💡Okay, so I have these four integrals to solve. Let me take them one by one and think through each step carefully. I remember that integration can sometimes be tricky, especially with trigonometric functions, square roots, and substitutions. I need to recall some integration techniques like substitution, trigonometric identities, and maybe even some geometry if needed.Starting with the first integral: (int_{0}^{frac{pi}{2}} sin^{3} x , dx). Hmm, I remember that for integrals involving powers of sine or cosine, sometimes it's helpful to use trigonometric identities to simplify the expression. Specifically, for odd powers, we can factor out one sine or cosine term and use the Pythagorean identity. So, for (sin^3 x), I can write it as (sin x cdot sin^2 x). Then, using the identity (sin^2 x = 1 - cos^2 x), the integral becomes:[int_{0}^{frac{pi}{2}} sin x (1 - cos^2 x) , dx]Now, this looks like a good candidate for substitution. Let me set (u = cos x), which means (du = -sin x , dx). So, (sin x , dx = -du). I also need to change the limits of integration accordingly. When (x = 0), (u = cos 0 = 1), and when (x = frac{pi}{2}), (u = cos frac{pi}{2} = 0). Substituting, the integral becomes:[- int_{1}^{0} (1 - u^2) , du]Which is the same as:[int_{0}^{1} (1 - u^2) , du]This is straightforward to integrate. The integral of 1 is (u), and the integral of (u^2) is (frac{u^3}{3}). So, evaluating from 0 to 1:[left[ u - frac{u^3}{3} right]_0^1 = (1 - frac{1}{3}) - (0 - 0) = frac{2}{3}]Okay, that seems reasonable. I think I did that correctly. Let me double-check my substitution steps. Yes, everything looks good.Moving on to the second integral: (int_{0}^{ln 2} sqrt{e^{x}-1} , dx). This one has an exponential function inside a square root, which might be a bit challenging. I think substitution is the way to go here as well.Let me set (t = sqrt{e^x - 1}). Then, squaring both sides, (t^2 = e^x - 1), so (e^x = t^2 + 1). Taking the derivative of both sides with respect to x, (e^x dx = 2t dt), which means (dx = frac{2t}{t^2 + 1} dt).Now, I need to change the limits of integration. When (x = 0), (t = sqrt{e^0 - 1} = sqrt{1 - 1} = 0). When (x = ln 2), (t = sqrt{e^{ln 2} - 1} = sqrt{2 - 1} = 1).Substituting into the integral:[int_{0}^{1} t cdot frac{2t}{t^2 + 1} , dt = 2 int_{0}^{1} frac{t^2}{t^2 + 1} , dt]Hmm, this simplifies to:[2 int_{0}^{1} left(1 - frac{1}{t^2 + 1}right) , dt]Because (frac{t^2}{t^2 + 1} = 1 - frac{1}{t^2 + 1}). That makes it easier to integrate. Now, integrating term by term:[2 left[ int_{0}^{1} 1 , dt - int_{0}^{1} frac{1}{t^2 + 1} , dt right] = 2 left[ t bigg|_{0}^{1} - arctan t bigg|_{0}^{1} right]]Calculating each part:[2 left[ (1 - 0) - left( frac{pi}{4} - 0 right) right] = 2 left( 1 - frac{pi}{4} right) = 2 - frac{pi}{2}]That seems correct. Let me verify the substitution steps again. Yes, everything checks out.Third integral: (int_{-a}^{a} x^{2} sqrt{a^{2}-x^{2}} , dx). This looks like a standard integral that might relate to the area of a circle or something similar. The limits are symmetric around zero, so maybe I can use a substitution that takes advantage of that symmetry.Let me try the substitution (x = a sin theta). Then, (dx = a cos theta , dtheta), and (sqrt{a^2 - x^2} = sqrt{a^2 - a^2 sin^2 theta} = a cos theta).Changing the limits of integration: when (x = -a), (sin theta = -1), so (theta = -frac{pi}{2}). When (x = a), (sin theta = 1), so (theta = frac{pi}{2}).Substituting into the integral:[int_{-frac{pi}{2}}^{frac{pi}{2}} (a^2 sin^2 theta) (a cos theta) (a cos theta , dtheta) = a^4 int_{-frac{pi}{2}}^{frac{pi}{2}} sin^2 theta cos^2 theta , dtheta]Simplify the integrand. I remember that (sin^2 theta cos^2 theta) can be rewritten using double-angle identities. Specifically, (sin^2 theta cos^2 theta = frac{1}{4} sin^2 2theta). So:[a^4 cdot frac{1}{4} int_{-frac{pi}{2}}^{frac{pi}{2}} sin^2 2theta , dtheta = frac{a^4}{4} int_{-frac{pi}{2}}^{frac{pi}{2}} sin^2 2theta , dtheta]Again, using the identity (sin^2 x = frac{1 - cos 2x}{2}), so:[frac{a^4}{4} cdot frac{1}{2} int_{-frac{pi}{2}}^{frac{pi}{2}} (1 - cos 4theta) , dtheta = frac{a^4}{8} left[ int_{-frac{pi}{2}}^{frac{pi}{2}} 1 , dtheta - int_{-frac{pi}{2}}^{frac{pi}{2}} cos 4theta , dtheta right]]Calculating each integral:First integral: (int_{-frac{pi}{2}}^{frac{pi}{2}} 1 , dtheta = pi).Second integral: (int_{-frac{pi}{2}}^{frac{pi}{2}} cos 4theta , dtheta). The integral of (cos 4theta) is (frac{sin 4theta}{4}). Evaluating from (-frac{pi}{2}) to (frac{pi}{2}):[frac{sin 4left(frac{pi}{2}right)}{4} - frac{sin 4left(-frac{pi}{2}right)}{4} = frac{sin 2pi}{4} - frac{sin(-2pi)}{4} = 0 - 0 = 0]So, the second integral is zero. Therefore, the entire expression simplifies to:[frac{a^4}{8} cdot pi = frac{pi a^4}{8}]That makes sense. The integral is positive and relates to the area of a semicircle, which is consistent with the result.Finally, the fourth integral: (int_{1}^{2} frac{sqrt{x^{2}-1}}{x} , dx). This looks similar to the second integral, with a square root of (x^2 - 1). Maybe substitution will help here as well.Let me set (t = sqrt{x^2 - 1}). Then, (t^2 = x^2 - 1), so (x^2 = t^2 + 1). Differentiating both sides, (2x , dx = 2t , dt), which simplifies to (x , dx = t , dt). Therefore, (dx = frac{t}{x} dt).But since (x = sqrt{t^2 + 1}), we can write (dx = frac{t}{sqrt{t^2 + 1}} dt).Changing the limits of integration: when (x = 1), (t = sqrt{1 - 1} = 0). When (x = 2), (t = sqrt{4 - 1} = sqrt{3}).Substituting into the integral:[int_{0}^{sqrt{3}} frac{t}{sqrt{t^2 + 1}} cdot frac{t}{sqrt{t^2 + 1}} , dt = int_{0}^{sqrt{3}} frac{t^2}{t^2 + 1} , dt]This is similar to the second integral. Let's rewrite the integrand:[frac{t^2}{t^2 + 1} = 1 - frac{1}{t^2 + 1}]So, the integral becomes:[int_{0}^{sqrt{3}} left(1 - frac{1}{t^2 + 1}right) , dt = int_{0}^{sqrt{3}} 1 , dt - int_{0}^{sqrt{3}} frac{1}{t^2 + 1} , dt]Calculating each part:First integral: (int_{0}^{sqrt{3}} 1 , dt = sqrt{3}).Second integral: (int_{0}^{sqrt{3}} frac{1}{t^2 + 1} , dt = arctan t bigg|_{0}^{sqrt{3}} = frac{pi}{3} - 0 = frac{pi}{3}).Putting it all together:[sqrt{3} - frac{pi}{3}]That seems correct. Let me just verify the substitution steps once more. Yes, everything looks consistent.So, to recap:1. For the first integral, using substitution with (u = cos x) simplified the integral nicely, leading to (frac{2}{3}).2. The second integral required a substitution with (t = sqrt{e^x - 1}), which transformed the integral into a manageable form, resulting in (2 - frac{pi}{2}).3. The third integral was tackled using a trigonometric substitution (x = a sin theta), leveraging the symmetry of the integral and leading to (frac{pi a^4}{8}).4. The fourth integral used substitution (t = sqrt{x^2 - 1}), similar to the second one, and resulted in (sqrt{3} - frac{pi}{3}).All steps seem logical and consistent. I don't see any mistakes in my reasoning, so I think these are the correct answers.

