🧠:[<think>Okay, so I need to find sin x given that x is between -π/2 and 0, and that cos 2x equals a. The answer choices are A through D, each involving square roots of expressions with (1 - a)/2 or (1 + a)/2, either positive or negative. Hmm, let's start by recalling some trigonometric identities. I know that there are double-angle formulas for cosine. The one that comes to mind is cos 2x = 1 - 2 sin²x. Alternatively, it can also be written as cos 2x = 2 cos²x - 1. Since the problem gives me cos 2x = a, maybe I can use one of these identities to solve for sin x. Let me try the first one: cos 2x = 1 - 2 sin²x. If I substitute a for cos 2x, then the equation becomes a = 1 - 2 sin²x. Let's solve for sin²x. Subtract 1 from both sides: a - 1 = -2 sin²x. Multiply both sides by -1: 1 - a = 2 sin²x. Then divide both sides by 2: sin²x = (1 - a)/2. So, sin x would be the square root of (1 - a)/2, but since square roots can be positive or negative, I need to consider the sign based on the interval where x lies. The problem states that x is in (-π/2, 0), which is the fourth quadrant. In the fourth quadrant, sine is negative. Therefore, sin x should be negative, so we take the negative square root. That would make sin x = -√[(1 - a)/2], which is option B. Let me double-check this because sometimes I mix up the identities. Let's verify using another identity. Alternatively, using cos 2x = 2 cos²x - 1. If I use this, then a = 2 cos²x - 1. Solving for cos²x, we get (a + 1)/2 = cos²x. Then cos x = ±√[(a + 1)/2]. However, x is in the fourth quadrant where cosine is positive, so cos x = √[(a + 1)/2]. But we need sin x, not cos x. Alternatively, since we know sin²x + cos²x = 1, so sin x = ±√(1 - cos²x). Substituting cos x from above, sin x = ±√[1 - (a + 1)/2] = ±√[(2 - a - 1)/2] = ±√[(1 - a)/2]. Again, since x is in the fourth quadrant, sin x is negative, so sin x = -√[(1 - a)/2]. That still gives me option B. Wait, but let me check if there's another way to approach this. For example, using the identity sin x = ±√[(1 - cos 2x)/2]. Wait, but since 2x is in (-π, 0). So 2x is in the third or fourth quadrants? If x is between -π/2 and 0, then 2x is between -π and 0. So 2x is in the third or fourth quadrants. But cos 2x = a. The cosine in the third quadrant is negative, and in the fourth quadrant is positive. So depending on the value of a, 2x could be in different quadrants, but since we know that a is the cosine of 2x, which is in (-π, 0). Hmm, but a is given as a value, so a can be between -1 and 1. But x is in (-π/2, 0), so 2x is in (-π, 0). The cosine of angles in (-π, 0) is the same as cosine of angles in (0, π) because cosine is even, so cos 2x is equal to cos |2x|. Therefore, a would be in the range of (-1, 1), but perhaps depending on where 2x is. But maybe that's complicating things. Let me stick to the first approach. Using the double-angle formula, I found that sin x = -√[(1 - a)/2], which is option B. However, let me check if I made any mistake in the sign. x is in (-π/2, 0), so that's the fourth quadrant. In the fourth quadrant, sine is negative, cosine is positive. So yes, sin x should be negative, so we take the negative square root. Alternatively, maybe I can plug in a specific value for x to test. For example, let me choose x = -π/4. Then 2x = -π/2. What's cos 2x? cos(-π/2) = 0. So a = 0. Then sin x = sin(-π/4) = -√2/2. Let's check the answer choices when a = 0. Option B would be -√[(1 - 0)/2] = -√(1/2) = -√2/2, which matches. Option A would be √(1/2) which is positive, which is incorrect. Similarly, options C and D with (1 + a)/2 would be (1 + 0)/2 = 1/2, so sqrt(1/2) and -sqrt(1/2). D is -sqrt(1/2), which is also -√2/2. Wait, but D is -sqrt[(1 + a)/2], which when a=0 is also -√(1/2). So in this case, both B and D give the correct answer. Hmm, so maybe my test case is not sufficient. Let me try another value. Let me pick x = -π/6. Then 2x = -π/3. cos 2x = cos(-π/3) = cos(π/3) = 1/2. So a = 1/2. Then sin x = sin(-π/6) = -1/2. Let's compute the options: Option B: -sqrt[(1 - 1/2)/2] = -sqrt[(1/2)/2] = -sqrt(1/4) = -1/2. Correct. Option D: -sqrt[(1 + 1/2)/2] = -sqrt[(3/2)/2] = -sqrt(3/4) = -√3/2. Which is not equal to -1/2. So in this case, option B is correct. Another test case: Let x = -π/3. Then 2x = -2π/3. cos 2x = cos(-2π/3) = cos(2π/3) = -1/2. So a = -1/2. Then sin x = sin(-π/3) = -√3/2. Option B: -sqrt[(1 - (-1/2))/2] = -sqrt[(3/2)/2] = -sqrt(3/4) = -√3/2. Correct. Option D: -sqrt[(1 + (-1/2))/2] = -sqrt[(1/2)/2] = -sqrt(1/4) = -1/2. Not correct. So test cases confirm that option B is correct. Therefore, the answer should be B. Wait, but let me check when x approaches 0. If x is approaching 0 from the left, then 2x approaches 0, so cos 2x approaches 1, so a approaches 1. Then sin x approaches 0. Let's see option B: -sqrt[(1 - a)/2] as a approaches 1, the expression inside the sqrt becomes (1 - 1)/2 = 0, so sin x approaches 0, which is correct. Similarly, if x approaches -π/2, then 2x approaches -π, so cos 2x = cos(-π) = -1. Therefore, a = -1. Then sin x = sin(-π/2) = -1. Let's compute option B: -sqrt[(1 - (-1))/2] = -sqrt[(2)/2] = -sqrt(1) = -1. Correct. So all test cases work with option B. Therefore, I think the correct answer is B. Final Answerboxed{B}</think>Given ( x in left(-frac{pi}{2}, 0right) ) and (cos 2x = a), we need to find (sin x).First, we use the double-angle identity for cosine:[cos 2x = 1 - 2 sin^2 x]Substituting ( a ) for (cos 2x), we get:[a = 1 - 2 sin^2 x]Solving for (sin^2 x):[2 sin^2 x = 1 - a implies sin^2 x = frac{1 - a}{2}]Taking the square root of both sides, we consider the sign of (sin x). Since ( x ) is in the interval (left(-frac{pi}{2}, 0right)) (the fourth quadrant), (sin x) must be negative. Therefore:[sin x = -sqrt{frac{1 - a}{2}}]To verify, we checked specific values of ( x ) and confirmed that the result matches the given conditions and answer choices. All test cases confirmed that the correct answer is B.Thus, the final answer is (boxed{B}).,<think>Okay, let me try to solve this problem. So, the question is: Given that x is in the interval (-π/2, 0) and cos(2x) = a, we need to find sin x. The options are A to D, each involving square roots of expressions with (1 - a)/2 or (1 + a)/2, and some with negative signs.First, I remember that there are double-angle formulas for cosine. The one that might be helpful here is cos(2x) = 1 - 2sin²x. Alternatively, it can also be written as cos(2x) = 2cos²x - 1. Since we need to find sin x, maybe the first form is more useful here. Let me check.Given that cos(2x) = a, using the identity cos(2x) = 1 - 2sin²x, we can set up the equation:a = 1 - 2sin²xThen, solving for sin²x:2sin²x = 1 - asin²x = (1 - a)/2So, sin x would be either sqrt[(1 - a)/2] or -sqrt[(1 - a)/2]. But since the answer choices have these expressions, I need to figure out the correct sign based on the interval where x lies.The problem states that x is in (-π/2, 0). That is, x is between -π/2 and 0, which is in the fourth quadrant. In the fourth quadrant, sine is negative. So, sin x should be negative. Therefore, the correct value of sin x is -sqrt[(1 - a)/2].Looking at the options, that's option B: -sqrt[(1 - a)/2]. But wait, let me make sure I didn't mix up any formulas. Let me verify using another identity.Alternatively, using the identity cos(2x) = 2cos²x - 1. If I solve for cos²x:2cos²x = 1 + acos²x = (1 + a)/2But the question is asking for sin x, not cos x. However, since sin²x + cos²x = 1, we can relate them. But since we already used the first identity and got sin²x = (1 - a)/2, and considering the quadrant for x, which is the fourth quadrant where sine is negative, so we take the negative square root.Alternatively, perhaps I can express sin x in terms of cos(2x) through another identity? Let me think. There's also sin x = sqrt[(1 - cos(2x))/2], but again, considering the quadrant.Wait, but according to the double-angle formula, sin²x = (1 - cos(2x))/2, so sin x = ±sqrt[(1 - cos(2x))/2]. Since cos(2x) = a, substituting that in, sin x = ±sqrt[(1 - a)/2]. And as x is in (-π/2, 0), which is the fourth quadrant, sine is negative there. So sin x is negative sqrt[(1 - a)/2]. So the answer should be B.But wait, let me check if the answer options have B as that. The options are:A: sqrt[(1 - a)/2]B: -sqrt[(1 - a)/2]C: sqrt[(1 + a)/2]D: -sqrt[(1 + a)/2]So yes, B is the negative sqrt of (1 - a)/2, which matches our conclusion. Therefore, the answer is B.But just to be thorough, let me consider another approach. Suppose we use the identity for sin x in terms of cos 2x. Let's see:From the double angle formula, cos(2x) = 1 - 2sin²xSo, 2sin²x = 1 - cos(2x)sin²x = (1 - cos(2x))/2Therefore, sin x = ±sqrt[(1 - cos(2x))/2]Given that cos(2x) = a, then sin x = ±sqrt[(1 - a)/2]. Again, considering x is in (-π/2, 0), which is quadrant IV, sine is negative here. Therefore, sin x is negative sqrt[(1 - a)/2], which is option B.Alternatively, maybe there's a way to express sin x in terms of cos x, but I think the approach above is straightforward. Let me also check with an example. Suppose x is a specific angle in (-π/2, 0), let's say x = -π/4. Then 2x = -π/2, so cos(2x) = cos(-π/2) = 0. Therefore, a = 0. Then sin x = sin(-π/4) = -sqrt(2)/2. Let's check the options with a = 0:A: sqrt[(1 - 0)/2] = sqrt(1/2) = sqrt(2)/2 (positive)B: -sqrt[(1 - 0)/2] = -sqrt(1/2) = -sqrt(2)/2C: sqrt[(1 + 0)/2] = sqrt(1/2) = sqrt(2)/2D: -sqrt[(1 + 0)/2] = -sqrt(1/2) = -sqrt(2)/2So both B and D give -sqrt(2)/2, but in reality, sin(-π/4) is indeed -sqrt(2)/2. Wait, but if we use the formula from B: sqrt[(1 - a)/2], but with a = 0, we get sqrt(1/2), but with a negative sign. So B gives -sqrt(1/2). Similarly, D would be -sqrt(1/2) as well if a = 0. Wait, in this example, both B and D would be the same?Wait, no. If a = 0, then option B is -sqrt[(1 - 0)/2] = -sqrt(1/2), and option D is -sqrt[(1 + 0)/2] = -sqrt(1/2). So in this case, both B and D give the same result? Wait, that can't be. Wait, let me check the problem statement again.Wait, the problem says cos(2x) = a. If x = -π/4, then 2x = -π/2, so cos(2x) = 0. Therefore, a = 0. Then, sin x = sin(-π/4) = -sqrt(2)/2. So, substituting a = 0 into the options:A: sqrt[(1 - 0)/2] = sqrt(1/2) ≈ 0.707 (positive, incorrect)B: -sqrt(1/2) ≈ -0.707 (correct)C: sqrt[(1 + 0)/2] = sqrt(1/2) ≈ 0.707 (positive, incorrect)D: -sqrt[(1 + 0)/2] = -sqrt(1/2) ≈ -0.707 (also correct?)Wait, so in this case, both B and D would give the correct answer? But that's contradictory. So perhaps my example is not sufficient? Wait, but if x is in (-π/2, 0), let's pick another example. Let's take x = -π/6. Then 2x = -π/3, so cos(2x) = cos(-π/3) = 0.5. Therefore, a = 0.5. Then sin x = sin(-π/6) = -0.5.Now, check the options with a = 0.5:A: sqrt[(1 - 0.5)/2] = sqrt(0.5/2) = sqrt(0.25) = 0.5 (positive, incorrect)B: -sqrt[(1 - 0.5)/2] = -sqrt(0.25) = -0.5 (correct)C: sqrt[(1 + 0.5)/2] = sqrt(1.5/2) = sqrt(0.75) ≈ 0.866 (positive, incorrect)D: -sqrt[1.5/2] ≈ -0.866 (incorrect)So in this case, only B gives the correct answer. Similarly, for the previous example where a = 0, both B and D gave the same answer. But why?Wait, when a = 0, both (1 - a)/2 and (1 + a)/2 become 0.5. But for x = -π/4, sin x is -sqrt(0.5). However, in that case, both B and D evaluate to -sqrt(0.5). Wait, but (1 - a)/2 when a = 0 is 0.5, and (1 + a)/2 is 0.5 as well. So sqrt(0.5) and sqrt(0.5) under negative signs. So in that case, both B and D would be same. However, the original question must have distinct options. Let me check the original problem again.The options are:A: sqrt[(1 - a)/2]B: -sqrt[(1 - a)/2]C: sqrt[(1 + a)/2]D: -sqrt[(1 + a)/2]Ah, okay, so when a = 0, both (1 - a)/2 and (1 + a)/2 are 0.5. But in the general case, (1 - a)/2 and (1 + a)/2 are different. So perhaps when a = 0, both expressions under the square roots become same, but in other cases, they differ.So in the first example with x = -π/4, a = 0, so B and D both give -sqrt(0.5). But in reality, sin x is indeed -sqrt(0.5). So in that case, both B and D are correct? That seems conflicting, but probably because in that specific case, the two expressions coincide. But when a is not 0, then B and D differ.In the second example, x = -π/6, a = 0.5, so B gives -sqrt[(1 - 0.5)/2] = -sqrt(0.25) = -0.5, which is correct, while D gives -sqrt[(1 + 0.5)/2] = -sqrt(0.75) ≈ -0.866, which is incorrect. Therefore, only B is correct here. So, perhaps the answer is B.But why in the first example, both B and D gave the correct answer? Because when a = 0, (1 - a)/2 = (1 + a)/2. So, in general, when a is 0, these two expressions coincide. Therefore, for the problem, when a is arbitrary, as long as x is in (-π/2, 0), the correct expression is B. Let's confirm with another example.Take x = -π/3. Then 2x = -2π/3, so cos(2x) = cos(-2π/3) = cos(2π/3) = -0.5. So a = -0.5. Then sin x = sin(-π/3) = -sqrt(3)/2 ≈ -0.866.Now, check the options with a = -0.5:A: sqrt[(1 - (-0.5))/2] = sqrt(1.5/2) = sqrt(0.75) ≈ 0.866 (positive, incorrect)B: -sqrt(1.5/2) ≈ -0.866 (correct)C: sqrt[(1 + (-0.5))/2] = sqrt(0.5/2) = sqrt(0.25) = 0.5 (positive, incorrect)D: -sqrt(0.25) = -0.5 (incorrect)So here, B gives the correct answer. Therefore, in these examples, B is correct, while D is incorrect except when a = 0, where both B and D coincide. Therefore, in general, the answer must be B.Therefore, the correct answer is B: -sqrt[(1 - a)/2].Final Answerboxed{B}</think>Given ( x in left(-frac{pi}{2}, 0right) ) and (cos 2x = a), we need to find (sin x).First, we use the double-angle identity for cosine:[cos 2x = 1 - 2sin^2 x]Substituting (cos 2x = a) into the equation, we get:[a = 1 - 2sin^2 x]Solving for (sin^2 x), we have:[2sin^2 x = 1 - a implies sin^2 x = frac{1 - a}{2}]Taking the square root of both sides, we get:[sin x = pm sqrt{frac{1 - a}{2}}]Since (x) is in the interval (left(-frac{pi}{2}, 0right)), which is in the fourth quadrant where sine is negative, we take the negative root:[sin x = -sqrt{frac{1 - a}{2}}]Thus, the correct answer is (boxed{B}).]

