💡Okay, so I have this problem about the function ( f(x) = ln{x} - ax - 3 ) where ( a ) is a real number and not zero. There are two parts: first, finding the monotonic intervals of ( f(x) ), and second, a more complex problem involving the tangent line and another function ( g(x) ). Let me try to tackle each part step by step.Starting with part 1: finding the monotonic intervals of ( f(x) ). I remember that to find monotonic intervals, I need to look at the derivative of the function. So, let me compute ( f'(x) ).The derivative of ( ln{x} ) is ( frac{1}{x} ), the derivative of ( -ax ) is ( -a ), and the derivative of a constant like -3 is 0. So, putting it together, ( f'(x) = frac{1}{x} - a ). Hmm, wait, is that right? Let me double-check. Yeah, ( frac{d}{dx} ln{x} = frac{1}{x} ), and ( frac{d}{dx} (-ax) = -a ). So, yes, ( f'(x) = frac{1}{x} - a ).Now, to find the monotonic intervals, I need to see where ( f'(x) ) is positive or negative. Let me set ( f'(x) = 0 ) to find critical points.So, ( frac{1}{x} - a = 0 ) implies ( frac{1}{x} = a ), which gives ( x = frac{1}{a} ). But wait, ( a ) is a real number and not zero, so ( x ) must be positive because the domain of ( ln{x} ) is ( x > 0 ). So, ( x = frac{1}{a} ) is a critical point.Now, depending on the value of ( a ), the critical point ( x = frac{1}{a} ) will be in different positions. Let me consider two cases: when ( a > 0 ) and when ( a < 0 ).Case 1: ( a > 0 ). Then, ( x = frac{1}{a} ) is positive. Let me analyze the sign of ( f'(x) ) around this critical point.For ( x < frac{1}{a} ), let's pick a test point, say ( x = frac{1}{2a} ). Plugging into ( f'(x) ):( f'left(frac{1}{2a}right) = frac{1}{frac{1}{2a}} - a = 2a - a = a > 0 ).So, ( f'(x) ) is positive when ( x < frac{1}{a} ), meaning ( f(x) ) is increasing on ( (0, frac{1}{a}) ).For ( x > frac{1}{a} ), let's pick ( x = frac{2}{a} ). Then,( f'left(frac{2}{a}right) = frac{1}{frac{2}{a}} - a = frac{a}{2} - a = -frac{a}{2} < 0 ).So, ( f'(x) ) is negative when ( x > frac{1}{a} ), meaning ( f(x) ) is decreasing on ( (frac{1}{a}, infty) ).Case 2: ( a < 0 ). Then, ( x = frac{1}{a} ) is negative, but since the domain of ( f(x) ) is ( x > 0 ), the critical point is not in the domain. So, in this case, ( f'(x) = frac{1}{x} - a ). Since ( a ) is negative, ( -a ) is positive. So, ( f'(x) = frac{1}{x} + |a| ), which is always positive for all ( x > 0 ). Therefore, ( f(x) ) is increasing on its entire domain ( (0, infty) ).Wait, hold on. If ( a < 0 ), then ( f'(x) = frac{1}{x} - a = frac{1}{x} + |a| ). Since both ( frac{1}{x} ) and ( |a| ) are positive, their sum is positive. So, yes, ( f'(x) > 0 ) for all ( x > 0 ) when ( a < 0 ). So, ( f(x) ) is strictly increasing on ( (0, infty) ).So, summarizing part 1:- If ( a > 0 ), ( f(x) ) is increasing on ( (0, frac{1}{a}) ) and decreasing on ( (frac{1}{a}, infty) ).- If ( a < 0 ), ( f(x) ) is increasing on ( (0, infty) ).Alright, that seems solid. Let me move on to part 2.Part 2 is more involved. It says: If the tangent line of the graph of the function ( y = f(x) ) at the point ( (2, f(2)) ) has a slope of ( 45^{circ} ), and for any ( t in [1, 2] ), the function ( g(x) = x^3 + x^2[f'(x) + frac{m}{2}] ) is not always monotonic on the interval ( (t, 3) ), find the range of values for ( m ).Okay, let's parse this step by step.First, the tangent line at ( (2, f(2)) ) has a slope of ( 45^{circ} ). The slope of a tangent line is given by the derivative at that point. So, ( f'(2) = tan(45^{circ}) ). Since ( tan(45^{circ}) = 1 ), we have ( f'(2) = 1 ).From part 1, we know ( f'(x) = frac{1}{x} - a ). So, plugging in ( x = 2 ):( f'(2) = frac{1}{2} - a = 1 ).Solving for ( a ):( frac{1}{2} - a = 1 )( -a = 1 - frac{1}{2} )( -a = frac{1}{2} )( a = -frac{1}{2} ).Wait, hold on. Let me double-check that. If ( f'(2) = 1 ), then ( frac{1}{2} - a = 1 ). So, subtract ( frac{1}{2} ) from both sides: ( -a = frac{1}{2} ), so ( a = -frac{1}{2} ). Yes, that's correct.So, ( a = -frac{1}{2} ). Therefore, the function ( f(x) ) becomes:( f(x) = ln{x} - (-frac{1}{2})x - 3 = ln{x} + frac{1}{2}x - 3 ).Wait, hold on, the original function is ( f(x) = ln{x} - a x - 3 ). So, substituting ( a = -frac{1}{2} ), we get:( f(x) = ln{x} - (-frac{1}{2})x - 3 = ln{x} + frac{1}{2}x - 3 ).Yes, that's correct.Now, moving on to the second part: for any ( t in [1, 2] ), the function ( g(x) = x^3 + x^2[f'(x) + frac{m}{2}] ) is not always monotonic on the interval ( (t, 3) ). We need to find the range of ( m ).First, let me write down ( g(x) ). We have:( g(x) = x^3 + x^2 [f'(x) + frac{m}{2}] ).We already know ( f'(x) = frac{1}{x} - a ), and we found ( a = -frac{1}{2} ). So, substituting ( a ):( f'(x) = frac{1}{x} - (-frac{1}{2}) = frac{1}{x} + frac{1}{2} ).Therefore, substituting into ( g(x) ):( g(x) = x^3 + x^2 left( frac{1}{x} + frac{1}{2} + frac{m}{2} right ) ).Let me simplify this expression step by step.First, inside the brackets:( frac{1}{x} + frac{1}{2} + frac{m}{2} = frac{1}{x} + frac{1 + m}{2} ).So, ( g(x) = x^3 + x^2 left( frac{1}{x} + frac{1 + m}{2} right ) ).Now, distribute ( x^2 ):( g(x) = x^3 + x^2 cdot frac{1}{x} + x^2 cdot frac{1 + m}{2} ).Simplify each term:- ( x^2 cdot frac{1}{x} = x ).- ( x^2 cdot frac{1 + m}{2} = frac{(1 + m)}{2} x^2 ).So, putting it all together:( g(x) = x^3 + x + frac{(1 + m)}{2} x^2 ).Let me write this in standard polynomial form:( g(x) = x^3 + frac{(1 + m)}{2} x^2 + x ).Alternatively, to make it clearer, I can write:( g(x) = x^3 + left( frac{1 + m}{2} right ) x^2 + x ).Now, the problem states that for any ( t in [1, 2] ), ( g(x) ) is not always monotonic on ( (t, 3) ). So, ( g(x) ) is not monotonic on any interval ( (t, 3) ) where ( t ) is between 1 and 2.To determine when a function is monotonic, we look at its derivative. If the derivative does not change sign on the interval, the function is monotonic. If the derivative changes sign, the function is not monotonic.Therefore, for ( g(x) ) to not be always monotonic on ( (t, 3) ), its derivative ( g'(x) ) must change sign in that interval. So, ( g'(x) ) must have at least one critical point in ( (t, 3) ).Therefore, we need to ensure that ( g'(x) ) has at least one root in ( (t, 3) ) for any ( t in [1, 2] ).So, let me compute ( g'(x) ):( g'(x) = frac{d}{dx} left( x^3 + frac{(1 + m)}{2} x^2 + x right ) ).Calculating term by term:- ( frac{d}{dx} x^3 = 3x^2 ).- ( frac{d}{dx} left( frac{(1 + m)}{2} x^2 right ) = frac{(1 + m)}{2} cdot 2x = (1 + m)x ).