💡First, I identify the key components of the arithmetic sequence: the first term ( a = -6 ), the common difference ( d = 4 ), and the last term ( l = 38 ).Next, I use the formula for the ( n )-th term of an arithmetic sequence, ( a_n = a + (n-1)d ), and set ( a_n ) equal to the last term ( l ) to find the number of terms.Substituting the known values into the formula:[38 = -6 + (n-1) cdot 4]I then solve for ( n ):[44 = (n-1) cdot 4][n-1 = 11][n = 12]This calculation shows that there are 12 terms in the sequence.

💡Alright, so I've got this problem about a sequence {a_n} where the sum of the first 9 terms is 153, and there's a point P(a_n, a_{n+1}) that lies on the line x - y + 3 = 0. I need to find the general term formula for {a_n}, then form a new sequence {b_n} by taking specific terms from {a_n}, and finally prove an inequality involving the reciprocals of {b_n}.Starting with part (1). The point P(a_n, a_{n+1}) is on the line x - y + 3 = 0. That means if I plug in x = a_n and y = a_{n+1} into the equation, it should satisfy. So, substituting, I get a_n - a_{n+1} + 3 = 0, which simplifies to a_{n+1} = a_n + 3. Hmm, so this tells me that each term is 3 more than the previous term. That means {a_n} is an arithmetic sequence with a common difference of 3.Now, since it's an arithmetic sequence, the nth term can be written as a_n = a_1 + (n - 1)d, where d is the common difference. Here, d = 3, so a_n = a_1 + 3(n - 1).Next, I know the sum of the first 9 terms is 153. The formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 * (2a_1 + (n - 1)d). Plugging in n = 9, S_9 = 9/2 * (2a_1 + 8*3) = 153.Let me compute that: 9/2 * (2a_1 + 24) = 153. Multiplying both sides by 2 to eliminate the denominator: 9*(2a_1 + 24) = 306. Then, dividing both sides by 9: 2a_1 + 24 = 34. Subtracting 24: 2a_1 = 10, so a_1 = 5.Therefore, the general term is a_n = 5 + 3(n - 1) = 3n + 2. Let me check that: when n=1, a_1=5, which is correct. The common difference is 3, so each term increases by 3. The 9th term would be 3*9 + 2 = 29. The sum of the first 9 terms is (5 + 29)*9/2 = 34*9/2 = 153, which matches. So, part (1) is done, and the general term is a_n = 3n + 2.Moving on to part (2). I need to form a new sequence {b_n} by taking the 2nd term, 8th term, 24th term, etc., from {a_n}. The pattern seems to be n * 2^n. Let me see: for n=1, it's 1*2^1=2, so the 2nd term; for n=2, it's 2*2^2=8, the 8th term; for n=3, 3*2^3=24, the 24th term, and so on. So, in general, the k-th term of {b_n} is a_{k*2^k}.Given that a_n = 3n + 2, then b_n = a_{n*2^n} = 3*(n*2^n) + 2 = 3n*2^n + 2.Now, I need to find the sum of the first n terms of {b_n}, denoted as S_n. So, S_n = b_1 + b_2 + ... + b_n = sum_{k=1}^n (3k*2^k + 2).This can be split into two sums: 3*sum_{k=1}^n (k*2^k) + sum_{k=1}^n 2. The second sum is straightforward: sum_{k=1}^n 2 = 2n.The first sum, sum_{k=1}^n (k*2^k), is a bit trickier. I remember there's a formula for sum_{k=1}^n k*r^k. Let me recall. I think it's r*(1 - (n+1)*r^n + n*r^{n+1}) / (1 - r)^2, but let me verify.Let me denote T = sum_{k=1}^n k*2^k. To find T, I can use the method of generating functions or recurrence.Let me write T = 2 + 2*2^2 + 3*2^3 + ... + n*2^n.Multiply both sides by 2: 2T = 2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n+1}.Subtract the original T from this equation:2T - T = T = (2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n+1}) - (2 + 2*2^2 + 3*2^3 + ... + n*2^n).Simplify term by term:= (2^2 - 2) + (2*2^3 - 2*2^2) + (3*2^4 - 3*2^3) + ... + (n*2^{n+1} - n*2^n).Each term simplifies to:= (4 - 2) + (16 - 8) + (48 - 24) + ... + (n*2^{n+1} - n*2^n).Which is:= 2 + 8 + 24 + ... + n*2^n*(2 - 1) = 2 + 8 + 24 + ... + n*2^n.Wait, that doesn't seem to help much. Maybe I need a different approach.Alternatively, consider that sum_{k=1}^n k*r^k = r*(1 - (n+1)*r^n + n*r^{n+1}) / (1 - r)^2.Let me plug in r=2:sum_{k=1}^n k*2^k = 2*(1 - (n+1)*2^n + n*2^{n+1}) / (1 - 2)^2.Simplify denominator: (1 - 2)^2 = 1, so denominator is 1.Numerator: 2*(1 - (n+1)*2^n + n*2^{n+1}) = 2 - 2(n+1)*2^n + 2n*2^{n+1}.Wait, let me compute it step by step:sum_{k=1}^n k*2^k = 2*(1 - (n+1)*2^n + n*2^{n+1}) / (1 - 2)^2.Compute denominator: (1 - 2)^2 = 1.Numerator: 2*(1 - (n+1)*2^n + n*2^{n+1}) = 2 - 2(n+1)*2^n + 2n*2^{n+1}.Wait, 2n*2^{n+1} = 2n*2*2^n = 4n*2^n.Similarly, -2(n+1)*2^n = -2n*2^n - 2*2^n.So, numerator becomes:2 - 2n*2^n - 2*2^n + 4n*2^n = 2 + ( -2n*2^n + 4n*2^n ) + ( -2*2^n )Simplify:2 + 2n*2^n - 2*2^n = 2 + 2^n*(2n - 2).Factor out 2:2 + 2*(2^{n-1})*(2n - 2) = 2 + 2^{n}*(n - 1).Wait, that doesn't seem right. Maybe I made a mistake in expanding.Wait, let's re-examine:Numerator: 2*(1 - (n+1)*2^n + n*2^{n+1}) = 2*1 - 2*(n+1)*2^n + 2*n*2^{n+1}.Compute each term:2*1 = 2.-2*(n+1)*2^n = -2n*2^n - 2*2^n = -n*2^{n+1} - 2^{n+1}.2*n*2^{n+1} = n*2^{n+2}.So, combining all terms:2 - n*2^{n+1} - 2^{n+1} + n*2^{n+2}.Factor terms with 2^{n+1}:= 2 + (-n -1 + 2n)*2^{n+1}.Simplify coefficients:(-n -1 + 2n) = n -1.So, numerator becomes: 2 + (n -1)*2^{n+1}.Therefore, sum_{k=1}^n k*2^k = 2 + (n -1)*2^{n+1}.So, T = sum_{k=1}^n k*2^k = (n -1)*2^{n+1} + 2.Therefore, going back to S_n:S_n = 3*T + 2n = 3*((n -1)*2^{n+1} + 2) + 2n.Compute this:= 3*(n -1)*2^{n+1} + 6 + 2n.So, S_n = 3(n -1)2^{n+1} + 2n + 6.Let me check for n=1: b_1 = a_2 = 3*2 + 2 = 8. So, S_1 = 8. Plugging into the formula: 3(0)*4 + 2 + 6 = 0 + 2 +6=8. Correct.For n=2: b_1=8, b_2=a_8=3*8+2=26. So, S_2=8+26=34. Plugging into the formula: 3(1)*8 + 4 +6=24 +4 +6=34. Correct.For n=3: b_3=a_{24}=3*24+2=74. So, S_3=8+26+74=108. Formula: 3(2)*16 +6 +6=96 +6 +6=108. Correct.So, the formula seems to hold.Moving on to part (3). I need to prove that the sum of reciprocals of {b_n} up to n terms is less than 1/3. That is, 1/b_1 + 1/b_2 + ... + 1/b_n < 1/3.Given that b_n = 3n*2^n + 2. So, 1/b_n = 1/(3n*2^n + 2). I need to show that the sum of these from n=1 to n is less than 1/3.First, let's see if we can find an upper bound for 1/(3n*2^n + 2). Notice that 3n*2^n + 2 > 3n*2^n, so 1/(3n*2^n + 2) < 1/(3n*2^n).Therefore, sum_{k=1}^n 1/b_k < sum_{k=1}^n 1/(3k*2^k) = (1/3) sum_{k=1}^n 1/(k*2^k).Now, I need to evaluate sum_{k=1}^infty 1/(k*2^k). I recall that sum_{k=1}^infty x^k /k = -ln(1 - x) for |x| <1.So, if I set x=1/2, sum_{k=1}^infty (1/2)^k /k = -ln(1 - 1/2) = -ln(1/2) = ln(2).Therefore, sum_{k=1}^infty 1/(k*2^k) = ln(2). So, sum_{k=1}^n 1/(k*2^k) < ln(2).Thus, sum_{k=1}^n 1/b_k < (1/3) ln(2). Since ln(2) is approximately 0.693, so (1/3)*0.693 ≈ 0.231, which is less than 1/3 ≈ 0.333. So, the sum is indeed less than 1/3.But to make it more rigorous, let's note that sum_{k=1}^infty 1/(k*2^k) = ln(2), so sum_{k=1}^n 1/(k*2^k) < ln(2) < 1. So, (1/3) sum_{k=1}^n 1/(k*2^k) < (1/3) ln(2) < 1/3, since ln(2) < 1.5, so (1/3)*1.5=0.5, but actually ln(2)≈0.693, so (1/3)*0.693≈0.231 < 1/3.Alternatively, since 1/(k*2^k) is decreasing, we can compare it to an integral or use another method, but the key point is that the infinite sum is less than 1/3, so the finite sum is also less than 1/3.Therefore, the inequality holds.

