answer：Okay, so I have this problem here about a function f(x) which is given as -sin²x + a sinx + b cosx. It's said to be an even function, and also that f(π) = -1. I need to find f(x) first, and then solve another part involving some inequality with m. Let me try to break this down step by step.Starting with part (1): Finding f(x). Since f(x) is an even function, that means f(-x) = f(x) for all x. So, let me write down what f(-x) would be. f(-x) = -sin²(-x) + a sin(-x) + b cos(-x). I know that sin(-x) = -sinx and cos(-x) = cosx. So substituting these in, we get:f(-x) = -sin²x - a sinx + b cosx.But since f(x) is even, f(-x) must equal f(x). So:- sin²x - a sinx + b cosx = - sin²x + a sinx + b cosx.Hmm, let me subtract the right side from both sides to see what that gives:(- sin²x - a sinx + b cosx) - (- sin²x + a sinx + b cosx) = 0.Simplifying this:- sin²x - a sinx + b cosx + sin²x - a sinx - b cosx = 0.Wait, the sin²x terms cancel out, and the b cosx terms also cancel out. So we're left with:- a sinx - a sinx = 0 => -2a sinx = 0.This must hold for all x, so the coefficient must be zero. Therefore, -2a = 0 => a = 0.Okay, so a is zero. Now, let's use the other condition given: f(π) = -1.Compute f(π):f(π) = -sin²π + a sinπ + b cosπ.We know sinπ = 0, cosπ = -1. So substituting:f(π) = -0 + 0 + b*(-1) = -b.But f(π) is given as -1, so -b = -1 => b = 1.So now, we have a = 0 and b = 1. Therefore, the function f(x) is:f(x) = -sin²x + 0*sinx + 1*cosx = -sin²x + cosx.Alright, that seems straightforward. So part (1) is done.Moving on to part (2). It says: Given θ ∈ (0, π/2), and tanθ = √2. If for any x ∈ [-π/2, 0], the inequality a ≤ f(2x + θ) + m ≤ 4b always holds, find the range of values for m.Wait, from part (1), a = 0 and b = 1, so substituting these in, the inequality becomes:0 ≤ f(2x + θ) + m ≤ 4*1 => 0 ≤ f(2x + θ) + m ≤ 4.So, we need to find m such that for all x in [-π/2, 0], f(2x + θ) + m is between 0 and 4.First, let's express f(2x + θ). Since f(x) = -sin²x + cosx, then:f(2x + θ) = -sin²(2x + θ) + cos(2x + θ).I need to analyze this function over x ∈ [-π/2, 0]. Let me think about the range of 2x + θ.Since x ∈ [-π/2, 0], then 2x ∈ [-π, 0]. Adding θ, which is between 0 and π/2, so 2x + θ ∈ [-π + θ, θ]. Since θ is positive and less than π/2, the lower bound is -π + θ, which is less than -π/2, and the upper bound is θ, which is less than π/2.So, 2x + θ ranges from -π + θ to θ. Let me denote φ = 2x + θ, so φ ∈ [-π + θ, θ]. So, f(φ) = -sin²φ + cosφ.I need to find the range of f(φ) over φ ∈ [-π + θ, θ]. Then, since f(φ) + m must be between 0 and 4, I can set up inequalities to solve for m.But before that, maybe I can express f(φ) in a different form to make it easier to find its maximum and minimum.Let me write f(φ) = -sin²φ + cosφ.Recall that sin²φ = (1 - cos2φ)/2, so substituting:f(φ) = - (1 - cos2φ)/2 + cosφ = -1/2 + (cos2φ)/2 + cosφ.Hmm, not sure if that helps. Alternatively, maybe express f(φ) as a quadratic in terms of cosφ.Let me set t = cosφ. Then, sin²φ = 1 - t². So:f(φ) = -(1 - t²) + t = -1 + t² + t.So, f(φ) = t² + t - 1, where t = cosφ.Now, t = cosφ, and φ ∈ [-π + θ, θ]. Since cos is even, cosφ = cos(-φ), so the range of t can be determined by looking at the maximum and minimum of cosφ over φ ∈ [-π + θ, θ].But let's think about the interval [-π + θ, θ]. Since θ ∈ (0, π/2), then -π + θ ∈ (-π, -π/2). So, the interval