🧠:<think>Okay, so I have this problem where I need to find the least positive angle α such that the product of four terms equals 1/256. The terms are of the form (3/4 - sin²(kα)) where k is 1, 3, 9, and 27. Hmm, that seems a bit complicated, but maybe I can break it down step by step.First, I remember that sin²(x) can be rewritten using the double-angle identity. The identity is sin²(x) = (1 - cos(2x))/2. So, maybe I can substitute that into each term. Let me try that.For each term, 3/4 - sin²(x) becomes 3/4 - (1 - cos(2x))/2. Let me compute that:3/4 - (1 - cos(2x))/2 = 3/4 - 1/2 + (cos(2x))/2 = (3/4 - 2/4) + (cos(2x))/2 = 1/4 + (cos(2x))/2.So each term simplifies to 1/4 + (cos(2kα))/2. Therefore, the product becomes:(1/4 + cos(2α)/2)(1/4 + cos(6α)/2)(1/4 + cos(18α)/2)(1/4 + cos(54α)/2) = 1/256.Hmm, interesting. So each term is 1/4 plus something. Since the product is 1/256, which is (1/4)^4, maybe each term is equal to 1/4? Because if each term is 1/4, then their product would be (1/4)^4 = 1/256. That seems like a good possibility.So, if 1/4 + cos(2kα)/2 = 1/4 for each k, then cos(2kα)/2 must be zero. That implies cos(2kα) = 0 for each k. So, let's write that down:cos(2α) = 0,cos(6α) = 0,cos(18α) = 0,cos(54α) = 0.Hmm, so each of these cosine terms equals zero. When does cos(θ) = 0? That happens when θ is an odd multiple of 90 degrees, right? So θ = 90° + 180°n, where n is an integer.So, let's apply that to each equation:1. 2α = 90° + 180°n,2. 6α = 90° + 180°m,3. 18α = 90° + 180°p,4. 54α = 90° + 180°q,where n, m, p, q are integers.I need to find the smallest positive α that satisfies all these equations. Let's solve each for α:1. α = 45° + 90°n,2. α = 15° + 30°m,3. α = 5° + 10°p,4. α = 5°/3 + 10°q/3.Hmm, these are all different expressions. I need α to satisfy all four equations simultaneously. So, α must be a common solution to all four. Let me see if I can find such an α.Looking at the first equation, α must be 45° more than a multiple of 90°, which is quite restrictive. The second equation requires α to be 15° more than a multiple of 30°, the third requires 5° more than a multiple of 10°, and the fourth is a bit more complex.Wait, maybe I can find a common multiple or something. Let's consider the first equation: α = 45° + 90°n. Let's plug this into the second equation:45° + 90°n = 15° + 30°m.Simplify:45° - 15° = 30°m - 90°n,30° = 30°(m - 3n).So, 1 = m - 3n, which means m = 3n + 1.So, for each integer n, m is determined. Let's choose n = 0 for the smallest positive solution:α = 45°, m = 1.Check the third equation: α = 5° + 10°p.45° = 5° + 10°p,40° = 10°p,p = 4.That works. Now check the fourth equation: 54α = 90° + 180°q.54 * 45° = 2430°.2430° = 90° + 180°q,2430° - 90° = 180°q,2340° = 180°q,q = 13.Hmm, that's an integer, so that works. So, α = 45° is a solution.But wait, is it the least positive solution? Let me check if there's a smaller α that satisfies all four equations.Let me try n = -1 in the first equation:α = 45° - 90° = -45°, which is negative, so we discard it.n = 0 gives 45°, which is positive.But wait, maybe there's a smaller positive α if we choose higher n? Hmm, no, because n increases α, so 45° is the smallest positive solution from the first equation.But hold on, let's think again. Maybe α doesn't have to satisfy all four equations simultaneously in the same way. Because the product is 1/256, which is (1/4)^4, but maybe each term doesn't have to be exactly 1/4. Maybe some terms are larger than 1/4 and some are smaller, but their product is still 1/256.Wait, but 1/4 is the maximum value for each term because sin²(x) is between 0 and 1, so 3/4 - sin²(x) is between -1/4 and 3/4. But actually, sin²(x) can be up to 1, so 3/4 - sin²(x) can be as low as -1/4. But since all the terms are multiplied together, and the product is positive, each term must be positive. So, 3/4 - sin²(x) > 0, which implies sin²(x) < 3/4, so |sin(x)| < sqrt(3)/2, which is approximately 0.866. So, x must not be 60°, 120°, etc.But since the product is 1/256, which is positive, all four terms must be positive. So, each term is between 0 and 1/4 + something.Wait, no, 3/4 - sin²(x) can be as high as 3/4, but in our earlier substitution, it became 1/4 + cos(2x)/2, which ranges from 0 to 1/2, because cos(2x) ranges from -1 to 1. So, 1/4 + (cos(2x))/2 ranges from 0 to 1/2.So, each term is between 0 and 1/2, and their product is 1/256. Hmm, so each term could be 1/4, but maybe not necessarily all exactly 1/4. But since 1/4 * 1/4 * 1/4 * 1/4 = 1/256, perhaps the minimal solution is when all terms are 1/4, which gives α = 45°, but maybe there's a smaller angle where the product still equals 1/256.Wait, maybe I can think of it differently. Let me consider that the product is 1/256, which is (1/4)^4. If each term is 1/4, then the product is 1/256. So, if I can find α such that each term is 1/4, then that would satisfy the equation. So, setting each term equal to 1/4:1/4 + cos(2kα)/2 = 1/4,Which implies cos(2kα)/2 = 0,So cos(2kα) = 0.Which is what I had before. So, that leads to α = 45°, but is that the minimal solution? Maybe not, because perhaps the angles 3α, 9α, 27α could wrap around the unit circle and still satisfy the condition for smaller α.Wait, let's think about this. If α is 45°, then 3α is 135°, 9α is 405°, which is equivalent to 45°, and 27α is 1215°, which is equivalent to 135° (since 1215 - 3*360 = 1215 - 1080 = 135°). So, the angles cycle between 45° and 135°, which are both 90° apart, so their cosines are zero. So, that works.But maybe if α is smaller, say, α = 5°, then 3α = 15°, 9α = 45°, 27α = 135°. Let's check:For α = 5°,First term: 3/4 - sin²(5°). Compute sin(5°) ≈ 0.0872, so sin²(5°) ≈ 0.0076, so 3/4 - 0.0076 ≈ 0.7424.Second term: 3/4 - sin²(15°). sin(15°) ≈ 0.2588, sin² ≈ 0.06699, so 3/4 - 0.06699 ≈ 0.683.Third term: 3/4 - sin²(45°). sin(45°) ≈ 0.7071, sin² ≈ 0.5, so 3/4 - 0.5 = 0.25.Fourth term: 3/4 - sin²(135°). sin(135°) is same as sin(45°), so sin² is 0.5, so 3/4 - 0.5 = 0.25.So, the product is approximately 0.7424 * 0.683 * 0.25 * 0.25 ≈ 0.7424 * 0.683 ≈ 0.507, then 0.507 * 0.25 ≈ 0.1268, and 0.1268 * 0.25 ≈ 0.0317. Which is much larger than 1/256 ≈ 0.00390625. So, α = 5° doesn't work.Hmm, so maybe α needs to be larger. Let's try α = 15°.Compute each term:First term: 3/4 - sin²(15°). sin(15°) ≈ 0.2588, sin² ≈ 0.06699, so 3/4 - 0.06699 ≈ 0.683.Second term: 3/4 - sin²(45°) = 3/4 - 0.5 = 0.25.Third term: 3/4 - sin²(135°) = 0.25.Fourth term: 3/4 - sin²(405°). 405° is equivalent to 45°, so same as third term: 0.25.So, the product is 0.683 * 0.25 * 0.25 * 0.25 ≈ 0.683 * 0.015625 ≈ 0.01064, which is still larger than 1/256.Hmm, so α = 15° gives a product of about 0.01064, which is bigger than 1/256. So, we need a larger α? Or perhaps not necessarily?Wait, 45° gave us a product of (1/4)^4 = 1/256, which is exactly what we need. So, maybe 45° is the solution. But the question says the least positive angle. So, is there a smaller angle than 45° that satisfies the equation?Wait, let's think again. If I take α = 45°, then 3α = 135°, 9α = 405°, which is 45°, and 27α = 1215°, which is 135°. So, it cycles between 45° and 135°, both of which have sin² equal to 0.5, so 3/4 - 0.5 = 1/4. So, each term is 1/4, so the product is 1/256. That works.But maybe there's a smaller α where the terms don't all equal 1/4, but their product still equals 1/256. Perhaps some terms are larger than 1/4 and some are smaller, but overall, the product is 1/256.So, maybe I can set up equations where some terms are larger and some are smaller. Let me think about how to approach this.Alternatively, maybe I can use a substitution. Let me set θ = α, then the angles are θ, 3θ, 9θ, 27θ. So, the product is:(3/4 - sin²θ)(3/4 - sin²3θ)(3/4 - sin²9θ)(3/4 - sin²27θ) = 1/256.Hmm, this seems similar to a telescoping product or something with multiple angles. Maybe there's a pattern or identity that can help here.Wait, I remember that sin(3x) can be expressed in terms of sin(x). The identity is sin(3x) = 3sinx - 4sin³x. Maybe I can use that to relate the terms.But I'm not sure if that will help directly. Alternatively, maybe I can express sin²x in terms of multiple angles.Wait, another thought: since 3/4 - sin²x = 1/4 + (cos2x)/2, as I did earlier, so each term is 1/4 + (cos2x)/2.So, the product becomes:Product = [1/4 + (cos2θ)/2][1/4 + (cos6θ)/2][1/4 + (cos18θ)/2][1/4 + (cos54θ)/2] = 1/256.Hmm, so that's four terms each of the form 1/4 + (coskθ)/2, where k is 2, 6, 18, 54. So, 2, 6, 18, 54 are multiples of 3: 2*3^0, 2*3^1, 2*3^2, 2*3^3.Wait, that's interesting. So, each term is 1/4 + (cos(2*3^n θ))/2 for n = 0,1,2,3.So, maybe there's a pattern or a telescoping product here when considering multiple angles.Alternatively, perhaps I can consider setting each term equal to 1/4, which gives us the solution α = 45°, but as we saw, maybe there's a smaller angle where the product is still 1/256.Wait, another approach: let's consider that the product is 1/256, which is (1/4)^4. So, if each term is 1/4, the product is 1/256. But if the terms are not all 1/4, but some are larger and some are smaller, their product can still be 1/256.So, perhaps a smaller α exists where some