💡Alright, so I have this problem about vectors forming an n-gon, and I need to prove this equation involving the squares of their magnitudes and the cosines of the angles between them. Hmm, okay, let me try to break this down.First, let's understand what's given. We have vectors a₁, a₂, ..., aₙ, which are the sides of an n-gon. That means these vectors form a closed polygon, right? So, if I add all these vectors together, they should sum up to zero because the polygon is closed. That seems important.The problem wants me to prove that a₁² equals the sum of the squares of the other vectors plus twice the sum of the products of their magnitudes times the cosine of the angles between each pair. Mathematically, it's written as:a₁² = a₂² + a₃² + ... + aₙ² + 2 ∑_{i > j > 1} a_i a_j cos φ_{ij}Where φ_{ij} is the angle between vectors a_i and a_j. Okay, so I need to relate the square of the first vector to the squares of the others and the angles between them.Since the vectors form a closed polygon, their sum is zero. So, a₁ + a₂ + ... + aₙ = 0. That means a₁ = -(a₂ + a₃ + ... + aₙ). Maybe I can use this to express a₁ in terms of the other vectors and then square both sides.Let me write that down:a₁ = -(a₂ + a₃ + ... + aₙ)Now, if I square both sides, I get:a₁² = |a₂ + a₃ + ... + aₙ|²Expanding the right-hand side, I know that the square of the magnitude of a vector sum is equal to the sum of the squares of the magnitudes plus twice the sum of the dot products of each pair. So,|a₂ + a₃ + ... + aₙ|² = a₂² + a₃² + ... + aₙ² + 2 ∑_{i > j > 1} a_i · a_jBut the dot product a_i · a_j is equal to |a_i||a_j|cos φ_{ij}, which is a_i a_j cos φ_{ij} since a_i and a_j are the magnitudes. So, substituting that in, we get:|a₂ + a₃ + ... + aₙ|² = a₂² + a₃² + ... + aₙ² + 2 ∑_{i > j > 1} a_i a_j cos φ_{ij}Therefore, putting it all together:a₁² = a₂² + a₃² + ... + aₙ² + 2 ∑_{i > j > 1} a_i a_j cos φ_{ij}Wait, that seems to be exactly what we needed to prove! So, by using the fact that the vectors form a closed polygon and expanding the square of their sum, we arrived at the desired equation.But let me double-check to make sure I didn't miss anything. The key steps were recognizing that the vectors sum to zero, expressing a₁ in terms of the other vectors, squaring both sides, and then expanding the magnitude squared. The expansion naturally leads to the sum of squares and the sum of dot products, which translates to the sum involving cosines of angles.I think this makes sense. The only thing I might have overlooked is whether the angles φ_{ij} are correctly defined as the angles between vectors a_i and a_j. Since we're dealing with vectors in a polygon, the angles between them are indeed the internal angles of the polygon, which should correspond to the angles between the vectors as defined.Also, it's important to note that when we take the dot product a_i · a_j, we're considering the angle between them in the context of their placement in the polygon. Since the polygon is closed, these angles are consistent with the geometry of the shape.Another thing to consider is the range of the indices in the summation. The problem specifies i > j > 1, which means we're summing over all unique pairs of vectors from a₂ to aₙ, ensuring we don't double-count any pairs. This is crucial because each pair should only be considered once in the expansion.To summarize, the proof relies on the properties of vectors forming a closed polygon and the expansion of the magnitude squared of their sum. By expressing a₁ in terms of the other vectors and carefully expanding, we arrive at the equation we needed to prove. It all ties together neatly with vector addition and the dot product properties.I think I've covered all the necessary steps and verified that each part makes sense. This seems like a solid proof, but I'll present it in a clear, step-by-step manner to ensure it's understandable.

