💡Alright, so I have this problem with two sets, A and B. Let me try to understand what each part is asking.First, set A is defined as all real numbers x such that x squared is less than or equal to 3x plus 10. So, A = {x | x² ≤ 3x + 10}. I need to figure out what this set looks like. Maybe I should solve the inequality x² ≤ 3x + 10.Let me rearrange the inequality: x² - 3x - 10 ≤ 0. Hmm, this is a quadratic inequality. To solve it, I can factor the quadratic or find its roots. Let me try factoring. The quadratic is x² - 3x - 10. I need two numbers that multiply to -10 and add to -3. Let's see, 2 and -5: 2 * (-5) = -10, and 2 + (-5) = -3. So, the factors are (x + 2)(x - 5) ≤ 0.Okay, so the inequality is (x + 2)(x - 5) ≤ 0. To solve this, I can use a sign chart. The critical points are x = -2 and x = 5. These divide the number line into three intervals: (-∞, -2), (-2, 5), and (5, ∞).Testing each interval:1. For x < -2, say x = -3: (-3 + 2)(-3 - 5) = (-1)(-8) = 8 > 0.2. For -2 < x < 5, say x = 0: (0 + 2)(0 - 5) = (2)(-5) = -10 < 0.3. For x > 5, say x = 6: (6 + 2)(6 - 5) = (8)(1) = 8 > 0.Since the inequality is ≤ 0, the solution includes the intervals where the product is negative or zero. So, the solution is [-2, 5]. Therefore, set A is all real numbers from -2 to 5, including both endpoints.Next, set B is defined as {x | a + 1 ≤ x ≤ 2a + 1}. So, B is an interval whose endpoints depend on the value of a. For part (1), a is given as 3. Let me compute B when a = 3.Substituting a = 3 into B: a + 1 = 3 + 1 = 4, and 2a + 1 = 2*3 + 1 = 7. So, B = [4, 7].Now, part (1) asks for the complement of A in ℝ, union with B. The complement of A, denoted as ∁ℝA, is all real numbers not in A. Since A is [-2, 5], the complement is (-∞, -2) union (5, ∞).So, ∁ℝA = (-∞, -2) ∪ (5, ∞). Now, we need to find (∁ℝA) ∪ B. Let's write that out: (-∞, -2) ∪ (5, ∞) ∪ [4, 7].Looking at these intervals, [4, 7] overlaps with (5, ∞) starting at 5, but [4, 7] also includes 4 to 5. So, combining them, the union becomes (-∞, -2) ∪ [4, ∞). Wait, is that correct?Wait, no. Let me think again. The complement of A is (-∞, -2) ∪ (5, ∞). B is [4, 7]. So, when we take the union, we have (-∞, -2) from the complement, and then from B we have [4, 7]. But (5, ∞) from the complement and [4, 7] from B overlap between 5 and 7. So, combining these, the union would be (-∞, -2) ∪ [4, ∞). Because [4, 7] covers up to 7, and (5, ∞) covers beyond 5, so together they cover from 4 to infinity.Wait, but does [4, 7] and (5, ∞) together make [4, ∞)? Let me check:- From 4 to 5, we have [4,5) from B and nothing from the complement.- From 5 to 7, we have [5,7] from B and (5, ∞) from the complement.- Beyond 7, we have (7, ∞) from the complement.So, combining all these, it's (-∞, -2) ∪ [4, ∞). Hmm, that seems correct.But wait, in the complement of A, it's (5, ∞), so when we take the union with [4,7], it's [4,7] union (5, ∞) which is [4, ∞). So yes, the union is (-∞, -2) ∪ [4, ∞).But wait, is that right? Because [4,7] includes 4 to 7, and (5, ∞) includes everything beyond 5. So together, they cover from 4 to infinity. So, yes, the union is (-∞, -2) ∪ [4, ∞).But let me double-check. If I have (-∞, -2) and then [4, ∞), is that correct? Because the complement of A is (-∞, -2) and (5, ∞), and B is [4,7]. So, when we take the union, it's (-∞, -2) combined with [4,7] and (5, ∞). So, [4,7] and (5, ∞) together make [4, ∞). So, yes, the union is (-∞, -2) ∪ [4, ∞).Okay, so for part (1), the answer is (-∞, -2) ∪ [4, ∞).Now, moving on to part (2). It says if A ∩ B = B, determine the