💡Okay, so I have this problem here: The sum of three different numbers is 100. The two larger numbers differ by 10, and the two smaller numbers differ by 5. I need to find the value of the largest number. Hmm, let me try to figure this out step by step.First, I'll assign variables to the three numbers to make it easier. Let's say the smallest number is ( a ), the middle number is ( b ), and the largest number is ( c ). So, we have ( a ), ( b ), and ( c ) such that ( a < b < c ).The problem states that the sum of these three numbers is 100. So, I can write the equation:[ a + b + c = 100 ]Next, it says the two larger numbers differ by 10. That means the largest number ( c ) is 10 more than the middle number ( b ). So, I can write:[ c = b + 10 ]It also mentions that the two smaller numbers differ by 5. That means the middle number ( b ) is 5 more than the smallest number ( a ). So, I can write:[ b = a + 5 ]Now, I have three equations:1. ( a + b + c = 100 )2. ( c = b + 10 )3. ( b = a + 5 )I can substitute equation 3 into equation 2 to express ( c ) in terms of ( a ). Let's do that:[ c = (a + 5) + 10 = a + 15 ]Now, I can substitute both ( b ) and ( c ) in terms of ( a ) into equation 1. So, substituting ( b = a + 5 ) and ( c = a + 15 ) into ( a + b + c = 100 ), we get:[ a + (a + 5) + (a + 15) = 100 ]Let's simplify this equation:[ a + a + 5 + a + 15 = 100 ][ 3a + 20 = 100 ]Now, subtract 20 from both sides:[ 3a = 80 ]Then, divide both sides by 3:[ a = frac{80}{3} ][ a approx 26.6667 ]Hmm, so the smallest number is approximately 26.6667. That seems a bit unusual because it's not an integer, but maybe that's okay. Let's see if that makes sense.Now, let's find ( b ) using equation 3:[ b = a + 5 = frac{80}{3} + 5 = frac{80}{3} + frac{15}{3} = frac{95}{3} ][ b approx 31.6667 ]And then ( c ) using equation 2:[ c = b + 10 = frac{95}{3} + 10 = frac{95}{3} + frac{30}{3} = frac{125}{3} ][ c approx 41.6667 ]Let me check if these values satisfy the original equation:[ a + b + c = frac{80}{3} + frac{95}{3} + frac{125}{3} = frac{80 + 95 + 125}{3} = frac{300}{3} = 100 ]Okay, that checks out. So, the largest number is ( c = frac{125}{3} ), which is approximately 41.6667.Wait a second, the problem didn't specify that the numbers have to be integers. So, even though the numbers are not whole numbers, they still satisfy all the given conditions. Therefore, the largest number is ( frac{125}{3} ), which is approximately 41.67.I think that's the answer. But just to be thorough, let me try to see if there's another way to approach this problem, maybe by assuming the numbers are integers and see if that leads to a different solution.If I assume that ( a ), ( b ), and ( c ) are integers, then ( a = frac{80}{3} ) is not an integer, which would mean there's no solution with integer values. But since the problem doesn't specify that the numbers need to be integers, I think my original solution is valid.Alternatively, maybe I made a mistake in setting up the equations. Let me double-check.I set ( a ) as the smallest, ( b ) as the middle, and ( c ) as the largest. Then, ( c - b = 10 ) and ( b - a = 5 ). Substituting ( b = a + 5 ) into ( c = b + 10 ) gives ( c = a + 15 ). Then, substituting into the sum equation gives ( a + (a + 5) + (a + 15) = 100 ), which simplifies to ( 3a + 20 = 100 ), leading to ( a = frac{80}{3} ). That seems correct.So, unless there's a different interpretation of the problem, I think my answer is correct. The largest number is ( frac{125}{3} ), which is approximately 41.67.

💡To convert the radian measure of ( -frac{23}{12}pi ) to degrees, I start by recalling the conversion factor between radians and degrees, which is ( 1 text{ rad} = frac{180}{pi}^circ ).Next, I multiply the given radian measure by this conversion factor to find the equivalent angle in degrees. This involves multiplying ( -frac{23}{12}pi ) by ( frac{180}{pi} ).After simplifying the expression, the (pi) terms cancel out, leaving me with the numerical calculation. Finally, performing the multiplication gives the angle in degrees.