🔑:Since cos 2x = a, we have 1 - 2sin^2x = a, which gives sin^2x = frac{1-a}{2}, and since x in left(-frac{pi}{2}, 0right), we have sin x < 0, thus sin x = -sqrt{frac{1-a}{2}}. Therefore, the correct choice is: B By using the double-angle formula for cosine and given conditions, we calculate sin^2x = frac{1-a}{2}, and from sin x < 0 we find sin x = -sqrt{frac{1-a}{2}}, thus obtaining the answer. This problem provides the value of cos 2x and asks for sin x. It mainly tests knowledge of trigonometric functions of any angle and the double-angle formula for cosine, and is considered a basic question.The final answer is boxed{text{B}}.

🧠:[<think>Okay, so I need to find the remainder when X^n is divided by X² - 4X + 3. Hmm, let me remember how polynomial division works. When you divide a polynomial by a quadratic, the remainder should be a first-degree polynomial, right? So the remainder will look like aX + b, where a and b are constants. The problem is asking for the remainder of X^n divided by X² - 4X + 3. Let me think about how to approach this.Maybe I can use the division algorithm. For polynomials, we have X^n = (X² - 4X + 3) * Q(X) + R(X), where Q(X) is the quotient and R(X) is the remainder, which is of degree less than 2. So R(X) = aX + b. Then, if I plug in the roots of the divisor polynomial X² - 4X + 3 into the equation, the term with Q(X) will disappear, leaving me with X^n = R(X) at those roots. Let me find the roots of X² - 4X + 3. Let's factor that: X² - 4X + 3 = (X - 1)(X - 3). So the roots are X = 1 and X = 3. That means if I substitute X = 1 and X = 3 into the equation X^n = aX + b, I should get two equations. For X = 1: 1^n = a*1 + b ⇒ 1 = a + b.For X = 3: 3^n = a*3 + b ⇒ 3^n = 3a + b.Now I have a system of equations:1. a + b = 12. 3a + b = 3^nI can solve this system for a and b. Let's subtract the first equation from the second: (3a + b) - (a + b) = 3^n - 1 ⇒ 2a = 3^n - 1 ⇒ a = (3^n - 1)/2.Then, substitute a back into the first equation: (3^n - 1)/2 + b = 1 ⇒ b = 1 - (3^n - 1)/2 ⇒ b = (2 - 3^n + 1)/2 ⇒ b = (3 - 3^n)/2.So the remainder R(X) = aX + b = [(3^n - 1)/2] X + (3 - 3^n)/2. Let me check if this makes sense. Alternatively, maybe I can express the remainder using modular arithmetic. Since X² ≡ 4X - 3 mod (X² - 4X + 3), we can reduce higher powers of X by substituting X² with 4X - 3. So, for X^n, we can write X^n = X^{n-2} * X² = X^{n-2}*(4X - 3) = 4X^{n-1} - 3X^{n-2}. This recurrence relation might help.Let me try to find a pattern or recurrence relation here. Let’s denote R_n as the remainder of X^n divided by X² - 4X + 3. Then, R_n = 4X^{n-1} - 3X^{n-2} mod (X² - 4X + 3). But since we're working modulo X² - 4X + 3, any higher power can be reduced. So, actually, R_n = 4R_{n-1} - 3R_{n-2}, with initial conditions. Let me check that.For n=2: X² divided by X² -4X +3. The remainder is 4X - 3. So R_2 = 4X -3.For n=1: X, the remainder is X. So R_1 = X.For n=0: 1, remainder is 1. So R_0 = 1.But maybe starting from n=0. Let me verify the recurrence. Suppose R_n = 4R_{n-1} -3R_{n-2}. Then for n=2: R_2 = 4R_1 -3R_0 = 4*X -3*1 = 4X -3, which matches. For n=3: R_3 = 4R_2 -3R_1 = 4*(4X -3) -3X = 16X -12 -3X =13X -12. Let me compute directly: X^3 = X*(X^2) = X*(4X -3) =4X² -3X =4*(4X -3) -3X=16X -12 -3X=13X -12. So that works. So recurrence is valid.Therefore, R_n = 4R_{n-1} -3R_{n-2} with R_0 =1, R_1 = X. Alternatively, solving this linear recurrence relation. The characteristic equation is r² -4r +3=0. Which factors as (r -1)(r -3)=0. So roots r=1 and r=3. Therefore, general solution is R_n = A(1)^n + B(3)^n. But R_n is a linear polynomial, so perhaps expressed as R_n = a_n X + b_n. Let me see. Wait, maybe I need to model the coefficients a_n and b_n.Alternatively, since R_n = a_n X + b_n, then according to the recurrence R_n = 4R_{n-1} -3R_{n-2}, so:a_n X + b_n =4(a_{n-1}X + b_{n-1}) -3(a_{n-2}X + b_{n-2} )Which gives:a_n X + b_n = (4a_{n-1} -3a_{n-2})X + (4b_{n-1} -3b_{n-2}))Therefore, equating coefficients:a_n =4a_{n-1} -3a_{n-2}b_n =4b_{n-1} -3b_{n-2}So both a_n and b_n satisfy the same recurrence relation. Now, let's find initial conditions.For n=0: R_0 =1, so a_0 X + b_0 =1 ⇒ a_0=0, b_0=1.For n=1: R_1 =X ⇒ a_1=1, b_1=0.Then, for a_n:a_0=0, a_1=1a_n=4a_{n-1} -3a_{n-2}Similarly for b_n:b_0=1, b_1=0b_n=4b_{n-1} -3b_{n-2}So now, solving these linear recursions for a_n and b_n.First for a_n:Characteristic equation: r² -4r +3=0, roots 1 and 3.So general solution: a_n = C(1^n) + D(3^n)Using initial conditions:For n=0: a_0=0= C + DFor n=1: a_1=1= C +3DSolving:From first equation: C= -DSubstitute into second: -D +3D=1 ⇒2D=1⇒D=1/2, so C= -1/2Thus, a_n= (-1/2)(1^n) + (1/2)(3^n)= (3^n -1)/2Similarly for b_n:Same recurrence, so general solution b_n = E(1^n) + F(3^n)Initial conditions:For n=0: b_0=1= E + FFor n=1: b_1=0= E +3FSolving:From second equation: E= -3FSubstitute into first: -3F + F=1 ⇒ -2F=1 ⇒F= -1/2, so E= -3*(-1/2)= 3/2Thus, b_n= (3/2)(1^n) + (-1/2)(3^n)= (3 -3^n)/2Therefore, the remainder R_n= a_n X + b_n= [(3^n -1)/2] X + (3 -3^n)/2Which is the same as before. Let me check for n=2:R_2= (9 -1)/2 X + (3 -9)/2= (8/2)X + (-6/2)=4X -3. Which matches.For n=3:R_3= (27 -1)/2 X + (3 -27)/2=26/2 X + (-24)/2=13X -12. Which matches the earlier result.So the formula seems valid.Alternatively, using the initial method with plugging in roots:We had X=1 gives 1 = a + bX=3 gives 3^n= 3a + bThen solving:From 1: b=1 -aSubstitute into second equation:3a + (1 -a)=3^n ⇒2a +1=3^n ⇒2a=3^n -1 ⇒a=(3^n -1)/2Then b=1 - (3^n -1)/2=(2 -3^n +1)/2=(3 -3^n)/2. Yep, same result.Therefore, regardless of the method, we get the remainder as [(3^n -1)/2] X + (3 -3^n)/2. Let me simplify this expression.Factor out 1/2:R(X)= (1/2)[(3^n -1)X + (3 -3^n)] = (1/2)[3^n X -X +3 -3^n] = (1/2)[3^n (X -1) + (3 - X)]Alternatively, maybe not necessary. The answer can be written as:R(X) = frac{3^n -1}{2} X + frac{3 - 3^n}{2}But we can also combine terms:= frac{(3^n -1)X + 3 -3^n}{2}= frac{3^n X - X +3 -3^n}{2}= frac{3^n (X -1) + (3 - X)}{2}But perhaps leaving it as two separate terms is clearer.Alternatively, factor 3 -3^n = - (3^n -3), so:R(X) = frac{3^n -1}{2} X - frac{3^n -3}{2}But not sure if that's helpful.Alternatively, writing both terms over the same denominator:R(X) = frac{(3^n -1)X + 3 -3^n}{2}But maybe this is the simplest form. Let me verify with n=2 again:(9 -1)X +3 -9=8X -6, divided by 2:4X -3. Correct.Another check, n=1:(3 -1)X +3 -3=2X +0=2X. Wait, but for n=1, the remainder is X. Wait, this seems inconsistent. Wait, maybe my formula is incorrect?Wait, when n=1, R(X)= [(3^1 -1)/2] X + (3 -3^1)/2=(2/2)X + (0)/2= X +0= X. Which is correct. So n=1 gives X. That's right. So n=1 works. For n=0, it would give [(1 -1)/2]X + (3 -1)/2=0 +1=1. Correct.So the formula works for n>=0, but the problem states n>=2. However, even for n=0,1,2, it works.So conclusion: the remainder is [(3^n -1)/2] X + (3 -3^n)/2. Therefore, written as:R(X) = frac{3^n -1}{2} X + frac{3 - 3^n}{2}Alternatively, combining the terms:= frac{(3^n -1) X + 3 -3^n}{2}= frac{3^n X - X +3 -3^n}{2}= frac{3^n (X -1) + (3 - X)}{2}But the first expression is probably the clearest. So the remainder is ( (3^n -1)X + (3 -3^n) ) divided by 2. Hence, boxed, it would be:boxed{dfrac{(3^n - 1)X + (3 - 3^n)}{2}}Alternatively, maybe factor out 3^n -1? Wait, but that might not help. Alternatively, separate the terms:boxed{dfrac{3^n -1}{2} X + dfrac{3 -3^n}{2}}But both forms are correct, perhaps the first is better. Wait, in polynomial remainder, coefficients can be written as fractions. Let me check if the problem requires integer coefficients? Wait, the original polynomial has integer coefficients, and the remainder will have coefficients in fractions. But since 3^n -1 is even, because 3^n is odd, so 3^n -1 is even, so (3^n -1)/2 is integer. Similarly, 3 -3^n is -(3^n -3). For n>=2, 3^n -3 is divisible by 3(3^{n-1} -1). For example, when n=2, 9-3=6, which is 3*2, divisible by 2. (3^n -3)=3(3^{n-1}-1). For n>=2, 3^{n-1} -1 is even because 3^{n-1} is odd, so 3^{n-1} -1 is even. Therefore, 3(3^{n-1}-1) is divisible by 6. Therefore, 3 -3^n= -3(3^{n-1} -1). So (3 -3^n)/2= -3(3^{n-1} -1)/2. Since 3^{n-1} -1 is even, (3^{n-1} -1)/2 is integer, so (3 -3^n)/2 is -3 times integer. So coefficients are integers.Thus, the remainder has integer coefficients, as expected. Let me verify for n=2: (9-1)/2=4, (3-9)/2=-3. So 4X -3, which is correct.For n=3: (27-1)/2=13, (3-27)/2=-12. So 13X -12. Which is correct. So the coefficients are integers, as expected.Therefore, the remainder is ( (3^n -1)/2 ) X + ( (3 -3^n)/2 ). So in boxed form, that's:boxed{dfrac{3^n - 1}{2} X + dfrac{3 - 3^n}{2}}Alternatively, some might prefer writing it as frac{(3^n -1)X + (3 - 3^n)}{2} inside the box. Either form is acceptable. Let me check once more for n=4:Compute using the formula: (81 -1)/2 X + (3 -81)/2=80/2 X + (-78)/2=40X -39.Alternatively, compute using the recurrence:R_4=4R_3 -3R_2=4*(13X -12) -3*(4X -3)=52X -48 -12X +9=40X -39. Correct. So the formula holds. Therefore, confident that this is the correct remainder.Final AnswerThe remainder is boxed{dfrac{(3^n - 1)X + (3 - 3^n)}{2}}.</think>To determine the remainder of the Euclidean division of (X^n) by (X^2 - 4X + 3), we start by noting that the remainder will be a polynomial of degree less than 2, i.e., (aX + b). First, we find the roots of the polynomial (X^2 - 4X + 3), which are (X = 1) and (X = 3). Substituting these roots into the polynomial (X^n) gives us two equations:- For (X = 1): (1^n = a cdot 1 + b implies 1 = a + b)- For (X = 3): (3^n = a cdot 3 + b implies 3^n = 3a + b)We solve this system of equations:1. (a + b = 1)2. (3a + b = 3^n)Subtracting the first equation from the second gives:[2a = 3^n - 1 implies a = frac{3^n - 1}{2}]Substituting (a) back into the first equation:[frac{3^n - 1}{2} + b = 1 implies b = 1 - frac{3^n - 1}{2} = frac{2 - 3^n + 1}{2} = frac{3 - 3^n}{2}]Thus, the remainder (R(X)) is:[R(X) = left( frac{3^n - 1}{2} right) X + left( frac{3 - 3^n}{2} right)]Verification using the recurrence relation approach confirms the solution. The recurrence relation for the remainder (R_n = 4R_{n-1} - 3R_{n-2}) with initial conditions (R_0 = 1) and (R_1 = X) leads to the same result.Therefore, the remainder of the Euclidean division of (X^n) by (X^2 - 4X + 3) is:[boxed{dfrac{(3^n - 1)X + (3 - 3^n)}{2}}],<think>Okay, so I need to find the remainder when X^n is divided by X² - 4X + 3. Hmm, let me recall how polynomial division works. When you divide a polynomial by a quadratic, the remainder should be a first-degree polynomial, right? So, the remainder will look like aX + b, where a and b are constants. The task is to find these constants.First, let me note that X² - 4X + 3 can be factored. Let me check: the discriminant is 16 - 12 = 4, so roots are (4 ± 2)/2, which gives 3 and 1. So, the polynomial factors as (X - 1)(X - 3). That might be useful later.Now, when dividing X^n by X² - 4X + 3, the division algorithm tells us that there exists a quotient Q(X) and a remainder R(X) such that X^n = (X² - 4X + 3)Q(X) + R(X), where the degree of R(X) is less than 2. So, R(X) is of the form aX + b. To find a and b, maybe I can plug in the roots of the divisor polynomial into the equation. Since X = 1 and X = 3 make the divisor zero, substituting these into the equation should give me equations involving a and b.Let me try that. If I substitute X = 1 into the equation, I get:1^n = (0)Q(1) + R(1) => 1 = a*1 + b => a + b = 1.Similarly, substituting X = 3 gives:3^n = (0)Q(3) + R(3) => 3 = a*3 + b => 3a + b = 3^n.So now I have a system of two equations:1) a + b = 12) 3a + b = 3^nSubtracting the first equation from the second gives 2a = 3^n - 1, so a = (3^n - 1)/2. Then, substituting back into the first equation, b = 1 - a = 1 - (3^n - 1)/2 = (2 - 3^n + 1)/2 = (3 - 3^n)/2.Therefore, the remainder R(X) = [(3^n - 1)/2]X + (3 - 3^n)/2.Let me check if this makes sense for small n. Let's try n = 2. Then X² divided by X² - 4X + 3 should have remainder 4X - 3. Let's compute using our formula:a = (9 - 1)/2 = 4, b = (3 - 9)/2 = -3. So R(X) = 4X - 3. Which is correct, since X² = (X² -4X +3) + 4X -3. Yep, that works.Another test case: n=3. Then X³ divided by X² -4X +3. Let's perform the division manually.Divide X³ by X² -4X +3. The first term is X. Multiply divisor by X: X³ -4X² +3X. Subtract from X³: 0 +4X² -3X. Now, divide 4X² by X², which gives 4. Multiply divisor by 4: 4X² -16X +12. Subtract: (4X² -3X) - (4X² -16X +12) = 13X -12. So remainder is 13X -12.Using our formula: a = (3^3 -1)/2 = (27 -1)/2 =26/2=13, b=(3 -27)/2=(-24)/2=-12. So R(X)=13X -12. Correct. Nice.Another one, n=1. Wait, but n >=2. So maybe n=2 and n=3 are sufficient.Thus, the general formula seems to hold. Therefore, the remainder is [(3^n -1)/2]X + (3 -3^n)/2. To write this as a single expression, factor out 1/2:(1/2)( (3^n -1)X + 3 -3^n ) = (1/2)(3^n X - X +3 -3^n ) = (1/2)(3^n (X -1) + ( -X +3 )).But maybe leave it as is. So, the remainder is ( (3^n -1)X + (3 -3^n) ) / 2.Alternatively, factor out (X -1):But the question just asks for the remainder. So perhaps writing it as (3^n -1)/2 X + (3 -3^n)/2. Alternatively, we can write it as:[(3^n - 1)X + (3 - 3^n)] / 2.Alternatively, we can factor out (3^n -1)/2 from both terms:(3^n -1)/2 * X + (3 -3^n)/2 = (3^n -1)/2 (X - 1) + (3 -3^n)/2 + (3^n -1)/2 *1.Wait, maybe not necessary. The answer is probably expected in the form aX + b, which is ( (3^n -1)X + (3 -3^n) ) divided by 2.Alternatively, simplifying:(3^n -1)X/2 + (3 -3^n)/2 = (3^n X - X +3 -3^n)/2 = (3^n (X -1) + (3 - X))/2.But perhaps the first form is better.Alternatively, let me check for n=2:(3^2 -1)/2 X + (3 -3^2)/2 = (9 -1)/2 X + (3 -9)/2 = 8/2 X + (-6)/2 = 4X -3. Which is correct.Similarly, n=3: (27 -1)/2 X + (3 -27)/2 = 26/2 X + (-24)/2 =13X -12. Correct.So, the formula works. Therefore, the remainder is ( (3^n -1)X + (3 -3^n) ) / 2.Alternatively, we can factor 3^n:(3^n (X -1) + ( -X +3 )) / 2.But unless there's a simpler form, the first expression is fine.Hence, the remainder is ( (3^n -1)X + (3 -3^n) ) divided by 2. So, boxed form would be boxed{dfrac{(3^n - 1)X + (3 - 3^n)}{2}}.Alternatively, written as boxed{frac{3^n -1}{2} X + frac{3 -3^n}{2}}.But the problem says "determine the remainder", and since the remainder is a polynomial, both forms are correct. Maybe the first expression is preferable as a single fraction.Alternatively, since both terms have a denominator of 2, we can write:boxed{frac{(3^n - 1)X + (3 - 3^n)}{2}}.Yes, that should be the answer.Final AnswerThe remainder is boxed{dfrac{(3^n - 1)X + (3 - 3^n)}{2}}.</think>To determine the remainder of the Euclidean division of (X^n) by (X^2 - 4X + 3), we note that the remainder must be a first-degree polynomial of the form (aX + b). The polynomial (X^2 - 4X + 3) factors as ((X - 1)(X - 3)), so we use the roots (X = 1) and (X = 3) to set up a system of equations.Substituting (X = 1) into the equation (X^n = (X^2 - 4X + 3)Q(X) + aX + b) gives:[1^n = a cdot 1 + b implies a + b = 1.]Substituting (X = 3) into the same equation gives:[3^n = a cdot 3 + b implies 3a + b = 3^n.]We solve this system of equations:1. (a + b = 1)2. (3a + b = 3^n)Subtracting the first equation from the second, we get:[2a = 3^n - 1 implies a = frac{3^n - 1}{2}.]Substituting (a) back into the first equation:[b = 1 - a = 1 - frac{3^n - 1}{2} = frac{2 - 3^n + 1}{2} = frac{3 - 3^n}{2}.]Thus, the remainder is:[frac{3^n - 1}{2}X + frac{3 - 3^n}{2}.]The final answer is:[boxed{dfrac{(3^n - 1)X + (3 - 3^n)}{2}}.]]