- ( frac{d}{dx} x = 1 ).So, putting it together:( g'(x) = 3x^2 + (1 + m)x + 1 ).So, ( g'(x) = 3x^2 + (1 + m)x + 1 ).We need ( g'(x) ) to have at least one root in ( (t, 3) ) for any ( t in [1, 2] ). In other words, for every ( t ) between 1 and 2, the equation ( 3x^2 + (1 + m)x + 1 = 0 ) must have at least one solution in ( (t, 3) ).Alternatively, since ( g'(x) ) is a quadratic function, it can have at most two real roots. For ( g'(x) ) to have a root in ( (t, 3) ), either:1. The quadratic has one root in ( (t, 3) ) and another outside, or2. Both roots are in ( (t, 3) ).But since the problem states that ( g(x) ) is not always monotonic on ( (t, 3) ), it's sufficient that ( g'(x) ) has at least one root in ( (t, 3) ).Therefore, we need to ensure that for every ( t in [1, 2] ), the interval ( (t, 3) ) contains at least one root of ( g'(x) = 0 ).To find the range of ( m ), let's analyze the quadratic equation ( 3x^2 + (1 + m)x + 1 = 0 ).First, let me compute the discriminant to see when real roots exist.Discriminant ( D = (1 + m)^2 - 4 cdot 3 cdot 1 = (1 + m)^2 - 12 ).For real roots, ( D geq 0 ), so ( (1 + m)^2 geq 12 ), which implies ( 1 + m geq 2sqrt{3} ) or ( 1 + m leq -2sqrt{3} ). Therefore, ( m geq 2sqrt{3} - 1 ) or ( m leq -2sqrt{3} - 1 ).But we don't know yet if these roots lie in ( (t, 3) ). So, even if there are real roots, we need to ensure that at least one of them is in ( (t, 3) ) for any ( t in [1, 2] ).Alternatively, perhaps it's better to consider the behavior of ( g'(x) ) at the endpoints ( x = t ) and ( x = 3 ). If ( g'(t) ) and ( g'(3) ) have opposite signs, then by the Intermediate Value Theorem, there must be a root in ( (t, 3) ).So, let's compute ( g'(t) ) and ( g'(3) ):( g'(t) = 3t^2 + (1 + m)t + 1 ).( g'(3) = 3(9) + (1 + m)(3) + 1 = 27 + 3 + 3m + 1 = 31 + 3m ).We need ( g'(t) ) and ( g'(3) ) to have opposite signs for any ( t in [1, 2] ). That is, for any ( t in [1, 2] ), ( g'(t) cdot g'(3) < 0 ).So, ( [3t^2 + (1 + m)t + 1] cdot [31 + 3m] < 0 ).This inequality must hold for all ( t in [1, 2] ). Therefore, both factors must have opposite signs for all ( t ) in that interval.Let me denote ( A(t) = 3t^2 + (1 + m)t + 1 ) and ( B = 31 + 3m ). So, ( A(t) cdot B < 0 ) for all ( t in [1, 2] ).This implies that either:1. ( A(t) > 0 ) for all ( t in [1, 2] ) and ( B < 0 ), or2. ( A(t) < 0 ) for all ( t in [1, 2] ) and ( B > 0 ).But let's analyze ( A(t) ). ( A(t) = 3t^2 + (1 + m)t + 1 ). Since the coefficient of ( t^2 ) is positive (3), the parabola opens upwards. Therefore, ( A(t) ) will have a minimum at its vertex.The vertex occurs at ( t = -frac{b}{2a} = -frac{1 + m}{6} ). Since ( t in [1, 2] ), which is positive, and the vertex is at ( t = -frac{1 + m}{6} ), which is negative if ( 1 + m > 0 ), or positive if ( 1 + m < 0 ).Wait, let me compute the vertex:( t_v = -frac{1 + m}{2 cdot 3} = -frac{1 + m}{6} ).So, if ( 1 + m > 0 ), then ( t_v < 0 ), meaning the minimum of ( A(t) ) is at ( t < 0 ), so on ( [1, 2] ), ( A(t) ) is increasing.If ( 1 + m < 0 ), then ( t_v > 0 ), so the minimum is at ( t_v ), which could be within or outside [1, 2].But regardless, since ( A(t) ) is a quadratic opening upwards, the minimum value on [1, 2] will be at one of the endpoints or at the vertex if it's inside the interval.But let's consider both cases.Case 1: ( 1 + m > 0 ). Then, ( t_v < 0 ), so on [1, 2], ( A(t) ) is increasing. Therefore, the minimum of ( A(t) ) on [1, 2] is at ( t = 1 ).Compute ( A(1) = 3(1)^2 + (1 + m)(1) + 1 = 3 + 1 + m + 1 = 5 + m ).Similarly, ( A(2) = 3(4) + (1 + m)(2) + 1 = 12 + 2 + 2m + 1 = 15 + 2m ).Since ( A(t) ) is increasing, ( A(t) ) ranges from ( 5 + m ) to ( 15 + 2m ) on [1, 2].Case 2: ( 1 + m < 0 ). Then, ( t_v > 0 ). Let's compute ( t_v ):( t_v = -frac{1 + m}{6} ).Since ( 1 + m < 0 ), ( t_v = frac{|1 + m|}{6} ).We need to check if ( t_v ) is in [1, 2].So, ( t_v in [1, 2] ) implies ( 1 leq -frac{1 + m}{6} leq 2 ).Multiply all parts by -6 (remembering to reverse inequalities):( -6 leq 1 + m leq -12 ).But ( 1 + m < 0 ), so ( 1 + m leq -12 ) implies ( m leq -13 ).Wait, let me verify:If ( t_v geq 1 ), then ( -frac{1 + m}{6} geq 1 ) implies ( - (1 + m) geq 6 ) implies ( 1 + m leq -6 ) implies ( m leq -7 ).Similarly, ( t_v leq 2 ) implies ( -frac{1 + m}{6} leq 2 ) implies ( - (1 + m) leq 12 ) implies ( 1 + m geq -12 ) implies ( m geq -13 ).Therefore, if ( m in [-13, -7] ), then ( t_v in [1, 2] ).Otherwise, if ( m < -13 ), then ( t_v > 2 ), and if ( m > -7 ), ( t_v < 1 ).So, summarizing:- If ( m > -7 ), ( t_v < 1 ), so on [1, 2], ( A(t) ) is increasing.- If ( m in [-13, -7] ), ( t_v in [1, 2] ), so ( A(t) ) has its minimum at ( t_v ).- If ( m < -13 ), ( t_v > 2 ), so on [1, 2], ( A(t) ) is decreasing.This is getting complicated, but let's proceed.Our goal is to have ( A(t) cdot B < 0 ) for all ( t in [1, 2] ). So, either ( A(t) > 0 ) and ( B < 0 ) for all ( t in [1, 2] ), or ( A(t) < 0 ) and ( B > 0 ) for all ( t in [1, 2] ).But let's analyze both possibilities.First, suppose ( A(t) > 0 ) for all ( t in [1, 2] ) and ( B < 0 ).Then, ( A(t) > 0 ) for all ( t in [1, 2] ) implies that the minimum of ( A(t) ) on [1, 2] is positive.Similarly, ( B = 31 + 3m < 0 ) implies ( m < -frac{31}{3} approx -10.333 ).But let's see if ( A(t) > 0 ) for all ( t in [1, 2] ).From earlier, depending on ( m ), the behavior of ( A(t) ) changes.If ( m > -7 ), ( A(t) ) is increasing on [1, 2], so the minimum is at ( t = 1 ), which is ( 5 + m ). So, ( 5 + m > 0 ) implies ( m > -5 ).But if ( m > -5 ), then ( B = 31 + 3m ). If ( m > -5 ), ( 31 + 3m > 31 - 15 = 16 > 0 ). So, ( B > 0 ). But we wanted ( B < 0 ). Therefore, this case is not possible because ( A(t) > 0 ) and ( B > 0 ), which would make ( A(t) cdot B > 0 ), contradicting our requirement.If ( m in [-13, -7] ), then ( A(t) ) has its minimum at ( t_v in [1, 2] ). The minimum value is ( A(t_v) ). Let's compute ( A(t_v) ).The minimum value of a quadratic ( ax^2 + bx + c ) is ( c - frac{b^2}{4a} ).So, ( A(t_v) = 1 - frac{(1 + m)^2}{12} ).We need ( A(t_v) > 0 ):( 1 - frac{(1 + m)^2}{12} > 0 )( frac{(1 + m)^2}{12} < 1 )( (1 + m)^2 < 12 )( |1 + m| < 2sqrt{3} approx 3.464 )So, ( -3.464 < 1 + m < 3.464 )Which implies ( -4.464 < m < 2.464 ).But in this case, ( m in [-13, -7] ), which is outside the range ( -4.464 < m < 2.464 ). Therefore, ( A(t_v) ) is negative in this interval, meaning ( A(t) ) is negative somewhere in [1, 2]. Therefore, ( A(t) ) cannot be always positive on [1, 2] when ( m in [-13, -7] ).If ( m < -13 ), then ( A(t) ) is decreasing on [1, 2], so the minimum is at ( t = 2 ), which is ( 15 + 2m ). So, ( 15 + 2m > 0 ) implies ( m > -7.5 ). But ( m < -13 ), so ( 15 + 2m < 15 - 26 = -11 < 0 ). Therefore, ( A(t) ) is decreasing from ( A(1) = 5 + m ) to ( A(2) = 15 + 2m ). Since ( m < -13 ), ( A(1) = 5 + m < 5 -13 = -8 < 0 ), and ( A(2) = 15 + 2m < 15 -26 = -11 < 0 ). So, ( A(t) ) is negative throughout [1, 2].Therefore, in this case, ( A(t) < 0 ) for all ( t in [1, 2] ).So, summarizing:- If ( m > -7 ), ( A(t) ) is increasing on [1, 2], with minimum at ( t = 1 ). ( A(t) > 0 ) if ( m > -5 ), but ( B = 31 + 3m ) is positive, so ( A(t) cdot B > 0 ).- If ( m in [-13, -7] ), ( A(t) ) has a minimum in [1, 2], but ( A(t_v) < 0 ), so ( A(t) ) is negative somewhere in [1, 2].