answer：Okay, so I have this problem where I need to find the value of ( b ) for the function ( f(x) = x^3 + 2a x^2 + b x + a^2 ). The problem states that the function has an extreme value of 17 at ( x = 2 ). Hmm, okay. Let me try to figure this out step by step.First, I remember that extreme values of a function occur where the first derivative is zero or undefined. Since this is a polynomial function, its derivative will also be a polynomial, which is defined everywhere. So, I just need to find where the derivative is zero.Let me compute the first derivative of ( f(x) ). The derivative of ( x^3 ) is ( 3x^2 ), the derivative of ( 2a x^2 ) is ( 4a x ), the derivative of ( b x ) is ( b ), and the derivative of ( a^2 ) is 0 because it's a constant. So, putting it all together, the first derivative ( f'(x) ) is:[ f'(x) = 3x^2 + 4a x + b ]Okay, so at the extreme point ( x = 2 ), the derivative should be zero. That gives me the equation:[ f'(2) = 3(2)^2 + 4a(2) + b = 0 ]Let me compute that:[ 3(4) + 8a + b = 0 ][ 12 + 8a + b = 0 ]So, that's my first equation:[ 8a + b = -12 quad text{(Equation 1)} ]Now, the problem also says that the function has an extreme value of 17 at ( x = 2 ). That means when I plug ( x = 2 ) into the original function ( f(x) ), I should get 17. Let me write that out:[ f(2) = (2)^3 + 2a(2)^2 + b(2) + a^2 = 17 ]Calculating each term:- ( (2)^3 = 8 )- ( 2a(2)^2 = 2a times 4 = 8a )- ( b(2) = 2b )- ( a^2 ) remains as is.So, putting it all together:[ 8 + 8a + 2b + a^2 = 17 ]Let me rearrange this equation:[ a^2 + 8a + 2b + 8 = 17 ][ a^2 + 8a + 2b = 9 quad text{(Equation 2)} ]Now, I have two equations:1. ( 8a + b = -12 ) (Equation 1)2. ( a^2 + 8a + 2b = 9 ) (Equation 2)I need to solve this system of equations to find the values of ( a ) and ( b ). Let me see how to approach this.From Equation 1, I can express ( b ) in terms of ( a ):[ b = -12 - 8a ]Now, I can substitute this expression for ( b ) into Equation 2. Let me do that:[ a^2 + 8a + 2(-12 - 8a) = 9 ]Let me simplify this step by step.First, distribute the 2 into the parentheses:[ a^2 + 8a - 24 - 16a = 9 ]Combine like terms:- ( 8a - 16a = -8a )- So, the equation becomes:[ a^2 - 8a - 24 = 9 ]Now, subtract 9 from both sides to set the equation to zero:[ a^2 - 8a - 24 - 9 = 0 ][ a^2 - 8a - 33 = 0 ]Okay, so now I have a quadratic equation in terms of ( a ):[ a^2 - 8a - 33 = 0 ]I need to solve this quadratic equation. Let me try factoring first. I'm looking for two numbers that multiply to -33 and add up to -8. Hmm, let's see:- Factors of 33 are 1 and 33, 3 and 11.Looking for a pair that adds up to -8. Let's see:- 3 and -11: 3 + (-11) = -8. Yes, that works.So, the quadratic factors as:[ (a + 3)(a - 11) = 0 ]Setting each factor equal to zero gives the solutions:1. ( a + 3 = 0 ) => ( a = -3 )2. ( a - 11 = 0 ) => ( a = 11 )Okay, so ( a ) can be either -3 or 11. Now, let's find the corresponding ( b ) values using the expression we had earlier:[ b = -12 - 8a ]First, for ( a = -3 ):[ b = -12 - 8(-3) ][ b = -12 + 24 ][ b = 12 ]Next, for ( a = 11 ):[ b = -12 - 8(11) ][ b = -12 - 88 ][ b = -100 ]So, we have two possible solutions:1. ( a = -3 ), ( b = 12 )2. ( a = 11 ), ( b = -100 )But wait, the problem mentions that the function has an extreme value at ( x = 2 ). I need to make sure that this point is indeed an extreme value, which means it should be a local maximum or minimum. To confirm this, I can check the second