🧠:<think>Okay, so I have this problem where I need to show that the quadratic mean of a set of numbers is always greater than or equal to the arithmetic mean. And then I also need to find when they are equal. Hmm, let me think about how to approach this.First, let me recall what quadratic mean and arithmetic mean are. The quadratic mean, also known as the root mean square, is defined as the square root of the average of the squares of the numbers. So for numbers (a_1, a_2, ldots, a_n), it's (sqrt{frac{a_1^2 + a_2^2 + ldots + a_n^2}{n}}). On the other hand, the arithmetic mean is just the average of the numbers, which is (frac{a_1 + a_2 + ldots + a_n}{n}).I need to show that the quadratic mean is always greater than or equal to the arithmetic mean. That is, I need to prove that:[sqrt{frac{a_1^2 + a_2^2 + ldots + a_n^2}{n}} geq frac{a_1 + a_2 + ldots + a_n}{n}]Hmm, how can I prove this? Maybe I can square both sides to get rid of the square root. Let me try that.Squaring both sides gives:[frac{a_1^2 + a_2^2 + ldots + a_n^2}{n} geq left(frac{a_1 + a_2 + ldots + a_n}{n}right)^2]Okay, so now I have an inequality without the square root. Let me simplify the right-hand side.Expanding the right-hand side:[left(frac{a_1 + a_2 + ldots + a_n}{n}right)^2 = frac{(a_1 + a_2 + ldots + a_n)^2}{n^2}]So now the inequality becomes:[frac{a_1^2 + a_2^2 + ldots + a_n^2}{n} geq frac{(a_1 + a_2 + ldots + a_n)^2}{n^2}]To make it easier, let me multiply both sides by (n^2) to eliminate the denominators:[n(a_1^2 + a_2^2 + ldots + a_n^2) geq (a_1 + a_2 + ldots + a_n)^2]Alright, so now I have:[n(a_1^2 + a_2^2 + ldots + a_n^2) geq (a_1 + a_2 + ldots + a_n)^2]This looks familiar. I think this is related to the Cauchy-Schwarz inequality. Let me recall that the Cauchy-Schwarz inequality states that for any real numbers (x_1, x_2, ldots, x_n) and (y_1, y_2, ldots, y_n):[(x_1y_1 + x_2y_2 + ldots + x_ny_n)^2 leq (x_1^2 + x_2^2 + ldots + x_n^2)(y_1^2 + y_2^2 + ldots + y_n^2)]If I set all (y_i = 1), then the inequality becomes:[(x_1 + x_2 + ldots + x_n)^2 leq (x_1^2 + x_2^2 + ldots + x_n^2)(1^2 + 1^2 + ldots + 1^2)]Since there are (n) terms of 1, the sum (1^2 + 1^2 + ldots + 1^2 = n). So the inequality simplifies to:[(x_1 + x_2 + ldots + x_n)^2 leq n(x_1^2 + x_2^2 + ldots + x_n^2)]Which is exactly the inequality I derived earlier! So, by the Cauchy-Schwarz inequality, this holds true. Therefore, the quadratic mean is always greater than or equal to the arithmetic mean.Now, I need to find when equality holds. From the Cauchy-Schwarz inequality, equality occurs if and only if the sequences are proportional. That is, there exists a constant (k) such that (x_i = k y_i) for all (i). In our case, since we set all (y_i = 1), this implies that (x_i = k) for all (i). Therefore, all (x_i) must be equal.Translating back to our original problem, this means that equality holds if and only if all the numbers (a_1, a_2, ldots, a_n) are equal.Let me double-check this with a simple example. Suppose all (a_i = c) for some constant (c). Then the quadratic mean is:[sqrt{frac{c^2 + c^2 + ldots + c^2}{n}} = sqrt{frac{n c^2}{n}} = sqrt{c^2} = |c|]And the arithmetic mean is:[frac{c + c + ldots + c}{n} = frac{n c}{n} = c]So, if (c) is non-negative, both means are equal to (c). If (c) is negative, the quadratic mean is (|c|), which is greater than (c). Wait, that seems contradictory. Did I make a mistake?Hmm, actually, in the definition of quadratic mean, we take the square root, which is always non-negative. So if all (a_i) are equal and negative, the quadratic mean would still be the absolute value, which is greater than the arithmetic mean (which is negative). So, in that case, even if all (a_i) are equal but negative, the quadratic mean is greater than the arithmetic mean.But wait, according to the Cauchy-Schwarz inequality, equality holds when all (a_i) are equal, regardless of their sign. So, if