- If ( m < -13 ), ( A(t) ) is decreasing on [1, 2], with both ( A(1) ) and ( A(2) ) negative.Therefore, the only case where ( A(t) ) is negative for all ( t in [1, 2] ) is when ( m < -13 ). But wait, when ( m < -13 ), ( A(t) ) is negative on [1, 2], but ( B = 31 + 3m ). If ( m < -13 ), ( 3m < -39 ), so ( 31 + 3m < 31 - 39 = -8 < 0 ). Therefore, ( A(t) < 0 ) and ( B < 0 ), so ( A(t) cdot B > 0 ). Which is not what we want.Wait, we need ( A(t) cdot B < 0 ). So, if ( A(t) < 0 ) and ( B < 0 ), their product is positive. So, that doesn't satisfy the condition.Alternatively, if ( A(t) > 0 ) and ( B < 0 ), then ( A(t) cdot B < 0 ). But earlier, we saw that when ( m > -5 ), ( A(t) > 0 ) on [1, 2], but ( B = 31 + 3m ). If ( m > -5 ), ( 3m > -15 ), so ( 31 + 3m > 16 > 0 ). Therefore, ( A(t) > 0 ) and ( B > 0 ), so ( A(t) cdot B > 0 ).So, neither case seems to satisfy ( A(t) cdot B < 0 ) for all ( t in [1, 2] ). Hmm, maybe I need to think differently.Wait, perhaps instead of requiring ( A(t) cdot B < 0 ) for all ( t in [1, 2] ), it's sufficient that for each ( t in [1, 2] ), there exists some ( x in (t, 3) ) such that ( g'(x) = 0 ). That is, for each ( t ), the interval ( (t, 3) ) contains at least one root of ( g'(x) ).Alternatively, perhaps the quadratic ( g'(x) = 3x^2 + (1 + m)x + 1 ) must have at least one root in ( (t, 3) ) for every ( t in [1, 2] ).So, for each ( t in [1, 2] ), the equation ( 3x^2 + (1 + m)x + 1 = 0 ) must have a solution in ( (t, 3) ).This is equivalent to saying that the smallest root of ( g'(x) = 0 ) is less than 3, and the largest root is greater than 1.Wait, but actually, for every ( t in [1, 2] ), the interval ( (t, 3) ) must contain at least one root. So, the roots must be such that the larger root is greater than 1, and the smaller root is less than 3. But more specifically, for each ( t ), at least one root is in ( (t, 3) ).This is a bit tricky. Maybe another approach is to ensure that the quadratic ( g'(x) ) has one root in ( (1, 3) ), and the other root is either less than 1 or greater than 3. But since we need for every ( t in [1, 2] ), the interval ( (t, 3) ) contains at least one root, perhaps the quadratic must have one root in ( (1, 3) ), and the other root outside of ( [1, 3] ).But let me think.Suppose the quadratic has two real roots, say ( r_1 ) and ( r_2 ), with ( r_1 < r_2 ).We need that for every ( t in [1, 2] ), either ( r_1 in (t, 3) ) or ( r_2 in (t, 3) ). But since ( t ) can be as large as 2, to ensure that even when ( t = 2 ), there is a root in ( (2, 3) ), we need at least one root in ( (2, 3) ).Similarly, for ( t = 1 ), we need a root in ( (1, 3) ). But if we already have a root in ( (2, 3) ), then for ( t = 1 ), that root is still in ( (1, 3) ). So, perhaps it's sufficient to have one root in ( (2, 3) ), and the other root can be anywhere else.But let's verify.If the quadratic has one root in ( (2, 3) ), then for any ( t in [1, 2] ), ( (t, 3) ) will contain that root in ( (2, 3) ). So, that would satisfy the condition.Alternatively, if both roots are in ( (2, 3) ), then for any ( t in [1, 2] ), ( (t, 3) ) will contain both roots, so certainly at least one.But if one root is in ( (1, 2) ) and the other is in ( (2, 3) ), then for ( t in [1, 2] ), depending on ( t ), the interval ( (t, 3) ) may or may not contain the root in ( (1, 2) ). For example, if ( t ) is just above the root in ( (1, 2) ), then ( (t, 3) ) would not contain that root, but it would contain the other root in ( (2, 3) ). So, as long as there is a root in ( (2, 3) ), it will be included in ( (t, 3) ) for any ( t in [1, 2] ).Therefore, the key is to ensure that the quadratic ( g'(x) = 3x^2 + (1 + m)x + 1 ) has at least one root in ( (2, 3) ).So, let's use the Intermediate Value Theorem on ( g'(x) ) at ( x = 2 ) and ( x = 3 ).Compute ( g'(2) ) and ( g'(3) ):( g'(2) = 3(4) + (1 + m)(2) + 1 = 12 + 2 + 2m + 1 = 15 + 2m ).( g'(3) = 3(9) + (1 + m)(3) + 1 = 27 + 3 + 3m + 1 = 31 + 3m ).For there to be a root in ( (2, 3) ), ( g'(2) ) and ( g'(3) ) must have opposite signs. So, ( g'(2) cdot g'(3) < 0 ).Therefore:( (15 + 2m)(31 + 3m) < 0 ).Let me solve this inequality.First, find the roots of the quadratic in terms of ( m ):( 15 + 2m = 0 ) implies ( m = -frac{15}{2} = -7.5 ).( 31 + 3m = 0 ) implies ( m = -frac{31}{3} approx -10.333 ).So, the critical points are at ( m = -10.333 ) and ( m = -7.5 ).Now, let's analyze the sign of ( (15 + 2m)(31 + 3m) ):- For ( m < -10.333 ): - ( 15 + 2m < 15 + 2(-10.333) = 15 - 20.666 = -5.666 < 0 ). - ( 31 + 3m < 31 + 3(-10.333) = 31 - 31 = 0 ). - So, both factors are negative, their product is positive.- For ( -10.333 < m < -7.5 ): - ( 15 + 2m < 0 ) (since ( m < -7.5 )). - ( 31 + 3m > 0 ) (since ( m > -10.333 )). - So, one factor is negative, the other positive, their product is negative.- For ( m > -7.5 ): - ( 15 + 2m > 0 ). - ( 31 + 3m > 0 ). - So, both factors are positive, their product is positive.Therefore, ( (15 + 2m)(31 + 3m) < 0 ) when ( -10.333 < m < -7.5 ).So, ( m in left( -frac{31}{3}, -frac{15}{2} right ) ).But let me write it as fractions:( -frac{31}{3} approx -10.333 ) and ( -frac{15}{2} = -7.5 ).So, ( m in left( -frac{31}{3}, -frac{15}{2} right ) ).But let me verify this conclusion.If ( m in left( -frac{31}{3}, -frac{15}{2} right ) ), then ( g'(2) cdot g'(3) < 0 ), which implies that there is a root in ( (2, 3) ). Therefore, for any ( t in [1, 2] ), the interval ( (t, 3) ) will contain this root in ( (2, 3) ), ensuring that ( g(x) ) is not monotonic on ( (t, 3) ).Therefore, the range of ( m ) is ( -frac{31}{3} < m < -frac{15}{2} ).But let me express ( -frac{31}{3} ) and ( -frac{15}{2} ) as exact fractions:( -frac{31}{3} = -10 frac{1}{3} ) and ( -frac{15}{2} = -7 frac{1}{2} ).So, the range is ( -10 frac{1}{3} < m < -7 frac{1}{2} ).But let me check if this is the final answer or if there are additional constraints.Wait, earlier, I considered that as long as there is a root in ( (2, 3) ), it satisfies the condition for all ( t in [1, 2] ). But is that sufficient?Suppose ( m ) is such that there is a root in ( (2, 3) ). Then, for any ( t in [1, 2] ), ( (t, 3) ) will include ( (2, 3) ), hence the root in ( (2, 3) ) is included. Therefore, ( g'(x) ) changes sign in ( (t, 3) ), so ( g(x) ) is not monotonic on ( (t, 3) ).Therefore, the condition ( g'(2) cdot g'(3) < 0 ) is sufficient to ensure that there is a root in ( (2, 3) ), which in turn ensures that for any ( t in [1, 2] ), ( (t, 3) ) contains a root, making ( g(x) ) not monotonic on that interval.Therefore, the range of ( m ) is ( -frac{31}{3} < m < -frac{15}{2} ).But let me double-check the calculations:We had ( g'(2) = 15 + 2m ) and ( g'(3) = 31 + 3m ).Setting ( (15 + 2m)(31 + 3m) < 0 ):- The roots are at ( m = -15/2 = -7.5 ) and ( m = -31/3 approx -10.333 ).- The inequality is negative between these two roots.Yes, that seems correct.Therefore, the range of ( m ) is ( -frac{31}{3} < m < -frac{15}{2} ).Expressed as fractions, that's ( -frac{31}{3} < m < -frac{15}{2} ).But let me write it in the box as requested.**Final Answer**The range of values for ( m ) is boxed{-dfrac{31}{3} < m < -dfrac{15}{2}}.

💡First, I need to determine the least common multiple (LCM) of 13 and 9 to find the smallest number that both 13 and 9 can divide into without leaving a remainder. Since 13 and 9 are coprime (they have no common factors other than 1), the LCM is simply their product: 13 × 9 = 117.Next, I need to identify how many multiples of 117 fall within the range of 200 to 500. I'll start by finding the smallest multiple of 117 that is greater than 200. Dividing 200 by 117 gives approximately 1.71, so the next whole number is 2. Multiplying 117 by 2 gives 234, which is the first multiple within the range.Continuing this process, I'll calculate subsequent multiples of 117:- 117 × 3 = 351- 117 × 4 = 468- 117 × 5 = 585The multiple 585 exceeds the upper limit of 500, so it is excluded from consideration. Therefore, the multiples of 117 between 200 and 500 are 234, 351, and 468.In conclusion, there are three integers between 200 and 500 that have both 13 and 9 as factors.

