💡Alright, so I have this problem where I need to find positive integers C and D such that C, D, C - D, and C + D are all prime numbers. Then, I need to figure out something about the sum of these four primes. The options are whether the sum is even, divisible by 3, divisible by 5, divisible by 7, or prime. I have to choose one of these options.First, let me understand the problem. I need to find two prime numbers C and D such that when I subtract D from C and add D to C, both results are also prime numbers. So, all four numbers: C, D, C - D, and C + D must be prime.Since we're dealing with primes, I know that except for 2, all primes are odd numbers. So, if both C and D were odd primes, then C - D and C + D would be even numbers because subtracting or adding two odd numbers gives an even number. The only even prime number is 2. So, for C - D and C + D to be prime, they must be equal to 2 or another prime number. But since 2 is the only even prime, if C - D or C + D is even, it must be 2.Wait, but if both C and D are odd, then C - D and C + D are even, so they can only be 2. But C + D is definitely larger than C - D, so C + D can't be 2 because it's larger. Therefore, C - D must be 2, and C + D must be another prime number.Alternatively, if one of C or D is 2, the even prime, then the other must be odd. Let's explore both possibilities.Case 1: Both C and D are odd primes.As I thought earlier, C - D and C + D would both be even. Since they must be prime, they must be 2. So, C - D = 2 and C + D is another prime. Let's see if this is possible.If C - D = 2, then C = D + 2. Then, C + D = (D + 2) + D = 2D + 2 = 2(D + 1). Since C + D must be prime, and it's equal to 2 times (D + 1), the only way this can be prime is if D + 1 = 1, which would make C + D = 2. But D + 1 = 1 implies D = 0, which is not a prime number. Therefore, this case leads to a contradiction because D would have to be 0, which isn't prime. So, both C and D can't be odd primes.Case 2: One of C or D is 2.Let's consider D = 2, which is the even prime. Then, C must be an odd prime. So, C - D = C - 2 must be prime, and C + D = C + 2 must also be prime.So, we need to find a prime number C such that both C - 2 and C + 2 are also prime. These are called prime triplets or prime constellations. The most famous example is 3, 5, 7. Let's check:If C = 5, then C - 2 = 3 (prime), C = 5 (prime), and C + 2 = 7 (prime). So, that works. So, C = 5, D = 2.Let me check another one. If C = 7, then C - 2 = 5 (prime), C = 7 (prime), and C + 2 = 9, which is not prime. So, that doesn't work.If C = 11, then C - 2 = 9 (not prime), so that doesn't work either.If C = 3, then C - 2 = 1, which isn't prime. So, that doesn't work.So, the only C that works in this case is 5, with D = 2.Alternatively, what if C = 2? Then D would have to be an odd prime. So, C - D = 2 - D. But since D is positive and a prime, D must be at least 2. So, 2 - D would be non-positive, which can't be a prime. Therefore, C can't be 2.So, the only possibility is C = 5 and D = 2.Therefore, the four primes are C = 5, D = 2, C - D = 3, and C + D = 7.Now, let's find the sum of these four primes: 5 + 2 + 3 + 7.5 + 2 is 7, 7 + 3 is 10, 10 + 7 is 17.So, the sum is 17.Now, let's check the options:A. Even: 17 is odd, so no.B. Divisible by 3: 17 divided by 3 is about 5.666, so no.C. Divisible by 5: 17 divided by 5 is 3.4, so no.D. Divisible by 7: 17 divided by 7 is about 2.428, so no.E. Prime: 17 is a prime number, so yes.Therefore, the sum is prime, which is option E.Wait, but let me double-check if there are any other possible values of C and D that satisfy the conditions. Maybe I missed something.Earlier, I considered C = 5 and D = 2. Let's see if there are other primes C where C - 2 and C + 2 are also primes.C = 5: 3, 5, 7 – all primes.C = 7: 5, 7, 9 – 9 is not prime.C = 11: 9, 11, 13 – 9 is not prime.C = 13: 11, 13, 15 – 15 is not prime.C = 17: 15, 17, 19 – 15 is not prime.C = 19: 17, 19, 21 – 21 is not prime.C = 23: 21, 23, 25 – 21 and 25 are not primes.So, it seems that the only such prime C is 5. Therefore, the only possible set is C = 5, D = 2, with the four primes being 2, 3, 5, 7, and their sum is 17, which is prime.Therefore, the answer is E.

💡First, Georgie has 8 choices for the window to enter the mansion. Once inside, he must exit through a different window, leaving him with 7 options. To find the total number of ways, multiply the number of entry choices by the number of exit choices: 8 × 7 = 56.

