🧠:[<think>Okay, let's see. The problem is about finding the value of ω for the function f(x) = cos(ωx + π/4) such that the distance between two adjacent zero points is π/6. The options are A: 3, B: 6, C: 12, D: 24. Hmm, alright, let's start by recalling some basics about cosine functions and their zeros.First, the general form of a cosine function is f(x) = A cos(Bx + C) + D. In this case, the function is f(x) = cos(ωx + π/4). So, comparing to the general form, A is 1, B is ω, C is π/4, and D is 0. The amplitude is 1, the vertical shift is 0, and the horizontal shift is -C/B, which would be -π/(4ω). But maybe the horizontal shift isn't directly important here. We are concerned about the zeros of the function, which are the points where f(x) = 0.So, to find the zeros, we set cos(ωx + π/4) = 0. The cosine function equals zero at odd multiples of π/2. That is, cos(θ) = 0 when θ = π/2 + kπ, where k is any integer. So, setting ωx + π/4 = π/2 + kπ. Then, solving for x, we have:ωx = π/2 - π/4 + kπWait, no, let me do that again. Starting with:ωx + π/4 = π/2 + kπSubtract π/4 from both sides:ωx = π/2 - π/4 + kπSimplify π/2 - π/4 = π/4, so:ωx = π/4 + kπThen, x = (π/4 + kπ)/ωSo, the zeros of the function occur at x = (π/4 + kπ)/ω for integer k.Now, the problem states that the distance between two adjacent zero points is π/6. So, we need to find the difference between consecutive zeros. Let's take two consecutive values of k, say k and k+1, and find the difference in their x-values.For k, x_k = (π/4 + kπ)/ωFor k+1, x_{k+1} = (π/4 + (k+1)π)/ωSubtracting these two gives:x_{k+1} - x_k = [π/4 + (k+1)π - (π/4 + kπ)]/ωSimplify numerator:π/4 cancels with -π/4, (k+1)π - kπ = πSo, the difference is π/ωBut the problem says this distance is π/6. Therefore, π/ω = π/6Solving for ω, divide both sides by π:1/ω = 1/6Therefore, ω = 6Wait, but that seems straightforward, so the answer is B: 6? Let me double-check.Wait, but is the distance between adjacent zeros really π/ω? Let me think again. Cosine function usually has a period of 2π, but here it's compressed or stretched by ω. The period of f(x) = cos(ωx + φ) is 2π/ω. But zeros occur twice every period, right? Because cosine crosses zero twice in each period. So the distance between consecutive zeros should be half the period? Wait, but actually, the zeros are separated by half the period? Let me recall.In the standard cosine function, cos(x), zeros occur at π/2, 3π/2, 5π/2, etc., so the distance between them is π, which is half the period (2π). So yes, the distance between consecutive zeros is half the period.Therefore, if the period is 2π/ω, then the distance between zeros is π/ω. So that's consistent with what we found earlier. Therefore, if the problem states that the distance between adjacent zeros is π/6, then π/ω = π/6, so ω = 6. Therefore, the answer is B: 6.But let me check again with the phase shift. The function is cos(ωx + π/4). Does the phase shift affect the distance between zeros? Hmm. Let's see. Suppose we have a phase shift, does that change the period or the spacing between zeros?No, the phase shift shifts the graph left or right, but doesn't affect the period or the spacing between consecutive zeros. The period is still 2π/ω, and the zeros are still separated by π/ω. So regardless of the phase shift, the distance between consecutive zeros should be π/ω. Therefore, even with the phase shift π/4, the distance remains the same. Therefore, the answer should indeed be ω = 6, option B.Wait, but let me test with specific numbers. Suppose ω = 6. Then the function is cos(6x + π/4). Let's find two adjacent zeros.Set 6x + π/4 = π/2 + kπSo, 6x = π/2 - π/4 + kπ = π/4 + kπx = (π/4 + kπ)/6For k = 0, x = π/(4*6) = π/24For k = 1, x = (π/4 + π)/6 = (5π/4)/6 = 5π/24Difference is 5π/24 - π/24 = 4π/24 = π/6. Okay, that works. So the difference is π/6 as desired. So ω = 6 is correct.Alternatively, let's check another value. Suppose ω = 12. Then the difference would be π/12. Which is not π/6. So that's not the answer. Similarly, ω = 3 would give π/3, which is larger than π/6. ω = 24 would give π/24, which is smaller. So yes, 6 is the correct answer. Therefore, the answer is B: 6.Final Answerboxed{B}</think>To find the value of ( omega ) for the function ( f(x) = cos(omega x + frac{pi}{4}) ) such that the distance between two adjacent zero points is ( frac{pi}{6} ), we start by determining the zeros of the function. The cosine function equals zero at odd multiples of ( frac{pi}{2} ). Therefore, we set:[omega x + frac{pi}{4} = frac{pi}{2} + kpi quad text{for integer } k]Solving for ( x ), we get:[omega x = frac{pi}{2} - frac{pi}{4} + kpi][omega x = frac{pi}{4} + kpi][x = frac{pi/4 + kpi}{omega}]The distance between consecutive zeros (for ( k ) and ( k+1 )) is:[x_{k+1} - x_k = frac{pi/4 + (k+1)pi}{omega} - frac{pi/4 + kpi}{omega} = frac{pi}{omega}]Given that this distance is ( frac{pi}{6} ), we set up the equation:[frac{pi}{omega} = frac{pi}{6}]Solving for ( omega ), we find:[omega = 6]Thus, the value of ( omega ) is (boxed{B}).,<think>Okay, so I need to find the value of ω given that the distance between two adjacent zero points of the function f(x) = cos(ωx + π/4) is π/6. The options are A: 3, B: 6, C: 12, D: 24. Let me think through this step by step.First, let me recall that the zeros of the cosine function occur where the argument of the cosine is an odd multiple of π/2. That is, cos(θ) = 0 when θ = π/2 + kπ for any integer k. So, for the given function f(x) = cos(ωx + π/4), the zeros occur when ωx + π/4 = π/2 + kπ, where k is an integer.Let me write that equation down:ωx + π/4 = π/2 + kπTo find the x-values where the function is zero, I need to solve for x:ωx = π/2 + kπ - π/4Simplify the right side:π/2 - π/4 = π/4, so:ωx = π/4 + kπThen, solving for x:x = (π/4 + kπ) / ωSo the zeros of the function occur at x = (π/4 + kπ)/ω for any integer k.Now, the problem states that the distance between two adjacent zero points is π/6. Adjacent zeros would correspond to consecutive integers k and k+1. Let's compute the difference between x_{k+1} and x_k.x_{k+1} = (π/4 + (k+1)π)/ω = (π/4 + kπ + π)/ω = (π/4 + kπ)/ω + π/ωx_k = (π/4 + kπ)/ωSo, x_{k+1} - x_k = π/ωBut the problem says this distance is π/6. Therefore:π/ω = π/6Solving for ω, we can divide both sides by π:1/ω = 1/6Therefore, ω = 6.Wait, but hold on. Let me check if this is correct. Because when we consider the zeros of the cosine function, adjacent zeros are separated by π in the argument. But here, the function is cos(ωx + π/4). The period of this function is 2π/ω. However, the zeros of the cosine function are spaced by π in its argument. So, the distance between zeros in terms of x would be π/(ω). So, indeed, the distance between two adjacent zeros is π/ω.Given that the problem states this distance is π/6, so π/ω = π/6 ⇒ ω = 6. Therefore, the answer is B: 6.But let me confirm this with an example. Suppose ω = 6. Then the function is cos(6x + π/4). Let's find two adjacent zeros.Set 6x + π/4 = π/2 + kπSolving for x:6x = π/2 - π/4 + kπ = π/4 + kπx = (π/4 + kπ)/6 = π/(4*6) + kπ/6 = π/24 + kπ/6So, the zeros occur at x = π/24 + kπ/6. Let's take k = 0: x = π/24. For k = 1: x = π/24 + π/6 = π/24 + 4π/24 = 5π/24. The difference between these two zeros is 5π/24 - π/24 = 4π/24 = π/6. That's correct. So the distance between two adjacent zeros is indeed π/6 when ω = 6. Therefore, the answer is B:6.But wait, let me check if there's another interpretation. Maybe the problem is referring to consecutive zeros without skipping any? But cosine normally has a zero every π in its argument. So, between θ = π/2 and θ = 3π/2, which is a difference of π, so in the x variable, that would correspond to a distance of π/(ω). So yes, that's consistent. Therefore, the answer should be 6.Wait, but let me check another value to make sure. Let's take option C:12. If ω=12, then the distance between zeros would be π/12. Since the problem states that the distance is π/6, which is double π/12. So that's not correct. Similarly, ω=3 would give distance π/3, which is larger than π/6. ω=24 would give π/24, which is smaller. So indeed, the correct answer is 6.Therefore, the answer is B.Final Answerboxed{B}</think>To find the value of ( omega ) given that the distance between two adjacent zero points of the function ( f(x) = cos(omega x + frac{pi}{4}) ) is ( frac{pi}{6} ):1. The zeros of the cosine function occur where the argument is an odd multiple of ( frac{pi}{2} ). For the given function, the zeros occur when: [ omega x + frac{pi}{4} = frac{pi}{2} + kpi quad text{for integer } k ]2. Solving for ( x ): [ omega x = frac{pi}{2} - frac{pi}{4} + kpi = frac{pi}{4} + kpi ] [ x = frac{pi/4 + kpi}{omega} ]3. The distance between two adjacent zeros corresponds to consecutive integers ( k ) and ( k+1 ): [ x_{k+1} - x_k = frac{pi/4 + (k+1)pi}{omega} - frac{pi/4 + kpi}{omega} = frac{pi}{omega} ]4. Given this distance is ( frac{pi}{6} ): [ frac{pi}{omega} = frac{pi}{6} ] Solving for ( omega ): [ omega = 6 ]Thus, the value of ( omega ) is (boxed{B}).]

🔑:The distance between two adjacent zero points of the function f(x)=cos (ωx+ frac {π}{4})(ω > 0) is frac {π}{6},Hence, T=2× frac {π}{6}= frac {π}{3},Also, frac {2π}{ω}= frac {π}{3},Solving for ω, we get ω=6.Therefore, the answer is: boxed{B}.This is based on the property of the cosine function where the distance between two adjacent zero points is exactly half a period, which can be used to find the value of ω. This question primarily tests the application of the graph and properties of the cosine function, and is a basic question.

🧠:[<think>Okay, let's see. The question is asking which of the given options is a fractional expression. The options are A, B, C, and D. Hmm, first I need to remember what a fractional expression is. If I recall correctly, a fractional expression is a ratio of two polynomials where the denominator is not a constant, right? Or maybe it's any ratio where the denominator has variables? Wait, maybe I should double-check the definition.From what I remember, a fractional expression, also known as a rational expression, is a fraction in which the numerator and the denominator are polynomials. The key thing is that the denominator cannot be a constant unless the numerator is also not a polynomial? Wait, no. Let me think. If both numerator and denominator are polynomials, then it's a rational expression. So even if the denominator is a constant, as long as it's a polynomial, then it's a rational expression. But maybe the term "fractional expression" here is being used differently?Wait, the problem is in Chinese, and it's translated as "fractional expression." Maybe in Chinese math terminology, a fractional expression specifically refers to a fraction where the denominator contains variables? So unlike a polynomial, which has only constants and variables in the numerator, a fractional expression has variables in the denominator. So that would mean options where the denominator has variables would be fractional expressions.Let me look at each option:Option A: (8x)/(3π). The numerator is 8x, which is a polynomial (degree 1), and the denominator is 3π, which is a constant (pi is a constant, approximately 3.14). So since the denominator is a constant, this would just be a polynomial (8x divided by 3π is still a linear term). So maybe this is not considered a fractional expression here?Option B: (x² - y²)/(x - y). The numerator is x² - y², which is a polynomial, and the denominator is x - y, also a polynomial. However, x² - y² can be factored into (x - y)(x + y), so this simplifies to x + y, as long as x ≠ y. So in simplified form, this is just a polynomial. But before simplification, it's a rational expression. But does the question consider simplified forms or not? The problem says "expressions," so maybe as they are written. So B is a rational expression with variables in the denominator. But since it simplifies to a polynomial, maybe it's not considered a fractional expression? Hmm, this is confusing.Option C: (x - y)/5. The numerator is x - y, a polynomial, and the denominator is 5, a constant. So similar to option A, this is just a polynomial divided by a constant, which is still a polynomial. So maybe this is not a fractional expression either.Option D: 5/8. Both numerator and denominator are constants. That's just a simple fraction, a rational number. But in terms of expressions, since there are no variables, is this considered a fractional expression? If the definition requires variables, then no. But if any fraction counts, then yes. But I think fractional expressions here refer to ones with variables in the denominator.Wait, maybe the question is using "fractional expression" as opposed to an integer expression. But in that case, D is a fractional constant. But in the context of algebra, probably fractional expressions are those that involve variables in the denominator.Looking back at the options, A and C have denominators that are constants (3π and 5), so they are technically rational expressions but with constant denominators, which are equivalent to polynomials. B has a denominator that is a polynomial with variables, but it simplifies to a polynomial when simplified. D is a constant fraction.So the problem is from a Chinese math context, and maybe in their terminology, fractional expressions are those that cannot be simplified to a polynomial, i.e., they have variables in the denominator that can't be canceled out. So if B simplifies to x + y, then it's equivalent to a polynomial, so not a fractional expression? Then that leaves none of them except maybe A, B, C, or D. Wait, but D is just a constant fraction. Maybe the question is different.Wait, perhaps in Chinese math terminology, a fractional expression is simply any algebraic expression that is written as a fraction, regardless of the denominator. If that's the case, then all options A, B, C, D are fractional expressions. But the question is asking "which one is a fractional expression," implying that only one of them is. That can't be. So maybe the term is being used differently.Alternatively, maybe the question is distinguishing between fractions where the denominator is a numeral (i.e., a constant) versus those where the denominator includes variables. In some contexts, a fractional expression might require the denominator to have variables, otherwise, it's just a numerical coefficient. So if that's the case, then among the options, only B has a denominator with variables. Let me check:A: denominator is 3π, a constant.B: denominator is x - y, variables.C: denominator is 5, constant.D: denominator is 8, constant.So if fractional expression is defined as having variables in the denominator, then B is the only one. However, option B can be simplified to x + y, which is a polynomial. So does the simplification affect whether it's considered a fractional expression? If we consider the expression as given, without simplifying, then B is a fractional expression. But if we simplify it, it becomes a polynomial.Hmm, tricky. In algebra, when we talk about expressions, we usually consider them in their original form unless specified otherwise. So unless the question mentions simplifying, we take them as they are. Therefore, B is a fractional expression because it's written as a fraction with variables in the denominator. Even though it simplifies to a polynomial, the original expression is a fractional expression.But then, in that case, A and C are also fractions, but their denominators are constants. So perhaps in this terminology, fractional expressions are those with variables in the denominator. If that's the case, then B is the answer. But let's check the options again.Wait, the problem is presented in Chinese, so maybe the term used is "分式" which directly translates to "fractional expression" and in Chinese math textbooks, 分式 is defined as an expression of the form A/B where A and B are polynomials and B is not a constant polynomial (i.e., B contains variables). So according to that definition, only expressions where the denominator has variables are considered fractional expressions. So in that case, B is the answer.Alternatively, maybe 分式 includes any fractions, but in Chinese curriculum, I think 分式 specifically refers to rational expressions where the denominator has variables. For example, in Chinese math, 分式方程 are equations involving fractional expressions, which typically have variables in the denominator.So, if that's the case, then among the options, only B has a denominator with variables. The others have denominators that are constants. Therefore, the answer should be B.Wait, but let me check once more. If we consider the definition, a fractional expression (分式) is a ratio of two polynomials where the denominator is not a constant. So:A: Numerator is 8x (polynomial), denominator is 3π (constant). So not a 分式.B: Numerator is x² - y² (polynomial), denominator is x - y (polynomial with variables). So yes, 分式.C: (x - y)/5: denominator is 5 (constant). Not a 分式.D: 5/8: both constants. Not a 分式.So the answer should be B. But wait, but B can be simplified to x + y, so maybe the question is testing whether simplification affects the classification. But in definitions, the classification is based on the original form, not the simplified one. For example, (x² - y²)/(x - y) is considered a 分式, even though it simplifies to a polynomial, because in its original form, it's a ratio of two polynomials with denominator containing variables.Therefore, the answer is B. But let me check the options again. The options are A, B, C, D. So the answer is B. However, in some contexts, maybe A is also considered a fractional expression because it's written as a fraction. But according to the stricter definition where the denominator must have variables, then B is the only one. Therefore, the answer is B. But let me confirm with an example.For instance, in the Chinese textbook definition, 分式 is defined as 分母中含有字母的式子, which translates to "expressions where the denominator contains letters (variables)." So yes, only B has a denominator with variables. Therefore, the answer is B.But wait, looking back at the problem statement: "Among the following expressions, which one is a fractional expression?" The options are given. So according to the definition where fractional expressions require variables in the denominator, B is the answer.So I think the answer is B.Final Answerboxed{B}</think>To determine which of the given expressions is a fractional expression, we need to consider the definition of a fractional expression. A fractional expression (or rational expression) is a ratio of two polynomials where the denominator contains variables. - Option A: (frac{8x}{3pi}) has a denominator (3pi), which is a constant. Therefore, it is not a fractional expression.- Option B: (frac{x^2 - y^2}{x - y}) has a denominator (x - y), which is a polynomial containing variables. Even though it simplifies to (x + y), the original form is a fractional expression.- Option C: (frac{x - y}{5}) has a denominator (5), which is a constant. Therefore, it is not a fractional expression.- Option D: (frac{5}{8}) has both numerator and denominator as constants. Therefore, it is not a fractional expression.According to the definition where a fractional expression requires the denominator to contain variables, the correct answer is (boxed{B}).,<think>Alright, so I need to figure out which of the given options is a fractional expression. Let me start by recalling what a fractional expression is. If I remember correctly, a fractional expression, also known as a rational expression, is a ratio of two polynomials. That means both the numerator and the denominator should be polynomials, and the denominator cannot be a constant unless it's considered a polynomial of degree zero. But wait, even if the denominator is a constant, as long as it's a non-zero polynomial, it should still be a fractional expression. Hmm, maybe I need to double-check that.Let's look at the options one by one. Option A: (8x)/(3π). The numerator is 8x, which is a polynomial of degree 1. The denominator is 3π. Now, π is a constant, albeit an irrational one. So 3π is just a constant term. Since both numerator and denominator are constants or polynomials, does that make this a fractional expression? But wait, the denominator here is 3π, which is an irrational number. But polynomials typically have coefficients that are rational numbers, unless specified otherwise. However, the question doesn't specify whether we're considering coefficients in real numbers or just rationals. If we consider the denominator to be a polynomial, then 3π would be a constant polynomial. But π is transcendental, not algebraic. Hmm, this might complicate things. Maybe fractional expressions can still have constants in the denominator, even if they are transcendental? I'm a bit confused here. Let me hold off on this and check the other options.Option B: (x² - y²)/(x - y). This simplifies to (x + y)(x - y)/(x - y), right? So if x ≠ y, this simplifies to x + y. But does that mean it's not a fractional expression? Wait, but the original expression is a ratio of two polynomials. Even if it simplifies to a polynomial, the original form is still a fractional expression. However, sometimes people might argue that if it simplifies to a polynomial, then it's not a fractional expression. But technically, a fractional expression is just a ratio of polynomials, regardless of whether it can be simplified. So perhaps B is a fractional expression. But I need to be careful here.Option C: (x - y)/5. The numerator is x - y, which is a polynomial. The denominator is 5, which is a constant. So similar to option A, but here the denominator is a rational constant. If we consider coefficients in real numbers, then 5 is a polynomial of degree 0. So this should be a fractional expression. But again, if the denominator is just a constant, is that still considered a fractional expression? Wait, in the definition, it just needs to be a ratio of two polynomials. Since 5 is a polynomial (constant polynomial), then (x - y)/5 is a ratio of two polynomials, hence a fractional expression. But then why is option D: 5/8 also a candidate? Let me check that.Option D: 5/8. Both numerator and denominator are constants. So this is just a constant value, 5/8. But since both 5 and 8 are constants (polynomials of degree 0), then 5/8 is also a ratio of two polynomials, hence a fractional expression. But is a constant considered a fractional expression? Because if so, then both C and D would be fractional expressions. Wait, but the question is asking which one is a fractional expression. Maybe multiple are, but the options are A, B, C, D. Let me check the original question again.The problem states: "Among the following expressions, which one is a fractional expression?" with options A to D. So it's a multiple-choice question with only one correct answer. So perhaps the key here is to determine which one is considered a fractional expression, while the others are not. Maybe there's a trick here.Let's revisit each option:A: (8x)/(3π). As mentioned earlier, the denominator is 3π, which is an irrational constant. If we consider the field of coefficients, usually in algebra, polynomials are defined over a field, often the real numbers or complex numbers. But a fractional expression is a ratio of two polynomials. So if 3π is considered a valid coefficient (as a real number), then the denominator is a constant polynomial, so the whole expression is a fractional expression. However, if the problem is considering coefficients only in the rational numbers, then 3π would not be a valid coefficient, making the denominator not a polynomial over the rationals, hence not a fractional expression over the rationals. But the problem doesn't specify the field. Hmm.B: (x² - y²)/(x - y). As I mentioned, this simplifies to x + y when x ≠ y. But the original expression is a ratio of two polynomials. However, since (x - y) is a polynomial, this is a fractional expression. But perhaps the problem is considering whether it simplifies to a polynomial? If so, then B is equivalent to a polynomial, so it's not a "proper" fractional expression. But I think technically, fractional expressions include any ratio of polynomials, even if they simplify. So B would still be a fractional expression. Wait, but the question might be testing on the difference between a polynomial and a fractional expression. If B simplifies to a polynomial, then it's not considered a fractional expression. Hmm. This is a bit ambiguous.C: (x - y)/5. As discussed, this is a ratio of a polynomial to a constant polynomial. So it's a fractional expression. Similarly, D: 5/8 is a ratio of two constant polynomials, so technically a fractional expression. But maybe the question is expecting a fractional expression where the denominator is a non-constant polynomial. If that's the case, then only A, B, and C have non-constant denominators? Wait, A's denominator is 3π, which is a constant. B's denominator is x - y, which is a non-constant polynomial. C's denominator is 5, a constant. D's denominator is 8, a constant. So if the key is that the denominator must be a non-constant polynomial, then only B is a fractional expression. But then again, the definition says a ratio of two polynomials, regardless of their degrees. So even if the denominator is a constant, as long as it's a polynomial, it's a fractional expression. Wait, maybe the term "fractional expression" is being used differently here. In some contexts, a fractional expression might require that the variable is in the denominator. If that's the case, then only options A, B, and maybe C? Wait, A's denominator is 3π, which doesn't have a variable. B's denominator is x - y, which has variables. C's denominator is 5, constant. D's denominator is 8, constant. So if fractional expression is defined as an expression where the variable is in the denominator, then only B would qualify. But I need to confirm the exact definition.Looking back, the term "fractional expression" might be translated differently. In Chinese math education, for example, a fractional expression (分式) specifically requires that the denominator contains variables and is not a constant. So in that case, only expressions where the denominator has variables are considered fractional expressions. If that's the case here, then among the options, only B has variables in the denominator. The others have constants in the denominator, so they would be considered rational expressions but not fractional expressions. In this context, fractional expressions are distinct from simple fractions or constant denominators.Given that this is a multiple-choice question likely following such a definition, the answer would be B. But let me verify this.If we use the definition that a fractional expression must have variables in the denominator, then:A: Denominator is 3π (constant) – not a fractional expression.B: Denominator is x - y (variables) – fractional expression.C: Denominator is 5 (constant) – not a fractional expression.D: Denominator is 8 (constant) – not a fractional expression.Therefore, the correct answer is B.But wait, initially, I thought C and D could be fractional expressions since they're ratios of polynomials, but if the definition requires variables in the denominator, then only B is correct. This must be the intended answer. So even though B simplifies to a polynomial, the original form has a denominator with variables, making it a fractional expression. The simplification is only valid when x ≠ y, but the expression itself is still considered fractional because it's written as a ratio with variables in the denominator.Therefore, the answer is B.Final Answerboxed{B}</think>To determine which expression is a fractional expression, we need to recall that a fractional expression (or rational expression) is a ratio of two polynomials, where the denominator is not a constant (i.e., it contains variables). Analysis of each option:- Option A: (frac{8x}{3pi}) The numerator (8x) is a polynomial, and the denominator (3pi) is a constant (irrational). Since the denominator does not contain any variables, this is not a fractional expression.- Option B: (frac{x^2 - y^2}{x - y}) The numerator (x^2 - y^2) and the denominator (x - y) are both polynomials. Although this simplifies to (x + y) when (x neq y), the original form is a ratio of two polynomials with variables in the denominator. Therefore, this is a fractional expression.- Option C: (frac{x - y}{5}) The numerator (x - y) is a polynomial, and the denominator (5) is a constant. Since the denominator does not contain any variables, this is not a fractional expression.- Option D: (frac{5}{8}) Both the numerator and the denominator are constants. Since there are no variables in the denominator, this is not a fractional expression.Conclusion: The correct answer is (boxed{B}) because it is the only expression with variables in the denominator, making it a fractional expression.]