💡Alright, so I have this problem where I need to find all triples (a, b, p) of natural numbers such that p is a prime number and the equation (a + b)^p = p^a + p^b holds true. Hmm, okay, let me try to figure this out step by step.First, let me understand the equation: (a + b)^p equals p^a plus p^b. Since p is a prime number, that might give me some constraints to work with. I guess I should consider different cases for a and b, maybe when they're equal or not.Let me start by assuming that a equals b. If a = b, then the equation becomes (2a)^p = p^a + p^a, which simplifies to (2a)^p = 2p^a. Dividing both sides by 2 gives (2a)^p / 2 = p^a. Hmm, that seems a bit complicated. Maybe I can test small values for a and p.Let's try p = 2 first because it's the smallest prime. If p = 2, then the equation becomes (2a)^2 = 2*2^a. Simplifying that, we get 4a^2 = 2*2^a, which is 4a^2 = 2^(a+1). Dividing both sides by 2, we have 2a^2 = 2^a. So, 2a^2 = 2^a. Let's test small values of a:- If a = 1: 2*1^2 = 2^1 → 2 = 2. That works!- If a = 2: 2*2^2 = 2^2 → 8 = 4. Nope, doesn't work.- If a = 3: 2*3^2 = 2^3 → 18 = 8. Nope. So, the only solution when a = b is when a = b = 1 and p = 2. That gives us the triple (1, 1, 2). Okay, that's one solution.Now, let's consider the case where a ≠ b. Without loss of generality, let's assume a < b. Then, the equation becomes (a + b)^p = p^a + p^b. Maybe I can factor out p^a from the right-hand side: (a + b)^p = p^a(1 + p^{b - a}).Hmm, since p is a prime, I wonder if I can analyze the divisibility here. The left side, (a + b)^p, must be divisible by p. The right side is p^a times (1 + p^{b - a}). Since p is prime, (1 + p^{b - a}) is not divisible by p because p^{b - a} is a multiple of p, and adding 1 makes it not divisible by p. So, p must divide (a + b)^p, which means p divides (a + b). Therefore, a + b must be a multiple of p.Let me denote a + b = kp, where k is some natural number. Then, the equation becomes (kp)^p = p^a + p^b. Let's rewrite this as k^p * p^p = p^a + p^b. Dividing both sides by p^a, we get k^p * p^{p - a} = 1 + p^{b - a}.Hmm, this seems tricky. Let me think about the exponents. Since p is prime, and a and b are natural numbers with a < b, b - a is at least 1. So, p^{b - a} is at least p. Therefore, 1 + p^{b - a} is at least p + 1.On the left side, we have k^p * p^{p - a}. Since p^{p - a} is a large number unless a is close to p, maybe I can consider small values of a and p.Let's try p = 2 again. Then, the equation becomes (a + b)^2 = 2^a + 2^b. Since a < b, let's try a = 1:(1 + b)^2 = 2^1 + 2^b → (1 + b)^2 = 2 + 2^b.Let's test small values of b:- b = 1: (1 + 1)^2 = 2 + 2^1 → 4 = 4. Wait, but a < b, so b can't be 1 if a = 1.- b = 2: (1 + 2)^2 = 2 + 2^2 → 9 = 6. Nope.- b = 3: (1 + 3)^2 = 2 + 2^3 → 16 = 10. Nope.- b = 4: (1 + 4)^2 = 2 + 2^4 → 25 = 18. Nope. Hmm, doesn't seem to work for a = 1 and p = 2. Let's try a = 2:(2 + b)^2 = 2^2 + 2^b → (2 + b)^2 = 4 + 2^b.Testing b:- b = 3: (2 + 3)^2 = 4 + 8 → 25 = 12. Nope.- b = 4: (2 + 4)^2 = 4 + 16 → 36 = 20. Nope.- b = 5: (2 + 5)^2 = 4 + 32 → 49 = 36. Nope. Not working either. Maybe p = 3?Let's try p = 3. Then, the equation becomes (a + b)^3 = 3^a + 3^b. Again, a < b.Let's try a = 1:(1 + b)^3 = 3 + 3^b.Testing b:- b = 2: (1 + 2)^3 = 3 + 9 → 27 = 12. Nope.- b = 3: (1 + 3)^3 = 3 + 27 → 64 = 30. Nope.- b = 4: (1 + 4)^3 = 3 + 81 → 125 = 84. Nope. Not working. Let's try a = 2:(2 + b)^3 = 9 + 3^b.Testing b:- b = 3: (2 + 3)^3 = 9 + 27 → 125 = 36. Nope.- b = 4: (2 + 4)^3 = 9 + 81 → 216 = 90. Nope. Hmm, not working either. Maybe p = 5?Wait, this might not be the best approach. Maybe I should think about the growth rates of both sides. The left side is (a + b)^p, which grows polynomially, while the right side is p^a + p^b, which grows exponentially. So, for large a and b, the right side will dominate. Therefore, maybe the solutions are only for small values of a and b.Let me try p = 2 again, but this time with a = 1 and b = 1, which we already saw works. What about a = 1 and b = 2?(1 + 2)^2 = 2^1 + 2^2 → 9 = 2 + 4 → 9 = 6. Nope.a = 1, b = 3:(1 + 3)^2 = 2 + 8 → 16 = 10. Nope.a = 1, b = 4:(1 + 4)^2 = 2 + 16 → 25 = 18. Nope.a = 2, b = 3:(2 + 3)^2 = 4 + 8 → 25 = 12. Nope.a = 2, b = 4:(2 + 4)^2 = 4 + 16 → 36 = 20. Nope.a = 3, b = 4:(3 + 4)^2 = 8 + 16 → 49 = 24. Nope.Hmm, seems like p = 2 only works when a = b = 1.What about p = 3? Let's try a = 1, b = 2:(1 + 2)^3 = 3 + 9 → 27 = 12. Nope.a = 1, b = 3:(1 + 3)^3 = 3 + 27 → 64 = 30. Nope.a = 2, b = 3:(2 + 3)^3 = 9 + 27 → 125 = 36. Nope.a = 1, b = 4:(1 + 4)^3 = 3 + 81 → 125 = 84. Nope.a = 2, b = 4:(2 + 4)^3 = 9 + 81 → 216 = 90. Nope.Not working either. Maybe p = 5?a = 1, b = 2:(1 + 2)^5 = 5 + 25 → 243 = 30. Nope.a = 1, b = 3:(1 + 3)^5 = 5 + 125 → 1024 = 130. Nope.a = 2, b = 3:(2 + 3)^5 = 25 + 125 → 3125 = 150. Nope.This seems like a dead end. Maybe I need to think differently.Let me go back to the equation: (a + b)^p = p^a + p^b. If a ≠ b, let's assume a < b. Then, p^b is much larger than p^a, so maybe (a + b)^p is approximately p^b. So, (a + b)^p ≈ p^b. Taking logarithms, p log(a + b) ≈ b log p. So, b ≈ p log(a + b) / log p.But I'm not sure if this helps. Maybe I can consider the ratio of the two sides. Let's divide both sides by