answer：Alright, so I've got this problem here about a function ( f(x) = x^3 + a x^2 + b x + a^2 ). It says that this function has an extremum of 10 at ( x = 1 ), and I need to find ( f(2) ). The options are given as A: 11 or 18, B: 11, C: 18, D: 17 or 18. Hmm, okay, let me try to work through this step by step.First off, I remember that an extremum (which is a maximum or minimum point) occurs where the derivative of the function is zero. So, I should start by finding the derivative of ( f(x) ). Let's compute that.The function is ( f(x) = x^3 + a x^2 + b x + a^2 ). Taking the derivative with respect to ( x ), we get:( f'(x) = 3x^2 + 2a x + b ).Okay, so that's the derivative. Now, since there's an extremum at ( x = 1 ), plugging that into the derivative should give zero. So, let's set up the equation:( f'(1) = 3(1)^2 + 2a(1) + b = 0 ).Simplifying that, we have:( 3 + 2a + b = 0 ).Alright, so that's our first equation: ( 3 + 2a + b = 0 ). Let me note that down.Next, the problem also tells us that the function has an extremum of 10 at ( x = 1 ). That means when we plug ( x = 1 ) into the original function ( f(x) ), we should get 10. Let's write that out:( f(1) = (1)^3 + a(1)^2 + b(1) + a^2 = 10 ).Simplifying that, we get:( 1 + a + b + a^2 = 10 ).So, that's our second equation: ( a^2 + a + b + 1 = 10 ). Let me write that as ( a^2 + a + b = 9 ).Now, we have a system of two equations:1. ( 3 + 2a + b = 0 )2. ( a^2 + a + b = 9 )I need to solve this system for ( a ) and ( b ). Let me see how to approach this. Maybe I can express ( b ) from the first equation and substitute into the second equation.From equation 1:( 3 + 2a + b = 0 )Subtracting 3 and ( 2a ) from both sides:( b = -3 - 2a )Okay, so ( b = -3 - 2a ). Now, let's substitute this into equation 2.Equation 2 is:( a^2 + a + b = 9 )Substituting ( b ):( a^2 + a + (-3 - 2a) = 9 )Simplify this:( a^2 + a - 3 - 2a = 9 )Combine like terms:( a^2 - a - 3 = 9 )Subtract 9 from both sides:( a^2 - a - 12 = 0 )So, now we have a quadratic equation in terms of ( a ):( a^2 - a - 12 = 0 )Let me try to factor this quadratic. Looking for two numbers that multiply to -12 and add up to -1. Hmm, 3 and -4? Yes, because 3 * (-4) = -12 and 3 + (-4) = -1.So, factoring:( (a + 3)(a - 4) = 0 )Therefore, the solutions are:( a = -3 ) or ( a = 4 )Alright, so ( a ) can be either -3 or 4. Let's find the corresponding ( b ) for each case using ( b = -3 - 2a ).Case 1: ( a = -3 )( b = -3 - 2(-3) = -3 + 6 = 3 )Case 2: ( a = 4 )( b = -3 - 2(4) = -3 - 8 = -11 )So, we have two possible pairs: ( (a, b) = (-3, 3) ) and ( (a, b) = (4, -11) ).Now, before proceeding, I remember that just because the derivative is zero at a point doesn't necessarily mean it's an extremum. It could be a point of inflection. So, I need to check whether ( x = 1 ) is indeed an extremum for both cases.Let's analyze each case.**Case 1: ( a = -3 ), ( b = 3 )**Our function becomes:( f(x) = x^3 - 3x^2 + 3x + (-3)^2 = x^3 - 3x^2 + 3x + 9 )Compute the derivative:( f'(x) = 3x^2 - 6x + 3 )Let me factor this:( f'(x) = 3(x^2 - 2x + 1) = 3(x - 1)^2 )So, ( f'(x) = 3(x - 1)^2 ). Since this is a square term multiplied by 3, it's always non-negative. That means the function is increasing on both sides of ( x = 1 ). Therefore, ( x = 1 ) is not an extremum; it's a point of inflection. So, this case doesn't satisfy the condition of having an extremum at ( x = 1 ). Therefore, we can discard this solution.**Case 2: ( a = 4 ), ( b = -11 )**Our function becomes:( f(x) = x^3 + 4x^2 - 11x + (4)^2 = x^3 + 4x^2 - 11x + 16 )Compute the derivative:( f'(x) = 3x^2 + 8x - 11 )Let me factor this quadratic if possible. Let's see if it factors.Looking for two numbers that multiply to ( 3 * (-11) = -33 ) and add up to 8. Hmm, 11 and -3? 11 * (-3) = -33, and 11 + (-3) = 8. Perfect.So, we can write:( f'(x) = 3x^2 + 11x - 3x - 11 = x(3x + 11) - 1(3x + 11) = (x - 1)(3x + 11) )So, ( f'(x) = (x - 1)(3x + 11) ). Therefore, the critical points are at ( x = 1 ) and ( x = -11/3 ).Now, let's analyze the sign of the derivative around ( x = 1 ). Since ( x = 1 ) is a critical point, we can test intervals around it.