🧠:[<think>Okay, let's tackle these two problems one by one. First, problem (1): solving the equation (x)/(x - 1) = (x - 1)/(2x - 2). Hmm, let me think. Alright, so we have a rational equation here. The first step is probably to cross-multiply to eliminate the denominators. But before I do that, I should check if there are any restrictions on the variable x. Since the denominators are x - 1 and 2x - 2, I need to make sure that neither of them is zero. So, solving x - 1 ≠ 0 gives x ≠ 1, and 2x - 2 ≠ 0. Let's solve the second one: 2x - 2 ≠ 0 → 2x ≠ 2 → x ≠ 1. So both denominators are zero when x is 1, which means x cannot be 1. Got that. So x ≠ 1 is a restriction.Now, moving on to solving the equation. Let's cross-multiply. That would give:x * (2x - 2) = (x - 1) * (x - 1)Let me write that out:Left side: x*(2x - 2) = 2x² - 2xRight side: (x - 1)² = x² - 2x + 1So, setting them equal:2x² - 2x = x² - 2x + 1Now, subtract x² - 2x + 1 from both sides to bring everything to one side:2x² - 2x - x² + 2x - 1 = 0Simplify terms:(2x² - x²) + (-2x + 2x) - 1 = 0 → x² - 1 = 0So, x² - 1 = 0. This factors into (x - 1)(x + 1) = 0. Therefore, x = 1 or x = -1.Wait a minute, but earlier we found that x cannot be 1 because it makes the denominators zero. So x = 1 is an extraneous solution and must be rejected. Therefore, the only solution is x = -1.Let me check that. Plug x = -1 into the original equation:Left side: (-1)/(-1 - 1) = (-1)/(-2) = 1/2Right side: (-1 - 1)/(2*(-1) - 2) = (-2)/(-2 - 2) = (-2)/(-4) = 1/2Yes, both sides equal 1/2. So that works. So problem (1) solution is x = -1.Now, moving on to problem (2): solving the system of inequalities:5x - 1 > 3x - 4and-1/3 x ≤ 2/3 - xLet's solve each inequality separately first and then find the intersection.Starting with the first inequality: 5x - 1 > 3x - 4Subtract 3x from both sides: 5x - 3x -1 > -4 → 2x -1 > -4Add 1 to both sides: 2x > -3Divide by 2: x > -3/2So, solution to the first inequality is x > -1.5Now the second inequality: -1/3 x ≤ 2/3 - xLet me rewrite this for clarity: -(1/3)x ≤ (2/3) - xFirst, maybe multiply both sides by 3 to eliminate denominators. Since 3 is positive, the inequality sign remains the same.Multiplying both sides by 3: -x ≤ 2 - 3xNow, let's solve for x. Let's add 3x to both sides:(-x + 3x) ≤ 2 → 2x ≤ 2Divide both sides by 2: x ≤ 1So, the second inequality gives x ≤ 1Now, combining both inequalities: x > -1.5 and x ≤ 1. So the solution set is all real numbers greater than -1.5 and less than or equal to 1.Expressed in interval notation, that would be (-1.5, 1]. But since the problem asks for the solution set, we can write it in inequality terms or interval. Let me verify if I did all steps correctly.First inequality steps again:5x -1 > 3x -4Subtract 3x: 2x -1 > -4Add 1: 2x > -3Divide by 2: x > -3/2. Yes, that's correct.Second inequality steps:-1/3 x ≤ 2/3 - xMultiply by 3: -x ≤ 2 - 3xAdd 3x: 2x ≤ 2Divide by 2: x ≤ 1. Correct.So combining both, x must be greater than -1.5 and less than or equal to 1.Let me check if the second inequality was correctly transformed. Let's test a number in the solution set, say x = 0.Original second inequality: -1/3*0 ≤ 2/3 - 0 → 0 ≤ 2/3. True.Another number, x = 1: -1/3*1 ≤ 2/3 -1 → -1/3 ≤ -1/3. Since the inequality is ≤, equality is allowed, so that's okay.And x = -2, which is outside the solution set. Plug into the second inequality: -1/3*(-2) ≤ 2/3 - (-2) → 2/3 ≤ 2/3 + 2 → 2/3 ≤ 8/3. True. Wait, but x = -2 is less than -1.5, which is outside the first inequality. So the second inequality is true for x = -2, but the first inequality is not. So the intersection is indeed x > -1.5 and x ≤ 1.Another test for the second inequality with x = 2 (outside solution set): -1/3*2 ≤ 2/3 -2 → -2/3 ≤ -4/3. Which is false. So correct.Thus, the solution set is -1.5 < x ≤ 1. In box form, written as x ∈ (-3/2, 1]. But the problem says "Find the solution set", so maybe present it in interval notation or using inequalities. Since the original inequalities are in terms of fractions, maybe write using fractions instead of decimals.So, -3/2 < x ≤ 1.I think that's correct.Final Answer(1) boxed{-1}(2) boxed{left( -dfrac{3}{2}, 1 right]}</think>(1) To solve the equation (frac{x}{x-1} = frac{x-1}{2x-2}):First, note that (x neq 1) because the denominators would be zero. Cross-multiplying gives:[x(2x - 2) = (x - 1)^2]Expanding both sides:[2x^2 - 2x = x^2 - 2x + 1]Simplifying:[2x^2 - 2x - x^2 + 2x - 1 = 0 implies x^2 - 1 = 0]Factoring:[(x - 1)(x + 1) = 0]Since (x neq 1), the solution is (x = -1).Verification shows both sides equal (frac{1}{2}) when (x = -1). Thus, the solution is (boxed{-1}).(2) To find the solution set of the inequality system:[begin{cases}5x - 1 > 3x - 4 -frac{1}{3}x leq frac{2}{3} - xend{cases}]For the first inequality (5x - 1 > 3x - 4):[5x - 1 > 3x - 4 implies 2x - 1 > -4 implies 2x > -3 implies x > -frac{3}{2}]For the second inequality (-frac{1}{3}x leq frac{2}{3} - x):[-frac{1}{3}x leq frac{2}{3} - x implies -x leq 2 - 3x implies 2x leq 2 implies x leq 1]Combining both solutions:[-frac{3}{2} < x leq 1]Thus, the solution set is (boxed{left( -dfrac{3}{2}, 1 right]}).,<think>Okay, let's tackle these two problems one by one. Starting with problem (1): solving the equation (frac{x}{x-1} = frac{x-1}{2x-2}). Hmm, first, I need to remember how to solve rational equations. Usually, you cross-multiply to eliminate the denominators, but I have to be careful about the domain restrictions. The denominators here are (x-1) and (2x-2). Let's note that (x-1 neq 0) so (x neq 1), and (2x - 2) is the same as (2(x - 1)), so it also can't be zero, which again means (x neq 1). So, whatever solution we get, (x = 1) is excluded.Alright, moving on. Let's cross-multiply to solve the equation. So, multiply both sides by ((x - 1)(2x - 2)). But wait, since (2x - 2 = 2(x - 1)), this simplifies to multiplying by (2(x - 1)^2). Let me check:Left side: (frac{x}{x-1} times 2(x - 1)^2 = x times 2(x - 1)).Right side: (frac{x - 1}{2(x - 1)} times 2(x - 1)^2 = (x - 1) times (x - 1)).Wait, that seems right. Let's write it out step by step.Original equation:[frac{x}{x - 1} = frac{x - 1}{2x - 2}]First, note that (2x - 2 = 2(x - 1)), so rewrite the equation as:[frac{x}{x - 1} = frac{x - 1}{2(x - 1)}]Now, cross-multiplying:(x times 2(x - 1) = (x - 1) times (x - 1))Simplify both sides:Left side: (2x(x - 1))Right side: ((x - 1)^2)So the equation becomes:(2x(x - 1) = (x - 1)^2)Hmm, let's expand both sides.Left side: (2x^2 - 2x)Right side: (x^2 - 2x + 1)Bring all terms to one side:(2x^2 - 2x - x^2 + 2x - 1 = 0)Simplify:(x^2 - 1 = 0)Wait, that's straightforward. So (x^2 = 1), which gives (x = 1) or (x = -1). But hold on, earlier we established that (x neq 1) because the denominators would be zero. Therefore, the only solution is (x = -1).Let me double-check. Substitute (x = -1) into the original equation:Left side: (frac{-1}{-1 - 1} = frac{-1}{-2} = frac{1}{2})Right side: (frac{-1 - 1}{2(-1) - 2} = frac{-2}{-2 - 2} = frac{-2}{-4} = frac{1}{2})Yes, both sides are equal. So (x = -1) is the valid solution. Great.Now, moving on to problem (2): solving the system of inequalities:[begin{cases}5x - 1 > 3x - 4 -frac{1}{3}x leq frac{2}{3} - xend{cases}]Let's solve each inequality separately and then find the intersection of the solution sets.Starting with the first inequality: (5x - 1 > 3x - 4)Subtract (3x` from both sides: (2x - 1 > -4)Add 1 to both sides: (2x > -3)Divide by 2: (x > -frac{3}{2})So the first inequality gives (x > -1.5).Now the second inequality: (-frac{1}{3}x leq frac{2}{3} - x)Let's eliminate the fractions by multiplying both sides by 3 to make it simpler. Since 3 is positive, the inequality direction remains the same.Multiply all terms by 3: (-x leq 2 - 3x)Now, add (3x` to both sides to get rid of the variable on the right:(-x + 3x leq 2)Which simplifies to:(2x leq 2)Divide both sides by 2: (x leq 1)So the second inequality gives (x leq 1).Now, combining both inequalities: (x > -frac{3}{2}) and (x leq 1). So the solution set is all real numbers greater than (-frac{3}{2}) and up to and including 1. In interval notation, this would be ((-1.5, 1]).Let me check if I did each step correctly. For the first inequality:Original: (5x -1 > 3x -4)Subtract 3x: (2x -1 > -4)Add 1: (2x > -3) → (x > -3/2). That looks good.Second inequality: (-frac{1}{3}x leq frac{2}{3} - x)Multiply by 3: (-x ≤ 2 - 3x)Add 3x: (2x ≤ 2) → (x ≤ 1). Correct.Intersection is indeed (x) values greater than (-1.5) and less than or equal to 1. So the solution set is (-1.5 < x leq 1).Just to be thorough, let's pick a number in this interval, say x = 0.First inequality: (5(0) -1 = -1 > 3(0) -4 = -4). True.Second inequality: (-1/3*(0) = 0 ≤ 2/3 -0 = 2/3. True.Another test: x = 1.First inequality: (5(1)-1=4 > 3(1)-4=-1). True.Second inequality: (-1/3*1 = -1/3 ≤ 2/3 -1 = -1/3. So equality holds, which is allowed. Correct.Test x = -2 (outside the interval):First inequality: 5*(-2)-1=-11 > 3*(-2)-4=-10? -11 > -10 is false. Correctly excluded.Test x = 2 (outside the interval):Second inequality: -1/3*2 = -2/3 ≤ 2/3 -2 = -4/3. Is -2/3 ≤ -4/3? No. So x=2 is excluded. Correct.Therefore, the solution set is indeed all real numbers between -1.5 and 1, including 1 but not including -1.5.Final Answer(1) (boxed{-1})(2) (boxed{(-frac{3}{2}, 1]})</think>(1) To solve the equation (frac{x}{x-1} = frac{x-1}{2x-2}):First, note that (2x - 2 = 2(x - 1)), so rewrite the equation as:[frac{x}{x - 1} = frac{x - 1}{2(x - 1)}]Cross-multiplying gives:[x times 2(x - 1) = (x - 1)^2]Simplifying both sides:[2x(x - 1) = (x - 1)^2]Expanding and simplifying:[2x^2 - 2x = x^2 - 2x + 1][x^2 - 1 = 0][x^2 = 1]The solutions are (x = 1) or (x = -1). However, (x = 1) is excluded because it makes the denominators zero. Therefore, the only solution is (x = -1).(boxed{-1})(2) To find the solution set of the inequality system:[begin{cases}5x - 1 > 3x - 4 -frac{1}{3}x leq frac{2}{3} - xend{cases}]For the first inequality (5x - 1 > 3x - 4):[5x - 1 > 3x - 4][2x - 1 > -4][2x > -3][x > -frac{3}{2}]For the second inequality (-frac{1}{3}x leq frac{2}{3} - x):[-frac{1}{3}x leq frac{2}{3} - x]Multiplying by 3:[-x leq 2 - 3x][2x leq 2][x leq 1]Combining both inequalities, the solution set is:[-frac{3}{2} < x leq 1](boxed{(-frac{3}{2}, 1]})]

🔑: Solution:# Part 1: Solve the equation frac{x}{x-1}=frac{x-1}{2x-2}1. Start with the given equation: frac{x}{x-1}=frac{x-1}{2x-2}.2. Notice that 2x-2 can be factored as 2(x-1), so rewrite the equation as: frac{x}{x-1}=frac{x-1}{2(x-1)}.3. To eliminate the denominators, multiply both sides by 2(x-1): 2(x-1) cdot frac{x}{x-1} = 2(x-1) cdot frac{x-1}{2(x-1)}.4. Simplify both sides: 2x = x-1.5. Solve for x: 2x - x = -1 Rightarrow x = -1.6. Check the solution: When x=-1, the denominator 2(x-1) neq 0, so x=-1 is a valid solution.Therefore, the solution to the equation is boxed{x=-1}.# Part 2: Find the solution set of the inequality system left{begin{array}{l}5x-1>3x-4-frac{1}{3}xleqfrac{2}{3}-xend{array}right.1. Solve the first inequality 5x-1>3x-4: - Subtract 3x from both sides: 2x - 1 > -4. - Add 1 to both sides: 2x > -3. - Divide by 2: x > -frac{3}{2}.2. Solve the second inequality -frac{1}{3}x leq frac{2}{3} - x: - Add x to both sides: -frac{1}{3}x + x leq frac{2}{3}. - Combine like terms: frac{2}{3}x leq frac{2}{3}. - Divide by frac{2}{3}: x leq 1.Therefore, the solution set of the inequality system is boxed{-1.5 < x leq 1}.

🧠:[<think>Okay, so I need to find the value of the sum [log₂1] + [log₂2] + [log₂3] + ... + [log₂2002]. Hmm, let me start by recalling what the brackets mean here. Usually, in math problems, brackets can denote different things like the floor function, ceiling function, or maybe even the integer part. Since the problem is about logarithms, which can often result in non-integer values, I think it's likely that the brackets here represent the floor function. The floor function of a number is the greatest integer less than or equal to that number. So, for example, [3.7] would be 3, and [5] would just be 5. Let me confirm that assumption first. If the problem uses another notation, maybe I should check, but given the context, floor function makes sense here. So I'll proceed under that assumption.So, the problem is asking for the sum of the floor of log base 2 of each integer from 1 to 2002. That is, for each n from 1 to 2002, compute floor(log₂(n)) and add them all up. Let me write that out more formally. The sum S is:S = Σ_{n=1}^{2002} floor(log₂(n))Alright, so my goal is to compute this sum. Let's think about how to approach this. Maybe it's helpful to consider the values of log₂(n) and when the floor of those values increases by 1. Since log₂(n) is a monotonically increasing function, each time n crosses a power of 2, the value of log₂(n) increases by 1, and hence the floor value would also increase by 1.For example, log₂(1) = 0, so floor(log₂(1)) = 0. Then log₂(2) = 1, so floor(log₂(2)) = 1. Then log₂(3) is approximately 1.58496, so floor(log₂(3)) = 1. This continues until n=4, where log₂(4)=2, so floor(log₂(4))=2. Similarly, between 4 and 8, the floor value is 2, and so on.Therefore, the floor(log₂(n)) is equal to k when n is in the interval [2^k, 2^(k+1) - 1]. For each integer k, the numbers n from 2^k to 2^(k+1)-1 will have floor(log₂(n)) = k. So, if I can determine how many numbers fall into each interval [2^k, 2^(k+1)-1], then multiply that count by k, and sum over all relevant k, I can compute the total sum S.Let me formalize that idea. Let k be an integer such that 2^k ≤ n < 2^(k+1). Then floor(log₂(n)) = k. The number of integers n in this interval is (2^(k+1) - 1) - 2^k + 1 = 2^(k+1) - 1 - 2^k + 1 = 2^(k+1) - 2^k = 2^k. So, for each k, there are 2^k numbers n where floor(log₂(n)) = k. However, this is only true up until the point where 2^(k+1) -1 is less than or equal to 2002. The last interval might be shorter if 2002 is not exactly one less than a power of 2.So, first, let me find the maximum k such that 2^k ≤ 2002. That is, find k_max where k_max = floor(log₂(2002)). Let me compute log₂(2002). Since 2^10 = 1024, 2^11 = 2048. So log₂(2002) is slightly less than 11, because 2^11 = 2048. Therefore, floor(log₂(2002)) = 10. So the values of k will range from 0 up to 10. However, for k=10, the interval is [2^10, 2^11 -1] = [1024, 2047], but our upper limit is 2002, so the last interval will be [1024, 2002], which has fewer numbers than 2^10. Let me check:For each k from 0 to 10:- When k=0: [2^0, 2^1 -1] = [1,1], numbers 1. Count = 1- k=1: [2,3], count=2- k=2: [4,7], count=4- k=3: [8,15], count=8- ...- k=10: [1024, 2047], but since our upper limit is 2002, the count here is 2002 - 1024 + 1 = 979Wait, let me verify that. For k=10, the interval would normally be [2^10, 2^11 -1] = [1024, 2047]. However, our upper limit is 2002, which is less than 2047. Therefore, the count for k=10 is 2002 - 1024 +1 = 979. Let me compute that: 2002 - 1024 = 978, plus 1 is 979. Okay.For k from 0 to 9, the counts are each 2^k, and for k=10, it's 979. Therefore, the total sum S would be the sum from k=0 to 9 of (k * 2^k) ) + (10 * 979). Then, we can compute this.So, the formula is:S = Σ_{k=0}^9 k*2^k + 10*979Therefore, the problem reduces to computing this sum. Now, I need to compute Σ_{k=0}^9 k*2^k. There is a known formula for Σ_{k=0}^n k*r^k. The formula for the sum Σ_{k=0}^n k*r^k is r*(1 - (n+1)*r^n + n*r^(n+1)) / (1 - r)^2. Since here r=2, the formula becomes:Σ_{k=0}^n k*2^k = 2 - (n+1)*2^(n+1) + n*2^(n+2)) / (1 - 2)^2Wait, maybe I should recall the exact formula. Alternatively, there's another way. Let me recall that Σ_{k=0}^n k*2^k.Let me consider S = Σ_{k=0}^n k*2^kMultiply both sides by 2:2S = Σ_{k=0}^n k*2^(k+1) = Σ_{k=1}^{n+1} (k-1)*2^kSubtract original S from this:2S - S = S = Σ_{k=1}^{n+1} (k-1)*2^k - Σ_{k=0}^n k*2^kLet me write out the terms:Σ_{k=1}^{n+1} (k-1)*2^k = Σ_{k=1}^{n} (k-1)*2^k + (n)*2^{n+1}Σ_{k=0}^n k*2^k = 0*2^0 + Σ_{k=1}^n k*2^k = Σ_{k=1}^n k*2^kTherefore, S = [Σ_{k=1}^n (k-1)*2^k + n*2^{n+1}] - Σ_{k=1}^n k*2^k= Σ_{k=1}^n (k-1 - k)*2^k + n*2^{n+1}= Σ_{k=1}^n (-1)*2^k + n*2^{n+1}= -Σ_{k=1}^n 2^k + n*2^{n+1}Σ_{k=1}^n 2^k is a geometric series, which is 2^(n+1) - 2Therefore,S = - (2^(n+1) - 2) + n*2^(n+1) = -2^(n+1) + 2 + n*2^(n+1) = (n*2^(n+1) - 2^(n+1)) + 2 = 2^(n+1)(n - 1) + 2Alternatively, factor:= (n - 1)*2^(n+1) + 2Therefore, the formula for S = Σ_{k=0}^n k*2^k is (n - 1)*2^(n+1) + 2Let me verify this formula with a small n. Let's take n=1:Σ_{k=0}^1 k*2^k = 0*1 + 1*2 = 2Using the formula: (1 -1)*2^(2) + 2 = 0 + 2 = 2. Correct.n=2:Σ_{k=0}^2 k*2^k = 0 + 2 + 8 = 10Formula: (2 -1)*2^3 + 2 = 1*8 + 2 = 10. Correct.n=3:Sum: 0 + 2 + 8 + 24 = 34Formula: (3 -1)*2^4 + 2 = 2*16 + 2 = 34. Correct.Okay, so the formula works. Therefore, for our case, when n=9:Σ_{k=0}^9 k*2^k = (9 -1)*2^(10) + 2 = 8*1024 + 2 = 8192 + 2 = 8194Wait, let's compute that:(9 -1) = 82^10 = 10248*1024 = 81928192 + 2 = 8194Therefore, the sum from k=0 to 9 of k*2^k is 8194. Then, we have the term for k=10, which is 10*979 = 9790.So total sum S = 8194 + 9790 = 17984Wait, let me compute 8194 + 9790:8000 + 9000 = 17000194 + 790 = 98417000 + 984 = 17984Therefore, the total sum is 17,984. Hmm, but let me check this again step by step to make sure I didn't make a mistake.First, for the sum from k=0 to 9 of k*2^k, which is 8194. Then, the sum for k=10 is 10*979=9790. So total sum is 8194 + 9790 = 17984.But let me verify the formula once again for n=9. If the formula is (n -1)*2^(n+1) +2, then for n=9:(9 -1)*2^(10) +2 =8*1024 +2=8192 +2=8194. That's correct.Then the term for k=10 is 10*(2002 -1024 +1). Wait, how did we get 979? Let me verify that.For k=10, the interval should be [2^10, 2^11 -1] = [1024, 2047]. But since our upper limit is 2002, the number of terms is 2002 -1024 +1. Let's compute that:2002 - 1024 = 978, then 978 +1=979. Yes, correct. So 979 terms, each contributing 10. So 10*979=9790. Adding that to 8194 gives 17984. So the answer should be 17984.But let me test this with a smaller number to see if my reasoning holds. Suppose instead of 2002, we take n=10. Then the sum would be [log₂1] + [log₂2] + ... + [log₂10]. Let's compute this manually and see if the formula works.For n=10:floor(log₂1)=0floor(log₂2)=1floor(log₂3)=1floor(log₂4)=2floor(log₂5)=2floor(log₂6)=2floor(log₂7)=2floor(log₂8)=3floor(log₂9)=3floor(log₂10)=3So sum is 0 +1 +1 +2 +2 +2 +2 +3 +3 +3 = 0 +1*2 +2*4 +3*3=0 +2 +8 +9=19.Using the formula: first find k_max. floor(log₂10)=3, so k from 0 to 3. However, the upper limit is 10, which is less than 2^4 -1=15.For k=0: [1,1], count=1k=1: [2,3], count=2k=2: [4,7], count=4k=3: [8,15], but upper limit is 10, so count=10-8+1=3Therefore, sum=0*1 +1*2 +2*4 +3*3=0 +2 +8 +9=19. Correct.Using the formula:Σ_{k=0}^2 k*2^k + 3*3= ?First, Σ_{k=0}^2 k*2^k. Using the formula, (2 -1)*2^3 +2=1*8 +2=10. Then add 3*3=9, total sum=10 +9=19. Correct. So the formula works here.Another test case: n=7. Then k_max=2, since 2^3=8>7. So sum would be:[log₂1] + [log₂2] + [log₂3] + ... +[log₂7]Which is 0 +1 +1 +2 +2 +2 +2 =0 +1*2 +2*4= 0 +2 +8=10.Using formula: Σ_{k=0}^1 k*2^k + 2*(7 -4 +1)= ?Σ_{k=0}^1 k*2^k=(1 -1)*2^2 +2=0 +2=2. Then 2*(7 -4 +1)=2*4=8. Total=2 +8=10. Correct.So the formula seems to work. Therefore, applying this to the original problem, the sum S=17984. Therefore, the answer should be 17984. However, let me check once again for possible mistakes.Wait, let me check the calculation of 8*1024. 8*1000=8000, 8*24=192, so 8000+192=8192, then add 2 gives 8194. Correct.Then 2002 -1024 +1: 2002-1024=978, 978+1=979. 10*979=9790. Then 8194 +9790: 8000+9000=17000, 194+790=984, total 17000+984=17984. Correct.But let me check for k=0. Wait, when k=0, the interval is [1,1], which is just 1. The formula counts 2^0=1 term, which is correct. For k=1, [2,3], 2 terms, which is 2^1=2. Correct. So up to k=9, which would normally be [512,1023], which has 512 numbers (1023-512 +1=512). Then k=10: [1024,2047], but we have only up to 2002, so 2002-1024 +1=979. Correct.Therefore, all steps check out. Therefore, the answer should be boxed{17984}Final Answerboxed{17984}</think>To find the value of the sum (left[log _{2} 1right]+left[log _{2} 2right]+left[log _{2} 3right]+cdots+left[log _{2} 2002right]), we need to compute the floor values of logarithms base 2 for each integer from 1 to 2002 and sum them up.The floor function (leftlfloor log_2 n rightrfloor) gives the greatest integer (k) such that (2^k leq n < 2^{k+1}). For each (k), the integers (n) that satisfy (2^k leq n < 2^{k+1}) are in the interval ([2^k, 2^{k+1} - 1]). The number of integers in this interval is (2^k) except for the last interval, which might be shorter if 2002 is not a power of 2.First, determine the maximum (k) such that (2^k leq 2002). Since (2^{10} = 1024) and (2^{11} = 2048), the maximum (k) is 10. For (k) from 0 to 9, the intervals are ([2^k, 2^{k+1} - 1]) with (2^k) numbers each. For (k = 10), the interval is ([1024, 2002]) with (2002 - 1024 + 1 = 979) numbers.The sum can be split into two parts: the sum from (k = 0) to 9 and the term for (k = 10):[S = sum_{k=0}^{9} k cdot 2^k + 10 cdot 979]Using the formula for the sum (sum_{k=0}^{n} k cdot 2^k = (n - 1) cdot 2^{n+1} + 2), we compute the sum for (k = 0) to 9:[sum_{k=0}^{9} k cdot 2^k = (9 - 1) cdot 2^{10} + 2 = 8 cdot 1024 + 2 = 8192 + 2 = 8194]Adding the term for (k = 10):[10 cdot 979 = 9790]Thus, the total sum is:[S = 8194 + 9790 = 17984]The value of the sum is (boxed{17984}).,<think>Okay, so I need to find the value of the sum where each term is the floor of the logarithm base 2 of numbers from 1 up to 2002. That is, compute [log₂1] + [log₂2] + [log₂3] + ... + [log₂2002]. Hmm, floor functions can sometimes be tricky, but maybe there's a pattern here.First, let me recall that the floor of a logarithm base 2 of a number gives the highest power of 2 that is less than or equal to that number. For example, log₂8 is exactly 3, so the floor is 3. But log₂9 is approximately 3.1699, so the floor is 3. So, in general, [log₂n] = k if 2ᵏ ≤ n < 2ᵏ⁺¹. So for each integer k, all numbers n in the interval [2ᵏ, 2ᵏ⁺¹) will have [log₂n] = k.Therefore, to compute the sum, I can break the range from 1 to 2002 into intervals where the floor value is constant. For each k, determine how many numbers n satisfy 2ᵏ ≤ n < 2ᵏ⁺¹, and then multiply that count by k, and sum all those products.But first, I need to find the maximum k such that 2ᵏ ≤ 2002. Let me compute log₂2002. Since 2¹⁰ = 1024, 2¹¹ = 2048. So log₂2002 is between 10 and 11. Specifically, since 2¹⁰ = 1024, 2¹¹ = 2048, so 2002 is less than 2048, so [log₂2002] = 10. Therefore, the values of k in the sum will range from 0 up to 10.Wait, but when n=1, log₂1 = 0, so [log₂1] = 0. So starting from n=1, the first term is 0. Then from n=2, log₂2 = 1, floor is 1. So for n=2, the floor is 1. Then for n=3, log₂3 is approx 1.584, so floor is 1. So n=2 and 3 (which are 2 numbers) have floor 1. Then n=4, log₂4=2, so floor is 2. Then n=4,5,6,7: 4 to 7, which are 4 numbers, floor is 2. Similarly, n=8 to 15 (8 numbers) floor is 3, and so on.So in general, for each k ≥ 0, the numbers n from 2ᵏ to 2ᵏ⁺¹ - 1 will have [log₂n] = k. The count of numbers in each interval is 2ᵏ⁺¹ - 2ᵏ = 2ᵏ. Except for the last interval, which might be shorter if 2002 is not exactly a power of 2.So the plan is:1. For k from 0 to 10, compute the number of terms where [log₂n] = k.2. Multiply each count by k.3. Sum all these products.First, let's confirm the maximum k needed. Since 2¹⁰ = 1024, 2¹¹ = 2048. But 2002 is less than 2048, so the highest interval is from 2¹⁰ = 1024 up to 2002. So for k=10, the number of terms is 2002 - 1024 + 1 = 979. Wait, let me check that.Wait, the intervals are [2ᵏ, 2ᵏ⁺¹). So for k=10, the interval would be [1024, 2048), but since we're only going up to 2002, the actual interval is [1024, 2002]. Therefore, the number of terms here is 2002 - 1024 + 1 = 979. The +1 is because both endpoints are inclusive.For the other k values, from k=0 up to k=9, the intervals are [2ᵏ, 2ᵏ⁺¹ - 1], so the number of terms is 2ᵏ. For example:- k=0: [1,1], count=1- k=1: [2,3], count=2- k=2: [4,7], count=4- k=3: [8,15], count=8- ...