let's try to break this down. Since each dwarf is either a truth-teller or a liar, their responses depend on their type. But wait, the problem says that Snow White asked them to raise their hands if they love a certain ice cream. So when she asks about butter, if a dwarf loves butter, then if they're truthful, they should raise their hand. But if they're a liar, they would lie about it. So, if a liar doesn't love butter, they would raise their hand to lie, and if they do love butter, they wouldn't raise their hand? Wait, no. Wait, no. Let's think carefully.If a dwarf is a truth-teller, then when asked if they love butter, they will raise their hand if they actually do love butter. If they are a liar, they will lie about it, so they will raise their hand only if they do NOT love butter. Wait, no. Wait, the question is: when Snow White asks those who love butter to raise their hands. So, the truthful ones will raise their hand if they love butter. The liars will lie, so they will raise their hand if they do NOT love butter. Because the question is "if you love butter, raise your hand." So a liar would do the opposite. So if they love butter, they would not raise their hand, and if they don't love butter, they would raise their hand. So, when asked about butter, all dwarfs raised their hands. That means that all the truthful dwarfs who love butter raised their hands, and all the liars who do NOT love butter also raised their hands. Since everyone raised their hands, that implies that any dwarf who is a truth-teller must love butter (because otherwise they wouldn't have raised their hands), and any dwarf who is a liar must NOT love butter (because they lied about it). Wait, but then the truthful dwarfs must all love butter, and the liars must not love butter. But each dwarf loves exactly one type of ice cream. So the liars must love either chocolate or fruit. And the truth-tellers all love butter. But there are 10 dwarfs. Let's denote T as the number of truth-tellers and L as the number of liars. So T + L = 10.All truth-tellers love butter. So number of butter lovers is T (truth-tellers) plus any liars who love butter. Wait, but earlier conclusion was that liars cannot love butter because when asked about butter, all of them raised their hands. Wait, if a liar loved butter, he would have to lie when asked if he loves butter. So he would not raise his hand. But since everyone raised their hand, that means that all liars must have not loved butter, so they lied by raising their hands. So all liars do not love butter, and all truth-tellers do love butter. Therefore, all truth-tellers are butter lovers, and all liars are either chocolate or fruit lovers. So the number of butter lovers is exactly T. Then, when Snow White asked about chocolate, half of them (5) raised their hands. Similarly, when asked about fruit, only 1 raised his hand.But let's analyze each step. So first, when asked about butter, everyone raised their hands. So:- Truth-tellers: all T of them love butter, so they raised their hands truthfully.- Liars: all L = 10 - T of them do not love butter, but they lied by raising their hands.Therefore, total hands raised for butter: T (truth-tellers) + L (liars) = 10. Which matches the problem statement.Then, when asked about chocolate, half (5) raised their hands. Let's figure out who raised their hands here. The truthful dwarfs (T) will raise their hands if they love chocolate. But we already know that all truth-tellers love butter, so none of the truth-tellers can love chocolate (since each dwarf loves exactly one type). Therefore, the truth-tellers will not raise their hands for chocolate. The liars (L), on the other hand, when asked about chocolate, will lie. So if a liar actually loves chocolate, he would lie and not raise his hand. If a liar does not love chocolate (i.e., loves fruit), he would lie and raise his hand. Wait, let me clarify:A liar's response when asked if they love chocolate: If they do love chocolate, they would lie and not raise their hand. If they don't love chocolate (i.e., they love fruit), they would lie and raise their hand. Therefore, the number of liars who raised their hands for chocolate is equal to the number of liars who love fruit. Let’s denote:Let’s let the number of liars who love chocolate be C, and the number who love fruit be F. Then C + F = L. When asked about chocolate, the liars who love fruit (F) will lie and raise their hands, and the liars who love chocolate (C) will tell the truth (but they are liars, so they lie). Wait, no: when asked "do you love chocolate?" a liar will lie. So if a liar loves chocolate, he would say "no" (i.e., not raise his hand), and if he doesn't love chocolate, he would say "yes" (raise his hand). Therefore, the number of hands raised for chocolate is the number of liars who do NOT love chocolate (i.e., F) plus the number of truth-tellers who love chocolate (which is zero, since all truth-tellers love butter). Therefore, the number of hands raised for chocolate is F. The problem states that half the dwarfs (5) raised their hands. So F = 5.Then, when asked about fruit, only 1 dwarf raised his hand. Let's analyze that. The truthful dwarfs will raise their hands if they love fruit, but since all truth-tellers love butter, they don't love fruit. So the truth-tellers won't raise their hands. The liars, when asked about fruit, will lie. So if a liar loves fruit, he would lie and not raise his hand. If a liar does not love fruit (i.e., loves chocolate), he would lie and raise his hand. So the number of hands raised for fruit is the number of liars who love chocolate (C). Because those liars would be lying about not loving fruit. Wait, let me rephrase:When asked "do you love fruit?", a liar will lie. So:- Liars who love fruit: when asked, they would say "no" (lie), so they don't raise their hands.- Liars who love chocolate: when asked, they would say "yes" (lie), so they do raise their hands.Therefore, the number of hands raised for fruit is C. The problem states that only 1 dwarf raised his hand, so C = 1.But we know that C + F = L. Since F = 5 and C = 1, then L = 5 + 1 = 6. Therefore, there are 6 liars and T = 10 - L = 4 truth-tellers.Wait, but let's verify this. If T = 4, then there are 4 truth-tellers, all loving butter. Then liars are 6, with 1 loving chocolate and 5 loving fruit. Let's check the responses:First question: butter. All 4 truth-tellers raise hands. All 6 liars, who don't love butter, lie and raise hands. Total 10, which matches.Second question: chocolate. The truth-tellers (4) don't love chocolate, so they don't raise hands. The liars: 5 who love fruit lie and raise their hands, 1 who loves chocolate lies and doesn't raise his hand. So total raised hands: 5, which is half of 10. Correct.Third question: fruit. Truth-tellers (4) don't love fruit, so no hands. Liars: 1 who loves chocolate lies and raises his hand (since he doesn't love fruit), and 5 who love fruit lie and don't raise their hands. So only 1 hand raised. Correct.Therefore, the number of truthful dwarfs is 4. So the answer should be 4.But let me just check if there are any other possibilities. Suppose, for example, that when they asked about chocolate, maybe some truth-tellers did raise their hands. But wait, no, the truth-tellers all love butter, so they cannot love chocolate. Therefore, none of the truth-tellers would raise their hands for chocolate. Similarly, for fruit. So the only hands raised for chocolate and fruit come from the liars. So our reasoning holds. So T = 4, L = 6. Hence, the answer is 4.Final AnswerThe number of truthful dwarfs is boxed{4}.</think>Given that each of the 10 dwarfs either always tells the truth or always lies, and each loves exactly one type of ice cream (butter, chocolate, or fruit), we need to determine how many are truthful based on their responses to Snow White's questions.1. Butter Ice Cream: All dwarfs raised their hands. - Truth-tellers (T) love butter, so they raised their hands. - Liars (L = 10 - T) do not love butter but lied by raising their hands. - Therefore, T + L = 10, which is consistent since all raised their hands.2. Chocolate Ice Cream: Half (5) raised their hands. - Truth-tellers (T) do not love chocolate (they love butter), so none raised their hands. - Liars who love fruit (F) lied by raising their hands (since they do not love chocolate). - Therefore, F = 5 (since 5 hands were raised).3. Fruit Ice Cream: Only 1 raised his hand. - Truth-tellers (T) do not love fruit (they love butter), so none raised their hands. - Liars who love chocolate (C) lied by raising their hands (since they do not love fruit). - Therefore, C = 1 (since 1 hand was raised).Since the number of liars (L) is the sum of those who love chocolate (C) and fruit (F): - L = C + F = 1 + 5 = 6 - Therefore, the number of truth-tellers (T) is 10 - L = 10 - 6 = 4.Thus, the number of truthful dwarfs is (boxed{4}).,<think>Okay, so there's this problem with 10 dwarfs. Each one either always tells the truth or always lies. Each of them loves exactly one type of ice cream: butter, chocolate, or fruit. Snow White asked them three times to raise their hands depending on the ice cream they love. First, when she asked those who love butter ice cream to raise their hands, everyone raised their hand. Then, when she asked about chocolate, half of them (so 5) raised their hands. Lastly, for fruit, only one dwarf raised his hand. The question is, how many of the dwarfs are truthful?Hmm, let me start by breaking down the information. There are 10 dwarfs. Each is either a truth-teller or a liar. They each love one ice cream type: butter, chocolate, or fruit. First, when asked about butter, everyone raised their hand. That seems odd because if only the butter lovers are supposed to raise their hands, then either all dwarfs are butter lovers, but that can't be since some raised their hands for chocolate and fruit. But wait, wait. Wait, the dwarfs might be lying or telling the truth when they raise their hands. Wait, when Snow White asks a question, the truthful dwarfs would raise their hand if they actually love that ice cream, right? Because they tell the truth. But the liars, when asked if they love a certain ice cream, would lie. So if they are asked "Raise your hand if you love butter ice cream," a liar who does not love butter would lie and raise their hand anyway. Wait, hold on. If a liar is asked to raise their hand if they love butter, but they don't love butter, they would lie by not raising their hand? Wait, no, wait. Wait, the exact logic here needs to be clear.Wait, if a truthful dwarf loves butter, he raises his hand when asked. If he doesn't love butter, he doesn't raise. A liar, on the other hand, does the opposite. So if a liar loves butter, when asked, he lies and doesn't raise his hand. If he doesn't love butter, he lies and raises his hand. Wait, no. Wait, let's think carefully.Suppose the question is: "Raise your hand if you love butter ice cream."- A truthful dwarf: If he loves butter, he raises his hand. If he doesn't love butter, he doesn't raise.- A liar dwarf: If he loves butter, he would lie by not raising his hand. If he doesn't love butter, he would lie by raising his hand.So when asked about butter, the number of dwarfs raising hands would be equal to the number of truthful dwarfs who love butter plus the number of liar dwarfs who do NOT love butter. Because the truthful ones only raise if they love it, and the liars raise if they don't love it.But in the problem, when asked about butter, everyone raised their hands. So all 10 dwarfs raised their hands. That means:Number of truthful dwarfs who love butter + Number of liar dwarfs who do not love butter = 10.But there are 10 dwarfs total, so this equation must hold.Similarly, when asked about chocolate, half (5) raised their hands. So:Number of truthful dwarfs who love chocolate + Number of liar dwarfs who do not love chocolate = 5.And when asked about fruit, only one raised their hand:Number of truthful dwarfs who love fruit + Number of liar dwarfs who do not love fruit = 1.Now, each dwarf loves exactly one type of ice cream, so the total number of dwarfs who love butter, chocolate, or fruit must be 10. Let me denote:Let T be the number of truthful dwarfs, and L = 10 - T be the number of liars.Each truthful dwarf must love exactly one type of ice cream. Let’s denote:B_t: truthful dwarfs who love butter.C_t: truthful dwarfs who love chocolate.F_t: truthful dwarfs who love fruit.Similarly for liars:B_l: liar dwarfs who love butter.C_l: liar dwarfs who love chocolate.F_l: liar dwarfs who love fruit.We know that B_t + C_t + F_t = T (since each truthful dwarf is exactly one type), and B_l + C_l + F_l = L (same for liars). Also, the total number of butter lovers is B_t + B_l, chocolate lovers C_t + C_l, and fruit lovers F_t + F_l. These must satisfy:B_t + B_l + C_t + C_l + F_t + F_l = 10.But since each dwarf has exactly one favorite, this is already covered.Now, going back to the first equation when everyone raised their hand for butter. As per the logic:When asked about butter, the number of dwarfs raising hands is B_t (truthful butter lovers) + (L - B_l) (liars who don't love butter). Because liars who don't love butter will lie and raise their hands.But B_l is the number of liars who love butter. So the number of liars who do not love butter is L - B_l. Therefore:B_t + (L - B_l) = 10.But since L = 10 - T, this becomes:B_t + (10 - T - B_l) = 10.Simplify: B_t - B_l - T + 10 = 10 → B_t - B_l - T = 0 → B_t = B_l + T.Similarly, when asked about chocolate, the number of dwarfs raising hands is C_t + (L - C_l) = 5.So:C_t + (10 - T - C_l) = 5.Simplify: C_t - C_l - T + 10 = 5 → C_t - C_l - T = -5 → C_t = C_l + T -5.For fruit, the number of hands raised is F_t + (L - F_l) = 1.So:F_t + (10 - T - F_l) = 1 → F_t - F_l - T +10 =1 → F_t - F_l - T = -9 → F_t = F_l + T -9.Now, since B_t + C_t + F_t = T (all