🧠:[<think>Okay, so I need to find the smallest positive period of the function f(f(f(x))) where f(x) is defined as arcsin(cos x). Hmm, let's start by understanding the original function f(x) = arcsin(cos x). Maybe if I break it down step by step, I can figure out its periodicity and then see how composing it three times affects the period.First, let me recall some properties of the arcsin function. The domain of arcsin is [-1, 1], and its range is [-π/2, π/2]. Since cos x always outputs values between -1 and 1, the domain of f(x) should be all real numbers. That's good; no restrictions on x there.Now, let's analyze f(x) = arcsin(cos x). I wonder if this can be simplified. I know that cos x can be written as sin(π/2 - x), but I'm not sure if that helps here. Alternatively, maybe there's a relationship between arcsin and cos that I can exploit.Wait, arcsin(cos x)... Let me think. If I let θ = arcsin(cos x), then sin θ = cos x. But sin θ = cos x implies that θ = π/2 - x + 2πn or θ = π/2 + x + 2πn for some integer n, right? Because sin θ = cos x can be rewritten as sin θ = sin(π/2 - x), which has solutions θ = π/2 - x + 2πn or θ = π/2 + x + 2πn.But θ is restricted to the range [-π/2, π/2] because it's the result of arcsin. So we need to find θ in that interval such that sin θ = cos x. Let's consider different cases.Case 1: Suppose x is such that π/2 - x is in [-π/2, π/2]. Then θ = π/2 - x. But π/2 - x ∈ [-π/2, π/2] implies that x ∈ [0, π]. So for x between 0 and π, arcsin(cos x) = π/2 - x.Case 2: If x is such that π/2 + x is in [-π/2, π/2], but wait, π/2 + x ∈ [-π/2, π/2] implies x ∈ [-π, 0]. So for x between -π and 0, θ = π/2 + x. But x here is negative, so π/2 + x would be π/2 - |x|. Hmm, but is this correct? Let's check.Alternatively, maybe it's better to consider the identity cos x = sin(π/2 - x), but then arcsin(sin(π/2 - x)). The arcsin of sin y is equal to y if y ∈ [-π/2, π/2], otherwise it's equal to π - y if y ∈ [π/2, 3π/2], adjusted by periodicity. Wait, so arcsin(sin z) is equal to z - 2πk if z ∈ [π/2 + 2πk, 3π/2 + 2πk] for some integer k, but adjusted to stay within the principal branch.But here, z = π/2 - x. So arcsin(sin(π/2 - x)) would be equal to π/2 - x if π/2 - x ∈ [-π/2, π/2], which is equivalent to x ∈ [0, π]. If π/2 - x is outside that interval, then we need to adjust it.For example, if π/2 - x > π/2, that is, if x < 0, then sin(π/2 - x) = sin(x') where x' = π/2 - x > π/2. Then arcsin(sin x') = π - x'. But x' = π/2 - x, so arcsin(sin x') = π - (π/2 - x) = π/2 + x. But since x < 0 here, this is π/2 + x, which would be in the range [-π/2, π/2] because x is between -π and 0.Wait, let's check when x is in [π, 2π]. Then π/2 - x would be in [-3π/2, -π/2], which is outside the principal branch. So arcsin(sin(π/2 - x)) would be equal to... Hmm, let's think. The sine function is periodic with period 2π, so sin(π/2 - x) is the same as sin(π/2 - x + 2π) = sin(5π/2 - x). If x ∈ [π, 2π], then 5π/2 - x ∈ [5π/2 - 2π, 5π/2 - π] = [π/2, 3π/2]. Then arcsin(sin(5π/2 - x)) would be π - (5π/2 - x) if 5π/2 - x ∈ [π/2, 3π/2]. So π - (5π/2 - x) = x - 3π/2. But we need to check if this is in [-π/2, π/2]. If x ∈ [π, 2π], then x - 3π/2 ∈ [-π/2, π/2]. So indeed, for x ∈ [π, 2π], arcsin(cos x) = x - 3π/2.Wait, this is getting complicated. Let me summarize:arcsin(cos x) is equal to:- π/2 - x when x ∈ [0, π]- π/2 + x when x ∈ [-π, 0]- x - 3π/2 when x ∈ [π, 2π]But since the function is periodic, we can probably express it as a triangle wave or some piecewise linear function with period 2π. Let me check a few specific values.For x = 0: arcsin(cos 0) = arcsin(1) = π/2. Which matches π/2 - 0 = π/2.For x = π/2: arcsin(cos π/2) = arcsin(0) = 0. π/2 - π/2 = 0, correct.For x = π: arcsin(cos π) = arcsin(-1) = -π/2. But π/2 - π = -π/2, correct.For x = -π/2: arcsin(cos(-π/2)) = arcsin(0) = 0. π/2 + (-π/2) = 0, correct.For x = -π: arcsin(cos(-π)) = arcsin(-1) = -π/2. π/2 + (-π) = -π/2, correct.For x = 3π/2: cos(3π/2) = 0, so arcsin(0) = 0. According to the third case, x - 3π/2 = 3π/2 - 3π/2 = 0, correct.For x = 2π: cos(2π) = 1, arcsin(1) = π/2. x - 3π/2 = 2π - 3π/2 = π/2, correct.So, the function f(x) = arcsin(cos x) is a piecewise linear function with period 2π, consisting of three segments:1. For x ∈ [0, π], f(x) = π/2 - x2. For x ∈ [-π, 0], f(x) = π/2 + x3. For x ∈ [π, 2π], f(x) = x - 3π/2But since it's periodic with period 2π, this pattern repeats every 2π. So, the graph of f(x) is a sawtooth wave that goes from π/2 at x = 0 down to -π/2 at x = π, then back up to π/2 at x = 2π. Wait, but between π and 2π, according to the third case, it's x - 3π/2. At x = π, that gives π - 3π/2 = -π/2, and at x = 2π, it gives 2π - 3π/2 = π/2. So it's a linear increase from -π/2 to π/2 over [π, 2π]. So yes, the function is a triangle wave with amplitude π/2 and period 2π.Wait, but a triangle wave usually has a period, so in this case, over [0, 2π], it's decreasing from π/2 to -π/2 over [0, π], then increasing back to π/2 over [π, 2π]. So the period is 2π.But since it's a periodic function with period 2π, then composing it multiple times might change the period. The question is about f(f(f(x))) and its smallest positive period.So, perhaps we need to find the period of f(f(f(x))). Let's first find the period of f(x), then f(f(x)), then f(f(f(x))).But wait, since f(x) has period 2π, then f(f(x)) would have a period that is possibly a divisor of 2π. But composing periodic functions can sometimes result in a smaller period. For example, if f(x) is a function with period T, then f(f(x)) might have a period smaller than T, depending on the symmetries of f.Alternatively, sometimes the composition can double the period, but in this case, maybe the period becomes smaller. Let's check.First, let's confirm that f(x) is indeed periodic with period 2π. Since cos(x) has period 2π, and arcsin is applied to it, the entire function f(x) = arcsin(cos x) must also have period 2π. Additionally, is there a smaller period? Let's check if f(x + T) = f(x) for some T < 2π.Suppose T = π. Then f(x + π) = arcsin(cos(x + π)) = arcsin(-cos x) = -arcsin(cos x) = -f(x). So f(x + π) = -f(x). Therefore, the function is not periodic with period π, but it has a sort of anti-periodicity. Then f(x + 2π) = f(x), which is the fundamental period.Therefore, the fundamental period of f(x) is 2π.Now, let's consider f(f(x)). Let's denote g(x) = f(f(x)). What's the period of g(x)?We know that f(x + 2π) = f(x), so g(x + 2π) = f(f(x + 2π)) = f(f(x)) = g(x). Therefore, the period of g(x) divides 2π. But maybe it's smaller. Let's see.To check if the period is smaller, we can test T = π. Let's compute g(x + π) = f(f(x + π)) = f(-f(x)) = arcsin(cos(-f(x))) = arcsin(cos(f(x))), since cos is even. But cos(f(x)) = cos(arcsin(cos x)). Hmm, is this equal to something?Wait, let's compute cos(arcsin(z)). Let θ = arcsin(z), then cos θ = sqrt(1 - z²). So cos(arcsin(z)) = sqrt(1 - z²). Therefore, cos(f(x)) = sqrt(1 - (cos x)^2) = |sin x|. Therefore, f(-f(x)) = arcsin(|sin x|).Wait, but f(-f(x)) = arcsin(cos(-f(x))) = arcsin(cos(f(x))) because cos is even. Then, as above, arcsin(cos(f(x))) = arcsin(|sin x|). Hmm, is this equal to something?Wait, |sin x| is non-negative, so arcsin(|sin x|) is equal to arcsin(sin |x|) ?Wait, no, not exactly. Let me think.arcsin(|sin x|). Since |sin x| is between 0 and 1, arcsin(|sin x|) is between 0 and π/2. Also, |sin x| = sin |x| when x is in [0, π], but not necessarily otherwise.But perhaps we can find an expression for arcsin(|sin x|). Let me consider different intervals.For x in [0, π/2], |sin x| = sin x, so arcsin(sin x) = x.For x in [π/2, 3π/2], |sin x| = |sin x|, which is sin(π - x) for x in [π/2, π], and -sin x for x in [π, 3π/2]. Wait, maybe breaking it into intervals:For x in [0, π/2], arcsin(|sin x|) = arcsin(sin x) = x.For x in [π/2, π], |sin x| = sin x, but arcsin(sin x) = π - x.For x in [π, 3π/2], |sin x| = -sin x, and arcsin(-sin x) = -arcsin(sin x). But since arcsin is odd, arcsin(-sin x) = -arcsin(sin x). But sin x is negative here, so let me see.Wait, if x is in [π, 3π/2], then sin x is negative, so |sin x| = -sin x. So arcsin(|sin x|) = arcsin(-sin x) = -arcsin(sin x). But arcsin(sin x) for x in [π/2, 3π/2] is π - x if x in [π/2, π], and -x + π if x in [π, 3π/2]? Wait, maybe it's better to think step by step.Wait, let's take x in [π, 3π/2]. Let θ = x - π ∈ [0, π/2]. Then sin x = -sin θ. So |sin x| = sin θ. Then arcsin(|sin x|) = arcsin(sin θ) = θ = x - π.Similarly, for x in [3π/2, 2π], |sin x| = sin(2π - x) and arcsin(sin(2π - x)) = 2π - x, but 2π - x is in [0, π/2] here, so arcsin(|sin x|) = 2π - x.Wait, this might be getting complicated. Let me summarize:arcsin(|sin x|) is equal to:- x, for x ∈ [0, π/2]- π - x, for x ∈ [π/2, π]- x - π, for x ∈ [π, 3π/2]- 2π - x, for x ∈ [3π/2, 2π]But this seems to have a period of π, but let's check.Wait, if we plot arcsin(|sin x|), it's a triangular wave that goes from 0 to π/2 and back to 0, repeating every π. Wait, no. Let's check specific points.At x = 0: arcsin(|sin 0|) = 0.At x = π/4: arcsin(sin π/4) = π/4.At x = π/2: arcsin(1) = π/2.Wait, hold on. Wait, when x = π/2, |sin x| = 1, so arcsin(1) = π/2. But according to the previous breakdown, for x ∈ [π/2, π], arcsin(|sin x|) = π - x. But at x = π/2, π - x = π/2, which matches. For x = π, arcsin(|sin π|) = arcsin(0) = 0. π - π = 0, correct.For x = 3π/4: arcsin(|sin 3π/4|) = arcsin(√2/2) = π/4. But π - x = π - 3π/4 = π/4, correct.For x = π: arcsin(0) = 0.For x = 3π/2: arcsin(|sin 3π/2|) = arcsin(1) = π/2. According to x - π: 3π/2 - π = π/2, correct.For x = 2π: arcsin(|sin 2π|) = 0. 2π - 2π = 0, correct.So, arcsin(|sin x|) is a triangular wave that starts at 0, rises to π/2 at x = π/2, decreases back to 0 at x = π, then rises again to π/2 at x = 3π/2, and decreases back to 0 at x = 2π. Wait, but this seems to have a period of π, but looking at the graph from 0 to 2π, it goes up to π/2, down to 0, up to π/2, down to 0. So actually, it's a waveform with two peaks in the interval [0, 2π], meaning the period is π.Wait, let's check. If we shift x by π, does the function repeat?arcsin(|sin(x + π)|) = arcsin(|sin x cos π + cos x sin π|) = arcsin(|-sin x|) = arcsin(|sin x|). So yes, arcsin(|sin(x + π)|) = arcsin(|sin x|). Therefore, the function arcsin(|sin x|) has period π. Therefore, f(f(x)) = arcsin(|sin x|) is periodic with period π.But wait, we initially thought that g(x) = f(f(x)) = arcsin(|sin x|). Is that right? Wait, let's recap:We had f(x) = arcsin(cos x). Then f(f(x)) = arcsin(cos(arcsin(cos x))). Then we found that cos(arcsin(cos x)) = |sin x|. Therefore, f(f(x)) = arcsin(|sin x|). Which we just determined has period π.Therefore, the period of f(f(x)) is π. So composing f twice reduces the period from 2π to π.Now, moving on to f(f(f(x))) = f(arcsin(|sin x|)). Let's denote h(x) = f(f(f(x))) = arcsin(cos(arcsin(|sin x|))).Again, let's try to simplify this. Let me first compute cos(arcsin(|sin x|)). Let θ = arcsin(|sin x|). Then cos θ = sqrt(1 - (|sin x|)^2) = sqrt(1 - sin²x) = |cos x|.Therefore, cos(arcsin(|sin x|)) = |cos x|. Therefore, h(x) = arcsin(|cos x|).So f(f(f(x))) = arcsin(|cos x|). Now, we need to find the period of arcsin(|cos x|).Let's analyze arcsin(|cos x|). The function |cos x| has period π because cos x has period 2π and taking absolute value reflects the negative part, making it repeat every π. Then, applying arcsin to it, which is a non-linear operation, but since |cos x| is periodic with period π, arcsin(|cos x|) should also have period π.But wait, let's verify. arcsin(|cos(x + π)|) = arcsin(|cos x cos π - sin x sin π|) = arcsin(|-cos x|) = arcsin(|cos x|). So yes, the period is π.But is there a smaller period? Let's check.Suppose T is the period. We need arcsin(|cos(x + T)|) = arcsin(|cos x|) for all x. Since |cos x| has fundamental period π, then T must be a divisor of π. The fundamental period of |cos x| is π, so unless arcsin introduces some symmetry that reduces the period further, the fundamental period of arcsin(|cos x|) should also be π.But let's check specific values. Let's compute arcsin(|cos x|) at x = 0, π/4, π/2, etc.At x = 0: arcsin(|cos 0|) = arcsin(1) = π/2.At x = π/4: arcsin(√2/2) = π/4.At x = π/2: arcsin(0) = 0.At x = π: arcsin(|cos π|) = arcsin(1) = π/2.At x = 3π/4: arcsin(√2/2) = π/4.At x = π/2 + π/2 = π: same as above.So the function arcsin(|cos x|) goes from π/2 at 0, decreases to 0 at π/2, increases back to π/2 at π, then decreases to 0 at 3π/2, and back to π/2 at 2π. Wait, but if the period is π, then between 0 and π, it should repeat. But in reality, between π and 2π, it's a mirror image. Wait, perhaps not.Wait, let's plot the function arcsin(|cos x|):Since |cos x| is symmetric about every multiple of π/2. So, the function |cos x| over [0, π/2] is cos x, decreasing from 1 to 0. Then over [π/2, π], it's -cos x, increasing from 0 to 1. Then over [π, 3π/2], it's -cos x, decreasing from 1 to 0, and over [3π/2, 2π], cos x, increasing from 0 to 1. Wait, actually |cos x| is a series of "bumps" each of length π/2, but with period π.But arcsin(|cos x|) would take the values:- For x in [0, π/2], |cos x| = cos x, so arcsin(cos x) = π/2 - x.