🔑:Answer: A, frac{8x}{3pi} is a polynomial, so option A is incorrect; B, frac{x^2-y^2}{x-y} is a fractional expression, so option B is correct; C, the denominator does not contain any letters, which makes it a polynomial, so option C is incorrect; D, the denominator does not contain any letters, which makes it a polynomial, so option D is incorrect; Therefore, the correct answer is boxed{text{B}}.

🧠:[<think>Okay, so I have this expression to evaluate: (frac{left(sqrt{3} cdot tan 12^{circ} - 3right) cdot csc 12^{circ}}{4 cos ^{2} 12^{circ} - 2}). Hmm, let's break this down step by step. First, maybe I should recall some trigonometric identities that could simplify this expression. Let's see... The denominator is (4 cos^2 12^{circ} - 2). That looks similar to the double-angle identity for cosine. Remember that (cos 2theta = 2cos^2 theta - 1), so if I factor out a 2 from the denominator, it becomes (2(2cos^2 12^{circ} - 1)), which is (2 cos 24^{circ}) because (2cos^2 theta - 1 = cos 2theta). So the denominator simplifies to (2 cos 24^{circ}). That might help.Now for the numerator: (sqrt{3} cdot tan 12^{circ} - 3) multiplied by (csc 12^{circ}). Let's handle that part by part. First, (tan 12^{circ}) is (sin 12^{circ}/cos 12^{circ}), and (csc 12^{circ}) is (1/sin 12^{circ}). So if I substitute those in, the numerator becomes (sqrt{3} cdot (sin 12^{circ}/cos 12^{circ}) - 3) all multiplied by (1/sin 12^{circ}). Let me write that out:Numerator: (left( sqrt{3} cdot frac{sin 12^{circ}}{cos 12^{circ}} - 3 right) cdot frac{1}{sin 12^{circ}})Let me distribute the (1/sin 12^{circ}) into the parentheses:First term: (sqrt{3} cdot frac{sin 12^{circ}}{cos 12^{circ}} cdot frac{1}{sin 12^{circ}} = sqrt{3} cdot frac{1}{cos 12^{circ}} = sqrt{3} cdot sec 12^{circ})Second term: (-3 cdot frac{1}{sin 12^{circ}} = -3 csc 12^{circ})So the numerator simplifies to (sqrt{3} sec 12^{circ} - 3 csc 12^{circ}). Hmm, not sure if that helps immediately. Maybe there's another identity or a way to express this in terms of sine and cosine to combine terms. Let's see:(sqrt{3} sec 12^{circ} = sqrt{3}/cos 12^{circ}) and (3 csc 12^{circ} = 3/sin 12^{circ}). So numerator is (sqrt{3}/cos 12^{circ} - 3/sin 12^{circ}). Maybe we can combine these two terms into a single fraction. Let's find a common denominator, which would be (sin 12^{circ} cos 12^{circ}).So numerator becomes: (sqrt{3} sin 12^{circ} - 3 cos 12^{circ}) all over (sin 12^{circ} cos 12^{circ}).Wait, that's interesting. So the numerator is (frac{sqrt{3} sin 12^{circ} - 3 cos 12^{circ}}{sin 12^{circ} cos 12^{circ}}), and the denominator of the entire expression is (2 cos 24^{circ}). So putting it all together, the entire expression is:(frac{sqrt{3} sin 12^{circ} - 3 cos 12^{circ}}{sin 12^{circ} cos 12^{circ}} div (2 cos 24^{circ}) = frac{sqrt{3} sin 12^{circ} - 3 cos 12^{circ}}{sin 12^{circ} cos 12^{circ} cdot 2 cos 24^{circ}})Hmm. Maybe there's a way to simplify the numerator (sqrt{3} sin 12^{circ} - 3 cos 12^{circ}). It looks like a linear combination of sine and cosine, which can be expressed as a single sine function with a phase shift. The general form is (a sin theta + b cos theta = R sin(theta + phi)), where (R = sqrt{a^2 + b^2}) and (phi = arctan(b/a)) or something like that. Wait, actually, it's (R sin(theta + phi)) or (R cos(theta + phi)), depending on the formula. Let me recall: (a sin theta + b cos theta = R sin(theta + phi)), where (R = sqrt{a^2 + b^2}) and (phi = arctan(b/a)). Alternatively, sometimes it's written as (R cos(theta - phi)). Let me check.Actually, let's derive it. Let's suppose:( sqrt{3} sin 12^{circ} - 3 cos 12^{circ} = R sin(12^{circ} + phi) )Expanding the right side using sine addition formula:( R (sin 12^{circ} cos phi + cos 12^{circ} sin phi) )Comparing coefficients:( sqrt{3} = R cos phi )( -3 = R sin phi )So then, dividing the two equations: ( tan phi = (-3)/sqrt{3} = -sqrt{3} ), so (phi = -60^{circ}) or (300^{circ}). Then, ( R = sqrt{ (sqrt{3})^2 + (-3)^2 } = sqrt{3 + 9} = sqrt{12} = 2 sqrt{3} ).Therefore, the numerator can be written as (2 sqrt{3} sin(12^{circ} - 60^{circ}) = 2 sqrt{3} sin(-48^{circ}) = -2 sqrt{3} sin 48^{circ}). Wait, is that right? Let me check.Wait, if ( phi = -60^{circ} ), then:( R sin(12^{circ} + phi) = 2 sqrt{3} sin(12^{circ} - 60^{circ}) = 2 sqrt{3} sin(-48^{circ}) = -2 sqrt{3} sin 48^{circ} ). So yes, the numerator is (-2 sqrt{3} sin 48^{circ}). Alternatively, if I use cosine, maybe that would be better? Let me try expressing it as a cosine function:Alternatively, write (a sin theta + b cos theta = R cos(theta - phi)). Let's try that:( sqrt{3} sin 12^{circ} - 3 cos 12^{circ} = R cos(12^{circ} - phi) )Expanding the right side:( R (cos 12^{circ} cos phi + sin 12^{circ} sin phi) )Comparing coefficients:Coefficient of (sin 12^{circ}): ( R sin phi = sqrt{3} )Coefficient of (cos 12^{circ}): ( R cos phi = -3 )So again, ( tan phi = sqrt{3}/(-3) = -1/sqrt{3} ), so (phi = -30^{circ}) or (330^{circ}). Then ( R = sqrt{ (sqrt{3})^2 + (-3)^2 } = sqrt{12} = 2 sqrt{3} ), same as before. Thus,( sqrt{3} sin 12^{circ} - 3 cos 12^{circ} = 2 sqrt{3} cos(12^{circ} + 30^{circ}) = 2 sqrt{3} cos 42^{circ} ). Wait, because the formula is ( R cos(theta - phi) ), but here phi is -30 degrees? Hmm, maybe I messed up the angle. Let's see:Wait, if we have ( R cos(theta - phi) ), and we found that ( phi = -30^{circ} ), then it's ( R cos(theta - (-30^{circ})) = R cos(theta + 30^{circ}) ). So substituting theta = 12°, we get ( 2 sqrt{3} cos(12° + 30°) = 2 sqrt{3} cos 42° ). Alternatively, since cosine is even, ( cos(-42°) = cos 42° ), but not sure if that helps. Either way, perhaps both expressions are valid, but one might be more helpful.So either way, the numerator is either (-2 sqrt{3} sin 48°) or (2 sqrt{3} cos 42°). Hmm. Let me check with actual values to see which one is correct. Let's compute both expressions numerically.Original numerator: (sqrt{3} sin12° - 3 cos12°). Let me compute this:First, compute (sqrt{3} ≈ 1.732), (sin12° ≈ 0.2079), (cos12° ≈ 0.9781).So, 1.732 * 0.2079 ≈ 0.35983 * 0.9781 ≈ 2.9343So numerator ≈ 0.3598 - 2.9343 ≈ -2.5745Now, compute (-2 sqrt{3} sin48°):First, sqrt(3) ≈ 1.732, so 2.5745 ≈ 2 * 1.732 * sin48°. Wait, but let me compute:-2 * 1.732 * sin48°. Sin48° ≈ 0.7431, so 2 * 1.732 ≈ 3.464, 3.464 * 0.7431 ≈ 2.574, so with the negative sign, it's -2.574, which matches the original numerator. So that's correct.Similarly, compute 2 * sqrt(3) * cos42°: cos42° ≈ 0.7431, so 2 * 1.732 * 0.7431 ≈ 2.574, but positive. Which doesn't match. So probably, the sine expression is correct with the negative sign. So the numerator is -2√3 sin48°, which is equal to -2√3 sin48°. Hmm.So going back, the entire expression is:Numerator: -2√3 sin48° / (sin12° cos12° * 2 cos24° )Wait, the denominator was:Original denominator after simplifying: 2 cos24°, but when we combined with the numerator's denominator, sin12° cos12°, so the total denominator is 2 cos24° * sin12° cos12°.Therefore, the entire expression is:(-2√3 sin48°) / (2 sin12° cos12° cos24° )The 2 cancels in numerator and denominator: -√3 sin48° / (sin12° cos12° cos24° )Now, maybe we can simplify the denominator as well. Let's see: sin12° cos12° is equal to (1/2) sin24°, by the double-angle identity: sin2θ = 2 sinθ cosθ, so sinθ cosθ = (1/2) sin2θ. Therefore, sin12° cos12° = (1/2) sin24°. So substituting that, the denominator becomes (1/2 sin24°) * cos24° = (1/2) sin24° cos24°. Then again, sin24° cos24° = (1/2) sin48°, so the denominator is (1/2)*(1/2) sin48° = (1/4) sin48°. So putting it all together:Expression = (-√3 sin48° ) / ( (1/4) sin48° ) = -√3 / (1/4) = -4√3Wait, that seems too straightforward. Let me verify step by step:Original expression after substituting numerator and denominator:(-2√3 sin48°) divided by (2 sin12° cos12° cos24° )First, cancel the 2's: -√3 sin48° divided by ( sin12° cos12° cos24° )Then, sin12° cos12° = (1/2) sin24°, so denominator becomes (1/2 sin24°) * cos24° = (1/2) sin24° cos24°Now, sin24° cos24° = (1/2) sin48°, so denominator is (1/2)*(1/2) sin48° = (1/4) sin48°Therefore, the expression is (-√3 sin48° ) / (1/4 sin48° ) = (-√3 ) / (1/4 ) = -4√3Yes! The sin48° cancels out, and we are left with -4√3. So the value of the expression is -4√3. But wait, let me check the sign again. The numerator was -2√3 sin48°, and the denominator after all simplifications became (1/4) sin48°, so when we divide them, the sin48° cancels, and we have (-2√3 ) / (1/4) which is -2√3 *4 = -8√3? Wait, wait, hold on. Wait, wait, there's a miscalculation here.Wait, in the numerator after cancelling the initial 2's, we have (-√3 sin48° ) in the numerator. Then denominator is (1/4) sin48°. So the expression becomes (-√3 sin48° ) / ( (1/4) sin48° ) = (-√3 ) / (1/4 ) = -√3 *4 = -4√3. That's correct.But let me confirm with actual numerical values to ensure I didn't make a mistake in the sign. Let's compute the original expression numerically step by step.Original expression: [ (sqrt(3) tan12° - 3) * csc12° ] / [4 cos^2 12° -2 ]Compute each part:First, compute tan12°: tan12 ≈ 0.2126sqrt(3) ≈ 1.732, so sqrt(3)*tan12 ≈ 1.732 * 0.2126 ≈ 0.368Then subtract 3: 0.368 - 3 ≈ -2.632Multiply by csc12°: csc12 ≈ 1/sin12 ≈ 1/0.2079 ≈ 4.810, so -2.632 * 4.810 ≈ -12.66Denominator: 4 cos^2 12° -2. cos12 ≈ 0.9781, so cos^2 12 ≈ 0.9566. Multiply by 4: 3.8264, subtract 2: 1.8264.Therefore, the entire expression ≈ -12.66 / 1.8264 ≈ -6.928.Now, what is -4√3? sqrt(3) ≈ 1.732, so 4*1.732 ≈ 6.928, so -4√3 ≈ -6.928. Which matches the numerical calculation. So that's correct. Therefore, the value is -4√3.But let me just recap the steps to make sure I didn't skip anything. Starting with the numerator, expressed it in terms of sine and cosine, combined terms, recognized the linear combination of sine and cosine, converted it into a single sine function with phase shift, simplified using identities, then simplified the denominator using double-angle identities, and eventually all the angle terms canceled out, leaving a simple constant. Numerically verified that the answer is indeed -4√3, which matches the calculated result. Therefore, I'm confident that the value of the expression is -4√3.Final Answerboxed{-4sqrt{3}}</think>To find the value of the expression (frac{left(sqrt{3} cdot tan 12^{circ} - 3right) cdot csc 12^{circ}}{4 cos ^{2} 12^{circ} - 2}), we start by simplifying both the numerator and the denominator.Denominator Simplification:The denominator (4 cos^2 12^{circ} - 2) can be rewritten using the double-angle identity for cosine:[4 cos^2 12^{circ} - 2 = 2(2 cos^2 12^{circ} - 1) = 2 cos 24^{circ}]Numerator Simplification:The numerator (left(sqrt{3} cdot tan 12^{circ} - 3right) cdot csc 12^{circ}) is simplified by expressing (tan 12^{circ}) and (csc 12^{circ}) in terms of sine and cosine:[left(sqrt{3} cdot frac{sin 12^{circ}}{cos 12^{circ}} - 3right) cdot frac{1}{sin 12^{circ}}]Distributing (frac{1}{sin 12^{circ}}):[frac{sqrt{3}}{cos 12^{circ}} - frac{3}{sin 12^{circ}}]Combining these terms over a common denominator (sin 12^{circ} cos 12^{circ}):[frac{sqrt{3} sin 12^{circ} - 3 cos 12^{circ}}{sin 12^{circ} cos 12^{circ}}]Recognizing the linear combination (sqrt{3} sin 12^{circ} - 3 cos 12^{circ}) as a multiple of a sine function, we use the identity (a sin theta + b cos theta = R sin(theta + phi)) where (R = sqrt{a^2 + b^2}) and (phi = arctanleft(frac{b}{a}right)). Here, (R = 2sqrt{3}) and (phi = -60^{circ}), leading to:[sqrt{3} sin 12^{circ} - 3 cos 12^{circ} = -2sqrt{3} sin 48^{circ}]Combining Numerator and Denominator:The numerator simplifies to (frac{-2sqrt{3} sin 48^{circ}}{sin 12^{circ} cos 12^{circ}}). Using the double-angle identity (sin 12^{circ} cos 12^{circ} = frac{1}{2} sin 24^{circ}), the denominator becomes:[2 cos 24^{circ} cdot frac{1}{2} sin 24^{circ} = sin 24^{circ} cos 24^{circ}]Further simplifying (sin 24^{circ} cos 24^{circ} = frac{1}{2} sin 48^{circ}), the denominator is:[frac{1}{2} sin 48^{circ}]Thus, the entire expression becomes:[frac{-2sqrt{3} sin 48^{circ}}{frac{1}{2} sin 48^{circ}} = frac{-2sqrt{3}}{frac{1}{2}} = -4sqrt{3}]The value of the expression is (boxed{-4sqrt{3}}).,<think>Okay, so I have this problem here: I need to find the value of the expression (frac{left(sqrt{3} cdot tan 12^{circ} - 3right) cdot csc 12^{circ}}{4 cos ^{2} 12^{circ} - 2}). Hmm, let's see. Trigonometry can be tricky, but maybe I can simplify this step by step. First, let me recall some trigonometric identities that might be useful here. I know that (tan theta = frac{sin theta}{cos theta}) and (csc theta = frac{1}{sin theta}). Also, the denominator has (4 cos^2 12^{circ} - 2), which looks similar to the double-angle identity for cosine. Specifically, (cos 2theta = 2cos^2 theta - 1). If I rearrange that, it becomes (2cos^2 theta - 1 = cos 2theta), so multiplying both sides by 2 gives (4cos^2 theta - 2 = 2cos 2theta). Wait, let me check that:Starting with (cos 2theta = 2cos^2 theta - 1), if I solve for (2cos^2 theta), it's (cos 2theta + 1). So, multiplying both sides by 2: (4cos^2 theta = 2cos 2theta + 2). Then subtracting 2: (4cos^2 theta - 2 = 2cos 2theta + 2 - 2 = 2cos 2theta). Oh, right! So the denominator simplifies to (2cos 24^{circ}) since (2theta = 24^{circ}) when (theta = 12^{circ}). So denominator is (2cos 24^{circ}). That's a good start.Now, let's look at the numerator: (left(sqrt{3} cdot tan 12^{circ} - 3right) cdot csc 12^{circ}). Let's break this down. First, let's express (tan 12^{circ}) as (frac{sin 12^{circ}}{cos 12^{circ}}) and (csc 12^{circ}) as (frac{1}{sin 12^{circ}}). So substituting these in, the numerator becomes:(left(sqrt{3} cdot frac{sin 12^{circ}}{cos 12^{circ}} - 3right) cdot frac{1}{sin 12^{circ}}).Let me distribute the multiplication over the subtraction. So that would be:(sqrt{3} cdot frac{sin 12^{circ}}{cos 12^{circ}} cdot frac{1}{sin 12^{circ}} - 3 cdot frac{1}{sin 12^{circ}}).Simplify the first term: (sqrt{3} cdot frac{sin 12^{circ}}{cos 12^{circ}} cdot frac{1}{sin 12^{circ}}). The (sin 12^{circ}) cancels out, leaving (sqrt{3}/cos 12^{circ}). The second term is (-3/sin 12^{circ}). So the numerator simplifies to:(frac{sqrt{3}}{cos 12^{circ}} - frac{3}{sin 12^{circ}}).Hmm, now I have the numerator as (frac{sqrt{3}}{cos 12^{circ}} - frac{3}{sin 12^{circ}}) and the denominator as (2cos 24^{circ}). Let me write the entire expression now:(frac{frac{sqrt{3}}{cos 12^{circ}} - frac{3}{sin 12^{circ}}}{2cos 24^{circ}}).Maybe I can combine the two terms in the numerator over a common denominator. The common denominator would be (cos 12^{circ} sin 12^{circ}). Let's try that:Numerator becomes:(frac{sqrt{3} sin 12^{circ} - 3 cos 12^{circ}}{cos 12^{circ} sin 12^{circ}}).So the entire expression is:(frac{sqrt{3} sin 12^{circ} - 3 cos 12^{circ}}{cos 12^{circ} sin 12^{circ} cdot 2cos 24^{circ}}).Hmm. Let's see if the numerator can be expressed in terms of a sine or cosine difference or something. The expression (sqrt{3} sin theta - 3 cos theta)... Maybe factor out a common term? Let's factor out a (sqrt{3}):Wait, (sqrt{3} sin 12^{circ} - 3 cos 12^{circ} = sqrt{3} sin 12^{circ} - sqrt{3} cdot sqrt{3} cos 12^{circ}). Wait, (sqrt{3} times sqrt{3} = 3, yes. So factoring out (sqrt{3}):(sqrt{3} (sin 12^{circ} - sqrt{3} cos 12^{circ})).But is that helpful? Maybe. Alternatively, perhaps we can write the numerator as a multiple of a sine or cosine function. Let's recall that (a sin theta + b cos theta = R sin(theta + phi)) or something similar. Wait, but here we have ( sqrt{3} sin theta - 3 cos theta ).The general formula is (a sin theta + b cos theta = R sin(theta + phi)), where (R = sqrt{a^2 + b^2}) and (tan phi = frac{b}{a}). Wait, actually, depending on the phase shift. Let me recall:Alternatively, (a sin theta + b cos theta = R sin(theta + phi)), where (R = sqrt{a^2 + b^2}) and (phi = arctanleft(frac{b}{a}right)). Hmm, but here we have (a = sqrt{3}) and (b = -3). So (R = sqrt{ (sqrt{3})^2 + (-3)^2 } = sqrt{3 + 9} = sqrt{12} = 2sqrt{3}). Then (phi = arctanleft( frac{-3}{sqrt{3}} right) = arctan(-sqrt{3})). So (phi = -60^{circ}) since (tan(-60^{circ}) = -sqrt{3}).Therefore, the numerator (sqrt{3} sin 12^{circ} - 3 cos 12^{circ}) can be written as (2sqrt{3} sin(12^{circ} - 60^{circ}) = 2sqrt{3} sin(-48^{circ})). But sine is an odd function, so (sin(-48^{circ}) = -sin 48^{circ}). Therefore, the numerator is ( -2sqrt{3} sin 48^{circ} ).Alternatively, maybe I could write it as a cosine function. Let me check that as well. Using the identity (a sin theta + b cos theta = R cos(theta - phi)), where (R = sqrt{a^2 + b^2}) and (phi = arctanleft( frac{a}{b} right)). Hmm, not sure if that helps more. Let's stick with the sine version for now.So numerator is ( -2sqrt{3} sin 48^{circ} ). So substituting back into the expression:Numerator: ( -2sqrt{3} sin 48^{circ} )Denominator: ( cos 12^{circ} sin 12^{circ} cdot 2 cos 24^{circ} )So the entire expression becomes:(frac{ -2sqrt{3} sin 48^{circ} }{ 2 cos 12^{circ} sin 12^{circ} cos 24^{circ} } )Simplify the constants: The numerator has -2√3 and the denominator has 2, so cancels out to -√3:(frac{ -sqrt{3} sin 48^{circ} }{ cos 12^{circ} sin 12^{circ} cos 24^{circ} } )Hmm. Let's see if we can manipulate the denominator. The denominator is ( cos 12^{circ} sin 12^{circ} cos 24^{circ} ). Maybe use double-angle identities here. Remember that ( sin 2theta = 2 sin theta cos theta ), so ( sin 12^{circ} cos 12^{circ} = frac{1}{2} sin 24^{circ} ). Let's apply that:So denominator becomes ( frac{1}{2} sin 24^{circ} cdot cos 24^{circ} ). Again, using the same identity: ( sin 24^{circ} cos 24^{circ} = frac{1}{2} sin 48^{circ} ). Therefore, denominator is ( frac{1}{2} times frac{1}{2} sin 48^{circ} = frac{1}{4} sin 48^{circ} ).So now, the expression is:( frac{ -sqrt{3} sin 48^{circ} }{ frac{1}{4} sin 48^{circ} } = frac{ -sqrt{3} sin 48^{circ} times 4 }{ sin 48^{circ} } = -4sqrt{3} )Wait, because the (sin 48^{circ}) terms cancel out, leaving us with (-4sqrt{3}). But let me double-check these steps because that seems like a big jump.Starting from the denominator:Original denominator: ( cos 12^{circ} sin 12^{circ} cos 24^{circ} )First step: ( cos 12^{circ} sin 12^{circ} = frac{1}{2} sin 24^{circ} ). Correct, because ( sin 2theta = 2 sin theta cos theta ), so ( sin theta cos theta = frac{1}{2} sin 2theta ).So denominator becomes ( frac{1}{2} sin 24^{circ} cdot cos 24^{circ} ). Then, ( sin 24^{circ} cos 24^{circ} = frac{1}{2} sin 48^{circ} ). So denominator is ( frac{1}{2} times frac{1}{2} sin 48^{circ} = frac{1}{4} sin 48^{circ} ). Correct. So denominator is 1/4 sin48.So the numerator is -√3 sin48. So the ratio is (-√3 sin48) / ( (1/4) sin48 ) = (-√3) / (1/4) = -√3 * 4 = -4√3. Yes, that's correct. So the entire expression simplifies to -4√3.Wait, but let me confirm again. The numerator after all the simplifications was -2√3 sin48 divided by denominator, which after simplifying step by step, became 1/4 sin48. So (-2√3 sin48) / ( (1/4) sin48 ) would be (-2√3) / (1/4) ) = -8√3. Wait, wait, hold on, maybe I messed up here.Wait, going back to the previous step:After simplifying numerator to -2√3 sin48 and denominator to 2 cos12 sin12 cos24. Then we had:Numerator: -2√3 sin48Denominator: 2 * [cos12 sin12 cos24] = 2 * [ (1/2 sin24) cos24 ] = 2 * [ (1/2 sin24 cos24) ] = 2 * [ (1/4 sin48) ] = 2 * (1/4 sin48) = (1/2 sin48)Wait, so denominator is 1/2 sin48. Therefore, the expression is:( -2√3 sin48 ) / ( (1/2) sin48 ) = (-2√3) / (1/2) = -4√3. Yes, that's correct.Wait, maybe I miscalculated earlier steps when I thought denominator was 1/4 sin48, but actually, let's re-express:Original denominator after step 1: cos12 sin12 cos24First, cos12 sin12 = 1/2 sin24, so denominator becomes 1/2 sin24 * cos24Then, sin24 cos24 = 1/2 sin48, so denominator becomes 1/2 * 1/2 sin48 = 1/4 sin48But wait, but in the previous calculation when we had the numerator as -2√3 sin48 and denominator as 2 * [cos12 sin12 cos24], which would be 2 * (1/4 sin48) ) = 1/2 sin48. Wait, no. Wait, the initial denominator was 2 cos12 sin12 cos24, right?Wait, let's go back. Let's track the denominator properly.Original denominator of the entire expression: 4 cos²12 - 2, which we converted to 2 cos24. But wait, no. Wait, initial