p^b:(a + b)^p / p^b = p^{a - b} + 1.Let me denote c = b - a, so b = a + c. Then, the equation becomes:(a + a + c)^p / p^{a + c} = p^{-c} + 1.Simplifying:(2a + c)^p / p^{a + c} = 1 + p^{-c}.Hmm, since p^{-c} is less than 1, the left side must be slightly larger than 1. So, (2a + c)^p must be slightly larger than p^{a + c}.Let me try small values for c. Let's say c = 1, so b = a + 1.Then, the equation becomes:(2a + 1)^p / p^{a + 1} = 1 + p^{-1} = (p + 1)/p.So, (2a + 1)^p = p^{a} (p + 1).Let me test p = 2:(2a + 1)^2 = 2^{a} * 3.Testing small a:a = 1: (3)^2 = 2 * 3 → 9 = 6. Nope.a = 2: (5)^2 = 4 * 3 → 25 = 12. Nope.a = 3: (7)^2 = 8 * 3 → 49 = 24. Nope.Not working.p = 3:(2a + 1)^3 = 3^{a} * 4.Testing a = 1:3^3 = 3 * 4 → 27 = 12. Nope.a = 2:5^3 = 9 * 4 → 125 = 36. Nope.a = 3:7^3 = 27 * 4 → 343 = 108. Nope.Not working.p = 5:(2a + 1)^5 = 5^{a} * 6.Testing a = 1:3^5 = 5 * 6 → 243 = 30. Nope.a = 2:5^5 = 25 * 6 → 3125 = 150. Nope.Not working.Hmm, maybe c = 2?So, b = a + 2.Then, the equation becomes:(2a + 2)^p / p^{a + 2} = 1 + p^{-2}.Simplifying:(2(a + 1))^p / p^{a + 2} = (p^2 + 1)/p^2.So, (2(a + 1))^p = p^{a} (p^2 + 1).Let me try p = 2:(2(a + 1))^2 = 2^{a} * 5.Testing a = 1:(4)^2 = 2 * 5 → 16 = 10. Nope.a = 2:(6)^2 = 4 * 5 → 36 = 20. Nope.a = 3:(8)^2 = 8 * 5 → 64 = 40. Nope.Not working.p = 3:(2(a + 1))^3 = 3^{a} * 10.Testing a = 1:(4)^3 = 3 * 10 → 64 = 30. Nope.a = 2:(6)^3 = 9 * 10 → 216 = 90. Nope.a = 3:(8)^3 = 27 * 10 → 512 = 270. Nope.Not working.p = 5:(2(a + 1))^5 = 5^{a} * 26.Testing a = 1:(4)^5 = 5 * 26 → 1024 = 130. Nope.a = 2:(6)^5 = 25 * 26 → 7776 = 650. Nope.Not working.This approach doesn't seem to be yielding any solutions. Maybe I need to consider that when a ≠ b, there are no solutions other than the one we found earlier when a = b = 1 and p = 2.Wait, let me think about the case when one of a or b is zero. But the problem states natural numbers, which usually start from 1, so a and b can't be zero. So, that's not an option.Another thought: maybe p = 2 is the only prime that works because it's the smallest, and higher primes make the right side too large. Let me check p = 2 with a = 1 and b = 1, which we already did, and it works. What about a = 2 and b = 2?(2 + 2)^2 = 2^2 + 2^2 → 16 = 4 + 4 → 16 = 8. Nope, doesn't work.a = 3 and b = 3:(6)^2 = 8 + 8 → 36 = 16. Nope.So, only a = b = 1 works for p = 2.What about p = 2 with a = 1 and b = 2?(1 + 2)^2 = 2^1 + 2^2 → 9 = 2 + 4 → 9 = 6. Nope.a = 1, b = 3:(4)^2 = 2 + 8 → 16 = 10. Nope.a = 2, b = 3:(5)^2 = 4 + 8 → 25 = 12. Nope.So, no solutions for p = 2 when a ≠ b.What about p = 3? Let's try a = 1, b = 2:(3)^3 = 3 + 9 → 27 = 12. Nope.a = 1, b = 3:(4)^3 = 3 + 27 → 64 = 30. Nope.a = 2, b = 3:(5)^3 = 9 + 27 → 125 = 36. Nope.Not working.p = 5:a = 1, b = 2:(3)^5 = 5 + 25 → 243 = 30. Nope.a = 1, b = 3:(4)^5 = 5 + 125 → 1024 = 130. Nope.a = 2, b = 3:(5)^5 = 25 + 125 → 3125 = 150. Nope.No luck there either.Maybe I should consider that the only solution is when a = b = 1 and p = 2. Let me see if there are any other possibilities.Wait, what if a = 0? But natural numbers start at 1, so that's not allowed. Similarly, b can't be zero.Another angle: let's consider the equation modulo p. Since p is prime, Fermat's little theorem might help, but I'm not sure. Let me see:(a + b)^p ≡ a + b mod p.On the right side, p^a + p^b ≡ 0 + 0 ≡ 0 mod p.So, we have a + b ≡ 0 mod p. That means p divides a + b. So, a + b must be a multiple of p.We already considered that earlier when we set a + b = kp. So, that's consistent.Given that, and knowing that a and b are natural numbers, maybe k is small. Let's try k = 1, so a + b = p.Then, the equation becomes p^p = p^a + p^b.But since a + b = p, and a < b, let's say a = 1, then b = p - 1.So, p^p = p^1 + p^{p - 1}.Simplify: p^p = p + p^{p - 1}.Divide both sides by p^{p - 1}:p = p^{1 - (p - 1)} + 1 → p = p^{2 - p} + 1.Wait, that doesn't make sense because for p ≥ 2, 2 - p is negative, making p^{2 - p} a fraction. But p is an integer, so this can't hold. Therefore, k = 1 doesn't work.What about k = 2? Then, a + b = 2p.So, the equation becomes (2p)^p = p^a + p^b.Again, let's assume a < b, so a = 1, b = 2p - 1.Then, (2p)^p = p + p^{2p - 1}.Divide both sides by p:(2p)^{p - 1} = 1 + p^{2p - 2}.Hmm, for p = 2:(4)^1 = 1 + 4 → 4 = 5. Nope.p = 3:(6)^2 = 1 + 81 → 36 = 82. Nope.p = 5:(10)^4 = 1 + 5^8 → 10000 = 1 + 390625 → 10000 = 390626. Nope.Not working.What if k = 3? a + b = 3p.Then, (3p)^p = p^a + p^b.Assuming a = 1, b = 3p - 1:(3p)^p = p + p^{3p - 1}.Divide by p:(3p)^{p - 1} = 1 + p^{3p - 2}.For p = 2:(6)^1 = 1 + 64 → 6 = 65. Nope.p = 3:(9)^2 = 1 + 729 → 81 = 730. Nope.Not working.This seems like a dead end too. Maybe k is not an integer, but since a and b are natural numbers, k has to be an integer.Wait, maybe I should consider that when a ≠ b, the equation doesn't have any solutions because the right side grows too fast compared to the left side. So, the only solution is when a = b = 1 and p = 2.Let me check p = 2 with a = 1 and b = 1 again:(1 + 1)^2 = 2^1 + 2^1 → 4 = 2 + 2 → 4 = 4. Yep, that works.Is there any other possibility? Maybe a = 0 or b = 0, but since natural numbers start at 1, that's not allowed. So, I think that's the only solution.Therefore, the only triple is (1, 1, 2).