range of the real number a.So, A ∩ B = B means that B is a subset of A. Because the intersection of A and B is B itself, which implies every element of B is also in A.So, B must be a subset of A. Remember, A is [-2, 5], and B is [a + 1, 2a + 1]. So, for B to be a subset of A, the entire interval [a + 1, 2a + 1] must lie within [-2, 5].Therefore, we need to ensure that:1. The lower bound of B, which is a + 1, is greater than or equal to the lower bound of A, which is -2.2. The upper bound of B, which is 2a + 1, is less than or equal to the upper bound of A, which is 5.So, we have two inequalities:a + 1 ≥ -2and2a + 1 ≤ 5Let me solve these inequalities.First inequality: a + 1 ≥ -2Subtract 1 from both sides: a ≥ -3Second inequality: 2a + 1 ≤ 5Subtract 1 from both sides: 2a ≤ 4Divide both sides by 2: a ≤ 2So, combining these two inequalities, we have -3 ≤ a ≤ 2.But wait, is that all? Let me think. Because B is defined as [a + 1, 2a + 1]. For B to be a valid interval, the lower bound must be less than or equal to the upper bound. So, a + 1 ≤ 2a + 1.Let me check that:a + 1 ≤ 2a + 1Subtract a from both sides: 1 ≤ a + 1Subtract 1 from both sides: 0 ≤ aSo, a must be greater than or equal to 0 for B to be a valid interval (i.e., not empty). Otherwise, if a < 0, then a + 1 > 2a + 1, and B would be an empty set.But in the condition A ∩ B = B, if B is empty, then A ∩ B is also empty, which equals B. So, in that case, the condition is satisfied.Therefore, we have two cases:Case 1: B is empty, which happens when a + 1 > 2a + 1, i.e., a < 0. In this case, A ∩ B = B is trivially true because both sides are empty sets.Case 2: B is not empty, which happens when a ≥ 0. In this case, for B to be a subset of A, we need -3 ≤ a ≤ 2. But since a ≥ 0 in this case, the range is 0 ≤ a ≤ 2.Combining both cases, the range of a is a < 2. Wait, let me think again.Wait, in Case 1, a < 0 satisfies the condition, and in Case 2, 0 ≤ a ≤ 2 also satisfies the condition. So, combining both, the range is a < 2.But wait, in Case 1, a < 0 is allowed, and in Case 2, 0 ≤ a ≤ 2 is allowed. So, overall, a can be any real number less than 2.But let me double-check. If a = 2, then B = [3, 5]. Since A is [-2, 5], B is [3,5], which is entirely within A. So, a = 2 is allowed.If a = -1, then B would be [0, -1], which is an empty set because 0 > -1. So, B is empty, and A ∩ B = B holds because both are empty.If a = 3, which is greater than 2, then B = [4,7]. But A is [-2,5], so B would be [4,7], which is not entirely within A because 7 > 5. So, A ∩ B would be [4,5], which is not equal to B. Therefore, a cannot be greater than 2.Similarly, if a = -4, then B would be [-3, -7], which is an empty set because -3 > -7. So, B is empty, and A ∩ B = B holds.Wait, but earlier I thought that for B to be non-empty, a must be ≥ 0. But actually, B is non-empty when a + 1 ≤ 2a + 1, which simplifies to a ≥ 0. So, for a < 0, B is empty.Therefore, the range of a is all real numbers a such that a < 2. Because for a < 0, B is empty and the condition holds, and for 0 ≤ a ≤ 2, B is non-empty and a subset of A.So, the range of a is (-∞, 2).But let me confirm with a = 2. If a = 2, B = [3,5], which is within A = [-2,5]. So, A ∩ B = B, which is [3,5]. So, it works.If a = 1.9, then B = [2.9, 4.8], which is within A.If a = -0.5, then B is [0.5, 0], which is empty, so A ∩ B = B holds.If a = 2.1, then B = [3.1, 5.2]. But A is [-2,5], so A ∩ B = [3.1,5], which is not equal to B = [3.1,5.2]. Therefore, a cannot be greater than 2.Hence, the range of a is all real numbers less than 2.

