answer：First, I note that the value of ( x ) is given as 32.The equation to solve is:[35 - left[ 23 - left{ text{a number} - x right} right] = 12 times 2 div frac{1}{2}]I start by simplifying the right side of the equation:[12 times 2 div frac{1}{2} = 24 div frac{1}{2} = 24 times 2 = 48]Now, the equation becomes:[35 - left[ 23 - left( N - 32 right) right] = 48]Next, I simplify inside the brackets:[35 - left[ 23 - N + 32 right] = 48][35 - left[ 55 - N right] = 48]Then, I distribute the negative sign:[35 - 55 + N = 48][-20 + N = 48]Finally, I solve for ( N ):[N = 48 + 20][N = 68]

answer：Alright, so I need to figure out how many even integers are in the top 15 rows of Pascal's Triangle. Hmm, okay. Let me start by recalling what Pascal's Triangle is. It's a triangular array where each number is the sum of the two numbers directly above it. The top of the triangle is row 0, which just has the number 1. Then row 1 has two 1s, row 2 has 1, 2, 1, and so on.First, I should probably write out the first 15 rows to get a sense of where the even numbers are. But that might take a while, and I'm not sure if I'll have time to do that accurately. Maybe there's a pattern or a formula I can use instead.I remember something about the binomial coefficients and whether they're odd or even. There's a theorem called Lucas's theorem that might help here. Lucas's theorem says that a binomial coefficient C(n, k) is odd if and only if every bit in the binary representation of k is less than or equal to the corresponding bit in n. If that's the case, then the coefficient is odd; otherwise, it's even.Okay, so to find even numbers, I need to find cases where at least one bit in k's binary representation is greater than the corresponding bit in n's binary representation. That sounds a bit complicated, but maybe I can use it to count the number of even numbers in each row.Let me think about how to apply this. For each row n, I can look at the binary representation of n. Then, for each k from 0 to n, I can check if any bit in k is greater than the corresponding bit in n. If so, then C(n, k) is even.But doing this manually for each row up to 15 seems tedious. Maybe there's a pattern in the number of even numbers per row. Let me see:- Row 0: 1 (all odd)- Row 1: 1, 1 (all odd)- Row 2: 1, 2, 1 (one even)- Row 3: 1, 3, 3, 1 (all odd)- Row 4: 1, 4, 6, 4, 1 (three evens)- Row 5: 1, 5, 10, 10, 5, 1 (two evens)- Row 6: 1, 6, 15, 20, 15, 6, 1 (three evens)- Row 7: 1, 7, 21, 35, 35, 21, 7, 1 (all odd)- Row 8: 1, 8, 28, 56, 70, 56, 28, 8, 1 (seven evens)- Row 9: 1, 9, 36, 84, 126, 126, 84, 36, 9, 1 (four evens)- Row 10: 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1 (six evens)- Row 11: 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1 (four evens)- Row 12: 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1 (eleven evens)- Row 13: 1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1 (six evens)- Row 14: 1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1 (thirteen evens)Wait, that seems inconsistent. For example, row 4 has three evens, but row 5 only has two. Maybe I'm missing something. Let me double-check row 5:Row 5: 1, 5, 10, 10, 5, 1. So, 10 and 10 are even. That's two evens. Okay, that's correct.Row 6: 1, 6, 15, 20, 15, 6, 1. So, 6, 20, and 6 are even. That's three evens. Got it.Row 7: All odd, as expected.Row 8: 1, 8, 28, 56, 70, 56, 28, 8, 1. So, 8, 28, 56, 70, 56, 28, 8 are even. That's seven evens.Row 9: 1, 9, 36, 84, 126, 126, 84, 36, 9, 1. So, 36, 84, 126, 126, 84, 36 are even. That's six evens. Wait, I thought I had four earlier. Let me recount: 36, 84, 126, 126, 84, 36. That's six evens. Okay, I must have miscounted before.Row 10: 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. So, 10, 120, 210, 252, 210, 120, 10 are even. That's seven evens. But earlier I thought it was six. Hmm, maybe I need to be careful.Wait, let's list them:- 10 (even)- 45 (odd)- 120 (even)- 210 (even)- 252 (even)- 210 (even)- 120 (even)- 45 (odd)- 10 (even)So, that's 10, 120, 210, 252, 210, 120, 10. That's seven evens. Okay, so I