truthful dwarfs), and B_l + C_l + F_l = L =10 - T (all liars).Let me substitute the equations we found:B_t = B_l + T,C_t = C_l + T -5,F_t = F_l + T -9.Sum these up:B_t + C_t + F_t = (B_l + T) + (C_l + T -5) + (F_l + T -9).But the left side is T, and the right side is B_l + C_l + F_l + 3T -14.But B_l + C_l + F_l = 10 - T, so substituting:Left side: TRight side: (10 - T) + 3T -14 = 10 - T +3T -14 = 2T -4.Thus:T = 2T -4 → T -4 =0 → T=4.So according to this, there are 4 truthful dwarfs. Let me check if this makes sense.If T=4, then L=6.Now, let's compute B_t, C_t, F_t in terms of B_l, C_l, F_l.From B_t = B_l +4,C_t = C_l +4 -5 = C_l -1,F_t = F_l +4 -9 = F_l -5.But B_t + C_t + F_t =4,So substituting:(B_l +4) + (C_l -1) + (F_l -5) =4Simplify: B_l + C_l + F_l +4 -1 -5 =4 → (B_l + C_l + F_l) -2 =4.But B_l + C_l + F_l =6 (since L=6), so 6 -2 =4. Which holds. So that's good.Now, let's check the total ice cream counts.Total butter lovers: B_t + B_l = (B_l +4) + B_l = 2B_l +4.Total chocolate lovers: C_t + C_l = (C_l -1) + C_l = 2C_l -1.Total fruit lovers: F_t + F_l = (F_l -5) + F_l = 2F_l -5.Since all dwarfs have exactly one favorite, these must sum to 10:(2B_l +4) + (2C_l -1) + (2F_l -5) =10Simplify: 2B_l +4 +2C_l -1 +2F_l -5 =10 → 2(B_l + C_l + F_l) -2 =10.But B_l + C_l + F_l =6, so 2*6 -2 =12 -2=10. Perfect.So, equations are consistent.Now, we need to check if the numbers make sense. Let's see:Butter lovers: 2B_l +4.Chocolate lovers:2C_l -1.Fruit lovers:2F_l -5.These should all be non-negative integers. Let's check.First, since B_l, C_l, F_l are non-negative integers (number of dwarfs can't be negative), and they sum to 6.So, let's denote:Butter lovers: 2B_l +4 ≥0 → Since B_l is non-negative, this is at least 4.Chocolate lovers:2C_l -1 ≥0 → 2C_l ≥1 → C_l ≥1 (since C_l must be integer).Fruit lovers:2F_l -5 ≥0 → 2F_l ≥5 → F_l ≥3 (since F_l must be integer).So, C_l ≥1 and F_l ≥3. But B_l + C_l + F_l =6.If F_l ≥3, then minimum F_l=3, C_l≥1, so B_l ≤6 -3 -1=2.Similarly, let's see possible values.Let’s let F_l=3. Then C_l ≥1, so C_l=1,2,3, etc., but with B_l=6 -3 -C_l.Let’s check:If F_l=3, then C_l can be 1, 2, or 3 (since B_l=6-3-C_l=3 -C_l. So if C_l=1, B_l=2; C_l=2, B_l=1; C_l=3, B_l=0. Let’s check each case.Case 1: F_l=3, C_l=1, B_l=2.Butter lovers:2*2 +4=8.Chocolate lovers:2*1 -1=1.Fruit lovers:2*3 -5=1.Total:8+1+1=10. Okay, this works.Case 2: F_l=3, C_l=2, B_l=1.Butter:2*1 +4=6.Chocolate:2*2 -1=3.Fruit:2*3 -5=1.Total:6+3+1=10. Also works.Case3: F_l=3, C_l=3, B_l=0.Butter:2*0 +4=4.Chocolate:2*3 -1=5.Fruit:1.Total:4+5+1=10. Also works.Next, F_l=4. Then B_l + C_l=6 -4=2. C_l≥1, so C_l=1,2. Then B_l=1,0.Case4: F_l=4, C_l=1, B_l=1.Butter:2*1 +4=6.Chocolate:2*1 -1=1.Fruit:2*4 -5=3.Total:6+1+3=10. Works.Case5: F_l=4, C_l=2, B_l=0.Butter:0+4=4.Chocolate:2*2 -1=3.Fruit:2*4 -5=3.Total:4+3+3=10. Okay.F_l=5: Then B_l + C_l=1. C_l≥1, so C_l=1, B_l=0.Case6: F_l=5, C_l=1, B_l=0.Butter:0+4=4.Chocolate:2*1 -1=1.Fruit:2*5 -5=5.Total:4+1+5=10. Good.F_l=6: Then B_l + C_l=0. Not possible since C_l≥1. So F_l can't be 6.So possible scenarios where all counts are non-negative integers. Now, check if in these scenarios, the numbers of dwarfs who raised hands match.Wait, but actually, we need to check if when Snow White asked about chocolate, 5 dwarfs raised their hands. Let me recall: For each ice cream type, the number of dwarfs raising hands is the number of truthful dwarfs who love that type plus the number of liars who do NOT love that type.Wait, but in the equations earlier, we had:For butter: B_t + (L - B_l) =10. Which, with T=4, L=6, becomes B_t +6 - B_l=10 → B_t - B_l=4. Which is satisfied because B_t = B_l +4.Similarly, for chocolate: C_t + (6 - C_l)=5 → C_t - C_l= -1. Which we had C_t = C_l -1. Wait, but C_t is truthful dwarfs who love chocolate, which must be non-negative. So C_l -1 ≥0 → C_l ≥1. Which is already considered.Similarly, for fruit: F_t + (6 - F_l)=1 → F_t - F_l= -5. Which gives F_t = F_l -5. But F_t is the number of truthful dwarfs who love fruit, which must be non-negative. So F_l -5 ≥0 → F_l ≥5. Wait, but in our cases above, we had F_l=3,4,5, which would mean F_t= -2, -1,0. But F_t cannot be negative. Wait, this is a problem.Wait, hold on. Wait, when we had T=4, we derived F_t = F_l + T -9 = F_l +4 -9= F_l -5. But F_t must be non-negative, so F_l -5 ≥0 → F_l ≥5. But earlier when we considered F_l=3, that would make F_t= -2, which is impossible. So that suggests that our initial cases where F_l=3,4 are invalid because F_t can't be negative.Wait, so that's a mistake in my earlier analysis. Because when I considered F_l=3, I didn't check if F_t is non-negative. So actually, F_t must be ≥0, so from F_t = F_l -5, we get F_l ≥5.Therefore, F_l must be at least 5. So F_l=5,6.But F_l can't be 6, since then B_l + C_l=6 - F_l=0, which would require C_l=0, but from the chocolate equation, C_l ≥1. So F_l can't be6. Therefore, F_l=5. Then B_l + C_l=1. Since C_l≥1 (from chocolate lovers needing C_l≥1), then C_l=1 and B_l=0.So only possible case is F_l=5, C_l=1, B_l=0.Thus:Butter lovers: B_t + B_l = (B_l +4) + B_l =0+4 +0=4.Chocolate lovers: C_t + C_l=(C_l -1)+C_l=1-1 +1=1.Fruit lovers: F_t +F_l=(F_l -5) +F_l=5-5 +5=5.Wait, total:4+1+5=10. Correct.But wait, let's verify the hands raised.For butter: B_t + (L - B_l)=4 + (6 -0)=10. Correct.For chocolate: C_t + (L - C_l)=0 + (6 -1)=5. Correct.For fruit: F_t + (L - F_l)=0 + (6 -5)=1. Correct.So in this scenario, all numbers check out. So the only possible case is F_l=5, C_l=1, B_l=0.Thus, the truthful dwarfs are:B_t = B_l +4=0+4=4.C_t = C_l -1=1 -1=0.F_t = F_l -5=5 -5=0.So all 4 truthful dwarfs love butter. The 6 liars: B_l=0, C_l=1, F_l=5. So 1 liar loves chocolate, 5 liars love fruit.So when asked about butter, all 4 truthful dwarfs (who love butter) raised their hands, and all 6 liars (who don't love butter, since they love chocolate or fruit) also raised their hands (because liars lie). Wait, but wait: The liars don't love butter, so when asked if they love butter, they lie and raise their hands. Hence, all 10 raise their hands. Correct.For chocolate: The truthful dwarfs who love chocolate are 0, so no one truthful raised. The liars who don't love chocolate are 6 - C_l=6 -1=5. So 5 liars raised their hands. Hence total 5, which matches.For fruit: The truthful dwarfs who love fruit are 0. The liars who don't love fruit are 6 - F_l=6 -5=1. So 1 liar raised his hand. Hence total 1, which matches.So this all checks out. So the number of truthful dwarfs is 4.Therefore, the answer is 4.Final Answerboxed{4}</think>Given 10 dwarfs, each either a truth-teller or a liar, and each loving exactly one type of ice cream (butter, chocolate, or fruit), we need to determine the number of truthful dwarfs based on their responses to Snow White's questions:1. All 10 dwarfs raised their hands when asked about butter.2. 5 dwarfs raised their hands when asked about chocolate.3. 1 dwarf raised his hand when asked about fruit. Key Steps:1. Hand Raising Analysis: - Butter: All 10 dwarfs raised their hands. This means the number of truthful dwarfs who love butter plus the number of liars who do not love butter equals 10. - Chocolate: 5 dwarfs raised their hands. This means the number of truthful dwarfs who love chocolate plus the number of liars who do not love chocolate equals 5. - Fruit: 1 dwarf raised his hand. This means the number of truthful dwarfs who love fruit plus the number of liars who do not love fruit equals 1.2. Equations Setup: - Let ( T ) be the number of truthful dwarfs and ( L = 10 - T ) be the number of liars. - For butter: ( B_t + (L - B_l) = 10 ) - For chocolate: ( C_t + (L - C_l) = 5 ) - For fruit: ( F_t + (L - F_l) = 1 ) - Where ( B_t, C_t, F_t ) are truthful dwarfs and ( B_l, C_l, F_l ) are liar dwarfs loving each ice cream type.3. Solving the Equations: - From the butter equation: ( B_t = B_l + T ) - From the chocolate equation: ( C_t = C_l + T - 5 ) - From the fruit equation: ( F_t = F_l + T - 9 ) - Summing these equations and using ( B_t + C_t + F_t = T ) and ( B_l + C_l + F_l = L ), we find ( T = 4 ).4. Validation: - With ( T = 4 ) and ( L = 6 ), we check the consistency: - Butter lovers: ( 2B_l + 4 ) - Chocolate lovers: ( 2C_l - 1 ) - Fruit lovers: ( 2F_l - 5 ) - Ensuring non-negative integers, the only valid solution is ( B_l = 0 ), ( C_l = 1 ), ( F_l = 5 ), leading to 4 truthful dwarfs. Conclusion:The number of truthful dwarfs is (boxed{4}).]

🔑:# Answer. 4.Solution. The gnomes who always tell the truth raised their hands once, while the gnomes who always lie raised their hands twice. In total, 16 hands were raised (10+5+1). If all the gnomes had told the truth, 10 hands would have been raised. If one truthful gnome is replaced by one liar, the number of raised hands increases by 1. Since there were 6 "extra" hands raised, 6 gnomes lied, and 4 told the truth.## 8th Grade. Grading Recommendations.Each problem is scored out of 7 points. Each score is an integer from 0 to 7. Some guidelines for grading are provided below. Naturally, the creators cannot foresee all cases. When evaluating a solution, it should be determined whether the provided solution is generally correct (although it may have flaws) - in which case the solution should be scored at least 4 points. Or if it is incorrect (although it may have significant progress) - in which case the score should not exceed 3 points.

🧠:[<think>Alright, so I need to find the number of natural numbers that don't exceed 2022 and aren't part of either of two arithmetic progressions. The first progression is the odd numbers: 1, 3, 5, ..., and the second one is 1, 4, 7, ... Hmm, okay. Let me break this down step by step.First, let me understand what exactly is being asked. I need to count all natural numbers up to 2022 that are not in either of these two arithmetic sequences. So, essentially, total numbers from 1 to 2022 minus the numbers that are in either of these two progressions. But wait, I have to be careful here because some numbers might be in both progressions. If I just subtract the counts of each progression, I might be double-counting the overlap. So, I need to use the principle of inclusion-exclusion here. That is, the total number of excluded numbers is the number in the first progression plus the number in the second progression minus the number that's in both. Then, subtract that total from 2022 to get the answer.Let me write that down:Number of natural numbers not exceeding 2022 = 2022Number of numbers in the first progression (odds) = ?Number of numbers in the second progression (1,4,7,...) = ?Number of numbers in both progressions = ?So, desired count = 2022 - (number in first + number in second - number in both)Yes, that makes sense.First, let's handle the first progression: the odd numbers. The arithmetic progression here is starting at 1 with a common difference of 2. So, the nth term is 1 + (n-1)*2 = 2n -1. We need to find how many terms of this sequence are less than or equal to 2022.So, solving 2n -1 ≤ 2022Add 1: 2n ≤ 2023Divide by 2: n ≤ 1011.5Since n must be a natural number, the number of terms is 1011.Wait, let me check that again. If n=1011, then term is 2*1011 -1 = 2022 -1 = 2021. Then n=1012 would be 2*1012 -1 = 2024 -1 = 2023, which is over 2022. So yes, there are 1011 terms in the first progression.Now, the second progression: starting at 1, common difference 3. So the nth term is 1 + (n-1)*3 = 3n -2. We need to find how many terms are ≤ 2022.Set 3n -2 ≤ 2022Add 2: 3n ≤ 2024Divide by 3: n ≤ 2024/3 ≈ 674.666...So n=674 gives 3*674 -2 = 2022 -2 = 2020. Then n=675 would be 3*675 -2 = 2025 -2 = 2023, which is over. So there are 674 terms in the second progression.Okay, so now we have 1011 numbers in the first progression and 674 in the second. But now we need to find the overlap, numbers that are in both progressions. That is, numbers that are odd and congruent to 1 modulo 3. Wait, let me think.First progression: all numbers congruent to 1 mod 2 (since they're odd). Second progression: numbers congruent to 1 mod 3. So the overlap would be numbers congruent to 1 mod 2 and 1 mod 3. To find numbers that satisfy both congruences, we can use the Chinese Remainder Theorem. Since 2 and 3 are coprime, the solution is numbers congruent to 1 mod 6. Let me verify that.Looking for x ≡1 mod 2 and x ≡1 mod 3. So x ≡1 mod lcm(2,3)=6. So yes, x ≡1 mod 6. Therefore, the common terms are the arithmetic progression starting at 1 with difference 6: 1,7,13,... So the nth term is 1 + (n-1)*6 = 6n -5. Let's check when 6n -5 ≤ 2022.Solving 6n -5 ≤ 2022Add 5: 6n ≤ 2027Divide by 6: n ≤ 2027/6 ≈ 337.833...So n=337 gives 6*337 -5 = 2022 -5 = 2017. n=338 would be 6*338 -5 = 2028 -5 = 2023, which is over. So the number of overlapping terms is 337.Therefore, using inclusion-exclusion, the total number of numbers in either progression is 1011 + 674 - 337 = 1011 + 674 is 1685, minus 337 is 1348. Therefore, the numbers not in either progression would be 2022 - 1348 = 674.Wait, hold on. 2022 - 1348 is 674? Let me check that subtraction: 2022 - 1300 is 722, then subtract 48 more: 722 - 48 = 674. Yeah. Hmm.But wait, this answer seems a bit low. Let me verify all steps again. Let me go through each calculation carefully.First progression: odd numbers up to 2022. The last term is 2021, which is 2*1011 -1. So 1011 terms. That seems correct.Second progression: starting at 1, difference 3. The nth term is 3n -2. Setting 3n -2 ≤2022 gives 3n ≤2024, so n=674.666..., so 674 terms. So the last term is 3*674 -2=2020, which is indeed ≤2022, next term would be 2023. Correct.Overlap: numbers in both progressions. As we saw, the overlapping numbers are those congruent to 1 mod 6. The nth term is 6n -5. Solving 6n -5 ≤2022 gives n= (2022 +5)/6=2027/6≈337.833, so 337 terms. Last term is 6*337 -5=2022 -5=2017. Next term would be 2023. Correct.So inclusion-exclusion gives 1011 +674 -337= 1348. Then total numbers not in either progression: 2022 -1348=674. Hmm. But let me cross-verify this with another approach.Alternatively, we can compute the number of numbers not in the first progression (which are the even numbers) and subtract the numbers that are in the second progression. But wait, that might complicate. Alternatively, think in terms of sets.Set A: numbers in first progression (odd numbers). Size |A|=1011.Set B: numbers in second progression (1,4,7,...). Size |B|=674.Set A∩B: numbers in both. Size |A∩B|=337.So |A∪B|=|A| + |B| - |A∩B|=1011+674-337=1348.Therefore, numbers not in A∪B=Total numbers - |A∪B|=2022-1348=674. So same answer. So that seems correct.But let's test with smaller numbers to see if the reasoning holds.Suppose we take numbers up to 10.First progression (odds): 1,3,5,7,9 (5 numbers)Second progression (1,4,7,10): 1,4,7,10 (4 numbers)Overlap: numbers in both. Which are 1,7 (since 1 is in both, 4 is even, 7 is odd, 10 is even). So overlap size 2.So inclusion-exclusion: 5+4-2=7. Numbers not in either: 10 -7=3. Let's list them: 2,6,8. Wait, 2,6,8 are not in either progression. But 10 is in the second progression. So numbers up to 10: 1,2,3,4,5,6,7,8,9,10. Excluding 1,3,5,7,9 (odds) and 1,4,7,10. So the excluded numbers are 1,3,4,5,7,9,10. Therefore, remaining numbers: 2,6,8. Which are 3 numbers. So 10-7=3. So that works. So the formula works here.Another example: numbers up to 7.First progression: 1,3,5,7 (4 numbers)Second progression:1,4,7 (3 numbers)Overlap: 1,7 (2 numbers)Inclusion-exclusion:4+3-2=5. Numbers not in either:7-5=2. Which are 2,6. Correct. So that works.So seems like the inclusion-exclusion approach is valid.Therefore, returning to the original problem, the answer would be 674. But wait, 674? Let me just confirm once again.Wait, the total numbers up to 2022:2022Numbers in first progression:1011Numbers in second:674Overlap:337So total excluded:1011 +674 -337=13482022 -1348=674But let me check another way. How many numbers are even and not in the second progression? Since numbers not in the first progression are the even numbers. So even numbers up to 2022:2022/2=1011. So 1011 even numbers.But some even numbers might be in the second progression. The second progression is 1,4,7,10,... which alternates even and odd? Wait, no. The progression is 1,4,7,10,13,... Each term increases by 3. So starting at 1, which is odd, then 4 (even), 7 (odd), 10 (even), etc. So every other term in the second progression is even. So the even numbers in the second progression are 4,10,16,... up to <=2022.So how many even numbers are in the second progression? Let's see. The second progression is 1,4,7,10,... So the even terms are those where the term is congruent to 0 mod 2. Since the progression is 1 mod 3, so 1,4,7,10,... So 4 mod 6, 10 mod 6=4, etc. Wait, 1 mod 3 and even. So numbers congruent to 4 mod 6. Because 4 mod 6 is equivalent to 1 mod 3 and 0 mod 2.So the even numbers in the second progression are 4,10,16,... which is an arithmetic progression starting at 4, with difference 6. So the nth term is 4 + (n-1)*6=6n -2. Wait, 4=6*1 -2, 10=6*2 -2, 16=6*3 -2, etc. So yes, 6n -2.Find how many terms of this sequence are <=2022.6n -2 <=20226n <=2024n<=2024/6≈337.333...So n=337, term is 6*337 -2=2022 -2=2020. Then n=338 would be 2022 +4=2026, which is over. So 337 even numbers in the second progression.Therefore, the even numbers not in the second progression would be total even numbers (1011) minus even numbers in the second progression (337), which is 1011 -337=674. Which matches the previous result. So that's another way to see it: numbers not in the first progression (even numbers) minus those even numbers that are in the second progression gives the numbers that are neither in the first nor the second progression. So that also gives 674.Therefore, regardless of the approach, the answer is 674. So I think that's correct.Final AnswerThe number of natural numbers not exceeding 2022 and not belonging to either progression is boxed{674}.</think>To find the number of natural numbers not exceeding 2022 and not belonging to either the arithmetic progression (1, 3, 5, ldots) or the arithmetic progression (1, 4, 7, ldots), we use the principle of inclusion-exclusion.1. Count numbers in the first progression (odd numbers): - The nth term is (2n - 1). - Solving (2n - 1 leq 2022) gives (n leq 1011.5), so there are 1011 terms.2. Count numbers in the second progression (1, 4, 7, ...): - The nth term is (3n - 2). - Solving (3n - 2 leq 2022) gives (n leq 674.666), so there are 674 terms.3. Count numbers in both progressions (numbers congruent to 1 mod 6): - The nth term is (6n - 5). - Solving (6n - 5 leq 2022) gives (n leq 337.833), so there are 337 terms.Using inclusion-exclusion principle:- Total numbers in either progression: (1011 + 674 - 337 = 1348).Subtracting this from the total numbers up to 2022:- Numbers not in either progression: (2022 - 1348 = 674).Thus, the number of natural numbers not exceeding 2022 and not belonging to either progression is (boxed{674}).,<think>Okay, so I need to find how many natural numbers up to 2022 are not in either of two arithmetic progressions. The first one is the odd numbers: 1, 3, 5, … and the second one is 1, 4, 7, …. Hmm. Let me think step by step.First, let me recall what an arithmetic progression (AP) is. It's a sequence where each term after the first is obtained by adding a constant difference. For the first AP, the difference is 2 (since it's all odd numbers), and for the second AP, the difference is 3. The problem wants numbers that are not in either of these two APs. So, effectively, I need to subtract the numbers that are in either AP from the total numbers up to 2022, but I have to be careful not to double-count the numbers that are in both APs.Right, inclusion-exclusion principle. The formula for the number of elements not in either set is Total - (Number in first AP + Number in second AP - Number in both APs). So the formula is:Number = Total - |A| - |B| + |A ∩ B|Where A is the first AP and B is the second AP.First, let me figure out the total number of natural numbers not exceeding 2022. Well, natural numbers start from 1, so numbers from 1 to 2022. That's 2022 numbers in total.Next, find |A|, the number of terms in the first AP (1, 3, 5, ...) up to 2022. The nth term of an AP is given by a_n = a_1 + (n-1)d. Here, a_1 = 1, d = 2. So, setting a_n ≤ 2022:1 + (n-1)*2 ≤ 2022Subtract 1: (n-1)*2 ≤ 2021Divide by 2: n-1 ≤ 1010.5So n-1 ≤ 1010.5, so n ≤ 1011.5. Since n must be an integer, n = 1011. So there are 1011 terms in the first AP.Similarly, for |B|, the second AP (1, 4, 7, ...) with a_1 = 1, d = 3. Let's find how many terms are ≤ 2022:1 + (n-1)*3 ≤ 2022Subtract 1: (n-1)*3 ≤ 2021Divide by 3: n-1 ≤ 2021/3 ≈ 673.666...So n-1 ≤ 673.666, so n ≤ 674.666. Therefore, n = 674. So there are 674 terms in the second AP.Now, the tricky part is |A ∩ B|, the numbers that are in both APs. These are numbers that are congruent to 1 mod 2 and 1 mod 3. Wait, since the first AP is all odd numbers (1 mod 2) and the second AP is 1 mod 3. So we need numbers that are 1 mod 2 and 1 mod 3. To find numbers that satisfy both congruences, we can use the Chinese Remainder Theorem. Let's see.We need to solve:x ≡ 1 mod 2x ≡ 1 mod 3Find x such that both congruences hold. The solutions to these will be numbers congruent to 1 mod lcm(2,3) = 6. Wait, since 2 and 3 are coprime, the solution is x ≡ 1 mod 6. Let me verify:If x ≡ 1 mod 6, then x = 6k + 1. Then x mod 2 = 1 mod 2 = 1, and x mod 3 = 1 mod 3 = 1. So yes, the numbers congruent to 1 mod 6 satisfy both congruences. Therefore, the common terms of both APs form an AP with a_1 = 1, d = 6.So the number of terms in |A ∩ B| up to 2022 is the number of terms in the AP 1, 7, 13, ... with difference 6. Let's compute that.Again, nth term formula: 1 + (n-1)*6 ≤ 2022Subtract 1: (n-1)*6 ≤ 2021Divide by 6: n-1 ≤ 2021/6 ≈ 336.833...So n-1 ≤ 336.833, so n ≤ 337.833. Therefore, n = 337. So there are 337 numbers common to both APs.Now, applying inclusion-exclusion:Number of numbers not in A or B = Total - |A| - |B| + |A ∩ B|Plugging in the numbers:2022 - 1011 - 674 + 337Let me compute step by step:2022 - 1011 = 10111011 - 674 = 337337 + 337 = 674Wait, that seems a bit coincidental. Let me check again:Total is 2022Minus |A| (1011): 2022 - 1011 = 1011Minus |B| (674): 1011 - 674 = 337Plus |A ∩ B| (337): 337 + 337 = 674So the result is 674. Wait, but 2022 - 1011 - 674 + 337 = 674. That seems low? Let me verify each step.First, total numbers: 2022. Correct.Number in A: AP with difference 2, starting at 1. The number of terms is floor((2022 - 1)/2) + 1 = floor(2021/2) + 1 = 1010 + 1 = 1011. Correct.Number in B: AP with difference 3, starting at 1. Number of terms is floor((2022 -1)/3) +1 = floor(2021/3) +1 = 673 +1 = 674. Correct.Number in both A and B: AP with difference 6, starting at 1. Number of terms is floor((2022 -1)/6) +1 = floor(2021/6) +1 = 336 +1 = 337. Correct.Then, applying inclusion-exclusion: 2022 - 1011 - 674 + 337 = 2022 - 1685 + 337 = (2022 - 1685) + 337 = 337 + 337 = 674. So 674 numbers not in either AP. But wait, is that possible? Let's see.Wait, but if we have 1011 odd numbers, which are half the numbers up to 2022 (since every other number is odd). Then the second AP is numbers congruent to 1 mod 3, which is roughly a third of the numbers. The overlap is numbers congruent to 1 mod 6, which is a sixth of the numbers.But if we subtract the counts and add back the overlap, we get 2022 - 1011 - 674 + 337. Let me compute this again:1011 + 674 = 1685. Then 2022 - 1685 = 337. Then 337 + 337 = 674. So yes, 674 is the answer.But just to make sure, let's take a smaller example and test. Suppose N=10.Total numbers: 10.Numbers in A (odd): 1,3,5,7,9 => 5 numbers.Numbers in B (1,4,7,10): 4 numbers.Numbers in both: 1,7 => 2 numbers.Numbers not in A or B: 10 -5 -4 +2 = 3. The numbers are 2,6,8. Let's check:Numbers not in A (even) and not in B (not 1,4,7,10). The even numbers up to 10 are 2,4,6,8,10. Excluding 4,10 (since they are in B), we have 2,6,8. That's 3 numbers. Correct. So the formula works here.So applying the same logic, the answer seems to be 674. Hmm. But let me check another edge case.Wait, for N=1. Total numbers=1. A has 1, B has 1. Overlap is 1. So numbers not in A or B: 1 -1 -1 +1 =0. Correct.Another example, N=6.Total numbers:6.A (odds):1,3,5 =>3 numbers.B:1,4 =>2 numbers (since 1 +3=4, next would be 7 which is beyond 6).Overlap:1.So numbers not in A or B:6 -3 -2 +1=2. The numbers are 2,6. Let's see: evens not 2,4,6. But 4 is in B, so exclude 4, so 2 and6. Correct.So the formula seems to work. Therefore, trusting the calculation, the answer is 674.But just to make sure, let's think about another approach.Alternatively, numbers not in either AP are numbers that are even (not in A) and not congruent to 1 mod 3 (not in B). Wait, but in the first AP, the numbers are all odd, so numbers not in A are the even numbers. Then, among the even numbers, how many are not in B?Wait, B is numbers congruent to 1 mod 3, which are all odd or even? Let's check. 1 mod 3 is 1,4,7,10,... So alternately odd and even. For example, 1 is odd, 4 is even, 7 is odd, 10 is even, etc. So in B, numbers can be both odd and even. So if we take even numbers not in B, that is, even numbers not congruent to 1 mod 3. Wait, but 1 mod 3 is equivalent to 1,4,7,10,... So even numbers in B are those congruent to 1 mod 3 and even. So 4,10,16,... which is 4 mod 6. Because 1 mod 3 and even implies 4 mod 6. So numbers in B that are even are numbers congruent to 4 mod 6.Therefore, even numbers not in B are even numbers not congruent to 4 mod 6, i.e., congruent to 0,2 mod 6.Wait, but this approach might complicate. Let's see.Alternatively, since numbers not in A are even numbers (2,4,6,...,2022). Then, from these, remove numbers that are in B. B is numbers congruent to 1 mod 3. So in the even numbers, how many are congruent to 1 mod 3? Let's see. Solve x ≡ 1 mod 3, x even. So x ≡ 1 mod 3 and x ≡ 0 mod 2.Solutions are numbers congruent to 4 mod 6 (since 4 ≡1 mod3 and 4 ≡0 mod2). So numbers congruent to 4 mod6. So in the even numbers, the ones in B are 4,10,16,... up to ≤2022.So to compute the number of even numbers not in B: total even numbers - numbers in B that are even.Total even numbers up to 2022: floor(2022/2)=1011.Numbers in B that are even: numbers congruent to 4 mod6 up to 2022. Let's find how many:First term 4, difference 6. The nth term is 4 + (n-1)*6 ≤ 2022.So 4 +6(n-1) ≤2022 => 6(n-1) ≤2018 => n-1 ≤2018/6≈336.333, so n=337.Therefore, there are 337 even numbers in B. So the number of even numbers not in B is 1011 -337=674. Which matches the previous result.Therefore, the answer is 674. So the number of natural numbers not exceeding 2022 and not in either progression is 674.Final Answerboxed{674}</think>To find the number of natural numbers not exceeding 2022 and not belonging to either the arithmetic progression (1, 3, 5, ldots) or the arithmetic progression (1, 4, 7, ldots), we use the inclusion-exclusion principle.1. Total numbers: There are 2022 natural numbers from 1 to 2022.2. Numbers in the first AP (1, 3, 5, ...): - The nth term is given by (1 + (n-1) cdot 2 leq 2022). - Solving, we get (n = 1011). So, there are 1011 terms.3. Numbers in the second AP (1, 4, 7, ...): - The nth term is given by (1 + (n-1) cdot 3 leq 2022). - Solving, we get (n = 674). So, there are 674 terms.4. Numbers in both APs: - These numbers are congruent to 1 mod 6 (solving (x equiv 1 mod 2) and (x equiv 1 mod 3)). - The nth term is given by (1 + (n-1) cdot 6 leq 2022). - Solving, we get (n = 337). So, there are 337 common terms.Applying the inclusion-exclusion principle:[text{Number of numbers not in either AP} = 2022 - 1011 - 674 + 337]Calculating step-by-step:[2022 - 1011 = 1011][1011 - 674 = 337][337 + 337 = 674]Thus, the number of natural numbers not exceeding 2022 and not belonging to either arithmetic progression is (boxed{674}).]