💡Alright, so I have this problem here: I need to find the number of irrational numbers ( x ) for which the value of ( frac{x+1}{x^{2}-3x+3} ) is an integer. The options are (A) 0, (B) 2, (C) 4, (D) 6. Hmm, okay, let me try to figure this out step by step.First, I know that the expression ( frac{x+1}{x^{2}-3x+3} ) needs to be an integer. Let's call that integer ( k ). So, I can set up the equation:[ frac{x+1}{x^{2}-3x+3} = k ]To solve for ( x ), I should probably get rid of the fraction by multiplying both sides by the denominator. Let's do that:[ x + 1 = k(x^{2} - 3x + 3) ]Now, let's expand the right side:[ x + 1 = kx^{2} - 3kx + 3k ]Hmm, now I have a quadratic equation in terms of ( x ). Let me rearrange all terms to one side so that I can have it in standard quadratic form:[ kx^{2} - 3kx + 3k - x - 1 = 0 ]Wait, let me combine like terms. The ( x ) terms are ( -3kx ) and ( -x ), so combining those gives:[ kx^{2} - (3k + 1)x + (3k - 1) = 0 ]Okay, so now I have a quadratic equation:[ kx^{2} - (3k + 1)x + (3k - 1) = 0 ]Now, since ( x ) is a real number, the discriminant of this quadratic must be non-negative. The discriminant ( D ) of a quadratic ( ax^{2} + bx + c ) is ( D = b^{2} - 4ac ). So, let's compute the discriminant for our equation:Here, ( a = k ), ( b = -(3k + 1) ), and ( c = 3k - 1 ). So,[ D = [-(3k + 1)]^{2} - 4(k)(3k - 1) ]Simplifying that:[ D = (3k + 1)^{2} - 4k(3k - 1) ]Let me expand ( (3k + 1)^{2} ):[ (3k + 1)^{2} = 9k^{2} + 6k + 1 ]And ( 4k(3k - 1) = 12k^{2} - 4k )So, subtracting these:[ D = 9k^{2} + 6k + 1 - 12k^{2} + 4k ][ D = (9k^{2} - 12k^{2}) + (6k + 4k) + 1 ][ D = -3k^{2} + 10k + 1 ]So, the discriminant is ( D = -3k^{2} + 10k + 1 ). For the quadratic to have real solutions, ( D ) must be greater than or equal to zero:[ -3k^{2} + 10k + 1 geq 0 ]Hmm, okay, so I need to solve this inequality for ( k ). Let's treat this as a quadratic inequality. First, let's find the roots of the equation ( -3k^{2} + 10k + 1 = 0 ). I can use the quadratic formula for this:[ k = frac{-b pm sqrt{b^{2} - 4ac}}{2a} ]In this case, ( a = -3 ), ( b = 10 ), and ( c = 1 ). Plugging these into the formula:[ k = frac{-10 pm sqrt{(10)^{2} - 4(-3)(1)}}{2(-3)} ][ k = frac{-10 pm sqrt{100 + 12}}{-6} ][ k = frac{-10 pm sqrt{112}}{-6} ]Simplify ( sqrt{112} ). Since ( 112 = 16 times 7 ), ( sqrt{112} = 4sqrt{7} ). So,[ k = frac{-10 pm 4sqrt{7}}{-6} ]Let me simplify this further by dividing numerator and denominator by -2:[ k = frac{5 mp 2sqrt{7}}{3} ]So, the roots are ( k = frac{5 + 2sqrt{7}}{3} ) and ( k = frac{5 - 2sqrt{7}}{3} ).Since the coefficient of ( k^{2} ) in the quadratic equation ( -3k^{2} + 10k + 1 ) is negative, the parabola opens downward. Therefore, the quadratic is positive between its two roots. So, the inequality ( -3k^{2} + 10k + 1 geq 0 ) holds for:[ frac{5 - 2sqrt{7}}{3} leq k leq frac{5 + 2sqrt{7}}{3} ]Now, let's approximate the numerical values of these roots to understand the range better.First, ( sqrt{7} ) is approximately 2.6458.So,[ frac{5 - 2sqrt{7}}{3} approx frac{5 - 2(2.6458)}{3} ][ approx frac{5 - 5.2916}{3} ][ approx frac{-0.2916}{3} ][ approx -0.0972 ]And,[ frac{5 + 2sqrt{7}}{3} approx frac{5 + 5.2916}{3} ][ approx frac{10.2916}{3} ][ approx 3.4305 ]So, ( k ) must be between approximately -0.0972 and 3.4305. But since ( k ) is an integer, let's list the integer values within this interval.The integers between -0.0972 and 3.4305 are 0, 1, 2, 3.So, possible integer values for ( k ) are 0, 1, 2, 3.Now, for each of these ( k ) values, we need to solve the quadratic equation ( kx^{2} - (3k + 1)x + (3k - 1) = 0 ) and check if the solutions ( x ) are irrational.Let's