had miscounted before.Similarly, for row 11:1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1.So, 330, 462, 462, 330 are even. That's four evens.Row 12:1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1.So, 12, 66, 220, 792, 924, 792, 220, 66, 12 are even. That's nine evens. Wait, earlier I thought it was eleven. Hmm, let's recount:- 12 (even)- 66 (even)- 220 (even)- 495 (odd)- 792 (even)- 924 (even)- 792 (even)- 495 (odd)- 220 (even)- 66 (even)- 12 (even)So, that's 12, 66, 220, 792, 924, 792, 220, 66, 12. That's nine evens. Okay, so I had miscounted before.Row 13:1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1.So, 78, 286, 1716, 1716, 286, 78 are even. That's six evens.Row 14:1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1.So, 14, 364, 2002, 3432, 2002, 364, 14 are even. That's seven evens. Wait, earlier I thought it was thirteen. Hmm, let's recount:- 14 (even)- 91 (odd)- 364 (even)- 1001 (odd)- 2002 (even)- 3003 (odd)- 3432 (even)- 3003 (odd)- 2002 (even)- 1001 (odd)- 364 (even)- 91 (odd)- 14 (even)So, that's 14, 364, 2002, 3432, 2002, 364, 14. That's seven evens. Okay, so I had miscounted before.Now, let's list the number of evens per row:- Row 0: 0- Row 1: 0- Row 2: 1- Row 3: 0- Row 4: 3- Row 5: 2- Row 6: 3- Row 7: 0- Row 8: 7- Row 9: 6- Row 10: 7- Row 11: 4- Row 12: 9- Row 13: 6- Row 14: 7Now, let's add these up:0 + 0 + 1 + 0 + 3 + 2 + 3 + 0 + 7 + 6 + 7 + 4 + 9 + 6 + 7Let's compute step by step:Start with 0.Add 0: 0Add 1: 1Add 0: 1Add 3: 4Add 2: 6Add 3: 9Add 0: 9Add 7: 16Add 6: 22Add 7: 29Add 4: 33Add 9: 42Add 6: 48Add 7: 55Wait, that's 55. But earlier, I thought the answer was 49. Hmm, maybe I made a mistake in counting the number of evens per row.Let me double-check the counts:Row 0: 0Row 1: 0Row 2: 1 (correct)Row 3: 0Row 4: 3 (correct)Row 5: 2 (correct)Row 6: 3 (correct)Row 7: 0Row 8: 7 (correct)Row 9: 6 (correct)Row 10: 7 (correct)Row 11: 4 (correct)Row 12: 9 (correct)Row 13: 6 (correct)Row 14: 7 (correct)So, adding them up:0 + 0 + 1 + 0 + 3 + 2 + 3 + 0 + 7 + 6 + 7 + 4 + 9 + 6 + 7Let's add them in pairs to make it easier:(0 + 0) = 0(1 + 0) = 1(3 + 2) = 5(3 + 0) = 3(7 + 6) = 13(7 + 4) = 11(9 + 6) = 15(7) = 7Now, add these intermediate sums:0 + 1 = 11 + 5 = 66 + 3 = 99 + 13 = 2222 + 11 = 3333 + 15 = 4848 + 7 = 55So, the total number of even integers in the top 15 rows is 55.But wait, earlier I thought the answer was 49. Maybe I was mistaken before. Let me check the initial counts again.Looking back at row 12, I had 9 evens. Let's verify:Row 12: 1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1.Evens: 12, 66, 220, 792, 924, 792, 220, 66, 12. That's 9 evens. Correct.Row 14: 1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1.Evens: 14, 364, 2002, 3432, 2002, 364, 14. That's 7 evens. Correct.So, the total is indeed 55. Maybe the initial answer of 49 was incorrect. Let me see if there's another way to calculate this.I recall that the number of odd numbers in row n is 2 raised to the number of ones in the binary representation of n. So, for example, row 4 is 100 in binary, which has one '1', so there are 2^1 = 2 odd numbers. But wait, row 4 has 5 numbers, and 5 - 2 = 3 evens, which matches our earlier count.Similarly, row 5 is 101 in binary, which has two '1's, so 2^2 = 4 odd numbers. Row 5 has 6 numbers, so 6 - 4 = 2 evens. Correct.Row 6 is 110 in binary, which has two '1's, so 2^2 = 4 odd numbers. Row 6 has 7 numbers, so 7 - 4 = 3 evens. Correct.Row 7 is 111 in binary, which has three '1's, so 2^3 = 8 odd numbers. Row 7 has 8 numbers, so 8 - 8 = 0 evens. Correct.Row 8 is 1000 in binary, which has one '1', so 2^1 = 2 odd numbers. Row 8 has 9 numbers, so 9 - 2 = 7 evens. Correct.Row 9 is 1001 in binary, which has two '1's, so 2^2 = 4 odd numbers. Row 9 has 10 numbers, so 10 - 4 = 6 evens. Correct.Row 10 is 1010 in binary, which has two '1's, so 2^2 = 4 odd numbers. Row 10 has 11 numbers, so 11 - 4 = 7 evens. Correct.Row 11 is 1011 in binary, which has three '1's, so 2^3 = 8 odd numbers. Row 11 has 12 numbers, so 12 - 8 = 4 evens. Correct.Row 12 is 1100 in binary, which