🧠:[<think>Okay, so I need to figure out the possible values of the integral of f squared, where f is a concave, positive function defined on (0,1) with integral 1. The question is asking what values can ∫₀¹ [f(x)]² dx take under these conditions. Let me start by recalling some concepts.First, a concave function. The definition given is that for any x₁ < x₂ in [0,1] and λ in [0,1], f(λx₁ + (1-λ)x₂) ≥ λf(x₁) + (1-λ)f(x₂). So concave from below here means the usual concave function, right? Because sometimes people might say concave up or down, but in this case, the definition matches the standard concave function, which curves downward. So the chord between any two points lies below the function.Since f is concave, it's also continuous on (0,1), right? I remember that concave functions on an open interval are continuous. So that's good. And it's integrable, so the integral over (0,1) is 1. Also, f is positive everywhere in (0,1).Now, the integral of f squared. I need to find the possible values of this integral given the constraints. So it's like an optimization problem: find the minimum and maximum possible values of ∫f² dx over all such functions f.Wait, but how do I approach this? Maybe using some inequalities. For instance, the Cauchy-Schwarz inequality relates the integral of f squared to the square of the integral of f, but here the integral of f is 1, so Cauchy-Schwarz would say that ∫f² dx ≥ (∫f dx)² = 1. But that's just the lower bound. However, this is only when we consider all functions with integral 1. But here we have the additional constraint that f is concave. So maybe the lower bound is higher? Or maybe the upper bound is restricted.Wait, but actually, Cauchy-Schwarz gives that ∫f² dx ≥ (∫f dx)² = 1, but equality holds when f is constant. Since f is concave and has integral 1, the constant function f(x) = 1 is concave (since the inequality becomes equality for all λ, x₁, x₂). So the minimal value of ∫f² dx is 1, achieved by the constant function. But is that the case? Wait, no. Wait, if f is constant, then it's both concave and convex, right? Because the definition of concave requires the chord to lie below the function, but if the function is linear (constant is a linear function), then the chord is on the function. So constant functions are indeed concave (and convex). Therefore, f(x) = 1 is a valid function here, and ∫f² dx = ∫1 dx = 1. So that's the minimum.But wait, maybe the minimum is actually higher? Because sometimes constraints can affect this. But since Cauchy-Schwarz gives the lower bound of 1, and we can achieve it, then that's the minimum.Now, what about the maximum? The maximum of ∫f² dx given that ∫f dx = 1 and f is concave. That seems more challenging. For the maximum, without any constraints, the integral can be made arbitrarily large by making f very tall and narrow, but since f has to be concave, maybe there's a limit to how peaked it can be.For example, in the case of concave functions on (0,1), the maximum value of f would be at the endpoints or somewhere in the middle? Wait, no. Concave functions on an interval attain their maximum at the endpoints if they are linear, but if they are strictly concave, maybe they have a maximum in the interior? Wait, actually, for concave functions, the maximum is attained on the boundary of the interval. Wait, is that correct?Wait, let's think. Take a simple concave function, like f(x) = -x² + 1. That's concave (since the second derivative is -2, which is negative). Its maximum is at x=0. But another concave function, say f(x) = √x, which is concave on (0,1). Its maximum is at x=1. Wait, but if we have a function that's increasing and concave, the maximum is at x=1. If it's decreasing and concave, the maximum is at x=0. If it's first increasing then decreasing, like a triangle function, but concave, then maybe the maximum is somewhere in the middle. Wait, but triangle functions are not differentiable, but they can be concave.Wait, actually, for a concave function on an interval, the maximum is achieved at the endpoints or in the interior? Let me recall: concave functions on a convex set are such that any local maximum is a global maximum. But on a closed interval, the maximum must be attained at the boundary or in the interior. If it's attained in the interior, then the function must be constant in some neighborhood? Wait, no. For example, f(x) = -x² + c, which is concave, has its maximum at x=0. So that's on the boundary. Another example: f(x) = -|x - 0.5| + 1. This is concave, and the maximum is at x=0.5, in the interior. So yes, concave functions can have their maximum in the interior. So depending on the function, the maximum can be either at the endpoints or inside.But how does that help with the integral of f squared?I need to find the maximum of ∫₀¹ [f(x)]² dx given that f is concave, positive, and ∫f(x) dx = 1.Since f is concave, it's continuous on (0,1). So maybe the maximum occurs when f is a tent function, i.e., linear increasing up to a point and then linear decreasing. But tent functions are concave only if the slopes are non-increasing? Wait, tent functions are concave if they are piecewise linear with first increasing then decreasing, but the slopes have to be decreasing. Wait, maybe not. Let me think. If a function is piecewise linear with two segments, first increasing with slope a, then decreasing with slope -b, where a and b are positive. For it to be concave, the slope must be non-increasing. So the first slope is a, then the next slope is -b. Since a > -b, the slope is decreasing. So tent functions are concave.Alternatively, affine functions. But tent functions can be concave. So perhaps the maximum is achieved by a tent function.Alternatively, maybe the function that is triangular, with a peak at some point, but adjusted so that the integral is 1.Alternatively, maybe the function that is linear, increasing from 0 to some point, then decreasing. Wait, but if it's defined on (0,1), but if we take f(0) and f(1) as limits. Wait, but the function is defined on (0,1), so the endpoints aren't included. However, since f is concave on (0,1) and integrable, it can be extended to [0,1] with upper semicontinuity. Hmm, maybe not necessary.Alternatively, consider that the maximum of ∫f² dx would be achieved when f is as "peaked" as possible, given the concavity constraint.In general, for non-negative functions with integral 1, the maximum of ∫f² is unbounded, but with concavity, maybe it's bounded. Let's see.Suppose we have a function that is triangular, symmetric, with peak at 0.5. Let's say it goes up linearly from (0,0) to (0.5, h), then down linearly from (0.5, h) to (1,0). But wait, this function is concave, right? The slope decreases from positive to negative. However, this function is not positive on (0,1), because at the endpoints, it's zero. But the problem states that f is positive on (0,1). So maybe approaching zero at the endpoints but remaining positive. Wait, but if the function is concave and positive on (0,1), can it approach zero at the endpoints?Wait, for example, take f(x) = 2 - 2x on (0,1). Then it's linear, decreasing, concave. The integral is ∫₀¹ (2 - 2x) dx = [2x - x²]₀¹ = (2 - 1) - 0 = 1. So this is a valid function. Its square integral is ∫₀¹ (2 - 2x)² dx = ∫₀¹ 4 - 8x + 4x² dx = [4x - 4x² + (4/3)x³]₀¹ = 4 - 4 + 4/3 = 4/3 ≈ 1.333.But another function: suppose we take a more peaked function. Let's consider a symmetric tent function that peaks at x=0.5. Let's say f(x) increases linearly from (0, a) to (0.5, b) and then decreases linearly from (0.5, b) to (1, a). Wait, but if it's concave, the slopes have to be non-increasing. So from 0 to 0.5, the slope is (b - a)/0.5 = 2(b - a). From 0.5 to 1, the slope is (a - b)/0.5 = 2(a - b). Since 2(b - a) > 2(a - b) because b > a, but for concavity, the slope must be non-increasing. So if the first slope is 2(b - a) and the second slope is 2(a - b), then we need 2(b - a) ≥ 2(a - b), which would require that b - a ≥ a - b ⇒ 2(b - a) ≥ 0 ⇒ b ≥ a. But since we are assuming a tent function peaking at 0.5, then b > a. So in that case, the slope is decreasing (from positive to negative), but the slope itself is non-increasing. So it's okay. So such a tent function is concave.But in this case, the endpoints are a. But the problem says f is positive on (0,1). If a > 0, then the function is positive everywhere. So let's set a > 0.So let's parametrize this tent function. Let me define f(x) as follows:f(x) = a + 2(b - a)x for x ∈ [0, 0.5]f(x) = a + 2(b - a)(1 - x) for x ∈ [0.5, 1]But wait, this might not integrate to 1. Let's compute the integral.Wait, actually, to make it symmetric and peaked at 0.5, let's define f(x) as:From 0 to 0.5: f(x) = m x + cFrom 0.5 to 1: f(x) = -m (x - 0.5) + c + 0.5mWait, maybe another approach. Let's say the tent function has maximum at 0.5, and it's linear on [0, 0.5] and [0.5, 1]. Let’s let the value at 0.5 be h, and the value at 0 and 1 be k. Since it's linear, the function from 0 to 0.5 is k + (h - k)/(0.5 - 0) x = k + 2(h - k)x. Similarly, from 0.5 to 1, it's h - 2(h - k)(x - 0.5). But since the function must be positive on (0,1), k must be positive. However, if k is positive, then the function is positive everywhere. Wait, but if k approaches zero, the function becomes a triangle with base 1 and height h. The integral of a triangle is (base * height)/2. So if we want the integral to be 1, then (1 * h)/2 = 1 ⇒ h = 2. So the maximum height is 2. Then, in that case, the function would be f(x) = 2 - 4|x - 0.5|. Wait, let's check:From 0 to 0.5: f(x) = 2 - 4(0.5 - x) = 2 - 2 + 4x = 4xFrom 0.5 to 1: f(x) = 2 - 4(x - 0.5) = 2 - 4x + 2 = 4 - 4xSo integrating from 0 to 0.5: ∫₀^0.5 4x dx = 2x² from 0 to 0.5 = 2*(0.25) = 0.5Similarly, from 0.5 to 1: ∫₀.5^1 (4 - 4x) dx = [4x - 2x²] from 0.5 to 1 = (4 - 2) - (2 - 0.5) = 2 - 1.5 = 0.5Total integral is 1, which is correct. So the triangular function with peak 2 at 0.5 and base 1 has integral 1. So this is a valid function. Now, the integral of f squared is:From 0 to 0.5: ∫₀^0.5 (4x)^2 dx = ∫₀^0.5 16x² dx = (16/3)x³ from 0 to 0.5 = (16/3)(1/8) = 2/3From 0.5 to 1: ∫₀.5^1 (4 - 4x)^2 dx. Let me compute this. Let u = x - 0.5, then when x = 0.5, u = 0; x = 1, u = 0.5.So ∫₀^0.5 (4 - 4(u + 0.5))² du = ∫₀^0.5 (4 - 4u - 2)^2 du = ∫₀^0.5 (2 - 4u)^2 du = ∫₀^0.5 4 - 16u + 16u² du = [4u - 8u² + (16/3)u³] from 0 to 0.5 = 4*(0.5) - 8*(0.25) + (16/3)*(0.125) = 2 - 2 + (2/3) = 2/3So total integral of f squared is 2/3 + 2/3 = 4/3 ≈ 1.333.But earlier, the linear function f(x) = 2 - 2x also had ∫f² dx = 4/3. Wait, that's interesting. So both the tent function peaking at 0.5 and the linear function decreasing from 2 to 0 have the same integral of f squared. Hmm.But wait, perhaps the maximum is 4/3? But let's check.Wait, maybe there's a function with a higher integral. Let me consider another concave function.Suppose we take a function that is constant 1 except near the endpoints, where it increases. Wait, but concave functions can't have upward jumps. If it's concave, increasing near the endpoints would require the function to bend upwards, which would make it convex there. So maybe not. Alternatively, suppose we take a function that is higher in the middle and lower at the ends, but more peaked than the tent function. But since the tent function is already the most peaked triangular function with integral 1, maybe that's the maximum.Alternatively, think about the family of concave functions parameterized by their maximum value. Let's say the function is symmetric around 0.5, increasing up to 0.5 and then decreasing. The more peaked it is, the higher the maximum value h, but the integral is fixed at 1. The integral is the area under the curve, so for a given h, the width of the peak would adjust to keep the area 1.But in the case of the tent function, the maximum h is 2, and as we saw, the integral of f squared is 4/3. If we make the function more peaked, but since it's concave, the slope can't be steeper than linear, right? Wait, if it's concave, then the function cannot curve upwards, so the steepest it can be is linear. So the tent function is the most peaked possible concave function with integral 1, because any more peaked function would have to curve upwards, making it convex.Therefore, the tent function with h=2 gives the maximum ∫f² dx = 4/3. Is that correct?Wait, let's check another function. Suppose we have a function that is constant 1 in the middle and slopes down to zero at the endpoints. For example, let's say it's 1 from a to 1 - a, and slopes linearly from 0 to 1 on [0, a] and from 1 to 0 on [1 - a, 1]. Then, the integral would be 1*(1 - 2a) + 2*(area of the triangles). The area of each triangle is (1/2)*a*1 = a/2. So total integral is (1 - 2a) + 2*(a/2) = 1 - 2a + a = 1 - a. To make the integral 1, we need 1 - a = 1 ⇒ a = 0. So that's just the constant function. So this approach doesn't help.Alternatively, if we make the flat part higher than 1, but then the slopes would have to adjust. Wait, but if the function is concave, the slopes have to be decreasing. Let's suppose that the flat part is h > 1, and the slopes go down to some k > 0 at the endpoints. Wait, but then the integral would be h*(1 - 2a) + 2*(area of the trapezoids or triangles). Wait, perhaps this is getting too complicated. Maybe the maximum is indeed 4/3, achieved by the tent function.Alternatively, let's use optimization. Let's parameterize the concave function. Since f is concave on (0,1), it can be represented as the minimum of a family of linear functions. Wait, but perhaps for symmetric reasons, the maximum of ∫f² dx occurs for a function that is symmetric around 0.5.Assume f is symmetric, so f(x) = f(1 - x). Then, the function is determined by its behavior on [0, 0.5]. Let's model f on [0, 0.5] as linear: f(x) = a + bx. Since it's concave, the slope b must be non-increasing. Wait, but on [0, 0.5], if it's linear, then the slope is constant. Then, on [0.5, 1], it would be a + b(1 - x). Wait, let me check symmetry. If f(x) is symmetric, then f(x) = a + b(0.5 - |x - 0.5|). Wait, that's similar to the tent function.But let's formalize this. Let’s define f(x) as follows:f(x) = m x + c for x ∈ [0, 0.5]f(x) = -m (x - 1) + c for x ∈ [0.5, 1]This ensures symmetry around x = 0.5. Wait, but integrating this:The integral from 0 to 0.5 is ∫₀^0.5 (m x + c) dx = [0.5 m x² + c x]₀^0.5 = 0.5 m (0.25) + c (0.5) = m/8 + c/2Similarly, the integral from 0.5 to 1 is the same due to symmetry, so total integral is 2*(m/8 + c/2) = m/4 + cWe need this integral to be 1: m/4 + c = 1Also, at x = 0.5, the two linear pieces meet, so m*(0.5) + c = -m*(0.5 - 1) + c ⇒ m*0.5 + c = m*0.5 + c, which is consistent.Now, to ensure positivity on (0,1). At x=0: f(0) = c. At x=1: f(1) = -m*(1 - 1) + c = c. So c must be positive. Also, at x=0.5, the function value is m*(0.5) + c. Since m and c are positive, this is positive as well.So we have two parameters, m and c, with the constraint m/4 + c = 1.We need to compute ∫₀¹ [f(x)]² dx. Due to symmetry, this is 2*∫₀^0.5 (m x + c)² dxCompute ∫₀^0.5 (m x + c)^2 dx = ∫₀^0.5 m² x² + 2 m c x + c² dx = [m² x³ / 3 + m c x² + c² x] from 0 to 0.5= m²*(0.125)/3 + m c*(0.25) + c²*(0.5)= m² / 24 + (m c)/4 + c² / 2Multiply by 2 to get the total integral:Total ∫f² dx = 2*(m² / 24 + (m c)/4 + c² / 2) = m² / 12 + (m c)/2 + c²Now, from the constraint m/4 + c = 1, we can express c = 1 - m/4. Substitute into the integral expression:Total ∫f² dx = m² / 12 + m*(1 - m/4)/2 + (1 - m/4)^2Let's compute each term:First term: m² / 12Second term: m*(1 - m/4)/2 = (m/2) - (m²)/8Third term: (1 - m/4)^2 = 1 - m/2 + m²/16Adding all together:m²/12 + (m/2 - m²/8) + (1 - m/2 + m²/16)Combine like terms:m² terms:m²/12 - m²/8 + m²/16 = (4m²/48 - 6m²/48 + 3m²/48) = (1m²)/48m terms:m/2 - m/2 = 0Constants:1So total ∫f² dx = 1 + m²/48Wait, that's interesting. So the integral of f squared is 1 + m²/48. To maximize this, we need to maximize m², which is equivalent to maximizing |m|. But since m is a slope, and we have constraints that c = 1 - m/4 must be positive. Since c > 0, then 1 - m/4 > 0 ⇒ m < 4. So m can be as large as approaching 4, but must be less than 4. Wait, but m is the slope from 0 to 0.5. If m approaches 4, then c approaches 1 - 4/4 = 0. So as m approaches 4, the function becomes steeper, approaching a tent function with slope 4 on [0, 0.5] and -4 on [0.5, 1], but with c approaching 0. Wait, but let's check the function when m approaches 4.If m approaches 4, then c approaches 0, so the function near x=0 is approaching 4x, and near x=1, approaching 4(1 - x). But at x=0.5, the value is m*0.5 + c = 2 + c. If m approaches 4, c approaches 0, so the peak at x=0.5 approaches 2. That matches the tent function we considered earlier, with peak 2. Then, the integral ∫f² dx approaches 1 + (4)² /48 = 1 + 16/48 = 1 + 1/3 = 4/3 ≈ 1.333. So as m approaches 4, ∫f² dx approaches 4/3. But m cannot exceed 4 because c must remain positive. Therefore, the maximum value of ∫f² dx is 4/3, achieved in the limit as m approaches 4 (i.e., the tent function with peak 2). However, when m=4, c=0, which would make f(0)=0 and f(1)=0, but the problem states that f must be positive on (0,1). Therefore, the function with m=4 is not allowed since f(0)=0 and f(1)=0, which are not in (0,1), but the limit as m approaches 4 from below would give functions that are positive on (0,1) (since c approaches 0 from above, so f(0)=c approaches 0, but remains positive). However, in the open interval (0,1), even if f(0) and f(1) approach 0, as long as they are positive in (0,1), it's okay. Wait, but actually, the function is defined on (0,1), so the endpoints aren't included. Therefore, even if c=0, the function at x=0 isn't considered. Wait, but if c=0, then at x approaching 0 from the right, f(x) approaches 0. But the problem says f must be positive on (0,1). So for all x in (0,1), f(x) > 0. If c=0, then near x=0, f(x) = 4x, which is positive for x > 0. Similarly near x=1, f(x) = 4(1 - x), which is positive for x < 1. Therefore, even with m=4 and c=0, the function f(x) is positive on (0,1). But wait, in that case, f(x) = 4x on (0, 0.5) and f(x) = 4(1 - x) on (0.5,1). This is a valid function, positive on (0,1), concave (as it's a tent function), integrable with integral 1, as we saw earlier. So why did we have the constraint m < 4 before? Because when we set c = 1 - m/4, if m=4, then c=0, which is allowed since f(x) is still positive on (0,1). Therefore, actually, m can be exactly 4, leading to c=0, which is acceptable. Therefore, the maximum ∫f² dx is exactly 4/3, achieved by the tent function with m=4, c=0, which is f(x) = 4x on [0, 0.5] and 4(1 - x) on [0.5,1].Therefore, the integral of f squared can take values from 1 to 4/3. Wait, but the Cauchy-Schwarz lower bound is 1, achieved by the constant function. The upper bound is 4/3, achieved by the tent function. So the possible values are [1, 4/3].But let me check if there are functions with higher ∫f² dx. Suppose we take a function that is more peaked, but still concave. Wait, but the tent function is linear, so any more peaked function would have to curve inward, which would require a convex region. Since the function must be concave everywhere, it can't have any convex parts. Therefore, the tent function is indeed the most peaked concave function possible with integral 1. Thus, the maximum is 4/3.Alternatively, suppose we consider asymmetric concave functions. For example, a function that increases steeply from 0 to some point and then decreases more gradually. But would that give a higher ∫f² dx?Let’s consider an asymmetric tent function. Suppose the peak is not at 0.5. Let's say the function increases from (0, a) to (c, h) and then decreases from (c, h) to (1, b). Since it's concave, the slope from 0 to c is (h - a)/c, and from c to 1 is (b - h)/(1 - c). For concavity, the slope must be non-increasing. So (h - a)/c ≥ (b - h)/(1 - c). Also, the integral must be 1.But this seems complex. Let's assume a = b = 0 for simplicity. Wait, but the function must be positive on (0,1). If a = 0, then near 0, the function is approaching 0, which is allowed as long as it's positive on (0,1). Let's try an asymmetric tent function with a peak at c ∈ (0,1). Let’s say f(x) increases from 0 to c with slope m, then decreases from c to 1 with slope -n. Then, the integral of f is the area of the triangle, which is (c * m * c)/2 + ((1 - c) * n * (1 - c))/2. Wait, but for a tent function, the slopes m and n must satisfy m*c = n*(1 - c), since the height h = m*c = n*(1 - c). So h = m*c = n*(1 - c). The integral is (h * c)/2 + (h * (1 - c))/2 = h/2*(c + 1 - c) = h/2. To have integral 1, h/2 = 1 ⇒ h = 2. Therefore, regardless of where the peak is, the height must be 2. Then, the integral of f squared would be:From 0 to c: ∫₀^c (m x)^2 dx = m² ∫₀^c x² dx = m² * (c³)/3From c to 1: ∫_c^1 (n (1 - x))² dx = n² ∫_c^1 (1 - x)^2 dx = n² ∫₀^{1 - c} u² du = n² * ( (1 - c)^3 ) /3Total ∫f² dx = [m² c³ + n² (1 - c)^3 ] / 3But since h = m c = 2, m = 2/c. Similarly, n = 2/(1 - c). Therefore:∫f² dx = [ ( (4/c²) * c³ ) + ( (4/(1 - c)^2 ) * (1 - c)^3 ) ] / 3 = [4c + 4(1 - c)] / 3 = [4c + 4 - 4c]/3 = 4/3.So regardless of c, the integral of f squared is 4/3. Therefore, any asymmetric tent function with height 2 (peaked anywhere in (0,1)) will have ∫f² dx = 4/3. Therefore, the maximum is indeed 4/3, regardless of where the peak is located.This suggests that the maximum value of ∫f² dx is 4/3, achieved by any tent function with height 2 and integral 1. The constant function gives the minimum value of 1.Therefore, the possible values of ∫₀¹ f²(x) dx for such functions f are between 1 and 4/3, inclusive.But to confirm there's no function with higher ∫f² dx, suppose there exists a concave function which is not piecewise linear but has higher ∫f² dx. Let's consider a concave function that is curved, like a parabola opening downward.Suppose f(x) = -a x² + b x + c, which is concave if a > 0. We need f(x) > 0 on (0,1), ∫₀¹ f(x) dx = 1.Let’s compute the integral:∫₀¹ (-a x² + b x + c) dx = -a/3 + b/2 + c = 1.Also, f(x) > 0 for all x ∈ (0,1). To maximize ∫f² dx, perhaps such a quadratic function can give a higher value than 4/3.Let’s attempt to construct such a function. Let’s set f(x) = -a x² + b x + c.First, since it's concave, a > 0.To ensure positivity on (0,1), the minimum of f(x) must be greater than 0. The minimum of a concave function is at the endpoints or where the derivative is zero.Wait, but f is concave (since -a x² + b x + c is concave if the second derivative is negative. The second derivative is -2a < 0, so yes, concave). The maximum is at the critical point, which is where the first derivative is zero: f’(x) = -2a x + b = 0 ⇒ x = b/(2a). Since x must be in (0,1), so 0 < b/(2a) < 1.The minimum of f(x) occurs at the endpoints x=0 and x=1 (since it's concave, the minimum is at the endpoints). Therefore, f(0) = c > 0 and f(1) = -a + b + c > 0.So we have constraints:1. c > 02. -a + b + c > 03. ∫₀¹ (-a x² + b x + c) dx = -a/3 + b/2 + c = 1We need to maximize ∫₀¹ (-a x² + b x + c)^2 dx.This integral would be complicated, but let's see if we can find parameters a, b, c that satisfy the constraints and give a higher value than 4/3.Alternatively, let's consider symmetric functions. Let’s assume the function is symmetric around x=0.5, so f(x) = f(1 - x). Then, the quadratic function would have its vertex at x=0.5. So f(x) = -a(x - 0.5)^2 + k, where k is the maximum at x=0.5. Since it's symmetric, the minimums are at x=0 and x=1, which must be positive.So f(0) = -a(0.25) + k > 0 ⇒ k > a/4Similarly, the integral is ∫₀¹ [-a(x - 0.5)^2 + k] dx = -a ∫₀¹ (x - 0.5)^2 dx + k ∫₀¹ dx = -a*(1/12) + k*1 = 1.Thus, -a/12 + k = 1 ⇒ k = 1 + a/12.From the positivity condition: k > a/4 ⇒ 1 + a/12 > a/4 ⇒ 1 > a/4 - a/12 ⇒ 1 > (3a - a)/12 ⇒ 1 > (2a)/12 ⇒ 1 > a/6 ⇒ a < 6.So a can vary from 0 to 6. Now, compute ∫₀¹ [ -a(x - 0.5)^2 + k ]² dx.Let’s expand the square:[ -a(x - 0.5)^2 + k ]² = a²(x - 0.5)^4 - 2 a k (x - 0.5)^2 + k²Integrate term by term:∫₀¹ a²(x - 0.5)^4 dx = a² ∫_{-0.5}^{0.5} t^4 dt = a² * [ t^5 /5 ]_{-0.5}^{0.5} = a² * (2*(0.5)^5 /5 ) = a² * (2*(1/32)/5) )= a²*(1/80)∫₀¹ -2 a k (x - 0.5)^2 dx = -2 a k ∫_{-0.5}^{0.5} t² dt = -2 a k * [ t^3 /3 ]_{-0.5}^{0.5} = -2 a k * (2*(0.5)^3 /3 ) = -2 a k * (2*(1/8)/3) = -2 a k * (1/12) = - a k /6∫₀¹ k² dx = k²Therefore, total integral:a² / 80 - a k /6 + k²But k = 1 + a/12. Substitute:= a² / 80 - a*(1 + a/12)/6 + (1 + a/12)^2Expand each term:First term: a² / 80Second term: -a/6 - a² /72Third term: 1 + 2a/12 + a² /144 = 1 + a/6 + a² /144Combine all terms:a² / 80 - a/6 - a² /72 + 1 + a/6 + a² /144Simplify:The -a/6 and +a/6 cancel.For a² terms:a² /80 - a² /72 + a² /144Convert to common denominator, which is 720:= (9a²)/720 - (10a²)/720 + (5a²)/720 = (9 -10 +5)a² /720 = 4a² /720 = a² /180Therefore, total integral is 1 + a² /180To maximize this, we need to maximize a². Since a <6, a can approach 6. So as a approaches 6, k approaches 1 +6/12 = 1.5. Then, the integral approaches 1 + (36)/180 = 1 + 0.2 = 1.2, which is 6/5 = 1.2. But 6/5 is 1.2, which is less than 4/3 ≈1.333. Therefore, this quadratic function gives a lower ∫f² dx than the tent function.Hence, the quadratic concave function cannot exceed 1.2, which is still less than 4/3. Therefore, the tent function remains the maximum.Another example: suppose we take a higher-degree polynomial. For example, f(x) = -a(x - 0.5)^4 + k, but this would be convex near the peak if a >0, which is not allowed. Since we need the function to be concave everywhere, higher-degree polynomials may not help.Therefore, it seems that the maximum value is indeed 4/3, achieved by the tent function, and the minimum is 1, achieved by the constant function.Hence, the possible values of the integral ∫₀¹ f²(x) dx are in the interval [1, 4/3].Final AnswerThe integral can take values between boxed{1} and boxed{dfrac{4}{3}}.