derivative or analyze the behavior around ( x = 2 ).Let me compute the second derivative ( f''(x) ):The first derivative was ( f'(x) = 3x^2 + 4a x + b ), so the second derivative is:[ f''(x) = 6x + 4a ]Now, evaluate the second derivative at ( x = 2 ):[ f''(2) = 6(2) + 4a = 12 + 4a ]If ( f''(2) > 0 ), then ( x = 2 ) is a local minimum. If ( f''(2) < 0 ), it's a local maximum. If ( f''(2) = 0 ), the test is inconclusive.Let's compute ( f''(2) ) for both values of ( a ):1. For ( a = -3 ):[ f''(2) = 12 + 4(-3) = 12 - 12 = 0 ]Hmm, the second derivative is zero, which means the test is inconclusive. So, ( x = 2 ) might not be an extreme value in this case, or it could be a point of inflection.2. For ( a = 11 ):[ f''(2) = 12 + 4(11) = 12 + 44 = 56 ]Since 56 is greater than 0, this means ( x = 2 ) is a local minimum when ( a = 11 ).Therefore, the case where ( a = -3 ) and ( b = 12 ) does not guarantee an extreme value at ( x = 2 ), because the second derivative is zero, making the test inconclusive. On the other hand, when ( a = 11 ) and ( b = -100 ), the second derivative is positive, confirming a local minimum at ( x = 2 ).Hence, the valid solution is ( a = 11 ) and ( b = -100 ).Wait, just to be thorough, let me plug ( a = -3 ) and ( b = 12 ) back into the original function and see if ( x = 2 ) is indeed an extreme value.So, if ( a = -3 ) and ( b = 12 ), the function becomes:[ f(x) = x^3 + 2(-3)x^2 + 12x + (-3)^2 ][ f(x) = x^3 - 6x^2 + 12x + 9 ]Let me compute the first derivative:[ f'(x) = 3x^2 - 12x + 12 ]At ( x = 2 ):[ f'(2) = 3(4) - 12(2) + 12 = 12 - 24 + 12 = 0 ]So, the derivative is zero, which is consistent. But since the second derivative is zero, let's check the behavior around ( x = 2 ) to see if it's an extreme.Let me pick points just less than 2 and just more than 2 to see if the function changes direction.Let's choose ( x = 1.9 ) and ( x = 2.1 ).Compute ( f'(1.9) ):[ f'(1.9) = 3(1.9)^2 - 12(1.9) + 12 ]First, ( (1.9)^2 = 3.61 )So,[ 3(3.61) = 10.83 ][ 12(1.9) = 22.8 ]So,[ f'(1.9) = 10.83 - 22.8 + 12 = 0.03 ]Similarly, compute ( f'(2.1) ):[ f'(2.1) = 3(2.1)^2 - 12(2.1) + 12 ]First, ( (2.1)^2 = 4.41 )So,[ 3(4.41) = 13.23 ][ 12(2.1) = 25.2 ]So,[ f'(2.1) = 13.23 - 25.2 + 12 = 0.03 ]Hmm, both ( f'(1.9) ) and ( f'(2.1) ) are positive (approximately 0.03). That means the derivative is positive just before and just after ( x = 2 ), which suggests that ( x = 2 ) is not a local maximum or minimum, but rather a point of inflection. So, in this case, even though the first derivative is zero at ( x = 2 ), it's not an extreme value because the function doesn't change direction there.Therefore, the case ( a = -3 ), ( b = 12 ) does not satisfy the condition of having an extreme value at ( x = 2 ). Hence, the only valid solution is ( a = 11 ), ( b = -100 ).Just to double-check, let me plug ( a = 11 ) and ( b = -100 ) back into the original function and compute ( f(2) ) to ensure it equals 17.So, the function becomes:[ f(x) = x^3 + 2(11)x^2 + (-100)x + (11)^2 ][ f(x) = x^3 + 22x^2 - 100x + 121 ]Compute ( f(2) ):[ f(2) = (2)^3 + 22(2)^2 - 100(2) + 121 ][ f(2) = 8 + 22(4) - 200 + 121 ][ f(2) = 8 + 88 - 200 + 121 ][ f(2) = (8 + 88) + (-200 + 121) ][ f(2) = 96 - 79 ][ f(2) = 17 ]Perfect, that's correct. So, with ( a = 11 ) and ( b = -100 ), the function does have an extreme value of 17 at ( x = 2 ).Therefore, the value of ( b ) is -100.**Final Answer**The value of ( b ) is boxed{-100}.