🔑:Note that X^{2}-4 X+3=(X-1)(X-3). Let Q_{n} and R_{n} be the quotient and the remainder of X^{n} when divided by (X-1)(X-3), so we have X^{n}=Q_{n}(X-1)(X-3)+R_{n}. By evaluating at 1 and 3, we get R_{n}(1)=1^{n}=1 and R_{n}(3)=3^{n}.Furthermore, the degree of R_{n} is strictly less than the degree of (X-3)(X-1), which is 2. Therefore, we can write R_{n} in the form a_{n} X+b_{n}. Thus, we have 3 a_{n}+b_{n}=3^{n} and a_{n}+b_{n}=1. By subtracting these equations, we obtain a_{n}=frac{3^{n}-1}{2}, and then b_{n}=1-a_{n}=frac{3-3^{n}}{2}, so R_{n}=frac{3^{n}-1}{2} X+frac{3-3^{n}}{2}.

🧠:[<think>Okay, so I have this problem here where there are 12 white balls lying in a groove, and I need to figure out in how many ways I can place 8 black balls among them such that no two black balls end up next to each other. Hmm, let me think about how to approach this. First, the problem is about arranging black balls among white ones without having two black balls adjacent. That sounds like a classic combinatorics problem. I remember something about arranging objects with restrictions, maybe using combinations? Let me recall. If we want to place objects such that they are not adjacent, we can use the idea of "slots" between the existing objects. Wait, in this case, the white balls are already lying in a groove. So there are 12 white balls, which creates 13 gaps where the black balls could go: one before the first white ball, one between each pair of white balls, and one after the last white ball. But if we want to place the black balls so that no two are next to each other, each black ball must go into a separate gap. So if there are 13 gaps and we need to choose 8 of them to place a black ball, the number of ways would be the combination of 13 choose 8. But wait, let me check if that's correct. Let me visualize this. Imagine the 12 white balls as W W W W W W W W W W W W. The gaps would be like: _ W _ W _ W ... W _. So there are indeed 13 gaps. If we need to place 8 black balls in these gaps such that no two are adjacent, each black ball goes into a different gap. Therefore, the number of ways is C(13,8). Calculating that, C(13,8) is 13! / (8! * (13-8)!) = 1287. But wait, let me verify if this is the right approach. Sometimes these problems have a twist. For instance, if the groove is circular, the calculation would be different, but the problem says "along a groove," which I assume is linear, not circular. So that shouldn't affect the number of gaps. Another thing to consider: does the order of the black balls matter? Since the black balls are indistinct, right? They are all black, so we're just choosing positions, not arranging different objects. So yes, combinations are the way to go. Let me think again. If we have n white balls, the number of gaps is n+1. Then placing k black balls in those gaps without two being adjacent is C(n+1, k). So here, n=12, k=8. Therefore, the answer is C(13,8). But let me compute that. 13 choose 8 is equal to 13 choose 5 (since C(n,k) = C(n, n−k)), which is (13×12×11×10×9)/(5×4×3×2×1). Let's compute that step by step. 13×12 = 156156×11 = 17161716×10 = 1716017160×9 = 154440Then divide by 5×4×3×2×1=120154440 ÷ 120: Let's divide 154440 by 10 first, which is 15444, then divide by 12: 15444 ÷ 12 = 1287. So yes, 1287. Therefore, the number of ways should be 1287. But let me check if there's another way to approach this problem to confirm. Alternatively, maybe using the inclusion-exclusion principle. If we first calculate the total number of ways to place 8 black balls among 12+8=20 positions (since each black and white ball takes a position) and then subtract the arrangements where at least two black balls are adjacent. But that might be more complicated. Let's see. Total positions would be 12 + 8 = 20. The total number of ways to choose 8 positions for black balls is C(20,8). Then subtract the number of arrangements where at least one pair of black balls is adjacent. However, inclusion-exclusion for this can get messy because subtracting the cases with one pair, adding back the cases with two pairs, etc. Considering that 8 black balls with no two adjacent is a standard problem, the first method is more straightforward. In the standard problem, arranging k non-adjacent objects in n positions is equivalent to choosing k positions with at least one gap between them. The formula is C(n - k +1, k). Wait, but in our case, the white balls are fixed. So maybe it's equivalent to placing black balls in the gaps between white balls. So the number of gaps is 13, and choosing 8 of them, which is C(13,8). Alternatively, another way to model this is to consider the white balls as fixed and inserting black balls into the gaps. Each black ball must go into a separate gap, so it's like distributing 8 identical balls into 13 distinct boxes, each box holding at most one ball. That's exactly C(13,8). Alternatively, imagine arranging all the black balls first, then placing white balls in between. But since the white balls are already there, maybe that's not the right approach. Wait, let me think again. If we have to place 8 black balls among 12 white balls such that no two are adjacent, the standard formula for this is C(n - k +1, k) where n is the number of white balls. Wait, but that formula is when you have n objects and you want to place k non-adjacent objects. Wait, no. Let me recall. If you have n white balls, the number of ways to place k black balls so that no two are adjacent is C(n +1, k). Because you have n+1 gaps. So yes, that's the same as before. So n=12, k=8, so C(13,8). Therefore, 1287. So the answer should be 1287. But let me just confirm with a smaller case. Suppose there are 2 white balls. Then, how many ways to place 1 black ball without adjacent? There are 3 gaps: _ W _ W _. So C(3,1)=3. Which makes sense: before the first white, between them, after the last. If we have 3 white balls, and want to place 2 black balls without adjacency. Then gaps: 4 gaps. Choose 2: C(4,2)=6. Let's list them:Positions: 1 & 2, 1 &3, 1&4, 2&3, 2&4, 3&4. Each corresponds to placing black balls in different gaps. That works. So the formula seems correct. Therefore, for 12 white balls and 8 black balls, it's C(13,8)=1287. So the answer is 1287. Wait, but hold on. Let's check if we have enough gaps. For 12 white balls, there are 13 gaps. We need to place 8 black balls, each in a separate gap. So 13 choose 8. That works. If the number of black balls were more than 13, then it's impossible, but here 8 is less than 13, so it's possible. Therefore, the answer is 1287. Final AnswerThe number of ways is boxed{1287}.</think>To determine the number of ways to place 8 black balls among 12 white balls such that no two black balls are adjacent, we consider the gaps between the white balls. With 12 white balls lying in a groove, there are 13 possible gaps (one before each white ball, one between each pair of white balls, and one after the last white ball). We need to choose 8 of these 13 gaps to place the black balls. This is a combinatorial problem where we need to calculate the number of combinations of 13 gaps taken 8 at a time.The number of ways to choose 8 gaps out of 13 is given by the binomial coefficient:[binom{13}{8}]Calculating this, we get:[binom{13}{8} = frac{13!}{8!(13-8)!} = frac{13!}{8!5!}]Simplifying further, we recognize that (binom{13}{8} = binom{13}{5}), which is easier to compute:[binom{13}{5} = frac{13 times 12 times 11 times 10 times 9}{5 times 4 times 3 times 2 times 1} = 1287]Thus, the number of ways to place 8 black balls among 12 white balls such that no two black balls are adjacent is (boxed{1287}).,<think>Okay, so I need to figure out how many ways I can place 8 black balls among 12 white balls arranged in a groove such that no two black balls are next to each other. Hmm, let me think. First, the white balls are already lying in a groove, so they are in a straight line and fixed in position. The task is to place the black balls in between them or at the ends without having two black balls adjacent. I remember that this is similar to some combinatorics problems I've seen before, maybe about arranging objects with certain restrictions.Let me start by visualizing the situation. There are 12 white balls in a row. So, let's represent them as W W W W W W W W W W W W. Now, I need to place 8 black balls (let's denote them as B) in between these white balls such that no two B's are next to each other. To ensure that no two black balls are adjacent, each black ball must be placed in a separate gap between the white balls. Additionally, there are also the ends of the row where we can place black balls. How many gaps are there? Well, if there are 12 white balls, then there are 11 gaps between them. Plus the two ends, so total gaps are 11 + 2 = 13. Wait, is that right? Let me check. If you have n objects in a line, the number of gaps between them is n-1, and including the two ends, it's n+1. So here, n=12 white balls, so gaps = 12 + 1 = 13. Yes, that's correct. So there are 13 possible positions where we can place a black ball, each position being either between two white balls or at the very start or end of the row.Since we don't want any two black balls to be adjacent, each black ball must go into a separate gap. Therefore, the problem reduces to choosing 8 distinct gaps out of these 13 where we can place the black balls. Ah, so this is a combination problem. The number of ways to choose 8 positions out of 13 without any restrictions is C(13,8), which is the combination formula. Since combinations are used when the order doesn't matter, which is the case here because the black balls are indistinct (assuming they are identical). Wait, let me confirm. Are the black balls distinguishable? The problem says "8 black balls", but doesn't specify if they are different. In most combinatorics problems like this, unless specified, we assume the objects are identical. So yes, they are identical, so the number of ways is just the number of ways to choose positions, which is C(13,8). But let me make sure I didn't overlook anything. Let's re-read the problem: "In how many ways can 8 black balls be placed among them so that no two black balls end up next to each other?" The key points are that the white balls are lying in a groove (so fixed positions), and we need to place 8 black balls such that none are adjacent. Another way to think about it is using the stars and bars method, but with restrictions. Wait, stars and bars is for distributing identical objects into distinct bins. Here, if we consider the gaps between white balls as bins, and the black balls as stars, then each bin can hold at most one star (to prevent two black balls being adjacent). So the number of ways is C(13,8) as before. Alternatively, we can model this as arranging the black and white balls with the given restrictions. If we have 12 white balls, and we want to insert 8 black balls such that they are not adjacent, we can think of first placing the 12 white balls, which creates 13 gaps (as previously established), and then choosing 8 of those gaps to place a black ball each. Therefore, the formula is indeed C(n + 1, k), where n is the number of white balls and k is the number of black balls. But wait, let me check that. If we have n white balls, then the number of gaps is n + 1. To place k black balls such that they are not adjacent, the formula is C(n + 1, k). In this case, n=12, so C(13,8). Alternatively, another approach is to think of arranging all the balls in a line where no two black balls are adjacent. The total number of balls would be 12 white + 8 black = 20 balls. But since the white balls are fixed, maybe this approach isn't directly applicable. Wait, no. If the white balls are fixed, their positions are already determined. So we need to place the black balls in the gaps between them. So yes, the key idea is that the white balls are fixed, creating 13 possible slots for the black balls, and we need to choose 8 of these 13 slots. Therefore, the answer should be C(13,8). Let me compute that. C(13,8) is calculated as 13! / (8! * (13-8)!) = 13! / (8!5!) Calculating this: 13 × 12 × 11 × 10 × 9 × 8! / (8! × 5 × 4 × 3 × 2 × 1) The 8! cancels out, so we have:(13 × 12 × 11 × 10 × 9) / (5 × 4 × 3 × 2 × 1) Calculating numerator: 13 × 12 = 156; 156 × 11 = 1716; 1716 × 10 = 17160; 17160 × 9 = 154,440Denominator: 5 × 4 = 20; 20 × 3 = 60; 60 × 2 = 120; 120 × 1 = 120So 154,440 / 120 = let's divide step by step:154,440 ÷ 10 = 15,444; 15,444 ÷ 12 = 1,287. So wait, 15,444 ÷ 12: 12 × 1,200 = 14,400; 15,444 - 14,400 = 1,044; 1,044 ÷ 12 = 87. So total is 1,200 + 87 = 1,287. Therefore, C(13,8) = 1,287. So the answer should be 1,287. Wait, but