💡Okay, so I have this problem where I need to test whether the failure probabilities of components in two devices are equal. Let me try to understand what's given and what I need to do.First, Device 1 has 800 components, and during a shift, 15 of them failed. Device 2 has 1000 components, and 25 failed during the same shift. I need to test the null hypothesis that the failure probabilities for both devices are the same against the alternative that they are different. The significance level is 0.05.Alright, so I think this is a hypothesis test for two proportions. Since we're dealing with proportions, and the alternative is that they are not equal, it's a two-tailed test. I remember that for such tests, we can use a z-test for two proportions.Let me recall the steps for this test:1. **State the hypotheses:** - Null hypothesis (H₀): p₁ = p₂ - Alternative hypothesis (H₁): p₁ ≠ p₂2. **Calculate the sample proportions:** - For Device 1: p₁ = 15/800 - For Device 2: p₂ = 25/10003. **Find the pooled sample proportion (p̂):** - p̂ = (total failures) / (total components) = (15 + 25) / (800 + 1000)4. **Calculate the standard error (SE) of the difference in proportions:** - SE = sqrt[p̂(1 - p̂)(1/n₁ + 1/n₂)]5. **Compute the z-score:** - z = (p₁ - p₂) / SE6. **Determine the critical value or p-value:** - Since it's a two-tailed test at α = 0.05, the critical z-values are ±1.96.7. **Compare the z-score to the critical value:** - If |z| > 1.96, reject H₀; otherwise, fail to reject H₀.Let me do these calculations step by step.First, calculating the sample proportions:- p₁ = 15/800 = 0.01875- p₂ = 25/1000 = 0.025Next, the pooled proportion p̂:- p̂ = (15 + 25) / (800 + 1000) = 40 / 1800 ≈ 0.022222Now, the standard error SE:- SE = sqrt[p̂(1 - p̂)(1/800 + 1/1000)]- Let's compute each part: - p̂(1 - p̂) = 0.022222 * (1 - 0.022222) ≈ 0.022222 * 0.977778 ≈ 0.021753 - (1/800 + 1/1000) = 0.00125 + 0.001 = 0.00225- So, SE = sqrt[0.021753 * 0.00225] ≈ sqrt[0.00004889] ≈ 0.00699Now, the z-score:- z = (p₁ - p₂) / SE = (0.01875 - 0.025) / 0.00699 ≈ (-0.00625) / 0.00699 ≈ -0.893So, the z-score is approximately -0.893.Now, comparing this to the critical value of ±1.96. Since -0.893 is between -1.96 and 1.96, we fail to reject the null hypothesis.Therefore, at the 0.05 significance level, there's not enough evidence to conclude that the failure probabilities of the two devices are different.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, p₁ and p₂ seem correct: 15/800 is indeed 0.01875, and 25/1000 is 0.025.The pooled proportion p̂ = 40/1800 ≈ 0.022222. That looks right.Calculating p̂(1 - p̂): 0.022222 * 0.977778 ≈ 0.021753. That seems correct.Then, 1/800 + 1/1000 = 0.00125 + 0.001 = 0.00225. Correct.Multiplying p̂(1 - p̂) by 0.00225 gives 0.021753 * 0.00225 ≈ 0.00004889. Taking the square root gives approximately 0.00699. That seems right.Then, z = (0.01875 - 0.025) / 0.00699 ≈ -0.00625 / 0.00699 ≈ -0.893. Correct.Since the z-score is -0.893, which is greater than -1.96 and less than 1.96, we fail to reject the null hypothesis.So, my conclusion is that there's not enough evidence to suggest that the failure probabilities are different between the two devices at the 0.05 significance level.I think that's it. I don't see any mistakes in my calculations or reasoning.

💡Okay, so I have this problem where Andrea is working with a regular hexagon and Christine is working with a regular octagon. Both polygons have a side length of 2. They each inscribe a circle inside their polygon and circumscribe a circle around it. Then they calculate the area between these two circles, which are called H for the hexagon and C for the octagon. I need to figure out the relationship between H and C.First, I should recall some properties of regular polygons, especially hexagons and octagons. I know that for regular polygons, the radius of the circumscribed circle (the one around the polygon) is called the circumradius, and the radius of the inscribed circle (the one inside the polygon) is called the apothem.For a regular polygon with side length 's', the formulas for the circumradius (R) and apothem (a) depend on the number of sides. Specifically, for a polygon with 'n' sides, the formulas are:- Circumradius (R) = s / (2 * sin(π/n))- Apothem (a) = s / (2 * tan(π/n))So, for both the hexagon and the octagon, I can calculate R and a, then find the area between the two circles, which is the area of the circumscribed circle minus the area of the inscribed circle.Let me start with the hexagon. A regular hexagon has 6 sides, so n = 6.Calculating the circumradius (R_hex) for the hexagon:R_hex = s / (2 * sin(π/6))I know that sin(π/6) is 0.5, so:R_hex = 2 / (2 * 0.5) = 2 / 1 = 2So the circumradius is 2.Now, the apothem (a_hex) for the hexagon:a_hex = s / (2 * tan(π/6))I remember that tan(π/6) is 1/√3, so:a_hex = 2 / (2 * (1/√3)) = 2 / (2/√3) = √3So the apothem is √3.Now, the area between the two circles for the hexagon (H) is the area of the circumscribed circle minus the area of the inscribed circle.Area of circumscribed circle = π * R_hex² = π * (2)² = 4πArea of inscribed circle = π * a_hex² = π * (√3)² = π * 3 = 3πSo, H = 4π - 3π = πAlright, so H is π.Now, moving on to the octagon. A regular octagon has 8 sides, so n = 8.Calculating the circumradius (R_oct) for the octagon:R_oct = s / (2 * sin(π/8))I need to compute sin(π/8). I remember that π/8 is 22.5 degrees. I can use the half-angle formula for sine:sin(θ/2) = √[(1 - cosθ)/2]Let me compute sin(22.5°). Let θ = 45°, so θ/2 = 22.5°.sin(22.5°) = √[(1 - cos45°)/2] = √[(1 - √2/2)/2]Let me compute that:First, cos45° = √2/2 ≈ 0.7071So, 1 - √2/2 ≈ 1 - 0.7071 ≈ 0.2929Divide that by 2: 0.2929 / 2 ≈ 0.14645Take the