</think>To determine the possible values of the integral (int_{0}^{1} f^{2}(x) , dx) for functions (f(x)) that are positive, integrable, concave, and have an integral of 1 over the interval ((0,1)), we need to consider both the minimum and maximum values this integral can take.1. Minimum Value: - By the Cauchy-Schwarz inequality, we know that (int_{0}^{1} f^{2}(x) , dx geq left( int_{0}^{1} f(x) , dx right)^{2} = 1). - Equality holds when (f(x)) is a constant function. The constant function (f(x) = 1) is indeed concave (and convex) and satisfies all the given conditions. Therefore, the minimum value of the integral is (boxed{1}).2. Maximum Value: - To find the maximum value, we consider the most peaked concave function possible. A tent function, which is linear on ([0, 0.5]) and linear on ([0.5, 1]) with a peak at (x = 0.5), is a candidate. - This tent function can be defined as (f(x) = 4x) for (x in [0, 0.5]) and (f(x) = 4(1 - x)) for (x in [0.5, 1]). The integral of this function over ([0, 1]) is 1, and it is positive and concave. - Calculating the integral of (f^{2}(x)) for this tent function: [ int_{0}^{0.5} (4x)^2 , dx + int_{0.5}^{1} (4(1 - x))^2 , dx = 2 int_{0}^{0.5} 16x^2 , dx = 2 left[ frac{16x^3}{3} right]_{0}^{0.5} = 2 left( frac{16 cdot (0.125)}{3} right) = frac{4}{3} ] - This shows that the maximum value of the integral is (boxed{dfrac{4}{3}}).Thus, the integral (int_{0}^{1} f^{2}(x) , dx) can take values between (boxed{1}) and (boxed{dfrac{4}{3}}).,<think>Okay, so I need to figure out the possible values that the integral of f squared can take, given that f is a positive, integrable function on (0,1) with integral 1, and it's concave from below. Wait, concave from below... I think that's the same as concave down, right? So the function lies above any chord connecting two points on its graph. The integral in question is ∫₀¹ [f(x)]² dx. The problem is asking what values can this integral take? So, possible minimum and maximum values?First, let me recall some concepts. For functions that are concave (down), there are certain properties. For example, a concave function on an interval is continuous on the interior of the interval, so f is continuous on (0,1). Also, concave functions have second derivatives ≤ 0 if they are twice differentiable, but we might not need differentiability here.Since the function is concave, positive, and integrates to 1 over (0,1), we need to consider what kind of concave functions satisfy these conditions and how the integral of their squares behaves.I remember that for functions with a given integral, the integral of their square is related to their variance. If we think of f(x) as a probability density function (since it's positive and integrates to 1), then ∫₀¹ [f(x)]² dx is related to the "Fisher information" or maybe it's just the expectation of f(x) with respect to itself? Hmm, not sure. Alternatively, using the Cauchy-Schwarz inequality, maybe?Wait, but we have constraints here: concave, positive, integral 1. So maybe I need to use Jensen's inequality because concave functions have certain properties with respect to expectations.Jensen's inequality states that for a concave function φ, φ(E[X]) ≥ E[φ(X)]. But here, we're dealing with the integral of f squared. How does that relate?Alternatively, maybe I can use the fact that for concave functions, there's a maximum or minimum value for the integral of their square given the integral constraint. Let's think about extremal cases.First, what's the minimum possible value of ∫ f²? For any function with ∫ f = 1, the minimum of ∫ f² is achieved when f is constant, by Cauchy-Schwarz. Because the L² norm is minimized when the function is as "spread out" as possible. Since we have a fixed L¹ norm, the minimum L² norm is (∫ |f| )² / (1 - 0) = 1² /1 =1. So the minimum is 1, achieved when f is constant. Wait, but here f is concave. If the minimum is achieved by the constant function, which is concave (since a constant function is both concave and convex), then the minimum value is 1. But wait, can we get lower?Wait, no. For functions in L¹, the minimum L² norm under the constraint ∫ f =1 is indeed 1, achieved by the constant function f(x)=1. But is this function allowed here? Yes, because f(x)=1 is concave (and convex). So the minimum is 1.But then the problem is, what is the maximum possible value of ∫ f²? Because concave functions can be "peaked", but how peaked can they be while still being concave and integrating to 1 over (0,1)?Wait, if a function is concave on (0,1), it must be continuous on (0,1), and its maximum is achieved somewhere. But being concave, it can't have a sharp peak like a Dirac delta function, which is not even a function. So the maximum value of f(x) is constrained by the concavity.So to maximize ∫ f², we want the function to be as "peaked" as possible. For concave functions, the most peaked function would be a linear function decreasing from some point, but wait, if it's concave, it's concave down, so it's like a ∩ shape. Wait, no. Wait, concave down is the same as concave, which is the cap shape. So concave down functions have the property that the line segment between any two points is below the function. So if the function is concave, then it can have a maximum at some point, and it slopes downward on either side. But since the interval is (0,1), the maximum could be at the endpoints? Wait, no. Wait, for a concave function on an interval, the maximum is attained at the boundary if the function is non-decreasing or non-increasing. Wait, actually, no. If a concave function is defined on (0,1), and it's concave, then if it has a maximum, it can be anywhere. But if the function is concave and defined on an open interval, it might not attain its maximum, but since the function is integrable and positive, maybe it can't go to infinity.Wait, let me think. For example, take the tent function: f(x) increasing on (0, a) and decreasing on (a,1). But that's convex. Wait, concave functions have to be above their chords. So the tent function is convex. A concave function would be like a hill that is rounded on top. Wait, maybe a parabola opening downward? But restricted to (0,1). If we take the function f(x) = -x² + bx + c, adjusted so that it's positive on (0,1) and integrates to 1.Alternatively, since the function is concave, the most "peaked" function would be a triangle function? Wait, but the triangle function is convex, not concave. Wait, maybe the concave function can be affine linear near the peak.Wait, let's formalize this. Suppose f is concave on (0,1). Then for any x, y in (0,1) and λ ∈ [0,1], f(λx + (1-λ)y) ≥ λf(x) + (1-λ)f(y). So, the function lies above the chord between any two points. So, the graph of the function is a concave set, meaning it's shaped like a cap. So the maximum is achieved at the midpoint? Not necessarily.Wait, actually, for a concave function on an interval, the maximum is achieved at the endpoints or at a point where the derivative changes from positive to negative. But since it's concave, the derivative is non-increasing. So if the function is differentiable, the derivative starts high at 0 and decreases. Wait, but the interval is open, so endpoints are not included. Hmm.Alternatively, think of a function that is linear on (0,1), but that's both concave and convex. But we need to maximize ∫ f². So maybe to maximize ∫ f², we need the function to be as concentrated as possible. Since the function is concave, it can't have a sharp peak. So the most concentrated concave function would be a triangle function with the peak at the center? Wait, but a triangle function is convex. Wait, concave function must be above the chords. If I imagine a function that is highest at some point c in (0,1) and then decreases linearly to zero at the endpoints. Wait, let's check concavity.Suppose f(x) is linear from (0,0) to (c, h) and then linear from (c, h) to (1,0). Wait, but this would make a triangle shape. However, this function is convex, not concave. Because the line between any two points on the graph would be above the function. Wait, no. Let's check.Take two points on the left side of the peak, say x1 and x2 both less than c. The function is linear there, so the chord between x1 and x2 is the same as the function, so it's both concave and convex on that interval. Similarly on the right side. But if we take a point on the left and a point on the right of c, then the chord between them would pass below the function's peak. Wait, let's parametrize.Suppose c is the midpoint, 0.5. Let’s define f(0) = 0, f(0.5) = h, f(1) = 0, and linear in between. Then, the integral of f over (0,1) would be the area of the triangle, which is (1 * h)/2 = 1. So h = 2. Then, f(x) is 4x for x in (0, 0.5) and 4(1 - x) for x in (0.5, 1). Wait, but that's a triangle function with height 2, integrating to 1. But is this function concave?Wait, take two points on either side of 0.5. For example, x1 = 0.4 and x2 = 0.6. The function at x1 is 4*0.4 = 1.6, at x2 is 4*(1 - 0.6) = 1.6. The midpoint is 0.5, and f(0.5) = 2. The chord between x1 and x2 is the line connecting (0.4, 1.6) and (0.6, 1.6), which is a horizontal line at 1.6. But f(0.5) = 2, which is above the chord. Therefore, the function is concave at that midpoint. Similarly, if we take any two points, the function lies above the chord. Therefore, the triangle function is concave. Wait, but the triangle function is usually considered convex. Hmm, maybe the terminology is different here. Wait, maybe I'm confused.Wait, actually, a function that is linear on both sides of a peak and forms a tent shape is convex. But in our case, the function is above the chords, which would make it concave. Wait, this seems conflicting. Let me think again.Take two points on the function, say (0,0) and (0.5, h). The function is linear between them. So, the chord between these two points is the same as the function. Then, take another two points, say (0.5, h) and (1,0). Again, linear. But if I take a point on the left side and a point on the right side, say (0.25, 0.5h) and (0.75, 0.5h). The chord between them would be a horizontal line at 0.5h. The function at the midpoint between them, 0.5, is h, which is above the chord. So the function is concave. Therefore, the triangle function as defined (with peak at the center) is indeed concave. So that seems to be the case.Therefore, this triangle function with integral 1 is a concave function. Then, its square integral would be ∫₀¹ [f(x)]² dx. Let's compute that. For the triangle function centered at 0.5, with f(x) = 4x for x in [0, 0.5] and 4(1 - x) for x in [0.5, 1]. Then, [f(x)]² is 16x² on [0, 0.5] and 16(1 - x)² on [0.5, 1]. Integrating:∫₀⁰˙⁵ 16x² dx + ∫⁰˙⁵¹ 16(1 - x)² dxCompute first integral: 16 * [x³/3] from 0 to 0.5 = 16*(0.125/3) = 16*(1/24) = 2/3Second integral: same as first, due to symmetry. So total integral is 2*(2/3) = 4/3 ≈ 1.333.So the integral of the square for this triangle function is 4/3. Is this the maximum?Wait, maybe we can get higher by making the function more peaked. But since the function is concave, how peaked can it be?Suppose instead of a triangle function with peak at 0.5, we have a function that is constant 1 somewhere and then drops linearly to 0 at the endpoints. Wait, but a constant function is concave, but if you make it constant over an interval and then linear, does that preserve concavity?Suppose f(x) is equal to h on [a, b], and then linearly decreases to 0 at 0 and 1. Wait, let's check concavity. For example, take two points in the constant region and a point outside. The function must lie above the chord. If we take a point in the constant region and a point in the linear region, the chord between them should lie below the function. But if the function is constant h on [a,b] and then linearly decreasing to 0 at 0 and 1, then between a point in [a,b] and a point in [0,a], the function is linear. Wait, but the chord between (a, h) and (0,0) is the same as the linear part. So the function is linear there, hence concave. Similarly, between (a, h) and (b, h), the function is constant, so the chord is the same. Between (b, h) and (1,0), linear again. So this function is concave.This is called a "trapezoidal" function, but flat on top. Let's compute its integral and the integral of its square.Suppose the flat region is from a to b, with height h. The integral would be h*(b - a) + 0.5*h*a + 0.5*h*(1 - b) = h*(b - a) + 0.5*h*(a + 1 - b) = h*(b - a + 0.5a + 0.5 - 0.5b) = h*(0.5b - 0.5a + 0.5) = 0.5*h*(b - a +1). Set this equal to 1. Then, 0.5*h*(b - a +1) =1, so h = 2/(b -a +1).To maximize ∫ f² dx, we need to maximize the integral of h² over [a,b] plus the integrals over the linear parts.Compute ∫₀^a ( (h/a)x )² dx + ∫_a^b h² dx + ∫_b^1 ( (h/(1 - b))(1 - x) )² dxFirst integral: (h²/a²) ∫₀^a x² dx = (h²/a²)(a³/3) = h² a /3Second integral: h² (b - a)Third integral: (h²/(1 - b)^2) ∫_b^1 (1 - x)^2 dx = (h²/(1 - b)^2) ∫₀^{1 - b} t² dt = (h²/(1 - b)^2)( (1 - b)^3 /3 ) = h² (1 - b)/3So total integral is h² [ a/3 + (b - a) + (1 - b)/3 ] = h² [ (a + 1 - b)/3 + (b - a) ] = h² [ (1 + a - b)/3 + b - a ] = h² [ 1/3 + (a - b)/3 + b - a ] = h² [1/3 + ( -2a + 2b ) /3 ] = h² [1 + (-2a + 2b)/3 ] = h² [1 + (2(b - a))/3 ]But h = 2/(b - a +1). Therefore, substituting:Integral = [4/(b -a +1)^2] * [1 + (2(b - a))/3 ] = 4/(b -a +1)^2 * [ (3 + 2(b - a))/3 ]Let me set t = b - a. Since 0 ≤ a < b ≤1, t ∈ (0,1). Then, the integral becomes:4/(t +1)^2 * (3 + 2t)/3 = (4/3)(3 + 2t)/(t +1)^2We need to maximize this expression with respect to t ∈ (0,1). Let's denote f(t) = (3 + 2t)/(t +1)^2. Find the maximum of f(t) over t ∈ (0,1).Compute derivative f’(t):f’(t) = [2(t +1)^2 - (3 + 2t)(2)(t +1)] / (t +1)^4Wait, more carefully:Numerator: d/dt [3 + 2t] * (t +1)^2 - (3 + 2t) * d/dt [ (t +1)^2 ] = 2*(t +1)^2 - (3 + 2t)*2*(t +1)Factor out 2*(t +1):2*(t +1)[ (t +1) - (3 + 2t) ] = 2*(t +1)( t +1 -3 -2t ) = 2*(t +1)( -t -2 ) = -2*(t +1)(t +2 )Therefore, f’(t) = -2*(t +1)(t +2 ) / (t +1)^4 = -2(t +2)/(t +1)^3Since t ∈ (0,1), the denominator is positive, and the numerator is -2(t +2), which is negative. Therefore, f’(t) is negative for all t ∈ (0,1). Therefore, f(t) is decreasing on (0,1). Therefore, maximum occurs at t=0.But t =0 corresponds to b =a, which would collapse the flat region to a point. However, t ∈ (0,1), so approaching t=0, the maximum value of f(t) approaches (3 +0)/(0 +1)^2 =3. Then, the integral becomes (4/3)*3 =4. Wait, that can't be right. Wait, as t approaches 0, the height h = 2/(t +1) approaches 2. The function becomes a triangle function with height 2 at the center, which is what we computed before, and the integral of the square is 4/3. But according to this, when t approaches 0, the expression (4/3)*(3 +2t)/(1 + t)^2 approaches (4/3)*3/1 =4, which contradicts the previous result.Wait, there must be a mistake here. Let me check my substitution.Original integral after substitution with t = b -a:Integral = 4/(t +1)^2 * (3 + 2t)/3 = (4/3)*(3 + 2t)/(t +1)^2.But when t approaches 0, this becomes (4/3)*(3)/(1)^2 =4, which is higher than the triangle function's 4/3. But that's impossible because when t approaches 0, the function becomes a very narrow flat region near a point, but still, the integral of the square can't be 4. Wait, maybe there's a mistake in setting up the integral.Wait, let me check the calculations again.Original function: flat from a to b with height h, linear from 0 to a and from b to1.Integral of f(x) over [0,1]:Area of the trapezoid: the two triangles and the rectangle.Left triangle: base a, height h. Area = (a * h)/2.Rectangle: width (b - a), height h. Area = h*(b - a).Right triangle: base (1 - b), height h. Area = ((1 - b)*h)/2.Total integral: h*(a/2 + b - a + (1 - b)/2) = h*( (a/2 - a) + (b - a) + (1 - b)/2 ) = h*( (-a/2) + b - a + (1 - b)/2 ) = h*( - (3a)/2 + b + (1 - b)/2 ) = h*( -3a/2 + b + 1/2 - b/2 ) = h*( -3a/2 + (b/2) + 1/2 ). Wait, this is getting complicated. Maybe my initial expression was wrong.Wait, perhaps I made a mistake in the integral calculation. Let me recompute.The integral of f(x) over [0,1] should be:Left linear part: integral from 0 to a of (h/a)x dx = (h/a)*(a²/2) = h*a/2.Flat part: integral from a to b of h dx = h*(b - a).Right linear part: integral from b to1 of (h/(1 - b))(1 - x) dx = (h/(1 - b)) * integral from b to1 of (1 -x) dx = (h/(1 - b)) * [ (1 - b)^2 / 2 ] = h*(1 - b)/2.Total integral: h*(a/2 + b - a + (1 - b)/2 ) = h*( (a/2 - a) + b + ( (1 - b)/2 - b/2 ) ) = h*( (-a/2) + b + (1 - 2b)/2 ) = h*( (-a/2) + b + 1/2 - b ) = h*( (-a/2) + 1/2 ). Wait, that's different. So according to this, the total integral is h*(1/2 - a/2). But we set this equal to 1. So h*( (1 - a)/2 ) =1, so h = 2/(1 -a). Wait, but we also have the right linear part. Wait, this seems inconsistent. Maybe my parametrization is wrong.Alternatively, maybe there's a better way. Let's suppose that the function is linear from 0 to c with slope m, reaching a maximum at c, then linear with slope -m from c to1. Wait, but since the function is concave, the slopes must be non-increasing. So, if we have a function that increases with slope m up to c, then decreases with slope -n after c. For concavity, we need the slope after c to be less than the slope before c. So, if m >0 and -n ≤ m, but since it's concave, the slope must be non-increasing. So after c, the slope is -n, which must be ≤ m. So to make it as peaked as possible, we might set n as large as possible. Wait, but n can't be arbitrary because the function must stay positive.Wait, let's try this approach. Let the function increase linearly from 0 to c with slope m, reaching a maximum value of m*c at c, then decrease linearly to 0 at 1 with slope -m*(c/(1 - c)) to ensure that the function is continuous. Wait, maybe. Let's see.Wait, if we want the function to be continuous at c, then the slope after c must satisfy: the value at 1 is 0, so from c to1, the function decreases from m*c to 0 over an interval of length (1 - c), so the slope is -m*c/(1 - c). For concavity, the slope after c must be less than or equal to the slope before c. Wait, the slope before c is m, and after c is -m*c/(1 - c). For concavity, the slope must be non-increasing, so we need -m*c/(1 - c) ≤ m. That is, -c/(1 - c) ≤1. Multiplying both sides by (1 - c) (which is positive since c <1):-c ≤1 -c → -c +c ≤1 → 0 ≤1, which is true. So that's okay. So the slope after c is more negative than -m, but since we have to stay positive, the slope after c is -m*c/(1 -c).Wait, but the function must stay positive on (0,1). So after c, the slope is -m*c/(1 -c). So the function after c is m*c - m*c/(1 -c)*(x - c). Wait, at x=1, this becomes m*c - m*c/(1 -c)*(1 -c) = m*c - m*c=0, which is correct. So the function is positive on [0,1].Now, compute the integral of f(x) over [0,1]:Integral from 0 to c: area of triangle with base c and height m*c: (c * m*c)/2 = m c² /2.Integral from c to1: area of triangle with base (1 -c) and height m*c: ((1 -c)*m*c)/2.Total integral: m c² /2 + m c (1 - c)/2 = m c /2 [c + (1 - c)] = m c /2.Set this equal to 1: m c /2 =1 → m = 2/(c).So the function is:From 0 to c: f(x) = (2/c) x.From c to1: f(x) = 2/(c) *c - (2/(c))*(c/(1 -c))(x -c) = 2 - (2c/(1 -c))(x -c).Simplify: 2 - (2c/(1 -c))(x -c) = 2 - (2c/(1 -c))x + (2c²)/(1 -c) = [2 + (2c²)/(1 -c)] - (2c/(1 -c))x.But maybe not necessary to simplify further.Now, compute ∫₀¹ [f(x)]² dx.First part: ∫₀^c [(2/c)x]^2 dx = ∫₀^c (4/c²)x² dx = (4/c²)*(c³/3) = (4/3)c.Second part: ∫_c^1 [2 - (2c/(1 -c))(x -c)]² dx.Let’s make substitution t = x -c, then t goes from 0 to1 -c.Integral becomes ∫₀^{1 -c} [2 - (2c/(1 -c)) t]^2 dt.Let’s expand the square:[2 - (2c/(1 -c)) t]^2 = 4 - 8c/(1 -c) t + 4c²/(1 -c)^2 t².Integrate term by term:∫₀^{1 -c} 4 dt =4(1 -c).∫₀^{1 -c} -8c/(1 -c) t dt = -8c/(1 -c) * [ t²/2 ]₀^{1 -c} = -8c/(1 -c) * ( (1 -c)^2 /2 ) = -8c*(1 -c)/2 = -4c(1 -c).∫₀^{1 -c} 4c²/(1 -c)^2 t² dt =4c²/(1 -c)^2 * [ t³/3 ]₀^{1 -c} =4c²/(1 -c)^2 * ( (1 -c)^3 /3 ) =4c²(1 -c)/3.Adding all three terms:4(1 -c) -4c(1 -c) +4c²(1 -c)/3 = [4(1 -c) -4c(1 -c)] +4c²(1 -c)/3.Factor 4(1 -c):4(1 -c)[1 -c] +4c²(1 -c)/3. Wait, no:Wait, first part: 4(1 -c) -4c(1 -c) =4(1 -c)(1 -c) =4(1 -c)^2. Then adding the third term: 4(1 -c)^2 +4c²(1 -c)/3.Factor out 4(1 -c)/3:4(1 -c)/3 [3(1 -c) +c² ].Compute inside the brackets: 3(1 -c) +c² =3 -3c +c².So total integral of the second part:4(1 -c)/3 (3 -3c +c²).Therefore, the total integral ∫₀¹ [f(x)]² dx is:First part: (4/3)c.Second part:4(1 -c)/3 (3 -3c +c²).Thus total integral:(4/3)c +4(1 -c)/3 (3 -3c +c²) = (4/3)[c + (1 -c)(3 -3c +c²)].Let’s compute the expression inside the brackets:c + (1 -c)(3 -3c +c²).Expand (1 -c)(3 -3c +c²):=1*(3 -3c +c²) -c*(3 -3c +c²)=3 -3c +c² -3c +3c² -c³=3 -6c +4c² -c³.Therefore, total expression inside the brackets: c +3 -6c +4c² -c³ =3 -5c +4c² -c³.Therefore, total integral is (4/3)(3 -5c +4c² -c³).So, ∫₀¹ [f(x)]² dx = (4/3)( -c³ +4c² -5c +3 ). We need to maximize this expression over c ∈(0,1).Let’s denote F(c) = -c³ +4c² -5c +3. Then, the integral is (4/3) F(c). To find maximum, find derivative F’(c):F’(c) = -3c² +8c -5.Set F’(c)=0:-3c² +8c -5=0 →3c² -8c +5=0.Solve quadratic equation:c = [8 ±√(64 -60)] /6 = [8 ±√4]/6 = [8 ±2]/6.Thus, c=(8 +2)/6=10/6=5/3≈1.666… which is outside (0,1). Discard.c=(8 -2)/6=6/6=1. So c=1. But c must be in (0,1), so the critical point is at c=1, which is endpoint.But wait, F’(c) is -3c² +8c -5. Let's check the sign of F’(c) over (0,1). For c=0, F’(0)= -5 <0. For c=1, F’(1)= -3 +8 -5=0. Therefore, F’(c) increases from -5 at c=0 to 0 at c=1. So F(c) is decreasing on (0,1). Therefore, maximum of F(c) occurs at c=0. But c=0 is not allowed. Wait, but c ∈ (0,1). Then, as c approaches 0, F(c) approaches -0 +0 -0 +3=3. At c=1, F(1)= -1 +4 -5 +3=1. Therefore, F(c) decreases from 3 to1 as c goes from0 to1. Therefore, the maximum of F(c) is 3, achieved as c approaches0. Therefore, the maximum integral is (4/3)*3=4. But this contradicts the previous triangle function result.Wait, but if c approaches0, the function becomes a very narrow triangle near 0. Wait, let's see. If c approaches0, then the function from0 to c is (2/c)x, which becomes a steep rise to height 2 at x=c, and then from c to1, it decreases to0 with slope -2c/(1 -c). As c approaches0, the slope after c becomes very shallow. Therefore, the function becomes a spike near0, but due to concavity, it must stay above the chord between any two points. But if we have a spike at0, how is that concave?Wait, no, because concavity requires that the function lies above the chord between any two points. If the function spikes up at0, but then is almost zero elsewhere, then the chord between a point near0 and a point near1 would lie below the function? Wait, let's take two points: x1=ε (very small) and x2=1. The function at x1 is 2ε/c ≈2 (if ε is small but c is approaching0, but ε must be less than c). Wait, if c approaches0, then ε must be less than c. So as c approaches0, x1 approaches0 as well. Then f(x1)=2x1/c. If x1 is proportional to c, say x1=kc for some k<1, then f(x1)=2k. So as c approaches0, for fixed k, x1 approaches0, but f(x1)=2k.Wait, but this is confusing. Maybe we need to consider the function as c approaches0. The function is linear from0 toc with slope 2/c, so it goes from0 to2 in that interval. Then, from c to1, it's a very shallow decrease. But for concavity, the function must lie above the chord between any two points. Let's take x1=0 and x2=1. The chord between (0,0) and (1,0) is the x-axis. The function is above this chord except at the endpoints. But between x=0 andx=c, the function is rising to2, then decreasing to0. Wait, but concavity would require that for any two points, the function lies above the chord. If we take two points on either side of the spike, say x1=0.1 and x2=0.2, with c=0.05, then the function at x1=0.1 is in the decreasing part. Wait, but this is getting too complicated.Wait, perhaps there's a mistake in assuming that such a function is concave. If c approaches0, the function has a very steep increasing part near0 and a shallow decreasing part. But for concavity, the slope must be non-increasing. In this case, the slope starts at 2/c (very large) and then drops to -2c/(1 -c) (very small negative). Since 2/c is positive and then becomes negative, but the slope is decreasing from a large positive to a negative, which is a decrease. Therefore, the derivative is non-increasing, so the function is concave. Therefore, even with c approaching0, the function is concave.But then, as c approaches0, the integral of f² approaches4. However, this contradicts our previous understanding that the triangle function (c=0.5) gives 4/3. How can the integral approach4?Wait, let's compute the integral when c approaches0. If c is very small, then:First part: ∫₀^c (2/c x)^2 dx = (4/c²)∫₀^c x² dx = (4/c²)(c³/3)=4c/3≈0.Second part: ∫_c^1 [2 - (2c/(1 -c))(x -c)]² dx. When c approaches0, this becomes ∫₀^1 [2 -0]^2 dx= ∫₀^14 dx=4. So as c approaches0, the integral approaches4. But how can this be? The function is spiking at0, but still integrating to1. However, the square of a function that spikes would have a higher integral. But in reality, the function near0 is 2/c x, which for x in (0,c). The square is (4/c²)x². Integrating this from0 to c gives (4/c²)(c³/3)=4c/3, which approaches0 as c→0. Then, the remaining part fromc to1 is approximately2 - (2c/(1 -c))(x -c)≈2, since c is very small. So the integral fromc to1 of2² dx≈4*(1 -c)≈4. Therefore, the total integral approaches4 as c→0.But how is this possible? If the function is mostly2 over (c,1), which is almost the entire interval (0,1) when c is tiny. But the integral of2 over (c,1) is2*(1 -c). To make the total integral1, we need to have the spike near0 contribute the remaining integral. Specifically, the integral of the spike is4c/3≈0, and the integral of the flat part is2*(1 -c). Wait, but in our previous calculation, when c approaches0, the total integral is supposed to be1. Wait, this is a contradiction.Wait, no. Wait, in this parametrization, the total integral was set to1 by setting m*c/2=1, so m=2/c. Then, the integral of the spike part is m c² /2=2/c *c² /2= c. The integral of the decreasing part is m c (1 -c)/2=2/c *c*(1 -c)/2= (1 -c). Therefore, total integral isc + (1 -c)=1. Therefore, the spike near0 has integralc, which approaches0 as c→0, and the decreasing part has integral(1 -c), which approaches1. But the decreasing part is approximately2 over (c,1). Wait, the height of the decreasing part atx=c is m c=2, and it decreases to0 atx=1 with slope-2c/(1 -c). Therefore, the function fromc to1 is2 - (2c/(1 -c))(x -c). If c approaches0, then this function is approximately2 -0=2 over (0,1). Wait, but then the integral would be2*(1 -c)≈2, which is more than1. But no, the integral is set to1. Therefore, there's a mistake in the previous reasoning.Wait, no. The integral fromc to1 is2*(1 -c) - correction term. Wait, no, let me recast.Wait, the function fromc to1 is linear starting at2 and ending at0 over the interval (c,1). Therefore, the average height is(2 +0)/2=1, so the integral is1*(1 -c). Therefore, as c approaches0, the integral approaches1, which matches. But the function is2 atx=c and decreases linearly to0 atx=1. Therefore, ifc is approaching0, the function is2 atx=0+, then decreases to0 over(0,1). So it's a line from(0,2) to(1,0). Wait, but that's exactly the triangle function we considered earlier with c=0.5? No, wait, ifc approaches0, then the peak is atx=0, but our original definition had peak atx=c. Wait, no, whenc approaches0, the peak is atx=c, which approaches0. So the function is a very short spike near0, rising to2 atx=c≈0, then decreasing linearly to0 atx=1. But this is not a triangle function symmetric about0.5.But in this case, as c approaches0, the function becomes a right triangle with base1 and height2, which would have area1. But the integral of the square would be ∫₀¹ (2(1 -x))² dx =4∫₀¹ (1 -x)^2 dx=4*(1/3)=4/3. Wait, this contradicts our previous conclusion that as c approaches0, the integral approaches4.Wait, there's clearly a mistake here. Let me check the function when c approaches0.If c approaches0, then the function from0 to c is (2/c)x, which goes from0 to2 asx goes from0 toc. Then, fromc to1, the function is2 - (2c/(1 -c))(x -c). As c approaches0, this becomes2 -0=2 forx in (0,1). But then the function would be0 atx=0, jumps to2 atx=0+, and stays2 untilx=1, which is not possible because it's supposed to be continuous. Wait, no, because as c approaches0, the interval from0 toc becomes negligible, and the function fromc to1 is2 - (2c/(1 -c))(x -c)≈2 forx in (0,1). But the integral of2 over (0,1) is2, but our total integral must be1. Therefore, there's a mistake in the previous parametrization.Wait, no. Wait, when we set m=2/c, then the integral of the function is m*c/2=1. Therefore, the integral is fixed at1. However, the function fromc to1 is linear from2 to0, but whenc approaches0, this slope becomes-2c/(1 -c)≈-0, so the function decreases very slowly from2 to0 over the interval (0,1). Therefore, the function is roughly2 -ε*x, whereε is very small. Then, the integral of this approximate function over (0,1) would be2*1 -ε*(1/2)=2 -ε/2. But we need this to be1. So 2 -ε/2=1 → ε/2=1 → ε=2. But ε=2c/(1 -c)≈2c. So2c≈2 →c≈1. But we are consideringc approaching0. This is a contradiction.This suggests that there's a mistake in the earlier reasoning. It appears that whenc approaches0, the function can't maintain the integral1 unless the height is adjusted. But according to the parametrization, the height is2, and the integral is fixed at1. But in reality, the integral fromc to1 of the decreasing function is:∫_c^1 [2 - (2c/(1 -c))(x -c)] dx.Let’s compute this integral:Let’s set t =x -c, so t from0 to1 -c.Integral becomes∫₀^{1 -c} [2 - (2c/(1 -c)) t] dt=2(1 -c) - (2c/(1 -c))*( (1 -c)^2 /2 )=2(1 -c) -c(1 -c)=2(1 -c) -c +c²=2 -2c -c +c²=2 -3c +c².Then, total integral is∫₀^c (2/c x) dx +∫_c^1 [...] dx= [c] + [2 -3c +c²]=2 -2c +c².But we set total integral=1. So2 -2c +c²=1 →c² -2c +1=0→(c -1)^2=0→c=1. Butc=1 is not allowed. Therefore, there's a mistake here. Wait, this indicates that my earlier calculation was wrong.This suggests that the parametrization is flawed. Let me retrace. The original setup was that the function increases from0 to2 atx=c with slope2/c, then decreases to0 atx=1 with slope-2c/(1 -c). The integral of the function is supposed to be1, but according to the previous calculation, it's2 -2c +c². Setting that equal to1 gives c=1, which is not in (0,1). Therefore, this approach is invalid. Hence, the mistake was in the parametrization.Therefore, my previous approach was incorrect. Let's abandon that and try a different method.Alternative approach:For concave functions on (0,1) with integral1, we can use the fact that the extremal functions for ∫f² are the constant function (giving minimum) and the function which is as "peaked" as possible (giving maximum). For concave functions, the most peaked function would be the one that is linear decreasing from the maximum point to the endpoints.Wait, let's consider symmetric functions. Suppose the function is symmetric aboutx=0.5, concave, and integrates to1. The most peaked such function would be a triangle function with peak at0.5, as we considered earlier. This gives ∫f²=4/3.But earlier, there was an attempt to parametrize a function with a peak at an arbitrary pointc, but that led to inconsistencies. Perhaps the maximum is indeed4/3, achieved by the symmetric triangle function.Alternatively, maybe the maximum is higher. Let's consider a function that is constant over a small interval near0.5 and decreases linearly to0. For example, let’s take a function that is constant h on [0.5 -a,0.5 +a], and decreases linearly to0 on [0,0.5 -a] and [0.5 +a,1]. This is a trapezoidal function centered at0.5.Compute the integral:The area under the trapezoid is:Central rectangle:2a*h.Two side triangles: each with basea and heighth. Area per triangle: (a*h)/2. Total for two: a*h.Total integral:2a*h +a*h=3a*h. Set this equal to1→3a*h=1→h=1/(3a).The integral of the square is:Central rectangle:2a*h².Two side regions: each side is a triangle where f(x) goes from0 toh. The square of a linear function. For one side, integral from0 to0.5 -a of ( (h/(0.5 -a))x )² dx. Similarly for the other side.Wait, the left side: from0 to0.5 -a, the function is linear from0 toh. The equation is f(x)= [h/(0.5 -a)]x.Similarly, the right side: from0.5 +a to1, the function is linear fromh to0: f(x)= [h/(0.5 -a)](1 -x).Integral of square on left side:∫₀^{0.5 -a} [h/(0.5 -a) x]^2 dx= [h²/(0.5 -a)^2] ∫₀^{0.5 -a} x² dx= [h²/(0.5 -a)^2] * [ (0.5 -a)^3 /3 ]= h² (0.5 -a)/3.Similarly, integral on right side is the same: h² (0.5 -a)/3.Integral in central rectangle:2a*h².Total integral of square:2a*h² +2*(h² (0.5 -a)/3 )=2a*h² + (2/3)h² (0.5 -a ).Substitute h=1/(3a):Total integral=2a*(1/(3a))² + (2/3)*(1/(3a))²*(0.5 -a )=2a*(1/(9a²)) + (2/3)*(1/(9a²))*(0.5 -a )=2/(9a) + (2/27a²)(0.5 -a )Simplify:=2/(9a) + (1 -2a)/(27a² )To find the maximum of this expression with respect toa ∈(0,0.5).This seems complicated, but perhaps taking derivative and setting to zero.Let’s denote the integral as I(a)=2/(9a) + (1 -2a)/(27a² ).Compute derivative I’(a):dI/da= -2/(9a² ) + [ (-2)(27a² ) - (1 -2a)(54a ) ] / (27a² )².Wait, better to compute term by term.First term: d/da [2/(9a)]= -2/(9a² ).Second term: d/da [ (1 -2a)/(27a²) ]= [ (-2)(27a² ) - (1 -2a)(54a ) ] / (27a² )².Wait, no. Use quotient rule:d/da [ (1 -2a)/a² ]= [ (-2)a² - (1 -2a)(2a) ] /a^4= [ -2a² -2a +4a² ] /a^4= [2a² -2a]/a^4=2(a² -a)/a^4=2(a -1)/a³.Therefore, derivative of second term:(1/27)*2(a -1)/a³.Therefore, total derivative:-2/(9a² ) + (2(a -1))/(27a³ ).Set this equal to0:-2/(9a² ) +2(a -1)/(27a³ )=0.Multiply both sides by27a³:-6a +2(a -1)=0→-6a +2a -2=0→-4a -2=0→-4a=2→a= -0.5.Which is outside the domain a ∈(0,0.5). Therefore, no critical points in the interval. Therefore, the function I(a) is monotonic.Check behavior as a approaches0:I(a)=2/(9a) + (1 -2a)/(27a² )≈2/(9a) +1/(27a² )→∞.But this requires h=1/(3a)→∞. However, the function must be concave. If a approaches0, the function becomes a spike at0.5 of height h=1/(3a)→∞. But a concave function cannot have a vertical slope. The slope on the left side is h/(0.5 -a)≈2h for small a. As a→0, h→∞, so slope→∞, which is impossible for a concave function, because concave functions have non-increasing slopes. If the slope approaches infinity, it's not possible. Therefore, such functions are not valid concave functions.Therefore, there must be a constraint on the slope due to concavity.This suggests that our previous approach of taking a trapezoidal function is flawed because it doesn't account for the concavity condition properly. Specifically, the slope on the increasing side must be non-increasing. So if we have a function that increases to a peak and then decreases, the slopes must be non-increasing. Therefore, the left slope must be ≥ the right slope.Wait, for a concave function, the derivative is non-increasing. Therefore, if the function increases to a peak atc and then decreases, the left derivative atc must be ≥ the right derivative atc.But for a piecewise linear function, the left derivative is the slope beforec, and the right derivative is the slope afterc. For concavity, we need slope afterc ≤ slope beforec. But for a peak atc, the slope beforec is positive, and the slope afterc is negative. So the condition is automatically satisfied since a positive number is ≥ a negative number. Therefore, any piecewise linear function that increases to a peak and then decreases is concave.But in that case, such a function can have arbitrary slopes, as long as the left slope is ≥ the right slope. But to maintain positivity, the slopes must be chosen so that the function does not go negative.Therefore, the maximum possible ∫f² is unbounded? But that can't be, since all functions are positive, concave, and integrate to1. If we make the function very peaked, like a very tall spike aroundc with tiny width, then ∫f² can be made arbitrarily large. But for concave functions, can we do that?Wait, but a concave function cannot have an arbitrary spike. Because to have a very tall spike, the function would have to increase very steeply, but then decrease very steeply as well. However, for concavity, the slope after the peak must be ≤ the slope before the peak. So if the function increases with slopem to the peak, then decreases with slope≤m. Therefore, if you make the increasing slopem very large, the decreasing slope can be at mostm. But to keep the function positive, the decreasing slope must be such that the function reaches zero beforex=1.Wait, let's formalize this. Suppose the function increases from0 atx=0 tomc atx=c with slopem, then decreases to0 atx=d with slope-n, wheren ≤m. Then, the integral is0.5*mc² +0.5*n(d -c)^2=1. To maximize ∫f², we need to maximize the integral of the square, which would be higher if the function is more peaked.But this is getting too vague. Let me look for an authoritative approach.I recall that for concave functions on [a,b], there's an inequality related to the integral of the square. Alternatively, using the fact that concave functions lie above their chords, and using some integral inequality.Alternatively, consider that for a concave function f on (0,1) with ∫₀¹ f(x)dx=1, the maximum of ∫f² occurs at the function which is the tent function (triangle) symmetric aroundx=0.5, and the minimum occurs at the constant function1.But earlier, when I considered the symmetric triangle function, ∫f²=4/3. But there was a conflicting approach suggesting that as the peak becomes narrower, the integral approaches4. But that must be incorrect due to the integral constraint.Wait, let's verify the symmetric triangle function. Let’s say f(x) is2 -4|x -0.5| forx in (0,1). Then, f(x)=2 -4x +2 forx <0.5, and2 +4x -2 forx ≥0.5. Wait, no.Wait, the tent function centered at0.5 with height2: f(x)=2 -4|x -0.5|. Then, integrating over (0,1):Area under the curve: two triangles each with base0.5 and height2. Area=2*(0.5*0.5*2)=1, which is correct. Then, ∫f² dx= ∫₀^0.5(2 -4x +2)² dx + ∫⁰˙⁵¹(2 +4x -2)² dx. Wait, this seems confused.Wait, f(x) from0 to0.5: f(x)=2 +4(x -0.5). Wait, no. Let me correctly define the function.Let’s define f(x)=4x forx ∈[0,0.5] and f(x)=4(1 -x) forx ∈[0.5,1]. Then, ∫₀¹ f(x)dx=2*0.5*2=1. ∫₀¹ f(x)² dx= ∫₀^0.5 16x² dx + ∫⁰˙⁵¹ 16(1 -x)² dx=2*16*(0.5)^3 /3=2*16*(1/8)/3=2*2/3=4/3≈1.333.Therefore, the integral is4/3 for the symmetric triangle function.Now, suppose we take an asymmetric triangle function, peaked at some pointc ≠0.5. Let's compute ∫f² for such a function.Let’s define f(x) as follows: increases linearly from0 atx=0 toh atx=c, then decreases linearly to0 atx=1. The integral is0.5*h*c +0.5*h*(1 -c)=0.5*h*(c +1 -c)=0.5*h=1⇒h=2.So the function is f(x)= (2/c)x forx ∈[0,c] and f(x)=2*(1 -x)/(1 -c) forx ∈[c,1].Compute ∫₀¹ f(x)² dx= ∫₀^c (4/c²)x² dx + ∫_c^1 (4/(1 -c)^2)(1 -x)^2 dx.First integral:4/c² *c³/3=4c/3.Second integral:4/(1 -c)^2 * ∫_c^1 (1 -x)^2 dx=4/(1 -c)^2 * [ (1 -c)^3 /3 ]=4(1 -c)/3.Total integral:4c/3 +4(1 -c)/3=4/3.So, regardless of c, the integral is always4/3. Therefore, any triangle function, symmetric or not, has ∫f²=4/3. Therefore, the maximum value is4/3.But earlier attempt with trapezoidal function gave higher value, but that was due to a mistake in parametrization. Therefore, the correct maximum is4/3.Therefore, the possible values of ∫₀¹ f(x)² dx are between1 and4/3.Hence, the integral can take any value in[1,4/3].But we need to confirm that there aren't other concave functions with higher ∫f².Suppose we take a function that is more peaked than the triangle function. For example, a function that is quadratic, concave, and integrates to1.Let’s consider f(x)=a -b(x -0.5)^2. Choosea andb such that f is positive on (0,1) and ∫₀¹ f(x)dx=1.Compute ∫₀¹ (a -b(x -0.5)^2 )dx= a -b ∫₀¹(x -0.5)^2 dx= a -b*(1/12)=1.To ensure positivity, the minimum of f(x) occurs atx=0.5, which isa -b*(0)=a. Wait, no: the function is a downward-opening parabola, so the minimum is at the ends. Wait, no, f(x)=a -b(x -0.5)^2 is a downward-opening parabola (since coefficient ofx² is -b), so it's concave. The maximum is atx=0.5, and the minimums are atx=0 andx=1. To ensure positivity, f(0)=a -b*(0.25)≥0 and similarly f(1)=a -b*(0.25)≥0. So a≥b/4.The integral isa -b/12=1. So a=1 +b/12.To ensure positivity:1 +b/12 -b/4 ≥0→1 -b/6 ≥0→b≤6.Thus, b can vary from0 to6.Compute ∫₀¹ f(x)^2 dx= ∫₀¹ [a -b(x -0.5)^2]^2 dx.Expand the square: a² -2ab(x -0.5)^2 +b²(x -0.5)^4.Integrate term by term:a²*1 -2ab∫₀¹(x -0.5)^2 dx +b²∫₀¹(x -0.5)^4 dx.Compute each integral:First term:a².Second term:-2ab*(1/12).Third term:b²*( ∫₀¹(x -0.5)^4 dx ). Let’s compute:Let t=x -0.5, then t ∈[-0.5,0.5]. ∫_{-0.5}^{0.5} t^4 dt=2*∫₀^{0.5}t^4 dt=2*( (0.5)^5 /5 )=2*(1/160)=1/80.Therefore, third term:b²*(1/80).So total integral:a² - (2ab)/12 + b²/80.Substitute a=1 +b/12:= (1 +b/12)^2 - (2*(1 +b/12)*b)/12 +b²/80.Expand (1 +b/12)^2=1 +b/6 +b²/144.Second term: -(2b/12 +2b²/144)/12= -(b/6 +b²/72).Third term: +b²/80.Combine all terms:1 +b/6 +b²/144 -b/6 -b²/72 +b²/80.Simplify:1 + (b/6 -b/6) + (b²/144 -b²/72 +b²/80).=1 + b²(1/144 -1/72 +1/80).Convert to common denominator, which is 720:1 + b²*(5/720 -10/720 +9/720)=1 + b²*(4/720)=1 +b²/180.Therefore, ∫₀¹ f(x)^2 dx=1 +b²/180.To maximize this expression, we need to maximize b. Sinceb≤6, the maximum occurs atb=6:∫₀¹ f(x)^2 dx=1 +36/180=1 +0.2=1.2=6/5=1.2.Which is less than4/3≈1.333. Therefore, the quadratic concave function gives a lower integral than the triangle function. Therefore, the triangle function gives a higher value.Thus, it seems that the maximum value of the integral is4/3, achieved by the triangle function, and the minimum is1, achieved by the constant function.Therefore, the possible values of ∫₀¹ f(x)² dx are in the interval[1,4/3].Hence, the answer is that the integral can take any value between1 and4/3, inclusive.Final AnswerThe integral can take values between boxed{1} and boxed{dfrac{4}{3}}.</think>Given a function ( f(x) ) defined on the interval ( (0,1) ) that is positive, integrable, integrates to 1, and is concave ( concave from below), we need to determine the possible values of the integral ( int_{0}^{1} f^{2}(x) , dx ). Key Steps and Reasoning:1. Minimum Value: - By the Cauchy-Schwarz inequality, the minimum value of ( int_{0}^{1} f^{2}(x) , dx ) for a function ( f ) with ( int_{0}^{1} f(x) , dx = 1 ) is achieved by the constant function ( f(x) = 1 ). This gives: [ int_{0}^{1} 1^{2} , dx = 1 ] - Therefore, the minimum value is ( 1 ).2. Maximum Value: - For concave functions, the most "peaked" function (which maximizes ( int_{0}^{1} f^{2}(x) , dx )) is a symmetric triangle function centered at ( x = 0.5 ). This function is defined as: [ f(x) = begin{cases} 4x & text{if } 0 leq x leq 0.5, 4(1 - x) & text{if } 0.5 leq x leq 1. end{cases} ] - Calculating the integral of the square of this function: [ int_{0}^{0.5} (4x)^2 , dx + int_{0.5}^{1} (4(1 - x))^2 , dx ] [ = 16 int_{0}^{0.5} x^2 , dx + 16 int_{0.5}^{1} (1 - x)^2 , dx ] [ = 16 left[ frac{x^3}{3} right]_{0}^{0.5} + 16 left[ frac{(1 - x)^3}{3} right]_{0.5}^{1} ] [ = 16 left( frac{(0.5)^3}{3} right) + 16 left( frac{(0.5)^3}{3} right) ] [ = 16 left( frac{1}{24} right) + 16 left( frac{1}{24} right) ] [ = frac{2}{3} + frac{2}{3} = frac{4}{3} ] - Therefore, the maximum value is ( frac{4}{3} ). Conclusion:The integral ( int_{0}^{1} f^{2}(x) , dx ) can take any value between ( 1 ) and ( frac{4}{3} ), inclusive.[boxed{1} quad text{and} quad boxed{dfrac{4}{3}}]]

🔑:a) One of the famous inequalities in integral calculus states that if the squares of the functions f_{1}, f_{2} are also integrable, thenleft(int_{a}^{b} f_{1}(x) f_{2}(x) d xright)^{2} leqq int_{a}^{b} f_{1}^{2}(x) d x int_{a}^{b} f_{2}^{2}(x) d x(Cauchy-Schwarz inequality). Indeed, based on f_{1}, f_{2}, the functionI(lambda)=int_{a}^{b}left[f_{1}(x)+lambda f_{2}(x)right]^{2} d x=int_{a}^{b} f_{1}^{2}(x) d x+2 lambda int_{a}^{b} f_{1}(x) f_{2}(x) d x+lambda^{2} int_{a}^{b} f_{2}^{2}(x) d xis non-negative for all lambda. Since I(lambda) is a quadratic function of lambda, this can only be true if the discriminant is non-positive, which is precisely the inequality (3). Applying (3) to the functions f_{1}(x)=f(x), f_{2}(x)=1, we getint_{0}^{1} f^{2}(x) d x geqq int_{0}^{1} f(x) d x=1Thus, the smallest possible value of (2) is 1, which can be achieved (even under the constraint (1)) by choosing f(x)=1 (and only by this choice).b) According to our assumption, f(x) is concave from below, i.e., convex from above. Since f(x) geqq 0, this means that the region F bounded by the x-axis, the y-axis, and the graph of f(x) is a convex set. Consider the horizontal chords of this set and associate with each chord a triangle whose base is the segment of the x-axis between 0 and 1, and whose sides pass through the endpoints of the chord. Denote the triangle corresponding to the chord of height m by H(m), the area of the part of F not covered by H(m) by A(m), and the part of H(m) not covered by F by B(m). Since F is convex, if m_{1}<m_{2}, Hleft(m_{1}right) contains Hleft(m_{2}right), and Aleft(m_{1}right) leqq Aleft(m_{2}right), Bleft(m_{1}right) geqq Bleft(m_{2}right). The largest possible value of m is equal to the maximum of f(x), denoted by M. Clearly, B(M) is 0, and since the area of H(M) is less than the area of F, M leqq 2. The smallest possible value of m is the smaller of f(0) and f(1), and this value can be chosen as m. Here, the limit of A(m) is 0, and as is easily seen, the functions A and B are continuous. Therefore, there exists an m such thatA(m)=B(m)and for this m, H(m)=1, meaning the third vertex of H(m) lies on the line y=2. Denote the function formed by the two sides of H(m) by f_{0}. For this function,begin{gathered}int_{0}^{1} f_{0}^{2}(x) d x-int_{0}^{1} f^{2}(x) d x=int_{0}^{1}left[f_{o}(x)-f(x)right]left[f_{0}(x)+right. f(x)] d x geqq 2 m int_{0}^{1}left[f_{0}(x)-f(x)right] d x geqq 0end{gathered}becauseint_{0}^{1}left[f_{0}(x)-f(x)right]left[f_{0}(x)+f(x)-2 mright] d x geqq 0since here the integrand is non-negative: where f_{0} leqq f, both f_{0} and f are less than m, so the product of both factors is positive. Since f_{0} is also concave from below, and its integral is 1, f_{0} also belongs to the set of functions for which we are seeking the maximum value of (2). It is easily seen that the square integral of f_{0} is 4 / 3, so for the functions under consideration,1 leqq int_{0}^{1} f^{2}(x) d x leqq frac{4}{3}

🧠:[<think>Okay, let me try to work through this problem step by step. The function given is ( f(x) = a ln x + frac{1}{2}x^2 - a x ), and I need to discuss its monotonicity first. Then, there's a part about finding the minimum value of λ such that ( f(x_1) + f(x_2) < lambda(x_1 + x_2) ) holds when f has two extreme points x₁ and x₂.Starting with part (I): Monotonicity of f(x). Monotonicity usually refers to where the function is increasing or decreasing, which is determined by the first derivative. So I need to find f'(x), set it to zero to find critical points, and then analyze intervals around those points.Let's compute the derivative. The function has three terms: ( a ln x ), ( frac{1}{2}x^2 ), and ( -a x ). The derivative of ( a ln x ) is ( a/x ), the derivative of ( frac{1}{2}x^2 ) is x, and the derivative of ( -a x ) is -a. So putting that all together:( f'(x) = frac{a}{x} + x - a ).Simplify that expression: ( f'(x) = x - a + frac{a}{x} ). To find critical points, set f'(x) = 0:( x - a + frac{a}{x} = 0 ).Multiply both sides by x to eliminate the denominator (assuming x ≠ 0, which it can't be anyway since ln x is part of the original function):( x^2 - a x + a = 0 ).So the equation becomes ( x^2 - a x + a = 0 ). Let's denote this quadratic equation in x. The critical points are the solutions to this equation. To have real solutions, the discriminant must be non-negative. The discriminant D is:( D = (-a)^2 - 4 times 1 times a = a^2 - 4a ).So for real critical points, we need ( a^2 - 4a geq 0 ). Let's factor that:( a(a - 4) geq 0 ).This inequality holds when ( a leq 0 ) or ( a geq 4 ). However, since the original function includes ( ln x ), the domain of f(x) is x > 0. So we need to consider the critical points in x > 0. Also, the parameter a could be positive or negative. Hmm, but if a is negative, let's see.Wait, the quadratic equation is ( x^2 - a x + a = 0 ). If a is negative, then the constant term is negative, so the quadratic would have two real roots (since discriminant would be positive when a is negative), but one positive and one negative because the product of the roots is a (from x² - a x + a = 0, product is c/a = a/1 = a). So if a is negative, the product of roots is negative, meaning one root positive and one negative. But since x > 0, we only consider the positive root. So when a < 0, there's only one critical point. When a ≥ 4, discriminant is non-negative (since a(a - 4) ≥ 0 when a ≥4), and since a is positive, both roots would be positive because sum of roots is a (from the quadratic equation) and product is a. So if a ≥4, both roots are positive, so two critical points. If 0 < a <4, discriminant is negative, so no real roots, meaning f'(x) is always positive or always negative? Let's check.Wait, let's first analyze the derivative f'(x) = x - a + a/x. Let's consider different cases based on a.Case 1: a = 0. Then f(x) simplifies to (1/2)x², since the ln x term and -a x term disappear. Then f'(x) = x. So derivative is x, which is positive for x >0. So f is increasing on (0, ∞).But a is a constant, maybe not necessarily zero. So if a is zero, function is quadratic. But the problem says "where a is a constant", so a could be zero. But likely, in general, a ≠0. Let's proceed.Case 2: a ≠0. Let's consider the derivative again: f'(x) = x - a + a/x. Let's write this as f'(x) = (x^2 - a x + a)/x. Since x >0, the sign of f'(x) depends on the numerator N(x) = x² - a x + a.So when N(x) is positive, f is increasing; when N(x) is negative, f is decreasing.We already found that the roots of N(x) are x = [a ± sqrt(a² -4a)]/2. So, depending on the value of a, the quadratic can have 0, 1, or 2 real roots.If a² -4a >0, which is when a <0 or a >4, two distinct real roots.If a² -4a =0, which is when a=0 or a=4, one real root (double root).If a² -4a <0, which is when 0 <a <4, no real roots.But since x >0, let's analyze for different ranges of a.Subcase 1: a <0.Then discriminant a² -4a is positive (since a is negative, a² is positive, -4a is positive, so sum is positive). So N(x) has two real roots. Since the product of roots is a (from x² - a x +a), which is negative when a <0. Therefore, one root positive and one negative. Therefore, only one critical point in x >0.So for a <0, f'(x) =0 at one point x1 in x >0. Let's analyze the behavior of N(x) when a <0.Since the quadratic N(x) =x² -a x +a opens upwards (coefficient of x² is 1 >0). Since one root is positive and one is negative, N(x) crosses the x-axis at x1 positive and x2 negative. Therefore, for x >0:- When x <x1 (the positive root), N(x) is negative (since between the roots, the quadratic is below the x-axis if it opens upwards). Wait, but the roots are at x1 positive and x2 negative. So for x >0, between negative infinity and x2, and x1 to infinity. But since x >0, we can only consider x >0. So for x >0, when x <x1, N(x) is negative (since between x2 and x1, but x2 is negative, so from x=0 up to x1, N(x) is negative). For x >x1, N(x) is positive.Therefore, f'(x) is negative for 0 <x <x1, and positive for x >x1. Therefore, f(x) is decreasing on (0, x1) and increasing on (x1, ∞). So the function has a minimum at x1.Subcase 2: a =0. Then N(x) =x², so f'(x) =x²/x =x. So derivative is x, which is positive for x >0. Thus, f is increasing on (0, ∞).Subcase 3: 0 <a <4. Then discriminant a² -4a <0, so N(x) =x² -a x +a has no real roots. Since the quadratic opens upwards (coefficient 1 >0), it is always positive. Therefore, N(x) >0 for all x, so f'(x) =N(x)/x >0 for all x >0. Therefore, f is increasing on (0, ∞).Subcase 4: a=4. Then discriminant is 16 -16 =0, so N(x) =x² -4x +4 = (x -2)^2. Therefore, N(x) is non-negative for all x, and zero at x=2. Therefore, f'(x) = (x -2)^2 /x. So for x >0, f'(x) ≥0, and zero only at x=2. Therefore, f is increasing on (0, ∞), with a horizontal tangent at x=2.Subcase 5: a >4. Then discriminant a² -4a >0, so N(x) has two real roots. The product of the roots is a (from quadratic formula), which is positive since a >4. The sum of the roots is a (from quadratic formula). Therefore, both roots are positive. Let's denote them x1 and x2, with x1 <x2.Since the quadratic opens upwards, N(x) is negative between x1 and x2, and positive otherwise. Therefore, for x >0:- For 0 <x <x1, N(x) >0 (since to the left of the first root, quadratic is positive)- For x1 <x <x2, N(x) <0- For x >x2, N(x) >0But wait, actually, when you have a quadratic opening upwards, it's positive outside the roots and negative between them. So since both roots are positive, for x <x1, N(x) >0; x1 <x <x2, N(x) <0; x >x2, N(x) >0.Therefore, the derivative f'(x) is positive for 0 <x <x1, negative for x1 <x <x2, and positive for x >x2. Therefore, the function f(x) is increasing on (0, x1), decreasing on (x1, x2), and increasing again on (x2, ∞). Therefore, f has a local maximum at x1 and a local minimum at x2.Wait, but wait, when the derivative goes from positive to negative, that's a maximum, and from negative to positive, that's a minimum. So yes, x1 is a local maximum and x2 is a local minimum. Wait, hold on. Wait, when moving from left to right on the x-axis:- From 0 to x1: increasing (f' >0)- From x1 to x2: decreasing (f' <0)- From x2 to ∞: increasing (f' >0)Therefore, x1 is a local maximum and x2 is a local minimum. So in this case, when a >4, there are two critical points: x1 (max) and x2 (min). Therefore, for part (I), the monotonicity is as follows:- If a <0: f decreases on (0, x1), increases on (x1, ∞)- If a =0: f increases on (0, ∞)- If 0 <a <4: f increases on (0, ∞)- If a =4: f increases on (0, ∞) with a stationary point at x=2- If a >4: f increases on (0, x1), decreases on (x1, x2), increases on (x2, ∞)So that's part (I). Now part (II): If f(x) has two extreme points, denoted as x₁ and x₂, and the inequality f(x₁) + f(x₂) < λ(x₁ + x₂) always holds, find the minimum value of λ.First, from part (I), f has two extreme points when a >4. So we need to consider a >4. Then x₁ and x₂ are the two critical points, with x₁ being the local maximum and x₂ the local minimum.Given that, the problem requires that for all a >4, the inequality f(x₁) + f(x₂) < λ(x₁ + x₂) holds. We need to find the minimal λ such that this inequality is always true for any a >4.Therefore, we need to express f(x₁) + f(x₂) in terms of a and x₁ + x₂, then find λ such that [f(x₁) + f(x₂)] / (x₁ + x₂) < λ, and find the minimal such λ.Alternatively, find the supremum of [f(x₁) + f(x₂)] / (x₁ + x₂) over all a >4, and that supremum will be the minimal λ.So first, let's find expressions for x₁ and x₂. Since x₁ and x₂ are roots of the quadratic equation x² -a x +a =0. So from quadratic formula:x = [a ± sqrt(a² -4a)] /2.So x₁ and x₂ are [a - sqrt(a² -4a)] /2 and [a + sqrt(a² -4a)] /2. Let's denote x₁ as the smaller root and x₂ as the larger. Wait, but since a >4, sqrt(a² -4a) is real. Let's compute x₁ + x₂: sum of roots is a. So x₁ +x₂ =a. Also, product of roots x₁ x₂ =a.So x₁ +x₂ =a, x₁ x₂ =a.Therefore, perhaps we can express f(x₁) +f(x₂) in terms of a.Let's compute f(x₁) +f(x₂). Since f(x) =a ln x + (1/2)x² -a x.So f(x₁) + f(x₂) = a(ln x₁ + ln x₂) + (1/2)(x₁² +x₂²) -a(x₁ +x₂).Simplify:First, ln x₁ + ln x₂ = ln(x₁ x₂) = ln a (since x₁ x₂ =a).Second, x₁² +x₂² can be written as (x₁ +x₂)² -2x₁x₂ = a² -2a.Third, x₁ +x₂ =a.Therefore, substituting:f(x₁) +f(x₂) = a ln a + (1/2)(a² -2a) -a*aSimplify term by term:- a ln a- (1/2)(a² -2a) = (1/2)a² -a- -a²Therefore, combining all terms:a ln a + (1/2 a² -a) -a² = a ln a - (1/2)a² -aSo f(x₁) +f(x₂) = a ln a - (1/2)a² -a.Therefore, the inequality is:a ln a - (1/2)a² -a < λ aBecause x₁ +x₂ =a, so λ(x₁ +x₂) =λ a.Thus, divide both sides by a (since a >4, a is positive, so inequality direction remains):ln a - (1/2)a -1 < λTherefore, the inequality becomes λ > ln a - (1/2)a -1.Thus, to satisfy the inequality for all a >4, λ must be greater than the maximum value of ln a - (1/2)a -1 for a >4. The minimal such λ is the supremum (least upper bound) of ln a - (1/2)a -1 over a >4.Therefore, we need to find the maximum of the function g(a) = ln a - (1/2)a -1 for a >4, and then set λ to be that maximum. Since the question asks for the minimal λ such that the inequality holds for all a >4, λ must be at least the maximum of g(a), so the minimal λ is the maximum of g(a).Therefore, compute the maximum of g(a) = ln a - (1/2)a -1 for a >4.To find the maximum, take derivative of g(a):g'(a) = 1/a - (1/2).Set derivative to zero:1/a -1/2 =0 => 1/a =1/2 => a=2.But wait, a=2 is less than 4, which is outside our domain of a >4. Therefore, the critical point at a=2 is not in the interval we are considering. Therefore, on a >4, the function g(a) is decreasing, because g'(a) =1/a -1/2. For a >4, 1/a <1/4, so 1/a -1/2 <1/4 -1/2 = -1/4 <0. Therefore, g(a) is decreasing for a >4. Therefore, its maximum on a >4 is at a=4.Thus, compute g(4) = ln4 - (1/2)*4 -1 = ln4 -2 -1 = ln4 -3 ≈1.386 -3 ≈-1.614.Therefore, the maximum value of g(a) for a >4 is at a=4, which is ln4 -3. But since the function g(a) is decreasing for a >4, as a increases beyond 4, g(a) decreases. Therefore, the supremum of g(a) over a >4 is ln4 -3. Therefore, the minimal λ is ln4 -3.But wait, let's verify this. If the function g(a) is decreasing on a >4, then the maximum is at a=4. Therefore, for all a >4, g(a) ≤g(4) =ln4 -3. Therefore, to have λ >g(a) for all a >4, we can take λ ≥ln4 -3. But since the problem states that the inequality f(x₁) +f(x₂) <λ(x₁ +x₂) must always hold, λ needs to be greater than the supremum of g(a). But since at a approaching 4 from above, g(a) approaches ln4 -3. However, when a approaches 4 from above, does g(a) approach ln4 -3? Let's check.Wait, when a=4, even though a=4 is the boundary, but the original problem states that f(x) has two extreme points. From part (I), when a=4, f(x) has a double root at x=2, but in that case, it's a point of inflection (since the derivative touches zero but doesn't cross it). Wait, no. Wait, when a=4, the quadratic equation x² -4x +4=0 has a double root at x=2. Therefore, the derivative f'(x)= (x-2)^2 /x. So at x=2, the derivative is zero, and since the derivative is non-negative everywhere (as (x-2)^2 is non-negative and x>0), the function is increasing on (0, ∞) with a stationary point at x=2. Therefore, at a=4, the function does not have two extreme points; it has a single stationary point which is neither a maximum nor a minimum—it's a point of inflection for the derivative, meaning the concavity changes.Therefore, when a=4, f(x) does not have two extreme points, so a must be strictly greater than 4 for two extreme points. Therefore, the domain for a is (4, ∞). Then, we should check the behavior of g(a) as a approaches 4 from above and as a approaches infinity.Compute the limit as a approaches 4 from the right:lim_{a→4+} g(a) = ln4 - (1/2)(4) -1 = ln4 -2 -1 = ln4 -3 ≈1.386 -3≈-1.614.As a approaches infinity:lim_{a→∞} g(a) = lim_{a→∞} [ln a - (1/2)a -1] = -∞, since the - (1/2)a term dominates.Since g(a) is continuous on (4, ∞) and decreasing, the maximum of g(a) on (4, ∞) is indeed approaching a=4 from the right, which is ln4 -3. However, since a must be greater than 4, the supremum is ln4 -3, but g(a) never actually reaches ln4 -3; it approaches it as a approaches 4 from above. Therefore, to satisfy the inequality f(x₁) +f(x₂) <λ(x₁ +x₂) for all a >4, λ must be at least ln4 -3. However, since the limit as a approaches 4 from above is ln4 -3, we need λ to be greater than or equal to this value. But since for a >4, g(a) < ln4 -3 (since g(a) is decreasing), actually, the inequality ln a - (1/2)a -1 <λ would hold for all a >4 if λ ≥ ln4 -3. However, since the problem states that the inequality must always hold, and we need the minimal such λ, the minimal λ is ln4 -3.But wait, let's check with a specific example. Let's take a approaching 4 from above, say a=4.1. Compute g(a):g(4.1)= ln(4.1) - (1/2)(4.1) -1 ≈1.411 -2.05 -1 ≈-1.639.Which is less than ln4 -3≈-1.613. So as a approaches 4+, g(a) approaches -1.613 from below. Wait, wait, no. Wait, when a=4, even though a=4 is not in the domain, the limit as a approaches 4 from above would be:lim_{a→4+} [ln a - (1/2)a -1] = ln4 -2 -1=ln4 -3≈-1.613.But when a is slightly above 4, say a=4.1:ln(4.1)≈1.411, (1/2)(4.1)=2.05, so 1.411 -2.05 -1≈-1.639, which is less than -1.613. So actually, as a increases beyond 4, g(a) decreases. Therefore, the maximum value of g(a) on a >4 is the left-hand limit as a approaches 4 from the right, which is ln4 -3. However, since a cannot be 4, and for a >4, g(a) is always less than ln4 -3. Wait, but wait, if a approaches 4 from above, g(a) approaches ln4 -3 from below. Wait, but ln4 -3 is approximately -1.613, and when a approaches 4+, g(a) approaches that value. Wait, but let me compute more carefully:Wait, if a approaches 4 from above, then:g(a) = ln a - (1/2)a -1.At a=4, it's ln4 -2 -1=ln4 -3≈1.386-3≈-1.614.Wait, but as a approaches 4 from above, ln a approaches ln4, (1/2)a approaches 2, so the limit is ln4 -2 -1=ln4 -3. But if a approaches 4 from above, g(a) approaches ln4 -3 from above? Wait, no. Wait, let's take a=4.0001:g(4.0001)= ln(4.0001) - (1/2)(4.0001) -1 ≈1.3863 -2.00005 -1 ≈-1.61375.But ln4 -3≈1.386294 -3≈-1.613706. So ln4 -3≈-1.613706, and g(4.0001)=≈-1.61375, which is slightly less than ln4 -3. Wait, that suggests that as a approaches 4 from above, g(a) approaches ln4 -3 from below. Therefore, the left-hand limit is ln4 -3, but from below. Therefore, the supremum of g(a) over a >4 is ln4 -3, but it's not attained. Therefore, for all a >4, g(a) <ln4 -3. Therefore, to satisfy the inequality ln a - (1/2)a -1 <λ for all a >4, we can take λ=ln4 -3, since even though g(a) never actually reaches ln4 -3, it gets arbitrarily close as a approaches 4 from above. Therefore, the minimal such λ is ln4 -3.But let's check with another a. Take a=5:g(5)=ln5 - (1/2)(5) -1≈1.609 -2.5 -1≈-1.891.Which is less than ln4 -3≈-1.613.Another example, a=10:g(10)=ln10 -5 -1≈2.302 -6≈-3.698.Even smaller.Therefore, the maximum of g(a) is approached as a approaches 4 from above, but never actually reaches ln4 -3. Therefore, the supremum is ln4 -3, so the minimal λ is λ=ln4 -3.But let's check if this is correct. If we set λ=ln4 -3, then the inequality f(x₁) +f(x₂) <λ(x₁ +x₂) becomes:a ln a - (1/2)a² -a <(ln4 -3)a.Divide both sides by a (positive):ln a - (1/2)a -1 <ln4 -3.Rearranged:ln a - (1/2)a -1 -ln4 +3 <0ln(a/4) - (1/2)a +2 <0.But as a approaches 4 from above, a/4 approaches1, so ln(a/4) approaches0, and -(1/2)a +2 approaches-2 +2=0. So ln(a/4) - (1/2)a +2 approaches0. Therefore, the left-hand side approaches0 from below, since for a slightly greater than4, ln(a/4)≈ (a-4)/4 - (a-4)^2/(2*16) +...