- For x in [π/2, π], |cos x| = -cos x, so arcsin(|cos x|) = arcsin(-cos x) = arcsin(cos(x - π/2 + π/2))? Wait, no. Wait, cos x in [π/2, π] is negative, so |cos x| = -cos x. Then arcsin(|cos x|) = arcsin(-cos x) = arcsin(cos(x)) but with x in [π/2, π]. Wait, arcsin(|cos x|) is equal to arcsin(|cos x|), which is the same as arcsin(|cos(x)|), so since |cos x| is symmetric, arcsin(|cos x|) is equal to arcsin(|cos(x)|), which is the same as arcsin(|cos(|x|)|), so it's even function.But perhaps I should look at specific intervals:For x in [0, π/2], |cos x| = cos x, so arcsin(cos x) = π/2 - x.For x in [π/2, π], |cos x| = -cos x, which is equal to cos(π - x). So arcsin(cos(π - x)) = π/2 - (π - x) = x - π/2.Therefore, arcsin(|cos x|) is equal to:- π/2 - x for x ∈ [0, π/2]- x - π/2 for x ∈ [π/2, π]- Then, since |cos x| is symmetric with period π, the pattern repeats every π.Therefore, arcsin(|cos x|) is a triangular wave that goes from π/2 at 0 to 0 at π/2, then back to π/2 at π, and so on. Therefore, the function has a period of π.Wait, let's check at x = π/2 + π = 3π/2. arcsin(|cos 3π/2|) = arcsin(0) = 0. Which is the same as at x = π/2. So the function is indeed periodic with period π.But wait, let's check at x = π/4 and x = π/4 + π. At x = π/4: arcsin(√2/2) = π/4. At x = 5π/4: |cos 5π/4| = |−√2/2| = √2/2, so arcsin(√2/2) = π/4. So same value. Similarly, at x = 3π/4: arcsin(|cos 3π/4|) = arcsin(√2/2) = π/4. Wait, but according to the previous piecewise definition, for x ∈ [π/2, π], arcsin(|cos x|) = x - π/2. So at x = 3π/4, that would be 3π/4 - π/2 = π/4, which matches. So yes, the function is symmetric around π/2.Therefore, arcsin(|cos x|) is a triangular wave with period π, going from π/2 at 0 to 0 at π/2, back to π/2 at π, then repeating. So its fundamental period is π.But hold on, the problem states to find the smallest positive period of f(f(f(x))) which is arcsin(|cos x|). So according to this, the period should be π. But the answer is supposed to be something else? Wait, maybe not. Let me verify.Wait, but is there a smaller period? For example, if T is such that arcsin(|cos(x + T)|) = arcsin(|cos x|) for all x. Let's suppose T = π/2.Check if arcsin(|cos(x + π/2)|) = arcsin(|cos x|). cos(x + π/2) = -sin x, so |cos(x + π/2)| = |sin x|. Therefore, arcsin(|sin x|). But arcsin(|sin x|) is not equal to arcsin(|cos x|). For example, at x = 0: arcsin(|cos 0|) = π/2, arcsin(|sin 0|) = 0. Not equal. Therefore, T = π/2 is not a period.How about T = π/1? So π. We already saw that it works. Because adding π to x, |cos(x + π)| = |cos x|, so arcsin(|cos(x + π)|) = arcsin(|cos x|). So π is a period. And there's no smaller period than π because the function arcsin(|cos x|) completes a full cycle over [0, π], and then repeats. For example, between 0 and π/2, it decreases from π/2 to 0, and between π/2 and π, it increases back to π/2. So the period is indeed π.But wait, let's check the values again. For example, let's take x = 0 and x = π. arcsin(|cos 0|) = π/2, arcsin(|cos π|) = π/2. Similarly, x = π/2: 0, x = 3π/2: 0. So over the interval [0, π], the function goes from π/2 to 0 to π/2. Then from π to 2π, it's the same as from 0 to π. So actually, the function has period π. Therefore, the fundamental period is π.But the question is about f(f(f(x))) = arcsin(|cos x|). So according to this, the period is π. However, let me check the original problem again. It says "the smallest positive period of f(f(f(x)))". The answer might be π, but maybe there's a mistake in my reasoning.Wait, let's cross-validate.Given that f(x) = arcsin(cos x) has period 2π, f(f(x)) has period π, and f(f(f(x))) has period π. But maybe when we compose these functions, the period changes in a non-trivial way.Wait, another approach: since f(x) is 2π-periodic, composing it three times would result in a function whose period divides 2π. But perhaps due to the specific symmetries, the period becomes π/2 or something else. Wait, but we found that f(f(f(x))) is arcsin(|cos x|), which we believe has period π.But let me check the function arcsin(|cos x|) numerically at some points to see if it's indeed π-periodic.Compute arcsin(|cos(0)|) = arcsin(1) = π/2.Compute arcsin(|cos(π)|) = arcsin(1) = π/2.Compute arcsin(|cos(π/2)|) = arcsin(0) = 0.Compute arcsin(|cos(3π/2)|) = arcsin(0) = 0.Compute arcsin(|cos(π/4)|) = arcsin(√2/2) = π/4.Compute arcsin(|cos(3π/4)|) = arcsin(√2/2) = π/4.Compute arcsin(|cos(π/8)|) = arcsin(cos(π/8)) ≈ arcsin(0.9239) ≈ 1.178 (which is π/2 - π/8 ≈ 1.178, since π/2 ≈ 1.5708). So π/2 - π/8 = 3π/8 ≈ 1.178, correct.Similarly, arcsin(|cos(3π/8)|) = arcsin(cos(3π/8)) ≈ arcsin(0.3827) ≈ 0.3927, which is π/8 ≈ 0.3927. Wait, but 3π/8 ≈ 1.178, so π/2 - 3π/8 = π/8 ≈ 0.3927. Correct.If we take x = π/8 and x = π/8 + π, which is 9π/8. Then arcsin(|cos(9π/8)|) = arcsin(|cos(9π/8)|). cos(9π/8) = cos(π + π/8) = -cos(π/8), so |cos(9π/8)| = cos(π/8). Therefore, arcsin(cos(π/8)) = π/2 - π/8 = 3π/8. However, at x = 9π/8, the function arcsin(|cos x|) = 3π/8. But at x = π/8 + π = 9π/8, the value is the same as at x = π/8. Wait, no. Wait, arcsin(|cos(9π/8)|) = arcsin(cos(π/8)) = π/2 - π/8 = 3π/8. However, if the function had period π, then arcsin(|cos(9π/8)|) should be equal to arcsin(|cos(π/8 + π)|) = arcsin(|cos(π/8)|) = 3π/8. Which matches. So adding π to x gives the same value, which is consistent with period π.But if we check x = π/4 and x = π/4 + π/2 = 3π/4. arcsin(|cos(3π/4)|) = arcsin(√2/2) = π/4. Similarly, arcsin(|cos(π/4)|) = π/4. But π/4 + π/2 = 3π/4, which is within the same period π. So this doesn't test a smaller period. Let's check x = 0 and x = π/2. arcsin(|cos 0|) = π/2, arcsin(|cos π/2|) = 0. These are different, so the period cannot be smaller than π.Therefore, it seems that the period of f(f(f(x))) is π. But the problem states "the smallest positive period of f(f(f(x))) is...". However, wait, the original function f(x) had period 2π, but after composing three times, we might have a smaller period. Wait, but according to the analysis, f(f(f(x))) = arcsin(|cos x|) has period π. So the answer should be π.But let me check once more. The problem might be expecting a different answer. Let me consider alternative approaches.Alternatively, since f(x) = arcsin(cos x) can be written as f(x) = π/2 - x for x in [0, π], and then repeats with period 2π. So, f(x) is a sawtooth wave with period 2π. Then f(f(x)) would involve applying this sawtooth wave to itself. Which might create another periodic function.But earlier steps showed that f(f(x)) = arcsin(|sin x|), which has period π, and then f(f(f(x))) = arcsin(|cos x|), which also has period π. So the conclusion is that the period halves each time we compose the function? Wait, f(x) has period 2π, f(f(x)) has period π, f(f(f(x))) has period π. So composing thrice doesn't reduce it further. So the answer is π.But maybe I made a mistake in assuming that the period of arcsin(|cos x|) is π. Let's check the graph of arcsin(|cos x|).Between 0 and π/2: arcsin(|cos x|) = π/2 - x, decreasing from π/2 to 0.Between π/2 and π: arcsin(|cos x|) = x - π/2, increasing from 0 to π/2.Between π and 3π/2: arcsin(|cos x|) = π/2 - (x - π) = 3π/2 - x, decreasing from π/2 to 0.Between 3π/2 and 2π: arcsin(|cos x|) = x - 3π/2 + π/2 = x - π, increasing from 0 to π/2.Wait, no, that doesn't align. Wait, if we consider the function arcsin(|cos x|), over [π, 2π], we can substitute y = x - π. Then |cos x| = |cos(y + π)| = | - cos y | = |cos y|. So arcsin(|cos x|) = arcsin(|cos y|) where y = x - π. Therefore, the function over [π, 2π] is the same as over [0, π], meaning the period is π. Therefore, the fundamental period is indeed π.Therefore, the smallest positive period of f(f(f(x))) is π. So the answer is π, which is boxed{pi}.Wait, but let me check once again with numerical examples to confirm.Take x = 0:f(f(f(0))) = arcsin(|cos 0|) = arcsin(1) = π/2.x = π/2:f(f(f(π/2))) = arcsin(|cos π/2|) = arcsin(0) = 0.x = π:f(f(f(π))) = arcsin(|cos π|) = arcsin(1) = π/2.x = 3π/2:f(f(f(3π/2))) = arcsin(|cos 3π/2|) = arcsin(0) = 0.x = 2π:f(f(f(2π))) = arcsin(|cos 2π|) = arcsin(1) = π/2.Now, if we take x = π/4:f(f(f(π/4))) = arcsin(|cos π/4|) = arcsin(√2/2) = π/4.x = π/4 + π = 5π/4:f(f(f(5π/4))) = arcsin(|cos 5π/4|) = arcsin(√2/2) = π/4.So the value at 5π/4 is the same as at π/4, which is consistent with period π.Similarly, x = π/2 + π = 3π/2: f(f(f(3π/2))) = 0, same as at π/2.Therefore, the function repeats every π. Hence, the period is π.Therefore, the answer should be boxed{pi}.Final Answerboxed{pi}</think>To find the smallest positive period of ( f(f(f(x))) ) where ( f(x) = arcsin(cos x) ):1. Analyze ( f(x) ): - ( f(x) = arcsin(cos x) ). - The function ( cos x ) has a period of ( 2pi ). - ( arcsin(cos x) ) is a piecewise linear function with period ( 2pi ).2. Simplify ( f(f(x)) ): - ( f(f(x)) = arcsin(cos(arcsin(cos x))) ). - Using the identity ( cos(arcsin(z)) = sqrt{1 - z^2} ), we get ( cos(arcsin(cos x)) = |sin x| ). - Therefore, ( f(f(x)) = arcsin(|sin x|) ). - The function ( arcsin(|sin x|) ) has a period of ( pi ).3. Simplify ( f(f(f(x)) ): - ( f(f(f(x))) = arcsin(cos(arcsin(|sin x|))) ). - Using the identity ( cos(arcsin(z)) = sqrt{1 - z^2} ), we get ( cos(arcsin(|sin x|)) = |cos x| ). - Therefore, ( f(f(f(x))) = arcsin(|cos x|) ). - The function ( arcsin(|cos x|) ) has a period of ( pi ).4. Verify the period: - The function ( arcsin(|cos x|) ) is analyzed over intervals and found to repeat every ( pi ). - Numerical checks confirm that the function values repeat every ( pi ).Thus, the smallest positive period of ( f(f(f(x))) ) is (boxed{pi}).,<think>Okay, let's try to figure out the smallest positive period of f(f(f(x))) where f(x) = arcsin(cos x). Hmm, this seems a bit tricky, but I'll take it step by step.First, I need to understand what the function f(x) = arcsin(cos x) looks like. Arcsin is the inverse sine function, right? Its domain is [-1, 1] and its range is [-π/2, π/2]. But cos x has a range of [-1, 1], so that's good because it fits into the domain of arcsin. So f(x) is defined for all real numbers x.Now, let's think about the periodicity of f(x). The cosine function, cos x, has a period of 2π. But since arcsin is being applied to it, does that affect the period? Let me check. Suppose I shift x by 2π. Then cos(x + 2π) = cos x, so arcsin(cos(x + 2π)) = arcsin(cos x), so f(x) is 2π periodic. But maybe there's a smaller period? Let's see.What if I shift x by π? Then cos(x + π) = -cos x. So arcsin(cos(x + π)) = arcsin(-cos x) = -arcsin(cos x), because arcsin is an odd function. So f(x + π) = -f(x). So if I apply f twice, maybe that would affect the period. Hmm, interesting.So f(x + π) = -f(x). Let's verify that. Let's take f(x + π) = arcsin(cos(x + π)) = arcsin(-cos x). Since arcsin(-y) = -arcsin y, that's correct. So f(x + π) = -f(x). Therefore, if I apply f again, f(f(x + π)) = f(-f(x)). Let's see what f(-f(x)) is.Since f is arcsin(cos(-f(x))) = arcsin(cos(f(x))), because cosine is even. So f(-f(x)) = arcsin(cos(f(x))). But f(x) is arcsin(cos x), so cos(f(x)) = cos(arcsin(cos x)). Let me compute that.Let’s denote θ = arcsin(cos x). Then sin θ = cos x. We know that cos θ = sqrt(1 - sin²θ) = sqrt(1 - cos²x) = |sin x|. But θ is in the range of arcsin, which is [-π/2, π/2], so cos θ is non-negative. Therefore, cos(arcsin(cos x)) = |sin x|. Wait, but that seems like a useful identity. So cos(f(x)) = |sin x|.Therefore, f(-f(x)) = arcsin(|sin x|). Hmm, interesting. So f(f(x + π)) = arcsin(|sin x|). Let's note that.But maybe I need to consider the composition f(f(f(x))) step by step. Let's start by understanding f(x), then f(f(x)), then f(f(f(x))), and see their periods.Starting with f(x) = arcsin(cos x). As we saw, f(x) has period 2π, but f(x + π) = -f(x). So applying f again, let's compute f(f(x + π)) = f(-f(x)) = arcsin(cos(-f(x))) = arcsin(cos(f(x))) since cosine is even. As before, cos(f(x)) = |sin x|, so f(f(x + π)) = arcsin(|sin x|).But arcsin(|sin x|) is equal to |x - π/2 - 2πn| - π/2 for x in some interval? Wait, maybe not. Let me think. If we have arcsin(|sin x|), since |sin x| is non-negative, arcsin(|sin x|) will be in [0, π/2]. Because the range of arcsin is [-π/2, π/2], but since the argument is non-negative, it will be in [0, π/2].But how does arcsin(|sin x|) behave? Let's take x in [0, π/2]. Then sin x is in [0, 1], so |sin x| = sin x, so arcsin(sin x) = x. If x is in [π/2, 3π/2], sin x is in [-1, 1], so |sin x| is in [0, 1], but arcsin(|sin x|) would give us arcsin(|sin x|) which is equivalent to arcsin(sin(π - x)) if x is in [π/2, π], but maybe I need a better approach.Alternatively, since |sin x| is symmetric and has a period of π. So arcsin(|sin x|) would inherit some periodicity. Let's check the period of arcsin(|sin x|). Since |sin x| has period π, but arcsin(|sin x|) is a composition of functions. Let's consider x in [0, π]. Then |sin x| = sin x for x in [0, π], so arcsin(sin x). But in [0, π/2], arcsin(sin x) = x. In [π/2, π], arcsin(sin x) = π - x. Wait, let's confirm:For x in [π/2, 3π/2], sin x is non-negative in [π/2, π] and non-positive in [π, 3π/2]. But since we have |sin x|, in [π/2, π], |sin x| = sin x, and in [π, 3π/2], |sin x| = -sin x. But arcsin(|sin x|) in [π, 3π/2] would be arcsin(-sin x) = arcsin(|sin x|) since we take absolute value. Wait, maybe this is getting confusing.Alternatively, let's consider the function arcsin(|sin x|). Let's plot it mentally. Since |sin x| is a π-periodic function with peaks at π/2 + kπ. The arcsin of |sin x| would take each hump of |sin x| and map it to [0, π/2]. Specifically, when |sin x| increases from 0 to 1, arcsin(|sin x|) increases from 0 to π/2, and when |sin x| decreases back to 0, arcsin(|sin x|) decreases back to 0. So the graph of arcsin(|sin x|) would look like a series of triangles going from 0 to π/2 and back to 0, repeating every π.Wait, that seems plausible. Let's take x from 0 to π. Then |sin x| = sin x, which goes from 0 to 1 and back to 0. Then arcsin(sin x) would be equal to x for x in [0, π/2], and π - x for x in [π/2, π]. So in [0, π], arcsin(|sin x|) is a triangular wave going from 0 to π/2 and back to 0. Then for x in [π, 2π], |sin x| = |sin(x - π)|, so the function is shifted by π. Therefore, arcsin(|sin x|) is indeed π-periodic. So f(f(x)) = arcsin(|sin x|) is π-periodic.Wait, but hold on, f(f(x)) is π-periodic? Let's check. If we replace x with x + π, then arcsin(|sin(x + π)|) = arcsin(|-sin x|) = arcsin(|sin x|). So yes, it's π-periodic. Therefore, f(f(x)) has period π.But is π the fundamental period? Let's check if there's a smaller period. Suppose we take T = π/2. Then arcsin(|sin(x + π/2)|) = arcsin(|cos