💡First, I recognize that 210 degrees is in the third quadrant of the unit circle, where cosine values are negative.Next, I identify the reference angle for 210 degrees by subtracting 180 degrees, which gives me 30 degrees.I know that the cosine of 30 degrees is √3/2.Since cosine is negative in the third quadrant, I apply the negative sign to the reference angle's cosine value.Therefore, cos 210 degrees is -√3/2.

💡Okay, so I have this problem about a triangular array of coins. It says there are 3780 coins arranged in such a way that the first row has 1 coin, the second row has 2 coins, and so on up to N coins in the Nth row. I need to find the sum of the digits of N. The answer choices are from A to E, with numbers 13 to 17. Hmm, triangular numbers. I remember that the formula for the sum of the first N natural numbers is given by S = N(N + 1)/2. So, in this case, the total number of coins is 3780, which means:N(N + 1)/2 = 3780I need to solve for N. Let me rearrange this equation to make it easier to handle. Multiplying both sides by 2 gives:N(N + 1) = 7560So, now I have a quadratic equation: N² + N - 7560 = 0To solve this quadratic equation, I can use the quadratic formula, which is N = [-b ± sqrt(b² - 4ac)]/(2a). Here, a = 1, b = 1, and c = -7560. Plugging these values in:N = [-1 ± sqrt(1 + 4*7560)]/2First, let's compute the discriminant:sqrt(1 + 4*7560) = sqrt(1 + 30240) = sqrt(30241)Hmm, what's the square root of 30241? Let me think. I know that 170² is 28900 and 180² is 32400, so it's somewhere between 170 and 180. Let me try 174: 174² = 30276, which is a bit higher than 30241. Maybe 173: 173² = 29929, which is lower. So, it's between 173 and 174. Wait, maybe I can factor 30241. Let me try dividing by small primes. 30241 divided by 7: 7*4320 is 30240, so 30241 is 7*4320 +1, which isn't divisible by 7. How about 11? 11*2749 is 30239, so 30241 is 2 more, not divisible by 11. Maybe 13? 13*2326 is 30238, so 30241 is 3 more, not divisible by 13. This is getting complicated. Maybe I should approximate the square root. Since 173² = 29929 and 174² = 30276, the square root of 30241 is approximately 173.9. So, sqrt(30241) ≈ 173.9.Now, plugging back into the quadratic formula:N = [-1 ± 173.9]/2We can ignore the negative solution because N can't be negative, so:N ≈ (-1 + 173.9)/2 ≈ 172.9/2 ≈ 86.45So, N is approximately 86.45. Since N has to be an integer, let's check N = 86 and N = 87.For N = 86:Total coins = 86*87/2 = (86*87)/2Calculating 86*87: 80*80=6400, 80*7=560, 6*80=480, 6*7=42. So, 6400 + 560 + 480 + 42 = 7482. Then, 7482/2 = 3741.Hmm, that's less than 3780. Let's try N = 87:Total coins = 87*88/2 = (87*88)/2Calculating 87*88: 80*80=6400, 80*8=640, 7*80=560, 7*8=56. So, 6400 + 640 + 560 + 56 = 7656. Then, 7656/2 = 3828.Wait, that's more than 3780. So, N is between 86 and 87, but since N has to be an integer, and 86 gives 3741 which is less than 3780, and 87 gives 3828 which is more than 3780, it seems like there's no integer N that gives exactly 3780 coins. But the problem says there are 3780 coins, so maybe I made a mistake in my calculations. Let me double-check.Wait, when I calculated 86*87, I think I did that wrong. Let me recalculate 86*87:86*80 = 688086*7 = 602So, 6880 + 602 = 7482. Then, 7482/2 = 3741. That's correct.Similarly, 87*88:87*80 = 696087*8 = 696So, 6960 + 696 = 7656. Then, 7656/2 = 3828. That's also correct.Hmm, so neither 86 nor 87 gives exactly 3780. Maybe I need to check if 3780 is a triangular