💡Okay, so I have this algebra problem here: factorize ( xy^{2} - x ). Hmm, factorizing expressions is something I've been practicing, but I want to make sure I understand each step clearly. Let me think about how to approach this.First, I remember that factorizing means breaking down an expression into simpler parts that can be multiplied together to get the original expression. So, I need to find common factors or patterns in ( xy^{2} - x ).Looking at the terms ( xy^{2} ) and ( -x ), I notice that both have an ( x ) in them. That seems like a good starting point. If I factor out the ( x ), I can simplify the expression.So, factoring ( x ) out of both terms, I get:[ x(y^{2} - 1) ]Okay, that looks right. Now, I have ( x ) multiplied by ( (y^{2} - 1) ). I wonder if I can factor ( y^{2} - 1 ) further. It looks familiar; isn't that a difference of squares?Yes, ( y^{2} - 1 ) is indeed a difference of squares because it can be written as ( y^{2} - 1^{2} ). The formula for factoring a difference of squares is ( a^{2} - b^{2} = (a - b)(a + b) ). Applying that here, I can factor ( y^{2} - 1 ) as:[ (y - 1)(y + 1) ]So, putting it all together, the original expression ( xy^{2} - x ) becomes:[ x(y - 1)(y + 1) ]Let me double-check to make sure I didn't make any mistakes. If I expand ( x(y - 1)(y + 1) ), I should get back to the original expression.First, multiply ( (y - 1)(y + 1) ):[ (y - 1)(y + 1) = y^{2} + y - y - 1 = y^{2} - 1 ]Then, multiply by ( x ):[ x(y^{2} - 1) = xy^{2} - x ]Yes, that matches the original expression. So, my factorization seems correct.I think I've covered all the steps. I identified the common factor, factored it out, recognized the difference of squares, applied the formula, and then verified my work by expanding. It all makes sense now.

💡Okay, so I have this problem about Fibonacci numbers. It says that if I take 8 consecutive Fibonacci numbers, their sum isn't a Fibonacci number. Hmm, interesting. I remember that Fibonacci numbers are a sequence where each number is the sum of the two preceding ones, starting from 0 and 1. So, it goes like 0, 1, 1, 2, 3, 5, 8, 13, 21, and so on.First, I need to understand what it means for a sum of numbers to be a Fibonacci number. If I add up these 8 consecutive Fibonacci numbers, the result should not be another number in the Fibonacci sequence. That seems a bit tricky, but maybe there's a pattern or a property I can use.Let me denote the 8 consecutive Fibonacci numbers as ( F_n, F_{n+1}, F_{n+2}, ldots, F_{n+7} ). So, I need to find the sum ( S = F_n + F_{n+1} + F_{n+2} + ldots + F_{n+7} ) and show that ( S ) is not a Fibonacci number.I recall that there's a formula for the sum of consecutive Fibonacci numbers. Maybe I can use that. Let me see... I think the sum of ( k ) consecutive Fibonacci numbers starting from ( F_n ) can be expressed in terms of other Fibonacci numbers. Specifically, I think it relates to ( F_{n+k+1} ) minus some earlier Fibonacci number.Wait, let me try to derive it. If I consider the sum ( S = F_n + F_{n+1} + F_{n+2} + ldots + F_{n+7} ), I can use the property of Fibonacci numbers that ( F_{k} = F_{k-1} + F_{k-2} ). Maybe I can express the sum recursively.Let me write down the sum:( S = F_n + F_{n+1} + F_{n+2} + F_{n+3} + F_{n+4} + F_{n+5} + F_{n+6} + F_{n+7} )I know that ( F_{n+1} = F_n + F_{n-1} ), but I'm not sure if that helps directly. Maybe I can look for a pattern or a telescoping sum.Alternatively, I remember that the sum of Fibonacci numbers has a closed-form expression. Maybe I can use Binet's formula, which expresses Fibonacci numbers in terms of powers of the golden ratio. But that might be complicated for this problem.Wait, another approach: I can use induction or some known identity about Fibonacci sums. Let me check if there's an identity for