💡Okay, so I have this problem here about the Cartesian coordinate system, a line, and a curve given by parametric equations. It has two parts, (Ⅰ) and (Ⅱ). Let me try to tackle them one by one.Starting with part (Ⅰ): They mention a point P with polar coordinates (√2, π/4). I need to figure out where this point is in relation to the curve C. Hmm, okay. I remember that polar coordinates (r, θ) can be converted to Cartesian coordinates (x, y) using the formulas x = r cos θ and y = r sin θ. So, let me compute that.Given r = √2 and θ = π/4, so:x = √2 * cos(π/4)y = √2 * sin(π/4)I know that cos(π/4) and sin(π/4) are both √2/2. So,x = √2 * (√2/2) = (2)/2 = 1y = √2 * (√2/2) = (2)/2 = 1So, point P has Cartesian coordinates (1, 1). Got that down.Now, the curve C is given parametrically as:x = √3 cos αy = √2 sin αI think it might help to write the Cartesian equation of curve C. Since it's parametric, I can express cos α and sin α in terms of x and y.From x = √3 cos α, we get cos α = x / √3From y = √2 sin α, we get sin α = y / √2Since cos² α + sin² α = 1, substituting gives:(x / √3)² + (y / √2)² = 1Which simplifies to:x² / 3 + y² / 2 = 1So, curve C is an ellipse centered at the origin with semi-major axis √3 along the x-axis and semi-minor axis √2 along the y-axis.Now, to determine the positional relationship between point P(1,1) and curve C, I need to check whether P lies inside, on, or outside the ellipse.To do that, I can plug the coordinates of P into the ellipse equation and see if the left-hand side is less than, equal to, or greater than 1.Plugging in x = 1 and y = 1:(1)² / 3 + (1)² / 2 = 1/3 + 1/2Let me compute that:1/3 is approximately 0.333, and 1/2 is 0.5, so adding them together gives approximately 0.833, which is less than 1.Therefore, point P lies inside the ellipse C.Alright, that was part (Ⅰ). Now, moving on to part (Ⅱ): Finding the minimum distance from a moving point Q on curve C to the line l, whose equation is x - y + 4 = 0.So, point Q is on curve C, which we already know is the ellipse x²/3 + y²/2 = 1. So, any point Q on C can be represented as (√3 cos α, √2 sin α) for some parameter α.The distance from a point (x, y) to the line Ax + By + C = 0 is given by the formula:d = |Ax + By + C| / √(A² + B²)In this case, the line is x - y + 4 = 0, so A = 1, B = -1, C = 4.So, plugging in the coordinates of Q into the distance formula:d = |√3 cos α - √2 sin α + 4| / √(1 + 1) = |√3 cos α - √2 sin α + 4| / √2I need to find the minimum value of this distance as α varies.Hmm, so the expression inside the absolute value is √3 cos α - √2 sin α + 4. To find the minimum distance, I need to minimize the numerator |√3 cos α - √2 sin α + 4|.I remember that expressions of the form a cos α + b sin α can be rewritten using the amplitude-phase form: R cos(α + φ), where R = √(a² + b²) and tan φ = b/a.Wait, in this case, it's √3 cos α - √2 sin α, so a = √3 and b = -√2.So, let me compute R:R = √[(√3)² + (-√2)²] = √[3 + 2] = √5And tan φ = b/a = (-√2)/√3 = -√(2/3)So, φ = arctan(-√(2/3)). Since tan is negative, φ is in the fourth quadrant. But since we're dealing with cos(α + φ), the phase shift doesn't affect the maximum or minimum values, just shifts the angle.So, we can write √3 cos α - √2 sin α as √5 cos(α + φ), where φ is such that tan φ = -√(2/3).Therefore, the expression inside the absolute value becomes:√5 cos(α + φ) + 4So, the distance d is:|√5 cos(α + φ) + 4| / √2Now, since cos(α + φ) ranges between -1 and 1, the expression √5 cos(α + φ) ranges between -√5 and √5.Therefore, √5 cos(α + φ) + 4 ranges between 4 - √5 and 4 + √5.Since we are taking the absolute value, we need to consider whether 4 - √5 is positive or negative.Compute 4 - √5: √5 is approximately 2.236, so 4 - 2.236 ≈ 1.764, which is positive. So, the expression inside the absolute value is always positive, because the minimum value is 4 - √5 ≈ 1.764 > 0.Therefore, |√5 cos(α + φ) + 4| = √5 cos(α + φ) + 4So, the distance d simplifies to:(√5 cos(α + φ) + 4) / √2To find the minimum distance, we need to minimize this expression.Since cos(α + φ) has a minimum value of -1, the minimum value of the numerator is √5*(-1) + 4 = 4 - √5.Therefore, the minimum distance is (4 - √5)/√2.But let me rationalize the denominator to make it look nicer:(4 - √5)/√2 = (4√2 - √10)/2So, the minimum distance is (4√2 - √10)/2.Let me double-check my steps to make sure I didn't make any mistakes.1. Converted polar coordinates of P to Cartesian: (1,1). That seems correct.2. Found the Cartesian equation of curve C: x²/3 + y²/2 = 1. Yes, that's the standard form of an ellipse.3. Plugged in (1,1) into the ellipse equation: 1/3 + 1/2 = 5/6 < 1. So, P is inside C. That makes sense.4. For part (Ⅱ), expressed the distance from Q(√3 cos α, √2 sin α) to line l: Correctly applied the distance formula.5. Rewrote √3 cos α - √2 sin α as √5 cos(α + φ). Calculated R correctly as √5. Calculated tan φ as -√(2/3). That seems right.6. Noted that the expression inside the absolute value is always positive, so removed the absolute value. Correct, since 4 - √5 is positive.7. Found the minimum value of the numerator by setting cos(α + φ) = -1, leading to 4 - √5. Then divided by √2 and rationalized.Yes, that seems solid. So, the minimum distance is (4√2 - √10)/2.**Final Answer**Ⅰ. Point ( P ) is inside curve ( C ).Ⅱ. The minimum distance is boxed{dfrac{4sqrt{2} - sqrt{10}}{2}}.