number. Let me see.Alternatively, maybe I can solve N(N + 1)/2 = 3780 more accurately.Multiplying both sides by 2:N² + N = 7560So, N² + N - 7560 = 0Using the quadratic formula:N = [-1 ± sqrt(1 + 4*7560)]/2Which is:N = [-1 ± sqrt(30241)]/2Wait, earlier I approximated sqrt(30241) as 173.9, but let me see if 173.9 squared is 30241.173.9² = (173 + 0.9)² = 173² + 2*173*0.9 + 0.9² = 29929 + 311.4 + 0.81 = 29929 + 311.4 = 30240.4 + 0.81 = 30241.21Oh, so sqrt(30241) ≈ 173.9, but actually, 173.9² is approximately 30241.21, which is very close to 30241. So, sqrt(30241) ≈ 173.9.Therefore, N ≈ (-1 + 173.9)/2 ≈ 172.9/2 ≈ 86.45So, N is approximately 86.45, which is between 86 and 87. Since N must be an integer, and the total coins at N=86 is 3741, which is less than 3780, and at N=87 it's 3828, which is more than 3780, it seems like there's no integer N that gives exactly 3780 coins. But the problem states that there are 3780 coins, so maybe I made a mistake in my approach. Let me think again.Wait, maybe I should check if 3780 is a triangular number. Let me see.A triangular number is of the form T_n = n(n + 1)/2. So, if 3780 is a triangular number, then n(n + 1) = 7560.So, I need to find n such that n² + n - 7560 = 0.Let me try to factor this equation. Maybe 7560 can be factored into two consecutive integers.Let me see, 7560 divided by 80 is 94.5, so maybe around 86 or 87.Wait, earlier I saw that 86*87 = 7482 and 87*88 = 7656. So, 7560 is between these two. Therefore, there is no integer n such that n(n + 1) = 7560. Therefore, 3780 is not a triangular number. But the problem says there are 3780 coins arranged in a triangular array. That seems contradictory.Wait, maybe I made a mistake in the initial formula. Let me double-check.The formula for the sum of the first N natural numbers is indeed N(N + 1)/2. So, if the total number of coins is 3780, then N(N + 1)/2 = 3780. So, N(N + 1) = 7560.But as we saw, 86*87 = 7482 and 87*88 = 7656. So, 7560 is between these two products. Therefore, there is no integer N such that N(N + 1) = 7560. Therefore, 3780 is not a triangular number.But the problem says there are 3780 coins arranged in a triangular array. That implies that N must be an integer. So, perhaps the problem is misstated, or I'm misunderstanding something.Wait, maybe the triangular array is not necessarily starting from 1 coin in the first row. But the problem says "the same configuration as before: 1 coin in the first row, 2 in the second, following the natural number order up to N coins in the Nth row." So, it is starting from 1.Hmm, this is confusing. Maybe I need to check my calculations again.Wait, let me calculate 86*87/2 again. 86*87 = 7482, divided by 2 is 3741. 87*88/2 is 7656/2 = 3828. So, 3780 is between these two. Therefore, there is no integer N such that the sum is 3780. So, perhaps the problem is incorrect, or maybe I'm missing something.Alternatively, maybe the triangular array is arranged differently. Wait, perhaps it's a different kind of triangular array, not the standard one. But the problem says "the same configuration as before: 1 coin in the first row, 2 in the second, following the natural number order up to N coins in the Nth row." So, it's the standard triangular number.Wait, maybe I made a mistake in the quadratic formula. Let me try solving N(N + 1)/2 = 3780 again.Multiplying both sides by 2: N² + N = 7560So, N² + N - 7560 = 0Using the quadratic formula:N = [-1 ± sqrt(1 + 4*7560)]/2Which is:N = [-1 ± sqrt(30241)]/2Now, sqrt(30241). Let me see, 173² = 29929, 174² = 30276. So, sqrt(30241) is between 173 and 174.Let me calculate 173.5²:173.5² = (173 + 0.5)² = 173² + 2*173*0.5 + 0.5² = 29929 + 173 + 0.25 = 30102.25Hmm, that's still less than 30241. Let's try 173.8²:173.8² = (173 + 0.8)² = 173² + 2*173*0.8 + 0.8² = 29929 + 276.8 + 0.64 = 29929 + 276.8 = 30205.8 + 0.64 = 30206.44Still less than 30241. Let's try 173.9²:173.9² = (173 + 0.9)² = 173² + 2*173*0.9 + 0.9² = 29929 + 311.4 + 0.81 = 29929 + 311.4 = 30240.4 + 0.81 = 30241.21Ah, so sqrt(30241) ≈ 173.9, as I thought earlier. So, N ≈ (-1 + 173.9)/2 ≈ 172.9/2 ≈ 86.45So, N is approximately 86.45, which is not an integer. Therefore, there is no integer N such that the sum is exactly 3780. But the problem states that there are 3780 coins arranged in such a triangular array. This seems contradictory because, as per the formula, it's not a triangular number. Wait, maybe the problem is referring to a different kind of triangular array, not the standard one. Or perhaps it's a misprint, and the number of coins is supposed to be a triangular number. Alternatively, maybe I need to consider that the triangular array could have a different starting point, but the problem clearly states it starts with 1 coin in the first row and increases by 1 each row. Wait, perhaps the problem is asking for the nearest integer N such that the sum is just over 3780, and then take the sum of its digits. Let me see.At N=86, the sum is 3741, which is less than 3780. At N=87, the sum is 3828, which is more than 3780. So, if we take N=87, the sum is 3828, which is the closest triangular number greater than 3780. But the problem says there are exactly 3780 coins, so maybe it's a misprint, and the intended answer is N=87, with the sum of digits 8+7=15, which is option C.Alternatively, maybe the problem is correct, and I'm missing something. Let me try to see if 3780 can be expressed as N(N + 1)/2 for some N.Let me rearrange the equation:N² + N - 7560 = 0Using the quadratic formula:N = [-1 ± sqrt(1 + 4*7560)]/2 = [-1 ± sqrt(30241)]/2As we saw, sqrt(30241) ≈ 173.9, so N ≈ (172.9)/2 ≈ 86.45So, N is approximately 86.45, which is not an integer. Therefore, 3780 is not a triangular number. Wait, maybe I can check if 3780 is a triangular number by seeing if 8N + 1 is a perfect square. The formula for triangular numbers is T_n = n(n + 1)/2, so 8T_n + 1 = 4n² + 4n + 1 = (2n + 1)². So, if 8*3780 + 1 is a perfect square, then 3780 is a triangular number.Calculating 8*3780 + 1:8*3780 = 3024030240 + 1 = 30241Is 30241 a perfect square? Let me check. Earlier, I saw that 173² = 29929 and 174² = 30276. So, 30241 is between these two. Let me see if 173.9² is 30241.21, which is very close. But 30241 is not a perfect square because 173² = 29929 and 174² = 30276, so there's no integer whose square is 30241. Therefore, 3780 is not a triangular number.This is confusing because the problem states that there are 3780 coins arranged in a triangular array. Maybe the problem is referring to a different kind of triangular array, or perhaps it's a misprint. Alternatively, maybe the problem is asking for the smallest N such that the sum is at least 3780, which would be N=87, as we saw earlier. Then, the sum of the digits of 87 is 8 + 7 = 15, which is option C.Given that the answer choices are 13, 14, 15, 16, 17, and 15 is one of them, I think the intended answer is 15, corresponding to N=87.So, despite the confusion, I think the answer is 15.

