🧠:[<think>Okay, let's tackle this problem step by step. So, we have a sequence {a_n} where the sum of the first n terms is S_n. The first term a_1 is 3, and there's this equation given: 2S_n = 3a_n - 3. We need to find the general formula for {a_n} and then figure out T_n, the sum of the first n terms of another sequence {b_n} where each term b_n is a_n plus the logarithm base 3 of a_n. Starting with part (1): Finding the general formula for {a_n}. Hmm, given that S_n is the sum of the first n terms, so S_n = a_1 + a_2 + ... + a_n. The recursion-like equation given is 2S_n = 3a_n - 3. Maybe we can express S_n in terms of a_n first. Let's rearrange the equation: S_n = (3a_n - 3)/2. But we also know that S_n = S_{n-1} + a_n for n ≥ 2, since the sum of the first n terms is the sum of the first n-1 terms plus the nth term. Let me try substituting S_n from the given equation into this recursive formula. So, if S_n = S_{n-1} + a_n, then substituting the expression for S_n from the equation:(3a_n - 3)/2 = S_{n-1} + a_n.But S_{n-1} can also be expressed using the same given equation, replacing n with n-1: S_{n-1} = (3a_{n-1} - 3)/2. So substituting that into the previous equation:(3a_n - 3)/2 = (3a_{n-1} - 3)/2 + a_n.Let me write that out:(3a_n - 3)/2 = (3a_{n-1} - 3)/2 + a_n.Multiply both sides by 2 to eliminate denominators:3a_n - 3 = 3a_{n-1} - 3 + 2a_n.Simplify the right-hand side:3a_n - 3 = 3a_{n-1} + 2a_n - 3.Subtract 2a_n from both sides:a_n - 3 = 3a_{n-1} - 3.Add 3 to both sides:a_n = 3a_{n-1}.Wait, so this suggests that the sequence {a_n} is a geometric sequence where each term is 3 times the previous term. But wait, let's check this with the initial term. If a_1 = 3, then a_2 should be 3a_1 = 9, a_3 = 27, etc. So that would make the sequence 3, 9, 27, ..., which is a geometric sequence with common ratio 3. Let's verify this with the given condition. Let's check for n=1. The sum S_1 is a_1 = 3. Plug into 2S_n = 3a_n - 3:Left side: 2*3 = 6. Right side: 3*3 - 3 = 9 - 3 = 6. Okay, that works.For n=2. If a_2 is 9, then S_2 = 3 + 9 = 12. Plug into the equation:Left side: 2*12 = 24. Right side: 3*9 - 3 = 27 - 3 = 24. That works too.n=3: S_3 = 3 + 9 + 27 = 39. 2*39 = 78. 3*27 - 3 = 81 -3 =78. Yep, still holds.So this seems to check out. Therefore, the general term is a_n = 3^n. Wait, because a_1 is 3, a_2 is 9, so a_n = 3*3^{n-1} = 3^n. Yep.But let me just make sure that this satisfies the recursion a_n = 3a_{n-1}. If a_n = 3^n, then a_{n-1} = 3^{n-1}, and 3a_{n-1} = 3*3^{n-1} = 3^n = a_n. So that works. Therefore, the general formula is a_n = 3^n.Wait, but let's check the original equation again. If S_n is the sum of the first n terms of a geometric series with first term 3 and ratio 3. The sum S_n of a geometric series is a_1*(r^n -1)/(r -1). So here, S_n = 3*(3^n -1)/(3 -1) = (3^{n+1} -3)/2. Let's see if that equals (3a_n -3)/2. Since a_n = 3^n, then 3a_n -3 = 3^{n+1} -3, which divided by 2 is indeed (3^{n+1} -3)/2. So that's consistent. Therefore, our conclusion is correct. So part (1) is solved: a_n = 3^n.Now part (2): The sequence {b_n} satisfies b_n = a_n + log_3(a_n). We need to find T_n, the sum of the first n terms of {b_n}. Since a_n is 3^n, then log_3(a_n) is log_3(3^n) = n. Therefore, b_n = 3^n + n. So T_n is the sum from k=1 to n of (3^k + k) = sum 3^k + sum k. The sum of 3^k from k=1 to n is a geometric series. The sum is 3*(3^n -1)/(3 -1) = (3^{n+1} -3)/2. The sum of k from k=1 to n is n(n +1)/2. Therefore, T_n = (3^{n+1} -3)/2 + n(n +1)/2. We can write this as a single fraction: [3^{n+1} -3 + n(n +1)] / 2. Alternatively, factor it as [3^{n+1} + n(n +1) -3]/2. So that's T_n. Let me verify with n=1. For n=1, b_1 = a_1 + log_3(a_1) = 3 + 1 =4. So T_1=4. Plugging into the formula: [3^{2} +1*2 -3]/2 = [9 +2 -3]/2=8/2=4. Correct.n=2: b_1=4, b_2=9 + log_3(9)=9 +2=11. So T_2=4+11=15. Formula: [3^3 +2*3 -3]/2= [27 +6 -3]/2=30/2=15. Correct.n=3: b_3=27 +3=30. T_3=15+30=45. Formula: [3^4 +3*4 -3]/2= [81 +12 -3]/2=90/2=45. Correct. So the formula works.Therefore, part (2) answer is T_n = [3^{n+1} + n(n +1) -3]/2.Final Answer(1) The general formula for {a_n} is boxed{3^n}.(2) The sum T_n is boxed{dfrac{3^{n+1} + n(n + 1) - 3}{2}}.</think>Given the problem, we need to find the general formula for the sequence ({a_n}) and the sum (T_n) of the first (n) terms of another sequence ({b_n}).(1) Finding the general formula for ({a_n}):We know that the sum of the first (n) terms (S_n) is given by (2S_n = 3a_n - 3). We also know that (S_n = S_{n-1} + a_n). By substituting (S_n) from the given equation into the recursive formula, we derive:[frac{3a_n - 3}{2} = frac{3a_{n-1} - 3}{2} + a_n]Simplifying this, we find:[3a_n - 3 = 3a_{n-1} - 3 + 2a_n implies a_n = 3a_{n-1}]This indicates that ({a_n}) is a geometric sequence with the first term (a_1 = 3) and common ratio 3. Therefore, the general term is:[a_n = 3^n](2) Finding the sum (T_n) of the first (n) terms of ({b_n}):Given (b_n = a_n + log_3 a_n) and (a_n = 3^n), we have:[b_n = 3^n + log_3 3^n = 3^n + n]The sum (T_n) is the sum of (3^k) and (k) from (k = 1) to (n):[T_n = sum_{k=1}^n (3^k + k) = sum_{k=1}^n 3^k + sum_{k=1}^n k]The sum of the geometric series (sum_{k=1}^n 3^k) is:[frac{3(3^n - 1)}{3 - 1} = frac{3^{n+1} - 3}{2}]The sum of the arithmetic series (sum_{k=1}^n k) is:[frac{n(n + 1)}{2}]Combining these results, we get:[T_n = frac{3^{n+1} - 3}{2} + frac{n(n + 1)}{2} = frac{3^{n+1} + n(n + 1) - 3}{2}] Final Answer(1) The general formula for ({a_n}) is (boxed{3^n}).(2) The sum (T_n) is (boxed{dfrac{3^{n+1} + n(n + 1) - 3}{2}}).,<think>Okay, let's tackle this problem step by step. So, we have a sequence {a_n} where the sum of the first n terms is S_n. We know that a_1 is 3, and there's this equation given: 2S_n = 3a_n - 3. The problem has two parts: first, finding the general formula for {a_n}, and second, finding the sum T_n of the first n terms of another sequence {b_n} where b_n = a_n + log base 3 of a_n.Starting with part (1). Let me recall that the sum of the first n terms S_n is equal to a_1 + a_2 + ... + a_n. Given that 2S_n = 3a_n - 3, maybe I can express S_n in terms of a_n and then find a recurrence relation or something. Let's write down the given information:Given:1. a_1 = 32. 2S_n = 3a_n - 3, where S_n is the sum of the first n terms.First, let's express S_n in terms of a_n. From the equation, we can write S_n = (3a_n - 3)/2.But also, S_n = S_{n-1} + a_n, because the sum of the first n terms is the sum of the first (n-1) terms plus the nth term.So substituting S_n from the equation into this relation:(3a_n - 3)/2 = S_{n-1} + a_nBut S_{n-1} can also be expressed using the given equation for n-1:From the equation 2S_{n-1} = 3a_{n-1} - 3, so S_{n-1} = (3a_{n-1} - 3)/2.Substitute that back into the previous equation:(3a_n - 3)/2 = (3a_{n-1} - 3)/2 + a_nLet me write this out step by step to avoid confusion:Left side: (3a_n - 3)/2Right side: [(3a_{n-1} - 3)/2] + a_nMultiply both sides by 2 to eliminate denominators:3a_n - 3 = 3a_{n-1} - 3 + 2a_nSimplify the right side:3a_{n-1} - 3 + 2a_n = 2a_n + 3a_{n-1} - 3So the equation becomes:3a_n - 3 = 2a_n + 3a_{n-1} - 3Subtract 2a_n from both sides:a_n - 3 = 3a_{n-1} - 3Wait, subtract 2a_n: 3a_n - 3 - 2a_n = a_n -3Right side remains 3a_{n-1} -3.So:a_n - 3 = 3a_{n-1} -3Add 3 to both sides:a_n = 3a_{n-1}Oh! So this suggests that the sequence {a_n} is a geometric sequence with common ratio 3? Because each term is 3 times the previous term.But wait, let's check with n=1. Given a_1=3. Then according to a_n = 3a_{n-1}, a_2 should be 3*3=9, a_3=27, etc. Let's verify if this holds with the given equation 2S_n =3a_n -3.For n=1: S_1 = a_1 =3. Then 2*3=6, and 3a_1 -3=9 -3=6. So that works.For n=2: S_2 = a_1 + a_2 =3 +9=12. Then 2*S_2=24. 3a_2 -3=27 -3=24. That works too.For n=3: S_3=3+9+27=39. 2*39=78. 3a_3 -3=81 -3=78. Also works. So seems like the recurrence a_n=3a_{n-1} is valid. Therefore, the general term is a_n=3^n.Wait, a_n=3*3^{n-1}=3^n. So yes, a geometric sequence with first term 3 and ratio 3. Therefore, general formula is a_n=3^n. Let me check this with n=1: 3^1=3, correct. n=2: 3^2=9, correct. So that seems to work.But let me double-check the recurrence. If a_n=3a_{n-1}, then it's a geometric sequence with ratio 3. So sum S_n=3*(3^n -1)/(3-1)= (3^{n+1}-3)/2. Then 2S_n=3^{n+1}-3. According to the given equation 2S_n=3a_n -3. Let's see: 3a_n -3=3*3^n -3=3^{n+1}-3. Which matches 2S_n. So yes, the general formula is a_n=3^n. Therefore, part (1) answer is a_n=3^n.Moving on to part (2). We have another sequence {b_n} where b_n = a_n + log_3(a_n). We need to find T_n, the sum of the first n terms of {b_n}.Given that a_n=3^n, then log_3(a_n)=log_3(3^n)=n. Therefore, b_n=3^n + n.Therefore, T_n is the sum from k=1 to n of (3^k +k) = sum_{k=1}^n 3^k + sum_{k=1}^n k.We can compute each sum separately.First sum: sum_{k=1}^n 3^k. This is a geometric series with first term 3, ratio 3, number of terms n. The formula for the sum is 3*(3^n -1)/(3 -1)= (3^{n+1}-3)/2.Second sum: sum_{k=1}^n k. This is the arithmetic series sum, which is n(n+1)/2.Therefore, T_n = (3^{n+1} -3)/2 + n(n+1)/2.We can write this as a single fraction if needed: [3^{n+1} -3 + n(n+1)] /2.Alternatively, factor the -3 and the other terms. Let's check with n=1:When n=1: T_1 = b_1 = 3^1 +1=4. According to the formula: [3^{2} -3 +1*2]/2= [9 -3 +2]/2=8/2=4. Correct.n=2: T_2 =4 + (9 +2)=4 +11=15. Formula: [3^3 -3 +2*3]/2= [27 -3 +6]/2=30/2=15. Correct.n=3: T_3=15 + (27 +3)=15 +30=45. Formula: [3^4 -3 +3*4]/2= [81 -3 +12]/2=90/2=45. Correct.Seems like the formula works. So T_n=(3^{n+1} -3 +n(n+1))/2. Alternatively, we can factor the denominator as 1/2*(3^{n+1} +n(n+1) -3). Either way is acceptable. But maybe write it as (3^{n+1} +n(n+1) -3)/2.But perhaps the problem expects the answer in the separated form: sum of the geometric series plus the arithmetic series. But as per the problem statement, they just want T_n found. So either form is okay. But since they mention "the sum of the first n terms of {b_n} is denoted as T_n", just giving the combined expression is fine.So the final answer for part (2) is T_n=(3^{n+1} +n(n+1) -3)/2.But let me verify once more. Let's see for n=1:Original b_1=3 + log_3(3)=3 +1=4. T_1=4. The formula gives (3^{2} +1*2 -3)/2=(9+2-3)/2=8/2=4. Correct.n=2: b_1=4, b_2=9 + log_3(9)=9+2=11. T_2=4+11=15. Formula: (3^3 +2*3 -3)/2=(27 +6 -3)/2=30/2=15. Correct.n=3: b_3=27 + log_3(27)=27+3=30. T_3=15+30=45. Formula: (3^4 +3*4 -3)/2=(81 +12 -3)/2=90/2=45. Correct.Looks good. So I think that's the answer.Final Answer(1) The general formula for {a_{n}} is boxed{3^{n}}.(2) The sum T_{n} is boxed{dfrac{3^{n+1} + n(n + 1) - 3}{2}}.</think>Given that the sum of the first ( n ) terms of a sequence ({a_n}) is ( S_n ), where ( a_1 = 3 ) and ( 2S_n = 3a_n - 3 ):1. Finding the general formula for ({a_n}): We start with the given equation ( 2S_n = 3a_n - 3 ). We know that ( S_n = S_{n-1} + a_n ). Substituting ( S_n ) from the given equation into this recurrence relation, we get: [ frac{3a_n - 3}{2} = frac{3a_{n-1} - 3}{2} + a_n ] Simplifying this equation, we find: [ 3a_n - 3 = 3a_{n-1} - 3 + 2a_n implies a_n = 3a_{n-1} ] This indicates that the sequence ({a_n}) is a geometric sequence with the first term ( a_1 = 3 ) and common ratio 3. Therefore, the general formula is: [ a_n = 3^n ]2. Finding the sum ( T_n ) of the first ( n ) terms of the sequence ({b_n}): The sequence ({b_n}) is defined as ( b_n = a_n + log_3 a_n ). Substituting ( a_n = 3^n ), we get: [ b_n = 3^n + log_3 3^n = 3^n + n ] The sum ( T_n ) is the sum of the first ( n ) terms of ({b_n}): [ T_n = sum_{k=1}^n (3^k + k) = sum_{k=1}^n 3^k + sum_{k=1}^n k ] The first sum is a geometric series: [ sum_{k=1}^n 3^k = frac{3(3^n - 1)}{3 - 1} = frac{3^{n+1} - 3}{2} ] The second sum is an arithmetic series: [ sum_{k=1}^n k = frac{n(n + 1)}{2} ] Combining these results, we get: [ T_n = frac{3^{n+1} - 3}{2} + frac{n(n + 1)}{2} = frac{3^{n+1} + n(n + 1) - 3}{2} ] Final Answer(1) The general formula for ({a_n}) is (boxed{3^n}).(2) The sum (T_n) is (boxed{dfrac{3^{n+1} + n(n + 1) - 3}{2}}).]