- For ( x < 1 ), say ( x = 0 ): ( f'(0) = (0 - 1)(0 + 11) = (-1)(11) = -11 < 0 )- For ( x > 1 ), say ( x = 2 ): ( f'(2) = (2 - 1)(6 + 11) = (1)(17) = 17 > 0 )So, the derivative changes from negative to positive at ( x = 1 ), which means the function changes from decreasing to increasing. Therefore, ( x = 1 ) is indeed a local minimum, which is an extremum. So, this case is valid.Therefore, the correct values are ( a = 4 ) and ( b = -11 ).Now, we need to compute ( f(2) ) with these values.So, ( f(x) = x^3 + 4x^2 - 11x + 16 ). Plugging in ( x = 2 ):( f(2) = (2)^3 + 4(2)^2 - 11(2) + 16 )Calculating each term:- ( 2^3 = 8 )- ( 4(2)^2 = 4*4 = 16 )- ( -11*2 = -22 )- ( +16 )Adding them up:( 8 + 16 - 22 + 16 )Compute step by step:- ( 8 + 16 = 24 )- ( 24 - 22 = 2 )- ( 2 + 16 = 18 )So, ( f(2) = 18 ).Therefore, the correct answer is C: 18.Wait, just to make sure I didn't make any calculation mistakes. Let me double-check the computation of ( f(2) ):( f(2) = 8 + 16 - 22 + 16 )Yes, 8 + 16 is 24, 24 - 22 is 2, 2 + 16 is 18. Correct.Also, just to recap, we had two possible solutions for ( a ) and ( b ), but one of them didn't result in an extremum at ( x = 1 ), so we discarded it. The other one gave us ( f(2) = 18 ), which is option C.I think that's thorough. I considered both cases, checked for the nature of the critical point, and computed the required value. Seems solid.**Final Answer**boxed{C}

answer：First, I need to convert the binary number (101101_2) to its decimal equivalent. To do this, I'll analyze each digit from right to left, multiplying each digit by (2) raised to the power of its position index, starting from (0).Breaking down the binary number (101101_2):- The rightmost digit (first from the right) is (1) in the (0)-th position, so its value is (1 times 2^0 = 1).- The second digit from the right is (0) in the (1)-st position, so its value is (0 times 2^1 = 0).- The third digit from the right is (1) in the (2)-nd position, so its value is (1 times 2^2 = 4).- The fourth digit from the right is (1) in the (3)-rd position, so its value is (1 times 2^3 = 8).- The fifth digit from the right is (0) in the (4)-th position, so its value is (0 times 2^4 = 0).- The leftmost digit (sixth from the right) is (1) in the (5)-th position, so its value is (1 times 2^5 = 32).Adding these values together gives the decimal equivalent:[1 times 2^0 + 0 times 2^1 + 1 times 2^2 + 1 times 2^3 + 0 times 2^4 + 1 times 2^5 = 1 + 0 + 4 + 8 + 0 + 32 = 45.]Next, I'll subtract 5 from the conversion result:[45 - 5 = 40.]Thus, the result after subtracting 5 from the decimal equivalent of (101101_2) is (40).

answer：Let's solve the problem step by step.**Step 1: Define Variables**Let ( g ) be the number of girls and ( b ) be the number of boys.Given that the number of boys is 20% more than the number of girls:[b = g + 0.2g = 1.2g]**Step 2: Total Number of Students**The total number of students is 180:[g + b = 180]Substitute ( b = 1.2g ):[g + 1.2g = 180 implies 2.2g = 180]Solving for ( g ):[g = frac{180}{2.2} = 81.818 approx 82]Then, the number of boys is:[b = 1.2 times 82 = 98.4 approx 98]**Step 3: Define Mean Scores**Let ( m_b ) be the mean score of the boys and ( m_g ) be the mean score of the girls.Given that the girls' average score is 25% less than the boys':[m_g = 0.75m_b]**Step 4: Total Score Calculation**The total score for all students is:[180 times 90 = 16200]Expressing the total score in terms of boys and girls:[82m_g + 98m_b = 16200]Substitute ( m_g = 0.75m_b ):[82(0.75m_b) + 98m_b = 16200]Simplify:[61.5m_b + 98m_b = 16200 implies 159.5m_b = 16200]Solving for ( m_b ):[m_b = frac{16200}{159.5} approx 101.567]**Conclusion:**Considering the calculation and rounding, the average score of the boys is approximately 102.The final answer is (boxed{102}).