has two '1's, so 2^2 = 4 odd numbers. Row 12 has 13 numbers, so 13 - 4 = 9 evens. Correct.Row 13 is 1101 in binary, which has three '1's, so 2^3 = 8 odd numbers. Row 13 has 14 numbers, so 14 - 8 = 6 evens. Correct.Row 14 is 1110 in binary, which has three '1's, so 2^3 = 8 odd numbers. Row 14 has 15 numbers, so 15 - 8 = 7 evens. Correct.So, using this method, the number of evens in each row is n + 1 minus 2 raised to the number of ones in the binary representation of n.Therefore, to find the total number of evens in the top 15 rows, we can sum (n + 1 - 2^{number of ones in binary n}) for n from 0 to 14.Let's compute this:For n = 0 to 14:n | binary | number of ones | 2^{ones} | n + 1 | evens = (n + 1) - 2^{ones}---|-------|-------------|---------|-------|-------0 | 0 | 0 | 1 | 1 | 01 | 1 | 1 | 2 | 2 | 02 | 10 | 1 | 2 | 3 | 13 | 11 | 2 | 4 | 4 | 04 | 100 | 1 | 2 | 5 | 35 | 101 | 2 | 4 | 6 | 26 | 110 | 2 | 4 | 7 | 37 | 111 | 3 | 8 | 8 | 08 | 1000 | 1 | 2 | 9 | 79 | 1001 | 2 | 4 | 10 | 610| 1010 | 2 | 4 | 11 | 711| 1011 | 3 | 8 | 12 | 412| 1100 | 2 | 4 | 13 | 913| 1101 | 3 | 8 | 14 | 614| 1110 | 3 | 8 | 15 | 7Now, let's sum the evens column:0 + 0 + 1 + 0 + 3 + 2 + 3 + 0 + 7 + 6 + 7 + 4 + 9 + 6 + 7As before, this adds up to 55.Wait, but earlier I thought the answer was 49. Maybe I was confusing the top 15 rows with something else. Let me double-check the initial problem statement."Determine how many even integers are in the top 15 rows of Pascal's Triangle."Yes, top 15 rows, which are rows 0 to 14. So, according to my calculations, the total number of even integers is 55.But in the initial assistant's answer, it was 49. Maybe the assistant made a mistake in counting the number of evens per row.Let me check the assistant's counts:- Row 2: 1 even- Row 4: 3 evens- Row 6: 5 evens- Row 8: 7 evens- Row 10: 9 evens- Row 12: 11 evens- Row 14: 13 evensAdding these: 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49.But according to my detailed count, rows 5, 9, 11, and 13 also have evens, which the assistant didn't include. So, the assistant only considered even-numbered rows (0, 2, 4, 6, 8, 10, 12, 14) and counted the evens in those, but ignored the odd-numbered rows (1, 3, 5, 7, 9, 11, 13), which also have some evens.Wait, no, in the assistant's answer, they only listed rows 2, 4, 6, 8, 10, 12, 14 as having evens, and summed those. But in reality, rows 5, 9, 11, and 13 also have evens, so the assistant's count was incomplete.Therefore, the correct total should include evens from all rows, not just the even-numbered ones.So, my conclusion is that the total number of even integers in the top 15 rows of Pascal's Triangle is 55.But wait, let me cross-verify with another method. I recall that the total number of entries in the top 15 rows is the sum from n=0 to 14 of (n + 1), which is (15)(16)/2 = 120.If I can find the total number of odd integers, then subtracting from 120 will give me the number of evens.Using the formula: for each row n, the number of odd integers is 2^{number of ones in binary n}.So, let's compute the number of odds per row:n | binary | number of ones | 2^{ones} | odds---|-------|-------------|---------|-----0 | 0 | 0 | 1 | 11 | 1 | 1 | 2 | 22 | 10 | 1 | 2 | 23 | 11 | 2 | 4 | 44 | 100 | 1 | 2 | 25 | 101 | 2 | 4 | 46 | 110 | 2 | 4 | 47 | 111 | 3 | 8 | 88 | 1000 | 1 | 2 | 29 | 1001 | 2 | 4 | 410| 1010 | 2 | 4 | 411| 1011 | 3 | 8 | 812| 1100 | 2 | 4 | 413| 1101 | 3 | 8 | 814| 1110 | 3 | 8 | 8Now, let's sum the odds:1 + 2 + 2 + 4 + 2 + 4 + 4 + 8 + 2 + 4 + 4 + 8 + 4 + 8 + 8Let's compute step by step:Start with 1.Add 2: 3Add 2: 5Add 4: 9Add 2: 11Add 4: 15Add 4: 19Add 8: 27Add 2: 29Add 4: 33Add 4: 37Add 8: 45Add 4: 49Add 8: 57Add 8: 65So, total odds = 65.Total entries = 120.Therefore, total evens = 120 - 65 = 55.Yes, that confirms it. So, the correct number of even integers in the top 15 rows is 55.I think the initial assistant's answer of 49 was incorrect because they only considered even-numbered rows and missed the evens in odd-numbered rows like 5, 9, 11, and 13. Therefore, the accurate total is 55.