problem:The entire expression is [ (sqrt3 tan12 - 3) csc12 ] / [4 cos²12 -2]We converted the denominator 4 cos²12 -2 to 2 cos24. That part is correct because 4 cos²θ - 2 = 2 cos2θ. So denominator is 2 cos24°.Wait, but in the numerator, after converting, we had:Numerator became [ sqrt3 / cos12 - 3 / sin12 ]Then we combined over common denominator:[ sqrt3 sin12 - 3 cos12 ] / [ cos12 sin12 ]So the entire expression is:[ (sqrt3 sin12 - 3 cos12) / (cos12 sin12) ] / [2 cos24]Which is equal to [ (sqrt3 sin12 - 3 cos12) / (cos12 sin12) ] * [1 / (2 cos24) ]Then, we wrote the numerator sqrt3 sin12 -3 cos12 as -2√3 sin48°, so replacing that:[ -2√3 sin48 / (cos12 sin12) ] * [1 / (2 cos24) ]So here, the numerator is -2√3 sin48, denominator is cos12 sin12 * 2 cos24Therefore, the total is (-2√3 sin48) / [2 cos12 sin12 cos24 ]Simplify constants: -2√3 / 2 = -√3, so:-√3 sin48 / [ cos12 sin12 cos24 ]Then, as before, cos12 sin12 = 1/2 sin24, so denominator becomes 1/2 sin24 cos24Then sin24 cos24 = 1/2 sin48, so denominator is 1/2 * 1/2 sin48 = 1/4 sin48Therefore, overall: (-√3 sin48) / (1/4 sin48) = -√3 / (1/4) = -4√3. Yes, that's correct.Wait, but in my earlier step-by-step, I had numerator as -2√3 sin48 and denominator as 2 * ... but now after rechecking, it's -2√3 sin48 divided by 2 cos12 sin12 cos24. Then 2 cancels, and then denominator is cos12 sin12 cos24, which is 1/4 sin48. So numerator is -√3 sin48, denominator is 1/4 sin48. So ratio is -√3 / (1/4) = -4√3. So answer is -4√3.Wait, but let's check if that makes sense. Alternatively, maybe I made a mistake in converting the numerator into -2√3 sin48.Let me verify that part again. So we had sqrt3 sin12 -3 cos12, which we converted using the formula a sinθ + b cosθ = R sin(theta + phi). Wait, but in this case, it's a sinθ + b cosθ, but with b negative. Let me confirm the phase shift.We had sqrt3 sin12 -3 cos12. Let me compute R:R = sqrt( (sqrt3)^2 + (-3)^2 ) = sqrt(3 +9) = sqrt(12) = 2 sqrt3. Correct.Then, the angle phi is arctan(b/a) = arctan(-3 / sqrt3) = arctan(-sqrt3) = -60 degrees. Therefore, sqrt3 sin12 -3 cos12 = 2 sqrt3 sin(12 -60) = 2 sqrt3 sin(-48) = -2 sqrt3 sin48. Yes, that's correct.Therefore, the numerator is indeed -2 sqrt3 sin48. Then, proceeding as before, the expression simplifies to -4 sqrt3.But wait, just to be 100% sure, maybe plug in approximate values.Let me compute the original expression numerically and check if it equals approximately -4√3.First, compute sqrt3 ≈1.732, tan12° ≈0.21256, csc12°≈1 / sin12°≈1 /0.2079≈4.8107.Compute numerator: sqrt3 * tan12 -3 ≈1.732 *0.21256 -3 ≈0.368 -3≈-2.632. Then multiply by csc12°≈-2.632 *4.8107≈-12.66.Denominator:4 cos²12° -2. cos12≈0.9781, so cos²12≈0.9566. Then 4*0.9566≈3.8264 -2≈1.8264.So the entire expression≈-12.66 /1.8264≈-6.93.Now, compute -4√3≈-4*1.732≈-6.928≈-6.93. So the approximate value matches. Therefore, the exact value is -4√3. So the answer is boxed{-4sqrt{3}}.Final Answerboxed{-4sqrt{3}}</think>To find the value of the expression (frac{left(sqrt{3} cdot tan 12^{circ} - 3right) cdot csc 12^{circ}}{4 cos ^{2} 12^{circ} - 2}), we start by simplifying the denominator using the double-angle identity for cosine:[4 cos^2 12^circ - 2 = 2 cos 24^circ]Next, we simplify the numerator. Express (tan 12^circ) and (csc 12^circ) in terms of sine and cosine:[left(sqrt{3} cdot frac{sin 12^circ}{cos 12^circ} - 3right) cdot frac{1}{sin 12^circ}]Distribute and simplify:[left(frac{sqrt{3} sin 12^circ}{cos 12^circ} - 3right) cdot frac{1}{sin 12^circ} = frac{sqrt{3}}{cos 12^circ} - frac{3}{sin 12^circ}]Combine the terms over a common denominator:[frac{sqrt{3} sin 12^circ - 3 cos 12^circ}{cos 12^circ sin 12^circ}]Recognize the numerator as a multiple of a sine function using the identity (a sin theta + b cos theta = R sin(theta + phi)):[sqrt{3} sin 12^circ - 3 cos 12^circ = -2sqrt{3} sin 48^circ]Thus, the numerator becomes:[frac{-2sqrt{3} sin 48^circ}{cos 12^circ sin 12^circ}]The denominator simplifies using double-angle identities:[cos 12^circ sin 12^circ cos 24^circ = frac{1}{4} sin 48^circ]Combining the numerator and denominator:[frac{-2sqrt{3} sin 48^circ}{frac{1}{4} sin 48^circ} = -8sqrt{3}]However, correcting the simplification steps, we find the final value is:[frac{-sqrt{3} sin 48^circ}{frac{1}{4} sin 48^circ} = -4sqrt{3}]Thus, the value of the expression is:[boxed{-4sqrt{3}}]]

🔑:4.begin{aligned}text { Original expression } & =frac{left(frac{sqrt{3} sin 12^{circ}}{cos 12^{circ}}-3right) cdot frac{1}{sin 12^{circ}}}{4 cos ^{2} 12^{circ}-2} text { (convert tangent and cotangent to sine and cosine) } & =frac{sqrt{3} sin 12^{circ}-3 cos 12^{circ}}{2 sin 12^{circ} cos 12^{circ}left(2 cos ^{2} 12^{circ}-1right)} & =frac{2 sqrt{3}left(frac{1}{2} sin 12^{circ}-frac{sqrt{3}}{2} cos 12^{circ}right)}{sin 24^{circ} cos 24^{circ}} text { (reverse use of double angle formula) } & =frac{2 sqrt{3}left(sin 12^{circ} cos 60^{circ}-cos 12^{circ} sin 60^{circ}right)}{sin 24^{circ} cos 24^{circ}} text { (constant transformation) } & =frac{4 sqrt{3} sin left(12^{circ}-60^{circ}right)}{2 sin 24^{circ} cos 24^{circ}} text { (reverse use of difference angle formula) } & =frac{4 sqrt{3} sin left(-48^{circ}right)}{sin 48^{circ}}=-4 sqrt{3} text { (reverse use of double angle formula). }end{aligned}

🧠:[<think>Okay, let's try to figure out the minimum number of water sources Beto needs to guarantee that all platforms get filled, no matter how Ana arranges the heights. Hmm, so the garden is an n x n grid, each 1x1 platform has a unique height. Beto places water sources on some platforms, and water can only flow downhill—meaning from higher to lower adjacent platforms. Beto wants to place the fewest possible water sources such that, regardless of how Ana assigned the heights, all platforms will be watered.First, let me rephrase the problem. Since Ana assigns heights in a permutation of 1 to n², each platform has a distinct height. Beto needs to choose a set of platforms such that every possible permutation (height arrangement) will result in the entire grid being covered by water starting from those sources. The key here is that Beto's choice must work for any possible height arrangement. So, we need a set of cells that are "universal" in some sense—no matter how the heights are ordered, these cells will be the sources that can cover the entire grid through the water flow.Let me think about smaller cases. Maybe n=1? Then the garden is just 1x1. Beto needs only 1 water source, obviously. For n=2, a 2x2 grid. Let's see. If Ana arranges the heights in some order, Beto needs to place water sources such that regardless of the permutation, all cells get watered. What's the minimum number here?Suppose Beto places one water source. If that source is on the highest cell, then water will flow to all lower cells. But the problem is that Ana could arrange the heights such that the source Beto chose is not the highest. Wait, but Beto doesn't know Ana's arrangement. So if Beto picks any cell, Ana could make that cell not the highest, so water wouldn't flow to the higher cells. Therefore, Beto can't just pick one source because Ana could make that cell low, and there would be higher cells not covered.Wait, but water can't flow uphill. So if Beto places a water source on a cell, it can only water cells that are lower and connected via a path of decreasing heights. But if Ana arranges heights such that the water source is in a local minimum, then water can't flow anywhere. Therefore, Beto's sources need to be placed in such a way that, no matter the height arrangement, every cell is either a source or can be reached by a descending path from a source.So, the problem reduces to finding the minimum number of cells such that in any permutation of heights, every cell is either one of these sources or lies on a descending path from one of the sources. Hmm.This seems related to the concept of "hitting sets" in graph theory. Maybe in the context of directed graphs where edges go from higher to lower cells. But since the heights are arbitrary, the direction of the edges is arbitrary except that they must form a DAG (since all heights are distinct, there are no cycles). Wait, but in reality, for any permutation, the directed graph is a DAG where each edge goes from higher to lower cell. Beto needs to choose a set of nodes such that every node is reachable from at least one of these nodes, regardless of the DAG's structure.But since the permutation is arbitrary, the DAG could be any possible linear extension of the grid's poset. Wait, perhaps another way: the problem is equivalent to covering the grid with sources such that in any possible total order, the sources are positioned in such a way that every other element is below at least one source in the order and can be reached via a descending path in the grid.But I'm not sure. Maybe another approach: consider that in any permutation, the highest cell must be a source, otherwise, water cannot reach it. But if Beto doesn't choose the highest cell as a source, and Ana made that the highest cell, then Beto loses. But Beto doesn't know where Ana placed the highest cell. Therefore, Beto must place a source on every cell that could potentially be the highest in some permutation. But since Ana could choose any permutation, the highest cell could be any cell. Therefore, Beto would need to place a source on every cell, which is impossible because the question asks for the minimum number.Wait, that can't be right. There must be a smarter way. Because even if the highest cell is not a source, but if there is a path from a source to the highest cell through increasing heights, but water can't flow uphill. So actually, the highest cell cannot be watered unless it is itself a source. Therefore, in order to cover all possible permutations, Beto must place a source on every cell that could be the highest in some permutation. But since in any permutation, any cell can be the highest, does that mean Beto needs to place a source on every cell? That can't be the answer.Wait, no. Because the water flows downhill. So if you have a source at some cell, then it can water all cells that are lower than it and reachable via a descending path. However, the highest cell in the entire grid cannot be watered by any other cell, since all others are lower. Therefore, the highest cell must be a source itself. Similarly, the second-highest cell, if it's adjacent to the highest cell, could be watered by the highest cell. But if it's not adjacent, then it needs to be a source. Wait, but Ana could arrange the heights such that the second-highest cell is isolated from the highest cell, i.e., not adjacent. Therefore, Beto would need to place a source on that second-highest cell as well. But Ana could do this for any cell. Therefore, is the minimal number of sources n, the size of a maximal set of cells such that no two are adjacent? Wait, that might be a dominating set problem.Alternatively, maybe it's related to the concept of a "feedback vertex set" or something else. Let me think differently.Suppose that in any permutation, the set of sources must include all the peaks. A peak is a cell that is higher than all its neighbors. Because water cannot flow into a peak from any neighbor, so if a peak is not a source, it can't be watered. Therefore, Beto must place a source on every peak. But Ana could arrange the heights to create as many peaks as possible. Therefore, the minimal number of sources Beto needs is equal to the maximum number of peaks that Ana can create. So the problem reduces to: what's the maximum number of peaks possible in an n x n grid, where a peak is a cell higher than all four (or fewer, if on the edge) adjacent cells.Then, Beto needs to place at least that many sources to cover all peaks. But is this correct? Let's see. If Ana can create a permutation with k peaks, then Beto needs at least k sources. Conversely, if Beto places k sources, each being a peak in some permutation, then Ana could create a permutation where those k cells are peaks, hence Beto needs at least k sources. Therefore, the minimal number of sources Beto needs is equal to the maximum number of peaks that Ana can create in the grid.Therefore, the problem reduces to finding the maximum number of peaks in an n x n grid. Then, the minimal number of water sources Beto needs is equal to that maximum number. So, the question becomes: what's the maximum number of peaks possible in an n x n grid?A peak is a cell that is higher than all its adjacent cells. How can we arrange as many peaks as possible? To maximize the number of peaks, we need to arrange the heights such that as many cells as possible are higher than all their neighbors. Since peaks cannot be adjacent to each other (otherwise, one would be lower than the other), so they must be placed in a way that no two peaks are adjacent. So, the problem of finding the maximum number of peaks is equivalent to finding the maximum independent set in the grid graph. In graph theory, an independent set is a set of vertices with no two adjacent. For a grid graph (n x n), the maximum independent set is known. For a chessboard, it's the number of black or white squares, whichever is larger. For even n, it's n²/2. For odd n, it's (n² + 1)/2.Wait, for example, in a chessboard, the maximum independent set is 32 for an 8x8 grid. For a 3x3 grid, it's 5, since there are 9 squares, and the independent set can be 5. But wait, does this correspond to peaks?But in the case of peaks, can we actually have an independent set of peaks? Because each peak must be higher than its neighbors, so two peaks cannot be adjacent. Therefore, the maximum number of peaks is indeed the maximum independent set of the grid graph. Therefore, the maximum number of peaks is ⌈n²/2⌉. Wait, but wait, in a grid graph, the maximum independent set is actually known to be ⌈n²/2⌉. For even n, it's exactly n²/2. For odd n, it's (n² + 1)/2. For example, 3x3 grid has 9 cells, so (9 + 1)/2 = 5. So, maximum independent set is 5.But hold on, if that's the case, then the maximum number of peaks Ana can create is equal to the size of the maximum independent set. Therefore, Beto needs to place at least that many sources. But is this correct?Wait, but in reality, even if two cells are not adjacent, can they both be peaks? Let's take a 2x2 grid. The maximum independent set is 2. Let's check if Ana can make two peaks. Suppose the grid has four cells. If Ana assigns the highest value to the top-left cell, the second-highest to the bottom-right cell, then both of these cells would be peaks if their neighbors are lower. But in a 2x2 grid, each cell has two neighbors. If top-left is the highest, then its neighbors (top-right and bottom-left) must be lower. Similarly, bottom-right is the second-highest. But then the neighbors of bottom-right are top-right and bottom-left, which are already lower than bottom-right. Wait, but in a 2x2 grid, if top-left is 4, bottom-right is 3, then top-right and bottom-left could be 2 and 1. Then, top-left (4) is a peak, and bottom-right (3) is also a peak because its neighbors are 2 and 1. So yes, in a 2x2 grid, you can have two peaks. Therefore, maximum independent set size is 2, which is equal to n²/2. So Beto would need 2 sources. But wait, can Beto cover all cells with 2 sources? Let's see. Suppose he places sources on two diagonal cells. If Ana assigns those two as the highest and second-highest, then water from the highest will flow to its neighbors, and the second-highest (if placed on the other diagonal) would flow to its neighbors. But in the 2x2 case, if sources are on, say, top-left and bottom-right, then if those are the two highest, water from top-left (4) flows to top-right (2) and bottom-left (1). Water from bottom-right (3) flows to top-right (2) and bottom-left (1). So all cells get watered. But if Ana had arranged the heights such that the two sources are not the highest, then water wouldn't flow to the actual highest cells. Wait, but Beto's sources are fixed. Wait, no, the problem says Beto places water sources in some platforms after Ana has placed the heights. Wait, no—no, actually, let me check the problem statement again."Ana places all the platforms at a different height. Afterwards, Beto places water sources in some of the platforms."Wait, so first Ana arranges the heights, then Beto places the water sources knowing Ana's arrangement. Wait, but the problem says "Find the least number of water sources that Beto needs to win no matter how Ana places the platforms." So Beto has to choose the number of sources such that, for any Ana's arrangement, Beto can choose that number of sources (specific to Ana's arrangement) to flood everything. But the question is about the minimal number k such that, regardless of Ana's arrangement, Beto can choose k sources to flood everything. So Beto can choose the sources after seeing Ana's arrangement, but we need the minimal k such that for any arrangement, there exists a set of k sources that can flood the entire garden.Ah, this is different. Initially, I thought Beto had to choose the sources before Ana's arrangement, but actually, Beto can choose the sources after Ana sets the heights, but Beto must use a number k that works for any Ana's arrangement. So the minimal k such that for any height permutation, there exists a set of k cells which are sources, such that all other cells are downhill from some source.Therefore, in other words, we need the minimal k such that every permutation of the grid can be covered by k sources. So this is equivalent to the concept of the "dual" of the permutation's DAG. For any permutation (which is a linear order), the DAG has edges from higher to lower cells. Then, the minimal number of sources needed is the minimal number of nodes such that every node is reachable from at least one source. This is called the minimum number of "kernels" or "dominating sets" in some contexts, but here it's specifically for DAGs.In any DAG, the minimal number of sources required to reach all nodes is equal to the number of minimal elements in the DAG's poset. Wait, but in a permutation, the DAG is a total order, so there's only one minimal element (the lowest cell) and one maximal element (the highest cell). Wait, but that's not right. Wait, in the permutation, the DAG is actually a Hasse diagram where each cell is connected to its adjacent cells only if they are lower. So it's not a total order, but a grid where each cell is connected to its adjacent cells, and edges go from higher to lower.But in that case, the DAG is not necessarily a linear order. Wait, perhaps I need to think in terms of graph theory. The problem is equivalent to: given a graph (the grid), and a orientation of the edges where each edge is directed from higher to lower node (since all heights are distinct, it's a DAG), find the minimal k such that in any such orientation, the DAG has a set of k nodes with out-degree zero (sources) such that all other nodes are reachable from them.Wait, no. In our case, the water sources are nodes where Beto places the water, so they are the actual sources in the DAG (nodes with in-degree zero). Because water starts there and flows out. So in the DAG, the water sources are the nodes with no incoming edges (since water can't come into them from elsewhere). But in reality, the DAG can have multiple sources and sinks. The sinks would be the local minima.But Beto can choose any nodes as water sources, regardless of their in-degree. Wait, actually, when Beto places a water source on a node, it doesn't matter its in-degree; water will flow out from it to its neighbors if they are lower. So in terms of the DAG, placing a water source on a node allows all nodes reachable from it (following the directed edges) to be watered. Therefore, the problem reduces to: given any DAG which is a orientation of the grid graph (with edges going from higher to lower), find the minimum number k such that there exists k nodes (water sources) that can reach all other nodes in the DAG. Then, the answer is the maximum such k over all possible orientations (since Beto needs to have k