square root: √0.14645 ≈ 0.38268So, sin(22.5°) ≈ 0.38268Therefore, R_oct = 2 / (2 * 0.38268) = 2 / 0.76536 ≈ 2.6131So, the circumradius is approximately 2.6131.Now, the apothem (a_oct) for the octagon:a_oct = s / (2 * tan(π/8))Again, π/8 is 22.5°, so I need tan(22.5°). I can use the half-angle formula for tangent:tan(θ/2) = (1 - cosθ) / sinθLet θ = 45°, so θ/2 = 22.5°.tan(22.5°) = (1 - cos45°) / sin45° = (1 - √2/2) / (√2/2)Let me compute that:1 - √2/2 ≈ 1 - 0.7071 ≈ 0.2929Divide that by √2/2 ≈ 0.7071:0.2929 / 0.7071 ≈ 0.4142So, tan(22.5°) ≈ 0.4142Therefore, a_oct = 2 / (2 * 0.4142) = 2 / 0.8284 ≈ 2.4142So, the apothem is approximately 2.4142.Now, the area between the two circles for the octagon (C) is the area of the circumscribed circle minus the area of the inscribed circle.Area of circumscribed circle = π * R_oct² ≈ π * (2.6131)² ≈ π * 6.8284Area of inscribed circle = π * a_oct² ≈ π * (2.4142)² ≈ π * 5.8284So, C ≈ 6.8284π - 5.8284π ≈ (6.8284 - 5.8284)π ≈ πWait, that's interesting. So, C is approximately π as well.But let me double-check my calculations because that seems too clean. Maybe I made an approximation error.Looking back at the octagon:R_oct ≈ 2.6131, so R_oct² ≈ 6.8284a_oct ≈ 2.4142, so a_oct² ≈ 5.8284Subtracting: 6.8284 - 5.8284 = 1.0000So, C ≈ π * 1.0000 ≈ πSimilarly, H was exactly π.So, both H and C are equal to π. Therefore, H = C.But wait, let me check if my approximations are accurate enough. Maybe the exact values are slightly different.Let me compute R_oct exactly:R_oct = 2 / (2 * sin(π/8)) = 1 / sin(π/8)We know that sin(π/8) = sin(22.5°) = √(2 - √2)/2 ≈ 0.38268So, R_oct = 1 / (√(2 - √2)/2) = 2 / √(2 - √2)Multiply numerator and denominator by √(2 + √2):R_oct = 2√(2 + √2) / √((2 - √2)(2 + √2)) = 2√(2 + √2) / √(4 - 2) = 2√(2 + √2) / √2 = √2 * √(2 + √2) = √(2*(2 + √2)) = √(4 + 2√2)Similarly, a_oct = 2 / (2 * tan(π/8)) = 1 / tan(π/8)tan(π/8) = tan(22.5°) = √(2) - 1 ≈ 0.4142So, a_oct = 1 / (√2 - 1) = (√2 + 1) / ((√2 - 1)(√2 + 1)) = (√2 + 1)/1 = √2 + 1 ≈ 2.4142Therefore, R_oct² = (√(4 + 2√2))² = 4 + 2√2a_oct² = (√2 + 1)² = 2 + 2√2 + 1 = 3 + 2√2So, the area between the circles for the octagon is:π*(R_oct² - a_oct²) = π*(4 + 2√2 - 3 - 2√2) = π*(1) = πSimilarly, for the hexagon, we had H = π.Therefore, both H and C are exactly π. So, H = C.But wait, that seems counterintuitive because an octagon has more sides than a hexagon, so I would expect the area between the circles to be different. Maybe my exact calculations show they are equal, but let me think again.Wait, no, actually, the area between the circles depends on the difference between the squares of the circumradius and the apothem. For both polygons, this difference turned out to be 1, so both areas are π.But let me verify with exact values:For the hexagon:R_hex = 2a_hex = √3R_hex² - a_hex² = 4 - 3 = 1So, H = π*1 = πFor the octagon:R_oct² = 4 + 2√2a_oct² = 3 + 2√2R_oct² - a_oct² = (4 + 2√2) - (3 + 2√2) = 1So, C = π*1 = πTherefore, both H and C are equal to π. So, H = C.But wait, that seems surprising because the octagon is closer to a circle than the hexagon, so I would expect the area between the circles to be smaller for the octagon. But according to the calculations, they are equal.Wait, maybe I made a mistake in the exact calculations. Let me re-examine.For the octagon:R_oct = √(4 + 2√2)a_oct = √2 + 1So, R_oct² = 4 + 2√2a_oct² = (√2 + 1)² = 2 + 2√2 + 1 = 3 + 2√2Subtracting: 4 + 2√2 - 3 - 2√2 = 1Yes, that's correct.So, both areas are π. Therefore, H = C.But wait, that seems to contradict the intuition that more sides would make the area between the circles smaller. Maybe because the side length is fixed, the difference in radii is such that the areas end up being the same.Alternatively, perhaps the difference in the squares of the radii is the same for both polygons when the side length is 2.Wait, let me think about it differently. Maybe for regular polygons with the same side length, as the number of sides increases, the difference between the circumradius and apothem decreases, but in this case, the difference in their squares is the same.But according to the exact calculations, for both the hexagon and octagon with side length 2, the difference R² - a² is 1, leading to the same area between the circles.So, the conclusion is that H = C.But let me check if I made any mistake in the formulas.For a regular polygon with side length s, the circumradius R = s / (2 * sin(π/n))The apothem a = s / (2 * tan(π/n))So, R² - a² = (s²)/(4 * sin²(π/n)) - (s²)/(4 * tan²(π/n)) = s²/4 [1/sin²(π/n) - 1/tan²(π/n)]But 1/tan²(π/n) = cot²(π/n) = (cos²(π/n))/(sin²(π/n))So, R² - a² = s²/4 [1/sin²(π/n) - cos²(π/n)/sin²(π/n)] = s²/4 [ (1 - cos²(π/n)) / sin²(π/n) ) ] = s²/4 [ sin²(π/n) / sin²(π/n) ) ] = s²/4Wait, that can't be right because for the hexagon, s=2, so s²/4 = 4/4 = 1, which matches R² - a² =1.For the octagon, s=2, so s²/4 =1 as well, which also matches R² - a²=1.Wait, so actually, for any regular polygon with side length s, R² - a² = s²/4Is that a general formula?Let me verify:R = s / (2 * sin(π/n))a = s / (2 * tan(π/n))So, R² = s² / (4 * sin²(π/n))a² = s² / (4 * tan²(π/n)) = s² / (4 * (sin²(π/n)/cos²(π/n))) = s² * cos²(π/n) / (4 * sin²(π/n))So, R² - a² = s²/(4 sin²(π/n)) - s² cos²(π/n)/(4 sin²(π/n)) = s²/(4 sin²(π/n)) [1 - cos²(π/n)] = s²/(4 sin²(π/n)) * sin²(π/n) = s²/4Yes, so R² - a² = s²/4 for any regular polygon with side length s.Therefore, regardless of the number of sides, as long as the side length is the same, the area between the circumscribed and inscribed circles is π*(s²/4).In this case, s=2, so s²/4=1, so the area is π*1=π.Therefore, for both the hexagon and the octagon, H and C are equal to π, so H=C.That's why both areas are the same.

