🧠:[<think>Okay, let's try to tackle this problem step by step. First, I need to understand the function f(x) given here. It's defined piecewise: for x in (1, +∞), it's 2 times f(x - 2), and for x in [-1, 1], it's 1 - |x|. Then, there's an equation f(x) - log_a(x + 1) = 0, which has exactly 5 different roots in [0, 5]. We need to find the range of a that satisfies this condition. The options are given as A to D. Alright, let's start by understanding the function f(x).So, the function f is defined differently in different intervals. For x between -1 and 1, inclusive, it's 1 minus the absolute value of x. That part is straightforward. For x greater than 1, it's 2 times f(x - 2). That recursive definition suggests that the function is self-similar but scaled up as x increases. So, maybe it's a fractal-like function or something with periodic behavior but increasing amplitude? Let me check.Since f(x) = 2f(x - 2) for x > 1, this means that every time x increases by 2, the function value doubles. So, for example, on the interval (1, 3], which is x > 1 but x - 2 ≤ 1, so f(x) = 2f(x - 2). Since x - 2 is in [-1, 1] when x is in (1, 3], then f(x) = 2*(1 - |x - 2|). Similarly, for x in (3, 5], which would be x - 2 in (1, 3], so f(x) = 2f(x - 2) = 2*(2f(x - 4)) = 4f(x - 4). Since x - 4 is in [-1, 1] when x is in (3, 5], f(x) = 4*(1 - |x - 4|). So, each interval (2n + 1, 2n + 3] for n = 0, 1, 2,... would have f(x) = 2^n * (1 - |x - 2n - 2|). Wait, maybe not. Let me check again.Wait, for x in (1, 3], x - 2 is in (-1, 1], so f(x) = 2*(1 - |x - 2|). For x in (3, 5], x - 2 is in (1, 3], so f(x) = 2*f(x - 2) = 2*(2*(1 - |x - 4|)) = 4*(1 - |x - 4|). Similarly, for x in (5,7], it would be 8*(1 - |x - 6|), etc. So, in each interval (2k - 1, 2k + 1], the function f(x) is 2^k*(1 - |x - 2k|). Is that right?Wait, for k=0, the interval (-1,1], f(x) = 1 - |x|. For k=1, (1,3], f(x) = 2*(1 - |x - 2|). For k=2, (3,5], f(x) = 4*(1 - |x - 4|). For k=3, (5,7], f(x) = 8*(1 - |x - 6|), etc. So, the coefficient doubles each time, and the center of the triangle shifts by 2 each interval. So, each interval of length 2, centered at 2k, with height 2^k. So, the graph of f(x) on [0,5] would be:From 0 to 1: part of the original triangle wave, 1 - |x|. Then from 1 to 3: a triangle wave scaled by 2, peaking at x=2 with value 2*(1 - 0) = 2. Then from 3 to 5: scaled by 4, peaking at x=4 with value 4*(1 - 0) = 4. Wait, but in each interval, the function is a triangle that goes from 0 at the left end, up to the peak in the middle, then back to 0 at the right end. So, in (1,3], the function starts at x=1: 2*(1 - |1 - 2|) = 2*(1 - 1) = 0. Then peaks at x=2: 2*(1 - 0) = 2, then decreases back to 0 at x=3. Similarly, (3,5] starts at 0, peaks at x=4 with value 4, then back to 0 at x=5.Therefore, on [0,1], it's the original triangle, from (0,1) decreasing from 1 to 0. Then from 1 to 3, a triangle going from 0 up to 2 at x=2, then back to 0 at x=3. Then from 3 to 5, another triangle from 0 up to 4 at x=4, then back to 0 at x=5. So, in total, the function f(x) on [0,5] has peaks at x=0 (value 1), x=2 (value 2), x=4 (value 4), and at x=5 it's 0. Wait, but x=0 is part of the interval [-1,1], so the function is 1 - |x|. So at x=0, f(0) = 1 - 0 = 1. At x=1, f(1) = 1 - 1 = 0. Then from x=1 to x=3, the function is 2*(1 - |x - 2|). So at x=2, it's 2*1=2, and at x=1 and x=3, it's 0. Similarly, from x=3 to x=5, it's 4*(1 - |x - 4|). So at x=4, it's 4*1=4, and at x=3 and x=5, it's 0.So, the graph of f(x) over [0,5] is a series of triangles: from 0 to 1, descending from 1 to 0; from 1 to 3, ascending to 2 at x=2 then descending to 0 at x=3; from 3 to 5, ascending to 4 at x=4 then descending to 0 at x=5.Now, the equation f(x) - log_a(x + 1) = 0. We need this equation to have exactly 5 different roots in [0,5]. So, we need to find the number of intersections between f(x) and log_a(x + 1) in [0,5], which should be exactly 5. We need to find the values of a > 0, a ≠ 1, for which this is true.First, let's note that log_a(x + 1) is defined for x + 1 > 0, which is always true for x ≥ 0, since x + 1 ≥ 1. So, log_a(x + 1) is defined on [0,5]. The behavior of log_a(x + 1) depends on the base a:- If a > 1, log_a(x + 1) is increasing, concave down, starting at log_a(1) = 0 when x=0, and increasing to log_a(6) when x=5.- If 0 < a < 1, log_a(x + 1) is decreasing, concave up, starting at 0 when x=0 and decreasing to log_a(6), which is negative, but since x is in [0,5], x +1 is from 1 to 6, so log_a(6) is negative when a < 1.However, since f(x) is non-negative on [0,5] (as it's composed of triangles with peaks at 1,2,4 and zeros at the endpoints), and log_a(x + 1) is negative for a < 1 in x ∈ [0,5] (since log_a(6) < 0 when a < 1). Therefore, the equation f(x) - log_a(x +1) = 0 would have f(x) = log_a(x +1). If a < 1, log_a(x +1) is negative, but f(x) is non-negative. Therefore, the equation would have no solutions for a < 1. Wait, but when x=0, log_a(1) = 0, and f(0) = 1. So, at x=0, f(x) - log_a(x +1) = 1 - 0 = 1 > 0. For x >0, log_a(x +1) is negative (if a <1), so f(x) - log_a(x +1) would be positive (since f(x) ≥0 and subtracting a negative is adding a positive), so the equation would have no roots. Therefore, a must be greater than 1, because otherwise, if a <1, there are no solutions. So, we can eliminate any options that include a <1. But looking at the options, all options are intervals with a >1 or starting above 1. Wait, let's check the options:A: (1,√3)B: (⁴√5, +∞)C: (√3, +∞)D: (⁴√5, √3)So all options are intervals with a >1, since ⁴√5 is about 5^(1/4) ≈ 1.495, so all options are a >1. Therefore, we can restrict our analysis to a >1.Therefore, log_a(x +1) is an increasing function, starting at 0 when x=0, and increasing to log_a(6) when x=5.We need to find a such that log_a(x +1) intersects f(x) exactly 5 times in [0,5]. Let's analyze the graph of f(x) and log_a(x +1). Since f(x) is a series of triangles, and log_a(x +1) is a smooth increasing curve. The number of intersections depends on how the logarithmic curve interacts with the triangles. Let's break down the intervals:[0,1]: f(x) is 1 - x (since |x| = x here), decreasing from 1 to 0.[1,3]: f(x) is 2*(1 - |x -2|). So, it's a triangle increasing from 0 at x=1 to 2 at x=2, then decreasing back to 0 at x=3.[3,5]: f(x) is 4*(1 - |x -4|). So, increasing from 0 at x=3 to 4 at x=4, then decreasing back to 0 at x=5.Therefore, in each interval, f(x) has a triangular shape. The log function is increasing throughout. So, the intersections can occur in each of these intervals. Let's consider how many times the log curve can intersect each triangle.First, in [0,1]: f(x) is decreasing from 1 to 0, and log_a(x +1) is increasing from 0 to log_a(2). Since at x=0, f(0)=1 and log_a(1)=0, so f(x) starts above. At x=1, f(1)=0 and log_a(2). If log_a(2) >0, which it is since a >1. So, the log function is increasing from 0 to log_a(2) ≈ (ln2)/lna. Since f(x) is decreasing from 1 to 0, and log_a(x+1) is increasing from 0 to log_a(2). So, depending on the value of a, log_a(2) can be less than 1 or greater than 1. If log_a(2) <1, then the log curve goes from 0 to some value less than 1, while f(x) goes from 1 to 0. So, they can intersect once in [0,1]. If log_a(2) =1, then at x=1, log_a(2) =1 implies a=2. If log_a(2) >1, which would require a <2, since log_a(2) = (ln2)/(lna). If a <2, then log_a(2) >1. Wait, if a >1, then:log_a(2) >1 is equivalent to 2 > a^1, so a <2.Similarly, log_a(2) =1 when a=2, and log_a(2) <1 when a >2.Therefore:- If a <2, log_a(2) >1, so at x=1, log_a(2) >1, but f(1)=0. Wait, wait, no. Wait, at x=1, log_a(2) is log_a(1 +1)=log_a(2). But f(1)=0. So, regardless of a, at x=1, log_a(2) is some positive number, and f(1)=0. Therefore, in the interval [0,1], f(x) starts at 1 and decreases to 0, while log_a(x +1) starts at 0 and increases to log_a(2). So, if log_a(2) <1, then the log curve will cross f(x) once in [0,1]. If log_a(2) =1, then they touch at x=1. If log_a(2) >1, then the log curve would cross f(x) once in [0,1), but then at x=1, log_a(2) >1 while f(1)=0, so there might be another crossing near x=1? Wait, let me think.Wait, no. Let's consider the two functions on [0,1]. At x=0: f=1, log=0. At x=1: f=0, log=log_a(2). So:- If log_a(2) <1, which is equivalent to a >2, then log_a(2) is less than 1. So, since f(x) is decreasing and log is increasing, they cross exactly once in [0,1].- If log_a(2) =1 (a=2), then at x=1, they meet at 1. But f(1)=0 and log_2(2)=1, so f(1)=0 vs log_2(2)=1. Wait, no. If a=2, then log_2(1 + x) at x=1 is log_2(2)=1. But f(1)=0, so they don't meet at x=1. Wait, perhaps I made a mistake here. Wait, log_a(2)=1 would mean a=2. So, if a=2, log_2(2)=1. But f(1)=0. So, even if a=2, the log function at x=1 is 1, while f(1)=0. Therefore, even when a=2, log function at x=1 is 1, which is higher than f(1)=0. Therefore, if log_a(2) <1, which is when a >2, then log_a(2) <1, so the log function at x=1 is less than 1, but f(1)=0. Wait, I think there's confusion here. Wait, let's clarify:Wait, log_a(x +1) at x=1 is log_a(2). The value of f(1) is 0. So, when x approaches 1 from the left, f(x) approaches 0 from above, and log_a(x +1) approaches log_a(2). Therefore:- If log_a(2) >0, which it is for all a >1, but log_a(2) can be greater or less than 1.Wait, but in the interval [0,1], f(x) is 1 - x. So, let's solve f(x) = log_a(x +1) in [0,1].Set 1 - x = log_a(x +1). At x=0: 1 -0 = log_a(1) → 1=0, which is not true. Wait, but f(0)=1, log_a(1)=0, so f(0) - log_a(1)=1 -0=1≠0. So, x=0 is not a root. At x=1: f(1)=0, log_a(2). So, equation holds only if log_a(2)=0, which is impossible for a >0, a≠1. Therefore, in [0,1], the equation f(x) = log_a(x +1) could have one solution if the log curve crosses the line 1 - x once. Let's check for a >1.We can analyze the function g(x) = 1 - x - log_a(x +1) on [0,1]. We need to find when g(x) =0. Let's see:At x=0: g(0)=1 -0 -0=1 >0.At x=1: g(1)=0 - log_a(2) = - log_a(2) <0 (since log_a(2) >0 for a >1). So, since g is continuous, by Intermediate Value Theorem, there is at least one root in (0,1). To check if there's exactly one root, note that g'(x) = -1 - (1/( (x +1) ln a )). For a >1, ln a >0, so g'(x) = -1 - (1/( (x +1) ln a )) <0 for all x in [0,1]. Therefore, g is strictly decreasing on [0,1], so exactly one root in (0,1). Therefore, regardless of a >1, there is exactly one root in (0,1).Therefore, in [0,1], there is always 1 root. So, that's one root.Next, in [1,3]: f(x) is a triangle going from 0 at x=1 up to 2 at x=2 and back to 0 at x=3. The function f(x) here is 2*(1 - |x -2|). So, for x in [1,2], it's 2*(1 - (2 -x))=2*(x -1). For x in [2,3], it's 2*(1 - (x -2))=2*(3 -x). So, it's linear increasing from 0 to 2, then linear decreasing back to 0.Meanwhile, log_a(x +1) is increasing throughout. Let's analyze how many times they intersect in [1,3].At x=1: f(1)=0, log_a(2). Since a >1, log_a(2) >0. So, log_a(2) is above f(1). Then, as x increases from 1 to 2, f(x) increases to 2 at x=2, while log_a(x +1) increases to log_a(3). At x=2: f(2)=2, log_a(3). So, whether they cross depends on whether log_a(3) is less than or greater than 2.Similarly, from x=2 to x=3, f(x) decreases from 2 to 0, while log_a(x +1) continues to increase. At x=3: f(3)=0, log_a(4). Again, log_a(4) >0.So, in the interval [1,3], depending on a, the log curve might intersect f(x) twice: once while f is increasing (in [1,2]) and once while f is decreasing (in [2,3]). Or maybe once or not at all.To find the number of intersections, we need to check when log_a(x +1) crosses the line 2*(x -1) on [1,2] and 2*(3 -x) on [2,3].First, in [1,2], the equation is 2*(x -1) = log_a(x +1).In [2,3], the equation is 2*(3 -x) = log_a(x +1).So, potentially two intersections in [1,3]. But whether they occur depends on the value of a.Similarly, in [3,5], f(x) is 4*(1 - |x -4|). So, in [3,4], it's 4*(x -3), increasing from 0 to 4, and in [4,5], it's 4*(5 -x), decreasing back to 0. The log_a(x +1) is increasing to log_a(6). So, in [3,4], f(x) increases from 0 to 4, and log_a(x +1) increases from log_a(4) to log_a(5). In [4,5], f(x) decreases from 4 to 0, while log_a(x +1) increases from log_a(5) to log_a(6). So, in [3,4], equation 4*(x -3) = log_a(x +1), and in [4,5], equation 4*(5 -x) = log_a(x +1). Each could have one intersection, depending on a.So, in total, possible intersections:- [0,1]: 1- [1,3]: up to 2- [3,5]: up to 2Total up to 5. But we need exactly 5. So, the question is, for which a does the equation have exactly 5 roots. Therefore, we need all these intersections to occur except perhaps some, but in such a way that the total is 5.Wait, but let's think:If in [1,3], there are 2 intersections, and in [3,5], there are 2 intersections, plus the one in [0,1], that's 5. So, we need a such that in both [1,3] and [3,5], the log curve intersects f(x) twice each. But perhaps for certain a, in [3,5], there is only 1 intersection or none. Similarly, in [1,3], maybe only 1. So, we need to find a where in [1,3] and [3,5], each interval has two intersections. Wait, but how?Wait, maybe in some intervals, the log curve is tangent to f(x), leading to a double root. But the problem states "exactly 5 different roots", so tangency would count as a single root. So, we need to ensure that in each interval where intersections can occur, they cross exactly the number of times needed. Let's think step by step.First, [0,1]: always 1 root.[1,3]: Let's see. The maximum of f(x) here is 2. So, log_a(x +1) must reach 2 somewhere in [1,3]. The log_a(x +1) at x=3 is log_a(4). So, if log_a(4) <2, then the maximum value of log_a(x +1) in [1,3] is log_a(4). If log_a(4) >=2, then log_a(x +1) would reach 2 at some point in [1,3].Similarly, in [3,5], the maximum of f(x) is 4. The log_a(x +1) at x=5 is log_a(6). If log_a(6) <4, then log_a(x +1) never reaches 4 in [3,5], so the equation 4*(something) = log_a(x +1) may have 0,1, or 2 solutions.Wait, maybe a better approach is to analyze each interval and find the conditions on a that allow a certain number of intersections.Starting with [1,3]:In [1,2], equation: 2(x -1) = log_a(x +1)Let’s denote this as equation (1).In [2,3], equation: 2(3 -x) = log_a(x +1)Equation (2).We need each equation to have one solution in their respective intervals for two intersections. So, for equation (1) in [1,2], we can set h1(x) = 2(x -1) - log_a(x +1). We need h1(x)=0. At x=1: h1(1)=0 - log_a(2) = -log_a(2) <0. At x=2: h1(2)=2(1) - log_a(3) = 2 - log_a(3). So, if h1(2) >0, then by Intermediate Value Theorem, there is a solution in (1,2). Similarly, h1 is increasing or decreasing?h1’(x) = 2 - (1 / ((x +1) ln a)). Since a >1, ln a >0. So, h1’(x) = 2 - [1 / ((x +1) ln a)].Depending on the value of a, h1’(x) can be positive or negative. However, since x ∈ [1,2], x +1 ∈ [2,3], so 1/((x +1) ln a) is less than 1/(2 ln a). If a is large enough, ln a is large, so 1/(2 ln a) is small, so h1’(x) ≈ 2, which is positive. Therefore, if a is large, h1 is increasing from h1(1)= -log_a(2) to h1(2)=2 - log_a(3). For h1(2) >0, we need 2 > log_a(3) → a^2 >3 → a >√3. So, if a >√3, then log_a(3) = ln3 / ln a < ln3 / ln√3 = ln3 / (0.5 ln3) = 2, so log_a(3) <2. Therefore, h1(2)=2 - log_a(3) >0. Therefore, if a >√3, equation h1(x)=0 has exactly one solution in (1,2). If a =√3, log_a(3)=2, so h1(2)=0, so there is a root at x=2. But x=2 is in the interval [1,2]? Wait, x=2 is the boundary between [1,2] and [2,3]. But in the interval [1,2], x=2 is the endpoint. So, if a=√3, then at x=2, h1(2)=0, which is part of equation (1) at x=2, but equation (1) is defined on [1,2]. So, x=2 would be a solution here. However, at x=2, equation (2) (for [2,3]) would also have 2(3 -2) = log_a(3). So, 2 = log_a(3). If a=√3, then log_{√3}(3) = 2, so 2=2. Therefore, x=2 is a solution for both equations (1) and (2). But x=2 is only one point. Therefore, if a=√3, then x=2 is a root, but it's a single root (even though it's on the boundary). So, in this case, in [1,3], we would have one root at x=2. However, since the function is continuous, we need to check if there's another root in (1,2) or (2,3).Wait, if a=√3, then in equation (1) on [1,2], h1(x) at x=1 is -log_{√3}(2) ≈ - (ln2)/(0.5 ln3) ≈ - (0.693)/(0.549) ≈ -1.26. At x=2, h1(2)=0. So, h1(x) goes from -1.26 to 0, increasing. Therefore, since h1 is increasing, there is exactly one root in (1,2). But wait, h1(2)=0, so the root is at x=2. So, in [1,2], the root is at x=2. Similarly, in equation (2) on [2,3], h2(x)=2(3 -x) - log_a(x +1). At x=2, h2(2)=2(1) - log_{√3}(3)=2 -2=0. At x=3, h2(3)=0 - log_{√3}(4)= - (ln4)/(0.5 ln3)≈ - (1.386)/(0.549)≈ -2.52. The derivative h2’(x)= -2 - [1/((x +1) ln a)]. For a=√3, ln a =0.5 ln3. So h2’(x)= -2 - [1/((x +1)(0.5 ln3))]. Since x ∈ [2,3], x +1 ∈ [3,4], so denominator is at least 3*(0.549)=1.647, so 1/1.647≈0.607, so h2’(x)≈ -2 -0.607≈-2.607 <0. Therefore, h2(x) is decreasing from 0 to -2.52, so no roots in (2,3). Therefore, for a=√3, in [1,3], there is exactly one root at x=2.Therefore, for a=√3, total roots would be [0,1]:1, [1,3]:1, [3,5]: ?In [3,5], f(x) is 4*(1 - |x -4|). So, similar to previous intervals. Let's analyze:In [3,4], equation: 4(x -3) = log_a(x +1). Equation (3).In [4,5], equation: 4(5 -x) = log_a(x +1). Equation (4).At x=3: log_a(4) versus 0. So log_a(4) >0. At x=4: log_a(5) versus 4. So, if log_a(5) <4, then equation (3) has a solution in (3,4). Similarly, equation (4): at x=5, log_a(6) versus 0. log_a(6) >0. So, in [4,5], if log_a(5) <4, equation (4) could have a solution.But let's check for a=√3. Then log_{√3}(5) = ln5 / (0.5 ln3) ≈1.609/0.549≈2.93. Since 2.93 <4, then at x=4, log_a(5)≈2.93 <4. Therefore, in equation (3), h3(x)=4(x -3) - log_a(x +1). At x=3: h3(3)=0 - log_a(4) ≈ -2.52. At x=4: h3(4)=4(1) - log_a(5)≈4 -2.93≈1.07. Since h3 is increasing? Let's check derivative h3’(x)=4 - [1/((x +1) ln a)]. For a=√3, ln a=0.5 ln3≈0.549. So h3’(x)=4 - [1/((x +1)*0.549)]. For x ∈ [3,4], x +1 ∈ [4,5], so denominator is ≥4*0.549≈2.196. Therefore, [1/2.196]≈0.455. So h3’(x)=4 -0.455≈3.545>0. So h3 is increasing from -2.52 to 1.07. Therefore, crosses zero once in (3,4). Similarly, in equation (4): h4(x)=4(5 -x) - log_a(x +1). At x=4: h4(4)=4(1) - log_a(5)=4 -2.93≈1.07. At x=5: h4(5)=0 - log_a(6)≈- (ln6)/(0.5 ln3)≈- (1.792)/(0.549)≈-3.26. The derivative h4’(x)= -4 - [1/((x +1) ln a)]. Again, a=√3. So h4’(x)= -4 - [1/((x +1)*0.549)]. For x ∈ [4,5], x +1 ∈ [5,6], denominator ≈5*0.549≈2.745, so [1/2.745]≈0.364. Therefore, h4’(x)= -4 -0.364≈-4.364 <0. Therefore, h4 is decreasing from 1.07 to -3.26, so crosses zero once in (4,5). Therefore, for a=√3, in [3,5], there are two roots. So total roots:[0,1]:1[1,3]:1 (at x=2)[3,5]:2Total:4. But we need exactly 5. Therefore, a=√3 gives 4 roots, which is less than required. Therefore, a=√3 is not in the range. Wait, but this contradicts the earlier thought. Wait, maybe my analysis is wrong.Wait, for a=√3, in [1,3], we have 1 root (at x=2), but earlier, when a >√3, in [1,3], equations (1) and (2) both have solutions? Wait, let me recap.For a >√3:In equation (1) on [1,2], h1(x) = 2(x -1) - log_a(x +1). At x=1, h1(1)=-log_a(2). At x=2, h1(2)=2 - log_a(3). For a >√3, log_a(3) <2, so h1(2) >0. Since h1 is increasing (derivative h1’(x)=2 - [1/((x +1) ln a)]. For a >√3, ln a > ln√3 ≈0.549, so denominator (x +1) ln a >2*0.549≈1.098. So 1/( (x +1) ln a ) <1/1.098≈0.912. So h1’(x)=2 -0.912≈1.088 >0. Therefore, h1 is increasing. Therefore, h1(x) goes from -log_a(2) to 2 - log_a(3). Since a >√3, log_a(3) <2, so h1(2) >0. Since h1 is increasing and starts negative and ends positive, crosses zero once in (1,2). Similarly, in equation (2) on [2,3], h2(x)=2(3 -x) - log_a(x +1). At x=2: h2(2)=2(1) - log_a(3)=2 - log_a(3). For a >√3, log_a(3) <2, so h2(2) >0. At x=3: h2(3)=0 - log_a(4) <0. The derivative h2’(x)= -2 - [1/((x +1) ln a)]. Since ln a >0, h2’(x) <0. Therefore, h2(x) is decreasing from positive to negative, so crosses zero once in (2,3). Therefore, for a >√3, in [1,3], there are two roots: one in (1,2) and one in (2,3). Therefore, total roots:[0,1]:1[1,3]:2[3,5]: ?Now, for [3,5], with a >√3.At x=3: log_a(4). For a >√3, log_a(4) = ln4 / ln a. Since a >√3, ln a > ln√3 ≈0.549, so log_a(4) < ln4 /0.549 ≈1.386/0.549≈2.525. Then, in equation (3) on [3,4], h3(x)=4(x -3) - log_a(x +1). At x=3: h3(3)=0 - log_a(4)≈-2.525. At x=4: h3(4)=4(1) - log_a(5). For a >√3, log_a(5)=ln5 / ln a < ln5 /0.549≈1.609/0.549≈2.93. So h3(4)=4 - log_a(5) >4 -2.93≈1.07 >0. The derivative h3’(x)=4 - [1/((x +1) ln a)]. Since a >√3, ln a >0.549. For x ∈ [3,4], x +1 ∈ [4,5], so denominator (x +1) ln a >4*0.549≈2.196. Therefore, [1/(2.196)]≈0.456. So h3’(x)=4 -0.456≈3.544>0. Therefore, h3 is increasing from -2.525 to 1.07, crosses zero once in (3,4). Similarly, in equation (4) on [4,5], h4(x)=4(5 -x) - log_a(x +1). At x=4: h4(4)=4(1) - log_a(5)≈4 -2.93≈1.07. At x=5: h4(5)=0 - log_a(6)≈- (ln6)/ln a. Since a >√3, log_a(6)≈1.792/ln a <1.792/0.549≈3.26. Therefore, h4(5)≈-3.26. The derivative h4’(x)= -4 - [1/((x +1) ln a)] < -4 -0 (since [1/(...)] >0). So h4’(x) < -4, hence h4 is decreasing from 1.07 to -3.26, crosses zero once in (4,5). Therefore, in [3,5], two roots. Therefore, total roots:[0,1]:1[1,3]:2[3,5]:2Total:5.Therefore, for a >√3, we have exactly 5 roots. But the problem states that the equation has exactly 5 different roots. However, the answer options include C: (√3, +∞), but also D: (⁴√5, √3). Wait, this suggests that my analysis might be missing something. Wait, let's check for a=√3, as before:For a=√3, in [1,3], we have one root (at x=2), and in [3,5], two roots, and [0,1] one root: total 4. For a >√3, we get 5 roots. For a <√3, but greater than 1, what happens?Suppose a is between 1 and √3. Let's take a=2. Let's check.For a=2:In [0,1]: 1 root.In [1,3]:Equation (1) on [1,2]: h1(x)=2(x -1) - log_2(x +1). At x=1: h1(1)=0 - log_2(2)= -1. At x=2: h1(2)=2 - log_2(3)≈2 -1.585≈0.415>0. h1’(x)=2 - [1/((x +1) ln2)]. ln2≈0.693. For x ∈ [1,2], x +1 ∈ [2,3], so denominator≈2*0.693≈1.386. Therefore, 1/1.386≈0.722. So h1’(x)=2 -0.722≈1.278>0. Therefore, h1 is increasing from -1 to 0.415, crosses zero once in (1,2).Equation (2) on [2,3]: h2(x)=2(3 -x) - log_2(x +1). At x=2: h2(2)=2(1) - log_2(3)≈2 -1.585≈0.415>0. At x=3: h2(3)=0 - log_2(4)= -2. h2’(x)= -2 - [1/((x +1) ln2)]. For x ∈ [2,3], x +1 ∈ [3,4], denominator≈3*0.693≈2.079. 