💡Alright, so I'm trying to solve this geometry problem, and it's a bit tricky, but I'll take it step by step. Let me first understand what's given and what's being asked.We have a triangle ( ABC ), and point ( D ) lies on side ( AC ). There are two circles involved: ( S_1 ) inscribed in triangle ( ABD ) and ( S_2 ) inscribed in triangle ( BCD ). The circle ( S_1 ) touches segment ( BD ) at point ( M ), and circle ( S_2 ) touches segment ( DC ) at point ( N ). We're told that the ratio of the radii of circles ( S_1 ) and ( S_2 ) is ( frac{7}{4} ). Additionally, we know that ( BM = 3 ), ( MN = 1 ), and ( ND = 1 ). The goal is to find the lengths of the sides of triangle ( ABC ).Okay, let's break this down. First, I need to visualize the triangle and the points. I imagine triangle ( ABC ) with ( D ) somewhere on ( AC ). The circles ( S_1 ) and ( S_2 ) are the incircles of triangles ( ABD ) and ( BCD ), respectively. The points ( M ) and ( N ) are the points where these incircles touch ( BD ) and ( DC ), respectively.Given the ratio of the radii ( frac{r_1}{r_2} = frac{7}{4} ), and the lengths ( BM = 3 ), ( MN = 1 ), ( ND = 1 ), I need to find the lengths of ( AB ), ( BC ), and ( AC ).Let me recall that the radius of an incircle of a triangle is related to the area and the semi-perimeter of the triangle. Specifically, ( r = frac{A}{s} ), where ( A ) is the area and ( s ) is the semi-perimeter.Since ( S_1 ) is the incircle of ( ABD ), its radius ( r_1 ) can be expressed as ( r_1 = frac{A_1}{s_1} ), where ( A_1 ) is the area of ( ABD ) and ( s_1 ) is its semi-perimeter. Similarly, ( r_2 = frac{A_2}{s_2} ) for ( S_2 ).Given the ratio ( frac{r_1}{r_2} = frac{7}{4} ), we can write ( frac{A_1 / s_1}{A_2 / s_2} = frac{7}{4} ), which simplifies to ( frac{A_1}{A_2} cdot frac{s_2}{s_1} = frac{7}{4} ).Now, let's think about the points ( M ) and ( N ). Since ( M ) is the point where ( S_1 ) touches ( BD ), and ( N ) is where ( S_2 ) touches ( DC ), we can use the properties of tangents from a point to a circle being equal.In triangle ( ABD ), the lengths from ( B ) to the points of tangency on ( BD ) and ( AB ) should be equal. Similarly, in triangle ( BCD ), the lengths from ( C ) to the points of tangency on ( DC ) and ( BC ) should be equal.Given ( BM = 3 ), and ( MN = 1 ), ( ND = 1 ), we can deduce that ( BD = BM + MN + ND = 3 + 1 + 1 = 5 ). So, ( BD = 5 ).Let me denote the lengths as follows:- In triangle ( ABD ), let the tangents from ( A ) be ( x ), from ( B ) be ( 3 ), and from ( D ) be ( y ).- In triangle ( BCD ), let the tangents from ( B ) be ( 3 ), from ( C ) be ( z ), and from ( D ) be ( 1 ).Wait, that might not be accurate. Let me think again.Actually, in triangle ( ABD ), the incircle ( S_1 ) touches ( AB ) at some point, say ( P ), ( BD ) at ( M ), and ( AD ) at some point, say ( Q ). Similarly, in triangle ( BCD ), the incircle ( S_2 ) touches ( BC ) at some point, say ( R ), ( CD ) at ( N ), and ( BD ) at some point, say ( S ).Given that ( BM = 3 ), ( MN = 1 ), and ( ND = 1 ), we can see that ( BD = BM + MN + ND = 3 + 1 + 1 = 5 ). So, ( BD = 5 ).In triangle ( ABD ), the lengths from ( B ) to the points of tangency on ( AB ) and ( BD ) are equal. Let's denote this common length as ( BM = 3 ). Similarly, in triangle ( BCD ), the lengths from ( C ) to the points of tangency on ( BC ) and ( CD ) are equal. Let's denote this common length as ( CN = 1 ) because ( ND = 1 ).Wait, that might not be correct. Let me clarify.In triangle ( ABD ), the incircle touches ( AB ) at ( P ), ( BD ) at ( M ), and ( AD ) at ( Q ). The lengths from ( B ) to ( P ) and ( M ) are equal, so ( BP = BM = 3 ). Similarly, the lengths from ( D ) to ( M ) and ( Q ) are equal, so ( DM = DQ ). But we know ( DM = BD - BM = 5 - 3 = 2 ), so ( DQ = 2 ).Similarly, in triangle ( BCD ), the incircle touches ( BC ) at ( R ), ( CD ) at ( N ), and ( BD ) at ( S ). The lengths from ( C ) to ( R ) and ( N ) are equal, so ( CR = CN = 1 ). The lengths from ( D ) to ( N ) and ( S ) are equal, so ( DN = DS = 1 ). Therefore, ( BS = BD - DS = 5 - 1 = 4 ).Now, let's denote the sides of triangle ( ABC ) as follows:- ( AB = c )- ( BC = a )- ( AC = b )We need to find ( a ), ( b ), and ( c ).From triangle ( ABD ):- ( AB = c = BP + AP = 3 + AP )- ( AD = AQ + DQ = AQ + 2 )- ( BD = 5 )From triangle ( BCD ):- ( BC = a = BR + CR = BR + 1 )- ( CD = DN + CN = 1 + 1 = 2 )- ( BD = 5 )Wait, that can't be right because ( CD = 2 ), but ( AC = AD + DC ), so ( AC = AD + 2 ). But we don't know ( AD ) yet.Let me try to express the semi-perimeters and areas of triangles ( ABD ) and ( BCD ) to use the ratio of the radii.For triangle ( ABD ):- Sides: ( AB = c ), ( BD = 5 ), ( AD = x ) (let's denote ( AD = x ))- Semi-perimeter ( s_1 = frac{c + 5 + x}{2} )- Area ( A_1 = r_1 cdot s_1 )For triangle ( BCD ):- Sides: ( BC = a ), ( CD = 2 ), ( BD = 5 )- Semi-perimeter ( s_2 = frac{a + 2 + 5}{2} = frac{a + 7}{2} )- Area ( A_2 = r_2 cdot s_2 )Given ( frac{r_1}{r_2} = frac{7}{4} ), so ( frac{A_1 / s_1}{A_2 / s_2} = frac{7}{4} ), which simplifies to ( frac{A_1}{A_2} cdot frac{s_2}{s_1} = frac{7}{4} ).Now, let's find expressions for ( A_1 ) and ( A_2 ) using Heron's formula.For triangle ( ABD ):( A_1 = sqrt{s_1(s_1 - c)(s_1 - 5)(s_1 - x)} )For triangle ( BCD ):( A_2 = sqrt{s_2(s_2 - a)(s_2 - 2)(s_2 - 5)} )This seems complicated. Maybe there's a better approach.Let me consider the lengths from the points of tangency.In triangle ( ABD ):- ( BP = BM = 3 )- ( AP = AQ )- ( DQ = DM = 2 )So, ( AB = BP + AP = 3 + AP )( AD = AQ + DQ = AP + 2 )Similarly, in triangle ( BCD ):- ( CR = CN = 1 )- ( BR = BS = 4 )- ( DS = DN = 1 )So, ( BC = BR + CR = 4 + 1 = 5 )( CD = CN + DN = 1 + 1 = 2 )Wait, so ( BC = 5 ) and ( CD = 2 ). Therefore, ( AC = AD + DC = AD + 2 ).But we need to find ( AD ).From triangle ( ABD ), we have:( AB = 3 + AP )( AD = AP + 2 )Let me denote ( AP = y ). Then:( AB = 3 + y )( AD = y + 2 )Now, the semi-perimeter of ( ABD ) is:( s_1 = frac{AB + BD + AD}{2} = frac{(3 + y) + 5 + (y + 2)}{2} = frac{2y + 10}{2} = y + 5 )The area ( A_1 = r_1 cdot s_1 = r_1(y + 5) )Similarly, for triangle ( BCD ):Sides are ( BC = 5 ), ( CD = 2 ), ( BD = 5 )Semi-perimeter ( s_2 = frac{5 + 2 + 5}{2} = frac{12}{2} = 6 )Area ( A_2 = r_2 cdot 6 )Given ( frac{r_1}{r_2} = frac{7}{4} ), so ( r_1 = frac{7}{4} r_2 )Now, let's express ( A_1 ) and ( A_2 ) using Heron's formula.For triangle ( ABD ):( A_1 = sqrt{s_1(s_1 - AB)(s_1 - BD)(s_1 - AD)} )( A_1 = sqrt{(y + 5)(y + 5 - (3 + y))(y + 5 - 5)(y + 5 - (y + 2))} )Simplify:( A_1 = sqrt{(y + 5)(2)(y)(3)} )( A_1 = sqrt{6y(y + 5)} )For triangle ( BCD ):( A_2 = sqrt{6(6 - 5)(6 - 2)(6 - 5)} )( A_2 = sqrt{6 cdot 1 cdot 4 cdot 1} )( A_2 = sqrt{24} = 2sqrt{6} )So, ( A_2 = 2sqrt{6} ) and ( A_2 = r_2 cdot 6 ), so:( 2sqrt{6} = 6 r_2 )( r_2 = frac{sqrt{6}}{3} )Then, ( r_1 = frac{7}{4} r_2 = frac{7}{4} cdot frac{sqrt{6}}{3} = frac{7sqrt{6}}{12} )Now, ( A_1 = r_1 cdot s_1 = frac{7sqrt{6}}{12} cdot (y + 5) )But we also have ( A_1 = sqrt{6y(y + 5)} )So, equate the two expressions:( sqrt{6y(y + 5)} = frac{7sqrt{6}}{12} (y + 5) )Divide both sides by ( sqrt{6} ):( sqrt{y(y + 5)} = frac{7}{12} (y + 5) )Square both sides:( y(y + 5) = left( frac{7}{12} right)^2 (y + 5)^2 )( y(y + 5) = frac{49}{144} (y + 5)^2 )Assuming ( y + 5 neq 0 ), we can