terms are larger than 1/4 and others are smaller, but the product remains 1/256.But how can I find such an α?Alternatively, maybe I can use the fact that 3/4 - sin²x = (1/2 + cos2x/2). Wait, no, that's what I did earlier.Wait, another idea: since the product involves angles θ, 3θ, 9θ, 27θ, which are all multiples of θ by powers of 3, maybe there's a way to express this product in terms of a single angle through multiple-angle identities.But I'm not sure. Alternatively, maybe I can use logarithms to turn the product into a sum, but that might complicate things.Wait, perhaps I can consider that the product equals 1/256, so taking the fourth root, each term on average is 1/4. So, maybe the terms oscillate around 1/4, but their product is 1/256.Alternatively, perhaps I can consider that for the product to be 1/256, the sum of the logs is log(1/256), but that might not be helpful here.Wait, maybe I can consider that each term is 1/4 + (cos2kθ)/2, so if I denote each term as t_k = 1/4 + (cos2kθ)/2, then the product t0*t1*t2*t3 = 1/256.But I'm not sure if that helps.Wait, another thought: since 2kθ for k = 1,3,9,27, which are 2θ, 6θ, 18θ, 54θ, so 2θ, 6θ, 18θ, 54θ.Wait, 54θ is 3^3 * 2θ, so maybe there's a pattern here where each angle is three times the previous one.So, starting from 2θ, each subsequent angle is tripled. So, 2θ, 6θ, 18θ, 54θ.Hmm, so perhaps I can express this product in terms of multiple angles.Alternatively, maybe I can use the identity for multiple angles, such as cos(3x) = 4cos³x - 3cosx. But I'm not sure how that would help here.Wait, another approach: let's consider that if we let θ = α, then 3θ = 3α, 9θ = 9α, 27θ = 27α. So, each subsequent angle is tripled.So, the product becomes:(3/4 - sin²θ)(3/4 - sin²3θ)(3/4 - sin²9θ)(3/4 - sin²27θ) = 1/256.Hmm, maybe I can write this as a product over n=0 to 3 of (3/4 - sin²(3^n θ)).I wonder if there's a telescoping product formula for such terms.Alternatively, perhaps I can use induction or some recursive relation.Wait, let me think about the product:Let P = (3/4 - sin²θ)(3/4 - sin²3θ)(3/4 - sin²9θ)(3/4 - sin²27θ).I need to find θ such that P = 1/256.I recall that in some trigonometric products, especially those involving multiple angles, there might be a telescoping product or a product that simplifies to a single term.Alternatively, maybe I can express each term in terms of cos2θ, as I did earlier.So, each term is 1/4 + (cos2kθ)/2, where k is 1,3,9,27.So, P = [1/4 + (cos2θ)/2][1/4 + (cos6θ)/2][1/4 + (cos18θ)/2][1/4 + (cos54θ)/2].Hmm, that's four terms. Maybe I can multiply them step by step.Let me try multiplying the first two terms:[1/4 + (cos2θ)/2][1/4 + (cos6θ)/2] = (1/4)(1/4) + (1/4)(cos6θ)/2 + (cos2θ)/2*(1/4) + (cos2θ)/2*(cos6θ)/2.Simplify:= 1/16 + (cos6θ)/8 + (cos2θ)/8 + (cos2θ cos6θ)/4.Hmm, that's getting complicated. Maybe I can use the identity for cosA cosB = [cos(A+B) + cos(A-B)]/2.So, the last term is (cos2θ cos6θ)/4 = [cos(8θ) + cos(-4θ)]/8 = [cos8θ + cos4θ]/8.So, the first two terms multiplied together give:1/16 + (cos6θ)/8 + (cos2θ)/8 + [cos8θ + cos4θ]/8.Combine like terms:1/16 + [cos2θ + cos4θ + cos6θ + cos8θ]/8.Hmm, that's still complicated. Maybe I can do the same for the next two terms.Multiply the third and fourth terms:[1/4 + (cos18θ)/2][1/4 + (cos54θ)/2] = (1/4)(1/4) + (1/4)(cos54θ)/2 + (cos18θ)/2*(1/4) + (cos18θ)/2*(cos54θ)/2.Simplify:= 1/16 + (cos54θ)/8 + (cos18θ)/8 + (cos18θ cos54θ)/4.Again, use the identity for cosA cosB:(cos18θ cos54θ)/4 = [cos(72θ) + cos(-36θ)]/8 = [cos72θ + cos36θ]/8.So, the third and fourth terms multiplied together give:1/16 + (cos54θ)/8 + (cos18θ)/8 + [cos72θ + cos36θ]/8.Combine like terms:1/16 + [cos18θ + cos36θ + cos54θ + cos72θ]/8.So, now the entire product P is:[1/16 + (cos2θ + cos4θ + cos6θ + cos8θ)/8] * [1/16 + (cos18θ + cos36θ + cos54θ + cos72θ)/8].This is getting really messy. Maybe there's a better approach.Wait, another thought: perhaps I can use the identity for sin(3x) in terms of sinx, but I don't see how that directly helps.Alternatively, maybe I can consider that 3/4 - sin²x = (sqrt(3)/2)^2 - sin²x. Hmm, that's reminiscent of the identity for cos(A - B)cos(A + B) = cos²A - sin²B, but I'm not sure.Wait, let me try to think differently. Suppose I set x = sin²θ, then 3/4 - x is a term. Similarly, 3/4 - sin²3θ, which is 3/4 - (3sinθ - 4sin³θ)^2. Hmm, but that seems complicated.Alternatively, maybe I can express 3/4 - sin²x in terms of cos2x, as I did earlier, and then multiply all the terms together.Wait, let me write each term as 1/4 + (cos2kθ)/2, as before. So, the product is:(1/4 + (cos2θ)/2)(1/4 + (cos6θ)/2)(1/4 + (cos18θ)/2)(1/4 + (cos54θ)/2) = 1/256.Let me denote each term as T_k = 1/4 + (cos(2*3^k θ))/2 for k=0,1,2,3.So, T0*T1*T2*T3 = 1/256.I wonder if there's a way to express this product in terms of a single cosine term or something that can be simplified.Alternatively, maybe I can use the fact that each term is 1/4 + (cosφ)/2, where φ is 2θ, 6θ, 18θ, 54θ.So, each term is 1/4 + (cosφ)/2. Let me denote this as T = 1/4 + (cosφ)/2.I can write T = (1 + 2cosφ)/4.So, T = (1 + 2cosφ)/4.So, the product becomes:Product = [(1 + 2cos2θ)/4][(1 + 2cos6θ)/4][(1 + 2cos18θ)/4][(1 + 2cos54θ)/4] = 1/256.So, Product = [ (1 + 2cos2θ)(1 + 2cos6θ)(1 + 2cos18θ)(1 + 2cos54θ) ] / 256 = 1/256.Therefore, the numerator must be 1:(1 + 2cos2θ)(1 + 2cos6θ)(1 + 2cos18θ)(1 + 2cos54θ) = 1.Hmm, that's a simpler equation: the product of these four terms equals 1.So, (1 + 2cos2θ)(1 + 2cos6θ)(1 + 2cos18θ)(1 + 2cos54θ) = 1.Now, maybe this can be simplified. Let me see.I recall that for certain angles, the product of terms like (1 + 2cosφ) can be related to roots of unity or something similar.Wait, actually, I remember that for angles that are multiples of each other, especially in geometric progression, there might be a telescoping product or a known identity.In particular, the product (1 + 2cosφ)(1 + 2cos3φ)(1 + 2cos9φ)... can be simplified.Wait, yes! There's an identity that says that the product from k=0 to n of (1 + 2cos(3^k φ)) equals something.Wait, let me recall. I think the product (1 + 2cosφ)(1 + 2cos3φ)(1 + 2cos9φ)... up to n terms can be expressed in terms of sine functions.Wait, actually, I found a resource that says:The product from k=0 to n of (1 + 2cos(3^k x)) = (sin(3^{n+1} x))/(2^n sinx).Wait, let me check that.Yes, I think it's similar to the identity for the product of (2cosx + 1), but I'm not sure. Let me test it for small n.Let n=0: product is (1 + 2cosx). According to the formula, sin(3^{1}x)/(2^0 sinx) = sin3x/sinx. Is that equal to (1 + 2cosx)?Yes, because sin3x = 3sinx - 4sin³x, so sin3x/sinx = 3 - 4sin²x = 3 - 4(1 - cos²x) = 3 - 4 + 4cos²x = -1 + 4cos²x. Wait, that doesn't match 1 + 2cosx.Hmm, maybe I remembered the identity incorrectly. Let me think again.Wait, another approach: use complex numbers. Let me denote z = e^{iφ}, then cosφ = (z + z^{-1})/2.So, 1 + 2cosφ = 1 + z + z^{-1} = (z^2 + z + 1)/z.Similarly, 1 + 2cos3φ = (z^6 + z^3 + 1)/z^3.Wait, so the product becomes:(z^2 + z + 1)/z * (z^6 + z^3 + 1)/z^3 * (z^18 + z^9 + 1)/z^9 * (z^54 + z^27 + 1)/z^27.Multiply these together:Numerator: (z^2 + z + 1)(z^6 + z^3 + 1)(z^18 + z^9 + 1)(z^54 + z^27 + 1).Denominator: z * z^3 * z^9 * z^27 = z^{1+3+9+27} = z^{40}.Hmm, numerator is a product of terms of the form z^{2*3^k} + z^{3^k} + 1 for k=0 to 3.Wait, actually, each term is z^{2*3^k} + z^{3^k} + 1. So, the product is:(z^2 + z + 1)(z^6 + z^3 + 1)(z^18 + z^9 + 1)(z^54 + z^27 + 1).I wonder if this product simplifies to something.Wait, let me consider that each term is a factor of z^{3^{k+1}} - 1.Because z^{3^{k+1}} - 1 = (z^{3^k} - 1)(z^{2*3^k} + z^{3^k} + 1).So, z^{2*3^k} + z^{3^k} + 1 = (z^{3^{k+1}} - 1)/(z^{3^k} - 1).Therefore, the product becomes:Product = [(z^3 - 1)/(z - 1)] * [(z^9 - 1)/(z^3 - 1)] * [(z^27 - 1)/(z^9 - 1)] * [(z^81 - 1)/(z^27 - 1)].Notice that this is a telescoping product:(z^3 - 1)/(z - 1) * (z^9 - 1)/(z^3 - 1) * (z^27 - 1)/(z^9 - 1) * (z^81 - 1)/(z^27 - 1) = (z^81 - 1)/(z - 1).So, the numerator becomes z^81 - 1, and the denominator is z - 1.Therefore, the entire product is (z^81 - 1)/(z - 1).Therefore, going back, the product (1 + 2cosφ)(1 + 2cos3φ)(1 + 2cos9φ)(1 + 2cos27φ) = (z^81 - 1)/(z - 1).But z = e^{iφ}, so z^81 = e^{i81φ}.So, (z^81 - 1)/(z - 1) = e^{i81φ} - 1 / (e^{iφ} - 1) = [e^{i81φ} - 1]/[e^{iφ} - 1].But this is equal to 1 + e^{iφ} + e^{i2φ} + ... + e^{i80φ}.Wait, but we have that:(1 + 2cosφ)(1 + 2cos3φ)(1 + 2cos9φ)(1 + 2cos27φ) = (e^{i81φ} - 1)/(e^{iφ} - 1).So, the product equals (e^{i81φ} - 1)/(e^{iφ} - 1).But we have that this product equals 1, as per our earlier equation.So, (e^{i81φ} - 1)/(e^{iφ} - 1) = 1.Therefore, e^{i81φ} - 1 = e^{iφ} - 1.So, e^{i81φ} = e^{iφ}.Therefore, e^{i81φ} = e^{iφ}.Which implies that 81φ ≡ φ mod 360°, because e^{iθ} is periodic with period 360°.So, 81φ - φ = 80φ ≡ 0 mod 360°.Therefore, 80φ = 360°k, where k is an integer.So, φ = (360°k)/80 = (9°k)/2.But φ = 2θ, so 2θ = (9°k)/2.Therefore, θ = (9°k)/4.So, θ = (9/4)°k.Now, we need θ to be the least positive angle, so k=1.Therefore, θ = 9/4°, which is 2.25°.But wait, let me verify this.If θ = 9/4°, then φ = 2θ = 9/2°, and 81φ = 81*(9/2)° = 729/2° = 364.5°, which is equivalent to 4.5°, since 364.5° - 360° = 4.5°.Wait, but 81φ = 729/2°, which is 364.5°, which is 360° + 4.5°, so e^{i81φ} = e^{i4.5°}.Similarly, e^{iφ} = e^{i4.5°}.Wait, so e^{i81φ} = e^{iφ}, so indeed, they are equal. Therefore, the equation is satisfied.So, θ = 9/4° is a solution. Is it the least positive solution?Yes, because if k=0, θ=0°, which is not positive. k=1 gives θ=9/4°, which is positive and the smallest.Therefore, the least positive angle α is 9/4°, which is 9/4 degrees.So, m = 9, n = 4, which are coprime. Therefore, m + n = 13.Final AnswerThe value of ( m + n ) is boxed{13}.</think>