let me confirm the arithmetic again because it's easy to make a mistake here. Let's compute 13 × 12 × 11 × 10 × 9 divided by 120. Alternatively, compute step by step:13/1 = 1313 × 12 = 156156 × 11 = 1,7161,716 × 10 = 17,16017,160 × 9 = 154,440Then divide 154,440 by 120:Divide numerator and denominator by 10: 15,444 / 1215,444 divided by 12: 12 × 1,000 = 12,000; 15,444 - 12,000 = 3,4443,444 divided by 12: 12 × 200 = 2,400; 3,444 - 2,400 = 1,0441,044 divided by 12: 12 × 87 = 1,044So total is 1,000 + 200 + 87 = 1,287. Correct. Alternatively, 12 × 1,287 = 12 × 1,200 + 12 × 87 = 14,400 + 1,044 = 15,444. Yes, so 154,440 ÷ 120 = 1,287. So the combination is 1,287.Therefore, the number of ways is 1,287. But let me check once again if the approach is correct. Wait, another way to think about this problem: if the white balls are fixed, then to place the black balls such that they are not adjacent, we need to have at least one white ball between any two black balls. However, since we have 12 white balls, inserting 8 black balls with at least one white between them would require how many white balls? Let's see: If we have 8 black balls, then to separate them, we need at least 7 white balls between them. So total white balls needed would be 7. But here, we have 12 white balls, which is more than enough. Therefore, the remaining 12 - 7 = 5 white balls can be distributed as additional spacers. Wait, this might be a different approach. Let's think of it as arranging the 8 black balls with at least one white ball between each. So first, place 8 black balls with one white ball between each. That would take up 8 black + 7 white = 15 balls. But we have 12 white balls, so we are short by 12 - 7 = 5 white balls. These 5 white balls can be distributed as additional spacers in the gaps between the black balls, including the ends. The number of gaps here is 8 + 1 = 9 gaps (since there are 8 black balls, there are 9 gaps: one before the first black, one between each pair, and one after the last). But wait, this approach might not be applicable here because the white balls are already fixed in position. Wait, maybe I confused the problem. Let me clarify. The original problem states that there are 12 white balls lying along a groove, and we need to place 8 black balls among them such that no two black balls are adjacent. So the white balls are fixed; their positions are not to be changed. Therefore, the approach where we consider the gaps between white balls is correct. The black balls are to be placed in these gaps, with each gap containing at most one black ball. Since there are 13 gaps, choosing 8 of them gives the required number of arrangements. Alternatively, if we were to arrange both black and white balls with the restriction, the formula is C(n - k + 1, k) where n is the total number of balls? Wait, no. Let me think again. Suppose we have to arrange 12 white and 8 black balls in a line such that no two black balls are adjacent. The number of ways would be C(12 + 1, 8) = C(13,8), which matches our previous result. Because the formula for arranging k non-adjacent objects among n objects is C(n + 1, k). So in this case, since the white balls are fixed, the number of ways is C(13,8). Therefore, the answer is 1,287. But just to ensure I haven't mixed up the formula, let me recall that when arranging k non-adjacent objects among n fixed objects, the number of ways is indeed C(n + 1, k). For example, if you have n white balls, the number of ways to place k black balls so that no two are adjacent is C(n + 1, k). This is because you have n + 1 gaps (including the ends) and you need to choose k of them. Yes, that makes sense. For example, if you have 3 white balls, there are 4 gaps (including ends), and the number of ways to place 2 black balls without adjacency is C(4,2) = 6. Let's verify:Positions: [W1] [W2] [W3]Gaps: before W1, between W1-W2, between W2-W3, after W3. So 4 gaps. Choosing 2: possible placements:1. Before W1 and between W1-W22. Before W1 and between W2-W33. Before W1 and after W34. Between W1-W2 and between W2-W35. Between W1-W2 and after W36. Between W2-W3 and after W3Which gives 6 ways, matching C(4,2)=6. So the formula works here.Therefore, returning to the original problem, with 12 white balls, there are 13 gaps, and we need to choose 8. So the answer is C(13,8) = 1,287. But let me consider another perspective to double-check. Suppose we first place the 8 black balls in the groove. To ensure they are not adjacent, we need to have at least one white ball between each pair. So between the 8 black balls, there must be at least 7 white balls. Since there are 12 white balls in total, subtracting these 7 gives 12 - 7 = 5 white balls remaining. These 5 can be freely distributed in the 8 + 1 = 9 available gaps (before the first black ball, between each pair, and after the last black ball). So the problem reduces to distributing 5 identical white balls into 9 distinct gaps, which is a stars and bars problem. The formula for this is C(5 + 9 - 1, 5) = C(13,5). Wait a minute, this is the same as C(13,8), because C(n, k) = C(n, n - k). Here, C(13,5) = C(13,8). Since 13 - 5 = 8. So 13 choose 5 is equal to 13 choose 8, which is 1,287. Therefore, both approaches give the same result, confirming that the answer is indeed 1,287. So, to recap: when placing 8 black balls among 12 fixed white balls such that no two are adjacent, we can either:1. Consider the 13 gaps created by the 12 white balls and choose 8 of them to place a black ball each, resulting in C(13,8) ways.2. Model the problem as distributing the remaining 5 white balls into the gaps around the 8 black balls, which also results in C(13,5) = C(13,8) ways.Either way, we arrive at the same answer, which gives me confidence that 1,287 is the correct number of ways.Therefore, the final answer is 1,287, which can be written as boxed{1287}.Final Answerboxed{1287}</think>To determine the number of ways to place 8 black balls among 12 white balls such that no two black balls are adjacent, we start by noting that the 12 white balls create 13 gaps (11 between the white balls and 2 at the ends). We need to choose 8 of these 13 gaps to place the black balls.This is a combination problem, where we need to calculate the number of ways to choose 8 positions out of 13. The formula for combinations is given by:[binom{n}{k} = frac{n!}{k!(n-k)!}]Substituting ( n = 13 ) and ( k = 8 ):[binom{13}{8} = frac{13!}{8! cdot 5!}]Calculating the factorial values step-by-step:[binom{13}{8} = frac{13 times 12 times 11 times 10 times 9}{5 times 4 times 3 times 2 times 1} = frac{154440}{120} = 1287]Thus, the number of ways to place 8 black balls among 12 white balls such that no two black balls are adjacent is (boxed{1287}).]

🔑:85. Among 12 white balls, there are 13 positions where black balls can be placed. This can be done in C_{13}^{8}= =1287 ways.

🧠:[<think>Okay, so I have this problem here. Let's see. It's about a function f(x) = 2|x - 1| - |2x + m|, and it's given that the maximum value of this function is 3. Then part (Ⅰ) asks to find the value of the real number m, and part (Ⅱ) is about finding the minimum value of a² + b² + c² given that a - 2b + c = m. Hmm, okay. Let's tackle part (Ⅰ) first.First, the function is f(x) = 2|x - 1| - |2x + m|. Since it's a combination of absolute value functions, the maximum value likely occurs at one of the "corner" points where the expressions inside the absolute values change their sign, that is, when x - 1 = 0 or 2x + m = 0. So the critical points are x = 1 and x = -m/2.So, I should probably analyze the function in intervals determined by these critical points. Depending on the value of m, the critical point x = -m/2 could be to the left or right of x = 1. Let's consider different cases based on the position of -m/2 relative to 1.But first, since m is a real number and m ≥ 0, as given. So m is non-negative. Then -m/2 is non-positive. So x = -m/2 is less than or equal to 0. Therefore, x = -m/2 is always to the left of x = 0, and x = 1 is to the right. So the critical points are at x = -m/2 (≤0) and x = 1. Therefore, the intervals we need to consider are:1. x < -m/22. -m/2 ≤ x < 13. x ≥ 1In each of these intervals, the expressions inside the absolute values can be simplified because we know their sign. Let's analyze each interval.Case 1: x < -m/2In this interval, both x - 1 and 2x + m are negative.So |x - 1| = -(x - 1) = -x + 1|2x + m| = -(2x + m) = -2x - mTherefore, f(x) = 2(-x + 1) - (-2x - m) = -2x + 2 + 2x + m = ( -2x + 2x ) + (2 + m) = 0x + (2 + m) = 2 + mSo in this interval, f(x) is constant 2 + m.Case 2: -m/2 ≤ x < 1In this interval, 2x + m is non-negative (since x ≥ -m/2), but x - 1 is still negative (since x < 1).So |x - 1| = -(x - 1) = -x + 1|2x + m| = 2x + mTherefore, f(x) = 2(-x + 1) - (2x + m) = -2x + 2 - 2x - m = (-2x - 2x) + (2 - m) = -4x + (2 - m)So here, f(x) is a linear function with slope -4, decreasing as x increases.Case 3: x ≥ 1In this interval, both x - 1 and 2x + m are non-negative.So |x - 1| = x - 1|2x + m| = 2x + mTherefore, f(x) = 2(x - 1) - (2x + m) = 2x - 2 - 2x - m = (2x - 2x) + (-2 - m) = 0x + (-2 - m) = -2 - mSo in this interval, f(x) is constant -2 - m.Now, let's summarize the behavior of f(x):- For x < -m/2, f(x) = 2 + m (constant)- For -m/2 ≤ x < 1, f(x) = -4x + (2 - m) (linear decreasing)- For x ≥ 1, f(x) = -2 - m (constant)We need to find the maximum value of f(x). The maximum occurs either in the first interval or at the point where the linear function starts in the second interval. Let's check.In the first interval, the value is constant 2 + m. In the third interval, it's constant -2 - m. Since m ≥ 0, 2 + m is positive, and -2 - m is negative, so the maximum must be in the first interval or in the second interval.In the second interval, the function is linear decreasing from x = -m/2 to x = 1. So the maximum in the second interval would occur at x = -m/2. Let's compute the value there.At x = -m/2, since it's the start of the interval, we can compute f(-m/2). But wait, in the first interval, x < -m/2, so approaching from the left, f(x) is 2 + m. At x = -m/2, we enter the second interval. Let's compute f(-m/2):From the second interval's expression: f(-m/2) = -4*(-m/2) + (2 - m) = 2m + 2 - m = m + 2Which matches the first interval's value. So at x = -m/2, the function is continuous and has value 2 + m.Then, as x increases from -m/2 to 1, f(x) decreases from 2 + m to f(1). Let's compute f(1):From the second interval's expression, at x = 1 (but x = 1 is not included in the second interval; it's the start of the third interval). But approaching x = 1 from the left:f(1-) = -4*(1) + (2 - m) = -4 + 2 - m = -2 - mWhich matches the third interval's constant value. So the function is decreasing from 2 + m to -2 - m over the interval [-m/2, 1). Therefore, the maximum value of f(x) is 2 + m, achieved for all x ≤ -m/2, and the minimum is -2 - m, achieved for x ≥ 1.But the problem states that the maximum value of the function is 3. So 2 + m = 3. Solving for m:2 + m = 3 => m = 1Wait, is that all? Let me check again.Wait, but hold on. The function is 2|x - 1| - |2x + m|. Let me verify if there's another critical point or maybe the slope in the second interval could lead to a higher maximum.Wait, in the second interval, the function is linear with slope -4, so it's decreasing. Therefore, its maximum in that interval is at x = -m/2, which is 2 + m, and then it decreases. So the only maximum is 2 + m. Therefore, if the maximum value is given as 3, then m = 1.But let's check for m = 1. If m = 1, then the critical points are at x = -1/2 and x = 1. Then in the intervals:x < -1/2: f(x) = 2 + 1 = 3-1/2 ≤ x < 1: f(x) = -4x + 2 - 1 = -4x + 1At x = -1/2, f(x) = -4*(-1/2) + 1 = 2 + 1 = 3At x approaching 1 from the left, f(x) approaches -4*(1) + 1 = -3And for x ≥ 1: f(x) = -2 -1 = -3So the maximum value is indeed 3 when m = 1. So m = 1 is the answer for part (Ⅰ).Wait, but maybe there's a case where m could be different? Let's see. Suppose that the maximum could be achieved somewhere else. Let's check.Suppose m is such that when we analyze the function in another way. For example, maybe if m is negative, but the problem says m ≥ 0, so m is non-negative. So we don't need to consider m negative.Alternatively, maybe we can use another approach. Let's consider the function f(x) = 2|x - 1| - |2x + m|.We can write f(x) as follows:f(x) = 2|x - 1| - |2x + m|Let me try to analyze this function by considering different expressions. Alternatively, maybe express it in terms of y = 2x.Wait, not sure. Alternatively, perhaps express 2|x - 1| as |2x - 2|. Wait, but 2|x - 1| is not the same as |2x - 2|. Because 2|x - 1| = |2x - 2|, yes. Because for any real number a, 2|a| = |2a|. So, 2|x -1| = |2x - 2|.So f(x) = |2x - 2| - |2x + m|So f(x) = |2x - 2| - |2x + m|This might be useful. So f(x) can be written as the difference of two absolute values of linear functions in x. The maximum of such a function can be found by considering the points where the expressions inside the absolute values change sign, which is at x = 1 and x = -m/2 as before.Alternatively, let's consider the function f(x) = |2x - 2| - |2x + m|. Let me denote t = 2x. Then f(x) can be written as |t - 2| - |t + m|. Let's define g(t) = |t - 2| - |t + m|. Then f(x) = g(2x).Since x can be any real number, t can be any real number as well (since t = 2x covers all real numbers as x does). Therefore, the maximum of f(x) is the same as the maximum of g(t). So we can analyze g(t) instead.So, g(t) = |t - 2| - |t + m|We need to find the maximum value of g(t) over all real t, given m ≥ 0, and set it equal to 3, then solve for m.Let's analyze g(t). The function g(t) is the difference of two absolute values. Let's find its maximum.The function g(t) can be considered in different intervals based on the critical points t = 2 and t = -m.Case 1: t < -mIn this case, |t - 2| = -(t - 2) = -t + 2|t + m| = -(t + m) = -t - mSo g(t) = (-t + 2) - (-t - m) = (-t + 2) + t + m = ( -t + t ) + (2 + m ) = 0 + 2 + m = 2 + mCase 2: -m ≤ t < 2In this case, |t - 2| = -(t - 2) = -t + 2|t + m| = t + mSo g(t) = (-t + 2) - (t + m) = -t + 2 - t - m = -2t + (2 - m)Case 3: t ≥ 2In this case, |t - 2| = t - 2|t + m| = t + mSo g(t) = (t - 2) - (t + m) = t - 2 - t - m = -2 - mTherefore, similar to before:- For t < -m: g(t) = 2 + m (constant)- For -m ≤ t < 2: g(t) = -2t + (2 - m) (linear decreasing)- For t ≥ 2: g(t) = -2 - m (constant)The maximum value of g(t) is 2 + m, achieved when t < -m. Then, as t increases, g(t) decreases from 2 + m at t = -m to g(2) = -2 - m. Since the maximum of g(t) is 2 + m, and this is equal to the maximum of f(x), which is given as 3. Therefore:2 + m = 3 ⇒ m = 1So, this confirms the previous result. Therefore, m = 1 is indeed the answer for part (Ⅰ).Okay, moving on to part (Ⅱ). Given real numbers a, b, c satisfying a - 2b + c = m (which is 1), find the minimum value of a² + b² + c².So, the problem is now: minimize a² + b² + c² subject to the constraint a - 2b + c = 1.This is a classic problem of finding the minimum distance from the origin to the plane a - 2b + c = 1 in three-dimensional space. The minimum value of a² + b² + c² is the square of the distance from the origin to the plane.The formula for the distance from a point (x₀, y₀, z₀) to the plane Ax + By + Cz + D = 0 is |Ax₀ + By₀ + Cz₀ + D| / sqrt(A² + B² + C²). In our case, the plane is a - 2b + c - 1 = 0, so A = 1, B = -2, C = 1, D = -1. The origin is (0, 0, 0), so substituting:Distance = |1*0 + (-2)*0 + 1*0 -1| / sqrt(1² + (-2)² + 1²) = | -1 | / sqrt(1 + 4 + 1) = 1 / sqrt(6)Therefore, the minimum distance is 1/sqrt(6), so the minimum value of a² + b² + c² is (1/sqrt(6))² = 1/6.Alternatively, using the method of Lagrange multipliers. Let me confirm.Let’s consider the function F(a, b, c) = a² + b² + c² with the constraint G(a, b, c) = a - 2b + c - 1 = 0.The method of Lagrange multipliers tells us that at the minimum, the gradient of F is proportional to the gradient of G. Therefore:∇F = λ∇GCompute gradients:∇F = (2a, 2b, 2c)∇G = (1, -2, 1)Thus:2a = λ*1 → a = λ/22b = λ*(-2) → b = -λ2c = λ*1 → c = λ/2Now, substitute a, b, c into the constraint equation:a - 2b + c = 1(λ/2) - 2*(-λ) + (λ/2) = 1λ/2 + 2λ + λ/2 = 1Combine terms:(λ/2 + λ/2) + 2λ = 1 → λ + 2λ = 1 → 3λ = 1 → λ = 1/3Therefore, a = (1/3)/2 = 1/6, b = -1/3, c = (1/3)/2 = 1/6Compute a² + b² + c²:(1/6)² + (-1/3)² + (1/6)² = 1/36 + 1/9 + 1/36 = (1 + 4 + 1)/36 = 6/36 = 1/6So, the minimum value is indeed 1/6. Therefore, the answer to part (Ⅱ) is 1/6.Alternatively, another way to think about it is that the minimum of a² + b² + c² under the linear constraint a - 2b + c = 1 is given by the square of the distance from the origin to the plane, which is (|1| / sqrt(1² + (-2)² + 1²))² = (1 / sqrt(6))² = 1/6. So same result.Therefore, part (Ⅱ) answer is 1/6.I think that's all. Let me just check if there's any possible mistake here. For part (Ⅰ), we considered the function and broke it down into intervals, checked the maximum, found m=1. For part (Ⅱ), used either Lagrange multipliers or distance formula. Both methods gave 1/6. Seems solid.Final AnswerⅠ. boxed{1}Ⅱ. boxed{dfrac{1}{6}}</think>Given the function ( f(x) = 2|x - 1| - |2x + m| ) with a maximum value of 3, we need to find the value of ( m ) and the minimum value of ( a^2 + b^2 + c^2 ) given ( a - 2b + c = m ). Part (Ⅰ)1. Analyze the function ( f(x) ) in different intervals determined by the critical points ( x = 1 ) and ( x = -frac{m}{2} ).2. For ( x < -frac{m}{2} ), ( f(x) = 2 + m ).3. For ( -frac{m}{2} leq x < 1 ), ( f(x) = -4x + (2 - m) ), which is a linear function decreasing from ( 2 + m ) to ( -2 - m ).4. For ( x geq 1 ), ( f(x) = -2 - m ).5. The maximum value of ( f(x) ) is ( 2 + m ), which is given as 3. Solving ( 2 + m = 3 ) gives ( m = 1 ).Thus, the value of ( m ) is (boxed{1}). Part (Ⅱ)1. Given ( a - 2b + c = 1 ), we need to find the minimum value of ( a^2 + b^2 + c^2 ).2. This is equivalent to finding the minimum distance from the origin to the plane ( a - 2b + c = 1 ) in 3D space.3. Using the formula for the distance from a point to a plane, the distance is ( frac{|1|}{sqrt{1^2 + (-2)^2 + 1^2}} = frac{1}{sqrt{6}} ).4. The square of this distance is ( left( frac{1}{sqrt{6}} right)^2 = frac{1}{6} ).Thus, the minimum value of ( a^2 + b^2 + c^2 ) is (boxed{dfrac{1}{6}}).,<think>Okay, so I need to solve this problem where there's a function f(x) = 2|x - 1| - |2x + m|, and it's given that the maximum value of this function is 3. Then part (I) asks for the value of m, and part (II) is about finding the minimum value of a² + b² + c² given that a - 2b + c = m. Let's start with part (I).First, I need to analyze the function f(x). It's made up of absolute value terms, which can be tricky because of their piecewise nature. Absolute value functions have corners where the expression inside the absolute value changes sign. So, to handle f(x) = 2|x - 1| - |2x + m|, I should find the points where the expressions inside the absolute values are zero. That is, x - 1 = 0 and 2x + m = 0. Solving these gives x = 1 and x = -m/2. These are the critical points where the function might change its slope.So, the critical points are at x = 1 and x = -m/2. Depending on the value of m, these points could be in different orders on the number line. For example, if m is positive, then -m/2 is negative, so x = -m/2 is to the left of x = 0, and x = 1 is to the right. If m is zero, then both critical points are at x = 1 and x = 0. Wait, but m is given to be greater than or equal to 0. So, m ≥ 0. So, x = -m/2 will be less than or equal to 0. Therefore, the critical points are at x = -m/2 (which is ≤ 0) and x = 1. So, these divide the real line into three intervals:1. x < -m/22. -m/2 ≤ x < 13. x ≥ 1In each of these intervals, the expressions inside the absolute values can be resolved with their signs, so we can write f(x) as a linear function without absolute values.So, let's consider each interval:1. For x < -m/2:In this interval, x - 1 is negative (since x < -m/2 ≤ 0 < 1), so |x - 1| = -(x - 1) = -x + 1.Similarly, 2x + m is negative because x < -m/2, so 2x + m < 2*(-m/2) + m = -m + m = 0. Therefore, |2x + m| = -(2x + m) = -2x - m.Therefore, f(x) = 2*(-x + 1) - (-2x - m) = -2x + 2 + 2x + m = ( -2x + 2x ) + (2 + m) = 0x + (2 + m) = 2 + m.Wait, that simplifies to a constant value 2 + m in this interval. So, for all x < -m/2, f(x) is just 2 + m. Interesting.2. For -m/2 ≤ x < 1:Here, x is greater than or equal to -m/2 but less than 1.In this interval, x - 1 is still negative (since x < 1), so |x - 1| = -x + 1.However, 2x + m is non-negative because x ≥ -m/2. Let's check: 2x + m ≥ 2*(-m/2) + m = -m + m = 0. So, |2x + m| = 2x + m.Therefore, f(x) = 2*(-x + 1) - (2x + m) = -2x + 2 - 2x - m = (-2x - 2x) + (2 - m) = -4x + (2 - m).So, in this interval, the function is linear with slope -4.3. For x ≥ 1:Here, x - 1 is non-negative, so |x - 1| = x - 1.Also, 2x + m is positive because x ≥ 1 and m ≥ 0, so 2x + m ≥ 2*1 + 0 = 2. Thus, |2x + m| = 2x + m.Therefore, f(x) = 2*(x - 1) - (2x + m) = 2x - 2 - 2x - m = (2x - 2x) + (-2 - m) = 0x + (-2 - m) = -2 - m.So, in this interval, f(x) is constant at -2 - m.Now, putting it all together:- For x < -m/2: f(x) = 2 + m (constant)- For -m/2 ≤ x < 1: f(x) = -4x + (2 - m) (linear with slope -4)- For x ≥ 1: f(x) = -2 - m (constant)We need to find the maximum value of this function. The maximum value is given as 3. Let's analyze the function in each interval.First, in the interval x < -m/2, the function is a constant 2 + m. Similarly, in x ≥ 1, it's a constant -2 - m. The middle interval is a linear function with slope -4, so it's decreasing. So, the maximum value of the function will either be the constant in the left interval (2 + m) or the starting point of the middle interval. Let's check the middle interval.At x = -m/2 (the start of the middle interval), the value is:f(-m/2) = -4*(-m/2) + (2 - m) = 2m + 2 - m = m + 2.Wait, that's the same as the constant value in the left interval. So, when approaching x = -m/2 from the left, the function is 2 + m, and at x = -m/2, it's also m + 2. Then, from there, the function decreases with slope -4 as x increases towards 1. So, the maximum value of the function is m + 2. Then, when x approaches 1 from the left, the value of the function would be:At x approaching 1 from the left, f(x) = -4*(1) + (2 - m) = -4 + 2 - m = -2 - m.Which is the same as the constant in the right interval. So, the function starts at m + 2 when x = -m/2, then decreases to -2 - m at x = 1. So, the maximum value of the entire function is m + 2, and the minimum is -2 - m.But the problem states that the maximum value is 3. So, we set m + 2 = 3. Therefore, m = 1. Wait, that seems straightforward, but let me check if there are any other possibilities.Wait, maybe there's a case where the middle interval's maximum is higher? But in the middle interval, the function is decreasing, so its maximum is at x = -m/2, which is m + 2, same as the left interval. So, the overall maximum is m + 2. Therefore, m + 2 = 3 implies m = 1.But let's verify this. Let's plug m = 1 into the function and check its maximum.If m = 1, then f(x) = 2|x -1| - |2x +1|.Let's analyze this function.Critical points are at x = 1 and x = -1/2 (since m =1, so -m/2 = -0.5).So, intervals:1. x < -0.52. -0.5 ≤ x < 13. x ≥ 1In x < -0.5: f(x) = 2 + 1 = 3In -0.5 ≤ x <1: f(x) = -4x + (2 -1) = -4x +1. At x = -0.5, f(x) = -4*(-0.5) +1 = 2 +1 = 3. Then, it decreases to x =1: f(1) = -4*1 +1 = -3.In x ≥1: f(x) = -2 -1 = -3.So, indeed, the maximum value is 3, achieved at x ≤ -0.5 and at x = -0.5. So, this checks out. Therefore, m =1 is correct.Wait, but the problem says "the function f(x) has a maximum value of 3". So, we set m + 2 =3, leading to m=1. So, part (I) answer is m=1.But before finalizing, let me make sure there's no other scenario. For example, suppose m is such that the critical points x=-m/2 and x=1 coincide. If m= -2x, but since m ≥0, x=-m/2 would be non-positive. So, unless m=0, x=-m/2=0. If m=0, then the critical points are x=0 and x=1. Then, let's check m=0.If m=0, then f(x)=2|x-1| - |2x|.Critical points at x=0 and x=1.For x <0: f(x) = 2( -x +1 ) - (-2x) = -2x +2 +2x =2.For 0 ≤x <1: f(x)=2(-x +1) -2x= -2x +2 -2x= -4x +2.For x ≥1: f(x)=2(x -1) -2x= 2x -2 -2x= -2.So, the maximum value would be 2 (at x <0 and x=0). But the problem states maximum is 3, so m=0 is invalid. So, m=1 is correct.Another check: suppose m=2, then the maximum would be m +2=4, which is more than 3. So, to get maximum 3, m must be 1.Therefore, part (I) answer is m=1.Now, moving on to part (II): Given real numbers a, b, c satisfying a -2b +c = m, which is 1, find the minimum value of a² + b² + c².So, we need to minimize a² + b² + c² subject to the constraint a -2b +c =1.This is a classic problem of finding the minimum distance from the origin to the plane a -2b +c =1 in three-dimensional space. The minimum value of a² + b² + c² is the square of the distance from the origin to the plane.The formula for the distance from a point (x0, y0, z0) to the plane Ax + By + Cz + D =0 is |Ax0 + By0 + Cz0 + D| / sqrt(A² + B² + C²). In our case, the plane is a -2b +c =1, which can be written as 1*a + (-2)*b +1*c -1 =0. So, A=1, B=-2, C=1, D=-1.The distance from the origin (0,0,0) to this plane is |0 +0 +0 -1| / sqrt(1 +4 +1) = | -1 | / sqrt(6) = 1/sqrt(6). Therefore, the minimum value of a² + b² + c² is (1/sqrt(6))² =1/6.Alternatively, using Lagrange multipliers:We want to minimize f(a,b,c)=a² + b² + c² subject to g(a,b,c)=a -2b +c -1=0.Set up the Lagrangian: L = a² + b² + c² - λ(a -2b +c -1).Take partial derivatives:∂L/∂a = 2a - λ =0 --> 2a = λ∂L/∂b = 2b + 2λ =0 --> 2b + 2λ =0 --> b = -λWait, wait, let me check again. The derivative with respect to b:∂L/∂b = 2b - λ*(-2) =0. Wait, no. Wait, the derivative of L with respect to b is:d/db [a² + b² + c² - λ(a -2b +c -1)] = 2b - λ*(-2) = 2b +2λ =0Therefore, 2b +2λ =0 --> b = -λSimilarly, derivative with respect to c:∂L/∂c = 2c - λ*(1) =0 --> 2c = λDerivative with respect to λ: -(a -2b +c -1)=0 --> a -2b +c =1.Now, we have the equations:1. 2a = λ2. 2b +2λ =0 --> b = -λ3. 2c = λ4. a -2b +c =1Let's substitute λ from equation 1 into equations 2 and 3.From equation 2: b = -λ = -2aFrom equation 3: 2c = λ =2a --> c = aNow, substitute into equation 4:a -2b +c =1But b = -2a, c =a, so:a -2*(-2a) +a =1 --> a +4a +a =1 -->6a =1 -->a=1/6Then, b = -2*(1/6) = -1/3c =1/6Therefore, a=1/6, b=-1/3, c=1/6Then, compute a² + b² + c²:(1/6)^2 + (-1/3)^2 + (1/6)^2 = (1/36) + (1/9) + (1/36) = (1 +4 +1)/36 =6/36=1/6Therefore, the minimum value is 1/6.Alternatively, using projection: The minimum of a² + b² + c² under the constraint a -2b +c =1 is equal to the square of the distance from the origin to the plane, which is (1)^2 / (1^2 + (-2)^2 +1^2) =1/(1 +4 +1)=1/6. So same result.Hence, part (II) answer is 1/6.Final AnswerⅠ. boxed{1}Ⅱ. boxed{dfrac{1}{6}}</think>Given the function ( f(x) = 2|x - 1| - |2x + m| ) with a maximum value of 3, we need to find the value of ( m ) and the minimum value of ( a^2 + b^2 + c^2 ) given ( a - 2b + c = m ). Part (I)1. Critical Points Analysis: - The critical points are ( x = 1 ) and ( x = -frac{m}{2} ). - These points divide the real line into three intervals: ( x < -frac{m}{2} ), ( -frac{m}{2} leq x < 1 ), and ( x geq 1 ).2. Function Analysis in Each Interval: - For ( x < -frac{m}{2} ), ( f(x) = 2 + m ) (constant). - For ( -frac{m}{2} leq x < 1 ), ( f(x) = -4x + (2 - m) ) (linear with slope -4). - For ( x geq 1 ), ( f(x) = -2 - m ) (constant).3. Maximum Value: - The maximum value occurs at ( x = -frac{m}{2} ) and is ( m + 2 ). - Given the maximum value is 3, we solve ( m + 2 = 3 ) to get ( m = 1 ).Thus, the value of ( m ) is (boxed{1}). Part (II)1. Constraint and Minimization: - We need to minimize ( a^2 + b^2 + c^2 ) subject to ( a - 2b + c = 1 ). - Using the formula for the distance from the origin to the plane ( a - 2b + c = 1 ): - The distance is ( frac{|1|}{sqrt{1^2 + (-2)^2 + 1^2}} = frac{1}{sqrt{6}} ). - The minimum value of ( a^2 + b^2 + c^2 ) is the square of this distance: ( left( frac{1}{sqrt{6}} right)^2 = frac{1}{6} ).Thus, the minimum value of ( a^2 + b^2 + c^2 ) is (boxed{dfrac{1}{6}}).]