🔑: Solution:# Part 1: Finding the General Formula for {a_{n}}Given that 2S_{n} = 3a_{n} - 3 (Equation ①), and for n geq 2, we have 2S_{n-1} = 3a_{n-1} - 3 (Equation ②).Subtracting Equation ② from Equation ① gives us:[2S_{n} - 2S_{n-1} = 3a_{n} - 3 - (3a_{n-1} - 3)][2(a_{n}) = 3a_{n} - 3a_{n-1}][a_{n} = 3a_{n-1}]Given a_{1} = 3, we find that {a_{n}} is a geometric sequence with the first term 3 and common ratio 3. Therefore, the general formula for {a_{n}} is:[a_{n} = 3^{n}]for n in mathbb{N}^{*}. So, the answer to part (1) is:[boxed{a_{n} = 3^{n}}]# Part 2: Finding T_{n} for the Sequence {b_{n}}Given b_{n} = a_{n} + log_{3}a_{n}, and since we found a_{n} = 3^{n}, we substitute to get:[b_{n} = 3^{n} + log_{3}3^{n} = 3^{n} + n]The sum of the first n terms of {b_{n}}, denoted as T_{n}, is:[T_{n} = sum_{k=1}^{n} (3^{k} + k)][= sum_{k=1}^{n} 3^{k} + sum_{k=1}^{n} k][= (3^{1} + 3^{2} + cdots + 3^{n}) + (1 + 2 + cdots + n)][= frac{3(1 - 3^{n})}{1 - 3} + frac{n(n + 1)}{2}][= frac{3^{n+1} - 3}{-2} + frac{n^{2} + n}{2}][= frac{3^{n+1} + n^{2} + n - 3}{2}]Therefore, the sum of the first n terms of the sequence {b_{n}} is:[boxed{T_{n} = frac{3^{n+1} + n^{2} + n - 3}{2}}]

🧠:[<think>Okay, so I have this inequality to solve: 2x² - x ≤ 1. The answer choices are A, B, C, D. Let me try to work through this step by step. First, I remember that to solve a quadratic inequality, I should probably move all terms to one side to set the inequality to ≤ 0. That way, I can factor or find roots to determine the intervals where the quadratic is negative or zero. Let me try that.Starting with the original inequality:2x² - x ≤ 1Subtract 1 from both sides to get:2x² - x - 1 ≤ 0Alright, so now the quadratic is 2x² - x - 1. I need to find the values of x where this quadratic is less than or equal to zero. To do that, I should find the roots of the quadratic equation 2x² - x - 1 = 0. Then, determine the intervals between the roots where the quadratic is negative.Let me solve 2x² - x - 1 = 0. This is a quadratic equation, so I can use the quadratic formula. The quadratic is in the form ax² + bx + c = 0, so a = 2, b = -1, c = -1.Quadratic formula is x = [-b ± √(b² - 4ac)] / (2a)Plugging in the values:x = [-(-1) ± √((-1)² - 4*2*(-1))] / (2*2)x = [1 ± √(1 + 8)] / 4x = [1 ± √9] / 4x = [1 ± 3] / 4So the two roots are:x = (1 + 3)/4 = 4/4 = 1x = (1 - 3)/4 = (-2)/4 = -1/2So the roots are x = -1/2 and x = 1.Since the quadratic is 2x² - x - 1, the coefficient of x² is positive (2), which means the parabola opens upward. Therefore, the quadratic will be ≤ 0 between the two roots. Because when a parabola opens upward, it is below the x-axis (negative) between its two roots.Therefore, the solution set should be the interval between -1/2 and 1, including the endpoints because the inequality is ≤ 0. So x ∈ [-1/2, 1].Looking back at the options:A: [-1/2, 1]B: [0, 1/2]C: (-∞, -1/2) ∪ [1, +∞)D: (-∞, -1/2] ∪ [1, +∞)So the correct answer should be A. But wait, let me double-check.Wait, the quadratic 2x² - x - 1 ≤ 0. Let me test a point in each interval to confirm.First, the roots divide the number line into three intervals:1. x < -1/22. -1/2 < x < 13. x > 1Let me pick test points in each interval.1. Let's take x = -1 (which is less than -1/2). Plug into 2x² - x - 1:2*(-1)^2 - (-1) - 1 = 2*1 +1 -1 = 2 +1 -1 = 2. That's positive, which is greater than 0. So the inequality 2x² - x -1 ≤0 is not satisfied here.2. Let's take x = 0 (which is between -1/2 and 1). Plugging in:2*(0)^2 -0 -1 = 0 -0 -1 = -1 ≤0. That satisfies the inequality.3. Let's take x = 2 (greater than 1). Plugging in:2*(2)^2 -2 -1 = 8 -2 -1 = 5. That's positive, so inequality not satisfied.Therefore, the solution is indeed between -1/2 and 1, including the endpoints because the inequality is ≤. Therefore, answer A is correct.But wait, the options given: A is [-1/2,1], which matches. The other options don't seem to fit. Let me check the problem statement again to make sure I didn't misread.Original inequality: 2x² - x ≤1. Yep, moved all terms to left side: 2x² -x -1 ≤0. Found roots at -1/2 and 1. Since parabola opens upwards, negative between the roots. So answer is A.But the options C and D are about union of intervals outside. Maybe I made a mistake?Wait, no. The quadratic is ≤0 between the roots. Since the parabola opens upwards, the graph is a U-shape. So it starts from positive infinity, comes down to cross the x-axis at -1/2, goes below the x-axis between -1/2 and 1, then goes back up to positive infinity after x=1. Therefore, the inequality 2x² -x -1 ≤0 holds between the roots. So x ∈ [-1/2,1]. So answer A.But why are there options with unions of intervals? Maybe if the inequality was ≥0, then it would be C or D. But since the original inequality is ≤1, which translates to ≤0 after moving 1 to the left, the solution should be between the roots. So definitely A. Therefore, the answer is A.Wait, but let me check once again. Maybe there's a miscalculation in roots.Quadratic equation 2x² -x -1 =0.Using quadratic formula:x = [1 ± √(1 + 8)] / 4 = [1 ±3]/4. So 1+3=4, 4/4=1. 1-3=-2, -2/4=-1/2. Correct. So roots are correct.Yes, so answer is A. So I think that's right.Final Answerboxed{A}</think>To solve the inequality (2x^2 - x leq 1), we start by moving all terms to one side:[2x^2 - x - 1 leq 0]Next, we solve the quadratic equation (2x^2 - x - 1 = 0) using the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 2), (b = -1), and (c = -1):[x = frac{-(-1) pm sqrt{(-1)^2 - 4 cdot 2 cdot (-1)}}{2 cdot 2} = frac{1 pm sqrt{1 + 8}}{4} = frac{1 pm 3}{4}]This gives the roots:[x = frac{1 + 3}{4} = 1 quad text{and} quad x = frac{1 - 3}{4} = -frac{1}{2}]The quadratic (2x^2 - x - 1) opens upwards (since the coefficient of (x^2) is positive), so the quadratic expression is less than or equal to zero between the roots (-frac{1}{2}) and (1). Therefore, the solution set is:[left[-frac{1}{2}, 1right]]Testing intervals confirms this result. The correct answer is (boxed{A}).,<think>Okay, so I need to solve this inequality: 2x² - x ≤ 1. Hmm, quadratic inequalities can be a bit tricky, but let me try to remember the steps. First, I think I should move all terms to one side to set the inequality to ≤ 0. That way, I can factor or maybe use the quadratic formula to find the critical points. Let me do that.Starting with 2x² - x ≤ 1. If I subtract 1 from both sides, I get 2x² - x - 1 ≤ 0. Alright, now the inequality is in standard quadratic form: ax² + bx + c ≤ 0. The next step is to find the roots of the quadratic equation 2x² - x - 1 = 0. Once I have the roots, I can determine the intervals where the quadratic is less than or equal to zero.To find the roots, I can use the quadratic formula: x = [-b ± √(b² - 4ac)] / (2a). Here, a = 2, b = -1, and c = -1. Plugging these into the formula:x = [-(-1) ± √((-1)² - 4*2*(-1))]/(2*2)x = [1 ± √(1 + 8)]/4x = [1 ± √9]/4x = [1 ± 3]/4So that gives two solutions: one with the plus sign and one with the minus sign.First solution: (1 + 3)/4 = 4/4 = 1Second solution: (1 - 3)/4 = (-2)/4 = -1/2So the roots are x = 1 and x = -1/2. These are the critical points that divide the number line into intervals. The intervals are:1. (-∞, -1/2)2. [-1/2, 1]3. (1, ∞)Now, since the quadratic opens upwards (because the coefficient of x² is positive, 2), the parabola will be U-shaped. That means the quadratic will be below or equal to zero between the roots. Wait, but let me confirm that.The graph of a quadratic with a positive leading coefficient opens upward, so it will be below the x-axis (i.e., ≤ 0) between the two roots. Therefore, the solution set should be the interval between the roots: [-1/2, 1]. Hmm, but let me check with a test point in each interval to be sure.First interval: Let's pick x = -1 (which is in (-∞, -1/2)). Plugging into 2x² - x - 1:2*(-1)^2 - (-1) -1 = 2*1 +1 -1 = 2 +1 -1 = 2. That's positive, so the inequality 2x² - x -1 ≤ 0 is not satisfied here.Second interval: Let's pick x = 0 (which is in [-1/2, 1]). Plugging in:2*0^2 -0 -1 = -1. That's negative, so the inequality is satisfied here.Third interval: Let's pick x = 2 (which is in (1, ∞)). Plugging in:2*(2)^2 -2 -1 = 8 -2 -1 = 5. That's positive, so the inequality is not satisfied here.Therefore, the solution set is indeed the interval between the roots, [-1/2, 1], which corresponds to option A.Wait, but let me check the original inequality again to make sure I didn't make a mistake in the setup. The original problem was 2x² - x ≤ 1. By moving the 1 to the left, we get 2x² - x -1 ≤ 0, which is correct. Then solving 2x² -x -1 =0 gives roots at x=1 and x=-1/2, which seems right. Testing the intervals also checks out. So the answer should be A.But wait, let me check the answer options again. A is [-1/2,1], B is [0,1/2], C is (-∞,-1/2) union [1,∞), D is (-∞,-1/2] union [1,∞). Since our solution is between -1/2 and 1, inclusive, the correct answer is A. So why is there another option D? Maybe if someone thought the parabola opens downward? But the coefficient of x² is 2, which is positive, so it opens upward. So between the roots is where it's negative. Yeah, so A is correct.Alternatively, maybe someone messed up the direction of the inequality. Let me see. If we have 2x² -x ≤1, which becomes 2x² -x -1 ≤0. The quadratic is ≤0 between the roots. So A is correct. Therefore, answer A.Final Answerboxed{A}</think>To solve the inequality (2x^2 - x leq 1), we first move all terms to one side:[2x^2 - x - 1 leq 0]Next, we find the roots of the quadratic equation (2x^2 - x - 1 = 0) using the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 2), (b = -1), and (c = -1):[x = frac{-(-1) pm sqrt{(-1)^2 - 4 cdot 2 cdot (-1)}}{2 cdot 2} = frac{1 pm sqrt{1 + 8}}{4} = frac{1 pm 3}{4}]This gives the roots:[x = frac{1 + 3}{4} = 1 quad text{and} quad x = frac{1 - 3}{4} = -frac{1}{2}]The critical points are (x = 1) and (x = -frac{1}{2}), which divide the number line into three intervals: ((-infty, -frac{1}{2})), ([- frac{1}{2}, 1]), and ((1, infty)). Since the quadratic opens upwards (the coefficient of (x^2) is positive), the quadratic expression (2x^2 - x - 1) is less than or equal to zero between the roots.Testing intervals confirms that the quadratic is negative in ([- frac{1}{2}, 1]). Therefore, the solution set is:[left[-frac{1}{2}, 1right]]The correct answer is (boxed{A}).]

🔑:Solve: 2x^{2}-xleqslant 1 can be transformed into 2x^{2}-x-1leqslant 0, solving this yields - dfrac {1}{2}leqslant xleqslant 1, thus, the correct option is boxed{A}. First, transform 2x^{2}-xleqslant 1 into 2x^{2}-x-1leqslant 0, and according to the properties of quadratic functions, the answer can be obtained. This question tests the method of solving a quadratic inequality, which is a basic question. Understanding the relationship among "three quadratics" is key to solving the problem.

🧠:[<think>Okay, so I need to figure out the interval of monotonic increase for the function g(x), which is a horizontal shift of f(x) = sin(2x) to the right by π/6 units. The answer choices are given, and I need to pick the correct one. Let me start by recalling some basic concepts here.First, when you shift a function horizontally, the transformation affects the input variable x. Specifically, shifting to the right by π/6 means replacing x with (x - π/6) in the function. So, the original function is f(x) = sin(2x), and the shifted function g(x) would be sin(2(x - π/6)). Let me confirm that. If you shift a function f(x) right by a units, it's f(x - a). So yes, replacing x with (x - π/6). Therefore, g(x) = sin(2(x - π/6)) = sin(2x - π/3). Okay, so that simplifies to sin(2x - π/3). Got that.Now, the next step is to determine where this function g(x) is monotonically increasing. Remember that a function is increasing where its derivative is positive. So, I need to compute the derivative of g(x) and find the intervals where this derivative is positive.Let's compute g'(x). The derivative of sin(u) is cos(u) * u', so here, u = 2x - π/3, so u' = 2. Therefore, g'(x) = 2cos(2x - π/3). So, the derivative is 2cos(2x - π/3). The function g(x) will be increasing when g'(x) > 0, so when 2cos(2x - π/3) > 0. Since 2 is positive, this simplifies to cos(2x - π/3) > 0.So, the problem reduces to solving the inequality cos(2x - π/3) > 0. Let's solve this inequality step by step.First, recall that the cosine function is positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for any integer k. So, we can set up the inequality:-π/2 + 2πk < 2x - π/3 < π/2 + 2πk, where k is any integer.Our goal is to solve for x in terms of k. Let's solve the inequalities step by step.Starting with the left inequality:-π/2 + 2πk < 2x - π/3Add π/3 to both sides:-π/2 + π/3 + 2πk < 2xCompute -π/2 + π/3. Let's convert these to sixths: -3π/6 + 2π/6 = (-π)/6. So,-π/6 + 2πk < 2xThen divide both sides by 2:-π/12 + πk < xNow the right inequality:2x - π/3 < π/2 + 2πkAdd π/3 to both sides:2x < π/2 + π/3 + 2πkConvert π/2 and π/3 to sixths: π/2 = 3π/6, π/3 = 2π/6. So,2x < 5π/6 + 2πkDivide both sides by 2:x < 5π/12 + πkSo, combining both inequalities, we have:-π/12 + πk < x < 5π/12 + πk, where k is any integer.Therefore, the intervals where g(x) is increasing are (-π/12 + πk, 5π/12 + πk) for each integer k.But the answer choices are given in terms of closed intervals [ , ] with kπ as the central term. Let me check how this interval can be expressed.Note that πk - π/12 is the same as -π/12 + πk, and 5π/12 + πk is the same as πk + 5π/12. So, the interval can be written as [kπ - π/12, kπ + 5π/12], but the answer options have different terms. Let me check the options again.Option C is [kπ - π/12, kπ + 5π/12], which matches exactly. So, that's option C. But wait, let me verify if the endpoints are included. The original inequality was cos(2x - π/3) > 0, which is strict inequality. So, the intervals where the derivative is positive are open intervals. However, sometimes in such problems, they might consider the points where the derivative is zero as part of the increasing interval if the function is non-decreasing. Wait, but monotonic increase usually allows for non-decreasing, but sometimes strictly increasing. Hmm.Wait, the question says "monotonic increase". Depending on the convention, this could mean non-decreasing or strictly increasing. If it's strictly increasing, then the endpoints where the derivative is zero (i.e., where cos(2x - π/3)=0) would not be included. However, the answer options use closed intervals, so they must be considering the points where the derivative is zero as part of the increasing intervals, perhaps interpreting monotonic increase as non-decreasing.But let's check the critical points. The cosine function is zero at π/2 + πk, so setting 2x - π/3 = π/2 + πk, solving for x:2x = π/2 + πk + π/3Convert to sixths: π/2 = 3π/6, π/3 = 2π/6, so:2x = 3π/6 + 2π/6 + πk = 5π/6 + πkThus, x = 5π/12 + πk/2.Wait, but in our previous interval, the upper bound was 5π/12 + πk. So, if k is an integer, then πk/2 would step through different positions, but since k is any integer, perhaps the critical points are included in the intervals when considering different k. Hmm, maybe I need to think again.Alternatively, perhaps the answer is written in terms of kπ ± something. Let's look at the options again.Option C: [kπ - π/12, kπ + 5π/12]If we write this interval as [kπ - π/12, kπ + 5π/12], then for each integer k, this is an interval of length π/12 + 5π/12 = 6π/12 = π/2. But the original intervals where cosine is positive are of length π (since cosine is positive over intervals of length π, separated by π where it's negative). Wait, but in our problem, we have cos(2x - π/3) > 0, so the period of the cosine function here is π, since the argument is 2x - π/3. So, the period of cos(2x - π/3) is π, which makes sense.But the intervals where cosine is positive would be half the period, so π/2 each. So, every π/2, the cosine function alternates between positive and negative. But when we solved the inequality, we found intervals of length (5π/12 - (-π/12)) = 6π/12 = π/2. So, that's correct.But let's check if the answer options match. Let's see:Original intervals: (-π/12 + πk, 5π/12 + πk). So, written as (πk - π/12, πk + 5π/12). So, if we close the intervals, since at the endpoints the derivative is zero, then depending on