🧠:<think>Okay, so I have this problem where I need to find the absolute value of the product of two integers, a and b, given that the cubic polynomial x³ + a x² + b x + 9a has two roots that coincide, and all three roots are integers. Hmm, that sounds a bit complicated, but let's break it down step by step.First, let me recall that if a polynomial has two coinciding roots, it means that it has a double root and another distinct root. So, in this case, the cubic polynomial has roots that are m, m, and n, where m is the repeated root, and n is the distinct one. Since all roots are integers, m and n must be integers.Given that, I can express the polynomial as (x - m)²(x - n). If I expand this, I can then compare it to the given polynomial x³ + a x² + b x + 9a to find relationships between m, n, a, and b.Let me try expanding (x - m)²(x - n). First, (x - m)² is x² - 2mx + m². Then, multiplying by (x - n):(x² - 2mx + m²)(x - n) = x³ - n x² - 2m x² + 2m n x + m² x - m² n.Combining like terms, this becomes:x³ - (2m + n)x² + (m² + 2mn)x - m² n.Now, comparing this with the given polynomial x³ + a x² + b x + 9a, I can equate the coefficients:1. Coefficient of x²: -(2m + n) = a.2. Coefficient of x: m² + 2mn = b.3. Constant term: -m² n = 9a.So, from the first equation, I have a = -(2m + n). From the third equation, I have -m² n = 9a. Let me substitute a from the first equation into the third equation:-m² n = 9*(-(2m + n)) => -m² n = -9*(2m + n).Simplifying, I get:-m² n = -18m - 9n.Multiplying both sides by -1:m² n = 18m + 9n.Let me rearrange this equation:m² n - 18m - 9n = 0.Hmm, maybe I can factor this equation. Let me factor out n from the first and third terms:n(m² - 9) - 18m = 0.So, n(m² - 9) = 18m.Then, solving for n:n = (18m)/(m² - 9).So, n must be an integer because all roots are integers. Therefore, (m² - 9) must divide 18m.Let me write that as:(m² - 9) divides 18m.Since m is an integer, m² - 9 is also an integer. Let me factor m² - 9:m² - 9 = (m - 3)(m + 3).So, (m - 3)(m + 3) divides 18m. Therefore, (m - 3)(m + 3) must be a divisor of 18m.Hmm, so (m - 3)(m + 3) divides 18m. Let's think about what this implies.Since (m - 3) and (m + 3) are two integers that differ by 6, their product divides 18m. So, I need to find integer values of m such that (m - 3)(m + 3) divides 18m.Let me consider possible integer values of m. Since m is an integer, and m² - 9 must divide 18m, perhaps I can list possible divisors of 18m and see which ones can be expressed as (m - 3)(m + 3).Alternatively, since (m - 3)(m + 3) divides 18m, let's denote d = (m - 3)(m + 3). Then d divides 18m, so d must be a divisor of 18m. Let's also note that d = m² - 9.Given that, perhaps it's easier to find integer m such that m² - 9 divides 18m. Let me write that as:(m² - 9) | 18m.Which means that there exists an integer k such that 18m = k*(m² - 9).So, 18m = k*(m² - 9).Let's rearrange this equation:k*m² - 18m - 9k = 0.This is a quadratic equation in terms of m. Let me write it as:k*m² - 18m - 9k = 0.Since m is an integer, the discriminant of this quadratic equation must be a perfect square. The discriminant D is:D = (-18)² - 4*k*(-9k) = 324 + 36k².Wait, that seems a bit complicated. Maybe there's another approach.Let me think about possible small integer values of m and see if they satisfy the condition that (m² - 9) divides 18m.Let me start testing integer values for m.First, m = 0: m² - 9 = -9, which divides 18*0 = 0. So, technically, it divides, but m = 0 would make the roots 0, 0, n. Plugging into the polynomial, let's see:If m = 0, then the polynomial is (x - 0)²(x - n) = x²(x - n) = x³ - n x². Comparing to x³ + a x² + b x + 9a, we have:Coefficient of x²: -n = a.Coefficient of x: 0 = b.Constant term: 0 = 9a.From constant term, 0 = 9a => a = 0, but in the problem statement, a and b are nonzero integers. So, m = 0 is invalid.Next, m = 1:m² - 9 = 1 - 9 = -8. Does -8 divide 18*1 = 18? 18 divided by -8 is -2.25, which is not an integer. So, no.m = -1:m² - 9 = 1 - 9 = -8. Does -8 divide 18*(-1) = -18? -18 divided by -8 is 2.25, not integer. So, no.m = 2:m² - 9 = 4 - 9 = -5. Does -5 divide 18*2 = 36? 36 divided by -5 is -7.2, not integer.m = -2:m² - 9 = 4 - 9 = -5. Does -5 divide 18*(-2) = -36? -36 divided by -5 is 7.2, not integer.m = 3:m² - 9 = 9 - 9 = 0. Division by zero is undefined, so invalid.m = -3:Same as above, m² - 9 = 0, undefined.m = 4:m² - 9 = 16 - 9 = 7. Does 7 divide 18*4 = 72? 72 divided by 7 is about 10.2857, not integer.m = -4:7 divides 18*(-4) = -72. -72 / 7 ≈ -10.2857, not integer.m = 5:m² - 9 = 25 - 9 = 16. Does 16 divide 18*5 = 90? 90 / 16 = 5.625, not integer.m = -5:16 divides 18*(-5) = -90. -90 / 16 = -5.625, not integer.m = 6:m² - 9 = 36 - 9 = 27. Does 27 divide 18*6 = 108? 108 / 27 = 4, which is integer. So, m = 6 is a possible solution.Let me check m = 6:Compute n = (18m)/(m² - 9) = (18*6)/(36 - 9) = 108 / 27 = 4. So, n = 4.Wait, but n = 4, and m = 6. So the roots are 6, 6, 4.Now, let's compute a and b.From earlier, a = -(2m + n) = -(2*6 + 4) = -(12 + 4) = -16.Wait, but in the problem statement, it's stated that a and b are nonzero integers. So, a = -16, which is nonzero. Let's compute b:b = m² + 2mn = 6² + 2*6*4 = 36 + 48 = 84.So, b = 84.Then, the polynomial is x³ + (-16)x² + 84x + 9*(-16) = x³ -16x² +84x -144.Let me verify if this has a double root at 6 and a single root at 4.Compute (x - 6)²(x - 4) = (x² -12x +36)(x -4) = x³ -4x² -12x² +48x +36x -144 = x³ -16x² +84x -144. Yes, that's correct.So, m = 6, n = 4, a = -16, b = 84. So, |ab| = |-16 * 84| = | -1344 | = 1344. Hmm, but that doesn't seem to match the answer I saw earlier. Wait, maybe I made a mistake in calculation.Wait, no, let me double-check. Wait, m = 6, then n = 4, but let me check if n is indeed 4 or maybe I miscalculated.Wait, n = (18m)/(m² -9) = (18*6)/(36 -9) = 108 /27 = 4. That's correct.Then a = -(2m + n) = -(12 +4) = -16. Correct.b = m² + 2mn = 36 + 2*6*4 = 36 +48 = 84. Correct.So, |ab| = |-16 * 84|. Let me compute 16*84:16*80 = 1280, 16*4=64, so total is 1280 +64 = 1344. So, |ab| is 1344. But wait, the answer I saw earlier was 96. Hmm, maybe I made a mistake.Wait, maybe I missed another possible m value.Let me continue testing m values beyond 6.m = 7:m² - 9 = 49 - 9 = 40. Does 40 divide 18*7=126? 126 /40 = 3.15, not integer.m = -6:m² -9 = 36 -9 =27. Does 27 divide 18*(-6) = -108? -108 /27 = -4, which is integer.So, m = -6, n = (18*(-6))/(36 -9) = (-108)/27 = -4.So, n = -4.Then, a = -(2m +n) = -(2*(-6) + (-4)) = -(-12 -4) = -(-16) =16.b = m² + 2mn = (-6)^2 + 2*(-6)*(-4) = 36 + 48 =84.So, a =16, b=84. Then, |ab| = |16*84| = 1344. Same as before.Wait, so both m=6 and m=-6 give |ab|=1344.But earlier, I thought the answer was 96. Maybe I made a mistake.Wait, perhaps I missed another m value. Let me check m=3 and m=-3, but m=3 gives m²-9=0, which is undefined, so skip.m=9:m² -9=81-9=72. Does 72 divide 18*9=162? 162/72=2.25, not integer.m=-9:72 divides 18*(-9)=-162. -162/72=-2.25, not integer.m=12:m² -9=144-9=135. 135 divides 18*12=216? 216/135=1.6, not integer.m=-12:135 divides 18*(-12)=-216. -216/135=-1.6, not integer.Hmm, maybe I need to consider other divisors. Wait, when I set m² -9 divides 18m, perhaps I can factor 18m.Wait, 18m = 2*3²*m.So, m² -9 divides 2*3²*m. Let me think about possible factors.Alternatively, perhaps I can set m² -9 as a divisor of 54, since 18m is 18 times m, and m is an integer. Wait, but that might not be directly applicable.Wait, let me think differently. Since m² -9 divides 18m, and m² -9 = (m-3)(m+3), which are two factors that differ by 6. So, perhaps (m-3) and (m+3) are divisors of 54, because 18m can be expressed as 54 times something.Wait, 18m = (m-3)(m+3)*k, where k is integer. So, (m-3)(m+3) must divide 18m.Let me think of possible pairs (m-3, m+3) that are divisors of 18m.Wait, maybe it's better to think of possible m such that (m-3) divides 54, since m-3 and m+3 differ by 6.Wait, let me list all pairs of divisors of 54 that differ by 6.Divisors of 54 are: ±1, ±2, ±3, ±6, ±9, ±18, ±27, ±54.Looking for pairs (d, d+6) that are both divisors of 54.Looking at positive divisors:1 and 7: 7 not a divisor.2 and 8: 8 not a divisor.3 and 9: both divisors.6 and 12: 12 not a divisor.9 and 15: 15 not a divisor.18 and 24: 24 not a divisor.27 and 33: 33 not a divisor.54 and 60: 60 not a divisor.Negative divisors:-1 and -7: -7 not a divisor.-2 and -8: -8 not a divisor.-3 and -9: both divisors.-6 and -12: -12 not a divisor.-9 and -15: -15 not a divisor.-18 and -24: -24 not a divisor.-27 and -33: -33 not a divisor.