answer：Alright, so I have this problem here: find the derivative of y with respect to x, where y is equal to the natural logarithm of 1 over the square root of (1 plus x squared). That is, y = ln(1 / sqrt(1 + x²)). Hmm, okay, let's see how I can tackle this.First off, I remember that when taking derivatives of logarithmic functions, the chain rule is usually involved. The chain rule is when you have a function inside another function, and you need to take the derivative of the outer function and multiply it by the derivative of the inner function. So, in this case, the outer function is the natural logarithm, and the inner function is 1 over the square root of (1 + x²).Let me write that down to make it clearer. Let’s say u is equal to 1 over sqrt(1 + x²). Then y is ln(u). So, y = ln(u), and u = 1 / sqrt(1 + x²). Okay, that seems manageable.Now, to find dy/dx, I need to use the chain rule, which says that dy/dx is equal to dy/du multiplied by du/dx. So, first, I'll find dy/du, which is the derivative of ln(u) with respect to u. I remember that the derivative of ln(u) is 1/u. So, dy/du is 1/u.Next, I need to find du/dx, which is the derivative of 1 over sqrt(1 + x²) with respect to x. Hmm, okay, let's see. 1 over sqrt(1 + x²) can be rewritten as (1 + x²)^(-1/2). That might make it easier to differentiate using the power rule.So, if u = (1 + x²)^(-1/2), then du/dx is the derivative of that. The power rule says that the derivative of u^n is n*u^(n-1)*du/dx. So, applying that here, n is -1/2, so the derivative is (-1/2)*(1 + x²)^(-3/2)*d/dx(1 + x²).Now, the derivative of (1 + x²) with respect to x is 2x. So, putting it all together, du/dx is (-1/2)*(1 + x²)^(-3/2)*2x. Hmm, the 2 and the 1/2 should cancel out, right? So, that simplifies to -x*(1 + x²)^(-3/2).So, du/dx is -x over (1 + x²)^(3/2). Okay, got that.Now, going back to the chain rule, dy/dx is dy/du times du/dx. We already have dy/du as 1/u, and u is 1 over sqrt(1 + x²). So, 1/u is just sqrt(1 + x²).Therefore, dy/dx is sqrt(1 + x²) times (-x over (1 + x²)^(3/2)). Let me write that out: sqrt(1 + x²) * (-x) / (1 + x²)^(3/2).Hmm, simplifying that, sqrt(1 + x²) is the same as (1 + x²)^(1/2). So, we have (1 + x²)^(1/2) times (-x) over (1 + x²)^(3/2). When you multiply these together, you add the exponents. So, (1 + x²)^(1/2 - 3/2) is (1 + x²)^(-1).So, that simplifies to (-x) / (1 + x²). Therefore, the derivative dy/dx is -x over (1 + x²).Wait, let me double-check that. Starting from the beginning: y = ln(1 / sqrt(1 + x²)). Another way to write that is y = ln(1) - ln(sqrt(1 + x²)). Since ln(a/b) is ln(a) - ln(b). And ln(1) is 0, so y simplifies to -ln(sqrt(1 + x²)).Then, since sqrt(1 + x²) is (1 + x²)^(1/2), y is - (1/2) ln(1 + x²). Now, taking the derivative of that, dy/dx is - (1/2) times the derivative of ln(1 + x²). The derivative of ln(1 + x²) is (2x)/(1 + x²). So, dy/dx is - (1/2)*(2x)/(1 + x²), which simplifies to -x/(1 + x²). Okay, that matches what I got earlier. So, that’s reassuring.I think I’ve got it right then. The derivative of y with respect to x is -x divided by (1 + x²). It's always good to check using a different method to make sure I didn't make any mistakes. Both methods led me to the same answer, so I feel confident about it.Just to recap, the steps were:1. Recognize that y is a composition of functions, so the chain rule applies.2. Let u be the inner function, which is 1 over sqrt(1 + x²).3. Express y in terms of u, which is ln(u).4. Differentiate y with respect to u, getting 1/u.5. Differentiate u with respect to x, which involved rewriting u as (1 + x²)^(-1/2) and applying the power rule.6. Multiply the derivatives together using the chain rule.7. Simplify the resulting expression.8. Double-check the result by rewriting the original function and differentiating again.This process helped me ensure that I didn't make any errors and that the answer makes sense. It's also a good practice to verify results using alternative methods when possible, especially when dealing with calculus problems that can sometimes be tricky.Another thing I noticed is that the function y = ln(1 / sqrt(1 + x²)) can be simplified before taking the derivative, which might make the differentiation process a bit more straightforward. By using logarithmic properties, I was able to rewrite y as - (1/2) ln(1 + x²), which is simpler to differentiate.This experience reinforces the importance of being comfortable with algebraic manipulations and logarithmic properties when working on calculus problems. Sometimes, simplifying the expression first can save time and reduce the chances of making mistakes during differentiation.In summary, the key steps were applying the chain rule correctly, carefully handling the exponents and simplifying the expressions, and verifying the result through an alternative method. This problem was a good exercise in applying calculus concepts and attention to detail.

answer：First, I need to determine the relative speed between the two trains since they are moving in the same direction. The faster train is moving at 108 km/h, and the slower train is moving at 54 km/h. By subtracting the speed of the slower train from the speed of the faster train, I can find the relative speed.Next, I need to convert this relative speed from kilometers per hour to meters per second to match the time given in seconds. This conversion is necessary because the time taken to cross the man is provided in seconds, and I want to maintain consistent units throughout the calculation.Once I have the relative speed in meters per second, I can use the time taken to cross the man to calculate the length of the faster train. The formula for distance is speed multiplied by time, so by multiplying the relative speed in meters per second by the time in seconds, I will obtain the length of the faster train.Finally, after performing these calculations, I will arrive at the length of the faster train.