💡Okay, so I have this problem about a sequence {a_n} defined by a recurrence relation, and I need to find real numbers x and y (with x not equal to zero) such that another sequence involving b_n is constant. Let me try to unpack this step by step.First, let me write down the given information:- The sequence {a_n} starts with a_1 = a_2 = 1.- For n > 1, a_{n+1} is given by (n² a_n² + 5) / ((n² - 1) a_{n-1}).Then, we define another sequence {b_n} where each term is b_n = (n x + y) a_n. The question is whether there exist real numbers x and y (x ≠ 0) such that the sequence { (b_{n+2} + b_n) / b_{n+1} } is constant for all n.So, I need to find x and y such that (b_{n+2} + b_n) / b_{n+1} is the same for all n. Let's denote this constant as C. Therefore, for all n, we have:(b_{n+2} + b_n) / b_{n+1} = C.This implies that:b_{n+2} + b_n = C b_{n+1}.So, the sequence {b_n} satisfies a linear recurrence relation of order 2 with constant coefficients. That is, it's a second-order linear recurrence.Given that b_n = (n x + y) a_n, let's substitute this into the recurrence:(n+2)x + y) a_{n+2} + (n x + y) a_n = C ( (n+1)x + y ) a_{n+1}.So, we have:[(n+2)x + y] a_{n+2} + [n x + y] a_n = C [ (n+1)x + y ] a_{n+1}.Now, I need to express a_{n+2} in terms of a_{n+1} and a_n using the given recurrence relation for {a_n}.From the given, a_{n+1} = (n² a_n² + 5) / ( (n² - 1) a_{n-1} ). Similarly, a_{n+2} can be expressed as:a_{n+2} = ( (n+1)² a_{n+1}² + 5 ) / ( ( (n+1)² - 1 ) a_n ).Simplify the denominator: (n+1)² - 1 = n² + 2n + 1 - 1 = n² + 2n = n(n + 2).So, a_{n+2} = ( (n+1)² a_{n+1}² + 5 ) / ( n(n + 2) a_n ).Let me substitute this expression for a_{n+2} back into the equation involving b_n:[(n+2)x + y] * [ ( (n+1)² a_{n+1}² + 5 ) / ( n(n + 2) a_n ) ] + [n x + y] a_n = C [ (n+1)x + y ] a_{n+1}.Hmm, this looks a bit complicated. Let me see if I can simplify this equation.First, let's write out the first term:[(n+2)x + y] * [ ( (n+1)² a_{n+1}² + 5 ) / ( n(n + 2) a_n ) ].Let me factor out the constants:= [ (n+2)x + y ] / ( n(n + 2) a_n ) * [ (n+1)² a_{n+1}² + 5 ].Similarly, the second term is:[ n x + y ] a_n.So, putting it all together:[ (n+2)x + y ] / ( n(n + 2) a_n ) * [ (n+1)² a_{n+1}² + 5 ] + [ n x + y ] a_n = C [ (n+1)x + y ] a_{n+1}.This seems quite involved. Maybe I can multiply both sides by n(n + 2) a_n to eliminate the denominators:[ (n+2)x + y ] [ (n+1)² a_{n+1}² + 5 ] + [ n x + y ] a_n * n(n + 2) a_n = C [ (n+1)x + y ] a_{n+1} * n(n + 2) a_n.Simplify term by term.First term: [ (n+2)x + y ] [ (n+1)² a_{n+1}² + 5 ].Second term: [ n x + y ] * n(n + 2) a_n².Third term: C [ (n+1)x + y ] * n(n + 2) a_{n+1} a_n.So, the equation becomes:[ (n+2)x + y ] [ (n+1)² a_{n+1}² + 5 ] + [ n x + y ] n(n + 2) a_n² = C [ (n+1)x + y ] n(n + 2) a_{n+1} a_n.This is a quadratic equation in terms of a_{n+1} and a_n. It's quite complex, but perhaps we can find a relationship between x and y such that this equation holds for all n.Alternatively, maybe I can look for a pattern or a substitution that simplifies this.Wait, perhaps instead of substituting a_{n+2} directly, I can use the original recurrence relation to express a_{n+2} in terms of a_{n+1} and a_n, and then see if that helps.Given that a_{n+1} = (n² a_n² + 5) / ( (n² - 1) a_{n-1} ), perhaps I can find a relationship between a_{n+2} and a_{n+1}, a_n, and a_{n-1}.But that might complicate things further.Alternatively, maybe I can consider the ratio of consecutive terms or look for a telescoping product.Wait, let's think about the sequence {b_n}.Given that b_n = (n x + y) a_n, and we want (b_{n+2} + b_n) / b_{n+1} to be constant.Let me denote this constant as k. So, (b_{n+2} + b_n) / b_{n+1} = k.This implies that b_{n+2} + b_n = k b_{n+1}.So, the recurrence is b_{n+2} - k b_{n+1} + b_n = 0.This is a