-54 and -60: -60 not a divisor.So, the only pairs are (3,9) and (-3,-9).So, for (m-3, m+3) = (3,9):Then m-3=3 => m=6and m+3=9 => m=6.Consistent.Similarly, for (m-3, m+3)=(-3,-9):m-3=-3 => m=0, but m=0 is invalid as earlier.Alternatively, m+3=-3 => m=-6and m-3=-9 => m=-6. So, m=-6.So, m=6 and m=-6 are the only possible values.So, as I did earlier, m=6 gives n=4, a=-16, b=84.m=-6 gives n=-4, a=16, b=84.Thus, |ab|=|(-16)(84)|=1344 and |ab|=|16*84|=1344.Wait, but the problem statement says "two roots coincide", but in both cases, I have a double root at 6 or -6, which is correct.But the answer I saw earlier was 96, so maybe I made a mistake.Wait, let me check my calculations again.Wait, when m=6, n=4.Compute a = -(2m +n) = -(12 +4) = -16.Compute b = m² + 2mn = 36 + 2*6*4 = 36 +48=84.So, 9a = 9*(-16)=-144.But in the polynomial, the constant term is 9a, which should be equal to -m² n.So, -m² n = -36*4 = -144.Yes, that's correct.Similarly, for m=-6, n=-4.a = -(2*(-6) + (-4)) = -(-12 -4)=16.b = (-6)^2 + 2*(-6)*(-4)=36 + 48=84.9a=9*16=144.And -m² n= -36*(-4)=144. Correct.So, everything checks out. So, |ab|=1344.But earlier, I thought the answer was 96. Maybe I confused it with another problem.Wait, perhaps I made a mistake in the initial step.Wait, let me double-check the expansion of (x - m)^2 (x - n).(x - m)^2 = x² - 2mx + m².Multiply by (x - n):(x² - 2mx + m²)(x - n) = x³ -n x² -2m x² + 2m n x + m² x - m² n.Combine like terms:x³ + (-n -2m)x² + (2m n + m²)x - m² n.So, coefficients:x³: 1x²: - (2m +n)x: m² + 2m nconstant: -m² n.So, that's correct.Comparing to x³ + a x² +b x +9a,We get:- (2m +n) = a => a = - (2m +n)m² + 2m n = b- m² n = 9a => m² n = -9a.So, substituting a from the first equation into the third:m² n = -9*(- (2m +n)) =9*(2m +n).So, m² n =18m +9n.Which leads to m² n -18m -9n=0.Then, n(m² -9)=18m.Thus, n=18m/(m² -9).So, that's correct.Therefore, m=6 gives n=4, a=-16, b=84, |ab|=1344.m=-6 gives n=-4, a=16, b=84, |ab|=1344.Wait, but the problem states that a and b are nonzero integers, which is satisfied.So, perhaps the answer is 1344. But I thought earlier it was 96. Maybe I made a mistake in the earlier problem.Alternatively, perhaps I made a mistake in the calculation.Wait, let me check m=3: m²-9=0, which is invalid.m=4: m²-9=7, which doesn't divide 72.Wait, maybe I missed another value. Let me check m=9:m² -9=72, which divides 18*9=162? 162/72=2.25, not integer.m=-9: 72 divides 18*(-9)=-162. -162/72=-2.25, not integer.m=12: m²-9=135, divides 18*12=216? 216/135=1.6, not integer.m=-12: 135 divides -216? -216/135=-1.6, not integer.So, indeed, only m=6 and m=-6 give integer n.Thus, |ab|=1344.Wait, but the problem statement says "two roots coincide, and all three roots are integers." So, that's satisfied.Wait, perhaps I made a mistake in the original problem, or perhaps the answer is indeed 1344.But in the initial problem, I saw an answer of 96, but maybe that was a different problem.Wait, let me check the original problem again."Suppose a,b are nonzero integers such that two roots of x³ + a x² +b x +9a coincide, and all three roots are integers. Find |ab|."So, according to my calculations, |ab|=1344.But perhaps I made a mistake in the coefficients.Wait, let me check the polynomial when m=6, n=4:(x-6)^2(x-4) = (x² -12x +36)(x-4) = x³ -4x² -12x² +48x +36x -144 = x³ -16x² +84x -144.Which matches x³ +a x² +b x +9a with a=-16, b=84, 9a=-144.Yes, correct.Similarly, for m=-6, n=-4:(x+6)^2(x+4) = (x² +12x +36)(x+4) = x³ +4x² +12x² +48x +36x +144 = x³ +16x² +84x +144.Which matches x³ +a x² +b x +9a with a=16, b=84, 9a=144.Yes, correct.So, in both cases, |ab|=1344.But the initial problem's answer was 96, so perhaps I made a mistake in the problem statement.Wait, perhaps the polynomial was x³ +a x² +b x +9a, but maybe I misread it as x³ +a x² +b x +9a, but maybe it's x³ +a x² +b x +9a.Wait, no, that's what I used.Alternatively, maybe I have a mistake in the calculation of |ab|.Wait, in the case of m=6, a=-16, b=84. So, ab= (-16)(84)= -1344, |ab|=1344.Similarly, m=-6, a=16, b=84, ab=16*84=1344, |ab|=1344.So, that's correct.Wait, but maybe I made a mistake in the initial equation.Wait, when I set n=18m/(m² -9), for m=6, n=4, correct.But perhaps I can find another solution where n is different.Wait, let me think again about the equation n(m² -9)=18m.Perhaps I can write this as n=18m/(m² -9). So, for n to be integer, 18m must be divisible by m² -9.So, m² -9 must divide 18m.Another approach: Let me set k=m² -9, so k divides 18m, and k=m² -9.So, k divides 18m, and k = m² -9.So, m² ≡9 mod k.But maybe that's not helpful.Alternatively, since k divides 18m, and k=m² -9, so m² -9 divides 18m.Let me write that as m² -9 | 18m.So, m² -9 divides 18m.Let me think of m as a positive integer greater than 3, since m=1,2,3, etc., but m=3 gives k=0, which is invalid.Wait, m=6 we already saw.m=9: m² -9=72, which divides 18*9=162? 162/72=2.25, not integer.m=12: m² -9=135, divides 18*12=216? 216/135=1.6, not integer.m=15: m²-9=225-9=216, divides 18*15=270? 270/216=1.25, not integer.m=18: m²-9=324-9=315, divides 18*18=324? 324/315≈1.028, not integer.So, no solution beyond m=6.Similarly for negative m:m=-6: m²-9=27, divides 18*(-6)=-108? -108/27=-4, integer.m=-9:72 divides 18*(-9)=-162? -162/72=-2.25, not integer.m=-12:135 divides 18*(-12)=-216? -216/135=-1.6, not integer.So, only m=6 and m=-6.Thus, |ab|=1344.But wait, in the initial problem, the answer was 96, so maybe I made a mistake in the problem statement.Wait, let me check again: the polynomial is x³ +a x² +b x +9a, right?Yes, and the roots are m, m, n.So, I think my calculations are correct, and the answer should be 1344.But perhaps I made a mistake in the problem statement, maybe the polynomial is x³ +a x² +b x +9a, or perhaps x³ +a x² +b x +9.Wait, no, the problem says x³ +a x² +b x +9a.Hmm, I think I did everything correctly. So, perhaps the answer is 1344.But to be sure, let me try to see if there's another approach.Let me consider that since the polynomial has integer roots, by Rational Root Theorem, any rational root p/q must have p dividing the constant term and q dividing the leading coefficient. Since the leading coefficient is 1, all rational roots are integers dividing the constant term, which is 9a.So, possible roots are divisors of 9a, which are ±1, ±3, ±9, ±a, ±3a, ±9a, but since a is an integer, but a itself can be anything, so it's a bit broad.But since we have a double root, say m, then m must be a root of both the polynomial and its derivative.So, let me compute the derivative of the polynomial:f(x) =x³ +a x² +b x +9af'(x)=3x² +2a x +b.So, if m is a double root, then f(m)=0 and f'(m)=0.So, let's set up these equations:1. m³ +a m² +b m +9a =0.2. 3m² +2a m +b=0.Let me solve these two equations for a and b.From equation 2: b= -3m² -2a m.Substitute into equation 1:m³ +a m² + (-3m² -2a m) m +9a =0.Simplify:m³ +a m² -3m³ -2a m² +9a =0.Combine like terms:(1 -3)m³ + (a -2a)m² +9a=0 =>-2m³ -a m² +9a=0.Rearrange:-2m³ -a m² +9a=0 =>-2m³ =a m² -9a =>-2m³ =a(m² -9).Thus,a= (-2m³)/(m² -9).So, a must be an integer, so (m² -9) must divide -2m³.But since m² -9 divides 18m as before, and now it also divides -2m³, perhaps we can find m such that m² -9 divides both 18m and -2m³.But perhaps this approach is more complicated.Alternatively, let me use the earlier result: a= -2m³/(m² -9).But from earlier, a= -(2m +n).So, equate:-2m³/(m² -9) = -(2m +n).Multiply both sides by (m² -9):-2m³ = -(2m +n)(m² -9).Simplify:-2m³ = -2m(m² -9) -n(m² -9).Multiply both sides by -1:2m³ =2m(m² -9) +n(m² -9).So,2m³ =2m³ -18m +n(m² -9).Subtract 2m³ from both sides:0= -18m +n(m² -9).Which is the same as earlier: n(m² -9)=18m.So, consistent.Thus, the only solutions are m=6 and m=-6, leading to |ab|=1344.Therefore, I think the correct answer is 1344.But wait, perhaps I made a mistake in the initial step. Let me check m=3:Wait, m=3, then m² -9=0, which is invalid.m=4:n=18*4/(16-9)=72/7, not integer.m=5:n=90/16=5.625, not integer.m=6:n=108/27=4, integer.m=7:n=126/40=3.15, not integer.m=8:n=144/55≈2.618, not integer.m=9:n=162/72=2.25, not integer.m=10:n=180/91≈1.978, not integer.m=11:n=198/112≈1.767, not integer.m=12:n=216/135=1.6, not integer.So, only m=6 and m=-6.Thus, the answer is |ab|=1344.Wait, but in the initial problem, the answer was 96, so perhaps I made a mistake.Wait, maybe I misread the problem. Let me check again.The problem says: "two roots of x³ +a x² +b x +9a coincide, and all three roots are integers."So, the polynomial is x³ +a x² +b x +9a.So, in my calculations, I got |ab|=1344.But perhaps I made a mistake in the initial setup.Wait, let me try another approach.Let me denote the roots as m, m, n.Then, sum of roots: 2m +n = -a.Sum of products: m² + 2mn =b.Product: m² n= -9a.So, from product: m² n = -9a => a= -m² n /9.From sum of roots: 2m +n = -a => a= -2m -n.Thus,-2m -n = -m² n /9.Multiply both sides by -9:18m +9n =m² n.Which is the same as before.Thus, m² n -18m -9n=0 => n(m² -9)=18m.So, n=18m/(m² -9).Thus, same as before.Therefore, only m=6 and m=-6 give integer n.Thus, |ab|=1344.So, I think the correct answer is 1344.But perhaps the problem was different. Wait, maybe the polynomial was x³ +a x² +b x +9, which would change the calculations.But the problem says x³ +a x² +b x +9a, so 9a is correct.Thus, I think the answer is 1344.But since the initial problem had an answer of 96, perhaps I made a mistake.Wait, let me check m=3:m=3, m²-9=0, invalid.m=4: n=18*4/(16-9)=72/7, not integer.m=5:90/16, not integer.m=6:108/27=4.m=7:126/40, not integer.m=8:144/55, not integer.m=9:162/72=2.25, not integer.m=10:180/91, not integer.m=12:216/135=1.6, not integer.So, only m=6 and m=-6.Thus, |ab|=1344.Therefore, I think the correct answer is 1344.</think>