- k=9: [512,1023], count=512Then k=10: [1024,2002], count=2002 - 1024 +1 = 979.So the total sum would be:Sum = Σ (from k=0 to 9) [k * 2ᵏ] + 10 * 979Wait, let me check that.Wait, for k=0: [1,1], count=1, so term is 0*1=0k=1: [2,3], count=2, term is 1*2=2k=2: [4,7], count=4, term is 2*4=8k=3: [8,15], count=8, term is 3*8=24...k=9: [512,1023], count=512, term is 9*512=4608Then k=10: [1024,2002], count=979, term is 10*979=9790Therefore, the total sum is the sum from k=0 to 9 of k*2ᵏ plus 9790.But we need to compute the sum Σ (k*2ᵏ) from k=0 to 9.There is a formula for the sum Σ (k*r^k) from k=0 to n. The formula is r*(1 - (n+1)*r^n + n*r^(n+1)) / (1 - r)^2 when r ≠ 1. Since here r=2, we can use this.Alternatively, I can compute it manually, but since we need to compute up to k=9, maybe the formula is quicker.Let me recall the formula:Σ_{k=0}^n k*2^k = 2 - (n+2)*2^{n+1}Wait, let me check for a small n. For n=0: 0*2^0 = 0. The formula would be 2 - (0+2)*2^{0+1} = 2 - 2*2 = 2 - 4 = -2, which is not correct. Hmm, maybe the formula is different.Wait, perhaps the formula is Σ_{k=0}^n k*2^k = (n-1)*2^{n+1} + 2. Let's test for n=1:Sum is 0*1 +1*2 = 2. Formula gives (1-1)*2^{2} +2 = 0 +2 =2. Correct.For n=2: 0*1 +1*2 +2*4 = 0 +2 +8=10. Formula: (2-1)*2^{3} +2=1*8 +2=10. Correct.For n=3: Sum is 0 +2 +8 +3*8= 2 +8 +24=34. Formula: (3-1)*2^4 +2=2*16 +2=34. Correct. So the formula is (n-1)*2^{n+1} +2.Therefore, Σ_{k=0}^n k*2^k = (n -1)*2^{n+1} +2.Therefore, for n=9, Σ_{k=0}^9 k*2^k = (9 -1)*2^{10} +2 =8*1024 +2= 8192 +2=8194.But let me verify for n=3 again with the formula:(n-1)*2^{n+1} +2 = (3-1)*2^4 +2 = 2*16 +2=32 +2=34. Which matches. So formula seems correct.So for n=9, it's 8*2^{10} +2=8*1024 +2=8192 +2=8194.Therefore, the sum from k=0 to 9 of k*2ᵏ is 8194.But wait, let's check for n=0. If n=0, formula would be (0 -1)*2^{1} +2= -1*2 +2=0. Which matches because sum is 0*1=0. Correct.So yes, formula works.Therefore, our sum Σ_{k=0}^9 k*2^k =8194.But wait, hold on, in our problem, when k=0, the term is 0*1=0. When k=1, it's 1*2=2, etc. So the formula gives the sum from k=0 to n, which is exactly what we need.Therefore, the sum from k=0 to 9 is 8194. Then we have the last term for k=10, which is 10*979=9790. So total sum is 8194 + 9790 = 17984.Wait, but wait a second. Let me verify this because sometimes these formulas can be tricky.Alternatively, maybe compute the sum step by step:For k=0: 0*1=0k=1:1*2=2k=2:2*4=8k=3:3*8=24k=4:4*16=64k=5:5*32=160k=6:6*64=384k=7:7*128=896k=8:8*256=2048k=9:9*512=4608So adding these up:0 (k=0)+2 (k=1) = 2+8 (k=2) =10+24 (k=3) =34+64 (k=4) =98+160 (k=5)=258+384 (k=6)=642+896 (k=7)=1538+2048 (k=8)=3586+4608 (k=9)=8194Yes, that matches the formula's result. So the sum up to k=9 is indeed 8194.Then the last term is k=10, which is 9790. So total sum is 8194 + 9790 = 17984.Wait, but let me check the count for k=10 again. The interval is from 1024 to 2002 inclusive. So the number of terms is 2002 - 1024 +1.Calculating that:2002 - 1024 = 978978 +1 = 979. So 979 terms, each contributing 10, so 10*979=9790. Correct.Therefore, total sum is 8194 + 9790 = 17984.But let me double-check with another approach to be safe.Alternatively, we can note that each term [log₂n] is equal to the number of times 2 fits into n when taking powers. But another way is to think about for each number from 1 to 2002, compute the floor of log2(n) and add them all up.But given that the formula approach aligns with the step-by-step computation, and the counts for each k seem correct, 17984 should be the answer.But let me test with a smaller number to see if the method works. For example, compute the sum up to n=7.Compute manually:[log₂1]=0, [log₂2]=1, [log₂3]=1, [log₂4]=2, [log₂5]=2, [log₂6]=2, [log₂7]=2Sum: 0 +1 +1 +2 +2 +2 +2= 10.Using the formula:For k=0: [1,1], count=1, term=0*1=0k=1: [2,3], count=2, term=1*2=2k=2: [4,7], count=4, term=2*4=8But since n=7, which is in k=2, the interval [4,7] is 4 numbers. So sum is 0 +2 +8 =10. Which matches. So the formula works here.Another test case: n=8.Sum up to n=8:[log₂1]=0, [log₂2]=1, [log₂3]=1, [log₂4]=2, [log₂5]=2, [log₂6]=2, [log₂7]=2, [log₂8]=3Sum: 0 +1 +1 +2 +2 +2 +2 +3=13.Using the formula:k=0: 0*1=0k=1:1*2=2k=2:2*4=8k=3:3*1=3 (since n=8 is in k=3, which is [8,15], but since we stop at 8, count=1). So sum=0+2+8+3=13. Correct.Therefore, the method is validated. Therefore, the total sum for the original problem is 17984. Wait, but let me check once again the calculation:Sum up to k=9:8194Plus k=10:10*979=97908194 + 9790: 8000 + 9000 =17000; 194 +790=984; total 17000 + 984=17984. Yes.Therefore, the answer is 17984.But hold on, let me cross-verify with another approach. For each number from 1 to 2002, [log₂n] is the exponent of the highest power of 2 less than or equal to n. So another way to compute the sum is to note that for numbers in [2ᵏ, 2ᵏ⁺¹), their contribution is k*(2ᵏ⁺¹ - 2ᵏ) except for the last interval. So for k from 0 to 10.But for k=0: [1,1], count=1, which is 2⁰k=1: [2,3], count=2=2¹k=2: [4,7], count=4=2²...k=9: [512,1023], count=512=2⁹k=10: [1024,2002], count=2002 -1024 +1=979So the sum is Σ_{k=0}^9 k*2ᵏ +10*979. Which is exactly what we computed before. So that's consistent.Hence, confident that 17984 is the correct answer.Final Answerboxed{17984}</think>To find the value of the sum (left[log _{2} 1right]+left[log _{2} 2right]+left[log _{2} 3right]+cdots+left[log _{2} 2002right]), we need to consider the floor function of logarithms base 2. The floor value (left[log_{2} nright]) is the highest integer (k) such that (2^k leq n < 2^{k+1}).1. Identify the intervals: For each integer (k), the numbers (n) in the interval ([2^k, 2^{k+1})) contribute (k) to the sum. The number of terms in each interval is (2^k) for (k) from 0 to 9. For (k = 10), the interval is ([1024, 2002]) with (2002 - 1024 + 1 = 979) terms.2. Sum the contributions: - For (k) from 0 to 9, the sum of contributions is (sum_{k=0}^{9} k cdot 2^k). - For (k = 10), the contribution is (10 cdot 979).3. Calculate the sum: - Using the formula for the sum (sum_{k=0}^{n} k cdot 2^k = (n - 1) cdot 2^{n+1} + 2), we find: [ sum_{k=0}^{9} k cdot 2^k = (9 - 1) cdot 2^{10} + 2 = 8 cdot 1024 + 2 = 8192 + 2 = 8194 ] - For (k = 10), the contribution is (10 cdot 979 = 9790).4. Total sum: - Adding the contributions from all intervals: [ 8194 + 9790 = 17984 ]Thus, the value of the sum is (boxed{17984}).]

🔑:Solution. From the definition of the logarithm, it follows thatlog _{2} 1=0, quad log _{2} 2=1, quad log _{2} 4=2and so on, the function always takes the next integer value at the next power of 2. We also know that the log _{2} x function is monotonically increasing. The left[log _{2} xright] function takes the smaller value between two powers of 2 as many times as there are integers between the two powers of 2; for example, the value 2 is taken at log _{2} 4, log _{2} 5, log _{2} 6, log _{2} 7.The summation should be continued until we reach the value 2002. Since 2^{10}=1024<2002, the last value will be 10, which the function will take at 2003-1024=979 points.Thus, the value of the sought sum is:begin{aligned}& (0+2 cdot 1+4 cdot 2+8 cdot 3+ldots+512 cdot 9)+979 cdot 10= = & left(sum_{k=1}^{9} k cdot 2^{k}right)+9790=left(sum_{k=1}^{9} 2^{k}right)+left(sum_{k=2}^{9} 2^{k}right)+ldots+left(sum_{k=9}^{9} 2^{k}right)+9790= = & left(2^{10}-2right)+left(2^{10}-2^{2}right)+ldots+left(2^{10}-2^{9}right)+9790= = & 9 cdot 2^{10}-left(2^{10}-2right)+9790=2^{13}+9792=17984 .end{aligned}

🧠:[<think>Alright, let's try to tackle this problem. So, we have 2006 distinct complex numbers with the property that the product of any two of them (including multiplying a number by itself) is also one of these 2006 numbers. We need to find the sum of all these numbers. Hmm, interesting. Let me think step by step.First, let's parse the problem again. We have 2006 distinct complex numbers. Let's call this set S. The key property is that for any two elements a and b in S (they could be the same), the product a*b is also in S. So, the set is closed under multiplication, even when multiplying an element by itself. That seems like a multiplicative closure property. So, S is a finite set closed under multiplication, which makes me think of groups in abstract algebra. But I'm not sure if it's a group because we might not have inverses. Wait, but maybe inverses are required here? Let's see.If S is closed under multiplication, then for any element a in S, all its powers must be in S. But since S is finite, the powers of a must eventually repeat. So, a must be a root of unity. That makes sense because roots of unity are finite in number (there are exactly n nth roots of unity) and their products are also roots of unity. However, in this problem, the set S has 2006 elements, which is a specific number. So, maybe S is a group under multiplication? If so, then it's a finite multiplicative group of complex numbers, which must be cyclic. All finite multiplicative subgroups of the complex numbers are cyclic, generated by some root of unity. So, the group would consist of all the nth roots of unity for some n. But in that case, the number of elements would be n. However, here n is 2006. So, is S the set of 2006th roots of unity? But wait, the problem states that the product of any two elements is also in S. If S is the set of 2006th roots of unity, then multiplying any two elements would give another 2006th root of unity, right? Because (e^(2πik/2006)) * (e^(2πil/2006)) = e^(2πi(k+l)/2006), which is another 2006th root of unity. However, the set of 2006th roots of unity has exactly 2006 elements, all distinct. So, that seems to satisfy the problem's conditions. Then, the sum of all 2006th roots of unity is zero. Because the sum of all nth roots of unity is always zero for n ≥ 1. Is that correct?Wait, let me recall. The sum of all nth roots of unity is indeed zero. That's a well-known result. The roots of unity are the solutions to z^n = 1, and the sum is the coefficient of z^{n-1} in the polynomial z^n - 1, which is zero. Therefore, the sum of all nth roots of unity is zero. So, if S is the set of 2006th roots of unity, then the sum would be zero. Therefore, the answer might be zero. But is that the only possibility?Wait, the problem says there are 2006 distinct complex numbers such that the product of any two is one of these numbers. So, S is closed under multiplication. If S is closed under multiplication and contains 2006 elements, which are roots of unity, then it must be a multiplicative group. Because for closure, we need that if a and b are in S, then a*b is in S. Also, since all elements are roots of unity, each element has finite order, so every element has an inverse (which is also a root of unity, hence in S). Therefore, S is a finite multiplicative group, hence cyclic. Therefore, S is the group of nth roots of unity where n is 2006. Therefore, the sum is zero.But wait, let me confirm this. Suppose n is 2006. Then the group of nth roots of unity has exactly 2006 elements, all distinct. So, the problem states that the set has 2006 distinct complex numbers closed under multiplication. Then, such a set must be a multiplicative group, hence cyclic, hence the sum is zero. So, the answer is zero. But maybe there's another possibility? Maybe S is not the entire group but some other set? Hmm.Wait, suppose S is not a group but just closed under multiplication. Then, even if it's closed under multiplication, if it's finite, then it must contain the inverses. Because in a finite semigroup, every element has an idempotent power, but in this case, since all elements are complex numbers, and the set is closed under multiplication, inverses must be present. Because if a is in S, then a^k is in S for all k. But since S is finite, a^k must cycle, so a^m = 1 for some m, hence a^{m-1} is the inverse of a. Therefore, S is a group. So, S is a finite multiplicative group of complex numbers, hence cyclic. Therefore, S is the group of nth roots of unity, where n is 2006, so the sum is zero.Therefore, the answer should be zero. But let me check for a smaller case. Suppose n=2. If we have two complex numbers closed under multiplication. The set {1, -1} is closed under multiplication, and the sum is 1 + (-1) = 0. Similarly, for n=3, the cube roots of unity sum to zero. So, in general, the sum of all nth roots of unity is zero. Therefore, if S is such a group, the sum is zero. So, given that in the problem S has 2006 elements, which is the same as the group of 2006th roots of unity, the sum should be zero. Therefore, the answer is boxed{0}.Wait, but the problem says "the product of any two of them (including self-multiplication) is one of these 2006 numbers". So, does this mean that even if we take two same elements, their product is in S? So, for any a in S, a^2 must be in S. Similarly, a^3, etc. But if S is a group, then that's automatically satisfied because a^k would be in S for any integer k. However, if S is not a group, but just closed under multiplication, including self-multiplication, then it's a multiplicative subsemigroup of the complex numbers. However, in the complex numbers, a finite multiplicative subsemigroup is necessarily a group. Because in a finite semigroup, every element has an idempotent power, but in the multiplicative semigroup of complex numbers, the only idempotent elements are 0 and 1. Since all elements are non-zero (since 0 multiplied by anything is 0, which would require 0 to be in the set if there's any zero, but the set has 2006 distinct elements, so 0 can't be there because 0 multiplied by any other element is 0, which would have to be in the set, but 0 is only one element. So, if there's a 0, then the product of 0 and any other element is 0, which is in the set, but the other elements multiplied together must also stay in the set. But since 0 is in the set, and the rest are non-zero, but multiplying a non-zero element by 0 gives 0. However, since all elements must be distinct, and there's 2006 elements, if 0 is included, then the rest 2005 elements must be closed under multiplication among themselves as well as with 0. But the problem says "the product of any two of them (including self-multiplication) is one of these 2006 numbers". If 0 is one of them, then multiplying 0 by any other element gives 0, which is allowed. However, the rest 2005 elements must also be closed under multiplication. But if they are non-zero, then they form a finite multiplicative subsemigroup, hence a group. But 2005 is an odd number, and multiplicative groups of complex numbers must have even order if they include -1. Wait, actually, the order can be any number. For example, the nth roots of unity have order n. So, 2005th roots of unity would have order 2005, which is allowed. However, in the problem, the total number is 2006 elements, including 0. But if there's 0, then the rest 2005 elements must form a group under multiplication, which is possible only if 0 is not in the set. Wait, because if 0 is in S, then multiplying 0 by any element gives 0, so closure is maintained, but the other elements must be closed under multiplication as well. However, the problem states that all 2006 elements are distinct. If 0 is in the set, then the other 2005 elements must be non-zero and closed under multiplication, hence forming a group. But 2005 is an odd number. The multiplicative group of complex numbers must be cyclic, and the cyclic group of order 2005 exists (the 2005th roots of unity). So, in that case, S would consist of 0 and all 2005th roots of unity. But wait, 0 multiplied by any root of unity is 0, which is in S, and the product of any two roots of unity is another root of unity, which is in S. However, 0 is a distinct element. But the problem states that the product of any two elements is one of the 2006 numbers. So, in this case, the product of 0 and any element is 0, which is in S. But in that case, the sum would be 0 plus the sum of the 2005th roots of unity. The sum of the 2005th roots of unity is zero, so the total sum would be 0 + 0 = 0. Therefore, even if 0 is included, the sum is still zero. But wait, in this case, S has 1 + 2005 = 2006 elements. So, maybe S could be the union of 0 and the 2005th roots of unity. However, the problem says "the product of any two of them (including self-multiplication) is one of these 2006 numbers". If 0 is included, then multiplying 0 by itself is 0, which is okay. Multiplying 0 by any root of unity is 0, which is okay. Multiplying two roots of unity gives another root of unity, which is in S. So, this set is closed under multiplication. But are all these elements distinct? The roots of unity and 0 are all distinct because 0 is not a root of unity. So, yes, they are distinct. Therefore, such a set S could be {0} ∪ {2005th roots of unity}, which has 2006 elements. In that case, the sum would be 0 + sum of 2005th roots of unity, which is 0. So, regardless of whether 0 is included or not, the sum is zero.Wait, but if S includes 0, then there must be an element 0, but multiplying 0 by any element gives 0. However, all elements must be distinct. So, if we have 0 in S, then S is {0} ∪ G, where G is a multiplicative group of order 2005. But does such a group G exist? Yes, the 2005th roots of unity. So, in that case, the sum is 0 + 0 = 0. But wait, the problem says "2006 distinct complex numbers". If 0 is among them, then S has 0 and the 2005th roots of unity. Alternatively, if S is the 2006th roots of unity, which does not include 0, then the sum is also zero. Therefore, regardless of whether 0 is included or not, the sum is zero. However, is it possible for S to include 0?Wait, let's check if S can include 0. Suppose S includes 0. Then, since S is closed under multiplication, multiplying 0 with any element of S must result in 0, which is in S. But since all elements are distinct, 0 is the only zero element. Now, the remaining 2005 elements must be closed under multiplication as well. Because if you take any two non-zero elements, their product must be in S. But since 0 is already in S, the product of two non-zero elements must be either 0 or another non-zero element. But 0 can only be obtained if one of the elements is 0. So, actually, multiplying two non-zero elements must result in another non-zero element. Therefore, the non-zero elements form a closed multiplicative set, hence a group. Therefore, S is {0} ∪ G, where G is a multiplicative group of order 2005. But in such a group G, the product of any two elements is in G, and the sum of the elements of G is zero (since G is a group of roots of unity), so the total sum is 0 + 0 = 0.Alternatively, if S does not contain 0, then S is a multiplicative group of order 2006, which is the group of 2006th roots of unity, and the sum is zero. Therefore, in either case, the sum is zero. So, regardless of whether 0 is in the set or not, the sum is zero. Therefore, the answer is boxed{0}.But wait, there's a catch here. If S is a multiplicative group of 2006 elements, then it must be the 2006th roots of unity. But 2006 is equal to 2 × 17 × 59. So, the group of 2006th roots of unity is cyclic, and the sum is zero. If S is {0} ∪ G, where G is a multiplicative group of 2005 elements, which is also cyclic (the 2005th roots of unity), then the sum is 0 + 0 = 0. However, 2005 is equal to 5 × 401. So, such roots of unity exist. However, in the problem, we are told that there are 2006 distinct complex numbers. So, both possibilities exist: either S is a multiplicative group of 2006 elements (without 0), or S is {0} ∪ a multiplicative group of 2005 elements. However, in both cases, the sum is zero.But does the problem allow for both possibilities? Let me check. The problem states: "the product of any two of them (including self-multiplication) is one of these 2006 numbers". If S is {0} ∪ G, where G is a group of 2005 elements, then multiplying 0 by any element gives 0, which is in S. Multiplying two elements of G gives an element of G, which is in S. So, that works. However, is there a requirement that the multiplicative identity is in S? If S is a group, then yes, it contains 1. If S is {0} ∪ G, then G must contain 1, because G is a group. So, 1 is in G, hence 1 is in S. Therefore, in both cases, 1 is present in S.But wait, if S is {0} ∪ G, then 1 is in G, so 1 is in S. So, in that case, multiplying 1 by any element a gives a. Therefore, for 0, 1*0=0, which is in S. For 1*1=1, which is in S. For other elements, 1*g=g, which is in S. So, that's okay.But here's another thought: can we have S being a group without 0? Yes. If S is the group of 2006th roots of unity, which includes 1, and all other roots. So, in that case, the sum is zero. If S is {0} ∪ the 2005th roots of unity, which also includes 1, then the sum is zero. Therefore, in both cases, the sum is zero. Therefore, regardless of whether 0 is in S or not, the sum is zero. Therefore, the answer is zero.But wait, the problem states that there are 2006 distinct complex numbers. If 0 is included, then we have the 2005th roots of unity plus 0. If 0 is not included, then it's the 2006th roots of unity. So, both possibilities exist, but in both cases, the sum is zero. Therefore, regardless of the structure of S, the sum must be zero.Is there any other possible structure for S? For example, could S be a union of different groups? For instance, multiple groups along with 0? Wait, but if S is a union of different groups, then multiplying elements from different groups might result in something not in S. So, unless those groups are somehow compatible, which would complicate things. However, since the problem states that the product of any two elements is in S, then if S is a union of groups, then multiplying elements from different groups must land back in S. But unless all those groups are subgroups of a larger group. But this seems complicated. However, the simplest solution is that S is either a cyclic group of 2006th roots of unity or {0} union a cyclic group of 2005th roots of unity, both leading to the sum being zero. Therefore, the answer is zero.Alternatively, perhaps the problem forces S to be a group. Because if 0 were in S, then 0 multiplied by any element is 0, but does that affect the distinctness? If all elements are distinct, and 0 is one of them, then the rest are non-zero. But maybe having 0 complicates things. Wait, but the problem doesn't say anything about inverses or the identity element. It just says closure under multiplication. So, S could be {0} ∪ G, as above, or a group without 0.But the key point is that in both cases, the sum is zero. Therefore, regardless of the structure, the answer must be zero. Therefore, the answer is boxed{0}.Final Answerboxed{0}</think>Given 2006 distinct complex numbers such that the product of any two of them (including self-multiplication) is one of these numbers, we need to find the sum of these 2006 numbers.1. Closure under Multiplication: The set ( S ) of these 2006 complex numbers is closed under multiplication. This property suggests that ( S ) could be a multiplicative group or a set containing 0 and a multiplicative group.2. Roots of Unity: If ( S ) is a multiplicative group, it must be a finite subgroup of the complex numbers under multiplication. Such subgroups are cyclic and consist of roots of unity. The sum of all ( n )-th roots of unity is zero.3. Possibility of Including 0: If 0 is included in ( S ), the remaining 2005 elements must form a multiplicative group (since the product of any two non-zero elements must be non-zero and in ( S )). The sum of these 2005 roots of unity is zero, and adding 0 to this sum still results in zero.4. Verification: Whether ( S ) is the set of 2006th roots of unity (sum zero) or the union of 0 and the 2005th roots of unity (sum zero), the total sum in both cases is zero.Thus, the sum of these 2006 complex numbers is (boxed{0}).,<think>Okay, let's try to tackle this problem. So, we have 2006 distinct complex numbers, and the product of any two of them (including multiplying a number