answer：Okay, so I need to show that a convex polyhedron always has two faces with the same number of edges. Hmm, let me think about this step by step.First, what is a convex polyhedron? It's a three-dimensional shape where all its faces are flat polygons, and the shape doesn't have any indentations. So, if I imagine a cube or a tetrahedron, those are convex polyhedrons. Now, each face of a polyhedron is a polygon, right? And polygons can have different numbers of edges, like triangles have 3 edges, squares have 4 edges, pentagons have 5 edges, and so on.The problem is asking me to prove that in any convex polyhedron, there must be at least two faces that have the same number of edges. So, for example, in a cube, all faces are squares, so they all have 4 edges. That's a case where all faces have the same number of edges, so obviously, there are multiple faces with the same number of edges.But what about a more complex polyhedron, like a soccer ball pattern, which is a truncated icosahedron? It has 12 pentagonal faces and 20 hexagonal faces. So, in this case, there are two different types of faces, but within each type, all faces have the same number of edges. So, the pentagons all have 5 edges, and the hexagons all have 6 edges. So, again, there are multiple faces with the same number of edges.But wait, what if we have a polyhedron where each face has a different number of edges? Is that possible? Let me think. If I try to construct such a polyhedron, I would need each face to have a unique number of edges. So, one face has 3 edges, another has 4, another has 5, and so on. But is that possible?I recall something called Euler's formula, which relates the number of vertices (V), edges (E), and faces (F) of a convex polyhedron. It states that V - E + F = 2. Maybe I can use this formula to analyze the situation.Let's denote the number of faces as F. Each face has a certain number of edges, but each edge is shared by two faces. So, if I sum up all the edges of all the faces, I'll get twice the number of edges of the polyhedron. Let's denote the number of edges of the i-th face as e_i. Then, we have:Sum_{i=1 to F} e_i = 2ENow, if all faces had a different number of edges, then the number of edges per face would have to be unique. So, the smallest possible number of edges for a face is 3 (a triangle), then 4, 5, etc. So, if we have F faces, the minimum total number of edges would be the sum of the first F integers starting from 3.Wait, that might not be correct. Let me think again. If each face has a unique number of edges, then the number of edges per face would be 3, 4, 5, ..., up to F+2. Because if you start at 3, the next unique number is 4, then 5, and so on.So, the total number of edges would be:Sum_{k=3}^{F+2} k = [Sum_{k=1}^{F+2} k] - [Sum_{k=1}^{2} k] = [(F+2)(F+3)/2] - [3] = (F^2 + 5F + 6)/2 - 3 = (F^2 + 5F)/2But we also know that Sum_{i=1 to F} e_i = 2E, so:(F^2 + 5F)/2 = 2ETherefore,E = (F^2 + 5F)/4Now, let's plug this into Euler's formula:V - E + F = 2But we need to express V in terms of F as well. Each edge is shared by two faces, and each face has e_i edges, which also relates to the number of vertices. Each face with e_i edges has e_i vertices, but each vertex is shared by multiple faces.Wait, this is getting complicated. Maybe there's a simpler way to approach this.I remember that in any convex polyhedron, the number of edges must satisfy certain conditions. For example, each face must have at least 3 edges, and each vertex must have at least 3 edges meeting there.Let me think about the average number of edges per face. If all faces had a unique number of edges, then the average number of edges per face would be higher than if there were repetitions.But I'm not sure if that's the right direction. Maybe I should consider the Pigeonhole Principle. If there are more faces than the number of possible distinct edge counts, then at least two faces must share the same number of edges.What's the range of possible edge counts for the faces? The smallest is 3, and the largest depends on the polyhedron. But in a convex polyhedron, the maximum number of edges a face can have is limited by the total number of edges.Wait, but without knowing the exact number of edges, it's hard to say. Maybe I can find a lower bound on the number of faces.From Euler's formula, V - E + F = 2. Also, we know that each face has at least 3 edges, and each edge is shared by two faces, so 2E >= 3F, which implies E >= (3/2)F.Similarly, each vertex has at least 3 edges meeting there, so 2E >= 3V, which implies V <= (2/3)E.Plugging this into Euler's formula:V - E + F = 2(2/3)E - E + F = 2(-1/3)E + F = 2So,F = (1/3)E + 2But we also have E >= (3/2)FSubstituting F from above:E >= (3/2)((1/3)E + 2) = (1/2)E + 3So,E >= (1/2)E + 3Subtracting (1/2)E from both sides:(1/2)E >= 3So,E >= 6Which makes sense, as the simplest polyhedron, a tetrahedron, has 6 edges.But how does this help me? Maybe I can find a relationship between F and E that shows that F is large enough that the number of possible distinct edge counts is less than F, forcing at least two faces to have the same number of edges.Wait, let's think about the number of possible distinct edge counts. The smallest number of edges a face can have is 3. What's the maximum? It's not fixed, but in a convex polyhedron, it's limited by the total number of edges.But perhaps I can consider that the number of possible distinct edge counts is less than F, so by the Pigeonhole Principle, at least two faces must share the same number of edges.But I need to make this more precise.Suppose, for contradiction, that all faces have a distinct number of edges. Then, the number of faces F must be less than or equal to the number of possible distinct edge counts.What's the number of possible distinct edge counts? It's the number of integers from 3 up to some maximum m.But without knowing m, it's hard to say. However, in any convex polyhedron, the number of edges per face is bounded by the total number of edges.Wait, maybe I can use the fact that the sum of the edges of all faces is 2E, and if all faces have distinct edge counts, then the sum would be at least the sum of the first F integers starting from 3.So, Sum_{i=1 to F} e_i >= Sum_{k=3}^{F+2} k = (F+2)(F+3)/2 - 3But we also have Sum_{i=1 to F} e_i = 2ESo,2E >= (F+2)(F+3)/2 - 3Simplify:2E >= (F^2 + 5F + 6)/2 - 3Multiply both sides by 2:4E >= F^2 + 5F + 6 - 6So,4E >= F^2 + 5FBut from earlier, we have E >= (3/2)FSo,4*(3/2)F >= F^2 + 5FSimplify:6F >= F^2 + 5FSubtract 6F:0 >= F^2 - FSo,F^2 - F <= 0Which implies F(F - 1) <= 0Since F is the number of faces, it's at least 4 (for a tetrahedron). So, F >= 4, which makes F(F - 1) >= 4*3 = 12 > 0This is a contradiction because we have F(F - 1) <= 0, but F >= 4 implies F(F - 1) >= 12.Therefore, our assumption that all faces have distinct edge counts must be false. Hence, there must be at least two faces with the same number of edges.So, that's the proof. By assuming that all faces have distinct edge counts, we arrived at a contradiction using Euler's formula and the Pigeonhole Principle. Therefore, a convex polyhedron must have at least two faces with the same number of edges.