go through each ( k ):1. **Case 1: ( k = 0 )**Plugging ( k = 0 ) into the quadratic equation:[ 0x^{2} - (0 + 1)x + (0 - 1) = 0 ][ -x - 1 = 0 ][ -x = 1 ][ x = -1 ]So, ( x = -1 ) is a rational number. Therefore, this case doesn't contribute to irrational solutions.2. **Case 2: ( k = 1 )**Plugging ( k = 1 ) into the quadratic equation:[ 1x^{2} - (3(1) + 1)x + (3(1) - 1) = 0 ][ x^{2} - 4x + 2 = 0 ]Now, let's compute the discriminant ( D ):[ D = (-4)^{2} - 4(1)(2) ][ D = 16 - 8 ][ D = 8 ]Since ( D = 8 ) is positive and not a perfect square, the roots will be irrational. So, the solutions are:[ x = frac{4 pm sqrt{8}}{2} ][ x = frac{4 pm 2sqrt{2}}{2} ][ x = 2 pm sqrt{2} ]So, two irrational solutions here: ( 2 + sqrt{2} ) and ( 2 - sqrt{2} ).3. **Case 3: ( k = 2 )**Plugging ( k = 2 ) into the quadratic equation:[ 2x^{2} - (3(2) + 1)x + (3(2) - 1) = 0 ][ 2x^{2} - 7x + 5 = 0 ]Compute the discriminant ( D ):[ D = (-7)^{2} - 4(2)(5) ][ D = 49 - 40 ][ D = 9 ]Since ( D = 9 ) is a perfect square, the roots will be rational. Let's find them:[ x = frac{7 pm sqrt{9}}{4} ][ x = frac{7 pm 3}{4} ]So, the solutions are:[ x = frac{10}{4} = 2.5 ][ x = frac{4}{4} = 1 ]Both are rational numbers. Therefore, no irrational solutions in this case.4. **Case 4: ( k = 3 )**Plugging ( k = 3 ) into the quadratic equation:[ 3x^{2} - (3(3) + 1)x + (3(3) - 1) = 0 ][ 3x^{2} - 10x + 8 = 0 ]Compute the discriminant ( D ):[ D = (-10)^{2} - 4(3)(8) ][ D = 100 - 96 ][ D = 4 ]Again, ( D = 4 ) is a perfect square, so the roots are rational:[ x = frac{10 pm sqrt{4}}{6} ][ x = frac{10 pm 2}{6} ]So, the solutions are:[ x = frac{12}{6} = 2 ][ x = frac{8}{6} = frac{4}{3} ]Both are rational numbers. Therefore, no irrational solutions here either.So, summarizing:- ( k = 0 ): 1 rational solution- ( k = 1 ): 2 irrational solutions- ( k = 2 ): 2 rational solutions- ( k = 3 ): 2 rational solutionsTherefore, the only integer values of ( k ) that give irrational solutions for ( x ) are ( k = 1 ), which gives two irrational solutions: ( 2 + sqrt{2} ) and ( 2 - sqrt{2} ).Hence, the number of irrational numbers ( x ) is 2.**Final Answer**boxed{B}

💡Okay, so I've got this complex number problem here: z equals i raised to the 2017th power divided by (1 minus 2i). I need to find the imaginary part of z. Hmm, complex numbers can be a bit tricky, but let's take it step by step.First off, I remember that i is the imaginary unit, where i squared is -1. So, powers of i cycle every four exponents. That is, i^1 is i, i^2 is -1, i^3 is -i, and i^4 is 1, then it repeats. So, if I can figure out where 2017 falls in this cycle, I can simplify i^2017.To do that, I can divide 2017 by 4 and find the remainder. Let me calculate that: 2017 divided by 4. Well, 4 times 504 is 2016, so 2017 is 504 times 4 plus 1. That means the remainder is 1. So, i^2017 is the same as i^1, which is just i. Okay, that simplifies the numerator to i.So now, z is i divided by (1 minus 2i). To simplify this expression, I need to get rid of the imaginary unit in the denominator. I remember that to do this, I can multiply both the numerator and the denominator by the complex conjugate of the denominator. The complex conjugate of (1 - 2i) is (1 + 2i). So, let's do that.Multiplying the numerator and denominator by (1 + 2i), we get:z = [i * (1 + 2i)] / [(1 - 2i)(1 + 2i)]Let me compute the denominator first. Using the formula (a - b)(a + b) = a^2 - b^2, so here, a is 1 and b is 2i. Therefore, the denominator becomes:1^2 - (2i)^2 = 1 - 4i^2But i^2 is -1, so this becomes:1 - 4*(-1) = 1 + 4 = 5Okay, so the denominator simplifies to 5. Now, let's work on the numerator:i * (1 + 2i) = i*1 + i*2i = i + 2i^2Again, i^2 is -1, so:i + 2*(-1) = i - 2So, putting it all together, z is:(i - 2) / 5Which can be written as:(-2 + i) / 5Or, separating the real and imaginary parts:-2/5 + (1/5)iSo, the complex number z is -2/5 plus (1/5)i. The question asks for the imaginary part of z. The imaginary part is the coefficient of i, which is 1/5.Wait a second, let me double-check my steps to make sure I didn't make a mistake. Starting from i^2017, I correctly found the remainder when dividing by 4, which was 1, so i^2017 is i. Then, I multiplied numerator and denominator by the conjugate, which is the right approach. The denominator calculation gave me 5, which seems correct. The numerator, when expanded, gave me i - 2, which also seems right. So, the final expression is (-2 + i)/5, which breaks down into real part -2/5 and imaginary part 1/5.Therefore, the imaginary part of z is 1/5. Looking back at the options given:A: -2/5B: (1/5)iC: 1/5D: -1/5So, the correct answer should be option C: 1/5.Wait, hold on. The options include both numerical values and expressions with i. The question specifically asks for the imaginary part, which is a real number coefficient, not an imaginary number. So, option C is 1/5, which is correct, whereas option B is (1/5)i, which is the entire imaginary component but not just the coefficient. So, yeah, I think I got it right.Just to recap:1. Simplify i^2017: remainder 1 when divided by 4, so i^2017 = i.2. Multiply numerator and denominator by the conjugate (1 + 2i).3. Denominator becomes 5.4. Numerator becomes i - 2.5. Separate into real and imaginary parts: -2/5 + (1/5)i.6. Imaginary part is 1/5.Yep, that all checks out. I don't think I made any mistakes here. So, I'm confident that the answer is 1/5, which is option C.**Final Answer**boxed{C}

💡Alright, so I have this problem here: the equation of a line is given by y = x sin α + 1, where α is a real number. I need to determine the range of the angle of inclination of this line. The options are A, B, C, and D, each providing different intervals. Hmm, okay, let me think through this step by step.First, I remember that the angle of inclination of a line is the angle θ that the line makes with the positive direction of the x-axis, measured counterclockwise. This angle θ is between 0 and π radians (or 0 and 180 degrees). The slope of the line is equal to the tangent of this angle, so slope m = tan θ.Looking at the given equation, y = x sin α + 1, I can see that the slope of this line is sin α. So, m = sin α. Now, since α is a real number, sin α can take any value between -1 and 1. That means the slope m is in the interval [-1, 1].Okay, so m = tan θ, and m ∈ [-1, 1]. Therefore, tan θ ∈ [-1, 1]. Now, I need to find all angles θ between 0 and π where the tangent of θ is between -1 and 1.Let me recall the behavior of the tangent function. The tangent function, tan θ, is periodic with a period of π, and it has vertical asymptotes at θ = π/2 and θ = 3π/2, etc. But since θ is between 0 and π, I only need to consider θ in that interval.In the first quadrant, from 0 to π/2, tan θ increases from 0 to infinity. In the second quadrant, from π/2 to π, tan θ decreases from negative infinity to 0. So, tan θ is positive in the first quadrant and negative in the second quadrant.Given that tan θ ∈ [-1, 1], let's break this down:1. When tan θ is between 0 and 1 (inclusive), θ is between 0 and π/4. Because tan(π/4) = 1, and as θ increases from 0 to π/4, tan θ increases from 0 to 1.2. When tan θ is between -1 and 0 (inclusive), θ is between 3π/4 and π. Because tan(3π/4) = -1, and as θ increases from 3π/4 to π, tan θ increases