x|). Compare this with arcsin(|sin x|). Are these the same? No, unless |cos x| = |sin x|, which is only true at certain points. So the function is not π/2-periodic. So the fundamental period is π.Therefore, f(f(x)) has period π. Now, moving on to f(f(f(x))) = f(arcsin(|sin x|)). Let's denote g(x) = arcsin(|sin x|), which is π-periodic. Then f(g(x)) = arcsin(cos(g(x))).So let's compute cos(g(x)) where g(x) = arcsin(|sin x|). Let's denote θ = g(x) = arcsin(|sin x|). Then sin θ = |sin x|, and cos θ = sqrt(1 - sin²θ) = sqrt(1 - sin²x) = |cos x|. Wait, because sin²θ = sin²x, so 1 - sin²θ = 1 - sin²x = cos²x, so cos θ = |cos x|. But θ is in [0, π/2] because arcsin(|sin x|) is in [0, π/2]. Therefore, cos θ is non-negative, so cos(arcsin(|sin x|)) = |cos x|. Therefore, f(g(x)) = arcsin(|cos x|).Therefore, f(f(f(x))) = arcsin(|cos x|). Now, we need to find the smallest positive period of arcsin(|cos x|).Let's analyze arcsin(|cos x|). Let's first consider the function |cos x|. It has a period of π, since cos(x + π) = -cos x, and absolute value makes it π-periodic. Then, arcsin(|cos x|) would be the composition of arcsin with |cos x|.Similarly to the previous step, arcsin(|cos x|) would be a function that takes |cos x|, which ranges between 0 and 1, and maps it via arcsin to [0, π/2]. Let's see how this behaves.Take x in [0, π/2]. Then |cos x| = cos x, decreasing from 1 to 0. So arcsin(cos x) would decrease from π/2 to 0. In x = 0, arcsin(1) = π/2. In x = π/2, arcsin(0) = 0.For x in [π/2, 3π/2], |cos x| = |cos(x)|, which is symmetric. Let's consider x in [π/2, π]. Here, cos x is negative, so |cos x| = -cos x. Then arcsin(|cos x|) = arcsin(-cos x) = arcsin(|cos x|). Wait, but arcsin(|cos x|) is the same as arcsin(|cos(x)|), which is the same as in [0, π/2], mirrored.Wait, perhaps it's better to see the periodicity. Since |cos x| is π-periodic, arcsin(|cos x|) would also be π-periodic. But let's check if it has a smaller period.Suppose we shift x by π/2. Then |cos(x + π/2)| = |sin x|. So arcsin(|cos(x + π/2)|) = arcsin(|sin x|). But arcsin(|sin x|) is different from arcsin(|cos x|). For example, at x = 0, arcsin(|cos 0|) = arcsin(1) = π/2, whereas arcsin(|sin 0|) = arcsin(0) = 0. So shifting by π/2 doesn't give the same function. Therefore, π is the period.But wait, maybe the function arcsin(|cos x|) has a period smaller than π? Let's check for T = π/2. Let's compute arcsin(|cos(x + π/2)|) = arcsin(|cos x cos π/2 - sin x sin π/2|) = arcsin(|-sin x|) = arcsin(|sin x|). Which is different from arcsin(|cos x|). So shifting by π/2 gives a different function. Therefore, the period is not π/2.But maybe the function is π-periodic. Let's check if arcsin(|cos(x + π)|) = arcsin(|-cos x|) = arcsin(|cos x|). Yes, because absolute value negates the negative sign. Therefore, arcsin(|cos(x + π)|) = arcsin(|cos x|), so it's π-periodic. Therefore, the period is π. But is that the fundamental period? Let's check if there's a smaller period.Suppose T = π/2. Then arcsin(|cos(x + π/2)|) = arcsin(|sin x|). Which is not the same as arcsin(|cos x|). Therefore, T = π is the fundamental period of arcsin(|cos x|). Wait, but hold on. Let's consider another approach.Let me graph arcsin(|cos x|) mentally. Since |cos x| is a π-periodic function, peaking at 0, π, 2π, etc., and reaching 0 at π/2, 3π/2, etc. The arcsin of |cos x| would take each of these peaks and convert them into a triangular wave similar to before, but shifted. Let's see:For x in [-π/2, π/2], |cos x| = cos x, which decreases from 0 to π/2, reaches 1 at x=0, and decreases back to 0. Then arcsin(cos x) would be equal to π/2 - |x|. Wait, for x in [-π/2, π/2], arcsin(cos x) = π/2 - |x|. Because when x = 0, arcsin(1) = π/2; when x approaches π/2, cos x approaches 0, so arcsin(0) = 0. So the graph is a V-shape between -π/2 and π/2, peaking at π/2 and going down to 0 at the ends.But since |cos x| is π-periodic, this pattern would repeat every π. Wait, but actually, if we consider x beyond π/2, say in [π/2, 3π/2], |cos x| = |cos(x)|, which is the same as |cos(x - π)|, so the function would be symmetric around π. So arcsin(|cos x|) would have this V-shape in each interval of length π, centered at multiples of π.Therefore, the function arcsin(|cos x|) is indeed π-periodic. But does it have a smaller period? Let's test T = π/2. If we take x = 0, arcsin(|cos 0|) = π/2. x = π/2, arcsin(|cos π/2|) = 0. x = π, arcsin(|cos π|) = π/2. x = 3π/2, arcsin(|cos 3π/2|) = 0. x = 2π, arcsin(1) = π/2. If we take T = π/2, then f(x + π/2) = arcsin(|cos(x + π/2)|) = arcsin(|sin x|). Comparing f(x) = arcsin(|cos x|) and f(x + π/2) = arcsin(|sin x|), these are different functions. For example, at x = 0, f(0) = π/2, f(0 + π/2) = arcsin(|sin 0|) = 0. At x = π/4, f(π/4) = arcsin(√2/2) = π/4, and f(π/4 + π/2) = arcsin(|sin 3π/4|) = arcsin(√2/2) = π/4. Wait, so at x = π/4, they are the same? Hmm.Wait, maybe the functions f(x) and f(x + π/2) are actually the same? Let me check more points.At x = π/2: f(π/2) = arcsin(|cos π/2|) = arcsin(0) = 0. f(π/2 + π/2) = f(π) = arcsin(|cos π|) = arcsin(1) = π/2.Wait, so f(x + π/2) is not equal to f(x). Therefore, shifting by π/2 does not give the same function, so the period is not π/2. Therefore, the fundamental period of arcsin(|cos x|) is π.But wait, let's check another point. Let’s take x = π/4. Then f(x) = arcsin(cos(π/4)) = arcsin(√2/2) = π/4. If we take x = π/4 + π = 5π/4. Then f(5π/4) = arcsin(|cos(5π/4)|) = arcsin(√2/2) = π/4. But cos(5π/4) = -√2/2, so |cos(5π/4)| = √2/2. So arcsin(√2/2) is π/4. So f(5π/4) = π/4. But f(π/4 + π) = π/4, same as f(π/4). So π is indeed a period.But is there a smaller period? Let's check T = π/2. For example, take x = 0: f(0) = π/2, f(0 + π/2) = 0, f(0 + π) = π/2. So over π/2, the function goes from π/2 to 0 and back to π/2. Wait, but if we take T = π, then the function repeats every π. If we take T = π/2, the function would need to satisfy f(x + π/2) = f(x), but as shown above, f(0 + π/2) = 0 ≠ f(0) = π/2. Therefore, π is indeed the fundamental period.Therefore, f(f(f(x))) = arcsin(|cos x|) is π-periodic. But the problem states that we need the smallest positive period of f(f(f(x))). But let's make sure that there isn't a smaller period. Let's suppose that there is a smaller period T, such that T divides π. Let's test T = π/2.Suppose T = π/2 is a period. Then arcsin(|cos(x + π/2)|) = arcsin(|sin x|) should equal arcsin(|cos x|). But as we saw before, these are different functions. For example, at x = 0, arcsin(|cos 0|) = π/2, and arcsin(|sin 0|) = 0. Hence, T = π/2 is not a period. How about T = π/3? Let's check x = 0: arcsin(|cos(0 + π/3)|) = arcsin(cos(π/3)) = arcsin(0.5) = π/6, which is not equal to arcsin(|cos 0|) = π/2. So T = π/3 isn't a period. Therefore, π is indeed the fundamental period.Wait, but hold on. Let's think again. Maybe arcsin(|cos x|) has a different fundamental period. Let's consider the function arcsin(|cos x|). Let's see its graph.Since |cos x| is π-periodic and even, arcsin(|cos x|) would also be π-periodic and even. Let's analyze the function over [0, π/2] and see how it behaves.For x in [0, π/2], |cos x| = cos x. So arcsin(cos x) = π/2 - x. For x in [π/2, π], |cos x| = -cos x (since cos x is negative here), so |cos x| = -cos x. Then arcsin(|cos x|) = arcsin(-cos x). But wait, arcsin(|cos x|) = arcsin(-cos x) only if we consider x in [π/2, π]. But arcsin is applied to |cos x|, which is a positive value. So arcsin(|cos x|) is equal to arcsin(|cos(π - x)|) because |cos x| = |cos(π - x)|. So for x in [π/2, π], let’s set y = π - x, which is in [0, π/2]. Then arcsin(|cos x|) = arcsin(|cos y|) = π/2 - y = π/2 - (π - x) = x - π/2. So in [π/2, π], arcsin(|cos x|) = x - π/2.Therefore, in [0, π/2], the function is π/2 - x, decreasing from π/2 to 0. In [π/2, π], it's x - π/2, increasing from 0 to π/2. Then in [π, 3π/2], since |cos x| = |cos(x - π)|, it's the same as the previous interval. So the function is a triangular wave going from π/2 to 0 to π/2 every π. Wait, no, in [0, π], it goes from π/2 to 0 to π/2, forming a V-shape. Wait, no:Wait, in [0, π/2], arcsin(|cos x|) = π/2 - x, which goes from π/2 to 0.In [π/2, π], arcsin(|cos x|) = x - π/2, which goes from 0 to π/2.So over [0, π], the function goes from π/2 down to 0 and back up to π/2. Then in [π, 2π], it repeats the same pattern. So the function is indeed π-periodic because after π, it repeats the same shape.But if we look at the graph, from 0 to π, it's a symmetric V-shape. Therefore, is the function also 2π periodic? Well, yes, because π divides 2π, but the fundamental period is π.However, wait, if the function is symmetric around π/2, does that introduce a smaller period? Let's see. For example, if we shift x by π/2, does the function repeat? Let's take x = 0: arcsin(|cos 0|) = π/2. x = 0 + π/2 = π/2: arcsin(|cos π/2|) = 0. x = π/2 + π/2 = π: arcsin(|cos π|) = π/2. x = π + π/2 = 3π/2: arcsin(|cos 3π/2|) = 0. x = 3π/2 + π/2 = 2π: arcsin(|cos 2π|) = π/2. So shifting by π/2 alternates between π/2 and 0. So the function doesn't repeat every π/2; instead, it alternates between high and low values. Therefore, π is still the fundamental period.Therefore, f(f(f(x))) = arcsin(|cos x|) has a fundamental period of π. Hence, the smallest positive period is π.Wait, but let me double-check. Let's suppose that T is the period, so that arcsin(|cos(x + T)|) = arcsin(|cos x|) for all x. We need the smallest T > 0 such that |cos(x + T)| = |cos x| for all x. Wait, but |cos(x + T)| = |cos x| implies that T is a period of |cos x|, which is π. Therefore, the minimal period is π. Therefore, f(f(f(x))) has period π.But wait, the question is about f(f(f(x))). Wait, but when we did f(f(f(x))), we ended up with arcsin(|cos x|). But is that correct?Let me recap:f(x) = arcsin(cos x). Then f(f(x)) = arcsin(|sin x|). Then f(f(f(x))) = arcsin(|cos x|). Wait, how did we get from f(f(x)) to f(f(f(x)))?Yes:f(f(f(x))) = f(arcsin(|sin x|)) = arcsin(cos(arcsin(|sin x|))). Then we found that cos(arcsin(|sin x|)) = |cos x|. Therefore, f(f(f(x))) = arcsin(|cos x|). So that's correct.Therefore, the final function is arcsin(|cos x|), which has period π.Therefore, the smallest positive period is π.But hold on, the original problem says "the smallest positive period of f(f(f(x)))". But according to our deduction, it's π. However, let me check an example. Take x = 0: f(f(f(0))) = arcsin(|cos 0|) = arcsin(1) = π/2. x = π: f(f(f(π))) = arcsin(|cos π|) = arcsin(1) = π/2. x = π/2: f(f(f(π/2))) = arcsin(|cos π/2|) = arcsin(0) = 0. x = 3π/2: f(f(f(3π/2))) = arcsin(|cos 3π/2|) = arcsin(0) = 0. So the function has the same value at x and x + π. For example, x = 0 and x = π both give π/2. x = π/2 and x = 3π/2 both give 0. So the period is indeed π.But let's check intermediate points. For example, x = π/4: f(f(f(π/4))) = arcsin(|cos π/4|) = arcsin(√2/2) = π/4. x = π/4 + π = 5π/4: f(f(f(5π/4))) = arcsin(|cos 5π/4|) = arcsin(√2/2) = π/4. So it's the same. x = π/3: arcsin(|cos π/3|) = arcsin(0.5) = π/6. x = π/3 + π = 4π/3: arcsin(|cos 4π/3|) = arcsin(0.5) = π/6. So same value. Therefore, it's indeed π-periodic.But is there a smaller period? Let's suppose T = π/2. Check x = 0: arcsin(|cos 0|) = π/2. x = 0 + π/2 = π/2: arcsin(|cos π/2|) = 0. x = π/2 + π/2 = π: arcsin(|cos π|) = π/2. x = π + π/2 = 3π/2: arcsin(|cos 3π/2|) = 0. x = 3π/2 + π/2 = 2π: arcsin(|cos 2π|) = π/2. So the function alternates between π/2 and 0 every π/2. Therefore, the function does not repeat every π/2; instead, it requires a full π to repeat. Therefore, the fundamental period is π.Therefore, the answer should be π. But wait, the problem statement says "the smallest positive period of f(f(f(x)))". So according to our analysis, it's π. But I need to make sure that there's no smaller period. Suppose someone claims that the period is π/2. Let's test x = 0 and x = π/2. f(f(f(0))) = π/2, f(f(f(π/2))) = 0. If the period were π/2, then f(f(f(π/2))) should equal f(f(f(0))), which it doesn't. Therefore, π/2 is not a period.Alternatively, maybe there's a period smaller than π but not dividing π? For example, T = π/3. Let's check x = 0: f(f(f(0))) = π/2. x = π/3: f(f(f(π/3))) = arcsin(|cos π/3|) = arcsin(0.5) = π/6. x = 2π/3: arcsin(|cos 2π/3|) = arcsin(0.5) = π/6. x = π: π/2. x = 4π/3: π/6. x = 5π/3: π/6. x = 2π: π/2. So the values are π/2, π/6, π/6, π/2, π/6, π/6, π/2. Clearly, there's no repetition every π/3. Therefore, the period is indeed π.Hence, the smallest positive period of f(f(f(x))) is π.Final AnswerThe smallest positive period of ( f(f(f(x))) ) is (boxed{pi}).</think>Given the function ( f(x) = arcsin(cos x) ), we need to determine the smallest positive period of ( f(f(f(x))) ).1. Understanding ( f(x) ): - ( f(x) = arcsin(cos x) ). - The cosine function has a period of ( 2pi ), but ( f(x) ) has a smaller period due to the properties of the arcsin function. - It is found that ( f(x + pi) = -f(x) ), indicating a period of ( pi ).2. Analyzing ( f(f(x)) ): - ( f(f(x)) = arcsin(|sin x|) ). - The function ( |sin x| ) has a period of ( pi ), so ( arcsin(|sin x|) ) also has a period of ( pi ).3. Analyzing ( f(f(f(x)) ): - ( f(f(f(x))) = arcsin(|cos x|) ). - The function ( |cos x| ) has a period of ( pi ), so ( arcsin(|cos x|) ) also has a period of ( pi ).4. Verification of the Fundamental Period: - Checking smaller periods like ( pi/2 ) and ( pi/3 ) shows they do not satisfy the periodicity condition. - The function ( arcsin(|cos x|) ) is verified to be ( pi )-periodic by evaluating key points and intervals.Thus, the smallest positive period of ( f(f(f(x))) ) is (boxed{pi}).]

🔑:-1 . pi.From the fact that cos x is an even function, we know that f(x) is an even function.begin{array}{l}text { and } f(x+pi)=arcsin [cos (x+pi)] =arcsin (-cos x)=-arcsin (cos x) =-f(x),end{array}Therefore, f(f(x+pi))=f(-f(x))=f(f(x)). Thus, the period of f(f(x)) is pi.Hence, the smallest positive period of f(f(f(x))) is pi.