🔑:Answer: 674.Solution. These two progressions define numbers of the form 1+2n and 1+3n. This indicates that the desired numbers are of the form 6n and 6n-4, n in mathbb{N}. Since 2022 is divisible by 6, the numbers of the form 6n will be frac{2022}{6}=337, and there will be as many numbers of the form 6n-4. Therefore, the total number of numbers is 337 cdot 2=674.

🧠:[<think>Okay, so I need to solve this trigonometry problem involving a triangle ABC. The given equation is cos A + sin A - 2/(cos B + sin B) = 0, and we need to find the value of (a + b)/c. The answer choices are 1, √2, √3, and 2. Hmm. Let's start by writing down what's given and what we need to find.First, in triangle ABC, the sides opposite angles A, B, and C are a, b, c respectively. The equation given is:cos A + sin A - 2/(cos B + sin B) = 0So rearranging that, we have:cos A + sin A = 2/(cos B + sin B)Our goal is to find (a + b)/c. Since the problem involves trigonometric functions of angles and sides of a triangle, I might need to use the Law of Sines or the Law of Cosines. Let me recall those:Law of Sines: a/sin A = b/sin B = c/sin C = 2R (where R is the radius of the circumscribed circle)Law of Cosines: c² = a² + b² - 2ab cos CAlternatively, for angle A: a² = b² + c² - 2bc cos ASimilarly for angle B.But since we have cos A + sin A and cos B + sin B in the equation, maybe there's a way to express these terms in a different form or relate angles A and B. Also, note that in a triangle, the sum of angles A + B + C = π. So maybe we can express angle C in terms of A and B if needed.Let me first look at the equation again:cos A + sin A = 2/(cos B + sin B)Let me denote cos A + sin A as X and cos B + sin B as Y. Then the equation is X = 2/Y. So X * Y = 2. Therefore, (cos A + sin A)(cos B + sin B) = 2.Hmm, that seems like a useful step. Let me compute that product:(cos A + sin A)(cos B + sin B) = cos A cos B + cos A sin B + sin A cos B + sin A sin BCombine terms:= cos A cos B + sin A sin B + cos A sin B + sin A cos BNotice that cos A cos B + sin A sin B = cos(A - B) by the cosine addition formula.Similarly, cos A sin B + sin A cos B = sin(A + B) by the sine addition formula.So the product becomes:cos(A - B) + sin(A + B) = 2Therefore, we have the equation:cos(A - B) + sin(A + B) = 2Now, since the maximum value of cos(A - B) is 1 and the maximum value of sin(A + B) is 1, the sum can only be 2 if both terms are equal to 1 individually. Because if one term is less than 1, the other term can't compensate to reach 2. So:cos(A - B) = 1andsin(A + B) = 1So let's solve these equations.First, cos(A - B) = 1. The cosine of an angle is 1 when the angle is 0 modulo 2π. But since A and B are angles in a triangle, they must each be between 0 and π, so A - B = 0. Therefore, A = B.Second, sin(A + B) = 1. The sine of an angle is 1 at π/2. So A + B = π/2.But wait, in a triangle, A + B + C = π. If A + B = π/2, then C = π - (A + B) = π - π/2 = π/2. So angle C is a right angle. Therefore, triangle ABC is a right-angled triangle at C.Also, since A = B from the first equation, and A + B = π/2, so 2A = π/2, which implies A = π/4, and therefore B = π/4 as well. So triangle ABC is a right-angled isoceles triangle with angles A = B = 45°, C = 90°.In such a triangle, the legs are equal, and the hypotenuse is leg * √2. Let me denote the sides as follows: since angle C is 90°, side c is opposite angle C, so c is the hypotenuse. Sides a and b are opposite angles A and B, which are both 45°, so a = b. Therefore, in this triangle, a = b, and c = a√2.Therefore, (a + b)/c = (a + a)/ (a√2) = (2a)/(a√2) = 2/√2 = √2. So the answer is √2, which is option B.Wait, but let me verify if this is correct. Let me check the steps again.First, starting from the equation cos A + sin A = 2/(cos B + sin B). Then we multiplied both sides by (cos B + sin B) to get (cos A + sin A)(cos B + sin B) = 2. Then we expanded it into cos(A - B) + sin(A + B) = 2. Then since both cos and sin terms can be at maximum 1, we concluded that each must be 1. Therefore, A = B and A + B = π/2, leading to a right-angled isoceles triangle. Then sides a = b, c = a√2, so (a + b)/c = √2.Yes, this seems correct. Let me check with actual numbers.Suppose a = b = 1, then c = √2. Then (1 + 1)/√2 = 2/√2 = √2. So that's correct.Alternatively, check if the original equation holds. Let's compute cos A + sin A. Since A = 45°, cos 45° = sin 45° = √2/2. So cos A + sin A = √2/2 + √2/2 = √2. Similarly, cos B + sin B is the same, since B = 45°, so also √2. Then 2/(cos B + sin B) = 2/√2 = √2. Therefore, cos A + sin A = √2 and 2/(cos B + sin B) = √2. Thus, √2 - √2 = 0, which satisfies the equation. So that's correct.Therefore, the answer is √2, which is option B.Final Answerboxed{B}</think>Given the equation (cos A + sin A - frac{2}{cos B + sin B} = 0) in (triangle ABC), we need to find the value of (frac{a + b}{c}).First, we rearrange the given equation:[cos A + sin A = frac{2}{cos B + sin B}]Multiplying both sides by (cos B + sin B), we get:[(cos A + sin A)(cos B + sin B) = 2]Expanding the left-hand side, we use trigonometric identities:[cos A cos B + sin A sin B + cos A sin B + sin A cos B = cos(A - B) + sin(A + B)]Thus, the equation becomes:[cos(A - B) + sin(A + B) = 2]Since the maximum values of (cos theta) and (sin theta) are 1, both terms must individually equal 1:[cos(A - B) = 1 quad text{and} quad sin(A + B) = 1]From (cos(A - B) = 1), we get (A = B). From (sin(A + B) = 1), we get (A + B = frac{pi}{2}). Therefore, (C = pi - (A + B) = frac{pi}{2}), making (triangle ABC) a right-angled triangle at (C).Since (A = B = frac{pi}{4}), the triangle is isosceles with legs (a = b) and hypotenuse (c = asqrt{2}). Thus:[frac{a + b}{c} = frac{a + a}{asqrt{2}} = frac{2a}{asqrt{2}} = frac{2}{sqrt{2}} = sqrt{2}]The value of (frac{a + b}{c}) is (boxed{B}).,<think>Okay, let me try to figure out this problem. So, we have triangle ABC with sides a, b, c opposite angles A, B, C respectively. The given equation is cos A + sin A - 2/(cos B + sin B) = 0. We need to find the value of (a + b)/c. The options are 1, √2, √3, or 2. Hmm, trigonometry in triangles. Maybe I can relate the trigonometric identities with the sides using the Law of Sines or Cosines?First, let me write down the given equation again: cos A + sin A - 2/(cos B + sin B) = 0. So, rearranging that, we get cos A + sin A = 2/(cos B + sin B). Maybe if I square both sides? But before that, perhaps simplify the expressions. Let's see. Let me denote cos A + sin A as something. I remember that cos θ + sin θ can be written as √2 sin(θ + 45°) or √2 cos(θ - 45°). Let me verify that. Yes, using the identity:cos θ + sin θ = √2 sin(θ + 45°) = √2 cos(θ - 45°).So maybe that could help. Let's try that for both sides. Let me write:cos A + sin A = √2 sin(A + 45°) = √2 cos(A - 45°)Similarly, cos B + sin B = √2 sin(B + 45°) = √2 cos(B - 45°)So substituting into the equation:√2 sin(A + 45°) = 2 / [√2 sin(B + 45°)]Simplify the right-hand side:2 / [√2 sin(B + 45°)] = (2 / √2) / sin(B + 45°) = √2 / sin(B + 45°)So the equation becomes:√2 sin(A + 45°) = √2 / sin(B + 45°)Divide both sides by √2:sin(A + 45°) = 1 / sin(B + 45°)Which implies that sin(A + 45°) * sin(B + 45°) = 1But the product of sines equals 1. Since the maximum value of sine is 1, the only way this can happen is if both sin(A + 45°) and sin(B + 45°) are equal to 1 or -1. But since angles in a triangle are between 0 and 180°, A + 45° would be between 45° and 225°, and same for B + 45°. However, sine of angles greater than 180° is negative, but since we have angles in a triangle, both A and B are less than 180°, so A + 45° and B + 45° would be less than 225°. The sine function reaches 1 at 90°, and -1 at 270°, but since A and B are less than 180°, their +45° would be less than 225°, so the maximum possible sine is 1 (at 90°), and the minimum is sin(225°) = -√2/2. However, the product of the sines has to be 1. Since the maximum value of sine is 1, the only possible way for their product to be 1 is if both sin(A + 45°) and sin(B + 45°) are 1. Because if one is greater than 1, it's impossible. So:sin(A + 45°) = 1 and sin(B + 45°) = 1Which implies:A + 45° = 90° + 360°kB + 45° = 90° + 360°kBut since A and B are angles in a triangle, they must be between 0° and 180°, so:A + 45° = 90° ⇒ A = 45°Similarly, B + 45° = 90° ⇒ B = 45°So angles A and B are both 45°, making triangle ABC an isosceles right-angled triangle? Wait, but if A and B are both 45°, then angle C would be 180° - 45° - 45° = 90°. So triangle ABC is a right-angled isoceles triangle with legs a and b, and hypotenuse c.But wait, in a right-angled triangle with angles 45°, 45°, 90°, sides a and b are equal, and c = a√2. So (a + b)/c = (a + a)/ (a√2) = 2a / (a√2) = 2/√2 = √2. So that would be answer B, √2. Hmm, but let me check if that's correct.Wait, but maybe I made an assumption too quickly. Let me verify again. The key step was when I said that sin(A + 45°) * sin(B + 45°) = 1. Since sine can't be more than 1, both factors have to be 1. So A + 45° = 90°, so A = 45°, and B + 45° = 90°, so B = 45°. Therefore, angles A and B are 45°, angle C is 90°. Then by the Law of Sines, a/sin A = b/sin B = c/sin C. Since A and B are 45°, and C is 90°, then a = b = c sin 45° = c*(√2/2). Therefore, a = b = c√2/2. Then (a + b)/c = (c√2/2 + c√2/2)/c = (c√2)/c = √2. So that's answer B.But let me check if there are other possibilities. For example, could sin(A + 45°) and sin(B + 45°) both be -1? But since angles A and B are between 0° and 180°, A + 45° is between 45° and 225°, so sin(A + 45°) is between -√2/2 and 1. So sin(A + 45°) can't be -1 because that would require A + 45° = 270°, but then A would be 225°, which is impossible. So the only possible solution is both sines equal to 1, leading to A = B = 45°, C = 90°. Therefore, (a + b)/c = √2, which is option B. Therefore, the answer is B.But wait, let me verify using another approach. Let's suppose angles A and B are both 45°, then cos A + sin A = cos 45° + sin 45° = √2/2 + √2/2 = √2. Similarly, cos B + sin B = √2. Then 2/(cos B + sin B) = 2/√2 = √2. So cos A + sin A = √2, which is equal to 2/(cos B + sin B). So the equation holds. So this checks out. So the conclusion is correct.Alternatively, maybe use Law of Cosines. If angle C is 90°, then c² = a² + b². If it's a right-angled isoceles triangle, then a = b, so c = a√2. Then (a + b)/c = 2a/(a√2) = √2. Same result.Alternatively, suppose we don't assume it's a right-angled triangle. Let me try using the Law of Sines. Since a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius.But maybe express cos A + sin A in terms of the sides. Hmm, not sure. Alternatively, since we have angles A and B related, maybe using the fact that A + B + C = π. If A and B are both 45°, then C = 90°, but if they are different, how would that affect? But according to the earlier reasoning, the equation only holds if both A and B are 45°, leading to C = 90°. So maybe that's the only solution.Wait, let's check if there's another possibility where sin(A + 45°) and sin(B + 45°) multiply to 1. For example, if sin(A + 45°) = 2, but sine can't be more than 1. Similarly, if one is 1 and the other is 1. So no other possibilities. Therefore, the only solution is A = B = 45°, C = 90°, leading to (a + b)/c = √2. So the answer is B.Final Answerboxed{B}</think>Given the equation (cos A + sin A - frac{2}{cos B + sin B} = 0) in (triangle ABC), we need to find the value of (frac{a + b}{c}).First, we rewrite the given equation as:[cos A + sin A = frac{2}{cos B + sin B}]Using the identity (cos theta + sin theta = sqrt{2} sin(theta + 45^circ)), we substitute:[sqrt{2} sin(A + 45^circ) = frac{2}{sqrt{2} sin(B + 45^circ)}]Simplifying the right-hand side:[sqrt{2} sin(A + 45^circ) = frac{sqrt{2}}{sin(B + 45^circ)}]Dividing both sides by (sqrt{2}):[sin(A + 45^circ) = frac{1}{sin(B + 45^circ)}]This implies:[sin(A + 45^circ) cdot sin(B + 45^circ) = 1]Since the maximum value of (sin theta) is 1, the only solution is:[sin(A + 45^circ) = 1 quad text{and} quad sin(B + 45^circ) = 1]Thus:[A + 45^circ = 90^circ implies A = 45^circ][B + 45^circ = 90^circ implies B = 45^circ]Therefore, (angle A = angle B = 45^circ) and (angle C = 90^circ), making (triangle ABC) a right-angled isosceles triangle with legs (a) and (b), and hypotenuse (c). In such a triangle, (a = b) and (c = asqrt{2}). Hence:[frac{a + b}{c} = frac{a + a}{asqrt{2}} = frac{2a}{asqrt{2}} = frac{2}{sqrt{2}} = sqrt{2}]The value of (frac{a + b}{c}) is (boxed{B}).]

🔑:2.B.From cos A+sin A-frac{2}{cos B+sin B}=0, we getsqrt{2} sin left(A+frac{pi}{4}right)-frac{2}{sqrt{2} sin left(B+frac{pi}{4}right)}=0,which means sin left(A+frac{pi}{4}right) cdot sin left(B+frac{pi}{4}right)=1.By the boundedness of the sine function and the fact that angle A, angle B are interior angles of a triangle, we havesin left(A+frac{pi}{4}right)=1 text { and } sin left(B+frac{pi}{4}right)=1 text {. }Thus, angle A=angle B=frac{pi}{4}. Therefore, angle C=frac{pi}{2}.Hence, frac{a+b}{c}=sin A+sin B=sqrt{2}.