spans from just below -π to just below π/2.The function cosφ is decreasing from 0 to π, increasing from π to 2π, but in the interval [-π, π], it's symmetric around 0.Wait, perhaps it's better to note that cosφ is decreasing on [0, π], so on [θ, π], it decreases from cosθ to -1. But our interval is from -π + θ to θ, which is symmetric around -π/2 + θ/2 or something? Maybe not.Alternatively, perhaps I can find the maximum and minimum of cosφ over φ ∈ [-π + θ, θ].Let me consider the function cosφ over φ ∈ [-π + θ, θ]. The maximum value occurs at the point where φ is closest to 0, which is θ, since θ is positive and less than π/2. So, cosθ is the maximum.The minimum value occurs at the point where φ is farthest from 0, which is -π + θ. So, cos(-π + θ) = cos(π - θ) = -cosθ.Wait, is that correct? Because cos(-π + θ) = cos(π - θ) because cos is even. And cos(π - θ) = -cosθ. So yes, the minimum value is -cosθ.So, t = cosφ ∈ [-cosθ, cosθ].But wait, is that accurate? Let me think again. When φ ranges from -π + θ to θ, cosφ starts at cos(-π + θ) = cos(π - θ) = -cosθ, then increases to cos0 = 1 at φ=0, but wait, does φ reach 0?Wait, φ = 2x + θ, and x ∈ [-π/2, 0], so φ = 2x + θ ∈ [2*(-π/2) + θ, 2*0 + θ] = [-π + θ, θ]. So, φ ranges from -π + θ to θ, which is a span of 2θ. But θ is less than π/2, so the total span is less than π.But in this interval, does φ pass through 0? Let's see: when x = -θ/2, then φ = 2*(-θ/2) + θ = -θ + θ = 0. So yes, φ does pass through 0 when x = -θ/2, which is within the interval x ∈ [-π/2, 0] since θ ∈ (0, π/2), so -θ/2 ∈ (-π/4, 0), which is within [-π/2, 0].Therefore, the maximum value of cosφ is 1, achieved at φ=0, and the minimum value is cos(-π + θ) = cos(π - θ) = -cosθ.Wait, hold on. At φ = -π + θ, cosφ = cos(-π + θ) = cos(π - θ) = -cosθ. But when φ increases from -π + θ to θ, cosφ increases from -cosθ to cosθ. But wait, when φ=θ, cosφ=cosθ.Wait, but if θ is in (0, π/2), then cosθ is positive, and -cosθ is negative. So, the range of t = cosφ is from -cosθ to cosθ. But wait, when φ=0, cosφ=1, which is greater than cosθ since θ is less than π/2, so cosθ is less than 1. So, actually, the maximum value of cosφ is 1, and the minimum is -cosθ.Wait, let's test with θ = π/4. Then, φ ranges from -π + π/4 = -3π/4 to π/4. So, cosφ at -3π/4 is -√2/2, and at π/4 is √2/2. But at φ=0, cosφ=1, which is higher than √2/2. So, yes, the maximum is 1, and the minimum is -√2/2.Therefore, in general, for θ ∈ (0, π/2), the range of t = cosφ is [-cosθ, 1].Wait, is that correct? Because when φ moves from -π + θ to θ, cosφ starts at -cosθ, increases to 1 at φ=0, then decreases to cosθ at φ=θ.Yes, that makes sense. So, the maximum value of t is 1, and the minimum is -cosθ.So, t ∈ [-cosθ, 1].Therefore, f(φ) = t² + t - 1, where t ∈ [-cosθ, 1].So, to find the range of f(φ), we can consider the function g(t) = t² + t - 1 over t ∈ [-cosθ, 1].We need to find the minimum and maximum of g(t) over this interval.First, let's find the critical points of g(t). The derivative g’(t) = 2t + 1. Setting this equal to zero gives t = -1/2.So, the critical point is at t = -1/2. We need to check if this is within our interval [-cosθ, 1].Since θ ∈ (0, π/2), cosθ ∈ (0, 1). So, -cosθ ∈ (-1, 0). Therefore, t = -1/2 is within [-cosθ, 1] because -cosθ > -1, so -1/2 > -cosθ (since cosθ < 1, so -cosθ > -1, but -1/2 is greater than -cosθ only if cosθ < 1/2. Wait, hold on.Wait, if θ ∈ (0, π/2), then cosθ ∈ (0, 1). So, -cosθ ∈ (-1, 0). So, t = -1/2 is in [-cosθ, 1] only if -1/2 ≥ -cosθ, which is equivalent to cosθ ≥ 1/2.But cosθ ≥ 1/2 implies θ ≤ π/3, since cos(π/3) = 1/2.But θ is in (0, π/2). So, if θ ≤ π/3, then -1/2 ≥ -cosθ, so t = -1/2 is within the interval. If θ > π/3, then -1/2 < -cosθ, so t = -1/2 is not in the interval.Wait, let's clarify:If θ ≤ π/3, then cosθ ≥ 1/2, so -cosθ ≤ -1/2. Therefore, the interval [-cosθ, 1] includes -1/2 because -cosθ ≤ -1/2.If θ > π/3, then cosθ < 1/2, so -cosθ > -1/2, meaning the interval [-cosθ, 1] does not include -1/2.So, depending on θ, the critical point t = -1/2 may or may not be inside the interval.But in our problem, θ is given such that tanθ = √2. Let's compute θ.Given tanθ = √2, so θ = arctan(√2). Let me compute arctan(√2). Since tan(π/3) = √3 ≈ 1.732, and tan(π/4) = 1. So, √2 ≈ 1.414, which is between 1 and √3, so θ is between π/4 and π/3. Specifically, θ ≈ 54.7 degrees, which is less than π/3 (≈60 degrees). Wait, actually, arctan(√2) is approximately 54.7 degrees, which is less than 60 degrees, so θ ≈ 54.7°, which is less than π/3 (≈60°). Wait, actually, no: 54.7° is less than 60°, so θ is less than π/3.Wait, let me confirm:tan(π/3) = √3 ≈ 1.732, tan(π/4) = 1, tan(54.7°) ≈ √2 ≈ 1.414. So, θ = arctan(√2) ≈ 54.7°, which is less than π/3 (≈60°). Therefore, cosθ = 1 / sqrt(1 + (√2)^2) = 1 / sqrt(3) ≈ 0.577, which is greater than 1/2 (0.5). So, cosθ ≈ 0.577 > 0.5.Therefore, since cosθ ≈ 0.577, which is greater than 1/2, so -cosθ ≈ -0.577, which is less than -1/2 (-0.5). Therefore, the interval [-cosθ, 1] includes -1/2 because -cosθ ≈ -0.577 < -0.5.So, in this specific case, the critical point t = -1/2 is within the interval [-cosθ, 1].Therefore, to find the extrema of g(t) = t² + t - 1 over t ∈ [-cosθ, 1], we need to evaluate g(t) at the critical point t = -1/2 and at the endpoints t = -cosθ and t = 1.So, let's compute g(-cosθ), g(-1/2), and g(1).First, compute g(-cosθ):g(-cosθ) = (-cosθ)^2 + (-cosθ) - 1 = cos²θ - cosθ - 1.Similarly, g(-1/2) = (-1/2)^2 + (-1/2) - 1 = 1/4 - 1/2 - 1 = (1 - 2 - 4)/4 = (-5)/4 = -5/4.g(1) = (1)^2 + 1 - 1 = 1 + 1 - 1 = 1.So, the maximum value of g(t) is 1, and the minimum is the lesser of g(-cosθ) and g(-1/2). Let's compute g(-cosθ):We have cosθ = 1 / sqrt(3), as tanθ = √2 implies that in a right triangle, opposite side is √2, adjacent is 1, hypotenuse is sqrt(1 + 2) = sqrt(3). So, cosθ = adjacent / hypotenuse = 1 / sqrt(3).Therefore, cos²θ = 1/3.So, g(-cosθ) = 1/3 - (1 / sqrt(3)) - 1.Compute this:1/3 - 1 - 1/sqrt(3) = (-2/3) - 1/sqrt(3).Which is approximately (-0.6667) - 0.577 ≈ -1.2437.Compare this with g(-1/2) = -5/4 = -1.25.So, g(-cosθ) ≈ -1.2437 and g(-1/2) = -1.25. So, the minimum is g(-1/2) = -5/4.Therefore, the range of f(φ) is [-5/4, 1].Wait, but let's compute g(-cosθ) exactly:g(-cosθ) = cos²θ - cosθ - 1 = (1/3) - (1/√3) - 1.Let me write it as:(1/3 - 1) - 1/√3 = (-2/3) - 1/√3.So, it's -2/3 - 1/√3. Let me compute this:-2/3 ≈ -0.6667, 1/√3 ≈ 0.577, so total ≈ -0.6667 - 0.577 ≈ -1.2437.Which is slightly greater than -1.25, which is -5/4. So, the minimum is indeed at t = -1/2, which is -5/4.Therefore, f(φ) ∈ [-5/4, 1].So, f(2x + θ) ∈ [-5/4, 1].Now, the inequality given is:0 ≤ f(2x + θ) + m ≤ 4.So, substituting the range of f(2x + θ):For the lower bound: f(2x + θ) + m ≥ 0 ⇒ m ≥ -f(2x + θ).Since f(2x + θ) can be as low as -5/4, the maximum of -f(2x + θ) is 5/4. Therefore, to satisfy m ≥ -f(2x + θ) for all x, m must be at least 5/4.Similarly, for the upper bound: f(2x + θ) + m ≤ 4 ⇒ m ≤ 4 - f(2x + θ).Since f(2x + θ) can be as high as 1, the minimum of 4 - f(2x + θ) is 4 - 1 = 3. Therefore, to satisfy m ≤ 4 - f(2x + θ) for all x, m must be at most 3.Therefore, combining both inequalities, m must satisfy:5/4 ≤ m ≤ 3.So, the range of m is [5/4, 3].Wait, let me double-check:We have f(φ) ∈ [-5/4, 1]. So, f(φ) + m ∈ [m - 5/4, m + 1].We need [m - 5/4, m + 1] ⊆ [0, 4].Therefore:1. m - 5/4 ≥ 0 ⇒ m ≥ 5/4.2. m + 1 ≤ 4 ⇒ m ≤ 3.Yes, that's correct. So, m must be between 5/4 and 3.But wait, 5/4 is 1.25, which is the same as 5/4, and 3 is 3. So, the range is [5/4, 3].But in the initial problem, part (2) says "the inequality a ≤ f(2x + θ) + m ≤ 4b". From part (1), a = 0 and b = 1, so it's 0 ≤ f(2x + θ) + m ≤ 4.So, yes, that's consistent.Therefore, the range of m is [5/4, 3].But wait, let me confirm the exact values:We found that f(φ) ∈ [-5/4, 1], so f(φ) + m ∈ [m - 5/4, m + 1].We need m - 5/4 ≥ 0 ⇒ m ≥ 5/4.And m + 1 ≤ 4 ⇒ m ≤ 3.So, m ∈ [5/4, 3].Yes, that seems correct.So, summarizing:(1) f(x) = -sin²x + cosx.(2) The range of m is [5/4, 3].**Final Answer**(1) boxed{-sin^2 x + cos x}(2) boxed{left[ dfrac{5}{4}, 3 right]}

answer：Okay, so I have this problem here: I need to find the equation of the curve that's symmetric to the curve ( y^2 = 4x ) with respect to the line ( x = 2 ). The options are A: ( y^2 = 8 - 4x ), B: ( y^2 = 4x - 8 ), C: ( y^2 = 16 - 4x ), and D: ( y^2 = 4x - 16 ). Hmm, let me try to figure this out step by step.First, I remember that when you reflect a point over a vertical line ( x = a ), the x-coordinate changes, but the y-coordinate stays the same. The formula for reflecting a point ( (x, y) ) over the line ( x = a ) is ( (2a - x, y) ). So, in this case, since we're reflecting over ( x = 2 ), the reflection of any point ( (x, y) ) would be ( (4 - x, y) ).Now, the original curve is ( y^2 = 4x ). Let's say there's a point ( (x, y) ) on the symmetric curve. Its reflection over ( x = 2 ) would be ( (4 - x, y) ), and this reflected point should lie on the original curve ( y^2 = 4x ). So, substituting ( (4 - x, y) ) into the original equation should give me the equation of the symmetric curve.Let me write that out. If ( (4 - x, y) ) is on ( y^2 = 4x ), then substituting gives:( y^2 = 4(4 - x) )Simplify that:( y^2 = 16 - 4x )So, that should be the equation of the symmetric curve. Let me check the options: Option C is ( y^2 = 16 - 4x ), so that must be the answer.Wait, just to make sure I didn't make a mistake. Let me think about it again. The reflection over ( x = 2 ) changes ( x ) to ( 4 - x ). So, if I plug ( x' = 4 - x ) into the original equation ( y^2 = 4x ), I get ( y^2 = 4(4 - x) ), which is ( y^2 = 16 - 4x ). Yeah, that seems right.Maybe I can also visualize it. The original curve ( y^2 = 4x ) is a parabola that opens to the right with vertex at the origin. Reflecting it over the line ( x = 2 ) should create another parabola that opens to the left, with its vertex at ( x = 4 ) (since the original vertex was at ( x = 0 ), and reflecting over ( x = 2 ) would place it 2 units on the other side). So, the vertex of the symmetric curve should be at ( (4, 0) ). Looking at the equation ( y^2 = 16 - 4x ), if I rewrite it as ( y^2 = -4x + 16 ) or ( y^2 = -4(x - 4) ), that makes sense because it's a parabola opening to the left with vertex at ( (4, 0) ). So, that confirms it.I think I'm confident now that option C is the correct answer.**Final Answer**boxed{C}