answer：Alright, so I have this sequence problem here. It's about a sequence defined by a recurrence relation. Let me try to understand what it's asking for.First, the sequence is given by ( a_1 = a ) and ( a_{n+1} = frac{5a_n - 8}{a_n - 1} ) for ( n in mathbb{N}^* ). So, it's a recursively defined sequence where each term depends on the previous one.Part (1) says: If ( a = 3 ), prove that the sequence ( left{ frac{a_n - 2}{a_n - 4} right} ) is a geometric sequence, and find the general term formula for the sequence ( {a_n} ).Okay, so starting with ( a = 3 ), I need to show that ( frac{a_n - 2}{a_n - 4} ) is a geometric sequence. That means each term is a constant multiple of the previous term. Then, using that, I can find a general formula for ( a_n ).Let me think about how to approach this. Maybe I can define a new sequence ( b_n = frac{a_n - 2}{a_n - 4} ) and try to find a recurrence relation for ( b_n ).So, ( b_n = frac{a_n - 2}{a_n - 4} ). Let's try to express ( b_{n+1} ) in terms of ( b_n ).Given ( a_{n+1} = frac{5a_n - 8}{a_n - 1} ), I can substitute this into the expression for ( b_{n+1} ):( b_{n+1} = frac{a_{n+1} - 2}{a_{n+1} - 4} = frac{frac{5a_n - 8}{a_n - 1} - 2}{frac{5a_n - 8}{a_n - 1} - 4} ).Hmm, that looks a bit complicated. Let me simplify the numerator and the denominator separately.First, the numerator:( frac{5a_n - 8}{a_n - 1} - 2 = frac{5a_n - 8 - 2(a_n - 1)}{a_n - 1} = frac{5a_n - 8 - 2a_n + 2}{a_n - 1} = frac{3a_n - 6}{a_n - 1} ).Similarly, the denominator:( frac{5a_n - 8}{a_n - 1} - 4 = frac{5a_n - 8 - 4(a_n - 1)}{a_n - 1} = frac{5a_n - 8 - 4a_n + 4}{a_n - 1} = frac{a_n - 4}{a_n - 1} ).So now, ( b_{n+1} = frac{frac{3a_n - 6}{a_n - 1}}{frac{a_n - 4}{a_n - 1}} = frac{3a_n - 6}{a_n - 4} ).Factor out the 3 in the numerator:( b_{n+1} = frac{3(a_n - 2)}{a_n - 4} = 3 cdot frac{a_n - 2}{a_n - 4} = 3b_n ).Oh, that's neat! So, ( b_{n+1} = 3b_n ), which means ( {b_n} ) is a geometric sequence with common ratio 3.Now, since ( b_n ) is geometric, I can write its general term. First, I need the initial term ( b_1 ).Given ( a_1 = 3 ), so:( b_1 = frac{a_1 - 2}{a_1 - 4} = frac{3 - 2}{3 - 4} = frac{1}{-1} = -1 ).Therefore, the general term for ( b_n ) is:( b_n = b_1 cdot 3^{n-1} = -1 cdot 3^{n-1} = -3^{n-1} ).But ( b_n = frac{a_n - 2}{a_n - 4} ), so:( frac{a_n - 2}{a_n - 4} = -3^{n-1} ).Let me solve for ( a_n ):Multiply both sides by ( a_n - 4 ):( a_n - 2 = -3^{n-1}(a_n - 4) ).Expand the right-hand side:( a_n - 2 = -3^{n-1}a_n + 4 cdot 3^{n-1} ).Bring all terms involving ( a_n ) to the left:( a_n + 3^{n-1}a_n = 2 + 4 cdot 3^{n-1} ).Factor out ( a_n ):( a_n(1 + 3^{n-1}) = 2 + 4 cdot 3^{n-1} ).Solve for ( a_n ):( a_n = frac{2 + 4 cdot 3^{n-1}}{1 + 3^{n-1}} ).Simplify numerator and denominator:Factor out 2 in the numerator:( a_n = frac{2(1 + 2 cdot 3^{n-1})}{1 + 3^{n-1}} ).Wait, that doesn't seem to simplify much further. Maybe I can write it as:( a_n = frac{4 cdot 3^{n-1} + 2}{3^{n-1} + 1} ).Yes, that looks better. So, that's the general term for ( a_n ).Alright, so part (1) seems done. I've shown that ( left{ frac{a_n - 2}{a_n - 4} right} ) is a geometric sequence with ratio 3, and found the general term for ( a_n ).Now, moving on to part (2): If for any positive integer ( n ), ( a_n > 3 ), determine the range of the real number ( a ).So, I need to find all real numbers ( a ) such that every term ( a_n ) in the sequence is greater than 3.Hmm, okay. Let's think about this. The sequence is defined recursively, so the behavior of the sequence depends on the initial term ( a ).I need to ensure that ( a_n > 3 ) for all ( n geq 1 ). So, starting from ( a_1 = a ), each subsequent term must also be greater than 3.Let me analyze the recurrence relation:( a_{n+1} = frac{5a_n - 8}{a_n - 1} ).I need ( a_{n+1} > 3 ) whenever ( a_n > 3 ). So, let's set up the inequality:( frac{5a_n - 8}{a_n - 1} > 3 ).Solve this inequality for ( a_n ):Multiply both sides by ( a_n - 1 ). But wait, I need to be careful about the sign of ( a_n - 1 ). Since ( a_n > 3 ), ( a_n - 1 > 2 > 0 ). So, multiplying