answer：Alright, so I have this problem about triangle ABC. The sides opposite angles A, B, and C are a, b, c respectively. The area of the triangle is S, and there's this equation given: 2S - √3 bc cos A = 0. Part (1) asks for the measure of angle A. Hmm, okay. I remember that the area of a triangle can be expressed in terms of two sides and the sine of the included angle. Specifically, S = (1/2) bc sin A. Maybe I can use that here.Let me write down the given equation: 2S - √3 bc cos A = 0. If I substitute S with (1/2) bc sin A, then it becomes 2*(1/2 bc sin A) - √3 bc cos A = 0. Simplifying that, the 2 and 1/2 cancel out, so I have bc sin A - √3 bc cos A = 0.Hmm, I can factor out bc since it's common to both terms: bc(sin A - √3 cos A) = 0. Now, bc can't be zero because they are sides of a triangle, so that means sin A - √3 cos A = 0. So, sin A = √3 cos A. If I divide both sides by cos A (assuming cos A ≠ 0), I get tan A = √3. I remember that tan 60° = √3, so A must be 60°, which is π/3 radians. That seems straightforward.Okay, so for part (1), angle A is π/3 radians or 60 degrees. Got that.Now, part (2) says if a = 7 and bc = 40, find the perimeter of triangle ABC. So, a is the side opposite angle A, which we've found to be 60 degrees. And bc is 40. The perimeter would be a + b + c, so I need to find b + c.I think I can use the Law of Cosines here. The Law of Cosines states that a² = b² + c² - 2bc cos A. We know a, bc, and angle A, so maybe we can find b² + c² first.Plugging in the values: 7² = b² + c² - 2*40*cos(60°). 7 squared is 49, and cos(60°) is 0.5, so 2*40*0.5 is 40. So, 49 = b² + c² - 40. Therefore, b² + c² = 49 + 40 = 89.Now, I need to find b + c. I remember that (b + c)² = b² + 2bc + c². We know b² + c² is 89 and bc is 40, so 2bc is 80. Therefore, (b + c)² = 89 + 80 = 169. Taking the square root of both sides, b + c = 13.So, the perimeter is a + b + c = 7 + 13 = 20. That should be the perimeter.Wait, let me double-check. If b + c is 13, and bc is 40, does that make sense? Let's see, if b and c are roots of the equation x² - 13x + 40 = 0, then the roots would be x = [13 ± √(169 - 160)] / 2 = [13 ± 3]/2, so x = 8 or x = 5. So, b and c could be 5 and 8. Then, checking with the Law of Cosines: 5² + 8² - 2*5*8*cos(60°) = 25 + 64 - 80*0.5 = 89 - 40 = 49, which is 7². Perfect, that works out.So, yeah, the perimeter is 7 + 5 + 8 = 20. That makes sense. I think I did everything correctly.**Final Answer**(1) The measure of angle ( A ) is (boxed{dfrac{pi}{3}}).(2) The perimeter of ( triangle ABC ) is (boxed{20}).