🔑:(Ⅰ) We have:f(x) = 2|x - 1| - |2x + m| = |2x - 2| - |2x + m| leq |(2x - 2) - (2x + m)| = |-m - 2|Since m geq 0, we get:f(x) leq |-m - 2| = m + 2Equality holds when x = 1.Therefore, we have:f_{text{max}} = m + 2Given that the maximum value of f(x) is 3, we have:m + 2 = 3Thus:boxed{m = 1}(Ⅱ) By applying the Cauchy-Schwarz inequality, we have:(a^2 + b^2 + c^2)(1^2 + (-2)^2 + 1^2) geq (a - 2b + c)^2Since a - 2b + c = m = 1, it follows that:a^2 + b^2 + c^2 geq frac{(a - 2b + c)^2}{1^2 + (-2)^2 + 1^2} = frac{1^2}{6} = frac{1}{6}The equality holds when a:1 = b:-2 = c:1, which occurs at a = frac{1}{6}, b = -frac{1}{3}, c = frac{1}{6}. Hence, the minimum value of a^2 + b^2 + c^2 is:boxed{frac{1}{6}}

🧠:[<think>Okay, so I need to find the area bounded by these two polar curves: r = 6 sin φ and r = 4 sin φ. Hmm, let's think. Both of these equations are in polar coordinates, and they look like circles. I remember that equations of the form r = a sin φ represent circles with diameter a along the y-axis. So, r = 6 sin φ should be a circle with radius 3 centered at (0, 3) in Cartesian coordinates, and r = 4 sin φ is a circle with radius 2 centered at (0, 2). Wait, but how do these two circles intersect? Maybe they are concentric? But the centers are at different y-values. Wait, actually, in polar coordinates, when you have r = a sin φ, it's a circle that sits above the origin, tangent to the origin, with diameter a. So, the first circle has a diameter of 6, so radius 3, centered at (0, 3), and the second one has diameter 4, radius 2, centered at (0, 2). So, they both are on the upper half-plane, touching the origin, but different sizes. Let me visualize this. The bigger circle (radius 3) would encompass the smaller circle (radius 2) if their centers are 1 unit apart. Wait, centers are at (0, 3) and (0, 2), so the distance between centers is 1. The radius of the bigger circle is 3, and the smaller is 2. Since 3 - 2 = 1, which is exactly the distance between the centers. That means the smaller circle is just touching the bigger circle internally. So, they are tangent to each other at one point. Therefore, the area bounded by them would be the region inside the bigger circle but outside the smaller circle. Wait, but since they are tangent, there's no overlapping area except at the point of tangency. But maybe I'm missing something. Wait, let me check again.Wait, actually, maybe the smaller circle is entirely inside the bigger circle? Let me see. The center of the bigger circle is at (0, 3), radius 3. The center of the smaller circle is at (0, 2), radius 2. The distance between centers is 1, and the radius of the bigger circle is 3. The smaller circle's center is 1 unit below the bigger circle's center. The topmost point of the smaller circle is at (0, 2 + 2) = (0, 4), and the bottommost point is at (0, 2 - 2) = (0, 0). The bigger circle's topmost point is at (0, 3 + 3) = (0, 6), and the bottommost point is (0, 3 - 3) = (0, 0). So, both circles touch the origin. But the smaller circle's top is at (0, 4), and the bigger circle's top is at (0, 6). So, the smaller circle is entirely inside the bigger circle? Wait, let's check a point. Take the center of the smaller circle at (0, 2). The distance from this point to the center of the bigger circle (0, 3) is 1. Since the radius of the bigger circle is 3, the distance from (0, 3) to (0, 2) is 1, which is less than 3, so the entire smaller circle is inside the bigger circle? Wait, but the smaller circle has radius 2. So, from its center at (0, 2), it extends from 0 to 4 on the y-axis. The bigger circle extends from 0 to 6 on the y-axis. So, the smaller circle is entirely inside the bigger circle. But wait, but the centers are only 1 unit apart. If the smaller circle has radius 2, then from (0, 2), going up 2 units gets to (0, 4), and down 2 units to (0, 0). The bigger circle goes from (0, 0) to (0, 6). So, the smaller circle is entirely within the bigger circle. But wait, if the centers are 1 unit apart, and the sum of the radius of the smaller circle (2) and the distance between centers (1) is 3, which is equal to the radius of the bigger circle. Therefore, they are tangent. That means the smaller circle touches the bigger circle exactly at one point. That point would be along the line connecting their centers, which is the vertical line x=0. So, the point of tangency is at (0, 3 + 1) = wait, no. Wait, the distance between centers is 1. If you have two circles, one with radius 3 centered at (0, 3), and the other with radius 2 centered at (0, 2). The distance between centers is 1. Then, the point of tangency is located along the line connecting the centers. So, starting from the bigger circle's center (0, 3), moving towards the smaller circle's center (0, 2), which is 1 unit down. The point of tangency would be in the direction from the bigger circle towards the smaller circle, at a distance equal to the radius of the bigger circle minus the distance between centers? Wait, maybe not. Let me recall the formula for two circles. If two circles are tangent internally, the distance between centers is equal to the difference of their radii. Wait, in this case, the bigger circle has radius 3, the smaller has radius 2. The difference is 1, which is equal to the distance between centers (1). Therefore, they are tangent internally. So, the smaller circle lies entirely inside the bigger circle, touching at one point. Therefore, the area bounded by both circles would be the area of the bigger circle minus the area of the smaller circle? But if they are tangent, there's no annular region between them, because the smaller one is just touching the bigger one. Wait, but that contradicts my previous thought. Wait, let's think again. If they are tangent internally, then yes, the smaller circle is entirely inside the bigger circle, except for the single point where they touch. Therefore, the area bounded by both curves would actually be the area inside the bigger circle but outside the smaller circle. However, since the smaller circle is only tangent, not overlapping, then the area between them is just the area of the bigger circle minus the area of the smaller circle. But wait, if the smaller circle is entirely inside, except for the tangent point, then subtracting its area would give the area of the region inside the bigger circle but outside the smaller circle, which is an annular region except that they are tangent. Wait, but in reality, if they are tangent, there is no area between them. Wait, no. If the circles are tangent internally, then the area between them is the area of the larger circle minus the area of the smaller circle. Because the smaller circle is entirely inside the larger one, except for the tangent point. Wait, but in that case, the area between them is the area of the larger circle minus the area of the smaller circle. For example, if you have a big circle and a small circle inside it, tangent at one point, the area between them is the area of the big circle minus the small circle. So, maybe that's the case here. Let me verify. The area of the bigger circle is π*(3)^2 = 9π, the smaller is π*(2)^2 = 4π. So, the area between them would be 5π. But wait, the problem says "the area of figures bounded by lines given in polar coordinates". So, maybe the bounded area is the region that is inside both circles? But if they are tangent, the only point common to both is the tangent point. So, the area of their intersection is zero. Therefore, perhaps the problem is not about the area between them, but maybe the area enclosed by both curves? Hmm, this is confusing. Let me double-check.Wait, maybe I made a mistake in assuming the smaller circle is entirely inside the bigger one. Let me plot these circles more carefully. The equation r = 6 sin φ is a circle with diameter 6, so radius 3, centered at (0, 3) in Cartesian. Similarly, r = 4 sin φ is a circle with diameter 4, radius 2, centered at (0, 2). The distance between centers is 1 (from (0, 3) to (0, 2)). The radius of the bigger circle is 3, and the radius of the smaller is 2. Since the distance between centers (1) is equal to the difference of the radii (3 - 2 = 1), they are internally tangent. Therefore, the smaller circle is entirely inside the larger circle, touching at one point. Therefore, the area bounded by both curves would be the area inside the larger circle and outside the smaller circle. So, that area is the area of the larger circle minus the area of the smaller circle: 9π - 4π = 5π. But wait, in polar coordinates, when you compute areas, you have to be careful about the limits of integration. Maybe it's not as straightforward as subtracting the areas because the curves are given in polar coordinates, and we need to set up the integral properly. Let me recall that the area in polar coordinates is given by (1/2) ∫[r]^2 dφ. So, perhaps the region bounded by both curves is between the two circles, but since they are tangent, maybe the limits of φ are the same? Wait, no, in polar coordinates, both circles are traced out as φ goes from 0 to π. Because r = a sin φ is a circle that is above the origin, and φ ranges from 0 to π to complete the circle. So, for both curves, they are defined from φ = 0 to φ = π. Therefore, to find the area between them, we can compute the integral from 0 to π of (1/2)( (6 sin φ)^2 - (4 sin φ)^2 ) dφ. Let me check if that makes sense. So, for each angle φ, the outer radius is 6 sin φ and the inner radius is 4 sin φ, so the area element is (1/2)(R^2 - r^2) dφ. Then, integrating from 0 to π would give the total area between the two circles. Let me compute that. So, expanding the squares: (6 sin φ)^2 - (4 sin φ)^2 = 36 sin²φ - 16 sin²φ = 20 sin²φ. Therefore, the integral becomes (1/2) * 20 ∫ sin²φ dφ from 0 to π. That simplifies to 10 ∫ sin²φ dφ from 0 to π. The integral of sin²φ dφ over 0 to π is (π/2), since ∫ sin²φ dφ from 0 to π is π/2. Wait, no. Wait, the integral of sin²φ over 0 to 2π is π, so over 0 to π, it's π/2. Therefore, 10*(π/2) = 5π. Which matches the previous result. So, the area between the two circles is 5π. Therefore, the answer is 5π. But let me confirm if the setup is correct. So, in polar coordinates, both circles are traced from φ = 0 to φ = π. At each angle φ, the radius of the outer circle is 6 sin φ, and the inner circle is 4 sin φ. Therefore, between φ = 0 and φ = π, the region between the two circles is indeed the area we want. Therefore, integrating (1/2)( (6 sin φ)^2 - (4 sin φ)^2 ) dφ from 0 to π gives the correct area, which is 5π. So, that should be the answer.But just to be thorough, let me consider if there's another region bounded by these two curves. Since they are tangent at the origin? Wait, wait, hold on. Wait, r = 6 sin φ and r = 4 sin φ. When φ = 0, both r = 0. When φ = π/2, r = 6 and 4 respectively. When φ = π, both r = 0 again. So, the two circles intersect only at the origin (when φ = 0 and φ = π), but since they are circles in Cartesian coordinates, they also intersect at the point of tangency. Wait, earlier I thought they were tangent at (0, 4), but actually, the point of tangency should be along the line connecting their centers. The centers are at (0, 3) and (0, 2). So, the line connecting them is the vertical line x = 0. The point of tangency should be 1 unit away from the center of the bigger circle towards the smaller circle. Wait, the distance between centers is 1, which is equal to the difference in radii. So, the point of tangency is located at (0, 3 - 3 + 2) = no, let me think. Wait, the bigger circle has center at (0, 3) and radius 3, so it goes from (0, 0) to (0, 6). The smaller circle has center at (0, 2) and radius 2, so it goes from (0, 0) to (0, 4). Therefore, both circles pass through the origin, and the smaller circle reaches up to (0, 4), while the bigger circle reaches up to (0, 6). The point of tangency is actually at (0, 4). Because from the smaller circle's perspective, moving up 2 units from its center (0, 2) gets to (0, 4). From the bigger circle's perspective, moving down 2 units from its center (0, 3) gets to (0, 1). Wait, no. Wait, the point of tangency should be where both circles meet. Let's compute it algebraically. Let's convert both equations to Cartesian coordinates and solve for intersection points.For r = 6 sin φ:Multiply both sides by r: r^2 = 6 r sin φConvert to Cartesian: x² + y² = 6 y=> x² + y² - 6 y = 0Complete the square for y:x² + (y² - 6 y + 9) = 9=> x² + (y - 3)^2 = 9So, that's a circle with center (0, 3) and radius 3.Similarly, for r = 4 sin φ:Multiply by r: r² = 4 r sin φConvert to Cartesian: x² + y² = 4 y=> x² + y² - 4 y = 0Complete the square:x² + (y² - 4 y + 4) = 4=> x² + (y - 2)^2 = 4So, a circle with center (0, 2) and radius 2.Now, to find the points of intersection, solve the two equations:1. x² + (y - 3)^2 = 92. x² + (y - 2)^2 = 4Subtract equation 2 from equation 1:[(y - 3)^2 - (y - 2)^2] = 9 - 4 = 5Expand the squares:(y² - 6 y + 9) - (y² - 4 y + 4) = 5Simplify:y² -6y +9 -y² +4y -4 = 5(-2y +5) = 5-2y = 0y = 0Then plug y = 0 into equation 2:x² + (0 - 2)^2 = 4x² + 4 = 4x² = 0x = 0So, the only point of intersection is (0, 0). Wait, that contradicts my earlier thought that they are tangent at (0, 4). Hmmm. So according to this, the only point of intersection is the origin. But how come? If they are supposed to be tangent, but the calculations show they only intersect at the origin. Wait, that must be wrong. Wait, let me check the algebra again.Wait, subtracting equation 2 from equation 1:(y - 3)^2 - (y - 2)^2 = 5Expand both:(y² -6y +9) - (y² -4y +4) = 5Simplify:y² -6y +9 -y² +4y -4 = 5(-2y +5) = 5-2y = 0y = 0So, y = 0. Then x² + (0 - 3)^2 = 9 => x² + 9 = 9 => x² = 0 => x=0.Similarly, equation 2 gives the same. So, only the origin. Therefore, the two circles intersect only at the origin. But that can't be. How is that possible? Wait, but when you have two circles, one centered at (0,3) with radius 3 and the other at (0,2) with radius 2. The distance between centers is 1. The sum of the radii is 5, the difference is 1. Since the distance between centers (1) is equal to the difference in radii (3 - 2 = 1), they are internally tangent. So, they should intersect at exactly one point. But according to the algebra, they intersect only at the origin. Wait, but the origin is (0,0). Let me check if that point is on both circles. For the first circle: (0 - 3)^2 + 0^2 = 9 = 3^2. Yes, the origin is on the first circle. For the second circle: (0 - 2)^2 + 0^2 = 4 = 2^2. Yes, the origin is on the second circle. But where is the other point of tangency? Wait, according to the equations, there is only one intersection point. That suggests that the circles are tangent at the origin. Wait, but how? The origin is on both circles, but if the distance between centers is 1, and radii are 3 and 2, then the circles should be tangent at another point. Wait, but maybe they are tangent only at the origin. Let me visualize. The first circle is centered at (0,3) with radius 3, so it touches the origin. The second circle is centered at (0,2) with radius 2, so it also touches the origin. The line connecting the centers is the vertical line x=0. The distance between centers is 1. The first circle has radius 3, the second has radius 2. If you move from the center of the first circle (0,3) towards the center of the second (0,2), which is 1 unit down, then from (0,3), moving 3 units in the direction towards (0,2) would reach (0,0). Similarly, from (0,2), moving 2 units towards (0,3) would reach (0,4). Wait, but (0,4) is not on the first circle. Let's check: (0 - 3)^2 + (4 - 3)^2 = 9 + 1 = 10 ≠ 9. So, (0,4) is not on the first circle. Therefore, the circles only intersect at the origin. So, they are tangent at the origin? Wait, if two circles touch at exactly one point, they are tangent. But in this case, the origin is the point where both circles meet, but since both circles pass through the origin and have their centers along the y-axis, maybe they cross each other at the origin but have another intersection point. Wait, but according to the algebra, they only intersect at the origin. That seems odd. Let me check with another method.Take the two equations in polar coordinates: r = 6 sin φ and r = 4 sin φ. Setting them equal: 6 sin φ = 4 sin φ => 2 sin φ = 0 => sin φ = 0 => φ = 0, π. Therefore, the only solutions are φ = 0 and φ = π, which both correspond to the origin (since r=0 at those angles). Therefore, in polar coordinates, the only points of intersection are at the origin. So, the two circles only intersect at the origin. Therefore, how can they bound a region? If they only intersect at the origin, then the area bounded by both curves would be... Hmm, maybe the area that is inside both curves. But since both curves pass through the origin and extend upwards, maybe the region where both are satisfied? But since one is always larger than the other. Wait, for any angle φ between 0 and π, 6 sin φ is greater than 4 sin φ (since sin φ is non-negative in [0, π]). Therefore, the circle r = 6 sin φ is always outside the circle r = 4 sin φ in the upper half-plane. Therefore, the area bounded by both curves would be the area between the two circles from φ = 0 to φ = π, but since they only meet at the origin, this area is like a crescent moon shape, but since they meet only at the origin, maybe it's the area inside the larger circle and outside the smaller circle. But since they don't cross elsewhere, this area would indeed be the region between the two circles from φ = 0 to φ = π. So, even though they only intersect at the origin, integrating from 0 to π of the difference between the squares of the radii would still give the area between them. Which we calculated as 5π. But let's see, since the circles only meet at the origin, the area between them is like a circular sector from 0 to π for both circles, subtracting the smaller from the larger. Since for each angle φ, the radius of the larger circle is 6 sin φ and the smaller is 4 sin φ, so the area between them is (1/2) ∫ (36 sin²φ - 16 sin²φ) dφ from 0 to π = 10 ∫ sin²φ dφ from 0 to π = 10*(π/2) = 5π. So, this seems to be the correct answer. But just to confirm, let's compute the areas of each circle in polar coordinates. The area of r = 6 sin φ is (1/2) ∫ (6 sin φ)^2 dφ from 0 to π. Which is (1/2)*36 ∫ sin²φ dφ from 0 to π = 18*(π/2) = 9π. Similarly, the area of r = 4 sin φ is (1/2)*16 ∫ sin²φ dφ from 0 to π = 8*(π/2) = 4π. Therefore, subtracting gives 5π, which matches. Therefore, the area between the two circles is 5π. So, even though they only intersect at the origin, the area bounded between them is still the difference of their areas. Therefore, the answer is 5π. I think that's solid.Final AnswerThe area of the region bounded by the given lines is boxed{5pi}.</think>To calculate the area bounded by the polar curves ( r = 6 sin phi ) and ( r = 4 sin phi ):1. Identify the curves: Both equations represent circles in polar coordinates. The circle ( r = 6 sin phi ) has a radius of 3 and is centered at ((0, 3)) in Cartesian coordinates. The circle ( r = 4 sin phi ) has a radius of 2 and is centered at ((0, 2)) in Cartesian coordinates.2. Determine the relationship between the circles: The distance between the centers of the circles is 1 unit. The radius of the larger circle (3) minus the radius of the smaller circle (2) equals the distance between their centers (1), indicating they are internally tangent at the origin.3. Set up the integral for the area between the curves: The area between two polar curves ( r = R(phi) ) and ( r = r(phi) ) from ( phi = a ) to ( phi = b ) is given by: [ frac{1}{2} int_{a}^{b} left( [R(phi)]^2 - [r(phi)]^2 right) dphi ] Here, ( R(phi) = 6 sin phi ) and ( r(phi) = 4 sin phi ), with ( phi ) ranging from 0 to ( pi ).4. Compute the integral: [ text{Area} = frac{1}{2} int_{0}^{pi} left( (6 sin phi)^2 - (4 sin phi)^2 right) dphi ] Simplify the integrand: [ (6 sin phi)^2 - (4 sin phi)^2 = 36 sin^2 phi - 16 sin^2 phi = 20 sin^2 phi ] Therefore, [ text{Area} = frac{1}{2} int_{0}^{pi} 20 sin^2 phi , dphi = 10 int_{0}^{pi} sin^2 phi , dphi ] Using the identity ( int_{0}^{pi} sin^2 phi , dphi = frac{pi}{2} ): [ text{Area} = 10 cdot frac{pi}{2} = 5pi ]The area of the region bounded by the given lines is (boxed{5pi}).,<think>Okay, so I need to find the area bounded by the two polar curves r = 6 sin φ and r = 4 sin φ. Hmm, let me think. Both of these equations are in polar coordinates, and they look like circles. I remember that equations of the form r = a sin φ are circles that are tangent to the origin. The radius should be a/2, right? So for r = 6 sin φ, the radius is 3, and for r = 4 sin φ, the radius is 2. Both circles are above the x-axis because they are in terms of sine, which is positive in the first and second quadrants. Now, to find the area between them. Wait, how do these two circles intersect? Let me visualize them. The first circle has a radius of 3 and is centered along the positive y-axis at (0, 3), and the second circle has a radius of 2 and is centered at (0, 2). So the smaller circle is entirely inside the larger one? Because the centers are both on the y-axis, separated by 1 unit (from y=3 to y=2), and the radii are 3 and 2. The distance between the centers is 1, and the sum of the radii is 5, which is more than the distance between centers, but since one radius is larger and the smaller one is closer to the origin, maybe they don't overlap? Wait, wait. Wait, if the center of the larger circle is at (0,3) with radius 3, so it touches the origin. The smaller circle is centered at (0,2) with radius 2, so it also touches the origin. But how do they relate? Let's see: the larger circle goes from y=0 to y=6 (since center is at (0,3) and radius 3, so top point is (0,6) and bottom is origin). The smaller circle is centered at (0,2), radius 2, so it goes from y=0 to y=4. So the smaller circle is entirely inside the larger one? Because from y=0 to y=4, the smaller circle is within the larger one which goes up to y=6. But wait, in polar coordinates, both circles are only drawn as φ goes from 0 to π, because r = a sin φ is 0 at φ=0 and φ=π, reaching maximum at φ=π/2.But maybe they overlap in some regions. Wait, but if the smaller circle is entirely within the larger one, then the area between them would just be the area of the larger circle minus the area of the smaller one. But let me confirm if they indeed don't overlap except at the origin.Wait, but if the centers are at (0,3) and (0,2), the distance between centers is 1. The radii are 3 and 2. So the distance between centers (1) is less than the difference of the radii (3 - 2 = 1). So actually, the smaller circle is tangent to the larger circle internally. That is, they touch at exactly one point. Wait, if the centers are 1 unit apart, and the radii differ by 1 unit, then yes, they are tangent. So they touch at one point. Where is that point? Let's see. The centers are along the y-axis. The larger circle is at (0,3) with radius 3, so the bottom point is the origin. The smaller circle is at (0,2) with radius 2, so the top point is (0,4). Wait, the larger circle goes up to (0,6), and the smaller one up to (0,4). Wait, but how are they tangent? Wait, distance between centers is 1. The radius of the larger circle is 3, the radius of the smaller is 2. So if you have two circles with centers 1 unit apart, radii 3 and 2. Then, the larger circle would enclose the smaller circle if the distance between centers plus the radius of the smaller is less than or equal to the radius of the larger. But here, distance between centers is 1, radius of smaller is 2, so 1 + 2 = 3, which is equal to the radius of the larger. Therefore, the smaller circle lies entirely within the larger circle, touching it at exactly one point: the point that is in the direction from the center of the larger circle towards the center of the smaller circle. Since the centers are along the y-axis, the point of tangency would be along the line connecting the centers, which is the y-axis. Specifically, starting from the center of the larger circle (0,3), moving towards the center of the smaller circle (0,2), a distance of 1 unit. But the radius of the larger circle is 3, so moving from (0,3) towards (0,2) by 3 units would reach (0,0). Wait, no, the direction from (0,3) to (0,2) is downward. So the point of tangency should be at (0,3) minus 3 units in the direction towards (0,2). Wait, but the distance between centers is 1, which is the same as the difference in radii. Therefore, the smaller circle is just touching the larger circle at the origin. Wait, that doesn't make sense because the smaller circle is centered at (0,2) with radius 2, so it touches the origin as well. Wait, both circles pass through the origin. Wait, if you plug φ = 0 into r = 6 sin φ, you get r = 0. Similarly, φ = π also gives r = 0. So both circles pass through the origin, but their other points?Wait, maybe I need to actually find where the two curves intersect. To find the area between them, I need to know the limits of integration where they intersect. So set 6 sin φ = 4 sin φ. Solving this equation: 6 sin φ = 4 sin φ => 2 sin φ = 0 => sin φ = 0 => φ = 0 or φ = π. So they only intersect at φ = 0 and φ = π, which are the same point (the origin). Therefore, the two circles only intersect at the origin, and the smaller circle is entirely inside the larger one. Wait, but how?Wait, if the smaller circle is centered at (0,2) with radius 2, then all points on the smaller circle are within 2 units from (0,2). The maximum y-coordinate on the smaller circle is 4, and the minimum is 0. The larger circle is centered at (0,3) with radius 3, so it goes from 0 to 6 in the y-axis. Therefore, the entire smaller circle is inside the larger circle. But they only touch at the origin. Wait, but if the smaller circle is centered at (0,2) with radius 2, then the distance from the origin to the center is 2, and the radius is 2, so the origin is on the circle. Similarly, the larger circle is centered at (0,3) with radius 3, so the origin is also on that circle. So both circles pass through the origin, but the smaller circle is entirely inside the larger circle except for the origin. So the area between them is the area of the larger circle minus the area of the smaller circle. But wait, in polar coordinates, the area calculation is a bit different. Let me recall the formula for the area in polar coordinates: (1/2) ∫ r^2 dφ over the interval. But since both circles are swept out as φ goes from 0 to π, and they only intersect at φ = 0 and φ = π, then between φ = 0 and φ = π, the radius of the larger circle is always greater than the radius of the smaller circle. Therefore, the area between them would be the integral from 0 to π of (1/2)( (6 sin φ)^2 - (4 sin φ)^2 ) dφ. Let me verify this. So the area between two polar curves between angles α and β is (1/2) ∫ (outer^2 - inner^2) dφ from α to β. In this case, since from 0 to π, the outer curve is r = 6 sin φ and the inner curve is r = 4 sin φ, so the area should indeed be (1/2) ∫₀^π [ (6 sin φ)^2 - (4 sin φ)^2 ] dφ. Let me compute that. First, simplify the integrand:(6 sin φ)^2 - (4 sin φ)^2 = 36 sin²φ - 16 sin²φ = 20 sin²φ.So the integral becomes (1/2) * 20 ∫₀^π sin²φ dφ = 10 ∫₀^π sin²φ dφ.Now, the integral of sin²φ dφ over 0 to π is (π/2). Because ∫ sin²φ dφ from 0 to π is equal to π/2. Let me recall that ∫ sin²x dx from 0 to π is (π/2). Since sin²x has an average value of 1/2 over the interval, integrating over π gives (1/2)*π.Therefore, 10 * (π/2) = 5π.Wait, so the area between the two circles is 5π? But let's check this another way. If the area of the larger circle is (1/2) * π * (6/2)^2? Wait, in polar coordinates, the area for r = a sin φ is (π a²)/4. Wait, no, wait. Wait, the standard area for a circle in polar coordinates when given as r = 2a sin φ is π a². Wait, let's think. The general equation for a circle in polar coordinates is r = 2a sin φ, which is a circle with radius a, centered at (0, a). So if we have r = 6 sin φ, that's equivalent to r = 2 * 3 sin φ, so the radius is 3, area is π * 3² = 9π. Similarly, r = 4 sin φ is r = 2 * 2 sin φ, radius 2, area π * 2² = 4π. So the area between them should be 9π - 4π = 5π. That matches the integral result. So the answer is 5π. So maybe that's it. But let me make sure. The key was realizing that both circles are symmetric about the y-axis, only existing from φ = 0 to φ = π, and the area between them is the difference of their areas. But since they're both circles in polar coordinates, their areas can be calculated as standard circle areas (πr²) but in polar coordinates, the formula when given as r = a sin φ, the area is (π a²)/4? Wait, no, wait. Wait, when you have a circle in polar coordinates of the form r = 2a sin φ, then the area is π a². Because that's a circle with radius a centered at (0, a). So if we have r = 6 sin φ, that's equivalent to 2a = 6, so a = 3, so area is π * 3² = 9π. Similarly, r = 4 sin φ is 2a = 4, so a = 2, area is π * 2² = 4π. So difference is 5π, which is the same as calculated through the integral. Therefore, the area bounded by the two lines (circles) is 5π. So I think that's the answer.Final AnswerThe area of the region bounded by the given lines is boxed{5pi}.</think>To calculate the area of the region bounded by the polar curves ( r = 6 sin phi ) and ( r = 4 sin phi ):1. Identify the curves: Both equations represent circles in polar coordinates. The circle ( r = 6 sin phi ) has a radius of 3 and is centered at ((0, 3)). The circle ( r = 4 sin phi ) has a radius of 2 and is centered at ((0, 2)).2. Determine the relationship between the circles: The distance between the centers of the circles is 1 unit. The radius of the larger circle is 3, and the radius of the smaller circle is 2. The smaller circle is entirely within the larger circle and touches it at the origin.3. Set up the integral for the area between the curves: The area between two polar curves ( r = f(phi) ) and ( r = g(phi) ) (where ( f(phi) geq g(phi) )) from ( phi = alpha ) to ( phi = beta ) is given by: [ text{Area} = frac{1}{2} int_{alpha}^{beta} left( [f(phi)]^2 - [g(phi)]^2 right) dphi ] Here, ( f(phi) = 6 sin phi ) and ( g(phi) = 4 sin phi ), with ( alpha = 0 ) and ( beta = pi ).4. Simplify the integrand: [ (6 sin phi)^2 - (4 sin phi)^2 = 36 sin^2 phi - 16 sin^2 phi = 20 sin^2 phi ]5. Compute the integral: [ text{Area} = frac{1}{2} int_{0}^{pi} 20 sin^2 phi , dphi = 10 int_{0}^{pi} sin^2 phi , dphi ] Using the identity ( int_{0}^{pi} sin^2 phi , dphi = frac{pi}{2} ): [ text{Area} = 10 cdot frac{pi}{2} = 5pi ]Thus, the area of the region bounded by the given lines is (boxed{5pi}).]

