🧠:<think>Okay, so I have this problem in front of me, and I need to solve it step by step. Let me read it again carefully.In the Cartesian coordinate system xOy, with the origin O as the pole and the non-negative half-axis of the x-axis as the polar axis, a polar coordinate system is established. The polar equation of curve C is ρ = 4 cos θ (0 ≤ θ ≤ π/2), and the parametric equation of line l is given as:x = -3 + t cos(π/6)y = t sin(π/6)where t is the parameter.Part (1) asks me to find the Cartesian equation of line l and the parametric equation of curve C.Alright, let's tackle part (1) first.Starting with line l. The parametric equations are given in terms of t, so I need to eliminate the parameter t to find the Cartesian equation.Given:x = -3 + t cos(π/6)y = t sin(π/6)First, I can solve for t from the second equation:y = t sin(π/6)So, t = y / sin(π/6)I know that sin(π/6) is 1/2, so t = y / (1/2) = 2y.Now, substitute this value of t into the first equation:x = -3 + (2y) cos(π/6)I remember that cos(π/6) is √3/2, so:x = -3 + 2y*(√3/2)Simplify that:x = -3 + y√3So, rearranging terms to get the standard Cartesian form:x - √3 y + 3 = 0Okay, so that's the Cartesian equation of line l.Now, moving on to finding the parametric equation of curve C. The polar equation is given as ρ = 4 cos θ, with θ between 0 and π/2.I recall that in polar coordinates, ρ = 4 cos θ is the equation of a circle. To convert this into Cartesian coordinates, I can use the relationships:x = ρ cos θy = ρ sin θand ρ² = x² + y².Given ρ = 4 cos θ, multiply both sides by ρ:ρ² = 4ρ cos θWhich translates to:x² + y² = 4xSo, rearranging:x² - 4x + y² = 0Completing the square for x:x² - 4x + 4 + y² = 4Which becomes:(x - 2)² + y² = 4So, this is a circle with center at (2, 0) and radius 2.But since θ is between 0 and π/2, the curve C is only the part of this circle in the first quadrant, right? Because θ from 0 to π/2 corresponds to the first quadrant in Cartesian coordinates.Therefore, the parametric equations for this circle can be written using an angle parameter, say α, which will range from 0 to π, since we're covering the upper half of the circle (from θ=0 to θ=π/2, but in Cartesian, it's a semicircle in the first quadrant).So, the parametric equations for a circle are generally:x = h + r cos αy = k + r sin αHere, the center (h, k) is (2, 0), and the radius r is 2. So, substituting:x = 2 + 2 cos αy = 0 + 2 sin αTherefore, the parametric equations are:x = 2 + 2 cos αy = 2 sin αAnd since θ ranges from 0 to π/2, α should range from 0 to π to cover the entire upper half of the circle in the first quadrant.Wait, hold on. If θ is from 0 to π/2, does that mean α should also be from 0 to π/2? Hmm, maybe I need to think about this.In polar coordinates, θ is the angle from the positive x-axis. So, when θ is 0, we're at (4, 0) in polar, which is (4, 0) in Cartesian. As θ increases to π/2, we move up along the circle to (0, 4) in Cartesian.But in the parametric equations, α is the parameter for the circle. So, when α is 0, we're at (2 + 2*1, 0) = (4, 0). When α is π/2, we're at (2 + 0, 2*1) = (2, 2). Wait, that's not (0, 4). Hmm, maybe I'm mixing up the angles here.Wait, perhaps I need to adjust the parameterization. Let me think.Alternatively, maybe I should express the parametric equations in terms of θ. Since the original equation is in terms