💡Okay, so I have this problem where I need to find the monotonically increasing interval of the function ( y = sinleft(2x - frac{pi}{3}right) ). There are four options given, labeled A to D. I need to figure out which one is correct.First, I remember that the sine function, ( sin(u) ), is increasing on the interval ( left[-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpiright] ) for any integer ( k ). This is because the sine function has a period of ( 2pi ) and its increasing behavior repeats every ( 2pi ).In this problem, the function inside the sine is ( 2x - frac{pi}{3} ). Let me denote this as ( u = 2x - frac{pi}{3} ). So, the function becomes ( y = sin(u) ).Since ( sin(u) ) is increasing when ( u ) is in ( left[-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpiright] ), I can set up the inequality:[-frac{pi}{2} + 2kpi leq 2x - frac{pi}{3} leq frac{pi}{2} + 2kpi]Now, I need to solve this inequality for ( x ). Let me do this step by step.First, add ( frac{pi}{3} ) to all parts of the inequality to isolate the term with ( x ):[-frac{pi}{2} + frac{pi}{3} + 2kpi leq 2x leq frac{pi}{2} + frac{pi}{3} + 2kpi]Next, I need to combine the constants ( -frac{pi}{2} + frac{pi}{3} ) and ( frac{pi}{2} + frac{pi}{3} ). To do this, I'll find a common denominator, which is 6.For the left side:[-frac{pi}{2} + frac{pi}{3} = -frac{3pi}{6} + frac{2pi}{6} = -frac{pi}{6}]For the right side:[frac{pi}{2} + frac{pi}{3} = frac{3pi}{6} + frac{2pi}{6} = frac{5pi}{6}]So now, the inequality becomes:[-frac{pi}{6} + 2kpi leq 2x leq frac{5pi}{6} + 2kpi]Now, divide all parts of the inequality by 2 to solve for ( x ):[-frac{pi}{12} + kpi leq x leq frac{5pi}{12} + kpi]This can be rewritten as:[left[ kpi - frac{pi}{12}, kpi + frac{5pi}{12} right], quad k in mathbb{Z}]Looking back at the options provided:A: ( left[ kpi - frac{pi}{12}, kpi + frac{5pi}{12} right], k in mathbb{Z} )B: ( left[ 2kpi - frac{pi}{12}, 2kpi + frac{5pi}{12} right], k in mathbb{Z} )C: ( left[ kpi - frac{pi}{6}, kpi + frac{5pi}{6} right], k in mathbb{Z} )D: ( left[ 2kpi - frac{pi}{6}, 2kpi + frac{5pi}{6} right], k in mathbb{Z} )The interval I found matches option A exactly. So, option A is the correct answer.Wait a second, let me double-check my steps to make sure I didn't make any mistakes. I started by setting ( u = 2x - frac{pi}{3} ), then used the increasing interval for sine, solved the inequality, and ended up with the interval in option A. It seems correct.Just to be thorough, let me consider the period of the function. The original sine function has a period of ( 2pi ), but since the argument is ( 2x - frac{pi}{3} ), the period is halved to ( pi ). Therefore, the increasing intervals should repeat every ( pi ), which is consistent with option A, where the interval is shifted by ( kpi ).Yes, I'm confident that option A is the correct answer.**Final Answer**boxed{A}

💡Okay, so I have this problem here with vectors, and I need to find the value of k. Let me read it again to make sure I understand.Given vectors a = (1, 2) and b = (-3, 2), if k times vector a plus vector b is perpendicular to vector a minus three times vector b, then what is the value of k? The options are A: -19, B: -1/3, C: 11/9, D: 19.Alright, so I remember that if two vectors are perpendicular, their dot product is zero. So, I need to compute the dot product of (k*a + b) and (a - 3b) and set it equal to zero. Then solve for k.Let me write down the vectors first. Vector a is (1, 2) and vector b is (-3, 2). So, let me compute k*a + b.k*a would