answer：Alright, so I have this problem where I need to convert the angle -1120 degrees into the form 2kπ + α, where k is an integer and α is between 0 and 2π. Then, I also need to find all angles β that have the same terminal side as α but lie within the interval [-4π, 0]. Hmm, okay, let's break this down step by step.First, I know that angles can be represented in different ways by adding or subtracting full rotations, which are 360 degrees or 2π radians. Since the given angle is negative, it means we're rotating clockwise from the positive x-axis. To find an equivalent positive angle, I can add multiples of 360 degrees until I get an angle between 0 and 360 degrees.So, starting with -1120 degrees, I need to figure out how many full rotations (360 degrees) fit into this angle. Let me divide 1120 by 360 to see how many times 360 goes into 1120.1120 ÷ 360 ≈ 3.111...So, 3 full rotations would be 3 × 360 = 1080 degrees. If I subtract that from 1120, I get 1120 - 1080 = 40 degrees. But since the original angle is negative, it's like rotating clockwise 1120 degrees, which is the same as rotating clockwise 3 full circles (1080 degrees) plus an additional 40 degrees. Therefore, the equivalent positive angle would be 360 - 40 = 320 degrees.Wait, is that right? Let me double-check. If I have -1120 degrees, adding 360 degrees multiple times should bring it within the 0 to 360 range. So, -1120 + 360 × 4 = -1120 + 1440 = 320 degrees. Yes, that matches. So, -1120 degrees is coterminal with 320 degrees.Now, I need to express this in radians. Since 360 degrees is 2π radians, 320 degrees would be (320/360) × 2π = (8/9) × 2π = 16π/9 radians. So, -1120 degrees is equal to 16π/9 radians.But the problem asks for the angle in the form 2kπ + α, where k is an integer and α is between 0 and 2π. Since 16π/9 is already between 0 and 2π (because 2π is approximately 6.283 and 16π/9 is approximately 5.585), I can write -1120 degrees as:-1120° = -4 × 360° + 320° = -8π + 16π/9.So, in the form 2kπ + α, it's -8π + 16π/9, where k = -4 and α = 16π/9.Okay, that seems to satisfy the first part. Now, moving on to the second part: finding all angles β that have the same terminal side as α (which is 16π/9) but lie within the interval [-4π, 0].Angles that have the same terminal side differ by integer multiples of 2π. So, the general form for β would be:β = 16π/9 + 2πk, where k is an integer.But we need β to be between -4π and 0. Let's find the appropriate values of k.First, let's express β in terms of π:β = 16π/9 + 2πk.We need β ≥ -4π and β < 0.So, let's solve for k:-4π ≤ 16π/9 + 2πk < 0.Divide all parts by π to simplify:-4 ≤ 16/9 + 2k < 0.Subtract 16/9 from all parts:-4 - 16/9 ≤ 2k < -16/9.Convert -4 to ninths: -4 = -36/9.So:-36/9 - 16/9 = -52/9 ≤ 2k < -16/9.Divide all parts by 2:-26/9 ≤ k < -8/9.Since k must be an integer, let's find the integers between -26/9 and -8/9.-26/9 ≈ -2.888...-8/9 ≈ -0.888...So, the integers k that satisfy this inequality are k = -2 and k = -1.Now, let's find β for these values of k.For k = -1:β = 16π/9 + 2π(-1) = 16π/9 - 2π = 16π/9 - 18π/9 = -2π/9.For k = -2:β = 16π/9 + 2π(-2) = 16π/9 - 4π = 16π/9 - 36π/9 = -20π/9.Let's check if these angles are within the interval [-4π, 0].-20π/9 ≈ -6.981, which is less than -4π ≈ -12.566? Wait, no, -20π/9 is approximately -6.981, which is greater than -4π (-12.566). So, it's within the interval.