answer：Okay, so I've got this function f(x) = ax - 1/x - (a + 1)ln x. There are two parts to the problem. First, when a is 0, I need to find the maximum value of f(x). Second, I need to figure out the range of a such that f(x) has exactly one zero. Hmm, let's take it step by step.Starting with part (1): When a = 0, the function simplifies. Let me substitute a = 0 into f(x). So, f(x) becomes 0*x - 1/x - (0 + 1)ln x, which simplifies to -1/x - ln x. So, f(x) = -1/x - ln x. Now, I need to find the maximum value of this function. Since it's a function of x, I should probably take its derivative and find critical points.Taking the derivative of f(x) with respect to x: f'(x) = d/dx (-1/x - ln x). The derivative of -1/x is 1/x², and the derivative of -ln x is -1/x. So, f'(x) = 1/x² - 1/x. Let me write that as (1 - x)/x². Okay, so f'(x) = (1 - x)/x².Now, to find critical points, I set f'(x) = 0. So, (1 - x)/x² = 0. The numerator must be zero, so 1 - x = 0, which gives x = 1. So, x = 1 is a critical point. Now, I need to determine if this is a maximum or a minimum. Let's check the sign of f'(x) around x = 1.For x < 1, say x = 0.5, f'(0.5) = (1 - 0.5)/(0.5)² = 0.5 / 0.25 = 2, which is positive. So, the function is increasing before x = 1. For x > 1, say x = 2, f'(2) = (1 - 2)/4 = -1/4, which is negative. So, the function is decreasing after x = 1. Therefore, x = 1 is a maximum point.So, the maximum value occurs at x = 1. Let's compute f(1): f(1) = -1/1 - ln 1 = -1 - 0 = -1. So, the maximum value is -1. That seems straightforward.Moving on to part (2): If f(x) has exactly one zero, find the range of values for a. Hmm, so we need to find all a such that the equation f(x) = 0 has exactly one solution. Let's first write down f(x) again: f(x) = ax - 1/x - (a + 1)ln x.To find when f(x) has exactly one zero, I think we need to analyze the behavior of f(x) and its derivative. Maybe we can use calculus to determine when the function crosses the x-axis only once.First, let's find the derivative of f(x) with respect to x. f'(x) = d/dx [ax - 1/x - (a + 1)ln x]. The derivative of ax is a, the derivative of -1/x is 1/x², and the derivative of -(a + 1)ln x is -(a + 1)/x. So, putting it all together, f'(x) = a + 1/x² - (a + 1)/x.Let me write that as f'(x) = a + (1 - (a + 1)x)/x². Wait, maybe it's better to factor it differently. Let's see:f'(x) = a + 1/x² - (a + 1)/x. Let's combine the terms over a common denominator, which would be x². So, f'(x) = (a x² + 1 - (a + 1)x)/x².Simplify the numerator: a x² - (a + 1)x + 1. Let me factor this quadratic in x. Let's see if it factors nicely.Looking at a x² - (a + 1)x + 1, let's try to factor it. Let me write it as a x² - (a + 1)x + 1. Let's see if we can factor this as (m x - n)(p x - q). Hmm, maybe.Alternatively, let's use the quadratic formula to find the roots of the numerator. The numerator is a x² - (a + 1)x + 1. So, discriminant D = [-(a + 1)]² - 4*a*1 = (a + 1)² - 4a = a² + 2a + 1 - 4a = a² - 2a + 1 = (a - 1)².So, the roots are x = [(a + 1) ± sqrt((a - 1)²)]/(2a). Since sqrt((a - 1)²) is |a - 1|. So, x = [(a + 1) ± |a - 1|]/(2a).Let me consider two cases: a >= 1 and a < 1.Case 1: a >= 1. Then |a - 1| = a - 1. So, the roots are:x = [(a + 1) + (a - 1)]/(2a) = (2a)/(2a) = 1,andx = [(a + 1) - (a - 1)]/(2a) = (2)/(2a) = 1/a.Case 2: a < 1. Then |a - 1| = 1 - a. So, the roots are:x = [(a + 1) + (1 - a)]/(2a) = (2)/(2a) = 1/a,andx = [(a + 1) - (1 - a)]/(2a) = (2a)/(2a) = 1.So, regardless of a, the critical points are x = 1 and x = 1/a. But we need to consider the domain of f(x). Since f(x) has ln x and 1/x, x must be positive, so x > 0.Now, let's analyze the behavior of f(x) as x approaches 0+ and as x approaches infinity.As x approaches 0+:- ax approaches 0,- 1/x approaches infinity,- ln x approaches -infinity.So, f(x) = ax - 1/x - (a + 1)ln x. Let's see:- ax is negligible,- -1/x approaches -infinity,- -(a + 1)ln x approaches positive infinity because ln x approaches -infinity and it's multiplied by a negative.So, which term dominates? Let's see: -1/x goes to -infinity, and -(a + 1)ln x goes to +infinity. Which one is stronger? Let's compare the rates.As x approaches 0+, 1/x grows faster