🧠:[<think>Okay, so I have this geometry problem here. Let me try to parse it step by step. The problem says: In triangle ABC, AB equals BC, and the cosine of angle B is -7/18. There's an ellipse with foci at points A and B that passes through point C. We need to find the eccentricity of this ellipse. Alright, let's break this down.First, triangle ABC is isosceles because AB equals BC. Wait, AB = BC? That means the triangle has two sides equal: AB and BC. So, the triangle is isosceles with base AC and equal sides AB and BC. Wait, but in a triangle, if two sides are equal, the angles opposite them are equal. So, if AB = BC, then angles at A and C are equal? Wait, but AB is a side, BC is another side. Wait, let me clarify.Wait, in triangle ABC, vertices are A, B, C. So AB is a side, BC is another side, and AC is the third side. So if AB = BC, then sides AB and BC are equal. Therefore, the triangle is isosceles with apex at point B? Because sides AB and BC are the two equal sides, so the base is AC. Therefore, the base angles at A and C would be equal. Hmm, okay. So angle at B is between sides AB and BC, which are equal. So angle B is the vertex angle, and angles at A and C are equal. Got it.Now, we are given that cos B = -7/18. So angle B has a cosine of -7/18. Since cosine is negative, angle B must be obtuse because cosine is negative in the second quadrant (angles between 90 and 180 degrees). So angle B is greater than 90 degrees. That makes sense because in an isosceles triangle, if the vertex angle is obtuse, then the base angles would be acute.Now, there's an ellipse with foci at points A and B that passes through point C. We need to find the eccentricity of the ellipse. Okay, so let's recall some properties of ellipses.An ellipse is defined as the set of points where the sum of the distances from any point on the ellipse to the two foci is constant. That constant is equal to the major axis length, 2a, where a is the semi-major axis. The distance between the two foci is 2c, where c is the focal distance. The eccentricity e of the ellipse is given by e = c/a. So our goal is to find e = c/a.Given that the ellipse passes through point C, the sum of the distances from C to A and C to B must equal 2a. So, in this case, CA + CB = 2a. But wait, in triangle ABC, AB = BC, so BC = AB. Let's denote AB = BC = x, and AC = y. Then, in triangle ABC, we know sides AB = BC = x, angle at B is angle B, with cos B = -7/18.So maybe we can use the Law of Cosines to relate the sides. Law of Cosines says that in a triangle, c² = a² + b² - 2ab cos C. So here, if we consider angle at B, then the sides adjacent to angle B are AB and BC, which are both equal to x, and the side opposite angle B is AC = y. Therefore, by the Law of Cosines:AC² = AB² + BC² - 2 * AB * BC * cos BWhich becomes:y² = x² + x² - 2 * x * x * (-7/18)Simplify:y² = 2x² - 2x²*(-7/18) = 2x² + (14/18)x² = 2x² + (7/9)x²Convert 2 to 18/9:y² = (18/9)x² + (7/9)x² = (25/9)x²Therefore, y = (5/3)x. So AC is (5/3)x. So the sides of the triangle are AB = BC = x, and AC = (5/3)x.Now, going back to the ellipse. The foci are at A and B, and the ellipse passes through point C. So for point C, the sum of distances to A and B is CA + CB. But CB is equal to AB, which is x. Wait, CB is equal to AB? Wait, AB = BC = x, so CB is x. Then CA is (5/3)x, so the sum CA + CB is (5/3)x + x = (8/3)x. Therefore, 2a = (8/3)x, so a = (4/3)x.Now, the distance between the foci is AB. Wait, the foci are at A and B, so the distance between them is AB, which is x. Therefore, 2c = x, so c = x/2.Therefore, the eccentricity e is c/a = (x/2)/(4x/3) = (1/2)/(4/3) = (1/2)*(3/4) = 3/8. Hmm, so e = 3/8. Wait, is this correct?Wait, hold on. Let me check my steps again.First, sides AB = BC = x. Then, by Law of Cosines, AC² = AB² + BC² - 2*AB*BC*cos B. Since cos B is -7/18, the formula becomes AC² = x² + x² - 2x²*(-7/18) = 2x² + (14/18)x² = 2x² + (7/9)x² = (18/9 + 7/9)x² = 25/9 x², so AC = 5x/3. That seems correct.Then, for the ellipse passing through C, sum of distances from C to A and C to B is 2a. So CA + CB = 5x/3 + x = 8x/3, so 2a = 8x/3, hence a = 4x/3. Then, the distance between foci is AB = x, so 2c = x, so c = x/2. Therefore, e = c/a = (x/2)/(4x/3) = 3/8. So 3/8. That seems straightforward, but let me verify once more.Wait, but in the problem statement, it's an ellipse with foci at A and B passing through C. So the sum CA + CB = 2a, which is 8x/3, so a = 4x/3. The distance between foci is AB = x, so c = x/2. Then e = c/a = (x/2)/(4x/3) = (1/2)/(4/3) = 3/8. Hmm. So 3/8 is the answer. That seems right.Wait, but perhaps I made an error in assuming that AB is the distance between foci? Yes, foci are at A and B, so the distance between them is AB, which is x, so 2c = x, hence c = x/2. Then, a is 4x/3, so e = c/a = (x/2)/(4x/3) = 3/8. Yes.But let me think again. Is there another way to approach this problem? Let's consider coordinate geometry. Maybe assign coordinates to the triangle ABC and compute the required parameters.Let me try setting up coordinates. Let me place point B at the origin (0,0). Since AB = BC = x, and angle at B is angle with cos B = -7/18. Let's set up the triangle in such a way that points A and C are placed symmetrically with respect to the axes.Since AB = BC = x, let's place point A at (d, e) and point C at (d, -e) so that BA = BC. But angle at B is given, so perhaps another coordinate system.Alternatively, place point B at the origin, and let’s have AB along the x-axis. But since angle B is between AB and BC, which are both length x. Wait, maybe this approach is better.Let me place point B at (0,0). Let’s place point A somewhere on the x-axis, say at (k, 0). Then, since AB = BC = x, point C should be somewhere such that the distance from B to C is x. However, angle at B is given, so perhaps using vectors or coordinate geometry.Alternatively, since AB = BC = x, and angle at B is θ with cos θ = -7/18, we can place point B at the origin, point A along the positive x-axis at (x, 0). Then point C will be somewhere in the plane such that BC = x and angle at B is θ.Using the Law of Cosines, as we did earlier, gives AC = 5x/3. Then, coordinates might help. Let me try.Let’s set point B at (0,0). Point A is at (x, 0). Then, point C is such that BC = x and angle at B is θ with cos θ = -7/18. Let's find coordinates of point C.Since angle at B is between BA and BC vectors. BA is from B to A: (x, 0) - (0,0) = (x, 0). BC is from B to C: let’s denote point C as (a, b). Then vector BC is (a, b). The angle between BA and BC is θ, so:cos θ = (BA · BC) / (|BA| |BC|) = (x * a + 0 * b) / (x * x) = a/x = -7/18Therefore, a = (-7/18)x.Since BC has length x, the coordinates of point C must satisfy a² + b² = x². So:(-7x/18)² + b² = x²Compute (49x²)/324 + b² = x²Therefore, b² = x² - 49x²/324 = (324x² - 49x²)/324 = (275x²)/324Thus, b = ± (sqrt(275)x)/18. Let's take positive b for simplicity, so point C is at (-7x/18, sqrt(275)x/18).Therefore, coordinates:A: (x, 0)B: (0, 0)C: (-7x/18, sqrt(275)x/18)Now, we need to compute CA and CB. Wait, CB is the distance from C to B, which is x, as given. CA is the distance from C to A.Compute CA:Distance between (x, 0) and (-7x/18, sqrt(275)x/18):Δx = x - (-7x/18) = x + 7x/18 = 25x/18Δy = 0 - sqrt(275)x/18 = -sqrt(275)x/18Therefore, CA = sqrt[(25x/18)^2 + (sqrt(275)x/18)^2]Calculate:(25x/18)^2 = (625x²)/324(sqrt(275)x/18)^2 = (275x²)/324Sum: (625 + 275)/324 x² = 900/324 x² = (900 ÷ 36)/(324 ÷ 36) x² = 25/9 x²Thus, CA = sqrt(25/9 x²) = 5x/3, which matches the previous result. Good.So, sum CA + CB = 5x/3 + x = 8x/3, so 2a = 8x/3, so a = 4x/3. Correct.Distance between foci is AB. AB is the distance between A(x, 0) and B(0,0), which is x. Therefore, 2c = x, so c = x/2.Therefore, eccentricity e = c/a = (x/2)/(4x/3) = (1/2)/(4/3) = 3/8. So e = 3/8.Therefore, the answer is 3/8, which in boxed form is boxed{dfrac{3}{8}}.Wait, let me just think if there's any possible mistake here. For example, is the sum CA + CB indeed equal to 2a? Yes, because in an ellipse, the sum of the distances from any point on the ellipse to the two foci is 2a. Since point C is on the ellipse, that sum is 2a. Therefore, that's correct.Also, the distance between the foci is 2c. Here, the distance between A and B is AB, which is x, so 2c = x, hence c = x/2. Then, using a = 4x/3, e = c/a = (x/2)/(4x/3) = 3/8. Correct. So seems like all steps check out. Therefore, the eccentricity is 3/8.Final Answerboxed{dfrac{3}{8}}</think>In (triangle ABC), (AB = BC) and (cos B = -dfrac{7}{18}). An ellipse with foci at (A) and (B) passes through (C). We need to find the eccentricity of the ellipse.1. Identify the triangle properties: - (triangle ABC) is isosceles with (AB = BC = x). - Using the Law of Cosines for (cos B = -dfrac{7}{18}): [ AC^2 = AB^2 + BC^2 - 2 cdot AB cdot BC cdot cos B ] [ AC^2 = x^2 + x^2 - 2x^2 left(-dfrac{7}{18}right) ] [ AC^2 = 2x^2 + dfrac{14x^2}{18} = dfrac{36x^2}{18} + dfrac{14x^2}{18} = dfrac{50x^2}{18} = dfrac{25x^2}{9} ] [ AC = dfrac{5x}{3} ]2. Ellipse properties: - The sum of distances from any point on the ellipse to the foci is (2a). - For point (C), the sum (CA + CB = dfrac{5x}{3} + x = dfrac{8x}{3}). Thus, (2a = dfrac{8x}{3}) and (a = dfrac{4x}{3}). - The distance between the foci (A) and (B) is (AB = x), so (2c = x) and (c = dfrac{x}{2}).3. Eccentricity calculation: - Eccentricity (e = dfrac{c}{a}): [ e = dfrac{dfrac{x}{2}}{dfrac{4x}{3}} = dfrac{x}{2} cdot dfrac{3}{4x} = dfrac{3}{8} ]The eccentricity of the ellipse is (boxed{dfrac{3}{8}}).,<think>Okay, let me try to figure out this problem step by step. So, we have triangle ABC where AB equals BC, and the cosine of angle B is -7/18. There's an ellipse with foci at A and B that passes through point C. We need to find the eccentricity of this ellipse. Hmm, okay. Let's start by recalling some properties of ellipses and triangles.First, in an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant and equal to the major axis length, 2a. The distance between the foci is 2c, and the eccentricity e is c/a. So, to find the eccentricity, we need to find c and a.Given that the ellipse passes through point C, the sum of the distances from C to A and C to B should be 2a. Also, since the foci are at A and B, the distance between A and B is 2c. Therefore, if we can find the lengths of AB, BC, and AC, we can compute both 2a (which is AC + BC) and 2c (which is AB). Then, e = c/a = (AB/2) / ((AC + BC)/2) = AB / (AC + BC). Wait, but since AB = BC, as given in the problem, that might simplify things. Let me check.Wait, the problem says AB = BC. So triangle ABC is isosceles with AB = BC. That means sides AB and BC are equal, so the triangle has two equal sides. Let me draw this triangle mentally. Vertex B is the vertex angle, and AB and BC are the equal sides. So angle B is the vertex angle. The cosine of angle B is given as -7/18. Since cosine is negative, angle B must be obtuse, right? Because cosine is negative in the second quadrant, which is angles between 90 and 180 degrees.So, let's denote AB = BC = x. Then, since the triangle is isosceles with AB = BC, the sides AB and BC are equal, and the base is AC. Let me assign coordinates to the triangle to make it easier. Maybe place point B at the origin, and points A and C on a coordinate plane such that BA = BC = x, and angle B is known. Let's see.Let me set point B at (0, 0). Since AB = BC = x, and angle B is between AB and BC, which are both length x. Let me place point A at (x, 0) on the x-axis. Then, since AB = BC = x, point C should be somewhere such that the distance from B(0,0) to C is x, and angle at B is given with cosine -7/18. Wait, but angle between BA and BC is angle B. So if BA is along the positive x-axis from B to A, then BC would make an angle of, say, θ with BA, where θ is angle B. But cosine θ is -7/18, which is negative, so θ is greater than 90 degrees.Alternatively, maybe coordinate system setup can help. Let me try to position point B at the origin (0,0). Let me have BA along the positive x-axis, so point A is at (x, 0). Then, since angle at B is θ with cos θ = -7/18, and BC = AB = x, point C would be located at a position such that the vector from B to C makes an angle θ with the vector from B to A. Since θ is obtuse, point C would be in the second quadrant relative to point B. Wait, but since BA is along the positive x-axis, and angle between BA and BC is θ, which is obtuse, then point C should be in the second quadrant. But BC is length x, so coordinates of C can be (x cos θ, x sin θ). Let me verify.Yes, if we take BA as the positive x-axis, then the coordinates of point C relative to B would be (x cos θ, x sin θ), where θ is angle ABC. Since cos θ = -7/18, sin θ would be sqrt(1 - cos²θ) = sqrt(1 - 49/324) = sqrt(275/324) = sqrt(275)/18. Since θ is in the second quadrant, sin θ is positive. So coordinates of C would be (x*(-7/18), x*(sqrt(275)/18)) = (-7x/18, sqrt(275)x/18).Then, point A is at (x, 0), point B is at (0,0), point C is at (-7x/18, sqrt(275)x/18). Then, we can compute the length AC. Let's compute AC.Coordinates of A: (x, 0)Coordinates of C: (-7x/18, sqrt(275)x/18)So, vector AC is (-7x/18 - x, sqrt(275)x/18 - 0) = (-25x/18, sqrt(275)x/18)Therefore, length AC is sqrt[ (-25x/18)^2 + (sqrt(275)x/18)^2 ].Let me compute that:(-25x/18)^2 = (625x²)/324(sqrt(275)x/18)^2 = (275x²)/324Total: (625 + 275)x² / 324 = 900x² / 324 = (900/324)x² = (25/9)x²So sqrt of that is (5/3)xTherefore, length AC is (5/3)x.So, AB = BC = x, and AC = (5/3)x.Now, the ellipse has foci at A and B, passing through point C. For an ellipse, the sum of distances from any point on the ellipse to the foci is 2a. Since point C is on the ellipse, then AC + BC = 2a.But BC = x, and AC = (5/3)x, so AC + BC = (5/3)x + x = (8/3)x. Therefore, 2a = (8/3)x, so a = (4/3)x.The distance between the foci is AB, which is x. Wait, AB is the distance between A and B. Since AB = x, then 2c = x, so c = x/2.Therefore, the eccentricity e = c/a = (x/2) / (4x/3) = (x/2) * (3/(4x)) = 3/8.Wait, so the eccentricity is 3/8. Let me check that again to make sure I didn't make a mistake.First, coordinates:Point B at (0,0), point A at (x,0), point C at (-7x/18, sqrt(275)x/18). Then AC is distance between (x,0) and (-7x/18, sqrt(275)x/18). The x-coordinate difference is x - (-7x/18) = x + 7x/18 = 25x/18. Wait, hold on! Wait, earlier I thought vector AC was (-7x/18 - x, ...), but actually, the coordinates of C are (-7x/18, sqrt(275)x/18), so the x-coordinate of C is -7x/18. The x-coordinate of A is x. So the difference in x-coordinates is x - (-7x/18) = x + 7x/18 = (18x + 7x)/18 = 25x/18. Similarly, the difference in y-coordinates is 0 - sqrt(275)x/18 = -sqrt(275)x/18. Therefore, the vector AC is (25x/18, -sqrt(275)x/18). Then the length AC is sqrt[(25x/18)^2 + (sqrt(275)x/18)^2].Wait, this contradicts what I had earlier. Wait, earlier I thought point C was at (-7x/18, sqrt(275)x/18), so when subtracting A's coordinates (x,0) from C's coordinates, it's (-7x/18 - x, sqrt(275)x/18 - 0) = (-25x/18, sqrt(275)x/18). Wait, no. Wait, point A is at (x, 0), point C is at (-7x/18, sqrt(275)x/18). So the vector from A to C is (-7x/18 - x, sqrt(275)x/18 - 0) = (-25x/18, sqrt(275)x/18). Therefore, the distance AC is sqrt[(-25x/18)^2 + (sqrt(275)x/18)^2]. So (-25x/18)^2 is (625x²)/324, and (sqrt(275)x/18)^2 is (275x²)/324. Adding them gives 900x²/324 = 25x²/9. sqrt of that is 5x/3. So AC is indeed 5x/3. Then, BC is x, since BC = AB = x. Therefore, AC + BC = 5x/3 + x = 8x/3, so 2a = 8x/3, hence a = 4x/3.Distance between foci is AB = x, so 2c = x, hence c = x/2. Then, eccentricity e = c/a = (x/2)/(4x/3) = (x/2)*(3/(4x)) = 3/8. So e = 3/8. That seems right.Wait, but let me check again. Since in an ellipse, the sum of distances from any point on the ellipse to the foci is 2a. Here, point C is on the ellipse, so AC + BC = 2a. Since BC = x, AC = 5x/3, so sum is x + 5x/3 = 8x/3 = 2a => a = 4x/3. Then, the distance between foci is AB = x, so 2c = x => c = x/2. Therefore, e = c/a = (x/2)/(4x/3) = 3/8. So yes, 3/8. That seems correct.But let me verify the triangle calculations again. In triangle ABC, AB = BC = x, angle at B is θ with cos θ = -7/18. Then, by the Law of Cosines, AC² = AB² + BC² - 2*AB*BC*cos θ. Wait, since AB = BC = x, then AC² = x² + x² - 2*x*x*cos θ = 2x² - 2x²*(-7/18) = 2x² + (14x²)/18 = 2x² + (7x²)/9 = (18x² + 7x²)/9 = 25x²/9. Therefore, AC = 5x/3. That's consistent with what I found earlier. So AC is indeed 5x/3. Therefore, sum AC + BC is 5x/3 + x = 8x/3. So 2a = 8x/3, so a = 4x/3. Distance between foci is AB = x, so 2c = x => c = x/2. Therefore, e = c/a = (x/2)/(4x/3) = 3/8. So the eccentricity is 3/8.Therefore, the answer is 3/8.Final AnswerThe eccentricity of the ellipse is boxed{dfrac{3}{8}}.</think>Given (triangle ABC) with (AB = BC) and (cos B = -dfrac{7}{18}), we need to find the eccentricity of an ellipse with foci at (A) and (B) that passes through (C).1. Identify triangle properties: - (AB = BC = x) - (cos B = -dfrac{7}{18}) implies (angle B) is obtuse. - Using the Law of Cosines to find (AC): [ AC^2 = AB^2 + BC^2 - 2 cdot AB cdot BC cdot cos B ] [ AC^2 = x^2 + x^2 - 2x^2 left(-dfrac{7}{18}right) ] [ AC^2 = 2x^2 + dfrac{14x^2}{18} = dfrac{36x^2 + 14x^2}{18} = dfrac{50x^2}{18} = dfrac{25x^2}{9} ] [ AC = dfrac{5x}{3} ]2. Ellipse properties: - The sum of distances from any point on the ellipse to the foci is (2a). For point (C): [ AC + BC = dfrac{5x}{3} + x = dfrac{8x}{3} implies 2a = dfrac{8x}{3} implies a = dfrac{4x}{3} ] - The distance between the foci (A) and (B) is (2c): [ AB = x implies 2c = x implies c = dfrac{x}{2} ]3. Eccentricity calculation: - Eccentricity (e) is given by (e = dfrac{c}{a}): [ e = dfrac{dfrac{x}{2}}{dfrac{4x}{3}} = dfrac{x}{2} cdot dfrac{3}{4x} = dfrac{3}{8} ]The eccentricity of the ellipse is (boxed{dfrac{3}{8}}).]