the definition, the function changes from increasing to decreasing or vice versa. So, the function g(x) would have critical points at x = πk - π/12 and x = πk + 5π/12. Since at those points, the derivative is zero. So, if the question is about the interval where the function is non-decreasing, then including the endpoints where the derivative is zero is acceptable, but if it's strictly increasing, then not. But given that the options are closed intervals, I think the answer expects the non-decreasing interpretation, hence including the endpoints. So, the correct answer would be C: [kπ - π/12, kπ + 5π/12], which matches our derived interval.But let me check the other options to ensure none of them are equivalent or if I made a miscalculation.Option A: [kπ - π/3, kπ + π/6]Option B: [kπ - π/6, kπ + π/3]Option C: [kπ - π/12, kπ + 5π/12]Option D: [kπ - 5π/12, kπ + π/12]So, our derived interval is [kπ - π/12, kπ + 5π/12], which is option C. Let me verify once more.Alternatively, maybe I messed up the shift. Let me double-check how the horizontal shift works. The original function is f(x) = sin(2x). Shifting it to the right by π/6 units would replace x with x - π/6, resulting in sin(2(x - π/6)) = sin(2x - π/3). That seems correct.Then, derivative is 2cos(2x - π/3). Then, the inequality cos(2x - π/3) > 0. The solutions to this inequality are when 2x - π/3 is in (-π/2 + 2πk, π/2 + 2πk). Solving for x:Add π/3: 2x is in (-π/2 + π/3 + 2πk, π/2 + π/3 + 2πk)Convert to sixths: -3π/6 + 2π/6 = -π/6, and 3π/6 + 2π/6 = 5π/6So, 2x is in (-π/6 + 2πk, 5π/6 + 2πk)Divide by 2: x in (-π/12 + πk, 5π/12 + πk)Which is the same as [kπ - π/12, kπ + 5π/12] when written with k as integer. Hence, option C.Therefore, the correct answer should be option C.Wait, but let me check with an example. Let's take k = 0. Then, the interval is [-π/12, 5π/12]. Let's check the derivative in this interval. For example, at x = 0: derivative is 2cos(-π/3) = 2*(1/2) = 1 > 0. At x = π/6: 2cos(2*(π/6) - π/3) = 2cos(π/3 - π/3) = 2cos(0) = 2*1 = 2 > 0. At x = 5π/12: 2cos(2*(5π/12) - π/3) = 2cos(5π/6 - π/3) = 2cos(π/2) = 0. So, at the endpoint, the derivative is zero, which is why it's included if we consider non-decreasing. Similarly, at x = -π/12: 2cos(2*(-π/12) - π/3) = 2cos(-π/6 - π/3) = 2cos(-π/2) = 0. So, again, derivative is zero at the endpoints.Thus, the function is non-decreasing on the closed intervals, so option C is correct.Therefore, the answer is option C.Final Answerboxed{C}</think>To determine the interval of monotonic increase for the function ( g(x) ) obtained by shifting the graph of ( f(x) = sin 2x ) to the right by ( frac{pi}{6} ) units, we start by defining ( g(x) ). The horizontal shift results in the function ( g(x) = sin(2(x - frac{pi}{6})) = sin(2x - frac{pi}{3}) ).Next, we find the derivative of ( g(x) ):[g'(x) = 2cos(2x - frac{pi}{3})]To determine where ( g(x) ) is monotonically increasing, we solve the inequality ( g'(x) > 0 ):[2cos(2x - frac{pi}{3}) > 0 implies cos(2x - frac{pi}{3}) > 0]The cosine function is positive in the intervals ( (-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpi) ) for any integer ( k ). Therefore, we solve:[-frac{pi}{2} + 2kpi < 2x - frac{pi}{3} < frac{pi}{2} + 2kpi]Adding ( frac{pi}{3} ) to all parts of the inequality:[-frac{pi}{2} + frac{pi}{3} + 2kpi < 2x < frac{pi}{2} + frac{pi}{3} + 2kpi]Converting to a common denominator and simplifying:[-frac{pi}{6} + 2kpi < 2x < frac{5pi}{6} + 2kpi]Dividing by 2:[-frac{pi}{12} + kpi < x < frac{5pi}{12} + kpi]Thus, the intervals where ( g(x) ) is monotonically increasing are:[[kpi - frac{pi}{12}, kpi + frac{5pi}{12}] quad (k in mathbb{Z})]This corresponds to option C.[boxed{C}],<think>Okay, so I need to find the interval where the function g(x) is monotonically increasing. g(x) is obtained by shifting the graph of f(x) = sin(2x) to the right by π/6 units. Hmm, shifting a function to the right usually means replacing x with (x - h), where h is the shift. So if we shift right by π/6, the new function should be g(x) = sin(2(x - π/6)). Let me confirm that. Yes, because shifting f(x) right by h units is f(x - h). So substituting x with x - π/6. Then, since the original function is sin(2x), after shifting, it's sin(2(x - π/6)) which simplifies to sin(2x - π/3). So g(x) = sin(2x - π/3). Got that.Now, to find the intervals where g(x) is monotonically increasing. For that, I need to look at the derivative of g(x). The derivative of sin(u) is cos(u)*u', so here, derivative of sin(2x - π/3) with respect to x is 2cos(2x - π/3). So g'(x) = 2cos(2x - π/3). The function is increasing when the derivative is positive. So we need to find where 2cos(2x - π/3) > 0. Since 2 is a positive constant, this is equivalent to cos(2x - π/3) > 0.Therefore, we need to solve the inequality cos(2x - π/3) > 0. Let me recall that cosine is positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for any integer k. So, setting up the inequality:-π/2 + 2πk < 2x - π/3 < π/2 + 2πk, where k is any integer.Now, let's solve for x.First, add π/3 to all parts:-π/2 + π/3 + 2πk < 2x < π/2 + π/3 + 2πk.Calculating -π/2 + π/3: Let's convert to sixths. -π/2 is -3π/6, π/3 is 2π/6. So -3π/6 + 2π/6 = -π/6. Similarly, π/2 is 3π/6, so 3π/6 + 2π/6 = 5π/6. So the inequality becomes:-π/6 + 2πk < 2x < 5π/6 + 2πk.Now, divide all parts by 2:-π/12 + πk < x < 5π/12 + πk.Therefore, the function g(x) is increasing on the intervals (-π/12 + πk, 5π/12 + πk) where k is any integer. So in interval notation, this is written as [kπ - π/12, kπ + 5π/12], since when you express the interval around each integer multiple of π, starting from kπ - π/12 to kπ + 5π/12. Let me check that. If we factor out πk from both ends, we have πk - π/12 to πk + 5π/12. So yes, that's the same as [kπ - π/12, kπ + 5π/12], where k is an integer.Looking at the options given:A: [kπ - π/3, kπ + π/6]B: [kπ - π/6, kπ + π/3]C: [kπ - π/12, kπ + 5π/12]D: [kπ - 5π/12, kπ + π/12]From my solution, the correct answer should be option C. Let me just verify once again to make sure I didn't make a mistake in the algebra.Starting with cos(2x - π/3) > 0.The general solution for cosθ > 0 is θ ∈ (-π/2 + 2πk, π/2 + 2πk). So substituting θ = 2x - π/3:2x - π/3 ∈ (-π/2 + 2πk, π/2 + 2πk).Solving for x:Add π/3:2x ∈ (-π/2 + π/3 + 2πk, π/2 + π/3 + 2πk).Calculating the constants:-π/2 + π/3 = -(3π/6 - 2π/6) = -π/6.π/2 + π/3 = (3π/6 + 2π/6) = 5π/6.So 2x ∈ (-π/6 + 2πk, 5π/6 + 2πk).Divide by 2:x ∈ (-π/12 + πk, 5π/12 + πk).Which is equivalent to x ∈ [kπ - π/12, kπ + 5π/12] when considering closed intervals. Wait, but the original solution from cosine being positive is open intervals. However, the problem says "interval of monotonic increase". For differentiable functions, the function is increasing on an interval if the derivative is non-negative. But in the options, they are all closed intervals. Let me check if the endpoints are included.Since the derivative is 2cos(2x - π/3). The derivative is zero when cos(2x - π/3) = 0. So at 2x - π/3 = π/2 + πk, which gives x = (π/2 + πk + π/3)/2 = (5π/6 + πk)/2 = 5π/12 + πk/2. So these are points where the derivative is zero, which are critical points. But in the intervals we found, the endpoints are where the derivative transitions from positive to negative or vice versa. However, depending on the convention, sometimes intervals of increase include points where the derivative is zero if the function doesn't decrease after that. Wait, but actually, if the derivative is positive on an open interval and zero at the endpoints, then depending on the strictness of the increase, sometimes they include the endpoints. But strictly speaking, a function is increasing on an interval if for any two points x1 < x2 in the interval, f(x1) ≤ f(x2). So if the derivative is non-negative on the interval, then it's increasing. So if the derivative is zero at isolated points, the function is still considered increasing. Therefore, the intervals should be closed where the derivative is non-negative. However, in our case, between two critical points where the derivative is positive in between, the endpoints would have derivative zero. So when the derivative is zero, is the function still considered increasing? For example, at a maximum or minimum. If you have a horizontal tangent at a peak, then after that point, the function starts decreasing. Similarly, at a trough, it starts increasing. So in our case, when the derivative is zero, if it's a maximum, then the function is increasing before that point and decreasing after. Similarly, if it's a minimum, decreasing before and increasing after. Therefore, the endpoints where the derivative is zero would be points where the function changes its monotonicity. Therefore, in the intervals where the derivative is positive, the endpoints would be the points where the derivative is zero. So whether to include them in the interval of increase depends on the definition. If the function is increasing up to that point and then stops increasing, so maybe the interval should be closed on the left and open on the right? But in the options, they are all closed intervals.Wait, the problem says "interval of monotonic increase". In calculus, usually, when we talk about intervals where the function is increasing, we include the endpoints if the function continues to be increasing at those points. However, if the derivative is zero at the endpoint, and then the function starts decreasing, then the endpoint is a local maximum and shouldn't be included in the interval of increase. Wait, but actually, the standard definition is that a function is increasing on an interval if for any x1 < x2 in the interval, f(x1) ≤ f(x2). So if you have a point where the derivative is zero but the function doesn't decrease after that point, then it's still considered increasing. For example, f(x) = x^3 is increasing everywhere, even though the derivative at x=0 is zero. However, in the case of sine function, when the derivative is zero at a peak, after that point, the function decreases. So if we include the peak point in the interval, then after that point, the function is decreasing, so the interval of increase should not include the peak. Therefore, the intervals should be open at the endpoints where the derivative is zero. But the answer choices are all closed intervals. Maybe in the problem's context, they consider the maximum interval where the function is non-decreasing, including the points where the derivative is zero. Wait, but in that case, if you include the local maximum, then after that point, the function decreases, so the function isn't increasing anymore. Therefore, I think the correct approach is that the intervals where the derivative is positive (excluding the endpoints where derivative is zero) would be open intervals. However, given that the answer choices are closed intervals, perhaps in this problem, they are considering the intervals between two consecutive critical points, including the endpoints. Wait, but let me check the original problem again. It says "interval of monotonic increase". Monotonic increase usually allows for the derivative to be non-negative, but in the case of smooth functions like sine, the increase would switch to decrease exactly at the critical points. Therefore, the maximum interval on which the function is increasing would be the closed interval including the starting critical point (a minimum) and ending at the next critical point (a maximum). But in reality, at the critical point (maximum), the function stops increasing. So if we take the definition that increasing allows for non-decreasing, then including the endpoint where it reaches the maximum would be incorrect because after that point, it decreases. Therefore, the intervals should be closed on the left (including the minimum) and open on the right (excluding the maximum). However, since all options are closed intervals, maybe in this problem's context, they are using the term "monotonic increase" to include intervals where the function is non-decreasing, so including the points where the derivative is zero. But in that case, the function would be non-decreasing on [a, b] if it never decreases on that interval. But if the derivative is positive on (a, b) and zero at a and b, but between a and b, it's increasing, then including a and b would still be okay if beyond a and b, the function continues. Wait, this is getting confusing. Let me think with an example. Take the standard sine function. After shifting, our function is sin(2x - π/3). Let's consider its behavior. The derivative is 2cos(2x - π/3). The critical points are at 2x - π/3 = π/2 + πk, so x = (π/2 + πk + π/3)/2 = (5π/6 + πk)/2 = 5π/12 + πk/2. So between these critical points, the function is either increasing or decreasing. Since cosine is positive in intervals of length π, separated by π. Therefore, the increasing intervals should be between a minimum and a maximum. So starting at a minimum, increasing to a maximum, then decreasing. Therefore, the increasing interval should start at a minimum point (where derivative is zero) and end at the next maximum point (derivative zero again). But at the maximum point, the function stops increasing. So if you include both endpoints, the interval would be from minimum to maximum, which is half a period. The period of sin(2x - π/3) is π, so half-period is π/2. Wait, let's check. The period of sin(2x) is π, so sin(2x - π/3) also has period π. So half-period is π/2. So the distance between a minimum and a maximum is π/2. But according to our previous calculation, the interval where the derivative is positive is (-π/12 + πk, 5π/12 + πk). So the length of this interval is 5π/12 - (-π/12) = 6π/12 = π/2, which is half a period. So this interval is indeed between two critical points: starting at -π/12 + πk, which is a minimum, and ending at 5π/12 + πk, which is a maximum. Therefore, if we include both endpoints, even though the derivative is zero there, the function is non-decreasing on the closed interval because it rises from the minimum to the maximum without decreasing. So in that case, the answer should be closed intervals [kπ - π/12, kπ + 5π/12], which is option C. Therefore, I think option C is the correct answer. Let me check the options again to make sure.Option C: [kπ - π/12, kπ + 5π/12] for k ∈ ℤ. Yes, that matches our result. So the answer is C.Final Answerboxed{C}</think>To determine the interval of monotonic increase for the function ( g(x) ) obtained by shifting the graph of ( f(x) = sin 2x ) to the right by ( frac{pi}{6} ) units, we first identify the shifted function. The shifted function is ( g(x) = sin(2(x - frac{pi}{6})) = sin(2x - frac{pi}{3}) ).Next, we find the derivative of ( g(x) ):[g'(x) = 2cos(2x - frac{pi}{3})]The function ( g(x) ) is monotonically increasing when its derivative is positive:[2cos(2x - frac{pi}{3}) > 0 implies cos(2x - frac{pi}{3}) > 0]The cosine function is positive in the intervals ( (-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpi) ) for any integer ( k ). Setting up the inequality:[-frac{pi}{2} + 2kpi < 2x - frac{pi}{3} < frac{pi}{2} + 2kpi]Solving for ( x ):1. Add ( frac{pi}{3} ) to all parts:[-frac{pi}{2} + frac{pi}{3} + 2kpi < 2x < frac{pi}{2} + frac{pi}{3} + 2kpi]2. Simplify the constants:[-frac{pi}{6} + 2kpi < 2x < frac{5pi}{6} + 2kpi]3. Divide by 2:[-frac{pi}{12} + kpi < x < frac{5pi}{12} + kpi]Thus, the intervals of monotonic increase for ( g(x) ) are:[[kpi - frac{pi}{12}, kpi + frac{5pi}{12}] quad (k in mathbb{Z})]The correct answer is (boxed{C}).]

🔑:By shifting the graph of the function f(x) = sin 2x to the right by frac{pi}{6} units, we obtain the graph of the function g(x) = sin 2(x - frac{pi}{6}) = sin (2x - frac{pi}{3}). Let 2kpi - frac{pi}{2} leq 2x - frac{pi}{3} leq 2kpi + frac{pi}{2}, we find kpi - frac{pi}{12} leq x leq kpi + frac{5pi}{12}. Thus, the interval of increase for the function is left[kpi - frac{pi}{12}, kpi + frac{5pi}{12}right], where k in mathbb{Z}.Therefore, the correct answer is boxed{C}.This problem mainly examines the transformation rules of the graph of the function y = Asin(omega x + varphi) and the monotonicity of the sine function, which is a basic question.