the sum of ( k ) consecutive Fibonacci numbers.After a quick search in my mind, I recall that the sum of ( k ) consecutive Fibonacci numbers starting from ( F_n ) is equal to ( F_{n+k} - F_{n-1} ). Let me verify that.Let's test it for a small ( k ). Suppose ( k = 1 ): Sum is ( F_n ), and according to the formula, it should be ( F_{n+1} - F_{n-1} ). But ( F_{n+1} - F_{n-1} = (F_n + F_{n-1}) - F_{n-1} = F_n ). Okay, that works.For ( k = 2 ): Sum is ( F_n + F_{n+1} ). The formula gives ( F_{n+2} - F_{n-1} ). Let's compute ( F_{n+2} - F_{n-1} = (F_{n+1} + F_n) - F_{n-1} ). But ( F_{n+1} = F_n + F_{n-1} ), so substituting, we get ( (F_n + F_{n-1} + F_n) - F_{n-1} = 2F_n ). Wait, but ( F_n + F_{n+1} = F_n + (F_n + F_{n-1}) = 2F_n + F_{n-1} ). Hmm, that doesn't match. Maybe my recollection is incorrect.Alternatively, perhaps the formula is ( F_{n+k} - F_{n} ). Let's test that.For ( k = 1 ): ( F_{n+1} - F_n = F_{n-1} ), which is not equal to ( F_n ). So that doesn't work either.Maybe I need to think differently. Let's try to express the sum recursively.Let ( S_k = F_n + F_{n+1} + ldots + F_{n+k-1} ). Then, ( S_{k+1} = S_k + F_{n+k} ). Also, since ( F_{n+k} = F_{n+k-1} + F_{n+k-2} ), maybe I can relate ( S_{k+1} ) to previous sums.Alternatively, I can use generating functions or matrix exponentiation, but that might be overkill.Wait, maybe I can use the fact that the sum of Fibonacci numbers has a relationship with Fibonacci numbers themselves. Let me look for a pattern.Let's compute the sum for small ( n ) and ( k = 8 ) to see if I can spot a pattern.Take ( n = 1 ): The numbers are ( F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, F_6 = 8, F_7 = 13, F_8 = 21 ). Sum = 1+1+2+3+5+8+13+21 = 54.Is 54 a Fibonacci number? Let's see: Fibonacci numbers around that range are 21, 34, 55, 89. 54 is between 34 and 55, so it's not a Fibonacci number.Another example: ( n = 2 ): Numbers are ( F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, F_6 = 8, F_7 = 13, F_8 = 21, F_9 = 34 ). Sum = 1+2+3+5+8+13+21+34 = 87.Is 87 a Fibonacci number? The Fibonacci sequence goes 21, 34, 55, 89, 144. 87 is between 55 and 89, so no.Another example: ( n = 3 ): Numbers are ( F_3 = 2, F_4 = 3, F_5 = 5, F_6 = 8, F_7 = 13, F_8 = 21, F_9 = 34, F_{10} = 55 ). Sum = 2+3+5+8+13+21+34+55 = 141.Is 141 a Fibonacci number? The sequence around there is 89, 144, 233. 141 is between 89 and 144, so no.Hmm, so in these examples, the sum is not a Fibonacci number. But I need to prove it in general, not just for specific cases.Let me try to find a general expression for the sum ( S = F_n + F_{n+1} + ldots + F_{n+7} ).I recall that the sum of Fibonacci numbers from ( F_1 ) to ( F_k ) is ( F_{k+2} - 1 ). Maybe I can use that.So, the sum from ( F_n ) to ( F_{n+7} ) can be expressed as the sum from ( F_1 ) to ( F_{n+7} ) minus the sum from ( F_1 ) to ( F_{n-1} ).Using the formula, that would be ( (F_{n+9} - 1) - (F_{n+1} - 1) = F_{n+9} - F_{n+1} ).Wait, so ( S = F_{n+9} - F_{n+1} ).Is that correct? Let me verify with my earlier example where ( n = 1 ):( F_{1+9} - F_{1+1} = F_{10} - F_2 = 55 - 1 = 54 ). Yes, that matches.Another example, ( n = 2 ):( F_{2+9} - F_{2+1} = F_{11} - F_3 = 89 - 2 = 87 ). Correct.And ( n = 3 ):( F_{3+9} - F_{3+1} = F_{12} - F_4 = 144 - 3 = 141 ). Correct.Okay, so the sum ( S = F_{n+9} - F_{n+1} ).Now, I need to show that ( S ) is not a Fibonacci number. So, I need to show that ( F_{n+9} - F_{n+1} ) cannot be equal to any ( F_k ) for some integer ( k ).Let me think about the properties of Fibonacci numbers. Each Fibonacci number is the sum of the two preceding ones, so they grow exponentially. Also, consecutive Fibonacci numbers are coprime, meaning they share no common divisors other than 1.Given that ( S = F_{n+9} - F_{n+1} ), I can try to express this in terms