🔑:Let's set (AB=BC=1). Then, since (cos B=-dfrac{7}{18}), we have (AC^2=AB^2+BC^2-2ABcdot BCcdotcos B=dfrac{25}{9}), thus (AC=dfrac{5}{3}), so (2a=1+dfrac{5}{3}=dfrac{8}{3}), since (2c=1), we get (e=dfrac{2c}{2a}=dfrac{3}{8}).Therefore, the eccentricity of the ellipse is boxed{dfrac{3}{8}}.

🧠:[<think>Okay, so I have this problem here: If all vertices of a rectangular solid are on the surface of a sphere with a radius of 3, then the maximum surface area of the rectangular solid is ______. Hmm, interesting. Let me try to figure this out step by step.First off, a rectangular solid, also known as a rectangular prism, has all its vertices on the sphere. The sphere has a radius of 3, so the diameter is 6. That means the maximum distance between any two vertices of the rectangular solid can't exceed the diameter of the sphere, right? But wait, in a rectangular solid, the space diagonal is the longest distance between any two vertices. So, the space diagonal of the rectangular solid must be equal to the diameter of the sphere. Let me confirm that. If all vertices are on the sphere, then the center of the sphere is the midpoint of the space diagonal. Therefore, the space diagonal should be twice the radius, which is 6. Yep, that makes sense. So, the space diagonal of the rectangular solid is 6.Let me recall the formula for the space diagonal of a rectangular solid with side lengths a, b, and c. The space diagonal d is given by d = sqrt(a² + b² + c²). Since the space diagonal here is 6, we have sqrt(a² + b² + c²) = 6. Squaring both sides gives a² + b² + c² = 36. Got that.Now, the problem asks for the maximum surface area of such a rectangular solid. The surface area S of a rectangular solid is given by S = 2(ab + bc + ac). So, our goal is to maximize S = 2(ab + bc + ac) given that a² + b² + c² = 36.Hmm, okay, so this is an optimization problem with a constraint. I remember that Lagrange multipliers are useful for such problems. But maybe there's another way too. Alternatively, maybe using the method of symmetry or some inequality.Let me think. If we need to maximize ab + bc + ac given that a² + b² + c² is fixed. Is there a known maximum for this expression?Wait, there's the Cauchy-Schwarz inequality or AM-QM inequality. But maybe not directly. Alternatively, since ab + bc + ac is related to (a + b + c)^2. Let me recall: (a + b + c)^2 = a² + b² + c² + 2(ab + bc + ac). So, if we can express ab + bc + ac in terms of (a + b + c)^2 and a² + b² + c². Since a² + b² + c² is fixed at 36, then ab + bc + ac = [(a + b + c)^2 - 36]/2. Therefore, to maximize ab + bc + ac, we need to maximize (a + b + c)^2. So, the maximum of ab + bc + ac occurs when (a + b + c)^2 is maximized. So, we need to maximize a + b + c given that a² + b² + c² = 36.Wait, is that correct? Let me check. If ab + bc + ac = [(a + b + c)^2 - (a² + b² + c²)] / 2, then since a² + b² + c² is fixed, maximizing ab + bc + ac is equivalent to maximizing (a + b + c)^2, which in turn is equivalent to maximizing |a + b + c|. Since a, b, c are lengths, they are positive, so a + b + c is positive. Therefore, maximizing (a + b + c)^2 is the same as maximizing a + b + c.So, now the problem reduces to maximizing a + b + c given that a² + b² + c² = 36. Hmm, how do we maximize the sum of variables given the sum of their squares is fixed?This is a classic optimization problem. The maximum of a + b + c under the constraint a² + b² + c² = 36 occurs when all variables are equal. Wait, is that true? Wait, no, actually, the maximum occurs when one variable is as large as possible, and the others are zero. Wait, but if we have a² + b² + c² = 36, then if we set two variables to zero, the third would be 6. Then a + b + c would be 6. But if all variables are equal, then each would be sqrt(12) ≈ 3.464, so a + b + c ≈ 10.392. Wait, but 10.392 is larger than 6. Wait, that contradicts my previous thought.Wait, no, actually, if you have three variables, the Cauchy-Schwarz inequality tells us that (a + b + c)^2 ≤ 3(a² + b² + c²). So, (a + b + c)^2 ≤ 3*36 = 108, so a + b + c ≤ sqrt(108) = 6*sqrt(3) ≈ 10.392. This maximum is achieved when a = b = c. So, when a = b = c, which would be a cube. Then each edge would be sqrt(36/3) = sqrt(12) = 2*sqrt(3). Then a + b + c would be 6*sqrt(3), which squared is 108, matching the inequality.So, the maximum of a + b + c is 6*sqrt(3), achieved when a = b = c. Therefore, the maximum of ab + bc + ac would be [ (6*sqrt(3))² - 36 ] / 2 = (108 - 36)/2 = 72/2 = 36. Therefore, the maximum surface area is 2*36 = 72. So, the maximum surface area is 72.Wait, but wait a second. Let me verify this. If the rectangular solid is a cube with edge length 2*sqrt(3), then the space diagonal is sqrt( (2√3)^2 + (2√3)^2 + (2√3)^2 ) = sqrt(12 + 12 + 12) = sqrt(36) = 6, which matches the sphere's diameter. The surface area would be 6*(2√3)^2 = 6*12 = 72. So that checks out.But wait, the problem is about a rectangular solid, not necessarily a cube. So, is the cube the one with the maximum surface area? Hmm. Let me think. Suppose we have a different rectangular solid, say, a very long and thin one. For example, if a approaches 6, and b and c approach 0. Then, the surface area would be 2(ab + bc + ac) ≈ 2(6*0 + 0*0 + 6*0) = 0. So, that's worse. If we make two sides large and one small, say, a = b = sqrt(18), c approaching 0. Then a² + b² = 18 + 18 = 36, so c = 0. Then the surface area would be 2(ab + bc + ac) = 2(18 + 0 + 0) = 36, which is less than 72. So, that's worse. So, making it more cube-like gives a larger surface area.Alternatively, if we set two variables equal and solve. Let's say a = b, then c = sqrt(36 - 2a²). Then the surface area is 2(a² + 2a*sqrt(36 - 2a²)). Let me compute this for some a. Let's take a = 3. Then c = sqrt(36 - 18) = sqrt(18) ≈ 4.242. Then surface area is 2(9 + 2*3*4.242) ≈ 2(9 + 25.452) ≈ 2*34.452 ≈ 68.904, which is less than 72. Hmm.If I take a = b = sqrt(12) ≈ 3.464, then c = sqrt(36 - 24) = sqrt(12) ≈ 3.464, so it's a cube. The surface area is 72. If I take a = 4, then c = sqrt(36 - 16 - 16) = sqrt(4) = 2. Then surface area is 2(16 + 4*2 + 4*2) = 2(16 + 8 + 8) = 2*32 = 64. Still less than 72. So, it seems like the cube indeed gives the maximum surface area.But let me think again. Is there a way to have a higher surface area? Let's see. Let's suppose that the maximum occurs when two variables are equal, and the third is different. Let me set a = b, and then find c. Then the surface area is 2(a² + 2ac). But we have 2a² + c² = 36. Let's express c in terms of a: c = sqrt(36 - 2a²). Then surface area S = 2(a² + 2a*sqrt(36 - 2a²)). To find the maximum, take derivative of S with respect to a.Let me compute dS/da. Let's denote f(a) = a² + 2a*sqrt(36 - 2a²). Then df/da = 2a + 2*sqrt(36 - 2a²) + 2a*(1/(2*sqrt(36 - 2a²)))*(-4a) = 2a + 2*sqrt(36 - 2a²) - (4a²)/sqrt(36 - 2a²).Set derivative equal to zero:2a + 2*sqrt(36 - 2a²) - (4a²)/sqrt(36 - 2a²) = 0.Multiply both sides by sqrt(36 - 2a²):2a*sqrt(36 - 2a²) + 2(36 - 2a²) - 4a² = 0.Let me expand:2a*sqrt(36 - 2a²) + 72 - 4a² - 4a² = 0Simplify:2a*sqrt(36 - 2a²) + 72 - 8a² = 0Move terms:2a*sqrt(36 - 2a²) = 8a² - 72Divide both sides by 2:a*sqrt(36 - 2a²) = 4a² - 36Square both sides:a²(36 - 2a²) = (4a² - 36)^2Left side: 36a² - 2a^4Right side: 16a^4 - 2*4a²*36 + 36² = 16a^4 - 288a² + 1296Set equation:36a² - 2a^4 = 16a^4 - 288a² + 1296Bring all terms to left side:36a² - 2a^4 -16a^4 + 288a² -1296 = 0Combine like terms:(36a² + 288a²) + (-2a^4 -16a^4) -1296 = 0324a² -18a^4 -1296 = 0Divide all terms by 18:18a² - a^4 -72 = 0Rearranged:-a^4 +18a² -72 = 0Multiply both sides by -1:a^4 -18a² +72 = 0Let me set y = a², then:y² -18y +72 = 0Quadratic equation: y = [18 ± sqrt(324 - 288)]/2 = [18 ± sqrt(36)]/2 = [18 ±6]/2So, y = (18 +6)/2 = 24/2 =12, or y=(18-6)/2=12/2=6Thus, a²=12 or a²=6. Then, since a is positive, a= sqrt(12)=2*sqrt(3) or a= sqrt(6). Let's check these solutions.First, a²=12, so a=2*sqrt(3). Then c= sqrt(36 -2a²)=sqrt(36 -24)=sqrt(12)=2*sqrt(3). So, a=b=2*sqrt(3), c=2*sqrt(3). Which is the cube case. Then surface area is 72.Second, a²=6, so a= sqrt(6). Then c= sqrt(36 -2*6)=sqrt(36-12)=sqrt(24)=2*sqrt(6). Then surface area would be 2(a² + 2ac)=2(6 +2*sqrt(6)*2*sqrt(6))=2(6 + 2*12)=2(6 +24)=2*30=60. Which is less than 72.So, the critical points are at a=2√3 (cube) and a=√6 (another case). The cube gives higher surface area. So, seems like the cube is the maximum.But let me check if squaring both sides introduced any extraneous solutions. Let's go back to when we squared both sides of the equation:a*sqrt(36 - 2a²) =4a² -36If a=2√3, left side: 2√3 * sqrt(36 - 2*(12))=2√3*sqrt(36-24)=2√3*sqrt(12)=2√3*2√3=12Right side:4*(12) -36=48 -36=12. So equality holds.If a=√6, left side: √6*sqrt(36 -2*6)=√6*sqrt(24)=√6*2√6=12Right side:4*(6) -36=24 -36= -12. But left side is 12, right side is -12. So, they are not equal. Therefore, this is an extraneous solution introduced by squaring. Therefore, the only valid critical point is a=2√3. Hence, the maximum occurs at the cube.Therefore, the maximum surface area is indeed 72.But let me think again, is there a non-cube rectangular solid that could have a larger surface area?Suppose we consider a different approach. Let's parameterize the rectangular solid in terms of variables x, y, z such that x² + y² + z² = 36 (space diagonal squared). We need to maximize 2(xy + yz + zx). Alternatively, since the surface area is 2 times the sum of products of edges. So, to maximize xy + yz + zx.Is there a way to use Lagrange multipliers here? Let me set up the Lagrangian. Let f(x, y, z) = xy + yz + zx, and the constraint g(x, y, z) = x² + y² + z² -36 =0.The Lagrangian is L = xy + yz + zx - λ(x² + y² + z² -36)Take partial derivatives:∂L/∂x = y + z - 2λx =0∂L/∂y = x + z - 2λy =0∂L/∂z = x + y - 2λz =0∂L/∂λ = -(x² + y² + z² -36) =0So, we have the system of equations:1. y + z = 2λx2. x + z = 2λy3. x + y = 2λz4. x² + y² + z² =36Let me try to solve this system. From equations 1,2,3, we can try to find relationships between x, y, z.Subtract equation 1 and 2: (y + z) - (x + z) = 2λx - 2λy => y - x = 2λ(x - y)So, y - x = -2λ(y - x). If y ≠ x, then we can divide both sides by (y - x):1 = -2λ => λ = -1/2. But let's check if that's possible.Alternatively, if y = x, then from equation 1 and 2:From equation 1: x + z = 2λxFrom equation 2: x + z = 2λx (same as equation 1)From equation 3: 2x = 2λz => x = λzSo, if x = y, then from equation 3: x + x = 2λz => 2x = 2λz => x = λzFrom equation 1: x + z = 2λxSubstitute λz for x: λz + z = 2λ*(λz) => z(λ +1) = 2λ² zIf z ≠0 (which it can't be, since all edges can't be zero), then:λ +1 = 2λ²So, 2λ² - λ -1=0Solving: λ = [1 ± sqrt(1 +8)]/4 = [1 ±3]/4Thus, λ=1 or λ= -1/2.If λ=1:Then from x = λz => x = z. But since x=y, then x=y=z. So all variables equal. Then from equation 4: 3x²=36 =>x²=12 =>x=2√3. So, cube. Which gives surface area 72.If λ=-1/2:From x = λz => x = (-1/2)zFrom equation 1: x + z = 2λx => x + z = 2*(-1/2)x =>x + z = -x => z = -2xBut since x, y, z are lengths, they should be positive. However, z = -2x would imply z is negative if x is positive, which is impossible. So, this solution is invalid.Therefore, the only valid solution when x=y is the cube solution.Alternatively, suppose all variables are different. Let me assume that x ≠ y ≠ z. But given the symmetry of the equations, this might not hold. Let's see.From equations 1,2,3:y + z = 2λxx + z = 2λyx + y = 2λzLet me try to express each variable in terms of another. For example, from equation 1: z = 2λx - yFrom equation 2: z = 2λy -xThus, 2λx - y = 2λy -x => 2λx +x = 2λy + y => x(2λ +1) = y(2λ +1)If 2λ +1 ≠0, then x = y. Which brings us back to the previous case. If 2λ +1=0, then λ= -1/2. Then from equation 1: y + z = 2*(-1/2)x => y + z = -x. But x, y, z positive, so y + z = -x is impossible. Hence, the only solution is when x = y = z. Therefore, the cube is the only critical point, which gives the maximum surface area.Therefore, through Lagrange multipliers, we also arrive at the conclusion that the maximum surface area is achieved by the cube, with surface area 72.So, after going through both the algebraic approach and Lagrange multipliers, it seems the maximum surface area is indeed 72.But just to be thorough, let me consider another perspective. Suppose we fix the space diagonal and try to maximize the surface area. In 3D geometry, for a rectangular prism with a given space diagonal, is the cube the one with the maximum surface area?Intuitively, yes. Because the surface area depends on the sum of the products of the sides. To maximize the sum of the products given the sum of the squares is fixed, the case where all variables are equal gives the maximum sum. This is similar to the inequality that the arithmetic mean is maximized when all variables are equal, given a fixed sum of squares.Alternatively, using the Cauchy-Schwarz inequality, we can consider that (ab + bc + ca) ≤ (a² + b² + c²) * (something). Wait, but Cauchy-Schwarz might not directly apply here. However, we know from the earlier relation that ab + bc + ca = [(a + b + c)^2 - (a² + b² + c²)] / 2. Since a² + b² + c² is fixed at 36, to maximize ab + bc + ca, we need to maximize (a + b + c)^2. The maximum of (a + b + c)^2 under a² + b² + c² =36 is achieved when a = b = c, which again gives the cube.Therefore, all approaches converge to the conclusion that the maximum surface area is 72.Hence, the answer should be 72.Final AnswerThe maximum surface area of the rectangular solid is boxed{72}.</think>Given a rectangular solid with all vertices on the surface of a sphere of radius 3, we need to find the maximum surface area of the rectangular solid. 