both sides by ( a_n - 1 ) preserves the inequality:( 5a_n - 8 > 3(a_n - 1) ).Simplify:( 5a_n - 8 > 3a_n - 3 ).Subtract ( 3a_n ) from both sides:( 2a_n - 8 > -3 ).Add 8 to both sides:( 2a_n > 5 ).Divide by 2:( a_n > frac{5}{2} ).But we already have ( a_n > 3 ), which is stronger than ( a_n > frac{5}{2} ). So, the inequality ( a_{n+1} > 3 ) holds as long as ( a_n > 3 ).Wait, but this seems a bit circular. Because we assumed ( a_n > 3 ) to derive that ( a_{n+1} > 3 ). So, if we can ensure that ( a_1 > 3 ) and that the function ( f(x) = frac{5x - 8}{x - 1} ) maps ( x > 3 ) to ( f(x) > 3 ), then by induction, all ( a_n > 3 ).But let's check the behavior of ( f(x) ). Let me analyze the function ( f(x) = frac{5x - 8}{x - 1} ).First, find its fixed points by solving ( f(x) = x ):( frac{5x - 8}{x - 1} = x ).Multiply both sides by ( x - 1 ):( 5x - 8 = x(x - 1) ).Expand:( 5x - 8 = x^2 - x ).Bring all terms to one side:( x^2 - 6x + 8 = 0 ).Factor:( (x - 2)(x - 4) = 0 ).So, fixed points at ( x = 2 ) and ( x = 4 ).Hmm, interesting. So, 2 and 4 are fixed points. Let me see the behavior around these points.Compute the derivative of ( f(x) ) to analyze stability:( f'(x) = frac{(5)(x - 1) - (5x - 8)(1)}{(x - 1)^2} = frac{5x - 5 - 5x + 8}{(x - 1)^2} = frac{3}{(x - 1)^2} ).So, ( f'(x) = frac{3}{(x - 1)^2} ). At ( x = 2 ), ( f'(2) = 3/(1)^2 = 3 ), which is greater than 1, so the fixed point at 2 is unstable.At ( x = 4 ), ( f'(4) = 3/(3)^2 = 1/3 ), which is less than 1, so the fixed point at 4 is attracting.So, if the sequence starts near 4, it will converge to 4. If it starts near 2, it will move away from 2.But in our case, we need ( a_n > 3 ) for all ( n ). So, starting above 3, let's see where the sequence goes.Let me consider different cases for ( a ):Case 1: ( a > 4 )If ( a > 4 ), let's compute ( a_2 = frac{5a - 8}{a - 1} ).Since ( a > 4 ), denominator ( a - 1 > 3 ), numerator ( 5a - 8 > 20 - 8 = 12 ). So, ( a_2 > 12 / 3 = 4 ). So, ( a_2 > 4 ).Similarly, ( a_3 = frac{5a_2 - 8}{a_2 - 1} ). Since ( a_2 > 4 ), same logic applies: ( a_3 > 4 ).So, if ( a > 4 ), the sequence remains above 4, which is certainly above 3.Case 2: ( a = 4 )If ( a = 4 ), then ( a_2 = frac{5*4 - 8}{4 - 1} = frac{20 - 8}{3} = frac{12}{3} = 4 ). So, all terms are 4. Thus, ( a_n = 4 > 3 ) for all ( n ).Case 3: ( 3 < a < 4 )Here, starting with ( a ) between 3 and 4, let's compute ( a_2 ).( a_2 = frac{5a - 8}{a - 1} ).Since ( 3 < a < 4 ), denominator ( a - 1 ) is between 2 and 3.Numerator ( 5a - 8 ): For ( a = 3 ), it's 15 - 8 = 7; for ( a = 4 ), it's 20 - 8 = 12. So, numerator is between 7 and 12.Thus, ( a_2 ) is between ( 7/3 approx 2.33 ) and ( 12/3 = 4 ).Wait, but we need ( a_n > 3 ). So, if ( a_2 ) is less than 3, that would violate the condition.Wait, let me compute ( a_2 ) when ( a = 3.5 ):( a_2 = frac{5*3.5 - 8}{3.5 - 1} = frac{17.5 - 8}{2.5} = frac{9.5}{2.5} = 3.8 ).So, ( a_2 = 3.8 > 3 ).Wait, but earlier I thought ( a_2 ) could be as low as 7/3 ≈ 2.33, but that's when ( a ) approaches 3 from above.Wait, let me check ( a ) approaching 3 from above.Let ( a = 3 + epsilon ), where ( epsilon ) is a small positive number.Then, ( a_2 = frac{5(3 + epsilon) - 8}{(3 + epsilon) - 1} = frac{15 + 5epsilon - 8}{2 + epsilon} = frac{7 + 5epsilon}{2 + epsilon} ).As ( epsilon ) approaches 0, ( a_2 ) approaches ( 7/2 = 3.5 ).So, even when ( a ) is just above 3, ( a_2 ) is about 3.5, which is still above 3.Wait, so maybe my earlier thought that ( a_2 ) could be below 3 was incorrect. Let me recast that.Wait, when ( a ) is just above 3, ( a_2 ) is just above 3.5, which is still above 3. So, perhaps for ( 3 < a < 4 ), ( a_n ) remains above 3.But let me test with ( a = 3.1 ):( a_2 = frac{5*3.1 - 8}{3.1 - 1} = frac{15.5 - 8}{2.1} = frac{7.5}{2.1} ≈ 3.571 ).Still above 3.Similarly, ( a = 3.01 ):( a_2 = frac{5*3.01 - 8}{3.01 - 1} = frac{15.05 - 8}{2.01} ≈ frac{7.05}{2.01} ≈ 3.507 ).Still above 3.So, it seems that even when ( a ) is just above 3, ( a_2 ) is still above 3.5, which is above 3.Therefore, perhaps for ( a > 3 ), the sequence remains