🧠:[<think>Okay, so I need to find the value of N given this complicated expression. Let me write it down again to make sure I have it right:N = [sqrt(sqrt5 + 2) + sqrt(sqrt5 - 2)] / [sqrt(sqrt5 + 1)] - sqrt(3 - 2sqrt2)Hmm, okay. The problem gives me multiple-choice answers, so maybe simplifying each part step by step will help. Let me break it into two main parts: the first fraction and then subtracting that square root term. Let's start with the fraction:Numerator: sqrt(sqrt5 + 2) + sqrt(sqrt5 - 2)Denominator: sqrt(sqrt5 + 1)I wonder if I can simplify the numerator first. Both terms in the numerator have square roots with sqrt5 inside. Maybe if I let x = sqrt5, then the numerator becomes sqrt(x + 2) + sqrt(x - 2), and the denominator is sqrt(x + 1). Let's see if that substitution helps.So, x = sqrt5. Then, numerator is sqrt(x + 2) + sqrt(x - 2), denominator is sqrt(x + 1). Maybe squaring the numerator to see if it simplifies?Let me denote the numerator as A = sqrt(x + 2) + sqrt(x - 2). Then A squared would be:A^2 = (sqrt(x + 2) + sqrt(x - 2))^2 = (x + 2) + (x - 2) + 2*sqrt{(x + 2)(x - 2)}}Simplify the terms: x + 2 + x - 2 = 2x. Then the cross term is 2*sqrt{(x^2 - 4)}.So A^2 = 2x + 2*sqrt(x^2 - 4)Since x = sqrt5, x^2 = 5. Then x^2 - 4 = 1. So sqrt(x^2 - 4) = 1.Therefore, A^2 = 2*sqrt5 + 2*1 = 2sqrt5 + 2Therefore, A = sqrt(2sqrt5 + 2). Wait, but A is the original numerator: sqrt(x + 2) + sqrt(x - 2) which is sqrt(sqrt5 + 2) + sqrt(sqrt5 - 2). So we have A squared equals 2sqrt5 + 2. Hence, A is sqrt(2sqrt5 + 2).But then the denominator is sqrt(x + 1) = sqrt(sqrt5 + 1). So the entire fraction is sqrt(2sqrt5 + 2) / sqrt(sqrt5 + 1).Hmm, can I simplify this ratio? Let's see. Let's write it as sqrt(2sqrt5 + 2) divided by sqrt(sqrt5 + 1). Maybe combine the square roots:sqrt[(2sqrt5 + 2)/(sqrt5 + 1)]Let me compute the expression inside the square root:(2sqrt5 + 2)/(sqrt5 + 1). Maybe factor numerator and denominator.Numerator: 2sqrt5 + 2 = 2(sqrt5 + 1)Denominator: sqrt5 + 1So the ratio is 2(sqrt5 + 1)/(sqrt5 + 1) = 2. Therefore, sqrt[2] ?Wait, that seems too easy. So if I have sqrt[(2sqrt5 + 2)/(sqrt5 + 1)] = sqrt(2) because (2sqrt5 + 2) is 2*(sqrt5 + 1), so dividing by (sqrt5 + 1) gives 2, then sqrt(2). So the entire fraction simplifies to sqrt(2). That seems correct? Let me check.Original numerator squared: A^2 = 2sqrt5 + 2. Then denominator squared is (sqrt(sqrt5 + 1))^2 = sqrt5 + 1. Then (A / denominator)^2 = (2sqrt5 + 2)/(sqrt5 + 1) = 2(sqrt5 + 1)/(sqrt5 + 1) = 2. So (A / denominator) = sqrt(2). So yes, the fraction simplifies to sqrt(2). Okay, that's good. So the first part of N is sqrt(2).Now, the second part is minus sqrt(3 - 2sqrt2). Let me see if I can simplify sqrt(3 - 2sqrt2). This looks like a nested square root. Maybe express it as sqrt(a) - sqrt(b) or something like that.Suppose sqrt(3 - 2sqrt2) = sqrt(a) - sqrt(b). Then squaring both sides gives:3 - 2sqrt2 = a + b - 2sqrt(ab)Comparing the terms, the rational parts and the irrational parts:Rational parts: a + b = 3Irrational parts: -2sqrt(ab) = -2sqrt2Therefore, sqrt(ab) = sqrt2 => ab = 2So we have a + b = 3 and ab = 2. Let's solve for a and b.The solutions to t^2 - 3t + 2 = 0 are t = 1 and t = 2. So a = 2, b = 1 or vice versa.Therefore, sqrt(3 - 2sqrt2) = sqrt(2) - sqrt(1) = sqrt(2) - 1Check: (sqrt2 - 1)^2 = 2 - 2sqrt2 + 1 = 3 - 2sqrt2. Correct.So sqrt(3 - 2sqrt2) = sqrt2 - 1. Therefore, the second part is sqrt2 - 1.Therefore, N = sqrt2 - (sqrt2 - 1) = sqrt2 - sqrt2 + 1 = 1So N equals 1. Looking at the options, A is 1. So the answer should be A.Wait, let me double-check all steps to make sure.First part: sqrt(sqrt5 + 2) + sqrt(sqrt5 - 2) over sqrt(sqrt5 + 1). Squared the numerator, simplified, got 2sqrt5 + 2. Then divided by denominator squared (sqrt5 + 1), which gave 2. So sqrt(2). Then subtracted sqrt(3 - 2sqrt2) which is sqrt2 - 1. Therefore sqrt2 - (sqrt2 -1 ) = 1. All steps seem correct. So the answer is A) 1. So I think that's correct.Final Answerboxed{A}</think>To find the value of ( N ), we start by simplifying the given expression step by step.First, consider the fraction:[frac{sqrt{sqrt{5} + 2} + sqrt{sqrt{5} - 2}}{sqrt{sqrt{5} + 1}}]Let ( x = sqrt{5} ). The numerator becomes ( sqrt{x + 2} + sqrt{x - 2} ). Squaring the numerator:[(sqrt{x + 2} + sqrt{x - 2})^2 = (x + 2) + (x - 2) + 2sqrt{(x + 2)(x - 2)}]Simplifying, we get:[2x + 2sqrt{x^2 - 4}]Since ( x = sqrt{5} ), we have ( x^2 = 5 ), so:[2sqrt{5} + 2sqrt{1} = 2sqrt{5} + 2]Thus, the numerator squared is ( 2sqrt{5} + 2 ). The denominator squared is ( sqrt{sqrt{5} + 1} )^2 = ( sqrt{5} + 1 ). Therefore, the fraction simplifies to:[sqrt{frac{2sqrt{5} + 2}{sqrt{5} + 1}} = sqrt{frac{2(sqrt{5} + 1)}{sqrt{5} + 1}} = sqrt{2}]Next, simplify ( sqrt{3 - 2sqrt{2}} ). Assume ( sqrt{3 - 2sqrt{2}} = sqrt{a} - sqrt{b} ). Squaring both sides:[3 - 2sqrt{2} = a + b - 2sqrt{ab}]Matching rational and irrational parts, we get ( a + b = 3 ) and ( ab = 2 ). Solving, we find ( a = 2 ) and ( b = 1 ). Thus:[sqrt{3 - 2sqrt{2}} = sqrt{2} - 1]Combining the results, we have:[N = sqrt{2} - (sqrt{2} - 1) = 1]Thus, the value of ( N ) is (boxed{A}).,<think>Okay, let me try to figure out the value of N. The problem looks a bit complicated with all those nested square roots, but maybe if I break it down step by step, I can simplify each part. Let me write down the expression again to make sure I have it right:N = [√(√5 + 2) + √(√5 - 2)] / √(√5 + 1) - √(3 - 2√2)Hmm. So there are two main parts here: the first fraction involving √5 terms, and then subtracting another square root expression. Let me handle them one at a time.Starting with the first part: [√(√5 + 2) + √(√5 - 2)] / √(√5 + 1). This seems the more complex part. Maybe I can simplify the numerator first. Let me denote A = √(√5 + 2) + √(√5 - 2). Let me try to compute A squared to see if it simplifies.So, A² = [√(√5 + 2) + √(√5 - 2)]²= (√5 + 2) + (√5 - 2) + 2*√( (√5 + 2)(√5 - 2) )Simplify the terms:First, (√5 + 2) + (√5 - 2) = 2√5.Then, the cross term: 2*√[ (√5 + 2)(√5 - 2) ]Let me compute (√5 + 2)(√5 - 2). That's a difference of squares: (√5)² - (2)² = 5 - 4 = 1. So the cross term is 2*√1 = 2*1 = 2.Therefore, A² = 2√5 + 2 = 2(√5 + 1). Interesting. So A = √[2(√5 + 1)]. Wait, but A is the sum of two square roots, which is positive, so taking the square root gives a positive value.Therefore, A = √[2(√5 + 1)]. Now, the original expression is A divided by √(√5 + 1). So:[√(√5 + 2) + √(√5 - 2)] / √(√5 + 1) = √[2(√5 + 1)] / √(√5 + 1)Simplifying that, since √a / √b = √(a/b):= √[2(√5 + 1) / (√5 + 1)] = √2.Wait, really? Because 2(√5 +1) divided by (√5 +1) is 2. So √2. So the entire first part simplifies to √2. That seems surprising but let me check again.Yes, A² = 2(√5 +1), so A = √[2(√5 +1)]. Then A / √(√5 +1) = √[2(√5 +1)] / √(√5 +1) = √2. Because √(2(√5 +1)) / √(√5 +1) = √2 * √(√5 +1) / √(√5 +1) = √2. That makes sense. So the first part is √2. Okay, that's simpler than I thought.Now moving on to the second part: √(3 - 2√2). Let me see if I can simplify this. Remember that expressions of the form √(a - 2√b) can sometimes be written as √c - √d if c + d = a and cd = b. Let's try that.So suppose √(3 - 2√2) = √c - √d. Then, squaring both sides:3 - 2√2 = (√c - √d)² = c + d - 2√(cd)So, we have:c + d = 3and2√(cd) = 2√2 ⇒ √(cd) = √2 ⇒ cd = 2So, solving for c and d. We have c + d = 3 and cd = 2. This is a system of equations. Let me solve for c and d.The solutions are roots of the quadratic equation x² - 3x + 2 = 0. Factoring:x² - 3x + 2 = (x - 1)(x - 2) = 0 ⇒ x = 1 or x = 2.Therefore, c = 2 and d = 1, or vice versa. But since we have √c - √d, and c > d to keep the result positive. Let's check:If c = 2 and d = 1, then √2 - √1 = √2 - 1. Let's square that: (√2 - 1)² = 2 + 1 - 2√2 = 3 - 2√2. Perfect, that's exactly the expression under the square root. Therefore, √(3 - 2√2) = √2 - 1. Wait, but hold on: √2 - 1 is approximately 1.414 - 1 = 0.414. Let me check the original expression √(3 - 2√2). Compute 3 ≈ 3, 2√2 ≈ 2.828, so 3 - 2.828 ≈ 0.172. Then √0.172 ≈ 0.415. So that matches. So √(3 - 2√2) = √2 - 1. Therefore, the second part is √2 - 1.So putting it all together, N = first part - second part = √2 - (√2 - 1) = √2 - √2 + 1 = 1.Wait, that would mean N = 1. Which is option A. But let me double-check everything because sometimes when simplifying radicals, especially with nested roots, there can be steps where signs are tricky.First, check the first part:We had A = √(√5 + 2) + √(√5 - 2). Then A² = 2√5 + 2 + 2*1 = 2√5 + 2. Wait, no, wait. Wait, when I squared A, that was [√a + √b]² = a + b + 2√(ab). So in this case, a = √5 + 2, b = √5 - 2. Then ab = (√5 + 2)(√5 - 2) = 5 - 4 = 1, as before. So the cross term is 2*√1 = 2. Then A² = (√5 + 2) + (√5 - 2) + 2 = 2√5 + 0 + 2 = 2√5 + 2. Wait, that's different from what I thought before. Wait, wait, in my first calculation, I thought that (√5 + 2) + (√5 - 2) is 2√5, which is correct, and then the cross term is 2*√1=2. So A² = 2√5 + 2. Therefore, A² = 2(√5 + 1). Therefore, A = sqrt(2(√5 +1)), which is what I had earlier. Then, when I divided by sqrt(√5 +1), that becomes sqrt(2(√5 +1)) / sqrt(√5 +1) = sqrt(2). Correct, because sqrt(2)*sqrt(√5 +1)/sqrt(√5 +1) = sqrt(2). So that part is correct.Then the second term: √(3 - 2√2) = √2 - 1. Let's verify:(√2 -1)^2 = 2 +1 - 2√2 = 3 - 2√2. Yes, correct. So that's correct.Therefore, N = sqrt(2) - (sqrt(2) -1) = 1. So the answer is A) 1. But wait, let me check if there is a miscalculation here. Because when you subtract (sqrt(2) -1), it's sqrt(2) - sqrt(2) +1 =1. So that's correct. Therefore, N=1. So the answer should be A.But wait, the answer options include 1 as option A. Let me check again if the problem is written correctly. The original problem:N = [√(√5 +2) + √(√5 -2)] / √(√5 +1) - √(3 -2√2)So yes, according to the calculation, that's sqrt(2) - (sqrt(2)-1) =1. So answer A. But wait, let me check once again with another approach to be safe. Maybe I can compute numerically.Compute N numerically.First, compute each part.Compute sqrt5: approx 2.23607.First part numerator: sqrt(sqrt5 +2) + sqrt(sqrt5 -2)sqrt5 +2 ≈ 2.23607 + 2 ≈ 4.23607. sqrt(4.23607) ≈ 2.05817.sqrt5 -2 ≈ 2.23607 -2 ≈ 0.23607. sqrt(0.23607) ≈ 0.48587.Sum: 2.05817 + 0.48587 ≈ 2.54404.Denominator: sqrt(sqrt5 +1). sqrt5 +1 ≈ 3.23607. sqrt(3.23607) ≈ 1.799.So first part: 2.54404 / 1.799 ≈ approximately 1.414, which is sqrt(2) ≈ 1.4142. That checks out.Second part: sqrt(3 - 2*sqrt2). Compute sqrt2 ≈1.4142. So 2*sqrt2≈2.8284. 3 -2.8284≈0.1716. sqrt(0.1716)≈0.414. Which is sqrt2 -1≈1.4142 -1≈0.4142. So that checks out.Then N≈1.4142 -0.4142≈1. So N≈1. So the answer is A)1. So seems correct.But wait, let me check if there's an alternative way where maybe the first term is miscalculated. Let me think again. If someone thought that sqrt(a) + sqrt(b) over sqrt(c) is not sqrt(2), but maybe another value. But according to the algebra, it's correct. Squared, got 2(√5 +1), divided by (√5 +1) gives 2, sqrt(2). Then subtract sqrt(2)-1. So indeed 1. So seems correct.Alternatively, maybe the problem is written with brackets in a different way? Let me check:N = [√(√5 +2) + √(√5 -2)] / √(√5 +1) - √(3 -2√2)Yes, the division is only over the first fraction, and then subtract the second square root. So the order of operations is correct as I interpreted.Therefore, I think the answer is A)1.Final Answerboxed{A}</think>To find the value of ( N ), we start by simplifying the given expression step by step:[ N = frac{sqrt{sqrt{5} + 2} + sqrt{sqrt{5} - 2}}{sqrt{sqrt{5} + 1}} - sqrt{3 - 2sqrt{2}} ]First, consider the numerator of the fraction:Let ( A = sqrt{sqrt{5} + 2} + sqrt{sqrt{5} - 2} ). Squaring ( A ):[A^2 = left( sqrt{sqrt{5} + 2} + sqrt{sqrt{5} - 2} right)^2][= (sqrt{5} + 2) + (sqrt{5} - 2) + 2 sqrt{(sqrt{5} + 2)(sqrt{5} - 2)}][= 2sqrt{5} + 2 sqrt{1} quad text{(since ((sqrt{5} + 2)(sqrt{5} - 2) = 1))}][= 2sqrt{5} + 2][= 2(sqrt{5} + 1)]Thus, ( A = sqrt{2(sqrt{5} + 1)} ).Next, we divide ( A ) by ( sqrt{sqrt{5} + 1} ):[frac{sqrt{2(sqrt{5} + 1)}}{sqrt{sqrt{5} + 1}} = sqrt{frac{2(sqrt{5} + 1)}{sqrt{5} + 1}} = sqrt{2}]Now, consider the second term ( sqrt{3 - 2sqrt{2}} ). We simplify it by assuming it can be written as ( sqrt{c} - sqrt{d} ):[sqrt{3 - 2sqrt{2}} = sqrt{c} - sqrt{d}][3 - 2sqrt{2} = (sqrt{c} - sqrt{d})^2 = c + d - 2sqrt{cd}]Solving for ( c ) and ( d ):[c + d = 3 quad text{and} quad 2sqrt{cd} = 2sqrt{2} implies sqrt{cd} = sqrt{2} implies cd = 2]The roots of the quadratic equation ( x^2 - 3x + 2 = 0 ) are ( x = 1 ) and ( x = 2 ). Thus, ( c = 2 ) and ( d = 1 ):[sqrt{3 - 2sqrt{2}} = sqrt{2} - 1]Combining both parts:[N = sqrt{2} - (sqrt{2} - 1) = sqrt{2} - sqrt{2} + 1 = 1]Thus, the value of ( N ) is (boxed{A}).]

🔑:Let x=frac{sqrt{sqrt{5}+2}+sqrt{sqrt{5}-2}}{sqrt{sqrt{5}+1}} and y=sqrt{3-2sqrt{2}}.Note thatbegin{align*} x^2&=frac{left(sqrt{sqrt{5}+2}+sqrt{sqrt{5}-2}right)^2}{left(sqrt{sqrt{5}+1}right)^2} &=frac{left(sqrt{5}+2right)+2cdotleft(sqrt{sqrt{5}+2}cdotsqrt{sqrt{5}-2}right)+left(sqrt{5}-2right)}{sqrt{5}+1} &=frac{left(sqrt{5}+2right)+2+left(sqrt{5}-2right)}{sqrt{5}+1} &=frac{2sqrt{5}+2}{sqrt{5}+1} &=2. end{align*}Since x>0, we have x=sqrt{2}.On the other hand, note thatbegin{align*} y^2&=3-2sqrt{2} &=2-2sqrt{2}+1 &=left(sqrt{2}-1right)^2. end{align*}Since y>0, we have y=sqrt{2}-1.Finally, the answer is [N=x-y=boxed{textbf{(A) }1}.]~Someonenumber011 (Fundamental Logic)~MRENTHUSIASM (Reconstruction)

🧠:[<think>Okay, so I need to solve the equation (2^{x} + 2^{y} = 2^{x y - 1}) where (x) and (y) are integers. Hmm, let's see. First, I should probably consider the properties of exponents here. Since all terms are powers of 2, maybe I can express them in terms of the same base and compare exponents. But the equation has two variables, so it might be tricky. Let me start by checking small integer values for (x) and (y) to see if any solutions pop up.First, let's try some simple cases. If (x = y), then the equation becomes (2 times 2^{x} = 2^{x^2 - 1}). Simplifying, that's (2^{x + 1} = 2^{x^2 - 1}), so (x + 1 = x^2 - 1). Rewriting, (x^2 - x - 2 = 0). Factoring, ((x - 2)(x + 1) = 0), so (x = 2) or (x = -1). Therefore, if (x = y), the solutions are (2, 2) and (-1, -1). Let me check these.For (2, 2): Left-hand side (LHS) is (2^2 + 2^2 = 4 + 4 = 8). Right-hand side (RHS) is (2^{2 times 2 - 1} = 2^{3} = 8). That works. For (-1, -1): LHS is (2^{-1} + 2^{-1} = frac{1}{2} + frac{1}{2} = 1). RHS is (2^{(-1)(-1) - 1} = 2^{1 - 1} = 2^0 = 1). That also works. So (2,2) and (-1,-1) are solutions.But maybe there are other solutions where (x neq y). Let's consider that. Without loss of generality, assume (x geq y). Then, we can write (2^{y}(2^{x - y} + 1) = 2^{x y - 1}). Since (2^{x - y} + 1) is an odd integer (since (2^{x - y}) is even when (x > y), so even +1 is odd), the left-hand side has (2^{y}) times an odd number. The right-hand side is (2^{x y - 1}), which is a power of 2. Therefore, the odd factor on the left must be 1. Otherwise, the left-hand side would have an odd factor greater than 1, which the right-hand side doesn't. Therefore, we must have (2^{x - y} + 1 = 1). Solving this, (2^{x - y} = 0), but that's impossible since 2 to any integer power is positive. Wait, but if (2^{x - y} + 1 = 1), then (2^{x - y} = 0), which isn't possible. Hmm, that suggests that if (x neq y), then there are no solutions? But that can't be right. Wait, maybe my assumption that (x geq y) was wrong, but even if I assume (x < y), the same logic would apply. Let me think again.Wait, maybe the problem arises because if (x neq y), then the left-hand side has a factor that is an odd number greater than 1, which cannot be part of a power of 2. Therefore, unless the odd factor is 1, which would require (2^{|x - y|} + 1 = 1), which is impossible as above, there's no solution. So perhaps the only solutions are when (x = y). Therefore, the solutions are (2,2) and (-1,-1). But let's check some specific cases where (x neq y) just to be thorough.Suppose (x = 0). Then the equation becomes (2^{0} + 2^{y} = 2^{0 times y - 1}), which is (1 + 2^y = 2^{-1}). But (2^{-1} = 0.5), and (1 + 2^y = 0.5) implies (2^y = -0.5), which is impossible since (2^y) is always positive. Similarly, if (y = 0), same result.What if (x = 1)? Then the equation is (2^1 + 2^y = 2^{y - 1}). That is, (2 + 2^y = 2^{y - 1}). Let me rearrange this: (2^{y - 1} - 2^y = -2). Factor out (2^{y - 1}): (2^{y - 1}(1 - 2) = -2). So (-2^{y - 1} = -2), which simplifies to (2^{y - 1} = 2), so (y - 1 = 1), hence (y = 2). Let's check: (x = 1), (y = 2). LHS: (2^1 + 2^2 = 2 + 4 = 6). RHS: (2^{(1)(2) - 1} = 2^{1} = 2). 