1. Space Diagonal Constraint: The space diagonal of the rectangular solid must be equal to the diameter of the sphere, which is 6. For a rectangular solid with side lengths (a), (b), and (c), the space diagonal is given by: [ sqrt{a^2 + b^2 + c^2} = 6 ] Squaring both sides, we get: [ a^2 + b^2 + c^2 = 36 ]2. Surface Area Objective: The surface area (S) of the rectangular solid is given by: [ S = 2(ab + bc + ac) ] We need to maximize (ab + bc + ac) under the constraint (a^2 + b^2 + c^2 = 36).3. Maximizing the Sum of Products: Using the identity: [ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac) ] Substituting (a^2 + b^2 + c^2 = 36), we get: [ ab + bc + ac = frac{(a + b + c)^2 - 36}{2} ] To maximize (ab + bc + ac), we need to maximize ((a + b + c)^2).4. Applying Cauchy-Schwarz Inequality: By the Cauchy-Schwarz inequality: [ (a + b + c)^2 leq 3(a^2 + b^2 + c^2) ] Substituting (a^2 + b^2 + c^2 = 36), we get: [ (a + b + c)^2 leq 3 times 36 = 108 ] The maximum value is achieved when (a = b = c), i.e., when the rectangular solid is a cube.5. Cube Dimensions: If (a = b = c), then: [ 3a^2 = 36 implies a^2 = 12 implies a = 2sqrt{3} ] The surface area of the cube is: [ S = 6a^2 = 6 times 12 = 72 ]6. Verification Using Lagrange Multipliers: Solving the system of equations derived from the Lagrangian method confirms that the maximum surface area occurs when (a = b = c), leading to the cube.Thus, the maximum surface area of the rectangular solid is (boxed{72}).,<think>Okay, so I have this problem here: If all vertices of a rectangular solid are on the surface of a sphere with a radius of 3, then the maximum surface area of the rectangular solid is ______. Hmm, I need to find the maximum surface area of such a rectangular solid. Let me think about how to approach this.First, I know that a rectangular solid has 8 vertices, and all of them are on the sphere's surface. The sphere has a radius of 3, so its diameter is 6. The rectangular solid is probably something like a box, with length, width, and height. Let me denote the edges as x, y, z. So the surface area of the rectangular solid is 2(xy + yz + xz). That's the formula I remember.But how does the sphere come into play here? Since all the vertices are on the sphere, that means the sphere is the circumscribed sphere (circum sphere) of the rectangular solid. The center of the sphere is the center of the rectangular solid as well. So the distance from the center of the sphere to any vertex is equal to the radius, which is 3. The space diagonal of the rectangular solid would then be equal to the diameter of the sphere. Wait, the radius is 3, so the diameter is 6. Therefore, the space diagonal of the rectangular solid is 6. The space diagonal of a rectangular solid with edges x, y, z is sqrt(x² + y² + z²). So sqrt(x² + y² + z²) = 6. Squaring both sides, we get x² + y² + z² = 36. Okay, that's a key equation.So, the problem reduces to maximizing the surface area 2(xy + yz + xz) given that x² + y² + z² = 36. Hmm, how do I maximize this expression? Maybe using some optimization techniques. I remember that for optimization problems with constraints, Lagrange multipliers can be useful. But maybe there's a simpler way, perhaps using inequalities.I recall that for real numbers, the square of sums: (x + y + z)² = x² + y² + z² + 2(xy + yz + xz). So, if I rearrange that, 2(xy + yz + xz) = (x + y + z)² - (x² + y² + z²). Since we know x² + y² + z² = 36, then 2(xy + yz + xz) = (x + y + z)² - 36. Therefore, to maximize the surface area, we need to maximize (x + y + z)². Since the surface area is proportional to (x + y + z)² minus 36, the maximum surface area occurs when (x + y + z)² is as large as possible.But we have the constraint x² + y² + z² = 36. So, how do I maximize (x + y + z)² given that x² + y² + z² is fixed? This is a classic optimization problem. Let me recall that for fixed x² + y² + z², the expression x + y + z is maximized when x, y, z are all equal. Wait, but that would be the case if we want to maximize x + y + z given x² + y² + z² fixed. But actually, if x, y, z are equal, then x = y = z, so x² + y² + z² = 3x² = 36, so x² = 12, x = 2*sqrt(3). Then x + y + z = 6*sqrt(3), so (x + y + z)^2 = 36*3 = 108. Then surface area would be 108 - 36 = 72. So surface area would be 72? But wait, is this the maximum?But wait, perhaps that's not the case. Wait, if we set two variables to be as large as possible and the third as small as possible. For example, if we set x = y and z to be different, but I need to think carefully.Alternatively, maybe using the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality states that (x + y + z)^2 <= 3(x² + y² + z²). Which in this case would mean (x + y + z)^2 <= 3*36 = 108. So maximum (x + y + z)^2 is 108, which is achieved when x = y = z. So then in that case, the surface area is 108 - 36 = 72. But wait, if that's the case, then the maximum surface area would be 72. But that seems counterintuitive because when I think of a cube versus a very elongated rectangular solid, the surface area might be larger when stretched out. Wait, is that true?Wait, let's test with an example. Suppose we have a cube with edges 2*sqrt(3). Then each face is a square with area (2*sqrt(3))² = 12. There are 6 faces, so surface area is 72. Now, if we take a rectangular solid with edges, say, 6, 0, 0. But wait, edges can't be zero. Let's take edges approaching 6, 0, 0. Then the surface area would approach 2*(6*0 + 0*0 + 6*0) = 0. That's bad. But if we take edges like 5, 1, sqrt(36 - 25 -1) = sqrt(10). Then surface area would be 2*(5*1 + 1*sqrt(10) + 5*sqrt(10)) = 2*(5 + 6*sqrt(10)). Let's compute that: 10 + 12*sqrt(10). Since sqrt(10) is approx 3.16, so 12*3.16 = 37.92, so total approx 47.92. That's way less than 72. Hmm, so that's worse.Alternatively, if we take a rectangular solid with two edges equal and the third different. For instance, set x = y, then x² + x² + z² = 36 => 2x² + z² = 36. Then surface area is 2(x^2 + 2xz). Let me express z in terms of x: z = sqrt(36 - 2x²). Then surface area becomes 2(x² + 2x*sqrt(36 - 2x²)). Let's compute derivative to maximize this.Let me denote f(x) = x² + 2x*sqrt(36 - 2x²). Then df/dx = 2x + 2*sqrt(36 - 2x²) + 2x*(1/(2*sqrt(36 - 2x²)))*(-4x). Simplifying:df/dx = 2x + 2*sqrt(36 - 2x²) - (4x²)/sqrt(36 - 2x²)Set derivative to zero:2x + 2*sqrt(36 - 2x²) - (4x²)/sqrt(36 - 2x²) = 0Multiply through by sqrt(36 - 2x²):2x*sqrt(36 - 2x²) + 2(36 - 2x²) - 4x² = 0Let me compute term by term:First term: 2x*sqrt(36 - 2x²)Second term: 2*(36 - 2x²) = 72 - 4x²Third term: -4x²Combine second and third terms: 72 - 4x² - 4x² = 72 - 8x²So entire equation:2x*sqrt(36 - 2x²) + 72 - 8x² = 0Bring the 72 -8x² to the other side:2x*sqrt(36 - 2x²) = 8x² - 72Divide both sides by 2:x*sqrt(36 - 2x²) = 4x² - 36Square both sides to eliminate sqrt:x²*(36 - 2x²) = (4x² - 36)^2Left side: 36x² - 2x^4Right side: 16x^4 - 2*4x²*36 + 36² = 16x^4 - 288x² + 1296Set equation:36x² - 2x^4 = 16x^4 - 288x² + 1296Bring all terms to left side:36x² - 2x^4 -16x^4 + 288x² -1296 =0Combine like terms:(36x² + 288x²) + (-2x^4 -16x^4) -1296=0324x² -18x^4 -1296 =0Divide all terms by 18:18x² - x^4 -72=0Rearranged:-x^4 +18x² -72=0Multiply both sides by -1:x^4 -18x² +72=0Let me set y = x², then equation becomes:y² -18y +72=0Solve quadratic:y = [18 ± sqrt(324 - 288)]/2 = [18 ± sqrt(36)]/2 = [18 ±6]/2So y= (18+6)/2=24/2=12 or y=(18-6)/2=12/2=6Therefore, x²=12 or x²=6. Then x=2*sqrt(3) or sqrt(6). Let's check which solutions are valid.If x²=12, then z²=36 -2*12=12, so z=2*sqrt(3). Then x=z=2*sqrt(3). So the rectangular solid is a cube. Then surface area is 72. On the other hand, if x²=6, then z²=36 -2*6=24, z=2*sqrt(6). Then x=sqrt(6). Then surface area would be 2(x² + 2xz)= 2*(6 + 2*sqrt(6)*2*sqrt(6))=2*(6 + 2*12)=2*(6+24)=2*30=60. Which is less than 72. So the maximum in this case is the cube.But this seems to contradict my initial thought that stretching the box might increase surface area, but according to this, the cube gives the maximum surface area. Hmm. Wait, but this is under the assumption that two variables are equal. Maybe if all three variables are different, we can get a higher surface area. Wait, but the Cauchy-Schwarz inequality told us that (x + y + z)^2 <=3(x² + y² + z²) =108, so the maximum of (x + y + z)^2 is 108, which would be achieved when x=y=z. Therefore, surface area is 108 -36=72. So according to that, 72 is the maximum. So maybe the cube is indeed the one with maximum surface area.But let me check another approach. Maybe using Lagrange multipliers. Let me set up the problem: maximize f(x,y,z)=2(xy + yz + xz) subject to g(x,y,z)=x² + y² + z²=36.The Lagrangian is L=2(xy + yz + xz) - λ(x² + y² + z² -36).Taking partial derivatives:dL/dx=2(y + z) - 2λx =0dL/dy=2(x + z) - 2λy =0dL/dz=2(x + y) - 2λz =0And the constraint x² + y² + z²=36.So we have the equations:2(y + z) - 2λx =0 => y + z = λxSimilarly,x + z = λyx + y = λzSo from the first equation: y + z = λxSecond: x + z = λyThird: x + y = λzLet me try to solve these equations.Let me subtract the first and second equations: (y + z) - (x + z) = λx - λy => y - x = λ(x - y) => (y - x) = -λ(y - x).If y ≠x, then we can divide both sides by (y -x), getting 1 = -λ. So λ= -1. Let's see if that works.If λ=-1, then from first equation: y + z = -x. Similarly, second equation: x + z = -y. Third equation: x + y = -z.Let me check. From first equation: y + z = -x. From second: x + z = -y. Let's solve for z from first equation: z = -x - y. Substitute into second equation: x + (-x - y) = -y => x -x - y = -y => -y = -y. Which is always true. Similarly, third equation: x + y = -z. But z = -x - y, so x + y = -(-x - y) => x + y = x + y. Also holds. So in this case, we can choose any x, y, z such that z = -x - y. But they must also satisfy x² + y² + z²=36. Let's substitute z = -x - y into the constraint:x² + y² + (x + y)² =36 => x² + y² + x² + 2xy + y²=36 => 2x² + 2y² + 2xy=36 => x² + y² + xy=18.Hmm, this is a quadratic in x and y. Let's see if this can yield real solutions. For example, suppose x=y. Then x² + x² + x²=3x²=18 => x²=6 =>x=sqrt(6). Then z = -2x = -2sqrt(6). But then z²=24, so x² + y² + z²=6 +6 +24=36, which works. So in this case, we have x=y=sqrt(6), z=-2sqrt(6). Then surface area is 2(xy + yz + xz)=2*(6 + sqrt(6)*(-2sqrt(6)) + sqrt(6)*(-2sqrt(6)))=2*(6 -12 -12)=2*(-18)= -36. Wait, but surface area can't be negative. Hmm, something's wrong here.Wait, z is negative? But in the problem, the edges of the rectangular solid can be negative? No, lengths are positive. Wait, but in our Lagrangian method, we considered variables x, y, z as real numbers, but in reality, the edges must be positive. So if we get solutions where z is negative, they are not feasible. So the case where λ=-1 might lead to negative dimensions, which are invalid. Therefore, this solution is not acceptable. Hence, perhaps the only valid solution is when x=y=z. Let's check that.Assume x=y=z. Then the constraint gives 3x²=36 =>x²=12 =>x=2*sqrt(3). Then surface area is 2(xy + yz + xz)=2*(12 +12 +12)=72, which is positive. So that's valid.Alternatively, if we consider the case where λ is positive, maybe that gives us the solution.From the equations:y + z = λxx + z = λyx + y = λzIf we suppose that x, y, z are positive, then λ must be positive as well.Let me try to solve these equations. Let's subtract the first equation from the second: (x + z) - (y + z) = λy - λx => x - y = λ(y - x) => x - y = -λ(x - y). So either x = y or λ = -1. But λ is positive, so we must have x = y.Similarly, subtract the second equation from the third: (x + y) - (x + z) = λz - λy => y - z = λ(z - y) => y - z = -λ(y - z). Again, either y = z or λ = -1. But since λ is positive, y = z.Therefore, the only solution in positive variables is x = y = z. Hence, the cube is the only critical point, which gives the maximum surface area of 72.Therefore, the maximum surface area is 72.But wait, the problem says "rectangular solid", which is a box with all edges different? Or can it be a cube? Since a cube is a special case of a rectangular solid, so it's allowed. Therefore, the maximum surface area is achieved when the rectangular solid is a cube.Therefore, the answer is 72. Let me verify once again.Wait, another thought: maybe using the method of symmetry. If we consider that the maximum occurs when all variables are equal, that's due to the symmetry of the sphere. Since the sphere is symmetric in all directions, the most symmetric rectangular solid inscribed in the sphere would be the cube. And due to the symmetry, the cube should either maximize or minimize the surface area. Since when we stretch the box, the surface area decreases as tested earlier, the cube is indeed the maximum.Alternatively, consider that for a fixed diagonal (i.e., fixed x² + y² + z²), the maximum surface area is achieved when the edges are equal. That seems to hold here.Yes, this makes sense. So the answer is 72. Therefore, the maximum surface area is boxed{72}.Final AnswerThe maximum surface area of the rectangular solid is boxed{72}.</think>Given a rectangular solid with all vertices on the surface of a sphere of radius 3, we need to find the maximum surface area of the rectangular solid. First, we note that the sphere's radius is 3, so its diameter is 6. The space diagonal of the rectangular solid must be equal to this diameter, giving us the equation:[sqrt{x^2 + y^2 + z^2} = 6]Squaring both sides, we get:[x^2 + y^2 + z^2 = 36]The surface area of the rectangular solid is given by:[2(xy + yz + xz)]We aim to maximize this surface area under the constraint (x^2 + y^2 + z^2 = 36). Using the identity:[(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz)]we can express the surface area as:[2(xy + yz + xz) = (x + y + z)^2 - (x^2 + y^2 + z^2)]Substituting the constraint (x^2 + y^2 + z^2 = 36), we get:[2(xy + yz + xz) = (x + y + z)^2 - 36]To maximize the surface area, we need to maximize ((x + y + z)^2). By the Cauchy-Schwarz inequality:[(x + y + z)^2 leq 3(x^2 + y^2 + z^2)]Given (x^2 + y^2 + z^2 = 36), this inequality becomes:[(x + y + z)^2 leq 3 times 36 = 108]The maximum value of ((x + y + z)^2) is 108, achieved when (x = y = z). Substituting (x = y = z) into the constraint (x^2 + y^2 + z^2 = 36), we get:[3x^2 = 36 implies x^2 = 12 implies x = 2sqrt{3}]Thus, the surface area is:[2(xy + yz + xz) = 108 - 36 = 72]Therefore, the maximum surface area of the rectangular solid is (boxed{72}).]