🧠:<think>Okay, so I've got this problem with two functions, f(x) and g(x), and I need to tackle two parts. Let me break it down step by step.Starting with part (1): The function is f(x) = (x)/(1+x) - a ln(1+x). It says that f(x) has an extreme value at x=0, and I need to find the maximum value of f(x). Hmm, okay. So, extreme value means it's either a maximum or a minimum. Since the problem is asking for the maximum value, I suppose we need to confirm whether it's a maximum and then find that value.First, to find extreme values, we look at the derivative. So, I should compute f'(x). Let's do that.f(x) = x/(1+x) - a ln(1+x)So, f'(x) is the derivative of x/(1+x) which is [1*(1+x) - x*1]/(1+x)^2 = (1+x -x)/(1+x)^2 = 1/(1+x)^2.Then, the derivative of -a ln(1+x) is -a/(1+x). So, putting it together:f'(x) = 1/(1+x)^2 - a/(1+x)We are told that f(x) has an extreme value at x=0. That means f'(0) = 0.Let's compute f'(0):f'(0) = 1/(1+0)^2 - a/(1+0) = 1 - aSet this equal to zero because it's an extreme value:1 - a = 0 => a = 1Okay, so a is 1. Therefore, the function becomes f(x) = x/(1+x) - ln(1+x).Now, to find the maximum value, since it's an extreme value at x=0, we can check the second derivative or analyze the behavior around x=0.Let me compute the second derivative f''(x):f''(x) is the derivative of f'(x) = 1/(1+x)^2 - 1/(1+x)So, derivative of 1/(1+x)^2 is -2/(1+x)^3, and derivative of -1/(1+x) is 1/(1+x)^2.So, f''(x) = -2/(1+x)^3 + 1/(1+x)^2Simplify: f''(x) = (-2 + (1+x))/(1+x)^3 = (x -1)/(1+x)^3Wait, let me double-check that derivative:Wait, f'(x) = (1/(1+x)^2) - (1/(1+x))So, f''(x) = d/dx [ (1/(1+x)^2) ] - d/dx [ (1/(1+x)) ] = -2/(1+x)^3 + 1/(1+x)^2So, f''(x) = (-2/(1+x)^3) + (1/(1+x)^2) = (-2 + (1+x))/(1+x)^3Yes, that's correct.So, f''(x) = (x - 1)/(1+x)^3At x=0, f''(0) = (0 -1)/(1+0)^3 = -1. Since f''(0) is negative, that means the function is concave down at x=0, so it's a local maximum.Therefore, the maximum value is f(0). Let's compute f(0):f(0) = 0/(1+0) - ln(1+0) = 0 - 0 = 0.So, the maximum value is 0. That seems straightforward.Now, moving on to part (2):① We have g(x) = ln(1+x) - b x. The question is whether there exists a real number b such that g(x) < 0 for all x in (0, +∞). If so, find the range of b; if not, explain why.Alright, so we need to find if there's a b such that ln(1+x) < b x for all x > 0.Let me think about this. The function ln(1+x) is concave and grows slower than linear functions as x increases. So, for large x, ln(1+x) is much smaller than linear functions. But near x=0, ln(1+x) behaves like x - x^2/2 + x^3/3 - ..., so it's approximately x when x is near 0.So, for small x, ln(1+x) is roughly x, so to have ln(1+x) < b x for all x >0, b must be at least 1, because when x approaches 0, ln(1+x)/x approaches 1. So, if b is less than 1, near x=0, ln(1+x) would be greater than b x, making g(x) positive. So, maybe b has to be at least 1.But let's analyze it more carefully.Let me define h(x) = ln(1+x)/x. I want to know the supremum of h(x) over x >0 because if I can find the maximum value of h(x), then any b greater than or equal to that maximum would satisfy ln(1+x) < b x.But, what's the maximum of h(x) = ln(1+x)/x?Let's compute its derivative to find critical points.h(x) = ln(1+x)/xh'(x) = [ (1/(1+x)) * x - ln(1+x) * 1 ] / x^2Simplify numerator:x/(1+x) - ln(1+x)So, h'(x) = [x/(1+x) - ln(1+x)] / x^2We need to find when h'(x) = 0:x/(1+x) - ln(1+x) = 0So, x/(1+x) = ln(1+x)Let me define k(x) = x/(1+x) - ln(1+x). We need to find x where k(x) = 0.Compute k(0): 0 - 0 = 0.Compute k(x) for x >0:Let me compute k(1): 1/2 - ln(2) ≈ 0.5 - 0.693 ≈ -0.193 <0Compute k(0.5): (0.5)/(1.5) - ln(1.5) ≈ 0.333 - 0.405 ≈ -0.072 <0Compute k(0.1): 0.1/1.1 - ln(1.1) ≈ 0.0909 - 0.0953 ≈ -0.0044 <0So, k(x) is negative for x >0.Wait, but k(0) =0, and for x>0, k(x) <0. So, h'(x) is negative for all x>0, meaning h(x) is decreasing on (0, ∞).Therefore, the maximum value of h(x) is at x approaching 0, which is lim_{x→0} ln(1+x)/x =1.Therefore, h(x) is decreasing from 1 to 0 as x goes from 0 to ∞.Therefore, the supremum of h(x) is 1, and the infimum is 0.Therefore, for g(x) = ln(1+x) - b x <0 for all x>0, we need b > h(x) for all x>0, which means b must be greater than or equal to the supremum of h(x), which is 1.But wait, if b=1, then g(x) = ln(1+x) -x. Let's see what happens to g(x) when b=1.Compute lim_{x→0} (ln(1+x) -x)/x^2: It's a standard limit which is -1/2, so ln(1+x) -x ≈ -x^2/2 as x→0. So, g(x) approaches 0 from below as x approaches 0.Wait, but at x=0, g(0)=0. For x>0, if b=1, is g(x) <0?Compute the derivative of g(x) when b=1: g'(x) = 1/(1+x) -1 = -x/(1+x). So, for x>0, g'(x) <0, meaning g(x) is decreasing on (0, ∞). Since g(0)=0 and it's decreasing, then for x>0, g(x) <0.Hence, if b=1, g(x) <0 for all x>0.If b>1, then g(x)=ln(1+x)-b x. Since for x>0, ln(1+x) <x, so ln(1+x) -b x <x -b x = (1 -b)x <0 because b>1.Therefore, for b≥1, g(x) <0 for all x>0.If b<1, then near x=0, ln(1+x) ≈x, so ln(1+x) -b x ≈x -b x =x(1 -b) >0, since b<1. Therefore, g(x) >0 near x=0, which contradicts g(x)<0 for all x>0.Therefore, the range of b is [1, ∞).So, the answer is b ∈ [1, ∞).Now, part ②: Prove the inequality -1 < sum_{k=1}^n [k/(k² +1) - ln n] ≤ 1/2 for n=1,2,...Wait, the original problem says:Prove the inequality -1 < sum_{k=1}^{n} [k/(k² +1) - ln n] ≤ 1/2 for n=1,2,...Wait, actually, the sum is sum_{k=1}^n [k/(k² +1)] - ln n.So, let me denote S_n = sum_{k=1}^n [k/(k² +1)] - ln n.We need to show that -1 < S_n ≤ 1/2.Hmm, okay. So, first, let me see if I can express this sum in a telescoping manner or relate it to integrals or something.Wait, the term k/(k² +1) can be written as (1/2) * [ (2k)/(k² +1) ] = (1/2) * derivative of ln(k² +1). Hmm, but I'm not sure that helps directly.Alternatively, perhaps we can compare k/(k² +1) to ln(1 + 1/k) or something similar.Wait, from part (1), we had that for f(x) = x/(1+x) - ln(1+x), which had a maximum of 0 at x=0. Wait, that function f(x) was equal to x/(1+x) - ln(1+x), and we found that it's maximum at 0 is 0, and it's decreasing for x>0.So, for x>0, x/(1+x) - ln(1+x) ≤0, so x/(1+x) ≤ ln(1+x).Wait, but in part (2) ①, we saw that ln(1+x) < b x for b≥1. So, perhaps these inequalities can help.Wait, perhaps a better approach is to use the inequality that for x >0, x/(1+x) < ln(1+x) <x.Wait, let me test x=1: 1/2 < ln(2) <1, yes, that holds.So, for x>0, we have x/(1+x) < ln(1+x) <x.So, if we set x=1/k, then:(1/k)/(1 +1/k) < ln(1 +1/k) <1/kSimplify the left side:(1/k)/( (k+1)/k ) = 1/(k+1)So, 1/(k+1) < ln(1 +1/k) <1/kTherefore, for each k≥1,1/(k+1) < ln(1 +1/k) <1/kTherefore, if we consider the term k/(k² +1) = [k]/[k² +1] = [1]/[k +1/k] ≈ 1/(k + something). Wait, maybe not directly.Wait, perhaps instead, let's consider the sum S_n = sum_{k=1}^n [k/(k² +1)] - ln n.Let me see if I can relate this sum to telescoping sums or integrals.Alternatively, perhaps express k/(k² +1) as (1/2)(2k)/(k² +1) which is the derivative of ln(k² +1). Hmm, that might be useful for integrals.Wait, if I think of the sum as a Riemann sum approximation for the integral of 1/(x² +1), but I'm not sure.Alternatively, perhaps consider that:sum_{k=1}^n [k/(k² +1)] = (1/2) sum_{k=1}^n [ (2k)/(k² +1) ] = (1/2) sum_{k=1}^n [ d/dk ln(k² +1) ]But integrating 1/(x² +1) is arctan(x), but I don't think that directly helps.Alternatively, perhaps using the inequality 1/(k+1) < ln(1 +1/k) <1/k.Let me think:sum_{k=1}^n ln(1 +1/k) = sum_{k=1}^n [ln(k+1) - ln k] = ln(n+1) - ln1 = ln(n+1)Therefore, sum_{k=1}^n ln(1 +1/k) = ln(n+1)But from the inequality, we have:sum_{k=1}^n [1/(k+1)] < sum_{k=1}^n ln(1 +1/k) < sum_{k=1}^n [1/k]Which becomes:sum_{k=2}^{n+1} [1/k] < ln(n+1) < sum_{k=1}^n [1/k]So, sum_{k=2}^{n+1} [1/k] = H_{n+1} -1, where H_n is the nth harmonic number.And sum_{k=1}^n [1/k] = H_n.So, H_{n+1} -1 < ln(n+1) < H_nBut H_{n} = gamma + ln n + 1/(2n) - 1/(12n^2) + ..., but maybe that's too advanced.Alternatively, perhaps relate S_n to the harmonic series.Wait, our S_n is sum_{k=1}^n [k/(k² +1)] - ln n.Let me see if I can express k/(k² +1) in terms of 1/k or something.Wait, k/(k² +1) = [1/k] / [1 +1/k²]Hmm, maybe not helpful.Alternatively, perhaps write k/(k² +1) = [k +1 -1]/(k² +1) = [k +1]/(k² +1) -1/(k² +1)But [k +1]/(k² +1) = [k +1]/[(k +1)(k -1) + 2] Hmm, not sure.Alternatively, perhaps partial fractions.Wait, maybe not. Let me think differently.From the inequality 1/(k+1) < ln(1 +1/k) <1/k, as we had earlier.But we need to relate k/(k² +1) to ln(1 + something). Let me see:Wait, let me consider x =1/k.Then, from the inequality x/(1+x) < ln(1+x) <x.So, (1/k)/(1 +1/k) < ln(1 +1/k) <1/kWhich simplifies to:1/(k +1) < ln(1 +1/k) <1/kSo, that's the same as before.Now, let's write S_n = sum_{k=1}^n [k/(k² +1)] - ln n.Let me see if I can relate k/(k² +1) to ln(1 +1/k) or something.Wait, k/(k² +1) = [1]/[k +1/k]. Hmm, not directly.Alternatively, perhaps note that k/(k² +1) = [1]/[k +1/k] <1/k, since k +1/k >k, so 1/(k +1/k) <1/k.Similarly, 1/(k +1/k) >1/(k +1), since 1/k +1 <k +1/k for k≥1?Wait, let me check:For k≥1, k +1/k ≥k +1? Is that true?Wait, k +1/k - (k +1) =1/k -1.For k=1: 1 +1 - (1+1)=1 -1=0.For k=2: 2 +0.5 -3= -0.5 <0.So, actually, k +1/k <k +1 for k≥2.Therefore, 1/(k +1/k) >1/(k +1).So, k/(k² +1) =1/(k +1/k) >1/(k +1).Wait, so we have:1/(k +1) <k/(k² +1) <1/kBut from earlier, we have 1/(k +1) < ln(1 +1/k) <1/kSo, comparing, we have:1/(k +1) <k/(k² +1) <1/kAnd 1/(k +1) < ln(1 +1/k) <1/kSo, k/(k² +1) and ln(1 +1/k) are both between 1/(k +1) and1/k.But how are they related? Let's see:Compute k/(k² +1) - ln(1 +1/k):We need to see if this is positive or negative.Wait, let's compute for k=1:1/(1 +1) - ln(2) ≈0.5 -0.693≈-0.193 <0For k=2:2/(4 +1)=2/5=0.4, ln(1 +1/2)=ln(1.5)≈0.405, so 0.4 -0.405≈-0.005 <0For k=3:3/(9 +1)=0.3, ln(1 +1/3)=ln(4/3)≈0.2877, so 0.3 -0.2877≈0.0123 >0Hmm, so for k=1,2, the term is negative, for k=3, it becomes positive.Interesting. So, the difference alternates? Or perhaps after a certain k, it becomes positive.Wait, for k=4:4/(16 +1)=4/17≈0.2353, ln(1 +1/4)=ln(5/4)≈0.2231, so 0.2353 -0.2231≈0.0122 >0Similarly, for k=5:5/(25 +1)=5/26≈0.1923, ln(6/5)=ln(1.2)≈0.1823, so 0.1923 -0.1823≈0.01 >0So, it seems that for k≥3, k/(k² +1) > ln(1 +1/k)But for k=1,2, it's the other way.So, how can we use this to bound the sum S_n.Wait, our S_n is sum_{k=1}^n [k/(k² +1)] - ln n.Note that sum_{k=1}^n ln(1 +1/k)= ln(n+1), as we saw earlier.So, S_n = sum_{k=1}^n [k/(k² +1)] - ln n.But sum_{k=1}^n [k/(k² +1)] = sum_{k=1}^n [k/(k² +1)] = let's see if we can relate this to ln(n+1) or something.Wait, perhaps express S_n as sum_{k=1}^n [k/(k² +1) - ln(1 +1/k)] + sum_{k=1}^n ln(1 +1/k) - ln n.But sum_{k=1}^n ln(1 +1/k)=ln(n+1), so S_n = sum_{k=1}^n [k/(k² +1) - ln(1 +1/k)] + ln(n+1) - ln n.Thus, S_n = sum_{k=1}^n [k/(k² +1) - ln(1 +1/k)] + ln(1 +1/n)Hmm, okay, so S_n is equal to this sum plus ln(1 +1/n).Now, from earlier, we saw that for k=1,2, [k/(k² +1) - ln(1 +1/k)] is negative, and for k≥3, it's positive.So, let's write S_n = [k=1 term] + [k=2 term] + sum_{k=3}^n [k/(k² +1) - ln(1 +1/k)] + ln(1 +1/n)Compute the first two terms:k=1: 1/2 - ln(2) ≈-0.193k=2: 2/5 - ln(3/2)≈0.4 -0.405≈-0.005So, sum of first two terms≈-0.198Then, for k≥3, each term [k/(k² +1) - ln(1 +1/k)] is positive, as we saw.So, the total sum is -0.198 + positive terms + ln(1 +1/n)But ln(1 +1/n) is positive and less than 1/n.We need to bound S_n.Let me try to bound the sum.First, note that for k≥3,k/(k² +1) - ln(1 +1/k) = [k/(k² +1)] - [ln(k+1) - ln k]But we can write this as [k/(k² +1) + ln k] - ln(k+1)But not sure.Alternatively, perhaps bound each term.From earlier, we have for k≥3:k/(k² +1) - ln(1 +1/k) =k/(k² +1) - ln((k+1)/k)=k/(k² +1) - [ln(k+1) - ln k]But this seems to complicate.Alternatively, perhaps use the inequality that for x>0, x -x^2/2 < ln(1+x) <x.Wait, for x=1/k, we have 1/k -1/(2k²) < ln(1 +1/k) <1/kSo, ln(1 +1/k) >1/k -1/(2k²)Therefore, [k/(k² +1)] - ln(1 +1/k) < [k/(k² +1)] - [1/k -1/(2k²)] = [k/(k² +1)] -1/k +1/(2k²)Simplify:[k/(k² +1) -1/k] +1/(2k²) = [k^2 - (k² +1)]/[k(k² +1)] +1/(2k²)= (-1)/[k(k² +1)] +1/(2k²)Which is negative because (-1)/[k(k² +1)] is negative and 1/(2k²) is positive.Wait, but we have:[k/(k² +1)] - ln(1 +1/k) < [k/(k² +1)] - [1/k -1/(2k²)] = negative.But we know that for k≥3, [k/(k² +1)] - ln(1 +1/k) is positive.Hmm, perhaps a different approach.Alternatively, let's consider the function f(x) =x/(x² +1) - ln(1 +1/x) for x≥3.We can analyze its behavior.Compute derivative:f(x)=x/(x² +1) - ln(1 +1/x)=x/(x² +1) - ln((x+1)/x)=x/(x² +1) - ln(x+1) + ln xBut derivative f’(x)= [ (1)(x² +1) -x(2x) ]/(x² +1)^2 - [1/(x+1) -1/x]Simplify numerator of first term: (x² +1 -2x²)= -x² +1So, f’(x)= (-x² +1)/(x² +1)^2 - [1/(x+1) -1/x]Simplify the second part:1/(x+1) -1/x= [x - (x+1)]/[x(x+1)]= -1/[x(x+1)]So, f’(x)= (-x² +1)/(x² +1)^2 +1/[x(x+1)]Now, for x≥3, let's see:(-x² +1)/(x² +1)^2 is negative because numerator is negative.1/[x(x+1)] is positive.So, f’(x)= negative + positive.We need to see if f’(x) is positive or negative.Compute for x=3:(-9 +1)/(9 +1)^2 +1/(3*4)= (-8)/100 +1/12≈-0.08 +0.083≈0.003>0For x=4:(-16 +1)/(16 +1)^2 +1/(4*5)= (-15)/289 +1/20≈-0.0519 +0.05≈-0.0019<0So, at x=3, f’(x)≈0.003>0, and at x=4,≈-0.0019<0.So, f’(x) changes from positive to negative between x=3 and x=4.Therefore, f(x) has a maximum somewhere around x=3.5.So, the function f(x)=x/(x² +1) - ln(1 +1/x) increases up to x≈3.5, then decreases.But since f(x) is positive for k≥3, and the terms are decreasing after a certain point.Hmm, maybe this is getting too complicated.Alternatively, perhaps approximate the sum S_n.Let me consider that S_n = sum_{k=1}^n [k/(k² +1)] - ln n.Let me write k/(k² +1)= [1/2][2k/(k² +1)] = [1/2][ (k² +1 - (k² -1)) / (k² +1) ]= [1/2][1 - (k² -1)/(k² +1)].Wait, that might not help.Alternatively, perhaps note that k/(k² +1)= [1/k]/[1 +1/k²].Hmm, which is similar to the expansion for ln(1+x).Wait, maybe not.Alternatively, perhaps relate k/(k² +1) to an integral.Consider that k/(k² +1)= integral from 0 to1 of x^{k²} dx? No, that seems off.Wait, perhaps use the fact that k/(k² +1)= integral from 0 to1 of x^{k²} dx? Hmm, not directly.Alternatively, perhaps write k/(k² +1)= [1/2][ (k +1)/(k² +1) + (k -1)/(k² +1) ]Wait, not sure.Alternatively, perhaps use the fact that sum_{k=1}^n [k/(k² +1)] can be approximated by an integral.Compute integral from 1 to n of x/(x² +1) dx.Let u=x² +1, du=2x dx, so integral becomes (1/2) ln(u)= (1/2) ln(x² +1)So, integral from1 ton is (1/2)(ln(n² +1) - ln2)But our sum is sum_{k=1}^n [k/(k² +1)]≈ integral from1 ton+1 of x/(x² +1) dx= (1/2) ln((n+1)^2 +1) - (1/2) ln2.But this is an approximation.Alternatively, perhaps use the Euler-Maclaurin formula, but that might be overkill.Alternatively, consider that sum_{k=1}^n [k/(k² +1)]≈ (1/2) ln(n² +1) + CBut I'm not sure.Alternatively, let's compute the difference between S_n and the integral.But this might be too involved.Alternatively, let me note that:sum_{k=1}^n [k/(k² +1)] = sum_{k=1}^n [1/(k +1/k)]Hmm, since k/(k² +1)=1/(k +1/k)Now, 1/(k +1/k) can be compared to 1/(k +1).Since k +1/k ≤k +1 for k≥1? Wait, no, because 1/k ≤1.Wait, for k≥1, 1/k ≤1, so k +1/k ≤k +1.Therefore, 1/(k +1/k) ≥1/(k +1)Thus, sum_{k=1}^n [1/(k +1/k)] ≥sum_{k=1}^n [1/(k +1)]=H_{n+1} -1Similarly, 1/(k +1/k) ≤1/k, since k +1/k ≥k.Thus, sum_{k=1}^n [1/(k +1/k)] ≤sum_{k=1}^n [1/k]=H_nTherefore, H_{n+1} -1 ≤sum_{k=1}^n [k/(k² +1)] ≤H_nTherefore, S_n = sum_{k=1}^n [k/(k² +1)] - ln n ≤H_n - ln nAnd H_n - ln n approaches gamma≈0.577 as n→∞.But we need to show S_n ≤1/2.Wait, for n=1:S_1=1/(1+1) - ln1=0.5 -0=0.5So, S_1=0.5For n=2:sum_{k=1}^2 [k/(k² +1)]=1/2 +2/5=0.5 +0.4=0.9ln2≈0.693So, S_2=0.9 -0.693≈0.207 <0.5For n=3:sum=1/2 +2/5 +3/10=0.5 +0.4 +0.3=1.2ln3≈1.0986S_3=1.2 -1.0986≈0.1014 <0.5Similarly, as n increases, sum_{k=1}^n [k/(k² +1)] grows like (1/2) ln(n² +1), which is roughly ln n + (1/2) ln(1 +1/n²)≈ln n + negligible.Thus, S_n≈(1/2) ln(n² +1) - ln n≈(1/2)(2 ln n + negligible) - ln n≈ln n - ln n=0But actually, it's a bit more precise.Wait, let me compute:sum_{k=1}^n [k/(k² +1)] ≈ integral from1 ton of x/(x² +1) dx= (1/2) ln(n² +1) - (1/2) ln2Thus, S_n≈(1/2) ln(n² +1) - (1/2) ln2 - ln n= (1/2)(ln(n² +1) - 2 ln n) - (1/2) ln2= (1/2) ln( (n² +1)/n² ) - (1/2) ln2= (1/2) ln(1 +1/n²) - (1/2) ln2As n→∞, this approaches (1/2)(0) - (1/2) ln2≈-0.3466But our earlier terms show that S_n is positive for small n and approaches approximately -0.3466 as n increases.Wait, but the problem states -1 < S_n ≤1/2.From our calculations, S_n starts at 0.5 when n=1, then decreases, approaching approximately -0.3466.But we need to show that S_n >-1.Wait, so S_n is always greater than -1.Let me see:From the sum:sum_{k=1}^n [k/(k² +1)] ≥H_{n+1} -1So, S_n= sum - ln n ≥H_{n+1} -1 - ln nBut H_{n+1}=H_n +1/(n+1)So, S_n ≥H_n +1/(n+1) -1 - ln n= (H_n - ln n) +1/(n+1) -1We know that H_n - ln n approaches gamma≈0.577, so S_n ≥gamma +1/(n+1) -1≈0.577 + something -1≈-0.423 + something.But 1/(n+1) is positive, so S_n ≥gamma -1 + something positive.Since gamma≈0.577, gamma -1≈-0.423.Thus, S_n ≥-0.423 +1/(n+1) >-0.423.But we need to show S_n >-1.Which is certainly true because -0.423 >-1.Thus, S_n >-1.Similarly, from earlier, S_n ≤H_n - ln n≈gamma≈0.577<1/2?Wait, but H_n - ln n approaches gamma≈0.577, which is less than 1/2=0.5? No, 0.577>0.5.Wait, that contradicts.Wait, actually, H_n - ln n approaches gamma≈0.577, which is greater than 1/2.But our earlier computation for n=1, S_1=0.5, for n=2,≈0.207, n=3≈0.1014, and as n increases, S_n approaches≈-0.3466.Wait, so actually, the maximum of S_n is at n=1, which is 0.5, and then it decreases.Thus, S_n ≤1/2 for all n≥1.And S_n >-1 because it approaches≈-0.3466, which is greater than -1.Therefore, the inequality holds: -1 < S_n ≤1/2.Thus, proved.So, overall, the maximum value of f(x) is 0, the range of b is [1, ∞), and the inequality holds as required.</think>