6 ≠ 2, so that doesn't work. Hmm, contradiction here. So even though we derived (y = 2), plugging back in gives a wrong result. Therefore, maybe the assumption that such a solution exists is wrong.Wait, let me check my steps again. If (x = 1), then equation is (2 + 2^y = 2^{y - 1}). Let's test (y = 1): LHS = 2 + 2 = 4, RHS = 2^{0} = 1. Nope. (y = 0): LHS = 2 + 1 = 3, RHS = 2^{-1} = 0.5. No. (y = 2): 2 + 4 = 6 vs 2^{1} = 2. Not equal. (y = 3): 2 + 8 = 10 vs 2^{2} = 4. Not equal. So no solution here.What about (x = 3)? Let's try (x = 3). Then the equation is (8 + 2^y = 2^{3y - 1}). Let's see if there's a possible (y). Let's try small (y). (y = 1): 8 + 2 = 10 vs (2^{2} = 4). No. (y = 2): 8 + 4 = 12 vs (2^{5} = 32). No. (y = 3): 8 + 8 = 16 vs (2^{8} = 256). No. (y = 0): 8 + 1 = 9 vs 0.5. No. Negative (y): (y = -1): 8 + 0.5 = 8.5 vs (2^{-4} = 0.0625). No. So, no solution here either.How about (x = -1)? Let's set (x = -1), then the equation becomes (2^{-1} + 2^y = 2^{-y - 1}). So (0.5 + 2^y = 2^{-y - 1}). Let's test (y = -1): 0.5 + 0.5 = 1 vs (2^{0} = 1). That works, but that's the case (x = y = -1) which we already have. What about (y = 0): 0.5 + 1 = 1.5 vs (2^{-1} = 0.5). No. (y = 1): 0.5 + 2 = 2.5 vs (2^{-2} = 0.25). No. (y = -2): 0.5 + 0.25 = 0.75 vs (2^{1} = 2). No. So no other solutions here.Wait, maybe trying (x) and (y) with different signs? Let's try (x = 2), (y = -1). Then LHS: (4 + 0.5 = 4.5). RHS: (2^{(2)(-1) -1} = 2^{-3} = 0.125). Not equal. How about (x = 1), (y = -1): LHS: 2 + 0.5 = 2.5. RHS: (2^{-1 -1} = 2^{-2} = 0.25). No. Maybe (x = 3), (y = -1): 8 + 0.5 = 8.5 vs (2^{-3 -1} = 2^{-4} = 0.0625). Nope. Doesn't seem like mixed signs are working.Another approach: Let's consider the equation (2^x + 2^y = 2^{xy - 1}). Without loss of generality, assume (x leq y). Let me set (x leq y), then we can write (2^x(1 + 2^{y - x}) = 2^{xy - 1}). So similar to before, (1 + 2^{y - x}) must be a power of two, because the left-hand side is (2^x) times that term, and the right-hand side is a power of two. So (1 + 2^{y - x}) must be equal to (2^k) for some integer (k). Then, the equation becomes (2^x times 2^k = 2^{xy - 1}), so (x + k = xy - 1). Therefore, (k = xy - x - 1).But since (1 + 2^{y - x} = 2^k), then (2^{y - x} = 2^k - 1). So (2^k - 1) must be a power of two. But when is (2^k - 1) a power of two? Let's see. If (k = 1), then (2^1 - 1 = 1 = 2^0). If (k = 2), (2^2 - 1 = 3), not a power of two. (k = 3): 8 - 1 = 7. Not a power. (k = 0): 1 - 1 = 0, which isn't a power of two (unless we consider 0, but 2^m is always positive). So the only possibility here is (k = 1), which gives (2^{y - x} = 1), so (y - x = 0), hence (y = x). Therefore, this case brings us back to (x = y). Hence, the only solutions would be when (x = y), which we already solved. Therefore, the only solutions are (x = y = 2) and (x = y = -1).Wait, but let me confirm this logic. If (1 + 2^{y - x} = 2^k), then (2^{y - x} = 2^k - 1). So (2^k - 1) must be a power of two. The only time this is true is when (k = 1), since (2^1 - 1 = 1 = 2^0). For any other (k), (2^k - 1) is not a power of two. For example, (k = 2): 3, which is not a power of two. (k = 3): 7, same. (k = 0): 0, which isn't a power. Therefore, only (k = 1) works, leading to (y - x = 0). So yes, that seems to confirm that (x = y) is necessary.Therefore, the only solutions are when (x = y), leading to the quadratic equation we solved earlier, giving (x = 2) or (x = -1). Therefore, the solutions are (2,2) and (-1,-1). Let me check another case where maybe (x) or (y) is zero. Wait, earlier when I tried (x = 0), it didn't work. Let me check (x = 0), (y) arbitrary: (2^0 + 2^y = 2^{-1}). So 1 + 2^y = 0.5, which is impossible because 2^y is positive. Similarly for (y = 0). How about negative exponents? Let's suppose (x = -2). Then, equation becomes (2^{-2} + 2^y = 2^{-2y - 1}). So (0.25 + 2^y = 2^{-2y -1}). Let's try (y = -1): 0.25 + 0.5 = 0.75 vs 2^{2 - 1} = 2. Not equal. (y = -2): 0.25 + 0.25 = 0.5 vs 2^{4 -1} = 8. No. (y = 0): 0.25 + 1 = 1.25 vs 2^{-1} = 0.5. No. Not working.Alternatively, maybe one of the variables is 1 and the other is different. Let's take (x = 1), (y = 3). Then LHS: 2 + 8 = 10. RHS: (2^{3 -1} = 4). Not equal. (x = 1), (y = 2): 2 + 4 = 6 vs (2^{2 -1} = 2). No. Maybe (x = 3), (y = 2): 8 + 4 = 12 vs (2^{6 -1} = 32). No.Alternatively, maybe one variable is positive and the other is negative. Let's try (x = 2), (y = -1). Then LHS: 4 + 0.5 = 4.5 vs RHS: (2^{-2 -1} = 2^{-3} = 0.125). Not equal. (x = -1), (y = 2): 0.5 + 4 = 4.5 vs (2^{-2 -1} = 0.125). Still no.Wait, perhaps we can consider the equation in terms of symmetry. If we swap (x) and (y), the equation remains the same. So any solution ((x, y)) implies the solution ((y, x)). But in our earlier analysis, we found that unless (x = y), there are no solutions. So all solutions must be symmetric. Hence, the only solutions are indeed the ones where (x = y), which are (2,2) and (-1,-1).Another angle: Let's suppose (x) and (y) are both non-negative integers. Then, we can analyze the growth of both sides. For (x, y geq 0), the left-hand side (2^x + 2^y) grows exponentially in (x) and (y), but the right-hand side (2^{xy - 1}) grows even faster because it's a double exponential in terms of (x) and (y). Therefore, except for small values of (x) and (y), the RHS will be much larger. So we can check small non-negative integers.Testing (x = 0): as before, no solution. (x = 1): tried earlier. (x = 2): (y = 2) works. (x = 3): RHS would be (2^{3y -1}), which for (y = 3) is 2^8 = 256, while LHS is 8 + 8 = 16. So way smaller. Similarly, higher values will have RHS growing much faster.For negative integers: When (x) and (y) are negative, (2^{x}) and (2^{y}) become fractions, and the RHS is (2^{xy -1}). Let's see. For example, take (x = y = -2). Then LHS: (2^{-2} + 2^{-2} = 0.25 + 0.25 = 0.5). RHS: (2^{(-2)(-2) -1} = 2^{4 -1} = 8). Not equal. (x = y = -3): LHS: 0.125 + 0.125 = 0.25. RHS: 2^{9 -1} = 256. Still no. So only (x = y = -1) works there.Wait, another thought: suppose one of (x) or (y) is 1. Let me see. Let’s set (x = 1). Then the equation becomes (2 + 2^y = 2^{y - 1}). Rearranging, (2^{y - 1} = 2 + 2^y). Multiply both sides by 2: (2^y = 4 + 2^{y + 1}). Subtract (2^y) from both sides: 0 = 4 + 2^{y +1} - 2^y = 4 + 2^y(2 - 1) = 4 + 2^y. So (2^y = -4), which is impossible. Therefore, no solution when (x = 1). Similarly for (y = 1).Alternatively, maybe (x = 2), (y = 3). Then LHS: 4 + 8 = 12. RHS: (2^{6 -1} = 32). Nope. (x = 3), (y = 3): 8 + 8 = 16 vs 512. No.Alternatively, what if one of the variables is 0? But (2^0 + 2^y = 2^{-1}), so 1 + 2^y = 0.5, which is impossible as 2^y is positive.What about (x = 1), (y = -1): LHS: 2 + 0.5 = 2.5. RHS: (2^{-1 -1} = 0.25). Not equal.Wait, maybe there's a solution where one variable is 3 and the other is 1? Let me check: (x = 3), (y = 1). Then LHS: 8 + 2 = 10. RHS: (2^{3 -1} = 4). No.Alternatively, maybe (x = -1), (y = 1). LHS: 0.5 + 2 = 2.5. RHS: (2^{-1 -1} = 0.25). No.Another approach: Let’s take logarithms. Taking log base 2 of both sides:(log_2(2^x + 2^y) = xy - 1).But log of a sum is tricky. However, if we let (x geq y), then (2^x + 2^y = 2^y(1 + 2^{x - y})). So,(log_2(2^y(1 + 2^{x - y})) = xy - 1)Which simplifies to:(y + log_2(1 + 2^{x - y}) = xy - 1).But since (1 + 2^{x - y}) is an integer, and a power of two only when (x = y) as discussed earlier, leading to (x = y) and the quadratic equation. Hence, this again leads to the conclusion that (x = y) is necessary.Alternatively, if we suppose that one of (x) or (y) is much larger than the other. For example, suppose (x) is much larger than (y). Then (2^x) dominates (2^y), so (2^x approx 2^{xy -1}), which implies (x approx xy -1). Solving for (x), (x(1 - y) approx -1). So unless (y = 0) or (y = 1), but (y = 0) gives (x approx -1), which conflicts with (x) being much larger. (y = 1): Then equation becomes (x(1 - 1) = -1), which is 0 = -1. Contradiction. So no solution in this case.Similarly, if (y) is much larger than (x), same reasoning applies.Therefore, considering all these cases, the only solutions seem to be when (x = y), leading to (x = 2) or (x = -1), so the integer solutions are ((2, 2)) and ((-1, -1)).But just to be absolutely thorough, let's check if there might be other solutions where (x neq y). Suppose (x = 3), (y = 2). Then LHS: 8 + 4 = 12. RHS: (2^{6 -1} = 32). Not equal. (x = 2), (y = 3): same. (x = 4), (y = 1): 16 + 2 = 18 vs (2^{4 -1} = 8). No. (x = 1), (y = 3): 2 + 8 = 10 vs (2^{3 -1} = 4). Nope. (x = -1), (y = 0): 0.5 + 1 = 1.5 vs (2^{0 -1} = 0.5). No. (x = 0), (y = -1): same as above. (x = -2), (y = -1): 0.25 + 0.5 = 0.75 vs (2^{2 -1} = 2). No. (x = -1), (y = -2): same as above.Hmm. Maybe there's a case where one of (x) or (y) is 1 and the other is negative. Let's try (x = 1), (y = -2). Then LHS: 2 + 0.25 = 2.25. RHS: (2^{(1)(-2) -1} = 2^{-3} = 0.125). No. (x = -2), (y = 1): same. (x = 1), (y = -3): 2 + 0.125 = 2.125 vs (2^{-4} = 0.0625). No.Alternatively, maybe (x = 2), (y = 1). LHS: 4 + 2 = 6. RHS: (2^{2 -1} = 2). Not equal. (x = 1), (y = 2): same. Not equal.What if (x = 4), (y = 2). LHS: 16 + 4 = 20. RHS: (2^{8 -1} = 128). No.Alternatively, (x = 3), (y = 3): 8 + 8 = 16 vs (2^{9 -1} = 256). No.I think I've tried a lot of different combinations, and the only ones that work are when (x = y = 2) and (x = y = -1). Therefore, the solutions must be these pairs. To ensure completeness, let me attempt to formalize this.From the equation (2^x + 2^y = 2^{xy - 1}), assume (x leq y). Then, as before, write LHS as (2^x(1 + 2^{y - x})). Since RHS is a power of two, (1 + 2^{y - x}) must also be a power of two. Let (1 + 2^{d} = 2^k), where (d = y - x geq 0). Then, (2^d = 2^k - 1). The only time this is possible is when (k = 1) and (d = 0), because for (k geq 2), (2^k - 1) is odd and greater than 1, which cannot be a power of two. If (k = 1), then (2^d = 1), so (d = 0), meaning (y = x). Therefore, the only solutions occur when (x = y). Then, substituting back, we get the quadratic equation leading to (x = 2) or (x = -1). Therefore, the solutions are ((2, 2)) and ((-1, -1)).Thus, after thorough checking and considering various cases, these are the only integer solutions.Final AnswerThe integer solutions are boxed{(2, 2)} and boxed{(-1, -1)}.</think>To solve the equation (2^x + 2^y = 2^{xy - 1}) in integers, we start by considering the case where (x = y). Substituting (x = y) into the equation, we get:[2^x + 2^x = 2^{x^2 - 1}]Simplifying, this becomes:[2 cdot 2^x = 2^{x^2 - 1} implies 2^{x + 1} = 2^{x^2 - 1}]Equating the exponents, we get the quadratic equation:[x + 1 = x^2 - 1 implies x^2 - x - 2 = 0]Factoring, we find:[(x - 2)(x + 1) = 0 implies x = 2 text{ or } x = -1]Thus, the solutions when (x = y) are ((2, 2)) and ((-1, -1)). Checking these solutions:- For ((2, 2)), both sides of the equation are (8).- For ((-1, -1)), both sides of the equation are (1).Next, we consider the case where (x neq y). Without loss of generality, assume (x geq y). The left-hand side can be written as (2^y(2^{x - y} + 1)). Since the right-hand side is a power of 2, the term (2^{x - y} + 1) must also be a power of 2. However, (2^{x - y} + 1) is odd and greater than 1 unless (x = y). This implies no solutions exist for (x neq y).Testing other specific values for (x) and (y) (including negative values and mixed signs) confirms that no additional solutions exist. Therefore, the only solutions are when (x = y). Final AnswerThe integer solutions are boxed{(2, 2)} and boxed{(-1, -1)}.,<think>Okay, so I need to solve the equation (2^{x} + 2^{y} = 2^{x y - 1}) where (x) and (y) are integers. Hmm, let's see. First, since the equation involves exponents with base 2, maybe I can use properties of exponents or logarithms to simplify it. But since we're dealing with integers, perhaps there are only a few cases to check. Let me start by thinking about possible values for (x) and (y).First, note that (2^{x}) and (2^{y}) are both positive, no matter what integer values (x) and (y) take. However, if (x) or (y) is negative, then (2^{x}) or (2^{y}) would be fractions. Since the right-hand side is (2^{xy - 1}), which is also positive, but depending on the exponent, it could be a large number or a fraction. So, maybe some cases where (x) and (y) are non-negative integers, and other cases where one or both are negative.But dealing with negative exponents might complicate things because we’d have fractions. Let me first consider the case where both (x) and (y) are non-negative integers. That might be simpler.Also, the equation is symmetric in (x) and (y), so any solution ((x, y)) would imply that ((y, x)) is also a solution. That might help reduce the number of cases I need to check.Let me try small non-negative integer values for (x) and (y). Let's start with (x = y). If (x = y), then the equation becomes (2 cdot 2^{x} = 2^{x^2 - 1}). Simplifying the left side: (2^{x + 1} = 2^{x^2 - 1}). Since the bases are equal, the exponents must be equal: (x + 1 = x^2 - 1). Rearranging: (x^2 - x - 2 = 0). Factoring: ((x - 2)(x + 1) = 0). So, solutions (x = 2) or (x = -1). But we're considering non-negative integers first. So (x = 2), which gives (y = 2). Let's check this: Left side: (2^2 + 2^2 = 4 + 4 = 8). Right side: (2^{2 cdot 2 - 1} = 2^{3} = 8). It works! So ((2, 2)) is a solution. The other root (x = -1) would give (y = -1). Let me check that even though they are negative. Left side: (2^{-1} + 2^{-1} = frac{1}{2} + frac{1}{2} = 1). Right side: (2^{(-1)(-1) - 1} = 2^{1 - 1} = 2^{0} = 1). So ((-1, -1)) is also a solution. Interesting. So when (x = y), we have two solutions: ((2, 2)) and ((-1, -1)).But are there solutions where (x neq y)? Let's check. Suppose (x > y). Then, by symmetry, if there is a solution, swapping (x) and (y) will also be a solution. So maybe I can assume (x geq y) without loss of generality and find solutions, then the others would be their swaps.First, let's check if either (x) or (y) is 0. Suppose (x = 0). Then the equation becomes (2^{0} + 2^{y} = 2^{0 cdot y - 1}), which simplifies to (1 + 2^{y} = 2^{-1}). But the right side is (0.5), and the left side is at least 1 (since (2^{y} geq 1) if (y geq 0), and even larger if (y) is positive). If (y) is negative, then (2^{y}) is a fraction, but (1 + 2^{y} > 1), which is still more than 0.5. So no solution when (x = 0). Similarly, if (y = 0), same reasoning. So (x) and (y) can't be 0.Next, check (x = 1). Let's set (x = 1). Then the equation becomes (2^{1} + 2^{y} = 2^{y - 1}). That is, (2 + 2^{y} = 2^{y - 1}). Let's rearrange: (2^{y - 1} - 2^{y} = 2). Factor out (2^{y - 1}): (2^{y - 1}(1 - 2) = 2), so (-2^{y - 1} = 2), which implies (2^{y - 1} = -2). But exponentials are always positive, so no solution here. So (x = 1) gives no solution. Similarly, (y = 1) would be the same.Now, check (x = 2). Let (x = 2). Then the equation becomes (2^{2} + 2^{y} = 2^{2y - 1}), which is (4 + 2^{y} = 2^{2y - 1}). Let me write this as (2^{2y - 1} - 2^{y} = 4). Maybe factor out (2^{y}): (2^{y}(2^{y - 1} - 1) = 4). Let's let (k = y). Then equation: (2^{k}(2^{k - 1} - 1) = 4). Let's test small integer values for (k):If (k = 2): Left side: (4*(2^{1} - 1) = 4*(2 - 1) = 4*1 = 4). So this works. So (y = 2). So when (x = 2), (y = 2) is a solution, which we already found.If (k = 3): Left side: (8*(2^{2} - 1) = 8*(4 - 1) = 8*3 = 24 ≠ 4).If (k = 1): Left side: (2*(2^{0} - 1) = 2*(1 - 1) = 0 ≠ 4).If (k = 0): Left side: (1*(2^{-1} - 1) = 1*(0.5 - 1) = -0.5 ≠ 4).If (k = -1): Left side: (0.5*(2^{-2} - 1) = 0.5*(0.25 - 1) = 0.5*(-0.75) = -0.375 ≠ 4).So the only solution here is (k = 2). So (x = 2), (y = 2).Alternatively, maybe for (x = 3), let's try. (x = 3): (2^{3} + 2^{y} = 2^{3y - 1}), so (8 + 2^{y} = 2^{3y - 1}). Let's rearrange: (2^{3y - 1} - 2^{y} = 8). Let me factor (2^{y}): (2^{y}(2^{2y - 1} - 1) = 8). Let's try small integers (y):If (y = 2): Left side: (4*(2^{3 - 1} - 1) = 4*(4 - 1) = 12 ≠ 8).If (y = 1): Left side: (2*(2^{1 - 1} - 1) = 2*(1 - 1) = 0 ≠ 8).If (y = 3): Left side: (8*(2^{5 - 1} - 1) = 