than ln x. So, -1/x dominates, so f(x) approaches -infinity.As x approaches infinity:- ax approaches infinity,- 1/x approaches 0,- ln x approaches infinity, but slower than x.So, f(x) = ax - 1/x - (a + 1)ln x. The dominant term is ax, which goes to infinity. So, f(x) approaches infinity.So, f(x) goes from -infinity at x approaches 0+ to infinity at x approaches infinity. So, by the Intermediate Value Theorem, f(x) must cross zero at least once.But the question is when does it have exactly one zero. So, we need to ensure that f(x) doesn't have more than one zero. That would happen if f(x) is monotonic, or if it has a critical point but doesn't cross zero more than once.Wait, but from the derivative, we have critical points at x = 1 and x = 1/a. So, depending on the value of a, the function could have different behaviors.Let me consider different cases for a.Case 1: a = 0. Wait, in part (1), a was 0, and f(x) had a maximum at x = 1, f(1) = -1. So, f(x) approaches -infinity as x approaches 0+ and approaches infinity as x approaches infinity. But with a maximum at x = 1 of -1, which is below zero. So, the function starts at -infinity, goes up to -1 at x = 1, then goes back down to -infinity? Wait, no, as x approaches infinity, f(x) approaches infinity. Wait, no, when a = 0, f(x) = -1/x - ln x. As x approaches infinity, -1/x approaches 0, and -ln x approaches -infinity. So, f(x) approaches -infinity as x approaches infinity. Wait, that contradicts my earlier thought.Wait, when a = 0, f(x) = -1/x - ln x. As x approaches infinity, -1/x approaches 0, and -ln x approaches -infinity. So, f(x) approaches -infinity as x approaches infinity. So, f(x) goes from -infinity at x approaches 0+ to -infinity at x approaches infinity, with a maximum at x = 1 of -1. So, it never crosses zero. So, when a = 0, f(x) has no zeros.But in part (1), we found the maximum value is -1, which is negative. So, f(x) is always negative, so no zeros.So, for a = 0, f(x) has no zeros. So, to have exactly one zero, a cannot be zero.Case 2: a > 0.Let me consider a > 0. Then, the critical points are x = 1 and x = 1/a.If a > 1, then 1/a < 1.If a = 1, then 1/a = 1.If 0 < a < 1, then 1/a > 1.So, let's consider these subcases.Subcase 2a: a > 1.So, critical points at x = 1/a < 1 and x = 1.Let me analyze the behavior of f(x):- As x approaches 0+, f(x) approaches -infinity.- At x = 1/a, which is less than 1, f(x) has a critical point.- At x = 1, another critical point.- As x approaches infinity, f(x) approaches infinity.So, let's see the behavior around the critical points.First, let's compute f(1/a):f(1/a) = a*(1/a) - 1/(1/a) - (a + 1)ln(1/a) = 1 - a - (a + 1)(-ln a) = 1 - a + (a + 1)ln a.Similarly, f(1) = a*1 - 1/1 - (a + 1)ln 1 = a - 1 - 0 = a - 1.Since a > 1, f(1) = a - 1 > 0.Now, let's see the behavior of f(x):- As x approaches 0+, f(x) approaches -infinity.- At x = 1/a, f(x) = 1 - a + (a + 1)ln a.- At x = 1, f(x) = a - 1 > 0.- As x approaches infinity, f(x) approaches infinity.Now, we need to determine if f(x) crosses zero only once.Given that f(x) approaches -infinity as x approaches 0+, and f(x) approaches infinity as x approaches infinity, and f(1) > 0, we need to see if f(x) has a minimum somewhere that is above zero, or if it dips below zero.Wait, but f(x) has critical points at x = 1/a and x = 1.Let me compute f(1/a):f(1/a) = 1 - a + (a + 1)ln a.Since a > 1, ln a > 0, so (a + 1)ln a > 0. So, f(1/a) = 1 - a + positive term.But 1 - a is negative because a > 1. So, f(1/a) could be positive or negative depending on the magnitude.Wait, let's see. Let me take a specific example. Let a = 2.Then f(1/a) = f(1/2) = 1 - 2 + (2 + 1)ln 2 = -1 + 3*0.693 ≈ -1 + 2.079 ≈ 1.079 > 0.So, f(1/a) is positive.Similarly, f(1) = 2 - 1 = 1 > 0.So, in this case, f(x) starts at -infinity, increases to f(1/a) ≈ 1.079, then decreases to f(1) = 1, then increases to infinity. So, f(x) crosses zero once between x approaching 0+ and x = 1/a, and then remains positive. So, only one zero.Wait, but f(x) goes from -infinity to positive at x = 1/a, then decreases to positive at x = 1, then increases to infinity. So, it crosses zero once between 0 and 1/a, and then stays positive. So, only