by itself) is also one of these 2006 numbers. We need to find the sum of all these numbers. Hmm, interesting.First, let me parse the problem again to make sure I understand. We have a set S of 2006 distinct complex numbers. The condition is that for any two elements a and b in S (they can be the same), the product a*b is also in S. So, the set is closed under multiplication, even when multiplying an element by itself. The question is to find the sum of all these numbers.Alright, so closure under multiplication is a key property here. That usually makes me think of groups in abstract algebra, but since we're dealing with complex numbers, maybe there's a simpler structure here. Let me think. In group theory terms, a set closed under an associative operation (like multiplication) with an identity element and inverses would be a group. But here, the problem doesn't mention inverses, only closure under multiplication, so it might be a monoid. But complex numbers under multiplication do form a group if we exclude zero, but if zero is included, it's a monoid because zero doesn't have an inverse.Wait, but the problem says "distinct complex numbers," so could zero be one of them? If zero were in the set, then multiplying zero by any other element would have to give another element in the set. But 0 * a = 0, so if zero is in the set, then 0 must be one of the elements. But if 0 is an element, then multiplying any element by 0 gives 0. But since all elements are distinct, 0 can only appear once. So, the presence of zero would require that 0 is in the set, and that multiplying any element by 0 gives 0, which is already accounted for. However, if there's a zero, then the product of two non-zero elements must also be in the set, but the problem states there are 2006 distinct complex numbers. Let me see. If zero is included, then the rest of the 2005 elements must be closed under multiplication among themselves and when multiplied by zero, but zero times anything is zero. So, the non-zero elements must form a closed set under multiplication. Wait, but if there's a zero, then the set S would consist of zero and a multiplicatively closed subset of 2005 elements. However, the problem states that the product of any two elements, including self-multiplication, must be in S. So, if zero is in S, then multiplying any element by zero must be in S. So zero must be included. But then, if the non-zero elements are closed under multiplication, but they need to include all products. However, since 2005 is an odd number, which might complicate things because, for example, multiplicative groups of complex numbers often have even order if they are roots of unity. Wait, maybe this is a group? If we can show that all the non-zero elements form a group under multiplication, then since the set is finite, it must be a cyclic group, right?But 2005 is the number of non-zero elements if we include zero, but the problem says there are 2006 elements. Wait, 2006 is even. If the set is closed under multiplication and all elements are distinct, then perhaps all elements are roots of unity? Because roots of unity are closed under multiplication. But how many roots of unity would give a set of 2006 elements closed under multiplication?Wait, the set of all nth roots of unity for some n would form a group under multiplication. But if n is such that the number of roots is 2006, but the number of roots of unity is equal to n, so n would have to be 2006. But the multiplicative group of the 2006th roots of unity is cyclic of order 2006. But 2006 factors into 2 * 17 * 59. Hmm, not sure if that's useful. Wait, but if our set S is exactly the 2006th roots of unity, then the product of any two roots of unity is another root of unity, but the 2006th roots of unity are closed under multiplication. However, the product of two nth roots of unity is another nth root of unity only if n divides the exponents. Wait, actually, if you multiply two roots of unity, their product is another root of unity, but not necessarily of the same order. However, if the set is the group of all roots of unity of order dividing 2006, then they would be closed under multiplication. But that would include all roots of unity of orders dividing 2006, which would be more than 2006 elements. Wait, no, the group of 2006th roots of unity has exactly 2006 elements, since each root is e^(2πik/2006) for k from 0 to 2005. So, multiplying two such roots would give another root of unity of order dividing 2006. So, the set of 2006th roots of unity is closed under multiplication and contains 2006 elements. Moreover, they are distinct. So, if the set S is the 2006th roots of unity, then the product of any two elements is another element in the set. That seems to fit. Then, the sum of all 2006th roots of unity is zero. Because the sum of all nth roots of unity is zero for any n ≥ 1. That's a well-known result. So, if that's the case, then the answer is zero.But wait, the problem states that there are 2006 distinct complex numbers with the given property. The 2006th roots of unity satisfy this property. But is this the only possibility? Let me check. Suppose that S includes zero. Then, as we thought earlier, S would consist of zero and 2005 non-zero elements closed under multiplication. But if S includes zero, then when we take the sum, the sum would be the sum of the non-zero elements plus zero, which is the sum of the non-zero elements. However, the sum of the non-zero elements in such a set might not necessarily be zero. But if S is the 2006th roots of unity including zero, but wait, the 2006th roots of unity do not include zero. So that can't be. So if S includes zero, then the other elements must form a multiplicative closed set of 2005 elements. But 2005 is a prime number? Wait, 2005 is 5*401. So, if there is a group of order 2005, which is a prime, then it would be cyclic. But 2005 is not prime. Wait, 2005 is 5*401. So, if the non-zero elements form a group under multiplication, then the order of the group is 2005, which is composite, so the group would be cyclic if it's abelian, which it is because multiplication in complex numbers is commutative. So, a finite abelian group is cyclic if and only if it's order is a product of distinct primes? No, actually, every finite abelian group is a product of cyclic groups. But in our case, since we have 2005 elements, which is 5*401, both primes, then the multiplicative group would be isomorphic to Z_5 × Z_401. But unless 5 and 401 are coprime, which they are, so by the Chinese Remainder Theorem, Z_5 × Z_401 is isomorphic to Z_(5*401) = Z_2005. So, the multiplicative group of order 2005 would be cyclic. Therefore, if such a group exists, it's cyclic. Therefore, the non-zero elements would be the 2005th roots of unity, plus zero. Then, the sum would be the sum of the 2005th roots of unity plus zero, which is zero. But wait, but 2005th roots of unity are different from zero. However, if the set includes zero, then the product of zero and any element is zero, which is in the set. However, the product of two non-zero elements would have to be in the set. So, if the non-zero elements are the 2005th roots of unity, then they form a group under multiplication, closed, and when multiplied by zero, gives zero. So, the total set would be the union of zero and the 2005th roots of unity. Then, the sum would be zero (sum of roots of unity) plus zero (the element zero), which is zero. Wait, but the problem says there are 2006 elements. If we include zero and the 2005th roots of unity, that's 1 + 2005 = 2006 elements. So that would satisfy the problem's conditions.But then, in this case, the sum is zero. Alternatively, if the entire set is the 2006th roots of unity, which do not include zero, then their sum is zero. So, either way, the sum is zero. So, regardless of whether zero is included or not, the sum is zero. Wait, but how can we be sure that the set must be either the 2006th roots of unity or the union of zero and the 2005th roots of unity?Wait, the problem states that there are 2006 distinct complex numbers. So, depending on whether zero is included or not, the structure changes. If zero is not included, then all elements must be roots of unity forming a multiplicative group of order 2006. If zero is included, then the non-zero elements must form a multiplicative group of order 2005. However, the problem is asking for the sum, which in both cases is zero. Because the sum of all nth roots of unity is zero, and if you include zero, it's still zero. Wait, hold on. If we have the 2005th roots of unity plus zero, then the sum is sum(roots of unity) + zero = 0 + 0 = 0. If we have the 2006th roots of unity, sum is zero. So in either case, the sum is zero. Therefore, regardless of whether zero is included or not, the sum is zero. Therefore, the answer is zero.But wait, the problem says "the product of any two of them (including self-multiplication) is one of these 2006 numbers." If the set includes zero, then multiplying any element by zero gives zero, which is in the set. If the set is the 2006th roots of unity, then multiplying any two elements gives another element in the set. So, both possibilities satisfy the problem's condition. However, in both cases, the sum is zero. Therefore, regardless of the structure, the sum must be zero. Therefore, the answer is zero.But wait, is there a case where the sum isn't zero? Let's think. Suppose the set S is something else. For example, suppose S is a finite multiplicative subgroup of the complex numbers. It's known that all finite multiplicative subgroups of the complex numbers are cyclic, consisting of roots of unity. Therefore, if S is a finite multiplicative group, it's cyclic, so the sum is zero. If S includes zero, then the sum of the roots of unity is zero, plus zero is still zero. Therefore, regardless of whether zero is included or not, the sum must be zero.Alternatively, suppose that S is not a group but just a multiplicatively closed set. But in that case, since S is finite, the multiplicative closure would imply that it's a group. Wait, in finite semigroups, if you have cancellation property (which in complex numbers you do, except for zero), then it's a group. So, if S is multiplicatively closed and finite, and all elements are non-zero, then it's a group. If there is zero, then as we saw earlier, the non-zero elements form a group. So, in either case, the non-zero elements form a finite multiplicative subgroup of the complex numbers, hence cyclic, so sum is zero. Therefore, the total sum is zero. Therefore, the answer is zero.But let's check with smaller numbers. For example, suppose n = 1. Then, the only complex number is 0? Wait, but if n=1, then the set must contain a single element a such that a*a = a. So, a^2 = a, which implies a(a - 1) = 0, so a = 0 or a = 1. Since it's distinct, there's only one element. So, either {0} or {1}. The sum is 0 or 1. But in our problem, n=2006. However, in the case of n=1, the problem's answer would depend on the element. But in our problem, since the set is closed under multiplication, for n=2006, the set must be a group (if excluding zero) or a group with zero adjoined. Since the sum of the roots of unity is zero, the answer is zero.Another check: Let's take n=2. Suppose we have two complex numbers. If they are closed under multiplication, including self-multiplication. Let's say the elements are 1 and -1. Then, 1*1=1, 1*(-1)=-1, (-1)*1=-1, (-1)*(-1)=1. So, closed. The sum is 1 + (-1) = 0. Alternatively, if the set includes 0 and another element a. Then, 0*0=0, 0*a=0, a*0=0, a*a must be in the set. So, a*a must be either 0 or a. If a*a = a, then a(a - 1)=0, so a=1. So, the set {0, 1}. Then, products: 0*0=0, 0*1=0, 1*0=0, 1*1=1. So, closed. The sum is 0 + 1 = 1. But in this case, the sum is 1, not zero. Wait, so in this case, the sum is 1. But according to our previous reasoning, if the set includes zero, the non-zero elements form a group. But here, the non-zero element is 1, which is a trivial group. So, the sum is 1. But this contradicts the previous conclusion. So, this shows that if the set includes zero and a trivial group (just {1}), then the sum is 1. So, in that case, the answer isn't zero. But in the problem statement, n=2006. So, maybe the answer isn't necessarily zero?Wait, so there's a contradiction here. For n=2, there are two possibilities: {1, -1} with sum 0, or {0, 1} with sum 1. Therefore, the sum isn't uniquely determined unless there's an additional condition. However, in the problem statement, it's given that all 2006 complex numbers are distinct. So, in the n=2 case, both possibilities are allowed. Therefore, the problem might require the sum in the specific case, but we need more information.But in the original problem, the number is 2006. So, perhaps the key is that 2006 is even, and in the case when we have a group without zero, the sum is zero. If we include zero, then the non-zero elements form a group of odd order (2005). But in that case, does the sum of a group of odd order roots of unity also equal zero?Wait, the sum of all nth roots of unity is zero, regardless of n. For example, the sum of the 3rd roots of unity: 1 + ω + ω² = 0, where ω is a primitive 3rd root. Similarly, for 5th roots, sum is zero. So, even if n is odd, the sum is zero. So, if the non-zero elements are 2005th roots of unity, then their sum is zero. Then, adding zero gives total sum zero. Therefore, even if zero is included, the sum is zero. Therefore, for n=2006, regardless of whether zero is included or not, the sum is zero.But in the n=2 case, if the non-zero elements form a group of order 1 (i.e., {1}), then the sum is 1. However, in that case, the group is trivial. So, perhaps when the group is non-trivial, the sum is zero? Wait, no, even for n=3, if we have the 3rd roots of unity, the sum is zero. If we have {0, 1, -1}, then the sum is 0 + 1 + (-1) = 0. Wait, but {0, 1, -1} isn't closed under multiplication. Because 1*(-1) = -1, which is in the set; 0*anything is 0. But 1*1=1, (-1)*(-1)=1, so that's okay. Wait, but the set {0, 1, -1} is closed under multiplication. Because:0*0=0, 0*1=0, 0*(-1)=0,1*0=0, 1*1=1, 1*(-1)=-1,(-1)*0=0, (-1)*1=-1, (-1)*(-1)=1.Yes, all products are in the set. Then, the sum is 0 + 1 + (-1) = 0. So, even when zero is included, if the non-zero elements form a group (in this case, {1, -1}, which is a group under multiplication), then the sum is zero.Wait a second, so in the previous n=2 case, if we have {0,1}, the sum is 1, but {0,1} is closed under multiplication. But in that case, the non-zero elements do not form a group, because {1} is a trivial group, but if the set {0,1} is considered, the product 1*1=1, which is in the set, but since 1 is the only non-zero element, it's closed. However, in that case, the sum is 1. So, in that case, the non-zero elements form a trivial group, but the sum is not zero. Therefore, our previous reasoning that the sum is zero relies on the non-zero elements forming a group of roots of unity with sum zero. However, in the trivial group case, the sum is 1. Therefore, perhaps the answer is zero only when the non-zero elements form a non-trivial group. But in the problem statement, since n=2006 is even, can we have a group of 2006 elements (without zero) which is the roots of unity, sum zero, or a group of 2005 elements (with zero) which would also sum to zero. Wait, 2005 is an odd number. The sum of the 2005th roots of unity is zero. Then, including zero, the sum is still zero. But in the case of n=2, if we have {0,1}, sum is 1. So, why is this different? Because in that case, the non-zero element is 1, which is a root of unity of order 1. The sum of the first roots of unity is 1. But for any n ≥ 2, the sum of the nth roots of unity is zero. Wait, yes. The sum of all nth roots of unity is zero for n ≥ 1. Wait, no. Wait, for n=1, the only root of unity is 1, sum is 1. For n=2, the roots are 1 and -1, sum is 0. For n=3, 1, ω, ω², sum is 0. So, for n ≥ 2, the sum is zero. For n=1, it's 1. Therefore, in the case where the non-zero elements form a group of order 2005, which is ≥ 2, the sum is zero. If the non-zero elements form a group of order 1, which is n=1, the sum is 1. Therefore, in the original problem, since we have 2006 elements, if zero is included, the non-zero elements would be 2005, which is ≥ 2, so their sum is zero. If zero is not included, the 2006 elements form a group of order 2006, which is also ≥ 2, so their sum is zero. Therefore, in both cases, the sum is zero. Therefore, regardless of the structure, the answer is zero.But in the n=2 case, if we have {0,1}, then the non-zero elements are of order 1, so sum is 1, but in our problem, since 2006 is large, and the number of non-zero elements would be 2005 or 2006, both of which are ≥ 2, their sums would be zero. Therefore, the answer must be zero.Therefore, the answer is zero.Final Answerboxed{0}</think>Given 2006 distinct complex numbers such that the product of any two of them (including self-multiplication) is one of these 2006 numbers, we need to find the sum of these numbers.1. Closure Under Multiplication: The set of complex numbers is closed under multiplication. This implies the set forms a multiplicative group or a monoid (if zero is included).2. Roots of Unity: If the set does not include zero, it must be a multiplicative group of 2006th roots of unity. The sum of all nth roots of unity is zero for any ( n geq 2 ).3. Inclusion of Zero: If zero is included, the remaining 2005 non-zero elements must form a multiplicative group. The sum of the 2005th roots of unity is also zero, and adding zero to this sum results in zero.4. Verification with Smaller Cases: For smaller sets, such as {1, -1} or {0, 1, -1}, the sum is zero if the non-zero elements form a group of order ( geq 2 ). For larger ( n ), the sum of roots of unity (whether including zero or not) is always zero.Thus, regardless of whether zero is included or not, the sum of the 2006 complex numbers, which are roots of unity, is zero.[boxed{0}]]

🔑:Three, 13. Let there be n distinct complex numbers alpha_{1}, alpha_{2}, cdots, alpha_{n}.If one of them is 0, without loss of generality, let alpha_{n}=0, then the setleft{alpha_{1}^{2}, alpha_{1} alpha_{2}, cdots, alpha_{1} alpha_{n-1}right}=left{alpha_{1}, alpha_{2}, cdots, alpha_{n-1}right} .Thus, by alpha_{1}^{2} cdot alpha_{1} alpha_{2} cdots cdot alpha_{1} alpha_{n-1}=alpha_{1} cdot alpha_{2} cdots cdot alpha_{n-1}, we get alpha_{1}^{n-1}=1.Similarly, a_{k}^{n-1}=1(k=1,2, cdots, n-1).Since alpha_{1}, alpha_{2}, cdots, alpha_{n-1} are distinct complex numbers, we have alpha_{1}+alpha_{2}+cdots+alpha_{n-1}=0, which meansalpha_{1}+alpha_{2}+cdots+alpha_{n}=0 .If these n complex numbers are all non-zero, similarly, we get alpha_{k}^{n}=1. Since alpha_{1}, alpha_{2}, cdots, alpha_{n} are distinct complex numbers, we have alpha_{1}+alpha_{2}+cdots+alpha_{n}=0

🧠:[<think>Alright, let me try to figure out this problem. So, we have the equation (36x^2 - 140[x] - 21 = 0), where [x] is the greatest integer less than or equal to x. We need to find how many real numbers x satisfy this equation. The options are from 2 to 5, so the answer is probably one of those. Let me think step by step.First, I remember that [x] is the floor function, which means for any real number x, [x] is the largest integer less than or equal to x. For example, if x is 3.7, then [x] is 3. If x is -1.2, then [x] is -2, because -2 is the greatest integer less than -1.2. So, floor function can sometimes be tricky with negative numbers, but maybe in this problem, x is positive? Let me check.The equation is quadratic in x^2, but linear in [x]. Maybe we can let [x] = n, where n is an integer, and then express x in terms of n? Then, solve for x in each interval [n, n+1). Because if [x] = n, then x is in [n, n+1). So, for each integer n, we can set up the equation (36x^2 - 140n - 21 = 0), and then solve for x in [n, n+1). The solutions would have to lie in that interval. So, the number of solutions would be the number of integers n for which the equation (36x^2 = 140n + 21) has a solution x in [n, n+1). So, x would be sqrt((140n +21)/36). Let me write that down.So, let me denote n as an integer. Then, (x = pm sqrt{(140n + 21)/36}). But since [x] = n, if x is positive, then n must be less than or equal to x. If x is negative, then [x] would be negative. Wait, but x is in [n, n+1), so depending on the sign of n, x could be positive or negative. However, the equation (36x^2 -140n -21 =0) is quadratic, so x^2 is positive, which implies that (140n +21) must be positive because 36x^2 is positive. Therefore, (140n +21 > 0), so (140n > -21), which means (n > -21/140). Calculating that, -21/140 is -0.15. So, n must be greater than -0.15. Since n is an integer, n >= 0. Therefore, n is a non-negative integer. So, x must be non-negative as well because if n >=0, then x is in [n, n+1), which is non-negative. Therefore, x is non-negative, so we can ignore the negative square root. So, x = sqrt((140n +21)/36). Let me simplify that. sqrt((140n +21)/36) = sqrt(140n +21)/6.So, x must be equal to sqrt(140n +21)/6, and x must lie in [n, n+1). Therefore, we have the inequality:n <= sqrt(140n +21)/6 < n +1.Our goal is to find all integers n >=0 such that this inequality holds. Then, for each such n, there is a solution x in [n, n+1). So, the number of solutions would be the number of such n's. Therefore, we need to find all integers n >=0 where:n <= sqrt(140n +21)/6 < n +1.Let me square all parts of the inequality to eliminate the square root. But before that, let's make sure that the expression inside the square root is non-negative. Since n >=0, 140n +21 >=21 >0, so that's okay.But when we square inequalities, we have to be careful because squaring is not a linear operation. However, in this case, since all terms are positive (n is non-negative, sqrt(...)/6 is non-negative), squaring should preserve the inequalities.Therefore, let's first handle the left inequality:n <= sqrt(140n +21)/6Multiply both sides by 6:6n <= sqrt(140n +21)Square both sides:(6n)^2 <= 140n +2136n^2 <= 140n +21Bring all terms to the left:36n^2 -140n -21 <=0Similarly, the right inequality:sqrt(140n +21)/6 < n +1Multiply both sides by 6:sqrt(140n +21) <6n +6Square both sides:140n +21 < (6n +6)^2140n +21 <36n^2 +72n +36Bring all terms to the left:-36n^2 +140n +21 -72n -36 <0Simplify:-36n^2 +68n -15 <0Multiply both sides by -1 (which reverses the inequality):36n^2 -68n +15 >0Therefore, combining both inequalities, we have:36n^2 -140n -21 <=0 (from the left inequality)and36n^2 -68n +15 >0 (from the right inequality)So, we need to find integers n >=0 such that both inequalities hold.So, let's solve the quadratic inequalities.First inequality: 36n^2 -140n -21 <=0Let's find the roots of 36n^2 -140n -21 =0.Using quadratic formula:n = [140 ± sqrt(140^2 -4*36*(-21))]/(2*36)Calculate discriminant:140^2 -4*36*(-21) = 19600 + 4*36*21Compute 4*36=144, 144*21=3024So discriminant =19600 +3024=22624sqrt(22624). Let's see, 150^2=22500, so sqrt(22624)=150.746..., approximately 150.75.Therefore, n = [140 ±150.75]/72Compute both roots:First root: (140 +150.75)/72 =290.75/72 ≈4.038Second root: (140 -150.75)/72 = (-10.75)/72≈-0.1493Therefore, the quadratic 36n^2 -140n -21 <=0 holds for n between -0.1493 and 4.038. Since n is integer >=0, possible n values are 0,1,2,3,4.But we also need the second inequality: 36n^2 -68n +15 >0Let's find the roots of 36n^2 -68n +15=0.Again, quadratic formula:n = [68 ± sqrt(68^2 -4*36*15)]/(2*36)Compute discriminant:68^2=46244*36*15=2160So discriminant=4624-2160=2464sqrt(2464). Let's see, 49^2=2401, 50^2=2500, so sqrt(2464)=49.64 approx.So, n = [68 ±49.64]/72Compute both roots:First root: (68 +49.64)/72≈117.64/72≈1.633Second root: (68 -49.64)/72≈18.36/72≈0.255Therefore, the quadratic 36n^2 -68n +15 is positive when n <0.255 or n>1.633. Since n is integer >=0, this inequality holds when n <=0 or n>=2.But n must be an integer. So, n=0 or n>=2.But combining with the first inequality which allows n=0,1,2,3,4.So, the intersection of both inequalities would be n=0,2,3,4.Because for the first inequality, n=0,1,2,3,4 and for the second inequality, n=0,2,3,4. So, the overlap is n=0,2,3,4.But wait, when n=1, check whether the second inequality holds. For n=1:36(1)^2 -68(1) +15=36 -68 +15= -17 <0. Therefore, the second inequality is not satisfied for n=1. So n=1 is excluded.So, possible n values are 0,2,3,4.Wait, but we need to check for each n in 0,2,3,4 whether the original inequalities hold. Because quadratic inequalities can sometimes include or exclude endpoints. Wait, but we already considered that. Let me verify.Alternatively, perhaps I should check each n=0,2,3,4 and see if the original inequalities hold, just to be safe.So, let's check for each n=0,2,3,4:First, n=0:Check if 0 <= sqrt(140*0 +21)/6 <0 +1So, sqrt(21)/6 ≈4.583/6≈0.764. So, 0 <=0.764 <1. Yes, that's true. So, x≈0.764, and [x]=0. So, x is in [0,1). So, valid.Next, n=2:Check if 2 <= sqrt(140*2 +21)/6 <3Compute sqrt(280 +21)/6 =sqrt(301)/6 ≈17.35/6≈2.892So, check if 2 <=2.892 <3. Yes, that's true. So, x≈2.892, and [x]=2. So, x is in [2,3). Valid.n=3:Check if 3 <= sqrt(140*3 +21)/6 <4Compute sqrt(420 +21)/6 =sqrt(441)/6=21/6=3.5Check 3 <=3.5 <4. Yes, true. So, x=3.5, and [x]=3. So, x is in [3,4). But wait, 3.5 is in [3,4), yes. Valid.n=4:Check if 4 <= sqrt(140*4 +21)/6 <5Compute sqrt(560 +21)/6=sqrt(581)/6≈24.11/6≈4.018Check 4 <=4.018 <5. Yes, true. So, x≈4.018, and [x]=4. So, x is in [4,5). Valid.So, n=0,2,3,4 are valid. But wait, let me check n=5 just in case. For n=5:sqrt(140*5 +21)/6= sqrt(700 +21)/6=sqrt(721)/6≈26.85/6≈4.475But 4.475 is less than 5, but n=5, so check if 5 <=4.475 <6? No, 4.475 <5. So, invalid. So, n=5 is not possible.Similarly, check n=1:sqrt(140*1 +21)/6=sqrt(161)/6≈12.69/6≈2.115Check if 1 <=2.115 <2? No, 2.115 is not less than 2. So, invalid. So, n=1 is invalid.Therefore, the possible n are 0,2,3,4. Each of these gives exactly one solution in their respective intervals. So, total solutions are 4. Wait, but n=0 gives x≈0.764, n=2 gives≈2.892, n=3 gives 3.5, and n=4 gives≈4.018. So, four solutions.But wait, let me check for n=0: x≈0.764. Then [x]=0, which is correct. For n=2: x≈2.892, [x]=2, correct. For n=3: x=3.5, [x]=3, correct. For n=4: x≈4.018, [x]=4, correct. So, all four solutions are valid. So, the answer should be 4, which is option C.But hold on, the problem says "the number of real numbers x", so maybe some n might give two solutions? Wait, but since we took the positive square root only. Because earlier, we concluded that x must be non-negative. Wait, let's think again.Original equation: 36x^2 -140[x] -21=0. If x is negative, then [x] would be less than or equal to x, so if x is negative, [x] <=x <0. But in the equation, 36x^2 is positive, 140[x] is negative (since [x] is negative if x is negative), and -21 is negative. So, positive term minus negative minus negative. Wait, 36x^2 is positive, -140[x] would be positive if [x] is negative, because [x] is negative, so -140*negative is positive. Then, -21. So, total equation would be positive + positive -21 =0. So, possible.Wait, so maybe x can be negative as well? Hmm. Did we miss negative solutions?Because when we set [x] =n, n must be an integer less than or equal to x. If x is negative, n is less than or equal to x, which is negative. So, n is a negative integer. Earlier, I thought that n must be non-negative because 140n +21 must be positive, but let's check.Because 36x^2 =140n +21. Since 36x^2 is always non-negative, so 140n +21 must be non-negative. Therefore, 140n +21 >=0 => n >= -21/140 => n >= -0.15. Since n is an integer, n >=-0.15 => n >=0 because n is integer. Wait, if x is negative, then n is the floor of x, so n <=x <0. But n must satisfy 140n +21 >=0, so 140n >=-21 => n >= -0.15. But n is integer less than x, which is negative, so n <=x <0. Therefore, n is integer in [-0.15, 0). But the integers in that interval are n=-0.15 to 0, so n=-0.15 is approximately, but n must be integer. So, n must be -0.15 <=n <0. But n is integer. The only integer in that interval is n=0. But n=0, but x is negative, so x is in [0,1). But if x is in [0,1), then x is non-negative. Therefore, there are no negative solutions.Wait, this seems a bit confusing. Let me clarify.If x is negative, then [x] =n where n is a negative integer. For example, x=-0.5, then [x]=-1. But in that case, n would be -1. Let me see.Suppose x is in [-1,0), so n=-1. Then, let's check if 140n +21 >=0. For n=-1: 140*(-1) +21= -140 +21= -119 <0. So, 36x^2 =-119, which is impossible because 36x^2 is non-negative. Therefore, no solutions here.Similarly, for n=-2: 140*(-2) +21= -280 +21= -259 <0. So, same problem. Therefore, negative n would lead to 140n +21 negative, which would require 36x^2 to be negative, which is impossible. Therefore, there are no negative solutions. Therefore, x must be non-negative, and n must be non-negative integers. So, our initial conclusion that n=0,2,3,4 are the only possible solutions is correct, and each gives one x in their respective intervals. Therefore, 4 solutions. So, answer is C)4.But let me check once again for n=0. x≈sqrt(21)/6≈0.764. [x]=0, which is correct. For n=2: x≈sqrt(301)/6≈17.35/6≈2.892, [x]=2, but x is approximately 2.892, which is in [2,3), so [x]=2. Correct.n=3: x= sqrt(140*3 +21)/6= sqrt(441)/6=21/6=3.5. But [3.5]=3, so x=3.5 is in [3,4). Correct.n=4: x≈sqrt(581)/6≈24.11/6≈4.018. [x]=4, so x is in [4,5). Correct.Thus, four solutions. Therefore, the answer is C)4.Final Answerboxed{C}</think>Given the equation (36x^2 - 140[x] - 21 = 0), where ([x]) denotes the greatest integer less than or equal to (x), we need to find the number of real solutions (x).1. Let ([x] = n), where (n) is an integer. Then, (x) must satisfy (n leq x < n + 1).2. Substitute ([x] = n) into the equation: (36x^2 - 140n - 21 = 0). Solving for (x), we get (x = pm sqrt{frac{140n + 21}{36}}). Since (x) must be non-negative (as shown in the detailed reasoning), we consider the positive root (x = sqrt{frac{140n + 21}{36}}).3. For (x) to lie in the interval ([n, n + 1)), the following inequalities must hold: [ n leq sqrt{frac{140n + 21}{36}} < n + 1 ]4. Squaring and simplifying these inequalities, we derive: - For the left inequality: (36n^2 - 140n - 21 leq 0) - For the right inequality: (36n^2 - 68n + 15 > 0)5. Solving these quadratic inequalities, we find the valid integer values of (n) are 0, 2, 3, and 4. Each value of (n) corresponds to a unique solution (x) in the interval ([n, n + 1)): - For (n = 0): (x approx 0.764) - For (n = 2): (x approx 2.892) - For (n = 3): (x = 3.5) - For (n = 4): (x approx 4.018)Thus, there are 4 real solutions to the equation. The answer is (boxed{C}).,<think>Okay, so I need to solve this equation: 36x² - 140[x] - 21 = 0, where [x] is the greatest integer less than or equal to x. The question is asking how many real numbers x satisfy this equation, and the options are 2, 3, 4, or 5. Alright, let's break this down step by step.First, I remember that [x] is the floor function, which means for any real number x, [x] is the largest integer less than or equal to x. So, for example, if x is 3.7, then [x] is 3. If x is -1.2, then [x] is -2 because -2 is the greatest integer less than -1.2. Got that.So the equation involves both x² and [x]. That seems a bit tricky because x is both squared and inside a floor function. But maybe I can approach this by considering different intervals for x. Since [x] is an integer, let's let n = [x]. Then, by definition, n ≤ x < n + 1, where n is an integer. Then, substituting n into the equation, we can write 36x² - 140n - 21 = 0. So for each integer n, we can solve for x and check if the solution x lies within the interval [n, n + 1). If it does, then that x is a valid solution; if not, then we discard that solution for that n.So the strategy is: iterate over possible integer values of n, solve the quadratic equation 36x² - 140n - 21 = 0 for each n, and check if the solutions x lie within [n, n + 1). The number of such valid x's across all n will be the answer.But first, we need to figure out the possible range of n. Since x² is non-negative (because squared term), we can use the equation 36x² = 140n + 21. Therefore, 140n + 21 must be non-negative because the left side is 36x², which is always non-negative. So 140n + 21 ≥ 0 → 140n ≥ -21 → n ≥ -21/140 → n ≥ -0.15. Since n is an integer, n ≥ -0.15 → n ≥ 0. Wait, but n is an integer. So the smallest possible n is 0? Because -0.15 is between -1 and 0. But n must be an integer less than or equal to x. If x is negative, say x is between -1 and 0, then [x] is -1. So even though n must be ≥ -0.15, but n is an integer. Wait, hold on.Wait, let's re-examine that. The equation is 36x² - 140n -21 = 0. So 36x² = 140n + 21. Since x² is non-negative, 140n +21 must be ≥ 0. Therefore, 140n +21 ≥0 → 140n ≥ -21 → n ≥ -21/140 → n ≥ -0.15. Since n is an integer, n must be greater than or equal to the next integer greater than -0.15, which is 0. So n ≥ 0. Therefore, n can be 0, 1, 2, 3, etc. So possible values for n start at 0. But wait, hold on. If x is negative, then [x] is negative. For example, if x is -0.5, [x] is -1. But according to the inequality 140n +21 ≥0, if n is -1, then 140*(-1) +21 = -140 +21 = -119, which is negative, which would imply 36x² = -119, which is impossible because x² is non-negative. Therefore, if x is negative, then [x] is at least -1 (if x is between -1 and 0), but substituting n = -1 would lead to 140*(-1) +21 = -119, which is negative, so the equation cannot hold. Therefore, x cannot be negative because that would require 36x² to equal a negative number, which is impossible. Therefore, x must be non-negative. Therefore, [x] is non-negative. Hence, n is a non-negative integer: 0, 1, 2, etc.So, n ≥0. Therefore, x must be ≥0, since [x] = n ≥0. So x is in [n, n +1), where n is 0, 1, 2, etc. Therefore, we can start checking for n starting at 0, 1, 2, ... and see for each n, whether the solutions x of the quadratic equation lie in [n, n +1).Let me formalize this:For each integer n ≥0:1. Solve 36x² -140n -21 =0 → 36x² =140n +21 → x² = (140n +21)/36 → x = ±√[(140n +21)/36]But since x must be in [n, n +1) and n ≥0, and x must be non-negative (as established earlier), so we can ignore the negative roots. Therefore, x = √[(140n +21)/36] = √(140n +21)/6.Therefore, for each n, compute x = √(140n +21)/6, and check whether x ∈ [n, n +1). If yes, then that x is a solution; otherwise, it's not.Therefore, the number of solutions is the number of integers n for which √(140n +21)/6 ∈ [n, n +1).So, let's write the inequality:n ≤ √(140n +21)/6 < n +1.Multiply all parts by 6 to eliminate the denominator:6n ≤ √(140n +21) < 6(n +1)Now square all parts to eliminate the square root (since all terms are non-negative):(6n)² ≤ 140n +21 < [6(n +1)]²Which simplifies to:36n² ≤ 140n +21 < 36(n +1)²Therefore, we have two inequalities:1. 36n² ≤140n +212. 140n +21 <36(n +1)²We can solve each inequality to find the possible values of n.Let's start with the first inequality:36n² -140n -21 ≤0.This is a quadratic inequality. Let's find its roots:36n² -140n -21 =0Using quadratic formula:n = [140 ±√(140² +4*36*21)]/(2*36)Compute discriminant D:D = 140² + 4*36*21 = 19600 + 3024 = 22624√D ≈ √22624. Let's compute this. 150² =22500, so √22624 is a bit more than 150. 150²=22500, 151²=22801, which is higher. So √22624 ≈150.41 (since 150.41² ≈150² + 2*150*0.41 +0.41²≈22500 +123 +0.168≈22623.168, which is very close). So approximately 150.41.Therefore, n ≈ [140 ±150.41]/72Compute both roots:First root: (140 +150.41)/72 ≈290.41/72≈4.03Second root: (140 -150.41)/72≈-10.41/72≈-0.144So the quadratic 36n² -140n -21 is ≤0 between the roots n≈-0.144 and n≈4.03. Since n is a non-negative integer, this inequality holds for integers n=0,1,2,3,4.Now the second inequality:140n +21 <36(n +1)²Expand the right side: 36(n² +2n +1)=36n² +72n +36Bring all terms to the left:140n +21 -36n² -72n -36 <0Simplify:-36n² +68n -15 <0Multiply both sides by -1 (which reverses the inequality):36n² -68n +15 >0Again, solve the quadratic inequality 36n² -68n +15 >0.First, find roots:n = [68 ±√(68² -4*36*15)]/(2*36)Compute discriminant D:D = 4624 - 2160 = 2464√2464. Let's approximate this: 49²=2401, 50²=2500, so √2464≈49.64Thus,n = [68 ±49.64]/72Compute both roots:First root: (68 +49.64)/72≈117.64/72≈1.633Second root: (68 -49.64)/72≈18.36/72≈0.255So the quadratic 36n² -68n +15 is positive when n <0.255 or n >1.633. Since n is a non-negative integer, this inequality holds for integers n=0,1,2,3,4,... except n=1. Because for integers:- n=0: 0 <0.255, so inequality holds.- n=1: 1 is between 0.255 and1.633, so inequality does not hold.- n=2,3,4,...: n >1.633, so inequality holds.Therefore, combining both inequalities:From the first inequality, n can be 0,1,2,3,4.From the second inequality, n can be 0,2,3,4,... (excluding n=1).Therefore, the overlapping integer values are n=0,2,3,4. Wait, but first inequality allows n=0,1,2,3,4, second inequality allows n=0,2,3,4,5,... So the intersection is n=0,2,3,4.Wait, but hold on. The first inequality was 36n² -140n -21 ≤0, which holds for n between approximately -0.144 and 4.03. So n=0,1,2,3,4.The second inequality is 36n² -68n +15 >0, which holds for n <0.255 or n>1.633, so in integers, n=0,2,3,4,...Therefore, the overlap is n=0,2,3,4.Therefore, possible n's are 0,2,3,4. Wait, but n=1 is excluded. So n=0,2,3,4. That's four values. Hmm, but let's check if for each of these n, the original inequality holds.Wait, but maybe there's a mistake here. Let me verify. So the first inequality gives n=0,1,2,3,4. The second inequality (from the upper bound) requires that 140n +21 <36(n +1)^2. Which in integers n=0,2,3,4,... So combining both, n=0,2,3,4. So n=0,2,3,4. So four possible values of n. Then for each of these n, we need to check if x=√(140n +21)/6 is in [n, n +1). Wait, but maybe n=1 is also possible? Because even if the second inequality excludes n=1, but perhaps the quadratic solutions could still lie in [1,2)? Let me check.Wait, perhaps my approach is flawed. Let me explain.The first inequality 36n² ≤140n +21 and the second inequality 140n +21 <36(n +1)^2 are derived from the original inequality n ≤ √(140n +21)/6 <n +1. So, solving both inequalities gives the range of n where the original inequality holds. Therefore, combining these two inequalities, n must satisfy both 36n² ≤140n +21 and 140n +21 <36(n +1)^2. So the possible n's are integers in the intersection of these two conditions.From the first inequality, n ∈ [ -0.144, 4.03 ]From the second inequality, n ∈ (-∞, 0.255) ∪ (1.633, ∞)So the intersection is n ∈ [ -0.144, 0.255 ) ∪ (1.633, 4.03 ]Since n is a non-negative integer (n ≥0), the valid n's are n=0 and n=2,3,4.Therefore, n=0,2,3,4. So four possible n's.But let's check n=1 as well. Maybe even though it's excluded by the second inequality, perhaps due to approximation errors.Wait, let's re-examine the second inequality for n=1. Let's plug n=1 into the original inequality:6*1 ≤ √(140*1 +21) <6*(1 +1)Which is 6 ≤ √(161) <12Calculate √161: 12.69 approximately. So 6 ≤12.69 <12? Wait, 12.69 is not less than 12. So the upper bound fails. Therefore, n=1 does not satisfy the original inequality. Therefore, n=1 is invalid.Therefore, the valid n's are 0,2,3,4.So four values of n. But wait, the answer options are 2,3,4,5. So possible answer is 4? But let's check each n.For each n=0,2,3,4, we can compute x and verify if x is in [n, n+1).Let's do that.First, n=0:x = √(140*0 +21)/6 = √21 /6 ≈4.5837/6≈0.76395Check if x ∈ [0,1). Yes, 0.76395 is between 0 and1. So x≈0.764 is a valid solution. So one solution here.Next, n=2:x = √(140*2 +21)/6 =√(280 +21)/6 =√301/6 ≈17.3496/6≈2.8916Check if x ∈ [2,3). 2.8916 is between2 and3. Yes. So another valid solution. So second solution.n=3:x=√(140*3 +21)/6 =√(420 +21)/6=√441/6=21/6=3.5Check if x ∈ [3,4). 3.5 is between3 and4. Yes. So third solution.n=4:x=√(140*4 +21)/6=√(560 +21)/6=√581/6≈24.1039/6≈4.0173Check if x ∈ [4,5). 4.0173 is between4 and5. Yes. So fourth solution.Therefore, four solutions: approximately 0.764, 2.892, 3.5, and4.017. So four solutions. Therefore, the answer is C) 4.Wait, but let me check if there are more n's. Wait, n=5:Wait, even though in the first inequality, n was up to 4.03, so n=4 is the last integer. But let's check n=5 for safety.Compute x=√(140*5 +21)/6=√(700 +21)/6=√721/6≈26.8514/6≈4.4752Check if x ∈ [5,6). 4.4752 is not in [5,6). So invalid.n=4: x≈4.0173 is in [4,5). Valid.n=3: 3.5 in [3,4). Valid.n=2: 2.8916 in [2,3). Valid.n=0: 0.764 in [0,1). Valid.What about n= -1? But n has to be ≥0, as we saw earlier. So n= -1 is invalid.Wait, but let's just confirm again with n=1. Even though our inequalities say n=1 is invalid, perhaps due to calculation errors?n=1: x=√(140*1 +21)/6=√161/6≈12.69/6≈2.115Check if x ∈ [1,2). But 2.115 is not in [1,2); it's in [2,3). Therefore, x≈2.115 is in [2,3), but n=1. So inconsistency here. Therefore, n=1 is invalid because x is not in [1,2). Therefore, n=1 is invalid.But wait, x≈2.115, but n=1. So x is supposed to be in [n, n +1)= [1,2), but 2.115 is not in [1,2). Therefore, n=1 is invalid.Therefore, the solutions are only n=0,2,3,4, each giving one x. So total four solutions. Therefore, the answer is C) 4.But wait, let me check for n=4. x≈4.0173 is in [4,5). That's valid. But let's check if the quadratic equation for n=4 gives x exactly or approximately.140*4 +21=560 +21=581. So x=√581/6. √581 is approximately 24.1039, divided by6≈4.0173. So yes, approximately 4.0173, which is just barely in [4,5). So valid.Similarly, for n=3: x=21/6=3.5. Exactly 3.5, which is in [3,4). Valid.For n=2: x≈2.8916, which is in [2,3). Valid.For n=0: x≈0.764, in [0,1). Valid.Therefore, four solutions. So answer is 4, which is option C. So I think the answer is C) 4.Wait, but let me check again. The problem says "the number of real numbers x". So each n gives one x? But quadratic equation is degree two, so could have two solutions, but since we take the positive square root, only one solution per n. So yes, each valid n gives one x. So four x's. Therefore, answer is 4. So C.But hold on. Wait, let me verify the case when n=0. x=√(140*0 +21)/6=√21 /6≈0.764. So x≈0.764, and [x]=0. Which is correct because 0.764 is between0 and1, so floor(x)=0. So substituting back into the equation: 36*(0.764)^2 -140*0 -21 ≈36*0.583 -0 -21≈21 -21=0. So approximately zero. That works.Similarly, n=2: x≈2.8916. [x]=2. So check 36*(2.8916)^2 -140*2 -21≈36*8.36 -280 -21≈301 -280 -21=0. Correct.n=3: x=3.5. [x]=3. So 36*(3.5)^2 -140*3 -21=36*12.25 -420 -21=441 -420 -21=0. Correct.n=4: x≈4.0173. [x]=4. So 36*(4.0173)^2 -140*4 -21≈36*(16.14) -560 -21≈581 -560 -21=0. Correct.Therefore, all four solutions are valid. Hence, the answer is 4. Option C.But wait, I just thought—maybe there's a case when the equation could have two solutions for a particular n. But in this case, since we have x= sqrt((140n +21)/36), which is a positive root, and the negative root is discarded because x must be non-negative. So each n gives at most one x. Therefore, four n's give four x's. Therefore, the answer is 4. So option C.But let me check once more to ensure I didn't miss any possible n.Wait, when n=0, x≈0.764. What if there is another x in the same interval [0,1)? For example, if the quadratic equation had two solutions in [0,1). But x is calculated as sqrt((140n +21)/36). For n=0, it's sqrt(21)/6≈0.764, which is the only positive solution. The other solution is negative, which is discarded. So only one x per n. Therefore, four solutions.Therefore, I think the correct answer is C) 4.Final Answerboxed{C}</think>Given the equation (36x^2 - 140[x] - 21 = 0) where ([x]) denotes the greatest integer less than or equal to (x), we need to find the number of real solutions (x).1. Let (n = [x]), then (n) is an integer and (n leq x < n + 1).2. Substitute (n) into the equation: (36x^2 - 140n - 21 = 0).3. Solve for (x): (x = pm sqrt{frac{140n + 21}{36}}). Since (x geq 0), we consider only the positive root: (x = frac{sqrt{140n + 21}}{6}).4. Check if (x) lies within the interval ([n, n + 1)).We derive the inequalities:- (6n leq sqrt{140n + 21} < 6(n + 1))- Squaring these inequalities: - (36n^2 leq 140n + 21) - (140n + 21 < 36(n + 1)^2)Solving these inequalities:1. (36n^2 - 140n - 21 leq 0) gives roots approximately (-0.144) and (4.03), so (n) can be (0, 1, 2, 3, 4).2. (36(n + 1)^2 - 140n - 21 > 0) gives roots approximately (0.255) and (1.633), so (n) can be (0, 2, 3, 4, ldots).Intersection of these ranges gives (n = 0, 2, 3, 4).Checking each (n):- (n = 0): (x = frac{sqrt{21}}{6} approx 0.764) (valid in ([0, 1)))- (n = 2): (x = frac{sqrt{301}}{6} approx 2.892) (valid in ([2, 3)))- (n = 3): (x = frac{21}{6} = 3.5) (valid in ([3, 4)))- (n = 4): (x = frac{sqrt{581}}{6} approx 4.017) (valid in ([4, 5)))Thus, there are 4 real solutions. The answer is (boxed{C}).]

🔑:5.C.Let x=k+alpha(0 leqslant alpha < 1, k geqslant 0), i.e., the solution of the equation is a positive number.According to equation (1), we havebegin{array}{l}21=36(k+alpha)^{2}-140 k>36 k^{2}-140 k =36 kleft(k-frac{35}{9}right)=36left(k-frac{35}{18}right)^{2}-36left(frac{35}{18}right)^{2} .end{array}Therefore, when k>frac{35}{18}, the right side of the above inequality is an increasing function of k.If k geqslant 5, thenthe right side of equation (2) >36 times 5left(5-frac{35}{9}right)=200,which is a contradiction.Thus, 0 leqslant k leqslant 4.From equation (1), we getx=k+alpha=frac{sqrt{140 k+21}}{6} text {. }By taking k=0,1,2,3,4, we obtainx=frac{sqrt{21}}{6}, frac{sqrt{161}}{6}, frac{sqrt{301}}{6}, frac{7}{2}, frac{sqrt{581}}{6} text {. }Upon verification, x=frac{sqrt{161}}{6} does not satisfy the problem's conditions.Therefore, there are 4 real numbers that satisfy the original equation.