above 3.But wait, what if ( a = 3 )?Then, ( a_2 = frac{5*3 - 8}{3 - 1} = frac{15 - 8}{2} = frac{7}{2} = 3.5 ).So, ( a_2 = 3.5 > 3 ).Similarly, ( a_3 = frac{5*3.5 - 8}{3.5 - 1} = frac{17.5 - 8}{2.5} = frac{9.5}{2.5} = 3.8 ).And so on, approaching 4.So, it seems that as long as ( a > 3 ), the sequence ( a_n ) remains above 3.But wait, what if ( a = 3 )?Then, ( a_1 = 3 ), which is equal to 3, not greater than 3. So, ( a = 3 ) is excluded.What if ( a < 3 )?Let me try ( a = 2 ):( a_2 = frac{5*2 - 8}{2 - 1} = frac{10 - 8}{1} = 2 ).So, ( a_2 = 2 < 3 ). Thus, ( a_n ) is not greater than 3 for all ( n ).Similarly, ( a = 1 ):( a_2 = frac{5*1 - 8}{1 - 1} ). Wait, division by zero. So, undefined.So, ( a = 1 ) is not allowed.What about ( a = 5 ):( a_2 = frac{5*5 - 8}{5 - 1} = frac{25 - 8}{4} = frac{17}{4} = 4.25 ).Then, ( a_3 = frac{5*4.25 - 8}{4.25 - 1} = frac{21.25 - 8}{3.25} = frac{13.25}{3.25} ≈ 4.077 ).Approaching 4.So, seems like for ( a > 3 ), the sequence ( a_n ) stays above 3.But let me check the boundary case when ( a = 3 ). As we saw, ( a_1 = 3 ), which is not greater than 3, so ( a = 3 ) is excluded.Therefore, the range of ( a ) is ( a > 3 ).But wait, let me think again. Suppose ( a = 4 ), which is allowed, as ( a_n = 4 ) for all ( n ), which is greater than 3.If ( a > 4 ), the sequence converges to 4 from above, so all terms are above 4, hence above 3.If ( 3 < a < 4 ), the sequence starts at ( a ) and increases towards 4, so all terms are above 3.If ( a = 3 ), the first term is exactly 3, which doesn't satisfy ( a_n > 3 ).If ( a < 3 ), as we saw with ( a = 2 ), the sequence can drop below 3.Therefore, the range of ( a ) is ( a > 3 ).So, in conclusion, for part (2), the range of ( a ) is all real numbers greater than 3.**Final Answer**(1) The general term is (boxed{dfrac{4 cdot 3^{n-1} + 2}{3^{n-1} + 1}}).(2) The range of (a) is (boxed{(3, +infty)}).

answer：Alright, so I have this triangle ABC, and the angles are A, B, and C. The sides opposite these angles are a, b, and c respectively. The problem gives me an equation: a cos C + c cos A = -2b cos A. I need to find the value of angle A first, and then, given some specific values for a and b + c, find the area of the triangle.Starting with part 1: Find the value of angle A.Hmm, okay. So, I have this equation: a cos C + c cos A = -2b cos A. Let me write that down again to make sure I have it right:a cos C + c cos A = -2b cos A.I need to find angle A. I remember that in triangles, the Law of Cosines relates sides and angles. Maybe that can help here. The Law of Cosines says that for any triangle,a² = b² + c² - 2bc cos A,b² = a² + c² - 2ac cos B,c² = a² + b² - 2ab cos C.But I'm not sure if that's directly applicable here. Alternatively, maybe the Law of Sines can help. The Law of Sines states that:a / sin A = b / sin B = c / sin C = 2R,where R is the radius of the circumscribed circle. Maybe I can express a and c in terms of sine functions.Let me try that. From the Law of Sines, I can write:a = 2R sin A,b = 2R sin B,c = 2R sin C.So, substituting these into the equation:a cos C + c cos A = -2b cos A,we get:(2R sin A) cos C + (2R sin C) cos A = -2(2R sin B) cos A.Simplify this equation:2R sin A cos C + 2R sin C cos A = -4R sin B cos A.I can factor out 2R from the left side:2R (sin A cos C + sin C cos A) = -4R sin B cos A.Divide both sides by 2R (assuming R ≠ 0, which it can't be in a triangle):sin A cos C + sin C cos A = -2 sin B cos A.Wait a minute, sin A cos C + sin C cos A is equal to sin(A + C). Because sin(x + y) = sin x cos y + cos x sin y. So, that's a trigonometric identity.So, sin(A + C) = -2 sin B cos A.But in a triangle, the sum of angles is π radians (or 180 degrees). So, A + B + C = π. Therefore, A + C = π - B.So, sin(A + C) = sin(π - B) = sin B, because sin(π - x) = sin x.Therefore, sin B = -2 sin B cos A.So, sin B = -2 sin B cos A.Hmm, let's write that equation:sin B = -2 sin B cos A.Let me bring all terms to one side:sin B + 2 sin B cos A = 0.Factor out sin B:sin B (1 + 2 cos A) = 0.So, either sin B = 0 or 1 + 2 cos