divide both sides by ( y + 5 ):( y = frac{49}{144} (y + 5) )Multiply both sides by 144:( 144y = 49(y + 5) )( 144y = 49y + 245 )( 144y - 49y = 245 )( 95y = 245 )( y = frac{245}{95} = frac{49}{19} )So, ( y = frac{49}{19} )Now, recall that:( AB = 3 + y = 3 + frac{49}{19} = frac{57}{19} + frac{49}{19} = frac{106}{19} )( AD = y + 2 = frac{49}{19} + 2 = frac{49}{19} + frac{38}{19} = frac{87}{19} )Therefore, ( AC = AD + DC = frac{87}{19} + 2 = frac{87}{19} + frac{38}{19} = frac{125}{19} )Wait, but earlier we found ( BC = 5 ). So, the sides are:- ( AB = frac{106}{19} )- ( BC = 5 )- ( AC = frac{125}{19} )But this seems a bit messy. Let me check my calculations.Wait, when I calculated ( BC ), I assumed ( BC = BR + CR = 4 + 1 = 5 ). That seems correct because in triangle ( BCD ), the incircle touches ( BC ) at ( R ) and ( CD ) at ( N ), so ( BR = BS = 4 ) and ( CR = CN = 1 ).Similarly, in triangle ( ABD ), ( AP = AQ = y ), ( BP = BM = 3 ), and ( DQ = DM = 2 ). So, ( AB = 3 + y ), ( AD = y + 2 ), and ( BD = 5 ).Then, the semi-perimeter ( s_1 = y + 5 ), and the area ( A_1 = sqrt{6y(y + 5)} ). Also, ( A_1 = r_1 s_1 = frac{7sqrt{6}}{12} (y + 5) ).Equating these gives ( sqrt{6y(y + 5)} = frac{7sqrt{6}}{12} (y + 5) ), which simplifies to ( y = frac{49}{19} ).So, ( AB = 3 + frac{49}{19} = frac{106}{19} ), ( AD = frac{49}{19} + 2 = frac{87}{19} ), and ( AC = frac{87}{19} + 2 = frac{125}{19} ).But ( AC ) is ( frac{125}{19} ), which is approximately 6.5789, and ( AB ) is approximately 5.5789. However, ( BC = 5 ), so the sides are ( AB approx 5.58 ), ( BC = 5 ), and ( AC approx 6.58 ).This seems plausible, but let me check if the ratio of the radii is indeed ( frac{7}{4} ).We found ( r_2 = frac{sqrt{6}}{3} ) and ( r_1 = frac{7sqrt{6}}{12} ). So, ( frac{r_1}{r_2} = frac{frac{7sqrt{6}}{12}}{frac{sqrt{6}}{3}} = frac{7}{4} ), which matches the given ratio. So, that checks out.Therefore, the sides of triangle ( ABC ) are:- ( AB = frac{106}{19} )- ( BC = 5 )- ( AC = frac{125}{19} )But these fractions can be simplified or expressed as decimals, but since the problem doesn't specify, I'll leave them as fractions.Wait, but ( frac{106}{19} ) is approximately 5.5789, and ( frac{125}{19} ) is approximately 6.5789. However, the problem might expect integer lengths. Maybe I made a mistake somewhere.Let me re-examine my steps.When I calculated ( y ), I got ( y = frac{49}{19} ). Let me verify that.From the equation:( y = frac{49}{144} (y + 5) )Multiply both sides by 144:( 144y = 49y + 245 )( 95y = 245 )( y = frac{245}{95} = frac{49}{19} )That seems correct.Alternatively, maybe there's a different approach to find integer lengths.Let me consider that ( AC = AD + DC = y + 2 + 2 = y + 4 ). Wait, no, ( DC = 2 ), so ( AC = AD + DC = (y + 2) + 2 = y + 4 ). Wait, no, ( AD = y + 2 ), and ( DC = 2 ), so ( AC = y + 2 + 2 = y + 4 ). Wait, no, ( AC = AD + DC = (y + 2) + 2 = y + 4 ). But earlier, I had ( AC = frac{125}{19} ), which is approximately 6.5789, so ( y + 4 = frac{125}{19} ), so ( y = frac{125}{19} - 4 = frac{125 - 76}{19} = frac{49}{19} ), which matches.So, perhaps the sides are indeed fractions. Alternatively, maybe I made a mistake in assuming the lengths.Wait, in triangle ( BCD ), I assumed ( BC = 5 ), but let me double-check.In triangle ( BCD ), the incircle touches ( BC ) at ( R ) and ( CD ) at ( N ). The lengths from ( C ) to ( R ) and ( N ) are equal, so ( CR = CN = 1 ). Similarly, the lengths from ( B ) to ( R ) and ( S ) (the point where the incircle touches ( BD )) are equal, so ( BR = BS = 4 ). Therefore, ( BC = BR + CR = 4 + 1 = 5 ). That seems correct.Similarly, in triangle ( ABD ), ( AB = 3 + y ), ( AD = y + 2 ), and ( BD = 5 ). So, the sides are ( AB = 3 + y ), ( BD = 5 ), and ( AD = y + 2 ).Given that, and the ratio of the radii, we found ( y = frac{49}{19} ), leading to ( AB = frac{106}{19} ), ( AD = frac{87}{19} ), and ( AC = frac{125}{19} ).Alternatively, perhaps the problem expects integer lengths, and I might have made an error in the initial setup.Let me consider another approach. Maybe using the lengths ( BM = 3 ), ( MN = 1 ), ( ND = 1 ), so ( BD = 5 ). Then, in triangle ( ABD ), the incircle touches ( BD ) at ( M ), so ( BM = 3 ), and in triangle ( BCD ), the incircle touches ( DC ) at ( N ), so ( DN = 1 ).Let me denote the semi-perimeters and areas again.For triangle ( ABD ):- Let ( AB = c ), ( BD = 5 ), ( AD = x )- Semi-perimeter ( s_1 = frac{c + 5 + x}{2} )- Area ( A_1 = r_1 s_1 )For triangle ( BCD ):- ( BC = a ), ( CD = 2 ), ( BD = 5 )- Semi-perimeter ( s_2 = frac{a + 2 + 5}{2} = frac{a + 7}{2} )- Area ( A_2 = r_2 s_2 )Given ( frac{r_1}{r_2} = frac{7}{4} ), so ( frac{A_1 / s_1}{A_2 / s_2} = frac{7}{4} ), which simplifies to ( frac{A_1}{A_2} cdot frac{s_2}{s_1} = frac{7}{4} ).Now, using Heron's formula for both triangles.For ( ABD ):( A_1 = sqrt{s_1(s_1 - c)(s_1 - 5)(s_1 - x)} )For ( BCD ):( A_2 = sqrt{s_2(s_2 - a)(s_2 - 2)(s_2 - 5)} )But this seems too complex. Maybe I can use the fact that the ratio of the areas is related to the ratio of the radii and semi-perimeters.Alternatively, perhaps using the formula for the radius in terms of the area and semi-perimeter.Given ( r_1 = frac{A_1}{s_1} ) and ( r_2 = frac{A_2}{s_2} ), and ( frac{r_1}{r_2} = frac{7}{4} ), so ( frac{A_1}{s_1} / frac{A_2}{s_2} = frac{7}{4} ), which is ( frac{A_1}{A_2} cdot frac{s_2}{s_1} = frac{7}{4} ).Now, let's find ( A_1 ) and ( A_2 ) in terms of ( y ).From earlier, we have:- ( AB = 3 + y )- ( AD = y + 2 )- ( BD = 5 )So, ( s_1 = frac{(3 + y) + 5 + (y + 2)}{2} = frac{2y + 10}{2} = y + 5 )( A_1 = sqrt{(y + 5)(y + 5 - (3 + y))(y + 5 - 5)(y + 5 - (y + 2))} )Simplify:( A_1 = sqrt{(y + 5)(2)(y)(3)} = sqrt{6y(y + 5)} )For triangle ( BCD ):- ( BC = 5 )- ( CD = 2 )- ( BD = 5 )So, ( s_2 = frac{5 + 2 + 5}{2} = 6 )( A_2 = sqrt{6(6 - 5)(6 - 2)(6 - 5)} = sqrt{6 cdot 1 cdot 4 cdot 1} = sqrt{24} = 2sqrt{6} )So, ( A_2 = 2sqrt{6} )Now, ( r_2 = frac{A_2}{s_2} = frac{2sqrt{6}}{6} = frac{sqrt{6}}{3} )Given ( frac{r_1}{r_2} = frac{7}{4} ), so ( r_1 = frac{7}{4} r_2 = frac{7}{4} cdot frac{sqrt{6}}{3} = frac{7sqrt{6}}{12} )Now, ( A_1 = r_1 s_1 = frac{7sqrt{6}}{12} cdot (y + 5) )But we also have ( A_1 = sqrt{6y(y + 5)} )So, equate the two:( sqrt{6y(y + 5)} = frac{7sqrt{6}}{12} (y + 5) )Divide both sides by ( sqrt{6} ):( sqrt{y(y + 5)} = frac{7}{12} (y + 5) )Square both sides:( y(y + 5) = left( frac{7}{12} right)^2 (y + 5)^2 )( y(y + 5) = frac{49}{144} (y + 5)^2 )Assuming ( y + 5 neq 0 ), divide both sides by ( y + 5 ):( y = frac{49}{144} (y + 5) )Multiply both sides by 144:( 144y = 49y + 245 )( 95y = 245 )( y = frac{245}{95} = frac{49}{19} )So, ( y = frac{49}{19} )Therefore:- ( AB = 3 + y = 3 + frac{49}{19} = frac{57}{19} + frac{49}{19} = frac{106}{19} )- ( AD = y + 2 = frac{49}{19} + frac{38}{19} = frac{87}{19} )- ( AC = AD + DC = frac{87}{19} + 2 = frac{87}{19} + frac{38}{19} = frac{125}{19} )So, the sides of triangle ( ABC ) are:- ( AB = frac{106}{19} )- ( BC = 5 )- ( AC = frac{125}{19} )These are the exact values, but they are fractions. If we convert them to decimals:- ( AB approx 5.5789 )- ( BC = 5 )- ( AC approx 6.5789 )This seems consistent with the given ratio of the radii and the lengths provided. Therefore, the sides of triangle ( ABC ) are ( AB = frac{106}{19} ), ( BC = 5 ), and ( AC = frac{125}{19} ).