8*(16 - 1) = 8*15 = 120 ≠ 8).If (y = 0): Left side: (1*(2^{-1 -1} -1 ) = 1*(2^{-2} -1) = 1*(0.25 - 1) = -0.75 ≠ 8).Negative (y): (y = -1): Left side: (0.5*(2^{-2 -1} -1 ) = 0.5*(2^{-3} -1 ) = 0.5*(1/8 - 1) = 0.5*(-7/8) = -7/16 ≠ 8).So seems like no solution for (x = 3). Similarly, trying (x = 4) might get even larger terms on RHS.Alternatively, maybe try (x = -1). Let's see. If (x = -1), then the equation is (2^{-1} + 2^{y} = 2^{(-1)y - 1}). Simplify left side: (0.5 + 2^{y}). Right side: (2^{-y - 1}). So (0.5 + 2^{y} = 2^{- y - 1}). Let's rearrange: (2^{y} + 0.5 - 2^{- y - 1} = 0). Multiply both sides by (2^{y + 1}) to eliminate denominators: (2^{2y + 1} + 2^{y} - 1 = 0). Let me denote (z = 2^{y}), then the equation becomes (2 z^{2} + z - 1 = 0). Solving quadratic: (z = [-1 pm sqrt{1 + 8}]/4 = [-1 pm 3]/4). So solutions: (z = (-1 + 3)/4 = 0.5) or (z = (-1 - 3)/4 = -1). Since (z = 2^{y} > 0), discard negative solution. So (z = 0.5 = 2^{-1}), so (2^{y} = 2^{-1}) implies (y = -1). Therefore, when (x = -1), (y = -1), which is the solution we already found. So nothing new here.How about (x = -2)? Let's try (x = -2): Left side: (2^{-2} + 2^{y} = 0.25 + 2^{y}). Right side: (2^{(-2)y - 1} = 2^{-2y - 1}). So equation: (0.25 + 2^{y} = 2^{-2y - 1}). Multiply both sides by (2^{2y + 1}): (0.25 cdot 2^{2y + 1} + 2^{y} cdot 2^{2y + 1} = 1). Simplify:(0.25 cdot 2^{2y + 1} = 2^{2y -1}), and (2^{y} cdot 2^{2y + 1} = 2^{3y + 1}). So equation becomes (2^{2y -1} + 2^{3y + 1} = 1). Let me denote (w = 2^{y}). Then (2^{2y} = w^2), (2^{3y} = w^3). Then:Equation: ( (w^2)/2 + 2 w^3 = 1 ).Multiply both sides by 2: ( w^2 + 4 w^3 = 2 ).Rearranged: (4 w^3 + w^2 - 2 = 0).This is a cubic equation. Let's check possible integer solutions. If (w) is a power of 2 (since (w = 2^{y}), (y) integer). Let's test (w = 1): (4 + 1 - 2 = 3 ≠ 0). (w = 2): (32 + 4 - 2 = 34 ≠ 0). (w = 1/2): (4*(1/8) + 1/4 - 2 = 0.5 + 0.25 - 2 = -1.25 ≠ 0). (w = 1/4): (4*(1/64) + 1/16 - 2 ≈ 0.0625 + 0.0625 - 2 ≈ -1.875 ≠ 0). Negative (w) not possible. So no solution here. Therefore, (x = -2) gives no solution.Hmm, maybe trying some other approach. Let's consider the original equation: (2^{x} + 2^{y} = 2^{xy - 1}). Let me divide both sides by (2^{xy - 1}) to get (2^{x - (xy - 1)} + 2^{y - (xy - 1)} = 1). Let me denote exponents:First term exponent: (x - xy + 1 = x(1 - y) + 1)Second term exponent: (y - xy + 1 = y(1 - x) + 1)So equation becomes: (2^{x(1 - y) + 1} + 2^{y(1 - x) + 1} = 1). Hmm, not sure if this helps. Alternatively, perhaps factor the left side. But the left side is a sum of two terms, which might be hard to factor. Let me think.Alternatively, set (a = 2^{x}) and (b = 2^{y}). Then the equation becomes (a + b = 2^{(x y) - 1}). But (a = 2^{x}), so (x = log_2 a), similarly (y = log_2 b). Then substituting into the exponent: (a + b = 2^{(log_2 a)(log_2 b) - 1}). Not sure if that helps. Maybe take logarithms? But since variables are in exponents, that might complicate things.Alternatively, think about cases where one term on the left is much larger than the other. For example, suppose (2^{x}) is much larger than (2^{y}). Then (2^{x} approx 2^{xy - 1}), so (x approx xy - 1), which would imply (x(1 - y) approx -1). Since (x) and (y) are integers, perhaps (x(1 - y) = -1). So possibilities: (x = 1), (1 - y = -1) → (y = 2); or (x = -1), (1 - y = 1) → (y = 0). Let's check these possibilities.First, (x = 1), (y = 2). Plug into original equation: (2^1 + 2^2 = 2 + 4 = 6). Right side: (2^{1*2 - 1} = 2^{1} = 2). Not equal. So no.Second, (x = -1), (y = 0). Left side: (2^{-1} + 2^{0} = 0.5 + 1 = 1.5). Right side: (2^{(-1)(0) - 1} = 2^{-1} = 0.5). Not equal. So this approach may not work.Alternatively, maybe both terms on the left are equal. Then (2^{x} = 2^{y}) ⇒ (x = y), which is the case we already considered, leading to solutions ((2,2)) and ((-1, -1)).Alternatively, consider when the right-hand side is a power of two, and the left side is a sum of two powers of two. So 2^{xy -1} is a power of two, and 2^{x} + 2^{y} is a sum of two powers of two. So the sum of two powers of two must equal another power of two. When does that happen?In general, the sum of two distinct powers of two is another power of two only if one of them is 1 and the other is 1, but 1 + 1 = 2, which is 2^1. Wait, more precisely:If the sum of two powers of two is a power of two, then they must be equal. Wait, no. For example, 2^0 + 2^0 = 2^1. So if two equal terms: 2^k + 2^k = 2^{k+1}. But if they are different exponents, then the sum is not a power of two. For example, 2^1 + 2^0 = 3, which is not a power of two. Similarly, 2^2 + 2^1 = 6, not a power of two. So the only way the sum of two powers of two is a power of two is when they are equal, i.e., 2^{k} + 2^{k} = 2^{k+1}. Therefore, in our equation, (2^{x} + 2^{y}) is a power of two only if (x = y), which leads to (2^{x+1} = 2^{xy -1}), which gives (x + 1 = xy -1). That equation simplifies to (xy - x - 2 = 0), which factors as (x(y - 1) = 2). So integer solutions for (x) and (y):Possible pairs for (x(y - 1) = 2):x and (y - 1) are integer divisors of 2. So possible divisors: (1, 2), (2, 1), (-1, -2), (-2, -1).Thus:1. (x = 1), (y - 1 = 2) ⇒ (y = 3). Check this solution: Left side (2^1 + 2^3 = 2 + 8 = 10). Right side (2^{1*3 -1} = 2^{2} = 4). Not equal. Doesn't work.2. (x = 2), (y -1 = 1) ⇒ (y = 2). Check: (2^2 + 2^2 = 8), right side (2^{4 -1} = 8). Works. So that's the (2,2) solution.3. (x = -1), (y -1 = -2) ⇒ (y = -1). Check: (2^{-1} + 2^{-1} = 1), right side (2^{(-1)(-1) -1} = 2^{0} = 1). Works. That's the (-1,-1) solution.4. (x = -2), (y -1 = -1) ⇒ (y = 0). Check: (2^{-2} + 2^{0} = 0.25 + 1 = 1.25), right side (2^{(-2)(0) -1} = 2^{-1} = 0.5). Doesn't work.So only the cases where (x = 2, y = 2) and (x = -1, y = -1) work. Therefore, the only solutions when (x = y) are these.But we need to check if there are solutions when (x ≠ y). So in cases where (2^{x} + 2^{y} = 2^{xy -1}) with (x ≠ y). Let's consider one of them being greater than the other. Let's assume (x > y). Then (2^{x} + 2^{y} = 2^{xy -1}). Since (x > y), (2^{x}) is larger than (2^{y}). So the left side is approximately (2^{x}), so we can approximate (2^{x} ≈ 2^{xy -1}), leading to (x ≈ xy -1). So (x(1 - y) ≈ -1). Since (x) and (y) are integers, this suggests that (x(1 - y) = -1). Therefore, possible values for (x) and (y) such that (x) divides -1. So possible (x = 1) or (x = -1).Case 1: (x = 1). Then (1(1 - y) = -1) ⇒ (1 - y = -1) ⇒ (y = 2). So check (x = 1), (y = 2). Left side: (2 + 4 = 6), right side: (2^{1*2 -1} = 2^{1} = 2). Not equal. So no solution.Case 2: (x = -1). Then (-1(1 - y) = -1) ⇒ (1 - y = 1) ⇒ (y = 0). Check (x = -1), (y = 0). Left side: (0.5 + 1 = 1.5), right side: (2^{0 -1} = 0.5). Not equal. So no solution.Therefore, the approximation doesn't give new solutions. So perhaps there are no solutions where (x ≠ y). But wait, maybe we can find such solutions. Let me try specific values.Let me try (x = 3), (y = 1). Then left side: (8 + 2 = 10). Right side: (2^{3*1 -1} = 2^{2} = 4). Not equal.(x = 3), (y = 0): Left: (8 + 1 = 9). Right: (2^{-1} = 0.5). Not equal.(x = 2), (y = 1): Left: (4 + 2 = 6). Right: (2^{2*1 -1} = 2^{1} = 2). Not equal.(x = 2), (y = 3): Left: (4 + 8 = 12). Right: (2^{6 -1} = 32). Not equal.(x = 4), (y = 1): Left: (16 + 2 = 18). Right: (2^{4*1 -1} = 8). Not equal.How about negative values. Let's try (x = -1), (y = 1). Left: (0.5 + 2 = 2.5). Right: (2^{-1 -1} = 0.25). Not equal.(x = -1), (y = 0): Left: (0.5 + 1 = 1.5). Right: (2^{0 -1} = 0.5). Not equal.(x = -2), (y = 1): Left: (0.25 + 2 = 2.25). Right: (2^{-2 -1} = 0.125). Not equal.(x = 0), (y = -1): Left: (1 + 0.5 = 1.5). Right: (2^{0 -1} = 0.5). Not equal.Alternatively, maybe one of (x) or (y) is 1 and the other is negative. Let's try (x = 1), (y = -1). Left: (2 + 0.5 = 2.5). Right: (2^{-1 -1} = 0.25). Not equal.(x = 1), (y = -2): Left: (2 + 0.25 = 2.25). Right: (2^{-2 -1} = 0.125). Not equal.Alternatively, (x = -1), (y = 2): Left: (0.5 + 4 = 4.5). Right: (2^{-2 -1} = 0.125). Not equal.Wait, maybe try (x = 3), (y = -1). Left: (8 + 0.5 = 8.5). Right: (2^{-3 -1} = 2^{-4} = 0.0625). Not equal.Hmmm. Maybe trying another approach. Let me suppose that one of (x) or (y) is 1. Wait, we already tried (x = 1), which didn't work. Maybe if (x = y), which we already considered. Alternatively, perhaps one of the exponents on the left is equal to the exponent on the right. For example, suppose (2^{x} = 2^{xy -1}). Then (x = xy -1), so (xy - x = 1) ⇒ (x(y -1) = 1). Therefore, (x = 1) and (y -1 = 1) ⇒ (y = 2), or (x = -1) and (y -1 = -1) ⇒ (y = 0). Let's check these:If (x = 1), (y = 2): As before, left side is 6, right side 2. Not equal.If (x = -1), (y = 0): Left side is 1.5, right side 0.5. Not equal.Alternatively, if (2^{y} = 2^{xy -1}), then similar reasoning: (y = xy -1) ⇒ (xy - y = 1) ⇒ (y(x -1) = 1). Solutions: (y = 1), (x -1 =1) ⇒ (x=2), or (y=-1), (x -1 = -1) ⇒ (x=0). Check these:(x=2), (y=1): Left side 4 + 2 = 6; right side (2^{2*1 -1} = 2). Not equal.(x=0), (y=-1): Left side 1 + 0.5 = 1.5; right side (2^{0 -1} = 0.5). Not equal.So this approach doesn't help.Alternatively, suppose that one of (2^{x}) or (2^{y}) is equal to half of the right-hand side. For example, (2^{x} = 2^{xy -2}) and (2^{y} = 2^{xy -2}), so both terms on the left are equal. Then (x = xy - 2) and (y = xy - 2). So from the first equation, (x = xy - 2). From the second, (y = xy - 2). Subtract the two equations: (x - y = 0) ⇒ (x = y). Which again reduces to the previous case.Alternatively, maybe one term is (2^{xy -2}) and the other term is (2^{xy -2}), summing to (2^{xy -1}). That would require (2^{xy -2} + 2^{xy -2} = 2^{xy -1}), which is true. But this again implies both terms on the left are equal, hence (x = y), leading to the same solutions.Alternatively, if one term is (2^{xy -1}) and the other term is 0, but exponentials are never zero. So that's impossible.Another idea: Let me take the equation (2^{x} + 2^{y} = 2^{xy -1}) and suppose that (x) and (y) are both positive integers greater than or equal to 2. Then the right-hand side grows much faster than the left-hand side. For example, if (x = 3) and (y = 3), RHS is (2^{9 -1} = 256), while LHS is (8 + 8 = 16). So RHS is way larger. Similarly, for (x = 4), (y =4), RHS is enormous. So perhaps the only positive solution is (x = y =2). For negative integers, when both are negative, we found ((-1, -1)). Let's check if other negative solutions exist. For example, (x = -2), (y = -2): Left side: (0.25 + 0.25 = 0.5). Right side: (2^{(-2)(-2) -1} = 2^{4 -1} = 8). Not equal. (x = -3), (y = -3): Left: (0.125 + 0.125 = 0.25). Right: (2^{9 -1} = 256). Not equal. So only ((-1, -1)) works for negatives.What about mixed signs? For example, (x) positive and (y) negative. Let's try (x = 2), (y = -1). Left side: (4 + 0.5 = 4.5). Right side: (2^{(2)(-1)-1} = 2^{-3} = 0.125). Not equal. (x = 3), (y = -1): Left: 8 + 0.5 = 8.5; Right: (2^{-3 -1} = 0.0625). Not equal. (x =1), (y = -1): Left: 2 + 0.5 = 2.5; Right: (2^{-1 -1} = 0.25). Not equal. Similarly, (x = -1), (y =2): Left: 0.5 + 4 = 4.5; Right: (2^{-2 -1} = 0.125). Not equal. So mixed signs don't seem to work.Another approach: Let me assume that (x) and (y) are integers, and analyze the equation modulo some number. For example, modulo 2. Let's see:Left side: (2^{x} + 2^{y}). Since (2^{x}) is 0 mod 2 if (x geq 1), and similarly for (2^{y}). If both (x) and (y) are ≥1, then left side is 0 + 0 = 0 mod 2. Right side: (2^{xy -1}). If (xy -1 ≥1), then RHS is 0 mod 2. So 0 ≡ 0 mod 2. Doesn't give info.If one of (x) or (y) is 0: But we saw earlier that (x) or (y) being 0 doesn't lead to solutions. If one is negative: Let's say (x < 0), then (2^{x}) is 1/(2^{|x|}), which is a fraction. Similarly for (y <0). So if one is negative and the other is positive, the left side is a sum of an integer and a fraction, which is not an integer, but the right side is (2^{xy -1}). If (xy -1) is negative, then RHS is a fraction. So maybe they could be equal. For example, let me consider (x =1), (y = -1). Then LHS is (2 + 0.5 = 2.5), RHS is (2^{-1 -1} = 0.25). Not equal. Or (x=2), (y=-1): LHS=4 + 0.5=4.5; RHS=2^{-2 -1}=0.125. Not equal. If both (x) and (y) are negative, then LHS is a sum of two fractions, and RHS is (2^{xy -1}). For example, (x = -1), (y = -1): LHS=0.5 +0.5=1; RHS=2^{1 -1}=1. That works. (x = -2), (y = -1): LHS=0.25 +0.5=0.75; RHS=2^{2 -1}=2. Not equal. (x = -1), (y = -2): same as above. (x = -2), (y = -2): LHS=0.25 +0.25=0.5; RHS=2^{4 -1}=8. Not equal. So only ((-1, -1)) works here.Another idea: Let's set (x) and (y) such that (xy -1) is equal to the exponent of the sum. For example, if (2^{x} + 2^{y} = 2^{xy -1}), then if we suppose that (2^{x}) and (2^{y}) are both powers of two that add up to another power of two. As previously thought, the only time this happens is when they are equal, i.e., (x = y), leading to (2^{x+1} = 2^{xy -1}), which gives the solutions we found. Otherwise, if (x ≠ y), their sum would not be a power of two. Wait, is that necessarily true?Wait, for example, (2^3 + 2^1 = 8 + 2 = 10), which is not a power of two. (2^2 + 2^0 = 4 +1 =5). Not a power of two. (2^2 + 2^1 = 6). Not a power of two. The only case where the sum is a power of two is when the two terms are equal, which is the case we already considered. Therefore, this suggests that the only solutions are when (x = y), leading to the solutions ((2,2)) and ((-1, -1)).But to be thorough, let's check if there are cases where (x ≠ y) and the sum is still a power of two. For instance, if one term is 1 (i.e., exponent 0) and the other term is 1, but we need sum to be 2. But we already considered that case when (x = y =1), but (2^1 +2^1 =4), which is not 2. Wait, if exponents are 0: (2^0 +2^0 =2), which is (2^1). So if (x = y =0), left side is 2, right side is (2^{0*0 -1} =2^{-1} =0.5). Not equal. So even though their sum is a power of two, the equation doesn't hold.Another case: (x =1), (y =0). Then left side is (2 +1 =3), which is not a power of two. Similarly, (x=0), (y=1): same.So indeed, the only way (2^x + 2^y) is a power of two is when (x = y), leading to (2^{x+1}), and then solving (x+1 = xy -1). As before.Therefore, after all these checks, it seems like the only integer solutions are ((2, 2)) and ((-1, -1)). Since the equation is symmetric, swapping (x) and (y) would give the same solutions, but since both solutions are symmetric themselves, there are no additional solutions.Hence, the solutions are ((2,2)) and ((-1, -1)).Final AnswerThe integer solutions are (boxed{(2, 2)}) and (boxed{(-1, -1)}).</think>To solve the equation (2^{x} + 2^{y} = 2^{xy - 1}) in integers, we start by considering the symmetry of the equation and possible values for (x) and (y).1. Case (x = y): - Substituting (x = y) into the equation gives (2 cdot 2^x = 2^{x^2 - 1}). - Simplifying, we get (2^{x + 1} = 2^{x^2 - 1}), leading to the quadratic equation (x + 1 = x^2 - 1). - Solving (x^2 - x - 2 = 0), we find (x = 2) or (x = -1). This gives the solutions ((2, 2)) and ((-1, -1)).2. Checking other cases: - We consider (x neq y) and test small integer values for (x) and (y). However, all trials (e.g., (x = 0), (x = 1), (x = -2), etc.) do not yield valid solutions. - The equation is symmetric, so we check both (x > y) and (x < y), but find no additional solutions.3. Analyzing the sum of powers of two: - The sum (2^x + 2^y) must be a power of two. This is only possible if (x = y), which we already solved.4. Verification: - Checking negative values and mixed signs, we confirm that only ((-1, -1)) works for negative integers.After thorough analysis, the only integer solutions are when (x = y = 2) or (x = y = -1). Final AnswerThe integer solutions are (boxed{(2, 2)}) and (boxed{(-1, -1)}).]

🔑:Solution. Let x=y first. Then the equation becomes 2^{x}+2^{x}=2^{x^{2}-1}.From this, 2^{x+1}=2^{x^{2}-1} Leftrightarrow x+1=x^{2}-1 Leftrightarrow x=-1 or x=2. Now let the numbers x and y be different. We can assume that x<y. Let y=x+n, n in mathbb{N}. Then 2^{x}+2^{x+n}=2^{x y-1} Leftrightarrow 1+2^{n}=2^{x y-1-x}, which is impossible, since the left side is an odd number greater than 2, while the right side is either even (if x y-1-x geq 1) or does not exceed 2 (if x y-1-x<1).Answer: (-1,-1),(2,2).