≈(a-4)/4 - small positive term, and -(1/2)a +2≈-2 - (a-4)/2 +2≈- (a-4)/2. Therefore, ln(a/4)≈(a-4)/4, so overall expression:(a-4)/4 - (a-4)/2≈- (a-4)/4. So for a >4, this is negative. Therefore, the inequality holds. But when a approaches4 from above, the expression approaches0 from below. So the difference between the two sides approaches zero. Therefore, λ must be at least ln4 -3. If we choose λ=ln4 -3, then the inequality f(x₁) +f(x₂) <λ(x₁ +x₂) will hold for all a >4, since for a >4, the left-hand side is less than ln4 -3 times a, and approaches it as a approaches4. Therefore, the minimal λ is ln4 -3.But let's compute ln4. Since ln4=2 ln2≈2*0.693≈1.386, so ln4 -3≈1.386 -3≈-1.614.Wait, but λ is supposed to satisfy the inequality f(x₁) +f(x₂) <λ(x₁ +x₂). If λ is negative, but x₁ +x₂ =a >4, so λ(x₁ +x₂) would be negative if λ is negative. But f(x₁) +f(x₂) is also negative? Let's check with a=5.For a=5, f(x₁)+f(x₂)=5 ln5 - (1/2)(25) -5≈5*1.609 -12.5 -5≈8.045 -17.5≈-9.455.λ(x₁ +x₂)=(ln4 -3)*5≈(-1.614)*5≈-8.07.But -9.455 < -8.07 is false, because -9.455 is less than -8.07. Wait, but -9.455 is less than -8.07, which would mean that the inequality -9.455 < -8.07 is true. Wait, but if both sides are negative, a more negative left-hand side is less than a less negative right-hand side.Wait, but f(x₁) +f(x₂)= -9.455, λ(x₁ +x₂)= -8.07. Then, -9.455 < -8.07 is true. So even though λ is negative, the inequality holds.But when a approaches4 from above, f(x₁) +f(x₂) approaches (ln4 -3)*4≈-1.614*4≈-6.456.But let me compute f(x₁) +f(x₂) when a approaches4 from above:f(x₁)+f(x₂)=a ln a -0.5a² -a. When a approaches4, this becomes4 ln4 -0.5*16 -4=4 ln4 -8 -4=4 ln4 -12≈4*1.386 -12≈5.544 -12≈-6.456.λ(x₁ +x₂)= (ln4 -3)*a≈(-1.614)*4≈-6.456.So as a approaches4 from above, f(x₁) +f(x₂) approaches -6.456, and λ(x₁ +x₂) approaches -6.456. Therefore, the inequality f(x₁)+f(x₂) <λ(x₁ +x₂) becomes approaching -6.456 < -6.456, which is not true. But since a approaches4 from above, but a is strictly greater than4, then f(x₁)+f(x₂) is slightly less than -6.456 (since g(a)=ln a -0.5 a -1 is decreasing, so as a increases past4, g(a) decreases past ln4 -3≈-1.614, so multiplying by a, which is increasing, but the product's limit is -6.456. But since a is approaching4 from above, a=4+ε, then:f(x₁)+f(x₂)= (4+ε)ln(4+ε) -0.5(4+ε)^2 - (4+ε).Expanding this:=4 ln4 + ε ln4 +4*(ε/4) -0.5*(16 +8ε +ε²) -4 -ε (using Taylor expansion for ln(4+ε)≈ln4 + ε/4)Approximately:≈4 ln4 + ε ln4 + ε -0.5*(16 +8ε +ε²) -4 -ε=4 ln4 -8 -4 + ε ln4 + ε -4ε -0.5ε² -ε= (4 ln4 -12) + ε(ln4 +1 -4 -1) -0.5ε²= (4 ln4 -12) + ε(ln4 -4) -0.5ε²As ε→0+, the term ε(ln4 -4) is negative since ln4≈1.386 <4, so ln4 -4≈-2.614. Therefore, the next term is negative, so f(x₁)+f(x₂)≈(4 ln4 -12) + negative term, which is less than 4 ln4 -12. Therefore, f(x₁)+f(x₂) <4 ln4 -12. But 4 ln4 -12=4*(ln4 -3)=λ*(x₁ +x₂) when λ=ln4 -3 and a=4. However, since a is approaching4 from above, x₁ +x₂=a approaches4 from above, so λ*(x₁ +x₂)= (ln4 -3)*(4+ε)≈(ln4 -3)*4 + (ln4 -3)*ε≈-6.456 + (-1.614)*ε. Meanwhile, f(x₁)+f(x₂)≈-6.456 + (-2.614)*ε. Therefore, f(x₁)+f(x₂) -λ*(x₁ +x₂)≈[-6.456 + (-2.614)ε] -[-6.456 + (-1.614)ε]= (-2.614 +1.614)ε= -1ε <0.Therefore, f(x₁)+f(x₂) <λ*(x₁ +x₂) as ε→0+. Therefore, even though both sides approach -6.456, the left-hand side approaches it from below, and the right-hand side approaches it from above (since λ*(x₁ +x₂)= (ln4 -3)*a, with a approaching4 from above and λ=ln4 -3≈-1.614). Therefore, for a slightly above4, the left-hand side is slightly less than -6.456, and the right-hand side is slightly more than -6.456, so the inequality holds.Therefore, the minimal λ is ln4 -3. But ln4 is 2 ln2, so we can write it as 2 ln2 -3. However, the problem might prefer the answer in terms of ln4. Let me check if that's acceptable.Alternatively, compute ln4 -3≈1.386 -3≈-1.614. However, the problem asks for the minimum value of λ, and λ is a real number. The fact that it's negative doesn't matter, as long as it's the minimal value required. Therefore, the answer is λ=ln4 -3. But let's check if there's a simpler form.Alternatively, note that ln4 -3=2 ln2 -3. Either form is acceptable, but perhaps the problem expects the answer in terms of ln2. Let's see.Alternatively, since ln4=2 ln2, so λ=2 ln2 -3. Either form is okay, but likely the answer is written as 2 ln2 -3. However, we should verify.But let's check again the steps to ensure no miscalculations.We had:f(x₁)+f(x₂)=a ln a - (1/2)a² -a.Divide by a: ln a - (1/2)a -1 <λ.Thus, λ must be greater than the supremum of ln a - (1/2)a -1 over a >4.We found that this supremum is ln4 -3, since as a approaches4 from above, ln a approaches ln4, (1/2)a approaches2, so ln a - (1/2)a -1 approaches ln4 -2 -1=ln4 -3.Since the function g(a)=ln a - (1/2)a -1 is decreasing for a >4, its supremum on a >4 is indeed ln4 -3, hence the minimal λ is ln4 -3.Therefore, the minimal value of λ is ln4 -3, which can also be written as 2 ln2 -3. Depending on the required form, both are correct, but in terms of simplest form, maybe 2 ln2 -3 is preferable. However, the answer might need to be boxed as boxed{2ln 2 -3} or boxed{ln 4 -3}. Both are correct.Wait, let's check with a=4, even though a=4 is excluded. If we plug a=4 into λ=ln4 -3, then λ=ln4 -3≈-1.613. For a=4.1, f(x₁)+f(x₂)=4.1 ln4.1 -0.5*(4.1)^2 -4.1≈4.1*1.411 -0.5*16.81 -4.1≈5.785 -8.405 -4.1≈-6.72, while λ*(x₁ +x₂)= (ln4 -3)*4.1≈-1.613*4.1≈-6.613. So -6.72 < -6.613, which is true.Similarly, for a=5: f(x₁)+f(x₂)=5 ln5 -0.5*25 -5≈5*1.609 -12.5 -5≈8.045 -17.5≈-9.455, and λ*(5)= (ln4 -3)*5≈-1.614*5≈-8.07. So -9.455 < -8.07 is true.Therefore, the inequality holds for all a >4 with λ=ln4 -3. Therefore, the minimal λ is indeed ln4 -3.Final AnswerThe minimum value of (lambda) is (boxed{2ln 2 - 3}).</think>Given the function ( f(x) = a ln x + frac{1}{2}x^2 - ax ): (I) Monotonicity of ( f(x) )1. First Derivative: [ f'(x) = frac{a}{x} + x - a = frac{x^2 - ax + a}{x} ]2. Critical Points: Solve ( x^2 - ax + a = 0 ). The discriminant is ( a^2 - 4a ). - Discriminant Analysis: - ( a leq 0 ) or ( a geq 4 ): Two real roots (one positive if ( a < 0 ), both positive if ( a geq 4 )). - ( 0 < a < 4 ): No real roots.3. Monotonicity: - ( a < 0 ): One critical point ( x_1 ). ( f(x) ) decreases on ( (0, x_1) ) and increases on ( (x_1, infty) ). - ( 0 leq a < 4 ): ( f(x) ) is increasing on ( (0, infty) ). - ( a = 4 ): ( f(x) ) has a stationary point at ( x = 2 ), increasing on ( (0, infty) ). - ( a > 4 ): Two critical points ( x_1 ) (local maximum) and ( x_2 ) (local minimum). ( f(x) ) increases on ( (0, x_1) ), decreases on ( (x_1, x_2) ), and increases on ( (x_2, infty) ). (II) Minimum Value of ( lambda )1. Extreme Points: - For ( a > 4 ), roots ( x_1 ) and ( x_2 ) of ( x^2 - ax + a = 0 ) with ( x_1 + x_2 = a ) and ( x_1 x_2 = a ).2. Expression for ( f(x_1) + f(x_2) ): [ f(x_1) + f(x_2) = a ln a - frac{1}{2}a^2 - a ]3. Inequality: [ a ln a - frac{1}{2}a^2 - a < lambda a ] Simplifies to: [ ln a - frac{1}{2}a - 1 < lambda ]4. Finding ( lambda ): - The function ( g(a) = ln a - frac{1}{2}a - 1 ) is decreasing for ( a > 4 ). - The supremum of ( g(a) ) as ( a to 4^+ ) is ( ln 4 - 3 ).Thus, the minimum value of ( lambda ) is (boxed{2ln 2 - 3}).,<think>Okay, let's try to tackle this problem step by step. The function given is ( f(x) = a ln x + frac{1}{2}x^2 - a x ). We need to discuss its monotonicity first, which means figuring out where it's increasing or decreasing. Then, part II asks about finding the minimum value of λ such that ( f(x_1) + f(x_2) < lambda(x_1 + x_2) ) always holds when there are two extreme points ( x_1 ) and ( x_2 ).Starting with part (I): Monotonicity. To find where the function is increasing or decreasing, I need to compute the first derivative ( f'(x) ). Let's do that.The function is composed of three terms: ( a ln x ), ( frac{1}{2}x^2 ), and ( -a x ). The derivative of ( a ln x ) with respect to x is ( frac{a}{x} ). The derivative of ( frac{1}{2}x^2 ) is ( x ). The derivative of ( -a x ) is ( -a ). So putting it all together:( f'(x) = frac{a}{x} + x - a )Now, to discuss monotonicity, we need to find the critical points where ( f'(x) = 0 ) or where the derivative doesn't exist. The derivative ( f'(x) ) exists for ( x > 0 ) because of the ( ln x ) term. So, the domain of f(x) is ( x > 0 ).Setting ( f'(x) = 0 ):( frac{a}{x} + x - a = 0 )Let's multiply both sides by x to eliminate the denominator:( a + x^2 - a x = 0 )Rearranged:( x^2 - a x + a = 0 )Hmm, so the critical points are the solutions to this quadratic equation. The number of real solutions depends on the discriminant. The discriminant D is ( (-a)^2 - 4 times 1 times a = a^2 - 4a ).So, when ( a^2 - 4a > 0 ), which is when ( a(a - 4) > 0 ), meaning ( a < 0 ) or ( a > 4 ), there are two distinct real roots. When ( a = 0 ) or ( a = 4 ), the discriminant is zero, so there's exactly one real root. For ( 0 < a < 4 ), the discriminant is negative, so no real roots, meaning the derivative doesn't cross zero.Wait, but the original function when a = 0 would be ( f(x) = 0 + frac{1}{2}x^2 - 0 = frac{1}{2}x^2 ), whose derivative is x, which is zero only at x=0, but x=0 isn't in the domain. So actually, when a=0, the derivative is x, which is always positive for x > 0, so the function is increasing on its entire domain.Similarly, when a=4, discriminant is zero, so the quadratic equation has one real root. Let me check.So, for the critical points:- If a < 0: The quadratic equation ( x^2 - a x + a = 0 ). If a is negative, then -a is positive, so the equation is ( x^2 + |a|x + a = 0 ). Wait, but a is negative here, so the equation is x^2 - a x + a = x^2 + |a|x + a. Since a is negative, the constant term a is negative. So the quadratic equation with a positive coefficient on x^2, a positive coefficient on x, and a negative constant term. The product of roots is a, which is negative, so one positive and one negative root. But since x > 0, only the positive root is relevant. Wait, but discriminant for a < 0 would be a^2 - 4a. If a is negative, then a^2 is positive and -4a is positive, so discriminant is positive. So two real roots. However, since the product of the roots is a (from x1x2 = c/a = a/1 = a). Since a is negative, the product of roots is negative. Hence, one positive and one negative. Therefore, for a < 0, only one critical point in the domain x > 0.Wait, this seems conflicting with the previous statement. Let me recast.Given quadratic equation ( x^2 - a x + a = 0 ). The product of roots is a (from x1x2 = c). If a is positive, then product of roots is positive, so both roots positive or both negative. The sum of roots is a (from -b/a = a). So if a is positive, sum of roots is positive, product is positive, so both roots positive. If a is negative, sum of roots is a (negative) and product is a (negative), so one positive and one negative root. Therefore, when a is positive, there are two positive roots if discriminant is positive, one if discriminant is zero, none if discriminant is negative. When a is negative, there's one positive root and one negative root.But the domain is x > 0, so for a > 0:- If a > 4: discriminant is positive (since a^2 -4a >0 when a >4 or a <0). So two positive critical points.- If a =4: discriminant zero, one positive critical point.- If 0 < a <4: discriminant negative, no critical points.For a <0: discriminant positive (since a^2 -4a = a(a -4), a negative times (a -4) which is also negative if a <0, so a(a -4) positive). So discriminant positive, but as product of roots is a (negative), so one positive and one negative root. Therefore, only one critical point at x >0.So in summary:- When a <0: one critical point (x1 positive)- When 0 ≤ a ≤4: for a=0, derivative is x, always positive, so no critical points. For 0 <a <4, no critical points. For a=4, one critical point.- When a >4: two critical points.But the problem says in part (II) "if f(x) has two extreme points", so that corresponds to a >4.But let me verify. When a >4, two critical points, both positive. Therefore, for a >4, there are two extreme points x1 and x2. Then, for part (II), the question is under the condition that a is such that there are two extreme points, so a >4, and find the minimum λ such that f(x1) + f(x2) < λ(x1 +x2).So first, part (I): Discuss monotonicity.For different ranges of a:Case 1: a <0Then, derivative f'(x) = a/x +x -a. Since a is negative, let's analyze the behavior.First, as x approaches 0+, a/x approaches negative infinity (since a is negative), x approaches 0, so f'(x) approaches -infty. As x increases, the term a/x becomes less negative, x increases, so f'(x) increases. The critical point occurs at x1 (the positive root). After x1, the derivative becomes positive and continues to increase since x keeps increasing. Wait, but if there's only one critical point, which is where derivative crosses from negative to positive (since as x approaches infinity, f'(x) = a/x +x -a ≈ x -a, which tends to infinity since x dominates). Therefore, for a <0, the function first decreases until x1 and then increases after x1.Case 2: a =0Derivative f'(x) = 0 +x -0 =x. So f'(x) =x, which is always positive for x >0. So the function is increasing on (0, ∞).Case 3: 0 <a <4No critical points. Let's check the derivative's sign. f'(x) = a/x +x -a. Let's see as x approaches 0+, a/x dominates (positive since a >0) so f'(x) approaches +infty. As x increases, the derivative is a/x +x -a. Let's check at x =a. Then f'(a) = a/a +a -a =1 +a -a =1 >0. As x approaches infinity, f'(x) ≈x -a, which tends to infinity. Therefore, derivative is always positive. Therefore, the function is increasing on (0, ∞).Case 4: a =4Derivative f'(x) =4/x +x -4. The critical point is where x^2 -4x +4=0, which is (x-2)^2=0, so x=2. So the derivative is zero at x=2. Let's check around x=2. For x <2, say x=1, f'(1)=4 +1 -4=1>0. For x=3, f'(3)=4/3 +3 -4≈1.33 +3 -4≈0.33>0. Wait, but if the quadratic equation has a double root at x=2, then the derivative touches zero at x=2 but doesn't cross it. Let's check the derivative's sign. For x approaching 2 from the left, the derivative approaches zero from above. For x approaching 2 from the right, derivative approaches zero from above as well. Wait, but let's compute derivative around x=2. Let’s take x=2 - h and x=2 + h for small h>0.At x=2 - h:f'(2 - h) =4/(2 - h) + (2 - h) -4 ≈ [2 + h/2] + [2 - h] -4 = 2 + h/2 +2 -h -4= (-h/2). So as h approaches 0 from positive side, derivative approaches 0 from below? Wait, this contradicts. Wait, maybe my approximation is wrong.Wait, let me compute more accurately.First, expand 4/(2 - h):4/(2 - h) = 4/(2(1 - h/2)) = 2/(1 - h/2) ≈ 2(1 + h/2 + (h/2)^2 + ...) ≈ 2 + h + h^2/2.Then, (2 - h):= 2 - h.So total f'(2 - h) ≈ (2 + h + h^2/2) + (2 - h) -4 = 2 + h + h^2/2 +2 - h -4 = (2 +2 -4) + (h -h) + h^2/2 = 0 + 0 + h^2/2 ≈ h^2/2 >0.Similarly, at x=2 + h:4/(2 + h) = 4/(2(1 + h/2)) = 2/(1 + h/2) ≈ 2(1 - h/2 + (h/2)^2 - ...) ≈ 2 - h + h^2/2.Then, (2 + h) = 2 + h.So f'(2 + h) ≈ (2 - h + h^2/2) + (2 + h) -4 = 2 - h + h^2/2 +2 + h -4 = (2 +2 -4) + (-h +h) + h^2/2 = 0 +0 + h^2/2 ≈ h^2/2 >0.So around x=2, the derivative is positive on both sides. Hence, at a=4, the derivative has a double root at x=2, but the function does not change its monotonicity. Therefore, for a=4, the function is always increasing on (0, ∞), with a stationary point at x=2 where the derivative touches zero but doesn't cross it. So technically, the function is increasing on (0, ∞) even when a=4.Wait, that's interesting. So even though there is a critical point at x=2, the function doesn't change its increasing behavior. Therefore, for a=4, the function is increasing on the entire domain.Wait, but that contradicts with the idea that the derivative is zero at x=2. However, since the derivative is positive on both sides, the function is increasing, but momentarily has a horizontal tangent at x=2. So, in terms of monotonicity, it's increasing throughout, just with a point where the slope is zero. So in that case, the function is non-decreasing? But since the derivative is positive everywhere except at x=2 where it's zero, then it's increasing on (0, ∞). Because to be strictly increasing, the derivative needs to be non-negative and not zero on any interval. Since the derivative is zero only at a single point, it's still considered increasing (some call it non-decreasing, but in calculus, if derivative is non-negative and only zero at isolated points, it's still increasing). So for a=4, the function is increasing on (0, ∞).Case 5: a >4Then, there are two critical points, x1 and x2, both positive. Let's denote them as x1 < x2. The derivative f'(x) is a quadratic in x, opening upwards (coefficient of x^2 is 1 >0). Therefore, the derivative is positive when x <x1, negative between x1 and x2, and positive again when x >x2.Wait, wait, the quadratic is x^2 -a x +a. Since it's opening upwards, the derivative f'(x) = (x^2 -a x +a)/x (if we rearrange the original derivative equation). Wait, actually, the quadratic is x^2 -a x +a. The roots are x1 and x2 with x1 <x2. Since the quadratic is positive outside the roots and negative between them. Therefore, the derivative f'(x) = (x^2 -a x +a)/x. Therefore, since x>0, the sign of f'(x) is the same as the sign of x^2 -a x +a. Therefore, when x <x1 or x >x2, x^2 -a x +a >0, so f'(x) >0. When x1 <x <x2, x^2 -a x +a <0, so f'(x) <0. Therefore, the function f(x) is increasing on (0, x1), decreasing on (x1, x2), and increasing again on (x2, ∞).Therefore, summarizing part (I):- If a ≤4: f(x) is increasing on (0, ∞).Wait, but when a <0, as we saw earlier, there's a critical point, so the function first decreases then increases. Wait, hold on, there's inconsistency here. Earlier analysis for a <0 said that for a <0, the quadratic equation x^2 -a x +a =0 has one positive root and one negative root, so in the domain x>0, only x1 (positive root). Then, derivative goes from -infty (as x approaches 0+) to crossing zero at x1 and then tending to +infty as x approaches infinity. Therefore, the function is decreasing on (0, x1) and increasing on (x1, ∞). But in the previous analysis for a >4, the function has two critical points and is increasing, decreasing, increasing.Therefore, part (I) answer should be:- For a <0: f(x) is decreasing on (0, x1) and increasing on (x1, ∞), where x1 is the positive critical point.- For 0 ≤a ≤4: f(x) is increasing on (0, ∞).- For a >4: f(x) is increasing on (0, x1), decreasing on (x1, x2), and increasing on (x2, ∞), where x1 and x2 are the critical points with 0 <x1 <x2.Therefore, that's the monotonicity discussion for part (I).Now moving on to part (II). If f(x) has two extreme points, meaning a >4, as established. Then, we need to find the minimum λ such that f(x1) +f(x2) < λ(x1 +x2) always holds, where x1 and x2 are the two critical points.First, let's note that x1 and x2 are roots of the quadratic equation x^2 -a x +a =0. So, from quadratic formula:x1, x2 = [a ± sqrt(a^2 -4a)]/2Also, since x1 and x2 are roots, by Vieta's formula:x1 +x2 = ax1 x2 = aWait, let's check that. The quadratic equation is x^2 -a x +a =0. So sum of roots is a, product is a. Correct.Therefore, x1 +x2 =a, and x1 x2 =a.So, we need to compute f(x1) +f(x2) and relate it to λ(x1 +x2) = λ a. So, the inequality is f(x1) +f(x2) < λ a, and we need to find the minimum λ such that this holds for all a >4.So, first, let's compute f(x1) +f(x2).Given f(x) =a ln x + (1/2)x^2 -a x.Therefore, f(x1) +f(x2) =a (ln x1 + ln x2) + (1/2)(x1^2 +x2^2) -a (x1 +x2)Simplify:First, ln x1 + ln x2 = ln(x1 x2) = ln(a) (since x1 x2 =a)Second, x1^2 +x2^2 = (x1 +x2)^2 - 2x1 x2 = a^2 - 2aThird, x1 +x2 =aTherefore:f(x1) +f(x2) =a ln a + (1/2)(a^2 - 2a) -a *aSimplify term by term:a ln a + (1/2)a^2 -a -a^2Combine like terms:(1/2 a^2 -a^2) + a ln a -a= (-1/2 a^2) +a ln a -aThus, f(x1) +f(x2) = - (1/2)a^2 +a ln a -aSo the inequality becomes:- (1/2)a^2 +a ln a -a < λ aDivide both sides by a (since a >4, a is positive, so inequality sign remains):- (1/2)a + ln a -1 < λSo we have:λ > - (1/2)a + ln a -1Therefore, the inequality must hold for all a >4. Thus, the minimum value of λ is the supremum (least upper bound) of the expression - (1/2)a + ln a -1 for a >4. Hence, λ must be greater than the maximum value of the right-hand side. Wait, but the expression is - (1/2)a + ln a -1. As a increases, the term - (1/2)a dominates, so the expression tends to -infty as a approaches infinity. However, we need to check if the expression has a maximum over a >4. Let's find the critical points.Let’s define g(a) = - (1/2)a + ln a -1. Find its derivative:g’(a) = -1/2 + 1/aSet derivative equal to zero:-1/2 +1/a =0 → 1/a =1/2 → a=2But a >4, so the critical point at a=2 is outside our domain. Therefore, on the interval a >4, the derivative g’(a) = -1/2 +1/a. For a >4, 1/a <1/4, so -1/2 +1/a < -1/2 +1/4 = -1/4 <0. Therefore, g(a) is decreasing for all a >4. Therefore, the maximum of g(a) on a >4 occurs at a=4.Therefore, the supremum of g(a) is g(4). Compute g(4):g(4) = - (1/2)(4) + ln 4 -1 = -2 + ln 4 -1 = -3 + ln 4 ≈ -3 +1.386 ≈-1.614But wait, if g(a) is decreasing for a >4, then as a increases beyond 4, g(a) decreases. Therefore, the maximum value of g(a) for a >4 is at a approaching 4 from the right. However, at a=4, the function f(x) has only one critical point (double root), but the problem states "if f(x) has two extreme points", so strictly a >4. Therefore, the supremum of g(a) is the limit as a approaches 4 from the right.Compute lim_{a→4+} g(a):lim_{a→4} [ - (1/2)a + ln a -1 ] = - (1/2)(4) + ln4 -1 = -2 + ln4 -1 = -3 + ln4 ≈ -3 +1.386≈ -1.614However, since a must be greater than 4, the expression g(a) approaches -1.614 but is less than that for a >4. Wait, but since g(a) is decreasing, the maximum value is achieved as a approaches 4 from the right. Therefore, the supremum is -3 + ln4, which is approximately -1.614. Therefore, λ must be greater than this value. But since the problem states "the inequality f(x1) +f(x2) < λ(x1 +x2) always holds", and we need the minimum λ such that this is true for all a >4. Since for all a >4, - (1/2)a + ln a -1 < λ, and the supremum of the left-hand side is approaching -3 + ln4 as a approaches 4 from the right. But since a cannot be 4 (as that gives only one extreme point), the expression g(a) is always less than -3 + ln4 for a >4. Wait, but let's check when a approaches 4 from above.Suppose a=4.1:g(4.1)= -0.5*4.1 + ln4.1 -1 ≈ -2.05 +1.411 -1≈-2.05 +0.411≈-1.639Which is less than -3 + ln4 ≈ -1.614. Wait, no, -1.639 is less than -1.614. Wait, that's contradictory. Wait, wait, ln4≈1.386, so -3 +1.386≈-1.614. So when a approaches 4 from above, the limit is -1.614. But at a=4.1, g(a)≈-1.639, which is lower. So the function g(a) is decreasing, so its maximum on a >4 is indeed at a approaches 4+, which is -3 +ln4≈-1.614. But for any a >4, g(a) is less than that. Wait, but as a approaches 4 from above, g(a) approaches -3 + ln4, so this is the supremum. However, when a=5, for example:g(5)= -2.5 + ln5 -1≈-2.5 +1.609 -1≈-2.5 +0.609≈-1.891, which is less than -1.614.Wait, but why is the limit as a approaches4+ equal to -3 + ln4, but when a=4, the expression is:g(4) = -2 + ln4 -1 = -3 + ln4≈-1.614But at a=4, technically the function f(x) has only one critical point. But in part (II), we are to consider when f(x) has two extreme points, which is when a>4, so approaching a=4 from above, the expression approaches -3 + ln4. But for any a>4, the expression g(a) is less than -3 + ln4. Wait, no. Let me check with a=4.0001:g(4.0001)= -0.5*4.0001 + ln(4.0001) -1≈-2.00005 +1.3863 -1≈-2.00005 +0.3863≈-1.61375Which is very close to -1.614. So as a approaches 4 from above, g(a) approaches -3 + ln4≈-1.614. Therefore, the supremum (least upper bound) of g(a) over a >4 is -3 + ln4. Therefore, in order for the inequality to hold for all a >4, λ must be greater than or equal to this supremum. Since the inequality is strict (f(x1)+f(x2) < λ(x1+x2)), λ can be as small as this supremum, but since the supremum is approached but not achieved, λ can be set as -3 + ln4, because for all a >4, g(a) < -3 + ln4. Wait, but wait, when approaching a=4 from the right, g(a) approaches -3 + ln4. So for any ε>0, there exists a >4 such that g(a) > -3 + ln4 -ε. Therefore, in order for the inequality to hold for all a >4, λ must be at least -3 + ln4, as otherwise, for λ slightly less than that, there would be a values of a >4 where g(a) >λ.Therefore, the minimal λ is -3 + ln4. However, we can write ln4 as 2 ln2, so alternatively, it's -3 + 2 ln2.But let me verify this conclusion.Given that:We have f(x1) + f(x2) = - (1/2)a^2 + a ln a -aThe inequality is:- (1/2)a^2 + a ln a -a < λ aDivide both sides by a (a >4):- (1/2)a + ln a -1 < λTherefore, λ must be greater than the supremum of the left-hand side over a >4.We found that as a approaches 4 from above, the left-hand side approaches - (1/2)*4 + ln4 -1 = -2 + ln4 -1 = -3 + ln4And since for a >4, the left-hand side is decreasing, so the supremum is indeed -3 + ln4.Therefore, the minimal λ is -3 + ln4. To check if this is correct, let's compute:Compute -3 + ln4:ln4≈1.386-3 +1.386≈-1.614So λ must be greater than approximately -1.614. But since the problem asks for the minimum value of λ, it's exactly equal to the supremum, which is -3 + ln4.But maybe the problem expects an exact form. Since ln4 can be written as 2 ln2, so -3 +2 ln2. Let's verify with an example. Take a approaching 4 from above, say a=4+ε, ε→0+. Then:f(x1) +f(x2) = - (1/2)a^2 +a ln a -aAt a=4, this would be - (1/2)(16) +4 ln4 -4= -8 +4 ln4 -4= -12 +4 ln4=4(ln4 -3). Wait, but we had earlier f(x1)+f(x2) = - (1/2)a^2 +a ln a -a. So when a approaches4 from above, it's approaching - (1/2)(16) +4 ln4 -4= -8 +4 ln4 -4= -12 +4 ln4=4(ln4 -3). Then the inequality f(x1)+f(x2) < λ a becomes 4(ln4 -3) <4λ (since a=4), so λ>ln4 -3≈1.386 -3≈-1.614, which matches our previous result. However, when a approaches4 from above, the expression (f(x1)+f(x2))/a approaches (4(ln4 -3))/4=ln4 -3≈-1.614. Therefore, for the inequality to hold for all a>4, λ must be at least ln4 -3. Therefore, the minimal λ is ln4 -3.Alternatively, expressed as 2 ln2 -3. But ln4=2 ln2, so yes. However, maybe they prefer it in terms of ln2. Let me check if 2 ln2 -3 is the same as ln4 -3. Yes. Alternatively, some might write it as -3 + ln4. Both are correct. But to present the answer neatly, perhaps factor it as -3 + 2 ln2 or 2 ln2 -3. Both are acceptable. Let me check if this is the case.Alternatively, compute the exact value. Let me check with a=4. Then, even though a=4 is not allowed (since there's only one critical point), the limit as a approaches4 from above is λ=ln4 -3≈-1.6137. Since λ is supposed to be a real number, and the problem asks for the minimal λ such that the inequality holds for all a>4. Since the supremum is ln4 -3, λ must be at least this value, so the minimum λ is ln4 -3. Therefore, the answer is boxed{2 ln 2 - 3}But to check, let's verify with another a. Take a=5, which is greater than4.Compute f(x1)+f(x2)= - (1/2)(25) +5 ln5 -5= -12.5 +5*1.609 -5≈-12.5 +8.045 -5≈-19.5 +8.045≈-11.455Then, λ(x1 +x2)=λ*5. The inequality is -11.455 <5λ → λ> -11.455/5≈-2.291. Since our found λ is≈-1.614, which is greater than -2.291. So for a=5, any λ>≈-1.614 will satisfy the inequality. But as a increases, the required λ can be smaller (since the left-hand side becomes more negative). However, the minimal λ is determined by the a approaching4 from above, where the left-hand side is closest to λ. Therefore, regardless of larger a requiring smaller λ, the minimal λ is set by the maximum of the left-hand side, which occurs near a=4. Therefore, the answer is indeed 2 ln2 -3.Let me compute 2 ln2 -3:2*0.6931≈1.3862, 1.3862 -3≈-1.6138, which matches our approximation.Therefore, the minimal λ is 2 ln2 -3. So boxed, it's boxed{2 ln 2 - 3}Final AnswerThe minimum value of (lambda) is boxed{2 ln 2 - 3}.</think>Given the function ( f(x) = a ln x + frac{1}{2}x^2 - ax ):(I) Monotonicity of ( f(x) ):1. First Derivative: ( f'(x) = frac{a}{x} + x - a ).2. Critical Points: Solving ( f'(x) = 0 ) gives the quadratic equation ( x^2 - ax + a = 0 ). The discriminant is ( a^2 - 4a ). - For ( a < 0 ): One positive critical point. ( f(x) ) decreases on ( (0, x_1) ) and increases on ( (x_1, infty) ). - For ( 0 leq a leq 4 ): No critical points (except ( a = 4 ) with a single critical point where the function is still increasing). ( f(x) ) is increasing on ( (0, infty) ). - For ( a > 4 ): Two positive critical points ( x_1 ) and ( x_2 ). ( f(x) ) increases on ( (0, x_1) ), decreases on ( (x_1, x_2) ), and increases again on ( (x_2, infty) ).(II) Minimum value of ( lambda ):1. Extreme Points: For ( a > 4 ), the critical points ( x_1 ) and ( x_2 ) are roots of ( x^2 - ax + a = 0 ). By Vieta's formulas, ( x_1 + x_2 = a ) and ( x_1 x_2 = a ).2. Expression for ( f(x_1) + f(x_2) ): [ f(x_1) + f(x_2) = a ln a - frac{1}{2}a^2 - a ]3. Inequality: We need ( f(x_1) + f(x_2) < lambda (x_1 + x_2) ), which simplifies to: [ - frac{1}{2}a + ln a - 1 < lambda ]4. Supremum Analysis: The left-hand side function ( g(a) = - frac{1}{2}a + ln a - 1 ) is decreasing for ( a > 4 ). The supremum as ( a ) approaches 4 from the right is ( -3 + ln 4 ).Thus, the minimum value of ( lambda ) is (boxed{2 ln 2 - 3}).]

🔑:Solution:- (I) f'(x)= frac {a}{x}+x-a= frac {x^2-ax+a}{x} (x > 0), - When a 4, Delta =a^{2}-4a > 0, solving f'(x)=0 gives x_{1,2}= frac {apm sqrt {a^{2}-4a}}{2}, - The increasing intervals of f(x) are (0, frac {a- sqrt {a^{2}-4a}}{2}) and ( frac {a+ sqrt {a^{2}-4a}}{2},+infty), - The decreasing interval is ( frac {a- sqrt {a^{2}-4a}}{2}, frac {a+ sqrt {a^{2}-4a}}{2}).- (II) From (I), we know that f(x) has two extreme points when a > 4, and x_{1}+x_{2}=a, x_{1}x_{2}=a, - Thus, f(x_{1})+f(x_{2})=aln x_{1}+ frac {1}{2}x_{1}^{2}-ax_{1}+aln x_{2}+ frac {1}{2}x_{2}^{2}-ax_{2}=aln (x_{1}x_{2})+ frac {1}{2}(x_{1}^{2}+x_{2}^{2})-a(x_{1}+x_{2}) - =a(ln a- frac {1}{2}a-1) - Hence, frac {f(x_{1})+f(x_{2})}{x_{1}+x_{2}}=ln a- frac {1}{2}a-1, where ain(4,+infty). - Let varphi(a)=ln a- frac {1}{2}a-1, then varphi'(a)= frac {1}{a}- frac {1}{2}. - Since a > 4, varphi'(a) 0, the inequality f(x_{1})+f(x_{2}) < lambda(x_{1}+x_{2}) is equivalent to frac {f(x_{1})+f(x_{2})}{x_{1}+x_{2}} < lambda, - Therefore, the minimum value of lambda is boxed{ln 4-3}.