🧠:[<think>Okay, let's see. I need to find sin(theta + pi/6) given that sin(theta) is -5/13 and theta is in the third quadrant. Hmm. Alright, so first, I remember that there's a sine addition formula. Let me recall that. It's sin(a + b) = sin a cos b + cos a sin b. Right? So in this case, a is theta and b is pi/6. So sin(theta + pi/6) = sin(theta)cos(pi/6) + cos(theta)sin(pi/6). I know sin(theta) is given as -5/13. But I need cos(theta) as well. Since theta is in the third quadrant, both sine and cosine are negative there. So I can use the Pythagorean identity to find cos(theta). The identity is sin²(theta) + cos²(theta) = 1. Let's plug in sin(theta) = -5/13. Calculating sin²(theta): (-5/13)² = 25/169. So cos²(theta) = 1 - 25/169 = (169 - 25)/169 = 144/169. Therefore, cos(theta) is sqrt(144/169) = 12/13. But since theta is in the third quadrant, cos(theta) must be negative. So cos(theta) = -12/13. Alright, so now I have sin(theta) = -5/13 and cos(theta) = -12/13. Now let's plug these into the addition formula. First, sin(theta)cos(pi/6): that's (-5/13) times cos(pi/6). Cos(pi/6) is sqrt(3)/2. Then, cos(theta)sin(pi/6): that's (-12/13) times sin(pi/6). Sin(pi/6) is 1/2. So putting it all together: sin(theta + pi/6) = (-5/13)(sqrt(3)/2) + (-12/13)(1/2). Let's compute each term separately. First term: (-5/13)(sqrt(3)/2) = (-5 sqrt(3))/26. Second term: (-12/13)(1/2) = (-12)/26 = -6/13. Wait, 12 divided by 2 is 6, so it's -6/13? Wait, no. Wait, 12/13 times 1/2 is 6/13, so with the negative sign, it's -6/13. But 26 is the denominator? Wait, no. Wait, 12/13 multiplied by 1/2 is 6/13. So denominator is 13. So the second term is -6/13. So now we have (-5 sqrt(3)/26) + (-6/13). To add these, they need a common denominator. 26 is a common denominator. So convert -6/13 to -12/26. Therefore, sin(theta + pi/6) = (-5 sqrt(3) - 12)/26. Let me double-check my steps. Starting with the addition formula, correct. Calculated cos(theta) correctly using the Pythagorean identity, and accounted for the quadrant, so cos(theta) is negative. Then substituted both into the formula. Calculated each term, converted to common denominator. Yes, seems right. Wait, let me verify the calculation again. First term: (-5/13)(sqrt(3)/2) = (-5 sqrt(3))/26. Second term: (-12/13)(1/2) = (-12/26) = (-6/13). Wait, no. 12/13 times 1/2 is 6/13, so -12/13 * 1/2 is -6/13. But 6/13 is equal to 12/26. So yes, to get a common denominator of 26, -6/13 is equivalent to -12/26. Therefore, adding them gives (-5 sqrt(3) -12)/26. So the final answer should be (-5 sqrt(3) -12)/26. But let me check if I can factor anything out. Not really, unless they want it factored as - (5 sqrt(3) +12)/26, but both numerator terms are negative, so combining them gives a negative numerator. Alternatively, you can write it as (-5 sqrt(3) -12)/26. Either way is fine. Wait, but let me check once more if I messed up the signs. Since theta is in the third quadrant, both sin and cos are negative, which we accounted for. Then multiplying by cos(pi/6) which is positive, and sin(pi/6) which is positive. So sin(theta) is negative, multiplied by positive gives negative. cos(theta) is negative multiplied by positive gives negative. So both terms are negative, which matches the result. So yes, that seems correct. Alternatively, if I factor out the negative sign, it's - (5 sqrt(3) +12)/26. But the question doesn't specify the form, so both ways are acceptable. Probably best to write it as (-5√3 -12)/26. Wait, let me check the denominators again. The first term is over 26, the second term is over 13. So converting 6/13 to 12/26, then adding -5√3/26 + (-12/26) is correct. So numerator is -5√3 -12, denominator 26. Yep, that seems right. So the final answer is (-5√3 -12)/26. Let me write that in LaTeX: boxed{dfrac{-5sqrt{3} - 12}{26}}.But wait, sometimes they prefer not having the negative sign in the numerator. So maybe factor out a negative sign? So boxed{-dfrac{5sqrt{3} + 12}{26}}. Either form is correct, but perhaps the second one is more standard. Let me check. Let me see. If I write it as (-5√3 -12)/26, that's the same as -(5√3 +12)/26. So either way. Maybe the problem expects the answer with the negative sign in the numerator. Hmm. But in the third quadrant, when adding pi/6, which is 30 degrees, would that take the angle to... Let me see. Theta is in the third quadrant, so between pi and 3pi/2. Adding pi/6 to that would be between 7pi/6 and 5pi/3. So 7pi/6 is 210 degrees, which is in the third quadrant, 5pi/3 is 300 degrees, which is in the fourth quadrant. So depending on where exactly theta is. But since theta is in the third quadrant, theta + pi/6 could be in the third or fourth quadrant. But we don't need to worry about that because we just calculate the sine of that angle. However, sine can be positive or negative. Let me check with the answer. Our result is negative. Let's see. If theta is in the third quadrant, say between pi and 3pi/2. Adding pi/6 to it would make it between 7pi/6 and 5pi/3. 7pi/6 to 3pi/2 is still in the third quadrant, and 3pi/2 to 5pi/3 is in the fourth quadrant. So sin(theta + pi/6) would be negative in the third quadrant and positive in the fourth quadrant. So depending on theta's exact position. But since theta is just in the third quadrant, we can't know exactly where theta + pi/6 is. However, the problem just asks for the value, so even if it's in different quadrants, the calculation holds. But since we calculated it using the formula, regardless of the quadrant, the formula is valid. So our answer is negative, which is possible. So the answer is correct. Alternatively, maybe there's a miscalculation. Let me verify again. Given sin(theta) = -5/13, theta in QIII. So cos(theta) = -12/13. Then sin(theta + pi/6) = sin(theta)cos(pi/6) + cos(theta)sin(pi/6). Plugging in: (-5/13)(sqrt(3)/2) + (-12/13)(1/2). First term: (-5 sqrt(3))/26. Second term: (-12)/26. So total numerator: -5 sqrt(3) -12. Denominator: 26. Correct. So yes, that's right. Therefore, the answer should be boxed{-dfrac{5sqrt{3} + 12}{26}} or boxed{dfrac{-5sqrt{3} - 12}{26}}. Both are equivalent. I think the first form is preferable because it factors out the negative sign. So maybe go with that. Let me check the problem statement again. It says "put your final answer within boxed{}". The problem doesn't specify the form, so both are acceptable. But perhaps the first one is better. Let me confirm with standard conventions. Usually, if possible, you factor out the negative sign. So I'll go with -dfrac{5sqrt{3} + 12}{26}. Yeah, that's cleaner. So I think that's the better answer.Final Answerboxed{-dfrac{5sqrt{3} + 12}{26}}</think>Given (sin theta = -frac{5}{13}) and (theta) is an angle in the third quadrant, we need to find (sin left(theta + frac{pi}{6}right)).First, we use the Pythagorean identity to find (cos theta):[sin^2 theta + cos^2 theta = 1][left(-frac{5}{13}right)^2 + cos^2 theta = 1][frac{25}{169} + cos^2 theta = 1][cos^2 theta = 1 - frac{25}{169} = frac{144}{169}]Since (theta) is in the third quadrant, (cos theta) is negative:[cos theta = -frac{12}{13}]Next, we use the sine addition formula:[sin left(theta + frac{pi}{6}right) = sin theta cos frac{pi}{6} + cos theta sin frac{pi}{6}]Substituting the known values:[sin left(theta + frac{pi}{6}right) = left(-frac{5}{13}right) left(frac{sqrt{3}}{2}right) + left(-frac{12}{13}right) left(frac{1}{2}right)]Calculating each term separately:[left(-frac{5}{13}right) left(frac{sqrt{3}}{2}right) = -frac{5sqrt{3}}{26}][left(-frac{12}{13}right) left(frac{1}{2}right) = -frac{6}{13} = -frac{12}{26}]Combining the terms with a common denominator:[sin left(theta + frac{pi}{6}right) = -frac{5sqrt{3}}{26} - frac{12}{26} = frac{-5sqrt{3} - 12}{26}]Factoring out the negative sign for clarity:[sin left(theta + frac{pi}{6}right) = -frac{5sqrt{3} + 12}{26}]Thus, the final answer is:[boxed{-dfrac{5sqrt{3} + 12}{26}}],<think>Okay, so I need to find the value of sin(θ + π/6) given that sinθ is -5/13 and θ is in the third quadrant. Hmm, let me start by recalling some trigonometric identities. The sine of a sum formula is sin(A + B) = sinA cosB + cosA sinB. Right, so I can use that here. That means I need to find both sinθ and cosθ, then plug them into the formula. But wait, I already know sinθ is -5/13. However, since θ is in the third quadrant, I should figure out the cosine of θ as well. In the third quadrant, both sine and cosine are negative, right? Because in the third quadrant, x and y coordinates are both negative, so cosine (which is adjacent/hypotenuse, or x/r) and sine (opposite/hypotenuse, or y/r) are both negative. So, since sinθ = -5/13, that gives me the opposite side as -5 and the hypotenuse as 13. Then, using the Pythagorean theorem, the adjacent side would be sqrt(r² - y²) = sqrt(13² - (-5)²) = sqrt(169 - 25) = sqrt(144) = 12. But since we're in the third quadrant, the adjacent side (which is the cosine) should also be negative. Therefore, cosθ = -12/13.Okay, so now I have sinθ = -5/13 and cosθ = -12/13. Now, plugging into the sine addition formula: sin(θ + π/6) = sinθ cos(π/6) + cosθ sin(π/6). I need to compute each part. First, sinθ is -5/13, and cos(π/6) is sqrt(3)/2. Then, cosθ is -12/13, and sin(π/6) is 1/2. Let me compute each term separately:Term 1: sinθ cos(π/6) = (-5/13)(sqrt(3)/2) = (-5 sqrt(3))/26Term 2: cosθ sin(π/6) = (-12/13)(1/2) = (-12)/26 = -6/13Wait, hold on, let me check that. So, (-12/13)*(1/2) is (-12/26), which simplifies to (-6/13). Yeah, that's right.So, adding the two terms together: (-5 sqrt(3)/26) + (-6/13). Hmm, but I need a common denominator here. The first term is over 26 and the second term is over 13, which is equivalent to 26. So, converting -6/13 to -12/26. Therefore, combining:(-5 sqrt(3) - 12)/26.So, sin(θ + π/6) equals (-5 sqrt(3) - 12)/26. Let me check if that simplifies further. I don't think the numerator can be simplified, so that's the answer. But let me verify step by step again to ensure I didn't make a mistake.First, verifying the quadrant: third quadrant, so sine and cosine both negative. Correct. Then, using Pythagoras to find adjacent: 13² - 5² = 169 -25 = 144, sqrt(144)=12. Since in third quadrant, adjacent is negative, so cosθ = -12/13. Correct.Then, using the sine addition formula: sin(A + B) = sinA cosB + cosA sinB. Yes. So A is θ, B is π/6. So sinθ cos(π/6) + cosθ sin(π/6). Plugging in the values: (-5/13)(sqrt(3)/2) + (-12/13)(1/2). Calculating each term: (-5 sqrt(3))/26 + (-12/26). Yes, that combines to (-5 sqrt(3) -12)/26. I don't see any mistakes here. So, the final answer should be (-5 sqrt(3) -12)/26. Alternatively, it can be written as - (5 sqrt(3) +12)/26. Either way is acceptable. Since the problem says to put the answer inside a box, I should present it in the required format. Wait, but let me confirm once again. If θ is in the third quadrant, then π < θ < 3π/2. Then θ + π/6 would be between π + π/6 = 7π/6 and 3π/2 + π/6 = 10π/6 = 5π/3. So θ + π/6 is between 7π/6 and 5π/3, which is in the third and fourth quadrants. So the sine of θ + π/6 could be negative or positive. But since we computed the value as negative (-5 sqrt(3) -12)/26 ≈ (-8.66 -12)/26 ≈ -20.66/26 ≈ -0.795, which is negative. Let's check if that makes sense. If θ is in the third quadrant, adding π/6 (30 degrees) to it would take us to somewhere in the third or fourth quadrants. For example, if θ is 180 + 45 degrees (third quadrant), then θ + 30 degrees is 255 degrees, which is in the third quadrant, where sine is negative. If θ is approaching 270 degrees, like 270 - 15 = 255, adding 30 degrees would be 285, which is in the fourth quadrant where sine is negative. Wait, 285 degrees is in the fourth quadrant, but 255 is still in the third. Wait, 255 is 180 + 75, which is third quadrant, and 285 is 270 + 15, which is fourth. So depending on where θ is, θ + π/6 could be in the third or fourth. But in both quadrants, sine is negative (third and fourth). So the result being negative makes sense. Therefore, the answer being negative is correct. Alternatively, if I miscalculated and got a positive result, that would be wrong. But since both terms in the sum are negative, their sum is negative, which aligns with the expected quadrant. So, I think the answer is correct. Therefore, sin(θ + π/6) = (-5 sqrt(3) -12)/26.Final AnswerThe value of sin left(theta + frac{pi}{6}right) is boxed{-dfrac{5sqrt{3} + 12}{26}}.</think>Given (sin theta = -frac{5}{13}) and (theta) is an angle in the third quadrant, we need to find (sin left(theta + frac{pi}{6}right)).1. Determine (cos theta): - Since (theta) is in the third quadrant, both (sin theta) and (cos theta) are negative. - Using the Pythagorean theorem: [ cos theta = -sqrt{1 - sin^2 theta} = -sqrt{1 - left(-frac{5}{13}right)^2} = -sqrt{1 - frac{25}{169}} = -sqrt{frac{144}{169}} = -frac{12}{13} ]2. Apply the sine addition formula: - The formula for (sin(a + b)) is (sin a cos b + cos a sin b). - Here, (a = theta) and (b = frac{pi}{6}): [ sin left(theta + frac{pi}{6}right) = sin theta cos left(frac{pi}{6}right) + cos theta sin left(frac{pi}{6}right) ] - Substitute the known values: [ sin left(theta + frac{pi}{6}right) = left(-frac{5}{13}right) left(frac{sqrt{3}}{2}right) + left(-frac{12}{13}right) left(frac{1}{2}right) ] - Calculate each term: [ left(-frac{5}{13}right) left(frac{sqrt{3}}{2}right) = -frac{5sqrt{3}}{26} ] [ left(-frac{12}{13}right) left(frac{1}{2}right) = -frac{6}{13} = -frac{12}{26} ] - Combine the terms: [ sin left(theta + frac{pi}{6}right) = -frac{5sqrt{3}}{26} - frac{12}{26} = -frac{5sqrt{3} + 12}{26} ]The value of (sin left(theta + frac{pi}{6}right)) is (boxed{-dfrac{5sqrt{3} + 12}{26}}).]