of θ, perhaps I can use θ as the parameter.Given ρ = 4 cos θ, so in Cartesian:x = ρ cos θ = 4 cos θ * cos θ = 4 cos² θy = ρ sin θ = 4 cos θ * sin θ = 4 cos θ sin θSo, the parametric equations can be written as:x = 4 cos² θy = 4 cos θ sin θBut θ is between 0 and π/2, so that's the parameter range.Alternatively, I can express this using double-angle identities. Let me see:x = 4 cos² θ = 2(1 + cos 2θ)y = 4 cos θ sin θ = 2 sin 2θSo, another way to write the parametric equations is:x = 2 + 2 cos 2θy = 2 sin 2θBut θ is from 0 to π/2, so 2θ is from 0 to π. So, if I let α = 2θ, then α ranges from 0 to π, and the parametric equations become:x = 2 + 2 cos αy = 2 sin αWhich is the same as before. So, whether I use θ or α, it's the same result. So, I think that's correct.Therefore, the parametric equation of curve C is:x = 2 + 2 cos αy = 2 sin αwhere α is from 0 to π.So, that's part (1) done.Now, moving on to part (2). It asks for the range of distances from a moving point M on curve C to line l.So, I need to find the minimum and maximum distances from any point M on curve C to the line l.Given that curve C is the circle (x - 2)^2 + y^2 = 4, with y ≥ 0 (since θ is between 0 and π/2), and line l is x - √3 y + 3 = 0.So, to find the distance from a point (x, y) on curve C to line l, I can use the distance formula from a point to a line.The distance d from a point (x0, y0) to the line Ax + By + C = 0 is given by:d = |Ax0 + By0 + C| / sqrt(A² + B²)In our case, the line is x - √3 y + 3 = 0, so A = 1, B = -√3, C = 3.So, the distance from point M(x, y) on curve C to line l is:d = |x - √3 y + 3| / sqrt(1 + (√3)^2) = |x - √3 y + 3| / sqrt(1 + 3) = |x - √3 y + 3| / 2Since curve C is parameterized as x = 2 + 2 cos α, y = 2 sin α, I can substitute these into the distance formula.So, substituting x and y:d = |(2 + 2 cos α) - √3*(2 sin α) + 3| / 2Simplify the numerator:2 + 2 cos α - 2√3 sin α + 3 = (2 + 3) + 2 cos α - 2√3 sin α = 5 + 2 cos α - 2√3 sin αSo, d = |5 + 2 cos α - 2√3 sin α| / 2Since the absolute value is always non-negative, and the denominator is positive, the distance is non-negative.Now, to find the range of d, I need to find the minimum and maximum values of |5 + 2 cos α - 2√3 sin α| as α varies from 0 to π.But since the expression inside the absolute value is linear in cos α and sin α, it can be rewritten in the form R cos(α + φ) or R sin(α + φ), which will help in finding the maximum and minimum.Let me consider the expression:2 cos α - 2√3 sin αThis is of the form A cos α + B sin α, where A = 2 and B = -2√3.I can write this as R cos(α + φ), where R = sqrt(A² + B²) and tan φ = B/A.Calculating R:R = sqrt(2² + (-2√3)^2) = sqrt(4 + 12) = sqrt(16) = 4Calculating φ:tan φ = B/A = (-2√3)/2 = -√3So, φ is the angle whose tangent is -√3. Since tan(π/3) = √3, tan(-π/3) = -√3.But since A is positive and B is negative, the angle φ is in the fourth quadrant. However, since we're dealing with cos(α + φ), we can adjust accordingly.Alternatively, we can express it as R sin(α + φ'), but let's stick with cos.So, 2 cos α - 2√3 sin α = 4 cos(α + φ), where φ = π/3. Wait, let me verify.Wait, if tan φ = -√3, then φ = -π/3. So, cos φ = cos(-π/3) = 1/2, sin φ = sin(-π/3) = -√3/2.So, R cos(α + φ) = 4 [cos α cos φ - sin α sin φ] = 4 [cos α*(1/2) - sin α*(-√3/2)] = 4 [ (1/2) cos α + (√3/2) sin α ] = 2 cos α + 2√3 sin αBut our