🧠:<think>Okay, so I have this problem here: I need to find the maximum value of the expression cos(θ/2) times (1 + sinθ) for θ between 0 and π. Hmm, okay. Let's think about how to approach this.First, I remember that sometimes trigonometric expressions can be simplified using identities. Maybe I can rewrite this expression in a way that makes it easier to handle. Let me recall some half-angle and double-angle formulas. Wait, I know that sinθ can be written using a double-angle identity. Specifically, sinθ is equal to 2 sin(θ/2) cos(θ/2). So maybe I can substitute that into the expression. Let's try that:cos(θ/2) * (1 + sinθ) becomes cos(θ/2) * (1 + 2 sin(θ/2) cos(θ/2)).Hmm, that looks a bit better. Let me write that out:cos(θ/2) * [1 + 2 sin(θ/2) cos(θ/2)].Maybe I can factor this or find a substitution to make it simpler. Let's let x be sin(θ/2). Then, cos(θ/2) would be sqrt(1 - x²), since sin² + cos² = 1. So substituting that in, the expression becomes:sqrt(1 - x²) * [1 + 2x sqrt(1 - x²)].Okay, so now I have an expression in terms of x. Let me denote this as y:y = sqrt(1 - x²) * [1 + 2x sqrt(1 - x²)].Now, if I want to find the maximum value of y, I can consider taking the derivative of y with respect to x and setting it equal to zero to find critical points. But since y is a product of functions, taking the derivative might get a bit messy. Maybe squaring y will make it easier to handle?Let me square both sides to get y²:y² = [sqrt(1 - x²) * (1 + 2x sqrt(1 - x²))]² = (1 - x²) * [1 + 2x sqrt(1 - x²)]².Hmm, that still looks complicated, but let's try expanding it step by step. First, let me compute [1 + 2x sqrt(1 - x²)]²:[1 + 2x sqrt(1 - x²)]² = 1² + 2 * 1 * 2x sqrt(1 - x²) + [2x sqrt(1 - x²)]² = 1 + 4x sqrt(1 - x²) + 4x² (1 - x²).So now, y² becomes:y² = (1 - x²) * [1 + 4x sqrt(1 - x²) + 4x² (1 - x²)].This is still a bit complicated, but let's see if we can expand it further. Let me distribute (1 - x²) across the terms inside the brackets:y² = (1 - x²) * 1 + (1 - x²) * 4x sqrt(1 - x²) + (1 - x²) * 4x² (1 - x²) = (1 - x²) + 4x (1 - x²)^(3/2) + 4x² (1 - x²)^2.Wow, that's quite involved. Now, to find the maximum of y, which is sqrt(y²), I need to find the maximum of y². That means I should take the derivative of y² with respect to x, set it equal to zero, and solve for x.Let me denote f(x) = y², so f(x) = (1 - x²) + 4x (1 - x²)^(3/2) + 4x² (1 - x²)^2.Taking the derivative f’(x) will require the use of the product rule and chain rule for each term.Let's compute the derivative term by term.First term: d/dx [1 - x²] = -2x.Second term: d/dx [4x (1 - x²)^(3/2)]. Let me use the product rule here. Let u = 4x and v = (1 - x²)^(3/2).Then, du/dx = 4, and dv/dx = (3/2)(1 - x²)^(1/2) * (-2x) = -3x (1 - x²)^(1/2).So, the derivative is u dv/dx + v du/dx:4x * (-3x (1 - x²)^(1/2)) + (1 - x²)^(3/2) * 4= -12x² (1 - x²)^(1/2) + 4 (1 - x²)^(3/2).Third term: d/dx [4x² (1 - x²)^2]. Again, using the product rule. Let u = 4x² and v = (1 - x²)^2.Then, du/dx = 8x, and dv/dx = 2(1 - x²)(-2x) = -4x (1 - x²).So, the derivative is u dv/dx + v du/dx:4x² * (-4x (1 - x²)) + (1 - x²)^2 * 8x= -16x³ (1 - x²) + 8x (1 - x²)^2.Putting it all together, f’(x) is:-2x + [-12x² (1 - x²)^(1/2) + 4 (1 - x²)^(3/2)] + [-16x³ (1 - x²) + 8x (1 - x²)^2].This is a very complicated expression. Maybe I can factor out some common terms to simplify it.Looking at the second term, -12x² (1 - x²)^(1/2) + 4 (1 - x²)^(3/2), I can factor out 4 (1 - x²)^(1/2):4 (1 - x²)^(1/2) [ -3x² + (1 - x²) ]= 4 (1 - x²)^(1/2) [ -3x² + 1 - x² ]= 4 (1 - x²)^(1/2) (1 - 4x²).Similarly, in the third term, -16x³ (1 - x²) + 8x (1 - x²)^2, I can factor out 8x (1 - x²):8x (1 - x²) [ -2x² + (1 - x²) ]= 8x (1 - x²) [ -2x² + 1 - x² ]= 8x (1 - x²) (1 - 3x²).So, now, f’(x) becomes:-2x + 4 (1 - x²)^(1/2) (1 - 4x²) + 8x (1 - x²) (1 - 3x²).Hmm, still quite complicated, but perhaps we can set f’(x) = 0 and try to solve for x.So, setting f’(x) = 0:-2x + 4 (1 - x²)^(1/2) (1 - 4x²) + 8x (1 - x²) (1 - 3x²) = 0.This equation looks really difficult to solve algebraically. Maybe I can try substituting some values or look for patterns.Alternatively, perhaps there's a smarter substitution or another approach to maximize the original expression without going through all this calculus.Let me go back to the original expression:cos(θ/2) * (1 + sinθ).I wonder if I can express this in terms of a single trigonometric function. Let me see.We already wrote sinθ as 2 sin(θ/2) cos(θ/2). So, the expression becomes:cos(θ/2) * [1 + 2 sin(θ/2) cos(θ/2)].Let me denote t = θ/2, so θ = 2t, and since 0 < θ < π, we have 0 < t < π/2.Then, the expression becomes:cos(t) * [1 + 2 sin(t) cos(t)].Simplify inside the brackets:1 + 2 sin(t) cos(t) = 1 + sin(2t).So, the expression is cos(t) * (1 + sin(2t)).Hmm, that might be easier to handle. Let me write that:y = cos(t) * (1 + sin(2t)).Now, maybe I can write this as:y = cos(t) + cos(t) sin(2t).Hmm, not sure if that helps. Alternatively, perhaps use another identity for sin(2t). Wait, sin(2t) is 2 sin(t) cos(t). So, substituting back:y = cos(t) * [1 + 2 sin(t) cos(t)] = cos(t) + 2 sin(t) cos²(t).Hmm, maybe that's helpful. Let me write it as:y = cos(t) + 2 sin(t) cos²(t).Now, perhaps factor out cos(t):y = cos(t) [1 + 2 sin(t) cos(t)].Wait, that's the same as before. Maybe another approach is needed.Alternatively, let me consider using substitution. Let me set u = sin(t). Then, cos(t) = sqrt(1 - u²), and du = cos(t) dt, but I'm not sure if that helps.Alternatively, perhaps write everything in terms of sin(t):y = cos(t) + 2 sin(t) cos²(t) = sqrt(1 - u²) + 2u (1 - u²),where u = sin(t).So, y = sqrt(1 - u²) + 2u (1 - u²).Now, if I denote this as y(u), I can take the derivative with respect to u and set it equal to zero to find the maximum.Let me compute dy/du:dy/du = (1/(2 sqrt(1 - u²))) * (-2u) + 2(1 - u²) + 2u * (-2u) = (-u)/sqrt(1 - u²) + 2(1 - u²) - 4u² = (-u)/sqrt(1 - u²) + 2 - 2u² - 4u² = (-u)/sqrt(1 - u²) + 2 - 6u².Set dy/du = 0:(-u)/sqrt(1 - u²) + 2 - 6u² = 0.Let me rearrange:(-u)/sqrt(1 - u²) = 6u² - 2.Hmm, this equation still looks tricky. Let me square both sides to eliminate the square root, but I have to be careful because squaring can introduce extraneous solutions.So, squaring both sides:[ (-u)/sqrt(1 - u²) ]² = (6u² - 2)²=> u² / (1 - u²) = 36u⁴ - 24u² + 4.Multiply both sides by (1 - u²):u² = (36u⁴ - 24u² + 4)(1 - u²) = 36u⁴ - 24u² + 4 - 36u⁶ + 24u⁴ - 4u² = -36u⁶ + (36u⁴ + 24u⁴) + (-24u² - 4u²) + 4 = -36u⁶ + 60u⁴ - 28u² + 4.Bring all terms to one side:0 = -36u⁶ + 60u⁴ - 28u² + 4 - u² = -36u⁶ + 60u⁴ - 29u² + 4.So, we have a quartic equation in terms of u². Let me let v = u². Then, the equation becomes:-36v³ + 60v² - 29v + 4 = 0.Hmm, solving a cubic equation. Maybe I can factor this or find rational roots.Using the Rational Root Theorem, possible roots are factors of 4 over factors of 36, so ±1, ±2, ±4, ±1/2, ±1/3, etc. Let me test v=1:-36(1) + 60(1) - 29(1) + 4 = (-36 + 60 - 29 + 4) = (-36 + 60) + (-29 + 4) = 24 -25 = -1 ≠ 0.v=1/2:-36*(1/8) + 60*(1/4) -29*(1/2) +4= -4.5 + 15 -14.5 +4= (-4.5 -14.5) + (15 +4)= (-19) +19=0.Oh, v=1/2 is a root. So, we can factor (v - 1/2) from the cubic.Using polynomial division or synthetic division:Divide -36v³ + 60v² -29v +4 by (v - 1/2).Let me use synthetic division with v=1/2:Coefficients: -36, 60, -29, 4Bring down -36.Multiply -36 by 1/2: -18.Add to next coefficient: 60 + (-18)=42.Multiply 42 by 1/2:21.Add to next coefficient: -29 +21= -8.Multiply -8 by 1/2: -4.Add to last coefficient:4 + (-4)=0.So, the cubic factors as (v - 1/2)(-36v² +42v -8).Now, we can factor the quadratic: -36v² +42v -8.Let me factor out a -2: -2(18v² -21v +4).Now, factor 18v² -21v +4.Looking for two numbers a and b such that a*b=18*4=72 and a + b= -21. Hmm, factors of 72: 8 and 9, but 8+9=17, not 21. Maybe 12 and 6, but 12+6=18. Hmm, not quite. Alternatively, maybe it's not factorable, so let's use the quadratic formula:v = [21 ± sqrt(441 - 4*18*4)] / (2*18) = [21 ± sqrt(441 - 288)] /36 = [21 ± sqrt(153)] /36 = [21 ± 3 sqrt(17)] /36 = [7 ± sqrt(17)] /12.So, the roots are v=1/2, v=(7 + sqrt(17))/12, and v=(7 - sqrt(17))/12.Now, v = u², and u = sin(t), with t between 0 and π/2, so u is between 0 and 1. So, v must be between 0 and 1.Let's compute the numerical values:1/2 = 0.5(7 + sqrt(17))/12 ≈ (7 + 4.123)/12 ≈ 11.123/12 ≈ 0.927(7 - sqrt(17))/12 ≈ (7 - 4.123)/12 ≈ 2.877/12 ≈ 0.23975So, all three roots are between 0 and1, so we need to consider all of them.But remember, when we squared the equation earlier, we might have introduced extraneous solutions, so we need to check each root in the original derivative equation.Original equation after rearrangement:(-u)/sqrt(1 - u²) = 6u² - 2.Let me test each v:First, v=1/2, so u= sqrt(1/2)=√2/2≈0.707.Compute LHS: (-0.707)/sqrt(1 - 0.5)= (-0.707)/sqrt(0.5)= (-0.707)/(0.707)= -1.Compute RHS:6*(0.5) -2=3 -2=1.So, LHS=-1, RHS=1. Not equal. So, v=1/2 is an extraneous solution.Next, v=(7 + sqrt(17))/12≈0.927.Compute u≈sqrt(0.927)≈0.963.Compute LHS: (-0.963)/sqrt(1 -0.927)= (-0.963)/sqrt(0.073)= (-0.963)/0.270≈-3.567.Compute RHS:6*(0.927) -2≈5.562 -2=3.562.So, LHS≈-3.567, RHS≈3.562. Not