🔑:Since sin theta = -frac{5}{13}, and theta is an angle in the third quadrant, we have cos theta = -sqrt{1-sin^2theta} = -frac{12}{13}, therefore, sin left(theta + frac{pi}{6}right) = frac{sqrt{3}}{2}sin theta + frac{1}{2}cos theta = frac{sqrt{3}}{2} times left(-frac{5}{13}right) + frac{1}{2} times left(-frac{12}{13}right) = -frac{5sqrt{3}+12}{26}. Hence, the answer is boxed{-frac{5sqrt{3}+12}{26}}. By using the fundamental trigonometric identity and the given information, we can find cos theta, and then use the special angle trigonometric values and the sine of sum formula to solve the problem. This question mainly examines the application of the fundamental trigonometric identity, special angle trigonometric values, and the sine of sum formula in simplifying and solving trigonometric functions. It tests computational skills and the ability to transform concepts, making it a basic question.

🧠:[<think>Okay, so I need to find the largest constant c such that for all positive real numbers x and y with x² + y² = 1, the inequality x⁶ + y⁶ ≥ cxy holds. Hmm, let's see. First, since x and y are positive and their squares add up to 1, they must be between 0 and 1. Maybe I can use some inequalities like Cauchy-Schwarz or AM-GM here? But I'm not sure yet. Let me think.The problem is to minimize x⁶ + y⁶ given x² + y² = 1 and x, y > 0, and then find the minimal value which will be equal to c times the maximum of xy? Wait, no. Wait, the inequality is x⁶ + y⁶ ≥ c xy. So for the minimal value of x⁶ + y⁶ over all possible x and y, divided by xy (but depending on x and y), the smallest ratio would determine the largest c. So maybe c is the minimum of (x⁶ + y⁶)/xy over all x, y >0 with x² + y² =1. If I can find that minimum value, then that would be c.Alternatively, to put it another way, the largest c such that c ≤ (x⁶ + y⁶)/xy for all x, y in the given domain. Therefore, c is the infimum of (x⁶ + y⁶)/xy over x, y >0 with x² + y² =1. So I need to compute that infimum.To approach this, maybe parameterize x and y. Since x² + y² =1, we can set x = cosθ and y = sinθ where θ ∈ (0, π/2) because x and y are positive. Then substitute into the expression:(x⁶ + y⁶)/xy = (cos⁶θ + sin⁶θ)/(cosθ sinθ)So now, the problem reduces to finding the minimum of (cos⁶θ + sin⁶θ)/(cosθ sinθ) for θ in (0, π/2). Let's compute this.First, let's simplify cos⁶θ + sin⁶θ. Recall that cos⁶θ + sin⁶θ can be written using the identity a³ + b³ = (a + b)(a² - ab + b²). Let me try that:cos⁶θ + sin⁶θ = (cos²θ)³ + (sin²θ)³ = (cos²θ + sin²θ)(cos⁴θ - cos²θ sin²θ + sin⁴θ)But cos²θ + sin²θ =1, so this becomes cos⁴θ - cos²θ sin²θ + sin⁴θ. Then, we can write cos⁴θ + sin⁴θ as (cos²θ + sin²θ)² - 2cos²θ sin²θ = 1 - 2cos²θ sin²θ. Therefore,cos⁶θ + sin⁶θ = 1*(1 - 2cos²θ sin²θ - cos²θ sin²θ) ? Wait, no. Wait, let me recast:Wait, cos⁶θ + sin⁶θ = (cos²θ + sin²θ)(cos⁴θ - cos²θ sin²θ + sin⁴θ) = 1*( (cos⁴θ + sin⁴θ) - cos²θ sin²θ )And since cos⁴θ + sin⁴θ = (cos²θ + sin²θ)^2 - 2cos²θ sin²θ = 1 - 2cos²θ sin²θ. Therefore,cos⁶θ + sin⁶θ = (1 - 2cos²θ sin²θ) - cos²θ sin²θ = 1 - 3cos²θ sin²θ.So, cos⁶θ + sin⁶θ = 1 - 3cos²θ sin²θ.Therefore, (cos⁶θ + sin⁶θ)/(cosθ sinθ) = [1 - 3cos²θ sin²θ]/(cosθ sinθ)Hmm, that seems manageable. Let's let t = cosθ sinθ. Since θ ∈ (0, π/2), t ∈ (0, 1/2], because the maximum of cosθ sinθ is 1/2 when θ=π/4. So t ∈ (0, 1/2].Express the expression in terms of t:First, note that cosθ sinθ = t, and cos²θ sin²θ = t². So the numerator is 1 - 3t², denominator is t. Therefore, the expression becomes (1 - 3t²)/t = 1/t - 3t.So we need to minimize f(t) = 1/t - 3t over t ∈ (0, 1/2].So now, it's a single variable calculus problem. Let's compute the derivative of f(t):f(t) = 1/t - 3tf’(t) = -1/t² - 3Set derivative equal to zero:-1/t² - 3 = 0 → -1/t² = 3 → t² = -1/3But t is positive, so there's no critical point in t >0. Therefore, the function f(t) is decreasing on (0, ∞), because derivative is always negative: -1/t² -3 is always less than zero.Therefore, since f(t) is decreasing on (0, 1/2], its minimum occurs at the right endpoint, t=1/2.Therefore, the minimal value of f(t) is f(1/2) = 1/(1/2) - 3*(1/2) = 2 - 1.5 = 0.5.Wait, 0.5? So the minimal value is 1/2. Therefore, the minimal value of (x⁶ + y⁶)/xy is 1/2, achieved when t=1/2, which corresponds to θ=π/4, i.e., x=y=√2/2.Wait, but if θ=π/4, then cosθ=sinθ=√2/2, so cosθ sinθ = (√2/2)(√2/2)= 2/4=1/2. Then, x⁶ + y⁶ = 2*( (√2/2)^6 ) = 2*( (2^(1/2))^6 / 2^6 ) = 2*(2^3 / 2^6 ) = 2*(8/64) = 2*(1/8) = 1/4. Then, xy = (√2/2)(√2/2)=1/2. So (x⁶ + y⁶)/xy = (1/4)/(1/2) = 1/2. Yes, that checks out.But wait, if f(t) is decreasing on (0, 1/2], then the minimal value is achieved at t=1/2, which is 1/2. Therefore, c=1/2 is the largest constant such that x⁶ + y⁶ ≥ c xy for all x, y>0 with x² + y²=1.But wait, let me check another point to make sure. Let's take θ approaching 0, so x approaches 1, y approaches 0. Then x⁶ + y⁶ approaches 1, and xy approaches 0. So (x⁶ + y⁶)/xy approaches infinity. Similarly, when θ approaches π/2, same thing. So the expression (x⁶ + y⁶)/xy becomes large as either x or y approaches 0. Therefore, the minimal value is indeed achieved at θ=π/4, giving c=1/2.But let me confirm with another angle. Suppose θ=π/6. Then cosθ=√3/2, sinθ=1/2. Then x=√3/2, y=1/2.Compute x⁶ + y⁶ = ( (√3/2)^6 + (1/2)^6 ) = ( (27/64) + (1/64) )=28/64=7/16≈0.4375Compute xy= (√3/2)(1/2)=√3/4≈0.4330Thus, (x⁶ + y⁶)/xy≈0.4375/0.4330≈1.01, which is greater than 1/2≈0.5. So that's consistent with the minimum being 1/2 at θ=π/4.Similarly, θ=π/4, which we already did: 1/4 divided by 1/2 is 1/2.Another test: θ=π/3, then cosθ=1/2, sinθ=√3/2.x⁶ + y⁶ = (1/2)^6 + (√3/2)^6=1/64 + 27/64=28/64=7/16≈0.4375xy=(1/2)(√3/2)=√3/4≈0.4330Same as θ=π/6, so ratio≈1.01. Still higher than 1/2.So seems like the minimal ratio is indeed 1/2. Therefore, the largest constant c is 1/2.But just to be thorough, let me check another angle. Maybe θ=π/8.θ=π/8, cosθ≈0.9239, sinθ≈0.3827x⁶≈0.9239^6≈approx 0.9239^2=0.8536, 0.8536^3≈0.8536*0.8536≈0.7286, *0.8536≈0.622. Similarly, y⁶≈0.3827^6≈approx (0.3827^2)^3≈0.1464^3≈0.0031. So x⁶ + y⁶≈0.622 +0.0031≈0.625xy≈0.9239*0.3827≈0.3535Thus, ratio≈0.625/0.3535≈1.77, which is still greater than 1/2.Wait, but if θ=π/4, ratio is 1/2, which is less than 1.77. Wait, no, in θ=pi/4, ratio is 1/2, which is lower. Wait, but theta=pi/4 is giving ratio=0.5, but theta=pi/8 is giving ratio≈1.77. Wait, that can't be. Wait, but when theta approaches 0, the ratio goes to infinity, so pi/8 is still giving a ratio higher than 1. So the minimal ratio is at pi/4. Hmm. So that must be the minimal. So yes, so the minimal ratio is 0.5, so c=1/2.Therefore, the answer is c=1/2.Final AnswerThe largest constant is boxed{dfrac{1}{2}}.</think>To find the largest constant ( c ) such that for real numbers ( x, y ) satisfying ( x > 0 ), ( y > 0 ), and ( x^2 + y^2 = 1 ), it always holds that ( x^6 + y^6 geq c xy ), we start by parameterizing ( x ) and ( y ) using trigonometric identities. Let ( x = cos theta ) and ( y = sin theta ) where ( theta in (0, pi/2) ).We need to minimize the expression ( frac{x^6 + y^6}{xy} ). Substituting ( x = cos theta ) and ( y = sin theta ), we get:[frac{cos^6 theta + sin^6 theta}{cos theta sin theta}]Using the identity for ( cos^6 theta + sin^6 theta ):[cos^6 theta + sin^6 theta = 1 - 3 cos^2 theta sin^2 theta]Thus, the expression becomes:[frac{1 - 3 cos^2 theta sin^2 theta}{cos theta sin theta}]Let ( t = cos theta sin theta ). Then ( t in (0, 1/2] ) and the expression simplifies to:[frac{1 - 3t^2}{t} = frac{1}{t} - 3t]We need to minimize ( f(t) = frac{1}{t} - 3t ) over ( t in (0, 1/2] ). The derivative of ( f(t) ) is:[f'(t) = -frac{1}{t^2} - 3]Since ( f'(t) ) is always negative for ( t > 0 ), ( f(t) ) is a decreasing function. Therefore, the minimum occurs at the right endpoint ( t = 1/2 ):[fleft(frac{1}{2}right) = frac{1}{1/2} - 3 left(frac{1}{2}right) = 2 - frac{3}{2} = frac{1}{2}]This minimum value corresponds to ( theta = pi/4 ), where ( x = y = frac{sqrt{2}}{2} ). Therefore, the largest constant ( c ) is ( frac{1}{2} ).The final answer is (boxed{dfrac{1}{2}}).,<think>Okay, so I need to find the largest constant c such that for all positive real numbers x and y with x² + y² = 1, the inequality x⁶ + y⁶ ≥ cxy holds. Hmm, let's start by understanding what the problem is asking. We have to determine the maximum value of c so that this inequality is always true under the given conditions. First, I recall that when dealing with inequalities under a constraint like x² + y² = 1, methods like Lagrange multipliers might be useful. Alternatively, maybe using substitution based on trigonometric identities since x and y are on the unit circle. But since x and y are positive, we can parameterize them using sine and cosine. Let me think.Let me try substituting x = cosθ and y = sinθ, where θ is in the first quadrant (0 < θ < π/2). Then x² + y² = 1 is automatically satisfied. Then the expression becomes (cos⁶θ + sin⁶θ) ≥ c cosθ sinθ. So, we need to find the minimal value of (cos⁶θ + sin⁶θ)/(cosθ sinθ) over θ in (0, π/2), and then c would be that minimal value. Because if the inequality holds for the minimal value of the left-hand side divided by the right-hand side, then it holds for all θ. So c is the minimum of (cos⁶θ + sin⁶θ)/(cosθ sinθ).Alternatively, maybe we can express x⁶ + y⁶ in terms of x² and y². Let's see. Since x² + y² = 1, perhaps we can use some algebraic manipulation. For example, x⁶ + y⁶ can be written as (x²)³ + (y²)³, which is a sum of cubes. The formula for a³ + b³ is (a + b)(a² - ab + b²). So applying that here:x⁶ + y⁶ = (x² + y²)(x⁴ - x²y² + y⁴) = 1*(x⁴ - x²y² + y⁴) = x⁴ + y⁴ - x²y².But we can also express x⁴ + y⁴ in terms of (x² + y²)² - 2x²y². Let's do that:x⁴ + y⁴ = (x² + y²)² - 2x²y² = 1 - 2x²y².Therefore, substituting back into x⁶ + y⁶:x⁶ + y⁶ = (1 - 2x²y²) - x²y² = 1 - 3x²y².So the inequality becomes 1 - 3x²y² ≥ c xy. So, 1 - 3x²y² ≥ c xy. Our goal is to find the largest c such that this holds for all x, y >0 with x² + y² =1.So, rearranging, we have c ≤ (1 - 3x²y²)/(xy). Therefore, the maximum possible c is the minimum of (1 - 3x²y²)/(xy) over all x, y >0 with x² + y² =1.Alternatively, since x² + y² =1, perhaps we can let t = xy. Then, we need to express 1 - 3t² in terms of t. But first, we need to relate t to the constraint.Note that for x, y >0 with x² + y² =1, the product xy is maximized when x = y = 1/√2, by AM-GM inequality. Because xy ≤ (x² + y²)/2 = 1/2. So the maximum value of xy is 1/2, achieved at x = y = 1/√2. The minimum value of xy would be approaching 0, when either x approaches 0 or y approaches 0.So t = xy ∈ (0, 1/2]. Therefore, our expression becomes (1 - 3t²)/t = 1/t - 3t. So we need to find the minimum of 1/t - 3t over t ∈ (0, 1/2]. Let's denote f(t) = 1/t - 3t. Then f'(t) = -1/t² - 3. Setting derivative to zero: -1/t² - 3 = 0 → -1/t² = 3 → t² = -1/3. Which is not possible. So there's no critical point in the real numbers. Therefore, the function f(t) is decreasing on (0, 1/2], since its derivative is always negative. Therefore, the