1/2.079≈0.481. So h2’(x)= -2 -0.481≈-2.481 <0. Therefore, h2 is decreasing from 0.415 to -2, crosses zero once in (2,3). Therefore, in [1,3], two roots.In [3,5]:Equation (3) on [3,4]: h3(x)=4(x -3) - log_2(x +1). At x=3: h3(3)=0 - log_2(4)= -2. At x=4: h3(4)=4 - log_2(5)≈4 -2.322≈1.678>0. h3’(x)=4 - [1/((x +1) ln2)]. For x ∈ [3,4], denominator≈4*0.693≈2.772. 1/2.772≈0.361. So h3’(x)=4 -0.361≈3.639>0. Therefore, h3 is increasing from -2 to1.678, crosses zero once in (3,4).Equation (4) on [4,5]: h4(x)=4(5 -x) - log_2(x +1). At x=4: h4(4)=4(1) - log_2(5)≈4 -2.322≈1.678>0. At x=5: h4(5)=0 - log_2(6)≈-2.585. h4’(x)= -4 - [1/((x +1) ln2)]. For x ∈ [4,5], denominator≈5*0.693≈3.465. 1/3.465≈0.289. So h4’(x)= -4 -0.289≈-4.289 <0. Therefore, h4 is decreasing from 1.678 to -2.585, crosses zero once in (4,5). Therefore, in [3,5], two roots. Total roots:1+2+2=5.But wait, a=2 is less than √3 (since √3≈1.732). Wait, but a=2 is greater than √3? Wait, 2≈2.0 and √3≈1.732. So a=2 is greater than √3. So, according to previous analysis, for a >√3, we have 5 roots, which includes a=2. But if we take a=1.5 (which is between 1 and √3), let's see:a=1.5.In [0,1]:1 root.In [1,3]:Equation (1): h1(x)=2(x -1) - log_{1.5}(x +1). At x=1: h1(1)=0 - log_{1.5}(2)= - (ln2)/(ln1.5)≈-0.693/0.405≈-1.71. At x=2: h1(2)=2 - log_{1.5}(3)=2 - (ln3)/(ln1.5)≈2 -1.585/0.405≈2 -3.915≈-1.915. So h1(2) is negative. Since h1 is increasing (derivative h1’(x)=2 - [1/((x +1) ln1.5)]). Let's compute h1’(x):ln1.5≈0.405. For x ∈ [1,2], x +1 ∈ [2,3]. So denominator≈2*0.405≈0.81. So [1/0.81≈1.23]. Therefore, h1’(x)=2 -1.23≈0.77>0. So h1 is increasing from -1.71 to -1.915. Wait, that can't be. If derivative is positive, but starting at -1.71 and going to -1.915, which is decreasing. Contradiction. Wait, my calculation must be wrong. Wait, h1(x)=2(x -1) - log_{1.5}(x +1). At x=2, h1(2)=2(1) - log_{1.5}(3). log_{1.5}(3)=ln3 / ln1.5≈1.0986/0.4055≈2.7095. Therefore, h1(2)=2 -2.7095≈-0.7095.Wait, earlier I must have miscalculated. Let's correct:log_{1.5}(3) = ln3 / ln1.5 ≈1.0986 /0.4055≈2.7095. Therefore, h1(2)=2 -2.7095≈-0.7095.Similarly, at x=1: h1(1)=0 - log_{1.5}(2)= - (ln2)/ln1.5≈-0.693/0.405≈-1.71.Derivative h1’(x)=2 - [1/((x +1) ln1.5)]. For x ∈ [1,2], x +1 ∈ [2,3]. So denominator ∈ [2*0.405≈0.81, 3*0.405≈1.215]. So [1/denominator] ∈ [0.823, 1.235]. So h1’(x)=2 - [0.823 to1.235] ∈ [0.765,1.177]. So h1’(x) >0. Therefore, h1 is increasing from -1.71 to -0.7095. Since it's increasing but stays negative, no roots in (1,2). Therefore, equation (1) has no solution.Equation (2) on [2,3]: h2(x)=2(3 -x) - log_{1.5}(x +1). At x=2: h2(2)=2(1) - log_{1.5}(3)=2 -2.7095≈-0.7095. At x=3: h2(3)=0 - log_{1.5}(4)= - (ln4)/(ln1.5)≈-1.386/0.405≈-3.423. The derivative h2’(x)= -2 - [1/((x +1) ln1.5)]. For x ∈ [2,3], denominator ∈ [3*0.405≈1.215,4*0.405≈1.62]. So [1/denominator] ∈ [0.617,0.823]. Therefore, h2’(x)= -2 - [0.617 to0.823]≈-2.617 to-2.823 <0. So h2 is decreasing from -0.7095 to-3.423. Therefore, no roots in (2,3). Therefore, in [1,3], no roots. Therefore, total roots so far:1.In [3,5]:Equation (3) on [3,4]: h3(x)=4(x -3) - log_{1.5}(x +1). At x=3: h3(3)=0 - log_{1.5}(4)≈-3.423. At x=4: h3(4)=4(1) - log_{1.5}(5)≈4 - (ln5)/ln1.5≈4 -1.609/0.405≈4 -3.973≈0.027. The derivative h3’(x)=4 - [1/((x +1) ln1.5)]. For x ∈ [3,4], x +1 ∈ [4,5], denominator≈4*0.405≈1.62. [1/1.62≈0.617]. So h3’(x)=4 -0.617≈3.383>0. Therefore, h3 is increasing from -3.423 to0.027. Therefore, crosses zero once in (3,4).Equation (4) on [4,5]: h4(x)=4(5 -x) - log_{1.5}(x +1). At x=4: h4(4)=4(1) - log_{1.5}(5)≈4 -3.973≈0.027. At x=5: h4(5)=0 - log_{1.5}(6)≈- (ln6)/ln1.5≈-1.792/0.405≈-4.425. The derivative h4’(x)= -4 - [1/((x +1) ln1.5)]. For x ∈ [4,5], denominator≈5*0.405≈2.025. [1/2.025≈0.494]. So h4’(x)= -4 -0.494≈-4.494 <0. Therefore, h4 is decreasing from0.027 to-4.425, crosses zero once in (4,5). Therefore, in [3,5], two roots. Total roots:1+0+2=3. Which is less than 5.Therefore, for a=1.5, which is between 1 and √3, total roots are 3. So, somewhere between a=1.5 and a=√3, the number of roots increases from 3 to 4. But the problem states that the equation has exactly 5 roots. Therefore, there must be another interval where the number of roots increases.Wait, perhaps my previous analysis is missing some intersections. Wait, when a decreases below √3, what happens in [1,3]?When a decreases, log_a(x +1) increases faster. For example, if a is just slightly larger than 1, log_a(x +1) increases rapidly.Wait, suppose a=⁴√5 ≈1.495. Let's check.a=⁴√5≈1.495.In [0,1]:1 root.In [1,3]:Equation (1): h1(x)=2(x -1) - log_a(x +1). At x=1: h1(1)=0 - log_a(2)= - (ln2)/(ln a). Since a=⁴√5, ln a=(1/4) ln5≈(1/4)*1.609≈0.402. So log_a(2)=ln2/0.402≈0.693/0.402≈1.724. Therefore, h1(1)= -1.724.At x=2: h1(2)=2 - log_a(3)=2 - (ln3)/0.402≈2 -1.0986/0.402≈2 -2.732≈-0.732.The derivative h1’(x)=2 - [1/((x +1) ln a)]. For x ∈ [1,2], x +1 ∈ [2,3], ln a≈0.402. So denominator≈2*0.402≈0.804. Therefore, [1/0.804≈1.244]. So h1’(x)=2 -1.244≈0.756>0. Therefore, h1 is increasing from -1.724 to -0.732. No roots in (1,2).Equation (2): h2(x)=2(3 -x) - log_a(x +1). At x=2: h2(2)=2(1) - log_a(3)≈2 -2.732≈-0.732. At x=3: h2(3)=0 - log_a(4)= - (ln4)/0.402≈-1.386/0.402≈-3.448. Derivative h2’(x)= -2 - [1/((x +1) ln a)]. For x ∈ [2,3], denominator≈3*0.402≈1.206. So [1/1.206≈0.829]. So h2’(x)= -2 -0.829≈-2.829<0. Therefore, h2 is decreasing from -0.732 to-3.448. No roots in (2,3). Therefore, in [1,3], no roots.In [3,5]:Equation (3): h3(x)=4(x -3) - log_a(x +1). At x=3: h3(3)=0 - log_a(4)≈-3.448. At x=4: h3(4)=4(1) - log_a(5)≈4 - (ln5)/0.402≈4 -1.609/0.402≈4 -4.00≈0. So approximately zero. Derivative h3’(x)=4 - [1/((x +1) ln a)]. For x ∈ [3,4], x +1 ∈ [4,5], denominator≈4*0.402≈1.608. [1/1.608≈0.622]. So h3’(x)=4 -0.622≈3.378>0. Therefore, h3 is increasing from -3.448 to0. If h3(4)=0, then there's a root at x=4. But since x=4 is in [3,4]? No, x=4 is the endpoint. So h3(x) reaches zero at x=4. But equation (3) is on [3,4], so x=4 is included. So, at x=4, h3(4)=0. Therefore, one root at x=4.Equation (4): h4(x)=4(5 -x) - log_a(x +1). At x=4: h4(4)=4(1) - log_a(5)=4 -4≈0. At x=5: h4(5)=0 - log_a(6)= - (ln6)/0.402≈-1.792/0.402≈-4.458. Derivative h4’(x)= -4 - [1/((x +1) ln a)]. For x ∈ [4,5], denominator≈5*0.402≈2.01. So [1/2.01≈0.497]. So h4’(x)= -4 -0.497≈-4.497<0. Therefore, h4 is decreasing from0 to-4.458. Therefore, equation h4(x)=0 has a root at x=4. Therefore, in [3,5], x=4 is a root for both equations (3) and (4). But x=4 is a single point. Therefore, in [3,5], we have one root at x=4. Therefore, total roots:1 (from [0,1]) +0 (from [1,3]) +1 (from [3,5])=2. But this contradicts our earlier analysis. Wait, but a=⁴√5≈1.495.Wait, perhaps my approximation is off. Let's do more precise calculation.For a=⁴√5, ln a=(ln5)/4≈1.6094/4≈0.40235.log_a(5)=ln5 / ln a=1.6094/0.40235≈4.0. Exactly. Because a=5^{1/4}, so a^4=5. Therefore, log_a(5)=4. Therefore, at x=4, log_a(5)=4. So equation (3) at x=4: h3(4)=4*(1) - log_a(5)=4 -4=0. Similarly, equation (4) at x=4: h4(4)=4*(1) - log_a(5)=0. Therefore, x=4 is a root for both equations. So in [3,5], x=4 is a root. But what about near x=4?For equation (3) on [3,4], h3(x)=4(x -3) - log_a(x +1). At x=4: h3(4)=0. The derivative at x=4: h3’(4)=4 - [1/(5*0.40235)]≈4 - [1/2.01175]≈4 -0.497≈3.503>0. So near x=4, h3 is increasing through zero. Therefore, x=4 is the only root in [3,4]. For equation (4) on [4,5], h4(x)=4(5 -x) - log_a(x +1). At x=4: h4(4)=0. The derivative h4’(4)= -4 - [1/(5*0.40235)]≈-4 -0.497≈-4.497<0. So near x=4, h4 is decreasing through zero. Therefore, x=4 is the only root in [4,5]. But since x=4 is the boundary between [3,4] and [4,5], it's counted as one root. Therefore, in [3,5], only one root at x=4. Therefore, total roots:1+0+1=2.But the problem states that the equation has exactly 5 roots. Therefore, there must be a critical value of a where in [3,5], the number of roots changes from 2 to1, and in [1,3], the number changes from0 to2. Therefore, to have exactly 5 roots, a must be such that in [1,3], there are two intersections, and in [3,5], two intersections. From previous analysis, this occurs when a >√3. However, when a=⁴√5≈1.495, which is less than √3≈1.732, but log_a(5)=4. This suggests that when log_a(5)=4, i.e., a=⁴√5, then equation (3) has a root exactly at x=4. For a >⁴√5, log_a(5) <4, so in equation (3), h3(x) at x=4 is 4 - log_a(5) >0, and since h3 is increasing, crosses zero once in (3,4). Similarly, in equation (4), h4(x)=4(5 -x) - log_a(x +1). At x=4: h4(4)=4 - log_a(5) >0. At x=5: h4(5)=0 - log_a(6). Since a >⁴√5, log_a(6)=ln6 / ln a. For example, if a=2, log_a(6)=ln6 / ln2≈1.792/0.693≈2.585. So h4(5)= -2.585. Therefore, h4(x) is decreasing from positive to negative, so crosses zero once in (4,5). Therefore, for a >⁴√5, in [3,5], two roots.But in [1,3], for a between ⁴√5 and √3, we need to check if there are two roots. For example, take a=⁵√10≈10^(1/5)≈1.5848, which is between ⁴√5≈1.495 and √3≈1.732.Let's take a=1.5848.In [1,3]:Equation (1): h1(x)=2(x -1) - log_a(x +1). At x=1: h1(1)= -log_a(2)= - (ln2)/ln a≈-0.693/(0.462)≈-1.5. At x=2: h1(2)=2 - log_a(3)=2 - (ln3)/0.462≈2 -1.0986/0.462≈2 -2.378≈-0.378. The derivative h1’(x)=2 - [1/((x +1) ln a)]. For x ∈ [1,2], denominator≈(2)*0.462≈0.924. [1/0.924≈1.082]. So h1’(x)=2 -1.082≈0.918>0. Therefore, h1 is increasing from -1.5 to-0.378. No roots.Equation (2): h2(x)=2(3 -x) - log_a(x +1). At x=2: h2(2)=2 - log_a(3)≈-0.378. At x=3: h2(3)=0 - log_a(4)= - (ln4)/0.462≈-1.386/0.462≈-3.0. Derivative h2’(x)= -2 - [1/((x +1) ln a)]. For x ∈ [2,3], denominator≈3*0.462≈1.386. [1/1.386≈0.722]. So h2’(x)= -2 -0.722≈-2.722 <0. Therefore, h2 is decreasing from -0.378 to-3.0. No roots. Therefore, in [1,3], no roots.In [3,5]:Equation (3): h3(x)=4(x -3) - log_a(x +1). At x=3: h3(3)=0 - log_a(4)≈-3.0. At x=4: h3(4)=4 - log_a(5)=4 - (ln5)/0.462≈4 -1.609/0.462≈4 -3.48≈0.52>0. Derivative h3’(x)=4 - [1/((x +1) ln a)]. For x ∈ [3,4], denominator≈4*0.462≈1.848. [1/1.848≈0.541]. So h3’(x)=4 -0.541≈3.459>0. Therefore, h3 is increasing from -3.0 to0.52, crossing zero once.Equation (4): h4(x)=4(5 -x) - log_a(x +1). At x=4: h4(4)=4 - log_a(5)≈0.52. At x=5: h4(5)=0 - log_a(6)= - (ln6)/0.462≈-1.792/0.462≈-3.88. Derivative h4’(x)= -4 - [1/((x +1) ln a)]. For x ∈ [4,5], denominator≈5*0.462≈2.31. [1/2.31≈0.433]. So h4’(x)= -4 -0.433≈-4.433<0. Therefore, h4 is decreasing from0.52 to-3.88, crossing zero once. Therefore, in [3,5], two roots. Total roots:1+0+2=3. Still less than 5.This suggests that even for a between ⁴√5 and √3, the total roots are still 3. Therefore, my previous analysis must be incorrect. Perhaps there's a different interval where the roots appear.Wait, maybe in [3,5], when a=⁴√5, the equation has one root at x=4, but for a >⁴√5, it has two roots. For a <⁴√5, in [3,5], there might be two roots. Wait, let's test a=1.4, which is less than ⁴√5≈1.495.a=1.4.In [3,5]:Equation (3): h3(x)=4(x -3) - log_{1.4}(x +1). At x=3: log_{1.4}(4)=ln4/ln1.4≈1.386/0.3365≈4.12. So h3(3)=0 -4.12≈-4.12. At x=4: h3(4)=4 - log_{1.4}(5)=4 - (ln5)/ln1.4≈4 -1.609/0.3365≈4 -4.78≈-0.78. Derivative h3’(x)=4 - [1/((x +1) ln1.4)]. For x ∈ [3,4], x +1 ∈ [4,5], ln1.4≈0.3365. Denominator≈4*0.3365≈1.346. [1/1.346≈0.743]. So h3’(x)=4 -0.743≈3.257>0. Therefore, h3 is increasing from -4.12 to-0.78. No roots.Equation (4): h4(x)=4(5 -x) - log_{1.4}(x +1). At x=4: h4(4)=4 - log_{1.4}(5)≈4 -4.78≈-0.78. At x=5: h4(5)=0 - log_{1.4}(6)= - (ln6)/ln1.4≈-1.792/0.3365≈-5.325. Derivative h4’(x)= -4 - [1/((x +1) ln1.4)]. For x ∈ [4,5], denominator≈5*0.3365≈1.6825. [1/1.6825≈0.594]. So h4’(x)= -4 -0.594≈-4.594 <0. Therefore, h4 is decreasing from -0.78 to-5.325. No roots. Therefore, in [3,5], no roots. Total roots:1+0+0=1.Therefore, for a <⁴√5, roots decrease further. So, it seems that the only way to get 5 roots is when a >√3. But wait, the answer options include D: (⁴√5, √3). Therefore, there must be an error in my previous analysis.Wait, maybe when a is in (⁴√5, √3), in [3,5], there are two roots, and in [1,3], there are two roots, and in [0,1], one root. Let's check for a=2^(1/2)=√2≈1.414, which is between ⁴√5≈1.495 and √3≈1.732. Wait, no, √2≈1.414 <⁴√5≈1.495. So maybe take a=1.6.Wait, let's try a=1.6.ln a≈0.47.In [1,3]:Equation (1): h1(x)=2(x -1) - log_{1.6}(x +1). At x=1: h1(1)=0 - log_{1.6}(2)= - (ln2)/ln1.6≈-0.693/0.47≈-1.475. At x=2: h1(2)=2 - log_{1.6}(3)=2 - (ln3)/0.47≈2 -1.0986/0.47≈2 -2.338≈-0.338. h1’(x)=2 - [1/((x +1) ln1.6)]. For x ∈ [1,2], denominator≈2*0.47≈0.94. [1/0.94≈1.064]. So h1’(x)=2 -1.064≈0.936>0. Therefore, h1 is increasing from -1.475 to-0.338. No roots.Equation (2): h2(x)=2(3 -x) - log_{1.6}(x +1). At x=2: h2(2)=2 - log_{1.6}(3)≈-0.338. At x=3: h2(3)=0 - log_{1.6}(4)= - (ln4)/0.47≈-1.386/0.47≈-2.95. Derivative h2’(x)= -2 - [1/((x +1) ln1.6)]. For x ∈ [2,3], denominator≈3*0.47≈1.41. [1/1.41≈0.709]. So h2’(x)= -2 -0.709≈-2.709 <0. Therefore, h2 is decreasing from -0.338 to-2.95. No roots. Therefore, [1,3] no roots.In [3,5]:Equation (3): h3(x)=4(x -3) - log_{1.6}(x +1). At x=3: h3(3)=0 - log_{1.6}(4)≈-2.95. At x=4: h3(4)=4 - log_{1.6}(5)=4 - (ln5)/0.47≈4 -1.609/0.47≈4 -3.424≈0.576>0. Derivative h3’(x)=4 - [1/((x +1) ln1.6)]. For x ∈ [3,4], denominator≈4*0.47≈1.88. [1/1.88≈0.532]. So h3’(x)=4 -0.532≈3.468>0. Therefore, h3 is increasing from -2.95 to0.576, crossing zero once.Equation (4): h4(x)=4(5 -x) - log_{1.6}(x +1). At x=4: h4(4)=4 - log_{1.6}(5)≈0.576. At x=5: h4(5)=0 - log_{1.6}(6)= - (ln6)/0.47≈-1.792/0.47≈-3.813. Derivative h4’(x)= -4 - [1/((x +1) ln1.6)]. For x ∈ [4,5], denominator≈5*0.47≈2.35. [1/2.35≈0.426]. So h4’(x)= -4 -0.426≈-4.426 <0. Therefore, h4 is decreasing from0.576 to-3.813, crossing zero once. Therefore, in [3,5], two roots. Total roots:1+0+2=3.So, even for a=1.6, total roots are 3. Therefore, I must have made a mistake in my initial analysis. It seems that only when a >√3, we get 5 roots. Therefore, the answer should be C: (√3, +∞). However, the options include D: (⁴√5, √3). Therefore, there must be another perspective.Wait, perhaps when a is in (⁴√5, √3), the equation f(x)=log_a(x +1) has two roots in [3,5] and two roots in [1,3], but the total would be 1+2+2=5. But my previous examples with a=1.6 didn't yield that. Maybe the key is that when a >⁴√5, log_a(5) <4, allowing two roots in [3,5], and when a <√3, log_a(3) >2, allowing two roots in [1,3]. Therefore, the overlap where both conditions are met is a ∈ (⁴√5, √3). Let's verify.Condition for two roots in [1,3]:log_a(3) >2 → a^2 <3 → a <√3.Condition for two roots in [3,5]:log_a(5) <4 → a^4 >5 → a >⁴√5.Therefore, the overlap is a ∈ (⁴√5, √3). Therefore, for these values of a, we have two roots in [1,3] and two roots in [3,5], plus one in [0,1], totaling 5. Therefore, the correct answer should be D: (⁴√5, √3). But earlier examples with a=2 (which is >√3) gave 5 roots. So there must be a miscalculation.Wait, if a >√3, then log_a(3) <2, so in [1,3], two roots. If a >⁴√5, log_a(5) <4, so two roots in [3,5]. Therefore, for a >max(√3, ⁴√5), which is a >√3 since √3≈1.732 and ⁴√5≈1.495, then a >√3 satisfies both conditions. However, if a is between ⁴√5 and √3, then log_a(3) >2 (since a <√3) but log_a(5) <4 (since a >⁴√5). Therefore, in [1,3], the equations h1 and h2 would have two roots (since log_a(3) >2, which is necessary for h1 and h2 to have two roots), but wait, no. If log_a(3) >2, then in equation (1) on [1,2], h1(2)=2 - log_a(3,<think>Okay, let's try to tackle this problem step by step. First, I need to understand the function f(x) given here. It's defined piecewise: for x in (1, +∞), it's 2 times f(x-2), and for x in [-1, 1], it's 1 - |x|. Then, we have the equation f(x) - log_a(x + 1) = 0, which we need to find for which values of a this equation has exactly 5 different roots in [0,5]. The answer choices are given, so maybe after solving, I can check which interval corresponds.First, let me try to sketch or understand the function f(x) on the interval [0,5]. Since the definition is recursive, I need to figure out f(x) in different intervals by breaking it down.Starting with the base case: for x in [-1,1], which includes [0,1], f(x) = 1 - |x|. So on [0,1], it's 1 - x because x is positive. So from 0 to 1, it's a linear function starting at (0,1) and decreasing to (1,0).Then for x in (1, +∞), f(x) = 2f(x-2). So for x in (1,3], since x-2 is in (-1,1]. But since we are dealing with x >=0, x in (1,3], x-2 is in (-1,1]. However, f is defined on [-1,1], so for x in (1,3], f(x) = 2f(x-2). But x-2 here would be in (-1,1]. However, since we are considering x in [0,5], maybe I should split the intervals into [0,1], (1,3], (3,5]. Let's see:For x in (1,3], x - 2 is in (-1,1], so f(x) = 2*(1 - |x - 2|). But since x is in (1,3], x - 2 is in (-1,1]. But since x is greater than 1, x - 2 is greater than -1. So in (1,3], x - 2 ranges from -1 to 1. Wait, but x is in (1,3], so x - 2 is in (-1,1]. So f(x) = 2*(1 - |x - 2|). So on (1,3], the function is 2*(1 - |x - 2|). That would be a triangle wave, peaking at x = 2 with value 2*1 = 2, decreasing linearly to 0 at x =1 and x=3. But wait, when x approaches 1 from the right, f(x) approaches 2*(1 - |1 - 2|) = 2*(1 - 1) = 0. And at x=2, it's 2*(1 - 0) = 2. Then at x=3, it's 2*(1 - |3 -2|) = 2*(1 -1) =0. So between 1 and 3, it's a triangle going from 0 up to 2 and back to 0.Similarly, for x in (3,5], x - 2 is in (1,3], so f(x) = 2*f(x - 2). So f(x) = 2* [2*(1 - |(x - 2) - 2|)] = 4*(1 - |x -4|). So here, x in (3,5], x -4 ranges from (-1,1]. Therefore, f(x) = 4*(1 - |x -4|). So again, a triangular function, peaking at x=4 with value 4*(1 -0) =4, and decreasing to 0 at x=3 and x=5.Therefore, putting it all together:On [0,1]: f(x) =1 - x.On (1,3]: f(x)=2*(1 - |x -2|).On (3,5]: f(x)=4*(1 - |x -4|).So the graph of f(x) on [0,5] is three linear segments:1. From (0,1) to (1,0).2. From (1,0) up to (2,2) and back down to (3,0).3. From (3,0) up to (4,4) and back down to (5,0).So it's a series of triangles doubling in height each time the interval increases by 2. So from 0 to1, height 1; 1 to3, height 2; 3 to5, height4.Now, the equation f(x) - log_a(x +1) =0, so f(x) = log_a(x +1). We need to find a >0, a ≠1 such that this equation has exactly 5 roots in [0,5].So, let's first note that x +1 ranges from 1 (when x=0) to 6 (when x=5). So log_a(x +1) is defined for x in [0,5], since x +1 >0.We need to analyze where the graphs of f(x) and log_a(x +1) intersect 5 times in [0,5].To do this, we can consider the behavior of log_a(x +1) relative to f(x). Depending on the base a, the logarithm function can be increasing or decreasing. Since a>0 and a≠1, if a>1, log_a is increasing; if 0<a<1, it's decreasing. However, the answer choices all have a>1 (since options B, C, D have lower bounds greater than 1 except option A which is (1, sqrt(3)), but maybe not). Wait, the options are:A: (1, sqrt(3))B: (4th root of5, +infty)C: (sqrt(3), +infty)D: (4th root of5, sqrt(3))Hmm, so all options except A and D have upper or lower bounds. So maybe both a>1 and 0<a<1 could be possible, but the answer choices only include a>1. Let me check the options again.Wait, the problem states "a >0 and a ≠1". So a can be greater than 1 or between 0 and 1. But the answer choices A is (1, sqrt(3)), B is (4th root of5, +infty), C is (sqrt(3), +infty), D is (4th root of5, sqrt(3)). So these are all intervals where a>1. So maybe only a>1 are considered here? Wait, but the problem didn't restrict a>1; it's just a>0 and a≠1. However, maybe in the options, all are for a>1. Let me check.Wait, the options are:A: (1, sqrt(3))B: (⁴√5, +infty)C: (sqrt(3), +infty)D: (⁴√5, sqrt(3))So these are all intervals where a>1. Because ⁴√5 is about 5^(1/4) ≈ 1.495, sqrt(3)≈1.732. So the options are intervals above 1. So perhaps even though the problem allows a<1, the correct answer is among these, so maybe we can focus on a>1.But let's proceed.First, let's analyze the function f(x) on [0,5]:- [0,1]: linear from (0,1) to (1,0).- (1,3]: triangle wave peaking at 2 with maximum 2.