💡Alright, so I have this geometry problem here. Let me read it again to make sure I understand it properly."In triangle ABC, angle A is 60 degrees, BC is 3 units long, and AB is less than AC. The length of side AB is... and then there are four options: A. 2, B. 3, C. 4, D. 6."Okay, so I need to find the length of side AB given these conditions. Let me visualize triangle ABC. Angle A is 60 degrees, which is a nice angle because it's one of those special angles that often come up in geometry problems. BC is the side opposite angle A, right? So BC is 3 units. Also, it's given that AB is less than AC. So, AB < AC.Hmm, so in triangle ABC, side opposite angle A is BC, which is 3. Since AB is less than AC, that means that side AC is longer than AB. In a triangle, the longer side is opposite the larger angle. So, if AB is less than AC, then angle C must be larger than angle B. That makes sense because in a triangle, the larger angles are opposite the longer sides.Now, I need to find the length of AB. Let me think about the tools I have for solving triangles. Since I know one angle and its opposite side, maybe the Law of Sines could help here. The Law of Sines says that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant. So, that is:[frac{AB}{sin C} = frac{AC}{sin B} = frac{BC}{sin A}]Given that angle A is 60 degrees and BC is 3, I can write:[frac{BC}{sin A} = frac{3}{sin 60^circ}]Calculating (sin 60^circ), I remember that (sin 60^circ = frac{sqrt{3}}{2}). So,[frac{3}{sqrt{3}/2} = frac{3 times 2}{sqrt{3}} = frac{6}{sqrt{3}} = 2sqrt{3}]So, the ratio is (2sqrt{3}). That means:[frac{AB}{sin C} = 2sqrt{3} quad text{and} quad frac{AC}{sin B} = 2sqrt{3}]But I don't know angles B and C yet. However, I do know that the sum of angles in a triangle is 180 degrees. So, angle B + angle C + angle A = 180 degrees. Since angle A is 60 degrees, angle B + angle C = 120 degrees.Also, since AB < AC, as given, angle C > angle B, as I thought earlier. So, angle C is larger than angle B.Let me denote angle B as x and angle C as y. So, x + y = 120 degrees, and y > x.From the Law of Sines, I have:[AB = 2sqrt{3} sin y][AC = 2sqrt{3} sin x]Given that AB < AC, so:[2sqrt{3} sin y < 2sqrt{3} sin x]Dividing both sides by (2sqrt{3}), which is positive, so the inequality remains the same:[sin y < sin x]But wait, since y > x, and both angles are between 0 and 180 degrees, how can (sin y < sin x)? Hmm, that seems contradictory because usually, if y > x, then (sin y) is greater than (sin x) if both angles are acute. But in this case, since angle A is 60 degrees, angles B and C must add up to 120 degrees. So, both angles B and C could be acute or one could be obtuse.Wait, if angle C is greater than angle B, and their sum is 120 degrees, then angle C could be greater than 60 degrees, but less than 120 degrees, because if angle C were 120 degrees, angle B would be 0, which isn't possible. So, angle C is between 60 and 120 degrees.Similarly, angle B is between 0 and 60 degrees.So, if angle C is between 60 and 120 degrees, and angle B is between 0 and 60 degrees, then (sin y) (where y is angle C) could be greater or less than (sin x) (where x is angle B). Because sine increases from 0 to 90 degrees and decreases from 90 to 180 degrees.So, if angle C is less than 90 degrees, then (sin y) would be greater than (sin x) because y > x. But if angle C is greater than 90 degrees, then (sin y) could be less than (sin x) if angle B is such that (sin x) is larger.Wait, that seems complicated. Maybe I should approach this differently.Alternatively, I can use the Law of Cosines. The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. The formula is:[c^2 = a^2 + b^2 - 2ab cos C]In this case, if I consider angle A, which is 60 degrees, then:[BC^2 = AB^2 + AC^2 - 2 cdot AB cdot AC cdot cos 60^circ]We know BC is 3, so:[3^2 = AB^2 + AC^2 - 2 cdot AB cdot AC cdot cos 60^circ][9 = AB^2 + AC^2 - 2 cdot AB cdot AC cdot frac{1}{2}][9 = AB^2 + AC^2 - AB cdot AC]So, that's one equation. But I have two variables here: AB and AC. I need another equation to solve for both.But I also know that AB < AC. So, maybe I can express AC in terms of AB or vice versa.Let me denote AB as c and AC as b, following the standard notation where a is BC, opposite angle A, which is 60 degrees. So, a = 3, angle A = 60 degrees, side opposite is a, so side BC is a.Then, sides AB and AC are sides c and b respectively.So, from the Law of Cosines, we have:[a^2 = b^2 + c^2 - 2bc cos A][3^2 = b^2 + c^2 - 2bc cos 60^circ][9 = b^2 + c^2 - 2bc cdot frac{1}{2}][9 = b^2 + c^2 - bc]So, that's the same equation as before.Now, since AB < AC, that means c < b.I need another equation or a way to relate b and c. Maybe I can use the Law of Sines again.From the Law of Sines:[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}]We already calculated that (frac{a}{sin A} = 2sqrt{3}), so:[frac{b}{sin B} = 2sqrt{3} quad text{and} quad frac{c}{sin C} = 2sqrt{3}]So,[b = 2sqrt{3} sin B][c = 2sqrt{3} sin C]Since angle B + angle C = 120 degrees, as we established earlier, we can express angle C as 120 - angle B.So, angle C = 120° - B.Therefore,[c = 2sqrt{3} sin (120° - B)]Using the sine of a difference identity:[sin (120° - B) = sin 120° cos B - cos 120° sin B]We know that (sin 120° = frac{sqrt{3}}{2}) and (cos 120° = -frac{1}{2}). So,[sin (120° - B) = frac{sqrt{3}}{2} cos B - (-frac{1}{2}) sin B = frac{sqrt{3}}{2} cos B + frac{1}{2} sin B]Therefore,[c = 2sqrt{3} left( frac{sqrt{3}}{2} cos B + frac{1}{2} sin B right ) = 2sqrt{3} cdot frac{sqrt{3}}{2} cos B + 2sqrt{3} cdot frac{1}{2} sin B][c = 3 cos B + sqrt{3} sin B]So, now we have expressions for both b and c in terms of angle B.Recall that:[b = 2sqrt{3} sin B][c = 3 cos B + sqrt{3} sin B]Now, let's substitute these into the equation from the Law of Cosines:[9 = b^2 + c^2 - bc]Substituting b and c:[9 = (2sqrt{3} sin B)^2 + (3 cos B + sqrt{3} sin B)^2 - (2sqrt{3} sin B)(3 cos B + sqrt{3} sin B)]Let me compute each term step by step.First, compute (b^2):[(2sqrt{3} sin B)^2 = 4 cdot 3 sin^2 B = 12 sin^2 B]Next, compute (c^2):[(3 cos B + sqrt{3} sin B)^2 = 9 cos^2 B + 2 cdot 3 cdot sqrt{3} cos B sin B + 3 sin^2 B = 9 cos^2 B + 6sqrt{3} cos B sin B + 3 sin^2 B]Now, compute the product (bc):[(2sqrt{3} sin B)(3 cos B + sqrt{3} sin B) = 2sqrt{3} cdot 3 sin B cos B + 2sqrt{3} cdot sqrt{3} sin^2 B = 6sqrt{3} sin B cos B + 6 sin^2 B]Now, plug all these back into the equation:[9 = 12 sin^2 B + (9 cos^2 B + 6sqrt{3} cos B sin B + 3 sin^2 B) - (6sqrt{3} sin B cos B + 6 sin^2 B)]Let me simplify term by term.First, expand the equation:[9 = 12 sin^2 B + 9 cos^2 B + 6sqrt{3} cos