all (a_i) are equal, whether positive or negative, the quadratic mean equals the arithmetic mean in absolute value, but if they are negative, the arithmetic mean is negative while the quadratic mean is positive. So, in terms of magnitude, they are equal, but in terms of actual values, the quadratic mean is always non-negative, while the arithmetic mean can be negative.Hmm, maybe I need to clarify this. Let me think again.If all (a_i) are equal, say (a_i = c), then the arithmetic mean is (c), and the quadratic mean is (|c|). So, if (c) is positive, both are equal. If (c) is negative, the quadratic mean is greater than the arithmetic mean because (|c| > c). So, in the case where all (a_i) are equal and positive, the quadratic mean equals the arithmetic mean. If they are equal and negative, the quadratic mean is greater.But wait, the problem didn't specify whether the numbers are positive or not. So, does the equality hold only when all numbers are equal and non-negative? Or does it hold when all numbers are equal, regardless of their sign?Looking back at the Cauchy-Schwarz inequality, equality holds when the vectors are proportional. In our case, the vectors are ((a_1, a_2, ldots, a_n)) and ((1, 1, ldots, 1)). So, for equality, we need (a_i = k cdot 1) for some constant (k), meaning all (a_i) are equal to (k). So, regardless of the sign of (k), the equality holds in the Cauchy-Schwarz inequality.But when we take the square root in the quadratic mean, it becomes (|k|), while the arithmetic mean is (k). So, if (k) is positive, both are equal. If (k) is negative, the quadratic mean is greater. So, in terms of the original inequality, equality holds only when all (a_i) are equal and non-negative.Wait, but the Cauchy-Schwarz inequality doesn't care about the sign; it's about the proportionality. So, maybe I need to consider that in the context of the original inequality, the quadratic mean is always non-negative, while the arithmetic mean can be negative. So, if all (a_i) are equal and negative, the quadratic mean is still greater than the arithmetic mean because it's positive.Therefore, the equality holds only when all (a_i) are equal and non-negative. If they are equal and negative, the quadratic mean is greater. So, the cases of equality are when all the numbers are equal and non-negative.Wait, but the problem didn't specify that the numbers are non-negative. So, maybe I need to consider that the quadratic mean is always greater than or equal to the arithmetic mean, regardless of the signs of the numbers, and equality holds only when all numbers are equal and non-negative.Alternatively, perhaps the problem assumes that the numbers are non-negative, as quadratic mean is often discussed in the context of non-negative numbers.Let me check the original problem statement again. It just says "the quadratic mean of (a_1, ldots, a_n)", without specifying that the numbers are non-negative. So, I think I need to consider that the numbers can be any real numbers.In that case, the quadratic mean is always greater than or equal to the arithmetic mean, and equality holds if and only if all the numbers are equal. Because even if they are negative, the quadratic mean is the absolute value of the common value, which is greater than or equal to the arithmetic mean, which is the same value (negative). So, in terms of magnitude, they are equal, but in terms of actual values, the quadratic mean is greater unless all numbers are zero.Wait, no. If all numbers are equal to zero, then both means are zero, so they are equal. If all numbers are equal to a positive number, both means are equal. If all numbers are equal to a negative number, the quadratic mean is positive, and the arithmetic mean is negative, so they are not equal. Therefore, equality holds only when all numbers are equal and non-negative, including zero.Wait, but if all numbers are equal to a negative