🔑:Let the lengths of the three dimensions of the rectangular solid be a, b, and c. The diameter of the sphere is equal to the length of the diagonal of the rectangular solid. From the given information we have: a^2 + b^2 + c^2 = 6^2 = 36. The surface area of the rectangular solid is given by: 2ab + 2ac + 2bc. Using the inequality of arithmetic and geometric means, we can write: 2ab + 2ac + 2bc leq 2(a^2 + b^2 + c^2) = 72. Equality holds when a = b = c, which means that when the rectangular solid is a cube, the surface area is maximized.Therefore, the maximum surface area is: boxed{72}. In this problem, we defined the dimensions of the rectangular solid and found the expression for its surface area. We then determined the maximum value using the relationship between the dimensions and the diagonal, which corresponds to the diameter of the sphere. The problem tests knowledge of the circumscribed sphere of a rectangular solid, how to find the maximum surface area, and the application of basic inequalities. It's important to note the conditions for equality when using basic inequalities to find maximum values.

🧠:[<think>Okay, so I need to find the imaginary part of (1 + i)^5. Hmm, complex numbers, right? Let me recall how to work with powers of complex numbers. Maybe using binomial expansion? Or perhaps converting to polar form would be easier? Let me think. First, if I try binomial expansion, (1 + i)^5 can be expanded using the binomial theorem. That would involve calculating combinations like 5 choose k, multiplied by 1^(5-k) and i^k. Since 1 raised to any power is still 1, it simplifies to the sum from k=0 to 5 of (5 choose k) * i^k. The imaginary part would be the terms where i^k is imaginary, which happens when k is odd. Because i^1 = i, i^3 = -i, i^5 = -i, etc. Wait, but in the expansion, each term's coefficient is (5 choose k) multiplied by i^k. So the imaginary parts would come from k=1, 3, 5. Let's see:For k=1: (5 choose 1)*i^1 = 5iFor k=3: (5 choose 3)*i^3 = 10*(-i) = -10iFor k=5: (5 choose 5)*i^5 = 1*(-i) = -iAdding those up: 5i -10i -i = (5 -10 -1)i = (-6i). So the imaginary part is -6? Wait, but that can't be right. Because (1 + i)^2 is 2i, then (1 + i)^4 would be (2i)^2 = -4, and (1 + i)^5 = (1 + i)*(-4) = -4 -4i. So the imaginary part would be -4. Wait, that contradicts the previous result. Hmm, so clearly I made a mistake in the binomial expansion. Let me check again.Alternatively, maybe converting to polar form is better. Let me try that. The complex number 1 + i can be represented in polar form as r*(cos θ + i sin θ). The modulus r is sqrt(1^2 + 1^2) = sqrt(2). The angle θ is arctangent of (1/1) which is 45 degrees or π/4 radians. Therefore, (1 + i)^5 would be (sqrt(2))^5 * [cos(5*π/4) + i sin(5*π/4)]. Let's compute that.First, (sqrt(2))^5. sqrt(2) is 2^(1/2), so (2^(1/2))^5 = 2^(5/2) = 2^2 * 2^(1/2) = 4*sqrt(2). Then, cos(5π/4) is cos(π + π/4) which is -sqrt(2)/2, and sin(5π/4) is also -sqrt(2)/2. Therefore, the expression becomes 4*sqrt(2) * [ -sqrt(2)/2 + i*(-sqrt(2)/2) ]. Let's compute the real and imaginary parts.For the real part: 4*sqrt(2) * (-sqrt(2)/2) = 4 * (- (sqrt(2)*sqrt(2))/2 ) = 4 * (-2/2) = 4*(-1) = -4.For the imaginary part: 4*sqrt(2) * (-sqrt(2)/2) = same calculation, so also -4. Therefore, the imaginary part is -4. So that's different from the -6 I got earlier. Which one is correct?Wait, let's check using another method. Let's compute (1 + i)^2 first. (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i -1 = 2i. Then, (1 + i)^4 = (2i)^2 = -4. Then, (1 + i)^5 = (1 + i)^4 * (1 + i) = (-4)*(1 + i) = -4 -4i. So the imaginary part is -4. That matches the polar form result. So where did I go wrong with the binomial expansion?Let me redo the binomial expansion step by step. (a + b)^5 expansion is:a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5.Here, a = 1 and b = i. So:1^5 + 5*1^4*i + 10*1^3*(i)^2 + 10*1^2*(i)^3 + 5*1*(i)^4 + (i)^5.Simplify each term:1 + 5i + 10*(i^2) + 10*(i^3) + 5*(i^4) + i^5.Compute the powers of i:i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i.So substituting:1 + 5i + 10*(-1) + 10*(-i) + 5*(1) + i.Simplify term by term:1 + 5i -10 -10i +5 + i.Combine like terms:Real parts: 1 -10 +5 = -4.Imaginary parts: 5i -10i +i = (-4i).Therefore, the total expression is -4 -4i. So the imaginary part is -4. Wait, so earlier when I thought the binomial expansion gave -6i, I must have made a mistake. Let me check my previous binomial approach.I had written:"For k=1: (5 choose 1)*i^1 = 5iFor k=3: (5 choose 3)*i^3 = 10*(-i) = -10iFor k=5: (5 choose 5)*i^5 = 1*(-i) = -iAdding those up: 5i -10i -i = (-6i). So the imaginary part is -6."Ah, here's the mistake. When k=5, i^5 is actually i, not -i. Wait, no. Wait, i^5 is i^(4+1) = (i^4)*i = 1*i = i. So in the expansion, the term for k=5 is (5 choose 5)*i^5 = 1*i = i, not -i. So that term should be +i, not -i. So then the sum would be 5i -10i +i = (-4i). Therefore, the imaginary part is -4. Which matches the other methods. So my mistake was in thinking that i^5 is -i, but actually i^5 is i. Let me confirm:i^1 = ii^2 = -1i^3 = -ii^4 = 1i^5 = i^4 * i = 1 * i = iYes, so i^5 is i. Therefore, the term for k=5 is +i. So the correct sum is 5i -10i +i = (-4i). So the imaginary part is -4. So all methods concur that the imaginary part is -4. Therefore, the answer is boxed{-4}. But wait, the problem says "the imaginary part of (1+i)^5". The imaginary part is the coefficient of i, which here is -4. So -4 is the answer.But let me verify once more through another method for thoroughness. Let's compute (1 + i)^5 step by step:First, (1 + i)^1 = 1 + i(1 + i)^2 = (1 + i)(1 + i) = 1 + 2i + i^2 = 1 + 2i -1 = 2i(1 + i)^3 = (1 + i)^2 * (1 + i) = 2i * (1 + i) = 2i + 2i^2 = 2i - 2 = -2 + 2i(1 + i)^4 = (1 + i)^3 * (1 + i) = (-2 + 2i)(1 + i) = -2(1) -2(i) + 2i(1) + 2i(i) = -2 -2i + 2i + 2i^2 = -2 + 0i + 2*(-1) = -2 -2 = -4(1 + i)^5 = (1 + i)^4 * (1 + i) = (-4)(1 + i) = -4 -4iSo, indeed, the imaginary part is -4. Therefore, the answer is -4.I think initially I made a mistake in the binomial expansion by miscalculating i^5 as -i instead of i, which led to an incorrect result. But after correcting that, all methods agree. So the imaginary part is -4.Final AnswerThe imaginary part of ((1+i)^5) is boxed{-4}.</think>To find the imaginary part of ((1 + i)^5), we can use multiple methods:1. Binomial Expansion: [ (1 + i)^5 = sum_{k=0}^{5} binom{5}{k} 1^{5-k} i^k ] The imaginary part comes from the terms where (k) is odd: - For (k = 1): (binom{5}{1} i = 5i) - For (k = 3): (binom{5}{3} i^3 = 10(-i) = -10i) - For (k = 5): (binom{5}{5} i^5 = 1(i) = i) Adding these terms: (5i - 10i + i = -4i). Thus, the imaginary part is (-4).2. Polar Form: - Convert (1 + i) to polar form: (r = sqrt{2}), (theta = frac{pi}{4}) - ((1 + i)^5 = (sqrt{2})^5 left( cos frac{5pi}{4} + i sin frac{5pi}{4} right)) - Calculate modulus: ((sqrt{2})^5 = 4sqrt{2}) - Calculate angles: (cos frac{5pi}{4} = -frac{sqrt{2}}{2}), (sin frac{5pi}{4} = -frac{sqrt{2}}{2}) - Result: (4sqrt{2} left( -frac{sqrt{2}}{2} - i frac{sqrt{2}}{2} right) = -4 - 4i) - Imaginary part is (-4).3. Step-by-Step Expansion: - ((1 + i)^2 = 2i) - ((1 + i)^4 = (2i)^2 = -4) - ((1 + i)^5 = (1 + i)(-4) = -4 - 4i) - Imaginary part is (-4).All methods confirm that the imaginary part of ((1 + i)^5) is (boxed{-4}).,<think>Okay, so I need to find the imaginary part of (1 + i)^5. Hmm, let's see. I remember that when dealing with complex numbers raised to powers, there are a couple of methods. One is using the binomial theorem to expand the expression, and another is converting the complex number to polar form and then applying De Moivre's theorem. Let me try both methods to confirm the answer, just to be sure.First, let's try the binomial theorem. The binomial theorem says that (a + b)^n = sum from k=0 to n of (n choose k) * a^(n - k) * b^k. So applying that to (1 + i)^5:(1 + i)^5 = Σ (5 choose k) * 1^(5 - k) * i^k for k = 0 to 5.Let's compute each term step by step.For k = 0:(5 choose 0) * 1^5 * i^0 = 1 * 1 * 1 = 1For k = 1:(5 choose 1) * 1^4 * i^1 = 5 * 1 * i = 5iFor k = 2:(5 choose 2) * 1^3 * i^2 = 10 * 1 * (-1) = -10For k = 3:(5 choose 3) * 1^2 * i^3 = 10 * 1 * (-i) = -10iFor k = 4:(5 choose 4) * 1^1 * i^4 = 5 * 1 * 1 = 5For k = 5:(5 choose 5) * 1^0 * i^5 = 1 * 1 * i^5 = i^5. Wait, what's i^5? Since i cycles every four powers: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1, so i^5 = i^(4 + 1) = i^4 * i = 1 * i = i. So the last term is i.Now, adding all the terms together:1 + 5i - 10 -10i + 5 + iLet's combine like terms. The real parts are 1 -10 +5 = (1 +5) -10 = 6 -10 = -4The imaginary parts are 5i -10i +i = (5 -10 +1)i = (-4)iSo the entire expression simplifies to -4 -4i. Therefore, the imaginary part is -4. Wait, but the problem asks for the imaginary part. So the imaginary part is -4. But let me check if I did that correctly.Wait, let's recount the terms:k=0: 1 (real)k=1:5i (imaginary)k=2:-10 (real)k=3:-10i (imaginary)k=4:5 (real)k=5:i (imaginary)So imaginary parts: 5i -10i +i = (5 -10 +1)i = (-4)i. So the coefficient is -4. So imaginary part is -4. So the answer should be -4. Hmm. But wait, let me check with the other method to make sure.Alternatively, using polar form. The complex number 1 + i can be represented in polar form as r*(cos θ + i sin θ). Let's compute r and θ.The modulus r is sqrt(1^2 + 1^2) = sqrt(2)The argument θ is arctangent of (1/1) which is π/4 radians.Therefore, 1 + i = sqrt(2)*(cos(π/4) + i sin(π/4))Then, raising to the 5th power using De Moivre's theorem:(1 + i)^5 = [sqrt(2)]^5 * [cos(5π/4) + i sin(5π/4)]Compute [sqrt(2)]^5. Since sqrt(2) is 2^(1/2), so [2^(1/2)]^5 = 2^(5/2) = 2^2 * 2^(1/2) = 4 * sqrt(2)Then, 5π/4 is 225 degrees, which is in the third quadrant. The cosine and sine there are both negative, equal to -sqrt(2)/2.Therefore:cos(5π/4) = -sqrt(2)/2sin(5π/4) = -sqrt(2)/2Therefore:(1 + i)^5 = 4*sqrt(2) * [ -sqrt(2)/2 + i*(-sqrt(2)/2) ]Let's compute this:Multiply 4*sqrt(2) with each term inside the brackets.First term: 4*sqrt(2)*(-sqrt(2)/2) = 4*(- (sqrt(2)*sqrt(2))/2 ) = 4*(- (2)/2 ) = 4*(-1) = -4Second term: 4*sqrt(2)*( -sqrt(2)/2 ) * i = 4*(- (sqrt(2)*sqrt(2))/2 ) *i = 4*(-2/2)*i = 4*(-1)*i = -4iTherefore, (1 + i)^5 = -4 -4i. So the imaginary part is -4, which matches the previous result.Therefore, the imaginary part is -4. So the answer should be boxed{-4}. Wait, but the problem says "the imaginary part of (1+i)^5". Imaginary part is the coefficient of i, which in the expression a + bi is b. So in -4 -4i, it's -4. So yes, -4. But let me just check once again. Maybe I made a mistake in the binomial expansion.Wait, let me recast the expansion:(1 + i)^5 = 1 + 5i + 10i^2 + 10i^3 + 5i^4 + i^5Wait, but maybe my coefficients are wrong? Wait, the coefficients for (a + b)^5 are 1, 5, 10, 10, 5, 1. So:(1)^5 + 5*(1)^4*(i) + 10*(1)^3*(i)^2 + 10*(1)^2*(i)^3 + 5*(1)*(i)^4 + (i)^5So:1 + 5i + 10i^2 + 10i^3 + 5i^4 + i^5Now, compute each term:1 is 1.5i is 5i.10i^2 is 10*(-1) = -10.10i^3 is 10*(-i) = -10i.5i^4 is 5*(1) = 5.i^5 is i.So adding them all up:1 + 5i -10 -10i +5 +iCombine real parts: 1 -10 +5 = -4Combine imaginary parts: 5i -10i +i = (-4i)So total: -4 -4i. So imaginary part is -4. Yes, correct. So the answer is -4. But hold on, maybe the problem is expecting the answer as a positive number? But no, the imaginary part is the coefficient, which can be negative. So -4 is the correct answer. But let me check online, maybe. Wait, maybe I can calculate (1 + i)^2 first and then build up.(1 + i)^1 = 1 + i(1 + i)^2 = (1 + i)(1 + i) = 1 + 2i + i^2 = 1 + 2i -1 = 2i(1 + i)^3 = (1 + i)^2 * (1 + i) = 2i*(1 + i) = 2i + 2i^2 = 2i -2 = -2 + 2i(1 + i)^4 = (1 + i)^3*(1 + i) = (-2 + 2i)(1 + i) = -2(1 + i) + 2i(1 + i) = -2 -2i + 2i + 2i^2 = -2 + 0i -2 = -4Wait, (1 + i)^4 is -4? Wait, but (1 + i)^4 should be [(1 + i)^2]^2 = (2i)^2 = -4. Yes, that's correct. So (1 + i)^4 = -4. Then (1 + i)^5 = (1 + i)^4*(1 + i) = (-4)*(1 + i) = -4 -4i. Therefore, the imaginary part is -4. So same result. So this seems correct.Therefore, the imaginary part is -4. So the answer is boxed{-4}. But wait, the problem says "the imaginary part". Sometimes people get confused between the imaginary part and the imaginary coefficient. The imaginary part of a complex number a + bi is b, which is a real number. So in this case, it's -4, not -4i. So yeah, the answer is -4.I think that's solid. Both methods give the same answer. Binomial expansion, polar form with De Moivre's theorem, and step-by-step multiplication all lead to -4 -4i. So the imaginary part is -4. Definitely, the answer is boxed{-4}.Final AnswerThe imaginary part of ((1+i)^5) is boxed{-4}.</think>To find the imaginary part of ((1 + i)^5), we can use both the binomial theorem and polar form with De Moivre's theorem.Using the Binomial Theorem:Expand ((1 + i)^5) using the binomial theorem:[(1 + i)^5 = sum_{k=0}^{5} binom{5}{k} 1^{5-k} i^k]Calculating each term:- For (k = 0): (binom{5}{0} cdot 1^5 cdot i^0 = 1)- For (k = 1): (binom{5}{1} cdot 1^4 cdot i^1 = 5i)- For (k = 2): (binom{5}{2} cdot 1^3 cdot i^2 = 10 cdot (-1) = -10)- For (k = 3): (binom{5}{3} cdot 1^2 cdot i^3 = 10 cdot (-i) = -10i)- For (k = 4): (binom{5}{4} cdot 1^1 cdot i^4 = 5 cdot 1 = 5)- For (k = 5): (binom{5}{5} cdot 1^0 cdot i^5 = i)Combining all terms:[1 + 5i - 10 - 10i + 5 + i]Separating real and imaginary parts:- Real parts: (1 - 10 + 5 = -4)- Imaginary parts: (5i - 10i + i = -4i)Thus, the imaginary part is (-4).Using Polar Form and De Moivre's Theorem:Convert (1 + i) to polar form:- Modulus: (sqrt{1^2 + 1^2} = sqrt{2})- Argument: (frac{pi}{4})Raising to the 5th power:[(1 + i)^5 = (sqrt{2})^5 left( cosleft(5 cdot frac{pi}{4}right) + i sinleft(5 cdot frac{pi}{4}right) right)]Calculating:- ((sqrt{2})^5 = 4sqrt{2})- (5 cdot frac{pi}{4} = frac{5pi}{4})- (cosleft(frac{5pi}{4}right) = -frac{sqrt{2}}{2})- (sinleft(frac{5pi}{4}right) = -frac{sqrt{2}}{2})Thus:[4sqrt{2} left( -frac{sqrt{2}}{2} - i frac{sqrt{2}}{2} right) = -4 - 4i]The imaginary part is (-4).Both methods confirm that the imaginary part of ((1 + i)^5) is (boxed{-4}).]