-2π/9 ≈ -0.698, which is also within [-4π, 0].Wait, but I thought -4π is about -12.566, so -20π/9 is about -6.981, which is greater than -4π. So, it's within the interval.But let me double-check the calculations.For k = -1:β = 16π/9 - 2π = 16π/9 - 18π/9 = -2π/9 ≈ -0.698 radians.For k = -2:β = 16π/9 - 4π = 16π/9 - 36π/9 = -20π/9 ≈ -6.981 radians.Yes, both are within [-4π, 0] since -4π ≈ -12.566 and 0 is 0.So, the angles β are -2π/9 and -20π/9.Wait, but let me check if there are more values of k that could give angles within this interval.If k = -3:β = 16π/9 + 2π(-3) = 16π/9 - 6π = 16π/9 - 54π/9 = -38π/9 ≈ -12.832 radians.But -38π/9 ≈ -12.832, which is less than -4π ≈ -12.566, so it's outside the interval.Similarly, k = 0:β = 16π/9 + 0 = 16π/9 ≈ 5.585, which is positive and outside the interval.So, only k = -1 and k = -2 give angles β within [-4π, 0].Therefore, the set of angles β is {β | β = 2kπ + 16π/9, k ∈ ℤ} intersected with [-4π, 0], which gives β = -2π/9 and β = -20π/9.Let me summarize:1. -1120° = -4 × 360° + 320° = -8π + 16π/9.2. The angles β in [-4π, 0] with the same terminal side as α = 16π/9 are β = -2π/9 and β = -20π/9.I think that's it. I should probably double-check my calculations to make sure I didn't make any mistakes.First, converting -1120° to radians:-1120° × (π/180) = -1120π/180 = -56π/9 ≈ -6.283π.Wait, that's not matching with 16π/9. Hmm, maybe I confused something.Wait, no, when I converted -1120° to a positive angle, I got 320°, which is 16π/9 radians. So, -1120° is coterminal with 320°, which is 16π/9 radians.But in terms of adding multiples of 2π, I have to express -1120° as 2kπ + α.So, -1120° in radians is -1120 × π/180 = -56π/9.Now, to express -56π/9 as 2kπ + α, where α is between 0 and 2π.Let's find k such that α = -56π/9 - 2kπ is between 0 and 2π.Let me solve for k:0 ≤ -56π/9 - 2kπ < 2π.Divide all parts by π:0 ≤ -56/9 - 2k < 2.Add 56/9 to all parts:56/9 ≤ -2k < 56/9 + 2.Convert 2 to ninths: 2 = 18/9.So:56/9 ≤ -2k < 74/9.Divide all parts by -2, remembering to reverse the inequalities:-56/18 ≥ k > -74/18.Simplify:-28/9 ≥ k > -37/9.Convert to decimals:-3.111... ≥ k > -4.111...So, k must be integers between -4.111 and -3.111, which means k = -4.So, k = -4.Then, α = -56π/9 - 2(-4)π = -56π/9 + 8π = -56π/9 + 72π/9 = 16π/9.Yes, that matches what I had before. So, -1120° = 2(-4)π + 16π/9.Okay, that checks out.Now, for the second part, finding β in [-4π, 0] with the same terminal side as α = 16π/9.As I did earlier, β = 16π/9 + 2πk.We need β ∈ [-4π, 0].So, let's solve for k:-4π ≤ 16π/9 + 2πk < 0.Divide by π:-4 ≤ 16/9 + 2k < 0.Subtract 16/9:-4 - 16/9 ≤ 2k < -16/9.Convert -4 to ninths: -36/9.So:-36/9 - 16/9 = -52/9 ≤ 2k < -16/9.Divide by 2:-26/9 ≤ k < -8/9.Which is approximately:-2.888... ≤ k < -0.888...So, k can be -2 and -1.For k = -1:β = 16π/9 + 2π(-1) = 16π/9 - 18π/9 = -2π/9.For k = -2:β = 16π/9 + 2π(-2) = 16π/9 - 36π/9 = -20π/9.Both -2π/9 and -20π/9 are within [-4π, 0], since -20π/9 ≈ -6.981 and -4π ≈ -12.566, so -20π/9 is greater than -4π.Wait, but -20π/9 is approximately -6.981, which is greater than -4π (-12.566), so it's within the interval.Similarly, -2π/9 ≈ -0.698, which is also within [-4π, 0].If I try k = -3:β = 16π/9 + 2π(-3) = 16π/9 - 6π = 16π/9 - 54π/9 = -38π/9 ≈ -12.832, which is less than -4π (-12.566), so it's outside the interval.And k = 0:β = 16π/9, which is positive and outside the interval.So, only k = -1 and k = -2 give valid β within [-4π, 0].Therefore, the angles β are -2π/9 and -20π/9.I think that's correct. I don't see any mistakes in the calculations now.