🧠:[<think>Okay, let's tackle these problems one by one. First, problem (1). So, the taxi starts at the company and picks up and drops off 5 batches of passengers. Each time, they move a certain distance, either positive (south) or negative (north). I need to find out where the driver ends up relative to the company after the fifth batch.Let me write down the distances for each batch:1. 5 km (south)2. 2 km (south)3. -4 km (north)4. -3 km (north)5. 6 km (south)To find the final position, I should add all these distances together. Let's do that step by step.Starting at 0 (the company).After batch 1: 0 + 5 = 5 km south.After batch 2: 5 + 2 = 7 km south.After batch 3: 7 + (-4) = 3 km south.After batch 4: 3 + (-3) = 0 km. Back to the company?Wait, that can't be right. Wait, after dropping off the fourth batch, the driver is at 0 km, which is the company. Then the fifth batch is 6 km south. So after the fifth batch, the driver is 6 km south of the company.Wait, but let me check the addition again:5 + 2 + (-4) + (-3) + 6.Let me compute this:5 + 2 = 77 + (-4) = 33 + (-3) = 00 + 6 = 6.Yes, that's correct. So the total displacement from the company is 6 km south. So direction is south, distance is 6 km.Okay, problem (1) seems straightforward. Just adding all the distances with their signs. Answer: 6 km south.Now problem (2). Fuel consumption. The taxi uses 0.3 liters per kilometer. Total fuel consumed is based on the total distance traveled, regardless of direction. So I need to sum the absolute values of each batch's distance and multiply by 0.3 liters/km.Let me compute the total distance:Batch 1: 5 kmBatch 2: 2 kmBatch 3: |-4| = 4 kmBatch 4: |-3| = 3 kmBatch 5: 6 kmTotal distance = 5 + 2 + 4 + 3 + 6 = let's add them up:5 + 2 = 77 + 4 = 1111 + 3 = 1414 + 6 = 20 km total.So total fuel consumed is 20 km * 0.3 liters/km = 6 liters.Wait, that's straightforward. So answer is 6 liters.Problem (3). Fare calculation. The fare is 8 yuan for up to 3 km, and additional 1.6 yuan per km beyond 3 km. So for each batch, we need to compute the fare based on the distance of that batch.Wait, the problem says "batches of passengers". So each batch is a separate trip? So each distance given is the distance for that individual trip? So each batch is a separate fare calculation.So for each of the 5 batches, compute the fare based on the distance of that batch, then sum all fares.Let me check. The problem states: "continuously picks up and drops off 5 batches of passengers". Hmm, does "continuously" mean that these are sequential trips, so each batch is a separate trip where the taxi picks up and drops off, then picks up the next batch. So each distance is the distance from the previous drop-off to the next drop-off? Wait, but if the taxi is moving from one drop-off to the next pickup, that would be considered as the next trip. However, the problem doesn't mention if the taxi is returning to the company after each batch. Wait, the problem says "picks up and drops off 5 batches of passengers on the north-south Beijing road". It might mean that each batch is a separate fare, starting from where the previous one was dropped off. So the distances given are the net movement for each batch. Therefore, each distance is the trip distance for that batch. So each batch is a separate fare.Therefore, for each batch (1 to 5), we compute the fare based on the distance of that batch, then sum all the fares.Therefore, for each distance, if the distance is <=3 km, the fare is 8 yuan. If the distance is >3 km, the fare is 8 + 1.6*(distance -3).But wait, the distance is in km. But the fare is for the trip. However, the problem says "distance of up to 3 kilometers" is 8 yuan. So for each trip (each batch), if the distance of that trip is <=3 km, charge 8 yuan. If more than 3 km, charge 8 + 1.6*(distance -3). So we need to compute each batch's fare and sum them up.But note that the distance given for each batch is the net distance from the previous location. So for example, batch 1 is 5 km. So the taxi takes the passenger 5 km south, which is a 5 km trip, so fare is 8 + 1.6*(5-3) = 8 + 3.2 = 11.2 yuan.Similarly for each batch. Let's compute each batch's fare:Batch 1: 5 km. Since 5 >3, fare = 8 + 1.6*(5-3) = 8 + 3.2 = 11.2 yuan.Batch 2: 2 km. 2 <=3, so fare is 8 yuan.Batch 3: -4 km. The fare is based on the absolute distance, right? Because distance can't be negative. So even though the taxi went north 4 km, the passenger is charged based on the 4 km trip. So fare is 8 +1.6*(4-3)=8 +1.6=9.6 yuan.Batch 4: -3 km. Similarly, absolute distance is 3 km. So fare is 8 yuan.Batch 5: 6 km. Fare is 8 +1.6*(6-3)=8 +4.8=12.8 yuan.Now sum all these fares:11.2 +8 +9.6 +8 +12.8.Let's compute step by step:11.2 +8 =19.219.2 +9.6=28.828.8 +8=36.836.8 +12.8=49.6 yuan.So total money received is 49.6 yuan.Wait, but let me double-check each fare:Batch 1: 5 km: 8 +1.6*(5-3)=8 +3.2=11.2. Correct.Batch 2: 2 km: 8 yuan. Correct.Batch 3: 4 km (absolute value): 8 +1.6*(4-3)=9.6. Correct.Batch 4: 3 km: 8 yuan. Correct.Batch 5: 6 km: 8 +1.6*(6-3)=8 +4.8=12.8. Correct.Sum: 11.2+8=19.2; 19.2+9.6=28.8; 28.8+8=36.8; 36.8+12.8=49.6. Yes.So the answer for part (3) is 49.6 yuan.But wait, the problem says "how much money does the driver receive in total during this process?" So unless there's something missing, like waiting time or other charges, but based on the problem statement, the fare is calculated purely on distance as described. So 49.6 yuan is correct.So summarizing:1. 6 km south.2. 6 liters.3. 49.6 yuan.Final Answer(1) The driver is boxed{6} km south of the company.(2) The total fuel consumed is boxed{6} liters.(3) The total money received is boxed{49.6} yuan.</think>To solve the given problems, we follow these steps:1. Determining the final position relative to the company: - We sum the distances traveled for each batch, considering the direction (south as positive, north as negative). - Calculation: (5 + 2 - 4 - 3 + 6 = 6) - Result: The driver is 6 km south of the company.2. Calculating the total fuel consumed: - We sum the absolute values of the distances traveled for each batch. - Total distance: (5 + 2 + 4 + 3 + 6 = 20) km - Fuel consumption: (20 text{ km} times 0.3 text{ liters/km} = 6) liters3. Calculating the total fare received: - For each batch, the fare is calculated as follows: - 8 yuan for distances up to 3 km. - 8 yuan plus 1.6 yuan per km for distances exceeding 3 km. - Batch 1 (5 km): (8 + 1.6 times (5 - 3) = 11.2) yuan - Batch 2 (2 km): (8) yuan - Batch 3 (4 km): (8 + 1.6 times (4 - 3) = 9.6) yuan - Batch 4 (3 km): (8) yuan - Batch 5 (6 km): (8 + 1.6 times (6 - 3) = 12.8) yuan - Total fare: (11.2 + 8 + 9.6 + 8 + 12.8 = 49.6) yuan Final Answers(1) The driver is boxed{6} km south of the company.(2) The total fuel consumed is boxed{6} liters.(3) The total money received is boxed{49.6} yuan.,<think>Alright, let's tackle this problem step by step. There are three parts to it, so I'll go through each one methodically. Starting with part (1): We need to find out where the driver is relative to the company after dropping off the fifth batch of passengers. The directions are given with south as positive and north as negative. The distances for each batch are 5, 2, -4, -3, and 6 km. Okay, so the total displacement from the company will just be the sum of all these distances. Let me add them up. First batch: +5 km (south)Second batch: +2 km (more south)Third batch: -4 km (north)Fourth batch: -3 km (north)Fifth batch: +6 km (south)Adding these together: 5 + 2 = 7; then 7 - 4 = 3; 3 - 3 = 0; 0 + 6 = 6. So total displacement is +6 km. That means the driver is 6 km south of the company. That seems straightforward.Wait, but displacement is the straight-line distance from the starting point, right? So even though the driver might have gone back and forth, the final position is just the sum of all movements. So yes, 6 km south. Got it.Moving on to part (2): Fuel consumption. The taxi uses 0.3 liters per kilometer. But here, we need the total distance traveled, not just the displacement. Because fuel is consumed for every kilometer driven, regardless of direction. So we need to calculate the total distance for each batch and sum them up, then multiply by 0.3 liters/km.Let's compute each leg:1st batch: 5 km (south)2nd batch: 2 km (south)3rd batch: 4 km (north)4th batch: 3 km (north)5th batch: 6 km (south)Adding all these distances: 5 + 2 + 4 + 3 + 6. Let's calculate that. 5+2=7; 7+4=11; 11+3=14; 14+6=20. Total distance traveled is 20 km. Therefore, fuel consumed is 20 km * 0.3 L/km = 6 liters. That seems right. But wait, let me double-check the additions. 5, 2, 4, 3, 6. 5+2=7, plus 4=11, plus3=14, plus6=20. Yep. 20 total km. 20*0.3=6 liters. Correct.Now part (3): Calculating the total fare. The fare structure is 8 yuan for up to 3 km, and an extra 1.6 yuan per km for distances exceeding 3 km. So for each batch, we need to compute the fare. The problem mentions "batches of passengers," so each batch is a separate trip? Wait, actually, the problem says "continuously picks up and drops off 5 batches of passengers." Hmm. Does that mean each batch is a separate fare, or is it one continuous trip with multiple stops? But since they are referred to as batches, and each has a distance, I think each batch is a separate fare. So each batch is a separate passenger group, and the fare is calculated per batch. So we need to compute the fare for each batch's distance and sum them up.Wait, but the fare standard is for the taxi, so maybe each batch is a separate ride. So each of the 5 batches is a separate ride, each with their own distance. So for each distance, compute the fare as per the standard: 8 yuan for up to 3 km, plus 1.6 yuan for each km beyond 3 km.So for each batch:1. Batch 1: 5 km. The first 3 km are 8 yuan. The remaining 2 km (5-3=2) are charged at 1.6 yuan/km. So 8 + 2*1.6 = 8 + 3.2 = 11.2 yuan.2. Batch 2: 2 km. This is within 3 km, so just 8 yuan.3. Batch 3: -4 km. Wait, distance can't be negative. The direction is just for displacement, but the actual distance traveled for the fare would be the absolute value. So even though it's -4 km (north), the fare is based on 4 km.So Batch 3: 4 km. 8 yuan for first 3 km, then 1 km over, so 8 + 1*1.6 = 9.6 yuan.4. Batch 4: -3 km. Absolute distance is 3 km. So exactly 3 km, so 8 yuan.5. Batch 5: 6 km. 8 yuan for first 3 km, then 3 km over. 3*1.6 = 4.8. Total 8 + 4.8 = 12.8 yuan.Now sum all these up:Batch 1: 11.2Batch 2: 8Batch 3: 9.6Batch 4: 8Batch 5: 12.8Adding them together:11.2 + 8 = 19.219.2 + 9.6 = 28.828.8 + 8 = 36.836.8 + 12.8 = 49.6 yuan total.But let me verify each calculation again:1. 5 km: 3 km @8, 2 km @1.6: 8 + 3.2=11.2. Correct.2. 2 km: 8 yuan. Correct.3. 4 km: 8 +1.6=9.6. Correct.4. 3 km: 8 yuan. Correct.5. 6 km: 3 km @8, 3 km @1.6: 8 +4.8=12.8. Correct.Sum: 11.2 +8=19.2; 19.2+9.6=28.8; 28.8+8=36.8; 36.8+12.8=49.6. Yes, total 49.6 yuan.Therefore, the answers are:1) 6 km south.2) 6 liters.3) 49.6 yuan.Wait, but I should check if the fare is calculated per trip or if there's any waiting time or other charges. The problem states the fare standard is based solely on distance: 8 yuan up to 3 km, 1.6 yuan per km beyond. So no extra charges mentioned. So as per given info, my calculation is correct.Also, for part (2), total distance is 5+2+4+3+6=20 km. 20*0.3=6 liters. Correct.And part (1), displacement is 5+2-4-3+6=6 km. Correct.So all three parts are sorted.Final Answer(1) The driver is boxed{6} km south of the company.(2) The total fuel consumed is boxed{6} liters.(3) The total fare received is boxed{49.6} yuan.</think>To determine the driver's position after dropping off the fifth batch of passengers, we sum the distances traveled in each batch, considering south as positive and north as negative:[5 + 2 - 4 - 3 + 6 = 6 text{ km (south)}]For fuel consumption, we calculate the total distance traveled regardless of direction:[5 + 2 + 4 + 3 + 6 = 20 text{ km}]Fuel consumed is then:[20 text{ km} times 0.3 text{ liters/km} = 6 text{ liters}]For the total fare, we compute the fare for each batch using the given fare standard (8 yuan for up to 3 km, plus 1.6 yuan per km for distances exceeding 3 km):- Batch 1 (5 km): (8 + 1.6 times 2 = 11.2) yuan- Batch 2 (2 km): (8) yuan- Batch 3 (4 km): (8 + 1.6 times 1 = 9.6) yuan- Batch 4 (3 km): (8) yuan- Batch 5 (6 km): (8 + 1.6 times 3 = 12.8) yuanSumming these fares:[11.2 + 8 + 9.6 + 8 + 12.8 = 49.6 text{ yuan}]Final answers:(1) The driver is boxed{6} km south of the company.(2) The total fuel consumed is boxed{6} liters.(3) The total fare received is boxed{49.6} yuan.]

🔑: Solution:# Part 1: Direction and Distance from the CompanyTo find the final position of the taxi relative to the company, we sum up all the distances traveled:[5 + 2 + (-4) + (-3) + 6 = 6 , text{km}]This means the taxi is 6 km to the south of the company.Final answer for part (1): boxed{text{South, 6 km}}# Part 2: Total Fuel ConsumptionTo calculate the total fuel consumption, we first find the absolute value of each distance to account for the total distance traveled, regardless of direction:[|5| + |2| + |-4| + |-3| + |6| = 5 + 2 + 4 + 3 + 6 = 20 , text{km}]Then, we multiply the total distance by the fuel consumption rate:[0.3 times 20 = 6 , text{liters}]Final answer for part (2): boxed{6 , text{liters}}# Part 3: Total Fare ReceivedTo calculate the total fare received, we calculate the fare for each batch of passengers:- For the first batch: 8 + 1.6 times (5 - 3) = 11.2 yuan- For the second batch: 8 yuan (since it's within 3 km)- For the third batch: 8 + 1.6 times (4 - 3) = 9.6 yuan- For the fourth batch: 8 yuan (since it's within 3 km)- For the fifth batch: 8 + 1.6 times (6 - 3) = 12.8 yuanAdding these fares together:[11.2 + 8 + 9.6 + 8 + 12.8 = 49.6 , text{yuan}]Final answer for part (3): boxed{49.6 , text{yuan}}

🧠:[<think>Okay, so I need to prove that for a positive odd integer n, the sum of the floor of log base 2 of n/k, where k ranges over the odd numbers from 1 to n, equals (n-1)/2. Hmm, let me start by understanding the problem better.First, let me restate it. The left-hand side (LHS) is a sum of terms like floor(log₂(n/k)), where k is each odd integer from 1 up to n. Since n is odd, all the divisors k here are odd. The right-hand side (RHS) is (n-1)/2, which is an integer because n is odd, so n-1 is even. So, I need to show that when you take all these floor values of log₂(n/k) and add them up, you get (n-1)/2.Maybe I should test this with a small odd integer first. Let's try n=1. Then the sum is just floor(log₂(1/1)) = floor(0) = 0. The RHS is (1-1)/2 = 0. So it works for n=1.Next, n=3. The sum is floor(log₂(3/1)) + floor(log₂(3/3)). Let's compute each term:floor(log₂(3)) = floor(1.58496) = 1floor(log₂(1)) = floor(0) = 0So the sum is 1 + 0 = 1. The RHS is (3-1)/2 = 1. That works.n=5. The sum is floor(log₂5) + floor(log₂(5/3)) + floor(log₂1). Let's calculate:floor(log₂5) = 2 since 2²=4 <5 <8=2³floor(log₂(5/3)) = floor(log₂(1.666...)) = floor(0.73696) = 0floor(log₂1)=0So sum is 2 + 0 + 0 = 2. RHS is (5-1)/2 = 2. Good.n=7. Let's check:Terms are floor(log₂7) + floor(log₂7/3) + floor(log₂7/5) + floor(log₂7/7).Compute each:floor(log₂7) = 2 (since 2²=4 <7 <8=2³)floor(7/3)=2.333..., log₂(2.333)≈1.22, floor is 1floor(7/5)=1.4, log₂1.4≈0.485, floor is 0floor(log₂1)=0So sum is 2 +1 +0 +0=3. RHS is (7-1)/2=3. Correct.n=9: Let's see.Sum is floor(log₂9) + floor(log₂9/3) + floor(log₂9/5) + floor(log₂9/7) + floor(log₂9/9)Compute each:floor(log₂9)=3 (since 2³=8 <9 <16=2⁴)9/3=3, log₂3≈1.584, floor=19/5=1.8, log₂1.8≈0.847, floor=09/7≈1.285, log₂1.285≈0.36, floor=0floor(log₂1)=0Sum: 3+1+0+0+0=4. RHS is (9-1)/2=4. Perfect.So small cases check out. Now, how to generalize?First, note that k ranges over the odd divisors of n? Wait, no, the problem says "Let n be a positive odd integer," and the sum is over terms where k is from 1 to n, but only odds? Wait, the problem statement says:floor(log₂(n)) + floor(log₂(n/3) ) + floor(log₂(n/5)) + ... + floor(log₂(n/n))So the denominators are 1, 3, 5, ..., n. Since n is odd, all the denominators k are odd numbers from 1 to n. So the sum is over all odd integers k from 1 to n, each term being floor(log₂(n/k)).We need to sum these terms and show that the total is (n-1)/2.Let me think about properties of floor functions and logarithms. The floor of log₂(n/k) is the exponent of the highest power of 2 less than or equal to n/k. So if we write n/k as m, then floor(log₂m) is the largest integer t such that 2^t ≤ m <2^{t+1}.Alternatively, this is equivalent to the number of times you can divide m by 2 before it's less than 1. But not exactly, since it's just the exponent. Wait, perhaps another approach.Alternatively, note that floor(log₂(n/k)) = the highest integer t where 2^t ≤ n/k. So 2^t ≤ n/k <2^{t+1}, which can be rearranged as n/(2^{t+1}) <k ≤n/(2^t).Thus, for each t, the number of k's such that floor(log₂(n/k)) = t is equal to the number of odd integers k in (n/(2^{t+1}), n/(2^t)].So if I can count the number of odd k's in each interval (n/(2^{t+1}), n/(2^t)] and multiply by t, then sum over t, that should give the total sum. Then maybe this total sum equals (n-1)/2.Alternatively, maybe there's a bijection or a combinatorial interpretation here. Since the result is (n-1)/2, which is reminiscent of the number of pairs or something similar.Alternatively, maybe considering binary representations. Since n is odd, its binary representation doesn't end with a 0. The terms floor(log₂(n/k)) relate to the exponents in the binary expansions.Wait, another idea: if we pair each term k with n/k. But since n is odd and k is odd, n/k is not necessarily integer. Hmm, maybe not helpful.Wait, perhaps another approach. Let's consider all the terms floor(log₂(n/k)) for k odd from 1 to n. For each odd k, n/k can be written as some real number, and floor(log₂(n/k)) is the exponent t where 2^t ≤ n/k <2^{t+1}.Alternatively, for each k, let's write n/k = 2^{t} + f, where 0 ≤f <1. Then t is floor(log₂(n/k)). But I'm not sure if that helps.Wait, maybe think in terms of binary expansions. Since n is odd, when we divide n by an odd k, n/k is a rational number with denominator dividing k. But again, not sure.Alternatively, think of each term floor(log₂(n/k)) as the number of times you can divide n/k by 2 before it's less than 1. Which is equivalent to the number of times you can multiply k by 2 before it exceeds n. Wait, that might be a useful interpretation.For example, floor(log₂(n/k)) is the largest integer t such that 2^t ≤ n/k. So 2^t ≤ n/k => k ≤n/2^t. So t is the maximum number such that k ≤n/2^t. So for each k, t is the maximum number where k ≤n/2^t. Alternatively, t is the number of times you can double k before it exceeds n. So if you start with k and keep doubling it, t is the number of times you can do this without exceeding n.Therefore, for each odd k, the term floor(log₂(n/k)) counts how many times you can double k (i.e., multiply by 2) before exceeding n. So, for example, if k=1, you can double it log₂n times, but since n is odd, log₂n is not integer. Wait, but floor(log₂n) is the exponent of the highest power of 2 less than n. For k=1, the number of doublings before exceeding n is floor(log₂n). Similarly, for k=3, you can double it floor(log₂(n/3)) times, and so on.Alternatively, if we think of all the numbers of the form k*2^t ≤n, where k is odd and t ≥0, then for each such number, the exponent t is counted in the term floor(log₂(n/k)) for that particular k. But how does this relate to the total sum?Wait, the total sum over all odd k of floor(log₂(n/k)) is equal to the total number of pairs (k, t) where k is odd, 1 ≤k ≤n, and t is a non-negative integer such that k*2^t ≤n.Because for each k, the number of t's is exactly floor(log₂(n/k)), since t can be 0,1,...,floor(log₂(n/k)). So the total sum counts the number of such pairs (k, t).Therefore, the original sum is equal to the number of pairs (k, t) with k odd, 1 ≤k ≤n, and k*2^t ≤n.So, perhaps we can interpret this count in another way. Let's consider all numbers m =k*2^t where k is odd, 1 ≤k ≤n, t ≥0, and m ≤n. Each such m can be uniquely written as k*2^t, where k is odd. Therefore, the number of such pairs (k, t) is equal to the number of integers m from 1 to n. Wait, no. Each m can be written uniquely as k*2^t, so the total number of such pairs is exactly n. But wait, that can't be right because if m=1, it can be written as 1*2^0; m=2 is 1*2^1; m=3 is 3*2^0; m=4 is 1*2^2, etc. So in general, each m from 1 to n corresponds to exactly one pair (k, t) where k is odd and t ≥0. Therefore, the total number of pairs (k, t) is exactly n. But according to the original sum, the total is (n-1)/2, which is different. So this seems contradictory. Therefore, my previous reasoning must be wrong.Wait, hold on. If each m from 1 to n corresponds to exactly one pair (k, t) with k odd and t ≥0, then the number of such pairs is n. But according to the problem, the original sum equals (n-1)/2, which is less than n. Therefore, my interpretation must be incorrect.Wait, but in the original sum, k is not arbitrary. Wait, no. Wait, the original sum counts for each odd k, the number of t such that k*2^t ≤n. So for each odd k, how many multiples of k by powers of 2 are ≤n. So for example, if k=1, how many t with 2^t ≤n, which is floor(log₂n) +1. Wait, but in the original sum, the term is floor(log₂(n/k)). Wait, when k=1, floor(log₂n). But if you consider t such that 2^t ≤n, then t can go from 0 to floor(log₂n), so there are floor(log₂n)+1 values. But the term in the sum is floor(log₂n), not floor(log₂n)+1. So my earlier assertion was wrong. Therefore, the total number of pairs (k, t) is the sum over k of (floor(log₂(n/k)) +1). Therefore, the total number of such pairs would be the original sum plus the number of terms in the sum, which is equal to the number of odd numbers from 1 to n. Since n is odd, there are (n+1)/2 odd numbers. Therefore, total pairs would be original_sum + (n+1)/2. But according to unique factorization, each m corresponds to exactly one pair (k, t), so total pairs is n. Therefore:original_sum + (n+1)/2 = nTherefore, original_sum = n - (n+1)/2 = (2n -n -1)/2 = (n -1)/2Which is exactly the result we need to prove!So, that's the key. By realizing that the original sum counts the number of pairs (k, t) with k odd, 1 ≤k ≤n, t ≥0, such that k*2^t ≤n, but subtracts 1 for each k (since floor(log₂(n/k)) = number of t's minus 1). Therefore, the total number of such pairs is original_sum + number of k's. Since each number m from 1 to n can be uniquely expressed as k*2^t with k odd, the total number of pairs is n. Therefore:original_sum + (number of k's) = nSince the number of k's is the number of odd numbers from 1 to n, which is (n +1)/2 (since in 1 to n, half are odd, rounded up). Wait, for n odd, the number of odd numbers from 1 to n is (n +1)/2. For example, n=3: 1,3 => 2 numbers; (3+1)/2=2. Correct.Therefore:original_sum + (n +1)/2 = nHence, original_sum = n - (n +1)/2 = (2n -n -1)/2 = (n -1)/2.Therefore, the identity is proven.So, the key idea was to relate the sum to the number of pairs (k, t) where k is odd and k*2^t ≤n. Each such pair corresponds to a unique integer m =k*2^t in [1, n]. The total number of such pairs is n. However, the original sum counts for each k, the number of t's such that t ≥1 (since floor(log₂(n/k)) is the maximum t where 2^t ≤n/k, which is equivalent to k*2^t ≤n. Wait, no:Wait, floor(log₂(n/k)) = the largest integer t where 2^t ≤n/k. Therefore, the number of t's where k*2^t ≤n is t from 0 to floor(log₂(n/k)). Therefore, the number of t's is floor(log₂(n/k)) +1. Therefore, the total number of pairs (k, t) is the sum over k of (floor(log₂(n/k)) +1) = original_sum + number of k's. Since the total is n, then original_sum + number_of_k = n => original_sum = n - number_of_k. Since number_of_k is (n+1)/2, original_sum =n - (n+1)/2=(n-1)/2. Which is exactly what we needed to prove. Therefore, this combinatorial interpretation proves the identity.That's a neat argument. So, the key was to recognize that each term floor(log₂(n/k)) counts the number of times you can double k (starting from t=1) before exceeding n, but since the unique factorization gives each number as k*2^t, the total count is n. Then subtracting the number of k's gives the original sum.Therefore, the proof is as follows:For each odd integer k from 1 to n, the term floor(log₂(n/k)) counts the number of integers t ≥1 such that k*2^t ≤n. The total number of pairs (k, t) with k odd, t ≥0, and k*2^t ≤n is equal to n, since each integer from 1 to n can be uniquely expressed as k*2^t where k is odd. The number of pairs with t ≥0 is the sum over k of (floor(log₂(n/k)) +1), which equals the original sum plus the number of odd k's. The number of odd k's from 1 to n is (n +1)/2. Therefore, original_sum + (n +1)/2 = n, leading to original_sum = (n -1)/2. Hence, proved.Final Answerboxed{dfrac{n-1}{2}}</think>To prove that for a positive odd integer ( n ), the sumleftlfloor log_{2} n rightrfloor + leftlfloor log_{2} left( frac{n}{3} right) rightrfloor + leftlfloor log_{2} left( frac{n}{5} right) rightrfloor + cdots + leftlfloor log_{2} left( frac{n}{n} right) rightrfloorequals (frac{n-1}{2}), we consider the following key steps:1. Interpretation of Floor Functions: Each term (leftlfloor log_{2} left( frac{n}{k} right) rightrfloor) for odd ( k ) counts the largest integer ( t ) such that ( 2^t leq frac{n}{k} ). This is equivalent to the number of times ( k ) can be doubled before exceeding ( n ).2. Counting Pairs ((k, t)): For each odd ( k ), the term (leftlfloor log_{2} left( frac{n}{k} right) rightrfloor) gives the number of valid ( t ) values (starting from ( t = 1 )) such that ( k cdot 2^t leq n ). The total count of pairs ((k, t)) where ( k ) is odd, ( t geq 0 ), and ( k cdot 2^t leq n ) is exactly ( n ), as each integer from 1 to ( n ) can be uniquely expressed as ( k cdot 2^t ) with ( k ) odd.3. Summing the Terms: The sum of (leftlfloor log_{2} left( frac{n}{k} right) rightrfloor) for all odd ( k ) from 1 to ( n ) is equivalent to the total number of pairs ((k, t)) minus the number of such ( k ). The number of odd ( k ) values from 1 to ( n ) is (frac{n+1}{2}).4. Final Calculation: The total number of pairs ((k, t)) is ( n ). Subtracting the number of odd ( k ) values from this total gives: [ n - frac{n+1}{2} = frac{2n - (n + 1)}{2} = frac{n - 1}{2} ]Thus, the sum of the floor functions is proven to be (frac{n-1}{2}).[boxed{dfrac{n-1}{2}}],<think>Okay, let me try to figure out this problem. So, we need to prove that for a positive odd integer n, the sum of these floor functions of logarithms equals (n-1)/2. The sum is written as the floor of log base 2 of n, plus the floor of log base 2 of n/3, plus the same for n/5, n/7, and so on up to n/n. Hmm.First, let me parse the problem again. The terms in the sum are floor(log₂(n/k)) where k ranges over the odd integers from 1 to n. Wait, actually, the denominators are 1, 3, 5, 7, ..., n? Since n is odd, n/k would be n divided by each odd integer up to n. So the denominators are the odd integers 1, 3, 5, ..., n. So the sum is over all odd divisors k of n? Wait, no, the denominators are just odd integers up to n, regardless of whether they divide n or not. Because it's written as n/1, n/3, n/5, etc., up to n/n. So k runs through all odd integers from 1 to n, stepping by 2 each time. So for example, if n is 7, then k is 1,3,5,7. Then the sum is floor(log₂7) + floor(log₂7/3) + floor(log₂7/5) + floor(log₂7/7).So maybe the first step is to compute this sum for a small odd n, say n=1, n=3, n=5, n=7, and check if the formula holds. Let's try that.For n=1: The sum is floor(log₂1/1) = floor(log₂1) = floor(0) = 0. The right-hand side is (1-1)/2 = 0. So 0=0, which works.For n=3: The terms are floor(log₂3) + floor(log₂3/3). So log₂3 is approximately 1.584, so floor is 1. Then 3/3=1, log₂1=0, floor is 0. Sum is 1+0=1. The right-hand side is (3-1)/2 = 1. So that works.For n=5: Terms are floor(log₂5) + floor(log₂5/3) + floor(log₂5/5). log₂5≈2.32, floor is 2. 5/3≈1.666, log₂≈0.70, floor 0. 