💡Alright, so I have this problem about triangle ABC, and I need to prove two inequalities involving sine and cosine functions of the angles and their halves. Let me try to wrap my head around this.First, let me write down what I need to prove:1. sin A sin(A/2) + sin B sin(B/2) + sin C sin(C/2) ≤ 4/√32. sin A cos(A/2) + sin B cos(B/2) + sin C cos(C/2) ≤ 4/√3Hmm, okay. Both inequalities involve sums of products of sine and cosine functions of the angles and their halves. Since it's a triangle, I know that A + B + C = π radians, or 180 degrees. That might come in handy later.Let me tackle the first inequality first.Starting with sin A sin(A/2). I remember that there are some trigonometric identities that relate sine of an angle and sine of half that angle. Maybe I can use a double-angle identity or something similar.I recall that sin A = 2 sin(A/2) cos(A/2). Let me try substituting that into sin A sin(A/2):sin A sin(A/2) = [2 sin(A/2) cos(A/2)] sin(A/2) = 2 sin²(A/2) cos(A/2)Okay, so sin A sin(A/2) simplifies to 2 sin²(A/2) cos(A/2). That seems manageable. Maybe I can find a way to bound this expression.Similarly, sin B sin(B/2) = 2 sin²(B/2) cos(B/2), and the same for sin C sin(C/2).So the entire sum becomes:2 sin²(A/2) cos(A/2) + 2 sin²(B/2) cos(B/2) + 2 sin²(C/2) cos(C/2)Hmm, maybe I can factor out the 2:2 [sin²(A/2) cos(A/2) + sin²(B/2) cos(B/2) + sin²(C/2) cos(C/2)]Now, I need to find an upper bound for this expression. Since all the terms are positive (because angles in a triangle are between 0 and π, so their halves are between 0 and π/2, where sine and cosine are positive), I can consider maximizing each term individually.Wait, but since the angles are related (A + B + C = π), I can't just maximize each term independently. I need to consider the constraint that A + B + C = π.Maybe I can use some inequality like AM-GM or Cauchy-Schwarz. Let me think about AM-GM.But before that, perhaps it's useful to express everything in terms of a single variable. Since it's a triangle, maybe I can parameterize the angles in terms of one variable, but that might complicate things.Alternatively, maybe I can consider substituting variables. Let me set x = A/2, y = B/2, z = C/2. Then, since A + B + C = π, we have 2x + 2y + 2z = π, so x + y + z = π/2.So, the expression becomes:2 [sin²x cosx + sin²y cosy + sin²z cosz]And we need to maximize this expression under the constraint x + y + z = π/2, where x, y, z > 0.Hmm, okay. So, now the problem is to maximize sin²x cosx + sin²y cosy + sin²z cosz, given that x + y + z = π/2.This seems more manageable. Maybe I can use Lagrange multipliers or some method from calculus to find the maximum.But since it's a symmetric expression, perhaps the maximum occurs when x = y = z. Let me test that.If x = y = z, then each is π/6. Let's compute sin²(π/6) cos(π/6):sin(π/6) = 1/2, so sin²(π/6) = 1/4.cos(π/6) = √3/2.So, sin²(π/6) cos(π/6) = (1/4)(√3/2) = √3/8.Therefore, the sum would be 3*(√3/8) = 3√3/8.Then, multiplying by 2, we get 2*(3√3/8) = 3√3/4 ≈ 1.299.But the upper bound we need to prove is 4/√3 ≈ 2.309, which is larger. So, if all angles are equal, the expression is about 1.299, which is less than 4/√3. So, maybe the maximum occurs when one angle is larger and the others are smaller.Wait, perhaps when one angle approaches π/2 and the others approach 0. Let me test that.Suppose x approaches π/2, then y and z approach 0.Compute sin²x cosx: sin²(π/2) cos(π/2) = 1*0 = 0.Similarly, sin²y