answer：Okay, so I have this trigonometry problem here: Given that tan(θ - π) equals 2, I need to find the value of sin²θ + sinθ cosθ - 2cos²θ. Hmm, let me think about how to approach this step by step.First, I remember that the tangent function has a period of π, which means that tan(θ - π) is the same as tanθ. So, if tan(θ - π) = 2, then tanθ must also equal 2. That simplifies things a bit because now I can work with tanθ = 2 instead of dealing with θ - π.Now, I need to find sin²θ + sinθ cosθ - 2cos²θ. This expression involves both sine and cosine functions, but since I know tanθ, which is the ratio of sine to cosine, maybe I can express everything in terms of tanθ.I recall that sin²θ can be written as (tan²θ) / (1 + tan²θ) and cos²θ can be written as 1 / (1 + tan²θ). Also, sinθ cosθ can be expressed as (tanθ) / (1 + tan²θ). Let me write these down:- sin²θ = tan²θ / (1 + tan²θ)- cos²θ = 1 / (1 + tan²θ)- sinθ cosθ = tanθ / (1 + tan²θ)Since tanθ is 2, let's substitute that into each of these expressions.First, calculate sin²θ:sin²θ = (2)² / (1 + (2)²) = 4 / (1 + 4) = 4/5.Next, calculate cos²θ:cos²θ = 1 / (1 + (2)²) = 1 / 5.Then, calculate sinθ cosθ:sinθ cosθ = 2 / (1 + 4) = 2/5.Now, plug these back into the original expression:sin²θ + sinθ cosθ - 2cos²θ = (4/5) + (2/5) - 2*(1/5).Let's compute each term:- 4/5 is straightforward.- 2/5 is also straightforward.- 2*(1/5) is 2/5.So, putting it all together:4/5 + 2/5 - 2/5.Hmm, let's add them up step by step. First, 4/5 + 2/5 is 6/5. Then, subtracting 2/5 from that gives 6/5 - 2/5 = 4/5.Wait, so the entire expression simplifies to 4/5. That seems straightforward. Let me double-check my steps to make sure I didn't make a mistake.1. I used the identity tan(θ - π) = tanθ, which is correct because tangent has a period of π.2. I expressed sin²θ, cos²θ, and sinθ cosθ in terms of tanθ. Let me verify those identities: - sin²θ = tan²θ / (1 + tan²θ): Yes, because sinθ = tanθ / sqrt(1 + tan²θ), so squaring both sides gives sin²θ = tan²θ / (1 + tan²θ). - cos²θ = 1 / (1 + tan²θ): Correct, since cosθ = 1 / sqrt(1 + tan²θ), squaring gives cos²θ = 1 / (1 + tan²θ). - sinθ cosθ = tanθ / (1 + tan²θ): Let me check: sinθ cosθ = (tanθ / sqrt(1 + tan²θ)) * (1 / sqrt(1 + tan²θ)) = tanθ / (1 + tan²θ). Yes, that's correct.3. Substituted tanθ = 2 into each expression: - sin²θ = 4/5 - cos²θ = 1/5 - sinθ cosθ = 2/54. Plugged into the original expression: 4/5 + 2/5 - 2*(1/5) = 4/5 + 2/5 - 2/5 = 4/5.Everything seems to check out. I don't see any mistakes in my calculations or reasoning. So, the final answer should be 4/5.Just to be thorough, let me consider another approach. Maybe using the identity for tan(θ - π) directly.I know that tan(θ - π) = (tanθ - tanπ) / (1 + tanθ tanπ). Since tanπ = 0, this simplifies to tanθ / 1 = tanθ. So, tan(θ - π) = tanθ, which is consistent with what I used earlier. Therefore, tanθ = 2 is correct.Alternatively, I could express the original expression in terms of tanθ by dividing numerator and denominator by cos²θ. Let's try that.The expression is sin²θ + sinθ cosθ - 2cos²θ. If I divide each term by cos²θ, I get:(sin²θ / cos²θ) + (sinθ cosθ / cos²θ) - 2(cos²θ / cos²θ) = tan²θ + tanθ - 2.So, the expression simplifies to tan²θ + tanθ - 2. Since tanθ = 2, substitute that in:(2)² + 2 - 2 = 4 + 2 - 2 = 4.Wait a minute, that gives me 4, but earlier I got 4/5. That's conflicting. What did I do wrong here?Oh, I see. When I divide the expression by cos²θ, I have to remember that I'm effectively rewriting the expression as (tan²θ + tanθ - 2) * (cos²θ). So, the entire expression is (tan²θ + tanθ - 2) * cos²θ.Since tanθ = 2, tan²θ = 4, so tan²θ + tanθ - 2 = 4 + 2 - 2 = 4. Then, multiply by cos²θ, which is 1/5. So, 4 * (1/5) = 4/5.Ah, okay, so I had to remember that dividing by cos²θ changes the expression, and I need to multiply back by cos²θ to get the original expression. That makes sense now. So, both methods lead to the same result, 4/5.Therefore, I'm confident that the answer is 4/5.**Final Answer**The value of the expression is boxed{dfrac{4}{5}}.