answer：Okay, so I have this geometry problem here. It says that the straight line 3x - 4y + 5 = 0 intersects with the circle x² + y² = r² at points A and B. The angle AOB is 120 degrees, where O is the origin. I need to find the value of r.Hmm, let me visualize this. There's a circle centered at the origin with radius r, and a straight line that cuts through the circle at two points A and B. The angle between the two radii OA and OB is 120 degrees. So, triangle AOB is an isosceles triangle with OA and OB both equal to r, and the angle at O is 120 degrees.I remember that in such cases, the distance from the center of the circle to the line can be related to the radius and the angle. Maybe I can use some trigonometry here.First, I should find the distance from the origin to the line 3x - 4y + 5 = 0. The formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / sqrt(A² + B²). In this case, the point is (0,0), so plugging in, the distance d is |3*0 - 4*0 + 5| / sqrt(3² + (-4)²) = |5| / 5 = 1. So, the distance from the origin to the line is 1 unit.Now, how does this distance relate to the radius and the angle? I think in the triangle AOB, the distance d is the length of the perpendicular from O to the line AB. Since triangle AOB is isosceles with OA = OB = r and angle AOB = 120 degrees, maybe I can find the relationship between d, r, and the angle.Let me draw triangle AOB. OA and OB are both length r, and the angle between them is 120 degrees. The line AB is the chord of the circle, and the distance from O to AB is d = 1.I recall that in a circle, the length of the chord can be related to the radius and the central angle. The formula for the length of chord AB is 2r sin(theta/2), where theta is the central angle. So, AB = 2r sin(120°/2) = 2r sin(60°) = 2r*(√3/2) = r√3.But I also know that the distance from the center to the chord is d = 1, and there's a relationship between the radius, the chord length, and the distance. The formula is d = r cos(theta/2). Let me check that.Yes, in an isosceles triangle, the distance from the vertex to the base is equal to the length of the equal sides times the cosine of half the vertex angle. So, d = r cos(theta/2). Plugging in the values, d = r cos(60°) = r*(1/2). So, 1 = (1/2)r, which means r = 2.Wait, that seems straightforward. Let me verify if this makes sense. If the distance from the center to the chord is 1, and the radius is 2, then the chord length should be 2*2*sin(60°) = 4*(√3/2) = 2√3. Alternatively, using the chord length formula, AB = 2*sqrt(r² - d²) = 2*sqrt(4 - 1) = 2*sqrt(3), which matches. So, that seems consistent.Alternatively, I can use the area of triangle AOB. The area can be calculated in two ways: using the formula (1/2)*OA*OB*sin(theta) and also using the formula (1/2)*AB*d.First method: Area = (1/2)*r*r*sin(120°) = (1/2)*r²*(√3/2) = (√3/4)r².Second method: Area = (1/2)*AB*d = (1/2)*(r√3)*1 = (√3/2)r.Setting these equal: (√3/4)r² = (√3/2)r. Dividing both sides by √3/4, we get r² = 2r, so r = 2. That's the same result as before.So, both methods confirm that r = 2. I think that's the correct answer.