from -1 to 0.So, combining these two intervals, θ must be in [0, π/4] ∪ [3π/4, π). Wait, why is it [3π/4, π) and not [3π/4, π]? Let me think. At θ = π, tan θ is undefined because tan π = 0, but actually, tan π is 0, which is within our range. Hmm, but π is the upper limit of the angle of inclination, so it's included. Wait, no, at θ = π, the line would be pointing directly to the left, which would have a slope of 0? Wait, no, tan π is 0, but that's actually at θ = 0 and θ = π, the line is horizontal. Wait, no, at θ = 0, it's pointing to the right, and at θ = π, it's pointing to the left, both with slope 0. But in our case, the slope can be negative, so when θ is approaching π from below, tan θ approaches 0 from the negative side.Wait, maybe I'm overcomplicating. Let me visualize the unit circle. For θ between 0 and π/2, tan θ is positive, increasing from 0 to infinity. For θ between π/2 and π, tan θ is negative, decreasing from 0 to negative infinity. So, when tan θ is between -1 and 0, θ is between 3π/4 and π. Because tan(3π/4) = -1, and as θ increases from 3π/4 to π, tan θ increases from -1 to 0.Therefore, the angles θ where tan θ is between -1 and 1 are θ ∈ [0, π/4] ∪ [3π/4, π). So, that would correspond to option C.Wait, but let me double-check. If θ is exactly π/4, tan θ is 1, which is included. If θ is exactly 3π/4, tan θ is -1, which is included. And at θ = 0, tan θ is 0, which is included. At θ approaching π, tan θ approaches 0 from the negative side, which is still within our range.So, the range of θ is [0, π/4] union [3π/4, π). That matches option C.But wait, let me think again. The problem says "the range of the angle of inclination." The angle of inclination is typically defined as the smallest non-negative angle between the positive x-axis and the line, measured counterclockwise. So, it's always between 0 and π, inclusive. So, our previous conclusion holds.Alternatively, sometimes people define the angle of inclination as strictly between 0 and π, not including π. But in our case, since tan θ approaches 0 as θ approaches π, and 0 is included in our slope range, I think θ = π is included because the line can be horizontal to the left, which would correspond to θ = π.But looking back at the options, option C is [0, π/4] ∪ [3π/4, π), which includes π/4 and 3π/4 but not π. Wait, why is that? Because at θ = π, the line is horizontal to the left, which would have a slope of 0. But in our case, the slope is sin α, which can be 0. So, θ = π is possible when sin α = 0, which occurs at α = 0, π, 2π, etc.Wait, but in the given equation, y = x sin α + 1, when sin α = 0, the equation becomes y = 1, which is a horizontal line. The angle of inclination for a horizontal line is 0 or π, depending on the direction. But in the definition, the angle of inclination is the smallest non-negative angle, so it's 0 for a horizontal line to the right, and π for a horizontal line to the left. But in reality, both correspond to the same line, just traversed in opposite directions.Hmm, this is getting a bit confusing. Maybe I should refer back to the standard definition. The angle of inclination is the angle between 0 and π such that tan θ equals the slope. So, for a horizontal line, tan θ = 0, which occurs at θ = 0 and θ = π. But since θ is between 0 and π, both are valid, but in reality, they represent the same line. So, perhaps θ = π is included as a valid angle of inclination.But in our case, the slope can be 0, so θ can be 0 or π. But in the options, option C includes [0, π/4] and [3π/4, π), which includes π/4 and 3π/4 but not π. Wait, but if θ can be π, then the interval should include