🧠:[<think>Okay, so I need to find the range of the dot product of vectors PA and PB, where P is the left vertex of the ellipse Γ: x²/3 + y² = 1, and A and B are moving points on Γ. Let me break this down step by step.First, let me recall that the standard form of an ellipse is x²/a² + y²/b² = 1. In this case, a² = 3 and b² = 1, so a = √3 and b = 1. The vertices of the ellipse are at (±a, 0) and (0, ±b). Therefore, the left vertex P is at (-√3, 0).Next, points A and B are moving on the ellipse. Let me parametrize their positions. For an ellipse, a common parametrization is using angles, similar to circles, but adjusted for the ellipse's axes. So, a point on the ellipse can be written as (a cosθ, b sinθ) where θ is the parameter angle. So, in this case, points A and B can be represented as:A: (√3 cosθ₁, sinθ₁)B: (√3 cosθ₂, sinθ₂)Since both A and B are moving, θ₁ and θ₂ can vary independently from 0 to 2π.Now, vectors PA and PB. Vector PA is from point P to point A, so it's (Ax - Px, Ay - Py). Similarly for PB. Since P is (-√3, 0), then:PA = (√3 cosθ₁ - (-√3), sinθ₁ - 0) = (√3 (cosθ₁ + 1), sinθ₁)PB = (√3 cosθ₂ - (-√3), sinθ₂ - 0) = (√3 (cosθ₂ + 1), sinθ₂)The dot product PA · PB is then:[√3 (cosθ₁ + 1)][√3 (cosθ₂ + 1)] + [sinθ₁][sinθ₂]Let me compute that:First term: √3 * √3 (cosθ₁ + 1)(cosθ₂ + 1) = 3 (cosθ₁ + 1)(cosθ₂ + 1)Second term: sinθ₁ sinθ₂So the dot product is 3 (cosθ₁ + 1)(cosθ₂ + 1) + sinθ₁ sinθ₂.Now, the problem is to find the range of this expression as θ₁ and θ₂ vary. Since θ₁ and θ₂ are independent variables, this is a function of two variables, and we need to find its maximum and minimum values.Alternatively, maybe we can find a way to express this in terms of sum and differences of angles? Let me see.Alternatively, perhaps consider θ₁ and θ₂ as variables, and try to find the extrema.But since the expression is symmetric in θ₁ and θ₂, maybe we can set θ₁ = θ₂ and see if that gives some extremum? Not sure. Wait, but the problem doesn't state that A and B are the same point, so they can vary independently.Alternatively, maybe fix θ₁ and then find the maximum and minimum over θ₂, then find the overall max and min over θ₁. But that might be complicated.Alternatively, perhaps we can write the expression as:3 (cosθ₁ + 1)(cosθ₂ + 1) + sinθ₁ sinθ₂Let me expand the first term:3 [cosθ₁ cosθ₂ + cosθ₁ + cosθ₂ + 1] + sinθ₁ sinθ₂= 3 cosθ₁ cosθ₂ + 3 cosθ₁ + 3 cosθ₂ + 3 + sinθ₁ sinθ₂So the entire expression is:3 cosθ₁ cosθ₂ + sinθ₁ sinθ₂ + 3 cosθ₁ + 3 cosθ₂ + 3Hmm. Maybe we can group terms. Let's see:3 cosθ₁ cosθ₂ + sinθ₁ sinθ₂ can be written as:cosθ₁ cosθ₂ (3) + sinθ₁ sinθ₂ (1)Which resembles the formula for cos(θ₁ - θ₂) but scaled. Wait, cos(θ₁ - θ₂) = cosθ₁ cosθ₂ + sinθ₁ sinθ₂. So here we have 3 cosθ₁ cosθ₂ + sinθ₁ sinθ₂. So that's 2 cosθ₁ cosθ₂ + [cosθ₁ cosθ₂ + sinθ₁ sinθ₂] = 2 cosθ₁ cosθ₂ + cos(θ₁ - θ₂). Hmm, maybe not helpful.Alternatively, think of it as a linear combination. Alternatively, perhaps factor terms with cosθ₁ and terms with cosθ₂:= 3 cosθ₁ cosθ₂ + 3 cosθ₁ + 3 cosθ₂ + sinθ₁ sinθ₂ + 3Hmm, maybe factor 3 cosθ₁ (cosθ₂ + 1) + 3 cosθ₂ + sinθ₁ sinθ₂ + 3. Not sure.Alternatively, perhaps consider θ₁ and θ₂ as variables and take partial derivatives with respect to θ₁ and θ₂ to find critical points.Let me denote the expression as f(θ₁, θ₂) = 3 cosθ₁ cosθ₂ + sinθ₁ sinθ₂ + 3 cosθ₁ + 3 cosθ₂ + 3.To find the extrema, we can set the partial derivatives to zero.First, compute ∂f/∂θ₁:∂f/∂θ₁ = -3 sinθ₁ cosθ₂ + cosθ₁ sinθ₂ - 3 sinθ₁Similarly, ∂f/∂θ₂ = -3 cosθ₁ sinθ₂ + sinθ₁ cosθ₂ - 3 sinθ₂Set these partial derivatives to zero:-3 sinθ₁ cosθ₂ + cosθ₁ sinθ₂ - 3 sinθ₁ = 0-3 cosθ₁ sinθ₂ + sinθ₁ cosθ₂ - 3 sinθ₂ = 0This seems a system of two equations with two variables θ₁ and θ₂. Solving this system might be complex, but maybe we can find some symmetry.Alternatively, suppose θ₁ = θ₂ = θ. Let's see if that gives a critical point.Then, substituting θ₁ = θ₂ = θ into the equations:First equation:-3 sinθ cosθ + cosθ sinθ - 3 sinθ = (-3 sinθ cosθ + sinθ cosθ) - 3 sinθ = (-2 sinθ cosθ) - 3 sinθ = -sinθ (2 cosθ + 3) = 0Similarly, the second equation would be the same as the first since θ₁ = θ₂:-3 cosθ sinθ + sinθ cosθ - 3 sinθ = (-3 cosθ sinθ + cosθ sinθ) - 3 sinθ = (-2 cosθ sinθ) -3 sinθ = -sinθ (2 cosθ + 3) = 0So setting θ₁ = θ₂ = θ, we get -sinθ (2 cosθ + 3) = 0Solutions are sinθ = 0 or 2 cosθ + 3 = 0. But 2 cosθ + 3 = 0 implies cosθ = -3/2, which is impossible since cosθ ranges between -1 and 1. So the only solutions are sinθ = 0, i.e., θ = 0 or π.So critical points at θ = 0 and θ = π.Let me compute f(θ, θ) when θ = 0:f(0,0) = 3*1*1 + 0 + 3*1 + 3*1 + 3 = 3 + 0 + 3 + 3 + 3 = 12When θ = π:f(π, π) = 3*(-1)(-1) + 0 + 3*(-1) + 3*(-1) + 3 = 3*1 + 0 -3 -3 +3 = 3 -3 -3 +3 = 0So when θ₁ = θ₂ = 0, the value is 12; when θ₁ = θ₂ = π, the value is 0. But maybe these are only critical points when θ₁ = θ₂. There might be other critical points when θ₁ ≠ θ₂.Alternatively, maybe consider when θ₁ = -θ₂. Let me try θ₁ = -θ₂ = θ.Then f(θ, -θ) = 3 cosθ cos(-θ) + sinθ sin(-θ) + 3 cosθ + 3 cos(-θ) + 3= 3 cos²θ - sin²θ + 3 cosθ + 3 cosθ + 3= 3 cos²θ - sin²θ + 6 cosθ + 3But this might not necessarily lead to extrema. Maybe not the right approach.Alternatively, perhaps use trigonometric identities to express f(θ₁, θ₂) in a different form.Looking back at the expression:f(θ₁, θ₂) = 3 cosθ₁ cosθ₂ + sinθ₁ sinθ₂ + 3 cosθ₁ + 3 cosθ₂ + 3Let me write this as:cosθ₁ cosθ₂ *3 + sinθ₁ sinθ₂ *1 + 3 (cosθ₁ + cosθ₂) + 3Hmm, this looks similar to the dot product of two vectors. Suppose we have vectors u = (sqrt(3) cosθ₁, sinθ₁) and v = (sqrt(3) cosθ₂, sinθ₂). Then their dot product would be 3 cosθ₁ cosθ₂ + sinθ₁ sinθ₂. But in our case, f(θ₁, θ₂) is exactly this dot product plus 3 (cosθ₁ + cosθ₂) + 3.So, f(θ₁, θ₂) = u · v + 3 (cosθ₁ + cosθ₂) + 3.But u and v are position vectors of points A and B on the ellipse. Wait, actually, the ellipse is x²/3 + y² =1, so points A and B are exactly (sqrt(3) cosθ, sinθ). So u and v are the position vectors of A and B.But in the problem, vectors PA and PB are from point P to points A and B. So maybe there's a relation between PA, PB and the position vectors of A and B.Given that P is (-sqrt(3), 0), then PA = A - P = (sqrt(3) cosθ₁ + sqrt(3), sinθ₁) = sqrt(3)(cosθ₁ +1, 0) + (0, sinθ₁). Wait, no: PA is (sqrt(3) cosθ₁ - (-sqrt(3)), sinθ₁ - 0) = sqrt(3)(cosθ₁ +1), sinθ₁). Similarly for PB.But earlier, we already established that PA · PB = 3 (cosθ₁ +1)(cosθ₂ +1) + sinθ₁ sinθ₂.Wait, but if u and v are the position vectors of A and B, then u · v = 3 cosθ₁ cosθ₂ + sinθ₁ sinθ₂.So PA · PB = u · v + 3 (cosθ₁ + cosθ₂) + 3.Alternatively, maybe express PA and PB in terms of the position vectors:PA = A - P = (sqrt(3) cosθ₁, sinθ₁) - (-sqrt(3), 0) = (sqrt(3)(cosθ₁ +1), sinθ₁)Similarly, PB = (sqrt(3)(cosθ₂ +1), sinθ₂)Therefore, PA · PB = 3 (cosθ₁ +1)(cosθ₂ +1) + sinθ₁ sinθ₂.Which is the same as before.Alternatively, maybe let’s denote cosθ₁ = c1, sinθ₁ = s1, cosθ₂ = c2, sinθ₂ = s2.Then PA · PB = 3(c1 +1)(c2 +1) + s1 s2.We need to find the maximum and minimum of this expression with the constraints that c1² + s1² =1 and c2² + s2² =1.But since θ₁ and θ₂ are independent, we can treat c1, s1, c2, s2 as variables on the unit circle. So, perhaps we can use Lagrange multipliers for two variables.But this might get complex. Alternatively, perhaps fix θ₁ and then find the maximum over θ₂, then see what the minimum and maximum over θ₁ would be.Let me try that approach. Suppose θ₁ is fixed. Then PA · PB can be considered as a function of θ₂:f(θ₂) = 3 (c1 +1)(c2 +1) + s1 s2.Which can be rewritten as:3(c1 +1) c2 + 3(c1 +1) + s1 s2.So:f(θ₂) = [3(c1 +1)] c2 + [s1] s2 + 3(c1 +1)This is a linear combination of c2 and s2, plus a constant term. The maximum and minimum of this expression over θ₂ (i.e., over c2 and s2 with c2² + s2²=1) can be found using the formula for the maximum of A c2 + B s2 + C, which is C ± sqrt(A² + B²).Yes, because A c + B s has maximum sqrt(A² + B²) and minimum -sqrt(A² + B²). So in this case:A = 3(c1 +1)B = s1C = 3(c1 +1)Therefore, the maximum of f(θ₂) is 3(c1 +1) + sqrt([3(c1 +1)]² + s1²)and the minimum is 3(c1 +1) - sqrt([3(c1 +1)]² + s1²)So, the range of PA · PB for a fixed θ₁ is [3(c1 +1) - sqrt(9(c1 +1)^2 + s1^2), 3(c1 +1) + sqrt(9(c1 +1)^2 + s1^2)]Now, we need to find the overall maximum and minimum over θ₁.Let’s compute the expression inside the square root:9(c1 +1)^2 + s1^2.But since c1² + s1² =1, s1² =1 -c1².Therefore:9(c1 +1)^2 + (1 -c1²) =9(c1² +2c1 +1) +1 -c1² =9c1² +18c1 +9 +1 -c1²=8c1² +18c1 +10.Therefore, the maximum for a fixed θ₁ is 3(c1 +1) + sqrt(8c1² +18c1 +10)and the minimum is 3(c1 +1) - sqrt(8c1² +18c1 +10)So now, we need to find the maximum and minimum of these expressions over c1 ∈ [-1,1].First, let me denote c1 as x for simplicity, where x ∈ [-1,1].So, we have:Maximum value: 3(x +1) + sqrt(8x² +18x +10)Minimum value: 3(x +1) - sqrt(8x² +18x +10)So, the overall maximum of PA · PB is the maximum of 3(x +1) + sqrt(8x² +18x +10) over x ∈ [-1,1]Similarly, the overall minimum is the minimum of 3(x +1) - sqrt(8x² +18x +10) over x ∈ [-1,1]So, let's compute these.First, let's tackle the maximum:f_max(x) = 3(x +1) + sqrt(8x² +18x +10)We need to find the maximum of f_max(x) over x ∈ [-1,1]Similarly, for the minimum:f_min(x) = 3(x +1) - sqrt(8x² +18x +10)Find the minimum of f_min(x) over x ∈ [-1,1]To find the extrema, we can take the derivatives of these functions with respect to x and set them to zero.First, compute f_max(x):f_max(x) = 3x +3 + sqrt(8x² +18x +10)Derivative f_max’(x) = 3 + (16x +18)/(2 sqrt(8x² +18x +10)) = 3 + (8x +9)/sqrt(8x² +18x +10)Set this equal to zero:3 + (8x +9)/sqrt(8x² +18x +10) = 0=> (8x +9)/sqrt(8x² +18x +10) = -3Square both sides:(8x +9)^2 / (8x² +18x +10) = 9Multiply both sides by denominator:(8x +9)^2 = 9(8x² +18x +10)Expand left side: 64x² + 144x +81Right side: 72x² + 162x +90Bring all terms to left:64x² +144x +81 -72x² -162x -90 = 0=> -8x² -18x -9 =0Multiply by -1:8x² +18x +9 =0Solve quadratic:x = [-18 ± sqrt(324 - 288)] /16 = [-18 ± sqrt(36)] /16 = [-18 ±6]/16So,x = (-18 +6)/16 = (-12)/16 = -3/4x = (-18 -6)/16 = (-24)/16 = -3/2But x must be in [-1,1]. So x = -3/4 is valid, x=-3/2 is outside the interval.So critical point at x = -3/4.Check if this is a maximum. Compute second derivative or check endpoints.Compute f_max(-3/4):First compute 3*(-3/4 +1) + sqrt(8*(9/16) +18*(-3/4) +10)Compute 3*(1/4) + sqrt( (72/16) - (54/4) +10 )= 3/4 + sqrt( (9/2) - (27/2) +10 )= 3/4 + sqrt( (-18/2) +10 )= 3/4 + sqrt(-9 +10) = 3/4 + sqrt(1) = 3/4 +1 = 7/4 = 1.75Now compute f_max at endpoints x=-1 and x=1.At x=-1:f_max(-1) = 3*(-1 +1) + sqrt(8*1 +18*(-1) +10) = 0 + sqrt(8 -18 +10) = sqrt(0) =0Wait, but 8*(-1)^2 +18*(-1) +10 =8*1 -18 +10=8-18+10=0. So sqrt(0)=0. Thus, f_max(-1)=0.At x=1:f_max(1)=3*(1 +1) + sqrt(8*1 +18*1 +10)=6 + sqrt(8 +18 +10)=6 + sqrt(36)=6+6=12So comparing the values:At x=-3/4: 1.75At x=-1:0At x=1:12Thus, the maximum of f_max(x) is 12 at x=1.Wait, but when x=1, that corresponds to θ₁=0, so point A is at (√3,0). Similarly, when θ₁=0, the maximum over θ₂ is achieved at θ₂=0, as we saw earlier.But wait, when we set θ₁=0, then the expression becomes:PA · PB = 3(1 +1)(cosθ₂ +1) +0*sinθ₂= 6(cosθ₂ +1)The maximum of 6(cosθ₂ +1) is when cosθ₂=1, i.e., θ₂=0, giving 6*2=12. Similarly, minimum when cosθ₂=-1, giving 6*0=0. So that checks out. So when θ₁=0, the maximum PA · PB is 12 and minimum is 0.But when θ₁ varies, we need to check if there's a higher value. But in this analysis, the maximum over x=1 (i.e., θ₁=0) gives the highest value of 12. So overall maximum is 12.Now, let's look at the minimum of f_min(x) = 3(x +1) - sqrt(8x² +18x +10). We need to find the minimum of this over x ∈ [-1,1].Again, take derivative:f_min’(x) = 3 - (16x +18)/(2 sqrt(8x² +18x +10)) = 3 - (8x +9)/sqrt(8x² +18x +10)Set derivative equal to zero:3 - (8x +9)/sqrt(8x² +18x +10) =0=> (8x +9)/sqrt(8x² +18x +10) =3Square both sides:(8x +9)^2 / (8x² +18x +10) =9Same as before:(8x +9)^2 =9(8x² +18x +10)64x² +144x +81 =72x² +162x +90Left side minus right side:-8x² -18x -9=0 => 8x² +18x +9=0Which we solved earlier, giving x=-3/4 and x=-3/2. Again, x=-3/4 is in [-1,1].Thus, critical point at x=-3/4.Compute f_min(-3/4):3*(-3/4 +1) - sqrt(8*(9/16) +18*(-3/4) +10)=3*(1/4) - sqrt( (72/16) - (54/4) +10 )=3/4 - sqrt( (9/2) - (27/2) +10 )=3/4 - sqrt( (-18/2) +10 )=3/4 - sqrt(-9 +10) =3/4 -1= -1/4= -0.25Now check endpoints:At x=-1:f_min(-1)=3*(-1 +1) - sqrt(8*1 +18*(-1)+10)=0 - sqrt(8 -18 +10)=0 - sqrt(0)=0At x=1:f_min(1)=3*(1 +1) - sqrt(8*1 +18*1 +10)=6 - sqrt(36)=6 -6=0So, comparing the values:At x=-3/4: -0.25At x=-1:0At x=1:0Thus, the minimum of f_min(x) is -0.25 at x=-3/4. Wait, but -0.25 is -1/4. But let me check the calculation again.Wait, f_min(-3/4):3*(-3/4 +1) - sqrt(8*(-3/4)^2 +18*(-3/4) +10)First, compute 3*(1/4) = 3/4 =0.75Then compute the sqrt part:8*(9/16) +18*(-3/4) +10= (72/16) - (54/4) +10= (9/2) - (27/2) +10= (-18/2)+10= -9 +10=1Therefore, sqrt(1)=1. So, f_min=0.75 -1= -0.25. Yes, that's correct.Therefore, the minimum value is -1/4.But wait, but when x=-3/4, then θ₁ corresponds to cosθ₁=-3/4. Let me check if that's possible.But cosθ₁ can be -3/4, since cosine ranges between -1 and 1. So yes, valid.Therefore, the overall minimum of PA · PB is -1/4, and the maximum is 12.Wait, but in the earlier analysis when θ₁=θ₂=π, we got PA · PB=0, which is higher than -1/4. So maybe the minimum is indeed -1/4. Let me verify with θ₁ and θ₂ corresponding to x=-3/4.Wait, x here is c1=cosθ₁=-3/4. Then, s1=sinθ₁=±sqrt(1 - (9/16))=±sqrt(7/16)=±√7/4.So, for θ₁ with cosθ₁=-3/4 and sinθ₁=√7/4 or -√7/4. Let's take sinθ₁=√7/4 first.Then, when θ₁ is such that cosθ₁=-3/4 and sinθ₁=√7/4, then the minimum value of PA · PB is achieved when θ₂ is chosen to minimize the expression.From earlier, for fixed θ₁, the minimum is 3(c1 +1) - sqrt(8c1² +18c1 +10)Plugging c1=-3/4:3*(-3/4 +1) - sqrt(8*(9/16) +18*(-3/4) +10)=3*(1/4) - sqrt(72/16 -54/4 +10)=3/4 - sqrt(9/2 -27/2 +10)=3/4 - sqrt(-18/2 +10)=3/4 - sqrt(-9 +10)=3/4 -1= -1/4So, when θ₁ is such that cosθ₁=-3/4 and sinθ₁=√7/4, then there exists a θ₂ that gives PA · PB=-1/4.Similarly, if sinθ₁=-√7/4, then the same result since sinθ₁ is squared in the sqrt term.Therefore, the minimum value is indeed -1/4.But let me check this with actual points.Let me take θ₁ such that cosθ₁=-3/4, sinθ₁=√7/4. Then point A is (√3*(-3/4), √7/4)= (-3√3/4, √7/4). Point P is (-√3,0). So vector PA is A - P = (-3√3/4 - (-√3), √7/4 -0)= (√3/4, √7/4)Similarly, for θ₂, we need to find the θ₂ that minimizes PA · PB. Wait, but according to our earlier analysis, for fixed θ₁, the minimum PA · PB is achieved when θ₂ is such that [3(c1 +1)]c2 + s1 s2 is minimized. But we have already computed that the minimum is -sqrt( [3(c1 +1)]² + s1² )So with c1=-3/4 and s1=√7/4,[3(c1 +1)] = 3*(-3/4 +1)=3*(1/4)=3/4s1=√7/4Therefore, the minimum PA · PB is 3(c1 +1) - sqrt( [3(c1 +1)]² + s1² )=3/4 - sqrt( (9/16) + (7/16) )=3/4 - sqrt(16/16)=3/4 -1= -1/4So when θ₂ is chosen such that [3(c1 +1)]c2 + s1 s2 = -sqrt([3(c1 +1)]² + s1² )This occurs when the vector (3(c1 +1), s1) is in the opposite direction of (c2, s2). Therefore, (c2, s2) = - (3(c1 +1), s1)/sqrt( [3(c1 +1)]² + s1² )In our case, 3(c1 +1)=3/4, s1=√7/4. So, the vector is (3/4, √7/4). Its norm is sqrt( (9/16) + (7/16) )=sqrt(16/16)=1.Therefore, (c2, s2) must be -(3/4, √7/4), so c2=-3/4, s2=-√7/4. Therefore, θ₂ is such that cosθ₂=-3/4, sinθ₂=-√7/4.Therefore, point B is (√3*(-3/4), -√7/4). Then vector PB is B - P = (√3*(-3/4) - (-√3), -√7/4 -0)= (√3*(-3/4 +1), -√7/4)= (√3*(1/4), -√7/4)Then PA · PB = (√3/4, √7/4) · (√3/4, -√7/4)= (3/16) - (7/16)= (-4)/16= -1/4. Which matches the calculation.Therefore, this point indeed gives PA · PB=-1/4. So the minimum is -1/4.Therefore, the range of PA · PB is from -1/4 to 12.Wait, but earlier when we had θ₁=θ₂=π, we calculated PA · PB=0. But according to this analysis, the minimum is -1/4, which is less than 0. So that seems correct.Therefore, the final answer should be that the range of PA · PB is [-1/4, 12].But let me check once more for errors.Wait, when we considered θ₁=θ₂=0, PA · PB=12, which is correct. When θ₁=θ₂=π, PA · PB=0, but when we choose specific θ₁ and θ₂, we can get negative values, down to -1/4.Another check: take θ₁=θ such that cosθ= -3/4, sinθ=√7/4, and θ₂ such that cosθ₂=-3/4, sinθ₂=-√7/4. Then PA · PB=-1/4. This seems correct.Therefore, the range is from -1/4 to 12.But the problem states "the range of PA · PB". So, in boxed form, boxed{[-frac{1}{4}, 12]}.But let me check once again if there could be lower values. Suppose we choose θ₁ and θ₂ such that PA and PB are in opposite directions, but given the ellipse, maybe there's a more negative dot product. But according to the analysis, the minimum is -1/4.Alternatively, consider θ₁=π, θ₂=0.PA is vector from P(-√3,0) to A(√3*cosπ, sinπ)=(-√3,0). So PA is (-√3 - (-√3),0 -0)=(0,0). Wait, no. Wait, if θ₁=π, then point A is (√3 cosπ, sinπ)=(-√3, 0). Then PA is (-√3 - (-√3),0 -0)=(0,0). So PA is the zero vector. Then PA · PB=0 · PB=0. So in this case, PA · PB=0. So when A coincides with P, the dot product is zero regardless of B.Similarly, if B coincides with P, then PB is zero vector, and the dot product is zero.But according to our calculation, the minimum is achieved when both A and B are not at P, but at specific points.But the key point is that when A is near P, PA is small, but when B is in a certain position, the dot product can be negative.But according to the analysis, -1/4 is the minimum. Let me see.Another test case: Let’s take θ₁=π/2, so A is (0,1). Then PA is (0 - (-√3),1 -0)=(√3,1). Let’s take θ₂=3π/2, so B is (0,-1). Then PB is (0 - (-√3), -1 -0)=(√3,-1). Then PA · PB= (√3)(√3) + (1)(-1)=3 -1=2. Which is positive.Alternatively, take θ₁=π/3, θ₂=2π/3.Wait, perhaps this is getting too time-consuming. According to the mathematical analysis, the minimum is -1/4 and maximum is 12. The critical points check out. The test case with specific angles gives the calculated minimum. Therefore, I think the answer is correct.Final Answerboxed{left[ -dfrac{1}{4}, 12 right]}</think>The ellipse (Gamma: frac{x^2}{3} + y^2 = 1) has its left vertex at (P(-sqrt{3}, 0)). Points (A) and (B) are moving points on (Gamma). We need to find the range of (overrightarrow{PA} cdot overrightarrow{PB}).1. Parametrization of Points (A) and (B): - Points (A) and (B) can be parametrized as (A(sqrt{3} cos theta_1, sin theta_1)) and (B(sqrt{3} cos theta_2, sin theta_2)).2. Vectors (overrightarrow{PA}) and (overrightarrow{PB}): - (overrightarrow{PA} = (sqrt{3} (cos theta_1 + 1), sin theta_1)) - (overrightarrow{PB} = (sqrt{3} (cos theta_2 + 1), sin theta_2))3. Dot Product Calculation: - The dot product (overrightarrow{PA} cdot overrightarrow{PB}) is: [ 3 (cos theta_1 + 1)(cos theta_2 + 1) + sin theta_1 sin theta_2 ] - Expanding this, we get: [ 3 cos theta_1 cos theta_2 + 3 cos theta_1 + 3 cos theta_2 + 3 + sin theta_1 sin theta_2 ]4. Finding the Range: - We analyze the expression by considering (theta_1) and (theta_2) as independent variables. - Using trigonometric identities and optimization techniques, we find the maximum and minimum values of the expression.5. Critical Points and Extrema: - By setting partial derivatives to zero and solving, we find critical points. - The maximum value is found when (theta_1 = theta_2 = 0), giving (12). - The minimum value is found when (cos theta_1 = -frac{3}{4}) and (cos theta_2 = -frac{3}{4}) with specific sine values, giving (-frac{1}{4}).6. Verification: - The critical points and endpoints are verified to ensure correctness.The range of (overrightarrow{PA} cdot overrightarrow{PB}) is (boxed{left[ -dfrac{1}{4}, 12 right]}).,<think>Alright, let's tackle this problem step by step. So, we have an ellipse Γ defined by the equation x²/3 + y² = 1. The left vertex is point P. Then there are two moving points A and B on Γ. We need to find the range of the dot product of vectors PA and PB. First, let me recall some basics about ellipses. The standard form of an ellipse is x²/a² + y²/b² = 1, where a is the semi-major axis and b is the semi-minor axis. In this case, comparing with the given equation, a² = 3 and b² = 1, so a = √3 and b = 1. Since the denominator under x² is larger, the major axis is along the x-axis. The vertices are at (±a, 0), so the left vertex P is at (-√3, 0). Got that.Now, points A and B are moving points on the ellipse. So their coordinates must satisfy x²/3 + y² = 1. The problem asks for the range of the dot product of vectors PA and PB. Let me recall the formula for the dot product. If we have two vectors u = (u₁, u₂) and v = (v₁, v₂), their dot product is u₁v₁ + u₂v₂. So, PA ⋅ PB is the sum of the products of their corresponding components.First, I need to express vectors PA and PB in terms of coordinates of points A and B. Since P is (-√3, 0), vector PA would be A - P, right? So if A is (x₁, y₁), then PA = (x₁ - (-√3), y₁ - 0) = (x₁ + √3, y₁). Similarly, vector PB would be (x₂ + √3, y₂) if B is (x₂, y₂). Therefore, the dot product PA ⋅ PB is (x₁ + √3)(x₂ + √3) + y₁y₂.So the expression we need to analyze is (x₁ + √3)(x₂ + √3) + y₁y₂. Let's expand that:(x₁ + √3)(x₂ + √3) = x₁x₂ + √3 x₁ + √3 x₂ + 3Adding the y₁y₂ term, the total becomes:x₁x₂ + √3 x₁ + √3 x₂ + 3 + y₁y₂So, the dot product PA ⋅ PB equals x₁x₂ + √3(x₁ + x₂) + 3 + y₁y₂.Now, since A and B are points on the ellipse Γ, their coordinates satisfy x²/3 + y² = 1. So for point A: x₁²/3 + y₁² = 1, and similarly for point B: x₂²/3 + y₂² = 1.Our goal is to find the maximum and minimum values of the expression x₁x₂ + √3(x₁ + x₂) + 3 + y₁y₂, given that x₁²/3 + y₁² = 1 and x₂²/3 + y₂² = 1.Hmm, this seems a bit complicated with two moving points. Maybe there's a way to parametrize points A and B on the ellipse. Since it's an ellipse, parametric equations can be written using trigonometric functions. Let's recall that for an ellipse x²/a² + y²/b² = 1, a parametrization is (a cos θ, b sin θ). So, for our ellipse, a = √3 and b = 1, so point A can be written as (√3 cos θ, sin θ) and point B as (√3 cos φ, sin φ), where θ and φ are angles parameterizing the positions of A and B on the ellipse.Using this parametrization, let's substitute into our expression for the dot product.First, compute x₁x₂: (√3 cos θ)(√3 cos φ) = 3 cos θ cos φNext, √3(x₁ + x₂) = √3(√3 cos θ + √3 cos φ) = √3 * √3 (cos θ + cos φ) = 3 (cos θ + cos φ)Then, the constant term is +3.Then, y₁y₂ = (sin θ)(sin φ) = sin θ sin φPutting it all together:PA ⋅ PB = 3 cos θ cos φ + 3 (cos θ + cos φ) + 3 + sin θ sin φSimplify:= 3 cos θ cos φ + 3 cos θ + 3 cos φ + 3 + sin θ sin φHmm. Let's see if we can combine terms. The 3 cos θ cos φ and sin θ sin φ can be grouped together. Let's write them as:= [3 cos θ cos φ + sin θ sin φ] + 3 cos θ + 3 cos φ + 3Hmm, maybe we can use some trigonometric identities here. Recall that cos(θ - φ) = cos θ cos φ + sin θ sin φ. But here we have 3 cos θ cos φ + sin θ sin φ. It's not quite the same. Maybe factor out something?Alternatively, perhaps express the entire expression in terms of cos(θ - φ) or something similar. Let's see.Alternatively, maybe we can write 3 cos θ cos φ + sin θ sin φ as a combination of cos(θ - φ) and something else. Let's try:3 cos θ cos φ + sin θ sin φ = (3 cos θ cos φ) + (sin θ sin φ)Let me see. Let me denote that term as C. Then:C = 3 cos θ cos φ + sin θ sin φPerhaps factor out cos θ cos φ?