🔑:## SolutionS=frac{1}{2} int_{alpha}^{beta}left(r_{1}^{2}(varphi)-r_{2}^{2}(varphi)right) d varphiWhere alpha=-frac{pi}{2} ; beta=frac{pi}{2}, and then we get:begin{aligned}& S=frac{1}{2} int_{-pi / 2}^{pi / 2}left(36 sin ^{2} varphi-16 sin ^{2} varphiright) d varphi= & =10 int_{-pi / 2}^{pi / 2} sin ^{2} varphi d varphi= & =10 int_{-pi / 2}^{pi / 2} frac{1-cos 2 varphi}{2} d varphi= & =left.5left(varphi-frac{sin 2 varphi}{2}right)right|_{-pi / 2} ^{pi / 2}= & =5left(frac{pi}{2}-frac{1}{2} sin piright)-5left(-frac{pi}{2}-frac{1}{2} sin (-pi)right)= & =5 cdot frac{pi}{2}+5 cdot frac{pi}{2}=5 piend{aligned}Source — «http://pluspi.org/wiki/index.php/������������ %D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD % mathrm{D} 1 % 82 % mathrm{D} 0 % mathrm{~B} 5 % mathrm{D} 0 % mathrm{~B} 3 % mathrm{D} 1 % 80 % mathrm{D} 0 % mathrm{~B} 0 % mathrm{D} 0 % mathrm{BB} % mathrm{D} 1 % 8 mathrm{~B}+16-31 »Categories: Kuznetsov's Problem Book Integrals Problem 16 | Integrals | Problems for CheckingUkrainian Banner Network- Last edited on this page: 11:11, 9 July 2010.- Content is available under CC-BY-SA 3.0.

🧠:[<think>Okay, let's see. So, the problem is about a game activity where everyone gets a prize: first, second, or third. The probabilities of winning each prize form a geometric sequence with the first term 'a' and a common ratio of 2. The prizes themselves form an arithmetic sequence with the first term 700 yuan and a common difference of -140. We need to find the expected prize money for participating.First, let me parse this step by step. There are three prizes, and all participants receive one of them, so the probabilities must add up to 1. The probabilities are in a geometric sequence where each term is double the previous one. Wait, geometric sequence with first term a and common ratio 2. So, the probabilities would be a, 2a, 4a? But since there are only three prizes, it should be first term a, second term 2a, third term 4a? But wait, if the common ratio is 2, then each subsequent term is multiplied by 2. However, since there are three terms, the total probability should be a + 2a + 4a = 7a. But since all participants receive a prize, the total probability must be 1. So, 7a = 1 => a = 1/7. Therefore, the probabilities would be 1/7, 2/7, 4/7 for first, second, and third prizes respectively. Hmm, that seems straightforward.Next, the prizes form an arithmetic sequence with the first term 700 yuan and a common difference of -140. Arithmetic sequence with first term 700, so the prizes are 700, 700 - 140, 700 - 2*140. Let me compute those. First prize: 700, second prize: 700 - 140 = 560, third prize: 560 - 140 = 420. So the prizes are 700, 560, and 420 yuan.Now, the expected prize money would be the sum of each prize multiplied by its probability. So, Expected Value (EV) = (700 * 1/7) + (560 * 2/7) + (420 * 4/7). Let me compute each term step by step.First term: 700 * (1/7) = 100. Second term: 560 * (2/7) = (560/7)*2 = 80*2 = 160. Third term: 420 * (4/7) = (420/7)*4 = 60*4 = 240. Then, adding them up: 100 + 160 + 240. Let's compute that: 100 + 160 = 260, 260 + 240 = 500. So the expected prize money is 500 yuan.Wait a second, that seems straightforward, but let me check again to make sure I didn't make a mistake.First, confirming the probabilities: geometric sequence with first term a, ratio 2. So a, 2a, 4a. Sum is 7a = 1 => a = 1/7. So probabilities 1/7, 2/7, 4/7. Correct.Prizes: arithmetic sequence with first term 700, common difference -140. So each subsequent prize is 140 less. First: 700, second: 700 - 140 = 560, third: 560 - 140 = 420. Correct.Then, EV calculation: 700*(1/7) is 100, 560*(2/7) = (560*2)/7 = 1120/7 = 160, 420*(4/7) = (420*4)/7 = 1680/7 = 240. 100 + 160 + 240 = 500. That seems right.But let me consider: Is there any possibility that the geometric sequence is in the reverse order? For example, maybe the highest probability is for the third prize, so the sequence is 4a, 2a, a? But the problem says "the corresponding winning probabilities form a geometric sequence with the first term a and a common ratio of 2". So the first term is a, which would correspond to the first prize. Then the second term is a*2, third term is a*4. So the first prize has probability a, second 2a, third 4a, adding to 7a=1. So yes, first prize has the lowest probability, third has the highest. That seems counterintuitive, but the problem states that the probabilities form a geometric sequence with first term a and common ratio 2. So unless specified otherwise, the order is first, second, third. So first prize has probability a, second 2a, third 4a. So the third prize is the most probable. But in real-life scenarios, the highest prize usually has the lowest probability. But the problem doesn't mention anything about that, so we have to go with the given.Similarly, the prizes are an arithmetic sequence with first term 700 and common difference -140, so decreasing by 140 each time. So first prize 700, second 560, third 420. So the arithmetic sequence is ordered first to third.Therefore, the calculation seems correct. The expected value is 500 yuan.Just to confirm once again:EV = (700 * 1/7) + (560 * 2/7) + (420 * 4/7)Calculate each term:700 / 7 = 100560 * 2 = 1120; 1120 / 7 = 160420 * 4 = 1680; 1680 / 7 = 240100 + 160 + 240 = 500.Yep, that's consistent. So I think that's the correct answer. The expected prize money is 500 yuan.Final AnswerThe expected prize money for participating in this game activity is boxed{500} yuan.</think>In a certain game activity, the rewards are divided into first, second, and third prizes, with all participants receiving a prize. The winning probabilities form a geometric sequence with the first term (a) and a common ratio of 2. The corresponding prizes form an arithmetic sequence with the first term of 700 yuan and a common difference of -140. First, we determine the probabilities. The geometric sequence for the probabilities is (a), (2a), and (4a). Since the total probability must be 1, we have:[a + 2a + 4a = 7a = 1 implies a = frac{1}{7}]Thus, the probabilities are (frac{1}{7}), (frac{2}{7}), and (frac{4}{7}) for the first, second, and third prizes respectively.Next, we determine the prize amounts. The arithmetic sequence for the prizes is 700, 700 - 140, and 700 - 2*140, which results in 700 yuan, 560 yuan, and 420 yuan for the first, second, and third prizes respectively.The expected prize money is calculated by multiplying each prize by its corresponding probability and summing the results:[text{Expected Value} = left(700 times frac{1}{7}right) + left(560 times frac{2}{7}right) + left(420 times frac{4}{7}right)]Calculating each term:[700 times frac{1}{7} = 100][560 times frac{2}{7} = frac{1120}{7} = 160][420 times frac{4}{7} = frac{1680}{7} = 240]Summing these results:[100 + 160 + 240 = 500]The expected prize money for participating in this game activity is (boxed{500}) yuan.,<think>Okay, let me try to figure out this problem step by step. So, the problem is about a game activity where everyone gets a prize, either first, second, or third. The probabilities of winning these prizes form a geometric sequence with the first term 'a' and a common ratio of 2. The prizes themselves form an arithmetic sequence starting at 700 yuan with a common difference of -140. We need to find the expected prize money for participating in this game.First, let me recall what geometric and arithmetic sequences are. A geometric sequence is one where each term is multiplied by a common ratio to get the next term. So, if the first term is 'a' and the common ratio is 2, then the probabilities would be a, 2a, 4a for first, second, and third prizes respectively. Wait, but since there are only three prizes, does that mean the probabilities are a, 2a, 4a? But hold on, the problem says all participants receive a prize, so the total probability should sum up to 1. That makes sense. So, the sum of the probabilities must be 1. So, if they form a geometric sequence with first term a and common ratio 2, then the three probabilities are a, 2a, and 4a. Therefore, the sum is a + 2a + 4a = 7a. Since all participants must receive a prize, the total probability is 1, so 7a = 1, which implies a = 1/7. Therefore, the probabilities are 1/7, 2/7, and 4/7 for first, second, and third prizes respectively.Now, the prizes form an arithmetic sequence with first term 700 yuan and common difference -140. An arithmetic sequence is one where each term is obtained by adding a common difference. Since the common difference is -140, the prizes would decrease by 140 each time. So, the first prize is 700, the second prize is 700 - 140 = 560 yuan, and the third prize is 560 - 140 = 420 yuan. Let me confirm that: 700, 560, 420. The common difference is indeed -140.Now, the expected prize money is the sum of each prize multiplied by its corresponding probability. So, we need to calculate (First prize * probability of first prize) + (Second prize * probability of second prize) + (Third prize * probability of third prize).Substituting the numbers:First prize: 700 yuan, probability 1/7.Second prize: 560 yuan, probability 2/7.Third prize: 420 yuan, probability 4/7.Calculating each term:700 * (1/7) = 100560 * (2/7) = Let's compute 560 divided by 7 first. 560 / 7 = 80. Then 80 * 2 = 160.420 * (4/7) = 420 divided by 7 is 60. Then 60 * 4 = 240.Now, adding up these three terms: 100 + 160 + 240 = 500.So, the expected prize money is 500 yuan.Wait, let me check if I did everything correctly. Let's verify the arithmetic again.First term: 700 * 1/7 = 700/7 = 100. Correct.Second term: 560 * 2/7. Let's compute 560/7 = 80. Then 80*2 = 160. Correct.Third term: 420 * 4/7. 420/7 = 60. 60*4 = 240. Correct.Sum: 100 + 160 = 260. 260 + 240 = 500. Yes, that adds up to 500.Therefore, the expected prize money is 500 yuan. Hmm. That seems straightforward. Let me just check if I interpreted the sequences correctly.The problem states the probabilities form a geometric sequence with first term 'a' and common ratio 2. So, the terms are a, 2a, 4a. Summing to 7a = 1. Therefore, a = 1/7. That's correct. The prizes are arithmetic with first term 700 and common difference -140. So, 700, 560, 420. Correct.Calculating the expectation: each prize times its probability. Then adding them up. The calculation steps seem right. So, I think 500 yuan is the correct expected value.I don't see any mistakes in the reasoning. Therefore, the answer should be 500 yuan.Final AnswerThe expected prize money for participating in this game activity is boxed{500} yuan.</think>In a certain game activity, the rewards are divided into first, second, and third prizes, with all participants receiving a prize. The winning probabilities form a geometric sequence with the first term (a) and a common ratio of 2. The corresponding prizes form an arithmetic sequence with the first term 700 yuan and a common difference of -140.First, we determine the probabilities. The geometric sequence for the probabilities is (a), (2a), (4a). Since the total probability must be 1, we have:[a + 2a + 4a = 7a = 1 implies a = frac{1}{7}]Thus, the probabilities are (frac{1}{7}), (frac{2}{7}), and (frac{4}{7}) for the first, second, and third prizes respectively.Next, we determine the prize amounts. The arithmetic sequence for the prizes is 700, 560, 420 (starting at 700 yuan with a common difference of -140).The expected prize money is calculated as follows:[text{Expected prize} = left(700 times frac{1}{7}right) + left(560 times frac{2}{7}right) + left(420 times frac{4}{7}right)]Calculating each term:[700 times frac{1}{7} = 100][560 times frac{2}{7} = 80 times 2 = 160][420 times frac{4}{7} = 60 times 4 = 240]Summing these results:[100 + 160 + 240 = 500]Thus, the expected prize money for participating in this game activity is (boxed{500}) yuan.]

🔑:(7) 500 Hint: From the problem, we know the probabilities of winning the first, second, and third prizes areP_{1}=a, P_{2}=2 a, P_{3}=4 a .From P_{1}+P_{2}+P_{3}=1, we get a=frac{1}{7}.So P_{1}=frac{1}{7}, P_{2}=frac{2}{7}, P_{3}=frac{4}{7}.The prizes for winning the first, second, and third prizes are xi_{1}=700, xi_{2}=560, xi_{3}=420.Therefore, E xi=700 times frac{1}{7}+560 times frac{2}{7}+420 times frac{4}{7}=500 (yuan).