linear homogeneous recurrence relation with constant coefficients. The characteristic equation would be r² - k r + 1 = 0.The roots of this equation are r = [k ± sqrt(k² - 4)] / 2.Depending on the discriminant, the roots can be real and distinct, real and equal, or complex.But since we are dealing with sequences, unless the roots are real, the solutions can involve oscillations or exponential growth, which might not necessarily align with the given a_n sequence.But perhaps this is a detour. Maybe I can instead look for specific values of x and y that make the expression (b_{n+2} + b_n) / b_{n+1} independent of n.Let me try plugging in small values of n and see if I can find a pattern or equations for x and y.First, let's compute b_1, b_2, b_3, etc., using the given a_n.Given a_1 = 1, a_2 = 1.Compute a_3:a_3 = (2² a_2² + 5) / ( (2² - 1) a_1 ) = (4*1 + 5) / (3*1) = 9 / 3 = 3.Similarly, a_4:a_4 = (3² a_3² + 5) / ( (3² - 1) a_2 ) = (9*9 + 5) / (8*1) = (81 + 5)/8 = 86/8 = 43/4 = 10.75.a_5:a_5 = (4² a_4² + 5) / ( (4² - 1) a_3 ) = (16*(43/4)^2 + 5) / (15*3).Compute (43/4)^2 = (1849)/16.So, 16*(1849/16) = 1849.Thus, numerator is 1849 + 5 = 1854.Denominator is 15*3 = 45.So, a_5 = 1854 / 45 = 41.2.Wait, 1854 divided by 45: 45*41 = 1845, so 1854 - 1845 = 9, so 41 + 9/45 = 41 + 1/5 = 41.2. Yes.So, a_5 = 41.2, which is 206/5.Similarly, a_6:a_6 = (5² a_5² + 5) / ( (5² - 1) a_4 ) = (25*(206/5)^2 + 5) / (24*(43/4)).Compute (206/5)^2 = (206)^2 / 25 = 42436 / 25.25*(42436 / 25) = 42436.So, numerator is 42436 + 5 = 42441.Denominator: 24*(43/4) = 6*43 = 258.Thus, a_6 = 42441 / 258.Let me compute that: 258*164 = 258*(160 + 4) = 258*160 + 258*4 = 41280 + 1032 = 42312.Subtract from 42441: 42441 - 42312 = 129.So, 129 / 258 = 1/2.Thus, a_6 = 164 + 1/2 = 164.5.So, a_6 = 329/2.Alright, so we have:a_1 = 1a_2 = 1a_3 = 3a_4 = 43/4 = 10.75a_5 = 206/5 = 41.2a_6 = 329/2 = 164.5Now, let's compute b_n = (n x + y) a_n for n=1,2,3,4,5,6.Compute b_1 = (1 x + y) a_1 = (x + y)*1 = x + y.b_2 = (2 x + y) a_2 = (2x + y)*1 = 2x + y.b_3 = (3x + y) a_3 = (3x + y)*3 = 9x + 3y.b_4 = (4x + y) a_4 = (4x + y)*(43/4) = (4x + y)*(10.75).Similarly, b_5 = (5x + y)*(206/5) = (5x + y)*41.2.b_6 = (6x + y)*(329/2) = (6x + y)*164.5.Now, let's compute the ratios (b_{n+2} + b_n) / b_{n+1} for n=1,2,3,4.First, for n=1:(b_3 + b_1)/b_2 = (9x + 3y + x + y)/(2x + y) = (10x + 4y)/(2x + y).Similarly, for n=2:(b_4 + b_2)/b_3 = [ (4x + y)*(43/4) + (2x + y) ] / (9x + 3y).For n=3:(b_5 + b_3)/b_4 = [ (5x + y)*(206/5) + (9x + 3y) ] / [ (4x + y)*(43/4) ].For n=4:(b_6 + b_4)/b_5 = [ (6x + y)*(329/2) + (4x + y)*(43/4) ] / [ (5x + y)*(206/5) ].Since we want all these ratios to be equal to the same constant k, we can set up equations:(10x + 4y)/(2x + y) = k,[ (4x + y)*(43/4) + (2x + y) ] / (9x + 3y) = k,and so on.Let me compute these expressions step by step.First, for n=1:(10x + 4y)/(2x + y) = k.Let me denote this as Equation (1):10x + 4y = k (2x + y).Similarly, for n=2:Compute numerator: (4x + y)*(43/4) + (2x + y).Compute (4x + y)*(43/4) = (43/4)(4x + y) = 43x + (43/4)y.Add (2x + y): 43x + (43/4)y + 2x + y = (43x + 2x) + (43/4 y + y) = 45x + (47/4)y.Denominator: 9x + 3y.So, the ratio is (45x + (47/4)y)/(9x + 3y) = k.Simplify numerator and denominator:Numerator: 45x + (47/4)y = (180x + 47y)/4.Denominator: 9x + 3y = 3(3x + y).So, ratio becomes (180x + 47y)/4 divided by 3(3x + y) = (180x + 47y)/(12(3x + y)).Thus, Equation (2):(180x + 47y)/(12(3x + y)) = k.Similarly, for n=3:Compute numerator: (5x + y)*(206/5) + (9x + 3y).Compute (5x + y)*(206/5) = (206/5)(5x + y) = 206x + (206/5)y.Add (9x + 3y): 206x + (206/5)y + 9x + 3y = (206x + 9x) + (206/5 y + 3y) = 215x + (206/5 + 15/5)y = 215x + (221/5)y.Denominator: (4x + y)*(43/4) = (43/4)(4x + y) = 43x + (43/4)y.So, the ratio is (215x + (221/5)y)/(43x + (43/4)y) = k.Simplify numerator and denominator:Numerator: 215x + (221/5)y = (1075x + 221y)/5.Denominator: 43x + (43/4)y = (172x + 43y)/4.So, ratio becomes (1075x + 221y)/5 divided by (172x + 43y)/4 = (1075x + 221y)/5 * 4/(172x + 43y) = (4300x + 884y)/(860x + 215y).Simplify numerator and denominator by dividing numerator and denominator by 5:Numerator: 4300x + 884y = 5*(860x + 176.8y). Wait, maybe not helpful.Alternatively, factor numerator and denominator:Numerator: 4300x + 884y = 4*(1075x + 221y).Denominator: 860x + 215y = 5*(172x + 43y).Wait, that's the same as before. Maybe not helpful.Alternatively, let's write the ratio as:(4300x + 884y)/(860x + 215y) = k.Let me note this as Equation (3):4300x + 884y = k (860x + 215y).Similarly, for n=4:Compute numerator: (6x + y)*(329/2) + (4x + y)*(43/4).Compute (6x + y)*(329/2) = (329/2)(6x + y) = 987x + (329/2)y.Compute (4x + y)*(43/4) = (43/4)(4x + y) = 43x + (43/4)y.Add these together: 987x + (329/2)y + 43x + (43/4)y = (987x + 43x) + (329/2 y + 43/4 y) = 1030x + (658/4 + 43/4)y = 1030x + (701/4)y.Denominator: (5x + y)*(206/5) = (206/5)(5x + y) = 206x + (206/5)y.So, the ratio is (1030x + (701/4)y)/(206x + (206/5)y) = k.Simplify numerator and denominator:Numerator: 1030x + (701/4)y = (4120x + 701y)/4.Denominator: 206x + (206/5)y = (1030x + 206y)/5.So, ratio becomes (4120x + 701y)/4 divided by (1030x + 206y)/5 = (4120x + 701y)/4 * 5/(1030x + 206y) = (20600x + 3505y)/(4120x + 824y).Simplify numerator and denominator:Numerator: 20600x + 3505y = 5*(4120x + 701y).Denominator: 4120x + 824y = 4*(1030x + 206y).Wait, that's the same as before. Alternatively, factor:Numerator: 20600x + 3505y = 5*(4120x + 701y).Denominator: 4120x + 824y = 4*(1030x + 206y).Hmm, not sure if that helps.Alternatively, write as:(20600x + 3505y)/(4120x + 824y) = k.Let me denote this as Equation (4):20600x + 3505y = k (4120x + 824y).Now, we have four equations:Equation (1): 10x + 4y = k (2x + y).Equation (2): 180x + 47y = 12k (3x + y).Equation (3): 4300x + 884y = k (860x + 215y).Equation (4): 20600x + 3505y = k (4120x + 824y).This seems like a lot, but perhaps we can solve for k from the first equation and substitute into the others.From Equation (1):10x + 4y = k (2x + y).Solve for k:k = (10x + 4y)/(2x + y).Similarly, from Equation (2):180x + 47y = 12k (3x + y).Substitute k from Equation (1):180x + 47y = 12*(10x + 4y)/(2x + y)*(3x + y).Let me write this as:180x + 47y = [12*(10x + 4y)*(3x + y)] / (2x + y).Multiply both sides by (2x + y):(180x + 47y)(2x + y) = 12*(10x + 4y)*(3x + y).Let me expand both sides.Left side:(180x + 47y)(2x + y) = 180x*2x + 180x*y + 47y*2x + 47y*y = 360x² + 180xy + 94xy + 47y² = 360x² + 274xy + 47y².Right side:12*(10x + 4y)*(3x + y) = 12*[10x*3x + 10x*y + 4y*3x + 4y*y] = 12*(30x² + 10xy + 12xy + 4y²) = 12*(30x² + 22xy + 4y²) = 360x² + 264xy + 48y².So, set left side equal to right side:360x² + 274xy + 47y² = 360x² + 264xy + 48y².Subtract 360x² from both sides:274xy + 47y² = 264xy + 48y².Bring all terms to left side:274xy - 264xy + 47y² - 48y² = 0 => 10xy - y² = 0.Factor:y(10x - y) = 0.So, either y = 0 or 10x - y = 0 => y = 10x.But x ≠ 0, so possible solutions are y = 0 or y = 10x.Let me consider both cases.Case 1: y = 0.Then, from Equation (1):k = (10x + 4*0)/(2x + 0) = 10x / 2x = 5.So, k = 5.Now, let's check if this holds for Equation (3) and Equation (4).From Equation (3):4300x + 884y = k (860x + 215y).With y=0, this becomes:4300x = 5*(860x).Compute RHS: 5*860x = 4300x.So, 4300x = 4300x, which holds.Similarly, Equation (4):20600x + 3505y = k (4120x + 824y).With y=0, this becomes:20600x = 5*(4120x).Compute RHS: 5*4120x = 20600x.So, 20600x = 20600x, which holds.Therefore, y=0 and k=5 is a solution.Case 2: y = 10x.Then, from Equation (1):k = (10x + 4y)/(2x + y) = (10x + 4*10x)/(2x + 10x) = (10x + 40x)/(12x) = 50x / 12x = 50/12 = 25/6 ≈ 4.1667.So, k = 25/6.Now, let's check if this holds for Equation (3) and Equation (4).From Equation (3):4300x + 884y = k (860x + 215y).Substitute y=10x and k=25/6:Left side: 4300x + 884*10x = 4300x + 8840x = 13140x.Right side: (25/6)*(860x + 215*10x) = (25/6)*(860x + 2150x) = (25/6)*(3010x) = (25*3010x)/6 = 75250x/6 ≈ 12541.6667x.But 13140x ≈ 12541.6667x? No, 13140 ≠ 12541.6667.Therefore, this does not hold. Hence, y=10x is not a valid solution.Thus, the only solution is y=0 and k=5.Therefore, x can be any non-zero real number, and y=0.But wait, the problem asks for real numbers x and y with x ≠ 0. So, x can be any non-zero real, and y=0.But let me check if this works for the original recurrence.Given that y=0, b_n = n x a_n.Then, the ratio (b_{n+2} + b_n)/b_{n+1} = ( (n+2)x a_{n+2} + n x a_n ) / ( (n+1)x a_{n+1} ) = [ (n+2) a_{n+2} + n a_n ] / ( (n+1) a_{n+1} ).We need this to be constant, which we found to be 5.So, let's verify this with the values we have.For n=1:[ (1+2)a_3 + 1*a_1 ] / (2 a_2 ) = (3*3 + 1*1)/(2*1) = (9 + 1)/2 = 10/2 = 5. Correct.For n=2:[ (2+2)a_4 + 2 a_2 ] / (3 a_3 ) = (4*(43/4) + 2*1)/ (3*3) = (43 + 2)/9 = 45/9 = 5. Correct.For n=3:[ (3+2)a_5 + 3 a_3 ] / (4 a_4 ) = (5*(206/5) + 3*3)/ (4*(43/4)) = (206 + 9)/43 = 215/43 = 5. Correct.For n=4:[ (4+2)a_6 + 4 a_4 ] / (5 a_5 ) = (6*(329/2) + 4*(43/4))/ (5*(206/5)) = (987 + 43)/206 = 1030/206 = 5. Correct.So, it works for these values. Therefore, y=0 and x can be any non-zero real number.But the problem asks for real numbers x and y (with x ≠ 0). So, x can be any non-zero real, and y=0.But perhaps the simplest solution is x=1 and y=0, as any scalar multiple would work, but since x is arbitrary, we can set x=1 for simplicity.Thus, the answer is x=1 and y=0.

💡First, I'll define the variables for the ages of Michael (M), his older brother (O), and his younger brother (Y).Given that the younger brother is 5 years old, I'll set Y = 5.The combined age of all three brothers is 28, so I can write the equation:M + O + 5 = 28Simplifying this, I get:M + O = 23Next, I'll use the information that the older brother is 1 year older than twice Michael's age when Michael was a year younger. This translates to:O = 2(M - 1) + 1Simplifying this, I get:O = 2M - 1Now, I'll substitute the expression for O into the equation M + O = 23:M + (2M - 1) = 23Combining like terms, I get:3M - 1 = 23Adding 1 to both sides:3M = 24Dividing by 3:M = 8With Michael's age known, I'll find the older brother's age:O = 2(8) - 1 = 15Finally, to find the fraction of the older brother's age that represents the younger brother's age, I'll calculate:Fraction = Y / O = 5 / 15 = 1/3

💡First, I need to calculate the average price for each month by adding the minimum and maximum prices and dividing by two.For January:[text{Average}_{text{Jan}} = frac{30 + 34}{2} = 32]For February:[text{Average}_{text{Feb}} = frac{37 + 42}{2} = 39.5]For March:[text{Average}_{text{Mar}} = frac{32 + 34}{2} = 33]For April:[text{Average}_{text{Apr}} = frac{35 + 39}{2} = 37]For May:[text{Average}_{text{May}} = frac{33 + 33}{2} = 33]For June:[text{Average}_{text{Jun}} = frac{29 + 31}{2} = 30]Next, I'll calculate the differences in average prices between consecutive months to identify the price drops.From January to February:[Delta_{text{Jan-Feb}} = 39.5 - 32 = 7.5 quad (text{Increase})]From February to March:[Delta_{text{Feb-Mar}} = 33 - 39.5 = -6.5 quad (text{Drop})]From March to April:[Delta_{text{Mar-Apr}} = 37 - 33 = 4 quad (text{Increase})]From April to May:[Delta_{text{Apr-May}} = 33 - 37 = -4 quad (text{Drop})]From May to June:[Delta_{text{May-Jun}} = 30 - 33 = -3 quad (text{Drop})]Among these differences, the largest price drop is -6.5, which occurs from February to March.