💡Alright, so I have this problem where I need to arrange three expressions, a, b, and c, in ascending order. The expressions are given as:a = (2 tan 70°) / (1 + tan² 70°)b = √[(1 + cos 109°)/2]c = (√3/2) cos 81° + (1/2) sin 99°Hmm, okay. I need to figure out the values of a, b, and c and then compare them. Let me take it step by step.Starting with a. The expression for a is (2 tan 70°) / (1 + tan² 70°). I remember there's a trigonometric identity that relates tan and sin. Specifically, I think it's something like sin(2θ) = 2 tan θ / (1 + tan² θ). Let me verify that.Yes, indeed, the double-angle formula for sine is sin(2θ) = 2 sin θ cos θ. But another way to write this is using tan θ. Since tan θ = sin θ / cos θ, then 2 tan θ / (1 + tan² θ) would be 2 (sin θ / cos θ) / (1 + (sin² θ / cos² θ)).Simplifying the denominator: 1 + (sin² θ / cos² θ) = (cos² θ + sin² θ) / cos² θ = 1 / cos² θ.So, 2 tan θ / (1 + tan² θ) = 2 (sin θ / cos θ) / (1 / cos² θ) = 2 sin θ / cos θ * cos² θ = 2 sin θ cos θ = sin(2θ). So, yes, that identity holds.Therefore, a = sin(2 * 70°) = sin(140°). Now, sin(140°) is the same as sin(180° - 40°) = sin(40°) because sine is positive in the second quadrant and sin(180° - θ) = sin θ. So, a = sin(40°). I can note that down.Moving on to b. The expression is √[(1 + cos 109°)/2]. I recall that there's a half-angle identity for cosine: cos(θ/2) = √[(1 + cos θ)/2]. So, if I let θ = 109°, then cos(109°/2) = √[(1 + cos 109°)/2]. Therefore, b = cos(54.5°).But wait, 54.5° is in the first quadrant, so cosine is positive there. Alternatively, since cos(θ) = sin(90° - θ), we can write cos(54.5°) as sin(35.5°). So, b = sin(35.5°). That might be helpful for comparison later.Now, onto c. The expression is (√3/2) cos 81° + (1/2) sin 99°. Hmm, this looks like a combination of sine and cosine terms. Maybe I can express this as a single sine or cosine function using the formula for sine or cosine addition.Let me recall that A cos θ + B sin θ can be written as C sin(θ + φ) or C cos(θ - φ), where C = √(A² + B²) and φ is some angle. Let me try that.First, let's identify A and B. Here, A = √3/2 and B = 1/2. So, C = √[(√3/2)² + (1/2)²] = √[(3/4) + (1/4)] = √[1] = 1. Interesting, so C = 1.Now, to find φ, we can use tan φ = A/B. Wait, actually, depending on the formula. Let me think. If I write it as sin(θ + φ), then:sin(θ + φ) = sin θ cos φ + cos θ sin φ.Comparing with our expression, which is (√3/2) cos 81° + (1/2) sin 99°, but wait, the angles are different. 81° and 99°. Hmm, that complicates things.Alternatively, maybe I can use cofunction identities or angle addition formulas. Let me see.Wait, 99° is 90° + 9°, so sin 99° = sin(90° + 9°) = cos 9°. Similarly, cos 81° = cos(90° - 9°) = sin 9°. So, substituting these into c:c = (√3/2) sin 9° + (1/2) cos 9°Now, this looks like a standard sine addition formula. Let me recall that sin(A + B) = sin A cos B + cos A sin B. But here, we have coefficients √3/2 and 1/2, which correspond to cos 30° and sin 30°, respectively, since cos 30° = √3/2 and sin 30° = 1/2.So, c = sin 9° cos 30° + cos 9° sin 30° = sin(9° + 30°) = sin(39°). Therefore, c = sin(39°).Alright, so now I have:a = sin(40°)b = sin(35.5°)c = sin(39°)Now, I need to compare these three values. Since sine is an increasing function in the interval [0°, 90°], the larger the angle, the larger the sine value. So, sin(35.5°) < sin(39°) < sin(40°).Therefore, b < c < a.Wait, let me double-check to make sure I didn't make any mistakes in simplifying.For a: (2 tan 70°)/(1 + tan² 70°) = sin(140°) = sin(40°). That seems correct.For b: √[(1 + cos 109°)/2] = cos(54.5°) = sin(35.5°). That also seems correct.For c: (√3/2) cos 81° + (1/2) sin 99° = (√3/2) sin 9° + (1/2) cos 9° = sin(39°). Yes, that looks right.So, the order is b < c < a.I think that's the correct arrangement.

💡Okay, so I have this problem where a child walks to school at two different speeds and either arrives late or early. I need to find out how far the school is from the child's home. Let me try to understand what's going on here.First, when the child walks at 5 meters per minute, he's 6 minutes late. That means if he walked at a certain ideal speed, he'd be on time, but at 5 m/min, it takes him 6 extra minutes. On the other hand, when he walks faster, at 7 meters per minute, he arrives 30 minutes early. So, walking faster gets him to school much sooner than needed.I think I need to set up some equations to model this situation. Let's denote the distance from home to school as D (in meters) and the time it should ideally take to reach school on time as T (in minutes). When he walks at 5 m/min and is late, the time he takes is T + 6 minutes. So, the distance D can be expressed as:[ D = 5 times (T + 6) ]That's because distance equals speed multiplied by time.Similarly, when he walks at 7 m/min and arrives early, the time he takes is T - 30 minutes. So, the distance D can also be expressed as:[ D = 7 times (T - 30) ]Now I have two equations:1. ( D = 5(T + 6) )2. ( D = 7(T - 30) )Since both expressions equal D, I can set them equal to each other:[ 5(T + 6) = 7(T - 30) ]Let me expand both sides:[ 5T + 30 = 7T - 210 ]Now, I'll solve for T. Subtract 5T from both sides:[ 30 = 2T - 210 ]Then, add 210 to both sides:[ 240 = 2T ]Divide both sides by 2:[ T = 120 ]So, the ideal time to reach school on time is 120 minutes. Now, I can plug this back into one of the original equations to find D. Let's use the first equation:[ D = 5(T + 6) ][ D = 5(120 + 6) ][ D = 5 times 126 ][ D = 630 ]Therefore, the distance from the child's home to the school is 630 meters.Wait, let me double-check to make sure I didn't make any mistakes. If T is 120 minutes, then walking at 5 m/min takes 126 minutes, and the distance is 5 * 126 = 630 meters. Walking at 7 m/min takes 90 minutes, and the distance is 7 * 90 = 630 meters. Both give the same distance, so that checks out.I think that makes sense. The child needs to cover 630 meters to get to school. At 5 m/min, it takes him 126 minutes, which is 6 minutes more than the ideal 120 minutes. At 7 m/min, it takes him 90 minutes, which is 30 minutes less than the ideal time. So, everything seems consistent.I don't see any errors in my calculations, so I'm confident that the distance is indeed 630 meters.

💡Okay, so I have this problem about the Yi Rong ferry traveling between points A and B. Let me try to understand what's going on here.First, the ferry's speed is 40 km/h. On odd-numbered days, it goes downstream from A to B, and on even-numbered days, it goes upstream from B to A. The water speed is 24 km/h. So, when it's going downstream, the current helps it, and when it's going upstream, it's against the current.On one odd-numbered day, the ferry lost power at the midpoint to point C and drifted to point B. The captain recorded the travel time as being 43/18 times the usual time for an odd-numbered day. Hmm, okay. So, normally, it takes a certain time to go from A to B downstream. But when it lost power halfway, it had to drift the rest of the way, and that took longer—43/18 times longer.Then, on another even-numbered day, the ferry also lost power at the midpoint to point C. The repair crew worked for 1 hour while drifting, and then the ferry continued to point A at double its normal speed. The captain found that the travel time was exactly the same as the usual time for an even-numbered day. Interesting. So, despite losing power and having a repair, the total time was the same as usual.I need to find the distance between A and B.Let me define some variables:Let’s let x be the distance between A and B in kilometers.Let’s let y be the distance from A to C, which is the midpoint where the ferry lost power. So, if the total distance is x, then y should be x/2, right? Because it's the midpoint. Wait, but in the problem, it says "the midpoint to point C." So, maybe point C is the midpoint? So, y = x/2. Hmm, but in the solution above, y was 32 km, and x was 192 km. So, 32 is not x/2, which would be 96. Hmm, maybe I need to think differently.Wait, maybe point C isn't necessarily the midpoint. Maybe it's a point somewhere along the route, but the problem says "the midpoint to point C." Hmm, maybe it's the midpoint between A and B, so y = x/2. But in the solution, y was 32, and x was 192, so y would be 96. But in the solution, y was 32. Hmm, maybe I need to clarify.Wait, the problem says "the midpoint to point C." Maybe that means point C is the midpoint between A and B, so y = x/2. So, if x is 192, then y is 96. But in the solution, y was 32. Hmm, maybe I'm misunderstanding.Wait, let me read the problem again."On one odd-numbered day, the ferry lost power at the midpoint to point C and drifted to point B. The captain recorded the travel time as being 43/18 times the usual time for an odd-numbered day."So, it lost power at the midpoint to point C. So, maybe point C is the midpoint, so y = x/2. Then, it drifted from C to B, which is the other half of the distance, so x/2.Similarly, on the even-numbered day, it lost power at the midpoint to point C, so again, point C is the midpoint, so y = x/2.Wait, but in the solution above, y was 32, and x was 192, so y would be 96, but in the solution, y was 32. So, maybe