of other Fibonacci numbers to see if it can be a Fibonacci number.Let me write ( F_{n+9} ) in terms of earlier Fibonacci numbers. Using the Fibonacci recurrence, ( F_{n+9} = F_{n+8} + F_{n+7} ).Similarly, ( F_{n+8} = F_{n+7} + F_{n+6} ), and so on.But I'm not sure if that helps directly. Maybe I can find a relationship between ( F_{n+9} ) and ( F_{n+1} ).Alternatively, I can consider the Fibonacci sequence modulo some number to see if ( S ) can be a Fibonacci number. But that might be too abstract.Wait, another idea: I can use the fact that Fibonacci numbers are strictly increasing. So, if I can show that ( S ) lies strictly between two consecutive Fibonacci numbers, then it cannot be a Fibonacci number itself.Let me see. Since ( S = F_{n+9} - F_{n+1} ), and ( F_{n+9} = F_{n+8} + F_{n+7} ), then ( S = F_{n+8} + F_{n+7} - F_{n+1} ).But ( F_{n+7} = F_{n+6} + F_{n+5} ), so substituting, ( S = F_{n+8} + F_{n+6} + F_{n+5} - F_{n+1} ).This seems to be getting more complicated. Maybe I need a different approach.Let me think about the indices. If ( S = F_{n+9} - F_{n+1} ), and I want to see if ( S ) is a Fibonacci number, say ( F_k ), then ( F_k = F_{n+9} - F_{n+1} ).I can rearrange this as ( F_{n+9} = F_k + F_{n+1} ).But in the Fibonacci sequence, each number is the sum of the two preceding ones. So, unless ( F_k ) and ( F_{n+1} ) are consecutive Fibonacci numbers, their sum won't be a Fibonacci number.Wait, but ( F_{n+9} ) is much larger than ( F_{n+1} ). Specifically, ( F_{n+9} ) is several steps ahead of ( F_{n+1} ).Let me consider the ratio of consecutive Fibonacci numbers. As ( n ) increases, the ratio ( frac{F_{n+1}}{F_n} ) approaches the golden ratio ( phi approx 1.618 ).So, ( F_{n+9} ) is roughly ( phi^8 ) times ( F_n ), which is a significant increase.Given that ( S = F_{n+9} - F_{n+1} ), and ( F_{n+9} ) is much larger than ( F_{n+1} ), ( S ) is still a large number, but I need to see if it can be exactly a Fibonacci number.Alternatively, maybe I can find bounds for ( S ) in terms of Fibonacci numbers.I know that ( F_{n+8} < F_{n+9} < F_{n+8} + F_{n+7} ), but that doesn't directly help.Wait, let's consider the Fibonacci sequence:( F_{n}, F_{n+1}, F_{n+2}, ldots, F_{n+9} )I can see that ( F_{n+9} = F_{n+8} + F_{n+7} ), and ( F_{n+8} = F_{n+7} + F_{n+6} ), and so on.But I'm not sure if that helps in expressing ( S ) as a Fibonacci number.Another approach: Let's assume, for contradiction, that ( S = F_k ) for some ( k ). Then, ( F_{n+9} - F_{n+1} = F_k ).I need to see if this equation can hold for some ( k ).Let me consider the possible values of ( k ). Since ( F_{n+9} ) is much larger than ( F_{n+1} ), ( k ) must be close to ( n+9 ).Specifically, ( F_{n+9} - F_{n+1} ) is less than ( F_{n+9} ), so ( k ) must be less than ( n+9 ).But ( ( F_{n+8} < F_{n+9} - F_{n+1} < F_{n+9} ) ) because ( F_{n+9} - F_{n+1} = F_{n+8} + F_{n+7} - F_{n+1} ), and since ( F_{n+7} > F_{n+1} ) for ( n geq 1 ), this sum is greater than ( F_{n+8} ).Wait, let me verify that:( F_{n+9} - F_{n+1} = F_{n+8} + F_{n+7} - F_{n+1} )Since ( F_{n+7} > F_{n+1} ) for ( n geq 1 ), then ( F_{n+8} + F_{n+7} - F_{n+1} > F_{n+8} ).Also, ( F_{n+9} - F_{n+1} = F_{n+8} + F_{n+7} - F_{n+1} < F_{n+8} + F_{n+7} ) because ( F_{n+7} - F_{n+1} < F_{n+7} ).But ( F_{n+8} + F_{n+7} = F_{n+9} ), so ( F_{n+9} - F_{n+1} < F_{n+9} ).Therefore, ( F_{n+8} < F_{n+9} - F_{n+1} < F_{n+9} ).This means that ( S = F_{n+9} - F_{n+1} ) lies strictly between ( F_{n+8} ) and ( F_{n+9} ).But in the Fibonacci sequence, there are no numbers between ( F_{n+8} ) and ( F_{n+9} ) except for ( F_{n+8} ) and ( F_{n+9} ) themselves. Therefore, ( S ) cannot be a Fibonacci number because it's strictly between two consecutive Fibonacci numbers.Thus, the sum of 8 consecutive Fibonacci numbers is not a Fibonacci number.