= cos θ cos φ (3 + tan θ tan φ)But that might not be helpful. Alternatively, write it as:C = 2 cos θ cos φ + (cos θ cos φ + sin θ sin φ)= 2 cos θ cos φ + cos(θ - φ)But I don't know if that helps. Alternatively, maybe use multiple angle formulas.Alternatively, think of θ and φ as variables and try to find maxima and minima by taking derivatives. But since there are two variables, θ and φ, this might get complicated.Alternatively, notice that the expression is symmetric in θ and φ in some terms. Let's check:The terms involving θ and φ:3 cos θ cos φ + sin θ sin φ + 3 cos θ + 3 cos φ + 3Is there a way to separate variables? Let's see. Let me rearrange the terms:= [3 cos θ cos φ + 3 cos θ + 3 cos φ] + sin θ sin φ + 3Factor out 3 cos θ:= 3 cos θ (cos φ + 1) + 3 cos φ + sin θ sin φ + 3Not sure. Alternatively, factor out 3 from some terms:= 3(cos θ cos φ + cos θ + cos φ + 1) + sin θ sin φ - 3 cos θ cos φ ?Wait, that might not be helpful. Let me check the original expression again:PA ⋅ PB = 3 cos θ cos φ + 3 cos θ + 3 cos φ + 3 + sin θ sin φHmm, maybe factor 3(cos θ cos φ + cos θ + cos φ + 1) + sin θ sin φ - 3 cos θ cos φ. No, this seems messy.Alternatively, let's group the terms as follows:= 3(cos θ cos φ + cos θ + cos φ + 1) + sin θ sin φ - 3 cos θ cos φWait, that's not correct. Wait, original expression is:3 cos θ cos φ + 3 cos θ + 3 cos φ + 3 + sin θ sin φSo, 3 cos θ cos φ + 3 cos θ + 3 cos φ + 3 = 3 (cos θ cos φ + cos θ + cos φ + 1) Then plus sin θ sin φ. So:PA ⋅ PB = 3 (cos θ cos φ + cos θ + cos φ + 1) + sin θ sin φAlternatively, note that cos θ cos φ + cos θ + cos φ + 1 can be written as (cos θ + 1)(cos φ + 1) - sin θ sin φ. Wait:Let's check: (cos θ + 1)(cos φ + 1) = cos θ cos φ + cos θ + cos φ + 1. So yes, exactly. Therefore:PA ⋅ PB = 3 [(cos θ + 1)(cos φ + 1) - sin θ sin φ] + sin θ sin φ= 3 (cos θ + 1)(cos φ + 1) - 3 sin θ sin φ + sin θ sin φ= 3 (cos θ + 1)(cos φ + 1) - 2 sin θ sin φHmm, not sure if this helps, but maybe. Alternatively, maybe we can use substitution variables.Alternatively, let's denote u = θ + φ and v = θ - φ, but since θ and φ are independent variables, this might complicate things. Alternatively, fix one angle and express the other in terms of it.Alternatively, treat θ and φ as independent variables and try to find the extrema. Let's consider PA ⋅ PB as a function f(θ, φ) = 3 cos θ cos φ + 3 cos θ + 3 cos φ + 3 + sin θ sin φ.To find its extrema, take partial derivatives with respect to θ and φ, set them to zero, and solve.First, compute ∂f/∂θ:Derivative of 3 cos θ cos φ with respect to θ is -3 sin θ cos φDerivative of 3 cos θ is -3 sin θDerivative of 3 cos φ is 0Derivative of 3 is 0Derivative of sin θ sin φ is cos θ sin φSo ∂f/∂θ = -3 sin θ cos φ - 3 sin θ + cos θ sin φSimilarly, ∂f/∂φ:Derivative of 3 cos θ cos φ with respect to φ is -3 cos θ sin φDerivative of 3 cos θ is 0Derivative of 3 cos φ is -3 sin φDerivative of 3 is 0Derivative of sin θ sin φ is sin θ cos φSo ∂f/∂φ = -3 cos θ sin φ - 3 sin φ + sin θ cos φSet both partial derivatives equal to zero:1. -3 sin θ cos φ - 3 sin θ + cos θ sin φ = 02. -3 cos θ sin φ - 3 sin φ + sin θ cos φ = 0These are two equations with variables θ and φ. This seems quite complicated. Maybe there's some symmetry here. Suppose that θ = φ. Let's check if this is possible. If θ = φ, then points A and B coincide. Wait, but the problem says A and B are two moving points; they can coincide? The problem doesn't specify they are distinct. So maybe the maximum or minimum occurs when θ = φ. Let's test this.Set θ = φ. Then equations become:1. -3 sin θ cos θ - 3 sin θ + cos θ sin θ = 0Simplify:(-3 sin θ cos θ + cos θ sin θ) - 3 sin θ = (-2 sin θ cos θ) - 3 sin θ = 0Factor out sin θ:sin θ (-2 cos θ - 3) = 0Similarly, equation 2:-3 cos θ sin θ - 3 sin θ + sin θ cos θ = (-3 cos θ sin θ + sin θ cos θ) - 3 sin θ = (-2 cos θ sin θ) - 3 sin θ = 0Same as equation 1. So the equations reduce to sin θ (-2 cos θ - 3) = 0. Therefore, sin θ = 0 or -2 cos θ -3 =0. But -2 cos θ -3 =0 implies cos θ = -3/2, which is impossible since cos θ ranges between -1 and 1. Therefore, the only solution is sin θ = 0. So θ = 0 or π.If θ = 0, then point A is (√3, 0), which is the right vertex. Similarly, point B would also be (√3, 0). Let's compute PA ⋅ PB in this case.PA is from P(-√3, 0) to A(√3, 0): (√3 + √3, 0 - 0) = (2√3, 0). Similarly, PB is (2√3, 0). The dot product is (2√3)(2√3) + 0*0 = 12. Plus, the other terms when θ = φ =0:Original expression: 3*1*1 + 3*1 + 3*1 + 3 + 0 = 3 + 3 + 3 + 3 = 12. So that matches.If θ = π, point A is (-√3, 0), which is point P itself. Similarly, point B would be P. But vectors PA and PB would be zero vectors, so their dot product is 0. Let's check with the expression:3 cos π cos π + 3 cos π + 3 cos π + 3 + sin π sin π = 3*(-1)(-1) + 3*(-1) + 3*(-1) + 3 + 0 = 3 -3 -3 +3 = 0. Correct.But θ = φ = π gives PA ⋅ PB = 0. So that's a critical point. But we need to check if these are maxima or minima. When θ = φ = 0, the dot product is 12. When θ = φ = π, it's 0. But there might be other critical points where θ ≠ φ.Alternatively, maybe the maximum and minimum occur when A and B are at specific points, such as the vertices, co-vertices, or other points. Let's try some specific points.For example, let’s take A and B both at the right vertex (√3, 0). Then PA ⋅ PB = (2√3, 0) ⋅ (2√3, 0) = 12, as above. If A is at (√3, 0) and B is at (-√3, 0), then PA is (2√3, 0) and PB is (0, 0), so the dot product is 0. If A is at (0, 1) and B is at (0, 1), then PA is (√3, 1) and PB is (√3, 1). Their dot product is (√3)^2 + 1*1 = 3 + 1 = 4. If A is at (0,1) and B is at (0,-1), then PA is (√3,1) and PB is (√3, -1). The dot product is (√3)(√3) + (1)(-1) = 3 -1 = 2.Hmm, so we have different values here. Maybe 12 is the maximum, but what's the minimum? Let's see. When A and B are at (0,1) and (0,-1), the dot product is 2. Is there a case where it's lower?Wait, when A is at (√3, 0) and B is at (-√3, 0), the dot product is 0, which is lower. But when A and B are both at (-√3, 0), the dot product is 0 as well. So maybe 0 is the minimum? But when we had θ = φ = π, the points are overlapping at P, so the vectors PA and PB are zero vectors. But if the points are allowed to coincide, then the minimum is 0. However, if the points must be distinct, maybe the minimum is higher. But the problem says "two moving points", so I think they can coincide. Hence, 0 might be attainable. Wait, but when A and B are both P, then PA and PB are zero vectors, so their dot product is 0. However, if A and B are required to be on Γ, then P is on Γ (since it's the left vertex), so they can indeed coincide at P, resulting in 0. So perhaps the minimum is 0 and the maximum is 12. But earlier, when A and B are at (0,1) and (0,-1), the dot product was 2, which is higher than 0. So maybe 0 is the minimum, 12 is the maximum. But let's check more points.Suppose A is at (√3, 0) and B is at (0,1). Then PA is (2√3, 0) and PB is (√3, 1). The dot product is (2√3)(√3) + 0*1 = 6 + 0 = 6. Another value. So, the dot product varies between 0 and 12? Maybe. But maybe there's a lower value than 0.Wait, let's take A at (-√3, 0) and B somewhere else. Wait, if A is P, then PA is (0,0), so PA ⋅ PB is 0 regardless of B. Similarly, if B is P, then PB is (0,0), so the dot product is 0. So 0 is achieved when either A or B is P. But if neither A nor B is P, can we get negative values?Suppose A is at (√3/2, y) and B is at (√3/2, -y). Let's compute PA ⋅ PB.PA = (√3/2 + √3, y) = ( (3√3)/2, y )PB = (√3/2 + √3, -y ) = ( (3√3)/2, -y )Dot product: ( (3√3)/2 )^2 + y*(-y) = (27/4) - y²But since A is on the ellipse, x²/3 + y² = 1. So ( (√3/2)^2 )/3 + y² = 1 => ( 3/4 ) /3 + y² = 1 => 1/4 + y² =1 => y² = 3/4. So y = ±√3/2.Therefore, the dot product is 27/4 - 3/4 = 24/4 = 6. So in this case, the dot product is 6. Positive.What if we take A and B such that their coordinates lead to a negative contribution? For example, take A and B in such a way that their x-coordinates are negative. Let's try.Let A be ( -√3/2, y ) and B be ( -√3/2, y ). Since they are on the ellipse, x²/3 + y² =1 => ( 3/4 ) /3 + y² =1 => 1/4 + y² =1 => y² = 3/4. So y=±√3/2.Compute PA ⋅ PB. PA is ( -√3/2 + √3, y ) = ( √3/2, y ). Similarly, PB is ( √3/2, y ). Their dot product is ( (√3/2)^2 + y^2 ) = ( 3/4 + 3/4 ) = 6/4 = 3/2. Still positive.Hmm. Maybe it's challenging to get a negative value. Wait, let's think algebraically.Looking at the expression PA ⋅ PB = x₁x₂ + √3(x₁ + x₂) + 3 + y₁y₂.Given that x₁ and x₂ can be negative or positive, and y₁ and y₂ as well. But the term √3(x₁ + x₂) could be negative if x₁ + x₂ is negative. Similarly, x₁x₂ could be positive or negative. Let's see.Suppose both A and B are on the left side of the ellipse, left of the origin. So x₁ and x₂ are negative. Then x₁x₂ is positive (negative times negative). √3(x₁ + x₂) is negative. The term 3 is positive. y₁y₂ could be positive or negative. Let's see.For example, take A and B at (-√3, 0), which is P. Then PA ⋅ PB =0 as before.If A is at (-√3/2, √3/2) and B is at (-√3/2, -√3/2). Let's compute PA ⋅ PB.First, coordinates of A: x₁ = -√3/2, y₁ = √3/2Coordinates of B: x₂ = -√3/2, y₂ = -√3/2PA = ( -√3/2 + √3, √3/2 ) = ( √3/2, √3/2 )PB = ( -√3/2 + √3, -√3/2 ) = ( √3/2, -√3/2 )Dot product: (√3/2)(√3/2) + (√3/2)(-√3/2) = (3/4) - (3/4) = 0.Hmm, still zero. Let's try another point. Let’s take A at (0,1) and B at (0,-1). Then PA = (√3,1), PB = (√3, -1). Dot product: (√3)(√3) + (1)(-1) = 3 -1 = 2.Positive again. Let me try A at ( √3/2, √3/2 ) and B at ( √3/2, -√3/2 ). Wait, first check if these points are on the ellipse.For A: x²/3 + y² = ( (3/4)/3 ) + ( 3/4 ) = (1/4) + (3/4) = 1. Yes. Similarly for B.Compute PA ⋅ PB.PA = ( √3/2 + √3, √3/2 ) = ( (3√3)/2, √3/2 )PB = ( √3/2 + √3, -√3/2 ) = ( (3√3)/2, -√3/2 )Dot product: (3√3/2)(3√3/2) + (√3/2)(-√3/2) = (27/4) - (3/4) = 24/4 = 6. Positive again.Hmm. Maybe it's impossible to get negative values? Let's think again.Looking at the expression PA ⋅ PB = x₁x₂ + √3(x₁ + x₂) + 3 + y₁y₂.Suppose x₁ and x₂ are both large negative numbers. Wait, but the ellipse only allows x to be between -√3 and √3. So x₁ and x₂ can be as low as -√3. Let's take x₁ = x₂ = -√3. Then PA and PB would be zero vectors (since A and B are P). The dot product is 0. If x₁ = -√3 and x₂ is something else, then PA is (0, y₁) and PB is (x₂ + √3, y₂). Then the dot product is 0*(x₂ + √3) + y₁y₂ = y₁y₂. So if A is P, then the dot product is y₁y₂. Since A is P, y₁ =0, so the dot product is 0. Similarly, if B is P, then the dot product is 0.But if neither A nor B is P, then PA and PB have both components. Let's see if y₁y₂ can be negative. If y₁ and y₂ have opposite signs, then y₁y₂ is negative. So maybe we can get negative contributions from y₁y₂.But let's see. For example, take A at (0,1) and B at (0,-1). Then PA ⋅ PB = (√3,1) ⋅ (√3,-1) = (√3)^2 + (1)(-1) = 3 -1 = 2. Still positive.Wait, maybe even if y₁y₂ is negative, the other terms dominate. Let's try to find a case where y₁y₂ is negative and x₁x₂ + √3(x₁ + x₂) + 3 is less than the absolute value of y₁y₂.Wait, for example, let’s take A at (x, y) and B at (x, -y). Then PA ⋅ PB would be [x + √3]^2 + y*(-y) = (x + √3)^2 - y². Since A is on the ellipse, x²/3 + y² =1 => y² =1 - x²/3.Substitute into the expression: (x + √3)^2 - (1 - x²/3) = x² + 2√3 x + 3 -1 + x²/3 = (4x²)/3 + 2√3 x + 2.Now, to see if this can be negative. Let’s set this equal to some value and see if it can be negative.(4x²)/3 + 2√3 x + 2 = 0.Multiply both sides by 3: 4x² + 6√3 x + 6 =0.Discriminant: (6√3)^2 - 4*4*6 = 108 - 96 = 12 >0. So real roots exist. So x = [-6√3 ± √12]/8 = [-6√3 ± 2√3]/8 = [(-6√3 ± 2√3)]/8.So x = [(-6√3 + 2√3)/8] = (-4√3)/8 = -√3/2, or x = (-6√3 -2√3)/8 = (-8√3)/8 = -√3.So x = -√3/2 and x = -√3.But x must be in [-√3, √3]. So x = -√3 is allowed, but when x = -√3, then y² =1 - ( (-√3)^2 )/3 =1 -1=0, so y=0. Therefore, point is P, so PA ⋅ PB =0.For x = -√3/2, we can compute y² =1 - ( (3/4)/3 )=1 -1/4=3/4, so y=±√3/2. Therefore, taking A=(-√3/2, √3/2) and B=(-√3/2, -√3/2):PA ⋅ PB = ( (-√3/2 + √3 )^2 - (√3/2)^2 ) = ( (√3/2)^2 - 3/4 ) = ( 3/4 - 3/4 ) =0.So PA ⋅ PB=0. So in this case, the expression equals zero. Therefore, even when A and B are not P but symmetric about the x-axis at x=-√3/2, the dot product is zero. Therefore, zero is attainable even when neither A nor B is P.But earlier, when we tried A=(-√3/2, √3/2) and B=(-√3/2, -√3/2), the dot product was zero. So, perhaps the minimum value is zero, and the maximum is 12.But wait, the problem says "the range of PA ⋅ PB". So maybe it's [0, 12]. But wait, earlier when we took A and B at (0,1) and (0,-1), the dot product was 2, which is positive. But zero is lower than 2. So is zero the minimum? If so, then the range is [0,12]. But we need to verify if negative values are possible.Wait, let's consider another scenario. Let’s take A and B such that their y-coordinates are negative and positive in a way that y₁y₂ is negative, and see if the total expression can be negative.Let’s take A=(x, y) and B=(x', y') with y positive and y' negative. Let's see:PA ⋅ PB = x x' + √3(x + x') + 3 + y y'Suppose we can make this negative. Let's try to find such points.Let’s pick A=(0,1), B=(0,-1). Then PA ⋅ PB = 0*0 + √3(0+0) +3 + (1)(-1) =0 +0 +3 -1=2>0.Still positive. Let's pick A=(√3,0), B=(0,-1). Then PA=(2√3,0), PB=(√3,-1). Dot product: 2√3*√3 + 0*(-1)=6 +0=6>0.Another example: A=(√3/2, √3/2), B=(√3/2, -√3/2). PA=( (3√3)/2, √3/2 ), PB=( (3√3)/2, -√3/2 ). Dot product: (9*3)/4 + (-3/4)=27/4 -3/4=24/4=6>0.What if we take A=(x, y) and B=(x, -y) with x negative. Let’s compute:PA ⋅ PB = x^2 + √3*2x +3 - y^2.But since x²/3 + y²=1, y²=1 -x²/3. Substitute:PA ⋅ PB =x² + 2√3 x +3 - (1 -x²/3)=x² +2√3 x +3 -1 +x²/3= (4x²)/3 +2√3 x +2.This is the same expression as before, which equals zero when x=-√3/2 or x=-√3, but it's quadratic in x, opening upwards (since coefficient of x² is positive). Therefore, the minimum value is at x = -b/(2a) = -(2√3)/(2*(4/3)) )= -(2√3)/(8/3) )= - (6√3)/8 = - (3√3)/4 ≈ -1.3.But x must be ≥ -√3 ≈ -1.732. So x=-3√3/4 ≈-1.3 is within the valid range. Plugging into the expression:(4x²)/3 +2√3 x +2.Compute x= -3√3/4:x²= (9*3)/16=27/16So (4*(27/16))/3 +2√3*(-3√3/4) +2 = (108/16)/3 + 2*(-9/4) +2 = (108/48) - 18/4 +2 = (9/4) - (9/2) +2 = (9/4 -18/4 +8/4)= (-1/4). So the expression equals -1/4 here. Wait, negative!But wait, x=-3√3/4 is within the ellipse? Let's check x²/3 + y²=1.x=-3√3/4, x²= (9*3)/16=27/16. x²/3=9/16. Then y²=1 -9/16=7/16. So y=±√7/4. Therefore, the points exist. So, if we take A=(-3√3/4, √7/4) and B=(-3√3/4, -√7/4), then PA ⋅ PB should be -1/4.Wait, let me compute PA ⋅ PB for these points.Coordinates of A: x=-3√3/4, y=√7/4.PA = ( -3√3/4 + √3, √7/4 ) = ( (√3/4), √7/4 )Similarly, PB = ( -3√3/4 + √3, -√7/4 ) = ( √3/4, -√7/4 )Dot product: (√3/4)(√3/4) + (√7/4)(-√7/4 ) = (3/16) - (7/16 )= -4/16 = -1/4.Yes! So PA ⋅ PB = -1/4. So we have a negative value. Therefore, the minimum is at least -1/4.But earlier, when we set the derivative to zero, we found critical points at θ=0, π, etc., but here we have another critical point where the dot product is negative. Therefore, our initial assumption that the minimum is 0 was incorrect.This suggests that we need a more thorough analysis. Let's return to the parametrization approach.We had PA ⋅ PB expressed in terms of θ and φ as:3 cos θ cos φ + 3 cos θ + 3 cos φ + 3 + sin θ sin φ.We need to find the extrema of this function over θ and φ.Alternatively, perhaps use Lagrange multipliers, considering the constraints x₁²/3 + y₁² =1 and x₂²/3 + y₂²=1. But with two variables, this would involve four variables (x₁, y₁, x₂, y₂) and two constraints, leading to a complicated system.Alternatively, since the problem is symmetric in A and B, maybe we can assume θ = φ or some relation between θ and φ.Wait, in the example where PA ⋅ PB reached -1/4, points A and B had the same x-coordinate but opposite y-coordinates. So maybe in general, the extremum occurs when φ = -θ or some relation. Let's suppose that φ = -θ. Let's set φ = -θ and see what the expression becomes.Set φ = -θ. Then:PA ⋅ PB = 3 cos θ cos(-θ) + 3 cos θ + 3 cos(-θ) + 3 + sin θ sin(-θ)Since cos(-θ)=cos θ and sin(-θ)=-sin θ:= 3 cos²θ + 3 cos θ + 3 cos θ + 3 - sin²θSimplify:= 3 cos²θ + 6 cos θ + 3 - sin²θBut sin²θ =1 - cos²θ, so substitute:= 3 cos²θ +6 cos θ +3 - (1 - cos²θ) =3 cos²θ +6 cos θ +3 -1 +cos²θ=4 cos²θ +6 cos θ +2.So PA ⋅ PB becomes 4 cos²θ +6 cos θ +2. Let’s denote this as f(θ) =4 cos²θ +6 cos θ +2.To find extrema, take derivative f’(θ)= -8 cos θ sin θ -6 sin θ.Set to zero: -8 cos θ sin θ -6 sin θ=0.Factor out -sin θ: -sin θ (8 cos θ +6)=0.Solutions: sin θ=0 or 8 cos θ +6=0.If sin θ=0, then θ=0 or π.If 8 cos θ +6=0 => cos θ= -6/8= -3/4.So θ= arccos(-3/4) or θ=2π - arccos(-3/4). Let's compute the values.For θ=0: f(0)=4*1 +6*1 +2=4+6+2=12.For θ=π: f(π)=4*1 +6*(-1) +2=4-6+2=0.For θ= arccos(-3/4): compute f(θ)=4*(9/16) +6*(-3/4) +2= (36/16) - (18/4) +2= (9/4) - (9/2) +2= (9/4 -18/4 +8/4)= (-1/4). So minimum of -1/4.Therefore, when φ=-θ= arccos(-3/4), we get PA ⋅ PB= -1/4. So this is the minimum value when θ= arccos(-3/4) and φ= -θ.Therefore, under the assumption that φ= -θ, we can attain a minimum of -1/4 and maximum of12. But we need to check if there are other extrema when θ and φ are not related by φ=-θ.Alternatively, maybe the global maximum and minimum occur under this assumption.But to confirm, let's consider another scenario. Suppose θ and φ are independent. Let's try θ=0, φ=π.PA ⋅ PB=3*1*(-1)+3*1+3*(-1)+3 +0= -3 +3 -3 +3=0. So zero.Another case, θ=π/2, φ=π/2. Points A and B at (0,1). PA=(√3,1), PB=(√3,1). Dot product=3+1=4.θ=π/2, φ=-π/2. PA=(√3,1), PB=(√3,-1). Dot product=3-1=2.Alternatively, θ= arccos(-3/4), φ=arccos(-3/4). Let's compute:PA ⋅ PB=3*(-3/4)^2 +3*(-3/4) +3*(-3/4) +3 + sin θ sin φ.Wait, this might be complex, but perhaps substituting θ=φ= arccos(-3/4):cos θ= -3/4, sin θ= sqrt(1 - (9/16))=sqrt(7/16)=√7/4.Then PA ⋅ PB=3*(-3/4)^2 +3*(-3/4) +3*(-3/4) +3 + (√7/4)^2.Compute:3*(9/16) + (-9/4) + (-9/4) +3 +7/16=27/16 -18/4 +3 +7/16Convert to sixteenths:27/16 -72/16 +48/16 +7/16= (27 -72 +48 +7)/16=(10)/16=5/8≈0.625. Which is positive.So not an extremum. Therefore, the minimum of -1/4 occurs when φ=-θ= arccos(-3/4), and the maximum of12 occurs when θ=φ=0.But is -1/4 the global minimum? To confirm, we can consider other possibilities. For example, take θ= arccos(-3/4), φ= some other angle.Alternatively, use the method of Lagrange multipliers for two variables.But this might be time-consuming. Alternatively, notice that when we set φ=-θ, we found a minimum of -1/4, and other configurations give higher values. Therefore, it's plausible that -1/4 is the minimum and12 is the maximum.But to ensure this, let's analyze the original function f(θ, φ)=3 cosθ cosφ +3 cosθ +3 cosφ +3 +sinθ sinφ.Let’s rewrite this function as:3 cosθ cosφ + sinθ sinφ +3 (cosθ + cosφ) +3.Notice that 3 cosθ cosφ + sinθ sinφ can be rewritten using trigonometric identities. Recall that cos(θ - φ) = cosθ cosφ + sinθ sinφ. Here, we have 3 cosθ cosφ + sinθ sinφ = 2 cosθ cosφ + (cosθ cosφ + sinθ sinφ) = 2 cosθ cosφ + cos(θ - φ). Therefore:f(θ, φ)=2 cosθ cosφ + cos(θ - φ) +3 (cosθ + cosφ) +3.Not sure if this helps. Alternatively, consider expressing the function in terms of sum and difference angles.Let’s set α=θ + φ and β=θ - φ. Then θ=(α + β)/2, φ=(α - β)/2.But this substitution might complicate the expression further.Alternatively, use the method of considering f(θ, φ) as a quadratic form.Alternatively, fix θ and optimize over φ, then optimize over θ.Let’s fix θ and treat f as a function of φ:f(φ)=3 cosθ cosφ + sinθ sinφ +3 cosθ +3 cosφ +3.Group terms with cosφ and sinφ:= [3 cosθ +3] cosφ + [sinθ] sinφ +3 cosθ +3.Let’s write this as A cosφ + B sinφ + C, where:A=3 cosθ +3,B= sinθ,C=3 cosθ +3.Then, f(φ)=A cosφ +B sinφ +C.The maximum and minimum of A cosφ +B sinφ are ±√(A² +B²). Therefore, for fixed θ, the range of f(φ) is [C -√(A² +B²), C +√(A² +B²)].Therefore, the maximum of f(θ,φ) over φ is C +√(A² +B²), and the minimum is C -√(A² +B²).Thus, to find the global maximum and minimum, we need to compute these expressions for all θ.First, compute A=3 cosθ +3, B= sinθ, C=3 cosθ +3.Compute √(A² +B²):√[(3 cosθ +3)^2 + sin²θ]= √[9 cos²θ +18 cosθ +9 + sin²θ]= √[9 cos²θ + sin²θ +18 cosθ +9]But since sin²θ=1 - cos²θ:= √[9 cos²θ +1 - cos²θ +18 cosθ +9]= √[8 cos²θ +18 cosθ +10]So the maximum value for fixed θ is C +√(8 cos²θ +18 cosθ +10), and the minimum is C -√(8 cos²θ +18 cosθ +10).But C=3 cosθ +3. Therefore:Maximum over φ for fixed θ: 3 cosθ +3 +√(8 cos²θ +18 cosθ +10)Minimum over φ for fixed θ:3 cosθ +3 -√(8 cos²θ +18 cosθ +10)We need to find the global maximum and minimum of these two expressions over θ.This seems quite involved, but let's attempt to analyze it.First, let's find the global maximum of the maximum expression:F_max(θ)=3 cosθ +3 +√(8 cos²θ +18 cosθ +10)Similarly, the global minimum of the minimum expression:F_min(θ)=3 cosθ +3 -√(8 cos²θ +18 cosθ +10)We need to find the maximum of F_max(θ) and the minimum of F_min(θ).Let’s first tackle F_max(θ).Let’s denote t=cosθ, where t ∈ [-1,1].Then F_max(t)=3t +3 +√(8t² +18t +10)We need to find the maximum of F_max(t) over t ∈ [-1,1].Similarly, F_min(t)=3t +3 -√(8t² +18t +10), find its minimum.This is a one-variable calculus problem. Let's compute the derivative of F_max(t) with respect to t:d/dt [F_max(t)] =3 + [ (16t +18) / (2√(8t² +18t +10)) ) ]Set derivative to zero:3 + (8t +9)/√(8t² +18t +10) =0Multiply both sides by √(8t² +18t +10):3√(8t² +18t +10) +8t +9=0Square both sides to eliminate square root:9(8t² +18t +10) + (8t +9)^2 + 2*3√(8t² +18t +10)*(8t +9) =0Wait, this is getting complicated. Alternatively, move terms:3√(8t² +18t +10) = - (8t +9)Since the left side is a square root, it's non-negative, while the right side is - (8t +9). Therefore, - (8t +9) ≥0 =>8t +9 ≤0 =>t ≤ -9/8. But t ∈ [-1,1], so t ≤-9/8≈-1.125, but the minimum t is -1. Therefore, this equation can only hold if t ≤-9/8, which is outside the domain. Therefore, no solution. Therefore, the derivative does not vanish in the domain t ∈ [-1,1]. Therefore, F_max(t) is either increasing or decreasing throughout the interval. Let's check the derivative at t=1:d/dt [F_max(1)] =3 + (8*1 +9)/√(8*1 +18*1 +10)=3 +17/√36=3 +17/6≈3+2.83≈5.83>0At t=-1:d/dt [F_max(-1)]=3 + (8*(-1) +9)/√(8*1 +18*(-1) +10)=3 + (1)/√(8 -18 +10)=3 +1/0 undefined. But as t approaches -1 from the right, denominator approaches √(8 -18 +10)=√0=0, and numerator approaches1. Therefore, derivative approaches 3 + (1)/0^+ = +∞. Therefore, derivative is always positive in t ∈ [-1,1]. Therefore, F_max(t) is increasing on [-1,1]. Therefore, maximum at t=1: F_max(1)=3*1 +3 +√(8 +18 +10)=6 +√36=6+6=12.Similarly, minimum at t=-1: F_max(-1)=3*(-1)+3 +√(8*1 +18*(-1)+10)= -3 +3 +√(8 -18 +10)=0 +√0=0.But wait, at t=-1, the expression under the sqrt is 8(-1)^2 +18*(-1) +10=8 -18 +10=0. So F_max(-1)=3*(-1)+3 +0=0. So the maximum of F_max(t) is12 at t=1, and the minimum is0 at t=-1.Now consider F_min(t)=3t +3 -√(8t² +18t +10). We need to find its minimum over t ∈[-1,1].Compute derivative:d/dt [F_min(t)]=3 - [ (16t +18)/(2√(8t² +18t +10)) ) ]=3 - (8t +9)/√(8t² +18t +10)Set derivative to zero:3 - (8t +9)/√(8t² +18t +10)=0 => (8t +9)/√(8t² +18t +10)=3Square both sides:(8t +9)^2=9(8t² +18t +10)Expand left side:64t² +144t +81Right side:72t² +162t +90Bring all terms to left:64t² +144t +81 -72t² -162t -90= -8t² -18t -9=0Multiply both sides by -1:8t² +18t +9=0Solve: t=(-18 ±√(324 -288))/16=(-18 ±√36)/16=(-18 ±6)/16Thus, t=(-18 +6)/16=(-12)/16=-3/4≈-0.75 or t=(-18 -6)/16=(-24)/16=-3/2≈-1.5.But t ∈ [-1,1]. So t=-3/4 is valid, t=-3/2 invalid.Therefore, critical point at t=-3/4.Now, check value of F_min at t=-3/4 and endpoints t=-1 and t=1.First, compute F_min(-3/4)=3*(-3/4)+3 -√(8*(9/16)+18*(-3/4)+10)= -9/4 +3 -√( (72/16) -54/4 +10 )Convert to quarters:-9/4 +12/4=3/4.Inside sqrt:72/16=9/2, 54/4=27/2, 10=40/4.So:9/2 -27/2 +40/4= (-18/2) +10= -9 +10=1.Thus sqrt(1)=1.Therefore, F_min(-3/4)=3/4 -1= -1/4.At t=1:F_min(1)=3*1 +3 -√(8 +18 +10)=6 -√36=6-6=0.At t=-1:F_min(-1)=3*(-1)+3 -√(8*1 +18*(-1)+10)= -3+3 -√0=0.Therefore, the minimum of F_min(t) is -1/4 at t=-3/4, and maximum is0 at t=-1 and t=1.Therefore, combining the results:The global maximum of PA ⋅ PB is12, achieved when θ=φ=0 (points A and B at (√3,0)).The global minimum of PA ⋅ PB is-1/4, achieved when t=cosθ=-3/4, which corresponds to θ=arccos(-3/4), and φ=-θ, as we saw earlier.Therefore, the range of PA ⋅ PB is[-1/4,12].But wait, the question asks for the range. Should we write it as [-¼,12]? Yes.But in the problem statement, coordinates are given in fractions, so probably express -1/4 as -0.25, but in the boxed answer, we should use fractions.But let me verify the calculation for the minimum one more time.When t=cosθ=-3/4, θ=arccos(-3/4), φ=-θ.PA ⋅ PB=3 cosθ cosφ +3 cosθ +3 cosφ +3 +sinθ sinφ.Since φ=-θ, this becomes 3 cosθ cosθ +3 cosθ +3 cos(-θ) +3 +sinθ sin(-θ)=3 cos²θ +3 cosθ +3 cosθ +3 -sin²θ=3 cos²θ +6 cosθ +3 -sin²θ.But sin²θ=1 -cos²θ, so substitute:=3 cos²θ +6 cosθ +3 -1 +cos²θ=4 cos²θ +6 cosθ +2.With cosθ=-3/4:4*(9/16) +6*(-3/4) +2= 36/16 -18/4 +2=9/4 -9/2 +2= (9/4 -18/4 +8/4)=(-1/4). Correct.Therefore, the minimum is indeed -1/4.Therefore, the range is from -1/4 to12.Thus, the final answer is boxed{left[ -dfrac{1}{4}, 12 right]}.Final AnswerThe range of overrightarrow{PA} cdot overrightarrow{PB} is boxed{left[ -dfrac{1}{4}, 12 right]}.</think>The ellipse (Gamma: frac{x^2}{3} + y^2 = 1) has its left vertex at (P(-sqrt{3}, 0)). Points (A) and (B) are moving points on (Gamma). We need to find the range of the dot product (overrightarrow{PA} cdot overrightarrow{PB}).1. Parametrization of Points (A) and (B): - Points (A) and (B) can be parametrized as (A(sqrt{3} cos theta, sin theta)) and (B(sqrt{3} cos phi, sin phi)).2. Expression for (overrightarrow{PA} cdot overrightarrow{PB}): - Vectors (overrightarrow{PA}) and (overrightarrow{PB}) are ((sqrt{3} cos theta + sqrt{3}, sin theta)) and ((sqrt{3} cos phi + sqrt{3}, sin phi)). - The dot product is: [ (sqrt{3} cos theta + sqrt{3})(sqrt{3} cos phi + sqrt{3}) + sin theta sin phi ] - Expanding and simplifying: [ 3 cos theta cos phi + 3 cos theta + 3 cos phi + 3 + sin theta sin phi ]3. Using Trigonometric Identities and Critical Points: - The expression is analyzed using trigonometric identities and critical points are found by setting partial derivatives to zero. - Critical points are found when (theta = phi = 0) (maximum value) and when (theta = arccos(-frac{3}{4})) and (phi = -theta) (minimum value).4. Calculating Extremes: - Maximum Value: When (A) and (B) are at the right vertex ((sqrt{3}, 0)), the dot product is (12). - Minimum Value: When (theta = arccos(-frac{3}{4})) and (phi = -theta), the dot product is (-frac{1}{4}).5. Verification: - The calculations are verified using substitution and checking the range of the function over the interval ([-1, 1]) for (cos theta) and (cos phi).The range of (overrightarrow{PA} cdot overrightarrow{PB}) is (boxed{left[ -dfrac{1}{4}, 12 right]}).]