🔑:To find the imaginary part of (1+i)^5, we can first expand (1+i)^2:(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i.Then, we continue to expand (1+i)^5:begin{align*}(1+i)^5 &= (1+i)^2 times (1+i)^2 times (1+i) &= 2i times 2i times (1+i) &= -4 times (1+i) &= -4 - 4i.end{align*}The imaginary part of (1+i)^5 is the coefficient of i, which is -4.Therefore, the imaginary part of (1+i)^5 is boxed{-4}.

🧠:[<think>Okay, let's see. I need to find the intersection of sets M and N. First, let me make sure I understand the problem correctly. Set M is given as {0, 2, 3, 4}. Set N is defined as all elements x such that x = 2a where a is in M. So essentially, N is created by taking each element of M, multiplying it by 2, and collecting those results. Then, the intersection of M and N would be the elements that are common to both sets. The options given are A through D, with different subsets. Alright, let me break this down step by step.First, let me write down set M again to make sure I have it right. M = {0, 2, 3, 4}. Got that. Now, N is the set of elements x where x is twice some element a from M. So I need to compute N first. Let me list out the elements of N by taking each element of M and multiplying by 2.Starting with the first element of M, which is 0. So 2 times 0 is 0. Next element is 2. 2 times 2 is 4. Then 3. 2 times 3 is 6. Then 4. 2 times 4 is 8. So putting all those together, set N is {0, 4, 6, 8}. Let me verify that: 0 from 0, 4 from 2, 6 from 3, 8 from 4. Yes, that seems right. So N = {0, 4, 6, 8}.Now, the intersection of M and N, written as M ∩ N, is the set of elements that are in both M and N. Let me list out the elements of both sets again to compare.Set M: {0, 2, 3, 4}Set N: {0, 4, 6, 8}Looking for common elements. Let's check each element of M and see if it's also in N.Starting with 0: Is 0 in N? Yes, N has 0. So 0 is in the intersection.Next, 2: Is 2 in N? Let's check N's elements: 0, 4, 6, 8. No, 2 is not there. So 2 isn't in the intersection.Then 3: Is 3 in N? N has 0, 4, 6, 8. Nope, so 3 isn't in the intersection.Next, 4: Is 4 in N? Yes, N has 4. So 4 is in the intersection.So the common elements are 0 and 4. Therefore, the intersection M ∩ N is {0, 4}.Now, let's check the answer options. Option C is {0, 4}, so that should be the correct answer. But wait, let me double-check everything to make sure I didn't make a mistake.First, confirming set N again. Each element of M multiplied by 2:0*2 = 02*2 = 43*2 = 64*2 = 8Yes, so N is {0, 4, 6, 8}. Then comparing with M {0, 2, 3, 4}, the common elements are indeed 0 and 4. So the intersection is {0, 4}, which is option C. Therefore, the answer must be C.But just to be thorough, let me check if the problem might have any trick or if I misread something. For instance, sometimes there might be a question about whether the multiplication is done in another way, but here it's straightforward: x = 2a where a is in M. So each a in M is multiplied by 2. So 0, 2, 3, 4 become 0, 4, 6, 8. That seems correct. Then intersection is elements present in both. So yes, 0 and 4 are in both. The other options: A is {0}, but 4 is also in both. B is {0, 2}, but 2 is not in N. D is {2,4}, but 2 isn't in N. So definitely C is the right one. I think that's solid. I don't see any errors in my reasoning. So the answer should be option C.Final Answerboxed{C}</think>Given the sets ( M = {0, 2, 3, 4} ) and ( N = {x mid x = 2a, a in M} ), we need to find the intersection of sets ( M ) and ( N ).First, we determine the elements of set ( N ). Each element of ( M ) is multiplied by 2:- For ( 0 in M ), ( 2 times 0 = 0 )- For ( 2 in M ), ( 2 times 2 = 4 )- For ( 3 in M ), ( 2 times 3 = 6 )- For ( 4 in M ), ( 2 times 4 = 8 )Thus, set ( N ) is ( {0, 4, 6, 8} ).Next, we find the intersection of sets ( M ) and ( N ). The intersection consists of elements common to both sets:- Set ( M ): ( {0, 2, 3, 4} )- Set ( N ): ( {0, 4, 6, 8} )The common elements are ( 0 ) and ( 4 ).Therefore, the intersection of sets ( M ) and ( N ) is ( {0, 4} ), which corresponds to option C.[boxed{C}],<think>Okay, let's see. The problem is about sets M and N. I need to find their intersection. Let me start by understanding what each set is.First, set M is given as {0, 2, 3, 4}. So that's straightforward. The elements are 0, 2, 3, and 4.Now, set N is defined as {x | x = 2a, a ∈ M}. Hmm, okay. That notation means N is the set of all elements x such that x is equal to 2 times a, where a is an element of M. So, essentially, for every element in M, I multiply it by 2, and collect those results into set N.Let me write out what N would be. Let's take each element of M one by one and apply the operation.Starting with the first element of M: 0. Multiply by 2: 0*2=0. So 0 is in N.Next element: 2. Multiply by 2: 2*2=4. So 4 is in N.Third element: 3. Multiply by 2: 3*2=6. So 6 is in N.Fourth element: 4. Multiply by 2: 4*2=8. So 8 is in N.Therefore, set N is {0, 4, 6, 8}.Now, the problem asks for the intersection of sets M and N. The intersection of two sets is the set of elements that are common to both sets.So, set M is {0, 2, 3, 4} and set N is {0, 4, 6, 8}. Let's compare these.Looking for elements that are in both:- 0 is in both M and N.- 2 is in M but not in N (since N has 4, 6, 8).- 3 is in M but not in N.- 4 is in both M and N.- The other elements of N (6, 8) aren't in M.Therefore, the intersection should be {0, 4}.Looking at the options:A: {0}B: {0, 2}C: {0, 4}D: {2, 4}So option C is {0,4}, which matches what I found. Let me double-check to make sure I didn't make a mistake.First, verifying set N again. Multiplying each element of M by 2:0*2=0, 2*2=4, 3*2=6, 4*2=8. So N={0,4,6,8}. Correct.Intersection with M: elements common to both {0,2,3,4} and {0,4,6,8} are indeed 0 and 4. So yes, the intersection is {0,4}, which is option C. The answer should be C.Final Answerboxed{C}</think>Given the sets ( M = {0, 2, 3, 4} ) and ( N = {x mid x = 2a, a in M} ), we need to find their intersection.First, we determine the elements of set ( N ). For each element ( a ) in ( M ), we calculate ( 2a ):- When ( a = 0 ), ( 2a = 0 )- When ( a = 2 ), ( 2a = 4 )- When ( a = 3 ), ( 2a = 6 )- When ( a = 4 ), ( 2a = 8 )Thus, set ( N ) is ( {0, 4, 6, 8} ).Next, we find the intersection of sets ( M ) and ( N ). The intersection consists of elements common to both sets:- Set ( M ): {0, 2, 3, 4}- Set ( N ): {0, 4, 6, 8}The common elements are 0 and 4. Therefore, the intersection of sets ( M ) and ( N ) is ( {0, 4} ).The correct answer is (boxed{C}).]

🔑:Since M={0, 2, 3, 4},N={x|x=2a, ain M}={0, 4, 6, 8},Therefore, the intersection of sets M and N is {0, 4}.Hence, the correct option is boxed{text{C}}.

🧠:[<think>Okay, so I need to figure out which of the given inequalities is incorrect if we know that a is greater than b. Let's go through each option step by step. First, the original inequality is a > b. So any correct inequality should preserve this relationship when manipulated properly. Let's check each one.Option A: 3a > 3b. Hmm, if we multiply both sides of the original inequality by 3, which is a positive number, the direction of the inequality should stay the same. Since 3 is positive, multiplying both a and b by 3 should maintain the greater than relationship. So 3a should indeed be greater than 3b. So A seems correct. So A is not the answer.Option B: -a/3 < -b/3. Let's see. If we take the original inequality a > b, and multiply both sides by -1/3. But wait, when you multiply or divide an inequality by a negative number, the inequality sign flips. So starting with a > b, multiplying both sides by -1/3 would give -a/3 < -b/3. That's exactly what option B says. So B is correct. So B is not the answer.Option C: 4a - 3 > 4b - 3. Let's break this down. If we start with a > b, multiply both sides by 4, which is positive, so 4a > 4b. Then subtract 3 from both sides. Subtracting the same number from both sides doesn't change the inequality, so 4a - 3 > 4b - 3. That's exactly what option C states. So C is correct. Therefore, C is not the answer.Option D: ac² > bc². Hmm. Let's think here. If we have a > b, and we multiply both sides by c². However, c² is always non-negative because any real number squared is non-negative. But the problem is, if c² is zero, then multiplying both sides by c² would result in 0 > 0, which is not true. So if c is zero, then ac² = bc², so the inequality ac² > bc² would not hold. Therefore, the inequality in D is not necessarily true if c is zero. But wait, the problem says "which of the following inequalities is incorrect." So if there exists a case where the inequality doesn't hold, then D is incorrect. Since c could be zero, then D is not always true. The other options (A, B, C) are always true given a > b. Therefore, D is the incorrect inequality.Wait, but let me double-check. The problem states "if a > b", but it doesn't specify any restrictions on c. So c could be any real number, including zero. Therefore, unless there's a constraint that c is non-zero, D might not hold. So if c is zero, then c² is zero, so ac² = bc², regardless of a and b. Therefore, in that case, ac² is not greater than bc²; they are equal. Therefore, D is not necessarily true, making it incorrect. So the answer must be D.Final Answerboxed{D}</think>Given the inequality (a > b), we need to determine which of the provided options is incorrect.- Option A: (3a > 3b). Multiplying both sides of (a > b) by 3 (a positive number) preserves the inequality. Therefore, this is correct. - Option B: (-frac{a}{3} < -frac{b}{3}). Multiplying both sides of (a > b) by (-frac{1}{3}) (a negative number) reverses the inequality. Therefore, this is correct. - Option C: (4a - 3 > 4b - 3). First, multiply both sides by 4 (a positive number) to get (4a > 4b), then subtract 3 from both sides, which preserves the inequality. Therefore, this is correct. - Option D: (ac^2 > bc^2). Since (c^2) is non-negative, if (c = 0), then (ac^2 = bc^2 = 0), which does not satisfy the inequality. Therefore, this is not necessarily true.Thus, the incorrect inequality is (boxed{D}).,<think>Okay, let's see. The problem is asking which of the given inequalities is incorrect if we know that a is greater than b. The options are A, B, C, and D. Alright, let's go through each one step by step.Starting with option A: 3a > 3b. Hmm, since a > b, if we multiply both sides by 3, which is a positive number, the inequality direction should stay the same. So 3 times a should indeed be greater than 3 times b. That makes sense. So A is correct, not the answer we're looking for.Moving on to option B: -a/3 < -b/3. Let me think. If a > b, then dividing both sides by 3 (which is positive) would keep the inequality the same, so a/3 > b/3. But then we have a negative sign in front. So if we multiply both sides of a/3 > b/3 by -1, that would reverse the inequality, right? Multiplying or dividing by a negative number flips the inequality. So -a/3 should be less than -b/3. Wait, so that's exactly what option B says: -a/3 < -b/3. So that's correct too. So B is not the incorrect one. Hmm, maybe I need to check again. Let's take actual numbers to verify. Suppose a = 4 and b = 2. Then -a/3 = -4/3 ≈ -1.333 and -b/3 = -2/3 ≈ -0.666. So -1.333 < -0.666 is true. So yes, B is correct. So B is okay.Option C: 4a - 3 > 4b - 3. Let's see. Starting from a > b, multiply both sides by 4 (positive number), so 4a > 4b. Then subtract 3 from both sides: 4a - 3 > 4b - 3. That's straightforward. So C is definitely correct. So not the answer.Option D: ac² > bc². Hmm, okay. So if a > b, then multiplying both sides by c². Wait, but c² is a square, so it's non-negative. If c² is positive, then the inequality remains the same. But if c² is zero, then both sides become zero, so the inequality would not hold. So the problem here is that c² could be zero. If c is zero, then ac² = bc² regardless of a and b. So in that case, ac² is not greater than bc². Therefore, the inequality D is not necessarily true. So D is incorrect. Let me check with numbers. Suppose a = 5, b = 3, and c = 0. Then ac² = 5*0 = 0 and bc² = 3*0 = 0. So 0 > 0 is false. Therefore, D is incorrect when c is zero. So D is the incorrect inequality. So the answer must be D.But wait, let me double-check. The question says "which of the following inequalities is incorrect". So if there's a case where the inequality doesn't hold, then that option is incorrect. Since c could be zero, D isn't always true. So D is the incorrect one. So the answer is D. Yeah, I think that's right.Final Answerboxed{D}</think>Given ( a > b ), we need to determine which inequality is incorrect.- Option A: ( 3a > 3b ) Multiplying both sides of ( a > b ) by 3 (a positive number) preserves the inequality. Therefore, ( 3a > 3b ) is correct.- Option B: ( -frac{a}{3} < -frac{b}{3} ) Dividing both sides of ( a > b ) by 3 gives ( frac{a}{3} > frac{b}{3} ). Multiplying by -1 reverses the inequality, resulting in ( -frac{a}{3} < -frac{b}{3} ). This is correct.- Option C: ( 4a - 3 > 4b - 3 ) Multiplying both sides of ( a > b ) by 4 gives ( 4a > 4b ). Subtracting 3 from both sides preserves the inequality, so ( 4a - 3 > 4b - 3 ) is correct.- Option D: ( ac^2 > bc^2 ) If ( c = 0 ), then ( ac^2 = bc^2 = 0 ), which means the inequality ( 0 > 0 ) is false. Since ( c^2 ) can be zero, this inequality is not necessarily true. Therefore, it is incorrect.The incorrect inequality is (boxed{D}).]

🔑:Given a > b, we need to evaluate which of the following inequalities is incorrect.Option A: 3a > 3b- Starting from a > b- Multiply both sides by 3: 3a > 3b- This shows that 3a > 3b is boxed{text{correct}}.Option B: -frac{a}{3} b- Divide both sides by -3 (Note: Dividing or multiplying by a negative number reverses the inequality): -frac{a}{3} 4b-3- Starting from a > b- Multiply both sides by 4: 4a > 4b- Subtract 3 from both sides: 4a - 3 > 4b - 3- This shows that 4a-3 > 4b-3 is boxed{text{correct}}.Option D: ac^{2} > bc^{2}- This inequality depends on the value of c. If c = 0, then ac^{2} = bc^{2} = 0, which does not satisfy ac^{2} > bc^{2}.- Therefore, without additional information about c (specifically, ensuring c neq 0), this statement can be incorrect.- This shows that ac^{2} > bc^{2} is boxed{text{incorrect}} when c=0.Therefore, the incorrect inequality is boxed{D}.