sources that work for any orientation).But this seems too abstract. Maybe there's a pattern or known result.Alternatively, think about the problem as similar to the concept of "covering" the grid with sources such that every possible height permutation allows some set of k sources to cover the grid. So we need to determine the minimal k such that in every possible labeling of the grid with distinct heights, there exists a set of k cells whose descending paths cover the entire grid.This is similar to the concept of the "Grundy number" or "coloring number" but in a different context. Alternatively, think in terms of maximum matching.Wait, perhaps another angle: For any permutation of the grid, the minimum number of sources required is equal to the number of "basins" in the topography, where a basin is a region that drains to a particular local minimum. But since Beto can choose the sources, he can strategically pick cells that are high enough to cover multiple basins. However, since Ana can arrange the heights adversarially, she can create as many basins as possible, forcing Beto to use more sources. But how many basins can Ana create?In the worst case, Ana can arrange the grid to have as many local minima as possible. Each local minimum would require a source unless there's a path from another source to it. But a local minimum is a cell that is lower than all its neighbors, so water cannot flow into it from any neighbor. Therefore, each local minimum must be a source itself. Therefore, the maximum number of local minima that Ana can create is equivalent to the maximum number of sources Beto might need.Wait, but this contradicts the previous thought about peaks. Wait, no: local minima are cells that are lower than all neighbors, so they must be sources because water can't flow into them. Conversely, peaks (local maxima) are cells that can't be watered unless they are sources. Wait, but the problem is about water flowing downhill, so water sources spread to lower cells. Therefore, a local maximum can only be watered if it is a source, because all its neighbors are lower. Therefore, if Ana makes a local maximum not a source, it can't be watered. Similarly, a local minimum must be a source because otherwise, you can't get water into it. Wait, but water sources are placed by Beto. If Beto places a source in a local minimum, water can only stay there. But local minima are lower than all neighbors, so water can't flow out. Therefore, if Beto places a source in a local minimum, it only waters that cell. To water other cells, he needs sources in higher cells.Therefore, perhaps the key is that to cover the entire grid, Beto must place sources in a set of cells such that every cell is either a source or is reachable via a descending path from a source. Given that Ana can arrange the heights adversarially, Beto needs to choose a set of sources that are "independent" in some way.Alternatively, think of the grid as a graph where each node must be either a source or downstream from a source. Since Ana can choose the orientation of edges (by permuting heights), Beto needs a set S of nodes such that, for any orientation (which is a DAG), every node is either in S or reachable from S. This is called a "universal covering set" for all possible orientations. The minimal size of such a set is what we need.This seems related to the concept of "king nodes" in tournaments. In a tournament (a complete graph with oriented edges), a king is a node that can reach every other node in at most two steps. It's known that every tournament has a king. However, our problem is different because the graph is a grid, not a complete graph, and orientations are constrained to be DAGs with edges going from higher to lower.Alternatively, maybe we can model this as a graph where each node has a potential to cover its "lower closure" (all nodes that can be reached by descending paths from it). Beto needs to choose a set of nodes whose lower closures cover the entire graph, regardless of the height permutation (i.e., regardless of the orientation). So the question is: what's the minimal size of a set S such that, for any possible orientation (DAG) of the grid where edges go from higher to lower, the union of the lower closures of S covers all nodes.This is equivalent to saying that S is a hitting set for all possible upper sets in the grid's poset under any permutation. Wait, maybe not. Let me think.Alternatively, since the orientations correspond to all possible linear extensions of the grid poset, the problem reduces to finding a set of elements that intersect every maximal chain in the poset. But I need to be careful here.Wait, another idea: In order for Beto to cover all platforms, his set of sources must form a "dominating set" in every possible orientation. However, since orientations can vary, this seems too vague.Wait, here's a different approach inspired by the permutation argument. Since Ana can arrange the heights in any order, we need that for any permutation, there exists a set of k cells such that every cell is either one of the k or has a higher neighbor. Because if a cell has a higher neighbor, then water can flow from that neighbor down to it. Therefore, the problem reduces to: find the smallest k such that in any permutation of the grid, there exists a set S of k cells where every cell not in S has a neighbor in S with a higher value.This is equivalent to saying that S is a "dominating set" for the grid graph, but with the domination condition being that each cell not in S is adjacent to a cell in S that is higher in the permutation. The question then becomes: what is the minimal k such that in any labeling of the grid, there exists a dominating set S of size k where every cell not in S is adjacent to a higher cell in S.This seems related to the concept of "alpine" graphs or "uphill dominating sets," but I'm not sure if such a term exists. Alternatively, maybe this is similar to the idea that in any graph labeling, there exists a small dominating set where each dominated node has a higher neighbor in the set.Alternatively, consider that in any permutation, the highest cell must be in S, otherwise, its neighbors cannot be higher than it. Wait, but if the highest cell is not in S, then it must have a neighbor in S that is higher, but the highest cell is the maximum, so it can't have a higher neighbor. Therefore, the highest cell must be in S. Similarly, the second-highest cell: if it's adjacent to the highest cell, then it can be dominated by the highest cell (since the highest cell is in S and is higher). If it's not adjacent to the highest cell, then the second-highest cell must be in S. Wait, but Ana could arrange the heights such that the second-highest cell is not adjacent to the highest cell. Therefore, S must include the highest cell and any cell not adjacent to it that is the second-highest.But this seems recursive. Wait, let's formalize it. In any permutation, the maximum element (highest cell) must be in S. Then, consider the remaining grid. The next highest element, if it's adjacent to the maximum, is already dominated. If it's not adjacent, it must be in S. But Ana can arrange the heights such that multiple high cells are not adjacent to the maximum or to each other. Therefore, S would need to include all these non-adjacent high cells.This seems similar to the problem of selecting a set of cells such that every cell is either in S or adjacent to S, and S is an independent set. Wait, but in our case, S doesn't need to be independent. However, Ana can arrange heights such that S must include all the local maxima.But perhaps another angle: the problem requires that in any permutation, the set S must contain all local maxima. Because a local maximum is a cell higher than all its neighbors, so unless it's in S, it cannot be dominated by any neighbor (since all neighbors are lower). Therefore, the minimal S must include all local maxima. However, Ana can create as many local maxima as possible. The maximum number of local maxima in an n x n grid is equal to the size of the maximum independent set, which is ⌈n²/2⌉ as discussed before. Therefore, if Ana creates that many local maxima, then Beto must place a source on each, which would require ⌈n²/2⌉ sources. But this seems too large, and contradicts the example for n=2.Wait, for n=2, the maximum independent set is 2, so Beto would need 2 sources. Let's verify. In a 2x2 grid, if Ana assigns the heights such that two diagonal cells are local maxima. Then, Beto must place sources on both. However, in reality, in a 2x2 grid, you can only have at most two local maxima if they are on opposite corners. For example, heights arranged as:4 23 1Here, 4 and 1 are local maxima? Wait, 4 is higher than its neighbors 2 and 3, so it's a local maximum. 1 is lower than its neighbors 2 and 3, so it's a local minimum. Wait, maybe my previous reasoning was wrong. Let's see.Wait, a local maximum is higher than all adjacent cells. So in a 2x2 grid, the highest cell is a local maximum, and the second-highest cell could be adjacent to it or not. If Ana places the two highest cells on opposite corners, then each is a local maximum. For example:4 23 1Here, 4 is a local maximum (adjacent to 2 and 3). 1 is a local minimum. 2 is adjacent to 4 and 1; since 2 < 4 and 2 > 1, so 2 is not a local maximum. 3 is adjacent to 4 and 1; 3 < 4 and 3 > 1, so not a local maximum. Therefore, only one local maximum. Wait, so maybe in a 2x2 grid, you can only have one local maximum. Wait, that contradicts the previous thought.Wait, suppose Ana arranges the heights as:3 42 1Then, 4 is a local maximum (adjacent to 3 and 2). 1 is a local minimum. 3 and 2 are not local maxima. So only one local maximum.Alternatively, if Ana arranges:4 32 1Again, 4 is the only local maximum.Wait, perhaps in a 2x2 grid, you can have at most one local maximum. Because if the highest cell is in one corner, then its two neighbors are lower, so no other cell can be a local maximum. If the highest cell is in the center, but in 2x2 there is no center cell. Wait, 2x2 grid has four cells. So maybe in 2x2 grid, the maximum number of local maxima is 1. Then, in this case, Beto needs only 1 source. But wait, in the initial arrangement where the highest cell is the only local maximum, Beto can just place a source there. However, Ana could arrange the heights such that there's a local maximum that's not the highest cell. Wait, no. A local maximum is higher than all its neighbors. So if Ana makes a cell a local maximum, it has to be higher than its neighbors. But the highest cell in the grid is higher than all cells, including its neighbors. Therefore, the highest cell is always a local maximum. Any other local maximum must be lower than the highest cell but higher than their neighbors. However, if the highest cell is adjacent to another local maximum candidate, then that candidate cannot be a local maximum because it's adjacent to a higher cell. Therefore, in a 2x2 grid, the only local maximum is the highest cell, and the number of local maxima is 1. Therefore, Beto only needs 1 source. But earlier, I thought that in a 2x2 grid, Beto needs 2 sources. Which is correct?Wait, let's take an example. Suppose Ana arranges the 2x2 grid as follows:4 32 1Here, 4 is the highest, a local maximum. 3 is adjacent to 4 and 1. Since 3 < 4, it's not a local maximum. 2 is adjacent to 4 and 1; 2 < 4, not a local maximum. 1 is a local minimum. So only 4 is a local maximum. If Beto places a source on 4, water flows to 3 and 2, which then flow to 1. So all cells are watered.Alternatively, suppose Ana arranges the grid as:3 42 1Here, 4 is still the highest, local maximum. 3 is adjacent to 4 and 2; 3 < 4, so not a local maximum. Same result.Alternatively, what if Ana arranges:2 43 1Here, 4 is the highest, local maximum. 2 is adjacent to 4 and 3. 2 < 4 and 2 < 3, so not a local maximum. 3 is adjacent to 2 and 1. 3 > 1 and 3 > 2, but 3 < 4. Wait, but 3 is adjacent to 2, which is lower, and 1, which is lower. Wait, 3 is higher than its neighbors 2 and 1. But 3 is not the highest cell in the grid. So 3 is a local maximum. Similarly, 4 is a local maximum. So here, in this arrangement, there are two local maxima: 3 and 4. Therefore, Beto needs to place sources on both 3 and 4 to water the entire grid.Wait, this is a critical point. If Ana can arrange the grid such that there are two local maxima, then Beto needs to place sources on both. Let's verify.In the arrangement:2 43 1Here, 4 is a local maximum (higher than neighbors 2 and 1). 3 is a local maximum (higher than neighbors 2 and 1). Wait, 3 is adjacent to 2 and 1. 3 is higher than both, so yes, it's a local maximum. 4 is adjacent to 2 and 1. 4 is higher than both, so it's a local maximum. 2 is adjacent to 4 and 3, both of which are higher. 1 is adjacent to 4 and 3, both higher. Therefore, yes, two local maxima: 3 and 4. Therefore, Beto must place sources on both 3 and 4. Otherwise, if he only places a source on 4, then cell 3 will not get watered because it's a local maximum. Similarly, if he places a source only on 3, cell 4 won't get watered. Therefore, Beto needs two sources here. So in a 2x2 grid, the maximum number of local maxima is 2. Therefore, the minimal number of sources Beto needs is 2.But wait, how is this possible? In the 2x2 grid, if two diagonal cells are local maxima, then they are not adjacent. Because in the example above, cells (1,2) and (2,1) in a 2x2 grid (assuming matrix coordinates) are diagonal, so not adjacent. Therefore, local maxima can be non-adjacent, so you can have two in a 2x2 grid. Hence, the maximum number of local maxima is 2, which is equal to the size of the maximum independent set (2 in a 2x2 grid). Therefore, this suggests that Beto needs to place sources equal to the size of the maximum independent set. For 2x2, that's 2. For 3x3, maximum independent set is 5, so Beto needs 5 sources. But let's check.In a 3x3 grid, can Ana arrange the heights to have 5 local maxima? Let's see. The maximum independent set in a 3x3 grid is 5 cells, arranged such that no two are adjacent. For example, the four corners and the center cell. Wait, but center cell is adjacent to all four edges. Alternatively, a checkerboard pattern where we take all the black squares. In a 3x3 grid, there are 5 black squares if we start with black at the corners. Wait, no: 3x3 grid has 9 cells. A checkerboard pattern would have 5 of one color and 4 of the other. For example, (1,1), (1,3), (2,2), (3,1), (3,3) are black, totaling 5. These cells are not adjacent. If Ana assigns the highest 5 values to these cells, making each a local maximum, then Beto would need to place sources on all 5. But can Ana actually do that?Let's try to create a 3x3 grid where the 5 black cells are local maxima. Assign the highest value, say 9, to the center cell (2,2). Then assign the next four highest values (8,7,6,5) to the four corner cells. The remaining four cells (edges) get the lower values (4,3,2,1). Now, check if the corners are local maxima. Take corner (1,1). Its neighbors are (1,2) and (2,1). If (1,1) is 8, and its neighbors are 4 and 3, then yes, (1,1) is a local maximum. Similarly, (1,3) is 7, neighbors are (1,2)=4 and (2,3)=2. Local maximum. (3,1)=6, neighbors (3,2)=1 and (2,1)=3. Local maximum. (3,3)=5, neighbors (3,2)=1 and (2,3)=2. Local maximum. Center cell (2,2)=9, neighbors all around are 4,3,2,1. So also a local maximum. Therefore, all 5 black cells are local maxima. Thus, Beto would need to place sources on all 5. Therefore, the minimal number of sources needed is equal to the size of the maximum independent set.Therefore, generalizing, the maximum number of local maxima Ana can create is equal to the size of the maximum independent set of the grid graph, which is ⌈n²/2⌉. Therefore, Beto needs at least ⌈n²/2⌉ sources. But is this tight? In other words, can Beto always cover the grid with ⌈n²/2⌉ sources, regardless of Ana's arrangement?Wait, let's think. If Beto places sources on all cells of a maximum independent set, which are non-adjacent, then in any permutation, those cells are either local maxima or not. But if Ana has arranged the heights such that those cells are not local maxima, then their neighbors are higher, so water can flow into them from the neighbors. But since Beto can choose any cells as sources, he can choose the maximum independent set cells. But regardless of how Ana permutes the heights, Beto can choose a set of sources such that every local maximum is included. However, since Ana can choose the local maxima to be anywhere, but Beto can choose the sources after seeing Ana's arrangement. Wait, no—the problem states that Beto must choose the number of sources such that no matter how Ana arranges the heights, Beto can choose that number of sources to cover all platforms. So the number k must be such that, for any Ana's permutation, there exists a set of k cells that are sources, covering the entire grid.Therefore, k is the minimal number such that, in any permutation, there exists a dominating set of size k where each cell not in the set has a neighbor in the set with a higher value. This is equivalent to the minimal size k such that every permutation has a feedback vertex set of size k, which intersects every descending path. But this is unclear.Alternatively, think of it as a two-player game. Ana picks a permutation to maximize the number of required sources, and Beto picks the minimal k that works for all permutations. The previous examples suggest that the required k is the size of the maximum independent set, as Ana can make each cell in the independent set a local maximum, forcing Beto to place a source on each. However, if Beto can choose sources after seeing the permutation, perhaps he can do better.Wait, in the 2x2 example, if Ana makes two local maxima, Beto needs to place two sources. If Ana makes one local maximum, Beto can get away with one source. But the problem requires that Beto can guarantee a win no matter how Ana arranges the heights. Therefore, Beto must assume the worst-case scenario, which is Ana creating as many local maxima as possible. Hence, the minimal k is equal to the maximum number of local maxima possible in the grid, which is the size of the maximum independent set.Therefore, for an n x n grid, the maximum independent set has size ⌈n²/2⌉. Hence, the minimal number of sources Beto needs is ⌈n²/2⌉. But wait, let's check for n=1: 1, which matches. For n=2: 2, which matches the earlier example. For n=3: 5, which is (9+1)/2=5. So this seems to hold.But wait, let's think again. If Beto can choose the sources after seeing Ana's arrangement, maybe he doesn't need to use the maximum independent set. For instance, if Ana arranges the grid with only one local maximum, Beto can just use one source. However, if Ana can arrange it with, say, 5 local maxima (as in 3x3), then Beto needs 5 sources. Since the problem requires that Beto can win no matter how Ana places the platforms, he needs to have enough sources to cover the worst-case scenario, which is when Ana makes as many local maxima as possible. Therefore, the minimal number of sources required is equal to the maximum possible number of local maxima, which is the size of the maximum independent set.Therefore, the answer should be the size of the maximum independent set for the n x n grid graph, which is ⌈n²/2⌉. For even n, it's n²/2, for odd n, it's (n² + 1)/2. Hence, the least number of water sources Beto needs is the ceiling of n² divided by 2.But wait, let's verify with n=3. If Beto uses 5 sources, then in any permutation, those 5 sources can cover the grid. But how? Suppose Ana arranges the grid such that there are 5 local maxima. Then, Beto has to place sources on all 5. But if Ana arranges it with fewer local maxima, Beto can choose fewer sources. But the problem says "no matter how Ana places the platforms," so Beto must be able to cover all possibilities, including the case where Ana makes the maximum number of local maxima. Hence, the minimal k is indeed the maximum number of local maxima, which is the maximum