- (3,5]: triangle wave peaking at4 with maximum4.So f(x) has three segments:1. Decreasing from 1 to 0 on [0,1].2. Increasing from 0 to 2 on (1,2], then decreasing back to 0 on (2,3].3. Increasing from 0 to4 on (3,4], then decreasing back to 0 on (4,5].Now, log_a(x +1) is a logarithmic function. Since x +1 increases from1 to6, log_a(x +1) will behave depending on a.If a>1, log_a(x +1) is increasing, concave, starting at log_a(1)=0, going to log_a(6). If 0<a<1, log_a(x +1) is decreasing, convex, starting at0 and going to log_a(6) <0. But in the interval [0,5], since x+1 ≥1, log_a(x+1) is positive when a>1 and negative when 0<a<1. However, f(x) is non-negative on [0,5], as 1 - |x| is non-negative on [-1,1], and the recursive definition multiplies by 2 each time, so f(x) is non-negative. Therefore, if 0 < a <1, log_a(x +1) is negative on [0,5], so f(x) - log_a(x +1) =0 would imply f(x) = log_a(x +1), but since f(x) is non-negative and log_a(x+1) is negative, there are no solutions. Therefore, the equation f(x) - log_a(x +1) =0 can only have solutions when a>1. So perhaps only a>1 are considered here, which is why all answer options are intervals where a>1.Therefore, we can restrict to a>1.So log_a(x +1) is increasing, concave, starting at0 when x=0, going to log_a(6).So we need to find a>1 such that the equation f(x) = log_a(x +1) has exactly5 roots in [0,5].Now, to find the number of intersections between f(x) and log_a(x +1), we need to consider the shape of both functions.f(x) has peaks at x=0 (1), x=2 (2), x=4 (4). Wait, actually:Wait, at x=0, f(x)=1 -0=1.At x=1, f(x)=0.From x=1 to2, f(x) increases to2 at x=2, then decreases back to0 at x=3.From x=3 to4, f(x) increases to4 at x=4, then decreases back to0 at x=5.So the graph of f(x) has peaks at x=0 (1), x=2 (2), x=4 (4). So three peaks, but the first one is at x=0, which is the starting point. Then at x=2 and x=4.But the interval [0,5] includes x=0 to x=5.So the function f(x) is 1 at x=0, then decreases to0 at x=1, then goes up to2 at x=2, back to0 at x=3, up to4 at x=4, back to0 at x=5.So the graph is like a series of triangles with increasing height.Now, log_a(x +1) is an increasing function starting at x=0 with log_a(1)=0, then increasing to log_a(6) at x=5.So at x=0, f(x)=1 and log_a(1)=0. So f(x) starts above log_a(x +1).At x=1, f(x)=0, log_a(2). Depending on a, log_a(2) can be greater or less than0. But since a>1, log_a(2) is positive. So at x=1, f(x)=0, log_a(2) >0, so log_a(x+1) is above f(x) here.Between x=1 andx=3, f(x) rises to2 atx=2, then falls back to0 atx=3. log_a(x +1) continues to increase.Similarly, between x=3 andx=5, f(x) rises to4 atx=4, then falls back to0 atx=5. log_a(x +1) continues to increase to log_a(6).So we need to see how the log curve intersects with these triangles.The number of intersections will depend on the steepness of the log function, which is controlled by the base a.If a is very large, log_a(x +1) grows very slowly. For example, if a is large, log_a(6) is close to0, since 6 is much smaller than a. Wait, no: if a is large, log_a(6) is small. Wait, log_a(6) = ln(6)/ln(a). If a increases, log_a(6) decreases. So when a is large, log_a(x +1) is a slowly increasing function that stays low. Conversely, if a is slightly larger than1, log_a(x +1) increases more rapidly.So, the key is to adjust a such that log_a(x +1) intersects f(x) exactly5 times.To have exactly5 roots, we need intersections in each segment where f(x) and log_a(x +1) cross.Let me try to break down the possible regions where intersections can occur.First, on [0,1]: f(x) =1 -x. log_a(x +1) is increasing from0 to log_a(2). At x=0, f(x)=1 and log_a(1)=0, so f(x) is above. At x=1, f(x)=0 and log_a(2). Since log_a(2) >0 (because a>1), log_a(2) is positive. So f(x) goes from1 to0, while log_a(x +1) goes from0 to log_a(2). So they must cross once in (0,1). Because f(x) is decreasing from1 to0, and log_a(x +1) is increasing from0 to log_a(2). Since at x=0, f(x) > log_a(x +1), and at x=1, f(x)=0 < log_a(2). Therefore, exactly1 crossing in (0,1).So that's one root.Then, betweenx=1 andx=3: f(x) is a triangle from0 up to2 atx=2 and back to0. log_a(x +1) is increasing from log_a(2) to log_a(4).Depending on the value of a, log_a(4) could be above or below2. Similarly, log_a(2) could be above or below the minimum of f(x) in this interval.But since log_a(x +1) is increasing, and f(x) first increases to2 then decreases. So potentially, there could be0,1, or2 intersections here.Similarly, betweenx=3 andx=5: f(x) is a triangle from0 up to4 atx=4 and back to0. log_a(x +1) is increasing from log_a(4) to log_a(6). Depending on a, log_a(6) could be above or below4. So again, potentially0,1, or2 intersections here.In total, to have exactly5 roots, we need1 root in [0,1], and then2 roots in (1,3], and2 roots in (3,5]. So total 5. Alternatively, maybe different distributions, but given the options, likely this.But let's check:In [0,1]:1 root.In (1,3]: since f(x) goes from0 to2 and back to0, and log_a(x +1) is increasing from log_a(2) to log_a(4). So if log_a(x +1) intersects the rising part and the falling part of the triangle, there would be2 intersections here. Similarly, if log_a(x +1) is entirely above or below f(x), there would be0 or maybe tangent, but we need exactly2.Similarly, in (3,5], f(x) goes from0 to4 and back to0, while log_a(x +1) is increasing from log_a(4) to log_a(6). Again, if log_a(x +1) intersects both the rising and falling parts, there would be2 intersections here. If it's tangent at the peak, that's1 intersection. If it's entirely above or below, then0.So total roots would be1 (from [0,1]) + number in (1,3] + number in (3,5].To get exactly5 roots, we need 1 +2 +2=5. So we need both intervals (1,3] and (3,5] to have2 intersections each. Therefore, we need log_a(x +1) to intersect the triangle in (1,3] twice and the triangle in (3,5] twice.Alternatively, maybe in one interval it intersects twice and in the other once, but that would give total 1+2+1=4 or 1+1+2=4. So to get5, need 2+2.Therefore, the key is to find a such that log_a(x +1) intersects each of the two upper triangles twice.To have two intersections in each triangle, log_a(x +1) must cross the ascending and descending parts of the triangle. Therefore, log_a(x +1) must be below the peak of the triangle but high enough to cross both sides.For the triangle in (1,3], which peaks at2, log_a(4) must be greater than2? Wait, no. Wait, log_a(x +1) atx=3 is log_a(4). If log_a(4) >2, then at x=3, log_a(4) >2, but the triangle at x=3 is0. So if log_a(4) >2, then the log function would cross the descending part of the triangle (from2 to0) once. But we need two intersections. Wait, maybe not. Let me think.Wait, in the interval (1,3], f(x) is a triangle starting at0, rising to2 atx=2, then falling back to0 atx=3. log_a(x +1) is increasing from log_a(2) to log_a(4). To have two intersections here, the log function must cross the ascending part (from0 to2) and the descending part (from2 to0). Therefore, log_a(x +1) must be less than2 at x=2, but greater than0 at x=1. Wait, but log_a(2) is already greater than0. Hmm.Wait, let's formalize this. To have two intersections in (1,3], log_a(x +1) must intersect the ascending slope and the descending slope. So there must be some x in (1,2) where log_a(x +1) = f(x), and some x in (2,3) where log_a(x +1) = f(x). Therefore, at x=2, log_a(3) must be less than2 (since f(x)=2 atx=2). If log_a(3) <2, then the log function atx=2 is below the peak, so it would have crossed the ascending part once before x=2, and then cross the descending part once after x=2. Similarly, if log_a(4) >0, which it is since a>1, but we need log_a(4) to be less than the value of f(x) at some point in (2,3). Wait, but f(x) in (2,3) is decreasing from2 to0. So log_a(x +1) is increasing. So for log_a(x +1) to intersect f(x) in (2,3), log_a(x +1) must cross from below to above? Wait, f(x) is decreasing, log_a is increasing. So at x=2, f(x)=2, log_a(3). If log_a(3) <2, then at x=2, log_a(3) <2. Then as x increases from2 to3, f(x) decreases from2 to0, and log_a(x +1) increases from log_a(3) to log_a(4). So if log_a(4) >0, which it is, but when does log_a(x +1) cross f(x) in (2,3)? The question is whether log_a(x +1) can reach up to meet the descending f(x). If log_a(3) <2 and log_a(4) >0, but the question is whether log_a(4) is less than or greater than f(x) atx=3. Wait, f(x) atx=3 is0, and log_a(4) is positive. Therefore, atx=3, log_a(4) > f(x)=0. Therefore, since log_a(x +1) is increasing and f(x) is decreasing, there must be exactly one crossing in (2,3) if log_a(3) <2. Because atx=2, log_a(3) <2, and atx=3, log_a(4) >0 =f(3). So since log_a(x +1) is increasing and f(x) is decreasing, they cross exactly once in (2,3). But in (1,2), since log_a(x +1) is increasing and f(x) is increasing, if log_a(2) < f(x) at some point.Wait, atx=1, log_a(2) >0 =f(1). Atx=2, log_a(3) <2. So log_a(x +1) goes from0 atx=1 to log_a(3) <2 atx=2, while f(x) goes from0 to2. So they start atx=1 with log_a(2) >0 and f(x)=0, then atx=2, log_a(3) <2. So in the interval (1,2), log_a(x +1) starts above f(x) atx=1 (since f(x)=0, log_a(2) >0) and ends atx=2 with log_a(3) <2. But f(x) is increasing from0 to2. Therefore, depending on how log_a(x +1) and f(x) behave, there could be one crossing in (1,2). Because atx=1, log_a(2) >0 =f(1); atx=2, log_a(3) <2 =f(2). Therefore, since log_a(x +1) is increasing and f(x) is increasing, but log_a(x +1) starts above and ends below. Therefore, they must cross exactly once in (1,2). Therefore, in total, in (1,3], there are two crossings: one in (1,2), one in (2,3).Similarly, for the interval (3,5], f(x) goes from0 up to4 atx=4, then back to0 atx=5. log_a(x +1) is increasing from log_a(4) to log_a(6). To have two crossings here, log_a(x +1) must cross the ascending part (from0 to4) and the descending part (from4 to0). For this, log_a(6) must be less than4, and log_a(4) must be less than4? Wait, let's analyze.Atx=3, f(x)=0, log_a(4). Since log_a(4) >0, and f(x)=0, so log_a(4) >f(x) atx=3. Then, log_a(x +1) increases from log_a(4) to log_a(6). f(x) increases from0 to4 atx=4, then decreases to0 atx=5. So similar to the previous interval, in (3,4], f(x) increases from0 to4, and log_a(x +1) increases from log_a(4) to log_a(5). Atx=4, log_a(5). If log_a(5) <4, then atx=4, log_a(5) <4. Then, since atx=3, log_a(4) >0 =f(3), and atx=4, log_a(5) <4 =f(4). So since log_a(x +1) is increasing and f(x) is increasing, starting above and ending below, there is exactly one crossing in (3,4). Then, in (4,5], f(x) decreases from4 to0, while log_a(x +1) increases from log_a(5) to log_a(6). If log_a(6) <4, then atx=5, log_a(6) <0? No, log_a(6) is positive. Wait, if log_a(6) <4, then since f(x) is decreasing from4 to0 and log_a(x +1) is increasing from log_a(5) to log_a(6), which is less than4. Therefore, if log_a(5) <4 and log_a(6) <4, then in (4,5], since log_a(x +1) is increasing towards log_a(6) <4 and f(x) is decreasing from4 to0, there could be a crossing if log_a(6) is greater than0, which it is. Wait, but f(x) starts at4 and goes to0, log_a(x +1) starts atlog_a(5) <4 and goes to log_a(6) <4. So if log_a(5) <4, then atx=4, log_a(5) <4, and atx=5, log_a(6) <4. So the log function is below4 for the entire interval (3,5], but f(x) goes up to4. So for intersections in (3,5], we need log_a(x +1) to intersect f(x) both on the way up and on the way down.Wait, perhaps similar to the previous interval. Atx=3, log_a(4) >0 =f(3). As x increases from3 to4, f(x) increases from0 to4, and log_a(x +1) increases from log_a(4) to log_a(5). If log_a(5) <4, then atx=4, log_a(5) <4. So f(x) goes from0 to4, log_a(x +1) goes from log_a(4) to log_a(5). So if log_a(4) <4 and log_a(5) <4, then the log function is below f(x) atx=4. But since log_a(x +1) starts at log_a(4) >0 =f(3) and increases. So betweenx=3 andx=4, f(x) starts at0 and goes up to4, log_a(x +1) starts atlog_a(4) and goes up to log_a(5). If log_a(4) <4, then there will be a crossing in (3,4). Wait, similar to the previous interval.Wait, atx=3: f(x)=0, log_a(4) >0. So log_a(x +1) is above f(x).Atx=4: f(x)=4, log_a(5). If log_a(5) <4, then log_a(x +1) is below f(x) atx=4. Therefore, since log_a(x +1) is increasing and f(x) is increasing from0 to4, they cross once in (3,4). Then, betweenx=4 andx=5, f(x) decreases from4 to0, and log_a(x +1) increases fromlog_a(5) to log_a(6). If log_a(6) <4, then log_a(x +1) is still below4. Therefore, since log_a(x +1) is increasing and f(x) is decreasing, there might be another crossing in (4,5) if log_a(x +1) catches up to f(x). So if log_a(6) <4, but log_a(6) is some value less than4, then depending on the slopes, there might be one crossing. Alternatively, if log_a(6) >=4, then atx=5, log_a(6) >=4, but f(x)=0, so log_a(6) >0, but that would not necessarily guarantee a crossing.Wait, let me think again. Betweenx=4 andx=5:Atx=4, f(x)=4 and log_a(5) <4 (assuming log_a(5) <4). log_a(x +1) is increasing. f(x) decreases from4 to0. So, atx=4, log_a(x +1)=log_a(5) <4, and atx=5, log_a(6). So if log_a(6) <4, then log_a(x +1) is increasing from log_a(5) to log_a(6) <4, while f(x) decreases from4 to0. So there must be a point where log_a(x +1) crosses f(x) from below. So exactly one crossing in (4,5). Therefore, total two crossings in (3,5].Therefore, to have two crossings in (3,5], we need log_a(5) <4 and log_a(6) <4? Wait, but even if log_a(6) is greater than4, then in (4,5), log_a(x +1) would exceed4 and then decrease? No, log_a(x +1) is always increasing. Wait, if log_a(6) >4, then atx=5, log_a(6) >4, but f(x)=0. So log_a(x +1) increases past4 somewhere in (4,5). Therefore, if log_a(6) >4, then log_a(x +1) crosses4 at some point in (4,5). But f(x) is decreasing from4 to0. So in that case, there could be two crossings: one when log_a(x +1) crosses the descending f(x) from below to above at some point afterx=4, but since log_a(x +1) is increasing, and f(x) is decreasing, there would be exactly one crossing in (4,5) if log_a(6) >4. Wait, if log_a(6) >4, then log_a(x +1) would cross f(x) once in (4,5). However, in (3,4), if log_a(5) <4, then there's one crossing in (3,4). If log_a(5) >=4, then log_a(x +1) would be above f(x) in (3,4). But we need two crossings in (3,5]. So, to have two crossings, log_a(x +1) must intersect both in (3,4) and (4,5). Therefore, log_a(5) <4 and log_a(6) <4? Or log_a(5) <4 and log_a(6) >4?Wait, if log_a(6) >4, then log_a(x +1) crosses f(x) once in (4,5), and if log_a(5) <4, log_a(x +1) crosses once in (3,4), totaling two crossings. If log_a(6) <4, then log_a(x +1) is always below4 in (4,5), so crosses once in (4,5) because f(x) is decreasing from4 to0, and log_a(x +1) is increasing fromlog_a(5) <4 tolog_a(6) <4. So atx=4, log_a(5) <4 =f(4), and atx=5, log_a(6) <4 and f(x)=0. Therefore, since log_a(x +1) is increasing and f(x) is decreasing, they cross once in (4,5). Therefore, even if log_a(6) <4, there is still one crossing in (4,5). Wait, but log_a(x +1) is below4 in (4,5), and f(x) is decreasing from4 to0. So atx=4, log_a(5) <4, and atx=5, log_a(6) <4. Therefore, log_a(x +1) is below4 throughout (4,5). So f(x) starts at4 and decreases to0, log_a(x +1) is increasing fromlog_a(5) to log_a(6). So, if log_a(5) <4 and log_a(6) <4, then in (4,5), log_a(x +1) is increasing from below4 to below4, while f(x) is decreasing from4 to0. So atx=4, log_a(5) <4, so log_a(x +1) is below4. As x increases, log_a(x +1) increases but stays below4, while f(x) decreases. Therefore, they cross exactly once in (4,5). So regardless of whether log_a(6) is greater than or less than4, as long as log_a(5) <4, there is one crossing in (3,4) and one crossing in (4,5). So total two crossings in (3,5]. Therefore, to have two crossings in each of (1,3] and (3,5], we need:For (1,3]:- log_a(3) <2 and log_a(4) >0 (which is always true since a>1).For (3,5]:- log_a(5) <4 and log_a(6) >0 (also always true). So the main constraints are log_a(3) <2 and log_a(5) <4.Converting these into inequalities with a:log_a(3) <2 => 3 <a^2 =>a > sqrt(3).Similarly, log_a(5) <4 =>5 <a^4 =>a >5^(1/4) ≈1.495.So combining these, a must be greater than sqrt(3) and greater than5^(1/4). Since sqrt(3)≈1.732 and5^(1/4)≈1.495, the stricter condition is a>sqrt(3). However, this would suggest that a must be greater thansqrt(3), so the range is (sqrt(3), +infty). However, wait, but in the answer choices, option C is (sqrt(3), +infty). But there's also option D: (⁴√5, sqrt(3)).Wait, perhaps I made a mistake here. Let me check again.If we need both log_a(3) <2 and log_a(5) <4, which translate to:log_a(3) <2 => a^2 >3 =>a >sqrt(3).log_a(5) <4 =>a^4 >5 =>a >5^(1/4).Therefore, the two conditions are a>sqrt(3) and a>5^(1/4). Since sqrt(3)≈1.732 >5^(1/4)≈1.495, the combined condition is a>sqrt(3). Therefore, the range should be (sqrt(3), +infty), which is option C.But wait, let's verify with the answer options. Option C is (sqrt(3), +infty). However, the options also have D: (⁴√5, sqrt(3)). But according to our deduction, the stricter condition is a>sqrt(3). So why is there a discrepancy?Wait, perhaps my reasoning is flawed. Let's consider when a is between5^(1/4) andsqrt(3). Let's say a=2, which is greater thansqrt(3). Then log_2(3)≈1.585 <2, and log_2(5)≈2.32 <4, so both conditions satisfied, so two crossings in each upper interval, totaling5 roots.But if a is between5^(1/4) andsqrt(3), say a=1.5, which is less thansqrt(3)≈1.732. Then, check the conditions:log_a(3) <2 => a^2 >3? If a=1.5, then a^2=2.25 <3. Therefore, log_a(3) = ln3 / ln a. If a=1.5, ln3 / ln1.5 ≈1.0986 /0.4055≈2.708 >2. So log_a(3) >2, which violates the condition for having two crossings in (1,3]. So in this case, log_a(x +1) would not cross f(x) twice in (1,3], but maybe only once. Therefore, in this case, the total number of roots would be less than5.Therefore, the real condition is that a must satisfy both a>5^(1/4) and a>sqrt(3). Since sqrt(3) is larger, then the combined condition is a>sqrt(3). So answer is C.But the options have D: (⁴√5, sqrt(3)), which is between≈1.495 and1.732. But according to our analysis, when a is in this interval, log_a(3) >2, so we have only one crossing in (1,3], leading to total roots1+1+2=4, which is less than5. Therefore, the correct answer should be C.But the answer choices were given as A: (1,sqrt(3)), B: (⁴√5, +infty), C: (sqrt(3), +infty), D: (⁴√5, sqrt(3)). So the correct answer should be C.But wait, the original problem states that the equation has exactly5 roots. However, when a>sqrt(3), then log_a(3) <2, so two crossings in (1,3], and log_a(5) <4 (since a>sqrt(3) >5^(1/4)), so two crossings in (3,5], total 1+2+2=5. Therefore, the range is a>sqrt(3). So answer C.But the answer might be D if there are other constraints. Wait, let's check when a is between⁴√5 andsqrt(3). For example, a=1.6, which is greater than⁴√5≈1.495 but less thansqrt(3)≈1.732.Check log_a(3):log_1.6(3) = ln3 / ln1.6 ≈1.0986 /0.4700≈2.338 <2? No, 2.338>2. Therefore, log_a(3) >2, so in the interval (1,3], the log function exceeds the peak of2, so instead of crossing twice, it would intersect only once. Wait, if log_a(3) >2, then atx=2, log_a(3) >2, but f(x)=2 atx=2. So log_a(x +1) atx=2 is log_a(3). If log_a(3) >2, then atx=2, log_a(x +1) >f(x). Then, since log_a(x +1) is increasing, in (1,2], f(x) increases from0 to2, log_a(x +1) increases fromlog_a(2) tolog_a(3). If log_a(3) >2, then log_a(x +1) crosses f(x) once in (1,2), since it starts below atx=1 (log_a(2) >0) and ends above atx=2. Then, in (2,3), log_a(x +1) continues to increase to log_a(4). Since f(x) decreases from2 to0. If log_a(4) >2, then log_a(x +1) remains above f(x) in (2,3), so no crossings. If log_a(4) <2, then log_a(x +1) crosses f(x) once in (2,3). So in this case, if a is between⁴√5 andsqrt(3), then log_a(3) >2 (if a<sqrt(3)), but log_a(5) <4 (if a>⁴√5). Wait, let's take a=1.6 again.log_a(3)=log_1.6(3)=≈2.338>2.log_a(4)=log_1.6(4)=ln4 / ln1.6≈1.386 /0.470≈2.95 <4.So in (1,3], log_a(x +1) starts atlog_a(2)≈1.473, increases tolog_a(3)≈2.338 atx=2, then tolog_a(4)≈2.95 atx=3.f(x) in (1,3] is from0 up to2 and down to0.So in (1,2], log_a(x +1) starts at≈1.473 (which is greater than0) and goes to≈2.338. f(x) goes from0 to2. So they cross once in (1,2) where log_a(x +1) overtakes f(x). Then atx=2, log_a(3)≈2.338>2, so log_a(x +1) is above f(x) atx=2, and continues to increase to≈2.95 atx=3, while f(x) decreases to0. So in (2,3), log_a(x +1) is always above f(x), so no crossings. Therefore, in (1,3], only1 crossing.In (3,5], log_a(x +1) is from≈2.95 tolog_a(6)=ln6/ln1.6≈1.7918/0.470≈3.81 <4. So log_a(6)≈3.81 <4. f(x) in (3,5] is from0 up to4 and down to0. So in (3,4], f(x) increases from0 to4, log_a(x +1) increases from≈2.95 to≈3.81. Therefore, log_a(x +1) crosses f(x) once in (3,4). Then in (4,5], f(x) decreases from4 to0, log_a(x +1) increases from≈3.81 to≈3.81 (wait, no, log_a(6)≈3.81). So log_a(x +1) is increasing from≈3.81 to≈3.81 (wait, atx=5, log_a(6)=≈3.81). Wait, no, x=5 corresponds to log_a(6). So log_a(x +1) increases fromlog_a(4)≈2.95 tolog_a(6)≈3.81. So in (3,5], log_a(x +1) increases from≈2.95 to≈3.81.In (3,4], f(x) increases from0 to4. So log_a(x +1) starts at≈2.95 and increases to≈3.81, while f(x) goes from0 to4. So log_a(x +1) crosses f(x) once in (3,4). Then in (4,5], f(x) decreases from4 to0, log_a(x +1) increases to≈3.81. So atx=4, log_a(5)=≈3.81 <4, and f(x)=4. So in (4,5], since log_a(x +1) is increasing towards≈3.81 and f(x) decreases from4 to0, they cross once. So total two crossings in (3,5].Therefore, total roots would be1 (from [0,1]) +1 (from (1,3]) +2 (from (3,5]) =4 roots. Therefore, when a is in(⁴√5, sqrt(3)), we have4 roots. But the problem requires exactly5 roots. Therefore, a cannot be in this interval. Therefore, only when a>sqrt(3), which makes log_a(3) <2, leading to two crossings in (1,3], and log_a(5) <4 (since a>sqrt(3) >⁴√5), leading to two crossings in (3,5], totaling5 roots. Therefore, the correct answer is C: (sqrt(3), +infty).But wait, the options given include B: (⁴√5, +infty). Why is that? If a is in (⁴√5, sqrt(3)), we get4 roots, but if a>sqrt(3), we get5 roots. Therefore, the range is a>sqrt(3). Therefore, answer C.But let's confirm with another example. Suppose a=2 (>sqrt(3)).log_2(3)≈1.585 <2. So in (1,3], log_a(x +1) crosses f(x) twice. In (3,5], log_a(5)=log2(5)≈2.32 <4, log_a(6)=log2(6)≈2.58 <4. Therefore, in (3,5], log_a(x +1) crosses twice. So total roots1+2+2=5. Correct.If a=⁴√5≈1.495, log_a(5)=4, because a^4=5. So log_a(5)=4. Then in (3,5], atx=4, log_a(5)=4, which is equal to f(x)=4. So they touch atx=4. Then in (4,5], log_a(x +1) increases beyond4, but f(x) decreases to0. So atx=4, it's a tangent point, so maybe counts as one root, and then another root in (4,5). So total two roots in (3,5], but atx=4, it's a double root? If the function touches atx=4, then it's a single root with multiplicity two? But the problem states "exactly5 different roots". So if there's a tangent point, it's still one root. Therefore, if a=⁴√5, then log_a(5)=4, so atx=4, they touch, so one root there, and one root in (4,5). So total two roots in (3,5]. But in (1,3], log_a(3)=log_{⁴√5}(3). Let's compute:log_{⁴√5}(3)= ln3 / ( (ln5)/4 )=4*ln3 / ln5≈4*1.0986/1.6094≈4*0.682≈2.728>2. Therefore, log_a(3)=2.728>2, so in (1,3], similar to previous case, only one crossing. Therefore total roots1+1+2=4. Therefore, at a=⁴√5, we have4 roots. Therefore, the lower bound is not⁴√5. Therefore, the correct range is a>sqrt(3). Therefore, the answer is C.But why is option D: (⁴√5, sqrt(3)) present? Maybe there is a miscalculation. Alternatively, perhaps there's another constraint I missed.Wait, let's re-examine the (3,5] interval. If a=⁴√5, then log_a(5)=4. So atx=4, log_a(5)=4=f(4). So they touch atx=4. This could count as one root. Then in (4,5], log_a(x +1) increases beyond4, but f(x) decreases from4 to0. However, since a=⁴√5, log_a(6)=log_{⁴√5}(6)= ln6 / (ln5/4)=4*ln6/ln5≈4*1.7918/1.6094≈4*1.113≈4.453>4. So log_a(6)=4.453>4. Therefore, in (4,5], log_a(x +1) crosses f(x) once. So atx=4, tangent (one root), and one root in (4,5). However, if it's tangent atx=4, does that count as one root or two roots? In the context of the problem, it should count as one root. Therefore, total in (3,5], two roots. But in (1,3], when a=⁴√5, log_a(3)=4*ln3/ln5≈4*1.0986/1.6094≈2.728>2. So in (1,3], log_a(x +1) crosses once. So total roots1+1+2=4. But if a is slightly greater than⁴√5, say a=1.5, which is greater than⁴√5≈1.495 but less thansqrt(3)≈1.732, then log_a(5)=log_{1.5}(5)=ln5 / ln1.5≈1.6094/0.4055≈3.97≈4. So close to4. Then log_a(6)=ln6/ln1.5≈1.7918/0.4055≈4.42>4. So in this case, in (3,5], log_a(x +1) crosses f(x) twice (once in (3,4), once in (4,5)), but in (1,3], log_a(3)=log_{1.5}(3)=ln3/ln1.5≈1.0986/0.4055≈2.708>2, so crosses once. So total roots1+1+2=4. Therefore, even when a is slightly above⁴√5, we still get4 roots.Therefore, the correct interval must be a>sqrt(3), leading to5 roots. Hence, option C.But in the given options, option B is (⁴√5, +infty). If the correct answer is C, then why is B an option? Maybe there's a mistake in my analysis.Wait, let's think differently. Maybe the number of roots depends not only on log_a(3) <2 and log_a(5) <4 but also on log_a(2) <1. Because in the interval [0,1], we have f(x)=1 -x, which starts at1 and decreases to0. log_a(x +1) starts at0 and increases to log_a(2). For there to be exactly one crossing in [0,1], we need log_a(2) <1. Because if log_a(2) >=1, then log_a(2) >=1, which would mean that log_a(x +1) reaches1 atx=1, but f(x)=0 atx=1. So if log_a(2) >=1, then log_a(x +1) would cross f(x) somewhere in [0,1). Wait, but if log_a(2) >=1, then log_a(x +1) is increasing from0 to>=1, and f(x) is decreasing from1 to0. Therefore, they would cross exactly once regardless. Because atx=0, f(x)=1 >0=log_a(1), and atx=1, f(x)=0 <log_a(2). Therefore, even if log_a(2)>=1, there's still exactly one crossing in (0,1). So the number of roots in [0,1] is always1, regardless of a>1.Therefore, the previous analysis holds: total roots are1 + number in (1,3] + number in (3,5]. To have exactly5 roots, need2 in each of the other intervals. Therefore, constraints are log_a(3) <2 and log_a(5) <4, which givesa>sqrt(3) and a>⁴√5. Since sqrt(3) >⁴√5, the combined constraint isa>sqrt(3). Therefore, answer C.Therefore, the correct answer is C.Final Answerboxed{C}</think>Given the function ( f(x) ) defined as:[f(x) = begin{cases} 2f(x-2) & text{if } x in (1, +infty), 1 - |x| & text{if } x in [-1, 1],end{cases}]we need to determine the range of ( a ) such that the equation ( f(x) - log_a(x+1) = 0 ) has exactly 5 different roots in the interval ([0, 5]).First, we analyze the function ( f(x) ) on the interval ([0, 5]):- On ([0, 1]), ( f(x) = 1 - x ).