answer：Okay, so I have this problem here: Let ( f(x) = 4x^3 + mx^2 + (m - 3)x + n ) where ( m, n ) are real numbers. It says that ( f(x) ) is a monotonically increasing function on ( mathbb{R} ), and I need to find the value of ( m ).Hmm, okay. Monotonically increasing means that the function never decreases as ( x ) increases. So, for all real numbers ( x ), the derivative of ( f(x) ) should be non-negative. That makes sense because if the derivative is positive, the function is increasing, and if it's zero, it's constant at that point but not decreasing.Alright, so first step: find the derivative of ( f(x) ). Let me do that.The derivative of ( 4x^3 ) is ( 12x^2 ). The derivative of ( mx^2 ) is ( 2mx ). The derivative of ( (m - 3)x ) is ( m - 3 ). And the derivative of a constant ( n ) is zero. So putting it all together, the derivative ( f'(x) ) is:( f'(x) = 12x^2 + 2mx + (m - 3) )Okay, so ( f'(x) = 12x^2 + 2mx + (m - 3) ). Since ( f(x) ) is monotonically increasing, ( f'(x) geq 0 ) for all ( x in mathbb{R} ).Now, ( f'(x) ) is a quadratic function in terms of ( x ). For a quadratic function ( ax^2 + bx + c ), it is always non-negative if and only if two conditions are satisfied:1. The leading coefficient ( a ) is positive.2. The discriminant ( D = b^2 - 4ac ) is less than or equal to zero.So, let's check these conditions for our derivative ( f'(x) = 12x^2 + 2mx + (m - 3) ).First, the leading coefficient is 12, which is positive. That's good because it means the parabola opens upwards, so it has a minimum point. If the discriminant is less than or equal to zero, that minimum point will be at or above the x-axis, meaning the quadratic is always non-negative.So, let's compute the discriminant ( D ):( D = (2m)^2 - 4 times 12 times (m - 3) )Calculating that:( D = 4m^2 - 48(m - 3) )Let me simplify this:First, expand the second term:( 48(m - 3) = 48m - 144 )So, substituting back:( D = 4m^2 - 48m + 144 )Hmm, that's a quadratic in terms of ( m ). Let me see if I can factor this or simplify it further.Looking at ( 4m^2 - 48m + 144 ), I notice that all coefficients are divisible by 4. Let me factor out a 4:( D = 4(m^2 - 12m + 36) )Now, looking at the quadratic inside the parentheses: ( m^2 - 12m + 36 ). Hmm, that looks like a perfect square. Let me check:( (m - 6)^2 = m^2 - 12m + 36 ). Yes, that's exactly it.So, substituting back:( D = 4(m - 6)^2 )Alright, so the discriminant is ( 4(m - 6)^2 ). Since we need ( D leq 0 ) for the quadratic ( f'(x) ) to be non-negative for all ( x ), let's set up the inequality:( 4(m - 6)^2 leq 0 )Now, ( 4(m - 6)^2 ) is always non-negative because any real number squared is non-negative, and multiplied by 4, it's still non-negative. So, the only way this can be less than or equal to zero is if ( (m - 6)^2 = 0 ).Therefore:( (m - 6)^2 = 0 )Taking the square root of both sides:( m - 6 = 0 )So,( m = 6 )Alright, so ( m ) must be 6. Let me just double-check my steps to make sure I didn't make a mistake.1. Found the derivative correctly: ( f'(x) = 12x^2 + 2mx + (m - 3) ). That seems right.2. Identified that for ( f'(x) geq 0 ) for all ( x ), the discriminant must be ( leq 0 ). That's correct because a quadratic with a positive leading coefficient is always non-negative only if it doesn't cross the x-axis, which happens when the discriminant is non-positive.3. Calculated the discriminant: ( D = (2m)^2 - 4 times 12 times (m - 3) ) Which simplifies to: ( D = 4m^2 - 48m + 144 ) Factored out a 4: ( D = 4(m^2 - 12m + 36) ) Recognized that ( m^2 - 12m + 36 ) is ( (m - 6)^2 ). That's correct.4. So, ( D = 4(m - 6)^2 leq 0 ). Since ( 4(m - 6)^2 ) is non-negative, the only solution is ( m = 6 ). That makes sense.So, I think my steps are correct. Therefore, the value of ( m ) is 6.**Final Answer**The value of ( m ) is boxed{6}.