number, the quadratic mean is the absolute value, which is positive, and the arithmetic mean is negative, so they are not equal. Therefore, equality holds only when all numbers are equal and non-negative.But in the Cauchy-Schwarz inequality, equality holds when all numbers are equal, regardless of sign. So, perhaps the confusion arises because the quadratic mean is always non-negative, while the arithmetic mean can be negative. Therefore, when all numbers are equal and negative, the quadratic mean is greater than the arithmetic mean, not equal.Therefore, the cases of equality are when all numbers are equal and non-negative.Wait, but if all numbers are equal to zero, which is non-negative, then both means are zero, so they are equal. If all numbers are equal to a positive number, both means are equal. If all numbers are equal to a negative number, the quadratic mean is greater than the arithmetic mean.Therefore, the equality holds if and only if all numbers are equal and non-negative.But let me test this with an example. Suppose all (a_i = 2). Then quadratic mean is (sqrt{frac{2^2 + 2^2 + ldots + 2^2}{n}} = sqrt{frac{4n}{n}} = 2), and arithmetic mean is (frac{2 + 2 + ldots + 2}{n} = 2). So, equal.If all (a_i = -2), then quadratic mean is (sqrt{frac{(-2)^2 + (-2)^2 + ldots + (-2)^2}{n}} = sqrt{frac{4n}{n}} = 2), and arithmetic mean is (frac{-2 + (-2) + ldots + (-2)}{n} = -2). So, quadratic mean is greater.If all (a_i = 0), both means are zero, so equal.Therefore, equality holds only when all numbers are equal and non-negative, including zero.So, to summarize, the quadratic mean is always greater than or equal to the arithmetic mean, and equality holds if and only if all the numbers are equal and non-negative.Wait, but in the Cauchy-Schwarz inequality, equality holds when the vectors are proportional, regardless of the sign. So, in our case, when all (a_i) are equal, regardless of sign, the Cauchy-Schwarz inequality gives equality. But when we take the square root in the quadratic mean, it becomes the absolute value, which is always non-negative. Therefore, if all (a_i) are equal and negative, the quadratic mean is the absolute value, which is greater than the arithmetic mean, which is negative. So, in that case, equality does not hold.Therefore, the cases of equality are when all numbers are equal and non-negative.Alternatively, if we consider the numbers to be non-negative, then equality holds when all numbers are equal.But the problem didn't specify that the numbers are non-negative. So, I think the correct statement is that equality holds if and only if all numbers are equal and non-negative.Wait, but if all numbers are equal to zero, which is non-negative, then both means are zero, so they are equal. If all numbers are equal to a positive number, both means are equal. If all numbers are equal to a negative number, the quadratic mean is greater than the arithmetic mean.Therefore, the cases of equality are when all numbers are equal and non-negative.So, to conclude, the quadratic mean is always greater than or equal to the arithmetic mean, and equality holds if and only if all the numbers are equal and non-negative.But wait, let me think again. If all numbers are equal to a negative number, the quadratic mean is greater than the arithmetic mean. So, equality does not hold in that case. Therefore, equality holds only when all numbers are equal and non-negative.Yes, that seems correct.So, in summary, I have shown that the quadratic mean is always greater than or equal to the arithmetic mean by using the Cauchy-Schwarz inequality, and equality holds if and only if all the numbers are equal and non-negative.</think>

🧠:<think>First, I need to determine how many pencil boxes can be filled with 7,344 pencils when each box holds 7 pencils.To find this, I will divide the total number of pencils by the number of pencils per box.So, I divide 7,344 by 7.7,344 ÷ 7 equals 1,049.Therefore, you will fill 1,049 pencil boxes.</think>