A = 0.But in a triangle, angles are between 0 and π, so sin B = 0 would imply B = 0 or π, which is impossible because all angles in a triangle must be greater than 0 and less than π. So, sin B ≠ 0.Therefore, the other factor must be zero:1 + 2 cos A = 0.Solving for cos A:2 cos A = -1,cos A = -1/2.So, cos A = -1/2. Now, what angle A between 0 and π has cosine equal to -1/2?I remember that cos(2π/3) = -1/2, because 2π/3 is 120 degrees, and cosine of 120 degrees is -1/2.Therefore, angle A is 2π/3 radians, or 120 degrees.Okay, that seems solid. Let me recap:1. I used the Law of Sines to express a, b, c in terms of sine functions.2. Substituted into the given equation.3. Simplified using trigonometric identities, specifically sin(A + C) = sin B.4. Factored out sin B, leading to cos A = -1/2.5. Concluded that A = 2π/3.So, part 1 is solved. Angle A is 120 degrees or 2π/3 radians.Moving on to part 2: If a = 2√3, b + c = 4, find the area of triangle ABC.Alright, so we have side a opposite angle A, which is 120 degrees. We also know that b + c = 4. We need to find the area of the triangle.First, let me recall that the area of a triangle can be found in several ways. One common formula is:Area = (1/2)ab sin C,but in this case, I don't know angles B or C, or sides b and c individually. I only know their sum.Alternatively, I can use Heron's formula, which requires knowing all three sides. But I only know one side (a = 2√3) and the sum of the other two sides (b + c = 4). So, I don't have enough information for Heron's formula directly.Another approach is to use the formula involving two sides and the included angle. Since I know angle A and sides b and c, if I can find bc, then I can compute the area as:Area = (1/2)bc sin A.Yes, that seems promising because I can express bc in terms of known quantities.But how?I remember the Law of Cosines, which relates the sides of a triangle with one of its angles. Since I know angle A and side a, perhaps I can write an equation involving b and c.Law of Cosines states:a² = b² + c² - 2bc cos A.We know a, cos A, and we know that b + c = 4. Maybe we can express b² + c² in terms of (b + c)².Yes, because (b + c)² = b² + 2bc + c². So, b² + c² = (b + c)² - 2bc.Given that b + c = 4, then (b + c)² = 16. Therefore, b² + c² = 16 - 2bc.So, substituting into the Law of Cosines equation:a² = (16 - 2bc) - 2bc cos A.We know a = 2√3, so a² = (2√3)² = 4 * 3 = 12.Also, angle A is 120 degrees, so cos A = cos 120° = -1/2.Substituting these values into the equation:12 = (16 - 2bc) - 2bc*(-1/2).Simplify the right-hand side:First, compute -2bc*(-1/2). That is equal to bc.So, the equation becomes:12 = 16 - 2bc + bc.Simplify further:12 = 16 - bc.Subtract 16 from both sides:12 - 16 = -bc,-4 = -bc,Multiply both sides by -1:4 = bc.So, bc = 4.Great! Now, since we need the area, which is (1/2)bc sin A, and we know bc = 4, and sin A = sin 120°.Sin 120° is equal to sin(60°) = √3/2, because sin(180° - x) = sin x.So, sin 120° = √3/2.Therefore, the area is:Area = (1/2) * 4 * (√3/2).Simplify this:First, multiply 1/2 and 4: (1/2)*4 = 2.Then, multiply 2 by √3/2: 2*(√3/2) = √3.So, the area is √3.Wait, let me double-check that calculation:(1/2) * bc * sin A = (1/2) * 4 * (√3/2).Yes, 1/2 * 4 is 2, and 2 * √3/2 is √3. Correct.Alternatively, I can think of it as:(1/2) * 4 * (√3/2) = (4/2) * (√3/2) = 2 * (√3/2) = √3.Same result.So, the area is √3.Let me recap the steps for part 2:1. Recognized that the area can be expressed as (1/2)bc sin A.2. Needed to find bc.3. Used the Law of Cosines: a² = b² + c² - 2bc cos A.4. Expressed b² + c² in terms of (b + c)² - 2bc.5. Substituted known values: a² = 12, b + c = 4, cos A = -1/2.6. Solved for bc, getting bc = 4.7. Plugged bc and sin A into the area formula, resulting in √3.Everything seems consistent. I don't see any mistakes in the calculations.So, summarizing:1. Angle A is 120 degrees or 2π/3 radians.2. The area of triangle ABC is √3.**Final Answer**1. The value of angle ( A ) is (boxed{dfrac{2pi}{3}}).2. The area of (triangle ABC) is (boxed{sqrt{3}}).