be k*(1, 2) which is (k, 2k). Then adding vector b, which is (-3, 2), so adding component-wise: (k + (-3), 2k + 2). So that simplifies to (k - 3, 2k + 2).Next, I need to compute vector a - 3b. Vector a is (1, 2), and 3b is 3*(-3, 2) which is (-9, 6). So subtracting 3b from a: (1 - (-9), 2 - 6). That becomes (1 + 9, 2 - 6) which is (10, -4).So now, I have the two vectors: (k - 3, 2k + 2) and (10, -4). Their dot product should be zero because they're perpendicular.The dot product is calculated by multiplying corresponding components and then adding them together. So, (k - 3)*10 + (2k + 2)*(-4) = 0.Let me compute each part step by step.First, (k - 3)*10: that's 10k - 30.Second, (2k + 2)*(-4): let's distribute the -4. That would be -8k - 8.Now, adding these two results together: (10k - 30) + (-8k - 8) = 0.Combine like terms: 10k - 8k is 2k, and -30 - 8 is -38. So, 2k - 38 = 0.Solving for k: 2k = 38, so k = 19.Wait, that seems straightforward. Let me double-check my calculations to make sure I didn't make a mistake.Starting with k*a + b: (k, 2k) + (-3, 2) = (k - 3, 2k + 2). That looks right.Then a - 3b: (1, 2) - (-9, 6) = (1 + 9, 2 - 6) = (10, -4). That also looks correct.Dot product: (k - 3)*10 + (2k + 2)*(-4). Let's compute each term again.(k - 3)*10: 10k - 30.(2k + 2)*(-4): -8k - 8.Adding them: 10k - 30 -8k -8 = 2k -38. So, 2k = 38, k =19.Hmm, so k is 19, which is option D. Let me just check if I interpreted the vectors correctly.Vector a is (1,2), vector b is (-3,2). Yes, that's correct. So, k*a + b is (k -3, 2k + 2). a - 3b is (10, -4). Dot product is zero, so 10(k -3) + (-4)(2k + 2) = 0.Wait, another way to compute the dot product is 10*(k -3) + (-4)*(2k + 2). Let me compute that.10*(k -3) is 10k -30.-4*(2k +2) is -8k -8.Adding them: 10k -30 -8k -8 = 2k -38 =0. So, 2k=38, k=19.Yes, that's consistent. So, I think I did it right. So, the answer is D:19.**Final Answer**boxed{D}

💡First, I identify the coefficients of the quadratic equation (36u^2 - 24u - 120 = 0). Here, (a = 36), (b = -24), and (c = -120).Next, I recall that the product of the roots of a quadratic equation (au^2 + bu + c = 0) is given by the formula (frac{c}{a}).Applying this formula to the given equation:[frac{c}{a} = frac{-120}{36}]I simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 12:[frac{-120 div 12}{36 div 12} = frac{-10}{3}]Therefore, the product of the roots of the equation is (-frac{10}{3}).

💡Okay, so I have this problem here about a function f(x) = sin(ωx) - cos(ωx), where ω is greater than 1/4, and x is a real number. The question is about finding the range of ω such that the abscissa (which is the x-coordinate) of the intersection point of any axis of symmetry of f(x) with the x-axis does not belong to the interval (2π, 3π). The options are given as A, B, C, D, and I need to figure out which one is correct.First, I need to recall what an axis of symmetry is for a function. For trigonometric functions like sine and cosine, their graphs are periodic and have certain symmetries. For example, the sine function has symmetry about its midline, which is the x-axis if it's a standard sine function, but if it's shifted or scaled, the axis of symmetry might change.Given that f(x) is a combination of sine and cosine, I think it can be rewritten using the auxiliary angle formula. The auxiliary angle formula allows us to express a linear combination of sine and cosine as a single sine (or cosine) function with a phase shift. That might help me analyze the symmetry more easily.So, let me