minimum of f(t) occurs at the right endpoint, t =1/2. Therefore, the minimum value is f(1/2) = 1/(1/2) - 3*(1/2) = 2 - 3/2 = 1/2. Therefore, the minimal value of (1 - 3x²y²)/xy is 1/2. Therefore, the maximum c is 1/2. Therefore, the answer is boxed{dfrac{1}{2}}.Wait, but let's check this. If we substitute x = y =1/√2, then x⁶ + y⁶ = 2*(1/√2)^6 = 2*(1/(2^3)) = 2*(1/8) = 1/4. And xy = (1/√2)^2 = 1/2. Then c would have to satisfy 1/4 ≥ c*(1/2), so c ≤ 1/2. So that checks out. But is this the minimum? Wait, but according to our previous analysis, since f(t) is decreasing on (0,1/2], then the minimal value of f(t) is at t =1/2, which is 1/2. But let's test another point. For example, take x approaching 1 and y approaching 0. Then x⁶ approaches 1, y⁶ approaches 0, so x⁶ + y⁶ approaches 1. And xy approaches 0. So (1 - 3x²y²)/xy approaches 1/0, which tends to infinity. But since xy approaches 0, but x⁶ + y⁶ approaches 1, so 1 ≥ c*0, which is always true, but since we need the inequality to hold for all x,y, the critical case is where the left-hand side is minimized relative to the right-hand side. So the minimal value of (x⁶ + y⁶)/xy is 1/2, as when x = y =1/√2. So in that case, the ratio is (1/4)/(1/2) = 1/2. So that seems to be the minimal value. Therefore, c must be 1/2.But wait, let's verify with another point. Let's take θ = π/4, which gives x = y =1/√2. As before, x⁶ + y⁶ =1/4, and xy=1/2. So 1/4 = (1/2)*(1/2), so equality holds. Now take θ = π/6. Then x = cos(π/6)=√3/2, y = sin(π/6)=1/2. Then x⁶ = ( (√3/2)^2 )^3 = (3/4)^3 =27/64. y⁶=(1/2)^6=1/64. So x⁶ + y⁶=28/64=7/16. xy= (√3/2)(1/2)=√3/4. So the ratio is (7/16)/(√3/4)=7/(4√3)≈7/6.928≈1.01. Which is bigger than 1/2. Similarly, take θ=π/3. x=1/2, y=√3/2. Then x⁶=1/64, y⁶=27/64. Sum is 28/64=7/16. Same as before. xy=(1/2)(√3/2)=√3/4. So ratio is same, 7/(4√3). So approximately 1.01. So indeed, higher than 1/2. So the minimal ratio is at θ=π/4, which gives 1/2. So that's correct. Therefore, c=1/2.Wait, but let me check another angle. Let's take θ=π/8. Then x=cos(π/8)=√(2 + √2)/2≈0.9239, y=sin(π/8)=√(2 - √2)/2≈0.3827. Then x⁶≈0.9239^6≈approx 0.9239^2=0.8536, then ^3≈0.8536^3≈0.622. Similarly y⁶≈0.3827^6≈(0.3827^2)^3≈0.1464^3≈0.0031. So x⁶ + y⁶≈0.625. xy≈0.9239*0.3827≈0.3535. Then the ratio is 0.625 /0.3535≈1.768. Which is much larger than 1/2. So indeed, the minimal ratio is at θ=π/4. Therefore, the minimal value is 1/2. Thus, the largest c is 1/2. Therefore, the answer is boxed{dfrac{1}{2}}.But wait, just to make sure, let's try to do this using Lagrange multipliers. Let me set up the optimization problem. We need to minimize (x⁶ + y⁶)/(xy) subject to x² + y² =1. Let’s let’s set f(x, y) = (x⁶ + y⁶)/(xy) and the constraint g(x, y)=x² + y² -1=0. Then using Lagrange multipliers, the gradient of f should be proportional to the gradient of g.But first, let's compute f(x, y) = (x⁶ + y⁶)/(xy) = x⁵/y + y⁵/x. Then, compute the partial derivatives.Partial derivative with respect to x:df/dx = (5x⁴/y) - (y⁵/x²). Similarly, partial derivative with respect to y:df/dy = (5y⁴/x) - (x⁵/y²).The gradient of g is (2x, 2y). So, according to Lagrange multipliers, there exists λ such that:5x⁴/y - y⁵/x² = 2λ x,5y⁴/x - x⁵/y² = 2λ y.This seems complicated, but perhaps if we assume x = y, then we can check if that's a critical point.Let’s set x = y. Since x² + y² =1, then 2x² =1, so x = y =1/√2. Let’s check if the partial derivatives satisfy the equations.Compute df/dx at x = y =1/√2:5x⁴/y - y⁵/x² = 5*(1/√2)^4/(1/√2) - (1/√2)^5/( (1/√2)^2 )First term: 5*(1/4) / (1/√2) = (5/4) / (1/√2) = (5/4)*√2.Second term: (1/(2^(5/2))) / (1/2) = (1/(2^(5/2)))*(2) = 1/(2^(3/2)) = 1/(2*sqrt(2)).So first term: (5/4)*√2, second term: 1/(2*sqrt(2)). Let's convert to same denominator:First term: (5√2)/4 ≈5*1.414/4≈7.07/4≈1.7675Second term: 1/(2.828)≈0.3535So total: 1.7675 -0.3535≈1.414≈√2.Similarly, 2λ x = 2λ*(1/√2). So equation:√2 ≈2λ*(1/√2) → λ ≈ (√2)/(2*(1/√2)) )= (√2 * √2)/2= 2/2=1.Similarly, check the other partial derivative.df/dy =5y⁴/x -x⁵/y². At x = y =1/√2:5*(1/√2)^4/(1/√2) - (1/√2)^5/(1/√2)^2 = same as above. So same result: √2. And 2λ y =2λ*(1/√2). Which gives λ=1 as before. So this is consistent.Therefore, the point x = y =1/√2 is a critical point. Now, we need to check if this is a minimum. Since we saw earlier that as t approaches 0, f(t) approaches infinity, and the function is decreasing, so the critical point at t=1/2 (x=y=1/√2) must be the minimum. Therefore, this is indeed the minimal value. So c=1/2 is correct.Alternatively, another way to think about this is using substitution. Let’s let t = x/y. Since x and y are positive, t is positive. Then, from x² + y² =1, we can write x = t y. Substituting into the constraint: t² y² + y² =1 → y²(t² +1)=1 → y = 1/√(t² +1). Then x = t/√(t² +1). Then, x⁶ + y⁶ = (t^6)/(t² +1)^3 + 1/(t² +1)^3 = (t^6 +1)/(t² +1)^3. And xy = (t/√(t² +1))(1/√(t² +1)) = t/(t² +1). So the ratio is [ (t^6 +1)/(t² +1)^3 ] / [ t/(t² +1) ] = (t^6 +1)/(t(t² +1)^2 ). So we need to find the minimum of f(t) = (t^6 +1)/(t(t² +1)^2 ) for t >0.This seems more complicated, but maybe we can simplify. Let’s compute f(t):f(t) = (t^6 +1)/(t(t² +1)^2 ). Let’s substitute u = t + 1/t. Maybe? Not sure. Alternatively, note that t^6 +1 can be written as (t²)^3 +1, which factors as (t² +1)(t^4 - t² +1). Therefore:f(t) = (t² +1)(t^4 - t² +1)/(t(t² +1)^2 ) = (t^4 - t² +1)/(t(t² +1)).Simplify numerator and denominator:Numerator: t^4 - t² +1Denominator: t(t² +1) = t^3 +tSo f(t) = (t^4 - t² +1)/(t^3 + t). Let’s perform polynomial division:Divide t^4 - t² +1 by t^3 + t.t^4 / t^3 = t. Multiply divisor by t: t^4 + t². Subtract from dividend:(t^4 - t² +1) - (t^4 + t²) = -2t² +1. So the division gives t + (-2t² +1)/(t^3 + t). Therefore,f(t) = t + (-2t² +1)/(t^3 + t).Hmm, not sure if this helps. Alternatively, take derivative of f(t):f(t) = (t^6 +1)/(t(t² +1)^2 )Take natural log: ln f(t) = ln(t^6 +1) - ln t - 2 ln(t² +1)Differentiate:f’(t)/f(t) = [6t^5/(t^6 +1)] - [1/t] - [4t/(t² +1)]Set derivative to zero:6t^5/(t^6 +1) -1/t -4t/(t² +1) =0.Multiply through by t(t^6 +1)(t² +1) to eliminate denominators:6t^5 * t(t² +1) - (t^6 +1)(t² +1) -4t * t(t^6 +1) =0.Compute each term:First term: 6t^6(t² +1)Second term: -(t^6 +1)(t² +1)Third term: -4t²(t^6 +1)Expand first term: 6t^8 +6t^6Second term: -(t^8 + t^6 + t² +1)Third term: -4t^8 -4t²Combine all terms:6t^8 +6t^6 -t^8 -t^6 -t² -1 -4t^8 -4t² =0Combine like terms:(6t^8 - t^8 -4t^8) + (6t^6 -t^6) + (-t² -4t²) -1 =0(1t^8) + (5t^6) + (-5t²) -1=0Thus, the equation reduces to:t^8 +5t^6 -5t² -1=0.This seems difficult to solve. Maybe we can factor this equation. Let’s try to see if t=1 is a root:1 +5 -5 -1=0. Yes, t=1 is a root. Therefore, factor (t -1) from the polynomial. Let's perform polynomial division.Divide t^8 +5t^6 -5t² -1 by (t -1). Using synthetic division for t=1:Coefficients: 1 (t^8), 0 (t^7),5 (t^6),0 (t^5),0 (t^4),0 (t^3),-5 (t^2),0 (t), -1 (constant).Bring down 1. Multiply by 1:1. Add to next term:0 +1=1.Multiply by1:1. Add to next term:5 +1=6.Multiply by1:6. Add to next term:0 +6=6.Multiply by1:6. Add to next term:0 +6=6.Multiply by1:6. Add to next term:0 +6=6.Multiply by1:6. Add to next term:-5 +6=1.Multiply by1:1. Add to next term:0 +1=1.Multiply by1:1. Add to last term:-1 +1=0. So the division is exact. Therefore, the polynomial factors as (t -1)(t^7 +t^6 +6t^5 +6t^4 +6t^3 +6t^2 +t +1).So we have t=1 is a solution. Let’s check t=1 corresponds to x=y=1/√2, which we already considered. So that critical point. The other factor is t^7 +t^6 +6t^5 +6t^4 +6t^3 +6t^2 +t +1. This polynomial doesn't seem to have real roots for t>0. Let’s check t= -1 is not a root, but since t>0, we can ignore negative roots. Therefore, the only critical point in t>0 is t=1. Therefore, the minimal value occurs at t=1, which gives the ratio f(1)= (1 +1)/(1*(1 +1)^2 )=2/(1*4)=1/2. Therefore, confirming our previous result.Thus, the largest constant c is indeed 1/2.Final AnswerThe largest constant is boxed{dfrac{1}{2}}.</think>To find the largest constant ( c ) such that for real numbers ( x, y ) satisfying ( x > 0 ), ( y > 0 ), and ( x^2 + y^2 = 1 ), the inequality ( x^6 + y^6 geq cxy ) always holds, we start by expressing ( x^6 + y^6 ) in terms of ( x ) and ( y ).Using the identity for the sum of cubes, we have:[x^6 + y^6 = (x^2)^3 + (y^2)^3 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)]Given ( x^2 + y^2 = 1 ), this simplifies to:[x^6 + y^6 = x^4 + y^4 - x^2y^2]Next, we express ( x^4 + y^4 ) using the square of the sum of squares:[x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 = 1 - 2x^2y^2]Thus,[x^6 + y^6 = (1 - 2x^2y^2) - x^2y^2 = 1 - 3x^2y^2]We need to ensure:[1 - 3x^2y^2 geq cxy]Rewriting this inequality, we get:[c leq frac{1 - 3x^2y^2}{xy}]To find the largest ( c ), we need to minimize the expression ( frac{1 - 3x^2y^2}{xy} ) over all positive ( x ) and ( y ) satisfying ( x^2 + y^2 = 1 ).Let ( t = xy ). The maximum value of ( t ) occurs when ( x = y = frac{1}{sqrt{2}} ), giving ( t = frac{1}{2} ). Substituting ( t ) into the expression, we get:[frac{1 - 3t^2}{t} = frac{1 - 3 left( frac{1}{2} right)^2}{frac{1}{2}} = frac{1 - frac{3}{4}}{frac{1}{2}} = frac{frac{1}{4}}{frac{1}{2}} = frac{1}{2}]Thus, the minimum value of ( frac{1 - 3x^2y^2}{xy} ) is ( frac{1}{2} ), which occurs at ( x = y = frac{1}{sqrt{2}} ).Therefore, the largest constant ( c ) is (boxed{dfrac{1}{2}}).]

🔑:When x=y=frac{sqrt{2}}{2}, the inequality also holds, then,c leqslant frac{x^{6}+y^{6}}{x y}=frac{1}{2} text {. }Next, we prove that under the conditions x>0, y>0, x^{2}+y^{2}=1, the inequalityx^{6}+y^{6} geqslant frac{1}{2} x yalways holds.Equation (1) is equivalent toleft(x^{2}+y^{2}right)left[left(x^{2}+y^{2}right)^{2}-3 x^{2} y^{2}right] geqslant frac{1}{2} x y,which simplifies to 6(x y)^{2}+x y-2 leqslant 0.Let mu=x y, since 0<x y leqslant frac{x^{2}+y^{2}}{2}=frac{1}{2}, we have 0<mu leqslant frac{1}{2}.Let f(mu)=6 mu^{2}+mu-2,0<mu leqslant frac{1}{2}, then f(mu) is increasing on left(0, frac{1}{2}right], and fleft(frac{1}{2}right)=0. Therefore, on left(0, frac{1}{2}right], we have f(mu) leqslant 0. Thus,6(x y)^{2}+x y-2 leqslant 0 text {. }In summary, the maximum value of c is frac{1}{2}.Note: To find the maximum or minimum value of a parameter in an inequality, one can first estimate the upper or lower bound of the parameter, and then prove that the inequality always holds for this upper or lower bound.This problem can also be considered as follows: Find the minimum value of frac{x^{6}+y^{6}}{x y}, which is the maximum value of c. That is,begin{aligned}frac{x^{6}+y^{6}}{x y} & =frac{left(x^{2}+y^{2}right)left[left(x^{2}+y^{2}right)^{2}-3 x^{2} y^{2}right]}{x y} & =frac{1}{x y}-3 x y .end{aligned}Since 0<x y leqslant frac{1}{2}, we have frac{1}{x y} geqslant 2,-3 x y geqslant-frac{3}{2}. Therefore, frac{x^{6}+y^{6}}{x y} geqslant 2-frac{3}{2}=frac{1}{2}.When x=y=frac{sqrt{2}}{2}, the equality holds, thus the minimum value of frac{x^{6}+y^{6}}{x y} is frac{1}{2}, and the maximum value of c is frac{1}{2}.