independent set size.Therefore, the answer is ⌈n²/2⌉. For example, n=2 gives 2, which works. n=3 gives 5, which also works. So generalizing, the least number of water sources Beto needs is the ceiling of n squared over two.But wait, let me check another example. Take a 1x3 grid (though the problem is for n x n, but still). The maximum independent set is 2. If Ana arranges the heights as [3,1,2], then the local maxima are 3 and 2. So Beto needs two sources. If it's 1x3, the answer would be 2. For a 3x3 grid, it's 5. Hence, the pattern holds.Therefore, the answer should be boxed{leftlceil dfrac{n^2}{2} rightrceil}.Final Answerboxed{leftlceil dfrac{n^2}{2} rightrceil}</think>To determine the least number of water sources Beto needs to fill all platforms in an ( n times n ) garden, regardless of how Ana places the platforms, we need to consider the maximum number of local maxima (peaks) Ana can create. A local maximum is a cell higher than all its adjacent cells, and water cannot flow into it from any neighbor, so it must be a source.Key steps:1. Local Maxima and Independent Sets: The problem reduces to finding the maximum number of local maxima Ana can create. This is equivalent to finding the maximum independent set in the grid graph, where no two nodes are adjacent. For an ( n times n ) grid, the maximum independent set size is ( leftlceil frac{n^2}{2} rightrceil ).2. Verification with Examples: For smaller grids (e.g., ( 2 times 2 ) and ( 3 times 3 )), we verified that the number of local maxima matches the maximum independent set size. For example, a ( 2 times 2 ) grid can have 2 local maxima, and a ( 3 times 3 ) grid can have 5 local maxima.3. Generalization: The maximum number of local maxima (and thus the minimum number of sources required) for any ( n times n ) grid is given by the ceiling of half the total number of cells, which is ( leftlceil frac{n^2}{2} rightrceil ).Thus, the least number of water sources Beto needs to win no matter how Ana places the platforms is (boxed{leftlceil dfrac{n^2}{2} rightrceil}).,<think>Okay, let's try to figure out this problem. So, we have an n x n garden, right? Ana is going to place platforms of different heights on each 1x1 cell. Then Beto has to place water sources in some platforms such that water can flow to all platforms, moving only to adjacent (sharing a side) lower platforms. We need to find the minimum number of water sources Beto needs to guarantee that no matter how Ana arranges the heights, he can always flood the entire garden.Hmm, so first, the key points here are:1. Ana assigns a unique height to each platform. So every cell has a distinct height.2. Beto places water sources, and water flows from higher to lower adjacent platforms.3. We need the minimum number of water sources such that regardless of how Ana assigns the heights, Beto can cover the entire garden.So, this seems related to graph theory. If we model the garden as a graph where each cell is a node, and there's a directed edge from a higher platform to a lower adjacent platform. Then Beto needs to choose a set of nodes such that every node is reachable from at least one of these sources. This is called a dominating set in some contexts, but here it's specifically about the directed edges based on heights.But since the heights are assigned adversarially (Ana can choose any permutation), we need a strategy that works for any possible height arrangement.Wait, so in other words, for any possible directed acyclic graph (since heights are all different, there's no cycles) formed by the grid with edges pointing from higher to lower, we need the minimum number of sources such that every node is reachable from at least one source. The DAG here is a tournament in some sense because between any two adjacent cells, there is a direction determined by their heights. But actually, it's not a complete tournament since only adjacent cells have edges.But regardless, since Ana can choose the heights in any way, Beto has to choose sources that work for any possible DAG orientation. Therefore, the problem reduces to finding the minimum number of nodes needed such that every node is reachable from at least one of them in any DAG that is a permutation of the grid's adjacency.Hmm. Maybe another way to think about it: in any permutation of heights on the grid, what is the minimum number of starting points needed so that every cell is on a decreasing path from at least one starting point.This sounds similar to the concept of "greedy dominating sets" or something else. Wait, in permutation matrices, the longest increasing subsequence, but this is different.Alternatively, maybe we can model this as a graph where each node must be covered by a source or be downstream from a source. Since the heights are arbitrary, we need a set of sources such that in any possible orientation (assignment of directions based on heights), all nodes are reachable.This is similar to a universal covering set for all possible orientations. What's known about such sets?Alternatively, think about it in terms of the undirected graph. Each edge can be directed either way, depending on the heights. To cover all nodes, regardless of the edge directions, the sources must be such that in every possible orientation, the sources can reach all nodes.This is equivalent to saying that the set of sources must be a "hull" that intersects every possible "sink component" in any orientation. Wait, but how?Wait, in any DAG, there is at least one sink (a node with no outgoing edges). Similarly, in any DAG, the strongly connected components are single nodes, and the condensation is a DAG. So, in this case, the entire graph is a DAG, so it has a topological order.But how does that help?Alternatively, think about the problem as a game: Ana is trying to arrange heights to make Beto's job as hard as possible. Beto picks some number of sources, and Ana tries to arrange the heights so that the maximum number of sources is needed. But we need the minimal number such that no matter what Ana does, Beto's sources will cover everything.Another angle: in any permutation of numbers on a grid, what is the minimal number of cells needed such that every cell is part of a decreasing sequence starting from one of these cells. Decreasing here means each step is to a lower adjacent cell.Wait, in 1D, if you have a permutation of numbers 1 to n, the minimal number of sources needed so that every element is part of a decreasing sequence starting from a source. In 1D, the answer is 1, because you can just pick the highest element, and then the sequence will decrease from there. But in 2D, it's more complicated.But actually, in 1D, if Ana arranges the numbers in increasing order, then if Beto only puts a source at the leftmost element (the highest), then water can flow rightwards if the numbers are decreasing to the right. But if Ana arranges the numbers in a way that the maximum is somewhere in the middle, then water can flow both left and right. Wait, but in 1D, regardless of how Ana permutes the numbers, the maximum element is a single cell. If Beto places a source there, then water can flow to all other cells, since from the maximum, you can go to any adjacent cell (since they are lower), and from there, continue. So in 1D, the minimal number is 1, because the maximum element must be placed somewhere, and Beto can choose that as the source.But in 2D, it's different. For example, if Ana arranges the heights such that there are multiple "peaks" which are local maxima. Each local maximum would require a source, because water can't flow into a local maximum from a higher neighbor. Wait, but in Ana's arrangement, all heights are distinct, so there is a strict global maximum. However, besides the global maximum, there can be local maxima, which are cells higher than all their neighbors. Since all heights are distinct, each local maximum is only higher than its neighbors, but not necessarily higher than all other cells.Wait, but in any grid, the number of local maxima can be as high as... For example, in a chessboard pattern where every other cell is a local maximum, but of course, with distinct heights, you can't have adjacent local maxima. Actually, in a grid, the maximum number of local maxima (cells higher than all four neighbors) is roughly (n^2)/5, but I'm not sure. However, for our problem, since Ana can arrange the heights adversarially, she could potentially create as many local maxima as possible to force Beto to place a source in each. But maybe there's a smarter way.Alternatively, perhaps the problem relates to the concept of "feedback vertex set" or "hitting set" where you need to hit all cycles, but since the graph is a DAG, there are no cycles. Hmm.Wait, but since Ana can choose any height arrangement, which corresponds to any possible DAG where edges go from higher to lower. So Beto needs to select a set S of vertices such that in any DAG orientation (i.e., any linear extension of the grid poset?), S is a dominating set in the sense that all vertices are reachable from S.Wait, but in reality, the DAG is not any orientation, but orientations based on height assignments. Each cell's height is unique, so the DAG is a linear extension of the grid poset? Not exactly, because the grid's poset isn't a total order. Wait, maybe not.Alternatively, since all heights are distinct, the DAG is a transitive orientation where if there's a path from a to b, then a is higher than b. But actually, the DAG isn't necessarily transitive. For example, if a is higher than b, and b is higher than c, but a and c are not adjacent, so there might not be an edge between a and c. So transitivity isn't enforced unless there's a path.But regardless, the problem is that Beto needs to pick a set of nodes such that, regardless of the height permutation, every node is reachable from at least one node in the set via a decreasing path.So, in other words, the problem reduces to finding the minimum number k such that in any permutation of the grid, there exists k elements from which every other element can be reached by moving through adjacent cells with decreasing values.This seems related to the concept of "basins of attraction" in permutations or grid orientations, where each basin is a set of cells that drain to a particular sink. But here we are looking for the dual concept: sources that can cover the entire grid.Wait, in the context of sinks, the number of sinks (local minima) in a permutation can be as high as something, but here we are dealing with sources. Maybe the minimal number of sources required is equal to the maximum number of sinks that Ana can create? But I need to think carefully.Wait, but actually, if we consider the dual problem: instead of water flowing from high to low, think of it as something flowing from low to high. Then the number of sources Beto needs would correspond to the number of sinks (local minima) in Ana's arrangement. Because each local minimum would need to be a source if we were going upwards. But in our case, since water flows down, maybe it's the number of local maxima. But in Ana's arrangement, there is only one global maximum, but multiple local maxima.But wait, if Ana wants to make Beto's life hard, she would arrange the heights to have as many local maxima as possible. Each local maximum cannot be reached from any adjacent cell, so Beto must place a water source on each local maximum. Hence, the minimal number of sources Beto needs is equal to the maximum number of local maxima that Ana can create in the grid.Therefore, the problem reduces to: what's the maximum number of local maxima that can exist in an n x n grid with all distinct heights? Then Beto needs to place at least that many sources, and perhaps this is the minimal number required.But what is the maximum number of local maxima in an n x n grid? A local maximum is a cell that is higher than all four of its adjacent cells (assuming edge and corner cells have fewer neighbors). To maximize the number of local maxima, Ana can arrange the heights in a checkerboard pattern of high and low values. However, in such a pattern, each high cell is adjacent to low cells, but each high cell is not necessarily higher than all its neighbors. Wait, in a checkerboard pattern, each high cell is surrounded by low cells, so if all the high cells are arranged such that each is a local maximum, then you can have roughly n²/2 local maxima. But wait, in reality, adjacent high cells in the checkerboard can't both be local maxima because they are adjacent. Wait, no, in a checkerboard, the high cells are not adjacent to each other. Each high cell is surrounded by low cells, so if all high cells are higher than all low cells, then each high cell is a local maximum. Similarly, each low cell is a local minimum. So if Ana assigns the highest n²/2 values to the checkerboard cells, then each of those would be a local maximum. But in that case, the number of local maxima would be roughly (n²)/2. But that can't be right because in a 2x2 grid, the checkerboard pattern would have 2 local maxima, which is half of 4, so 2. But in a 2x2 grid, is the maximum number of local maxima 1? Wait, no. If you have:4 32 1Here, 4 is a local maximum, and 1 is a local minimum. But if you have:3 42 1Then 4 is the local maximum, and 1 is the local minimum. But if you arrange it as:4 23 1Then 4 and 3 are local maxima? No, 4 is higher than its neighbors 2 and 3, so 4 is a local maximum. 3 is higher than its neighbor 1 and 2, but 3 is adjacent to 4, which is higher. Wait, no. If 4 is in the top-left, then its neighbors are 2 (right) and 3 (below). If 4 is the highest, then 3 is adjacent to 4 (higher) and 1 (lower). So 3 is not a local maximum. So in 2x2 grid, maximum local maxima is 1.Wait, maybe my initial thought is wrong. Let's check 3x3 grid.Suppose we have a 3x3 grid. Can we have 5 local maxima? Probably not. Let's see. If we place the highest number in the center, then it's a local maximum. Then, place the next four highest numbers on the edges. Each edge cell has three neighbors: two adjacent edge cells and the center. If the center is the highest, then the edge cells cannot be higher than the center. So they can't be local maxima. Then, if we place the highest number in a corner, say, top-left corner. Then, its neighbors are right and down. If the top-left is the highest, then those neighbors are lower. Then, to create another local maximum, say, the top-right corner. But for the top-right corner to be a local maximum, it needs to be higher than its neighbors: left, down, and (if it's on the edge) but in 3x3, top-right has neighbors: left (top-middle), down (middle-right), and maybe diagonally? Wait, the problem says "sharing a side", so diagonals don't count. So top-right corner has two neighbors: left (top-middle) and down (middle-right). So if top-right is higher than those two, then it can be a local maximum. But if the top-left is already the highest, then the top-middle cell (neighbor of top-left) must be lower than top-left. Similarly, if top-right is higher than top-middle and middle-right, but since top-left is the highest, we can arrange the heights such that top-right is the second highest, then top-middle and middle-right are lower. Then top-right would be a local maximum. Similarly, maybe bottom-left and bottom-right can also be local maxima. So in a 3x3 grid, maybe we can have 4 local maxima: the four corners. Let's check:Assign the highest four numbers to the four corners. Each corner has two edge neighbors. So for example:9 2 83 1 47 5 6Here, the corners are 9, 8, 7, 6. Each corner is higher than its two neighbors. For instance, 9 is higher than 2 and 3; 8 is higher than 2 and 4; 7 is higher than 3 and 5; 6 is higher than 4 and 5. Wait, but in this case, 6 is adjacent to 5 and 4, which are lower. But the center is 1, which is lower than all. However, the middle cells (2,3,4,5) are lower than the corners. But for instance, the cell with 2 is adjacent to 9 and 8, which are higher, so 2 is not a local maximum. Similarly, 3 is adjacent to 9 and 7, which are higher. 4 is adjacent to 8 and 6, which are higher. 5 is adjacent to 7 and 6, which are higher. So in this arrangement, the four corners are local maxima, and the center is a local minimum. So in 3x3, we can have 4 local maxima. But is that the maximum?Alternatively, can we have 5 local maxima? Let's see. Suppose we try to place local maxima in all four corners and the center. The center has four neighbors. If the center is a local maximum, it has to be higher than all four adjacent cells. But if the corners are also local maxima, they need to be higher than their two neighbors. So for example, the top-left corner is higher than top-middle and middle-left. The center is higher than top-middle, middle-left, middle-right, and bottom-middle. But if top-left is higher than top-middle and middle-left, but the center is higher than top-middle and middle-left, there's no conflict as long as the center is higher than those. Wait, let's assign numbers:Let’s say center is 9, the highest. Then the four corners can be 8,7,6,5. Each corner needs to be higher than its two edge neighbors. Let's try:8 1 72 9 36 4 5Here, the corners: 8,7,6,5. Each corner needs to be higher than their edge neighbors. For top-left corner (8): neighbors are 1 (right) and 2 (down). 8 > 1 and 8 > 2: good. Top-right corner (7): neighbors are 1 (left) and 3 (down). 7 > 1 and 7 > 3: good. Bottom-left corner (6): neighbors are 2 (up) and 4 (right). 6 > 2 and 6 > 4: good. Bottom-right corner (5): neighbors are 3 (up) and 4 (left). 5 > 3 and 5 > 4: good. The center (9) is higher than all its neighbors: 1,2,3,4. So in this case, we have 5 local maxima: the four corners and the center. Wait, is that possible? Because the center is a local maximum, and the four corners are also local maxima. But the problem is that the edge cells adjacent to the corners (like the top-middle cell 1) are lower than both the corner and the center. Similarly, the middle-left cell 2 is lower than the corner and the center. So in this arrangement, yes, there are 5 local maxima. But is this a valid permutation? Let's check if all numbers are distinct. The numbers used are 1,2,3,4,5,6,7,8,9. Yes, in a 3x3 grid, that's all cells. So this works. So in a 3x3 grid, Ana can arrange the heights to have 5 local maxima. Therefore, Beto would need to place 5 water sources to cover all local maxima. Wait, but in this case, if Beto places a water source in the center (9), water can flow to all adjacent cells, which are 1,2,3,4. Then from those, water can flow to the edge cells. But the corners are local maxima, so water cannot flow into them from any adjacent cell. Therefore, Beto must also place water sources in each of the four corners. So in total, 5 sources: the four corners and the center. But wait, but in the example above, the center is the highest, so if Beto places a source there, water can flow to all adjacent cells. However, the corners are local maxima; they are not adjacent to any higher cells. Therefore, water cannot reach the corners unless Beto places a source there. So in this case, Beto needs to place a source at each local maximum. Hence, if Ana can create 5 local maxima, Beto needs 5 sources.But in a 3x3 grid, is 5 the maximum number of local maxima? Let's see. If we try to have more than 5, say 6. Let's attempt to have 6 local maxima. Each local maximum must be higher than all its neighbors. Let's see, in a 3x3 grid, each corner has two neighbors, each edge center has three neighbors, and the center has four neighbors. If we try to have local maxima at all four corners, the center, and one of the edges. Wait, but an edge center cell (like top-middle) has three neighbors: left, right, and down. If top-middle is a local maximum, it needs to be higher than left (which is a corner), right (another corner), and down (the center). But if the corners and center are already local maxima, they are higher than top-middle, so top-middle cannot be a local maximum. Therefore, it's impossible to have a local maximum on an edge if the adjacent corners and center are already local maxima. Therefore, the maximum number of local maxima in 3x3 is 5: four corners and the center.But this is specific to 3x3. For larger grids, how does this scale? For example, in a 4x4 grid, can Ana arrange the heights such that there are multiple local maxima? Maybe in a checkerboard-like pattern where every other cell is a local maximum.Wait, let's think of a general approach. If Ana wants to maximize the number of local maxima, she can arrange the grid such that local maxima are placed as many as possible, each separated by at least one cell. For example, in a chessboard pattern where every white square is a local maximum and every black square is a local minimum. But in this case, each local maximum (white square) is adjacent only to local minima (black squares), which are lower. Therefore, each white square would be a local maximum. In an n x n grid, approximately half of the cells can be local maxima. For even n, it would be n²/2, and for odd n, (n² + 1)/2.But wait, in a chessboard pattern, adjacent cells are alternating high and low. So each high cell is surrounded by low cells, hence being a local maximum. So in this case, the number of local maxima would be ceil(n² / 2). But in a 2x2 grid, this would be 2 local maxima, but earlier we saw that in 2x2, you can only have 1 local maximum. Wait, contradiction.Wait, in a 2x2 grid, a chessboard pattern would have two local maxima. For example:4 23 1Here, 4 and 1 are local maxima? Wait, no. 4 is higher than its neighbors 2 and 3, so 4 is a local maximum. 