B sin B + 3 sin^2 B - 6sqrt{3} sin B cos B - 6 sin^2 B]Now, combine like terms.Looking at the (sin^2 B) terms:12 sin²B + 3 sin²B - 6 sin²B = (12 + 3 - 6) sin²B = 9 sin²BLooking at the (cos^2 B) terms:9 cos²BLooking at the (sin B cos B) terms:6√3 sinB cosB - 6√3 sinB cosB = 0So, the cross terms cancel out.So, now the equation simplifies to:[9 = 9 sin^2 B + 9 cos^2 B]Factor out the 9:[9 = 9 (sin^2 B + cos^2 B)]But we know from the Pythagorean identity that (sin^2 B + cos^2 B = 1), so:[9 = 9 times 1][9 = 9]Hmm, that's interesting. So, this equation simplifies to an identity, which doesn't give me any new information. That means that my approach might not be the best way to solve this problem because it's leading me to a tautology.Maybe I need to try a different method. Let me think.Since I know angle A is 60 degrees and BC is 3, perhaps I can consider the triangle in terms of coordinates. Let me place point A at the origin, point B on the x-axis, and point C somewhere in the plane.So, let me assign coordinates:- Let point A be at (0, 0).- Let point B be at (c, 0), where c is the length of AB.- Let point C be at coordinates (d, e).Given that angle at A is 60 degrees, the angle between sides AB and AC is 60 degrees. So, the coordinates of C can be determined using trigonometry.Since AB is along the x-axis from (0,0) to (c, 0), then AC makes a 60-degree angle with AB. So, the coordinates of C can be expressed as (b cos 60°, b sin 60°), where b is the length of AC.So, point C is at (b * 0.5, b * (√3)/2).Now, the distance between points B and C is given as 3. So, the distance between (c, 0) and (b/2, (b√3)/2) is 3.Let me write the distance formula:[sqrt{(c - frac{b}{2})^2 + left(0 - frac{bsqrt{3}}{2}right)^2} = 3]Squaring both sides:[(c - frac{b}{2})^2 + left( frac{bsqrt{3}}{2} right)^2 = 9]Let me expand this:First, expand ((c - frac{b}{2})^2):[c^2 - c b + frac{b^2}{4}]Then, expand (left( frac{bsqrt{3}}{2} right)^2):[frac{3 b^2}{4}]So, adding them together:[c^2 - c b + frac{b^2}{4} + frac{3 b^2}{4} = 9]Simplify the terms:The (frac{b^2}{4} + frac{3 b^2}{4}) adds up to (b^2).So, the equation becomes:[c^2 - c b + b^2 = 9]Which is the same equation as before from the Law of Cosines. So, again, I end up with the same equation:[c^2 - c b + b^2 = 9]But I still have two variables, b and c, and only one equation. I need another equation or a way to relate b and c.Wait, but I also know that AB < AC, which is c < b. So, perhaps I can express b in terms of c or vice versa.Alternatively, maybe I can use the Law of Sines again to relate b and c.From the Law of Sines:[frac{b}{sin B} = frac{c}{sin C} = 2sqrt{3}]So,[b = 2sqrt{3} sin B][c = 2sqrt{3} sin C]But since angle C = 120° - B, as before, we can write:[c = 2sqrt{3} sin (120° - B)]Which we expanded earlier to:[c = 3 cos B + sqrt{3} sin B]So, now, if I can express b in terms of c or vice versa, maybe I can substitute into the equation (c^2 - c b + b^2 = 9).Let me see. From the Law of Sines, we have:[frac{b}{sin B} = frac{c}{sin (120° - B)}]So,[frac{b}{c} = frac{sin B}{sin (120° - B)}]Let me denote this ratio as k:[k = frac{sin B}{sin (120° - B)}]So,[b = k c]Therefore, I can express b in terms of c. Let's substitute this into the equation:[c^2 - c (k c) + (k c)^2 = 9][c^2 - k c^2 + k^2 c^2 = 9][c^2 (1 - k + k^2) = 9]So,[c^2 = frac{9}{1 - k + k^2}]But I need to find k in terms of known quantities or find another relation.Wait, k is (frac{sin B}{sin (120° - B)}). Let me compute this ratio.Let me denote angle B as x for simplicity. So, angle C = 120° - x.So,[k = frac{sin x}{sin (120° - x)}]Let me compute (sin (120° - x)):Using the sine of a difference identity:[sin (120° - x) = sin 120° cos x - cos 120° sin x][= frac{sqrt{3}}{2} cos x - (-frac{1}{2}) sin x][= frac{sqrt{3}}{2} cos x + frac{1}{2} sin x]Therefore,[k = frac{sin x}{frac{sqrt{3}}{2} cos x + frac{1}{2} sin x}]Let me factor out 1/2 from the denominator:[k = frac{sin x}{frac{1}{2} (sqrt{3} cos x + sin x)} = frac{2 sin x}{sqrt{3} cos x + sin x}]So,[k = frac{2 sin x}{sqrt{3} cos x + sin x}]Let me divide numerator and denominator by cos x:[k = frac{2 tan x}{sqrt{3} + tan x}]Let me denote t = tan x. So,[k = frac{2 t}{sqrt{3} + t}]So, now, k is expressed in terms of t, which is tan x.Now, going back to the equation:[c^2 = frac{9}{1 - k + k^2}]But I need to express this in terms of t.First, let's compute 1 - k + k^2:[1 - k + k^2 = 1 - frac{2 t}{sqrt{3} + t} + left( frac{2 t}{sqrt{3} + t} right)^2]Let me compute each term:First, compute k:[k = frac{2 t}{sqrt{3} + t}]Compute k^2:[k^2 = left( frac{2 t}{sqrt{3} + t} right)^2 = frac{4 t^2}{(sqrt{3} + t)^2}]Now, compute 1 - k + k^2:[1 - frac{2 t}{sqrt{3} + t} + frac{4 t^2}{(sqrt{3} + t)^2}]To combine these terms, let me find a common denominator, which is ((sqrt{3} + t)^2).So,[1 = frac{(sqrt{3} + t)^2}{(sqrt{3} + t)^2}][- frac{2 t}{sqrt{3} + t} = - frac{2 t (sqrt{3} + t)}{(sqrt{3} + t)^2}][frac{4 t^2}{(sqrt{3} + t)^2} = frac{4 t^2}{(sqrt{3} + t)^2}]So, combining all terms:[frac{(sqrt{3} + t)^2 - 2 t (sqrt{3} + t) + 4 t^2}{(sqrt{3} + t)^2}]Let me expand the numerator:First, expand ((sqrt{3} + t)^2):[(sqrt{3})^2 + 2 sqrt{3} t + t^2 = 3 + 2 sqrt{3} t + t^2]Next, expand (-2 t (sqrt{3} + t)):[-2 t sqrt{3} - 2 t^2]So, adding all terms together:[(3 + 2 sqrt{3} t + t^2) + (-2 t sqrt{3} - 2 t^2) + 4 t^2]Simplify term by term:- Constant term: 3- Terms with (sqrt{3} t): 2√3 t - 2√3 t = 0- Terms with t²: t² - 2 t² + 4 t² = (1 - 2 + 4) t² = 3 t²So, the numerator simplifies to:[3 + 3 t^2 = 3(1 + t^2)]Therefore,[1 - k + k^2 = frac{3(1 + t^2)}{(sqrt{3} + t)^2}]So, going back to the equation for c²:[c^2 = frac{9}{1 - k + k^2} = frac{9}{frac{3(1 + t^2)}{(sqrt{3} + t)^2}} = frac{9 (sqrt{3} + t)^2}{3(1 + t^2)} = 3 frac{(sqrt{3} + t)^2}{1 + t^2}]So,[c^2 = 3 frac{(sqrt{3} + t)^2}{1 + t^2}]But I also know that t = tan x, where x is angle B.Since x is an angle in a triangle, it's between 0 and 180 degrees, but in this case, since angle A is 60 degrees, x is between 0 and 120 degrees, but more specifically, since AB < AC, angle C > angle B, so x < 60 degrees.So, x is between 0 and 60 degrees, so t = tan x is between 0 and tan 60° = √3.So, t ∈ (0, √3).Now, let me compute ((sqrt{3} + t)^2):[(sqrt{3} + t)^2 = 3 + 2 sqrt{3} t + t^2]So,[c^2 = 3 frac{3 + 2 sqrt{3} t + t^2}{1 + t^2}]Let me split the fraction:[c^2 = 3 left( frac{3 + t^2}{1 + t^2} + frac{2 sqrt{3} t}{1 + t^2} right )]Simplify the first term:[frac{3 + t^2}{1 + t^2} = frac{(1 + t^2) + 2}{1 + t^2} = 1 + frac{2}{1 + t^2}]So,[c^2 = 3 left( 1 + frac{2}{1 + t^2} + frac{2 sqrt{3} t}{1 + t^2} right )][= 3 + frac{6}{1 + t^2} + frac{6 sqrt{3} t}{1 + t^2}]Hmm, this seems complicated. Maybe I can find another approach.Wait, perhaps instead of going through all this, I can consider specific values from the answer choices and see which one fits.The options are A. 2, B. 3, C. 4, D. 6.Given that AB < AC, and BC = 3, let's test each option.Let me start with option A: AB = 2.If AB = 2, then AC must be greater than 2.Using the Law of Cosines:[BC^2 = AB^2 + AC^2 - 2 cdot AB cdot AC cdot cos 60°][9 = 4 + AC^2 - 2 cdot 2 cdot AC cdot 0.5][9 = 4 + AC^2 - 2 AC][AC^2 - 2 AC + 4 - 9 = 0][AC^2 - 2 AC - 5 = 0]Solving this quadratic equation for AC:[AC = frac{2 pm sqrt{4 + 20}}{2} = frac{2 pm sqrt{24}}{2} = frac{2 pm 2 sqrt{6}}{2} = 1 pm sqrt{6}]Since length can't be negative, AC = 1 + √6 ≈ 1 + 2.45 ≈ 3.45.So, AC ≈ 3.45, which is greater than AB = 2. So, this satisfies AB < AC.Now, let's check if this is possible.Alternatively, let's check option B: AB = 3.Then, AC must be greater than 3.Using the Law of Cosines:[9 = 9 + AC^2 - 2 cdot 3 cdot AC cdot 0.5][9 = 9 + AC^2 - 3 AC][AC^2 - 3 AC = 0][AC (AC - 3) = 0]So, AC = 0 or AC = 3. Since AC can't be 0, AC = 3. But AB = 3 and AC = 3, so AB is not less than AC. It's equal. So, this violates the condition AB < AC. Therefore, option B is not correct.Option C: AB = 4.Then, AC must be greater than 4.Using the Law of Cosines:[9 = 16 + AC^2 - 2 cdot 4 cdot AC cdot 0.5][9 = 16 + AC^2 - 4 AC][AC^2 - 4 AC + 16 - 9 = 0][AC^2 - 4 AC + 7 = 0]Discriminant: 16 - 28 = -12 < 0. No real solution. So, AB cannot be 4.Option D: AB = 6.Then, AC must be greater than 6.Using the Law of Cosines:[9 = 36 + AC^2 - 2 cdot 6 cdot AC cdot 0.5][9 = 36 + AC^2 - 6 AC][AC^2 - 6 AC + 36 - 9 = 0][AC^2 - 6 AC + 27 = 0]Discriminant: 36 - 108 = -72 < 0. No real solution. So, AB cannot be 6.Therefore, the only valid option is A: AB = 2.Wait, but let me double-check my calculations for option A.When AB = 2:[9 = 4 + AC^2 - 2 cdot 2 cdot AC cdot 0.5][9 = 4 + AC^2 - 2 AC][AC^2 - 2 AC - 5 = 0]Solutions:[AC = frac{2 pm sqrt{4 + 20}}{2} = frac{2 pm sqrt{24}}{2} = 1 pm sqrt{6}]Since AC must be positive, AC = 1 + √6 ≈ 3.45, which is greater than AB = 2. So, this works.Additionally, let's check if the triangle inequality holds.AB + BC > AC?2 + 3 > 3.45? 5 > 3.45, yes.AB + AC > BC?2 + 3.45 > 3? 5.45 > 3, yes.AC + BC > AB?3.45 + 3 > 2? 6.45 > 2, yes.So, all triangle inequalities are satisfied.Therefore, AB = 2 is the correct answer.

💡Alright, so I have this problem here: I need to find the constant term in the expansion of ((frac{1}{x} + x^2)^3), which is denoted as (a). Then, using that value of (a), I have to find the area enclosed by the line (y = ax) and the curve (y = x^2). The options are given as A: (frac{27}{2}), B: (9), C: (frac{9}{2}), and D: (frac{27}{4}).Okay, let's start by understanding what it means to find the constant term in the expansion of ((frac{1}{x} + x^2)^3). I remember that when expanding expressions like ((A + B)^n), we can use the binomial theorem, which states that:[(A + B)^n = sum_{k=0}^{n} binom{n}{k} A^{n - k} B^{k}]In this case, (A = frac{1}{x}) and (B = x^2), and (n = 3). So, the expansion would be:[left(frac{1}{x} + x^2right)^3 = sum_{k=0}^{3} binom{3}{k} left(frac{1}{x}right)^{3 - k} left(x^2right)^{k}]Simplifying each term, let's compute the general term:[binom{3}{k} cdot left(frac{1}{x}right)^{3 - k} cdot left(x^2right)^{k} = binom{3}{k} cdot x^{-(3 - k)} cdot x^{2k} = binom{3}{k} cdot x^{-(3 - k) + 2k} = binom{3}{k} cdot x^{3k - 3}]So, each term in the expansion is (binom{3}{k} cdot x^{3k - 3}). Now, we need to find the constant term, which means the exponent of (x) should be zero. Therefore, we set the exponent equal to zero:[3k - 3 = 0 implies 3k = 3 implies k = 1]So, the constant term corresponds to (k = 1). Let's compute that term:[binom{3}{1} cdot x^{3(1) - 3} = 3 cdot x^{0} = 3]Therefore, the constant term (a) is 3.Now, moving on to the second part of the problem: finding the area enclosed by the line (y = ax) (which is now (y = 3x)) and the curve (y = x^2).First, I need to find the points where these two graphs intersect because the area between them will be bounded by these intersection points.Setting (3x = x^2), we can solve for (x):[x^2 - 3x = 0 implies x(x - 3) = 0]So, the solutions are (x = 0) and (x = 3). These are the x-coordinates of the points of intersection. Therefore, the area we need to find is between (x = 0) and (x = 3).To find the area between two curves, I can set up an integral of the top function minus the bottom function with respect to (x) over the interval of integration. In this case, between (x = 0) and (x = 3), I need to determine which function is on top.Looking at the two functions (y = 3x) and (y = x^2), let's analyze their behavior:- At (x = 0), both functions are zero.- At (x = 3), both functions are equal to 9.- Between (x = 0) and (x = 3), the line (y = 3x) is above the parabola (y = x^2). For example, at (x = 1), (y = 3(1) = 3) and (y = 1^2 = 1). So, (3x) is indeed above (x^2) in this interval.Therefore, the area (A) is given by the integral from (0) to (3) of the top function minus the bottom function:[A = int_{0}^{3} (3x - x^2) , dx]Now, let's compute this integral step by step.First, find the antiderivative of (3x - x^2):[int (3x - x^2) , dx = int 3x , dx - int x^2 , dx = frac{3}{2}x^2 - frac{1}{3}x^3 + C]Where (C) is the constant of integration, but since we're calculating a definite integral, we'll evaluate it at the bounds and subtract.Now, evaluate the antiderivative at the upper limit (x = 3):[frac{3}{2}(3)^2 - frac{1}{3}(3)^3 = frac{3}{2} times 9 - frac{1}{3} times 27 = frac{27}{2} - 9]Simplify this:[frac{27}{2} - 9 = frac{27}{2} - frac{18}{2} = frac{9}{2}]Next, evaluate the antiderivative at the lower limit (x = 0):[frac{3}{2}(0)^2 - frac{1}{3}(0)^3 = 0 - 0 = 0]Subtracting the lower limit evaluation from the upper limit evaluation gives:[frac{9}{2} - 0 = frac{9}{2}]Therefore, the area enclosed by the line (y = 3x) and the curve (y = x^2) is (frac{9}{2}).Looking back at the options provided:A: (frac{27}{2})B: (9)C: (frac{9}{2})D: (frac{27}{4})The correct answer is option C: (frac{9}{2}).I think I've covered all the steps thoroughly. I started by expanding the given expression to find the constant term, identified the value of (a), found the points of intersection between the line and the curve, set up the integral to calculate the area between them, and then evaluated the integral to find the area. Each step seems logical and follows from the previous one without any gaps or errors in reasoning. So, I feel confident that the answer is indeed (frac{9}{2}).**Final Answer**boxed{C}