cosy ≈ 0, and sin²z cosz ≈ 0.So, the sum approaches 0. That's not helpful.Wait, maybe when one angle is larger, but not approaching π/2. Let me try x = π/3, y = π/6, z = 0. But z can't be 0, but approaching 0.Compute sin²(π/3) cos(π/3) + sin²(π/6) cos(π/6) + sin²(0) cos(0).sin(π/3) = √3/2, so sin²(π/3) = 3/4.cos(π/3) = 1/2.So, sin²(π/3) cos(π/3) = (3/4)(1/2) = 3/8.sin(π/6) = 1/2, so sin²(π/6) = 1/4.cos(π/6) = √3/2.So, sin²(π/6) cos(π/6) = (1/4)(√3/2) = √3/8.sin²(0) cos(0) = 0*1 = 0.So, the sum is 3/8 + √3/8 + 0 ≈ 0.375 + 0.216 ≈ 0.591.Multiply by 2: ≈ 1.182, which is still less than 4/√3.Hmm, maybe I need a different approach.Wait, perhaps I can use the method of Lagrange multipliers to find the maximum of the function f(x, y, z) = sin²x cosx + sin²y cosy + sin²z cosz, subject to the constraint g(x, y, z) = x + y + z - π/2 = 0.Let me set up the Lagrangian:L = sin²x cosx + sin²y cosy + sin²z cosz - λ(x + y + z - π/2)Take partial derivatives with respect to x, y, z, and set them to zero.Compute ∂L/∂x:d/dx [sin²x cosx] = 2 sinx cosx * cosx + sin²x (-sinx) = 2 sinx cos²x - sin³xSimilarly, ∂L/∂y = 2 siny cos²y - sin³y∂L/∂z = 2 sinz cos²z - sin³zSet these equal to λ:2 sinx cos²x - sin³x = λ2 siny cos²y - sin³y = λ2 sinz cos²z - sin³z = λSo, all three expressions are equal. Therefore, 2 sinx cos²x - sin³x = 2 siny cos²y - sin³y = 2 sinz cos²z - sin³zThis suggests that the function h(t) = 2 sint cos²t - sin³t is equal for all three variables x, y, z.I need to find when h(t) is constant. Let me analyze h(t):h(t) = 2 sint cos²t - sin³t = sint (2 cos²t - sin²t)Using the identity cos²t = 1 - sin²t:h(t) = sint [2(1 - sin²t) - sin²t] = sint [2 - 2 sin²t - sin²t] = sint (2 - 3 sin²t)So, h(t) = sint (2 - 3 sin²t)We need h(x) = h(y) = h(z)This suggests that either x = y = z, or some other symmetric condition.If x = y = z, then x = y = z = π/6, as before. But we saw that gives a lower value than 4/√3.Alternatively, maybe two angles are equal, and the third is different. Let me suppose x = y ≠ z.Then, h(x) = h(z). So,sinx (2 - 3 sin²x) = sinz (2 - 3 sin²z)But since x + y + z = π/2, and x = y, then 2x + z = π/2, so z = π/2 - 2x.So, let me substitute z = π/2 - 2x into the equation:sinx (2 - 3 sin²x) = sin(π/2 - 2x) [2 - 3 sin²(π/2 - 2x)]We know that sin(π/2 - 2x) = cos2x, and sin²(π/2 - 2x) = cos²2x.So,sinx (2 - 3 sin²x) = cos2x [2 - 3 cos²2x]Let me compute cos2x [2 - 3 cos²2x]:cos2x * 2 - 3 cos³2xSo, the equation becomes:sinx (2 - 3 sin²x) = 2 cos2x - 3 cos³2xThis seems complicated. Maybe I can express everything in terms of sinx.Recall that cos2x = 1 - 2 sin²x.So, let me substitute:Left side: sinx (2 - 3 sin²x)Right side: 2(1 - 2 sin²x) - 3(1 - 2 sin²x)^(3/2)Wait, that might not be helpful. Alternatively, let me express cos2x in terms of sinx:cos2x = 1 - 2 sin²xSo, right side becomes:2(1 - 2 sin²x) - 3(1 - 2 sin²x)^(3/2)Hmm, this is getting messy. Maybe instead of trying to solve this analytically, I can consider specific values.Let me try x = π/6, then z = π/2 - 2*(π/6) = π/2 - π/3 = π/6. So, x = y = z = π/6, which is the case we already considered.Alternatively, try x = π/4, then z = π/2 - 2*(π/4) = π/2 - π/2 = 0. But z can't be 0.Alternatively, x = π/3, then z = π/2 - 2*(π/3) = π/2 - 2π/3 = -π/6, which is negative, so invalid.Hmm, maybe x = π/12, then z = π/2 - 2*(π/12) = π/2 - π/6 = π/3.So, x = π/12, y = π/12, z = π/3.Compute h(x) = sin(π/12) (2 - 3 sin²(π/12)).Compute sin(π/12) ≈ 0.2588sin²(π/12) ≈ 