point C is not the midpoint, but rather, the ferry lost power at a point C, which is the midpoint of the journey, but not necessarily the midpoint of the distance. Hmm, that might complicate things.Wait, maybe the midpoint is in terms of time, not distance. So, the ferry lost power halfway through the usual time, not halfway through the distance. Hmm, that could be.But the problem says "the midpoint to point C," which sounds like a point, so probably the midpoint in distance.Wait, maybe I need to clarify.Let me try to define variables again.Let x be the distance from A to B.Let y be the distance from A to C, which is the midpoint, so y = x/2.So, on the odd-numbered day, the ferry goes from A to C at downstream speed, then loses power and drifts from C to B.On the even-numbered day, the ferry goes from B to C at upstream speed, then loses power, drifts for 1 hour, then repairs and goes from wherever it is to A at double speed.Wait, but in the problem, on the even-numbered day, it lost power at the midpoint to point C, so maybe it was going from B to C, lost power, drifted for 1 hour, then repaired and went to A at double speed.Wait, but the problem says "the ferry also lost power at the midpoint to point C." So, maybe point C is the midpoint, so y = x/2.So, let's proceed with y = x/2.So, on the odd-numbered day, the ferry goes from A to C at downstream speed, then drifts from C to B.The usual time for an odd-numbered day is x / (40 + 24) = x / 64 hours.But on this day, it went from A to C, which is y = x/2, at downstream speed, then drifted from C to B, which is also y = x/2, at the speed of the current, which is 24 km/h.So, the time taken on this day is (y / 64) + (y / 24).And this time is equal to (43/18) times the usual time.So, equation 1: (y / 64) + (y / 24) = (43/18) * (x / 64)But since y = x/2, we can substitute that in.So, (x/2 / 64) + (x/2 / 24) = (43/18) * (x / 64)Simplify:(x / 128) + (x / 48) = (43x) / (18 * 64)Find a common denominator for the left side. 128 and 48 have a common denominator of 384.So, (3x / 384) + (8x / 384) = (43x) / 1152Combine terms:11x / 384 = 43x / 1152Multiply both sides by 1152 to eliminate denominators:11x * 3 = 43x33x = 43xWait, that can't be right. 33x = 43x implies 0 = 10x, which would mean x = 0, which doesn't make sense.Hmm, so maybe my assumption that y = x/2 is incorrect.Wait, maybe point C is not the midpoint in distance, but rather, the midpoint in time. So, the ferry lost power halfway through the usual time, not halfway through the distance.So, let's redefine y as the distance from A to C, which is such that the time taken to go from A to C is half the usual time.The usual time for A to B is x / 64.So, half the usual time is (x / 64) / 2 = x / 128.So, the distance y is the distance covered in x / 128 hours at downstream speed.So, y = 64 * (x / 128) = x / 2.Wait, so y is still x/2. So, same result.Hmm, but that leads to the same equation, which gives 33x = 43x, which is impossible.So, maybe my initial approach is wrong.Wait, let's look at the solution provided.In the solution, they defined y as the distance from A to C, and set up the equation:(x / 64) * (43 / 18) = (y / 64) + (x - y) / 24So, they didn't assume y = x/2.So, maybe point C is not the midpoint, but rather, a point such that the time taken to go from A to C is such that the total time is 43/18 times the usual time.So, let's proceed without assuming y = x/2.So, let me define:x = distance from A to B.y = distance from A to C.So, on the odd-numbered day, the ferry goes from A to C at downstream speed (40 + 24 = 64 km/h), then drifts from C to B at the speed of the current, which is 24 km/h.The usual time for A to B is x / 64.The time taken on this day is (y / 64) + ((x - y) / 24).And this time is equal to (43 / 18) * (x / 64).So, equation 1:(y / 64) + ((x - y) / 24) = (43 / 18) * (x / 64)Similarly, on the even-numbered day, the ferry goes from B to C at upstream speed (40 - 24 = 16 km/h), then loses power, drifts for 1 hour, then repairs and goes from wherever it is to A at double speed, which is 80 km/h.The usual time for B to A is x / 16.The time taken on this day is (y / 16) + 1 + ((x - y - 24 * 1) / 80)Wait, because when it loses power at C, it drifts for 1 hour, so it covers 24 km in that hour, so the remaining distance to A is x - y - 24.Then, it repairs and goes at double speed, which is 80 km/h, so the time is (x - y - 24) / 80.And the total time is equal to the usual time, which is x / 16.So, equation 2:(y / 16) + 1 + ((x - y - 24) / 80) = x / 16Now, we have two equations:1) (y / 64) + ((x - y) / 24) = (43 / 18) * (x / 64)2) (y / 16) + 1 + ((x - y - 24) / 80) = x / 16Let me simplify equation 1 first.Multiply both sides by 64 to eliminate denominators:y + (64 / 24)(x - y) = (43 / 18) xSimplify 64 / 24 = 8 / 3.So:y + (8/3)(x - y) = (43 / 18) xMultiply through:y + (8/3)x - (8/3)y = (43/18)xCombine like terms:(1 - 8/3)y + (8/3)x = (43/18)xConvert 1 to 3/3:(3/3 - 8/3)y + (8/3)x = (43/18)x(-5/3)y + (8/3)x = (43/18)xMultiply both sides by 18 to eliminate denominators:-30y + 48x = 43xSimplify:-30y + 48x = 43x-30y = -5xDivide both sides by -5:6y = xSo, x = 6yOkay, so equation 1 gives us x = 6y.Now, let's simplify equation 2.Equation 2:(y / 16) + 1 + ((x - y - 24) / 80) = x / 16Multiply all terms by 80 to eliminate denominators:5y + 80 + (x - y - 24) = 5xSimplify:5y + 80 + x - y - 24 = 5xCombine like terms:(5y - y) + (80 - 24) + x = 5x4y + 56 + x = 5xBring x to the right:4y + 56 = 4xDivide both sides by 4:y + 14 = xSo, equation 2 gives us x = y + 14Now, from equation 1, we have x = 6ySo, substitute x = 6y into equation 2:6y = y + 14Subtract y:5y = 14y = 14 / 5 = 2.8 kmWait, that can't be right because in the solution above, y was 32 km, and x was 192 km.Hmm, so I must have made a mistake.Wait, let me check my steps.In equation 2, I had:(y / 16) + 1 + ((x - y - 24) / 80) = x / 16Multiply all terms by 80:5y + 80 + (x - y - 24) = 5xWait, let me verify:(y / 16) * 80 = 5y1 * 80 = 80((x - y - 24) / 80) * 80 = x - y - 24x / 16 * 80 = 5xSo, yes, that's correct.Then, 5y + 80 + x - y - 24 = 5xCombine like terms:(5y - y) + (80 - 24) + x = 5x4y + 56 + x = 5xThen, 4y + 56 = 4xDivide by 4:y + 14 = xSo, x = y + 14From equation 1, x = 6ySo, 6y = y + 145y = 14y = 14 / 5 = 2.8 kmBut in the solution above, y was 32 km, x was 192 km.So, clearly, I have a mistake here.Wait, maybe I misapplied the equation for the even-numbered day.Let me re-examine the even-numbered day scenario.On the even-numbered day, the ferry goes from B to A, which is upstream, so speed is 16 km/h.It loses power at point C, which is y km from A, so it's (x - y) km from B.Wait, no, if y is the distance from A to C, then from B to C is x - y.So, the ferry goes from B to C at upstream speed, which is 16 km/h, so time is (x - y)/16.Then, it loses power at C, drifts for 1 hour, so it moves downstream at 24 km/h for 1 hour, covering 24 km.So, after drifting, it is at point D, which is 24 km downstream from C, so distance from A is y + 24 km.Then, it repairs and continues to A at double speed, which is 80 km/h.So, the remaining distance to A is y + 24 km.Wait, no, if it was at C, which is y km from A, and it drifts downstream for 1 hour, it moves towards B, so it's now at y - 24 km from A? Wait, no.Wait, if it's going downstream, it's moving away from A towards B. So, if it was at C, which is y km from A, drifting downstream for 1 hour at 24 km/h, it would be at y + 24 km from A.But if y + 24 km exceeds x, that would be beyond B, which doesn't make sense.Wait, but in the problem, it says "the ferry also lost power at the midpoint to point C." So, maybe point C is the midpoint, so y = x/2.Wait, but earlier, that led to a contradiction.Wait, maybe I need to clarify the direction.When going upstream from B to A, the ferry loses power at point C, which is y km from A, so it's (x - y) km from B.Then, it drifts downstream for 1 hour, so it moves towards B at 24 km/h, covering 24 km.So, after drifting, it's at (x - y) - 24 km from B, which is y + 24 km from A.Wait, no, if it was at C, which is y km from A, and it drifts downstream for 1 hour, it moves towards B, so it's now at y + 24 km from A, but that can't be because y + 24 might exceed x.Wait, perhaps I need to think differently.Wait, maybe when it loses power at C, it's y km from A, so (x - y) km from B.Then, it drifts downstream for 1 hour, so it moves towards B, covering 24 km, so it's now at (x - y) - 24 km from B, which is y + 24 km from A.But if y + 24 > x, that would mean it's beyond B, which is not possible.So, perhaps y + 24 <= x.Wait, but in the problem, it says "the repair crew worked for 1 hour while drifting, and then the ferry continued to point A at double its normal speed."So, after drifting for 1 hour, it's somewhere between C and B, but then it repairs and goes back to A.Wait, but if it's drifting downstream, it's moving towards B, so after drifting, it's further away from A, so to get back to A, it has to go upstream again, but at double speed.Wait, that seems complicated.Alternatively, maybe after losing power at C, it drifts for 1 hour, which takes it to some point D, then repairs and goes from D to A at double speed.So, the total time is time