π. But in option C, it's up to π, but not including π. Hmm, that's a bit conflicting.Wait, let me think again. If θ is exactly π, tan θ is 0, which is within our slope range. So, θ = π should be included. But in the interval [3π/4, π), π is not included. So, maybe the correct interval should be [0, π/4] ∪ [3π/4, π]. But that's not one of the options. Option C is [0, π/4] ∪ [3π/4, π), and option D is (0, π/4] ∪ [3π/4, π]. So, option C includes 0 and π/4, but not π, while option D excludes 0 but includes π/4 and 3π/4, and includes π.Wait, but if θ can be 0, which corresponds to a horizontal line to the right, and θ can be π, which corresponds to a horizontal line to the left, both with slope 0, then θ = 0 and θ = π should both be included. So, the interval should be [0, π/4] ∪ [3π/4, π]. But none of the options have that exactly. Option C is [0, π/4] ∪ [3π/4, π), which excludes π, and option D is (0, π/4] ∪ [3π/4, π], which excludes 0 but includes π.Hmm, this is tricky. Maybe I need to consider whether θ = 0 and θ = π are both included or not. If the line is horizontal, it can be considered to have an angle of inclination of 0 or π, depending on the direction. But in the definition, the angle of inclination is the smallest non-negative angle, so for a horizontal line, it's 0. However, if we consider the line extending in both directions, it can also be represented by θ = π. But in terms of the angle of inclination, it's typically defined as the smallest angle, so 0.But in our case, the slope can be 0, which corresponds to θ = 0 or θ = π. So, both should be included. But the options don't have both. Option C includes 0 but not π, and option D includes π but not 0. Wait, no, option C includes 0 and π/4, and up to π but not including π. Option D excludes 0 but includes π/4 and up to π.Wait, maybe I'm overcomplicating. Let me think about the possible values of θ. When sin α is positive, the slope is positive, so θ is between 0 and π/2. But since sin α is between -1 and 1, the slope is between -1 and 1. So, tan θ is between -1 and 1.So, tan θ ∈ [-1, 1]. Therefore, θ ∈ [0, π/4] ∪ [3π/4, π). Because tan θ is positive between 0 and π/4, and negative between 3π/4 and π.Wait, but at θ = π/4, tan θ = 1, which is included, and at θ = 3π/4, tan θ = -1, which is included. So, the intervals are closed at π/4 and 3π/4. But at θ = π, tan θ = 0, which is within our range, but in the interval [3π/4, π), π is not included. So, maybe the correct interval is [0, π/4] ∪ [3π/4, π). That would include all angles where tan θ is between -1 and 1, excluding θ = π because tan π is 0, which is within the range, but π is the upper limit.Wait, no, tan π is 0, which is within our slope range, so θ = π should be included. Therefore, the interval should be [0, π/4] ∪ [3π/4, π]. But that's not an option. The closest is option C: [0, π/4] ∪ [3π/4, π), which excludes π, and option D: (0, π/4] ∪ [3π/4, π], which excludes 0 but includes π.Hmm, maybe the answer is option C because at θ = π, the line is horizontal to the left, which is a valid angle of inclination, but in the interval [3π/4, π), π is excluded. Wait, but if θ can be π, then it should be included. Maybe the options have a typo, or perhaps I'm misunderstanding.Alternatively, perhaps the angle of inclination is defined as strictly less than π, so θ ∈ [0, π). In that case, θ = π is excluded, and the interval would be [0, π/4] ∪ [3π/4, π). That would make sense, and that's option C.Yes, I think that's it. The angle of inclination is typically defined as θ ∈ [0, π), meaning it doesn't include π. So, the interval would be [0, π/4] ∪ [3π/4, π), which is option C.