🔑:Let A(sqrt{3} cos alpha, sin alpha), B(sqrt{3} cos beta, sin beta), P(-sqrt{3}, 0),then overrightarrow{P A} cdot overrightarrow{P B}=(sqrt{3} cos alpha+sqrt{3})(sqrt{3} cos beta+sqrt{3})+sin alpha sin betabegin{array}{l}=3 cos alpha cos beta+3(cos alpha+cos beta)+sin alpha sin beta+3 =cos (alpha+beta)+2 cos (alpha-beta)+6 cos frac{alpha+beta}{2} cos frac{alpha-beta}{2}+3 =2 cos ^{2} frac{alpha+beta}{2}+4 cos ^{2} frac{alpha-beta}{2}+6 cos frac{alpha+beta}{2} cos frac{alpha-beta}{2},end{array}Let u=cos frac{alpha+beta}{2}, v=cos frac{alpha-beta}{2}, then u, v in[-1,1],overrightarrow{P A} cdot overrightarrow{P B}=2 u^{2}+6 u v+4 v^{2}=4left(v+frac{3}{4} uright)^{2}-frac{1}{4} u^{2} inleft[-frac{1}{4}, 12right].When overrightarrow{P A} cdot overrightarrow{P B}=-frac{1}{4}, Aleft(-frac{3 sqrt{3}}{4}, frac{sqrt{7}}{4}right), Bleft(-frac{3 sqrt{3}}{4},-frac{sqrt{7}}{4}right);When overrightarrow{P A} cdot overrightarrow{P B}=12, A(sqrt{3}, 0), B(sqrt{3}, 0). Therefore, the range of overrightarrow{P A} cdot overrightarrow{P B} isleft[-frac{1}{4}, 12right] text {. }

🧠:[<think>Okay, so I need to find all real numbers k such that the inequality a³ + b³ + c³ + d³ + 1 is greater than or equal to k times (a + b + c + d) for all a, b, c, d in the interval [-1, +∞). Hmm, this seems like an optimization problem where I need to determine the maximum possible value of k such that the inequality holds for all a, b, c, d in that range. First, maybe I should think about the structure of the inequality. The left-hand side (LHS) is a sum of cubes plus 1, and the right-hand side (RHS) is a linear combination of the variables multiplied by k. To find the maximum k for which this inequality holds, perhaps I can consider each variable separately? Or maybe use some inequality like AM-GM or Cauchy-Schwarz? Wait, but cubes might complicate things. Alternatively, since the variables are independent, maybe I can consider the inequality for each variable individually and then combine the results? Let me think.If the inequality has to hold for all a, b, c, d in [-1, ∞), then even if we set variables to their extreme values (like -1 or very large numbers), the inequality should still hold. So perhaps to find the minimal k, we need to check the cases where the LHS is minimized and the RHS is maximized. Wait, but actually, we need to ensure that LHS - k*(a + b + c + d) is always non-negative. So maybe the minimal value of LHS - k*(a + b + c + d) is non-negative. Therefore, we need to find k such that the minimum of a³ - k*a is at least -1/4? Wait, no. Wait, since there are four variables, each contributing a³ and a term with k, and the +1. Maybe if we can handle each variable separately. Let's suppose that each term a³ - k*a is non-negative, but that might not necessarily be the case. Because if we set three variables to -1, and one variable to something else, perhaps we can get a lower bound.Alternatively, maybe we can use the method of Lagrange multipliers to find the critical points where the function f(a,b,c,d) = a³ + b³ + c³ + d³ +1 - k(a + b + c + d) is minimized. Then, set the derivative with respect to each variable to zero. Let's see:The derivative of f with respect to a is 3a² - k. Setting this equal to zero gives 3a² - k = 0, so a = sqrt(k/3) or -sqrt(k/3). But the variables are in [-1, ∞). So critical points occur at these roots if they lie within [-1, ∞). Also, we need to check the boundaries, i.e., a = -1 and a approaching infinity.Wait, but if we use Lagrange multipliers here, we need to consider each variable's contribution. Since all variables are symmetric, the minimum would occur either when all variables are equal, or some at the boundary points. Maybe the minimal value occurs when some variables are set to -1 and others are set to some value where the derivative is zero. So perhaps we can assume some variables are equal and others are set to -1. For example, suppose three variables are set to -1, and the fourth variable is x. Then the function becomes:3*(-1)^3 + x³ +1 - k*(3*(-1) + x) = -3 + x³ +1 -k*(-3 + x) = x³ -2 + 3k - kx.We need this expression to be non-negative for x >= -1. So, we need x³ -kx + (3k -2) >=0 for all x >= -1. Similarly, if two variables are set to -1, and two variables are x, then the expression would be 2*(-1)^3 + 2x³ +1 -k*(2*(-1) + 2x) = -2 + 2x³ +1 -k*(-2 + 2x) = 2x³ -1 +2k -2kx. We need this to be non-negative for x >= -1. Hmm, but this seems complicated. Maybe the minimal case is when some variables are at -1 and others are at their critical points.Alternatively, maybe the minimal value occurs when all variables are equal? Let's suppose all variables are equal to x. Then the inequality becomes 4x³ +1 >= 4k x. Then, 4x³ -4k x +1 >=0. We need this inequality to hold for all x >= -1. But this might not capture the cases where variables are different.Alternatively, perhaps considering individual variables. Let's fix three variables and vary the fourth. Suppose a, b, c are fixed at some values, then d must satisfy d³ +1 -k d >= -(a³ + b³ + c³ -k(a + b + c)). Hmm, not sure.Alternatively, think of the inequality as the sum over each variable (a³ -k a) + (b³ -k b) + (c³ -k c) + (d³ -k d) +1 >=0. So the sum of (a³ -k a) for each variable plus 1 is non-negative. So if we can ensure that each term (a³ -k a) is bounded below by some value, then the total sum would be bounded below by 4*minimum(a³ -k a) +1 >=0. Wait, but that might not be the case because the minimum could be achieved at different points for different variables. However, if we can find the minimal value of (a³ -k a) for a >= -1, then since there are four variables, the total would be 4*min +1 >=0. So maybe this approach can work.Let’s try that. Let's find the minimal value of f(a) = a³ -k a for a in [-1, ∞). Then, the minimal value for each variable would be min_{a >= -1} (a³ -k a). Then, since we have four variables, the total sum would be 4*min +1 >=0. Therefore, the condition is 4*min +1 >=0. So, to find the maximum k such that 4*min +1 >=0.Therefore, first, we need to find min_{a >= -1} (a³ -k a). Let's compute this minimum. Let's find critical points by taking derivative of f(a):f’(a) = 3a² -k. Setting derivative to zero gives a = sqrt(k/3) or -sqrt(k/3). But since a >= -1, we need to check if -sqrt(k/3) >= -1. That is, sqrt(k/3) <=1, which implies k <=3. So, if k <=3, then the critical points are a = sqrt(k/3) and a = -sqrt(k/3). However, if k >3, then -sqrt(k/3) < -1, which is outside the domain. So in that case, the critical point a = -sqrt(k/3) is not in the domain, so the minimum would be at a = -1 or a = sqrt(k/3).Wait, hold on. Let's clarify:For f(a) = a³ -k a, on the interval [-1, ∞):Case 1: k <=3.Then sqrt(k/3) <=1, so the critical points are a = sqrt(k/3) and a = -sqrt(k/3). But since a >= -1, both critical points are within the domain if -sqrt(k/3) >= -1. Since sqrt(k/3) <=1, then -sqrt(k/3) >= -1. Therefore, for k <=3, the critical points are within the domain. Then, we need to evaluate f(a) at a = sqrt(k/3), a = -sqrt(k/3), and the endpoint a = -1.Wait, but a >= -1, so a = -sqrt(k/3) is in the domain only if -sqrt(k/3) >= -1, which is equivalent to sqrt(k/3) <=1, which is k <=3. So for k <=3, both critical points are in the domain. Then, we can compute the minimal value among these points.Wait, but for a >= -1, the function f(a) = a³ -k a. Let's compute f(-1) = (-1)^3 -k*(-1) = -1 +k.At a = sqrt(k/3), f(a) = (sqrt(k/3))^3 -k*sqrt(k/3) = (k^(3/2))/(3^(3/2)) -k^(3/2)/sqrt(3) = (k^(3/2)/3^(3/2)) - (k^(3/2)*3^(1/2))/3) = (k^(3/2)/3^(3/2)) - (k^(3/2)/3^(1/2)) = (1/3^(3/2) -1/3^(1/2))k^(3/2) = (-2/(3^(3/2)))k^(3/2).Similarly, at a = -sqrt(k/3), f(a) = (-sqrt(k/3))^3 -k*(-sqrt(k/3)) = - (k^(3/2))/(3^(3/2)) + k^(3/2)/sqrt(3) = (-1/3^(3/2) +1/3^(1/2))k^(3/2) = (2/3^(3/2))k^(3/2).Therefore, for k <=3, the critical points are at sqrt(k/3) and -sqrt(k/3). The function f(a) at a = sqrt(k/3) is negative, and at a = -sqrt(k/3) is positive. Therefore, the minimal value is at a = sqrt(k/3): f(a) = -2/(3^(3/2))k^(3/2). But also, we need to compare with the value at a = -1. Since when k <=3, the critical point a = -sqrt(k/3) is greater than or equal to -1, so the minimal value could be either the minimum at sqrt(k/3) or at a approaching infinity. Wait, but as a approaches infinity, f(a) = a³ -k a tends to infinity. So the minimal value is at a = sqrt(k/3) or at a = -1.Compare f(-1) = -1 +k and f(sqrt(k/3)) = -2/(3^(3/2))k^(3/2). Which is smaller?Let’s compute for k=3: f(-1) = -1 +3=2; f(sqrt(1))=f(1)=1 -3*1= -2. So for k=3, the minimal value is -2. But f(-1)=2, so the minimal value is at a=1 (sqrt(k/3)).Wait, but for k=3, sqrt(k/3)=1, so the critical point is at a=1, and f(1)=1 -3*1= -2. So for k=3, the minimum is -2. For k less than 3, say k=0, then f(a)=a³. The minimum is at a=-1, f(-1)=-1. So when k=0, the minimum is -1, which is less than f(sqrt(k/3))=0.So perhaps there is a threshold value of k where the minimal value transitions from being at a=-1 to a=sqrt(k/3). Let’s find when f(-1) = f(sqrt(k/3)):Set -1 +k = -2/(3^(3/2))k^(3/2). Let me compute 3^(3/2) = 3*sqrt(3) ≈5.196.So the equation is k -1 = (-2/(3*sqrt(3)))k^(3/2).But since the RHS is negative (because of the negative sign) and the LHS is k -1, for k <1, LHS is negative. Let's square both sides? Maybe not straightforward. Alternatively, let's set t = sqrt(k). Then k = t², and equation becomes t² -1 = (-2/(3*sqrt(3)))t³.Multiply both sides by 3*sqrt(3):3*sqrt(3)(t² -1) = -2t³.Bring all terms to left:2t³ +3*sqrt(3)t² -3*sqrt(3)=0.This is a cubic equation in t. Hmm, solving this might be complicated. Alternatively, perhaps there's a better approach. Let's test k=1:f(-1) = -1 +1=0; f(sqrt(1/3)) = -2/(3*sqrt(3))*(1)^(3/2)= -2/(3*sqrt(3))≈-0.385. So f(-1)=0 is greater than f(sqrt(1/3))≈-0.385. Therefore, the minimum is at sqrt(k/3). For k=1, the minimal value is -0.385. For k=0, the minimal value is -1. So maybe for all k in [0,3], the minimal value is either at a=-1 or at a=sqrt(k/3). Wait, but when k increases from 0 to 3, when does the minimal point switch from a=-1 to a=sqrt(k/3)?Wait, actually, the derivative at a=-1 is f’(-1) = 3*(-1)^2 -k = 3 -k. If 3 -k >0, that means the function is increasing at a=-1. So if the derivative is positive at a=-1, then the function is increasing to the right of a=-1, meaning that the minimal value is at a=-1. If the derivative is negative, then the function is decreasing at a=-1, meaning that the minimal value is somewhere to the right of a=-1.So, f’(-1)=3 -k. If 3 -k >0, i.e., k <3, then at a=-1, the function is increasing. Therefore, the minimum would be at a=-1. Wait, but that contradicts previous example for k=1 where the minimal value was at sqrt(1/3). Wait, maybe not. Let's think again.If the derivative at a=-1 is positive (i.e., function increasing at a=-1), that would mean that to the right of a=-1, the function increases. However, the critical point at a = sqrt(k/3) is a minimum or maximum?Wait, the second derivative is f''(a)=6a. At a=sqrt(k/3), the second derivative is 6*sqrt(k/3)=2*sqrt(3k). Which is positive, so that critical point is a local minimum. At a=-sqrt(k/3), the second derivative is -2*sqrt(3k), which is negative, so that's a local maximum. However, for k <=3, a=-sqrt(k/3) is >= -1. So in the interval [-1, ∞), the function f(a) has a local maximum at a=-sqrt(k/3) and a local minimum at a=sqrt(k/3). But since the function tends to infinity as a approaches infinity, the global minimum is at a=sqrt(k/3) if f(sqrt(k/3)) < f(-1). Otherwise, if f(-1) is less, then the minimum is at a=-1.But when is f(sqrt(k/3)) < f(-1)? That is, when -2/(3*sqrt(3))k^(3/2) < k -1. Let's check for k=1: -2/(3*sqrt(3)) ≈-0.385 < 0, which is true. So for k=1, the minimum is at sqrt(k/3). For k=0: f(sqrt(0))=0, f(-1)=-1. So 0 > -1, so the minimum is at a=-1. Wait, but when k=0, the function is f(a)=a³. Its minimum on [-1, ∞) is at a=-1, which is -1. But according to the previous formula, f(sqrt(0))=0, which is higher. So when k=0, the minimum is at a=-1.Therefore, perhaps the transition occurs when f(sqrt(k/3)) = f(-1). Let's solve the equation:-2/(3*sqrt(3))k^(3/2) = k -1.Multiply both sides by -3*sqrt(3)/2:k^(3/2) = (-3*sqrt(3)/2)(k -1).But the LHS is non-negative (since k is real and exponent 3/2 requires k >=0). The RHS is (-3*sqrt(3)/2)(k -1). For the RHS to be non-negative, (k -1) must be <=0, so k <=1. Therefore, possible solutions when k <=1.So set k <=1, and the equation is:k^(3/2) = (3*sqrt(3)/2)(1 -k).Let’s set t = sqrt(k), so k = t², then the equation becomes:t³ = (3*sqrt(3)/2)(1 - t²).Multiply both sides by 2:2t³ = 3*sqrt(3)(1 - t²).Bring all terms to left:2t³ +3*sqrt(3)t² -3*sqrt(3)=0.This is the same cubic equation as before. Let’s see if t= sqrt(3)/something is a solution. Let me try t= sqrt(3)/2 ≈0.866. Then:2*( (sqrt(3)/2)^3 ) +3*sqrt(3)*( (sqrt(3)/2)^2 ) -3*sqrt(3)First term: 2*( (3*sqrt(3))/8 ) = (6*sqrt(3))/8 = (3*sqrt(3))/4 ≈1.299Second term: 3*sqrt(3)*(3/4) = (9*sqrt(3))/4 ≈3.897Third term: -3*sqrt(3)≈-5.196Sum≈1.299 +3.897 -5.196≈0. So t= sqrt(3)/2 is a solution.Therefore, t= sqrt(3)/2, so k= t²= 3/4=0.75. Therefore, when k=3/4, the equation holds. Therefore, for k=3/4, the minimal value is achieved both at a=-1 and a=sqrt(k/3). Let's verify:For k=3/4, compute f(-1)= -1 +3/4= -1/4≈-0.25.Compute f(sqrt(k/3))= sqrt( (3/4)/3 )= sqrt(1/4)=1/2.Wait, wait, sqrt(k/3) when k=3/4 is sqrt( (3/4)/3 )= sqrt(1/4)=1/2.Then f(1/2)= (1/2)^3 - (3/4)(1/2)=1/8 - 3/8= -2/8= -1/4. So f(-1)= -1/4 and f(1/2)= -1/4. So both points give the same minimal value. Therefore, when k=3/4, the minimal value is -1/4, achieved at both a=-1 and a=1/2.Therefore, for k <=3/4, the minimal value is at a=-1, since when k <3/4, solving f(-1)=k -1 vs f(sqrt(k/3))=-2/(3*sqrt(3))k^(3/2). Let's check k=0.5:f(-1)=0.5 -1= -0.5.f(sqrt(0.5/3))=sqrt(0.5/3)=sqrt(1/6)≈0.408.Then f(0.408)= (0.408)^3 -0.5*0.408≈0.068 -0.204≈-0.136. Which is greater than -0.5. So for k=0.5, the minimal value is at a=-1, which is -0.5.Similarly, for k=1, f(-1)=0, f(sqrt(1/3))≈-0.385. So the minimal value is at sqrt(1/3). So the transition occurs at k=3/4. So for k <=3/4, the minimal value of f(a)=a³ -k a is at a=-1, which is k -1. For k >=3/4, the minimal value is at a=sqrt(k/3), which is -2/(3*sqrt(3))k^(3/2).Therefore, summarizing:min_{a >=-1} (a³ -k a) = { k -1, if k <=3/4 ; -2/(3*sqrt(3))k^(3/2), if k >=3/4 }But wait, when k=3/4, both expressions give -1/4, so it's continuous.Therefore, returning to the original problem: the sum over four variables of (a³ -k a) +1 >=0. So the minimal total sum would be 4*min +1 >=0. Therefore, if each variable's minimal contribution is m, then 4m +1 >=0.So, if k <=3/4, then each min is k -1, so 4(k -1) +1 >=0 =>4k -4 +1 >=0 =>4k -3 >=0 =>k >=3/4. Wait, but this is a contradiction because k <=3/4. Wait, this suggests that when k <=3/4, the condition 4(k -1) +1 >=0 must hold. However, solving 4k -3 >=0 gives k >=3/4. But we assumed k <=3/4. Therefore, the only possible k where both conditions hold is k=3/4. However, for k <3/4, 4(k -1)+1=4k -3 <0, which violates the inequality. Therefore, the minimal k is 3/4? Wait, but this is conflicting.Wait, no. Wait, if we have four variables, each contributing at least (k -1) when k <=3/4, then the total sum would be at least 4(k -1) +1. For this to be >=0, we need 4(k -1) +1 >=0 =>4k -4 +1 >=0 =>4k >=3 =>k >=3/4. But if we are considering k <=3/4, then this can only happen when k=3/4. Therefore, the minimal k where the inequality holds is 3/4. However, when k >3/4, the minimal contribution per variable is -2/(3*sqrt(3))k^(3/2). Then, the total sum would be 4*(-2/(3*sqrt(3))k^(3/2)) +1 >=0. Therefore, 1 -8/(3*sqrt(3))k^(3/2) >=0. Solving for k:8/(3*sqrt(3))k^(3/2) <=1 =>k^(3/2) <= (3*sqrt(3))/8 =>k <= ( (3*sqrt(3))/8 )^(2/3 ).Compute that:First, 3*sqrt(3) =3*1.732≈5.196. So 5.196/8≈0.6495. Then, (0.6495)^(2/3). Let's compute:Take natural log: ln(0.6495)≈-0.431, multiply by 2/3:≈-0.287, then exponentiate: e^(-0.287)≈0.75. So approximately 0.75. Wait, but 3/4=0.75. Hmm, interesting. Let me compute exactly:( (3*sqrt(3))/8 )^(2/3 ) = (3^(3/2)/8 )^(2/3 )= (3^(3/2))^(2/3)/8^(2/3)=3^(1)/8^(2/3)=3/(8^(2/3)).But 8^(2/3)= (8^(1/3))^2=2^2=4. Therefore, 3/4. So, exactly k <=3/4. Therefore, when k >=3/4, the condition 1 -8/(3*sqrt(3))k^(3/2) >=0 is equivalent to k <=3/4. Therefore, only k=3/4 satisfies this. Hence, regardless of whether k <=3/4 or k >=3/4, the only value that satisfies the inequality is k=3/4. Wait, that seems contradictory. Let me check again.Wait, when k >=3/4, the minimal total sum is 4*(-2/(3*sqrt(3))k^(3/2)) +1. We need this to be >=0. Therefore:1 - (8/(3*sqrt(3)))k^(3/2) >=0 => (8/(3*sqrt(3)))k^(3/2) <=1 =>k^(3/2) <= (3*sqrt(3))/8 =>k <= ( (3*sqrt(3))/8 )^(2/3 )=3/4. So even though we considered k >=3/4, the inequality only holds when k <=3/4. Therefore, combining both cases, the only k that satisfies the inequality is k=3/4. But wait, this seems confusing. Let me recapitulate:Case 1: k <=3/4. Then each variable contributes at least (k -1). Therefore, total sum >=4(k -1) +1=4k -3. We need 4k -3 >=0 =>k >=3/4. But since we are in the case k <=3/4, the only possibility is k=3/4.Case 2: k >=3/4. Then each variable contributes at least -2/(3*sqrt(3))k^(3/2). Therefore, total sum >=4*(-2/(3*sqrt(3))k^(3/2)) +1. We need this >=0, which reduces to k <=3/4. Therefore, again only k=3/4.Therefore, the only k that satisfies the inequality is k=3/4. Hence, the maximum possible k is 3/4. Therefore, the answer is boxed{dfrac{3}{4}}.But wait, let me test this with specific examples. For example, if a=b=c=d=-1, then LHS=4*(-1)^3 +1= -4 +1=-3. RHS=k*(4*(-1))= -4k. So the inequality becomes -3 >=-4k =>4k >=3 =>k >=3/4. So in this case, k must be at least 3/4. But we are supposed to find all k such that the inequality holds for all a,b,c,d. Therefore, k must be <=3/4 and >=3/4? Wait, but no. Wait, the problem says "Find all real numbers k such that a³ +b³ +c³ +d³ +1 >=k(a +b +c +d) for all a,b,c,d in [-1, +infty)".So when a=b=c=d=-1, the inequality becomes -3 >= -4k. Which implies k >= 3/4. But if we set a,b,c,d to be large positive numbers, then the LHS ~a³ and RHS ~k a. So for large a, a³ dominates k a, so the inequality holds for any k. However, when variables are set to -1, we get a lower bound on k. Therefore, k must be at least 3/4 to satisfy the case when all variables are -1. But earlier analysis suggested that k must be exactly 3/4. Wait, but there's a conflict here. Because when variables are set to -1, k must be >=3/4, but when variables are set to their critical points, we also require k <=3/4. Therefore, the only possible k is 3/4.But let's check another case. Suppose three variables are -1, and one variable is x. Then the inequality becomes:3*(-1)^3 +x³ +1 >=k*(3*(-1) +x).Which simplifies to:-3 +x³ +1 >= -3k +k x.Which is:x³ -kx + (-2 +3k) >=0.We need this to hold for all x >=-1.Take x= -1:(-1)^3 -k*(-1) + (-2 +3k) = -1 +k -2 +3k= -3 +4k >=0 =>4k >=3 =>k>=3/4.Take x approaching infinity: x³ dominates, so the LHS tends to infinity, which is okay.Take x= sqrt(k/3). Let’s substitute x= sqrt(k/3). Then:x³ -k x + (-2 +3k)= (k^(3/2))/(3^(3/2)) -k*(sqrt(k)/sqrt(3)) + (-2 +3k)= (k^(3/2)/3^(3/2) -k^(3/2)/sqrt(3)) + (-2 +3k)= (-2k^(3/2)/3^(3/2)) + (-2 +3k).Set this equal to zero for minimal case:-2k^(3/2)/3^(3/2) -2 +3k=0.Multiply both sides by 3^(3/2):-2k^(3/2) -2*3^(3/2) +9k*sqrt(3)=0.This seems complicated, but if we assume k=3/4:Left-hand side:-2*( (3/4)^(3/2) ) -2*3^(3/2) +9*(3/4)*sqrt(3).Compute each term:First term: -2*( (3*sqrt(3))/8 )= - (3*sqrt(3))/4 ≈-1.299Second term: -2*(5.196)≈-10.392Third term:9*(3/4)*1.732≈6.75*1.732≈11.691Total≈-1.299 -10.392 +11.691≈0. So yes, when k=3/4, the expression is zero. Therefore, at x=sqrt(k/3)=sqrt( (3/4)/3 )=sqrt(1/4)=1/2, substituting back:(1/2)^3 - (3/4)(1/2) + (-2 +3*(3/4))=1/8 -3/8 + (-2 +9/4)= (-2/8) + (-8/4 +9/4)= (-1/4) + (1/4)=0. So equality holds at x=1/2 and three variables at -1. Similarly, when all variables are -1, equality holds when k=3/4. Therefore, the minimal k is 3/4 and the maximum k is 3/4. Hence, the only possible k is 3/4. Therefore, the answer is boxed{dfrac{3}{4}}.Final AnswerThe maximum real number ( k ) is (boxed{dfrac{3}{4}}).</think>To find all real numbers ( k ) such that the inequality ( a^3 + b^3 + c^3 + d^3 + 1 geqslant k(a + b + c + d) ) holds for all ( a, b, c, d in [-1, +infty) ), we need to consider the function ( f(a, b, c, d) = a^3 + b^3 + c^3 + d^3 + 1 - k(a + b + c + d) ) and ensure it is non-negative for all ( a, b, c, d ) in the given interval.1. Critical Points Analysis: - For each variable ( a ), we analyze the function ( f(a) = a^3 - ka ). - The critical points are found by setting the derivative ( f'(a) = 3a^2 - k ) to zero, giving ( a = pm sqrt{frac{k}{3}} ). - For ( k leq 3 ), the critical points ( pm sqrt{frac{k}{3}} ) are within the interval ([-1, +infty)).2. Minimum Value Calculation: - For ( a geq -1 ), the minimum value of ( f(a) = a^3 - ka ) is evaluated at critical points and endpoints. - When ( k leq frac{3}{4} ), the minimum value at ( a = -1 ) is ( k - 1 ). - When ( k geq frac{3}{4} ), the minimum value at ( a = sqrt{frac{k}{3}} ) is ( -frac{2}{3sqrt{3}} k^{3/2} ).3. Sum of Minimum Values: - For four variables, the sum of the minimum values plus 1 must be non-negative. - When ( k leq frac{3}{4} ), the total sum is ( 4(k - 1) + 1 ), leading to ( 4k - 3 geq 0 ), hence ( k geq frac{3}{4} ). - When ( k geq frac{3}{4} ), the total sum is ( 1 - frac{8}{3sqrt{3}} k^{3/2} ), leading to ( k leq frac{3}{4} ).4. Verification: - Testing specific cases (e.g., all variables at -1 or critical points) confirms that ( k = frac{3}{4} ) is the only value satisfying the inequality for all ( a, b, c, d in [-1, +infty) ).Thus, the maximum real number ( k ) such that the inequality holds is (boxed{dfrac{3}{4}}).,<think>Alright, so I need to find all real numbers k such that the inequality a³ + b³ + c³ + d³ + 1 is greater than or equal to k times the sum of a, b, c, and d, for all a, b, c, d in the interval [-1, +∞). Hmm, okay. Let me start by understanding what the problem is asking. First, the inequality must hold for all real numbers a, b, c, d where each of them is greater than or equal to -1. So, regardless of what values a, b, c, d take in that interval, the left side (the sum of their cubes plus 1) must be at least as large as k times the sum of the variables. The task is to find all such k that make this true. Since the problem is about an inequality that has to hold for all a, b, c, d in the given interval, my approach should probably involve finding the minimal possible k such that the inequality holds. Because if we can find the smallest k for which the inequality is always true, then any larger k would automatically satisfy the inequality as well. So, the answer is likely going to be the maximum value of (a³ + b³ + c³ + d³ + 1)/(a + b + c + d) over all permissible a, b, c, d where the denominator is not zero. But we have to be careful about cases where the denominator could be zero or negative, which complicates things.Wait, actually, if we consider k as the minimal value such that the inequality holds, maybe we need to use some optimization techniques or calculus to find the maximum of the ratio (left side)/(right side). But since there are four variables, this might get complicated. Alternatively, maybe there's symmetry here, so we can assume that all variables are equal? Let me check if that's a valid approach.If we set a = b = c = d = t, then the inequality becomes 4t³ + 1 ≥ 4kt. Simplifying, we get 4t³ - 4kt + 1 ≥ 0. Then, perhaps we can analyze this single-variable inequality to find constraints on k. However, I need to ensure that this approach doesn't miss any cases where variables are not equal, which might give a higher required k. So, even if assuming symmetry gives a candidate for k, I need to verify whether this k works for all possible variables, not just when they are equal.Alternatively, maybe using Lagrange multipliers to find the extremum of the function f(a, b, c, d) = a³ + b³ + c³ + d³ + 1 - k(a + b + c + d), subject to the constraints a, b, c, d ≥ -1. But this might be complicated due to the four variables. Alternatively, maybe fixing three variables and varying one to find critical points.Another approach is to consider the inequality for each variable individually. Since the variables are independent, maybe the problem can be reduced to a single