0.06699So, h(x) ≈ 0.2588*(2 - 3*0.06699) ≈ 0.2588*(2 - 0.20097) ≈ 0.2588*1.799 ≈ 0.465Similarly, h(z) = sin(π/3) (2 - 3 sin²(π/3)).sin(π/3) ≈ 0.8660sin²(π/3) ≈ 0.75So, h(z) ≈ 0.8660*(2 - 3*0.75) ≈ 0.8660*(2 - 2.25) ≈ 0.8660*(-0.25) ≈ -0.2165But h(x) ≈ 0.465 and h(z) ≈ -0.2165, which are not equal. So, this doesn't satisfy h(x) = h(z).Hmm, maybe this approach isn't working. Perhaps I need to consider that the maximum occurs when two angles are equal, but not necessarily all three.Alternatively, maybe I can use Jensen's inequality, since the function might be concave or convex.Let me consider the function f(t) = sin²t cos t.Compute its second derivative to check concavity.First derivative: f'(t) = 2 sint cos t * cos t + sin²t (-sin t) = 2 sin t cos² t - sin³ tSecond derivative:f''(t) = 2 [cos t cos² t + sin t * 2 cos t (-sin t)] - [3 sin²t cos t]Wait, let me compute it step by step.f'(t) = 2 sin t cos² t - sin³ tDifferentiate f'(t):f''(t) = 2 [cos t * cos² t + sin t * 2 cos t (-sin t)] - 3 sin² t cos tWait, actually, let's use product rule:f'(t) = 2 sin t cos² t - sin³ tSo,f''(t) = 2 [cos t * cos² t + sin t * 2 cos t (-sin t)] - 3 sin² t cos tWait, no. Let me do it correctly.First term: d/dt [2 sin t cos² t] = 2 [cos t * cos² t + sin t * 2 cos t (-sin t)] = 2 [cos³ t - 2 sin² t cos t]Second term: d/dt [- sin³ t] = -3 sin² t cos tSo, overall:f''(t) = 2 cos³ t - 4 sin² t cos t - 3 sin² t cos t = 2 cos³ t - 7 sin² t cos tFactor out cos t:f''(t) = cos t (2 cos² t - 7 sin² t)Hmm, the sign of f''(t) depends on the term (2 cos² t - 7 sin² t). Let me see when this is positive or negative.Let me express it in terms of cos 2t:2 cos² t - 7 sin² t = 2*(1 + cos 2t)/2 - 7*(1 - cos 2t)/2 = (1 + cos 2t) - (7/2)(1 - cos 2t) = 1 + cos 2t - 7/2 + (7/2) cos 2t = (-5/2) + (9/2) cos 2tSo, f''(t) = cos t [ (-5/2) + (9/2) cos 2t ]This is complicated. It seems that f''(t) changes sign depending on t, so f(t) is neither convex nor concave over the entire interval. Therefore, Jensen's inequality might not be directly applicable.Hmm, maybe another approach. Let me think about substituting variables.Let me set u = sin t. Then, cos t = sqrt(1 - u²).So, f(t) = sin² t cos t = u² sqrt(1 - u²)Let me consider f(u) = u² sqrt(1 - u²), where u ∈ [0, 1]I can try to find the maximum of this function.Compute derivative:f'(u) = 2u sqrt(1 - u²) + u² * (1/2)(1 - u²)^(-1/2)(-2u) = 2u sqrt(1 - u²) - u³ / sqrt(1 - u²)Set f'(u) = 0:2u sqrt(1 - u²) - u³ / sqrt(1 - u²) = 0Multiply both sides by sqrt(1 - u²):2u (1 - u²) - u³ = 0Expand:2u - 2u³ - u³ = 0 => 2u - 3u³ = 0 => u(2 - 3u²) = 0Solutions: u = 0 or 2 - 3u² = 0 => u² = 2/3 => u = sqrt(2/3)So, maximum occurs at u = sqrt(2/3). Let's compute f(u):f(sqrt(2/3)) = (2/3) sqrt(1 - 2/3) = (2/3) sqrt(1/3) = (2/3)(1/√3) = 2/(3√3) ≈ 0.385So, the maximum of sin²t cos t is 2/(3√3), achieved when sin t = sqrt(2/3), i.e., t = arcsin(sqrt(2/3)) ≈ 54.7 degrees.Okay, so each term sin²t cos t is bounded above by 2/(3√3). Therefore, the sum sin²x cosx + sin²y cosy + sin²z cosz ≤ 3*(2/(3√3)) = 2/√3.Therefore, the original expression:2 [sin²x cosx + sin²y cosy + sin²z cosz] ≤ 2*(2/√3) = 4/√3Which is exactly what we needed to prove for the first inequality.Great, so that works out. So, the maximum of the sum is achieved when each term is maximized, which occurs when each angle is such that sin t = sqrt(2/3). But in a triangle, the angles are related, so it's not necessarily the case that all angles can achieve this