answer：Okay, so I have this problem where I need to find the number of pairs (a, b) where both a and b are natural numbers not exceeding 10. The condition is that the solution x to the equation ax = b must be less than 1/2 but greater than 1/3. Hmm, let me try to break this down step by step.First, let's understand what the equation ax = b means. If I solve for x, I get x = b/a. So, the problem is essentially asking for how many pairs (a, b) such that when I divide b by a, the result is between 1/3 and 1/2. That is, 1/3 < b/a < 1/2.Since both a and b are natural numbers from 1 to 10, I can approach this by checking each possible value of a and seeing which values of b satisfy the inequality 1/3 < b/a < 1/2.Let me write down the inequality again for clarity:1/3 < b/a < 1/2I can rewrite this inequality to make it easier to work with. If I multiply all parts of the inequality by a (since a is positive, the direction of the inequality won't change), I get:(1/3)a < b < (1/2)aSo, for each a from 1 to 10, I need to find the integer values of b that lie strictly between (1/3)a and (1/2)a.Let me go through each value of a one by one.Starting with a = 1:(1/3)*1 = 1/3 ≈ 0.333(1/2)*1 = 1/2 = 0.5So, b must be greater than 0.333 and less than 0.5. Since b must be a natural number, the only possible value is b = 1. But 1 is not less than 0.5, so there are no valid b for a = 1.Next, a = 2:(1/3)*2 ≈ 0.666(1/2)*2 = 1So, b must be greater than approximately 0.666 and less than 1. Again, b must be a natural number, so the only possible value is b = 1. But 1 is not less than 1, so no valid b for a = 2.a = 3:(1/3)*3 = 1(1/2)*3 = 1.5So, b must be greater than 1 and less than 1.5. The only natural number in this range is b = 2, but 2 is not less than 1.5. So, no valid b for a = 3.a = 4:(1/3)*4 ≈ 1.333(1/2)*4 = 2So, b must be greater than approximately 1.333 and less than 2. The only natural number in this range is b = 2, but 2 is not less than 2. So, no valid b for a = 4.a = 5:(1/3)*5 ≈ 1.666(1/2)*5 = 2.5So, b must be greater than approximately 1.666 and less than 2.5. The natural numbers in this range are b = 2. So, b = 2 is valid because 2 is greater than 1.666 and less than 2.5. Therefore, (5, 2) is a valid pair.a = 6:(1/3)*6 = 2(1/2)*6 = 3So, b must be greater than 2 and less than 3. The only natural number in this range is b = 3, but 3 is not less than 3. So, no valid b for a = 6.a = 7:(1/3)*7 ≈ 2.333(1/2)*7 = 3.5So, b must be greater than approximately 2.333 and less than 3.5. The natural numbers in this range are b = 3. So, b = 3 is valid because 3 is greater than 2.333 and less than 3.5. Therefore, (7, 3) is a valid pair.a = 8:(1/3)*8 ≈ 2.666(1/2)*8 = 4So, b must be greater than approximately 2.666 and less than 4. The natural numbers in this range are b = 3. So, b = 3 is valid because 3 is greater than 2.666 and less than 4. Therefore, (8, 3) is a valid pair.a = 9:(1/3)*9 = 3(1/2)*9 = 4.5So, b must be greater than 3 and less than 4.5. The natural numbers in this range are b = 4. So, b = 4 is valid because 4 is greater than 3 and less than 4.5. Therefore, (9, 4) is a valid pair.a = 10:(1/3)*10 ≈ 3.333(1/2)*10 = 5So, b must be greater than approximately 3.333 and less than 5. The natural numbers in this range are b = 4. So, b = 4 is valid because 4 is greater than 3.333 and less than 5. Therefore, (10, 4) is a valid pair.Let me summarize the valid pairs I found:- (5, 2)- (7, 3)- (8, 3)- (9, 4)- (10, 4)So, that's a total of 5 pairs. Wait, but the options given are (A) Five pairs, (B) Four pairs, (C) Three pairs, (D) Two pairs. So, according to my count, the answer should be (A) Five pairs.But let me double-check to make sure I didn't miss anything or count any extra pairs.Starting from a = 1 to a = 10:a = 1: No valid ba = 2: No valid ba = 3: No valid ba = 4: No valid ba = 5: b = 2 (Valid)a = 6: No valid ba = 7: b = 3 (Valid)a = 8: b = 3 (Valid)a = 9: b = 4 (Valid)a = 10: b = 4 (Valid)Yes, that's five pairs. So, the answer should be (A) Five pairs.Wait, but let me check if for a = 5, b = 2: 2/5 = 0.4, which is between 1/3 ≈ 0.333 and 1/2 = 0.5. Correct.For a = 7, b = 3: 3/7 ≈ 0.428, which is between 1/3 and 1/2. Correct.For a = 8, b = 3: 3/8 = 0.375, which is between 1/3 and 1/2. Correct.For a = 9, b = 4: 4/9 ≈ 0.444, which is between 1/3 and 1/2. Correct.For a = 10, b = 4: 4/10 = 0.4, which is between 1/3 and 1/2. Correct.So, all five pairs are valid. Therefore, the answer is (A) Five pairs.