try to rewrite f(x) = sin(ωx) - cos(ωx) using the auxiliary angle formula. The general form is A sin(θ) + B cos(θ) = C sin(θ + φ), where C = √(A² + B²) and φ is the phase shift. In this case, A is 1 (coefficient of sin) and B is -1 (coefficient of cos). So, C should be √(1² + (-1)²) = √2. Then, the phase shift φ can be found using tan(φ) = B/A, which is -1/1 = -1. So φ is -π/4 because tan(-π/4) = -1.Therefore, f(x) can be rewritten as √2 sin(ωx - π/4). That seems correct. Let me double-check: sin(ωx - π/4) = sin(ωx)cos(π/4) - cos(ωx)sin(π/4) = (sin(ωx) - cos(ωx)) / √2. So multiplying by √2 gives sin(ωx) - cos(ωx), which is our original function. Perfect.Now, the function is f(x) = √2 sin(ωx - π/4). The graph of this function is a sine wave with amplitude √2, angular frequency ω, and a phase shift of π/(4ω). The axes of symmetry for a sine function are the vertical lines passing through its maxima and minima. For the standard sine function, these are at x = π/2 + kπ for maxima and x = 3π/2 + kπ for minima, where k is an integer.But in our case, the function is shifted and scaled. So, the axes of symmetry will be at the points where the function reaches its maximum or minimum. Let me find the general form of these axes of symmetry.The maxima of f(x) occur where the derivative f'(x) is zero and the second derivative is negative. Similarly, minima occur where the derivative is zero and the second derivative is positive. Let me compute the derivative:f'(x) = d/dx [√2 sin(ωx - π/4)] = √2 ω cos(ωx - π/4).Setting this equal to zero:√2 ω cos(ωx - π/4) = 0.Since √2 ω ≠ 0 (because ω > 1/4), we have cos(ωx - π/4) = 0.The solutions to cos(θ) = 0 are θ = π/2 + kπ, where k is an integer. So,ωx - π/4 = π/2 + kπSolving for x:ωx = π/2 + π/4 + kπ = 3π/4 + kπx = (3π/4 + kπ) / ωSo, these are the x-coordinates where the function has maxima or minima, which are the axes of symmetry.Therefore, the abscissas (x-coordinates) of the axes of symmetry are x = (3π/4 + kπ)/ω for integers k.The problem states that these abscissas should not lie in the interval (2π, 3π). So, for all integers k, (3π/4 + kπ)/ω should not be in (2π, 3π).We need to find the range of ω such that for all k, (3π/4 + kπ)/ω ∉ (2π, 3π).Let me rephrase this condition. For each integer k, either (3π/4 + kπ)/ω ≤ 2π or (3π/4 + kπ)/ω ≥ 3π.So, for each k, either (3π/4 + kπ) ≤ 2πω or (3π/4 + kπ) ≥ 3πω.Let me solve these inequalities for ω.First inequality: (3π/4 + kπ) ≤ 2πωDivide both sides by π:3/4 + k ≤ 2ωSo, ω ≥ (3/4 + k)/2Second inequality: (3π/4 + kπ) ≥ 3πωDivide both sides by π:3/4 + k ≥ 3ωSo, ω ≤ (3/4 + k)/3Therefore, for each integer k, ω must satisfy either ω ≥ (3/4 + k)/2 or ω ≤ (3/4 + k)/3.But since ω > 1/4, we have to consider the possible values of k such that these inequalities make sense.Let me think about which k's are relevant. Since x must be in (2π, 3π), and x = (3π/4 + kπ)/ω, we can solve for k in terms of ω.But perhaps another approach is better. Let me consider the possible k's such that (3π/4 + kπ)/ω could lie in (2π, 3π). So, for each k, we can find the range of ω where (3π/4 + kπ)/ω is in (2π, 3π), and then exclude those ω's.So, for each k, if (3π/4 + kπ)/ω ∈ (2π, 3π), then ω must satisfy:2π < (3π/4 + kπ)/ω < 3πMultiply all parts by ω (since ω > 0, inequality signs don't change):2πω < 3π/4 + kπ < 3πωDivide all parts by π:2ω < 3/4 + k < 3ωSo, 2ω < 3/4 + k < 3ωWhich implies:2ω < 3/4 + k and 3/4 + k < 3ωSo, from the first inequality: 2ω < 3/4 + k => ω < (3/4 + k)/2From the second inequality: 3/4 + k < 3ω => ω > (3/4 + k)/3Therefore, for each k, if ω is in ((3/4 + k)/3, (3/4 + k)/2), then (3π/4 + kπ)/ω is in (2π, 3π). So, to ensure that (3π/4 + kπ)/ω is not in (2π, 3π), ω must not lie in any of these intervals for any integer k.Therefore, the range of ω is the complement of the union of all intervals ((3/4 + k)/3, (3/4 + k)/2) for all integers k, intersected with ω > 1/4.So, we need to find all ω > 1/4 such that ω is not in any ((3/4 + k)/3, (3/4 + k)/2) for any integer k.Now, let's find the values of k for which ((3/4 + k)/3, (3/4 + k)/2) overlaps with ω > 1/4.Let me compute ((3/4 + k)/3, (3/4 + k)/2) for different k.Start with k = 0:((3/4 + 0)/3, (3/4 + 0)/2) = (1/4, 3/8). So, ω must not be in (1/4, 3/8).But ω > 1/4, so the forbidden interval is (1/4, 3/8).Next, k = 1:((3/4 + 1)/3, (3/4 + 1)/2) = (7/12, 7/8). So, ω must not be in (7/12, 7/8).k = 2:((3/4 + 2)/3, (3/4 + 2)/2) = (11/12, 11/8). So, ω must not be in (11/12, 11/8).k = 3:((3/4 + 3)/3, (3/4 + 3)/2) = (15/12, 15/8) = (5/4, 15/8). But 15/8 is 1.875, which is beyond the options given, so maybe we don't need to consider k=3.Similarly, for negative k:k = -1:((3/4 + (-1))/3, (3/4 + (-1))/2) = (-1/12, -1/8). Since ω > 1/4, this interval doesn't affect our range.k = -2:Similarly, negative values will give negative intervals, which are irrelevant since ω > 1/4.Therefore, the forbidden intervals for ω are (1/4, 3/8), (7/12, 7/8), (11/12, 11/8), etc., but since the options don't go beyond 19/12, we can stop at k=2.Therefore, the allowed ω are:ω > 1/4, excluding (1/4, 3/8), (7/12, 7/8), (11/12, 11/8), etc.So, the allowed intervals are:[3/8, 7/12], [7/8, 11/12], [11/8, 15/12], etc.But looking at the options, the highest ω in the options is 19/12, so let's see:The first allowed interval is [3/8, 7/12], then [7/8, 11/12], then [11/8, 15/12], but 15/12 is 5/4, which is 1.25, and 19/12 is approximately 1.583.But in the options, option C is [3/8, 7/12] ∪ [7/8, 11/12], which seems to match our first two allowed intervals.Option A is [3/8, 11/12] ∪ [11/8, 19/12], which includes more, but our forbidden interval is (11/12, 11/8), so [11/12, 11/8] is actually forbidden, but in option A, it's [11/8, 19/12], which is allowed. Wait, maybe I need to think again.Wait, the forbidden intervals are (1/4, 3/8), (7/12, 7/8), (11/12, 11/8), etc. So, the allowed intervals are [3/8, 7/12], [7/8, 11/12], [11/8, 15/12], etc.But in the options, option C is [3/8, 7/12] ∪ [7/8, 11/12], which is exactly the first two allowed intervals. Option A includes [3/8, 11/12], which would include the forbidden interval (7/12, 7/8), so that's incorrect.Option B is (1/4, 5/12] ∪ [5/8, 3/4], which doesn't align with our allowed intervals.Option D is (1/4, 3/4] ∪ [9/8, 17/12], which also doesn't match.Therefore, the correct answer should be option C: [3/8, 7/12] ∪ [7/8, 11/12].Wait, but let me double-check. The forbidden intervals are (1/4, 3/8), (7/12, 7/8), (11/12, 11/8), etc. So, the allowed intervals are:From 3/8 to 7/12, then from 7/8 to 11/12, then from 11/8 to 15/12, etc.But in the options, option C only includes up to 11/12, which is correct for the first two allowed intervals. The next allowed interval would be [11/8, 15/12], but 15/12 is 5/4, which is 1.25, and the options don't go that high except for option A, which includes up to 19/12.But since the problem states ω > 1/4, and the options are given, I think option C is the correct one because it covers the first two allowed intervals without including any forbidden parts.So, I think the answer is C.