🧠:<think>Okay, so I have this problem about points on a plane and connecting them with segments. Let me try to understand it step by step.First, there's a finite set of points M, and no three points are on the same straight line. That means if I pick any three points, they form a triangle, right? No degenerate cases where they all lie on a line. Good.Now, some of these points are connected by segments, and importantly, no point is connected to more than one segment. So, each point is either isolated or connected to exactly one other point. This setup reminds me of a matching in graph theory, where each vertex has degree at most one. So, we're dealing with a matching on the point set M.The allowed operation is replacing a pair of intersecting segments AB and CD with the pair of opposite sides AC and BD of the quadrilateral ACBD. Hmm, okay. So, if two segments AB and CD cross each other, we can replace them with AC and BD. I need to visualize this. If I have four points A, B, C, D arranged such that AB and CD intersect, then replacing them with AC and BD would essentially "flip" the segments across the intersection point.The question is whether this sequence of replacements can be infinite. So, can we keep doing this operation forever, or will it eventually stop?Let me think about what happens when we perform this replacement. If AB and CD intersect, then AC and BD are the other two sides of the quadrilateral. I wonder if there's some measure that changes with each replacement. Maybe the total length of all segments?If I consider the lengths of AB, CD, AC, and BD, I recall something about the triangle inequality. In a quadrilateral, the sum of the lengths of any three sides must be greater than the fourth side. But here, we're replacing two intersecting sides with the other two. Is there a relationship between the lengths?Wait, maybe it's about the sum of the lengths. If I replace AB and CD with AC and BD, does the total length increase or decrease? Let me think. If AB and CD intersect, then AC and BD are the other two diagonals. In a convex quadrilateral, the sum of the lengths of the two diagonals is greater than the sum of the lengths of the two sides. But in our case, AB and CD are sides, and AC and BD are the other sides.Wait, no. Actually, in a convex quadrilateral, the sum of the lengths of the two diagonals is greater than the sum of the lengths of the two opposite sides. But here, we're replacing two sides with the other two sides. Hmm, maybe I need to think differently.Alternatively, maybe the key is that each replacement reduces some kind of potential function, ensuring that the process must terminate after a finite number of steps. Since the set of points is finite, the number of possible segments is finite, so maybe the number of possible replacements is also finite.But wait, the problem allows replacing any intersecting pair, and after each replacement, we might create new intersections. So, it's not immediately clear that the number of replacements is finite.Let me think about an invariant or something that strictly decreases with each replacement. Maybe the number of intersections? If each replacement removes one intersection and possibly creates others, but maybe the total number of intersections decreases.But I'm not sure. Maybe another approach. Since the set of points is finite, the number of possible segments is finite, so the number of possible configurations is also finite. If each replacement leads to a different configuration, then eventually, we must repeat a configuration, leading to a cycle. But the problem is whether the sequence can be infinite, meaning it doesn't terminate.Wait, but if each replacement strictly decreases some measure, like the total length of segments, then since the measure is bounded below (can't be negative), the