💡Okay, so I have this problem here: the line y = x + a is tangent to the curve y = ln x. I need to find the value of a. Hmm, let me think about how to approach this.First, I remember that for a line to be tangent to a curve at a point, two conditions must be satisfied. One, the line and the curve must intersect at that point. Two, the slopes of the line and the curve must be equal at that point. So, I need to find a point (x, y) where both the curve and the line pass through, and their derivatives at that point are the same.Alright, let's denote the point of tangency as (m, n). Since this point lies on the curve y = ln x, we can say that n = ln m. Also, since it lies on the line y = x + a, we can say that n = m + a. So, from these two equations, we have ln m = m + a. That's one equation.Now, the second condition is about the slopes. The slope of the line y = x + a is 1 because the coefficient of x is 1. The slope of the curve y = ln x at any point x is given by its derivative. Let me compute that derivative. The derivative of ln x is 1/x. So, at the point x = m, the slope of the curve is 1/m. Since the slopes must be equal for the line to be tangent, we have 1/m = 1. Wait, so 1/m equals 1? That means m must be 1 because 1 divided by m is 1 only when m is 1. So, m = 1. Now that I have m = 1, I can find n. Since n = ln m, and m is 1, ln 1 is 0. So, n = 0. Now, going back to the equation from the line: n = m + a. We know n is 0 and m is 1, so plugging those in: 0 = 1 + a. Solving for a, we subtract 1 from both sides: a = -1.Let me double-check to make sure I didn't make any mistakes. If a is -1, then the line is y = x - 1. Let's see if this line is tangent to y = ln x. We already found the point of tangency is at (1, 0). Let's verify that. Plugging x = 1 into y = ln x gives y = 0, which matches. Plugging x = 1 into y = x - 1 gives y = 0 as well, so the point (1, 0) is indeed on both the curve and the line. Now, checking the slopes. The derivative of y = ln x at x = 1 is 1/1 = 1, which is the same as the slope of the line. So, both conditions are satisfied. Therefore, I'm confident that a is -1.**Final Answer**The value of ( a ) is boxed{-1}.

💡Alright, so I have this function f(x) = 2cos²x - 2√3 sinx cosx - 1, and I need to do two things: first, calculate f(-π/12), and second, find the smallest positive period and the intervals where the function is monotonically increasing. Hmm, okay, let's start with part (1).First, I remember that cos²x can be rewritten using a double-angle identity. The identity is cos²x = (1 + cos2x)/2. So, if I substitute that into the function, I get:f(x) = 2*(1 + cos2x)/2 - 2√3 sinx cosx - 1.Simplifying that, the 2 and the denominator 2 cancel out, so it becomes:f(x) = (1 + cos2x) - 2√3 sinx cosx - 1.Then, the 1 and -1 cancel each other out, so we're left with:f(x) = cos2x - 2√3 sinx cosx.Wait, I also remember that sinx cosx can be expressed as (sin2x)/2. So, substituting that in:f(x) = cos2x - 2√3*(sin2x)/2.Simplifying further, the 2 and 2 cancel out, so:f(x) = cos2x - √3 sin2x.Hmm, this looks like a combination of cosine and sine terms with the same argument, 2x. I think I can write this as a single cosine function using the amplitude-phase form. The general form is A cos(θ + φ), where A is the amplitude and φ is the phase shift.To find A, I use the formula A = √(a² + b²), where a is the coefficient of cosθ and b is the coefficient of sinθ. Here, a = 1 and b = -√3, so:A = √(1² + (√3)²) = √(1 + 3) = √4 = 2.Okay, so the amplitude is 2. Now, to find the phase shift φ, I use the formula tanφ = b/a. Here, tanφ = (-√3)/1 = -√3. So, φ is the angle whose tangent is -√3. I know that tan(π/3) = √3, so tan(-π/3) = -√3. Therefore, φ = -π/3.So, putting it all together, the function can be written as:f(x) = 2 cos(2x - π/3).Wait, hold on, actually, when I use the formula A cos(θ + φ), if the coefficient of sin is negative, it might be better to write it as 2 cos(2x + π/3) with a