🧠:[<think>Okay, let's see. I have this quadratic equation: mx² + 2x + 1 = 0. The question is asking for the range of values for m so that the equation has real number solutions. The options are given, so I need to figure out which one is correct.First, I remember that for a quadratic equation ax² + bx + c = 0 to have real solutions, the discriminant must be non-negative. The discriminant is b² - 4ac. If that's greater than or equal to zero, then there are real roots. If it's negative, the roots are complex. So I need to calculate the discriminant here and set it to be ≥ 0.Let me write that out. Here, a is m, b is 2, and c is 1. So the discriminant would be (2)² - 4*(m)*(1). That simplifies to 4 - 4m. So 4 - 4m ≥ 0. Then, solving for m: 4 ≥ 4m, divide both sides by 4, so 1 ≥ m, which is the same as m ≤ 1. Wait, so m has to be less than or equal to 1. So then, looking at the options, B is m ≤ 1, D is m ≤1 and m ≠0. Hmm.But wait, hold on. Quadratic equation requires that a ≠ 0, right? Because if a is 0, then it's not a quadratic equation anymore, it's linear. So in this case, if m is 0, the equation becomes 2x + 1 = 0, which is linear, not quadratic. So even if m =0, it's a linear equation with a real solution. But the question specifically says "quadratic equation", so m cannot be 0. So even though when m=0, the equation still has a real solution, but since it's not quadratic anymore, we need to exclude m=0. Therefore, the correct range would be m ≤1 and m ≠0. Which is option D.Wait, but hold on. Let me double-check. The discriminant when m=0: If m is 0, the equation is linear, so discriminant isn't even considered here. But the problem states "quadratic equation", so m cannot be 0. Therefore, even though m=0 gives a real solution, we need to restrict m to non-zero values. Therefore, combining the discriminant condition m ≤1 with m ≠0, so the answer is D.But wait, let me confirm. The discriminant is 4 - 4m ≥0. So that gives m ≤1. But if m is negative, does that affect anything? Wait, no. Because even if m is negative, as long as the discriminant is non-negative, the quadratic equation will have real roots. So for example, if m is negative, say m = -1, then the equation becomes -x² + 2x +1 =0. The discriminant would be 4 -4*(-1)(1) = 4 +4 =8, which is positive. So real roots. So m can be negative, as long as m ≤1. So possible values for m are m ≤1, but m can't be 0. So the correct answer is D: m ≤1 and m ≠0.But wait, looking back at the options, option C is m ≥ -1 and m ≠0. Hmm. Wait, maybe I need to check if there's a lower bound on m. Let's see.Wait, the discriminant is 4 -4m ≥0. So 4 -4m ≥0 → 4 ≥4m → 1 ≥m → m ≤1. So there is no lower bound? Wait, if m is negative, does that affect the discriminant? Let's see, if m is -10, then discriminant is 4 -4*(-10) = 4 +40 =44, which is positive. So even if m is very negative, the discriminant is positive, so real roots. So there's no lower limit on m? Then why is there an option with m ≥ -1? Maybe someone might think that coefficient a (which is m) must be non-zero, but m can be any real number except zero. Wait, but the discriminant allows m to be any value less than or equal to 1, including negative numbers. So why is there an option with m ≥ -1? Maybe a miscalculation?Wait, let's check. Maybe the person who made the problem thought that the leading coefficient can't be negative? But no, quadratic equations can have negative coefficients. For example, mx² +2x +1=0 with m negative is still a quadratic equation. So as long as m ≠0, it's quadratic, and discriminant is 4-4m. Wait, hold on. Wait, discriminant is b² -4ac. So in this case, a is m, so discriminant is 4 -4*m*1=4 -4m. So discriminant is 4(1 -m). Therefore, discriminant ≥0 implies 1 -m ≥0 → m ≤1. So that's correct.So if m is less than or equal to 1, including negative values, but m cannot be zero. Therefore, the range is m ≤1 and m ≠0. So answer D. But the options include C: m ≥ -1 and m ≠0. So why is that there? Maybe the options are trying to trick you into considering something else?Wait, maybe there's a mistake in the discriminant calculation? Let me check again. Quadratic equation: mx² +2x +1=0. Discriminant is b² -4ac = (2)^2 -4*(m)*(1) =4 -4m. So discriminant is 4 -4m. To have real solutions, discriminant must be ≥0. So 4 -4m ≥0 → 4 ≥4m → 1 ≥m → m ≤1. Therefore, m must be less than or equal to 1. There's no lower bound, because as m approaches negative infinity, 4 -4m approaches positive infinity, so discriminant is always positive. Therefore, m can be any real number less than or equal to 1, except 0. So the answer should be D.But maybe the options are wrong? Let me check the options again:A: m ≥ -1B: m ≤1C: m ≥ -1 and m ≠0D: m ≤1 and m ≠0So according to my calculation, D is the correct answer. So why is option C there? Maybe someone thought that m has to be positive? But no, m can be negative. For example, m = -2, then equation is -2x² +2x +1=0. Discriminant is 4 -4*(-2)(1)=4 +8=12>0, real roots.Alternatively, maybe confusion between the coefficient of x² and the discriminant? For example, if someone thought that the quadratic equation must open upwards or downwards, but even if it opens downward (m negative), it can still cross the x-axis if discriminant is positive.Alternatively, maybe someone confused the quadratic formula. Wait, if the equation is mx² +2x +1=0, the solutions are x = [-2 ± sqrt(4 -4m)]/(2m). So if m is negative, the denominator is negative, but the numerator is also negative, but sqrt(4 -4m) is a real number as long as m ≤1. So even if m is negative, sqrt(4 -4m) is real. So solutions are real. Therefore, the answer is D.Therefore, the correct answer should be D. So why is option C there? Maybe a common mistake where people assume that the coefficient of x² must be positive? But in the problem statement, there's no such restriction. The problem just says "quadratic equation", which only requires that the coefficient of x² is non-zero. So m ≠0. So combining with discriminant, m ≤1 and m ≠0. So D.Therefore, the answer is D.Final Answerboxed{D}</think>To determine the range of values for ( m ) such that the quadratic equation ( mx^2 + 2x + 1 = 0 ) has real number solutions, we need to check the discriminant condition. For a quadratic equation ( ax^2 + bx + c = 0 ) to have real solutions, the discriminant ( b^2 - 4ac ) must be non-negative.Here, ( a = m ), ( b = 2 ), and ( c = 1 ). The discriminant is calculated as follows:[2^2 - 4 cdot m cdot 1 = 4 - 4m]For the equation to have real solutions, the discriminant must be non-negative:[4 - 4m geq 0]Solving for ( m ):[4 geq 4m implies 1 geq m implies m leq 1]Additionally, since the equation is quadratic, the coefficient ( m ) must not be zero. Therefore, combining these conditions, we get:[m leq 1 text{ and } m neq 0]Thus, the correct answer is (boxed{D}).,<think>Okay, so I have this quadratic equation: mx² + 2x + 1 = 0. The question is asking for the range of values for m such that the equation has real number solutions. The options are given as A to D. Let me think about how to approach this.First, I remember that for a quadratic equation ax² + bx + c = 0 to have real solutions, the discriminant must be non-negative. The discriminant is b² - 4ac. If that's greater than or equal to zero, then there are real roots. So maybe I need to compute the discriminant here and set it to be ≥ 0.Let me write that down. The given equation is mx² + 2x + 1 = 0. So, in standard form, a = m, b = 2, c = 1. Therefore, the discriminant D would be b² - 4ac = (2)² - 4*(m)*(1) = 4 - 4m.Wait, so the discriminant is 4 - 4m. For real solutions, this discriminant must be ≥ 0. So, 4 - 4m ≥ 0. Let's solve that inequality.4 - 4m ≥ 0 Subtract 4 from both sides: -4m ≥ -4 Now, divide both sides by -4. But wait, when you divide or multiply both sides of an inequality by a negative number, the inequality sign flips. So dividing by -4:m ≤ 1.Hmm, so m must be less than or equal to 1? But wait, quadratic equations also require that the coefficient of x² (which is m here) is not zero. Otherwise, it's not a quadratic equation anymore, it becomes a linear equation. So m ≠ 0. But in the options given, there are choices that mention m ≠ 0. Let me check the options again.Options:A: m ≥ -1B: m ≤ 1C: m ≥ -1 and m ≠ 0D: m ≤ 1 and m ≠ 0Wait, so according to my calculation, the discriminant is 4 - 4m ≥ 0 → m ≤ 1. But also, since it's a quadratic equation, m cannot be zero. So combining these, the answer should be m ≤ 1 and m ≠ 0, which is option D. But hold on, let me verify.Wait, but the discriminant here is 4 - 4m. So when m is 1, discriminant is 4 - 4*1 = 0, which is okay because discriminant zero still gives a real solution (a repeated root). So m can be 1, as long as it's ≤ 1. But also, m can't be zero. So if m is zero, the equation becomes 2x + 1 = 0, which is linear and has a real solution. Wait a minute, but the problem says "has real number solutions". If m is zero, it's a linear equation which does have a real solution. So does the problem consider m=0 as valid?Wait, this is a crucial point. The original question says "the quadratic equation mx² + 2x + 1 = 0 has real number solutions". So if m=0, it's not a quadratic equation anymore; it's linear. So even though it has a real solution, the problem specifies a quadratic equation, so m cannot be zero. Therefore, m must satisfy two things: discriminant ≥ 0 (for real solutions when it's quadratic) and m ≠ 0 (to ensure it's a quadratic equation). Therefore, the answer would be m ≤ 1 and m ≠ 0, which is option D.But let me double-check. Suppose m=0, then the equation is 2x +1 =0, which is linear, and x= -1/2 is a solution. However, the problem says "quadratic equation", so if m=0, it's not quadratic. Therefore, m cannot be zero. So even though m=0 would give a real solution, it's excluded because the equation is no longer quadratic. Therefore, m must be ≤1 and m≠0, so option D.Wait, but the discriminant calculation gave m ≤1. So if m is, say, -1, then the discriminant would be 4 -4*(-1)= 4 +4=8, which is positive, so real roots. If m is 1, discriminant is 0, so repeated real root. If m is 0.5, discriminant is 4 -2=2, positive. If m is 2, discriminant is 4 -8= -4, which is negative, so no real roots. So m needs to be ≤1 but also m≠0. So the correct answer is D.But looking at the options, option C is m ≥ -1 and m≠0. Why is that an option? Maybe I missed something. Let me check the discriminant again. The discriminant is 4 -4m. To have discriminant ≥0:4 -4m ≥0 → -4m ≥ -4 → m ≤1.Wait, so m can be any number less than or equal to 1. But m can be negative? If m is negative, then the quadratic equation would open downward. But does that affect the discriminant? Let's see. Let me take m=-1. Then the equation is -x² +2x +1=0. The discriminant would be 4 -4*(-1)(1)=4 +4=8, which is positive, so two real roots. So even if m is negative, as long as m ≤1, the discriminant is non-negative. So m can be negative. So why is there an option C with m ≥-1? Maybe someone confused the discriminant or miscalculated?But according to the discriminant, m can be any real number ≤1, including negative numbers, as long as m≠0. For example, m=-2, let's check: mx² +2x +1=0 becomes -2x² +2x +1=0. Discriminant is 4 -4*(-2)(1)=4 +8=12, which is positive. So two real roots. So m=-2 is allowed, as long as m ≠0. Therefore, m can be as small as possible (approaching negative infinity) as long as m ≤1 and m≠0. Wait, but discriminant is 4 -4m. If m approaches negative infinity, the discriminant becomes 4 -4*(-∞)= +∞, which is positive, so real roots. So actually, m can be any real number ≤1, except m=0.But none of the options say m ≤1 and m≠0 except option D. So that must be the correct answer. Then why is option C there? Maybe a trick to test if someone miscalculates the discriminant. Let me check my discriminant calculation again.Original equation: mx² +2x +1=0.Discriminant D = b² -4ac = (2)^2 -4*(m)*(1) = 4 -4m. Yes, that's correct.Set D ≥0: 4 -4m ≥0 → 4 ≥4m → 1 ≥m → m ≤1. So that's right.So m can be any number less than or equal to 1, except m=0. Therefore, the answer is D: m ≤1 and m≠0. So the correct answer is D.But wait, let me verify once more. Suppose m=0 is excluded, so m can be negative, zero, or positive up to 1. But if m=0 is excluded, then m can be negative (like -1, -2, etc.) or positive (up to 1). So the range is m ≤1, excluding 0. So that's exactly option D. So yeah, D is the answer.I think the confusion might be if someone thought that the coefficient m has to be positive, but in reality, quadratic equations can have negative coefficients as well. So as long as m is not zero and the discriminant is non-negative, the equation will have real solutions. So yeah, answer D.Final Answerboxed{D}</think>To determine the range of values for ( m ) such that the quadratic equation ( mx^2 + 2x + 1 = 0 ) has real number solutions, we need to check the discriminant of the quadratic equation. The discriminant for a quadratic equation ( ax^2 + bx + c = 0 ) is given by ( b^2 - 4ac ). For the equation to have real solutions, the discriminant must be non-negative (i.e., ( geq 0 )).Given the quadratic equation ( mx^2 + 2x + 1 = 0 ), we identify the coefficients:- ( a = m )- ( b = 2 )- ( c = 1 )The discriminant is calculated as follows:[Delta = b^2 - 4ac = 2^2 - 4 cdot m cdot 1 = 4 - 4m]For the equation to have real solutions, the discriminant must be non-negative:[4 - 4m geq 0]Solving this inequality:[4 - 4m geq 0 -4m geq -4 m leq 1]Additionally, since the equation must be quadratic, the coefficient ( m ) cannot be zero (i.e., ( m neq 0 )). Combining these conditions, we get:[m leq 1 quad text{and} quad m neq 0]Thus, the range of values for ( m ) is given by option D:[boxed{D}]

🔑:To determine the range of values for m that allows the quadratic equation mx^{2}+2x+1=0 to have real number solutions, we need to consider the discriminant Delta of the quadratic equation. The discriminant is given by Delta = b^{2} - 4ac, where a, b, and c are the coefficients of x^{2}, x, and the constant term, respectively. For real number solutions to exist, the discriminant must be greater than or equal to zero (Delta geqslant 0).Given the equation mx^{2}+2x+1=0, we have:- a = m- b = 2- c = 1Thus, the discriminant Delta is:[Delta = b^{2} - 4ac = 2^{2} - 4 cdot m cdot 1 = 4 - 4m]For the equation to have real number solutions, we require Delta geqslant 0:[4 - 4m geqslant 0]Solving for m, we get:[4 geqslant 4m implies 1 geqslant m implies m leqslant 1]However, since the equation is quadratic, we also need m neq 0 to ensure it does not degenerate into a linear equation. Therefore, the conditions for m are:[m leqslant 1 quad text{and} quad m neq 0]Hence, the correct answer is boxed{text{D}}.