process must terminate after a finite number of steps.So, maybe the total length of all segments decreases with each replacement. Let me check.Suppose we have two intersecting segments AB and CD. Let’s denote their intersection point as E. Then, by the triangle inequality, in triangle AEC, we have AC < AE + EC. Similarly, in triangle BED, we have BD < BE + ED. But AB = AE + EB and CD = CE + ED. So, AC + BD < (AE + EC) + (BE + ED) = AB + CD.Therefore, replacing AB and CD with AC and BD decreases the total length of the segments. So, each replacement reduces the total length.Since the total length is a positive real number and it decreases with each replacement, and it's bounded below by zero, the process must terminate after a finite number of steps. Therefore, the sequence of replacements cannot be infinite.Wait, but is the total length the only thing decreasing? What if after some replacements, the total length stops decreasing? But no, each replacement strictly decreases the total length, so it can't cycle or repeat configurations without decreasing.Therefore, the sequence must terminate, so it can't be infinite.I think that makes sense. Each operation strictly reduces the total length, and since we can't have an infinite decreasing sequence of positive real numbers, the process must stop after finitely many steps.Final Answerboxed{text{No}}</think>

🧠:<think>Alright, so I have this function ( f(x) = a ln x - x - frac{1}{2}x^2 ). The problem is divided into two parts: first, for ( a = 2 ), I need to find the extreme values of the function, and second, I have to discuss the monotonicity of ( f(x) ) in general. Let me try to tackle each part step by step.Starting with part (I): For ( a = 2 ), find the extreme values of ( f(x) ).Hmm, okay. To find extreme values, I remember that I need to find the critical points by taking the derivative of ( f(x) ) and setting it equal to zero. So, let me compute the derivative ( f'(x) ).Given ( f(x) = 2 ln x - x - frac{1}{2}x^2 ), the derivative ( f'(x) ) would be:- The derivative of ( 2 ln x ) is ( frac{2}{x} ).- The derivative of ( -x ) is ( -1 ).- The derivative of ( -frac{1}{2}x^2 ) is ( -x ).So, putting it all together, ( f'(x) = frac{2}{x} - 1 - x ).I can write this as ( f'(x) = frac{2}{x} - x - 1 ). To make it easier to solve for critical points, maybe I can combine the terms over a common denominator. Let me rewrite it:( f'(x) = frac{2 - x^2 - x}{x} ).So, ( f'(x) = frac{-x^2 - x + 2}{x} ). Hmm, that simplifies to ( f'(x) = frac{-(x^2 + x - 2)}{x} ).Wait, maybe I can factor the numerator. Let me try factoring ( x^2 + x - 2 ). Looking for two numbers that multiply to -2 and add to 1. That would be 2 and -1. So, ( x^2 + x - 2 = (x + 2)(x - 1) ).So, substituting back, ( f'(x) = frac{-(x + 2)(x - 1)}{x} ).Alright, so to find critical points, I set ( f'(x) = 0 ). That happens when the numerator is zero, so:( -(x + 2)(x - 1) = 0 ).This gives solutions ( x = -2 ) and ( x = 1 ). But since the domain of ( f(x) ) is ( x > 0 ) (because of the ( ln x ) term), ( x = -2 ) is not in the domain. So, the only critical point is at ( x = 1 ).Now, to determine if this critical point is a maximum or a minimum, I can use the second derivative test or analyze the sign changes of the first derivative around ( x = 1 ).Let me try the first derivative test. I'll pick points just less than 1 and just greater than 1 to see the sign of ( f'(x) ).First, let's choose ( x = 0.5 ) (which is less than 1):( f'(0.5) = frac{2}{0.5} - 0.5 - 1 = 4 - 0.5 - 1 = 2.5 ). So, positive.Next, choose ( x = 2 ) (which is greater than 1):( f'(2) = frac{2}{2} - 2 - 1 = 1 - 2 - 1 = -2 ). So, negative.Therefore, the derivative changes from positive to negative at ( x = 1 ), which means ( f(x) ) has a local maximum at ( x = 1 ).Now, let's compute the value of ( f(1) ):( f(1) = 2 ln 1 - 1 - frac{1}{2}(1)^2 = 0 - 1 - frac{1}{2} = -frac{3}{2} ).So, the function