5/5=1, log₂1=0. Sum is 2+0+0=2. RHS: (5-1)/2=2. Good.n=7: Terms: floor(log₂7)≈2.807, floor=2; floor(log₂7/3)=floor(log₂≈2.333)=1 (since 7/3≈2.333, log₂≈1.22, floor=1); floor(log₂7/5)=floor(log₂1.4)=0; floor(log₂7/7)=0. Sum: 2+1+0+0=3. RHS: (7-1)/2=3. Correct.n=9: Wait, but n must be odd. 9 is odd. Let's check. Terms: floor(log₂9)=3 (since 2^3=8<9<16=2^4). Then floor(log₂9/3)=floor(log₂3)=1. Then floor(log₂9/5)=floor(log₂1.8)=0. floor(log₂9/7)=floor(log₂≈1.285)=0. floor(log₂9/9)=0. Sum:3+1+0+0+0=4. RHS: (9-1)/2=4. Correct again.Hmm, so for small n it works. Now, how to generalize this.The sum is over k odd integers from 1 to n: floor(log₂(n/k)).We need to show that this sum is equal to (n-1)/2.So perhaps each term floor(log₂(n/k)) corresponds to some count, and when we sum over all odd k, we get the total count as (n-1)/2.Alternatively, maybe there's a bijection or combinatorial interpretation here. Let me think.Alternatively, let's analyze floor(log₂(n/k)). The floor of log base 2 of x is the exponent of the highest power of 2 less than or equal to x. So floor(log₂(n/k)) = m where 2^m ≤ n/k < 2^(m+1). So m is the largest integer such that 2^m ≤ n/k. So equivalently, k ≥ n/(2^{m+1}).So for each term in the sum, floor(log₂(n/k)) is the integer m such that n/k is in [2^m, 2^{m+1}). So k is in (n/2^{m+1}, n/2^m].But since k is odd, perhaps we can count the number of odd integers k in each interval (n/2^{m+1}, n/2^m] and multiply by m, then sum over m. That is, the total sum would be Σ m * (number of odd k in (n/2^{m+1}, n/2^m]).But maybe this is a way to approach it. Let me formalize this.Let’s consider that floor(log₂(n/k)) = m if and only if 2^m ≤ n/k < 2^{m+1}, which is equivalent to n/2^{m+1} < k ≤ n/2^m. So for each m ≥0, the number of odd integers k in (n/2^{m+1}, n/2^m] is equal to the number of terms in the sum where floor(log₂(n/k)) = m. Then the total sum is Σ m * (number of such k). So we need to compute this sum over m and show that it equals (n-1)/2.But how?Alternatively, maybe we can interpret (n-1)/2 as the number of odd integers less than n, but since n is odd, the number of odd integers from 1 to n is (n+1)/2. Wait, (n+1)/2. For example, n=7: 1,3,5,7: 4 terms, which is (7+1)/2=4. But the right-hand side is (n-1)/2. So (n-1)/2 is one less. Hmm.Wait, perhaps there's a different way. Let's think about binary representations. Since n is odd, its binary representation ends with a 1. Maybe the terms floor(log₂(n/k)) relate to the positions of 1s in the binary representation? Not sure.Alternatively, consider that when k divides n, floor(log₂(n/k)) is an integer. But k doesn't have to divide n here. Wait, but k ranges over all odd integers up to n, not just divisors.Alternatively, maybe express n in binary. Let's say n is a binary number with bits b_0 to b_t, where b_0=1 (since n is odd). Maybe each term floor(log₂(n/k)) corresponds to some bit? Not sure.Alternatively, perhaps note that the function floor(log₂(n/k)) counts the number of times you can divide n/k by 2 before it becomes less than 1. That is, floor(log₂(n/k)) is the largest integer m such that 2^m ≤ n/k. Which is equivalent to k ≤ n / 2^m. So for each k, floor(log₂(n/k)) is the exponent m where 2^m is the largest power of 2 less than or equal to n/k. So, if we fix m, then the number of k's (odd) such that floor(log₂(n/k))=m is equal to the number of odd integers k in (n / 2^{m+1}, n / 2^m].Therefore, the total sum would be the sum over m of m multiplied by the number of odd integers k in (n / 2^{m+1}, n / 2^m].If we can compute this sum and show it equals (n-1)/2, then we are done.Alternatively, perhaps instead of counting m * number of k's, we can think of each k contributing m = floor(log₂(n/k)). So, maybe change the order of summation. Let me think.Alternatively, maybe notice that each odd k from 1 to n can be written uniquely as k = n / (2^m * t), where t is an odd integer greater than or equal to 1, and m is a non-negative integer. Wait, not sure.Alternatively, think of each k as n divided by some number, which can be expressed as 2^m * s, where s is odd. But I need to formalize this.Wait, for each k, n/k can be expressed as 2^m * s, where s is an odd integer. Then m would be floor(log₂(n/k)) if s >=1. Wait, but n/k may not be an integer. Hmm.Alternatively, perhaps use generating functions. The sum over odd k of floor(log₂(n/k)). Hmm, not sure.Alternatively, let's consider that for each odd integer k, floor(log₂(n/k)) is the number of times you can divide n/k by 2 before it's less than 1. Which is equivalent to the exponent of the highest power of 2 in the integer part of n/k. But since n and k are both odd, n/k is a rational number with denominator dividing k. But perhaps not helpful.Wait, maybe instead of looking at each k, we can pair terms. For example, for each odd k, there's a corresponding term floor(log₂(n/k)). Maybe if we pair k and n/k, but not sure.Alternatively, think about the binary expansions. Since n is odd, let's write n as 2^m + 1 or some such. Wait, maybe not.Alternatively, use induction on n. Since n is odd, maybe step from n to n+2. Suppose the formula holds for n, then prove it for n+2.Let's try induction.Base case: n=1, which works as shown earlier.Assume it works for some odd integer n. Now, consider n+2. We need to show that the sum for n+2 is ((n+2)-1)/2 = (n+1)/2.The sum for n+2 is floor(log₂(n+2)) + floor(log₂((n+2)/3)) + ... + floor(log₂((n+2)/(n+2))).Comparing this to the sum for n, which is floor(log₂n) + floor(log₂(n/3)) + ... + floor(log₂(n/n)).So the difference between the two sums would be the terms for n+2 minus the terms for n. But the problem is that the denominators k for n+2 go up to n+2, which is two more than n. So when we go from n to n+2, the number of terms increases by 1 (since k goes up to n+2, which is two more than n, but since k must be odd, if n was odd, then n+2 is the next odd, so k increments by two terms: n+1 is even, so skipped, and n+2 is odd. Wait, no, n is odd, so n+1 is even, n+2 is odd. So the denominators for n+2 are 1,3,5,...,n, n+2. So one more term than in n. Wait, n had denominators up to n, which is odd, so k from 1 to n, step 2. Then n+2 would have k from 1 to n+2, step 2, so one more term: floor(log₂((n+2)/(n+2))) = 0.But also, the existing terms floor(log₂((n+2)/k)) versus floor(log₂(n/k)) might differ for some k.So the difference between the sums would be [sum_{k odd=1}^{n+2} floor(log₂((n+2)/k))] - [sum_{k odd=1}^{n} floor(log₂(n/k))] = floor(log₂(n+2)) + [sum_{k odd=1}^{n} (floor(log₂((n+2)/k)) - floor(log₂(n/k)))] + floor(log₂((n+2)/(n+2))). Since the last term is 0.But this seems complicated. Maybe induction is not straightforward here.Alternatively, perhaps there's a direct counting argument. Let's recall that (n-1)/2 is the number of even integers less than n. Wait, but n is odd, so numbers from 1 to n-1, half are even. But how does that relate to the sum?Alternatively, maybe each term floor(log₂(n/k)) counts the number of times a certain power of 2 divides into n/k. But since n/k isn't necessarily an integer.Alternatively, if we write each k as 2^m * t, where t is odd. But wait, k itself is odd, so m=0. So that might not help.Wait, but n/k might be even or odd. But n and k are both odd, so n/k is either odd (if k divides n) or a fraction with odd numerator and denominator. For example, n=7, k=3: 7/3 is not an integer. So floor(log₂(7/3)) = floor(log₂(2.333)) = 1.Hmm. Maybe another approach. Let's consider that for each odd k, floor(log₂(n/k)) is equal to the number of times you can divide n/k by 2 before it's less than 1. So for each k, we can write n/k = 2^{m} * r where 1 ≤ r < 2. Then m is floor(log₂(n/k)).But since n and k are odd, n/k is either an odd integer or a fraction with odd numerator and denominator. If n/k is an integer, then it's an odd integer, so m = log₂(n/k) if n/k is a power of 2. But since n/k is odd, the only way n/k is a power of 2 is if n/k=1, i.e., k=n. So in that case, floor(log₂(n/n)) = 0.For other k, n/k is either an odd integer greater than 1 or a fraction. If n/k is an odd integer greater than 1, then log₂(n/k) is log₂ of an odd integer, so not an integer, so floor(log₂(n/k)) is the exponent of the highest power of 2 less than that odd integer. For example, n/k=3, log₂3≈1.58, floor is 1. If n/k is a fraction, like 5/3≈1.666, log₂≈0.70, floor 0.Wait, but how does this help? Maybe we need to consider binary expansions.Alternatively, let's think of all the terms in the sum. For each odd k from 1 to n, we have floor(log₂(n/k)). Let's denote m_k = floor(log₂(n/k)). Then 2^{m_k} ≤ n/k < 2^{m_k +1}, which can be rearranged as n / 2^{m_k +1} < k ≤ n / 2^{m_k}. So for each k, m_k is the unique integer such that k is in (n / 2^{m_k +1}, n / 2^{m_k}].Therefore, each m_k corresponds to an interval where k lies. So if we can partition the interval of k from 1 to n into regions where m_k is constant, then the sum becomes the sum of m_k multiplied by the number of k's in each such interval.But since k must be odd, the number of k's in each interval (n / 2^{m+1}, n / 2^{m}] is equal to the number of odd integers in that interval.So our sum is Σ_{m=0}^{floor(log₂n)} m * (number of odd k in (n / 2^{m+1}, n / 2^{m}]).Our goal is to show that this sum equals (n-1)/2.Hmm. Let's try to compute the number of odd integers in each interval (n / 2^{m+1}, n / 2^{m}].The number of integers in (n / 2^{m+1}, n / 2^{m}] is floor(n / 2^{m}) - floor(n / 2^{m+1}). But we need the number of odd integers in that interval. The number of odd integers in an interval from a to b is approximately (b - a +1)/2, but adjusted depending on the parity of a and b.Specifically, the number of odd integers between a and b inclusive is floor((b +1)/2) - floor(a / 2). But here, the interval is (n / 2^{m+1}, n / 2^{m}], so from floor(n / 2^{m+1}) +1 to floor(n / 2^{m}).But since n is odd, and we are dividing by powers of 2, n / 2^{m} is a fraction. Wait, maybe more precise analysis.Let me denote L = n / 2^{m+1} and R = n / 2^{m}. The interval is (L, R]. The number of odd integers in this interval is equal to the number of integers k in (L, R] such that k is odd.But since n is odd, and we are dealing with divisions by powers of 2, maybe there's a pattern here.Alternatively, note that n is odd, so n = 2^0 * n. When we divide by 2^{m}, we get n / 2^{m} = (n) / 2^{m}, which is a fraction with denominator 2^{m}.But perhaps it's better to express n in binary. Let’s consider n as a binary number. Since n is odd, the least significant bit is 1. Suppose n is written in binary as b_tb_{t-1}...b_1b_0, with b_0=1.When we divide n by 2^{m}, we are effectively shifting the binary representation right by m digits. The floor(n / 2^{m}) is the integer obtained by removing the last m digits.But since n is odd, dividing by 2^{m} gives a number of the form (integer part) + (fraction part)/(2^{m}).But I'm not sure.Alternatively, perhaps it's helpful to think of the intervals (n / 2^{m+1}, n / 2^{m}] as doubling intervals. For each m, the interval (n / 2^{m+1}, n / 2^{m}] contains numbers k such that n/k is in [2^m, 2^{m+1}).But since k is odd, and n is odd, n/k is either an odd integer or a fraction in reduced form with an odd denominator.Wait, maybe for each m, the number of odd k in (n / 2^{m+1}, n / 2^{m}] is equal to the number of odd integers in that interval.But how to count them?Let’s denote that interval as (A, B], where A = n / 2^{m+1} and B = n / 2^m. The number of odd integers in (A, B] is equal to floor((B + 1)/2) - floor((A + 1)/2). Because the number of odd numbers up to x is floor((x + 1)/2). Therefore, the number in (A, B] is floor((B +1)/2) - floor((A +1)/2).But since A and B are fractions, we need to compute floor((B +1)/2) and floor((A +1)/2).But since n is odd, let's write n = 2p +1, where p is an integer.But not sure if this helps. Let's take a specific m and see.Suppose m=0. Then the interval is (n/2, n]. The number of odd integers in (n/2, n]. Since n is odd, n/2 is a half-integer. For example, if n=7, then n/2=3.5, so the interval (3.5, 7] includes integers 4,5,6,7. Among these, the odd integers are 5,7. So two numbers. Which is floor((7 +1)/2) - floor((3.5 +1)/2) = floor(8/2) - floor(4.5/2) = 4 - 2 = 2. Correct.Similarly, if m=1: interval is (n/4, n/2]. For n=7, n/4=1.75, n/2=3.5. So interval (1.75, 3.5] includes integers 2,3. Odd integers: 3. So 1 number. floor((3.5 +1)/2) - floor((1.75 +1)/2) = floor(4.5/2) - floor(2.75/2) = 2 -1=1. Correct.m=2: (n/8, n/4]. For n=7, n/8=0.875, n/4=1.75. Interval (0.875,1.75] includes integer 1. Odd integer:1. So 1 number. floor((1.75 +1)/2) - floor((0.875 +1)/2) = floor(2.75/2) - floor(1.875/2) =1 -0=1. Correct.m=3: (n/16, n/8]. For n=7, n/16≈0.4375, n/8≈0.875. Interval (0.4375,0.875] includes no integers. So 0 numbers. floor((0.875 +1)/2) - floor((0.4375 +1)/2)=floor(1.875/2)-floor(1.4375/2)=0 -0=0.So, in general, for each m, the number of odd integers in (n / 2^{m+1}, n / 2^{m}] is floor((n / 2^{m} +1)/2) - floor((n / 2^{m+1} +1)/2). Let's denote this as C(m).Therefore, the total sum is Σ_{m=0}^{∞} m * C(m). But since n is finite, this sum will terminate when n / 2^{m} <1.But how can we compute Σ m * C(m) ?Wait, let's compute Σ_{m=0}^{∞} m * [floor((n / 2^{m} +1)/2) - floor((n / 2^{m+1} +1)/2))]Maybe telescoping? Let's see.Let’s denote D(m) = floor((n / 2^{m} +1)/2). Then C(m) = D(m) - D(m+1). So the sum becomes Σ_{m=0}^{∞} m*(D(m)-D(m+1)).This is similar to Σ_{m=0}^{∞} m*D(m) - m*D(m+1) = Σ_{m=0}^{∞} m*D(m) - Σ_{m=0}^{∞} m*D(m+1).Change index in the second sum: Let m' = m+1, so when m=0, m'=1. So the second sum becomes Σ_{m'=1}^{∞} (m'-1)*D(m'). Therefore, the total sum is Σ_{m=0}^{∞} m*D(m) - Σ_{m'=1}^{∞} (m'-1)*D(m') = Σ_{m=0}^{∞} m*D(m) - Σ_{m=1}^{∞} (m -1)*D(m).This telescopes to:= 0*D(0) + Σ_{m=1}^{∞} [m*D(m) - (m -1)*D(m)] = Σ_{m=1}^{∞} D(m)Thus, the total sum is Σ_{m=0}^{∞} D(m) - D(0) ? Wait, no. Wait, original expression:Σ_{m=0}^{∞} m*(D(m)-D(m+1)) = Σ_{m=0}^{∞} [m*D(m) - m*D(m+1)]= Σ_{m=0}^{∞} m*D(m) - Σ_{m=0}^{∞} m*D(m+1)Second sum is Σ_{m=0}^{∞} m*D(m+1) = Σ_{k=1}^{∞} (k-1)*D(k) where k = m+1.Thus, total sum becomes:Σ_{m=0}^{∞} m*D(m) - Σ_{k=1}^{∞} (k-1)*D(k)= 0*D(0) + Σ_{k=1}^{∞} k*D(k) - Σ_{k=1}^{∞} (k-1)*D(k)= Σ_{k=1}^{∞} [k*D(k) - (k-1)*D(k)] = Σ_{k=1}^{∞} D(k)So the total sum is Σ_{k=1}^{∞} D(k), where D(k) = floor((n / 2^{k} +1)/2).Wait, but how does Σ_{k=1}^{∞} floor((n / 2^{k} +1)/2) relate to (n-1)/2?Hmm, maybe we can compute Σ_{k=1}^{∞} floor((n / 2^{k} +1)/2).Wait, let's compute this sum for n=7.n=7:Compute floor((7 / 2^k +1)/2) for k=1,2,3,...k=1: floor((7/2 +1)/2) = floor((3.5 +1)/2)=floor(4.5/2)=floor(2.25)=2k=2: floor((7/4 +1)/2)=floor((1.75 +1)/2)=floor(2.75/2)=1k=3: floor((7/8 +1)/2)=floor((0.875 +1)/2)=floor(1.875/2)=0k=4 onwards: 0.So sum is 2 +1 +0 +...=3. Which matches (7-1)/2=3.Similarly, for n=5:k=1: floor((5/2 +1)/2)=floor((2.5 +1)/2)=floor(3.5/2)=1k=2: floor((5/4 +1)/2)=floor((1.25 +1)/2)=floor(2.25/2)=1k=3: floor((5/8 +1)/2)=floor((0.625 +1)/2)=floor(1.625/2)=0Sum:1+1=2. Which is (5-1)/2=2.n=3:k=1: floor((3/2 +1)/2)=floor((1.5 +1)/2)=floor(2.5/2)=1k=2: floor((3/4 +1)/2)=floor((0.75 +1)/2)=floor(1.75/2)=0Sum:1+0=1=(3-1)/2=1.n=9:k=1: floor((9/2 +1)/2)=floor(4.5 +1)/2=floor(5.5/2)=2k=2: floor((9/4 +1)/2)=floor(2.25 +1)/2=floor(3.25/2)=1k=3: floor((9/8 +1)/2)=floor(1.125 +1)/2=floor(2.125/2)=1k=4: floor((9/16 +1)/2)=floor(0.5625 +1)/2=floor(1.5625/2)=0Sum:2+1+1=4=(9-1)/2=4.Yes, so it seems that Σ_{k=1}^{∞} floor((n / 2^{k} +1)/2) = (n -1)/2. Therefore, if we can prove that Σ_{k=1}^{∞} floor((n / 2^{k} +1)/2) = (n -1)/2, then the original identity holds.So our problem reduces to proving that Σ_{k=1}^{∞} floor((n / 2^{k} +1)/2) = (n -1)/2 for odd positive integers n.Let me try to compute this sum for general odd n.Let n be an odd integer. Let's express n in binary. Since n is odd, its binary representation ends with a 1. Let’s say n = 2^0 + a_1*2^1 + a_2*2^2 + ... + a_t*2^t, where a_t=1.Alternatively, write n as n = 2^{m_1} + 2^{m_2} + ... + 2^{m_k} +1, but not sure.Alternatively, note that n / 2^k is equal to the number n shifted right by k bits in binary. So floor((n / 2^k +1)/2) would correspond to taking the k-th shifted value, adding 1, and integer dividing by 2. Maybe there's a binary digit interpretation here.Alternatively, consider that Σ_{k=1}^{∞} floor((n + 2^{k}) / 2^{k+1}) )Wait, because (n / 2^{k} +1)/2 = (n + 2^{k}) / 2^{k+1}So floor((n + 2^{k}) / 2^{k+1})But floor((n + 2^{k}) / 2^{k+1}) is equal to floor(n / 2^{k+1} + 1/2). Wait, not exactly, because:(n + 2^{k}) / 2^{k+1} = n / 2^{k+1} + 2^{k} / 2^{k+1} = n / 2^{k+1} + 1/2.So floor(n / 2^{k+1} + 1/2). Therefore, floor((n / 2^{k} +1)/2) = floor(n / 2^{k+1} + 1/2).Therefore, the sum becomes Σ_{k=1}^{∞} floor(n / 2^{k+1} + 1/2).But how does this help?Alternatively, notice that floor(x + 1/2) is the nearest integer function rounding down when there's a tie. But since n is an integer, n / 2^{k+1} is a rational number with denominator 2^{k+1}. So adding 1/2 and flooring would give the integer closest to n / 2^{k+1}, rounding down on 0.5.But maybe not helpful.Alternatively, let's express n in binary. Let’s say n is written as b_tb_{t-1}...b_1b_0, with b_0=1.Then n / 2^{k+1} is the number represented by the binary digits shifted right by k+1 digits. Adding 1/2 is equivalent to adding 0.1 in binary. Then flooring would correspond to rounding to the nearest integer, but with ties broken down.But not sure.Alternatively, think of the sum Σ_{k=1}^{∞} floor(n / 2^{k+1} + 1/2). For each k, this is essentially counting how many times 2^{k+1} fits into n, adjusted by 1/2. Hmm.Alternatively, write n in binary. Let's take n=7 (111 in binary). Then:k=1: floor(7 / 4 + 0.5) = floor(1.75 + 0.5) = floor(2.25)=2k=2: floor(7 / 8 + 0.5) = floor(0.875 +0.5)=floor(1.375)=1k=3: floor(7 / 16 +0.5)=floor(0.4375 +0.5)=floor(0.9375)=0Which matches the earlier sum.Similarly, n=5 (101):k=1: floor(5/4 +0.5)=floor(1.25+0.5)=1.75 floor=1k=2: floor(5/8 +0.5)=floor(0.625+0.5)=1.125 floor=1k=3: floor(5/16 +0.5)=floor(0.3125+0.5)=0.8125 floor=0Sum:1+1=2.So the sum counts for each k>=1, the number of times 2^{k+1} fits into n plus a half, floored. But how to relate this to (n-1)/2.Wait, let's compute Σ_{k=1}^{∞} floor(n / 2^{k+1} + 1/2).Notice that this is similar to the sum Σ_{k=1}^{∞} floor(n / 2^{k} / 2 + 1/2).Alternatively, consider the standard identity that Σ_{k=1}^{∞} floor(n / 2^{k}) = n - s_2(n), where s_2(n) is the sum of the bits in the binary representation of n. But this is a different sum.Alternatively, maybe use induction. Suppose that for odd n, Σ_{k=1}^{∞} floor((n + 2^{k}) / 2^{k+1}) ) = (n -1)/2.Let’s check n=1:Σ_{k=1}^{∞} floor((1 + 2^{k}) / 2^{k+1} ). For k=1: floor((1+2)/4)=floor(3/4)=0. For k>=1, all terms are 0. Sum=0. (1-1)/2=0. Correct.Assume it's true for n. Now consider n+2.Σ_{k=1}^{∞} floor((n+2 + 2^{k}) / 2^{k+1}) = Σ_{k=1}^{∞} floor( (n +2 + 2^{k}) / 2^{k+1} )= Σ_{k=1}^{∞} floor( (n + 2^{k} + 2) / 2^{k+1} )But not sure how to relate this to the sum for n.Alternatively, maybe notice that adding 2 to n increments certain bits in the binary representation, but since n is odd, n+2 is next odd number. However, this might complicate the sum.Alternatively, another approach: Let’s observe that for an odd integer n, we can write n = 2m +1, where m is an integer.Then (n -1)/2 = m. So we need to show that the sum Σ_{k=1}^{∞} floor((n / 2^{k} +1)/2) = m.But since n = 2m +1,Σ_{k=1}^{∞} floor( (2m +1)/ 2^{k} +1)/2 )= Σ_{k=1}^{∞} floor( ( (2m +1)/ 2^{k} +1 ) / 2 )= Σ_{k=1}^{∞} floor( (2m +1 + 2^{k}) / 2^{k+1} )Hmm, not sure.Alternatively, maybe expand the binary representation.Let’s express m in binary. Since n=2m +1, m=(n-1)/2.Wait, perhaps if we consider the binary digits of m, each term in the sum corresponds to a digit.But this is vague.Alternatively, let’s note that for each k, floor((n / 2^{k} +1)/2) is equal to the number of times 2^{k+1} fits into n plus a correction term.Wait, let's denote that:floor((n / 2^{k} +1)/2) = floor(n / 2^{k+1} + 1/2)This is similar to counting the number of multiples of 2^{k+1} in n, rounded to the nearest integer. But since n is odd, maybe there's a pattern.Alternatively, consider writing n in binary. Let's say n is written as B, a binary number. Then dividing by 2^{k+1} shifts B right by k+1 bits. Adding 1/2 is like adding a 1 in the (k+1)th decimal place in binary. Then flooring is equivalent to truncating after the decimal point.But this might not directly help.Alternatively, think recursively. Let's suppose that n is odd. Divide n by 2: since n is odd, n=2m+1. Then floor(n / 2) = m. Then n / 2^{k} = m / 2^{k-1} + 1 / 2^{k}.But not sure.Alternatively, let's use the identity that for any integer x, floor(x / 2) + floor((x +1)/2) = x.Wait, let's test x=5: floor(5/2)=2, floor(6/2)=3. 2 +3=5. Yes.Similarly, x=4: 2 +2=4.Yes, this identity holds: floor(x/2) + floor((x+1)/2) = x.But how does this apply here?If we have Σ_{k=1}^{∞} floor((n + 2^{k}) / 2^{k+1}) )= Σ_{k=1}^{∞} floor( (n / 2^{k} +1)/2 )Let’s denote x_k = n / 2^{k}. Then floor(x_k / 2 + 1/2) = floor( (x_k +1)/2 )From the identity, floor(x_k / 2) + floor((x_k +1)/2) = x_k.But x_k is n / 2^{k}, which is not necessarily an integer. Wait, the identity floor(a) + floor(a + 1/2) = floor(2a). Not sure.Wait, let's step back. We need to find the sum Σ_{k=1}^{∞} floor( (n / 2^{k} +1)/2 ). Let's denote this sum as S(n).We need to show that S(n) = (n -1)/2 when n is odd.Let’s try to compute S(n) + S(n) = 2S(n).2S(n) = Σ_{k=1}^{∞} 2*floor( (n / 2^{k} +1)/2 )But 2*floor( (n / 2^{k} +1)/2 ) ≤ n / 2^{k} +1.But not sure.Alternatively, consider that S(n) = Σ_{k=1}^{∞} floor( (n + 2^{k}) / 2^{k+1} )= Σ_{k=1}^{∞} floor( n / 2^{k+1} + 1/2 )This is similar to adding up the rounded values of n / 2^{k+1}.But perhaps we can telescope this sum.Let’s compute Σ_{k=1}^{∞} floor(n / 2^{k+1} + 1/2)Notice that for each term floor(n / 2^{k+1} + 1/2), it is equal to the nearest integer to n / 2^{k+1}, rounded down on 0.5.But since n is odd, when k increases, n / 2^{k+1} becomes a fraction with denominator 2^{k+1}.Alternatively, note that the sum S(n) can be rewritten as Σ_{k=2}^{∞} floor(n / 2^{k} + 1/2), shifted index. Not sure.Alternatively, consider binary expansion again. Let's write n in binary as a t+1-bit number: n = b_0 + 2b_1 + 4b_2 + ... + 2^t b_t, where b_0=1.When we divide n by 2^{k+1}, we get a number whose binary representation is the original shifted right by k+1 digits. Adding 1/2 is like setting the (k+1)th bit after the decimal to 1 and flooring removes all bits after the decimal.For example, take n=7 (111 in binary):k=1: n / 2^{2} = 7 /4 = 1.75 (binary 1.11). Adding 0.5 (binary 0.1) gives 2.25 (binary 10.01). Floor is 2 (binary 10).k=2: n / 2^{3} = 7 /8 =0.875 (binary 0.111). Adding 0.5=1.375 (binary 1.011). Floor is 1.k=3: n / 16=0.4375 +0.5=0.9375, floor=0.So each term floor(n / 2^{k+1} +1/2) corresponds to taking the first k+1 bits of n, shifting, adding 0.1 in binary, and flooring.But how does this sum up to (n -1)/2.Alternatively, note that when we sum floor(n / 2^{k+1} +1/2) over k from 1 to infinity, we are effectively summing the bits of n in a certain way.Alternatively, consider that each term floor(n / 2^{k+1} +1/2) is equal to the number of times 2^{k+1} fits into n plus a correction.But maybe this is not helpful.Alternatively, let's use induction again. Suppose S(n) = (n-1)/2 holds for all odd numbers less than n. Then consider S(n). We can express n as n = 2m +1. Then maybe relate S(n) to S(m).But m could be even or odd. If n =2m +1, then m = (n -1)/2. If m is even, then n=2*(2k) +1=4k+1. If m is odd, then n=2*(2k+1)+1=4k+3.But not sure.Alternatively, express n in base 2. Let’s say n is written as 1b_1b_2...b_t. Then n=1 + 2b_1 +4b_2 + ... + 2^t b_t.Then S(n) = Σ_{k=1}^{∞} floor((1 + 2b_1 +4b_2 + ... + 2^t b_t)/2^{k+1} +1/2)But not sure.Alternatively, take the binary digits of n and consider each term in the sum S(n). For each k, the term floor(n / 2^{k+1} +1/2) is the value of the number formed by the digits from the (k+1)th bit onwards, plus half the k-th bit. But this might be too vague.Alternatively, consider that the sum S(n) counts the number of times each bit in the binary representation of n is set, weighted by some factor. But this might not directly lead to (n-1)/2.Alternatively, think of generating function for the sum.Let’s write S(n) = Σ_{k=1}^{∞} floor((n / 2^{k} +1)/2 )Let’s denote n as a binary number: n = sum_{i=0}^{t} b_i 2^i, with b_0=1.Then n / 2^{k} = sum_{i=0}^{t} b_i 2^{i -k} = sum_{i=0}^{k-1} b_i 2^{i -k} + sum_{i=k}^{t} b_i 2^{i -k}So n / 2^{k} = F_k + I_k, where F_k is the fractional part and I_k is the integer part.Then floor((n / 2^{k} +1)/2 ) = floor( (I_k + F_k +1)/2 )If F_k +1 >= 2, then this is floor(I_k / 2 + (F_k +1)/2 )= floor(I_k /2 + 1 + (F_k -1)/2 )= I_k /2 +1 + floor((F_k -1)/2 )But since F_k <1, F_k -1 <0, so floor((F_k -1)/2 )= -1.Wait, this seems complicated.Alternatively, note that I_k = floor(n / 2^{k}).Then floor((n / 2^{k} +1)/2 ) = floor( (I_k + F_k +1)/2 )If F_k +1 <2, which is always true since F_k <1, then F_k +1 <2, so (I_k + F_k +1)/2 = I_k/2 + (F_k +1)/2.Then floor(I_k/2 + (F_k +1)/2) = floor(I_k/2 + (F_k +1)/2).But since I_k is integer, I_k/2 is either integer or half-integer.If I_k is even, say I_k=2m, then floor(m + (F_k +1)/2 )= m + floor( (F_k +1)/2 ). Since F_k <1, (F_k +1)/2 <1, so floor is m.If I_k is odd, say I_k=2m+1, then floor( (2m+1)/2 + (F_k +1)/2 )= floor(m + 0.5 + (F_k +1)/2 )= m + floor(0.5 + (F_k +1)/2 ). Since F_k <1, 0.5 + (F_k +1)/2 <0.5 +1=1.5. So floor is m + 0 if 0.5 + (F_k +1)/2 <1, i.e., F_k <1, which is always true. Wait, 0.5 + (F_k +1)/2 = (F_k +2)/2. Since F_k <1, (F_k +2)/2 < (1 +2)/2 =1.5, so floor is m +0 if (F_k +2)/2 <1, which is F_k <0. But F_k >=0, so (F_k +2)/2 >=1. So floor is m +0 or m +1?Wait, let me compute:If I_k is even: floor(m + (F_k +1)/2) = m + floor( (F_k +1)/2 )Since (F_k +1)/2 is <1 (because F_k <1), so floor( (F_k +1)/2 ) =0. So total m.If I_k is odd: floor(m + 0.5 + (F_k +1)/2 )= floor(m + (F_k +2)/2 )Since (F_k +2)/2 =1 + F_k /2. Since F_k <1, this is <1.5. So floor(m +1 + F_k /2 )= m +1 if F_k /2 >=0.5, i.e., F_k >=1. But F_k <1, so F_k /2 <0.5. So floor(m +1 + F_k /2 )= m +1 if (F_k /2) >=0.5? No, since F_k is the fractional part, F_k <1, so F_k /2 <0.5. So (F_k +2)/2 =1 + F_k /2 <1 +0.5=1.5. So floor is m +1.Wait, for I_k odd, regardless of F_k, floor(m + (F_k +2)/2 )= floor(m +1 + F_k /2 ). Since F_k /2 is <0.5, m +1 + F_k /2 is < m +1.5. So floor is m +1.Wait, example:Let I_k =1 (m=0), F_k=0.5. Then floor(0 + (0.5 +2)/2 )=floor(1.25)=1.If F_k=0.9, floor(0 + (0.9 +2)/2 )=floor(1.45)=1.If F_k=0.1, floor(0 + (0.1 +2)/2 )=floor(1.05)=1.Thus, for I_k odd, floor( (n / 2^{k} +1)/2 )= m +1, where I_k=2m+1.For I_k even, floor( (n / 2^{k} +1)/2 )= m.Therefore, floor( (n / 2^{k} +1)/2 ) = floor(I_k /2 ) + c, where c=1 if I_k is odd, else 0.But floor(I_k /2 ) is equal to floor(I_k /2 ). Since I_k = floor(n / 2^{k} ), then floor(I_k /2 ) = floor(n / 2^{k+1} ).Thus, floor( (n / 2^{k} +1)/2 ) = floor(n / 2^{k+1} ) + c_k, where c_k=1 if I_k is odd, else 0.But since I_k = floor(n / 2^{k} ), so I_k is odd if and only if the (k)th bit in n's binary representation is 1.Wait, for example, n=7=111.I_1 = floor(7/2)=3 (binary 11), which is odd. So c_1=1.I_2 = floor(7/4)=1 (binary 1), odd, c_2=1.I_3 = floor(7/8)=0, even, c_3=0.So floor((n / 2^{k} +1)/2 )= floor(n / 2^{k+1} ) + c_k.But floor(n / 2^{k+1} ) = floor(I_k /2 ) + floor(F_k /2 ). But F_k /2 <0.5, so floor(F_k /2 )=0. Therefore, floor(n / 2^{k+1} ) = floor(I_k /2 ).Therefore, floor((n / 2^{k} +1)/2 ) = floor(n / 2^{k+1} ) + c_k, where c_k=1 if floor(n / 2^{k} ) is odd, else 0.Therefore, S(n) = Σ_{k=1}^{∞} [ floor(n / 2^{k+1} ) + c_k ] = Σ_{k=1}^{∞} floor(n / 2^{k+1} ) + Σ_{k=1}^{∞} c_k.But Σ_{k=1}^{∞} floor(n / 2^{k+1} ) = Σ_{k=2}^{∞} floor(n / 2^{k} ) = Σ_{k=1}^{∞} floor(n / 2^{k} ) - floor(n / 2^1 ).And we know that Σ_{k=1}^{∞} floor(n / 2^{k} ) = n - s_2(n), where s_2(n) is the number of 1s in the binary representation of n. But wait, no, actually, the standard identity is Σ_{k=1}^{∞} floor(n / 2^{k} ) = n - s_2(n), where s_2(n) is the sum of the digits in binary. However, I need to confirm.Actually, the identity is Σ_{k=1}^{∞} floor(n / 2^{k} ) = n - bitcount(n), where bitcount(n) is the number of 1s in binary expansion. Let's verify with n=7 (111):Σ_{k=1}^{∞} floor(7 /2^k ) = floor(7/2) + floor(7/4) + floor(7/8) + ... = 3 +1 +0+...=4. n - bitcount(n)=7 -3=4. Correct.Similarly, n=5 (101):floor(5/2)=2, floor(5/4)=1, floor(5/8)=0. Sum=3. n - bitcount(n)=5 -2=3. Correct.So Σ_{k=1}^{∞} floor(n / 2^{k} ) = n - s_2(n).Therefore, Σ_{k=2}^{∞} floor(n / 2^{k} ) = (n - s_2(n)) - floor(n / 2).But floor(n / 2) = (n -1)/2 since n is odd.So Σ_{k=2}^{∞} floor(n / 2^{k} ) = (n - s_2(n)) - (n -1)/2.Therefore, Σ_{k=1}^{∞} floor(n / 2^{k+1} ) = (n - s_2(n)) - (n -1)/2 - floor(n / 2^1 ) ?Wait, no, wait:Σ_{k=1}^{∞} floor(n / 2^{k+1} ) = Σ_{k=2}^{∞} floor(n / 2^{k} )= Σ_{k=1}^{∞} floor(n / 2^{k} ) - floor(n / 2^1 )= (n - s_2(n)) - floor(n / 2 )But since n is odd, floor(n / 2 ) = (n -1)/2.Thus, Σ_{k=1}^{∞} floor(n / 2^{k+1} ) = (n - s_2(n)) - (n -1)/2.And Σ_{k=1}^{∞} c_k is the number of times floor(n / 2^{k} ) is odd for k >=1.But floor(n / 2^{k} ) is odd if and only if the k-th bit in n's binary representation is 1. Wait, let's see.For example, n=7 (111):floor(7 /2^1)=3 (11) which is odd.floor(7 /2^2)=1 (1) which is odd.floor(7 /2^3)=0 which is even.So c_1=1, c_2=1, c_3=0,...Similarly, n=5 (101):floor(5 /2^1)=2 (10) even, c_1=0.floor(5 /2^2)=1 (1) odd, c_2=1.floor(5 /2^3)=0 even.Hmm, so floor(n /2^k) is odd when the k-th bit in n's binary representation is 1. But in the case of n=5, floor(5 /2^1)=2 which is even, but the first bit (k=1) in n=101 is 0.Wait, maybe there's a relationship between the bits of n and the parity of floor(n /2^k).Indeed, floor(n /2^k) is the number obtained by removing the last k bits of n. So if n's binary is b_tb_{t-1}...b_0, then floor(n /2^k) is b_tb_{t-1}...b_k.This number is odd if and only if its last bit (b_k) is 1.Therefore, floor(n /2^k) is odd if and only if the k-th bit of n is 1.Therefore, Σ_{k=1}^{∞} c_k = number of 1s in the binary representation of n, excluding the zeroth bit.But since n is odd, the zeroth bit is 1. Therefore, s_2(n) =1 + Σ_{k=1}^{∞} c_k.Therefore, Σ_{k=1}^{∞} c_k = s_2(n) -1.Putting it all together:S(n) = Σ_{k=1}^{∞} floor(n / 2^{k+1} ) + Σ_{k=1}^{∞} c_k= [ (n - s_2(n)) - (n -1)/2 ] + (s_2(n) -1 )= n - s_2(n) - (n -1)/2 + s_2(n) -1= n - (n -1)/2 -1= (2n - (n -1) -2)/2= (2n -n +1 -2)/2= (n -1)/2.Which is exactly what we needed to prove. Therefore, the sum S(n) = Σ_{k=1}^{∞} floor((n / 2^{k} +1)/2 ) = (n -1)/2.Therefore, returning to the original problem, the sum of floor(log₂(n/k)) over odd k from 1 to n is equal to S(n) = (n -1)/2. Hence, proved.Final Answerboxed{dfrac{n-1}{2}}</think>To prove that for a positive odd integer ( n ),leftlfloorlog _{2} nrightrfloor+leftlfloorlog _{2}(n / 3)rightrfloor+leftlfloorlog _{2}(n / 5)rightrfloor+leftlfloorlog _{2}(n / 7)rightrfloor+cdots+leftlfloorlog _{2}(n / n)rightrfloor=frac{n-1}{2},we analyze the sum of floor functions of logarithms.1. Understanding the Terms: - Each term ( leftlfloor log_2(n/k) rightrfloor ) for odd ( k ) corresponds to the largest integer ( m ) such that ( 2^m leq n/k < 2^{m+1} ). - This is equivalent to ( n/2^{m+1} < k leq n/2^m ).2. Counting Odd ( k ) in Intervals: - For each ( m ), count the number of odd integers ( k ) in the interval ( (n/2^{m+1}, n/2^m] ). - The number of odd integers in an interval ( (a, b] ) is given by ( leftlfloor frac{b+1}{2} rightrfloor - leftlfloor frac{a+1}{2} rightrfloor ).3. Summing Over All ( m ): - The total sum is ( sum_{m=0}^{infty} m cdot left( leftlfloor frac{n/2^m + 1}{2} rightrfloor - leftlfloor frac{n/2^{m+1} + 1}{2} rightrfloor right) ).4. Telescoping Series: - This sum telescopes to ( sum_{k=1}^{infty} leftlfloor frac{n/2^k + 1}{2} rightrfloor ).5. Binary Representation Analysis: - For an odd ( n ), the sum ( sum_{k=1}^{infty} leftlfloor frac{n/2^k + 1}{2} rightrfloor ) simplifies to ( frac{n-1}{2} ) by considering the binary representation of ( n ) and the properties of floor functions.6. Final Proof: - Using the identity that the sum of floor functions involving binary shifts and properties of odd integers, we show that the sum of the floor functions of logarithms equals ( frac{n-1}{2} ).Thus, the final answer is:boxed{dfrac{n-1}{2}}]

🔑:Solution: Note that leftlfloorlog _{2} krightrfloor is the cardinality of the set left{2,4,8, ldots, 2^{leftlfloorlog _{2} krightrfloor}right}, i.e., the number of powers of 2 that are even and are at most k. Then leftlfloorlog _{2}(n / k)rightrfloor is the number of even powers of 2 that are at most n / k, or equivalently (multiplying each such number by k ) the number of positive even numbers leq n whose greatest odd divisor is k. Summing over all odd k, we get the number of even numbers leq n, which is just (n-1) / 2.Let sigma(n) denote the sum of the (positive) divisors of n, including 1 and n itself.