- On ((1, 3]), ( f(x) = 2(1 - |x - 2|) ), which forms a triangle peaking at 2 when ( x = 2 ).- On ((3, 5]), ( f(x) = 4(1 - |x - 4|) ), which forms a triangle peaking at 4 when ( x = 4 ).Next, we consider the logarithmic function ( log_a(x + 1) ). Since ( a > 1 ), ( log_a(x + 1) ) is increasing and concave. We need to determine the values of ( a ) such that ( log_a(x + 1) ) intersects ( f(x) ) exactly 5 times.Key points of analysis:1. Interval ([0, 1]): ( f(x) = 1 - x ) starts at 1 and decreases to 0. ( log_a(x + 1) ) starts at 0 and increases to ( log_a(2) ). They intersect once.2. Interval ((1, 3]): ( f(x) ) forms a triangle from 0 to 2 and back to 0. For two intersections, ( log_a(3) < 2 ), which translates to ( a > sqrt{3} ).3. Interval ((3, 5]): ( f(x) ) forms a triangle from 0 to 4 and back to 0. For two intersections, ( log_a(5) < 4 ), which translates to ( a > sqrt[4]{5} ).Since ( sqrt{3} approx 1.732 ) is greater than ( sqrt[4]{5} approx 1.495 ), the stricter condition is ( a > sqrt{3} ).Thus, the range of ( a ) ensuring exactly 5 roots is ( (sqrt{3}, +infty) ).[boxed{C}]]

🔑:AnalysisThis problem examines the ability to judge the number of zeros of a function, combining numerical and graphical analysis to solve the problem. By drawing the graph of the function and using numerical-graphical analysis, we can derive the inequality and thus obtain the result.SolutionGiven the function f(x) = begin{cases}2f(x-2), & xin(1,+infty) 1-|x|, & xin[-1,1]end{cases}, for x in the interval [-1,5], Therefore, the equation about x, f(x)-log_{a}(x+1)=0 (a > 0 and aneq 1) has exactly 5 different roots in the interval [0,5],Therefore, f(x)=log_{a}(x+1) has exactly 5 different roots, Therefore, the function y=f(x) and the function y=log_{a}(x+1) have exactly 5 different intersection points, From the graph, we can derive: begin{cases}log_{a}3 sqrt{3}. Hence, the correct choice is boxed{C}.

🧠:[<think>Alright, let's try to figure out the smallest positive period of the function ( f(x) = sqrt{3}sin^{2}x + sin x cos x - frac{sqrt{3}}{2} ). The options given are A: 1, B: π, C: 2, D: 2π. Hmm, okay, so I need to determine the period of this function. Let me recall that the period of a function is the smallest positive number ( T ) such that ( f(x + T) = f(x) ) for all ( x ). First, let's analyze the given function. It has three terms: ( sqrt{3}sin^{2}x ), ( sin x cos x ), and a constant term ( -frac{sqrt{3}}{2} ). The constant term doesn't affect the period, so we can focus on the first two terms. Starting with the second term, ( sin x cos x ). I remember there's a double-angle identity for sine: ( sin 2x = 2sin x cos x ). So, if I rewrite ( sin x cos x ) as ( frac{1}{2}sin 2x ), that might simplify things. Let me do that:( sin x cos x = frac{1}{2}sin 2x ).Okay, so the second term becomes ( frac{1}{2}sin 2x ), which has a period of ( pi ), since the period of ( sin 2x ) is ( pi ). Now, the first term is ( sqrt{3}sin^{2}x ). To find its period, I should also try to express it using a double-angle identity. I recall that ( sin^2x = frac{1 - cos 2x}{2} ). Let me apply that:( sqrt{3}sin^{2}x = sqrt{3} cdot frac{1 - cos 2x}{2} = frac{sqrt{3}}{2} - frac{sqrt{3}}{2}cos 2x ).So substituting back into the original function:( f(x) = frac{sqrt{3}}{2} - frac{sqrt{3}}{2}cos 2x + frac{1}{2}sin 2x - frac{sqrt{3}}{2} ).Simplifying the constants: ( frac{sqrt{3}}{2} - frac{sqrt{3}}{2} = 0 ), so the function simplifies to:( f(x) = -frac{sqrt{3}}{2}cos 2x + frac{1}{2}sin 2x ).Now, this is a combination of sine and cosine terms with the same argument ( 2x ). Both ( cos 2x ) and ( sin 2x ) have periods of ( pi ), so their combination should also have a period of ( pi ), right? Because when you add two functions with the same period, the resulting function has the same period. However, I should check if there's a possibility of a smaller period. Let me think.Wait, sometimes when you combine sine and cosine terms, the amplitude and phase shift change, but the period remains the same as the original functions. Since both terms here have a period of ( pi ), the entire function ( f(x) ) should have a period of ( pi ). Therefore, the answer might be B: π. But let me verify.Alternatively, maybe we can combine the two terms into a single sine or cosine function. Let me try that. The expression ( -frac{sqrt{3}}{2}cos 2x + frac{1}{2}sin 2x ) can be written in the form ( Asin(2x + phi) ) or ( Acos(2x + phi) ).Let's use the formula for combining sine and cosine into a single sine function. Recall that ( asintheta + bcostheta = Rsin(theta + phi) ), where ( R = sqrt{a^2 + b^2} ) and ( phi = arctanleft(frac{b}{a}right) ). But here, our expression is ( -frac{sqrt{3}}{2}cos 2x + frac{1}{2}sin 2x ). Let me rearrange terms to match the formula:( frac{1}{2}sin 2x - frac{sqrt{3}}{2}cos 2x ).So, in this case, ( a = frac{1}{2} ) (coefficient of sine) and ( b = -frac{sqrt{3}}{2} ) (coefficient of cosine). Then:( R = sqrt{left(frac{1}{2}right)^2 + left(-frac{sqrt{3}}{2}right)^2} = sqrt{frac{1}{4} + frac{3}{4}} = sqrt{1} = 1 ).The phase angle ( phi ) is given by:( phi = arctanleft(frac{b}{a}right) = arctanleft(frac{-sqrt{3}/2}{1/2}right) = arctan(-sqrt{3}) ).Hmm, ( arctan(-sqrt{3}) ). The tangent of 60 degrees (π/3) is √3, so tangent of -60 degrees (-π/3) is -√3. Therefore, ( phi = -frac{pi}{3} ).So, the expression becomes:( Rsin(2x + phi) = sinleft(2x - frac{pi}{3}right) ).Therefore, ( f(x) = sinleft(2x - frac{pi}{3}right) ).Wait, is that right? Let me check the calculation again. So combining ( frac{1}{2}sin 2x - frac{sqrt{3}}{2}cos 2x ):If we use the identity ( asintheta + bcostheta = Rsin(theta + phi) ), but here it's ( asintheta + bcostheta ). Wait, perhaps another approach: using ( Asintheta + Bcostheta = Csin(theta + phi) ).Alternatively, perhaps using:( Asintheta + Bcostheta = sqrt{A^2 + B^2} sinleft(theta + arctanleft(frac{B}{A}right)right) ). Hmm, maybe I confused the formula earlier. Let me verify.Wait, actually, the formula is:( asintheta + bcostheta = sqrt{a^2 + b^2} sinleft(theta + phiright) ), where ( phi = arctanleft(frac{b}{a}right) ).But here, ( a = frac{1}{2} ), ( b = -frac{sqrt{3}}{2} ), so:( sqrt{a^2 + b^2} = 1 ), as calculated.Then, ( phi = arctanleft( frac{b}{a} right) = arctanleft( frac{ -sqrt{3}/2 }{ 1/2 } right) = arctan(-sqrt{3}) ), which is indeed ( -pi/3 ).Therefore, the expression is:( sinleft(2x - frac{pi}{3}right) ).So, ( f(x) = sinleft(2x - frac{pi}{3}right) ).The period of ( sin(2x - frac{pi}{3}) ) is the same as the period of ( sin(2x) ), which is ( pi ), because the period of ( sin(kx) ) is ( 2pi /k ), so here ( k = 2 ), so period is ( 2pi /2 = pi ). Therefore, the period of ( f(x) ) is ( pi ). Therefore, the answer is B: π.But wait, the original function was written as the combination of terms with period π. However, sometimes when functions are combined, even with the same period, the resulting function can have a smaller period if there's some symmetry. For example, if you have ( sin x ) and ( cos x ), each has period ( 2pi ), but if you have ( sin x + cos x ), the period is still ( 2pi ). However, in this case, our combined function is ( sin(2x - pi/3) ), which clearly has period ( pi ). So unless there is a further simplification or another identity that can reduce the period, the period should be ( pi ).Wait, but let's check if the phase shift affects the period. No, the phase shift doesn't affect the period; it's just a horizontal shift. So even if it's shifted, the period remains the same. Therefore, the function ( sin(2x - pi/3) ) has period ( pi ), so the original function ( f(x) ) also has period ( pi ). Therefore, the smallest positive period is π, so the answer is B.But just to be thorough, let's check if maybe the function repeats earlier. Let's suppose that ( f(x + T) = f(x) ) for some T less than π. Let's test T = π/2. Let's compute ( f(0) ) and ( f(0 + π/2) ):First, compute ( f(0) ):( f(0) = sqrt{3}sin^2 0 + sin 0 cos 0 - sqrt{3}/2 = 0 + 0 - sqrt{3}/2 = -sqrt{3}/2 ).Now, compute ( f(0 + π/2) = f(π/2) ):( f(π/2) = sqrt{3}sin^2(π/2) + sin(π/2)cos(π/2) - sqrt{3}/2 = sqrt{3}(1) + (1)(0) - sqrt{3}/2 = sqrt{3} - sqrt{3}/2 = sqrt{3}/2 ).So ( f(0) = -sqrt{3}/2 ) and ( f(π/2) = sqrt{3}/2 ). If the period were π/2, then ( f(π/2) ) should equal ( f(0) ), but it doesn't. So π/2 is not a period. How about T = π? Let's check f(x + π) vs f(x). Let's pick x = 0:f(0 + π) = f(π) = sqrt(3) sin²π + sinπ cosπ - sqrt(3)/2 = 0 + 0 - sqrt(3)/2 = -sqrt(3)/2, which is equal to f(0). Let's check another point, say x = π/4:f(π/4) = sqrt(3) sin²(π/4) + sin(π/4)cos(π/4) - sqrt(3)/2.Compute each term:sin²(π/4) = (sqrt(2)/2)^2 = 1/2,sin(π/4)cos(π/4) = (sqrt(2)/2)(sqrt(2)/2) = 1/2,So f(π/4) = sqrt(3)*(1/2) + 1/2 - sqrt(3)/2 = (sqrt(3)/2 - sqrt(3)/2) + 1/2 = 0 + 1/2 = 1/2.Now, f(π/4 + π) = f(5π/4):Compute f(5π/4):sin²(5π/4) = (sin(5π/4))² = (-sqrt(2)/2)^2 = 1/2,sin(5π/4)cos(5π/4) = (-sqrt(2)/2)(-sqrt(2)/2) = 1/2,Thus f(5π/4) = sqrt(3)*(1/2) + 1/2 - sqrt(3)/2 = same as before, 1/2. Which is equal to f(π/4). So it seems that T = π is indeed a period. But is there a smaller period? Let's check T = π/2, which we saw didn't work for x = 0 and x = π/4. Let's see another point. Let's take x = π/6:Compute f(π/6):sin²(π/6) = (1/2)^2 = 1/4,sin(π/6)cos(π/6) = (1/2)(sqrt(3)/2) = sqrt(3)/4,So f(π/6) = sqrt(3)*(1/4) + sqrt(3)/4 - sqrt(3)/2 = (sqrt(3)/4 + sqrt(3)/4) - sqrt(3)/2 = sqrt(3)/2 - sqrt(3)/2 = 0.Now, f(π/6 + π/2) = f(2π/3):Compute f(2π/3):sin²(2π/3) = (sqrt(3)/2)^2 = 3/4,sin(2π/3)cos(2π/3) = (sqrt(3)/2)(-1/2) = -sqrt(3)/4,So f(2π/3) = sqrt(3)*(3/4) + (-sqrt(3)/4) - sqrt(3)/2 = (3sqrt(3)/4 - sqrt(3)/4) - sqrt(3)/2 = (2sqrt(3)/4) - sqrt(3)/2 = sqrt(3)/2 - sqrt(3)/2 = 0.Which equals f(π/6). Hmm, so f(π/6 + π/2) = f(π/6). But f(0 + π/2) ≠ f(0), so T = π/2 is not a period for all x. Therefore, even though some points repeat after π/2, not all do, so the period can't be π/2.Therefore, the period must be π. Hence, the answer should be B: π.But just to make sure, let's recall that after simplifying, the function is ( sin(2x - pi/3) ), which has a period of π. So, the minimal period is π. Therefore, the correct answer is B.Final Answerboxed{B}</think>To determine the smallest positive period of the function ( f(x) = sqrt{3}sin^{2}x + sin x cos x - frac{sqrt{3}}{2} ), we start by simplifying the given function using trigonometric identities.First, rewrite the term ( sin x cos x ) using the double-angle identity:[sin x cos x = frac{1}{2}sin 2x]Next, rewrite the term ( sqrt{3}sin^{2}x ) using the identity ( sin^2x = frac{1 - cos 2x}{2} ):[sqrt{3}sin^{2}x = sqrt{3} cdot frac{1 - cos 2x}{2} = frac{sqrt{3}}{2} - frac{sqrt{3}}{2}cos 2x]Substituting these into the original function, we get:[f(x) = frac{sqrt{3}}{2} - frac{sqrt{3}}{2}cos 2x + frac{1}{2}sin 2x - frac{sqrt{3}}{2}]Simplifying the constants, we find:[f(x) = -frac{sqrt{3}}{2}cos 2x + frac{1}{2}sin 2x]This expression can be combined into a single sine function using the identity ( asintheta + bcostheta = Rsin(theta + phi) ):[-frac{sqrt{3}}{2}cos 2x + frac{1}{2}sin 2x = sinleft(2x - frac{pi}{3}right)]The period of ( sin(2x - frac{pi}{3}) ) is determined by the coefficient of ( x ), which is 2. The period of ( sin(kx) ) is ( frac{2pi}{k} ), so for ( k = 2 ), the period is:[frac{2pi}{2} = pi]Thus, the smallest positive period of ( f(x) ) is ( pi ).[boxed{B}],<think>Okay, so I need to find the smallest positive period of the function f(x) = √3 sin²x + sinx cosx - √3/2. The options are 1, π, 2, and 2π. Hmm, periods of trigonometric functions. Let me recall some basics. The basic periods of sinx and cosx are 2π, but when they are squared or have their arguments multiplied by something, the period can change. For example, sin²x has a period of π because sin²x = (1 - cos2x)/2, so the period is halved. Similarly, sinx cosx is equivalent to (sin2x)/2, so its period is π as well. But wait, let me check.First, let me break down each term in the function f(x):1. √3 sin²x: The sin²x term. As I thought, sin²x can be written using a double-angle identity. Specifically, sin²x = (1 - cos2x)/2. So this term becomes √3*(1 - cos2x)/2. That simplifies to √3/2 - (√3/2)cos2x. The period of cos2x is π, so this term has a period of π.2. sinx cosx: This term can be rewritten using the double-angle identity as well. sinx cosx = (sin2x)/2. So this term becomes (1/2) sin2x. The period of sin2x is π, so this term also has a period of π.3. -√3/2: This is just a constant term. Constants don't affect the period because they don't change with x.So both the first and second terms have a period of π. However, when combining functions with different periods, the overall period is the least common multiple (LCM) of their individual periods. But in this case, both terms have the same period π. So does that mean the entire function f(x) has a period of π? Wait, but sometimes when you combine functions with the same period, their combination could have a smaller period if there's some phase shift or other factor. Let me check that.Let's verify by substituting x with x + π/2 into f(x) and see if it's the same. Wait, but maybe it's easier to first simplify f(x) using trigonometric identities. Let me try to combine the terms into a single trigonometric function.Starting with f(x):f(x) = √3 sin²x + sinx cosx - √3/2First, replace sin²x with (1 - cos2x)/2:f(x) = √3*(1 - cos2x)/2 + sinx cosx - √3/2Simplify the first term:= √3/2 - (√3/2)cos2x + sinx cosx - √3/2Combine the constants:= - (√3/2)cos2x + sinx cosxNow, sinx cosx is (1/2) sin2x, so:= - (√3/2)cos2x + (1/2) sin2xSo f(x) simplifies to:(1/2) sin2x - (√3/2) cos2xThis looks like a single sinusoidal function. Remember that a sinθ + b cosθ can be written as R sin(θ + φ), where R = √(a² + b²) and φ is the phase shift. Let's apply that here.Let me denote θ = 2x. Then,f(x) = (1/2) sinθ - (√3/2) cosθCompute R:R = √[(1/2)² + (-√3/2)²] = √[1/4 + 3/4] = √[1] = 1So, f(x) = sin(θ - φ), where φ is such that:sinφ = (√3/2)/1 = √3/2, cosφ = (1/2)/1 = 1/2Therefore, φ = π/3, since sin(π/3) = √3/2 and cos(π/3) = 1/2.Wait, but in the formula, it's:a sinθ + b cosθ = R sin(θ + φ)But here, our coefficients are (1/2) and (-√3/2). So actually, it's:(1/2) sinθ - (√3/2) cosθ = sinθ * (1/2) + cosθ * (-√3/2)So comparing to a sinθ + b cosθ, a = 1/2, b = -√3/2.Alternatively, we can write it as:= R sin(θ + φ)But with R = √(a² + b²) = √( (1/2)^2 + (-√3/2)^2 ) = √(1/4 + 3/4) = √1 = 1Then,sin(θ + φ) = sinθ cosφ + cosθ sinφ = (1/2) sinθ - (√3/2) cosθTherefore,cosφ = 1/2, sinφ = -√3/2Wait, so that would make φ = -π/3, because cos(-π/3) = 1/2, sin(-π/3) = -√3/2.Therefore,f(x) = sin(2x - π/3)Alternatively, using the identity:a sinθ + b cosθ = R sin(θ + φ)But since b is negative here, φ is negative. So the expression becomes sin(2x - π/3). So f(x) is equivalent to sin(2x - π/3).Now, the period of sin(2x - π/3) is determined by the coefficient of x inside the sine function. The general period of sin(kx + φ) is 2π / |k|. Here, k = 2, so the period is 2π / 2 = π.Therefore, the period of f(x) is π, which is option B.But wait, the question is about the smallest positive period. Let me verify once again to make sure. If f(x) is written as sin(2x - π/3), then indeed its period is π. However, sometimes when functions are combined, even if their individual periods are the same, the combination might have a smaller period. But in this case, since we have already combined all terms into a single sine function with argument 2x shifted by π/3, the period should indeed be π. Therefore, the answer is π, which is option B.But let me check again by testing the function at some points. Let me compute f(0) and f(π), see if they are the same.Compute f(0):f(0) = √3 sin²0 + sin0 cos0 - √3/2 = 0 + 0 - √3/2 = -√3/2Compute f(π):f(π) = √3 sin²π + sinπ cosπ - √3/2 = 0 + 0 - √3/2 = -√3/2Now check f(π/2):f(π/2) = √3 sin²(π/2) + sin(π/2) cos(π/2) - √3/2 = √3 *1 + 1*0 - √3/2 = √3 - √3/2 = √3/2Similarly, f(π + π/2) = f(3π/2):= √3 sin²(3π/2) + sin(3π/2) cos(3π/2) - √3/2= √3*(1) + (-1)*0 - √3/2 = √3 - √3/2 = √3/2, which is the same as f(π/2). So f(x + π) = f(x) for these points.But let's check a point where x is not a multiple of π/2. Let's take x = π/6.Compute f(π/6):= √3 sin²(π/6) + sin(π/6) cos(π/6) - √3/2sin(π/6) = 1/2, cos(π/6) = √3/2So:= √3*(1/2)^2 + (1/2)(√3/2) - √3/2= √3*(1/4) + (√3/4) - √3/2= (√3/4 + √3/4) - √3/2= (√3/2) - √3/2 = 0Now compute f(π/6 + π) = f(7π/6):= √3 sin²(7π/6) + sin(7π/6) cos(7π/6) - √3/2sin(7π/6) = -1/2, cos(7π/6) = -√3/2So:= √3*(-1/2)^2 + (-1/2)(-√3/2) - √3/2= √3*(1/4) + (√3/4) - √3/2Same as before:= √3/4 + √3/4 - √3/2 = 0So f(7π/6) = f(π/6). Hence, adding π to x gives the same value. Thus, the period is π. Therefore, the answer is B: π.But let me check if there's a smaller period. Suppose someone suggests period π/2. Let's test if f(x + π/2) = f(x). Take x = 0:f(0 + π/2) = f(π/2) = √3/2 as beforef(0) = -√3/2, which is not equal. So π/2 is not a period. How about π/3?Take x = 0:f(0 + π/3) = f(π/3) = √3 sin²(π/3) + sin(π/3) cos(π/3) - √3/2sin(π/3) = √3/2, cos(π/3) = 1/2So:= √3*(3/4) + (√3/2)(1/2) - √3/2= (3√3/4) + (√3/4) - √3/2= (4√3/4) - 2√3/4 = (4√3 - 2√3)/4 = 2√3/4 = √3/2 ≠ f(0) = -√3/2. So π/3 is not a period.Similarly, check π/4. Let x = 0:f(π/4) = √3 sin²(π/4) + sin(π/4) cos(π/4) - √3/2= √3*(1/2) + (√2/2)(√2/2) - √3/2= (√3/2) + (2/4) - √3/2= (√3/2 - √3/2) + 1/2 = 1/2 ≠ f(0). So π/4 is not a period.Therefore, the smallest period is indeed π.Alternatively, since f(x) = sin(2x - π/3), the function is a sine wave with period π, and phase shift π/3. The phase shift doesn't affect the period. So the fundamental period is π.Hence, the correct answer is B: π.Final Answerboxed{B}</think>To find the smallest positive period of the function ( f(x) = sqrt{3} sin^2 x + sin x cos x - frac{sqrt{3}}{2} ), we start by simplifying the function using trigonometric identities.First, rewrite each term using double-angle identities:1. ( sqrt{3} sin^2 x ) can be rewritten using ( sin^2 x = frac{1 - cos 2x}{2} ): [ sqrt{3} sin^2 x = sqrt{3} left( frac{1 - cos 2x}{2} right) = frac{sqrt{3}}{2} - frac{sqrt{3}}{2} cos 2x ]2. ( sin x cos x ) can be rewritten using ( sin x cos x = frac{sin 2x}{2} ): [ sin x cos x = frac{sin 2x}{2} ]Substituting these into the original function:[f(x) = frac{sqrt{3}}{2} - frac{sqrt{3}}{2} cos 2x + frac{sin 2x}{2} - frac{sqrt{3}}{2}]Simplifying the constants:[f(x) = - frac{sqrt{3}}{2} cos 2x + frac{1}{2} sin 2x]This can be combined into a single sinusoidal function:[f(x) = frac{1}{2} sin 2x - frac{sqrt{3}}{2} cos 2x]Recognizing this as a form ( a sin theta + b cos theta ), we rewrite it using the amplitude-phase form:[f(x) = sin(2x - frac{pi}{3})]The period of ( sin(2x - frac{pi}{3}) ) is determined by the coefficient of ( x ), which is 2. The period of ( sin(kx) ) is ( frac{2pi}{k} ), so here the period is:[frac{2pi}{2} = pi]Thus, the smallest positive period of ( f(x) ) is ( pi ).[boxed{B}]]

🔑:To solve for the smallest positive period of the function f(x)=sqrt{3}sin^{2}x+sin x cos x-frac{sqrt{3}}{2}, we first simplify the function. Starting with the given function:f(x)=sqrt{3}sin^{2}x+sin x cos x-frac{sqrt{3}}{2}We can use trigonometric identities to simplify it. Recall that sin^{2}x = frac{1-cos 2x}{2} and sin x cos x = frac{sin 2x}{2}. Substituting these identities into the function gives:begin{align*}f(x) &= sqrt{3}left(frac{1-cos 2x}{2}right) + frac{sin 2x}{2} - frac{sqrt{3}}{2} &= frac{sqrt{3}}{2} - frac{sqrt{3}}{2}cos 2x + frac{1}{2}sin 2x - frac{sqrt{3}}{2} &= -frac{sqrt{3}}{2}cos 2x + frac{1}{2}sin 2x.end{align*}This can be rewritten using the sine of a sum formula, sin(A + B) = sin A cos B + cos A sin B, where A = 2x and B = -frac{pi}{3}. Thus, we have:f(x) = sinleft(2x - frac{pi}{3}right).The function sinleft(2x - frac{pi}{3}right) has a period of 2pi divided by the coefficient of x in the argument of the sine function, which is 2. Therefore, the smallest positive period of f(x) is:frac{2pi}{2} = pi.So, the correct answer is boxed{B}.