💡Okay, so I have this problem with two parts. Let me try to figure them out step by step.**Problem (1):** Find the range of positive real number ( a ) such that ( g(x) = ax - frac{a}{x} - 5ln x ) is an increasing function in its domain.Alright, so to determine if a function is increasing, I remember that I need to look at its derivative. If the derivative is positive for all ( x ) in the domain, then the function is increasing.So, let's compute the derivative of ( g(x) ).( g'(x) = frac{d}{dx}left(ax - frac{a}{x} - 5ln xright) )Calculating term by term:- The derivative of ( ax ) is ( a ).- The derivative of ( -frac{a}{x} ) is ( frac{a}{x^2} ) because ( frac{d}{dx}(-a x^{-1}) = a x^{-2} ).- The derivative of ( -5ln x ) is ( -frac{5}{x} ).Putting it all together:( g'(x) = a + frac{a}{x^2} - frac{5}{x} )Hmm, let me write that as a single fraction to make it easier to analyze:( g'(x) = frac{a x^2 + a - 5x}{x^2} )So, ( g'(x) = frac{a x^2 - 5x + a}{x^2} )Since ( x^2 ) is always positive for ( x > 0 ) (because the domain of ( g(x) ) is ( x > 0 ) due to the ( ln x ) term), the sign of ( g'(x) ) depends on the numerator:( a x^2 - 5x + a > 0 ) for all ( x > 0 ).So, we need ( a x^2 - 5x + a > 0 ) for all ( x > 0 ).This is a quadratic in terms of ( x ). Let me denote it as:( Q(x) = a x^2 - 5x + a )We need ( Q(x) > 0 ) for all ( x > 0 ).For a quadratic ( Q(x) = ax^2 + bx + c ), it is positive for all ( x ) if:1. The leading coefficient ( a > 0 ).2. The discriminant ( D = b^2 - 4ac < 0 ).Wait, but in our case, the quadratic is ( Q(x) = a x^2 - 5x + a ). So, the coefficients are:- Leading coefficient: ( a )- Linear coefficient: ( -5 )- Constant term: ( a )So, applying the conditions:1. ( a > 0 ) (which is already given in the problem)2. Discriminant ( D = (-5)^2 - 4(a)(a) = 25 - 4a^2 < 0 )So, ( 25 - 4a^2 < 0 )Solving for ( a ):( 25 < 4a^2 )( a^2 > frac{25}{4} )( a > frac{5}{2} ) or ( a < -frac{5}{2} )But since ( a ) is a positive real number, we discard the negative solution.Therefore, ( a > frac{5}{2} ).Wait, let me double-check. If ( a > frac{5}{2} ), then the discriminant is negative, so the quadratic never crosses zero, and since the leading coefficient is positive, it is always positive. That makes sense.So, the range of ( a ) is ( a > frac{5}{2} ).**Problem (2):** Given ( h(x) = x^2 - m x + 4 ), when ( a = 2 ), if there exists ( x_1 in (0,1) ) such that for all ( x_2 in [1,2] ), ( g(x_1) geq h(x_2) ), find the range of real number ( m ).Alright, so when ( a = 2 ), ( g(x) = 2x - frac{2}{x} - 5ln x ).We need to find ( m ) such that there exists some ( x_1 ) in ( (0,1) ) where ( g(x_1) ) is greater than or equal to ( h(x_2) ) for all ( x_2 ) in ( [1,2] ).So, in other words, the maximum value of ( h(x) ) on ( [1,2] ) must be less than or equal to the maximum value of ( g(x) ) on ( (0,1) ).Wait, but it's phrased as "there exists ( x_1 in (0,1) )" such that for all ( x_2 in [1,2] ), ( g(x_1) geq h(x_2) ). So, it's not necessarily the maximum of ( g(x) ), but there