has a maximum value of ( -frac{3}{2} ) at ( x = 1 ). Since the function tends to negative infinity as ( x ) approaches 0 and also tends to negative infinity as ( x ) approaches infinity (because the ( -frac{1}{2}x^2 ) term dominates), there is no minimum value.Alright, that takes care of part (I). Now, moving on to part (II): Discuss the monotonicity of the function ( f(x) ).Monotonicity refers to where the function is increasing or decreasing. To determine this, I need to analyze the sign of the derivative ( f'(x) ).Earlier, I found that ( f'(x) = frac{a - x - x^2}{x} ). Let me write that again:( f'(x) = frac{a - x - x^2}{x} ).Since ( x > 0 ), the sign of ( f'(x) ) depends on the numerator ( a - x - x^2 ). Let me denote the numerator as ( g(x) = a - x - x^2 ).So, ( g(x) = -x^2 - x + a ). This is a quadratic function in terms of ( x ). The quadratic is concave down because the coefficient of ( x^2 ) is negative.To find when ( g(x) = 0 ), we can solve the quadratic equation:( -x^2 - x + a = 0 ).Multiplying both sides by -1 to make it easier:( x^2 + x - a = 0 ).Using the quadratic formula, the roots are:( x = frac{-1 pm sqrt{1 + 4a}}{2} ).So, the roots are ( x = frac{-1 + sqrt{1 + 4a}}{2} ) and ( x = frac{-1 - sqrt{1 + 4a}}{2} ).Since ( x > 0 ), we can ignore the negative root because ( frac{-1 - sqrt{1 + 4a}}{2} ) will always be negative (as ( sqrt{1 + 4a} ) is non-negative, so subtracting 1 and dividing by 2 keeps it negative). Therefore, the only relevant root is ( x = frac{-1 + sqrt{1 + 4a}}{2} ).Now, the behavior of ( g(x) ) depends on the discriminant ( Delta = 1 + 4a ).Case 1: If ( Delta leq 0 ), which means ( 1 + 4a leq 0 ) or ( a leq -frac{1}{4} ). In this case, the quadratic ( g(x) ) does not cross the x-axis, meaning it is always negative (since the coefficient of ( x^2 ) is negative). Therefore, ( f'(x) = frac{g(x)}{x} ) is negative for all ( x > 0 ). So, ( f(x) ) is decreasing on ( (0, +infty) ).Case 2: If ( Delta > 0 ), which means ( a > -frac{1}{4} ). Here, the quadratic ( g(x) ) has two real roots. As mentioned earlier, only the positive root ( x = frac{-1 + sqrt{1 + 4a}}{2} ) is relevant.Now, let's analyze the sign of ( g(x) ):- For ( x < frac{-1 + sqrt{1 + 4a}}{2} ), since the quadratic opens downward, ( g(x) ) is positive.- For ( x > frac{-1 + sqrt{1 + 4a}}{2} ), ( g(x) ) is negative.Therefore, ( f'(x) ) is positive when ( x < frac{-1 + sqrt{1 + 4a}}{2} ) and negative when ( x > frac{-1 + sqrt{1 + 4a}}{2} ). This implies that ( f(x) ) is increasing on ( (0, frac{-1 + sqrt{1 + 4a}}{2}) ) and decreasing on ( (frac{-1 + sqrt{1 + 4a}}{2}, +infty) ).But wait, I need to consider the value of ( a ) more carefully. Specifically, when ( a ) is between ( -frac{1}{4} ) and 0, does the positive root ( x = frac{-1 + sqrt{1 + 4a}}{2} ) become positive or not?Let me compute ( frac{-1 + sqrt{1 + 4a}}{2} ) when ( a ) is in ( (-frac{1}{4}, 0] ).Take ( a = 0 ):( x = frac{-1 + sqrt{1 + 0}}{2} = frac{-1 + 1}{2} = 0 ).So, at ( a = 0 ), the positive root is 0. For ( a ) just above ( -frac{1}{4} ), say ( a = -frac{1}{8} ):( x = frac{-1 + sqrt{1 + 4*(-1/8)}}{2} = frac{-1 + sqrt{1 - 0.5}}{2} = frac{-1 + sqrt{0.5}}{2} approx frac{-1 + 0.707}{2} approx frac{-0.293}{2} approx -0.146 ).Wait, that's negative. Hmm, so for ( a ) in ( (-frac{1}{4}, 0) ), the positive root is actually negative? That can't be, because ( sqrt{1 + 4a} ) when ( a > -frac{1}{4} ) is greater than 0, but subtracting 1 might make it less than 1.Wait, let me compute it more carefully.For ( a = -frac{1}{8} ):( 1 + 4a = 1 + 4*(-1/8) = 1 - 0.5 = 0.5 ).So, ( sqrt{0.5} approx 0.707 ).Thus, ( x = frac{-1 + 0.707}{2} = frac{-0.293}{2} approx -0.146 ). So, it's negative.Therefore, for ( a ) in ( (-frac{1}{4}, 0) ), the positive root is actually negative, which is outside our domain ( x > 0 ). Therefore, in this case, ( g(x) = a - x - x^2 ) is always negative for ( x > 0 ), because the