answer：Okay, so I have this problem where I need to determine the number of pairs (a, b) of real numbers such that 12, a, b, ab form a geometric progression. Hmm, let me think about how to approach this.First, I remember that in a geometric progression, each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as r. So, if 12, a, b, ab are in geometric progression, then a should be 12 multiplied by r, right? So, a = 12r.Then, the next term, b, should be a multiplied by r, which would be 12r * r = 12r². So, b = 12r².Now, the fourth term in the progression is ab. Since it's a geometric progression, ab should be equal to b multiplied by r, which is 12r² * r = 12r³. So, ab = 12r³.But wait, ab is also the product of a and b. Since a = 12r and b = 12r², then ab = (12r)(12r²) = 144r³.So, from the geometric progression, we have ab = 12r³, and from the product, we have ab = 144r³. Therefore, setting these equal to each other: 12r³ = 144r³.Hmm, that seems interesting. If I subtract 12r³ from both sides, I get 0 = 132r³. So, 132r³ = 0. Dividing both sides by 132, I get r³ = 0, which implies r = 0.Wait, but if r = 0, then a = 12r = 0, and b = 12r² = 0. Then, ab = 0 * 0 = 0. So, the sequence would be 12, 0, 0, 0. Is that a valid geometric progression?Well, technically, a geometric progression requires a common ratio, but if the common ratio is zero, after the first term, all subsequent terms are zero. So, 12, 0, 0, 0 is a geometric progression with common ratio 0. But is that acceptable? I think in some contexts, it's considered a trivial geometric progression, but I'm not sure if it's considered valid here.Alternatively, maybe I made a mistake in my reasoning. Let me double-check.I said that since 12, a, b, ab is a geometric progression, then a = 12r, b = 12r², and ab = 12r³. Then, ab is also equal to a * b, which is 12r * 12r² = 144r³. So, equating the two expressions for ab: 12r³ = 144r³, which simplifies to 0 = 132r³, so r = 0.But if r = 0, then a = 0, b = 0, and ab = 0. So, the sequence is 12, 0, 0, 0. Is this acceptable? I think in some definitions, a geometric progression requires the common ratio to be non-zero, so that each term is non-zero. If that's the case, then r = 0 might not be acceptable.Alternatively, maybe there's another way to approach this problem. Let me think.Suppose the common ratio is r, so:a = 12rb = a * r = 12r²ab = b * r = 12r³But ab is also equal to a * b, which is (12r)(12r²) = 144r³So, 12r³ = 144r³Subtracting 12r³ from both sides: 0 = 132r³So, r³ = 0, which gives r = 0.So, seems like the only solution is r = 0, which gives a = 0, b = 0.But then, the sequence is 12, 0, 0, 0. Is this considered a geometric progression? I think it depends on the definition. Some sources say that in a geometric progression, the common ratio can be zero, but others might require the ratio to be non-zero to avoid all terms after the first being zero.If we allow r = 0, then there is exactly one such pair (a, b) = (0, 0). If we don't allow r = 0, then there are no solutions.But the problem says "real numbers," so r can be any real number, including zero. So, perhaps the answer is that there is exactly one such pair.Wait, but let me think again. Maybe I missed something. Is there another way to interpret the problem?Alternatively, maybe I can set up equations without assuming the common ratio.Let me denote the common ratio as r. Then:a = 12rb = a * r = 12r²ab = b * r = 12r³But ab is also a * b = (12r)(12r²) = 144r³So, 12r³ = 144r³Which simplifies to 12r³ - 144r³ = 0 => -132r³ = 0 => r³ = 0 => r = 0.So, again, r = 0 is the only solution, leading to a = 0, b = 0.Therefore, there is exactly one pair (a, b) = (0, 0).But wait, is there another possible common ratio? Maybe negative or something else?Wait, let's see. Suppose r ≠ 0, then from 12r³ = 144r³, we can divide both sides by r³ (since r ≠ 0), getting 12 = 144, which is not true. So, no solution for r ≠ 0.Therefore, the only solution is r = 0, leading to a = 0, b = 0.Hence, there is exactly one such pair.But wait, the initial problem says "pairs (a, b) of real numbers." So, if r = 0 is allowed, then (0, 0) is the only pair. If r = 0 is not allowed, then there are no solutions.But in the context of real numbers, r = 0 