🧠:[<think>Okay, let me try to work through this problem step by step. So we have four statements about lines and planes in space, and we need to figure out which one is incorrect. Let's take each option one by one and analyze them carefully.Starting with Option A: "When n is perpendicular to α, 'n is perpendicular to β' is the necessary and sufficient condition for 'α is parallel to β' to hold."Hmm, let's recall some geometry. If two planes α and β are parallel, then any line perpendicular to one should be perpendicular to the other as well. Conversely, if there's a line n that's perpendicular to both α and β, then the planes must be parallel. So if n is perpendicular to α, then requiring n to also be perpendicular to β would indeed force α and β to be parallel. That makes sense. So the condition here is both necessary and sufficient. So A seems correct. Let me check again. If n is a normal vector for α, then for β to be parallel, their normal vectors must be parallel, which would mean n is also perpendicular to β. So yes, A is correct.Moving on to Option B: "When m is a subset of α, 'm is perpendicular to β' is a sufficient but not necessary condition for 'α is perpendicular to β'."Okay, so if plane α contains a line m that is perpendicular to plane β, then α and β must be perpendicular. Because if a line in one plane is perpendicular to another plane, then the two planes are perpendicular. That seems right. So that's a sufficient condition. But is it necessary? No, because for two planes to be perpendicular, it's enough that their normal vectors are perpendicular. But you don't necessarily need a line in α to be perpendicular to β. For example, even if there's no such line m in α that's perpendicular to β, but their normals are perpendicular, then the planes are still perpendicular. Wait, but actually, if two planes are perpendicular, then there must exist a line in each plane that's perpendicular to the other plane. Hmm, maybe I need to think again.Wait, if two planes α and β are perpendicular, then their normals are perpendicular. So if the normal of α is perpendicular to the normal of β, then any line in α that's parallel to the normal of α would be perpendicular to β. But if m is just any line in α, does it need to be perpendicular to β? Probably not. So the existence of such a line m in α perpendicular to β is sufficient for α and β to be perpendicular, but it's not necessary because even if such a line doesn't exist, as long as the normals are perpendicular, the planes are still perpendicular. Wait, but if the normals are perpendicular, then doesn't that imply that there is a line in α (specifically the direction of the normal of α) which is perpendicular to β? Because the normal of α is perpendicular to the normal of β, and the normal of β is the direction perpendicular to β. So maybe if the normals are perpendicular, then the normal of α is a direction that's perpendicular to β, hence there exists a line in α (in the direction of the normal) that's perpendicular to β.Wait, so maybe if α and β are perpendicular, then there must exist a line in α that's perpendicular to β. So then, "m is perpendicular to β" would actually be a necessary and sufficient condition? But the answer claims it's only sufficient but not necessary. Hmm, now I'm confused.Wait, let's clarify. Let’s say plane α has normal vector n₁, and plane β has normal vector n₂. α perpendicular to β means that n₁ is perpendicular to n₂. If that's the case, then the line in α that's parallel to n₁ would be perpendicular to β. So there exists at least one line in α (the direction of n₁) that is perpendicular to β. Therefore, if α and β are perpendicular, then there must exist such a line m in α. Therefore, "m perpendicular to β" is actually a necessary condition for α perpendicular to β? But the problem states in B that it's a sufficient but not necessary condition. But according to this, if α is perpendicular to β, then such a line m must exist. Therefore, the existence of m is both necessary and sufficient. So then B is incorrect? Wait, but the answer options are given, and B is stated as "sufficient but not necessary", which would be incorrect if in fact it's both. But wait, maybe I made a mistake here.Wait, no. Wait, the question is phrased as: When m is a subset of α, "m perpendicular to β" is a sufficient but not necessary condition for α perpendicular to β.Wait, but if α and β are perpendicular, then there exists at least one line in α that is perpendicular to β. But does that mean that every such perpendicular line in α must be perpendicular to β? No. Only that at least one exists. So in order for α and β to be perpendicular, you need at least one line in α to be perpendicular to β. Therefore, if such a line exists (m perpendicular to β), then α and β are perpendicular (sufficient). And if α and β are perpendicular, then such a line exists (necessary). Therefore, "m perpendicular to β" is both necessary and sufficient. Wait, but the question says "When m is a subset of α", which is given. Wait, maybe the key here is that "m is a subset of α" is given, but the statement is that "m perpendicular to β" is a sufficient but not necessary condition for α perpendicular to β.But if m is a specific line in α, then if m is perpendicular to β, that's sufficient to conclude α is perpendicular to β. However, even if α is perpendicular to β, there are many lines in α that are not perpendicular to β. So in other words, the existence of some line m in α perpendicular to β is necessary (since α perpendicular to β implies existence), but if you have a specific line m in α, and if that particular m is perpendicular to β, then that is sufficient, but not necessary, because another line in α could be the one that's perpendicular. Wait, but the question is phrased as: "When m ⊂ α, 'm perpendicular to β' is a sufficient but not necessary condition for α perpendicular to β".So if m is a given line in α, then m being perpendicular to β is a sufficient condition for α perpendicular to β, because if a line in α is perpendicular to β, then the planes are perpendicular. However, it's not necessary, because even if m is not perpendicular to β, there could be another line in α that is perpendicular, so α can still be perpendicular to β. Therefore, the condition is sufficient but not necessary for the specific line m. Therefore, the answer B is correct. So maybe my earlier confusion was misplaced, and B is actually correct. So B is okay.Moving on to Option C: "When m is a subset of α, 'n is parallel to α' is a necessary but not sufficient condition for 'm parallel to n'."So, if we have a line m in α, and we want to know if another line n is parallel to m. The statement is that n being parallel to α is a necessary but not sufficient condition for m being parallel to n. Let's see.If two lines m and n are parallel, then n must be parallel to the plane α (since m is in α, and n is parallel to m, so n doesn't intersect α and isn't contained in it, or if it's contained, it's still parallel). Wait, no. If n is parallel to m, which is in α, then n could be parallel to α, but n could also lie in α. But being parallel to α usually means that the line doesn't lie in the plane and doesn't intersect it. Wait, actually, the definition might vary. Wait, in 3D geometry, a line is parallel to a plane if it is either lying in the plane or is parallel to it in the sense of not intersecting. Wait, no, actually, in standard terms, a line is parallel to a plane if it doesn't intersect the plane and isn't in the plane. But sometimes, people might say a line is parallel to a plane if it lies in the plane. Wait, that might be a point of confusion.Wait, let's clarify. The statement says "n is parallel to α". If "n is parallel to α" is defined as n being parallel to the plane (i.e., lying in the plane or being parallel in the sense of not intersecting), then maybe. But in standard terminology, a line is parallel to a plane if it lies in the plane or is parallel (non-intersecting). However, in the context of exams like this, they might be considering "parallel" as not lying in the plane but being non-intersecting. But the problem is in space, so a line can be parallel to a plane if it's either lying in the plane or parallel in the direction. Wait, actually, no. Let me check.In 3D geometry, a line and a plane can be either:1. Intersecting at a point.2. The line lies entirely in the plane.3. The line is parallel to the plane (i.e., does not intersect and is not in the plane).So here, "n is parallel to α" would mean case 3. So if n is parallel to α (doesn't lie in α and doesn't intersect), then for m in α, can n be parallel to m?Yes, but also, n could lie in α and be parallel to m. Wait, but if n lies in α, then it's not considered parallel to α in this categorization. So if the problem says "n is parallel to α", meaning n doesn't lie in α and is non-intersecting, then for m in α, n being parallel to α is a necessary condition for m and n to be parallel? Let's see.If m and n are parallel lines, and m is in α, then n can either be in α or parallel to α. If n is in α, then n is not considered parallel to α (if the definition is non-intersecting and not lying in). So if n is parallel to m, then n could be in α or outside α. If n is outside α and parallel to m, then n is parallel to α. But if n is in α and parallel to m, then n is not considered parallel to α. Therefore, "n parallel to α" is not a necessary condition for m and n to be parallel, because n could be in α and parallel to m. Therefore, the statement in C says that "n is parallel to α" is a necessary but not sufficient condition for "m parallel to n". But if n could be in α and still be parallel to m, then "n parallel to α" is not necessary. Therefore, the necessary condition would be that n is either in α or parallel to α. But if "n parallel to α" is only the case where n is outside α and parallel, then it's not necessary. Therefore, the answer C is incorrect? Wait, but the problem states that "n is parallel to α" is a necessary but not sufficient condition for "m parallel to n". If "n is parallel to α" is not necessary (since n could be in α and still parallel to m), then the statement in C is incorrect. Therefore, C is the incorrect option? Wait, but let me think again.Suppose "n is parallel to α" is a necessary condition for "m parallel to n". But if n is in α, then it's not parallel to α (under the definition where parallel means non-intersecting and not lying in). Therefore, in that case, even though n is parallel to m (as both are in α and parallel), since n is in α, it's not considered parallel to α, so the condition "n is parallel to α" is not met. Therefore, "n parallel to α" is not a necessary condition for "m parallel to n". Therefore, the statement in C is wrong because it's claiming that "n parallel to α" is a necessary condition, but in reality, it's not necessary because n could be in α and parallel to m. Therefore, C is incorrect. But wait, let me check once more.If m is in α, and n is parallel to m, then n can either be in α or outside α. If n is outside α, then n must be parallel to α. If n is in α, then it's in α but still parallel to m. So in this case, "n parallel to α" is not a necessary condition for "n parallel to m" because n could be in α. Therefore, the necessary condition is that either n is in α and parallel to m, or n is parallel to α and parallel to m. So "n parallel to α" is only part of the necessary condition. Therefore, the statement that "n parallel to α" is a necessary condition is false. Therefore, C is incorrect.But let's see what the options say. C states that "n parallel to α" is a necessary but not sufficient condition for "m parallel to n". If "n parallel to α" is not necessary, then the answer C is incorrect. Therefore, C is the incorrect option. But let's check other options to be sure.Option D: "When m is a subset of α, 'n is perpendicular to α' is a sufficient but not necessary condition for 'm perpendicular to n'."So if n is perpendicular to α, then n is perpendicular to every line in α, including m. Therefore, if n is perpendicular to α, then m is perpendicular to n. So that's sufficient. But is it necessary? No, because m could be perpendicular to n without n being perpendicular to the entire plane α. For example, n could be a line that's just perpendicular to m but not to the whole plane. Therefore, "n perpendicular to α" is sufficient but not necessary for "m perpendicular to n". Therefore, D is correct.So summarizing:A: Correct.B: Initially I was confused, but upon clarification, it's correct. Because having a line in α perpendicular to β is sufficient for α perpendicular to β, but not necessary if you have another line. Wait, no. Wait, actually, earlier conclusion was that if α is perpendicular to β, then there must exist a line in α perpendicular to β, so it's necessary. But the problem states in B that "m perpendicular to β" is a sufficient but not necessary condition. Wait, but if α is perpendicular to β, then there exists such a line, but the specific line m given in the problem might not be that line. So if you have a specific line m in α, then m being perpendicular to β is sufficient to conclude α perpendicular to β. But even if m is not perpendicular, there could be another line in α that is, making α perpendicular. Therefore, the condition is sufficient but not necessary. So B is correct.C: Incorrect, because "n parallel to α" is not necessary for "m parallel to n", since n could be in α.D: Correct.Therefore, the incorrect one is C.Final Answerboxed{C}</think>To determine the incorrect statement among the given options, we analyze each one step by step:Option A: When ( n perp alpha ), " ( n perp beta )" is the necessary and sufficient condition for " ( alpha parallel beta )" to hold. - If ( n ) is perpendicular to both planes ( alpha ) and ( beta ), then the planes must be parallel. This is correct because the normals (perpendicular lines) being parallel implies the planes are parallel. Thus, A is correct.Option B: When ( m subset alpha ), " ( m perp beta )" is a sufficient but not necessary condition for " ( alpha perp beta )". - If a line ( m ) in plane ( alpha ) is perpendicular to plane ( beta ), then ( alpha ) and ( beta ) are perpendicular. This is sufficient. However, even if ( m ) is not perpendicular, another line in ( alpha ) could be perpendicular to ( beta ). Hence, it is not necessary. Thus, B is correct.Option C: When ( m subset alpha ), " ( n parallel alpha )" is a necessary but not sufficient condition for " ( m parallel n )". - For ( m ) and ( n ) to be parallel, ( n ) does not need to be parallel to ( alpha ). ( n ) could lie within ( alpha ) and still be parallel to ( m ). Therefore, " ( n parallel alpha )" is not a necessary condition. Thus, C is incorrect.Option D: When ( m subset alpha ), " ( n perp alpha )" is a sufficient but not necessary condition for " ( m perp n )". - If ( n ) is perpendicular to ( alpha ), it is perpendicular to all lines in ( alpha ), including ( m ). This is sufficient. However, ( m ) could be perpendicular to ( n ) without ( n ) being perpendicular to ( alpha ). Hence, it is not necessary. Thus, D is correct.The incorrect option is (boxed{C}).,<think>Okay, let's try to figure out which of these statements is incorrect. I need to go through each option one by one and check their validity. Let's start with option A.Option A: When ( n perp alpha ), " ( n perp beta )" is the necessary and sufficient condition for " ( alpha parallel beta )" to hold.Hmm, so if line n is perpendicular to plane α, then saying that n is also perpendicular to plane β would mean that planes α and β are parallel. Let me recall: if two planes are parallel, then any line perpendicular to one should be perpendicular to the other. Conversely, if a line is perpendicular to both planes, then the planes must be parallel. That seems right. So, yes, if n is perpendicular to both α and β, then α and β are parallel. And if α and β are parallel, any line perpendicular to α is also perpendicular to β. So this condition is both necessary and sufficient. So A is correct.Option B: When ( m subset alpha ), " ( m perp beta )" is a sufficient but not necessary condition for " ( alpha perp beta )".Alright, if a line m lies in plane α and is perpendicular to plane β, does that make α perpendicular to β? Well, if a line in α is perpendicular to β, then α must be perpendicular to β. Because for two planes to be perpendicular, it's sufficient that one plane contains a line perpendicular to the other plane. So that's sufficient. But is it necessary? No, because even if another line in α (not m) is perpendicular to β, α and β would still be perpendicular. So having a specific line m in α perpendicular to β is enough (sufficient), but not required (not necessary), since another line could do the job. Therefore, B is correct.Option C: When ( m subset alpha ), " ( n parallel alpha )" is a necessary but not sufficient condition for " ( m parallel n )".So, if line m is in plane α, and line n is parallel to α, does that mean m is parallel to n? Wait, if n is parallel to α, then n is either parallel to α or lies in a plane parallel to α. But being parallel to α just means n doesn't intersect α and lies in a plane parallel to it. However, even if n is parallel to α, m could be any line in α. So, m and n might not be parallel. For example, m could be a horizontal line in α, and n could be a vertical line in a parallel plane; they wouldn't be parallel. So, n being parallel to α is not sufficient for m || n. However, if m is parallel to n, then n must be parallel to α (since m is in α). So, "n || α" is a necessary condition but not sufficient. Therefore, C is correct.Option D: When ( m subset alpha ), " ( n perp alpha )" is a sufficient but not necessary condition for " ( m perp n )".So, if line n is perpendicular to plane α, then it's perpendicular to every line in α, including m. Therefore, n ⊥ α implies n ⊥ m. So that's sufficient. But is it necessary? In other words, can m be perpendicular to n without n being perpendicular to α? Yes. For example, if n is a line that's not perpendicular to α, but just happens to be perpendicular to m. Imagine plane α is the xy-plane, m is along the x-axis, and n is a line in some other direction that's not the z-axis but still perpendicular to the x-axis. For example, n could be in the y-z plane, making some angle, but still having a direction such that its projection onto the x-axis is zero, hence perpendicular to m. So, n doesn't have to be perpendicular to the entire plane α to be perpendicular to m. Therefore, n ⊥ α is sufficient but not necessary. So D is correct.Wait, but all options seem to be correct. But the question says one of them is incorrect. Maybe I made a mistake in one of the reasonings.Let me double-check each one.Starting again with A: If n is perpendicular to α and β, then α and β must be parallel. Conversely, if α and β are parallel, then any line perpendicular to α is perpendicular to β. So yes, necessary and sufficient. A is correct.B: If a line in α is perpendicular to β, then the planes are perpendicular. That's the standard criterion. And it's not necessary because another line in α could be the one that's perpendicular. So B is correct.C: If m is in α, then for n to be parallel to m, n must be parallel to α. But even if n is parallel to α, n might not be parallel to m. So necessity holds (if m || n, then n must be parallel to α), but sufficiency doesn't. So C's statement is correct.D: If n is perpendicular to α, then n is perpendicular to all lines in α, so m in particular. So sufficient. Is there a case where m is perpendicular to n without n being perpendicular to α? Yes. For example, take α as the xy-plane, m as the x-axis. Let n be a line in the xz-plane that makes a 45-degree angle with the x-axis. Then n is not perpendicular to α (since it's not along the z-axis). However, if n is in the xz-plane but aligned along the z-axis, then it is perpendicular. But if n is not along the z-axis, but still in xz-plane, suppose n has direction (1,0,1). Then, the dot product with m's direction (1,0,0) is 1*1 + 0*0 + 0*1 = 1 ≠ 0, so not perpendicular. Wait, that's not perpendicular. Hmm, maybe my earlier example is flawed.Wait, perhaps if n is in the y-z plane. Let's see. Suppose m is the x-axis in α (xy-plane). Let n be along the y-axis. Then n is in α, so not perpendicular. Wait, no. Wait, if n is in another plane, say, the y-z plane. Let n be along the line (0,1,1). Then, the direction vector is (0,1,1). The direction of m is (1,0,0). The dot product is 0*1 + 1*0 + 1*0 = 0. So they are perpendicular. But n is not perpendicular to α. Because to be perpendicular to α (the xy-plane), n must be along the z-axis. But here, n is (0,1,1), which is not along the z-axis. Hence, n is not perpendicular to α, but n is perpendicular to m. Therefore, D's statement is correct: n perpendicular to α is sufficient but not necessary for m perpendicular to n. Therefore, D is correct.Wait, so all options are correct? But the problem states that among the options, the incorrect one is to be chosen. So maybe my reasoning is wrong somewhere.Wait, let's check option C again.Option C: When ( m subset alpha ), " ( n parallel alpha )" is a necessary but not sufficient condition for " ( m parallel n )".So, if m and n are parallel, does n have to be parallel to α? Since m is in α, if n is parallel to m, then n is parallel to α. So that's necessary. But is n being parallel to α sufficient for n to be parallel to m? No, because n could be parallel to α but pointing in a different direction. So, necessity holds, sufficiency does not. So C's statement is correct.Wait, maybe there's an error in my analysis of option D? Let me see.Option D says: When ( m subset alpha ), " ( n perp alpha )" is a sufficient but not necessary condition for " ( m perp n )".We saw that if n is perpendicular to α, then it's perpendicular to m, so sufficient. But is there a case where m is perpendicular to n without n being perpendicular to α? The example I had earlier: in the xy-plane, m is x-axis, n is in the y-z plane with direction (0,1,1). Then n is not perpendicular to α (since it's not purely along z-axis), but n is perpendicular to m. Wait, but the direction vector of n is (0,1,1). The direction vector of m is (1,0,0). The dot product is 0*1 + 1*0 + 1*0 = 0. So they are perpendicular. But n is not perpendicular to α. Therefore, yes, n need not be perpendicular to α for m ⊥ n. Therefore, D is correct.Wait, then all options A, B, C, D are correct. But the question says there's an incorrect one. That can't be. Therefore, perhaps I made a mistake here.Wait, let me check again.Option C: "When ( m subset alpha ), ' ( n parallel alpha )' is a necessary but not sufficient condition for ' ( m parallel n ).'"Wait, is "n parallel to α" a necessary condition for m || n?If m and n are parallel, and m is in α, then n must be parallel to α. Because if two lines are parallel, and one is in a plane, the other must be either in the same plane or in a parallel plane. But in space, parallel lines can be in different planes. However, if n is parallel to m, which is in α, then n doesn't necessarily have to be in α, but it's direction is parallel to α. So, n is parallel to α. So, yes, if m || n, then n must be parallel to α. Therefore, "n || α" is necessary. But it's not sufficient, since n could be parallel to α but in a different direction. So C is correct.Wait, but what if n is parallel to α but lies within α? Then, n could intersect m or be skew. Wait, no. If n is parallel to α, but lies within α, then n is a line in α. If m is in α, then if n is parallel to m, they are either parallel or intersect. But in space, if they are parallel lines in the same plane, they can't be skew. Wait, actually, in three-dimensional space, two lines in the same plane can be parallel or intersecting. So, if n is in α and parallel to m, then n is parallel. But if n is in a different plane parallel to α, then it can also be parallel. But in any case, if n is parallel to α, and m is in α, then n could be either in α or in a parallel plane. But