🧠:[<think>Okay, so I need to figure out the relationship between a, b, and c where a is 7 raised to the power of 0.3, b is 0.3 raised to the power of 7, and c is the logarithm of 0.3 with base 7. The options are A through D, each suggesting a different order. Let me start by analyzing each one step by step.First, let's recall some basic properties of exponents and logarithms. Exponents can vary widely depending on the base and the exponent. Logarithms, especially, have certain behaviors depending on whether the argument is greater than or less than the base. Let's break down each value:Starting with a = 7^{0.3}. Since 7 is a number greater than 1 and the exponent is 0.3, which is a positive number less than 1, this is going to be a root. Specifically, 0.3 is 3/10, so this is the 10th root of 7 cubed. But maybe I don't need to compute the exact value. Instead, I can estimate it. Since 7^1 = 7, and 7^{0.5} would be sqrt(7) which is about 2.645. Since 0.3 is less than 0.5, 7^{0.3} should be less than sqrt(7) but more than 1 because any positive number to a positive exponent is greater than 0, and since the base is greater than 1, the result is greater than 1. Let me check with a calculator in my mind. Maybe 7^{0.3} is approximately... Hmm, 7^0.3 is equal to e^{0.3 ln7}. Let's approximate ln7. ln7 is approximately 1.9459. So 0.3 * 1.9459 ≈ 0.5838. Then e^0.5838 is approximately... e^0.5 is about 1.6487, e^0.6 is about 1.8221. So 0.5838 is between 0.5 and 0.6, closer to 0.58. Maybe approximately 1.79? So a ≈ 1.79.Next, b = 0.3^7. Here, the base is 0.3, which is between 0 and 1, and the exponent is 7, a positive integer. Raising a number between 0 and 1 to a positive exponent makes it smaller. So 0.3^1 is 0.3, 0.3^2 is 0.09, 0.3^3 is 0.027, 0.3^4 is 0.0081, 0.3^5 is 0.00243, 0.3^6 is ~0.000729, and 0.3^7 is ~0.0002187. So b is approximately 0.0002187. That's a very small number, much less than 1.Now, c = log_7(0.3). Logarithms of numbers less than 1 with bases greater than 1 are negative. For example, log_7(1) is 0, and as the argument decreases below 1, the log becomes negative. Let me compute log_7(0.3). Using change of base formula, log_7(0.3) = ln(0.3)/ln(7). Let's compute ln(0.3). ln(0.3) is approximately -1.2039, and ln(7) is about 1.9459. So -1.2039 / 1.9459 ≈ -0.619. So c ≈ -0.619. So c is a negative number.So putting it all together: a is approximately 1.79, b is approximately 0.0002187, and c is approximately -0.619. Therefore, ordering them from smallest to largest: c (-0.619) < b (0.0002187) < a (1.79). Wait, but let me check that. Is 0.0002187 greater than -0.619? Yes, because any positive number is greater than a negative number. So c is the smallest, then b, then a. Therefore, the order is c < b < a. Looking at the options, that's option B: c < b < a. Wait, but hold on. Wait, let me double-check because the options are:A: a < b < cB: c < b < aC: c < a < bD: b < c < aWait, in my analysis, c is negative, b is positive but very small, and a is positive and around 1.79. So the order is c < b < a, which is option B. But wait, hold on. Let me check again. Maybe I made a mistake in the estimation of a.Wait, I approximated a as 1.79. Let me verify that again. 7^{0.3} is equal to e^{0.3 * ln7}. We said ln7 ≈ 1.9459, so 0.3*1.9459 ≈ 0.5838. Then e^0.5838. Let me compute this more accurately. e^0.5 is 1.64872, e^0.6 is 1.82211. The difference between 0.5 and 0.6 is 0.1 in the exponent, which corresponds to a multiplicative factor of about 1.82211 / 1.64872 ≈ 1.105. So each 0.01 increase in the exponent would lead to roughly 1.105^(0.1) ≈ e^(0.1*ln(1.105)) ≈ e^(0.1*0.0998) ≈ e^0.00998 ≈ 1.01005. So from 0.5 to 0.58 is 0.08, which is 8*0.01. So approximately multiplying 1.64872 by (1.01005)^8 ≈ 1.083. So 1.64872 * 1.083 ≈ 1.64872 + 1.64872*0.083 ≈ 1.64872 + 0.137 ≈ 1.7857. So yeah, around 1.7857. So 1.79 is a good approximation. So a ≈ 1.79, which is positive and greater than 1, b ≈ 0.0002, which is positive but very small, c ≈ -0.619. So order is c < b < a, so option B. But wait, let me check the options again.Wait, the options given are:A: a < b < cB: c < b < aC: c < a < bD: b < c < aSo yes, B is c < b < a. So my answer should be B? But wait, hold on. Wait, let me check once more.Alternatively, maybe my calculation for c is wrong. Let's re-examine c = log_7(0.3). Using change of base formula, log_7(0.3) = ln(0.3)/ln(7). ln(0.3) is ln(3/10) = ln(3) - ln(10) ≈ 1.0986 - 2.3026 ≈ -1.204. Then ln(7) ≈ 1.9459. So -1.204 / 1.9459 ≈ -0.619. That's correct. So c ≈ -0.619. So c is negative, b is a small positive number, and a is about 1.79, which is a positive number greater than 1. So yes, c is the smallest, then b, then a, so option B. Therefore, the answer should be B.But wait, let me check the problem statement again to ensure I didn't misread anything. The problem says: Let a = 7^{0.3}, b = 0.3^{7}, c = log_{7}0.3. The relationship among a, b, and c is (　　). Options are A to D. So yes, I think my reasoning is correct. So the answer is B: c < b < a.However, just to be thorough, let me check if there's any possible mistake. For example, perhaps I miscalculated the value of a? Let's see. Let's compute 7^{0.3} another way. Since 7^0.3 is the same as 10^(0.3 * log10(7)). log10(7) is approximately 0.8451. So 0.3 * 0.8451 ≈ 0.2535. So 10^0.2535. 10^0.25 is sqrt(sqrt(10)) ≈ 1.778, and 10^0.2535 is slightly higher. Let's say approximately 1.79. So that matches my previous calculation. So a is indeed approximately 1.79.Alternatively, maybe I can compare a and 2. 2 is 2. Since 7^0.3 is about 1.79, which is less than 2, so a is less than 2, but still greater than 1. So that's consistent.What about b? 0.3^7. Let me calculate it step by step. 0.3^2 = 0.09, 0.3^3 = 0.027, 0.3^4 = 0.0081, 0.3^5 = 0.00243, 0.3^6 = 0.000729, 0.3^7 = 0.0002187. So that's correct, approximately 0.0002187, which is 2.187 x 10^{-4}. So that's about 0.0002, which is positive but very small.And c is log base 7 of 0.3, which is definitely negative because 0.3 is less than 1. So the logarithm of a number less than 1 with base greater than 1 is negative. So c is negative. So putting it all together, c is negative, b is a small positive, and a is a larger positive. Therefore, c < b < a, which is option B.But wait, let me check once more. Maybe there's a trick here. For example, sometimes logarithms can be compared with exponents in a different way. Let me see.Alternatively, perhaps we can compare a and b without calculating their exact values. Let's see: a = 7^{0.3} and b = 0.3^7. Since 7 > 1 and 0.3 < 1, then 7^{0.3} > 1^{0.3} = 1, so a > 1. Meanwhile, 0.3^7 < 1^7 = 1, so b < 1. Therefore, a > 1 and b < 1, so a > b. Also, c is log_7(0.3). Since 0.3 < 7, and logarithm base 7 of a number less than 7 is less than 1. But more precisely, since 0.3 < 1, log_7(0.3) is negative, as we calculated. Therefore, c is negative. So since c is negative, b is positive but less than 1, and a is greater than 1, the order is c < b < a, which is option B. So this seems consistent.Alternatively, another way to compare a and b is to take logarithms. Let's take natural logarithm of a and b.ln(a) = ln(7^{0.3}) = 0.3 ln7 ≈ 0.3 * 1.9459 ≈ 0.5838ln(b) = ln(0.3^7) = 7 ln0.3 ≈ 7*(-1.2039) ≈ -8.4273So ln(a) ≈ 0.5838, ln(b) ≈ -8.4273. Therefore, ln(a) > ln(b), which means a > b. Since ln(c) is undefined here because c is already a logarithm. Wait, c itself is a logarithm. But we can compare c with a and b. Since c is negative, and both a and b are positive, but we already saw a > 1 and b < 1. So c is definitely the smallest. Then b is positive but very small, and a is the largest.Therefore, the correct ordering is c < b < a, which is option B. Therefore, the answer is B.Final Answerboxed{B}</think>To determine the relationship among (a = 7^{0.3}), (b = 0.3^7), and (c = log_{7}0.3):1. Calculating (a): - (a = 7^{0.3}) - Using the natural logarithm: (7^{0.3} = e^{0.3 ln 7}) - (ln 7 approx 1.9459), so (0.3 ln 7 approx 0.5838) - (e^{0.5838} approx 1.79) - Therefore, (a approx 1.79)2. Calculating (b): - (b = 0.3^7) - Calculating step-by-step: (0.3^2 = 0.09), (0.3^3 = 0.027), (0.3^4 = 0.0081), (0.3^5 = 0.00243), (0.3^6 = 0.000729), (0.3^7 = 0.0002187) - Therefore, (b approx 0.0002187)3. Calculating (c): - (c = log_{7}0.3) - Using the change of base formula: (log_{7}0.3 = frac{ln 0.3}{ln 7}) - (ln 0.3 approx -1.2039), so (frac{-1.2039}{1.9459} approx -0.619) - Therefore, (c approx -0.619)4. Comparing the values: - (c approx -0.619) (negative) - (b approx 0.0002187) (positive, very small) - (a approx 1.79) (positive, greater than 1)Thus, the order from smallest to largest is (c < b < a), which corresponds to option B.[boxed{B}],<think>Okay, so I need to figure out the relationship between a, b, and c where:a = 7^0.3b = 0.3^7c = log base 7 of 0.3And then choose the correct option from A to D. Let me start by recalling some logarithmic and exponential properties to compare these values.First, let's look at a = 7^0.3. Since 7 is greater than 1 and the exponent is 0.3, which is a positive number less than 1, the result should be between 1 and 7. Because any number greater than 1 raised to a power between 0 and 1 will be between 1 and the number itself. For example, 7^1 = 7, 7^0 = 1. So 7^0.3 is between 1 and 7. Maybe around 2? Let me check that later.Next, b = 0.3^7. Here, 0.3 is a number between 0 and 1, and raising it to the 7th power will make it even smaller. For example, 0.3^2 is 0.09, 0.3^3 is 0.027, and so on. So 0.3^7 must be a very small positive number, close to 0. Maybe around 0.0002 or something? I need to approximate this.Then c = log base 7 of 0.3. Since 0.3 is less than 1, the logarithm of a number less than 1 with a base greater than 1 is negative. So c will be negative. For example, log base 7 of 1 is 0, and log base 7 of 1/7 is -1. Since 0.3 is between 1/7 and 1, but closer to 1/7? Let me see. 1/7 is approximately 0.1429, so 0.3 is larger than that. So log base 7 of 0.3 is between -1 and 0. Maybe around -0.5? Hmm.So summarizing:- a is between 1 and 7, maybe around 2.- b is a very small positive number, close to 0.- c is a negative number, maybe around -0.5.So in order from smallest to largest: c (negative) < b (small positive) < a (around 2). So that would be c < b < a, which is option B. Wait, but let me verify my approximations because I might be wrong.First, let's get a better estimate of a = 7^0.3.I know that 7^0.3 is the same as e^(0.3 ln 7). Let's compute ln 7 ≈ 1.9459, so 0.3 * 1.9459 ≈ 0.5838. Then e^0.5838. What's e^0.5? Approximately 1.6487. e^0.6 is about 1.8221. So e^0.5838 is somewhere between 1.79 and 1.8. Let's say approximately 1.79. So a ≈ 1.79.Wait, so my initial thought that it's around 2 was a bit high. It's actually around 1.79. So a is less than 2.Next, b = 0.3^7. Let's compute that step by step:0.3^2 = 0.090.3^3 = 0.09 * 0.3 = 0.0270.3^4 = 0.027 * 0.3 = 0.00810.3^5 = 0.0081 * 0.3 = 0.002430.3^6 = 0.00243 * 0.3 = 0.0007290.3^7 = 0.000729 * 0.3 = 0.0002187So b ≈ 0.0002187. That's about 2.187 x 10^-4. So very small, as I thought.Then c = log base 7 of 0.3. Let's convert that to natural logarithm: c = ln(0.3)/ln(7). Compute ln(0.3) ≈ -1.2039, ln(7) ≈ 1.9459. So c ≈ -1.2039 / 1.9459 ≈ -0.619. So approximately -0.62.So putting these approximated values:a ≈ 1.79b ≈ 0.0002187c ≈ -0.62Therefore, c < b < a, which is option B.Wait, but let me check once more if I didn't miscalculate anything.For a: 7^0.3. Let me verify with logarithm.log10(7) ≈ 0.8451So log10(a) = 0.3 * 0.8451 ≈ 0.2535Therefore, a ≈ 10^0.2535 ≈ 10^(0.25 + 0.0035) ≈ 10^0.25 * 10^0.0035.10^0.25 is the fourth root of 10, which is approximately 1.7783.10^0.0035 ≈ 1 + 0.0035 * ln(10) ≈ 1 + 0.0035 * 2.3026 ≈ 1 + 0.008 ≈ 1.008.So multiplying them: 1.7783 * 1.008 ≈ 1.792. So that confirms a ≈ 1.792. So my previous calculation was correct.For b: 0.3^7. As calculated step by step, yes, 0.3^7 is 0.0002187. Correct.For c: log7(0.3) = ln(0.3)/ln(7) ≈ (-1.2039)/1.9459 ≈ -0.619. Correct.So the order is c ≈ -0.619 < b ≈ 0.0002 < a ≈ 1.792. Therefore, the correct order is c < b < a, which is option B. So the answer should be B.But wait, let me check the options again:A: a < b < cB: c < b < aC: c < a < bD: b < c < aSo yes, B is c < b < a, which matches our conclusion. So I think the answer is B.But let me just cross-verify one more time. Maybe using different methods.Another approach for a = 7^0.3. Let's think of 7^(1/3) is cube root of 7, which is about 1.913. Then 0.3 is slightly less than 1/3 (which is 0.333...). So 7^0.3 is slightly less than 7^(1/3) ≈ 1.913. So around 1.8 or so, which is consistent with previous calculation.For c = log7(0.3). Since 0.3 is 3/10, which is 3 * 7^(-1) * 10^(-0)? Wait, maybe not helpful. Alternatively, since log7(1/7) = -1, and 0.3 is greater than 1/7 (≈0.1428), so log7(0.3) is greater than -1. And since 0.3 is less than 1, the log is negative. So between -1 and 0, which we calculated as approximately -0.62.Alternatively, using change of base formula to base 10:log7(0.3) = log10(0.3)/log10(7) ≈ (-0.5229)/0.8451 ≈ -0.619, same as before.So yes, c ≈ -0.62.So the order is indeed c < b < a. Therefore, the answer is option B.Final Answerboxed{B}</think>To determine the relationship among (a = 7^{0.3}), (b = 0.3^7), and (c = log_{7}0.3):1. Calculating (a = 7^{0.3}): - Convert to natural logarithm: (a = e^{0.3 ln 7}). - (ln 7 approx 1.9459), so (0.3 ln 7 approx 0.5838). - (e^{0.5838} approx 1.79). Thus, (a approx 1.79).2. Calculating (b = 0.3^7): - Compute step-by-step: (0.3^2 = 0.09), (0.3^3 = 0.027), (0.3^4 = 0.0081), (0.3^5 = 0.00243), (0.3^6 = 0.000729), (0.3^7 = 0.0002187). Thus, (b approx 0.0002187).3. Calculating (c = log_{7}0.3): - Convert to natural logarithm: (c = frac{ln 0.3}{ln 7}). - (ln 0.3 approx -1.2039) and (ln 7 approx 1.9459). - (c approx frac{-1.2039}{1.9459} approx -0.619).Summarizing the values:- (a approx 1.79)- (b approx 0.0002187)- (c approx -0.619)Thus, the order is (c < b < a), corresponding to option B.(boxed{B})]

🔑:Since log_{7}0.3<log_{7}1=0, 0<0.3^{7}<0.3^{0}=1, 1=7^{0}<7^{0.3},Therefore, c<b<a,So, the correct option is boxed{text{B}}.This problem can be solved by utilizing the monotonicity of logarithmic, power, and exponential functions to compare their magnitudes.Mastering the monotonicity of logarithmic, power, and exponential functions is key to solving the problem. Pay attention to comparisons with 0 and 1.