simultaneously. However, since we used the fact that each term is individually bounded, and the sum is maximized when each term is as large as possible, the inequality holds.Now, moving on to the second inequality:sin A cos(A/2) + sin B cos(B/2) + sin C cos(C/2) ≤ 4/√3Again, let's start by expressing sin A cos(A/2).Using the double-angle identity: sin A = 2 sin(A/2) cos(A/2)So, sin A cos(A/2) = 2 sin(A/2) cos²(A/2)Similarly, sin B cos(B/2) = 2 sin(B/2) cos²(B/2)And sin C cos(C/2) = 2 sin(C/2) cos²(C/2)So, the entire sum becomes:2 sin(A/2) cos²(A/2) + 2 sin(B/2) cos²(B/2) + 2 sin(C/2) cos²(C/2)Factor out the 2:2 [sin(A/2) cos²(A/2) + sin(B/2) cos²(B/2) + sin(C/2) cos²(C/2)]Again, we can use the substitution x = A/2, y = B/2, z = C/2, so x + y + z = π/2.So, the expression becomes:2 [sinx cos²x + siny cos²y + sinz cos²z]We need to find the maximum of this expression under x + y + z = π/2.Again, since the expression is symmetric, perhaps the maximum occurs when x = y = z = π/6.Let me compute sin(π/6) cos²(π/6):sin(π/6) = 1/2, cos(π/6) = √3/2, so cos²(π/6) = 3/4.Thus, sinx cos²x = (1/2)(3/4) = 3/8.Sum of three such terms: 3*(3/8) = 9/8.Multiply by 2: 9/4 = 2.25.But 4/√3 ≈ 2.309, which is slightly larger. So, 2.25 < 4/√3, so the maximum might be higher.Wait, maybe when one angle is larger. Let me try x = π/3, y = π/6, z = 0.Compute sin(π/3) cos²(π/3) + sin(π/6) cos²(π/6) + sin(0) cos²(0)sin(π/3) = √3/2, cos(π/3) = 1/2, so cos²(π/3) = 1/4.Thus, sinx cos²x = (√3/2)(1/4) = √3/8 ≈ 0.216.sin(π/6) = 1/2, cos(π/6) = √3/2, so cos²(π/6) = 3/4.Thus, siny cos²y = (1/2)(3/4) = 3/8 ≈ 0.375.sinz cos²z = 0*1 = 0.Total sum: √3/8 + 3/8 ≈ 0.216 + 0.375 ≈ 0.591.Multiply by 2: ≈ 1.182, which is less than 4/√3.Hmm, maybe the maximum occurs when two angles are equal and the third is different. Let me try x = y = π/4, z = π/2 - 2*(π/4) = π/2 - π/2 = 0. But z = 0, which is not allowed.Alternatively, x = y = π/3, z = π/2 - 2*(π/3) = π/2 - 2π/3 = -π/6, which is negative. Not allowed.Wait, maybe x = π/2, y = z = 0. But then, sinx cos²x = sin(π/2) cos²(π/2) = 1*0 = 0. Similarly for y and z. So, sum is 0.Not helpful.Alternatively, maybe x = π/4, y = π/4, z = π/2 - π/2 = 0. Again, z = 0.Hmm, perhaps the maximum occurs when one angle is larger, but not too large.Wait, let me consider the function f(t) = sint cos²t.Compute its maximum.Take derivative:f'(t) = cost cos²t + sint * 2 cost (-sint) = cost cos²t - 2 sin²t costSet f'(t) = 0:cost cos²t - 2 sin²t cost = 0Factor out cost:cost (cos²t - 2 sin²t) = 0So, either cost = 0, which gives t = π/2, but then sint = 1, but cos²t = 0, so f(t) = 0.Or cos²t - 2 sin²t = 0 => cos²t = 2 sin²t => tan²t = 1/2 => tant = 1/√2 => t = arctan(1/√2) ≈ 35.26 degrees.So, maximum occurs at t ≈ 35.26 degrees.Compute f(t) at t = arctan(1/√2):sint = 1/√(1 + (√2)^2) = 1/√3cos t = √2/√3So, sint cos²t = (1/√3)( (√2/√3)^2 ) = (1/√3)(2/3) = 2/(3√3) ≈ 0.385So, the maximum of sint cos²t is 2/(3√3), same as before.Therefore, each term sint cos²t is bounded above by 2/(3√3). So, the sum sint cos²t + siny cos²y + sinz cos²z ≤ 3*(2/(3√3)) = 2/√3.Thus, the original expression:2 [sinx cos²x + siny cos²y + sinz cos²z] ≤ 2*(2/√3) = 4/√3Which is exactly what we needed to prove for the second inequality.So, both inequalities are proven by recognizing that each term can be individually bounded, and then summing those bounds.**Final Answer**Both inequalities are proven to hold, and the maximum value is boxed{dfrac{4}{sqrt{3}}}.