exists some ( x_1 ) where ( g(x_1) ) is greater than or equal to all ( h(x_2) ).But to satisfy this condition, ( g(x_1) ) must be at least as large as the maximum of ( h(x) ) on ( [1,2] ). Because if ( g(x_1) ) is greater than or equal to the maximum of ( h(x) ), then it's certainly greater than or equal to all other values of ( h(x) ) on that interval.Therefore, we need:( max_{x_2 in [1,2]} h(x_2) leq max_{x_1 in (0,1)} g(x_1) )So, first, let's find the maximum of ( h(x) ) on ( [1,2] ).( h(x) = x^2 - m x + 4 )This is a quadratic function opening upwards (since the coefficient of ( x^2 ) is positive). Its vertex is at ( x = frac{m}{2} ).So, depending on where ( frac{m}{2} ) is relative to the interval ( [1,2] ), the maximum will be at one of the endpoints.Case 1: If ( frac{m}{2} leq 1 ), i.e., ( m leq 2 ), then the function is increasing on ( [1,2] ), so the maximum is at ( x = 2 ).Case 2: If ( 1 < frac{m}{2} < 2 ), i.e., ( 2 < m < 4 ), then the function has its minimum at ( x = frac{m}{2} ), so the maximum will still be at the endpoints. Wait, but since it's a parabola opening upwards, the maximum on a closed interval will be at one of the endpoints regardless of where the vertex is.Wait, actually, for a quadratic opening upwards, the maximum on an interval is at one of the endpoints. So, regardless of where the vertex is, the maximum will be at either ( x = 1 ) or ( x = 2 ).Therefore, ( max_{x in [1,2]} h(x) = max{ h(1), h(2) } )Compute ( h(1) = 1 - m + 4 = 5 - m )Compute ( h(2) = 4 - 2m + 4 = 8 - 2m )So, ( max{5 - m, 8 - 2m} )We need to determine which of these is larger.Compare ( 5 - m ) and ( 8 - 2m ):Set ( 5 - m = 8 - 2m )Solving: ( 5 - m = 8 - 2m )Add ( 2m ) to both sides: ( 5 + m = 8 )Subtract 5: ( m = 3 )So, when ( m = 3 ), both are equal: ( 5 - 3 = 2 ) and ( 8 - 6 = 2 ).For ( m < 3 ):Let me test ( m = 2 ):( 5 - 2 = 3 ), ( 8 - 4 = 4 ). So, ( 8 - 2m ) is larger.For ( m = 4 ):( 5 - 4 = 1 ), ( 8 - 8 = 0 ). So, ( 5 - m ) is larger.Wait, that contradicts my earlier thought. Wait, let me see.Wait, when ( m < 3 ), which is larger between ( 5 - m ) and ( 8 - 2m )?Let me subtract them:( (8 - 2m) - (5 - m) = 3 - m )So, if ( 3 - m > 0 ), i.e., ( m < 3 ), then ( 8 - 2m > 5 - m ).If ( m > 3 ), then ( 5 - m > 8 - 2m ).At ( m = 3 ), they are equal.So, summarizing:- If ( m leq 3 ), ( max h(x) = 8 - 2m )- If ( m > 3 ), ( max h(x) = 5 - m )Okay, now we need to find the maximum of ( g(x) ) on ( (0,1) ).Given ( g(x) = 2x - frac{2}{x} - 5ln x )To find its maximum on ( (0,1) ), we can take its derivative and find critical points.Compute ( g'(x) ):( g'(x) = 2 + frac{2}{x^2} - frac{5}{x} )Simplify:( g'(x) = frac{2x^2 + 2 - 5x}{x^2} = frac{2x^2 - 5x + 2}{x^2} )Set ( g'(x) = 0 ):( 2x^2 - 5x + 2 = 0 )Solve the quadratic equation:Discriminant ( D = 25 - 16 = 9 )Solutions:( x = frac{5 pm 3}{4} )So,( x = frac{5 + 3}{4} = 2 )( x = frac{5 - 3}{4} = frac{1}{2} )So, critical points at ( x = frac{1}{2} ) and ( x = 2 )But our interval is ( (0,1) ), so only ( x = frac{1}{2} ) is