one zero.Similarly, if a = 3:f(1/3) = 1 - 3 + 4 ln 3 ≈ -2 + 4*1.0986 ≈ -2 + 4.394 ≈ 2.394 > 0.f(1) = 3 - 1 = 2 > 0.So, same behavior.Wait, but what if a is just slightly greater than 1, say a = 1.1.f(1/a) = 1 - 1.1 + (1.1 + 1)ln(1.1) ≈ -0.1 + 2.1*0.0953 ≈ -0.1 + 0.200 ≈ 0.1 > 0.So, still positive.So, in all cases where a > 1, f(1/a) > 0, and f(1) > 0. So, f(x) goes from -infinity to positive at x = 1/a, then decreases to positive at x = 1, then increases to infinity. So, it crosses zero once.Subcase 2b: a = 1.Then, f(x) = x - 1/x - 2 ln x.Critical points at x = 1 and x = 1 (since 1/a = 1). So, only one critical point at x = 1.Compute f(1) = 1 - 1 - 2*0 = 0.So, f(1) = 0.Now, let's see the behavior:As x approaches 0+, f(x) approaches -infinity.At x = 1, f(x) = 0.As x approaches infinity, f(x) approaches infinity.What about the derivative at x = 1?f'(x) = 1 + 1/x² - 2/x.At x = 1, f'(1) = 1 + 1 - 2 = 0.So, x = 1 is a critical point.Now, let's check the second derivative to see if it's a maximum or minimum.f''(x) = -2/x³ + 2/x².At x = 1, f''(1) = -2 + 2 = 0. Hmm, inconclusive.Wait, maybe we can analyze the behavior around x = 1.For x < 1, say x = 0.5:f'(0.5) = 1 + 1/(0.25) - 2/(0.5) = 1 + 4 - 4 = 1 > 0.For x > 1, say x = 2:f'(2) = 1 + 1/4 - 2/2 = 1 + 0.25 - 1 = 0.25 > 0.So, f'(x) is positive on both sides of x = 1. So, x = 1 is a point of inflection, not a maximum or minimum.Wait, but f(x) approaches -infinity as x approaches 0+, and f(1) = 0, and f(x) approaches infinity as x approaches infinity. So, f(x) is increasing throughout, except at x = 1 where the derivative is zero but doesn't change sign.So, f(x) is monotonically increasing on (0, ∞), with a critical point at x = 1 where the derivative is zero but doesn't change sign. So, f(x) crosses zero exactly once at x = 1.So, when a = 1, f(x) has exactly one zero at x = 1.Subcase 2c: 0 < a < 1.So, critical points at x = 1/a > 1 and x = 1.Let me analyze f(x):As x approaches 0+, f(x) approaches -infinity.At x = 1, f(1) = a - 1 < 0 because a < 1.At x = 1/a, f(1/a) = 1 - a + (a + 1)ln a.Since 0 < a < 1, ln a < 0, so (a + 1)ln a < 0.So, f(1/a) = 1 - a + negative term.But 1 - a is positive because a < 1. So, f(1/a) could be positive or negative.Wait, let's take a specific example. Let a = 0.5.f(1/a) = f(2) = 0.5*2 - 1/2 - (0.5 + 1)ln 2 ≈ 1 - 0.5 - 1.5*0.693 ≈ 0.5 - 1.0395 ≈ -0.5395 < 0.So, f(1/a) < 0.Similarly, f(1) = 0.5 - 1 = -0.5 < 0.As x approaches infinity, f(x) approaches infinity.So, f(x) starts at -infinity, increases to f(1/a) ≈ -0.5395, then decreases to f(1) = -0.5, then increases to infinity.Wait, but f(x) is increasing from x = 0+ to x = 1/a, then decreasing from x = 1/a to x = 1, then increasing again from x = 1 to infinity.But f(1/a) < 0 and f(1) < 0, so the function goes from -infinity, increases to a local maximum at x = 1/a which is still negative, then decreases to a local minimum at x = 1, which is also negative, then increases to infinity.Therefore, f(x) crosses zero exactly once between x = 1 and infinity, because it goes from negative at x = 1 to positive infinity. So, only one zero.Wait, but let me check another example. Let a = 0.25.f(1/a) = f(4) = 0.25*4 - 1/4 - (0.25 + 1)ln 4 ≈ 1 - 0.25 - 1.25*1.386 ≈ 0.75 - 1.7325 ≈ -0.9825 < 0.f(1) = 0.25 - 1 = -0.75 < 0.So, same behavior: f(x) goes from -infinity, increases to f(4) ≈ -0.9825, then decreases to f(1) = -0.75, then increases to infinity. So, crosses zero once between x = 1 and infinity.Wait, but what if a is very close to 1, say a = 0.9.f(1/a) = f(10/9) ≈ 0.9*(10/9) - 1/(10/9) - (0.9 + 1)ln(10/9).Compute:0.9*(10/9) = 1,1/(10/9) = 0.9,(0.9 + 1) = 1.9,ln(10/9) ≈ 0.10536.So, f(1/a) ≈ 1 - 0.9 - 1.9*0.10536 ≈ 0.1 - 0.200 ≈ -0.1 < 0.So, f(1/a) is still negative.Thus, in all cases where 0 < a < 1, f(x) has exactly one zero between x = 1 and infinity.Wait, but what about when a is negative? Let's consider that.Case 3: a < 0.So, a is negative. Let's see the critical points.From earlier, the critical points are x = 1 and x = 1/a. But since a is negative, 1/a is negative, which is not in the domain x > 0. So, the only critical point is x = 1.So, f(x) has only one critical point at x = 1.Let me