🧠:[<think>Okay, so I need to solve this rainfall problem from the ancient Chinese math book. Let me first understand the problem. There's this basin which is a frustum of a cone—that is, a cone with the top cut off, right? The basin has a top diameter, bottom diameter, and depth. They give all these measurements in chi and cun, and I remember that 1 chi is 10 cun. The question is, after a rainstorm, the basin has 9 cun of water accumulated. I need to find out how much rainfall there was on flat ground, measured in cun. The note says that rainfall is equal to the volume of water divided by the area of the basin's mouth. So, basically, the volume of water collected in the basin divided by the area of the top opening (the mouth) gives the rainfall, which makes sense because it's like spreading that volume over the area it was collected from.First, let's convert all measurements to cun since the answer needs to be in cun. The top diameter is 2 chi 8 cun. Since 1 chi is 10 cun, 2 chi is 20 cun, so 20 + 8 = 28 cun. The bottom diameter is 1 chi 2 cun, which is 10 + 2 = 12 cun. The depth of the basin is 1 chi 8 cun, which is 10 + 8 = 18 cun. The water accumulated is 9 cun deep. Hmm, so the water is filling the basin to half its depth? Wait, the basin is 18 cun deep, and the water is 9 cun. So, it's half full? Or is the depth measured differently? Wait, maybe not. Because a frustum's volume depends on the radii at the top and bottom. But when it's filled with water to a certain depth, does the water form a similar frustum, or is it a smaller frustum? Hmm, maybe I need to model the water as a frustum as well.Wait, the problem says "the accumulated water in the basin is 9 cun deep". So, the water itself forms a frustum with depth 9 cun. But does that frustum have the same top and bottom diameters as the original basin? Wait, no. Because if the basin is a frustum, then when you fill it partially, the surface of the water is a smaller circle. Wait, actually, if the basin is a frustum, then the sides are slanting. So when you fill it with water to a certain depth, the radius at the water's surface is going to be somewhere between the top and bottom radii. Hmm, this might require similar figures or linear interpolation.Alternatively, maybe the water forms a frustum similar to the basin but scaled down. Let me think. The original basin is a frustum with height 18 cun, top radius 14 cun (since diameter 28 cun), and bottom radius 6 cun (diameter 12 cun). If we fill it to a depth of 9 cun, which is half the height, does the water form a frustum with height 9 cun, and radii that are halfway between the top and bottom? Wait, but in a frustum, the radii change linearly along the height. So, if the original frustum has radii R1 = 14 cun at the top, R2 = 6 cun at the bottom, and height h = 18 cun, then at a depth of 9 cun from the top, the radius at that level would be R1 minus the difference over the height. Wait, actually, the radius decreases from R1 at the top to R2 at the bottom. So, the rate of decrease per cun of depth is (R1 - R2)/h = (14 - 6)/18 = 8/18 = 4/9 cun per cun. So, after 9 cun of depth from the top, the radius would be R1 - (4/9)*9 = 14 - 4 = 10 cun. Wait, that makes sense. So, the water forms a frustum with top radius 14 cun (the mouth), bottom radius 10 cun, and height 9 cun. Then, the volume of the water is the volume of this frustum. Then, the rainfall is that volume divided by the area of the mouth (which is π*14²). So, maybe that's the way to do it.Alternatively, maybe the problem is simpler. Since the basin is a frustum, and when you fill it with water, the water forms a frustum similar to the basin but scaled down. But in this case, since we filled it to half the depth, but the scaling factor might not be linear? Wait, actually, in similar figures, the scaling factor for linear dimensions is consistent. But in a frustum, the top and bottom radii are different, so scaling would require both radii and height to scale. However, in this case, the water is only filling part of the frustum. So, perhaps the water forms a frustum where the top radius is the same as the basin's top radius (since the mouth is open), and the bottom radius is determined by the depth of the water. Wait, but that doesn't make sense because the water's bottom radius would actually be the radius at the water's surface. Wait, no, when you pour water into the frustum, the bottom of the water is at the original basin's bottom, but since the basin is a frustum, the sides slope. Wait, perhaps I need to think of the basin as being a truncated cone. Let me visualize it. The original basin is a frustum with the larger radius at the top (14 cun) and smaller at the bottom (6 cun). If you pour water into it, the water will fill from the bottom up. So, the water forms a frustum with the bottom radius 6 cun and a top radius that depends on how high the water is filled. Wait, no, actually, if you have a frustum-shaped basin, when you put water into it, the water's shape is also a frustum, but inverted. Wait, maybe this is getting confusing.Wait, perhaps the key is to model the water as a frustum with the same slant sides as the basin, but with a height of 9 cun. So, the original basin has a height of 18 cun, top radius 14 cun, bottom radius 6 cun. The difference in radius over the height is 14 - 6 = 8 cun over 18 cun. So, the rate of change of radius per cun is 8/18 = 4/9 cun per cun. So, if the water is 9 cun deep, starting from the bottom, then the radius at the water's surface would be the bottom radius plus (4/9)*9 = 4 cun, so 6 + 4 = 10 cun. Wait, but if the water is 9 cun deep, is that measured from the bottom up? The problem says "accumulated water in the basin is 9 cun deep". So, I think it's the depth of the water, meaning from the surface of the water down to the bottom of the basin. So, if the total depth of the basin is 18 cun, and the water is 9 cun deep, then the water reaches halfway up the basin. Therefore, the water forms a frustum with height 9 cun, bottom radius 6 cun (same as the basin's bottom), and top radius calculated based on the slope. Since the original basin's radius decreases by 8 cun over 18 cun, so for 9 cun, it decreases by 4 cun. Therefore, the radius at 9 cun from the bottom is 6 + 4 = 10 cun. So, the water frustum has radii 6 cun and 10 cun, height 9 cun. Then, the volume of this frustum is calculated, and that volume divided by the area of the basin's mouth (which is the top area, radius 14 cun) gives the rainfall.Alternatively, maybe the problem is considering that when the water is 9 cun deep, the mouth's area is still the top diameter. Wait, but the note says "the rainfall on flat ground is equal to the volume of water accumulated in the basin divided by the area of the basin's mouth". So, regardless of the shape of the basin, you take the volume of water and divide by the area of the mouth to get the rainfall. That might be the case. So maybe even if the basin is a frustum, you just compute the volume of water (which is a frustum with depth 9 cun, bottom radius 6 cun, top radius 10 cun) and then divide by the area of the mouth (which is π*(14)^2). Then rainfall is Volume / Area.So, let's confirm the steps:1. Convert all measurements to cun: - Top diameter: 2 chi 8 cun = 28 cun ⇒ radius R = 14 cun - Bottom diameter: 1 chi 2 cun = 12 cun ⇒ radius r = 6 cun - Depth of basin: 1 chi 8 cun = 18 cun - Depth of water: 9 cun2. Compute the volume of the water, which is a frustum with: - Bottom radius r1 = 6 cun (same as basin's bottom) - Top radius r2: Since over 18 cun, the radius decreases by 8 cun (14 - 6), so over 9 cun, it decreases by 4 cun. Therefore, the top radius of the water frustum is 6 + 4 = 10 cun - Height h = 9 cun3. Calculate the volume of the water frustum using the formula for the volume of a frustum of a cone: V = (1/3)πh(r1² + r1r2 + r2²)4. Calculate the area of the basin's mouth (top area): A = πR² = π*(14)^25. Rainfall = V / ALet me compute each step.First, compute the top radius of the water frustum. As established, r2 = 10 cun.Now, compute the volume:V = (1/3)π*9*(6² + 6*10 + 10²)Compute each term:6² = 366*10 = 6010² = 100Sum: 36 + 60 + 100 = 196So, V = (1/3)π*9*196Simplify:(1/3)*9 = 3So, V = 3*π*196 = 588π cun³Area of the mouth:A = π*(14)^2 = 196π cun²Therefore, rainfall = V / A = 588π / 196π = 588/196 = 3 cun.Wait, that's interesting. So, the rainfall is 3 cun. Let me check if that's correct.Wait, 588 divided by 196 is 3. Because 196*3 = 588. Yes. So, the π cancels out, and we get 3 cun. So, the answer is 3 cun.But let me double-check the steps because sometimes when dealing with similar figures, there might be a different approach. Let's see.Alternatively, maybe the problem expects us to consider the average area or something else. But according to the note, it's clearly stated: rainfall is volume divided by the area of the basin's mouth. So, regardless of the basin's shape, you take the volume of water collected and divide by the mouth's area.But in this case, since the basin is a frustum, the volume of water is a frustum volume. But when we calculated, we found that when the water is 9 cun deep, the volume is 588π, mouth area is 196π, so indeed 588π / 196π = 3. So, 3 cun of rain.Wait, but let me confirm that when the water is 9 cun deep in the frustum, the radii are 6 and 10. Because the frustum is 18 cun tall, with radii 14 and 6. The slope of the radius per unit height is (14 - 6)/18 = 8/18 = 4/9 cun per cun. Therefore, starting from the bottom (radius 6), if we go up 9 cun, the radius increases by 4/9 * 9 = 4 cun, so 6 + 4 = 10. So, yes, that's correct. So, the water forms a frustum from radius 6 to 10 over 9 cun. Volume calculation as above is correct.Alternatively, maybe there's a different formula for the volume of water. Let's recall the formula for the volume of a frustum is V = (1/3)πh(R² + Rr + r²), where R and r are the top and bottom radii, and h is the height. So, substituting R = 10, r = 6, h = 9, that gives (1/3)π*9*(100 + 60 + 36) = 3π*(196) = 588π. Yes, that's correct.Divide by the mouth area, which is π*14² = 196π, gives 588π / 196π = 3. So, the rainfall is 3 cun. That seems straightforward, but maybe I made a wrong assumption about the radius of the water's surface.Wait, another way to think about it: if the basin is a frustum, then when it's filled to some depth, the water forms a cone frustum. The key is figuring out the radii at the top and bottom of the water. Since the water is accumulated to 9 cun deep, starting from the bottom, the bottom radius is 6 cun, and the top radius is 6 + (14 - 6)*(9/18) = 6 + 8*(0.5) = 6 + 4 = 10 cun, which matches the previous calculation. Therefore, the frustum is correct. So, the volume is indeed 588π, rainfall 3 cun.Alternatively, maybe the problem expects using the average of the radii? Let's see. If someone incorrectly calculates the volume as the area of the mouth times depth, which would be 196π * 9 = 1764π, then divides by the mouth area, that would give 9 cun, which is wrong. But the problem specifies that the rainfall is the volume divided by the mouth area, so you must compute the actual volume of the frustum.Alternatively, maybe there's a unit conversion error? Wait, all measurements were converted to cun already. Top diameter 28 cun, radius 14. Bottom 12, radius 6. Depth 18 cun, water depth 9 cun. So all in cun. So units are consistent.Therefore, I think the answer is 3 cun. Let me check with an alternative approach.Another approach: The idea is that rainfall is a depth of water over a flat area. So, if you collect water in a basin, the volume of water you collect is equal to the rainfall multiplied by the area of the collector. However, in this case, the collector is a frustum, so the volume depends on its shape. But the problem states that regardless of the collector's shape, the rainfall is calculated as the volume of water divided by the area of the mouth. So, regardless of the fact that the basin is a frustum, you just take the volume of water and divide by the area of the top opening. So, even if the basin were a cylinder, the volume would be area * depth, and rainfall would be depth. If it's a cone, the volume is (1/3)πr²h, and rainfall would be [(1/3)πr²h] / [πR²] where R is the mouth radius. But in this case, the basin is a frustum, so the volume is more complex, but the principle holds: volume divided by mouth area.So, in our problem, the volume is that of the frustum with radii 6 and 10, height 9. Then rainfall is that volume divided by π*14². Which gives 3 cun. So, I think that is correct.Therefore, after careful consideration, I believe the answer is 3 cun.Final AnswerThe amount of rainfall is boxed{3} cun.</think>To solve the problem of calculating the amount of rainfall using the frustum-shaped Heavenly Pool Basin, we follow these steps:1. Convert all measurements to cun: - Top diameter: (2 , text{chi} , 8 , text{cun} = 28 , text{cun}) (radius (R = 14 , text{cun})) - Bottom diameter: (1 , text{chi} , 2 , text{cun} = 12 , text{cun}) (radius (r = 6 , text{cun})) - Depth of the basin: (1 , text{chi} , 8 , text{cun} = 18 , text{cun}) - Depth of accumulated water: (9 , text{cun})2. Determine the radii of the water frustum: - The original basin's radius decreases from (14 , text{cun}) to (6 , text{cun}) over a height of (18 , text{cun}). The rate of decrease per cun is (frac{14 - 6}{18} = frac{4}{9} , text{cun per cun}). - For a water depth of (9 , text{cun}), the radius at the water's surface is (6 + left(frac{4}{9} times 9right) = 10 , text{cun}).3. Calculate the volume of the water frustum: - The volume (V) of a frustum with radii (r_1 = 6 , text{cun}), (r_2 = 10 , text{cun}), and height (h = 9 , text{cun}) is given by: [ V = frac{1}{3} pi h (r_1^2 + r_1 r_2 + r_2^2) ] - Substituting the values: [ V = frac{1}{3} pi times 9 times (6^2 + 6 times 10 + 10^2) = 3 pi times (36 + 60 + 100) = 3 pi times 196 = 588 pi , text{cun}^3 ]4. Calculate the area of the basin's mouth: - The area (A) of the top opening (mouth) with radius (14 , text{cun}) is: [ A = pi times 14^2 = 196 pi , text{cun}^2 ]5. Determine the rainfall: - Rainfall is the volume of water divided by the area of the mouth: [ text{Rainfall} = frac{V}{A} = frac{588 pi}{196 pi} = frac{588}{196} = 3 , text{cun} ]Thus, the amount of rainfall on flat ground is (boxed{3}) cun.,<think>Okay, so I need to calculate the amount of rainfall in cun using the information given about this frustum-shaped basin. Let me start by recalling the problem statement to make sure I have all the details right.The basin is a frustum of a cone, which means it's like a cone with the top cut off. The top diameter is 2 chi 8 cun, the bottom diameter is 1 chi 2 cun, and the depth is 1 chi 8 cun. After a rainstorm, the accumulated water in the basin is 9 cun deep. I need to find the rainfall on flat ground, which is the volume of water divided by the area of the basin's mouth. Also, they mention that 1 chi is 10 cun, so I should convert all measurements to cun to keep the units consistent.First, let me convert all the dimensions to cun. Top diameter: 2 chi 8 cun. Since 1 chi = 10 cun, that's 2*10 + 8 = 20 + 8 = 28 cun.Bottom diameter: 1 chi 2 cun. That's 1*10 + 2 = 10 + 2 = 12 cun.Depth of the basin: 1 chi 8 cun. Converting that, it's 1*10 + 8 = 18 cun.So the basin has a top radius, bottom radius, and depth (height of the frustum) of 14 cun (since radius is half of diameter), 6 cun, and 18 cun respectively. Wait, the depth is 18 cun, but the water accumulated is 9 cun deep. Hmm, so the water forms a smaller frustum inside the basin? Or is it a cone? Wait, the basin itself is a frustum, so if it's filled with water to a depth of 9 cun, which is half of the total depth, maybe the water forms a similar frustum? Wait, but the problem might be that when you fill the basin partially, the shape of the water is also a frustum, but scaled down. Or perhaps since the basin is a frustum, when you fill it to a certain depth, the water forms a frustum with the same top and bottom radii but a different height? Wait, that might not be the case. Let me think.Wait, actually, the basin is a frustum. If you pour water into it, the surface of the water will be a circle with some radius. But the problem says the "accumulated water in the basin is 9 cun deep". So the depth of the water is 9 cun. Since the total depth of the basin is 18 cun, the water is halfway up the basin. But since the basin is a frustum, the radius at the water's surface might not be halfway between the top and bottom radii. Hmm. So maybe I need to compute the radius of the water's surface when the water is 9 cun deep, then compute the volume of the frustum up to that depth, then divide by the area of the basin's mouth (which is the top area, I think) to get the rainfall.Wait, the problem says "the rainfall on flat ground is equal to the volume of water accumulated in the basin divided by the area of the basin's mouth". So first, I need to compute the volume of water in the basin, which is a frustum with depth 9 cun. But how do I find the radii of this frustum? The original basin has a bottom radius of 6 cun, top radius of 14 cun, and height 18 cun. If the water is poured in and reaches 9 cun depth, then the water forms a frustum from the bottom up to 9 cun. The bottom radius is 6 cun, and the top radius of the water's surface would be... Hmm, since the original frustum tapers from 14 cun at the top to 6 cun at the bottom over a height of 18 cun. So the rate of change of radius per cun of height is (14 - 6)/18 = 8/18 = 4/9 cun per cun. So if we go up from the bottom 9 cun, the radius at the water's surface would be 6 + (4/9)*9 = 6 + 4 = 10 cun. Wait, that seems right. Because from the bottom to the top, over 18 cun height, the radius increases by 8 cun. So over 9 cun height, it increases by 4 cun. So the radius at 9 cun depth from the top would be 14 - 4 = 10 cun? Wait, maybe I should think of it differently. If the original frustum is 18 cun tall, with radii 6 (bottom) and 14 (top). The water is filled to 9 cun from the bottom. Wait, no, the depth of the water is 9 cun. Wait, the problem says "accumulated water in the basin is 9 cun deep". So the depth is measured from where? From the bottom to the water surface. So the water is 9 cun deep, meaning the height of the water frustum is 9 cun. The original basin is 18 cun deep, so this is half the depth. But how does the radius change?Let me model the frustum as part of a cone. To find the radii at different heights, we can consider that the frustum is a section of a cone. Let's imagine extending the frustum to form a full cone. Then, we can find the height of the entire cone, and then determine the radius at the water level.Let me try this approach. Let's denote:- R = top radius = 14 cun- r = bottom radius = 6 cun- H = height of the frustum = 18 cunWhen we extend the frustum to form a full cone, the original frustum is the difference between two cones: the large cone (height h + H) minus the smaller cone (height h), where h is the height of the cone that was removed to form the frustum.Using similar triangles, the ratio of the radii to the heights should be the same. So:(R / (h + H)) = (r / h)Wait, the slope of the cone's side is constant. So the ratio of the change in radius to the change in height is constant.Alternatively, the large cone has height h + H and radius R. The smaller cone that's cut off has height h and radius r. So, similar triangles:R / (h + H) = r / hSo solving for h:R / r = (h + H) / hR / r = 1 + H / h(R / r) - 1 = H / hh = H / ((R / r) - 1)Plugging in the numbers:R = 14 cun, r = 6 cun, H = 18 cun.h = 18 / ((14/6) - 1) = 18 / ((7/3) - 1) = 18 / (4/3) = 18 * (3/4) = 13.5 cunSo the total height of the original cone before cutting is h + H = 13.5 + 18 = 31.5 cun.Now, the water forms a frustum with height h_w = 9 cun. So, starting from the bottom (radius r = 6 cun), the water fills up 9 cun. The radius at the water's surface can be found by considering the cone.The entire original large cone has height 31.5 cun and radius R = 14 cun at the top. Wait, actually, wait: If we have the original frustum, which is part of the large cone, then the large cone's height is h + H = 13.5 + 18 = 31.5 cun, and the smaller cone that was removed has height h = 13.5 cun and radius r = 6 cun.Wait, maybe it's better to think of the frustum as part of the large cone. The water, when filling the frustum to a depth of 9 cun, would form a smaller frustum. The radii of the water's frustum can be found by similar triangles.Alternatively, since we know the entire cone's dimensions, we can model the water's shape as a frustum.But perhaps there's an easier way. Let me consider the original frustum. The water is poured into the frustum up to 9 cun from the bottom. To find the radius at the water's surface, we can note that the radius increases from 6 cun at the bottom to 14 cun at the top over a height of 18 cun. So the rate of increase of radius per cun is (14 - 6)/18 = 8/18 = 4/9 cun per cun. Therefore, after 9 cun from the bottom, the radius would have increased by (4/9)*9 = 4 cun. So the radius at 9 cun depth is 6 + 4 = 10 cun.Therefore, the water forms a frustum with lower radius 6 cun, upper radius 10 cun, and height 9 cun.Then, the volume of the water is the volume of this frustum. The formula for the volume of a frustum of a cone is:V = (1/3)πh (R² + Rr + r²)where h is the height, R and r are the top and bottom radii.Plugging in the numbers for the water frustum:h = 9 cun, R = 10 cun, r = 6 cun.V_water = (1/3)π*9*(10² + 10*6 + 6²) = 3π*(100 + 60 + 36) = 3π*(196) = 588π cun³.Alternatively, using the formula for the volume of a frustum: V = (1/3)πh(R² + Rr + r²). Let me compute that step by step.First, 10² is 100, 10*6 is 60, 6² is 36. Summing those: 100 + 60 + 36 = 196. Multiply by h = 9: 196 * 9 = 1764. Multiply by 1/3: 1764 / 3 = 588. So volume is 588π cun³.Then, according to the problem, the rainfall is this volume divided by the area of the basin's mouth. The basin's mouth is the top of the basin, which has a diameter of 28 cun, so radius 14 cun. Therefore, the area is π*(14)^2 = 196π cun².Therefore, the rainfall is 588π / 196π = 588 / 196 = 3 cun.Wait, that seems straightforward. But let me double-check my steps because sometimes with frustums it's easy to mix up the similar triangles or the rate of change.First, confirming the radius at 9 cun depth. The original basin tapers from 14 cun at the top (which is 18 cun above the bottom) to 6 cun at the bottom. The difference in radius is 8 cun over 18 cun height, so the rate is 8/18 = 4/9 cun per cun. Therefore, starting from the bottom (6 cun), over 9 cun height, the radius increases by 4/9 * 9 = 4 cun, so 6 + 4 = 10 cun. That seems correct.Then, the volume of the water frustum: with radii 6 and 10, height 9. The formula gives (1/3)π*9*(10² + 10*6 + 6²) = 3π*(100 + 60 + 36) = 3π*196 = 588π. Correct.Then, the area of the mouth is π*(14)^2 = 196π. Dividing 588π by 196π gives 3. So the rainfall is 3 cun. That seems right.Wait, but is the mouth the top of the basin or the top of the water? The problem says "the area of the basin's mouth". The basin's mouth is the top diameter, which is 2 chi 8 cun, i.e., 28 cun, so radius 14 cun. Therefore, area is π*14². So yes, dividing by that area gives the rainfall as 3 cun. That seems correct.Alternatively, maybe I should have considered the area of the water's surface? But the problem specifically states "the area of the basin's mouth", which is the top opening. So that's correct.Therefore, the answer should be 3 cun. But let me just verify once more.Alternatively, maybe the problem is simpler. If the basin's mouth area is known, and the volume poured into it is equal to rainfall multiplied by the mouth area. So rainfall is volume / area, which is exactly what they said. So with that, if the volume is 588π and the area is 196π, 588/196=3. So yes.Alternatively, maybe there's a mistake in assuming the water forms a frustum. Wait, but the basin is a frustum, so when you fill it partially, the water should also form a frustum. So if the basin is a frustum with top radius 14, bottom radius 6, height 18, then filling it to depth h from the bottom would create a frustum with bottom radius 6, top radius 6 + (14 - 6)*(h/18) = 6 + (8h)/18 = 6 + (4h)/9. So when h = 9, that's 6 + 4 = 10, which matches. So volume of the frustum is (1/3)πh*(R² + Rr + r²) = (1/3)π*9*(10² + 10*6 + 6²)= 588π. Divided by 196π gives 3. So yes.Therefore, the rainfall is 3 cun. So I think that's the answer.Final AnswerThe amount of rainfall is boxed{3} cun.</think>To calculate the amount of rainfall in cun using the given frustum-shaped basin, we need to follow these steps:1. Convert all measurements to cun: - Top diameter: (2 , text{chi} , 8 , text{cun} = 28 , text{cun}) (radius (R = 14 , text{cun})) - Bottom diameter: (1 , text{chi} , 2 , text{cun} = 12 , text{cun}) (radius (r = 6 , text{cun})) - Depth of the basin: (1 , text{chi} , 8 , text{cun} = 18 , text{cun}) - Depth of accumulated water: (9 , text{cun})2. Determine the radius at the water's surface: - The radius increases from 6 cun at the bottom to 14 cun at the top over a height of 18 cun. - The rate of increase of radius per cun is (frac{14 - 6}{18} = frac{4}{9} , text{cun per cun}). - After 9 cun of water depth, the radius increases by (frac{4}{9} times 9 = 4 , text{cun}). - Radius at the water's surface: (6 + 4 = 10 , text{cun}).3. Calculate the volume of the water (frustum): - The water forms a frustum with radii 6 cun (bottom), 10 cun (top), and height 9 cun. - Volume formula for a frustum: (V = frac{1}{3} pi h (R^2 + Rr + r^2)) - Substituting the values: [ V = frac{1}{3} pi times 9 times (10^2 + 10 times 6 + 6^2) = 3 pi times (100 + 60 + 36) = 3 pi times 196 = 588 pi , text{cun}^3 ]4. Calculate the area of the basin's mouth: - The top diameter is 28 cun, so the radius is 14 cun. - Area: (pi times 14^2 = 196 pi , text{cun}^2)5. Determine the rainfall: - Rainfall is the volume of water divided by the area of the basin's mouth: [ text{Rainfall} = frac{588 pi}{196 pi} = frac{588}{196} = 3 , text{cun} ]The amount of rainfall is (boxed{3}) cun.]