🧠:[<think>Okay, so I need to find the equation of the parabola given by y = ax² + bx + c. The problem states that the parabola intersects the x-axis at points A and B, the vertex is at C(-1/4, -6 1/8), and the area of triangle ABC is 10 23/32. Hmm, let me break this down step by step.First, since the vertex is given, I know that the vertex form of a parabola is y = a(x - h)² + k, where (h, k) is the vertex. Here, the vertex C is at (-1/4, -6 1/8). Converting -6 1/8 to an improper fraction might be easier for calculations. Let's see: 6*8=48, plus 1 is 49, so -49/8. So the vertex form would be y = a(x + 1/4)² - 49/8. But the problem gives the equation in standard form y = ax² + bx + c. Maybe I can expand the vertex form to standard form. Let's try that. Expanding (x + 1/4)² gives x² + (1/2)x + 1/16. Multiply by a: a x² + (a/2)x + a/16. Then subtract 49/8: y = a x² + (a/2)x + (a/16 - 49/8). So in standard form, that would be y = ax² + (a/2)x + (a/16 - 49/8). So the coefficients are b = a/2 and c = a/16 - 49/8. So if I can find the value of a, I can get b and c as well.Now, the parabola intersects the x-axis at points A and B. The roots of the parabola are the x-intercepts, so we can call them A (x₁, 0) and B (x₂, 0). The vertex is at (-1/4, -49/8). The area of triangle ABC is given as 10 23/32. Let's convert that to an improper fraction too. 10*32 = 320, plus 23 is 343, so 343/32.So, triangle ABC has vertices at A, B, and C. The area of a triangle can be found using the formula: (1/2)*base*height. In this case, the base would be the distance between points A and B on the x-axis, which is |x₂ - x₁|. The height would be the vertical distance from the vertex C to the x-axis, which is the absolute value of the y-coordinate of C. Since the vertex is below the x-axis (y is negative), the height is 49/8.Wait, but hold on. The area of triangle ABC is (1/2)*base*height. So here, base is |x₂ - x₁| and height is 49/8. But the area is given as 343/32. So:(1/2)*|x₂ - x₁|*(49/8) = 343/32Let me compute that. Multiply both sides by 2:|x₂ - x₁|*(49/8) = 343/16Divide both sides by 49/8:|x₂ - x₁| = (343/16) / (49/8) = (343/16)*(8/49) = (343*8)/(16*49)Simplify 343 is 7³, 49 is 7², 8 is 2³, 16 is 2⁴.So:(7³ * 2³)/(2⁴ * 7²) )= (7/2)So |x₂ - x₁| = 7/2.So the distance between the roots is 7/2.But the roots of a quadratic are given by x = [-b ± sqrt(b² - 4ac)]/(2a). The difference between the roots is [sqrt(b² - 4ac)/a], because the ± would make the difference between the two roots 2*sqrt(b² - 4ac)/(2a) = sqrt(b² - 4ac)/a. Wait, actually, let me verify that.If the roots are x₁ and x₂, then x₁ + x₂ = -b/a and x₁x₂ = c/a. The difference |x₁ - x₂| can be computed as sqrt((x₁ + x₂)² - 4x₁x₂) = sqrt( (b²/a²) - 4c/a ) = sqrt(b² - 4ac)/|a|.But since we have the vertex at (-1/4, -49/8), and since the axis of symmetry is x = -b/(2a) = -1/4. So -b/(2a) = -1/4 => b = (2a)(1/4) = a/2. Which matches the earlier conclusion that b = a/2.So the difference between roots is sqrt(b² - 4ac)/|a|. But since the distance is 7/2, we can set sqrt(b² - 4ac)/|a| = 7/2.But since we already have expressions for b and c in terms of a, we can substitute those in.From earlier:b = a/2c = a/16 - 49/8So substitute into discriminant D = b² - 4ac:D = (a/2)^2 - 4a*(a/16 - 49/8)Compute term by term:(a/2)^2 = a²/44a*(a/16) = 4a*(a)/16 = a²/44a*(-49/8) = - (4a*49)/8 = - (49a)/2Therefore, D = a²/4 - (a²/4 - 49a/2) = a²/4 - a²/4 + 49a/2 = 49a/2So sqrt(D)/|a| = sqrt(49a/2)/|a| = (7 sqrt(a/2))/|a|But sqrt(a/2) is real only if a > 0, since the parabola intersects the x-axis at two points, so it opens upwards. Therefore, a > 0, so |a| = a. Therefore, (7 sqrt(a/2))/a = 7 / sqrt(2a)But we have this equal to 7/2, so:7 / sqrt(2a) = 7/2Divide both sides by 7:1 / sqrt(2a) = 1/2Take reciprocal:sqrt(2a) = 2Square both sides:2a = 4 => a = 2So a = 2. Then b = a/2 = 1, and c = a/16 - 49/8 = 2/16 - 49/8 = 1/8 - 49/8 = (-48)/8 = -6Therefore, the equation of the parabola is y = 2x² + x - 6.Wait, but let me verify this. Let's check if the vertex is indeed at (-1/4, -49/8). For the standard form y = 2x² + x -6, the vertex x-coordinate is -b/(2a) = -1/(4) = -1/4, which matches. Then the y-coordinate is f(-1/4). Let's compute that:2*(-1/4)^2 + (-1/4) -6 = 2*(1/16) - 1/4 -6 = 1/8 - 2/8 - 48/8 = (1 - 2 - 48)/8 = (-49)/8, which is -6 1/8. Correct. So the vertex is correct.Now, check the roots. The quadratic equation is 2x² + x -6 =0. Compute discriminant D = 1 + 48 = 49. So roots are x = [-1 ±7]/4. So (6/4)= 3/2 and (-8/4)= -2. So points A(-2, 0) and B(3/2, 0). The distance between them is 3/2 - (-2) = 3/2 + 2 = 7/2, which matches the earlier calculation. The area of triangle ABC would be (1/2)*7/2*49/8 = (7/4)*(49/8) = 343/32, which is 10 23/32. That's correct. So everything checks out.But wait, let me make sure I didn't skip any steps. The area calculation: base is 7/2, height is 49/8, so area is 1/2 * 7/2 * 49/8. Compute 1/2 *7/2 =7/4. 7/4 *49/8= (7*49)/(4*8)=343/32. Yes, which is the given area. So all correct.Therefore, the equation of the parabola is y = 2x² + x -6.Final AnswerThe equation of the parabola is boxed{y = 2x^2 + x - 6}.</think>Given the parabola ( y = ax^2 + bx + c ) with vertex ( Cleft(-frac{1}{4}, -6 frac{1}{8}right) ) and the area of (triangle ABC) as ( 10 frac{23}{32} ), we need to find the equation of the parabola.1. Vertex Form Conversion: The vertex form of the parabola is ( y = aleft(x + frac{1}{4}right)^2 - frac{49}{8} ). Expanding this, we get: [ y = aleft(x^2 + frac{1}{2}x + frac{1}{16}right) - frac{49}{8} ] Converting to standard form: [ y = ax^2 + frac{a}{2}x + left(frac{a}{16} - frac{49}{8}right) ] Thus, ( b = frac{a}{2} ) and ( c = frac{a}{16} - frac{49}{8} ).2. Distance Between Roots: The roots ( A ) and ( B ) are the x-intercepts. The distance between the roots is given by ( sqrt{frac{b^2 - 4ac}{a^2}} ). The area of (triangle ABC) is: [ frac{1}{2} times text{base} times text{height} = frac{343}{32} ] Here, the base is ( |x_2 - x_1| ) and the height is ( frac{49}{8} ). Solving for the base: [ frac{1}{2} times |x_2 - x_1| times frac{49}{8} = frac{343}{32} ] [ |x_2 - x_1| = frac{7}{2} ]3. Discriminant Calculation: Using the distance between roots: [ sqrt{frac{b^2 - 4ac}{|a|}} = frac{7}{2} ] Substituting ( b = frac{a}{2} ) and ( c = frac{a}{16} - frac{49}{8} ): [ sqrt{frac{left(frac{a}{2}right)^2 - 4aleft(frac{a}{16} - frac{49}{8}right)}{a}} = frac{7}{2} ] Simplifying the discriminant: [ sqrt{frac{49a}{2}} = frac{7}{2} ] Solving for ( a ): [ a = 2 ]4. Finding ( b ) and ( c ): Substituting ( a = 2 ): [ b = frac{2}{2} = 1 ] [ c = frac{2}{16} - frac{49}{8} = frac{1}{8} - frac{49}{8} = -6 ]The equation of the parabola is (boxed{y = 2x^2 + x - 6}).,<think>Okay, let's try to figure out the equation of the parabola. The problem states that the parabola intersects the x-axis at points A and B, has a vertex at C(-1/4, -6 1/8), and the area of triangle ABC is 10 23/32. Hmm, I need to find a, b, c such that the equation y = ax² + bx + c satisfies these conditions.First, let's recall that the vertex form of a parabola is y = a(x - h)² + k, where (h, k) is the vertex. Since the vertex is given as (-1/4, -6 1/8), which is (-0.25, -6.125) in decimal form. So maybe I can start by writing the equation in vertex form and then convert it to standard form.So, substituting h and k into the vertex form:y = a(x + 1/4)² - 6 1/8.But the problem gives the equation in standard form, y = ax² + bx + c, so I need to expand this vertex form to standard form. Let me do that.First, note that 6 1/8 is 49/8 as an improper fraction. So, the vertex form is:y = a(x + 1/4)² - 49/8.Expanding the squared term:(x + 1/4)² = x² + (1/2)x + 1/16.Multiply by a:y = a(x² + (1/2)x + 1/16) - 49/8.So, expanding:y = a x² + (a/2)x + a/16 - 49/8.Therefore, in standard form:y = ax² + (a/2)x + (a/16 - 49/8).So that gives us coefficients:b = a/2,c = a/16 - 49/8.So if I can find the value of a, then I can get b and c.Now, the parabola intersects the x-axis at points A and B. So, the roots are x_A and x_B. The distance between A and B on the x-axis is |x_B - x_A|, which is the length of the base of triangle ABC. The height would be the y-coordinate of the vertex, but since the vertex is below the x-axis (since y-coordinate is negative), the height is | -6 1/8 | = 6 1/8.Wait, but actually, the area of triangle ABC is given. Since the vertex is C, the triangle is formed by points A, B, and C. Points A and B are on the x-axis, and point C is the vertex. So, triangle ABC is a triangle with base AB and height equal to the vertical distance from C to the x-axis, which is |y-coordinate of C|. So, the area should be (1/2)*AB*height. The height is 6 1/8, which is 49/8. The area is given as 10 23/32, which is 10 + 23/32 = 320/32 + 23/32 = 343/32.Therefore, according to the area formula:Area = (1/2) * AB * (49/8) = 343/32.Let's solve for AB.Multiply both sides by 2:AB * (49/8) = 686/32.Simplify 686/32: divide numerator and denominator by 2: 343/16.Therefore:AB = (343/16) / (49/8) = (343/16) * (8/49) = (343 * 8) / (16 * 49).Simplify:343 is 7³ = 343, 49 is 7². So, 343 / 49 = 7. Then 8/16 = 1/2. Therefore:AB = 7 * (1/2) = 7/2.So, AB = 7/2. That is, the distance between points A and B is 7/2.But AB is also equal to the absolute difference between the roots of the quadratic equation. For a quadratic equation ax² + bx + c = 0, the difference between roots is sqrt(D)/|a|, where D is the discriminant. But wait, actually, the distance between roots is |x_B - x_A| = sqrt((x_A + x_B)^2 - 4x_Ax_B). Since for quadratic equation, x_A + x_B = -b/a and x_Ax_B = c/a. Therefore:|x_B - x_A| = sqrt( ( (-b/a)^2 - 4(c/a) ) ) = sqrt( (b² - 4ac)/a² ) = sqrt(b² - 4ac)/|a|.But since the parabola intersects the x-axis at two points, the discriminant must be positive. So, D = b² - 4ac > 0.We have AB = sqrt(D)/|a| = 7/2.So, sqrt(b² - 4ac)/|a| = 7/2.But we also have expressions for b and c in terms of a from earlier:b = a/2,c = a/16 - 49/8.So, let's substitute these into the discriminant.First, compute D = b² - 4ac.Substituting:D = (a/2)² - 4a*(a/16 - 49/8).Compute each term:(a/2)² = a²/4.4a*(a/16 - 49/8) = 4a*(a/16) - 4a*(49/8) = (4a²)/16 - (196a)/8 = a²/4 - 24.5a.Wait, 4a*(a/16) = (4a * a)/16 = 4a²/16 = a²/4. Correct.4a*(49/8) = (4a * 49)/8 = (196a)/8 = 24.5a. Yes.So, D = a²/4 - [a²/4 - 24.5a] = a²/4 - a²/4 + 24.5a = 24.5a.But 24.5a is equal to (49/2)a.Therefore, D = (49/2)a.Therefore, sqrt(D)/|a| = sqrt( (49/2)a ) / |a|.Since we have sqrt( (49/2)a ) / |a| = 7/2.Let me write this equation:sqrt( (49/2)a ) / |a| = 7/2.We can square both sides to eliminate the square root:( (49/2)a ) / a² = 49/4.Simplify numerator and denominator:(49/2)a / a² = (49/2)/a = 49/(2a).Set equal to 49/4:49/(2a) = 49/4.Divide both sides by 49:1/(2a) = 1/4.Therefore:2a = 4 => a = 2.Wait, that seems straightforward. So, a = 2.But let me check if this is correct. Let's verify step by step.First, discriminant D = 49/2 * a.Therefore, sqrt(D)/|a| = sqrt(49/2 * a)/|a|.But if a is positive, then |a| = a. So:sqrt( (49/2)*a ) / a = sqrt(49/(2a)) * sqrt(a) / a = sqrt(49/(2a)) / sqrt(a) ) = sqrt(49/(2a)) * 1/sqrt(a) ) = sqrt(49/(2a²)) = 7/(sqrt(2)*|a|).Wait, maybe I miscalculated earlier.Wait, sqrt(49/2 * a) divided by |a|:sqrt(49/2 * a) / |a| = sqrt( (49a)/2 ) / |a|.If a is positive, then this is sqrt(49a/2)/a = (7 sqrt(a/2)) ) / a = 7/(sqrt(2a)).Set equal to 7/2:7/(sqrt(2a)) = 7/2.Cancel the 7s:1/sqrt(2a) = 1/2.Take reciprocals:sqrt(2a) = 2.Square both sides:2a = 4 => a = 2.Yes, that checks out. So a = 2.Therefore, a = 2.Then, from earlier:b = a/2 = 2/2 = 1.c = a/16 - 49/8 = 2/16 - 49/8 = 1/8 - 49/8 = (-48/8) = -6.Therefore, the equation of the parabola is y = 2x² + x - 6.Wait, but let's verify this.First, check the vertex. The standard form is y = 2x² + x - 6.The x-coordinate of the vertex is -b/(2a) = -1/(4) which matches the given vertex at x = -1/4.Then, the y-coordinate is f(-1/4):f(-1/4) = 2*(-1/4)^2 + (-1/4) - 6 = 2*(1/16) - 1/4 - 6 = 1/8 - 2/8 - 48/8 = (1 - 2 - 48)/8 = (-49)/8 = -6.125, which is -6 1/8. Correct.So the vertex is correct.Now, check the area. The roots are the solutions to 2x² + x - 6 = 0.Using quadratic formula:x = [-1 ± sqrt(1 + 48)] / 4 = [-1 ± 7]/4.Thus, roots at x = (6)/4 = 3/2 and x = (-8)/4 = -2. So points A(-2, 0) and B(3/2, 0). The distance AB is 3/2 - (-2) = 3/2 + 2 = 7/2, which matches the earlier calculation.The height is 49/8. Therefore, area is (1/2)*(7/2)*(49/8) = (7/4)*(49/8) = 343/32, which is 10 23/32. Correct.Therefore, the equation y = 2x² + x - 6 satisfies all given conditions.Hence, the answer is y = 2x² + x - 6.Final AnswerThe equation of the parabola is boxed{y = 2x^2 + x - 6}.</think>Given the parabola ( y = ax^2 + bx + c ) with vertex ( Cleft(-frac{1}{4}, -6 frac{1}{8}right) ) and the area of (triangle ABC) as (10 frac{23}{32}), we need to find the equation of the parabola.1. Vertex Form Conversion: - The vertex form of the parabola is ( y = aleft(x + frac{1}{4}right)^2 - frac{49}{8} ). - Expanding this, we get: [ y = aleft(x^2 + frac{1}{2}x + frac{1}{16}right) - frac{49}{8} ] - Converting to standard form: [ y = ax^2 + frac{a}{2}x + frac{a}{16} - frac{49}{8} ] - Thus, ( b = frac{a}{2} ) and ( c = frac{a}{16} - frac{49}{8} ).2. Area of (triangle ABC): - The area is given as ( frac{343}{32} ). - The base ( AB ) is the distance between the roots, and the height is ( frac{49}{8} ). - Using the area formula: [ frac{1}{2} times AB times frac{49}{8} = frac{343}{32} ] - Solving for ( AB ): [ AB = frac{343}{16} div frac{49}{8} = frac{7}{2} ]3. Discriminant Calculation: - The distance between roots ( AB = frac{sqrt{b^2 - 4ac}}{|a|} = frac{7}{2} ). - Substituting ( b ) and ( c ): [ sqrt{left(frac{a}{2}right)^2 - 4aleft(frac{a}{16} - frac{49}{8}right)} / |a| = frac{7}{2} ] - Simplifying the discriminant: [ sqrt{frac{49}{2}a} / |a| = frac{7}{2} ] - Solving for ( a ): [ frac{7}{sqrt{2a}} = frac{7}{2} implies sqrt{2a} = 2 implies a = 2 ]4. Finding ( b ) and ( c ): - ( b = frac{a}{2} = 1 ) - ( c = frac{a}{16} - frac{49}{8} = -6 )Thus, the equation of the parabola is (boxed{y = 2x^2 + x - 6}).]

🔑:Solve for the vertex coordinates of y=a x^{2}+b x+c asmathrm{C}left(-frac{mathrm{b}}{2 mathrm{a}}, frac{4 mathrm{ac}-mathrm{b}^{2}}{4 mathrm{a}}right),thus -frac{mathrm{b}}{2 mathrm{a}}=-frac{1}{4},begin{array}{c}frac{4 mathrm{ac}-mathrm{b}^{2}}{4 mathrm{a}}=-6 frac{1}{8} . because mathrm{S}_{triangle mathrm{ABC}}=frac{1}{2}|mathrm{AB}| cdot mathrm{h}_{0}=frac{1}{2}|mathrm{AB}| cdot 6 frac{1}{8}=10 frac{23}{32} . therefore quad|mathrm{AB}|=frac{7}{2} . frac{sqrt{mathrm{b}^{2}-4 mathrm{ac}}}{|mathrm{a}|}=frac{7}{2} .end{array}Solving (1), (2), and (3) simultaneously, we getmathrm{a}=2, mathrm{~b}=1, mathrm{c}=-6 text {. }Therefore, the equation of the parabola is y=2 x^{2}+x-6.

🧠:[<think>Okay, let's see. The problem is about four children—Zhao, Qian, Sun, and Li—who all like playing chess. Some like Chinese chess, some like Go, and some might like both. We need to find out how many different ways their preferences can be arranged. Hmm, so the question is asking for the number of distinct ways these four people can have their preferences for Chinese chess and Go, considering that some may like both. First, I need to make sure I understand the problem correctly. Each child can have one or both of the two preferences: Chinese chess or Go. But the problem says "someone likes playing Chinese chess, someone likes playing Go, and someone likes both." So, does this mean that there must be at least one person in each category? Like, there must be at least one who likes Chinese chess, at least one who likes Go, and at least one who likes both? Or does it mean that among the four, there are some who like Chinese chess, some who like Go, and some who like both, but not necessarily all three categories must be non-empty? Wait, the wording is a bit ambiguous here. Let me read it again."Among them, someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess." So, the Chinese translation might be clearer, but in English, "someone" in each category. So, maybe it's required that all three categories are non-empty: at least one likes only Chinese chess, at least one likes only Go, and at least one likes both. But wait, maybe "someone" is just referring to the existence of people in each category. So, the total set must include people who like Chinese chess, people who like Go, and people who like both. But actually, in Chinese, the original problem might use terms that make it clearer. But since we're working with the translation, we need to be careful.Alternatively, maybe it's just saying that each child is in at least one of the categories: some like Chinese chess, some like Go, and some like both. But the problem states "someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess." So, the key here is that all three groups must be non-empty. That is, there must be at least one person who likes only Chinese chess, at least one who likes only Go, and at least one who likes both. Otherwise, if there were, say, no one who likes both, then the statement "someone likes both" wouldn't hold. Therefore, the problem is asking for the number of ways to assign the four children into these three categories (Chinese chess only, Go only, both), with each category having at least one person.Wait, but hold on. If we think of it as each child can be assigned to any subset of {Chinese chess, Go}, including both. But the problem says that in the group, there must be someone who likes Chinese chess, someone who likes Go, and someone who likes both. So, we need to count all possible assignments where all three categories are non-empty. But each child can be in one of three categories: C only, G only, or both. However, some might argue that "someone likes Chinese chess" could include those who like both. Similarly, "someone likes Go" could include those who like both. So, maybe the problem is just saying that there is at least one person who likes Chinese chess (could be only C or both), at least one who likes Go (could be only G or both), and at least one who likes both. In that case, the three conditions are:1. At least one person likes Chinese chess (either only C or both).2. At least one person likes Go (either only G or both).3. At least one person likes both.But that would mean that the overlap between the two sets (C and G) must be at least one. So, the total number of possible assignments is the number of ways to assign each of the four children to subsets such that:- The set of people who like Chinese chess (C or both) is non-empty.- The set of people who like Go (G or both) is non-empty.- The intersection (people who like both) is non-empty.But each child has three choices: C only, G only, or both. However, if we consider that each child can independently be in any subset of {C, G}, then each child has four possibilities: likes neither, likes only C, likes only G, or likes both. But the problem states that all four children like playing chess, so "likes neither" is not allowed. Therefore, each child must like at least one of the two types. So, each child has three choices: C only, G only, or both. So, in total, without any restrictions, there would be 3^4 = 81 possible assignments.But the problem says "someone likes Chinese chess, someone likes Go, and someone likes both." So, we need to exclude the cases where no one likes Chinese chess, or no one likes Go, or no one likes both. Wait, let's parse this again.The problem states that "someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess." So, in other words, all three categories must have at least one person. So, the assignments where:- All four children like only Chinese chess: invalid, because no one likes Go or both.- All four children like only Go: invalid, same reason.- All four children like both: invalid, because no one likes only Chinese chess or only Go.- Assignments where there are only people in two categories, for example, some like only Chinese chess and some like both, but none like only Go. Similarly for other combinations.Therefore, the problem is equivalent to finding the number of onto functions from the four children to the three categories {C only, G only, both}, with each category having at least one child. Wait, but actually, in standard combinatorics, this is similar to surjective functions. The number of ways to distribute n distinct objects into k distinct boxes with each box containing at least one object is given by k! * S(n, k), where S(n, k) is the Stirling numbers of the second kind. However, in this case, the boxes are distinct (C only, G only, both), and the children are distinct. So, yes, the formula would be applicable here.So, the number of ways is 3! * S(4, 3). The Stirling number S(4, 3) is equal to 6. Therefore, 3! * 6 = 6 * 6 = 36. But wait, is that correct?Wait, let me verify. The formula for the number of onto functions from a set of size n to a set of size k is k! * S(n, k). For n=4 and k=3, S(4,3) is 6. So, 6 * 6 = 36. But let's verify this another way. The number of surjective functions is also equal to the inclusion-exclusion principle: 3^4 - 3*2^4 + 3*1^4. Let's calculate that:3^4 = 813*2^4 = 3*16 = 483*1^4 = 3*1 = 3So, 81 - 48 + 3 = 36. Yes, that's correct. Therefore, the number of ways is 36.But hold on, in our problem, the three categories are C only, G only, and both. But wait, in standard surjective functions, the categories are distinct. However, in our case, the categories are:1. C only2. G only3. BothBut these are three distinct categories, so the formula applies. Therefore, the number of ways is 36. Therefore, the answer should be 36. But let me think again.Wait, but in our problem, each child is assigned to exactly one of the three categories: C only, G only, or both. But actually, "both" is not exclusive from C or G. Wait, actually, if we model each child's preference as a subset of {C, G}, where each child must choose at least one. Then, the possible subsets are {C}, {G}, {C, G}. So, each child has three choices. The problem states that in the entire set of four children, all three subsets must be present. That is, there is at least one child who likes only C, at least one who likes only G, and at least one who likes both. So, in other words, we need to count the number of functions from four children to the three subsets {C}, {G}, {C, G}, such that all three subsets are used. Which is exactly the same as surjective functions, hence 3! * S(4,3) = 36. So, the answer is 36.But let me confirm with another approach. Let's think of it as assigning each child to one of three categories: C only, G only, or both. Each child has three choices. The total number is 3^4 = 81. Now, subtract the assignments where one or more categories are missing.First, subtract the cases where C only is missing. If C only is missing, then all children are either G only or both. Similarly, if G only is missing, all are C only or both. If both is missing, all are C only or G only.But we need to use inclusion-exclusion here.Let A be the set of assignments where C only is missing.B be the set where G only is missing.C be the set where both is missing.We need to compute the total assignments minus |A ∪ B ∪ C|.By inclusion-exclusion:|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|.First, |A|: assignments where C only is missing. So, each child can be G only or both. That's 2^4 = 16.Similarly, |B| = 2^4 = 16 (C only or both).|C|: assignments where both is missing. So, each child can be C only or G only. That's 2^4 = 16.Now, |A ∩ B|: assignments where both C only and G only are missing. So, everyone must be in both. But since both is also a category, but if both C only and G only are missing, then everyone must be in the both category. However, the problem states that everyone must like at least one type of chess, so being in both is allowed. So, there's only one such assignment: all four children are in both. So, |A ∩ B| = 1.Similarly, |A ∩ C|: assignments where C only and both are missing. So, everyone must be in G only. That's only one assignment: all four in G only. Similarly, |B ∩ C|: assignments where G only and both are missing. Everyone must be in C only: one assignment.Finally, |A ∩ B ∩ C| is impossible, since all three categories can't be missing. So, that's 0.Therefore, |A ∪ B ∪ C| = 16 + 16 + 16 - 1 - 1 - 1 + 0 = 48 - 3 = 45.Therefore, the number of valid assignments is total assignments minus |A ∪ B ∪ C| = 81 - 45 = 36. Which matches the previous result. So, the answer should be 36.But let me check if there's another interpretation. Suppose the problem requires that each child can like Chinese chess, Go, or both, but the problem says "someone likes Chinese chess, someone likes Go, and someone likes both." So, if we interpret this as the sets of Chinese chess lovers and Go lovers must each be non-empty, and their intersection must also be non-empty. Then, this is another way to model the problem.In set theory terms, let S be the set of children who like Chinese chess, and T be the set who like Go. We need S ≠ ∅, T ≠ ∅, and S ∩ T ≠ ∅. Then, the number of such pairs (S, T) is the number we need.But how does this relate to the original problem? Each child can be in S, T, both, or neither. But the problem states all four children like playing chess, so neither is not allowed. Therefore, each child must be in at least one of S or T. So, the total number of possible pairs (S, T) where each child is in S, T, or both is 3^4 = 81, as each child has three choices.But we need the number of pairs (S, T) where S ≠ ∅, T ≠ ∅, and S ∩ T ≠ ∅. Let's compute that.Total pairs: 3^4 = 81.Subtract the cases where S = ∅: If S is empty, then all children must be in T only. But since S is empty, and T must be non-empty (because all children like chess), but in this case, T can be all four children. However, since we are subtracting S = ∅, which would mean that no one likes Chinese chess, but everyone must like Go (since they can't like neither). Wait, but if S = ∅, then T must include all children. But since all children must like at least one type, so if S is empty, then T must be all four. But in this case, does S ∩ T = ∅? Because S is empty, so yes. Therefore, to satisfy S ∩ T ≠ ∅, we need to subtract the cases where S = ∅ or T = ∅ or S ∩ T = ∅.Wait, no, the problem says "someone likes Chinese chess, someone likes Go, and someone likes both." So, S ≠ ∅, T ≠ ∅, and S ∩ T ≠ ∅. Therefore, the number of pairs (S, T) where S ≠ ∅, T ≠ ∅, and S ∩ T ≠ ∅.To compute this, we can use inclusion-exclusion again.Total number of pairs where S ≠ ∅ and T ≠ ∅: (2^4 - 1)(2^4 - 1) - ... Wait, no. Wait, actually, in this model, each child is assigned to be in S, T, both, or neither. But since neither is not allowed, each child has three choices: S only, T only, or both. So, the total number is 3^4 = 81. However, the number of pairs (S, T) where S ≠ ∅, T ≠ ∅, and S ∩ T ≠ ∅.Alternatively, the number of pairs (S, T) where S and T are subsets of the four children, such that S ≠ ∅, T ≠ ∅, and S ∩ T ≠ ∅, and every child is in at least one of S or T.This is equivalent to:Total number of such pairs = Total number of pairs (S, T) where every child is in S ∪ T, minus the pairs where S = ∅ or T = ∅, minus the pairs where S ∩ T = ∅.But this might be complicated. Alternatively, use inclusion-exclusion.First, total number of pairs (S, T) where every child is in S ∪ T: This is equivalent to each child being in S, T, or both. So, for each child, 3 choices. Total: 3^4 = 81.Now, subtract the cases where S = ∅. If S = ∅, then T must include all children (since they must be in S ∪ T). So, T = all four. Only one such pair. Similarly, subtract the cases where T = ∅: S must include all children. Only one such pair. So, subtract 1 + 1 = 2.Now, we have 81 - 2 = 79. But we also need to subtract the cases where S ∩ T = ∅. Wait, but S ∩ T = ∅ means that no child is in both S and T. So, each child is in S only or T only. However, since every child must be in at least one, each child is in exactly one of S or T. So, the number of such pairs is 2^4 = 16 (each child can be in S or T). But we have already subtracted the cases where S = ∅ or T = ∅. So, the number of pairs where S ∩ T = ∅ and S ≠ ∅ and T ≠ ∅ is 16 - 2 = 14.Therefore, the total number of valid pairs is 81 - 2 - 14 = 65.Wait, but this contradicts our previous answer of 36. So, which is correct?But here's the confusion: in the first approach, we considered each child being assigned to one of three categories: C only, G only, or both. We required all three categories to be non-empty, leading to 36 possibilities.In the second approach, we model it as subsets S and T (Chinese chess and Go lovers), with S ≠ ∅, T ≠ ∅, and S ∩ T ≠ ∅. But in this case, the number is 65. But these are different models.The discrepancy arises because the two models are different. In the first model, each child is assigned to exactly one category: C only, G only, or both. In the second model, we are considering all possible subsets S and T (with the constraints), which allows for more possibilities. For example, in the second model, multiple children can be in both S and T, but in the first model, "both" is a separate category.Wait, no. Actually, in both models, each child can be in S (C), T (G), or both. The difference is in how we count the assignments. In the first model, we count the number of assignments where each child is assigned to exactly one of the three categories (C only, G only, both), with all three categories present. This is 3^4 - cases where one or more categories are missing, leading to 36.In the second model, we are counting the number of pairs (S, T) where S and T are subsets of the children, with S ≠ ∅, T ≠ ∅, S ∩ T ≠ ∅, and every child is in S ∪ T. This would include all possible combinations where each child is in S only, T only, or both. The count here is different because when considering subsets S and T, the same pair (S, T) can be achieved in different ways. For example, if two children are in both S and T, the pair (S, T) is unique regardless of how many children are in both. However, in the first model, each child's individual assignment is considered.Wait, but actually, in the first model, each child is assigned to a category, so the total assignments are 3^4 = 81. The second model is also considering assignments where each child is assigned to S, T, or both, but counting distinct pairs (S, T). However, the same pair (S, T) can correspond to multiple assignments. For example, suppose S = {1,2}, T = {2,3}. This corresponds to child 2 being in both, child 1 in S only, child 3 in T only, and child 4 not being in either, but in our problem, child 4 must be in at least one. So, perhaps the second model is not the right way.But wait, in the problem statement, it says "someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess." This can be interpreted as:- There exists at least one child who likes Chinese chess (regardless of whether they also like Go).- There exists at least one child who likes Go (regardless of whether they also like Chinese chess).- There exists at least one child who likes both.So, in set terms, if S is the set of children who like Chinese chess (including those who like both), and T is the set who like Go (including those who like both), then we need S ≠ ∅, T ≠ ∅, and S ∩ T ≠ ∅.Therefore, the number of such pairs (S, T) where S ≠ ∅, T ≠ ∅, and S ∩ T ≠ ∅, with the additional constraint that every child is in S ∪ T (since they must like at least one type of chess). To compute this, we can use inclusion-exclusion. The total number of pairs (S, T) where every child is in S ∪ T is 3^4 = 81 (each child can be in S only, T only, or both).Now, subtract the cases where S = ∅ or T = ∅ or S ∩ T = ∅.But wait, we need to subtract the cases where S = ∅, T = ∅, or S ∩ T = ∅.First, the number of pairs where S = ∅: As before, S = ∅ implies T must include all children, which is 1 possibility.Similarly, pairs where T = ∅: 1 possibility.Pairs where S ∩ T = ∅: This means S and T are disjoint. Since every child is in S ∪ T, and S and T are disjoint, this is equivalent to partitioning the four children into S and T. Each child can be in S or T, but not both. So, 2^4 = 16 possibilities. However, this includes the cases where S = ∅ or T = ∅. So, to get the number of pairs where S ∩ T = ∅ and S ≠ ∅ and T ≠ ∅, we subtract 2, giving 14.Therefore, using inclusion-exclusion:Number of invalid pairs = |S=∅| + |T=∅| + |S ∩ T=∅| - |S=∅ and T=∅| - |S=∅ and S ∩ T=∅| - |T=∅ and S ∩ T=∅| + |S=∅ and T=∅ and S ∩ T=∅|.But this is getting complex. Let's use the standard inclusion-exclusion for three sets.But maybe it's simpler to consider:Total valid pairs = Total pairs - (pairs where S=∅ + pairs where T=∅ + pairs where S ∩ T=∅) + (pairs where S=∅ and T=∅ + pairs where S=∅ and S ∩ T=∅ + pairs where T=∅ and S ∩ T=∅) - pairs where all three conditions are violated.But pairs where all three are violated would be S=∅, T=∅, and S ∩ T=∅, which is impossible.So:Number of valid pairs = Total pairs - (A + B + C) + (D + E + F) - GWhere:A = pairs where S=∅:1B = pairs where T=∅:1C = pairs where S ∩ T=∅:16D = pairs where S=∅ and T=∅:1 (only possible if S=∅ and T=∅, but since children must like at least one chess, this is impossible. So D=0)E = pairs where S=∅ and S ∩ T=∅: If S=∅, then S ∩ T=∅ is automatically true. So this is the same as A=1F = pairs where T=∅ and S ∩ T=∅: Similarly, if T=∅, then S ∩ T=∅ is automatically true. So this is B=1G = pairs where S=∅, T=∅, and S ∩ T=∅: impossible, so 0Therefore:Number of valid pairs = 81 - (1 + 1 + 16) + (0 + 1 + 1) - 0 = 81 - 18 + 2 = 65.But according to this, the answer is 65. But this contradicts our first approach. Which one is correct?Wait, but in the problem statement, it says "the different ways these four people can like these two types of chess". So, if we consider that each child's preference is either Chinese chess only, Go only, or both, and we need to have at least one in each category, then the answer is 36. However, if we model it as subsets S and T where S and T can overlap, and we count distinct pairs (S, T) with the given conditions, then the answer is 65. But which interpretation is correct? Let's go back to the problem statement:"Among them, someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess."This can be read as:1. At least one person likes Chinese chess (could be only C or both).2. At least one person likes Go (could be only G or both).3. At least one person likes both.Therefore, the three conditions are:- S ≠ ∅ (someone likes Chinese chess, where S is the set of people who like Chinese chess, including those who like both).- T ≠ ∅ (someone likes Go).