equal. Close, but not exact. So, extraneous.Next, v=(7 - sqrt(17))/12≈0.23975.Compute u≈sqrt(0.23975)≈0.4896.Compute LHS: (-0.4896)/sqrt(1 -0.23975)= (-0.4896)/sqrt(0.76025)= (-0.4896)/0.872≈-0.561.Compute RHS:6*(0.23975) -2≈1.4385 -2≈-0.5615.So, LHS≈-0.561, RHS≈-0.5615. These are approximately equal. So, v≈0.23975 is a valid solution.Thus, the critical point occurs at v≈0.23975, which is u≈0.4896.So, u=sin(t)=0.4896, so t=arcsin(0.4896). Let me compute t≈arcsin(0.4896)≈29.3 degrees≈0.511 radians.So, t≈0.511 radians, which is θ/2, so θ≈1.022 radians≈58.6 degrees.Now, let's verify if this is indeed a maximum. Since we have only one valid critical point in the interval (0, π/2), and the function tends to zero at both ends (as t approaches 0 and π/2), this critical point must be the maximum.Now, let's compute the maximum value of y.Recall that y = cos(t) * (1 + sin(2t)).Compute cos(t)=sqrt(1 - u²)=sqrt(1 -0.23975)=sqrt(0.76025)=≈0.872.Compute sin(2t)=2 sin(t) cos(t)=2*0.4896*0.872≈2*0.427≈0.854.So, 1 + sin(2t)=1 +0.854≈1.854.Thus, y≈0.872 *1.854≈1.614.Wait, let me compute more accurately:cos(t)=sqrt(1 - (7 - sqrt(17))/12)=sqrt( (12 -7 + sqrt(17))/12 )=sqrt( (5 + sqrt(17))/12 ).Similarly, sin(t)=sqrt( (7 - sqrt(17))/12 ).But maybe it's better to compute y in terms of v.Wait, y = sqrt(1 - v) * (1 + 2u sqrt(1 - v)).But this seems complicated. Alternatively, since we have u=sin(t)=sqrt(v)=sqrt( (7 - sqrt(17))/12 ).So, let me compute y:y = cos(t) * (1 + sin(2t)).We have cos(t)=sqrt(1 - v)=sqrt(1 - (7 - sqrt(17))/12 )=sqrt( (12 -7 + sqrt(17))/12 )=sqrt( (5 + sqrt(17))/12 ).sin(2t)=2 sin(t) cos(t)=2 sqrt(v) * sqrt( (5 + sqrt(17))/12 ).So, sin(2t)=2 * sqrt( (7 - sqrt(17))/12 ) * sqrt( (5 + sqrt(17))/12 ).Multiply the terms inside the square roots:sqrt( (7 - sqrt(17))(5 + sqrt(17)) /144 ).Compute the numerator:(7 - sqrt(17))(5 + sqrt(17))=7*5 +7*sqrt(17) -5 sqrt(17) - (sqrt(17))²=35 +7 sqrt(17) -5 sqrt(17) -17=35 -17 + (7 sqrt(17) -5 sqrt(17))=18 + 2 sqrt(17).So, sin(2t)=2 * sqrt( (18 + 2 sqrt(17))/144 )=2 * sqrt( (9 + sqrt(17))/72 ).Simplify sqrt( (9 + sqrt(17))/72 )=sqrt( (9 + sqrt(17)) ) / (6 sqrt(2)).So, sin(2t)=2 * sqrt(9 + sqrt(17)) / (6 sqrt(2))= sqrt(9 + sqrt(17)) / (3 sqrt(2)).Therefore, 1 + sin(2t)=1 + sqrt(9 + sqrt(17))/(3 sqrt(2)).So, y=cos(t)*(1 + sin(2t))= sqrt( (5 + sqrt(17))/12 ) * [1 + sqrt(9 + sqrt(17))/(3 sqrt(2))].This seems quite complicated. Maybe it's better to rationalize or find a numerical approximation.Alternatively, perhaps we can express sqrt(9 + sqrt(17)) in terms of something else.Wait, let me compute sqrt(9 + sqrt(17)):Let me denote sqrt(9 + sqrt(17))=a + b sqrt(17), where a and b are rational numbers.Then, squaring both sides:9 + sqrt(17)=a² + 2ab sqrt(17) +17 b².Comparing the rational and irrational parts:Rational part:9 =a² +17 b².Irrational part:1=2ab.So, we have:From irrational part:2ab=1 => ab=1/2.From rational part:a² +17 b²=9.Let me solve for a in terms of b: a=1/(2b).Substitute into the rational part:(1/(2b))² +17 b²=9=>1/(4b²) +17 b²=9.Multiply both sides by 4b²:1 +68 b⁴=36 b².Rearrange:68 b⁴ -36 b² +1=0.Let me let z=b², so equation becomes:68 z² -36 z +1=0.Quadratic in z:z=(36 ±sqrt(1296 -272))/136=(36 ±sqrt(1024))/136=(36 ±32)/136.So,z=(36 +32)/136=68/136=0.5,or z=(36 -32)/136=4/136=1/34.So, z=1/2 or z=1/34.Since z=b², and b is positive (since sqrt(9 + sqrt(17)) is positive), so b=1/sqrt(2) or b=1/sqrt(34).But let's check:If z=1/2, then b=1/sqrt(2), then a=1/(2b)=1/(2*(1/sqrt(2)))=sqrt(2)/2.So, sqrt(9 + sqrt(17))=a +b sqrt(17)=sqrt(2)/2 + (1/sqrt(2)) sqrt(17)=sqrt(2)/2 + sqrt(17)/sqrt(2)= (sqrt(2) + sqrt(34))/2.Wait, but sqrt(17)/sqrt(2)=sqrt(34)/2, so yes, that's correct.Similarly, if z=1/34, then b=1/sqrt(34), a=1/(2b)=sqrt(34)/2.So, sqrt(9 + sqrt(17))=sqrt(34)/2 + (1/sqrt(34)) sqrt(17)=sqrt(34)/2 + sqrt(17)/sqrt(34)=sqrt(34)/2 + sqrt(17*34)/34= sqrt(34)/2 + sqrt(578)/34.Wait, sqrt(578)=sqrt(2*17²)=17 sqrt(2). So,sqrt(9 + sqrt(17))=sqrt(34)/2 +17 sqrt(2)/34= sqrt(34)/2 + sqrt(2)/2= (sqrt(34) + sqrt(2))/2.Wait, that's the same as the previous expression: (sqrt(2) + sqrt(34))/2. So, both roots give the same expression, just written differently.Thus, sqrt(9 + sqrt(17))=(sqrt(2) + sqrt(34))/2.So, going back to sin(2t)= sqrt(9 + sqrt(17))/(3 sqrt(2))= [ (sqrt(2) + sqrt(34))/2 ]/(3 sqrt(2))= (sqrt(2) + sqrt(34))/(6 sqrt(2)).Simplify:(sqrt(2))/(6 sqrt(2)) + sqrt(34)/(6 sqrt(2))=1/6 + sqrt(17)/6.Because sqrt(34)/sqrt(2)=sqrt(17*2)/sqrt(2)=sqrt(17).So, sin(2t)=1/6 + sqrt(17)/6=(1 + sqrt(17))/6.Therefore, 1 + sin(2t)=1 + (1 + sqrt(17))/6=(6 +1 + sqrt(17))/6=(7 + sqrt(17))/6.Now, cos(t)=sqrt( (5 + sqrt(17))/12 ).So, y=cos(t)*(1 + sin(2t))=sqrt( (5 + sqrt(17))/12 )*(7 + sqrt(17))/6.Let me compute this:sqrt( (5 + sqrt(17))/12 )=sqrt( (5 + sqrt(17)) )/(2 sqrt(3)).So, y= [sqrt(5 + sqrt(17))/(2 sqrt(3))] * (7 + sqrt(17))/6= [sqrt(5 + sqrt(17)) * (7 + sqrt(17))]/(12 sqrt(3)).Let me compute the numerator:sqrt(5 + sqrt(17))*(7 + sqrt(17)).Let me denote sqrt(5 + sqrt(17))=c.Then, c²=5 + sqrt(17).Let me compute c*(7 + sqrt(17))=7c +c sqrt(17).But I don't see an immediate simplification.Alternatively, maybe square y to see if it simplifies:y²= [ (5 + sqrt(17))/12 ] * [ (7 + sqrt(17))/6 ]².Compute [ (7 + sqrt(17))/6 ]²= (49 +14 sqrt(17) +17)/36=(66 +14 sqrt(17))/36=(33 +7 sqrt(17))/18.So, y²= [ (5 + sqrt(17))/12 ] * [ (33 +7 sqrt(17))/18 ].Multiply the numerators:(5 + sqrt(17))(33 +7 sqrt(17))=5*33 +5*7 sqrt(17) +33 sqrt(17) +7*(sqrt(17))²=165 +35 sqrt(17) +33 sqrt(17) +7*17=165 +68 sqrt(17) +119=284 +68 sqrt(17).Denominator:12*18=216.So, y²=(284 +68 sqrt(17))/216.Simplify numerator and denominator by dividing numerator and denominator by 4:Numerator:71 +17 sqrt(17).Denominator:54.So, y²=(71 +17 sqrt(17))/54.Thus, y= sqrt( (71 +17 sqrt(17))/54 ).We can rationalize this, but perhaps it's better to leave it as is or simplify further.Alternatively, let's compute the numerical value:Compute sqrt(17)≈4.123.Compute numerator inside sqrt:71 +17*4.123≈71 +70.091≈141.091.Divide by 54:141.091/54≈2.612.So, y≈sqrt(2.612)≈1.616.Which matches our earlier approximation of≈1.614.So, the maximum value is sqrt( (71 +17 sqrt(17))/54 ).Alternatively, we can write this as sqrt( (71 +17 sqrt(17))/54 ).But perhaps simplifying further:Note that (71 +17 sqrt(17))/54 can be written as (71 +17 sqrt(17))/(54)= (71/54) + (17/54)sqrt(17).But I don't think it simplifies more. Alternatively, factor numerator:71 +17 sqrt(17)=17 sqrt(17) +71. Not much to factor here.Alternatively, notice that 71 +17 sqrt(17)= (sqrt(17)*17 +71). Not particularly helpful.Alternatively, write as:sqrt( (71 +17 sqrt(17))/54 )=sqrt( (71 +17 sqrt(17))/54 ).But perhaps rationalizing or expressing differently.Alternatively, perhaps express in terms of sqrt(a) + sqrt(b):Suppose sqrt( (71 +17 sqrt(17))/54 )=sqrt(c) + sqrt(d).Then, squaring both sides:(71 +17 sqrt(17))/54= c + d + 2 sqrt(c d).So, we have:c + d=71/54≈1.3148,and 2 sqrt(c d)=17 sqrt(17)/54≈17*4.123/54≈69.091/54≈1.279.So, sqrt(c d)=1.279/2≈0.6395.Then, c + d≈1.3148,and sqrt(c d)≈0.6395.Let me denote c + d= S≈1.3148,and c*d= P≈(0.6395)²≈0.4088.So, we have a quadratic equation: x² - Sx + P=0,x² -1.3148x +0.4088=0.Solutions:x=(1.3148 ±sqrt(1.3148² -4*0.4088))/2≈(1.3148 ±sqrt(1.728 -1.635))/2≈(1.3148 ±sqrt(0.093))/2≈(1.3148 ±0.305)/2.So,x≈(1.3148 +0.305)/2≈1.6198/2≈0.8099,or x≈(1.3148 -0.305)/2≈1.0098/2≈0.5049.So, c≈0.8099, d≈0.5049.But 0.8099≈sqrt(0.656), and 0.5049≈sqrt(0.255). Not particularly helpful.Alternatively, perhaps express in terms of known constants, but it's probably not necessary. So, the exact value is sqrt( (71 +17 sqrt(17))/54 ).Alternatively, factor numerator and denominator:71 +17 sqrt(17) can be written as 17 sqrt(17) +71, and 54=6*9=6*3².But I don't see a way to simplify further. So, the maximum value is sqrt( (71 +17 sqrt(17))/54 ).Alternatively, rationalizing the denominator:sqrt( (71 +17 sqrt(17))/54 )=sqrt( (71 +17 sqrt(17))/54 )=sqrt( (71 +17 sqrt(17))/(6*9) )=sqrt( (71 +17 sqrt(17))/6 ) /3.But again, not much simpler.So, the exact maximum value is sqrt( (71 +17 sqrt(17))/54 ). Alternatively, rationalizing differently:Multiply numerator and denominator by sqrt(54):sqrt( (71 +17 sqrt(17)) *54 ) /54.But this doesn't seem helpful.Alternatively, perhaps factor numerator:71 +17 sqrt(17)=17 sqrt(17) +71=17(sqrt(17) +71/17)=17(sqrt(17) +4.176).Not helpful.Alternatively, perhaps approximate the exact value:sqrt( (71 +17 sqrt(17))/54 )≈sqrt( (71 +17*4.123)/54 )≈sqrt( (71 +70.091)/54 )≈sqrt(141.091/54 )≈sqrt(2.612 )≈1.616.So, approximately 1.616.But the problem asks for the maximum value, so we need to provide the exact value.Thus, the maximum value is sqrt( (71 +17 sqrt(17))/54 ).Alternatively, we can write this as sqrt(71 +17 sqrt(17))/sqrt(54)=sqrt(71 +17 sqrt(17))/(3 sqrt(6)).But that's as simplified as it gets.So, the maximum value is sqrt( (71 +17 sqrt(17))/54 ), which is approximately 1.616.Alternatively, we can rationalize the denominator:sqrt( (71 +17 sqrt(17))/54 )=sqrt(71 +17 sqrt(17))/(3 sqrt(6)).But I think the first form is acceptable.Therefore, the maximum value of the expression is sqrt( (71 +17 sqrt(17))/54 ).Final Answerboxed{sqrt{dfrac{71 + 17sqrt{17}}{54}}}</think>