expression is 2 cos α - 2√3 sin α, which is different. Hmm, so maybe I need to adjust the angle.Alternatively, perhaps it's better to write it as R sin(α + φ). Let's try that.Express 2 cos α - 2√3 sin α as R sin(α + φ).We know that R sin(α + φ) = R sin α cos φ + R cos α sin φComparing coefficients:R cos φ = -2√3 (coefficient of sin α)R sin φ = 2 (coefficient of cos α)So, R^2 = (-2√3)^2 + 2^2 = 12 + 4 = 16, so R = 4.Then, tan φ = (R sin φ)/(R cos φ) = (2)/(-2√3) = -1/√3So, φ is the angle whose tangent is -1/√3, which is -π/6 or 11π/6.Therefore, 2 cos α - 2√3 sin α = 4 sin(α - π/6)Wait, let me check:4 sin(α - π/6) = 4 [sin α cos(π/6) - cos α sin(π/6)] = 4 [sin α*(√3/2) - cos α*(1/2)] = 2√3 sin α - 2 cos αBut our expression is 2 cos α - 2√3 sin α, which is the negative of that. So, 2 cos α - 2√3 sin α = -4 sin(α - π/6)Alternatively, we can write it as 4 sin(α + 5π/6), but maybe it's simpler to stick with the negative.So, 2 cos α - 2√3 sin α = -4 sin(α - π/6)Therefore, the numerator becomes:5 + 2 cos α - 2√3 sin α = 5 - 4 sin(α - π/6)So, the distance d is:d = |5 - 4 sin(α - π/6)| / 2Now, since α ranges from 0 to π, let's find the range of sin(α - π/6).When α = 0, sin(-π/6) = -1/2When α = π, sin(π - π/6) = sin(5π/6) = 1/2But wait, actually, as α increases from 0 to π, α - π/6 increases from -π/6 to 5π/6.So, the angle α - π/6 ranges from -π/6 to 5π/6.Therefore, sin(α - π/6) ranges from sin(-π/6) = -1/2 to sin(π/2) = 1, and then back to sin(5π/6) = 1/2.So, the maximum value of sin(α - π/6) is 1, and the minimum is -1/2.Therefore, the expression inside the absolute value is:5 - 4 sin(α - π/6)So, when sin(α - π/6) is maximum (1), the expression is 5 - 4*1 = 1When sin(α - π/6) is minimum (-1/2), the expression is 5 - 4*(-1/2) = 5 + 2 = 7But wait, the expression is 5 - 4 sin(α - π/6). So, as sin(α - π/6) increases, the expression decreases, and vice versa.So, the maximum value of 5 - 4 sin(α - π/6) occurs when sin(α - π/6) is minimum, which is -1/2, giving 5 - 4*(-1/2) = 5 + 2 = 7The minimum value occurs when sin(α - π/6) is maximum, which is 1, giving 5 - 4*1 = 1But since we have the absolute value, |5 - 4 sin(α - π/6)|, we need to consider whether the expression inside can be negative.Looking at the expression 5 - 4 sin(α - π/6). The minimum value of sin(α - π/6) is -1/2, so 5 - 4*(-1/2) = 5 + 2 = 7, which is positive. The maximum value of sin(α - π/6) is 1, so 5 - 4*1 = 1, which is still positive. Therefore, the expression inside the absolute value is always positive, so |5 - 4 sin(α - π/6)| = 5 - 4 sin(α - π/6)Therefore, the distance d is (5 - 4 sin(α - π/6))/2So, d ranges from (5 - 4*1)/2 = (1)/2 = 1/2 to (5 - 4*(-1/2))/2 = (5 + 2)/2 = 7/2Therefore, the range of distances is from 1/2 to 7/2.Hence, the range of distances from a moving point M on curve C to line l is [1/2, 7/2]Wait, let me double-check my steps to make sure I didn't make a mistake.1. Converted the polar equation ρ = 4 cos θ to Cartesian, got (x - 2)^2 + y^2 = 4, correct.2. Parametrized the circle as x = 2 + 2 cos α, y = 2 sin α, with α from 0 to π, correct.3. Expressed the distance from (x, y) to line l as |x - √3 y + 3| / 2, correct.4. Substituted x and y from parametric equations, got |5 + 2 cos α - 2√3 sin α| / 2, correct.5. Expressed 2 cos α - 2√3 sin α as -4 sin(α - π/6), correct.6. Therefore, the expression inside the absolute value became 5 - 4 sin(α - π/6), correct.7. Determined the range of sin(α - π/6) as α varies from 0 to π, which is from -1/2 to 1, correct.8. Therefore, 5 - 4 sin(α - π/6) ranges from 1 to 7, correct.9. Divided by 2, got d ranging from 1/2 to 7/2, correct.So, yes, the range is [1/2, 7/2]Final Answer(1) The Cartesian equation of line l is boxed{x - sqrt{3}y + 3 = 0} and the parametric equation of curve C is begin{cases} x = 2 + 2cosalpha y = 2sinalpha end{cases} where (0 leq alpha leq pi).