positive phase shift. Let me double-check that.If I have f(x) = cos2x - √3 sin2x, and I want to write it as 2 cos(2x + φ), then expanding that gives:2 cos(2x + φ) = 2 cos2x cosφ - 2 sin2x sinφ.Comparing this to f(x) = cos2x - √3 sin2x, we can equate coefficients:2 cosφ = 1 => cosφ = 1/2,and-2 sinφ = -√3 => sinφ = √3/2.So, cosφ = 1/2 and sinφ = √3/2. That corresponds to φ = π/3. Therefore, f(x) = 2 cos(2x + π/3).Wait, that seems correct. So, f(x) = 2 cos(2x + π/3). Okay, so that's the simplified form.Now, for part (1), I need to calculate f(-π/12). Let's substitute x = -π/12 into the simplified function:f(-π/12) = 2 cos(2*(-π/12) + π/3).Simplify the argument inside the cosine:2*(-π/12) = -π/6,So, -π/6 + π/3 = (-π/6 + 2π/6) = π/6.Therefore, f(-π/12) = 2 cos(π/6).I know that cos(π/6) is √3/2, so:f(-π/12) = 2*(√3/2) = √3.Okay, that seems straightforward.Now, moving on to part (2). I need to find the smallest positive period and the intervals where f(x) is monotonically increasing.First, the period. The function f(x) is expressed as 2 cos(2x + π/3). The general form of a cosine function is A cos(Bx + C), and its period is 2π/|B|. Here, B is 2, so the period is 2π/2 = π. Therefore, the smallest positive period is π.Next, finding the intervals where f(x) is monotonically increasing. Since f(x) is a cosine function, it's periodic and has intervals where it increases and decreases. The function 2 cos(2x + π/3) will have its maximum and minimum points, and between those points, it will increase or decrease.To find where it's increasing, we can take the derivative and find where it's positive.Let's compute f'(x):f(x) = 2 cos(2x + π/3),f'(x) = -4 sin(2x + π/3).We need to find where f'(x) > 0, which means:-4 sin(2x + π/3) > 0,Divide both sides by -4 (remembering to flip the inequality sign):sin(2x + π/3) < 0.So, we need to find the intervals where sin(2x + π/3) < 0.The sine function is negative in the intervals (π + 2πk, 2π + 2πk) for integers k.So, let's solve for x:π + 2πk < 2x + π/3 < 2π + 2πk,Subtract π/3 from all parts:π - π/3 + 2πk < 2x < 2π - π/3 + 2πk,Simplify π - π/3 = 2π/3, and 2π - π/3 = 5π/3,So:2π/3 + 2πk < 2x < 5π/3 + 2πk,Divide all parts by 2:π/3 + πk < x < 5π/6 + πk.Therefore, the function f(x) is increasing on the intervals (π/3 + πk, 5π/6 + πk) for integers k.But wait, let me double-check that. Since the derivative is negative when sin(2x + π/3) is positive, and positive when sin(2x + π/3) is negative. So, f'(x) > 0 when sin(2x + π/3) < 0, which occurs in the intervals I found.But let me think about the standard intervals for sine being negative. Sine is negative in (π, 2π) plus multiples of 2π. So, solving for 2x + π/3 in that interval:π < 2x + π/3 < 2π,Subtract π/3:2π/3 < 2x < 5π/3,Divide by 2:π/3 < x < 5π/6.So, the function is increasing on (π/3, 5π/6), and since the period is π, this interval repeats every π units. So, the general solution is (π/3 + πk, 5π/6 + πk) for all integers k.Therefore, the intervals of monotonic increase are [π/3 + πk, 5π/6 + πk] for integers k.Wait, but in the original function, f(x) is continuous and differentiable everywhere, so the intervals where it's increasing are open intervals, but since the question asks for intervals of monotonic increase, it's acceptable to write them as closed intervals because the function is increasing on those closed intervals as well.So, summarizing:1. f(-π/12) = √3.2. The smallest positive period is π, and the function is monotonically increasing on intervals [π/3 + πk, 5π/6 + πk] for all integers k.I think that's it. Let me just recap to make sure I didn't make any mistakes.Starting with f(x) = 2cos²x - 2√3 sinx cosx -1, I used double-angle identities to rewrite it as 2 cos(2x + π/3). Then, for part (1), substituting x = -π/12 gave me √3. For part (2), the period is π, and by taking the derivative, I found the intervals where the function is increasing, which are [π/3 + πk, 5π/6 + πk] for integers k.Yeah, that seems correct.