from B to C upstream, plus 1 hour drifting, plus time from D to A at 80 km/h.And this total time equals the usual time from B to A, which is x / 16.So, let's define:Time from B to C: (x - y) / 16Time drifting: 1 hourDistance covered while drifting: 24 km downstream, so from C to D, which is 24 km towards B.So, distance from D to A: y + 24 kmTime from D to A: (y + 24) / 80So, total time:(x - y)/16 + 1 + (y + 24)/80 = x / 16So, equation 2:(x - y)/16 + 1 + (y + 24)/80 = x / 16Now, let's simplify this equation.Multiply all terms by 80 to eliminate denominators:5(x - y) + 80 + (y + 24) = 5xExpand:5x - 5y + 80 + y + 24 = 5xCombine like terms:5x - 4y + 104 = 5xSubtract 5x from both sides:-4y + 104 = 0-4y = -104y = 26So, y = 26 kmNow, from equation 1, we had x = 6ySo, x = 6 * 26 = 156 kmWait, but in the solution above, x was 192 km. So, something is still off.Wait, let me check equation 1 again.Equation 1:(y / 64) + ((x - y) / 24) = (43 / 18) * (x / 64)We had:y + (8/3)(x - y) = (43/18)xWhich led to x = 6yBut with y = 26, x = 156Let me check if this satisfies equation 1.Left side: (26 / 64) + ((156 - 26) / 24) = (26 / 64) + (130 / 24)Convert to decimals:26 / 64 ≈ 0.40625130 / 24 ≈ 5.416666...Total ≈ 5.82291666...Right side: (43 / 18) * (156 / 64)Calculate 156 / 64 ≈ 2.437543 / 18 ≈ 2.388888...Multiply: 2.4375 * 2.388888 ≈ 5.82291666...So, yes, it matches.So, x = 156 km, y = 26 kmBut in the solution above, x was 192 km, y was 32 kmSo, why the discrepancy?Wait, maybe I made a mistake in equation 2.Wait, in equation 2, I had:(x - y)/16 + 1 + (y + 24)/80 = x / 16But when I multiplied by 80, I got:5(x - y) + 80 + (y + 24) = 5xWhich simplifies to:5x - 5y + 80 + y + 24 = 5xWhich is:5x - 4y + 104 = 5xSo, -4y + 104 = 0y = 26But in the solution above, y was 32, leading to x = 192Wait, maybe the solution above had a different approach.Wait, in the solution above, they had:From equation 1:240y = 341xFrom equation 2:y = 32So, x = (240 * 32) / 341 ≈ 21.92But then they said that didn't make sense, so they assumed x = 192But that seems inconsistent.Wait, perhaps I need to re-examine the setup.Wait, in the problem, on the even-numbered day, after losing power at C, the ferry drifts for 1 hour, then repairs and continues to A at double speed.So, the total time is time from B to C upstream, plus 1 hour drifting, plus time from D to A at 80 km/h.But in my equation, I assumed that after drifting, the ferry is at y + 24 km from A, but that might not be correct.Wait, if the ferry is at C, which is y km from A, and it drifts downstream for 1 hour, it moves towards B, so it's now at y + 24 km from A, but if y + 24 > x, that's beyond B, which is not possible.So, perhaps y + 24 <= xBut in our case, y = 26, x = 156, so y + 24 = 50, which is less than 156, so it's okay.So, the ferry is at 50 km from A, then repairs and goes back to A at 80 km/h, so time is 50 / 80 = 0.625 hoursSo, total time:From B to C: (156 - 26)/16 = 130 / 16 ≈ 8.125 hoursDrifting: 1 hourFrom D to A: 50 / 80 = 0.625 hoursTotal: 8.125 + 1 + 0.625 = 9.75 hoursUsual time from B to A: 156 / 16 = 9.75 hoursSo, it matches.So, x = 156 km, y = 26 kmBut in the solution above, x was 192 km, y was 32 kmSo, why the difference?Wait, perhaps the solution above had a miscalculation.Wait, in the solution above, they had:From equation 1:240y = 341xFrom equation 2:y = 32So, x = (240 * 32) / 341 ≈ 21.92But that's inconsistent because with x = 192, y = 32, let's check equation 1.Equation 1:(y / 64) + ((x - y) / 24) = (43 / 18) * (x / 64)Plug in y = 32, x = 192Left side: 32 / 64 + (192 - 32)/24 = 0.5 + 160 / 24 ≈ 0.5 + 6.6667 ≈ 7.1667Right side: (43 / 18) * (192 / 64) = (43 / 18) * 3 = 43 / 6 ≈ 7.1667So, it matches.But in my approach, I got x = 156, y = 26, which also satisfies both equations.So, why the difference?Wait, perhaps the solution above had a different interpretation of the problem.Wait, in the problem, on the even-numbered day, after losing power at C, the repair crew worked for 1 hour while drifting, and then the ferry continued to point A at double its normal speed.So, the total time is time from B to C upstream, plus 1 hour drifting, plus time from D to A at 80 km/h.But in my equation, I assumed that after drifting, the ferry is at y + 24 km from A, but in reality, if the ferry is drifting for 1 hour, it's moving downstream, so it's moving away from A towards B.So, if it was at C, which is y km from A, after drifting for 1 hour, it's at y + 24 km from A.But then, to get back to A, it has to go upstream from y + 24 km to A, which is y + 24 km.But in the problem, it says "the ferry continued to point A at double its normal speed."So, the distance from D to A is y + 24 km, and the speed is 80 km/h, so time is (y + 24)/80.But in the solution above, they had:After drifting for 1 hour, the remaining distance is (x - y - 24) km.Wait, that would be if the ferry had drifted towards B, so the distance from D to A is x - (x - y - 24) = y + 24 km.Wait, no, if the ferry was at C, which is y km from A, and it drifts downstream for 1 hour, it's now at y + 24 km from A.So, the distance from D to A is y + 24 km.But in the solution above, they had:After drifting, the remaining distance is (x - y - 24) km.Wait, that would be if the ferry had drifted towards B, so the distance from D to A is x - (x - y - 24) = y + 24 km.Wait, no, that's the same as before.Wait, perhaps the solution above had a different interpretation.Wait, in the solution above, they had:From equation 1:240y = 341xFrom equation 2:y = 32So, x = (240 * 32) / 341 ≈ 21.92But that's inconsistent because with x = 192, y = 32, equation 1 is satisfied.Wait, perhaps the solution above had a miscalculation in equation 1.Wait, in the solution above, they had:From equation 1:43x = 48(8x - 5y)Which led to 43x = 384x - 240yThen, 240y = 341xBut in my approach, I had:From equation 1:(y / 64) + ((x - y) / 24) = (43 / 18) * (x / 64)Which led to x = 6ySo, which one is correct?Wait, let's re-examine equation 1.Equation 1:(y / 64) + ((x - y) / 24) = (43 / 18) * (x / 64)Multiply both sides by 64:y + (64 / 24)(x - y) = (43 / 18)xSimplify 64 / 24 = 8/3So:y + (8/3)(x - y) = (43/18)xMultiply through:y + (8/3)x - (8/3)y = (43/18)xCombine like terms:(1 - 8/3)y + (8/3)x = (43/18)xConvert 1 to 3/3:(3/3 - 8/3)y + (8/3)x = (43/18)x(-5/3)y + (8/3)x = (43/18)xMultiply both sides by 18 to eliminate denominators:-30y + 48x = 43xSo:-30y + 48x = 43x-30y = -5xDivide both sides by -5:6y = xSo, x = 6ySo, that's correct.In the solution above, they had:From equation 1:43x = 48(8x - 5y)Which led to 240y = 341xBut that seems incorrect.Wait, let me see.In the solution above, they had:From equation 1:43x = 48(8x - 5y)Which is:43x = 384x - 240yThen, 240y = 384x - 43x = 341xSo, 240y = 341xBut in my approach, I have x = 6ySo, 240y = 341x = 341 * 6y = 2046ySo, 240y = 2046yWhich implies 240 = 2046, which is not true.So, clearly, the solution above had a mistake in equation 1.Therefore, my approach is correct, leading to x = 6y, and from equation 2, y = 26, x = 156 km.But in the solution above, they had x = 192 km, y = 32 km, which also satisfies equation 1, but not equation 2.Wait, let me check equation 2 with x = 192, y = 32.Equation 2:(y / 16) + 1 + ((x - y - 24) / 80) = x / 16Plug in y = 32, x = 192Left side: 32 / 16 + 1 + (192 - 32 - 24)/80 = 2 + 1 + (136)/80 = 3 + 1.7 = 4.7Right side: 192 / 16 = 12So, 4.7 ≠ 12, so x = 192, y = 32 does not satisfy equation 2.Therefore, the solution above is incorrect.So, the correct answer is x = 156 km, y = 26 km.But wait, in the problem, on the even-numbered day, the ferry lost power at the midpoint to point C, which is y = x/2.But in my solution, y = 26, x = 156, so y = 26, which is not x/2.So, perhaps the problem states that point C is the midpoint, so y = x/2.But in that case, from equation 1, x = 6y = 6*(x/2) = 3x, which implies x = 0, which is impossible.So, there's a contradiction.Therefore, the problem must mean that point C is not the midpoint in distance, but rather, the midpoint in time.So, the ferry lost power halfway through the usual time, not halfway through the distance.So, let's redefine.Let’s let t be the usual time for A to B downstream, which is x / 64.So, halfway through the time is t / 2 = x / 128.So, the distance covered in t / 2 time at downstream speed is y = 64 * (x / 128) = x / 2.So, y = x / 2.Wait, so point C is the midpoint in distance.But then, as before, equation 1 leads to x = 6y = 6*(x/2) = 3x, which is impossible.So, this is a contradiction.Therefore, the problem must mean that point C is not the midpoint in time or distance, but rather, a specific point such that the time taken on the odd-numbered day is 43/18 times the usual time.Therefore, my initial approach is correct, leading to x = 156 km, y = 26 km.But the solution above had x = 192 km, y = 32 km, which does not satisfy equation 2.Therefore, the correct answer is x = 156 km.But let me check again.From equation 1, x = 6yFrom equation 2, y = 26, x = 156So, x = 156 kmTherefore, the distance between A and B is 156 km.But wait, in the solution above, they had x = 192 km, y = 32 km, which satisfies equation 1 but not equation 2.So, perhaps the solution above had a mistake.Therefore, the correct answer is 156 km.