💡Okay, so I have this problem about a quadratic function, f(x) = 2x² - 2ax + b. They told me that f(-1) = -8, and also that f(x) is always greater than or equal to f(-1) for all real numbers x. Then, there are two sets defined: set A is all x where f(x) > 0, and set B is all x such that |x - t| ≤ 1. Part I asks me to find (C_R A) ∪ B when t = 1. Hmm, okay. So first, I need to figure out what set A is. Since f(x) is a quadratic function, it's a parabola. The fact that f(x) ≥ f(-1) for all x means that the vertex of the parabola is at x = -1, and it's the minimum point because the coefficient of x² is positive (which makes the parabola open upwards). So, f(-1) is the minimum value of the function, which is -8.Given that, I can write the function in vertex form. The vertex form of a quadratic is f(x) = a(x - h)² + k, where (h, k) is the vertex. Here, the vertex is (-1, -8), so f(x) = 2(x + 1)² - 8. Let me check that: expanding it, it should be 2(x² + 2x + 1) - 8 = 2x² + 4x + 2 - 8 = 2x² + 4x - 6. Wait, but the original function is 2x² - 2ax + b. Comparing coefficients, 2x² + 4x -6 vs 2x² -2ax + b. So, 4x = -2a x, which means -2a = 4, so a = -2. And b is -6. So, f(x) = 2x² + 4x -6.Wait, but in the problem statement, it's f(x) = 2x² -2ax + b. So, if a is -2, then -2a is 4, which matches. And b is -6. So, f(x) = 2x² + 4x -6.Now, set A is where f(x) > 0. So, I need to solve 2x² + 4x -6 > 0. Let's factor this quadratic. 2x² +4x -6 can be factored as 2(x² + 2x -3). Then, x² + 2x -3 factors into (x + 3)(x - 1). So, 2(x + 3)(x - 1) > 0. To find where this is positive, we can look at the critical points x = -3 and x = 1. The quadratic opens upwards because the coefficient is positive. So, the function is positive when x < -3 or x > 1. Therefore, set A is (-∞, -3) ∪ (1, ∞).Now, set B is |x - t| ≤ 1. When t = 1, this becomes |x - 1| ≤ 1, which is the interval [0, 2]. So, B = [0, 2].Now, (C_R A) is the complement of A in R, which would be the set where f(x) ≤ 0. Since A is (-∞, -3) ∪ (1, ∞), the complement is [-3, 1]. So, C_R A = [-3, 1].Then, (C_R A) ∪ B is [-3, 1] ∪ [0, 2]. Since [0, 2] overlaps with [-3, 1] from 0 to 1, the union is [-3, 2]. So, the answer for part I is [-3, 2].Moving on to part II. Proposition P is A ∩ B ≠ empty set. So, if ¬P is true, that means A ∩ B is empty. So, sets A and B do not overlap. Set A is (-∞, -3) ∪ (1, ∞), and set B is [t - 1, t + 1]. For their intersection to be empty, B must be entirely contained within the complement of A, which is [-3, 1]. So, B must be within [-3, 1]. Therefore, we need t - 1 ≥ -3 and t + 1 ≤ 1. Solving these inequalities:t - 1 ≥ -3 ⇒ t ≥ -2t + 1 ≤ 1 ⇒ t ≤ 0So, combining these, t must be between -2 and 0, inclusive. Therefore, the range of t is [-2, 0].Wait, let me double-check. If t is in [-2, 0], then B = [t -1, t +1] would be from [-3, -1] when t = -2, and [ -1, 1] when t = 0. So, in both cases, B is within [-3, 1], which is the complement of A. So, A ∩ B would be empty because A is outside of [-3, 1]. That makes sense. If t were less than -2, say t = -3, then B would be [-4, -2], which overlaps with A's (-∞, -3). Similarly, if t were greater than 0, say t = 1, then B would be [0, 2], which overlaps with A's (1, ∞). So, indeed, t must be between -2 and 0 to prevent any overlap.So, I think my answers are correct.**Final Answer**(I) boxed{[-3, 2]}(II) boxed{[-2, 0]}

💡Okay, so I need to figure out how much a company should repay each year to fully pay off a million yuan loan in five years with an annual interest rate of r. Hmm, let me break this down step by step.First, the company borrows one million yuan at the beginning of the year. That means the loan starts right away, and interest will start accruing from the beginning. The interest rate is r, which I assume is a decimal, like 0.05 for 5%. The company plans to repay x million yuan each year at the end of each year for five years. So, the first repayment is at the end of the first year, the second at the end of the second year, and so on until the fifth year.I think this is an annuity problem because the company is making equal payments each period (yearly in this case) to pay off a loan. The key here is that each payment not only covers the interest that has accrued but also reduces the principal amount. So, over time, more of each payment goes toward the principal and less toward interest.Let me recall the formula for the present value of an ordinary annuity, which is what this seems to be since payments are made at the end of each period. The present value (PV) of an annuity is given by:PV = PMT × [(1 - (1 + r)^-n) / r]Where:- PV is the present value of the loan, which is 1 million yuan.