answer：Alright, so I have this complex number problem here, and I need to solve two parts. Let me try to figure it out step by step. First, the complex number is given as ( z = frac{(1 - i)^2 + 3(1 + i)}{2 - i} ). Okay, so I need to simplify this expression to find ( z ), and then find its conjugate ( overline{z} ). After that, in part (2), I have an equation involving ( z ): ( az + b = 1 - i ), and I need to find real numbers ( a ) and ( b ). Starting with part (1). Let me first compute the numerator: ( (1 - i)^2 + 3(1 + i) ). I think I should expand ( (1 - i)^2 ) first. So, ( (1 - i)^2 ) is ( 1^2 - 2 times 1 times i + (i)^2 ). That simplifies to ( 1 - 2i + (-1) ) because ( i^2 = -1 ). So, ( 1 - 2i - 1 ) which is ( -2i ). Okay, so the first part of the numerator is ( -2i ). Now, the second part is ( 3(1 + i) ). That should be straightforward: ( 3 times 1 + 3 times i = 3 + 3i ). So, combining both parts of the numerator: ( -2i + 3 + 3i ). Let me combine like terms. The real parts are just 3, and the imaginary parts are ( -2i + 3i = i ). So, the numerator simplifies to ( 3 + i ). Now, the denominator is ( 2 - i ). So, the entire expression for ( z ) is ( frac{3 + i}{2 - i} ). Hmm, I need to simplify this complex fraction. I remember that to simplify such expressions, I should multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of ( 2 - i ) is ( 2 + i ). So, multiplying numerator and denominator by ( 2 + i ): Numerator becomes ( (3 + i)(2 + i) ). Let me compute that. First, multiply 3 by 2: 6. Then, 3 by i: ( 3i ). Then, i by 2: ( 2i ). Finally, i by i: ( i^2 = -1 ). So, adding all these together: ( 6 + 3i + 2i - 1 ). Combine like terms: ( (6 - 1) + (3i + 2i) = 5 + 5i ). Denominator becomes ( (2 - i)(2 + i) ). That's a difference of squares, so ( (2)^2 - (i)^2 = 4 - (-1) = 4 + 1 = 5 ). So, now, ( z = frac{5 + 5i}{5} ). Simplifying that, divide both terms by 5: ( 1 + i ). Alright, so ( z = 1 + i ). That was part (1). Now, the conjugate of ( z ), which is ( overline{z} ), is simply the complex number with the imaginary part's sign changed. So, ( overline{z} = 1 - i ). Okay, that seems straightforward. Let me just double-check my steps to make sure I didn't make a mistake. First, expanding ( (1 - i)^2 ): 1 - 2i + i², which is 1 - 2i -1, so -2i. Then, 3(1 + i) is 3 + 3i. Adding them together: -2i + 3 + 3i = 3 + i. Then, dividing by 2 - i, I multiplied numerator and denominator by 2 + i, which gave me 5 + 5i over 5, simplifying to 1 + i. So, yes, that seems correct. Therefore, the conjugate is 1 - i. Moving on to part (2): If ( az + b = 1 - i ), find real numbers ( a ) and ( b ). Given that ( z = 1 + i ), let's substitute that into the equation: ( a(1 + i) + b = 1 - i ). Let me write that out: ( a(1 + i) + b = 1 - i ). Expanding the left side: ( a times 1 + a times i + b = a + ai + b ). Combine like terms: ( (a + b) + ai ). So, the equation becomes ( (a + b) + ai = 1 - i ). Since two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal, we can set up a system of equations. Real part: ( a + b = 1 ). Imaginary part: ( a = -1 ). So, from the imaginary part, we have ( a = -1 ). Substituting ( a = -1 ) into the real part equation: ( -1 + b = 1 ). Solving for ( b ): ( b = 1 + 1 = 2 ). Therefore, ( a = -1 ) and ( b = 2 ). Let me verify this solution. Plugging ( a = -1 ) and ( b = 2 ) back into the equation: ( (-1)(1 + i) + 2 = -1 - i + 2 = ( -1 + 2 ) + (-i) = 1 - i ). Yes, that matches the right side of the equation. So, the solution is correct. So, summarizing my findings: 1. The conjugate of ( z ) is ( 1 - i ). 2. The real numbers ( a ) and ( b ) are ( -1 ) and ( 2 ), respectively. I think that's all. I don't see any mistakes in my calculations, so I feel confident about these answers.**Final Answer**(1) (boxed{1 - i})(2) (boxed{a = -1}) and (boxed{b = 2})