is a real number, so I think it's acceptable.Therefore, the number of pairs is 1.Wait, but earlier I thought it might be infinite, but now I'm getting only one solution. Hmm.Wait, maybe I made a mistake in assuming that ab = 12r³. Let me double-check.Given the geometric progression: 12, a, b, ab.So, the ratio between a and 12 is r, so a = 12r.Then, the ratio between b and a is also r, so b = a * r = 12r².Then, the ratio between ab and b is r, so ab = b * r = 12r³.But ab is also equal to a * b, which is 12r * 12r² = 144r³.So, 12r³ = 144r³ => 12r³ - 144r³ = 0 => -132r³ = 0 => r³ = 0 => r = 0.So, yes, only r = 0 is the solution, leading to a = 0, b = 0.Therefore, only one pair.But wait, another thought: maybe the common ratio is not necessarily the same for all steps? No, in a geometric progression, the common ratio is consistent.So, the ratio between a and 12 is r, between b and a is r, and between ab and b is r. So, all must be equal to r.Therefore, the only solution is r = 0, leading to a = 0, b = 0.Hence, only one pair.But wait, let me think again. Maybe I can consider the case where r = 1? Let's see.If r = 1, then a = 12, b = 12, ab = 144. So, the sequence would be 12, 12, 12, 144. But 12, 12, 12 is a geometric progression with r = 1, but then the next term should be 12 * 1 = 12, but ab = 144 ≠ 12. So, that doesn't work.Similarly, if r = -1, then a = -12, b = 12, ab = -144. So, the sequence is 12, -12, 12, -144. Checking the ratios: -12 / 12 = -1, 12 / (-12) = -1, -144 / 12 = -12. So, the ratio changes from -1 to -12, which is not consistent. So, that doesn't work either.So, seems like r = 0 is the only possible common ratio that works, leading to a = 0, b = 0.Therefore, there is exactly one pair (a, b) = (0, 0).Wait, but earlier I thought it might be infinite, but now I'm convinced it's only one.Alternatively, maybe I can approach this problem differently. Let me set up equations without assuming the common ratio.Let me denote the common ratio as r. Then:a = 12rb = a * r = 12r²ab = b * r = 12r³But ab is also equal to a * b, which is (12r)(12r²) = 144r³So, 12r³ = 144r³Which simplifies to 12r³ - 144r³ = 0 => -132r³ = 0 => r³ = 0 => r = 0.So, again, r = 0 is the only solution, leading to a = 0, b = 0.Therefore, only one pair.Hence, the number of pairs is 1.

answer：First, I need to determine how many clarinets, trumpets, and pianists were accepted into the band based on the given percentages.For the clarinets, half of the 30 clarinets got in, which is 15 clarinets.For the trumpets, one-third of the 60 trumpets got in, which is 20 trumpets.For the pianists, one-tenth of the 20 pianists got in, which is 2 pianists.Adding these accepted members together gives a total of 37 non-flute players in the band.Since the total number of band members is 53, I subtract the 37 non-flute players to find out how many flutes were accepted: 53 - 37 = 16 flutes.Knowing that 80% of the flutes who tried out were accepted, I can set up the equation 0.80 * F = 16, where F is the total number of flutes that tried out.Solving for F, I divide both sides by 0.80: F = 16 / 0.80 = 20.Therefore, 20 flutes tried out for the band.

answer：First, I need to define the variables for the width and length of the rectangular land. Let ( w ) represent the width and ( l ) represent the length.Given that the length is twice the width, I can express this relationship as:[l = 2w]Next, I'll calculate the area of the rectangle using the formula:[text{Area} = l times w]Substituting the expression for ( l ) from the given relationship:[text{Area} = 2w times w = 2w^2]The requirement is that the area must be at least 800 square feet:[2w^2 geq 800]To find the minimum width, I'll solve for ( w ):[w^2 geq 400][w geq sqrt{400}][w geq 20]Now, I'll calculate the perimeter of the rectangle to determine the amount of fencing needed. The formula for the perimeter is:[P = 2l + 2w]Substituting ( l = 2w ) into the perimeter formula:[P = 2(2w) + 2w = 4w + 2w = 6w]To minimize the amount of fencing, I'll use the smallest possible width that satisfies the area requirement:[w = 20 text{ feet}]Then, the length will be:[l = 2w = 2 times 20 = 40 text{ feet}]Finally, the minimum perimeter is:[P = 6w = 6 times 20 = 120 text{ feet}]