1 is lower than its neighbors 2 and 3, so it's a local minimum. The other two cells, 2 and 3, each have two neighbors. 2 is adjacent to 4 (higher) and 1 (lower), so 2 is not a local maximum. 3 is adjacent to 4 (higher) and 1 (lower), so 3 is also not a local maximum. So in this 2x2 grid, only 4 is a local maximum. So the chessboard pattern doesn't actually give two local maxima in 2x2.Wait, maybe my initial thought is wrong. Let's try:3 14 2Here, 3 is higher than 1 and 4? No, 4 is higher than 3. So 4 is a local maximum. 3 is higher than 1 but lower than 4. 2 is higher than 1 but lower than 4. So only 4 is a local maximum. Wait, so in 2x2 grid, the maximum number of local maxima is 1.But in 3x3 grid, as we saw, it's possible to have 5 local maxima. So the chessboard idea doesn't hold. So maybe the maximum number of local maxima is more than n²/2 in some cases? Wait, 5 in 3x3 is more than half (which would be 4.5). So maybe the maximum number is roughly n²/2 + something.Alternatively, perhaps the maximum number of local maxima in an n x n grid is equal to the number of cells in a maximum independent set. Because in graph theory, an independent set is a set of vertices no two of which are adjacent. If we want to place local maxima, they cannot be adjacent, so they form an independent set. Therefore, the maximum number of local maxima is equal to the size of the maximum independent set in the grid graph.For a grid graph, which is bipartite, the maximum independent set is exactly half the number of vertices when the grid is even-sized. For example, in a 2x2 grid, the maximum independent set is 2, but we saw that only 1 local maximum is possible. Wait, contradiction again. Hmm.Wait, maybe it's not exactly the maximum independent set, because even though local maxima can't be adjacent, the independent set includes cells that are not adjacent, but local maxima also have the condition that all their neighbors are lower. However, in a chessboard pattern, even if you color the grid in black and white, and assign all white cells as high and black as low, then each white cell is a local maximum. Because each white cell is adjacent only to black cells, which are lower. So in that case, the number of local maxima would be equal to the size of the maximum independent set. But in a 2x2 grid, the maximum independent set is 2, but we can only have 1 local maximum. So something's wrong here.Wait, in the 2x2 grid, if you color it in chessboard fashion, white and black cells. If you assign the white cells as 4 and 1, and the black cells as 3 and 2. Then the white cells would be 4 and 1. But 1 is a local minimum, not a maximum. So the problem is that just coloring doesn't ensure that all cells of a color are higher than the other. So maybe the chessboard idea only works if you can assign the independent set cells to be all higher than the other partition. Which would require that the independent set is one partition of the bipartition. Since grid graphs are bipartite, their maximum independent set is exactly half the vertices (rounded up or down depending on parity). So in an even-sized grid, n even, the maximum independent set is n²/2. In odd-sized, it's (n² + 1)/2.But in the 2x2 grid, if we set white cells to 4 and 3, and black cells to 2 and 1. Then white cells are 4 and 3, which are both local maxima. Wait:4 23 1Here, 4 is higher than 2 and 3, so it's a local maximum. 3 is higher than 1, but 3 is adjacent to 4, which is higher. So 3 is not a local maximum. So only 4 is a local maximum. So even if we set white cells to the highest values, the lower white cell (3) is adjacent to a higher white cell (4), so it cannot be a local maximum. Therefore, in the 2x2 grid, the maximum number of local maxima is 1, which is less than the maximum independent set size of 2.So maybe the chessboard idea doesn't work because assigning the independent set to be the highest values doesn't guarantee that all of them are local maxima, since some may be adjacent diagonally? Wait, no, in the grid graph, adjacency is only via edges, not diagonally. So in the 2x2 grid, the two white cells are not adjacent (if we consider the chessboard coloring). Wait, in a 2x2 grid, the chessboard coloring would have two white and two black cells, with each white cell adjacent to two black cells. But the white cells are not adjacent to each other. So if you assign the two white cells to be the two highest values, then each white cell is higher than all its neighbors (which are black cells). Therefore, both white cells would be local maxima. Wait, but in the 2x2 grid, white cells are diagonal from each other, right? So in a 2x2 grid:W BB WIf we assign W cells as 4 and 3, and B cells as 2 and 1. Then:4 21 3Here, 4 is higher than 2 and 1, so it's a local maximum. 3 is higher than 1 and 2, but 3 is adjacent to 1 and 2, which are lower. Wait, but in this arrangement, 3 is adjacent to 1 (below) and 2 (right). If 3 is higher than both, then it's a local maximum. Similarly, 4 is higher than both its neighbors. So in this case, both 3 and 4 are local maxima. But in this arrangement, the two local maxima are diagonally opposite. So in a 2x2 grid, it's possible to have two local maxima. Wait, but earlier example had only 1. So it depends on the arrangement.Wait, let's take the 2x2 grid with cells:4 21 3Here, 4 is a local maximum (neighbors 2 and 1), 3 is a local maximum (neighbors 1 and 2). 2 is adjacent to 4 and 3, both higher, so not a local max. 1 is adjacent to 4 and 3, both higher. So yes, in this arrangement, there are two local maxima. Therefore, in a 2x2 grid, the maximum number of local maxima is 2.But then earlier example had a different arrangement where there was only 1 local maximum. So Ana can choose to arrange it to have 2 local maxima. Therefore, Beto must place 2 sources in the 2x2 grid to guarantee coverage. Because if Ana creates two local maxima, Beto has to place a source in each. So for 2x2 grid, the answer would be 2.But wait, let's verify. If Beto places sources in both local maxima (4 and 3 in the example), then water can flow from 4 to 2 and 1, and from 3 to 2 and 1. Therefore, all cells are covered. If Ana uses a different arrangement with only 1 local maximum, then Beto still needs to place 1 source. But since Ana can choose the worst-case arrangement (with 2 local maxima), Beto needs to have 2 sources.Therefore, in the 2x2 grid, the minimal number is 2. So perhaps in general, the minimal number of sources Beto needs is equal to the size of the maximum independent set of the grid graph, which for an n x n grid is ⌈n² / 2⌉. But wait, in the 3x3 grid, the maximum independent set is 5 (for 3x3, total cells 9, so (9 + 1)/2 = 5). And earlier, we saw that 5 local maxima are possible. So that seems to align. For 2x2, maximum independent set is 2, which matches. For 1x1 grid, it's 1. So maybe in general, the minimal number of sources Beto needs is equal to the size of the maximum independent set of the grid graph, which is ⌈n² / 2⌉.But wait, in the 3x3 grid, we saw an example with 5 local maxima. But the maximum independent set for 3x3 grid is indeed 5, as per the chessboard coloring (for odd n, it's (n² + 1)/2). So 3x3: (9 + 1)/2 = 5. So this matches. Similarly, for 4x4 grid, maximum independent set is 8, which is 16 / 2. So if Ana can create 8 local maxima in a 4x4 grid, then Beto needs 8 sources. But is that possible?Let's try a 4x4 grid with a chessboard pattern of local maxima. Assign the white cells as the highest 8 values and black cells as the lowest 8 values. Then, each white cell is adjacent only to black cells, which are all lower. Therefore, each white cell is a local maximum. Hence, there are 8 local maxima. Therefore, in a 4x4 grid, the maximum number of local maxima is 8, which is the size of the maximum independent set. Therefore, Beto needs 8 sources.Therefore, this seems to generalize. For an n x n grid, the maximum number of local maxima Ana can create is equal to the size of the maximum independent set of the grid graph, which is ⌈n² / 2⌉. Therefore, Beto needs to place that many sources to cover all local maxima, which are forced to be sources since water cannot flow into them.Therefore, the minimal number of water sources Beto needs is the size of the maximum independent set of the grid graph, which is ⎡n² / 2⎤ (ceil(n² / 2)).But let's verify with small cases:1x1 grid: ceil(1/2) = 1. Correct, since there's only one cell.2x2 grid: ceil(4 / 2) = 2. Which matches our earlier example.3x3 grid: ceil(9 / 2) = 5. Which matches the 5 local maxima example.4x4 grid: ceil(16 / 2) = 8. Which seems correct.Therefore, the minimal number of sources Beto needs is ⎡n² / 2⎤. But in mathematical terms, since n is an integer, n² is either even or odd. If n is even, then n² is even, so ⎡n² / 2⎤ = n² / 2. If n is odd, then n² is odd, so ⎡n² / 2⎤ = (n² + 1)/2. So the formula is:If n is even, then the number is n² / 2.If n is odd, then (n² + 1)/2.But both can be written as ⎡n² / 2⎤.But wait, the problem says "the least number of water sources that Beto needs to win no matter how Ana places the platforms." So regardless of Ana's arrangement, Beto needs to place ⎡n² / 2⎤ sources. But is this actually correct?Wait, hold on. If Ana arranges the grid so that half of the cells (in an independent set) are local maxima, then Beto has to place a source in each. But is there a way for Beto to cover all local maxima with fewer sources? For example, maybe some sources can cover multiple local maxima through water paths. But in the chessboard arrangement, each local maximum is isolated; water from one local maximum cannot reach another because they are separated by lower cells. For example, in the 2x2 grid with two local maxima diagonally opposite, water from one cannot reach the other because the adjacent cells are lower. Hence, Beto has to place a source in each local maximum.Therefore, in the worst case, the number of sources needed is equal to the maximum number of local maxima, which is the size of the maximum independent set. Therefore, the answer is ⎡n² / 2⎤.But let me check another example. Take the 3x3 grid with 5 local maxima. If Beto places sources in all 5 local maxima, then he can flood the grid. If he places fewer, say 4, then Ana could have placed a local maxima not covered by a source, and water cannot reach there. Therefore, Beto needs to place as many sources as the maximum number of local maxima Ana can create.Hence, the minimal number is the maximum number of local maxima, which is the size of the maximum independent set of the grid graph. For grid graphs, the maximum independent set is indeed known to be ⎡n² / 2⎤. Therefore, the answer should be ⎡n² / 2⎤, which is equal to (n² + 1) // 2 in integer division terms.Therefore, the least number of water sources Beto needs is the ceiling of n squared over two.So, final answer: ⎡n² / 2⎤, which can be written in LaTeX as boxed{leftlceil dfrac{n^2}{2} rightrceil}.Final AnswerThe least number of water sources Beto needs is boxed{leftlceil dfrac{n^2}{2} rightrceil}.</think>To determine the least number of water sources Beto needs to ensure he can fill all platforms in an (n times n) garden regardless of how Ana places the platforms, we need to consider the worst-case scenario where Ana arranges the heights to maximize the number of local maxima. A local maximum is a platform higher than all its adjacent platforms. Beto must place a water source at each local maxima since water cannot flow into a local maxima from a lower platform. The key insight is that the maximum number of local maxima Ana can create is equivalent to the size of the maximum independent set in the grid graph. For an (n times n) grid, the maximum independent set size is given by (leftlceil frac{n^2}{2} rightrceil). This is because the grid can be colored in a checkerboard pattern where each cell in one color class (either all black or all white) forms an independent set. In such a configuration, each cell in the independent set can be a local maximum if assigned the highest values.Thus, the minimum number of water sources Beto needs to guarantee covering all platforms, regardless of Ana's height arrangement, is the ceiling of half the total number of cells in the grid. This is mathematically represented as:[boxed{leftlceil dfrac{n^2}{2} rightrceil}]]

🔑:1. Claim and Initial Setup: We claim that the least number of water sources Beto needs is ( lceil frac{n^2}{2} rceil ). To justify this, we consider a chessboard coloring of the ( n times n ) garden, where there are ( lceil frac{n^2}{2} rceil ) black squares and ( lfloor frac{n^2}{2} rfloor ) white squares.2. Height Assignment: Assign distinct positive real numbers to the black squares in the interval ( (1, 2) ) and distinct positive real numbers to the white squares in the interval ( (0, 1) ). This ensures that no water can flow from a white square to a black square because the height of any black square is always greater than the height of any white square.3. Minimum Water Sources: Since water cannot flow from a white square to a black square, each black square must have its own water source. Therefore, at least ( lceil frac{n^2}{2} rceil ) water sources are needed in this configuration.4. Sufficiency of ( lceil frac{n^2}{2} rceil ) Water Sources: To show that ( lceil frac{n^2}{2} rceil ) water sources are sufficient for any configuration, partition the ( n times n ) garden into ( lceil frac{n^2}{2} rceil ) regions of size ( 2 times 1 ) (with possibly one ( 1 times 1 ) region if ( n ) is odd). In each of these regions, place a water source at the platform with the maximum height.5. Water Flow: In each ( 2 times 1 ) region, the water will flow from the platform with the maximum height to the platform with the minimum height. This ensures that all platforms in the garden will be filled with water.6. Conclusion: Therefore, the least number of water sources Beto needs to ensure that all platforms are filled with water, regardless of how Ana places the platforms, is ( lceil frac{n^2}{2} rceil ).The final answer is ( boxed{ lceil frac{n^2}{2} rceil } )

🧠:[<think>Okay, let me try to figure out this probability problem. So, the question is about a standard deck of 52 cards, without any jokers. After shuffling, four people each draw 13 cards. We need to find the probability that both red Aces (the Ace of Hearts and the Ace of Diamonds) end up in the same person's hand. Hmm, interesting. Let me break this down step by step.First, I know that probability problems often involve combinations because the order in which the cards are drawn usually doesn't matter. Since we're dealing with distributions of cards to different players, combinations will definitely come into play here. Let me recall that a combination C(n, k) is the number of ways to choose k items from n without considering the order.So, the total number of ways the two red Aces can be distributed among the four players is something we need to consider. But we want the specific case where both Aces are with the same person. Let me think: instead of looking at all possible distributions, maybe I can fix one Ace and see where the other Ace is. Wait, but since the deck is shuffled and the players are drawing cards in turn, maybe it's better to consider the positions of the two Aces in the deck.Alternatively, maybe we can think of dealing the cards one by one, but since each player gets 13 cards, perhaps the problem can be approached by considering the entire deck being partitioned into four groups of 13 each. Then, the two red Aces need to be in the same group. Therefore, the problem reduces to: given two specific cards in a 52-card deck, what's the probability they are in the same 13-card group when the deck is divided into four equal parts.Hmm, that might be a simpler way to think about it. Let me check if that approach is valid. Since the deck is shuffled randomly, the positions of the two red Aces are random. The four players each get 13 cards, so effectively, the deck is divided into four piles of 13. The two Aces can be in any of these piles. So, the probability that both are in the first pile, or both in the second, third, or fourth. So, total probability is 4 times the probability that both are in a specific pile.Therefore, the probability would be 4 multiplied by the probability that both Aces are in the first player's hand. Let's compute that. The total number of ways to choose 13 cards out of 52 is C(52, 13). The number of ways to choose 13 cards that include both red Aces is C(50, 11), since we need to include the two Aces and 11 more cards from the remaining 50. Therefore, the probability that both Aces are in the first player's hand is C(50, 11)/C(52, 13). Then, multiplying by 4 gives the total probability.Wait, let me verify this logic. If I fix the two Aces, then the first player's hand must include both. The number of possible hands that include both Aces is C(50, 11), as we need to choose 11 more cards from the remaining 50. The total number of possible hands for the first player is C(52, 13). Then, since there are four players, we multiply by 4. Is this correct?Alternatively, another way to think about it is: after placing the first Ace in a player's hand, the probability that the second Ace is in the same player's hand. Let me explore this approach.Suppose we fix one Ace, say the Ace of Hearts. The Ace of Hearts is in one of the four players' hands. Each player has 13 cards, so the Ace of Hearts is equally likely to be in any of the 52 positions, which correspond to each player's hands. But since each player has 13 cards, the probability that the Ace of Hearts is in any specific player's hand is 13/52 = 1/4. Now, given that the Ace of Hearts is in a particular player's hand, how many cards are left in that player's hand? Wait, no. Wait, when dealing the cards, each player gets exactly 13. So once the Ace of Hearts is in a player's hand, there are 12 remaining spots in that player's hand, and 51 remaining cards in total (since one card is already the Ace of Hearts). Therefore, the probability that the Ace of Diamonds is in the same player's hand is 12/51, because there are 12 remaining cards in that player's hand out of the 51 remaining cards.Therefore, the probability that both Aces are in the same hand would be 1 (for the first Ace) multiplied by 12/51. But since the first Ace can be in any of the four players' hands, but actually, once the first Ace is placed, the second Ace's probability is conditional. Wait, no. Let me think again.Actually, regardless of where the first Ace is, the second Ace has 51 remaining positions. Out of those 51 positions, 12 are in the same player's hand as the first Ace. Therefore, the probability that the second Ace is in the same hand as the first is 12/51. Therefore, the overall probability is 12/51 = 4/17 ≈ 0.235. So the probability is 4/17.Wait, but that seems conflicting with the previous approach. Let me check again.First approach: 4 * [C(50,11)/C(52,13)].Let's compute that.C(50,11) is 50 choose 11, and C(52,13) is 52 choose 13.The ratio is [50! / (11! * 39!)] / [52! / (13! * 39!)] = [50! * 13! * 39!] / [11! * 39! * 52!] = (13 * 12) / (52 * 51) = (156) / (2652) = 156 / 2652 = simplifying both by 12: 13 / 221. Wait, 156 ÷ 12 = 13; 2652 ÷12 = 221. So 13/221. Then multiplying by 4 gives 52/221. Simplify 52/221: 52 ÷13=4; 221 ÷13=17. So 4/17. So that matches the second approach. So both methods give 4/17. Therefore, the probability is 4/17. Therefore, the answer is 4/17, which is approximately 0.235.So both methods lead to the same answer. Therefore, the probability is 4/17. Let me just make sure I didn't miss anything.Wait, another way to think about it is hypergeometric distribution. The probability that both Aces are in the same hand. So, the total number of ways to choose 2 cards out of 52 is C(52,2). The number of ways that both are in the same player's hand: each player has C(13,2) combinations, and there are four players. So total favorable is 4*C(13,2). Therefore, probability is [4*C(13,2)] / C(52,2). Let's compute that.C(13,2) is (13*12)/2 = 78. So 4*78 = 312. C(52,2) is (52*51)/2 = 1326. So 312 / 1326. Simplify numerator and denominator by 6: 312 ÷6=52; 1326 ÷6=221. So 52/221. Then 52 divided by 221: 52 ÷13=4; 221 ÷13=17. So 4/17. Again, same result. Wow, so three different approaches all leading to 4/17. That gives me confidence that the answer is indeed 4/17.Therefore, the probability that both red Aces are in the same player's hand is 4/17.Final AnswerThe probability is boxed{dfrac{4}{17}}.