in ( (0,1) ).Now, let's analyze the behavior of ( g(x) ) around ( x = frac{1}{2} ).Compute the second derivative or test intervals.Let me test intervals around ( x = frac{1}{2} ):Choose ( x = frac{1}{4} ):( g'(1/4) = 2 + 2/(1/16) - 5/(1/4) = 2 + 32 - 20 = 14 > 0 )Choose ( x = 3/4 ):( g'(3/4) = 2 + 2/(9/16) - 5/(3/4) = 2 + (32/9) - (20/3) )Convert to common denominator, which is 9:( 2 = 18/9 ), ( 32/9 ), ( 20/3 = 60/9 )So, ( 18/9 + 32/9 - 60/9 = (18 + 32 - 60)/9 = (-10)/9 < 0 )Therefore, ( g(x) ) is increasing on ( (0, 1/2) ) and decreasing on ( (1/2, 1) ). So, ( x = 1/2 ) is a local maximum.Therefore, the maximum of ( g(x) ) on ( (0,1) ) is at ( x = 1/2 ).Compute ( g(1/2) ):( g(1/2) = 2*(1/2) - 2/(1/2) - 5ln(1/2) )Simplify:( 1 - 4 - 5ln(1/2) )( = -3 - 5ln(1/2) )But ( ln(1/2) = -ln 2 ), so:( = -3 + 5ln 2 )So, ( g(1/2) = 5ln 2 - 3 )So, the maximum of ( g(x) ) on ( (0,1) ) is ( 5ln 2 - 3 ).Now, we need this maximum to be greater than or equal to the maximum of ( h(x) ) on ( [1,2] ).So, depending on ( m ), the maximum of ( h(x) ) is either ( 8 - 2m ) or ( 5 - m ).Case 1: ( m leq 3 )Then, ( max h(x) = 8 - 2m )We need:( 5ln 2 - 3 geq 8 - 2m )Solve for ( m ):( 5ln 2 - 3 - 8 geq -2m )( 5ln 2 - 11 geq -2m )Multiply both sides by (-1), which reverses the inequality:( 11 - 5ln 2 leq 2m )( m geq frac{11 - 5ln 2}{2} )Compute ( frac{11 - 5ln 2}{2} ):First, approximate ( ln 2 approx 0.6931 )So,( 5 * 0.6931 = 3.4655 )( 11 - 3.4655 = 7.5345 )Divide by 2: ( 7.5345 / 2 ≈ 3.76725 )So, ( m geq 3.76725 )But in this case, ( m leq 3 ). So, ( m geq 3.76725 ) and ( m leq 3 ) cannot be satisfied simultaneously. Therefore, no solution in this case.Case 2: ( m > 3 )Then, ( max h(x) = 5 - m )We need:( 5ln 2 - 3 geq 5 - m )Solve for ( m ):( 5ln 2 - 3 - 5 geq -m )( 5ln 2 - 8 geq -m )Multiply both sides by (-1), reversing inequality:( 8 - 5ln 2 leq m )So, ( m geq 8 - 5ln 2 )Compute ( 8 - 5ln 2 ):Again, ( 5ln 2 ≈ 3.4655 )So, ( 8 - 3.4655 ≈ 4.5345 )So, ( m geq 4.5345 )But in this case, ( m > 3 ), so combining both, ( m geq 8 - 5ln 2 approx 4.5345 )Therefore, the range of ( m ) is ( m geq 8 - 5ln 2 )Expressed exactly, ( m geq 8 - 5ln 2 )So, that's the range.**Final Answer**(1) The range of ( a ) is ( boxed{left( dfrac{5}{2}, +infty right)} ).(2) The range of ( m ) is ( boxed{[8 - 5ln 2, +infty)} ).

💡First, I need to determine Ben's after-tax income. Since Ben spends 20% of his after-tax income on his car payment, which is 400, I can set up the equation 0.20 * A = 400, where A represents his after-tax income.Solving for A, I divide 400 by 0.20 to find that A = 2000. This means Ben's after-tax income is 2000.Next, I know that Ben pays 1/3 of his gross income in taxes. Let G represent his gross income. The relationship between gross income and after-tax income can be expressed as G - (1/3)G = A.Substituting the value of A (2000) into the equation, I get (2/3)G = 2000. To find G, I divide 2000 by (2/3), which simplifies to G = 3000.Therefore, Ben makes 3000 per month before taxes.