quadratic doesn't cross the x-axis in the positive domain.Wait, but hold on: when ( a > -frac{1}{4} ), the quadratic ( g(x) ) has two real roots, but both are negative or one positive and one negative? Wait, no. Actually, when ( a > -frac{1}{4} ), the discriminant is positive, so two real roots. But depending on ( a ), one root is positive and the other is negative.Wait, let's take ( a = 1 ):( x = frac{-1 + sqrt{1 + 4*1}}{2} = frac{-1 + sqrt{5}}{2} approx frac{-1 + 2.236}{2} approx 0.618 ).So, positive. So, for ( a > 0 ), the positive root is positive. For ( a = 0 ), the positive root is 0. For ( a ) between ( -frac{1}{4} ) and 0, the positive root is negative. So, in that case, ( g(x) ) doesn't cross the x-axis in the positive domain, so ( g(x) ) is always negative for ( x > 0 ).Therefore, summarizing:- If ( a leq -frac{1}{4} ), ( f'(x) leq 0 ) for all ( x > 0 ), so ( f(x) ) is decreasing on ( (0, +infty) ).- If ( -frac{1}{4} < a leq 0 ), ( f'(x) < 0 ) for all ( x > 0 ), so ( f(x) ) is decreasing on ( (0, +infty) ).- If ( a > 0 ), ( f'(x) ) changes sign from positive to negative at ( x = frac{-1 + sqrt{1 + 4a}}{2} ). Therefore, ( f(x) ) is increasing on ( (0, frac{-1 + sqrt{1 + 4a}}{2}) ) and decreasing on ( (frac{-1 + sqrt{1 + 4a}}{2}, +infty) ).Wait, but when ( a = 0 ), the critical point is at 0, which is the boundary of the domain. So, for ( a = 0 ), the function ( f(x) = -x - frac{1}{2}x^2 ), which is a downward opening parabola, so it's decreasing for all ( x > 0 ).So, that seems consistent.Therefore, putting it all together, the function ( f(x) ) is decreasing on ( (0, +infty) ) when ( a leq 0 ), and when ( a > 0 ), it first increases up to ( x = frac{-1 + sqrt{1 + 4a}}{2} ) and then decreases afterward.Let me just verify this with a quick example. Suppose ( a = 1 ), which is greater than 0. Then, the critical point is at ( x = frac{-1 + sqrt{5}}{2} approx 0.618 ). Let me compute ( f'(x) ) at ( x = 0.5 ) and ( x = 1 ).At ( x = 0.5 ):( f'(0.5) = frac{1}{0.5} - 0.5 - 1 = 2 - 0.5 - 1 = 0.5 ). Positive, so increasing.At ( x = 1 ):( f'(1) = 1 - 1 - 1 = -1 ). Negative, so decreasing.So, that aligns with the conclusion.Another example, ( a = -1 ), which is less than ( -frac{1}{4} ). Then, ( f'(x) = frac{-1 - x - x^2}{x} ). Let's compute at ( x = 1 ):( f'(1) = frac{-1 -1 -1}{1} = -3 ). Negative, so decreasing.At ( x = 0.5 ):( f'(0.5) = frac{-1 -0.5 -0.25}{0.5} = frac{-1.75}{0.5} = -3.5 ). Still negative.So, it's decreasing everywhere, as expected.Another test, ( a = -frac{1}{8} ), which is between ( -frac{1}{4} ) and 0. Let's compute ( f'(x) ) at ( x = 1 ):( f'(1) = frac{-1/8 -1 -1}{1} = frac{-1/8 -2}{1} = -frac{17}{8} ). Negative.At ( x = 0.5 ):( f'(0.5) = frac{-1/8 -0.5 -0.25}{0.5} = frac{-1/8 -0.75}{0.5} = frac{-0.875}{0.5} = -1.75 ). Still negative.So, even though ( a > -frac{1}{4} ), because ( a leq 0 ), the function is still decreasing everywhere.Therefore, my analysis seems consistent.So, to recap:(I) For ( a = 2 ), the function ( f(x) ) has a maximum value at ( x = 1 ) with ( f(1) = -frac{3}{2} ), and no minimum value.(II) The monotonicity of ( f(x) ) depends on ( a ):- If ( a leq -frac{1}{4} ), ( f(x) ) is decreasing on ( (0, +infty) ).- If ( -frac{1}{4} < a leq 0 ), ( f(x) ) is decreasing on ( (0, +infty) ).- If ( a > 0 ), ( f(x) ) is increasing on ( (0, frac{-1 + sqrt{1 + 4a}}{2}) ) and decreasing on ( (frac{-1 + sqrt{1 + 4a}}{2}, +infty) ).I think that covers both parts of the problem. I should double-check my calculations, especially the derivative and the critical points, but everything seems to add up.Final Answer(I) The maximum value of ( f(x) ) is (boxed{-frac{3}{2}}), and there is no minimum value.(II) The function ( f(x) ) is decreasing on ( (0, +infty) ) if ( a leq 0 ). If ( a > 0 ), it is increasing on ( left(0, frac{-1 + sqrt{1 + 4a}}{2}right) ) and decreasing on ( left(frac{-1 + sqrt{1 + 4a}}{2}, +inftyright) ).</think>