</think>To find the probability that the two red Aces (Heart A and Diamond A) are in the same person's hand when four players each draw 13 cards from a shuffled deck of 52 cards, we can use the following reasoning:1. First Approach (Combinations): - The total number of ways to choose 13 cards out of 52 is ( binom{52}{13} ). - The number of ways to choose 13 cards that include both red Aces is ( binom{50}{11} ) (choosing 11 more cards from the remaining 50). - The probability that both Aces are in a specific player's hand is ( frac{binom{50}{11}}{binom{52}{13}} ). - Multiplying by 4 (for the four players) gives the total probability: [ 4 times frac{binom{50}{11}}{binom{52}{13}} = 4 times frac{frac{50!}{11! cdot 39!}}{frac{52!}{13! cdot 39!}} = 4 times frac{13 times 12}{52 times 51} = 4 times frac{156}{2652} = frac{4}{17} ]2. Second Approach (Conditional Probability): - Once the first Ace is placed in a player's hand, the probability that the second Ace is in the same player's hand is ( frac{12}{51} ). - This gives the probability: [ frac{12}{51} = frac{4}{17} ]3. Third Approach (Hypergeometric Distribution): - The total number of ways to choose 2 cards out of 52 is ( binom{52}{2} ). - The number of ways both Aces are in the same player's hand is ( 4 times binom{13}{2} ). - The probability is: [ frac{4 times binom{13}{2}}{binom{52}{2}} = frac{4 times 78}{1326} = frac{312}{1326} = frac{52}{221} = frac{4}{17} ]All three approaches lead to the same result. Therefore, the probability that the two red Aces are in the same person's hand is (boxed{dfrac{4}{17}}).,<think>Okay, so I need to find the probability that both red Aces—Heart A and Diamond A—are in the same person's hand after a deck of 52 cards is shuffled and dealt evenly among four people, each getting 13 cards. Hmm, let's think about this step by step.First, let me understand the problem correctly. There are 52 cards, no jokers. Two of these cards are the red Aces: the Ace of Hearts and the Ace of Diamonds. The deck is shuffled, so the cards are randomly distributed. Four people each get 13 cards. I need to find the probability that both red Aces are in the same person's hand. So, regardless of which person it is, as long as both red Aces are with one person, that's the event we're interested in.Hmm, probability problems can sometimes be tricky, but maybe I can approach this using combinatorics. Let me recall that probability is often the number of favorable outcomes divided by the total number of possible outcomes. So, in this case, the total number of ways the two red Aces can be distributed among the four players, and the number of those ways where both are with the same person.Alternatively, maybe it's easier to think of it as the probability that the second red Ace is in the same hand as the first red Ace. Let me try that approach.Suppose we fix the position of the first red Ace. Since the deck is shuffled randomly, the first red Ace can be in any of the 52 positions. But perhaps it's better to imagine dealing the cards one by one. Wait, but actually, since all distributions are equally likely, maybe we can use the concept of conditional probability here.Once the first red Ace has been dealt to a particular person, the question is: what's the probability that the second red Ace is also dealt to the same person. Since the first Ace is in one of the four hands, and the second Ace must be in the same hand. Let me formalize this.Let's say we have four players: A, B, C, D. The first red Ace (say, Heart A) can be in any of the four players' hands. Once it's in someone's hand, how many cards are left in that hand? Wait, each player gets 13 cards. If one card is already the Heart A, then there are 12 remaining cards in that player's hand. But the deck is shuffled, so the rest of the cards are randomly distributed.Wait, but actually, when dealing the cards, once you fix one card in a player's hand, the other cards are dealt from the remaining 51 cards. So maybe the probability that the second Ace is in the same hand is the number of remaining slots in that player's hand divided by the remaining number of cards.Wait, so if one red Ace is already in a player's hand, which has 13 cards, then there are 12 remaining cards to be dealt to that player. The total number of remaining cards is 51. So the probability that the second red Ace is also dealt to the same player is 12/51. Therefore, the probability would be 12/51, which simplifies to 4/17. Hmm, is that correct?Wait, but let me check. Alternatively, maybe the answer is 4*(13 choose 2)/(52 choose 2). Wait, let's see. The total number of ways to choose the two red Aces is (52 choose 2). The number of favorable cases where both Aces are in the same hand: there are four hands, each with 13 cards. So for each hand, the number of ways to have both Aces in that hand is (13 choose 2). Therefore, total favorable is 4*(13 choose 2). So the probability is [4*(13*12)/2] / [52*51/2] = [4*13*12] / [52*51] = [4*12] / [4*51] since 52 is 4*13, so 52 cancels with 13, giving 4. So 4*12 / 4*51 = 12/51 = 4/17. So that's the same answer as before. So that seems to check out. Therefore, the probability is 4/17.But wait, let me think again to make sure. Is there another way to see this?Alternatively, suppose we consider that the two red Aces are among the 52 cards. The first Ace can be anywhere. The second Ace has to be in the same hand. Since the hands are dealt evenly, once the first Ace is in a particular hand, there are 12 remaining cards in that hand, out of the remaining 51 cards. Therefore, the probability that the second Ace is in the same hand is 12/51 = 4/17. So that's consistent.Alternatively, suppose you consider all possible pairs of cards. The total number of pairs is C(52,2). The number of pairs that are both in the first player's hand is C(13,2), similarly for the other three players. So total favorable pairs are 4*C(13,2). Then the probability is 4*C(13,2)/C(52,2) = 4*(13*12/2)/(52*51/2) = (4*13*12)/(52*51) = (12/51) = 4/17. So same result. So that's three different ways to get to 4/17, which gives me more confidence.But let me check if there's a different approach where maybe the answer is different. For example, sometimes in probability, especially with dealing cards to multiple players, it's useful to think of dealing the cards one by one, but in this case, since the cards are being dealt into four hands of 13 each, maybe it's equivalent to randomly assigning each card to a player, with each player getting exactly 13 cards.Alternatively, think of the two red Aces. The probability that both are assigned to the same player. Since the deck is shuffled and each player gets 13 cards, each card is equally likely to go to any player, but with the constraint that each player gets exactly 13. However, for the purpose of calculating the probability that two specific cards are in the same hand, we can consider the first card, assign it to any player, and then compute the probability that the second card is assigned to the same player, given that there are now 12 remaining slots in that player's hand and 51 remaining cards.Therefore, the probability would be 12/51, which is 4/17. So that's the same answer again.Therefore, regardless of the method, the probability seems to be 4/17. Therefore, the answer is 4/17. Wait, but let me think again to see if I made any hidden assumption here.Is there a different approach where the answer might differ? For example, suppose we fix one Ace in a particular player's hand. Then the number of ways the other Ace can be in the same hand is 12 (remaining cards), and the total number of possible positions for the second Ace is 51. So 12/51=4/17. That's correct.Alternatively, if you think of all possible ways to distribute the two Aces among the four players, the number of possible pairs of positions for the two Aces is 52*51/2. But since each player has 13 cards, the number of positions where both Aces are in the same player is 4*(13*12)/2. Therefore, the probability is [4*(13*12)/2]/[52*51/2] = [4*13*12]/[52*51] = [12/51] = 4/17.Alternatively, if we use hypergeometric distribution concepts. The probability that both Aces are in the same hand can be calculated as follows. The total number of ways to choose 13 cards out of 52 is C(52,13). For a specific player, the number of hands that include both red Aces is C(50,11), since we include the two Aces and 11 more cards from the remaining 50. Since there are four players, the total number of favorable hands is 4*C(50,11). Therefore, the probability is [4*C(50,11)] / C(52,13). Let's compute this.First, recall that C(n,k) = n!/(k!(n-k)! )So, 4*C(50,11)/C(52,13) = 4 * [50!/(11!39!)] / [52!/(13!39!)] = 4 * [50! *13!39! ] / [11!39! *52! ] = 4 * [13*12]/[52*51] = 4*(156)/(2652) = 624/2652 = divide numerator and denominator by 12: 52/221 = divide numerator and denominator by 13: 4/17. So same answer again. Therefore, this confirms the result once more.Therefore, after multiple approaches and verifications, the probability that both red Aces are in the same hand is 4/17. So I think that's the correct answer. Let me just check once more for any possible oversight.Wait, another way to think about it: when dealing the cards, the first Ace can be in any position. The second Ace needs to be in the same hand. Since each hand is 13 cards, once the first Ace is in a particular position, the hand it's in has 12 other cards. There are 51 remaining positions for the second Ace, 12 of which are in the same hand. Therefore, 12/51 = 4/17. That's the same reasoning again. So I think this is solid.Alternatively, if you think in terms of the four players, each has a 13/52 chance to get the first Ace. Then, given that a player has the first Ace, the chance they also get the second Ace is 12/51. Therefore, the total probability is the sum over all four players: 4*(13/52)*(12/51) = 4*(1/4)*(12/51) = 12/51 = 4/17. Exactly the same. So yes, this is consistent.Therefore, I feel confident that the answer is 4/17.Final AnswerThe probability that the two red Aces are in the same person's hand is boxed{dfrac{4}{17}}.</think>To find the probability that the two red Aces (Heart A and Diamond A) are in the same person's hand after a deck of 52 cards is shuffled and dealt evenly among four people (each getting 13 cards), we can use the following reasoning:1. Total number of ways to distribute the two red Aces: The total number of ways to choose 2 cards out of 52 is given by the combination ( binom{52}{2} ).2. Number of favorable outcomes: For each of the four players, the number of ways to have both red Aces in their 13-card hand is ( binom{13}{2} ). Therefore, the total number of favorable outcomes for all four players is ( 4 times binom{13}{2} ).3. Calculating the probability: - The total number of ways to choose 2 cards out of 52 is ( binom{52}{2} = frac{52 times 51}{2} ). - The total number of favorable outcomes is ( 4 times binom{13}{2} = 4 times frac{13 times 12}{2} ). - The probability is then the ratio of favorable outcomes to the total number of outcomes: [ frac{4 times frac{13 times 12}{2}}{frac{52 times 51}{2}} = frac{4 times 13 times 12}{52 times 51} = frac{4 times 12}{4 times 51} = frac{12}{51} = frac{4}{17} ]4. Alternative verification: Considering the conditional probability that the second red Ace is in the same hand as the first red Ace. After placing the first Ace in a hand, there are 12 remaining spots in that hand out of 51 remaining cards, leading to the same probability: [ frac{12}{51} = frac{4}{17} ]Thus, the probability that the two red Aces are in the same person's hand is (boxed{dfrac{4}{17}}).]

🔑:-1 . frac{4}{17}.Notice that, after the cards are shuffled, each person's cards are determined, i.e., the 52 cards are arranged in 52 positions. Let the four groups of card numbers be:begin{array}{l}1,5,9,13, cdots, 49 ; 2,6,10,14, cdots, 50 ; 3,7,11,15, cdots, 51 ; 4,8,12,16, cdots, 52 .end{array}Then the number of arrangements where the Ace of Hearts and the Ace of Diamonds are in the same group is M=4 mathrm{~A}_{13}^{2} mathrm{~A}_{50}^{50}.Thus, the required probability is P=frac{M}{52!}=frac{4}{17}.

🧠:[<think>Okay, so I need to figure out how many zeros the function f(x) = x² ln x + x² - 2 has. Hmm, zeros of the function mean the values of x where f(x) = 0. So I need to solve the equation x² ln x + x² - 2 = 0. Let me think about how to approach this. First, the function involves a logarithm, so the domain is x > 0 because ln x is only defined for positive x. Alright, so we only consider x > 0. Now, I should probably try to analyze the behavior of this function to see where it crosses the x-axis. Maybe I can check the limits as x approaches 0+ and as x approaches infinity, see how the function behaves at the extremes. Then, check if there are any critical points in between where the function might have a maximum or minimum, which could indicate the number of times it crosses the axis.Let me start by rewriting the function a bit to see if I can simplify it or factor something out. f(x) = x² ln x + x² - 2. I notice that both x² ln x and x² have a common factor of x². Let me factor that out: f(x) = x² (ln x + 1) - 2. Hmm, that might be useful. So the equation becomes x² (ln x + 1) = 2. Alternatively, maybe I can combine the terms: x² ln x + x² = x² (ln x + 1). So that's the same as before. So f(x) = x² (ln x + 1) - 2. So solving for x² (ln x + 1) = 2. Maybe I can write this as ln x + 1 = 2 / x². Then the equation becomes ln x = 2 / x² - 1. Not sure if that helps, but perhaps I can consider the functions y = ln x and y = 2 / x² - 1 and see where they intersect. Alternatively, maybe it's easier to analyze the original function f(x) by looking at its derivatives to find critical points, check increasing/decreasing intervals, and then see how many times it crosses zero. Let's try that approach.First, let's compute the first derivative f'(x). The function is f(x) = x² ln x + x² - 2. The derivative of x² ln x is 2x ln x + x²*(1/x) = 2x ln x + x. The derivative of x² is 2x. So combining these, f'(x) = 2x ln x + x + 2x = 2x ln x + 3x. Factor out x: f'(x) = x (2 ln x + 3). So critical points occur when f'(x) = 0. Since x > 0, the x term is always positive, so set 2 ln x + 3 = 0. Solving for x: 2 ln x = -3 → ln x = -3/2 → x = e^(-3/2). So the critical point is at x = e^(-3/2). That's approximately e^-1.5 ≈ 0.2231. Now, let's check the intervals around this critical point to see if the function is increasing or decreasing. For x < e^(-3/2), let's pick a value, say x = 0.1. Then ln x is very negative, so 2 ln x + 3 would be something like 2*(-2.3026) + 3 ≈ -4.6052 + 3 ≈ -1.6052, which is negative. But since x is positive, f'(x) is negative. So the function is decreasing when x < e^(-3/2). For x > e^(-3/2), let's pick x = 1. Then ln 1 = 0, so 2*0 + 3 = 3, which is positive. So f'(x) is positive here. Therefore, the function is increasing when x > e^(-3/2). Therefore, the function has a minimum at x = e^(-3/2). Let's compute the value of f at this critical point to see if the minimum is above or below the x-axis. First, compute x = e^(-3/2). Then ln x = -3/2. So f(x) = x² (ln x + 1) - 2. Let's substitute: f(e^(-3/2)) = [e^(-3/2)]² [(-3/2) + 1] - 2 = e^(-3) [(-1/2)] - 2 = (-1/2) e^(-3) - 2. Calculate that: e^(-3) ≈ 0.0498. So (-1/2)(0.0498) ≈ -0.0249. Then subtract 2: -0.0249 - 2 ≈ -2.0249. So the minimum value of the function is approximately -2.0249, which is below zero. Therefore, since the function has a minimum below zero, and considering the behavior at the endpoints (as x approaches 0+ and x approaches infinity), we need to check if the function crosses zero once or twice. If the function tends to positive infinity on both ends and has a minimum below zero, then it must cross the x-axis twice. But let's confirm the limits.First, as x approaches 0+:We have x² ln x. Let's analyze the limit of x² ln x as x approaches 0+. Let me recall that ln x approaches -infinity, but x² approaches 0. The product is 0 * (-infinity), which is an indeterminate form. Let's apply L’Hospital’s Rule. Rewrite x² ln x as (ln x) / (1/x²). Then as x approaches 0+, this is (-infinity)/(infinity). Applying L’Hospital’s Rule: derivative of numerator is (1/x), derivative of denominator is (-2)/x³. So the limit becomes (1/x) / (-2/x³) = (1/x) * (-x³/2) = -x²/2. As x approaches 0+, this limit is 0. Therefore, the term x² ln x approaches 0. Then, the entire function f(x) = x² ln x + x² - 2 approaches 0 + 0 - 2 = -2 as x approaches 0+. Wait, but earlier when I factored it as x²(ln x +1) -2, as x approaches 0+, x² approaches 0, ln x +1 approaches -infinity, so the product is 0*(-infty). Which is again indeterminate. But according to the earlier calculation, the limit is 0 -2 = -2. Wait, perhaps there was a miscalculation.Wait, no. Let me re-express. When I factored f(x) = x² (ln x +1) -2, and as x approaches 0+, x² approaches 0, ln x approaches -infty, so ln x +1 approaches -infty. So the product x² (ln x +1) is 0*(-infty). So it's indeterminate. But when I computed the limit of x² ln x as x approaches 0+ using L’Hospital, it was 0. Therefore, x² (ln x +1) = x² ln x + x². The limit of x² ln x is 0, the limit of x² is 0, so total limit is 0 + 0 = 0. Then f(x) approaches 0 -2 = -2. So that's consistent. Therefore, as x approaches 0+, f(x) approaches -2.But wait, that contradicts the earlier conclusion that the function approaches -2 as x approaches 0+. But in reality, the function f(x) approaches -2 from below? Wait, but when x approaches 0+, x² ln x + x² - 2. If x² ln x approaches 0 (from the negative side, since ln x is negative) and x² approaches 0, then f(x) approaches 0 + 0 -2 = -2. So the limit is -2. But when x is very small, is f(x) approaching -2 from above or below? Let's see. Let's take x approaching 0+.Take x = 0.1: ln 0.1 ≈ -2.3026. Then x² ln x ≈ 0.01*(-2.3026) ≈ -0.023, x² ≈ 0.01. Then f(x) ≈ -0.023 + 0.01 - 2 ≈ -2.013. So at x=0.1, f(x) is approximately -2.013, which is less than -2. Wait, but the limit as x approaches 0+ is -2. So the function approaches -2 from below? Wait, but when x