compute f(1) = a - 1. Since a < 0, f(1) = a - 1 < -1 < 0.Now, let's analyze the behavior:As x approaches 0+, f(x) approaches -infinity.At x = 1, f(1) = a - 1 < 0.As x approaches infinity, f(x) = ax - 1/x - (a + 1)ln x.Since a is negative, ax approaches -infinity, and -(a + 1)ln x: since a < 0, a + 1 could be positive or negative.Wait, if a < -1, then a + 1 < 0, so -(a + 1)ln x is positive because ln x is positive for x > 1, but multiplied by a negative coefficient.Wait, let's take a specific example. Let a = -2.f(x) = -2x - 1/x - (-2 + 1)ln x = -2x - 1/x + ln x.As x approaches infinity:-2x dominates, so f(x) approaches -infinity.As x approaches 0+, f(x) approaches -infinity.At x = 1, f(1) = -2 - 1 + 0 = -3 < 0.So, f(x) approaches -infinity at both ends, and has a critical point at x = 1 where f(1) = -3.Wait, but what about the derivative?f'(x) = a + 1/x² - (a + 1)/x.With a = -2, f'(x) = -2 + 1/x² - (-2 + 1)/x = -2 + 1/x² + 1/x.So, f'(x) = -2 + 1/x² + 1/x.Let me set f'(x) = 0: -2 + 1/x² + 1/x = 0.Multiply by x²: -2x² + 1 + x = 0 => -2x² + x + 1 = 0 => 2x² - x - 1 = 0.Solutions: x = [1 ± sqrt(1 + 8)] / 4 = [1 ± 3]/4.So, x = 1 or x = -0.5. Since x > 0, only x = 1 is valid.So, f(x) has a critical point at x = 1, which is a minimum or maximum.Compute f''(x) to check concavity.f''(x) = -2/x³ + 2/x².At x = 1, f''(1) = -2 + 2 = 0. Hmm, inconclusive.But let's check the behavior around x = 1.For x < 1, say x = 0.5:f'(0.5) = -2 + 1/(0.25) + 1/(0.5) = -2 + 4 + 2 = 4 > 0.For x > 1, say x = 2:f'(2) = -2 + 1/4 + 1/2 = -2 + 0.25 + 0.5 = -1.25 < 0.So, f(x) is increasing before x = 1 and decreasing after x = 1. So, x = 1 is a maximum.But f(1) = -3 < 0. So, the function increases from -infinity to x = 1, reaching a maximum of -3, then decreases back to -infinity as x approaches infinity.So, f(x) never crosses zero. So, for a = -2, f(x) has no zeros.Similarly, for any a < 0, f(x) approaches -infinity at both ends, has a maximum at x = 1 which is negative, so f(x) never crosses zero.Therefore, for a < 0, f(x) has no zeros.Putting it all together:- When a < 0: f(x) has no zeros.- When a = 0: f(x) has no zeros.- When 0 < a < 1: f(x) has exactly one zero.- When a = 1: f(x) has exactly one zero.- When a > 1: f(x) has exactly one zero.Wait, but earlier when a = 1, f(x) has exactly one zero at x = 1. For a > 1, f(x) has exactly one zero between 0 and 1/a, and then remains positive. Wait, no, actually, for a > 1, f(x) goes from -infinity, increases to a positive value at x = 1/a, then decreases to a positive value at x = 1, then increases to infinity. So, it crosses zero once between 0 and 1/a.Wait, but earlier when a = 2, f(x) crosses zero once between 0 and 1/a, and then remains positive. So, only one zero.Similarly, for a = 1, f(x) crosses zero exactly once at x = 1.For 0 < a < 1, f(x) crosses zero once between x = 1 and infinity.So, in all cases where a > 0, f(x) has exactly one zero.But wait, when a = 0, f(x) has no zeros, and when a < 0, f(x) has no zeros.Therefore, the range of a for which f(x) has exactly one zero is a > 0.But let me double-check for a = 1. When a = 1, f(x) has exactly one zero at x = 1.For a > 1, f(x) has exactly one zero between 0 and 1/a.For 0 < a < 1, f(x) has exactly one zero between 1 and infinity.So, yes, for all a > 0, f(x) has exactly one zero.Therefore, the range of a is (0, ∞).But wait, let me check a = 0.5 again. f(x) = 0.5x - 1/x - 1.5 ln x.As x approaches 0+, f(x) approaches -infinity.At x = 1, f(1) = 0.5 - 1 - 0 = -0.5.At x = 2, f(2) = 1 - 0.5 - 1.5 ln 2 ≈ 0.5 - 1.039 ≈ -0.539.Wait, but as x approaches infinity, f(x) approaches infinity because 0.5x dominates.So, f(x) goes from -infinity, increases to f(2) ≈ -0.539, then decreases to f(1) = -0.5, then increases to infinity. So, it must cross zero once between x = 1 and infinity.Yes, that's correct.Similarly, for a = 2, f(x) crosses zero once between 0 and 0.5, then remains positive.So, in all cases where a > 0, f(x) has exactly one zero.Therefore, the range of a is (0, ∞).**Final Answer**(1) The maximum value of ( f(x) ) when ( a = 0 ) is (boxed{-1}).(2) The range of values for ( a ) such that ( f(x) ) has exactly one zero is (boxed{(0, +infty)}).