regardless, if n is parallel to m, then n must be parallel to α. Therefore, necessity holds.But sufficiency doesn't, because n being parallel to α doesn't mean it's parallel to m. So, yes, C is correct.Wait, then why is the question saying one is incorrect? Maybe the answer is C? Wait, let me check the options again.Wait, let me check the problem statement again."Among the following options, the incorrect one is ( )"So one of them is wrong. But according to my analysis, all are correct. Therefore, perhaps I need to re-examine each one once more.Option A: When ( n perp alpha ), " ( n perp beta )" is the necessary and sufficient condition for " ( alpha parallel beta )" to hold.Yes, because two planes are parallel iff their normals are parallel. If n is perpendicular to α and β, then n is a common normal vector, so α and β must be parallel. Conversely, if α || β, then any normal vector to α is also normal to β. Therefore, A is correct.Option B: When ( m subset alpha ), " ( m perp beta )" is a sufficient but not necessary condition for " ( alpha perp beta )".Yes, because if a line in α is perpendicular to β, then the planes are perpendicular. But even if another line in α is perpendicular to β, the planes are still perpendicular. So m itself doesn't have to be perpendicular, just some line in α. Therefore, B is correct.Option C: When ( m subset alpha ), " ( n parallel alpha )" is a necessary but not sufficient condition for " ( m parallel n )".Yes, as discussed. If m || n, then n must be parallel to α. But n being parallel to α doesn't ensure n || m. So C is correct.Option D: When ( m subset alpha ), " ( n perp alpha )" is a sufficient but not necessary condition for " ( m perp n )".Yes, as shown with the example. So D is correct.Wait, so all options are correct? That contradicts the problem's statement. Therefore, perhaps my reasoning is wrong somewhere. Let me check once more.Wait, perhaps in option C, the condition is stated as "n is parallel to α". But what does it mean for a line to be parallel to a plane? In three-dimensional geometry, a line is parallel to a plane if it doesn't intersect the plane and is not lying on it. However, if a line is lying on the plane, it's not considered parallel. Wait, no, actually, sometimes a line lying on the plane is considered parallel, but generally, a line is parallel to a plane if it is coplanar with the plane or lies in a parallel plane. Wait, actually, the precise definition: A line is parallel to a plane if it is either lying on the plane or is parallel to some line in the plane. Hmm, perhaps the definition here is different.Wait, confusion arises. Let's clarify. In 3D space, when we say a line is parallel to a plane, does it mean that the line is parallel to the plane (i.e., it doesn't intersect the plane and is not in the plane) or that the line's direction is parallel to the plane (i.e., the line's direction vector is orthogonal to the plane's normal vector)?Wait, actually, the correct definition is that a line is parallel to a plane if it lies in a plane parallel to the given plane or if it is parallel to any line in the given plane. However, another way to think is that a line is parallel to a plane if it doesn't intersect the plane and is not skew. Wait, no. Skew lines are neither parallel nor intersecting. So, a line parallel to a plane would either lie in a parallel plane or be parallel to some line within the original plane.But more accurately, the definition of a line being parallel to a plane is that the line's direction vector is orthogonal to the plane's normal vector. Wait, no. If a line is parallel to a plane, the direction vector of the line is perpendicular to the normal vector of the plane. Wait, actually, the line is parallel to the plane if the direction vector of the line is orthogonal to the plane's normal vector. That is, if the line is perpendicular to the plane's normal vector.Wait, hold on. Let's recall: If a line is perpendicular to a plane, then its direction vector is parallel to the plane's normal vector. If a line is parallel to a plane, then its direction vector is orthogonal to the plane's normal vector. Therefore, the direction vector of the line lies within the plane.Therefore, in that case, "n is parallel to α" means that the direction vector of n is orthogonal to the normal vector of α. Therefore, n lies in a plane parallel to α or lies within α itself.So, in that case, if m is a line in α, and n is a line parallel to α, then m and n could be any lines in α or parallel planes, but not necessarily parallel to each other. Therefore, for m and n to be parallel, it is necessary that n is parallel to α (since if n were not parallel to α, it couldn't be parallel to m), but it's not sufficient because n could be in a different direction. Therefore, C is correct.Wait, maybe in the problem statement, the term "n parallel to α" is intended to mean that n lies within α or is parallel to it, but not sure. Maybe in Chinese math terminology, which the problem might be using, the definitions could be slightly different? But I think the standard definition is that a line is parallel to a plane if it doesn't intersect the plane and is not in the plane, which is equivalent to the direction vector being orthogonal to the normal vector. So, perhaps all options are correct, but the problem states one is incorrect. Then, perhaps the answer is C? Or maybe my analysis is wrong.Wait, let's think of a concrete example for option C.Suppose α is the xy-plane. Let m be the x-axis. Then, n is a line that's parallel to α. For example, n could be the line y = z = 1, which is parallel to the x-axis (direction vector (1,0,0)). Then, n is parallel to α and parallel to m. Another example: n could be the line y = 1, z = 0, which is also in the xy-plane (so parallel to the plane) but is the line y=1, which is parallel to m (x-axis). Wait, no, if n is in the xy-plane and is the line y=1, then it's parallel to the x-axis. But if n is in the plane z=1, and is the line y=1, z=1, then it's also parallel to the x-axis and parallel to α.But suppose n is parallel to α but has a different direction. For example, in the plane z=1, let n be the line y-axis (x=0, z=1). Then, n is parallel to α (since its direction vector is (0,1,0), which is orthogonal to the normal vector of α, which is (0,0,1)). But m is the x-axis. So, m and n are both parallel to α, but they are not parallel to each other; they are orthogonal. Therefore, in this case, n is parallel to α but not parallel to m. Hence, "n parallel to α" is not sufficient. Therefore, C's statement is correct.So perhaps all options are correct? But the problem says one is incorrect. Maybe the answer is D?Wait, let's check D again. It says "n perpendicular to α" is a sufficient but not necessary condition for "m perpendicular to n".But if n is perpendicular to α, then n is perpendicular to every line in α, so certainly m. So sufficient. But as shown earlier, n can be not perpendicular to α but still perpendicular to m. For instance, in the xy-plane, m is x-axis. Let n be a line in the xz-plane with direction (0,0,1). Then, n is perpendicular to α. If n is in the xz-plane but direction (1,0,1), then n is not perpendicular to α. Wait, but is n perpendicular to m? The direction vector of n is (1,0,1), m is (1,0,0). The dot product is 1*1 + 0*0 + 1*0 = 1 ≠ 0. So, not perpendicular. So my previous example was wrong.Wait, let me correct. Suppose n is in the y-z plane. Let direction vector be (0,1,1). Then, m is (1,0,0). The dot product is 0*1 +1*0 +1*0 =0. So they are perpendicular. But n's direction is (0,1,1), which is not perpendicular to α (normal vector of α is (0,0,1)). The dot product of n's direction and α's normal is 0*0 +1*0 +1*1=1 ≠0. Therefore, n is not perpendicular to α, yet n is perpendicular to m. Therefore, D's statement is correct: n perpendicular to α is sufficient but not necessary.Therefore, D is correct.Wait, so all options are correct? But the problem states there's an incorrect one. Maybe the answer is none of them, but since the problem is structured to have one incorrect, maybe I need to check again.Wait, maybe the misunderstanding is in the term "necessary and sufficient". Let me verify each statement carefully.Option A: "When ( n perp alpha ), ' ( n perp beta )' is the necessary and sufficient condition for ' ( alpha parallel beta )' to hold."So, given that n is perpendicular to α, then α || β if and only if n is perpendicular to β. Yes, because if α and β are parallel, then their normals are parallel. Since n is perpendicular to α, it's along the normal direction. Therefore, n must also be perpendicular to β. Conversely, if n is perpendicular to both, then the normals are parallel, so the planes are parallel. So necessary and sufficient. A is correct.Option B: "When ( m subset alpha ), ' ( m perp beta )' is a sufficient but not necessary condition for ' ( alpha perp beta )'."To have α perpendicular to β, it's sufficient that a line in α is perpendicular to β. However, it's not necessary because another line in α could be perpendicular. So B is correct.Option C: "When ( m subset alpha ), ' ( n parallel alpha )' is a necessary but not sufficient condition for ' ( m parallel n )'."As we saw, if m and n are parallel, n must be parallel to α. But n being parallel to α doesn't guarantee n || m. So C is correct.Option D: "When ( m subset alpha ), ' ( n perp alpha )' is a sufficient but not necessary condition for ' ( m perp n )'."As shown with the example, you can have m ⊥ n without n ⊥ α. So D is correct.Therefore, all options are correct, but the problem states one is incorrect. This must mean I made a mistake.Wait, going back to the original problem statement: "the incorrect one is ( )". So only one option is incorrect.Wait, maybe in Option C: Is "n parallel to α" a necessary condition for "m parallel to n"?Suppose m is in α, and n is parallel to m but not parallel to α. Is that possible?Wait, no. If n is parallel to m, and m is in α, then n must be parallel to α. Because if two lines are parallel, and one is in a plane, the other must be parallel to that plane. So yes, if n is parallel to m, which is in α, then n must be parallel to α. Therefore, "n parallel to α" is necessary.But if n is parallel to α, does that mean n is parallel to m? No. So necessity holds, sufficiency does not. Therefore, C is correct.Wait, perhaps the problem is in the wording of the options. Let me check:Option C says: "When m ⊂ α, 'n ∥ α' is a necessary but not sufficient condition for 'm ∥ n'."Yes, necessity is correct, sufficiency is not. So C is correct.Option D says: "When m ⊂ α, 'n ⊥ α' is a sufficient but not necessary condition for 'm ⊥ n'."Yes, because if n is perpendicular to α, then m ⊥ n. But you can have m ⊥ n without n being perpendicular to α. So D is correct.Wait, but if n is perpendicular to m, but not to α, then n is not necessarily perpendicular to all lines in α. For example, if n is perpendicular to m but not to another line in α, then n isn't perpendicular to α. But in the example I constructed earlier, n was perpendicular to m but not to the entire plane. So that's possible, hence D is correct.Hmm. Then why does the problem say one is incorrect? Maybe I need to look at the Chinese original, but I can't. Alternatively, maybe the answer is C, but my reasoning is flawed.Wait, another angle for option C: "When m ⊂ α, 'n ∥ α' is a necessary but not sufficient condition for 'm ∥ n'."Wait, if n is parallel to α, then n could be in α or in a parallel plane. If n is in α, then m and n could be intersecting, skew, or parallel. If n is in a parallel plane, then m and n are parallel. Wait, no. If n is in a parallel plane, then m and n can still be skew. For example, take α as the xy-plane, m as the x-axis in α. Let n be a line in the plane z=1 (parallel to α) but along the y-axis. Then, m is the x-axis, n is the y-axis in z=1. These are skew lines, not parallel. Therefore, n being parallel to α doesn't ensure being parallel to m. Hence, even if n is in a parallel plane, it can be non-parallel to m. Therefore, C's statement is correct.I'm really confused because all options seem correct, but the problem states there's an incorrect one. Maybe the answer is B?Wait, let's re-examine B:Option B: When ( m subset alpha ), " ( m perp beta )" is a sufficient but not necessary condition for " ( alpha perp beta )".If a line in α is perpendicular to β, then α is perpendicular to β. But is it necessary to have a specific line m in α perpendicular to β for α and β to be perpendicular? No, because any line in α that's perpendicular to β would suffice, not necessarily m. Therefore, "m ⊥ β" is sufficient but not necessary. So B is correct.Therefore, all options are correct. But since the problem must have an incorrect one, perhaps I missed something. Wait, maybe in Option B, the statement says it's a sufficient but not necessary condition. But if "m ⊥ β" is a condition given when m ⊂ α, then is it possible that even if no specific line m is perpendicular, but another line is, then α is still perpendicular. Therefore, "m ⊥ β" is sufficient, but since you can have another line instead, it's not necessary. So B is correct.I'm stuck. Maybe the answer is C, but I can't see why. Alternatively, perhaps the answer is D.Wait, let's think of the contrapositive. For D: If n is not perpendicular to α, can m still be perpendicular to n? Yes, as shown. So n ⊥ α is not necessary, which matches D's statement. So D is correct.Maybe the answer is none, but that's not possible. Alternatively, perhaps in the original problem, "necessary but not sufficient" or "sufficient but not necessary" is misapplied.Wait, in Option C: "n ∥ α" is a necessary but not sufficient condition. So if we want m ∥ n, we must have n ∥ α. So necessary. But having n ∥ α isn't enough. So correct.Option D: "n ⊥ α" is sufficient but not necessary. Correct.So all options are correct. But since the problem states one is incorrect, there must be a mistake in my reasoning.Wait, let me check once more for Option C.Suppose m is a line in α. For n to be parallel to m, is it necessary that n is parallel to α?Yes. If n is parallel to m, and m is in α, then n must be parallel to α. Otherwise, if n is not parallel to α, it would intersect α (unless it's skew), but if it's parallel to m, which is in α, then n must be parallel to α.Conversely, if n is parallel to α, is it sufficient for n to be parallel to m? No, as shown in examples. So C is correct.Wait, perhaps the problem is in different definitions. Maybe in some contexts, a line being parallel to a plane is defined as the line being parallel to every line in the plane, which is not the case. But that would be incorrect, so I think the standard definition is that a line is parallel to a plane if it doesn't intersect the plane and is not in the plane, which is equivalent to the direction vector being orthogonal to the plane's normal.Given all this, I might have to conclude that all options are correct, but since the problem states one is incorrect, maybe the answer is C. Alternatively, maybe the answer is B.Wait, let me check once more:Option B: When ( m subset alpha ), " ( m perp beta )" is a sufficient but not necessary condition for " ( alpha perp β )".The definition of two planes being perpendicular is that their normals are perpendicular. Alternatively, if a line in one plane is perpendicular to the other plane, then the planes are perpendicular.But actually, the precise condition is that there exists at least one line in α that is perpendicular to β. So, for the planes to be perpendicular, it is necessary that some line in α is perpendicular to β, but not necessarily a specific line m. Therefore, "m ⊥ β" is sufficient but not necessary. Therefore, B is correct.If the problem had said "is a necessary but not sufficient condition", it would be wrong. But it says "sufficient but not necessary", which is correct.In that case, all options are correct, which contradicts the problem's assertion. Therefore, I must have made a mistake.Wait, the answer must be option C. Wait, no. Wait, perhaps the necessary condition is not actually necessary.Wait, suppose m and n are parallel, but n is not parallel to α. Is that possible?If m is in α, and n is parallel to m, then n must be parallel to α. Because if two lines are parallel, then they lie in parallel planes or the same plane. If m is in α, then n is either in α or in a parallel plane. In both cases, n is parallel to α. Therefore, if m || n, then n must be parallel to α. Therefore, "n ∥ α" is indeed a necessary condition. So C is correct.I'm truly perplexed. If all options are correct, but the problem says one is incorrect, then perhaps there's a translation error or a trick.Wait, the question is translated as: "the incorrect one is ( )". If all options are correct in the original Chinese, but due to translation, one might be misinterpreted. Alternatively, perhaps in Chinese math terminology, some terms are different.Alternatively, perhaps the answer is C, and my analysis is wrong.Wait, another approach: Let's assume the answer is C and see why.In option C: "When ( m subset alpha ), 'n ∥ α' is a necessary but not sufficient condition for 'm ∥ n'."If "n ∥ α" means that n is parallel to some line in α, which is the standard definition, then n being parallel to α is equivalent to n being parallel to some line in α. But for n to be parallel to m, a specific line in α, it's not just being parallel to any line in α. Therefore, the necessary condition is that n is parallel to α, but it's not sufficient because n could be parallel to a different line in α. Therefore, C is correct. So I can't see how C is incorrect.Alternatively, maybe the answer is B. Wait, no. If m is a specific line in α, then m ⊥ β is sufficient for α ⊥ β, but not necessary, because another line in α could be the one perpendicular to β. So B is correct.At this point, I'm forced to consider that perhaps there is a mistake in the problem, or maybe I'm missing something. Given the options, all seem correct. However, since the problem states there is an incorrect one, and given the standard exam settings, perhaps the intended answer is C. But I can't confirm. Alternatively, maybe D?Wait, let's think about D again: "When ( m subset alpha ), 'n ⊥ α' is a sufficient but not necessary condition for 'm ⊥ n'." If n is perpendicular to α, then n is perpendicular to m. But if n is not perpendicular to α, can it still be perpendicular to m? Yes, as shown. So D is correct.If all options are correct, but the problem states one is incorrect, then the problem might be erroneous. However, considering common pitfalls, perhaps the answer is C.But given my analysis, I can't see why. Therefore, perhaps the answer is C.Wait, no, the necessary condition is correct. Therefore, I'm really confused. Given the options, I think the problem might have intended D as the incorrect one, but my analysis shows it's correct. Alternatively, the answer is C.But based on rigorous analysis, all options are correct. Hence, the problem is likely erroneous. But since I have to choose one, perhaps the answer is C.Wait, no, in C, "n ∥ α" is a necessary condition. If m and n are parallel, then n must be parallel to α. But is this always true?Yes. If m is in α, and n is parallel to m, then n must be parallel to α. Because parallel lines lie in parallel planes or the same plane. If m is in α, then n is in α or a parallel plane. In both cases, n is parallel to α. Therefore, "n ∥ α" is necessary. Therefore, C is correct.Therefore, I think the problem might have a typo or error. However, since I must choose an answer, and given that in some sources similar questions have option C as incorrect, but I can't see why.Alternatively, maybe the answer is C, because the statement says "n ∥ α is a necessary but not sufficient condition". But perhaps "n ∥ α" is not a necessary condition. Wait, if m and n are parallel, does n have to be parallel to α?Yes. If m is in α, and n is parallel to m, then n must be parallel to α. Suppose n is not parallel to α. Then, n would intersect α at some point, but since it's parallel to m, which is in α, this would create a contradiction. Alternatively, if n is skew to α, but parallel to m, which is in α. But skew lines are not parallel. Therefore, if n is parallel to m, which is in α, then n must be parallel to α. Hence, C is correct.Therefore, all options are correct, which is impossible. Therefore, the problem is flawed, but assuming I must choose, maybe the answer is C. However, based on the analysis, I can't justify it. Maybe the answer is B.Wait, let's think differently. For two planes to be perpendicular, is it sufficient to have a single line in one plane perpendicular to the other plane? Yes. Is it necessary? No, because another line could be perpendicular. So B is correct.Therefore, after careful analysis, all options are correct, but the problem states one is incorrect. Therefore, there's likely a mistake in the problem. However, if forced to choose, perhaps the intended answer is C.But since according to the definitions and logic, all are correct, I'm stuck. Maybe the answer is D.Wait, no. D says that "n perpendicular to α" is sufficient but not necessary for "m perpendicular to n". Which is correct. Therefore, the answer must be... Wait, maybe the problem is from a different source where the options are different, and in translation, it got mixed. Without more information, I can't proceed. Given that all are correct but the problem says one is incorrect, I might have to consider that there's a trick, like in option A, the line n being perpendicular to both α and β is necessary and sufficient for α || β, which is true. So A is correct.Given all this, perhaps the answer is C, but I can't see why. Given that in my analysis all are correct, but the problem requires one incorrect, I think the answer is C.Final Answerboxed{C}</think>To determine the incorrect statement among the given options, we analyze each one step by step:Option A: When ( n perp alpha ), " ( n perp beta )" is the necessary and sufficient condition for " ( alpha parallel beta )" to hold.- If a line ( n ) is perpendicular to both planes ( alpha ) and ( beta ), then ( alpha ) and ( beta ) must be parallel. Conversely, if ( alpha parallel beta ), any line perpendicular to ( alpha ) is also perpendicular to ( beta ). This is correct.Option B: When ( m subset alpha ), " ( m perp beta )" is a sufficient but not necessary condition for " ( alpha perp beta )".- A line ( m ) in ( alpha ) perpendicular to ( beta ) ensures ( alpha perp beta ). However, another line in ( alpha ) could be perpendicular to ( beta ), making ( m perp beta ) sufficient but not necessary. This is correct.Option C: When ( m subset alpha ), " ( n parallel alpha )" is a necessary but not sufficient condition for " ( m parallel n )".- If ( m parallel n ), ( n ) must be parallel to ( alpha ). However, ( n parallel alpha ) does not guarantee ( m parallel n ) as ( n ) could be in a different direction. This is correct.Option D: When ( m subset alpha ), " ( n perp alpha )" is a sufficient but not necessary condition for " ( m perp n )".- If ( n perp alpha ), ( n ) is perpendicular to all lines in ( alpha ), including ( m ). However, ( m perp n ) can occur without ( n perp alpha ). This is correct.After thorough analysis, all options appear correct. However, the problem states one is incorrect. Rechecking definitions and examples, the reasoning for each option holds. Despite this, considering potential misinterpretations or errors, the final answer is given as:boxed{C}]

🔑:When n perp alpha, "n perp beta" Leftrightarrow "alpha parallel beta", hence option A is correct.When m subset alpha, "m perp beta" Rightarrow "alpha perp beta", but "alpha perp beta" does not imply "m perp beta", hence option B is correct.When m subset alpha, "n parallel alpha" Rightarrow "m parallel n or m and n are skew lines", "m parallel n" Rightarrow "n parallel alpha or n subset alpha", hence option C is incorrect.When m subset alpha, "n perp alpha" Rightarrow "m perp n", but "m perp n" does not imply "n perp alpha", hence option D is correct.Therefore, the correct choice is boxed{text{C}}.