approaches 0+, x² ln x is approaching 0 from below (since ln x is negative), and x² approaches 0 from above. So x² ln x + x² is approaching 0 from below. Then subtracting 2, so the total function approaches -2 from below. Wait, but in the calculation at x=0.1, we had f(x)≈-2.013, which is less than -2, so approaching -2 from below. So the left-hand limit (as x approaches 0+) is approaching -2 from below. But since x can't be less than 0, it's just approaching from the right. Hmm, maybe the terminology here is a bit tricky, but the key point is that near x=0, f(x) is close to -2 but slightly less.But regardless, as x approaches 0+, the function approaches -2. As x approaches infinity, let's check the behavior. For x approaching infinity, x² ln x grows much faster than x². Since ln x tends to infinity, x² ln x will dominate over x², so f(x) = x² ln x + x² -2 ~ x² ln x, which tends to infinity. Therefore, as x approaches infinity, f(x) approaches infinity. So the function f(x) approaches -2 as x approaches 0+, has a minimum at x = e^(-3/2) of approximately -2.0249, which is even lower than the limit at 0, and then increases to infinity as x approaches infinity. Wait, hold on. Wait, the critical point is at x = e^(-3/2) ≈ 0.223, which is greater than 0. So between 0 and e^(-3/2), the function is decreasing, and at x = e^(-3/2), it reaches a minimum, then starts increasing. But as x approaches 0+, the function approaches -2, but at x = e^(-3/2), it's even lower, around -2.0249, and then as x increases past e^(-3/2), the function starts increasing towards infinity. So the function is decreasing from x=0 to x=e^(-3/2), reaching a minimum, then increasing from there onwards. Therefore, the graph starts near (0+, -2), goes down to a minimum at x≈0.223, then goes up to infinity as x increases. Since the minimum is below zero, and the function approaches -2 near zero and goes to infinity as x increases, the function must cross the x-axis exactly once, somewhere after the minimum point. Wait, but wait: when x approaches 0+, f(x) approaches -2, then decreases further to -2.0249, then increases. So if the function is going from -2 at x approaching 0, decreasing to -2.0249, then increasing to infinity. Since it's always negative until it starts increasing, but since the minimum is below zero, then as the function increases from the minimum, it goes from -2.0249 up to infinity. Therefore, it must cross zero exactly once in the interval (e^(-3/2), infinity). Because after the minimum, the function is increasing from -2.0249 to infinity, so it must cross zero once. However, before the minimum, the function is decreasing from -2 (as x approaches 0+) down to -2.0249. So it's decreasing and never crossing zero there. Hence, only one zero. Wait, but let me verify with some test points.Let's check the value of f(x) at x=1. f(1) = 1² ln 1 + 1² -2 = 0 +1 -2 = -1. Hmm, so at x=1, f(1) = -1. Still negative. Then check at x=2. f(2) = 4 ln 2 +4 -2 ≈ 4*0.6931 +2 ≈ 2.7724 +2 ≈ 4.7724. So positive. So between x=1 and x=2, the function crosses from -1 to 4.77, so crosses zero once. What about between x=e^(-3/2) ≈0.223 and x=1? Let's check at x=0.5. f(0.5) = 0.25 ln 0.5 +0.25 -2 ≈0.25*(-0.6931) +0.25 -2≈ -0.1733 +0.25 -2≈ -1.9233. Still negative. So from x=0.223 to x=1, the function goes from -2.0249 to -1. So still negative. Then from x=1 to x=2, it goes from -1 to positive. So crosses zero once. So only one zero. Therefore, the answer is 1.Wait, but let me check another point. Let's check x=e. At x=e, f(e) = e² ln e + e² -2 = e²*1 + e² -2 = 2e² -2 ≈2*(7.389) -2 ≈14.778 -2≈12.778. Positive. So between x=1 and x=e, function goes from -1 to ~12.778. So crosses zero once. How about between x=0.223 and x=1? At x=0.223, f(x)=~ -2.0249. At x=1, f(x)=-1. So it's increasing from -2.0249 to -1, but never reaching zero. So the function is always negative there. Then after x=1, it continues increasing, crosses zero at some point between 1 and 2. So only one zero. Hence, the number of zeros is 1. But wait, let me check another point. For example, x=1.5. f(1.5) = (2.25) ln(1.5) +2.25 -2. Let's compute ln(1.5) ≈0.4055. So 2.25*0.4055≈0.9124 +2.25 -2≈0.9124 +0.25≈1.1624. So positive. So between x=1 and x=1.5, it goes from -1 to ~1.16. Therefore, crosses zero once there. So only one crossing. So answer is 1. But let me check if there's another crossing between x approaching 0 and x= e^(-3/2). Wait, as x approaches 0, f(x) approaches -2, then decreases to -2.0249. So from -2 to -2.0249, it's getting more negative, so no crossing. Then from x=e^(-3/2) onwards, increasing to infinity. So crosses zero once. So total zeros:1. Wait, but let me think again. Is it possible that the function could cross zero more than once? For example, maybe the function starts at -2, goes down to -2.0249, then up to infinity. So if it's monotonic increasing after the minimum, then it would cross zero only once. But is the function indeed monotonic increasing after x=e^(-3/2)? Let's check the derivative. For x > e^(-3/2), f'(x) = x(2 ln x +3). Since x > e^(-3/2), let's check 2 ln x +3. For x > e^(-3/2), ln x > -3/2, so 2 ln x +3 > 2*(-3/2) +3 = -3 +3 =0. So for x > e^(-3/2), 2 ln x +3 is positive. Thus, f'(x) is positive. Therefore, the function is strictly increasing for x > e^(-3/2). Therefore, after the minimum point, the function is increasing all the way to infinity. Hence, it can cross zero at most once in that interval. Since f(x) at x=e^(-3/2) is ~-2.0249, which is less than zero, and as x increases, f(x) increases to infinity. Therefore, by the Intermediate Value Theorem, there must be exactly one c in (e^(-3/2), infinity) such that f(c)=0. But wait, before the critical point, from x approaching 0 to x=e^(-3/2), the function is decreasing from -2 to -2.0249, so it's always negative. Therefore, the function only crosses zero once, in the interval after the critical point. Hence, the number of zeros is 1.But wait, wait a second. Let me test at x=0.5. Wait, x=0.5 is greater than e^(-3/2)≈0.223. Wait, 0.5>0.223, so x=0.5 is in the increasing part. But we saw that f(0.5)≈-1.9233, which is greater than f(e^(-3/2))≈-2.0249. So indeed, the function is increasing from x=0.223 onwards. So as x increases from 0.223 to infinity, f(x) increases from -2.0249 to infinity. Therefore, since it's strictly increasing in that interval, and passes through f(1)=-1, f(1.5)=1.16, so crosses zero once between 1 and 1.5. Then in the interval x>e^(-3/2), exactly one zero. Therefore, total zeros:1.But earlier, I thought maybe there could be two zeros, but after analyzing, it's clear there's only one. So the answer is 1. But let me check another point. Wait, maybe there's another zero closer to x=0. Let me check x=0.1. f(0.1)≈(0.01)(ln 0.1) +0.01 -2≈0.01*(-2.3026)+0.01 -2≈-0.023 +0.01 -2≈-2.013. So f(0.1)≈-2.013. Then at x=0.223, it's ≈-2.0249. Then at x=0.5,≈-1.9233. So from x=0.223 to x=0.5, it goes from -2.0249 to -1.9233, which is increasing. Then from x=0.5 to 1, it goes from -1.9233 to -1. Still increasing. Then from x=1 to 2, it goes to positive. So yes, only crosses once. Therefore, the number of zeros is 1. So the answer is boxed{1}.Final Answerboxed{1}</think>To determine the number of zeros of the function ( f(x) = x^2 ln x + x^2 - 2 ), we analyze its behavior and critical points.1. Domain: The function is defined for ( x > 0 ) because ( ln x ) is only defined for positive ( x ).2. Rewriting the Function: The function can be factored as ( f(x) = x^2 (ln x + 1) - 2 ).3. First Derivative: The first derivative is ( f'(x) = x (2 ln x + 3) ). Setting ( f'(x) = 0 ) gives the critical point ( x = e^{-3/2} approx 0.2231 ).4. Behavior Around Critical Point: - For ( x < e^{-3/2} ), ( f'(x) < 0 ), so the function is decreasing. - For ( x > e^{-3/2} ), ( f'(x) > 0 ), so the function is increasing.5. Value at Critical Point: Evaluating ( f(x) ) at ( x = e^{-3/2} ): [ f(e^{-3/2}) = e^{-3} left( -frac{1}{2} right) - 2 approx -0.0249 - 2 approx -2.0249 ] This is a minimum value below zero.6. Limits: - As ( x to 0^+ ), ( f(x) to -2 ). - As ( x to infty ), ( f(x) to infty ).7. Intermediate Value Theorem: - The function is decreasing from ( x to 0^+ ) to ( x = e^{-3/2} ), reaching a minimum of approximately -2.0249. - Then, it increases to infinity as ( x to infty ). Since the function is strictly increasing after ( x = e^{-3/2} ) and passes through values from -2.0249 to infinity, it must cross the x-axis exactly once.8. Verification with Test Points: - At ( x = 1 ), ( f(1) = -1 ). - At ( x = 2 ), ( f(2) approx 4.7724 ). - The function crosses zero once between ( x = 1 ) and ( x = 2 ).Thus, the function ( f(x) ) has exactly one zero. The final answer is (boxed{1}).,<think>Okay, so I need to find the number of zeros for the function f(x) = x² ln x + x² - 2. Hmm, let's start by understanding what this function looks like and where it might cross the x-axis. First, the function involves a logarithm, so I should note the domain. Since ln x is only defined for x > 0, the domain of f(x) is (0, ∞). That's important because it tells me we don't have to consider any x values less than or equal to zero.Now, to find the zeros of f(x), I need to solve the equation x² ln x + x² - 2 = 0. Let's rewrite that equation to make it a bit simpler. Maybe factor out x² from the first two terms:x² (ln x + 1) - 2 = 0So, x² (ln x + 1) = 2Hmm, that might not be immediately helpful, but perhaps. Alternatively, maybe I can combine terms differently. Let's see:x² ln x + x² - 2 = x²(ln x + 1) - 2 = 0Yes, same as before. Maybe it's better to consider moving the 2 to the other side:x²(ln x + 1) = 2But I don't see an obvious algebraic way to solve this equation. So, maybe I should analyze the function's behavior to determine how many times it crosses the x-axis.To do that, I can look at the limits as x approaches the boundaries of the domain and check for monotonicity (whether the function is increasing or decreasing) by taking the derivative. First, let's check the behavior as x approaches 0+ and as x approaches infinity.Starting with x approaching 0+:As x approaches 0 from the right, ln x approaches -infinity. But x² approaches 0. So, we have a term that's 0 multiplied by -infinity. Let's analyze that limit:lim_{x->0+} x² ln xThis is a 0 * (-∞) indeterminate form. To resolve this, I can rewrite it as lim_{x->0+} (ln x) / (1/x²). Then, it becomes -∞ / ∞, which allows me to apply L'Hospital's Rule.Taking derivatives of numerator and denominator:d/dx (ln x) = 1/xd/dx (1/x²) = -2/x³So, applying L'Hospital's Rule:lim_{x->0+} (1/x) / (-2/x³) = lim_{x->0+} (1/x) * (-x³/2) = lim_{x->0+} (-x²/2) = 0Therefore, the limit of x² ln x as x approaches 0+ is 0. Then, the other terms in the function f(x) as x approaches 0+:x² ln x + x² - 2 approaches 0 + 0 - 2 = -2So, f(x) approaches -2 as x approaches 0 from the right.Next, as x approaches infinity:Let's see the behavior of each term. As x becomes very large, ln x grows to infinity, but much slower than x². So, x² ln x will dominate over x² and -2. Therefore, the entire function f(x) will behave like x² ln x, which goes to infinity. So, lim_{x->∞} f(x) = ∞.Now, we know that as x approaches 0+, f(x) approaches -2, and as x approaches infinity, f(x) approaches infinity. Therefore, by the Intermediate Value Theorem, there must be at least one zero somewhere between 0 and infinity.But the question is, how many zeros are there? To determine the exact number, we should check if the function is monotonic (always increasing or always decreasing) or if it has turning points where it could potentially cross the x-axis more than once.To check this, we can compute the derivative f'(x) and analyze its sign.Let's compute f'(x):f(x) = x² ln x + x² - 2First, differentiate term by term.The derivative of x² ln x:Use product rule: d/dx [x² ln x] = 2x ln x + x²*(1/x) = 2x ln x + xThe derivative of x² is 2x, and the derivative of -2 is 0.Therefore, f'(x) = 2x ln x + x + 2x = 2x ln x + 3xWait, let's verify that:Wait, hold on. Wait, f(x) is x² ln x + x² - 2. So:First term: x² ln x. Derivative is 2x ln x + x²*(1/x) = 2x ln x + x.Second term: x². Derivative is 2x.Third term: -2. Derivative is 0.So total derivative:f'(x) = (2x ln x + x) + 2x = 2x ln x + 3x.Yes, that's correct. So f'(x) = x(2 ln x + 3)Factor out x: x(2 ln x + 3). Since x > 0, the sign of f'(x) is determined by (2 ln x + 3).So, let's analyze when f'(x) is positive or negative.Set 2 ln x + 3 = 0:2 ln x + 3 = 0 => ln x = -3/2 => x = e^{-3/2} ≈ e^{-1.5} ≈ 0.2231So, for x < e^{-3/2}, 2 ln x + 3 < 0, so f'(x) = x*(negative) = negative.For x > e^{-3/2}, 2 ln x + 3 > 0, so f'(x) = x*(positive) = positive.Therefore, the function f(x) is decreasing on (0, e^{-3/2}) and increasing on (e^{-3/2}, ∞). So, there is a critical point at x = e^{-3/2}, which is a minimum because the function changes from decreasing to increasing there.Therefore, the function f(x) has a single minimum at x = e^{-3/2}. To determine the number of zeros, we need to check the value of the function at this critical point.If the minimum value is below zero, then since the function approaches -2 as x approaches 0 and goes to infinity as x approaches infinity, it must cross the x-axis twice: once between 0 and the critical point, and once after the critical point. Wait, but actually, since the function is decreasing before the critical point and increasing after, if the minimum is below zero, there might be two zeros. Wait, but wait: at x approaching 0, f(x) approaches -2, which is below zero, then the function decreases to the minimum. If the minimum is also below zero, then the function is always below zero from 0 to the critical point, and then increases from the critical point to infinity. Since it goes to infinity, it must cross zero once after the critical point. So, in that case, only one zero. Wait, no. Wait, hold on. Let's think again.Wait, as x approaches 0 from the right, f(x) approaches -2. Then, the function is decreasing until x = e^{-3/2}, so if it's already at -2 and decreasing, it goes more negative. Then, at x = e^{-3/2}, it's at its minimum. Then, it starts increasing. So, if the minimum value is less than zero, then after x = e^{-3/2}, the function increases from the minimum (which is below zero) to infinity, crossing the x-axis once. However, before the critical point, since the function was decreasing from -2 (approaching 0+) to the minimum, which is even lower, so it never crosses zero there. Therefore, only one zero after the critical point.But wait, that contradicts my initial thought. Wait, let's verify this with actual computation. Let's compute the value of f(x) at the critical point x = e^{-3/2}.Let me compute f(e^{-3/2}):f(x) = x² ln x + x² - 2So, substitute x = e^{-3/2}:First, compute ln x: ln(e^{-3/2}) = -3/2Then, x² = (e^{-3/2})² = e^{-3}Therefore, f(e^{-3/2}) = e^{-3}*(-3/2) + e^{-3} - 2Compute each term:First term: e^{-3}*(-3/2) = (-3/2)e^{-3}Second term: e^{-3}Third term: -2So, total:(-3/2)e^{-3} + e^{-3} - 2 = (-3/2 + 1)e^{-3} - 2 = (-1/2)e^{-3} - 2Since e^{-3} is positive, (-1/2)e^{-3} is negative, so f(e^{-3/2}) is negative minus 2, which is definitely negative. Therefore, the minimum value of the function is negative. Therefore, the function decreases from -2 at x approaching 0, reaches a minimum (lower than -2) at x = e^{-3/2}, then increases towards infinity. Therefore, since the function was already negative at x approaching 0, decreases to a more negative value, then increases to infinity, it must cross the x-axis exactly once after the critical point. Therefore, the function has only one zero.Wait, but wait. Let me double-check. If the function starts at -2, decreases to a lower value, then increases to infinity, crossing the x-axis once. So, only one zero. Therefore, the answer should be 1.But let me verify this with some test points. Let's compute f(1):f(1) = 1² ln 1 + 1² - 2 = 0 + 1 - 2 = -1So f(1) = -1, which is still negative.Compute f(2):f(2) = 4 ln 2 + 4 - 2 ≈ 4*0.6931 + 2 ≈ 2.7724 + 2 ≈ 4.7724 > 0So between x=1 and x=2, the function goes from -1 to +4.7724, so crosses zero once there.Now, check between x approaching 0 and x= e^{-3/2} (≈0.2231). Let's pick x=0.1:f(0.1) = (0.1)^2 ln 0.1 + (0.1)^2 - 2 = 0.01*(-2.3026) + 0.01 - 2 ≈ -0.023026 + 0.01 - 2 ≈ -2.013026 < 0So, at x=0.1, f(x) ≈ -2.013, which is even less than -2. So, the function is decreasing from -2 to about -2.013 at x≈0.2231, then increasing again. So, it never crosses zero before the critical point because it's moving from -2 to a lower value. Then, after the critical point, it increases past zero once. Therefore, only one zero.But wait, wait a second. The problem statement says "the number of zeros". If the function is continuous, starts at -2, goes lower, then comes up to infinity, crossing zero once, then yes, only one zero. But let me check another point after the critical point. Wait, the critical point is at x≈0.2231. Let's pick a point between 0.2231 and 1, say x=0.5.Compute f(0.5):x=0.5, x²=0.25, ln 0.5≈-0.6931f(0.5) = 0.25*(-0.6931) + 0.25 - 2 ≈ -0.173275 + 0.25 - 2 ≈ (-0.173275 + 0.25) -2 ≈ 0.076725 -2 ≈ -1.923275 < 0So, at x=0.5, f(x)≈-1.923, still negative.At x=1, f(1)=-1, still negative.At x=1.5:x=1.5, x²=2.25, ln 1.5≈0.4055f(1.5)=2.25*0.4055 + 2.25 - 2 ≈0.912375 + 2.25 -2≈3.162375 -2≈1.162375>0So, between x=1 and x=1.5, the function goes from -1 to +1.16, crossing zero once.But wait, we saw that at x=2, f(x)≈4.77, so continues increasing.So, even though the function starts increasing after x≈0.223, it's still negative until some point after x=1, where it becomes positive. Therefore, there is exactly one zero crossing.Therefore, the number of zeros of the function f(x)=x² ln x +x² -2 is 1.But wait, to make sure there's no other zero, let's check near x=0. Let's pick a very small x, say x=0.01:f(0.01)= (0.01)^2 ln 0.01 + (0.01)^2 -2 = 0.0001*(-4.605) + 0.0001 -2≈-0.0004605 +0.0001 -2≈-2.0003605 < -2So, even closer to zero, the function is approaching -2 from below. So, it's not crossing zero near zero. Hence, only one zero.Therefore, the answer should be 1.Wait, but wait. Let me confirm again. The function is continuous on (0, ∞). At x approaches 0+, f approaches -2. Then, it decreases to a minimum at x≈0.223, which is even lower (≈-2.000... something?), then increases to infinity. Since the function is always negative before the minimum, and then increases to cross zero once. Therefore, only one zero.Yes, that seems correct. So, the number of zeros is 1. Therefore, the answer is boxed{1}.Final AnswerThe number of zeros of the function is boxed{1}.</think>To determine the number of zeros of the function ( f(x) = x^2 ln x + x^2 - 2 ), we analyze its behavior and critical points.1. Domain: The function is defined for ( x > 0 ) because ( ln x ) is only defined for positive ( x ).2. Behavior at the boundaries: - As ( x to 0^+ ), ( x^2 ln x to 0 ) (using L'Hospital's Rule), so ( f(x) to -2 ). - As ( x to infty ), ( x^2 ln x ) dominates and ( f(x) to infty ).3. Derivative: Compute ( f'(x) ) to find critical points: [ f'(x) = 2x ln x + 3x = x(2 ln x + 3) ] Setting ( f'(x) = 0 ) gives ( x = e^{-3/2} approx 0.2231 ).4. Critical Point Analysis: - For ( x < e^{-3/2} ), ( f'(x) < 0 ) (function is decreasing). - For ( x > e^{-3/2} ), ( f'(x) > 0 ) (function is increasing). - The critical point ( x = e^{-3/2} ) is a minimum.5. Value at Critical Point: [ f(e^{-3/2}) = e^{-3} left( -frac{3}{2} right) + e^{-3} - 2 = -frac{1}{2}e^{-3} - 2 < 0 ] The minimum value is negative.6. Intermediate Value Theorem: - Since ( f(x) to -2 ) as ( x to 0^+ ) and increases to ( infty ) as ( x to infty ), and the minimum value is negative, the function must cross the x-axis once after the critical point.Thus, the function ( f(x) = x^2 ln x + x^2 - 2 ) has exactly one zero. The number of zeros is (boxed{1}).]

🔑:Let t=x^{2}, g(t)=frac{1}{2} t ln t+t-2, g^{prime}(t)=frac{1}{2}(ln t+1)+1=frac{1}{2}(ln t+3), then g(t) is monotonically decreasing on left(0, mathrm{e}^{-3}right) and monotonically increasing on left(mathrm{e}^{-3},+inftyright), and g(0)0, so f(x) has 1 zero point in (0,+infty).