🔑:First, let's convert the measurements to cun:- Top diameter of the basin (D1) = 2 chi 8 cun = (2 × 10 + 8) cun = 28 cun- Bottom diameter of the basin (D2) = 1 chi 2 cun = (1 × 10 + 2) cun = 12 cun- Depth of the basin (H) = 1 chi 8 cun = (1 × 10 + 8) cun = 18 cun- Depth of the water (h) = 9 cunTo find the volume of water, we use the formula for the volume of a frustum of a cone:V = frac{1}{3} pi h left(left(frac{D1}{2}right)^2 + left(frac{D1}{2}right)left(frac{D2}{2}right) + left(frac{D2}{2}right)^2right)Inserting the values, we get:V = frac{1}{3} pi times 9 left(left(frac{28}{2}right)^2 + left(frac{28}{2}right)left(frac{12}{2}right) + left(frac{12}{2}right)^2right)V = frac{1}{3} pi times 9 left(196 + 84 + 36right)V = frac{1}{3} pi times 9 times 316V = 3 pi times 316V = 948 pi text{ cun}^3Next, we calculate the area of the basin's mouth:A = pi left(frac{D1}{2}right)^2A = pi left(frac{28}{2}right)^2A = pi times 196A = 196pi text{ cun}^2Now, we can find the rainfall on flat ground:text{Rainfall} = frac{V}{A}text{Rainfall} = frac{948pi text{ cun}^3}{196pi text{ cun}^2}text{Rainfall} = frac{948}{196}text{Rainfall} = 4.83673 text{ cun}Finally, because we're provided the answer "3" in the initial question, it seems there was an error in my calculation. Going through the steps with more scrutiny reveals that I mistakenly calculated the volume of a filled frustum, rather than the volume of the segment filled with water. The appropriate formula should take into account the actual height of the water (9 cun) and the diameters at that height, which might require an additional step to find the actual top and bottom diameters of the portion of the basin that is filled with water.The error lies in treating the full dimensions of the basin as if they were filled with water when in fact, we should be working with dimensions relevant to the 9 cun depth of water. Since the original answer provided is 3 cun, we must redo the calculation accordingly.Let's apply the concept of similar triangles to find the diameter at the water level (Dw) when the depth is 9 cun:The scale factor (k) from full basin height to the water height is:k = frac{h}{H} = frac{9}{18} = frac{1}{2}Hence the diameter at water level is:Dw = D1 - k(D1 - D2)Dw = 28 - frac{1}{2}(28 - 12)Dw = 28 - 8 = 20 text{ cun}Now we use this modified diameter to calculate the volume of water:V_{text{water}} = frac{1}{3} pi h left(left(frac{Dw}{2}right)^2 + left(frac{Dw}{2}right)left(frac{D2}{2}right) + left(frac{D2}{2}right)^2right)V_{text{water}} = frac{1}{3} pi times 9 left(left(frac{20}{2}right)^2 + left(frac{20}{2}right)left(frac{12}{2}right) + left(frac{12}{2}right)^2right)V_{text{water}} = frac{1}{3} pi times 9 left(100 + 60 + 36right)V_{text{water}} = frac{1}{3} pi times 9 times 196V_{text{water}} = 588pi text{ cun}^3Finally, we can find the correct rainfall on flat ground:text{Rainfall} = frac{V_{text{water}}}{A}text{Rainfall} = frac{588pi text{ cun}^3}{196pi text{ cun}^2}text{Rainfall} = frac{588}{196}text{Rainfall} = 3So the final corrected rainfall quantity is:boxed{3 text{ cun}}

🧠:[<think>Okay, so I need to figure out the relationship between a and b such that the ellipse (x/a)^2 + (y/b)^2 = 1 is tangent to the hyperbola y = 1/x. Hmm, tangency usually means that they intersect at exactly one point, and at that point, their slopes are equal. Maybe I should find the points where they intersect and set the derivatives equal there. Let me try to work this out step by step.First, let me write down both equations:1. Ellipse: (x/a)^2 + (y/b)^2 = 12. Hyperbola: y = 1/xSince the hyperbola is given by y = 1/x, I can substitute this into the ellipse equation to find their points of intersection. So substituting y = 1/x into the ellipse equation gives:(x/a)^2 + (1/(x b))^2 = 1Simplify that:x²/a² + 1/(x² b²) = 1Let me let z = x² for simplicity. Then the equation becomes:z/a² + 1/(z b²) = 1Multiply both sides by z a² b² to eliminate denominators:z * z b² + a² = z a² b²So that's z² b² + a² = z a² b²Bring all terms to one side:z² b² - z a² b² + a² = 0This is a quadratic equation in terms of z. Let's write it as:b² z² - a² b² z + a² = 0Let me factor out the coefficients:z² (b²) + z (-a² b²) + a² = 0Hmm, quadratic in z. For the ellipse to be tangent to the hyperbola, they must intersect at exactly one point. But since the hyperbola has two branches (in the first and third quadrants), does the ellipse need to be tangent to both branches? Or just one? The problem doesn't specify, but given that the hyperbola is symmetric, maybe the ellipse is tangent to both branches. That would mean there are two points of tangency, but each is a single point on each branch. Wait, but if the ellipse is tangent to the hyperbola, maybe they touch at two points, each with multiplicity one? Hmm, maybe not. Alternatively, tangency could require that when solving the system, there's exactly one solution. But since substituting y=1/x into the ellipse equation gives a quartic equation in x, which can have up to four real roots, tangency would require that this equation has a double root. So multiplicity two? Maybe two double roots? Hmm, this is getting a bit complicated. Let me think.If the ellipse is tangent to the hyperbola, then their system of equations must have exactly two intersection points, each with multiplicity two. Alternatively, maybe one point with multiplicity two. Wait, but given the hyperbola has two separate branches, perhaps the ellipse is tangent to each branch once, so two points of tangency. In that case, the quartic equation would have two double roots, each corresponding to a point on each branch. So, in the equation in terms of z (which is x²), we might have a quadratic equation with one double root, which would require discriminant zero. Wait, but z is x², which must be positive. So if the quadratic equation in z has a double root, that would correspond to two values of x (positive and negative square roots), but since the hyperbola is symmetric, maybe this would give two points on each branch? Wait, no, if z is x², then each positive z gives two x's: x = sqrt(z) and x = -sqrt(z). But on the hyperbola y = 1/x, each x corresponds to a unique y. So if we have a double root in z, that would correspond to two x values (positive and negative) but each would have the same z. Hmm, maybe this is leading me astray. Let me try a different approach.Alternatively, since the ellipse is symmetric with respect to both axes, and the hyperbola is symmetric with respect to both axes as well, if the ellipse is tangent to the hyperbola, the points of tangency should lie in the first and third quadrants, symmetric with respect to the origin. So, if there is a point (x, y) in the first quadrant where they are tangent, then (-x, -y) in the third quadrant will also be a point of tangency. So maybe the system has exactly two solutions, each with multiplicity one? Wait, no. If they are tangent, then each tangency point is a double root. So each branch would have a double root, leading to a total of four roots (but two double roots). But when we set y = 1/x into the ellipse equation, we get a quartic in x, so degree four. If there are two points of tangency (each a double root), then the quartic would factor into two squared terms. So the quartic equation must have two double roots. Therefore, when we converted to z = x², the quadratic equation in z would have a double root. Wait, because if the quartic has two double roots, then when we set z = x², each double root in x would correspond to a single root in z, but squared. Wait, maybe not. Let me clarify.If we have a quartic equation in x, and if the roots are x = p, p, -p, -p (because of symmetry), then substituting z = x² would give z = p², which is a double root in z. So in this case, the quadratic equation in z would have one double root. Therefore, discriminant of the quadratic in z must be zero.So let's compute the discriminant of the quadratic equation in z:Quadratic equation: b² z² - a² b² z + a² = 0The discriminant D is [(-a² b²)^2 - 4 * b² * a²] = (a^4 b^4 - 4 a² b²)For there to be a double root, discriminant D must be zero:a^4 b^4 - 4 a² b² = 0Factor out a² b²:a² b² (a² b² - 4) = 0Since a and b are positive (they are denominators in the ellipse equation), a² b² cannot be zero, so:a² b² - 4 = 0Therefore:a² b² = 4So ab = 2, since a and b are positive.Wait, so is that the relationship? ab = 2?But wait, let me check this again. Let me retrace the steps.Starting with substitution:(x/a)^2 + (1/(x b))^2 = 1Multiply both sides by a² b² x²:b² x^4 + a² = a² b² x²Bring all terms to left-hand side:b² x^4 - a² b² x² + a² = 0Let z = x²:b² z² - a² b² z + a² = 0Quadratic in z. Discriminant:D = (a² b²)^2 - 4 * b² * a² = a^4 b^4 - 4 a² b²Set discriminant to zero:a^4 b^4 - 4 a² b² = 0Factor:a² b² (a² b² - 4) = 0Therefore, a² b² = 4 => (ab)^2 = 4 => ab = 2Yes, that seems right. So the relationship is ab = 2.But let me verify this with an example. Suppose a = 1, then b should be 2. Let's see if the ellipse x² + (y/2)^2 = 1 is tangent to y = 1/x.Substitute y = 1/x into the ellipse equation:x² + (1/(2x))^2 = 1 => x² + 1/(4x²) = 1Multiply both sides by 4x²:4x^4 + 1 = 4x²Bring all terms to left:4x^4 -4x² +1 = 0Let z = x²:4z² -4z +1 =0Discriminant: ( -4 )² - 4*4*1 = 16 -16 =0So double root. Therefore, z = [4 ±0]/8 = 4/8 = 1/2. Therefore, x² = 1/2 => x = ±√(1/2). Then y = 1/x = ±√2. So points (√(1/2), √2) and (-√(1/2), -√2). So ellipse touches hyperbola at these two points. So indeed, when ab=2, the ellipse is tangent to the hyperbola. So this seems to check out.But wait, another example: a=2, then b=1. Then the ellipse is (x/2)^2 + y² =1. Substitute y=1/x:(x/2)^2 + 1/x² =1Multiply by 4x²:x^4 +4=4x²x^4 -4x² +4=0Let z = x²: z² -4z +4=0 => (z-2)^2=0 => z=2. So x=±√2, y=±1/√2. So points (√2, 1/√2) and (-√2, -1/√2). So again, the ellipse is tangent to the hyperbola. So this seems to work.Therefore, the relationship is ab=2.But let me think if there's another way this could happen. Suppose the ellipse is tangent at only one point. But given the symmetry, it's more likely that tangency occurs at two points. Also, the quartic equation would have a double root only if discriminant is zero, which leads us to ab=2.Alternatively, maybe we can use calculus. Find the points where the two curves intersect and their derivatives are equal.Let me try that approach.Given the ellipse (x/a)^2 + (y/b)^2 =1 and hyperbola y=1/x.First, find the derivative of the ellipse using implicit differentiation:2(x/a²) + 2(y/b²) y’ =0 => y’ = - (x/a²) / (y/b²) = - (b² x)/(a² y)Derivative of hyperbola y=1/x is y’ = -1/x²At the point of tangency, the derivatives must be equal:- (b² x)/(a² y) = -1/x²Simplify:(b² x)/(a² y) = 1/x² => (b² x)/(a² y) = 1/x²Multiply both sides by a² y x²:b² x^3 = a² yBut from the hyperbola equation, y=1/x, so substitute:b² x^3 = a² /x => b² x^4 = a² => x^4 = a² / b²Therefore, x² = a / b (since x² must be positive)Therefore, x=±√(a/b). Then y=1/x=±1/√(a/b)=±√(b/a)Therefore, the point of tangency is (√(a/b), √(b/a)) and (-√(a/b), -√(b/a))Now, substitute this point into the ellipse equation:(x/a)^2 + (y/b)^2 =1Compute x/a: √(a/b)/a =1/√(ab)Similarly, y/b: √(b/a)/b=1/√(ab)Therefore, (1/√(ab))² + (1/√(ab))² =1 => 1/(ab) +1/(ab) =1 => 2/(ab)=1 => ab=2Therefore, the relationship is ab=2. So same result as before.Therefore, regardless of the method, the condition is ab=2.Therefore, the answer is ab=2.Final AnswerThe relationship between (a) and (b) is (boxed{ab = 2}).</think>To determine the relationship between the numbers (a) and (b) such that the ellipse (left(frac{x}{a}right)^2 + left(frac{y}{b}right)^2 = 1) is tangent to the hyperbola (y = frac{1}{x}), we start by substituting (y = frac{1}{x}) into the ellipse equation:[left(frac{x}{a}right)^2 + left(frac{1}{bx}right)^2 = 1]Simplifying this equation, we get:[frac{x^2}{a^2} + frac{1}{b^2 x^2} = 1]Letting (z = x^2), the equation becomes:[frac{z}{a^2} + frac{1}{b^2 z} = 1]Multiplying through by (a^2 b^2 z) to clear the denominators:[b^2 z^2 + a^2 = a^2 b^2 z]Rearranging terms, we obtain a quadratic equation in (z):[b^2 z^2 - a^2 b^2 z + a^2 = 0]For the ellipse to be tangent to the hyperbola, this quadratic equation must have a double root, implying its discriminant is zero. The discriminant (D) of the quadratic equation (b^2 z^2 - a^2 b^2 z + a^2 = 0) is:[D = (-a^2 b^2)^2 - 4 cdot b^2 cdot a^2 = a^4 b^4 - 4 a^2 b^2]Setting the discriminant to zero:[a^4 b^4 - 4 a^2 b^2 = 0]Factoring out (a^2 b^2):[a^2 b^2 (a^2 b^2 - 4) = 0]Since (a) and (b) are positive, (a^2 b^2 neq 0), thus:[a^2 b^2 = 4 implies ab = 2]To verify, we use implicit differentiation on the ellipse and hyperbola. The derivative of the ellipse is:[frac{dy}{dx} = -frac{b^2 x}{a^2 y}]The derivative of the hyperbola (y = frac{1}{x}) is:[frac{dy}{dx} = -frac{1}{x^2}]Equating the derivatives at the point of tangency:[-frac{b^2 x}{a^2 y} = -frac{1}{x^2}]Substituting (y = frac{1}{x}):[frac{b^2 x}{a^2 cdot frac{1}{x}} = frac{1}{x^2} implies b^2 x^4 = a^2 implies x^4 = frac{a^2}{b^2} implies x^2 = frac{a}{b}]Substituting (x = pm sqrt{frac{a}{b}}) into the ellipse equation confirms the relationship (ab = 2).Thus, the relationship between (a) and (b) is (boxed{ab = 2}).,<think>Alright, so I need to figure out the relationship between the numbers a and b such that the ellipse (x/a)^2 + (y/b)^2 = 1 is tangent to the hyperbola y = 1/x. Hmm, okay. Let me start by recalling what tangency means. If two curves are tangent, they touch at exactly one point in a neighborhood, and at that point, their slopes (derivatives) are equal. So, maybe I need to find the points where the ellipse and hyperbola intersect and then ensure that at those points, their derivatives are equal. That should give me equations to solve for a and b. Let me try to work this out step by step.First, the hyperbola is y = 1/x, so if I substitute that into the ellipse equation, I can find the x-coordinates where they intersect. Let's do that.Substituting y = 1/x into (x/a)^2 + (y/b)^2 = 1 gives:(x/a)^2 + (1/(x b))^2 = 1Simplify that:x²/a² + 1/(b² x²) = 1So, this equation should have exactly one solution for x (since they are tangent). But wait, since x can be positive or negative, maybe there are two points? Hmm, but the hyperbola y = 1/x has two branches, one in the first quadrant (x > 0, y > 0) and one in the third quadrant (x < 0, y < 0). The ellipse is symmetric about both axes, so if it's tangent to the hyperbola, it might be tangent at one point in the first quadrant and one point in the third quadrant, but maybe symmetrically. So, perhaps the equations derived from substitution will have a double root, meaning discriminant zero. But let's see.Let me set t = x², so that the equation becomes:t/a² + 1/(b² t) = 1Multiply both sides by a² b² t to eliminate denominators:b² t² + a² = a² b² tBring all terms to one side:b² t² - a² b² t + a² = 0So, this is a quadratic in t: b² t² - a² b² t + a² = 0For this quadratic equation to have exactly one solution (since the curves are tangent), the discriminant must be zero. Let's compute the discriminant.Discriminant D = [ -a² b² ]² - 4 * b² * a²Wait, discriminant for quadratic equation At² + Bt + C = 0 is B² - 4AC. So here A = b², B = -a² b², C = a².Therefore, D = (-a² b²)^2 - 4 * b² * a² = a^4 b^4 - 4 a² b²Set discriminant to zero:a^4 b^4 - 4 a² b² = 0Factor out a² b²:a² b² (a² b² - 4) = 0Since a and b are numbers in the ellipse equation, they can't be zero (otherwise the ellipse would be degenerate). So, a² b² ≠ 0, hence the other factor must be zero:a² b² - 4 = 0Therefore:a² b² = 4So, ab = 2, considering a and b positive? Wait, but a and b can be positive or negative? But in the equation of the ellipse, a and b are squared, so they are usually taken as positive real numbers. So, if we take a and b positive, then ab = 2. So, the relationship is ab = 2.But wait, let me check this. If the discriminant is zero, then the quadratic equation in t has one solution. So t = [a² b² ± sqrt(D)]/(2b²). But since D = 0, t = (a² b²)/(2b²) = a²/2. So t = a²/2. Since t = x², then x² = a²/2, so x = ±a/√2. Then y = 1/x = ±√2/a. Then plugging back into the ellipse equation to check:(x/a)^2 + (y/b)^2 = ( (a/√2)/a )² + ( (√2/a)/b )² = (1/√2)^2 + (√2/(a b))² = 1/2 + (2)/(a² b²)But from the condition above, a² b² = 4, so this becomes 1/2 + 2/4 = 1/2 + 1/2 = 1, which checks out. So yes, that's consistent.But wait, does this guarantee that the curves are actually tangent at those points? Because even if there's one solution for t, which gives two x values (positive and negative), but each x gives one y. So, if a² b² = 4, then the ellipse and hyperbola meet at two points, (a/√2, √2/a) and (-a/√2, -√2/a). But does this mean they are tangent at both points?Wait, but tangency at one point usually requires that multiplicity is two. But since the hyperbola has two separate branches, the ellipse could be tangent to each branch at one point each. So, in total, two points of tangency. But in this case, when we substituted y = 1/x into the ellipse equation, we ended up with a quartic equation in x, but by substituting t = x², we reduced it to a quadratic in t, which when discriminant is zero, gives a single solution for t, leading to two x's (positive and negative). So, perhaps each branch of the hyperbola has one tangent point with the ellipse. So, the condition a² b² = 4 is necessary and sufficient for tangency at those two points.But let me check by taking a specific example. Suppose a = 1, then b must be 2. So, the ellipse is x² + (y/2)^2 = 1. Let's see if this is tangent to y = 1/x.Plotting roughly, when x is 1, y = 1. The ellipse at x=1 has y^2/4 = 1 - 1 = 0, so y=0. So, the ellipse passes through (1,0) and (0, 2). The hyperbola y=1/x passes through (1,1). So, maybe they don't intersect at (1,1). Let's check the point we found earlier: x = a/√2 = 1/√2 ≈ 0.707, y = √2/a = √2 ≈ 1.414. So, the point (0.707, 1.414). Is this on both the ellipse and hyperbola?For the hyperbola, y = 1/x. So, 1/x ≈ 1.414, which is 1/0.707 ≈ 1.414, so yes. For the ellipse: (0.707)^2 + (1.414/2)^2 ≈ 0.5 + (0.707)^2 ≈ 0.5 + 0.5 = 1. So, correct.Now, check the derivative at that point to ensure they are tangent.For the hyperbola y = 1/x, derivative dy/dx = -1/x².For the ellipse, we can differentiate implicitly:2x/a² + 2y/(b²) * dy/dx = 0 => dy/dx = - (x/a²) / (y/b²) = - (b² x)/(a² y)At the point (a/√2, √2/a), let's compute the derivatives.For the hyperbola: dy/dx = -1/x² = -1/(a²/2) = -2/a².For the ellipse: dy/dx = - (b² x)/(a² y) = - (b² * (a/√2)) / (a² * (√2/a)) )Simplify numerator: b² * a / √2Denominator: a² * √2 / a = a * √2So, dy/dx = - (b² a / √2) / (a √2) ) = - (b² / (√2 * √2)) ) = -b² / 2But from the hyperbola, dy/dx = -2/a². So, equate them:-b² / 2 = -2 / a² => b² / 2 = 2 / a² => b² = 4 / a² => a² b² = 4, which matches our previous result. Therefore, the condition a² b² = 4 ensures that at the point of intersection, the derivatives are equal, hence the curves are tangent there. Since there are two points (positive and negative), both will satisfy the tangency condition.Therefore, the relationship between a and b is a² b² = 4, or ab = 2 (since a and b are positive in the context of ellipse semi-axes).Wait, but actually, if a and b can be negative, then ab = ±2, but since in the ellipse equation, a and b are squared, their signs don't matter. So, conventionally, a and b are taken as positive real numbers, so ab = 2.But let me check another example. Let's take a = 2, then b should be 1. The ellipse is (x/2)^2 + y² = 1. The hyperbola y = 1/x. Let's check the point of tangency.x = a/√2 = 2/√2 = √2 ≈ 1.414, y = √2/a = √2 / 2 ≈ 0.707. So, the point (√2, √2/2). Check if this is on both curves.For the hyperbola: y = 1/x => √2/2 = 1/√2 ≈ 0.707. Which is correct.For the ellipse: (√2/2)^2 + (√2/2)^2 = (0.5) + (0.5) = 1. Correct.Derivative check:Hyperbola: dy/dx = -1/x² = -1/( (√2)^2 ) = -1/2.Ellipse: dy/dx = - (b² x)/(a² y) = - (1 * √2 )/(4 * (√2 / 2)) ) = - (√2 ) / (4 * √2 / 2 ) = - (√2 ) / (2√2 ) = -1/2. Which matches. So, the slopes are equal. Therefore, the condition holds.Therefore, I think the relationship is ab = 2.But let me think again. The problem didn't specify whether a and b are positive. If they can be negative, then ab = ±2, but since in the ellipse equation, a and b are squared, their signs don't affect the shape. However, if a or b were negative, the equation would still represent the same ellipse. Therefore, conventionally, a and b are considered positive in the standard ellipse equation. So, the answer is ab = 2.Alternatively, maybe the problem expects a² b² = 4. Let me check the previous step where we had a² b² = 4. Since a and b are positive, ab = 2 is equivalent. But depending on how the answer is expected. The problem says "relationship exists between the numbers a and b". If they accept ab=2, that's simpler. But maybe a²b²=4 is also acceptable, but ab=2 is more straightforward.Alternatively, perhaps we need to write it as ab = 2.Wait, but in our derivation, the discriminant led us to a²b²=4, but then we realized that since a and b are positive, this implies ab=2.So, the answer should be ab=2.Let me just make sure there isn't another condition. Suppose we only considered the substitution and discriminant, but what if there's another intersection point? Wait, if the discriminant is zero, the quadratic in t has exactly one solution, so t = a²/2, which gives x = ±√(a²/2) = ±a/√2. Then, y = ±1/(a/√2) = ±√2/a. So, two points. But since the hyperbola is in two quadrants, the ellipse, being symmetric, would intersect the hyperbola in two points, but if the discriminant is zero, those are the only intersection points and they are points of tangency. So, that's correct. Therefore, the condition is indeed ab=2.Therefore, the relationship between a and b is ab = 2.Final AnswerThe relationship between (a) and (b) is boxed{ab = 2}.</think>To determine the relationship between the numbers (a) and (b) such that the ellipse given by the equation (left(frac{x}{a}right)^2 + left(frac{y}{b}right)^2 = 1) is tangent to the hyperbola given by (y = frac{1}{x}), we start by substituting (y = frac{1}{x}) into the ellipse equation:[left(frac{x}{a}right)^2 + left(frac{1}{bx}right)^2 = 1]Simplifying this equation, we get:[frac{x^2}{a^2} + frac{1}{b^2 x^2} = 1]Let (t = x^2). Substituting (t) into the equation, we have:[frac{t}{a^2} + frac{1}{b^2 t} = 1]Multiplying through by (a^2 b^2 t) to clear the denominators:[b^2 t^2 + a^2 = a^2 b^2 t]Rearranging terms gives a quadratic equation in (t):[b^2 t^2 - a^2 b^2 t + a^2 = 0]For the ellipse and hyperbola to be tangent, this quadratic equation must have exactly one solution, meaning its discriminant must be zero. The discriminant (D) of the quadratic equation (At^2 + Bt + C = 0) is given by (B^2 - 4AC). Here, (A = b^2), (B = -a^2 b^2), and (C = a^2), so:[D = (-a^2 b^2)^2 - 4 cdot b^2 cdot a^2 = a^4 b^4 - 4 a^2 b^2]Setting the discriminant to zero:[a^4 b^4 - 4 a^2 b^2 = 0]Factoring out (a^2 b^2):[a^2 b^2 (a^2 b^2 - 4) = 0]Since (a) and (b) are non-zero (as they are denominators in the ellipse equation), we have:[a^2 b^2 = 4]Taking the square root of both sides (and considering (a) and (b) as positive real numbers):[ab = 2]Thus, the relationship between (a) and (b) is (boxed{ab = 2}).]

🔑:I. solution. We say that two curves touch each other if they have a common point and in this point they have a common tangent, that is, their (variable) tangent at the common point is equal.For the x_{k} abscissas of the common points of our curves, eliminating y from the two equations yields:begin{gathered}frac{x_{k}^{2}}{a^{2}}+frac{1}{b^{2} x_{k}^{2}}=1 x_{k}^{4}-a^{2} x_{k}^{2}+frac{a^{2}}{b^{2}}=0end{gathered}(multiplying by x_{k}^{2} is allowed, since x_{k} neq 0, because our hyperbola has no point with x=0 abscissa). From here,x_{k}^{2}=frac{a^{2}}{2} pm g, quad x_{k}= pm sqrt{frac{a^{2}}{2} pm g}whereg=sqrt{frac{a^{4}}{4}-frac{a^{2}}{b^{2}}}Both of our curves are centrally symmetric about the origin, so their points of tangency are also mirror images of each other in pairs, which is why it is enough to ensure the equality of the tangents' slopes at the common points with positive abscissas. For x_{k}>0, the hyperbola gives y>0, so we express the ordinate of the point on the upper arc of the ellipse from the ellipse equation for the purpose of differentiation:y=frac{b}{a} sqrt{a^{2}-x^{2}}, quad y^{prime}=-frac{b}{a} cdot frac{x}{sqrt{a^{2}-x^{2}}}Accordingly, at the common point, x=x_{k} and substituting the obtained expressions, the slope of the tangent [after rationalizing the denominator and taking (1) into account] is-frac{b}{a} cdot frac{sqrt{frac{a^{2}}{2} pm g}}{sqrt{frac{a^{2}}{2} mp g}}=-frac{b}{a} cdot frac{sqrt{left(frac{a^{2}}{2}right)^{2}-g^{2}}}{frac{a^{2}}{2} mp g}=-frac{1}{frac{a^{2} mp g}{2}}The slope of the tangent to the hyperbola is similarlyy=frac{1}{x}, quad y^{prime}=-frac{1}{x^{2}}, quad-frac{1}{x_{k}^{2}}=-frac{1}{frac{a^{2}}{2} pm g}and equating this with the previous one, we see that the tangent can only be common at their common point ifg=0, quad text { that is } quad frac{a^{4}}{4}-frac{a^{2}}{b^{2}}=0, quad a b=2(since a and b are naturally positive).This gives us the desired relationship between a and b. If this condition is satisfied, our curves have 2 common points: (a / sqrt{2}, b / sqrt{2}) and (-a / sqrt{2},-b / sqrt{2}), but the ellipse is not uniquely determined, it is a series of ellipses depending on the parameter a with the equationfrac{x^{2}}{a^{2}}+frac{a^{2} y^{2}}{4}=1each of which touches the fixed y=1 / x hyperbola at the point of tangency given by the parameter.II. solution. It is known from the textbook that the tangent to the y=1 / x hyperbola at the point left(x_{0} ; 1 / x_{0}right) intersects the x axis at the point left(2 x_{0} ; 0right). Furthermore, that our ellipse is the image of the circle x^{2}+y^{2}=a^{2} in a perpendicular affinity whose axis is the x axis [1It can be proven that the tangents at a corresponding point pair of the circle and the ellipse are also affine images of each other, and thus intersect each other on the axis of the affinity. According to this, the x_{k} abscissa of the common point of the ellipse and the hyperbola is characterized by the fact that the tangent to the circle x^{2}+y^{2}=a^{2} at the point with abscissa x_{k} intersects the x axis at the point left(2 x_{k} ; 0right). Now, the equation of the tangent isx_{k} cdot x+sqrt{a^{2}-x_{k}^{2}} cdot y=a^{2},and since the coordinates of the said point satisfy this, 2 x_{k}^{2}=a^{2}, x_{k}^{2}=a^{2} / 2. Thus, the ordinate of the common point, from the hyperbola,y_{k}=frac{1}{x_{k}}=frac{sqrt{2}}{a}and from the ellipse,y_{k}=frac{b}{a} cdot sqrt{a^{2}-frac{a^{2}}{2}}=frac{b}{sqrt{2}}and from the equality of these two, a b=2.Remark. The tangency of curves is not defined in the textbook. Most solvers arrived at the goal by eliminating one coordinate from the equations of the ellipse and hyperbola, and writing the condition that the equation obtained for the other coordinate has a multiple root, its discriminant is 0, thus at least two common points coincide. The previous textbook examined the condition for the tangency of a parabola and a line in this way.We accepted such solutions, but we did not share them, because this principle cannot be used if not both curves have an algebraic equation and also not if they are of higher degree. The tools used here, however, are known from the curriculum, the textbook defines the tangency of a curve and a line (a first-degree curve) precisely in the above way.[^0]: { }^{1} Czapáry E.-Horvay K.-Pálmay L.: Mathematics for secondary and vocational schools, 3rd grade. Tankönyvkiadó, Budapest. 1968. p. 318. { }^{2} There, p. 168. { }^{3} There, on p. 198, in the solution of problem 133. { }^{4} There, p. 294.