- S ∩ T ≠ ∅ (someone likes both).So, in terms of subsets, the number of pairs (S, T) where S ≠ ∅, T ≠ ∅, S ∩ T ≠ ∅, and every child is in S ∪ T.According to our previous calculation, this is 65. However, this counts the number of distinct pairs (S, T). For example, if two different assignments of individual preferences result in the same sets S and T, they are considered the same in this count. But the problem says "the different ways these four people can like these two types of chess". If the question is about the individual preferences (i.e., how each child's preference is assigned), then the answer should be 36, as each child's choice is considered distinct. However, if it's about the overall distribution (i.e., how many people like C, G, or both), then it's a different count. But the problem states "the different ways these four people can like these two types of chess". This implies that each person's preference matters, so we need to count the number of assignments where each child is assigned a preference (C, G, or both), and the three conditions are met (at least one C, at least one G, at least one both).Therefore, this is the first model where each child is assigned to one of three categories, with all categories non-empty. Hence, the number is 3! * S(4, 3) = 36. Alternatively, via inclusion-exclusion: 3^4 - 3*2^4 + 3*1^4 = 81 - 48 + 3 = 36.But wait, let's verify this with an example. Suppose there are two children. How many ways would there be? Using the same logic, it would be 3^2 - 3*2^2 + 3*1^2 = 9 - 12 + 3 = 0, which is incorrect. Wait, no. If n=2, k=3, then S(2,3)=0, so 3! * 0 = 0. But in reality, if there are two children, you cannot have all three categories non-empty. So, that's correct. So, for two children, the answer is 0. Which makes sense. So, in our original problem with four children, the answer is 36.But let's see for three children: 3^3 - 3*2^3 + 3*1^3 = 27 - 24 + 3 = 6. Which is 3! * S(3,3)=6*1=6. So, yes, that's correct. For three children, there are six ways. For four children, as calculated, 36.But let's think differently. If we have four children, and we want to assign each to one of three categories (C, G, B), with all categories non-empty, then it's equivalent to the number of onto functions from four elements to three elements, which is 3! * S(4,3) = 6*6=36. So, that seems correct.Therefore, the answer should be 36. However, I need to reconcile this with the second approach where we got 65. The key difference is whether we are counting the number of assignments (36) or the number of subset pairs (65). Since the problem mentions "the different ways these four people can like these two types of chess", and since the people are distinct, we need to consider their individual assignments. Therefore, the correct answer is 36.But to make sure, let's take a small example. Suppose there are two children. According to the first approach, the number of ways is 0 because you can't have all three categories. According to the second approach, the number of pairs (S, T) where S ≠ ∅, T ≠ ∅, S ∩ T ≠ ∅, and each child is in S ∪ T. For two children:Total pairs where each child is in S ∪ T: 3^2=9.Subtract pairs where S=∅ (1), T=∅ (1), S ∩ T=∅ (2^2=4). But subtract overlaps:Number of valid pairs = 9 -1 -1 -4 +0 +1 +1 -0=9 -6 +2=5.But for two children, how many assignments are there where all three categories are present? Zero, because you can't have three categories with two children. But according to the second model, you get 5. This discrepancy shows that the two models are indeed different.Therefore, returning to the original problem, the correct interpretation depends on whether the problem requires each individual's preference to be accounted for (leading to 36) or the overall subsets (leading to 65). Given that the problem mentions four children and their individual preferences ("someone likes..."), but the answer is likely expecting the count of individual assignments where all three categories are covered. Therefore, the answer should be 36.However, I need to check the original Chinese problem's wording to be certain. Since I don't have the Chinese text, I'll proceed with the assumption that the problem wants the number of assignments where each child is assigned to C, G, or both, with all three categories present. Hence, the answer is boxed{36}.Final Answerboxed{36}</think>To determine the number of different ways Zhao, Qian, Sun, and Li can like Chinese chess (C), Go (G), or both, we need to ensure that each category (C only, G only, both) has at least one person. Each child can be assigned to one of three categories: C only, G only, or both. The total number of assignments without any restrictions is (3^4 = 81). We need to subtract the cases where at least one category is empty.Using the principle of inclusion-exclusion, we calculate the number of valid assignments:1. Subtract the cases where one category is empty: - Assignments missing C only: (2^4 = 16) - Assignments missing G only: (2^4 = 16) - Assignments missing both: (2^4 = 16) - Total: (16 + 16 + 16 = 48)2. Add back the cases where two categories are empty (since they were subtracted twice): - Assignments missing both C only and G only: (1) - Assignments missing both C only and both: (1) - Assignments missing both G only and both: (1) - Total: (1 + 1 + 1 = 3)3. Subtract the cases where all three categories are empty (impossible, so 0).Applying inclusion-exclusion:[81 - 48 + 3 = 36]Thus, the number of different ways these four people can like these two types of chess is (boxed{36}).,<think>Okay, so the problem is about four children—Zhao, Qian, Sun, and Li—who all like playing chess. Some like Chinese chess, some like Go, and some might like both. The question is asking how many different ways these four people can have preferences regarding these two types of chess. The answer needs to be the number of possible combinations, right?Let me start by understanding the problem. We have four children, each of whom can like Chinese chess, Go, or both. But wait, the problem says "someone likes playing Chinese chess, someone likes playing Go, and someone likes both." Hmm, does that mean that all three categories must be non-empty? In other words, there has to be at least one person in each category: Chinese chess only, Go only, and both. So, we can't have a situation where, say, all four like both chess types, because then there's no one who exclusively likes Chinese chess or Go. Similarly, if all liked only Chinese chess, then there would be no one in Go or both. So the condition is that all three groups (Chinese chess only, Go only, and both) must have at least one person each. So we need to count the number of ways to assign each of the four children to one of the three categories (Chinese chess, Go, both), such that none of the categories are empty.Wait, but is that the correct interpretation? The original problem says: "someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess." The way it's phrased might imply that there is at least one person in each of the three categories. So, each category must have at least one person. Therefore, we need to find the number of ways to distribute four distinct children into three distinct categories (Chinese chess only, Go only, both), with each category having at least one child. Then, the answer would be similar to the number of onto functions from a 4-element set to a 3-element set, but considering that the children are distinguishable and the categories are distinct as well.Wait, but actually, in inclusion-exclusion terms, the number of ways to distribute n distinguishable objects into k distinguishable boxes with each box non-empty is k! * S(n, k), where S(n, k) is the Stirling numbers of the second kind. But here, we have k=3 categories. Wait, but each child can be assigned to any of the three categories, but each category must have at least one child. So that is equivalent to 3^4 total assignments minus the assignments where one or more categories are empty. So using inclusion-exclusion.Alternatively, yes, the formula is 3! * S(4,3). Let me check: S(4,3) is the number of ways to partition 4 elements into 3 non-empty subsets, which is 6. Then 3! times that is 6*6=36. But wait, is that correct?Wait, if we have 3 categories, then the number of onto functions is 3^4 - 3*2^4 + 3*1^4. Let me compute that. So 3^4 is 81, 3*2^4 is 3*16=48, 3*1^4=3. So 81 - 48 + 3 = 36. So that matches. So 36 ways. Therefore, the answer is 36. But wait, but hold on. The problem says "someone likes Chinese chess, someone likes Go, and someone likes both." So does that mean that we have three categories: A (Chinese chess only), B (Go only), and C (both). Each must have at least one person. So this is exactly the number of onto functions from the four children to the three categories, which is 3! * S(4,3) = 6*6=36, as calculated. So that would be the answer.But let me think again. Maybe there's another way to interpret the problem. The problem states that "someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess." So maybe it's possible that a person is counted in both the "someone likes Chinese chess" and "someone likes Go" if they like both. But in terms of the categories, perhaps the problem is considering two sets: the set of people who like Chinese chess (which can include those who also like Go), and the set of people who like Go (which can include those who also like Chinese chess). The problem states that both sets are non-empty, and also that the intersection is non-empty (since someone likes both). So perhaps the problem is asking for the number of ordered pairs (C, G) where C is a subset of the four children (non-empty) who like Chinese chess, G is a subset (non-empty) who like Go, and the intersection C ∩ G is also non-empty.In that case, the problem becomes: how many pairs (C, G) with C ≠ ∅, G ≠ ∅, and C ∩ G ≠ ∅. So how many such pairs are there?First, the total number of ordered pairs (C, G) where C and G are any subsets of the four children is 2^4 * 2^4 = 256. But we need to subtract those pairs where C is empty, G is empty, or their intersection is empty. But we need to compute the number of pairs where C is non-empty, G is non-empty, and C ∩ G is non-empty.So using inclusion-exclusion again. The total number of pairs (C, G) with C non-empty and G non-empty is (2^4 -1)(2^4 -1)=15*15=225. But from these, we need to subtract the number of pairs where C and G are disjoint. So how many pairs (C, G) where C and G are non-empty and disjoint?For each child, they can be in C only, G only, or neither. But since C and G are disjoint and non-empty. So each child has three choices: C, G, or neither. However, we need at least one child in C and at least one child in G. So the total number is 3^4 - [number of ways where C is empty or G is empty]. The number of ways where C is empty is 2^4 (since each child can be in G or neither, but G can be empty). Similarly, the number where G is empty is 2^4. But then we subtract the overlap where both C and G are empty, which is 1. So using inclusion-exclusion, the number of pairs where C and G are disjoint and non-empty is 3^4 - 2*2^4 +1 = 81 - 32 +1 = 50. Wait, but hold on, this seems different. Wait, maybe I need to think again.Alternatively, the number of pairs (C, G) where C and G are disjoint is equal to the number of ways to assign each child to one of three possibilities: in C only, in G only, or in neither. So each child has three choices, so 3^4=81. But among these, some have C empty or G empty. So the number of pairs where C and G are disjoint and both non-empty is 3^4 - [number with C empty] - [number with G empty] + [number with both empty]. So that's 81 - 2^4 -2^4 +1=81-16-16+1=50. So 50 such pairs. Therefore, the number of pairs (C, G) where C and G are non-empty and their intersection is non-empty is total non-empty pairs (225) minus the disjoint non-empty pairs (50) = 225 -50=175. So 175.But wait, the problem says "the different ways these four people can like these two types of chess". So if the question is about the number of possible assignments where each child can like Chinese chess, Go, or both, with the constraints that there is at least one who likes Chinese chess, at least one who likes Go, and at least one who likes both. Then in terms of subsets, that would correspond to the assignments where the set of Chinese chess lovers (C) is non-empty, the set of Go lovers (G) is non-empty, and their intersection is non-empty. So each child can be in C only, G only, both, or neither. Wait, no. Wait, the problem states "someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess". So it doesn't say that everyone has to like at least one type. Wait, but the original problem says "four children, all like playing chess". So maybe "chess" here is ambiguous—does it mean that each child likes at least one type of chess? The problem states "all like playing chess", but then specifies "someone likes Chinese chess, someone likes Go, and someone likes both". So possibly, each child likes at least one of the two types (Chinese chess or Go), and additionally, all three categories (Chinese chess only, Go only, both) are non-empty.Therefore, in this case, the total number of assignments would be: each child must be assigned to at least one of Chinese chess or Go, and the assignments must result in at least one child in each of the three categories: Chinese chess only, Go only, and both. So the total number of such assignments.So, first, the total number of ways each child can like Chinese chess, Go, or both, with no restrictions except that they must like at least one. Since each child has 3 choices (C, G, both), but excluding the case where they like neither. Wait, but the problem says "all like playing chess", so each child must like at least one type. So each child has 3 choices: C only, G only, or both. Therefore, each child has 3 options, so the total number of assignments is 3^4=81. But we need to subtract those assignments where there are no C only, or no G only, or no both.So using inclusion-exclusion. Let’s define:Let A be the set of assignments where there are no Chinese chess only (i.e., every child is either Go only or both).B be the set where there are no Go only (every child is Chinese chess only or both).C be the set where there are no both (every child is Chinese chess only or Go only).We need to compute the total assignments (81) minus |A ∪ B ∪ C|.By inclusion-exclusion:|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|.Compute each term:|A|: No Chinese chess only. So each child can be Go only or both. That's 2 options per child, so 2^4=16. But we need to ensure that there is someone who likes Go only or both. Wait, but the problem requires that "someone likes playing Chinese chess", so if there are no Chinese chess only, but someone might still like both. Wait, no, in set A, assignments where there are no Chinese chess only. So in such assignments, every child is either Go only or both. However, the problem requires that "someone likes playing Chinese chess", which would mean that in the entire assignment, the set of Chinese chess players (both Chinese chess only and both) must be non-empty. Similarly, "someone likes Go" would require the set of Go players (Go only and both) must be non-empty. Additionally, "someone likes both" requires that the intersection (both) is non-empty.Wait, now I think my initial interpretation was wrong. Let me parse the problem again.Original problem:"Among them, someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess."So this can be interpreted as:- The set of Chinese chess players (C) is non-empty.- The set of Go players (G) is non-empty.- The set of those who like both (C ∩ G) is non-empty.So the problem is asking for the number of ordered pairs (C, G) where C and G are subsets of the four children, such that C ≠ ∅, G ≠ ∅, and C ∩ G ≠ ∅. Additionally, since "all like playing chess", does that mean each child is in at least one of C or G? The problem says "four children, all like playing chess", which might mean that each child likes at least one type of chess. So each child must be in C ∪ G. Therefore, the problem is asking for the number of ordered pairs (C, G) where:1. C ≠ ∅,2. G ≠ ∅,3. C ∩ G ≠ ∅,4. C ∪ G = {Zhao, Qian, Sun, Li} (all four children).So each child is in at least one of C or G, and the three conditions above. So how many such pairs (C, G) are there?First, let's compute the total number of ordered pairs (C, G) where C ∪ G = all four children. Each child must be in C, G, or both. Since each child has three choices (C only, G only, or both), and all must be covered, the total number is 3^4 = 81.Now, from these 81, we need to subtract those pairs where C is empty, G is empty, or C ∩ G is empty.But since C ∪ G must be all four children, C cannot be empty if G is empty, because if G is empty, then C must be all four. Similarly, if C is empty, G must be all four. So let's see:Wait, but if we require C ∪ G = all four children, then:- If C is empty, then G must be all four. But that would mean C ∩ G is empty (since C is empty). Similarly, if G is empty, then C must be all four, and C ∩ G is empty. So in this scenario, the only pairs where C ∩ G is empty are those where C and G are complements of each other. Wait, no. If C ∪ G = all four, and C ∩ G is empty, then C and G are partitions of the four children into two sets. So each child is in exactly one of C or G. So the number of such pairs (C, G) where C and G are disjoint and their union is all four is 2^4 = 16. Because for each child, you choose whether they are in C or in G. But since C and G cannot be empty (as C ∪ G is all four, so if C is empty, G is all four, but then C ∩ G would be empty). Wait, but in this problem, we have to exclude cases where C is empty, G is empty, or C ∩ G is empty. However, under the condition that C ∪ G is all four, the only way C is empty is if G is all four, and similarly for G empty. But in those cases, C ∩ G would be empty (if C is empty, C ∩ G is empty; if G is empty, same). So in the total of 81 assignments where C ∪ G is all four, we need to subtract the cases where C is empty, G is empty, or C ∩ G is empty.But since in the case where C ∪ G is all four, the only way C ∩ G is empty is if C and G form a partition of the four children (i.e., each child is in exactly one of C or G). So the number of such partitions is 2^4 - 2 = 14. Wait, why? Because for each child, you can assign them to C or G, so 2^4=16 ways. But we have to exclude the cases where all are in C (which would make G empty) or all are in G (which would make C empty). But since C ∪ G must be all four, if all are in C, G must be empty, which is already excluded. Wait, but in our problem, the total assignments where C ∪ G is all four is 3^4=81. But if we consider the assignments where C ∩ G is empty, that is, the partitions into C and G, then the number is 2^4=16. But among these 16, two of them have C empty or G empty (all in G or all in C). But since we already have the condition that C ∪ G is all four, those two cases (all in C or all in G) are included in the 3^4=81? Wait, no. If all are in C, then G is empty, but since C ∪ G is all four, that is allowed, but G is empty. But the problem requires that "someone likes Go", so G must be non-empty. Similarly, if all are in G, then C is empty, which violates "someone likes Chinese chess". So in the problem's constraints, we must have C ≠ ∅, G ≠ ∅, and C ∩ G ≠ ∅. Therefore, in the context where C ∪ G is all four (since all children like chess), the forbidden cases are:1. C is empty (G is all four) → violates C ≠ ∅.2. G is empty (C is all four) → violates G ≠ ∅.3. C ∩ G is empty → which is equivalent to C and G being a partition (each child in exactly one), except for the cases where C or G is empty.So the number of invalid assignments is:- Number of assignments where C is empty: 1 (since G must be all four).- Number of assignments where G is empty: 1 (C is all four).- Number of assignments where C ∩ G is empty (i.e., C and G partition the children, with neither empty): as calculated before, 2^4 - 2 = 14.So total invalid assignments: 1 + 1 + 14 = 16.Therefore, the valid assignments are total assignments (81) minus invalid (16) = 65.But wait, but inclusion-exclusion might be needed here. Let me check again.The formula is:Valid = Total - (A + B + C) + (AB + AC + BC) - ABCWhere:A: assignments where C is empty.B: assignments where G is empty.C: assignments where C ∩ G is empty.But in our scenario, since we already have C ∪ G = all four, the total is 81.Number of assignments where C is empty: 1 (G is all four).Number where G is empty: 1 (C is all four).Number where C ∩ G is empty: 14 (partitions into non-empty C and G).But A and B are each 1, and C is 14.However, the overlaps:A ∩ B: assignments where both C and G are empty, but since C ∪ G must be all four, this is impossible. So A ∩ B = 0.A ∩ C: assignments where C is empty and C ∩ G is empty. If C is empty, then C ∩ G is empty iff G is non-empty, which it is (since C ∪ G is all four and C is empty, G must be all four). But C ∩ G is empty because C is empty. So the number here is 1 (C empty, G all four).Similarly, B ∩ C: assignments where G is empty and C ∩ G is empty. Similarly, 1.A ∩ B ∩ C: impossible, as before.Therefore, using inclusion-exclusion:|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|= 1 + 1 + 14 - 1 - 1 + 0 = 14.Therefore, the number of invalid assignments is 14.Hence, the number of valid assignments is 81 - 14 = 67.Wait, this contradicts the previous calculation. So confusion arises. Let me step back.Given that all four children must be in C ∪ G, and we need C ≠ ∅, G ≠ ∅, and C ∩ G ≠ ∅.So total number of assignments where C ∪ G = all four: 3^4 = 81.From these, subtract:1. Assignments where C is empty: 1 (G = all four).2. Assignments where G is empty: 1 (C = all four).3. Assignments where C ∩ G = empty. But in this case, since C ∪ G = all four, C ∩ G = empty implies that C and G form a partition of the four children. The number of such partitions is 2^4 - 2 = 14 (each child assigned to C or G, minus the two cases where all are in C or all are in G).But these three sets (C empty, G empty, C ∩ G empty) may overlap. Specifically:- The assignments where C is empty or G is empty are already included in the 14 partitions where C ∩ G is empty (since when C is empty, G is all four, which is one of the 14 partitions; similarly for G empty). Wait, no. The 14 partitions are those where neither C nor G is empty. Because when you compute 2^4 - 2, you subtract the cases where all are in C or all are in G. So the 14 partitions are exactly the non-empty, disjoint C and G whose union is all four.Therefore, the total number of invalid assignments is:- 1 (C empty) + 1 (G empty) + 14 (C ∩ G empty, neither C nor G empty) = 16.Thus, valid assignments = 81 - 16 = 65.But earlier, using inclusion-exclusion considering overlaps, I got 14 invalid assignments. There's inconsistency here.Wait, another approach: The total number is 81.Invalid assignments are those where C is empty, G is empty, or C ∩ G is empty.But in the context where C ∪ G is all four:- If C is empty, then G = all four. This is 1 assignment.- If G is empty, then C = all four. This is 1 assignment.- If C ∩ G is empty, then C and G form a partition of the four children into two non-empty sets. The number of such partitions is equal to the number of ways to split 4 elements into two non-empty subsets, which is 2^4 - 2 = 14. Each partition corresponds to two assignments: (C, G) and (G, C). Wait, no. Each partition is an ordered pair (C, G). For unordered partitions, the number is 2^4 - 2 divided by 2, but since (C, G) is ordered, we need to count them all.Wait, no. If we consider ordered pairs, then for each non-empty subset C (not equal to the full set), G is the complement. But since we need C and G to be non-empty, the number of ordered pairs (C, G) where C and G are complements and non-empty is equal to the number of non-empty, proper subsets C, multiplied by 2 (since (C, G) and (G, C) are different). Wait, no. For each non-empty proper subset C, G is uniquely determined as the complement. So the number of such ordered pairs is equal to the number of non-empty proper subsets C, which is 2^4 - 2 = 14. Each such subset C corresponds to exactly one G (its complement). So there are 14 ordered pairs where C and G are non-empty, disjoint, and their union is all four.Therefore, the number of invalid assignments is 1 (C empty) +1 (G empty) +14 (C and G disjoint and non-empty) = 16.Thus, the valid assignments are 81 -16=65.But this contradicts the previous inclusion-exclusion result. What is the correct answer?Alternatively, perhaps the problem does not require that all four children like at least one type of chess. Let's re-examine the original problem statement.Original problem:"Zhao, Qian, Sun, and Li, four children, all like playing chess. Among them, someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess. Therefore, the different ways these four people can like these two types of chess are ______ kinds."The key is the translation of the original Chinese problem. The statement "all like playing chess" might mean that each child likes at least one type of chess (Chinese or Go). But it's possible that the translation is more literal, and "chess" here refers to the general concept, so they all like playing chess (either Chinese chess or Go). Therefore, it's necessary that each child likes at least one of the two types. Hence, the model where each child is assigned to C, G, or both, with all children assigned, i.e., C ∪ G = all four. So the total number of assignments is 3^4=81. From these, we must subtract the assignments where C is empty, G is empty, or C ∩ G is empty. As above, this gives 81 -16=65.But earlier, considering the problem as assigning each child to one of three categories (C only, G only, both), with each category non-empty, gives 3! * S(4,3)=6*6=36. So which is correct?Wait, these are two different interpretations. If the problem requires that:1. Each child is in exactly one category: C only, G only, or both. And each category must have at least one child. Then it's the number of onto functions to three categories, which is 3! * S(4,3)=36.But if the problem allows children to be in multiple categories (i.e., a child can be in both C and G), and the requirement is that the set C (Chinese chess players) is non-empty, the set G (Go players) is non-empty, and their intersection is non-empty, with no restriction on the children being in at least one, then the answer would be different. But the original problem states "four children, all like playing chess", which might mean that each child likes at least one type of chess. So in this case, each child must be in C, G, or both.In that case, the problem is equivalent to counting the number of ordered pairs (C, G) where C and G are subsets of the four children, such that:- C ≠ ∅,- G ≠ ∅,- C ∩ G ≠ ∅,- C ∪ G = all four children.Which we calculated as 65. However, there's a discrepancy here because when you consider each child being assigned to one of three categories (C only, G only, both), the count is different.Alternatively, the categories are:- A: C only,- B: G only,- C: both.Each child is in exactly one of these categories. The problem states that A, B, and C are all non-empty. Therefore, the number of ways is the number of ordered triples (A, B, C) where A, B, C are pairwise disjoint, A ∪ B ∪ C = all four, and A, B, C are all non-empty. This is equivalent to the number of ordered set partitions into three non-empty subsets, which is 3! * S(4,3)=6*6=36.But in this case, the categories are distinct: C only, G only, both. So if we think of it this way, it's 36.However, the other interpretation where a child can be in both C and G (i.e., the categories are not exclusive) leads to a different count. So which interpretation is correct?The problem says "someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess". This can be interpreted as there exists at least one person who likes only Chinese chess, at least one who likes only Go, and at least one who likes both. In this case, the three categories must each be non-empty. Therefore, the answer would be the number of ways to distribute the four children into three distinct categories (C only, G only, both), each non-empty, which is 3! * S(4,3)=36.But in another interpretation, if we just require that the sets C and G and their intersection are non-empty, without requiring that the "only" sets are non-empty, then the count is different. For example, if two children like both, one likes only C, and one likes only G, that's allowed. But if all children who like C also like G, then the "C only" category is empty, which would violate the problem's condition if we interpret it as requiring someone to like only C.The original problem statement in Chinese might clarify this, but since we don't have that, we have to rely on the translation given. The problem states: "someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess". The use of "someone" for each category suggests that each category must have at least one person. That is, there is at least one person who exclusively likes Chinese chess, at least one who exclusively likes Go, and at least one who likes both. If that's the case, then the three categories (C only, G only, both) must each be non-empty. Therefore, the number of ways is the number of ordered partitions of the four children into three non-empty subsets, which is 3! * S(4,3)=36.But let's verify with an example. Suppose there are two children. Then, how many ways? If we require each of the three categories (C only, G only, both) to have at least one child, but there are only two children, it's impossible. So the answer would be zero. Alternatively, if we only require that the sets C, G, and C ∩ G are non-empty, then with two children, the possibilities are that one child is in C only, and the other is in G only; or one is in both and the other is in C only; or one is in both and the other is in G only. So three possibilities. Wait, but how many ordered pairs (C, G) with C, G, and C ∩ G non-empty.For two children:Total assignments where each child is in at least one: 3^2=9.Subtract assignments where C is empty:1, G is empty:1, C ∩ G empty: partitions. For two children, the number of assignments where C and G are disjoint is 2^2=4 (each child in C or G but not both). But subtract the cases where C is empty or G is empty: 4 -2=2. So total invalid:1+1+2=4. Therefore, valid=9-4=5.But according to the category method, if we need all three categories (C only, G only, both) non-empty, it's impossible with two children, so zero. Contradiction.But according to the ordered pairs (C, G), requiring C, G, C ∩ G non-empty, it's possible. For two children:Possible assignments:1. Child 1 in C only, child 2 in G only.2. Child 1 in C only, child 2 in both.3. Child 1 in both, child 2 in G only.4. Child 1 in G only, child 2 in C only.5. Child 1 in G only, child 2 in both.6. Child 1 in both, child 2 in C only.But wait, this seems like 6 possibilities. However, in terms of sets:Each assignment corresponds to a pair (C, G). For example:1. C={1}, G={2}, both empty.2. C={1}, G={1,2}, both={1}.3. C={1,2}, G={2}, both={2}.4. Similarly for others.Wait, but if we require that C ∩ G is non-empty, then for two children, the possible pairs (C, G) where C, G, and C ∩ G are non-empty are:- C and G each have at least one child, and their intersection is non-empty.But since the union must be all children (if we consider that all children like chess), but the original problem doesn't specify whether they can dislike both. The problem says "all like playing chess", but if "chess" refers to either Chinese chess or Go, then yes, each child must like at least one. If "chess" is a general term, and the two types are Chinese chess and Go, then "all like playing chess" might mean they like at least one of the two types.Assuming that each child likes at least one type, for two children, the number of ordered pairs (C, G) with C, G, C ∩ G non-empty is:Each child can be in C only, G only, or both. Total assignments:3^2=9.Subtract assignments where C is empty (1), G is empty (1), or C ∩ G is empty. C ∩ G is empty means C and G are disjoint. With two children, the number of such assignments is 2^2=4 (each child in C or G). But subtract the cases where C or G is empty: 4 -2=2. So total invalid:1+1+2=4. Valid:9-4=5. But manually, the valid assignments are:1. Both children in both: C={1,2}, G={1,2}, intersection={1,2}. But this violates the requirement that someone likes only C or only G.Wait, no. If both children are in both, then the sets C only and G only are empty. Therefore, such an assignment is invalid. Therefore, the valid assignments are those where there is at least one in C only, at least one in G only, and at least one in both. But with two children, this is impossible. Hence, the answer should be zero. But according to the inclusion-exclusion, it's 5. Contradiction.This suggests that the interpretation determines the answer. If the problem requires that all three categories (C only, G only, both) are non-empty, then with two children, it's impossible, so zero. If the problem just requires that C, G, and their intersection are non-empty (regardless of the exclusivity), then with two children, it's possible. For example, one child in both, and the other in C only. Then C={1,2}, G={1}, so C is non-empty, G is non-empty, and their intersection is {1}, non-empty. Similarly, one in both and one in G only. So there are two such assignments. But according to inclusion-exclusion, we get 5 valid assignments, which includes cases where some children are in both. But if the problem requires that all three categories (C only, G only, both) are non-empty, then even if a child is in both, we still need someone in C only and someone in G only. Therefore, with two children, it's impossible. Hence, the correct interpretation is likely that the three categories must each be non-empty, leading to the answer 3! * S(n,3) for n=4, which is 36.Given the original problem mentions "someone likes playing Chinese chess, someone likes playing Go, and someone likes both types of chess", it's reasonable to interpret this as requiring at least one person in each of the three exclusive categories: Chinese chess only, Go only, and both. Therefore, the number of ways is the number of ordered partitions of the four children into three non-empty subsets, corresponding to these categories. This is calculated as 3! multiplied by the Stirling numbers of the second kind S(4,3). Since S(4,3)=6, the answer is 6*6=36.Therefore, the final answer is boxed{36}.Final Answerboxed{36}</think>The problem involves four children—Zhao, Qian, Sun, and Li—who all like playing chess. We need to determine the number of different ways these four people can like Chinese chess, Go, or both, given that there must be at least one person who likes Chinese chess, at least one who likes Go, and at least one who likes both.Key steps:1. Interpretation: The problem requires that each of the three categories (Chinese chess only, Go only, and both) must be non-empty.2. Set Partitions: The problem is equivalent to partitioning the four children into three non-empty subsets corresponding to Chinese chess only, Go only, and both.3. Stirling Numbers: The number of ways to partition 4 elements into 3 non-empty subsets is given by the Stirling numbers of the second kind, ( S(4, 3) ), which is 6.4. Permutations: Each partition can be permuted among the three categories (Chinese chess only, Go only, both), which is ( 3! ).Thus, the number of different ways is calculated as:[3! times S(4, 3) = 6 times 6 = 36]The final answer is (boxed{36}).]

🔑:【Analysis and Solution】Counting.4 people, some like playing Chinese chess, some like playing Go, and some like both; then exactly 2 people have the same hobbies;Considering the types where exactly 2 people have the same hobbies, there are C_{3}^{2}=3 cases.Case (1): 2 people like playing Chinese chess, 1 person likes playing Go, and 1 person likes both;Case (2): 1 person likes playing Chinese chess, 2 people like playing Go, and 1 person likes both;Case (3): 1 person likes playing Chinese chess, 1 person likes playing Go, and 2 people like both.Considering 4 different people, there are C_{4}^{2} times C_{2}^{1} times C_{1}^{1}=6 times 2 times 1=12 ways;Therefore, the different ways these four people can like these two types of chess are 3 times 12=36 ways.