- PMT is the annual payment, which is x million yuan.- r is the annual interest rate.- n is the number of periods, which is 5 years.So, plugging in the values we have:1 = x × [(1 - (1 + r)^-5) / r]I need to solve for x. Let me rearrange the equation:x = 1 / [(1 - (1 + r)^-5) / r]Simplifying that:x = r / [1 - (1 + r)^-5]Wait, that seems a bit off. Let me double-check. The present value formula is:PV = PMT × [(1 - (1 + r)^-n) / r]So, solving for PMT:PMT = PV / [(1 - (1 + r)^-n) / r]Which simplifies to:PMT = PV × r / [1 - (1 + r)^-n]Yes, that's correct. So, in this case:x = 1 × r / [1 - (1 + r)^-5]But wait, I think I might have missed something. The loan is taken out at the beginning of the year, so the first payment is at the end of the first year. Does that affect the calculation? Let me think. Since the payments are made at the end of each period, the formula I used should still apply because it's an ordinary annuity.Alternatively, if the payments were made at the beginning of each period, it would be an annuity due, and the formula would be slightly different. But in this case, it's at the end, so ordinary annuity is correct.Let me also consider the future value approach. The total amount owed after five years, considering the interest, should be equal to the sum of the future values of the payments. So, the future value of the loan is:FV = 1 × (1 + r)^5And the future value of the payments is:FV = x × [(1 + r)^5 - 1] / rSetting them equal:1 × (1 + r)^5 = x × [(1 + r)^5 - 1] / rSolving for x:x = [1 × (1 + r)^5 × r] / [(1 + r)^5 - 1]Hmm, that's a different expression than what I got earlier. Which one is correct?Wait, I think both approaches should give the same result. Let me see. From the present value approach, I had:x = r / [1 - (1 + r)^-5]From the future value approach, I have:x = r × (1 + r)^5 / [(1 + r)^5 - 1]Are these equivalent?Let me manipulate the first expression:x = r / [1 - (1 + r)^-5] = r / [1 - 1/(1 + r)^5] = r / [( (1 + r)^5 - 1 ) / (1 + r)^5 ] = r × (1 + r)^5 / [ (1 + r)^5 - 1 ]Yes, they are the same. So both methods lead to the same formula for x.Therefore, the annual repayment amount x is:x = [r × (1 + r)^5] / [ (1 + r)^5 - 1 ]This makes sense because as r increases, the denominator increases, making x increase, which is logical since higher interest rates mean higher repayments. Similarly, as the number of years increases, the denominator increases, making x decrease, which is also logical because spreading the repayments over more years reduces the annual payment.Let me test this formula with a simple example to see if it makes sense. Suppose r = 0 (no interest). Then:x = [0 × (1 + 0)^5] / [ (1 + 0)^5 - 1 ] = 0 / (1 - 1) = 0/0, which is undefined. But intuitively, if there's no interest, the company should repay 1 million yuan over 5 years, so x = 1/5 = 0.2 million yuan per year. So, the formula doesn't handle r = 0 well, but that's expected because the formula is derived for r > 0.Another test: suppose r = 100% per year, which is very high. Then:x = [1 × (1 + 1)^5] / [ (1 + 1)^5 - 1 ] = [1 × 32] / (32 - 1) = 32 / 31 ≈ 1.032 million yuan per year. That seems high, but with 100% interest, the debt doubles each year. So, the first year's payment would need to cover the interest for the first year, which is 1 million yuan, plus some principal. But since the payments are spread out, it's a bit more complicated. However, the result seems plausible.Another test: suppose r = 5% or 0.05. Then:x = [0.05 × (1.05)^5] / [ (1.05)^5 - 1 ]First, calculate (1.05)^5:1.05^1 = 1.051.05^2 = 1.10251.05^3 ≈ 1.1576251.05^4 ≈ 1.215506251.05^5 ≈ 1.2762815625So,x = [0.05 × 1.2762815625] / [1.2762815625 - 1] ≈ [0.063814078125] / [0.2762815625] ≈ 0.230975So, x ≈ 0.231 million yuan per year, or 231,000 yuan per year. That seems reasonable for a 5% interest rate over 5 years.Wait, but let me check with the present value formula:PV = PMT × [(1 - (1 + r)^-n) / r]So,1 = x × [(1 - (1.05)^-5) / 0.05]Calculate (1.05)^-5:1 / 1.2762815625 ≈ 0.7835261665So,1 = x × [(1 - 0.7835261665) / 0.05] = x × [0.2164738335 / 0.05] ≈ x × 4.32947667Therefore,x ≈ 1 / 4.32947667 ≈ 0.231 million yuan, which matches the previous result. So, the formula works correctly.Therefore, I'm confident that the formula for x is:x = [r × (1 + r)^5] / [ (1 + r)^5 - 1 ]This gives the annual repayment amount needed to pay off the loan in five years with annual interest rate r.