variable problem. If the inequality can be written as a sum over variables, each term being a³ + 1/4 ≥ k a, then by summing over four variables, we get the original inequality. Wait, but 1 is added in total, not 1/4 for each variable. So that approach might not directly work. But perhaps considering individual variables and then combining them.Suppose we fix three variables at their minimal value, which is -1, and vary the fourth. Let's see what happens. If we set a = b = c = -1 and vary d. Then the left side becomes (-1)³ + (-1)³ + (-1)³ + d³ + 1 = -1 -1 -1 + d³ + 1 = -2 + d³. The right side becomes k(-1 -1 -1 + d) = k(-3 + d). So the inequality is -2 + d³ ≥ k(-3 + d). Let's see for d ≥ -1. If d = -1: left side is -2 + (-1) = -3. Right side is k(-3 + (-1)) = -4k. So inequality becomes -3 ≥ -4k → multiplying both sides by -1 reverses inequality: 3 ≤ 4k → k ≥ 3/4.If d = 0: left side is -2 + 0 = -2. Right side is k(-3 + 0) = -3k. So -2 ≥ -3k → 2 ≤ 3k → k ≥ 2/3 ≈ 0.666...If d = 1: left side is -2 + 1 = -1. Right side is k(-3 + 1) = -2k. So -1 ≥ -2k → 1 ≤ 2k → k ≥ 1/2.If d approaches infinity: left side is d³ - 2, right side is k(d - 3). As d becomes very large, the dominant term on the left is d³ and on the right is kd. Since d³ grows much faster than kd, the left side will eventually be much larger than the right side, so for large d, the inequality holds for any k. So the critical cases are when d is small, particularly around d = -1, 0, 1.From d = -1, we get k ≥ 3/4. From d = 0, k ≥ 2/3. From d = 1, k ≥ 1/2. So the most restrictive here is k ≥ 3/4. But this is just one case where three variables are at -1 and one is varying. Maybe there are other cases where k needs to be larger.Alternatively, let's set two variables at -1 and two variables varying. For example, set a = b = -1, and c = d = t. Then the left side is 2*(-1)^3 + 2*t³ + 1 = -2 + 2t³ + 1 = -1 + 2t³. The right side is k*( -1 -1 + t + t ) = k*(-2 + 2t). So the inequality becomes -1 + 2t³ ≥ k*(-2 + 2t). Let's analyze this.First, if t = -1: left side is -1 + 2*(-1)^3 = -1 -2 = -3. Right side is k*(-2 + 2*(-1)) = k*(-4). So inequality: -3 ≥ -4k → 3 ≤ 4k → k ≥ 3/4.t = 0: left side -1 + 0 = -1. Right side k*(-2 + 0) = -2k. So -1 ≥ -2k → 1 ≤ 2k → k ≥ 1/2.t = 1: left side -1 + 2 = 1. Right side k*(-2 + 2*1) = 0. So inequality is 1 ≥ 0, which is always true, no constraint on k.Again, the most restrictive here is k ≥ 3/4. Hmm. So same as before.What if we set one variable at -1 and others varying? Let's say a = -1, b = c = d = t. Then left side is (-1)^3 + 3t³ + 1 = -1 + 3t³ + 1 = 3t³. Right side is k*(-1 + 3t). So inequality is 3t³ ≥ k*(-1 + 3t). Let's check for t ≥ -1.t = -1: left side 3*(-1)^3 = -3. Right side k*(-1 + 3*(-1)) = k*(-4). So -3 ≥ -4k → 3 ≤ 4k → k ≥ 3/4.t = 0: left side 0. Right side k*(-1 + 0) = -k. So 0 ≥ -k → k ≥ 0.t = 1: left side 3. Right side k*(-1 + 3) = 2k. So 3 ≥ 2k → k ≤ 3/2. But since we're looking for the lower bounds on k, this is an upper bound, which might not be necessary. Wait, but the problem is to find all k such that the inequality holds for all a, b, c, d. So, if k is too large, the inequality could fail. Wait, but in the original problem statement, the inequality is left side ≥ k*(sum). So, if k is too big, then for some a, b, c, d, the left side might not be large enough. Therefore, k can't be larger than some maximum value. Wait, but in the problem statement, it's asking for all real numbers k such that the inequality holds. So, actually, it might be a range of k from negative infinity up to some maximum value. Wait, no. Because if k is negative, then the right side becomes negative (since a, b, c, d are in [-1, ∞), their sum could be positive or negative). Hmm, actually, this is getting a bit more complicated.Wait, let's think again. The left side is a³ + b³ + c³ + d³ + 1, and the right side is k(a + b + c + d). The question is for which k does the left side always exceed the right side. If k is very large positive, then for some a, b, c, d, the right side might be too big. If k is negative, then the right side would be negative when a + b + c + d is positive, or positive when a + b + c + d is negative. Hmm, this seems complicated.But given that variables are in [-1, ∞), their sum can be as low as -4 (if all are -1) and as high as infinity. So, if k is positive, then when the sum a + b + c + d is negative, the right side becomes negative, and the left side is a³ + ... +1. Since each a, b, c, d ≥ -1, so each a³ ≥ -1, so the left side is at least -4 + 1 = -3. So if the right side is negative (when sum is negative and k is positive), then the left side is at least -3, and the right side is negative, so the inequality would hold. Similarly, if k is negative, then the right side would be positive when the sum is negative, but the left side is at least -3, so maybe not. Hmm, maybe the critical case is when the right side is positive. Wait, no, the problem says "for all a, b, c, d". So, even if the right side is positive or negative, the left side must be greater or equal. But when k is positive, and the sum a + b + c + d is positive, then the right side is positive, and the left side must be bigger. If the sum is negative, the right side is negative, so the inequality becomes left side ≥ a negative number, which might be easier because the left side can be as low as -3 (if all variables are -1), so if the right side is negative, then -3 ≥ negative number, which is true. Similarly, if k is negative, then when the sum is positive, the right side is negative, so left side ≥ negative number (which is true since left side can be as low as -3), but when the sum is negative, the right side becomes positive (since k is negative and sum is negative), so left side must be ≥ positive number. But the left side can be as low as -3, so it might not hold. So if k is negative, then for cases where sum is negative, the right side is positive, but left side is at least -3, which would not satisfy the inequality if the right side is positive. Therefore, k must be non-negative. Wait, let me check with k negative. Suppose k = -1. Then the inequality becomes a³ + b³ + c³ + d³ + 1 ≥ -1*(a + b + c + d). Let's set a = b = c = d = -1. Then left side is (-1)^3 *4 +1 = -4 +1 = -3. Right side is -1*(-4) = 4. So -3 ≥ 4? That's false. Therefore, k cannot be negative. So k must be non-negative.Therefore, k is non-negative, and the problem reduces to finding the maximum k such that the inequality holds for all a, b, c, d in [-1, ∞). So, k is the minimal upper bound for (a³ + b³ + c³ + d³ +1)/(a + b + c + d) over all a, b, c, d where a + b + c + d ≠ 0. But since when a + b + c + d = 0, the inequality becomes left side ≥ 0, which needs to be checked as well. So, first, when a + b + c + d = 0, we need to check if a³ + b³ + c³ + d³ +1 ≥ 0. Let me consider this case. For example, if all variables are -1, the sum is -4, so not zero. Let's try variables where the sum is zero. For instance, three variables at -1 and one at 3: a = b = c = -1, d = 3. Then sum is 0. Left side: (-1)^3*3 + 3³ +1 = -3 + 27 +1 = 25 ≥ 0. True. Another case: two variables at -1, two variables at 1. Then sum is 0. Left side: 2*(-1)^3 + 2*1³ +1 = -2 + 2 +1 =1 ≥0. True. Or maybe one variable at -1, three variables at (1/3). Then sum is -1 + 1 = 0. Left side: (-1)^3 + 3*(1/3)^3 +1 = -1 + 3*(1/27) +1 = 0 + 1/9 = 1/9 ≥0. True. So in these cases, the left side is non-negative. What if variables are such that sum is zero but left side is negative? Let's see. Suppose a = -1, b = c = d = t. Then sum is -1 + 3t = 0 → t = 1/3. Then left side is (-1)^3 + 3*(1/3)^3 +1 = -1 + 3*(1/27) +1 = 0 + 1/9 = 1/9 ≥0. Still positive. Hmm, maybe the left side is always non-negative when sum is zero. Let me check another case. Suppose two variables at x and two variables at y such that 2x + 2y = 0 → x = -y. Let x = t and y = -t. Then left side: 2t³ + 2(-t)^3 +1 = 2t³ - 2t³ +1 =1 ≥0. So yes, always 1. So in this case, left side is 1. So when the sum is zero, the left side is always at least 0, so the inequality holds. Therefore, in cases where the sum is zero, the inequality is satisfied regardless of k, as long as k is finite. Wait, but when the sum is zero, the right side is k*0 = 0, so the inequality becomes left side ≥0, which we saw holds. Therefore, the only critical cases are when the sum is non-zero, so we can focus on those. Therefore, the problem reduces to finding the maximum k such that (a³ + b³ + c³ + d³ +1)/(a + b + c + d) ≥ k for all a, b, c, d ∈ [-1, ∞) with a + b + c + d ≠ 0. Therefore, k is the infimum of (a³ + b³ + c³ + d³ +1)/(a + b + c + d) over all a, b, c, d in the domain with a + b + c + d >0 (since when a + b + c + d <0, the ratio would be negative, but we already saw that k must be non-negative, and in those cases, the inequality holds automatically). Wait, actually, if a + b + c + d is negative, then since k is non-negative, the right side is non-positive, and the left side is at least -3, as before. But in those cases, the left side can be as low as -3, but the right side is non-positive, so -3 ≥ non-positive number, which is not necessarily true. Wait, hold on. If a + b + c + d is negative, then the right side is k*(negative), which is non-positive if k is non-negative. The left side is a³ + b³ + c³ + d³ +1. The variables are each ≥ -1, so each a³ ≥ -1, so left side ≥ -4 +1 = -3. Therefore, if the right side is ≤0 (since k ≥0 and sum is negative), then the inequality becomes left side ≥ right side, which is equivalent to left side ≥ a non-positive number. Since left side can be as low as -3, but the right side is ≤0. So for example, if left side is -3, then we need -3 ≥ right side. But the right side is k*(sum). If sum is negative, then right side is k*(negative). So, if sum is negative and k is positive, right side is negative. So, the inequality becomes -3 ≥ negative number, which is true because -3 is less than any negative number. Wait, no. For example, if sum is -s (s positive), then right side is -k s. So the inequality is left side ≥ -k s. Since left side ≥ -3, and -k s can be any negative number (depending on s and k). So we need that -3 ≥ -k s. Which is equivalent to k s ≥3. But s is the absolute value of the sum, which can be as large as 4 (if all variables are -1). So, k s ≥3. The minimal s is 0 (if sum approaches zero from the negative side), but s can be up to 4. Wait, this is getting confusing. Let me think again.If a + b + c + d is negative, then right side is k*(negative). Since k is non-negative, right side ≤0. The left side is a³ + b³ + c³ + d³ +1. We need left side ≥ right side. Since right side is ≤0, and left side can be as low as -3 (when all variables are -1). So, for the inequality to hold in this case, the left side must be ≥ right side. So, for example, when all variables are -1, left side is -3, right side is k*(-4). So inequality is -3 ≥ -4k. Which simplifies to k ≥ 3/4. So, in this case, if k is at least 3/4, then -3 ≥ -4k holds. But if a + b + c + d is negative but not all variables are -1, then the left side might be higher than -3. For example, if three variables are -1 and the fourth is 0, sum is -3 +0 = -3. Left side is -3 +0 +1 = -2. Right side is k*(-3). So inequality is -2 ≥ -3k → k ≥ 2/3. If k is 3/4, then -2 ≥ -3*(3/4) → -2 ≥ -9/4 → -8/4 ≥ -9/4, which is true. So in this case, k=3/4 works.But if sum is very negative, for example, approaching negative infinity? Wait, but variables are in [-1, ∞), so the sum can't be less than -4. Therefore, the minimal sum is -4 (all variables at -1). So, for the case when sum is -4, left side is -3, so we need -3 ≥ -4k → k ≥ 3/4. For other sums between -4 and 0, say sum = -s where 0 < s ≤4, the inequality becomes left side ≥ -k s. Since left side can be as low as -3, and -k s can be as low as -4k. So, to ensure that even the minimal left side (-3) is ≥ -k s for all s in (0,4]. That is, -3 ≥ -k s for all s ∈ (0,4]. Which is equivalent to k s ≥3 for all s ∈ (0,4]. But s can be as small as approaching 0, so k s ≥3 would require k to be infinitely large, which is impossible. Therefore, this approach is flawed.Wait, perhaps I need to consider that when sum is negative, the inequality left side ≥ k*(sum) must hold. But sum is negative, so k*(sum) is negative if k is positive, as established. So left side ≥ negative number. Since left side is a³ + b³ + c³ + d³ +1. Each a, b, c, d ≥ -1, so each a³ ≥ -1. Therefore, left side ≥ -4 +1 = -3. So the left side is at least -3, and the right side is at most 0 (since it's k times a negative sum). So, for the inequality to hold, we need that -3 ≥ k*(sum) for all sum <0. Wait, but sum can be as negative as -4. So, when sum = -4, right side is -4k. So, -3 ≥ -4k → 4k ≥3 → k ≥3/4. For other sums, sum = -s, 0 < s <4. Then, right side is -k s. So the inequality becomes left side ≥ -k s. Since left side ≥ -3, we need -3 ≥ -k s for all s in (0,4]. Which is equivalent to k s ≥3 for all s ∈ (0,4]. But this is impossible because as s approaches 0, k s approaches 0, which is less than 3. Therefore, my previous reasoning is incorrect.Wait, no. Wait, we have left side ≥ right side. When sum is negative, right side is negative (if k is positive). Therefore, left side (which is ≥-3) must be ≥ a negative number. The most stringent case is when right side is the most negative (i.e., when sum is the most negative, which is -4). So, in that case, right side is -4k, and left side is -3. So we need -3 ≥ -4k → k ≥3/4. For sums that are less negative (closer to zero), right side is less negative (closer to zero), so left side (which is at least -3) will automatically be greater than right side (since right side is, say, -2k, which is greater than -4k). Therefore, the only critical case when sum is negative is when sum is minimal (-4), leading to k ≥3/4.Therefore, when sum is negative, the inequality imposes that k ≥3/4. For sum positive, we need to ensure that left side is at least k times sum. So, in that case, the ratio (left side)/sum ≥k must hold. Therefore, the minimal such k is the infimum of (left side)/sum over all a, b, c, d in [-1, ∞) with sum >0. So, the maximum k that works is the minimum between the lower bound from the negative sum case (k≥3/4) and the infimum from the positive sum case.Wait, but if in the positive sum case, the infimum of (left side)/sum is higher than 3/4, then the overall k can be 3/4. If the infimum in the positive case is lower than 3/4, then the maximum possible k would be that lower value. So, we need to check both cases.So, first, we already saw that when sum is negative, we need k ≥3/4. For sum positive, we need to find the minimal value of (left side)/sum, which would dictate the maximum k. If the minimal value is greater than or equal to 3/4, then k=3/4 works. Otherwise, k would have to be smaller. Let's check.Consider the positive sum case. Let's try to find the minimal value of (a³ + b³ + c³ + d³ +1)/(a + b + c + d). To minimize this ratio, we need to make the numerator as small as possible relative to the denominator. That is, make a³ + ... +1 small while making a + ... + d large. How can that happen? Let's see.Each variable is in [-1, ∞). The function x³ is increasing for x ≥ -1, since derivative 3x² is non-negative. Therefore, for variables in [-1, ∞), x³ is minimized at x = -1. Therefore, to minimize the numerator, we might want to set some variables to -1, which will make their cubes as small as possible (-1), while setting other variables to as small as possible (close to -1) but such that the denominator (sum) is as large as possible. Wait, but if we set variables to -1, the sum will be smaller. Hmm, this seems conflicting.Wait, perhaps if we set some variables to -1 (to reduce the numerator) and some variables to large positive numbers (to increase the denominator). Let's test this.Suppose we set three variables to -1 and one variable to t, which is large. Then, the sum is (-1)*3 + t = t -3. The numerator is (-1)^3*3 + t³ +1 = -3 + t³ +1 = t³ -2. Therefore, the ratio is (t³ -2)/(t -3). As t approaches infinity, this ratio behaves like t³/t = t², which goes to infinity. So, in this case, the ratio can be made arbitrarily large, so there's no upper bound in the positive sum case. But we are looking for the minimal ratio. So, perhaps the minimal ratio occurs when variables are set such that the numerator is minimized relative to the denominator. Let's consider setting some variables to -1 and others to some value.Alternatively, let's set two variables to -1 and two variables to t. Then, sum is (-1)*2 + 2t = 2t -2. The numerator is (-1)^3*2 + 2t³ +1 = -2 + 2t³ +1 = 2t³ -1. The ratio is (2t³ -1)/(2t -2) = (2t³ -1)/(2(t -1)). Let's analyze this for t ≥ -1.If t approaches 1 from above, denominator approaches 0 from the positive side, numerator approaches 2*1 -1 =1, so ratio approaches +infty. If t approaches 1 from below, denominator approaches 0 from the negative side, numerator approaches 1, so ratio approaches -infty. But since t ≥ -1, let's check t = -1: sum is 2*(-1) -2 = -4, which is negative. So we need t such that sum is positive, i.e., 2t -2 >0 → t >1. Wait, t must be greater than 1 for the sum to be positive. Then, as t increases beyond 1, the ratio (2t³ -1)/(2(t -1)) increases to infinity. So the minimal ratio in this case occurs at t approaching 1 from the right. As t approaches 1+, the ratio approaches (2 -1)/0+ = 1/0+ = +infty. So, actually, this case doesn't give a minimal ratio.Alternatively, set one variable to -1 and three variables to t. Then, sum is -1 +3t. The numerator is -1 +3t³ +1 =3t³. So ratio is 3t³/( -1 +3t ). To have sum positive: -1 +3t >0 → t >1/3. So, for t >1/3, ratio is 3t³/(3t -1). Let's compute derivative to find minima.Let f(t) = 3t³ / (3t -1). f’(t) = [9t²(3t -1) -3t³*3]/(3t -1)^2 = [27t³ -9t² -9t³]/(3t -1)^2 = (18t³ -9t²)/(3t -1)^2 = 9t²(2t -1)/(3t -1)^2.Setting derivative to zero: 9t²(2t -1) =0. Solutions t=0 or t=1/2. But t >1/3, so t=1/2 is critical point. Compute f(1/2): 3*(1/2)^3 / (3*(1/2) -1) = 3*(1/8)/(3/2 -1) = (3/8)/(1/2) = (3/8)*(2/1)= 3/4. Hmm, interesting. So, at t=1/2, the ratio is 3/4. Let's check behavior as t approaches 1/3 from the right: denominator approaches 0, numerator approaches 3*(1/3)^3 =3*(1/27)=1/9, so ratio approaches 1/9 /0+ = +infty. As t approaches infinity, ratio behaves like 3t³ /3t = t² → infinity. Therefore, the minimal ratio in this case is 3/4 at t=1/2.Therefore, when we set one variable to -1 and three variables to 1/2, the ratio is 3/4. Let's check this. a = -1, b = c = d =1/2. Then sum is -1 + 3*(1/2) = -1 + 3/2 =1/2. The left side is (-1)^3 + 3*(1/2)^3 +1 = -1 + 3*(1/8) +1 =0 + 3/8 = 3/8. Wait, 3/8 divided by 1/2 is (3/8)/(1/2)= 3/4. Yes, that's correct. So this gives a ratio of 3/4.Therefore, in this case, the minimal ratio in the positive sum case is 3/4. Therefore, combining both the negative sum case (which requires k ≥3/4) and the positive sum case (which also gives a minimal ratio of 3/4), the maximal k that satisfies the inequality for all a, b, c, d is 3/4. But wait, we need to confirm that this is indeed the minimal possible ratio in the positive sum case. Let's test another configuration. Suppose two variables are set to some t and two variables are set to s, such that sum is positive. Maybe symmetric cases. But in the previous analysis, setting one variable to -1 and three variables to 1/2 gave the minimal ratio. Let me check another case where two variables are set to -1 and two variables are set to t, but we saw earlier that in such cases, the minimal ratio approaches infinity as t approaches 1 from above, so that's not helpful. Alternatively, set a = b = c = d = t. Then, sum is 4t, and numerator is 4t³ +1. So ratio is (4t³ +1)/(4t). Let's see if this can be less than 3/4. For t=1/2: (4*(1/8) +1)/(2) = (0.5 +1)/2=1.5/2=0.75=3/4. So at t=1/2, the ratio is also 3/4. If t is larger than 1/2, say t=1: ratio is (4 +1)/4=5/4=1.25>3/4. If t is smaller than 1/2, but t must be ≥-1. Let's check t approaching -1: sum is -4, which is negative. So not in the positive sum case. For t between -1 and1/2, sum can be positive or negative. For example, t=0: sum=0. Not considered. t=1/4: sum=1. Numerator=4*(1/64) +1=1/16 +1=17/16≈1.06. Ratio≈1.06/1≈1.06>3/4. So in the symmetric case, the minimal ratio is also 3/4 at t=1/2. Therefore, both the symmetric case and the case with one variable at -1 and three at 1/2 give the ratio 3/4. Another test case: set three variables to 1/2 and one variable to -1, which we already did, giving ratio 3/4. So it seems that 3/4 is indeed the minimal ratio in the positive sum case, which coincides with the required k from the negative sum case. Therefore, k=3/4 is the maximal value such that the inequality holds for all a, b, c, d in [-1, ∞).To confirm, let's test k=3/4 in the original inequality. Let's take the case where three variables are 1/2 and one is -1. Then, left side: (-1)^3 +3*(1/2)^3 +1= -1 +3*(1/8)+1= -1 +3/8 +1= 3/8. Right side: (3/4)*( -1 + 3*(1/2) )= (3/4)*( -1 + 3/2 )= (3/4)*(1/2)=3/8. So 3/8 ≥3/8, which holds. Equality is achieved here.Another case where all variables are 1/2. Left side:4*(1/2)^3 +1=4*(1/8)+1=0.5 +1=1.5. Right side: (3/4)*(4*(1/2))= (3/4)*2=1.5. So 1.5≥1.5, equality holds.Another case where two variables are -1 and two variables are 1. Sum= -2 +2=0, but in that case, left side= 2*(-1)^3 +2*1 +1= -2 +2 +1=1. Right side=0, so 1≥0, which holds. If we take a case where variables are three at -1 and one at 3. Sum=0, left side= -3 +27 +1=25≥0, holds. If we take a case where sum is positive, say all variables=1, left side=4+1=5, right side=3/4*4=3, 5≥3 holds. If variables are two at 2 and two at -1. Sum=4 -2=2. Left side=2*(8) +2*(-1) +1=16 -2 +1=15. Right side=3/4*2=1.5. 15≥1.5 holds. If variables are set such that one variable is -1, three variables are 1/2, which we saw gives equality. Similarly, if all variables are 1/2, equality is achieved. So, this suggests that k=3/4 is indeed the maximal value where the inequality holds for all a, b, c, d in [-1, ∞). Therefore, the answer is k=3/4.Final Answerboxed{dfrac{3}{4}}</think>To find all real numbers ( k ) such that the inequality ( a^3 + b^3 + c^3 + d^3 + 1 geq k(a + b + c + d) ) holds for all ( a, b, c, d in [-1, +infty) ), we analyze the problem by considering different cases and configurations of the variables.1. Case when variables are set to their minimum value (-1): - Setting three variables to -1 and varying the fourth, we find that the inequality must hold for ( d geq -1 ). For ( d = -1 ), the inequality simplifies to ( -3 geq -4k ), leading to ( k geq frac{3}{4} ).2. Case when variables are set to other values: - Setting two variables to -1 and two variables to ( t ), we analyze the ratio ( frac{-1 + 2t^3}{-2 + 2t} ). For ( t = 1 ), the ratio gives ( k geq frac{1}{2} ), but the most restrictive case is when ( t = -1 ), leading to ( k geq frac{3}{4} ).3. Case when one variable is -1 and others are varied: - Setting one variable to -1 and three variables to ( t ), we analyze the ratio ( frac{3t^3}{-1 + 3t} ). For ( t = frac{1}{2} ), the ratio gives ( frac{3}{4} ), leading to ( k geq frac{3}{4} ).4. Symmetric case where all variables are equal: - Setting all variables to ( t ), we analyze the ratio ( frac{4t^3 + 1}{4t} ). For ( t = frac{1}{2} ), the ratio gives ( frac{3}{4} ), leading to ( k geq frac{3}{4} ).5. Verification of critical cases: - Testing configurations where equality holds, such as three variables at ( frac{1}{2} ) and one at -1, confirms that ( k = frac{3}{4} ) satisfies the inequality with equality.After analyzing all cases and verifying with critical configurations, we conclude that the maximum value of ( k ) such that the inequality holds for all ( a, b, c, d in [-1, +infty) ) is ( frac{3}{4} ).[boxed{dfrac{3}{4}}]]

🔑:When a=b=c=d=-1, we have 3 geqslant k(-4), so k geqslant 3. When a=b=c=d=frac{1}{2}, we have 4 times frac{1}{8}+1 geqslant kleft(4 times frac{1}{2}right), so k leqslant frac{3}{4}. Therefore, k=frac{3}{4}. Below is the proof thata^{3}+b^{3}+c^{3}+d^{3}+1 geqslant frac{3}{4}(a+b+c+d)holds for any a, b, c, d in (-1,+infty).First, we prove that 4 x^{3}+1 geqslant 3 x, x in (-1,+infty).In fact, from (x+1)(2 x-1) geqslant 0, we get4 x^{3}+1 geqslant 3 x, x in (-1,+infty)- Therefore, 4 a^{3}+1 geqslant 3 a, 4 b^{3}+1 geqslant 3 b, 4 c^{3}+1 geqslant 3 c, 4 d^{3}+1 geqslant 3 d; adding these four inequalities, we obtain the desired inequality (1).Thus, the real number k=frac{3}{4}.