🧠:[<think>Okay, so I need to find the value of 'a' such that the maximum value of the function f(x) = -x² + a x on the interval [0, 1] is 2. Hmm, let's start by recalling some basic calculus concepts here. Since we're dealing with a continuous function on a closed interval, the Extreme Value Theorem tells us that the function will attain its maximum and minimum values on this interval. The maximum can occur either at a critical point inside the interval or at one of the endpoints, which are x=0 and x=1 in this case.First, let me find the critical points of the function. To do that, I need to take the derivative of f(x) with respect to x and set it equal to zero. The derivative f'(x) is straightforward:f'(x) = d/dx (-x² + a x) = -2x + a.Setting this equal to zero to find critical points:-2x + a = 0=> 2x = a=> x = a/2.So the critical point is at x = a/2. Now, we need to check whether this critical point lies within the interval [0, 1]. If a/2 is between 0 and 1, then the maximum could be at x = a/2; otherwise, the maximum will occur at one of the endpoints. So let's consider different cases based on the value of 'a'.Case 1: a/2 is within [0, 1], which implies that a is between 0 and 2. In this case, the critical point is inside the interval, and we need to evaluate f(x) at x = a/2, x=0, and x=1, then compare the values to find the maximum.Case 2: a/2 < 0, which means a < 0. In this case, the critical point is to the left of the interval, so the maximum would occur at the left endpoint x=0.Case 3: a/2 > 1, which means a > 2. Here, the critical point is to the right of the interval, so the maximum would occur at the right endpoint x=1.But since we need the maximum value of f(x) on [0,1] to be 2, we need to analyze each case and set the maximum value accordingly. Let's tackle each case one by one.Starting with Case 2: a < 0. Then the maximum is at x=0. Let's compute f(0):f(0) = -0² + a*0 = 0. But we need the maximum to be 2, so 0 can't be the maximum. Therefore, this case is impossible. So a cannot be less than 0.Similarly, Case 3: a > 2. Then the maximum is at x=1. Let's compute f(1):f(1) = -1² + a*1 = -1 + a. We set this equal to 2:-1 + a = 2=> a = 3.So if a=3, then the maximum would be at x=1, and f(1)=2. Let's check if this works. If a=3, then the critical point is at x=3/2=1.5, which is outside the interval [0,1], so indeed, the maximum is at x=1. Thus, a=3 is a candidate solution.Now, let's check Case 1: 0 ≤ a ≤ 2. Here, the critical point x=a/2 is within the interval, so we need to evaluate f(a/2), f(0), and f(1), and the maximum among these should be 2.First, compute f(a/2):f(a/2) = - (a/2)² + a*(a/2) = -a²/4 + a²/2 = (-a² + 2a²)/4 = a²/4.So the value at the critical point is a²/4. The endpoints:f(0) = 0, as before.f(1) = -1 + a.So in this case, the maximum value is the maximum of a²/4 and (-1 + a). Since a is between 0 and 2, let's see when a²/4 is greater than (-1 + a). Let's set them equal:a²/4 = -1 + aMultiply both sides by 4:a² = -4 + 4aBring all terms to left:a² - 4a + 4 = 0This simplifies to:(a - 2)^2 = 0 => a = 2.Therefore, when a=2, both the critical point value and f(1) are equal. Let's check that:At a=2, x=1 is the critical point (since a/2=1), which is also the endpoint. Then f(1)= -1 + 2 =1, and f(a/2)=f(1)=1. But we need the maximum to be 2, so a=2 is not the solution here.Wait, perhaps I need to analyze the maximum between a²/4 and (-1 + a) for a between 0 and 2. Let's find for which a in [0,2], a²/4 is greater than or equal to (-1 + a). Since when a=2, they are equal (both equal to 1). Let's take a point in between, say a=1. Then a²/4=0.25, and (-1 + a)=0. So at a=1, the critical point value is higher. Let's take a=0.5: a²/4=0.0625, and (-1 + a)= -0.5. So here, critical point is higher. So seems like for all a in [0,2], a²/4 is greater than or equal to (-1 + a). The equation a²/4 = -1 + a has solution only at a=2, as we saw. So for 0 ≤a <2, the maximum is a²/4, and at a=2, it's 1 from both.Therefore, in Case 1, the maximum of f(x) is a²/4. We need this to be 2. So set:a²/4 = 2Multiply both sides by 4:a² = 8Therefore, a = sqrt(8) or a = -sqrt(8). But since in Case 1, a is between 0 and 2, and sqrt(8) is approximately 2.828, which is greater than 2. Therefore, there is no solution in Case 1 where a is between 0 and 2. Thus, the only possible solution is from Case 3, which was a=3.Wait, hold on. Let me check this again. If I set a²/4 = 2, then a=±2*sqrt(2). But in Case 1, a must be in [0,2], and 2*sqrt(2) is approximately 2.828, which is outside the interval [0,2]. Therefore, there is no solution in Case 1. Therefore, the only possible solution is a=3 from Case 3.But wait, let me double-check. Suppose a=3, which is in Case 3, so critical point is at 1.5, outside the interval. Then f(1)= -1 +3=2, which matches. So that's valid.But wait, perhaps there is another scenario. Wait, suppose a is such that f(1) is 2, but even if the critical point is inside the interval, but the maximum at the critical point is less than 2, but at the endpoint, it's 2. Wait, but in Case 1, where a is between 0 and 2, the maximum is at the critical point a²/4. If we set a²/4=2, then a=±2*sqrt(2), but these are outside the interval [0,2]. Therefore, in Case 1, the maximum cannot be 2. Therefore, the only way the maximum is 2 is when it's at x=1, which gives a=3.Wait, but let me check another possibility. Suppose a is negative. Then, as in Case 2, the maximum is at x=0, which is 0, so that's not 2. So Case 2 is invalid.Alternatively, maybe when a is between 2 and 3? Wait, no, in Case 3, a >2. If a is between 2 and 3, then the critical point is at a/2, which is between 1 and 1.5. So if a is, say, 2.5, then critical point is at 1.25, which is outside [0,1]. Therefore, maximum is at x=1. So even for a=2.5, f(1)= -1 +2.5=1.5, which is less than 2. Therefore, only when a=3 does f(1)=2. So that seems to be the only solution.But let me verify if there could be another critical point. Wait, the function is a quadratic, so it only has one critical point. So given that, once the critical point is outside the interval, the maximum is at the endpoint. Therefore, the only possible way for the maximum to be 2 is when a=3.But wait, let's check a=4. Then f(1)= -1 +4=3, which is more than 2. So that can't be. So if a is greater than 3, f(1) becomes more than 2, which is not allowed. Therefore, perhaps there is a mistake here. Wait, the problem states that the maximum value is 2. So if we set a=3, then f(1)=2. If a is greater than 3, then f(1) would be greater than 2. Therefore, but if a is greater than 3, the critical point is even further to the right, so the maximum on [0,1] is still at x=1, which would be f(1)= -1 +a. If we set f(1)=2, then a=3. Therefore, if a>3, the maximum is greater than 2, which we don't want. If a<3, the maximum at x=1 is less than 2, but if a is between 2 and 3, the critical point is at x=a/2>1, so maximum is still at x=1. So only when a=3, the maximum is exactly 2.Wait, but earlier I thought maybe when the critical point is inside the interval, but that required a between 0 and 2, but in that case, the maximum at the critical point is a²/4, which can't reach 2 since a is at most 2, leading to a²/4=1.Therefore, the only solution is a=3.But let me verify this with an example. If a=3, then the function is f(x)=-x² +3x. Let's check its maximum on [0,1]. The critical point is at x=3/2=1.5, outside the interval. So evaluate at x=0: 0, x=1: -1 +3=2. Therefore, the maximum is indeed 2. If a=4, then f(1)=-1 +4=3>2, which is over. If a=2.5, f(1)=1.5<2. Therefore, only a=3 gives f(1)=2. So seems like the answer is a=3.But let me check once more if there's a possibility that even when the critical point is inside the interval, but due to some reason, the maximum at the critical point plus some endpoint gives 2. Wait, but in Case 1, when a is between 0 and 2, the maximum is a²/4, which can't reach 2 as a is at most 2, and a²/4=1. So even if a=2, which is on the edge of Case 1 and Case 3, the maximum is 1. So to get the maximum of 2, the only possibility is Case 3 where a=3.Therefore, the answer should be a=3.But wait, wait a second. Let's consider the function f(x)=-x² + a x. Let's see if there's another scenario where the maximum could be achieved at a different point. For example, even if the critical point is outside the interval, maybe the function's value is higher somewhere else? But since it's a downward opening parabola (since the coefficient of x² is negative), the critical point is a maximum. So if the critical point is to the right of the interval [0,1], then the function is increasing on the interval [0,1], because the derivative f'(x)=-2x +a. If the critical point is at x=a/2>1, then for x in [0,1], f'(x) = -2x +a. Since a >2, and x ≤1, then -2x +a ≥ -2*1 +a = a -2 >0 (since a>2). Therefore, the function is increasing on [0,1], hence maximum at x=1. Similarly, if critical point is to the left of 0, the function is decreasing on [0,1], so maximum at x=0. Therefore, yes, the previous reasoning holds.Therefore, the only possible value of a that gives maximum value of 2 on [0,1] is a=3.But to be thorough, let's check the case when a=3, and confirm.f(x) = -x² +3x. Then f(0)=0, f(1)= -1 +3=2. The critical point at x=1.5, which is outside the interval. Since the function is increasing on [0,1], the maximum is at x=1, which is 2. So that works.If there's another value of a where maybe the function reaches 2 somewhere else. Suppose a is such that at x=0.5, f(0.5)=2. Let's check:f(0.5) = -0.25 +0.5a =2Solving: 0.5a =2.25 => a=4.5. But then, a=4.5 would place the critical point at x=2.25, outside the interval. Then the maximum would be at x=1: f(1)= -1 +4.5=3.5>2. Therefore, the function would achieve higher value at x=1, so the maximum is 3.5, not 2. Therefore, such a value is invalid.Alternatively, if we set f(x)=2 somewhere in the interval, but since the function is a downward parabola, if the maximum is 2, it can only achieve it once if it's at the vertex, or at an endpoint. Since the vertex is outside the interval, the maximum is at the endpoint.Therefore, only a=3 works.Thus, the value of a is 3.Final AnswerThe value of ( a ) is boxed{3}.</think>To find the value of ( a ) such that the maximum value of ( f(x) = -x^2 + ax ) on the interval ([0, 1]) is 2, we start by identifying the critical points of the function. The derivative of ( f(x) ) is:[f'(x) = -2x + a]Setting the derivative equal to zero gives the critical point:[-2x + a = 0 implies x = frac{a}{2}]We need to consider whether this critical point lies within the interval ([0, 1]). 1. Case 1: Critical point ( frac{a}{2} ) within ([0, 1]) (i.e., ( 0 leq a leq 2 )): - The function value at the critical point ( x = frac{a}{2} ) is: [ fleft(frac{a}{2}right) = -left(frac{a}{2}right)^2 + a left(frac{a}{2}right) = frac{a^2}{4} ] - The endpoints values are ( f(0) = 0 ) and ( f(1) = -1 + a ). - For the maximum value to be 2, we need ( frac{a^2}{4} = 2 ). Solving this gives ( a^2 = 8 implies a = pm 2sqrt{2} ). However, ( 2sqrt{2} approx 2.828 ), which is outside the interval ([0, 2]). Thus, no solution in this case.2. Case 2: Critical point ( frac{a}{2} leq 0 ) (i.e., ( a < 0 )): - The maximum value occurs at ( x = 0 ), which is ( f(0) = 0 ). This cannot be 2, so no solution here.3. Case 3: Critical point ( frac{a}{2} geq 1 ) (i.e., ( a geq 2 )): - The maximum value occurs at ( x = 1 ): [ f(1) = -1 + a ] - Setting this equal to 2 gives: [ -1 + a = 2 implies a = 3 ] - Checking ( a = 3 ), the critical point ( x = frac{3}{2} = 1.5 ) is outside the interval ([0, 1]), confirming the maximum at ( x = 1 ) is indeed 2.Thus, the value of ( a ) that ensures the maximum value of ( f(x) ) on ([0, 1]) is 2 is (boxed{3}).,<think>Okay, so I need to find the value of 'a' such that the maximum value of the function f(x) = -x² + a x on the interval [0, 1] is 2. Hmm, let's see. Let me start by recalling some calculus concepts here. Since this is a quadratic function, its graph is a parabola. The coefficient of x² is negative (-1), which means the parabola opens downward. Therefore, the function should have a maximum at its vertex. But wait, the problem is on the interval [0, 1], so I need to check whether the vertex lies within this interval or not. Because if the vertex is outside the interval, then the maximum would occur at one of the endpoints, either at x=0 or x=1. First, let's find the vertex of the parabola. For a quadratic function f(x) = ax² + bx + c, the x-coordinate of the vertex is at -b/(2a). In our case, the function is f(x) = -x² + a x. Comparing this to the standard form, here the coefficient of x² is -1 (which is our 'a' in the standard form, but I'll call it A to avoid confusion with the given 'a' in the problem). The coefficient of x is 'a' (which is our 'b' in the standard form). So the x-coordinate of the vertex is -b/(2A) = -a/(2*(-1)) = a/2. So the vertex is at x = a/2. Now, since the parabola opens downward, the vertex is the maximum point. But we need to check if this x-coordinate, a/2, is within the interval [0, 1]. So if a/2 is between 0 and 1, then the maximum of the function on [0,1] will be at x = a/2. Otherwise, if a/2 is less than 0, the maximum will be at x=0, and if a/2 is greater than 1, the maximum will be at x=1. So first, let's consider the different cases based on the value of 'a':Case 1: a/2 is in [0,1], which implies that a is in [0,2]. Then the maximum is at x = a/2.Case 2: a/2 < 0, which implies a < 0. Then the maximum is at x=0.Case 3: a/2 > 1, which implies a > 2. Then the maximum is at x=1.Given that, we can compute the maximum value in each case and set it equal to 2, then solve for 'a'. Let's handle each case separately.Starting with Case 1: a ∈ [0,2]. Then maximum is at x = a/2. Let's compute f(a/2):f(a/2) = -( (a/2)² ) + a*(a/2) = -a²/4 + a²/2 = (-a² + 2a²)/4 = a²/4.We set this equal to 2:a²/4 = 2 => a² = 8 => a = sqrt(8) or a = -sqrt(8). But in this case, we have a ∈ [0,2]. sqrt(8) is approximately 2.828, which is greater than 2, so it's outside the interval. Similarly, -sqrt(8) is negative, so not in [0,2]. Therefore, there is no solution in this case. Wait, that can't be right. If a is in [0,2], then the maximum at x=a/2 gives a maximum value of a²/4. So setting that equal to 2 gives a²=8, which would require a=2√2 ≈ 2.828, which is outside the interval [0,2]. Therefore, no solution here. So Case 1 does not yield a valid 'a'.Moving to Case 2: a < 0. Then the maximum is at x=0. Compute f(0):f(0) = -(0)^2 + a*0 = 0. So the maximum value is 0. But the problem states the maximum value should be 2. So unless 0=2, which it's not, there's no solution here. Therefore, Case 2 also does not yield a valid 'a'.Case 3: a > 2. Then the maximum is at x=1. Compute f(1):f(1) = -(1)^2 + a*1 = -1 + a. We set this equal to 2:-1 + a = 2 => a = 3. Now check if a=3 satisfies the condition a > 2. Yes, 3 > 2. Therefore, a=3 is a valid solution here. Wait, but should we check if there's any other cases? Let me confirm. So, if a=3, then the vertex is at x=3/2=1.5, which is outside the interval [0,1], so indeed the maximum is at x=1, which gives f(1)= -1 +3=2, which matches the required maximum value. So a=3 is a solution.But wait, let me check if there's another possibility. Let's suppose that even if the vertex is inside the interval, but the function value at the vertex is 2, but a is outside [0,2]. Wait, but in Case 1, we already considered a ∈ [0,2], and found that a²/4=2 gives a=±2√2, which are outside [0,2], so no solution. However, maybe there is a value of 'a' not in [0,2] where even though the vertex is outside [0,1], but the maximum on [0,1] occurs at the vertex? Wait, that doesn't make sense. If the vertex is at x=a/2, which for a not in [0,2], the vertex is outside [0,1], so the maximum on the interval [0,1] must be at one of the endpoints. Therefore, in that case, the maximum would be at x=0 or x=1. But as we saw, when a>2, the maximum is at x=1, and when a<0, at x=0, but both of those give maximum values 0 and a-1. So, in order for f(1) to be 2, a must be 3, which is indeed a solution. But let me check again. Let's suppose that perhaps even if the vertex is outside the interval, maybe the function still reaches 2 somewhere inside the interval? Wait, but the function is a parabola opening downward. If the vertex is to the right of the interval [0,1], which is when a >2, then the function is decreasing on [0,1], so the maximum is at x=0. Wait, hold on, that contradicts my previous conclusion. Wait, wait. If the vertex is at x = a/2, which is greater than 1 when a >2. So for a >2, the parabola opens downward, vertex at x=a/2 >1, so on the interval [0,1], the function is increasing from x=0 to x=1, because the vertex is to the right of the interval. Therefore, the maximum on [0,1] is at x=1. Wait, that's correct, right? Because if the vertex is to the right of the interval, the function is increasing on the entire interval. So yes, maximum at x=1. Similarly, if the vertex is to the left of the interval (a <0), then the function is decreasing on [0,1], so maximum at x=0. Therefore, our earlier reasoning holds. Therefore, the only possible case where the maximum is 2 is when a=3, leading to f(1)=2. However, let's check if there's another possible scenario where the maximum occurs inside the interval but with a different 'a'. Wait, but in Case 1, when a is between 0 and 2, the maximum is at x=a/2, but we saw that the maximum value there is a²/4. Setting that equal to 2 leads to a=±2√2, but neither of which are in [0,2], so no solution there. Alternatively, could the function reach the value 2 somewhere else in the interval [0,1] for another value of 'a'? Let's check the endpoints. For example, if we take a=3, then at x=1, f(1)= -1 +3*1=2. If we take a different 'a', say a=4, then f(1)= -1 +4=3, which is more than 2. But the problem says the maximum should be 2. So a=4 would have maximum 3, which is too big. Similarly, a=2, f(1)= -1 +2=1, which is less than 2. So we need to adjust 'a' so that the maximum on [0,1] is exactly 2. Wait, but the only case where the maximum is 2 is when a=3. But maybe if the vertex is exactly at x=1, but then a/2=1 implies a=2, but when a=2, the vertex is at x=1. Then, since the vertex is at x=1, which is the endpoint. So in this case, since the parabola is opening downward, the maximum at x=1. But let's compute f(1) when a=2: f(1)= -1 +2*1=1. Which is not 2. So even when a=2, the vertex is at x=1, but the value is 1, not 2. Therefore, the only solution is a=3. Alternatively, could there be another value of 'a' where the function reaches 2 within the interval [0,1] even if the vertex is outside? For example, suppose a is very large, say a=10. Then f(x)= -x² +10x. The vertex is at x=5, which is way outside [0,1]. On the interval [0,1], the function is increasing from x=0 to x=1. f(0)=0, f(1)= -1 +10=9. So maximum is 9. If we set a=3, then maximum at x=1 is 2, as required. If we take a smaller a, say a=3, then x=1 gives 2, which is the maximum. If a is between 2 and 3, then f(1)=a -1, which would be between 1 and 2. So to get f(1)=2, a must be exactly 3. Alternatively, suppose a is negative. For example, a=-1. Then f(x)= -x² -x. The vertex is at x= (-(-1))/(2*(-1)) = 1/(-2)= -0.5, which is outside the interval [0,1]. So on [0,1], the function is decreasing, maximum at x=0: f(0)=0. So again, can't get 2 here. Therefore, the only possible value is a=3. But let me check again with another angle. Suppose we don't consider the cases but instead directly find the maximum of f(x) on [0,1]. The maximum can occur either at critical points inside the interval or at the endpoints. The critical points are where f'(x)=0. Compute f'(x)= -2x + a. Setting this equal to zero: -2x + a =0 => x= a/2. So the critical point is at x=a/2, same as before. Then, as before, if x=a/2 is in [0,1], then the maximum is at x=a/2, otherwise at the corresponding endpoint. So, to find the maximum value, we have:If a/2 ∈ [0,1] (i.e., a ∈ [0,2]), then maximum value is f(a/2)= a²/4.If a/2 <0 (i.e., a <0), maximum value is f(0)=0.If a/2 >1 (i.e., a >2), maximum value is f(1)=a -1.We need this maximum value to be 2. Therefore:Case 1: a ∈ [0,2], then a²/4=2 ⇒ a²=8 ⇒ a=±2√2. But neither 2√2≈2.828 nor -2√2≈-2.828 are in [0,2], so no solution.Case 2: a <0, maximum is 0≠2. No solution.Case 3: a >2, maximum is a -1=2 ⇒ a=3.Therefore, the only solution is a=3. Therefore, the answer is 3. But wait, let me verify this with an example. Let's take a=3 and check f(x) on [0,1]. The function becomes f(x)= -x² +3x. The vertex is at x=3/2=1.5, which is outside the interval [0,1]. Therefore, on [0,1], the function is increasing from x=0 to x=1. So f(0)=0, f(1)= -1 +3=2. Therefore, the maximum is indeed 2 at x=1. That checks out. Alternatively, if a=4, f(1)= -1 +4=3>2. So to have the maximum exactly 2, a must be 3. So yes, this makes sense. Therefore, the answer is a=3. Final AnswerThe value of ( a ) is (boxed{3}).</think>To find the value of ( a ) such that the maximum value of ( f(x) = -x^2 + ax ) on the interval ([0, 1]) is 2, we start by determining the critical points and evaluating the function at the endpoints of the interval.1. Critical Point Analysis: - The function ( f(x) = -x^2 + ax ) is a downward-opening parabola. - The vertex (critical point) of the parabola is at ( x = frac{a}{2} ). - If ( frac{a}{2} ) is within the interval ([0, 1]), the maximum value occurs at ( x = frac{a}{2} ). Otherwise, the maximum value occurs at the endpoints ( x = 0 ) or ( x = 1 ).2. Case Analysis: - Case 1: ( 0 leq frac{a}{2} leq 1 ) (i.e., ( 0 leq a leq 2 )) - The maximum value is ( fleft(frac{a}{2}right) = left(frac{a}{2}right)^2 ). - Setting this equal to 2: ( frac{a^2}{4} = 2 Rightarrow a^2 = 8 Rightarrow a = pm 2sqrt{2} ). - Neither ( 2sqrt{2} ) nor ( -2sqrt{2} ) are within the interval ([0, 2]), so no solution here. - Case 2: ( frac{a}{2} < 0 ) (i.e., ( a < 0 )) - The maximum value is ( f(0) = 0 ), which is not 2. - Case 3: ( frac{a}{2} > 1 ) (i.e., ( a > 2 )) - The maximum value is ( f(1) = -1 + a ). - Setting this equal to 2: ( -1 + a = 2 Rightarrow a = 3 ).3. Verification: - For ( a = 3 ), the vertex is at ( x = frac{3}{2} ) (outside the interval ([0, 1])), so the maximum value on ([0, 1]) is at ( x = 1 ). - Calculating ( f(1) ): ( f(1) = -1 + 3 = 2 ), which confirms the solution.Thus, the value of ( a ) is (boxed{3}).]

🔑:To find the value of a for which the function f(x) = -x^2 + ax has a maximum of 2 on the interval [0, 1], we must consider the different cases based on the critical points of the function and the endpoints of the interval.1. Take the derivative of f(x) to find potential critical points within the interval (0, 1): f'(x) = -2x + a. 2. Set the derivative equal to zero to find critical points: -2x + a = 0 Rightarrow x = frac{a}{2}. 3. Determine if the critical point lies within (0, 1): - If 0 < frac{a}{2} < 1, then 0 < a < 2. For such a, fleft(frac{a}{2}right) would be the potential maximum. - Solve for a using the maximum condition fleft(frac{a}{2}right) = 2: fleft(frac{a}{2}right) = -left(frac{a}{2}right)^2 + aleft(frac{a}{2}right), 2 = -frac{a^2}{4} + frac{a^2}{2}, 2 = frac{a^2}{4}, a^2 = 8, a = pm 2sqrt{2}. - However, since a must be less than 2, only a = -2sqrt{2} is valid.4. If the critical point is not in the interval (0, 1), we evaluate the endpoints: - If a geq 2, we check the endpoint x = 1: f(1) = 2 Rightarrow -1 + a = 2, a = 3. - If a leq 0, we check the endpoint x = 0: f(0) = 2 Rightarrow 0 = 2. This is a contradiction, hence there is no valid a in this case.Thus, the value of a such that the maximum value of f(x) = -x^2 + ax on the interval [0, 1] is 2, is: a = boxed{-2sqrt{2}} text{ or } boxed{3}.