(2) The range of distances from a moving point M on curve C to line l is boxed{left[ frac{1}{2}, frac{7}{2} right]}.</think>

🧠:<think>To determine the ordered triple ((a, b, c)) for the function (h(x) = a f(bx) + c), we need to analyze the transformations applied to the graph of (y = f(x)).1. Horizontal Stretch by a Factor of 3: This transformation is achieved by replacing (x) with (frac{x}{3}) in the function (f(x)). Thus, (f(x)) becomes (fleft(frac{x}{3}right)).2. Shift Right by 2 Units: To shift the graph right by 2 units, we replace (x) with (x - 2) in the horizontally stretched function. This changes (fleft(frac{x}{3}right)) to (fleft(frac{x - 2}{3}right)).3. Shift Up by 5 Units: Finally, to shift the graph up by 5 units, we add 5 to the result. Therefore, the function becomes (fleft(frac{x - 2}{3}right) + 5).From these transformations, we identify the constants:- (a = 1) (no vertical scaling),- (b = frac{1}{3}) (horizontal stretch by a factor of 3),- (c = 5) (vertical shift upwards by 5 units).Thus, the ordered triple is ((1, frac{1}{3}, 5)).</think>

🧠:<think>Alright, so I have this problem here: I need to find the sum of all the zeros of the function ( y = 2(5 - x) sin(pi x) - 1 ) in the interval ( 0 leq x leq 10 ). Hmm, okay, let's break this down step by step.First, I know that finding zeros of a function means finding the values of ( x ) where ( y = 0 ). So, I need to solve the equation:[ 2(5 - x) sin(pi x) - 1 = 0 ]Let me rearrange this equation to make it a bit simpler. If I move the 1 to the other side:[ 2(5 - x) sin(pi x) = 1 ]And then divide both sides by 2:[ (5 - x) sin(pi x) = frac{1}{2} ]Okay, so now I have ( (5 - x) sin(pi x) = frac{1}{2} ). This seems a bit complicated because it's a product of two functions: ( (5 - x) ) and ( sin(pi x) ). I wonder if I can analyze this graphically or numerically.Let me think about the behavior of each part. The term ( (5 - x) ) is a linear function that decreases from 5 to -5 as ( x ) goes from 0 to 10. The term ( sin(pi x) ) is a sine wave with a period of 2, oscillating between -1 and 1. So, when I multiply these two, the amplitude of the sine wave will be scaled by ( (5 - x) ), which is decreasing.At ( x = 0 ), ( (5 - 0) sin(0) = 0 ), so the function starts at 0. At ( x = 1 ), ( (5 - 1) sin(pi) = 0 ). Similarly, at ( x = 2 ), ( (5 - 2) sin(2pi) = 0 ), and so on, up to ( x = 10 ), where ( (5 - 10) sin(10pi) = 0 ). So, the function crosses zero at every integer value of ( x ) from 0 to 10.But wait, the equation we have is ( (5 - x) sin(pi x) = frac{1}{2} ). So, we're looking for points where the product of ( (5 - x) ) and ( sin(pi x) ) equals ( frac{1}{2} ). This might happen between the integer zeros.Let me consider the behavior between each pair of integers. For example, between ( x = 0 ) and ( x = 1 ), ( (5 - x) ) is positive and decreasing, and ( sin(pi x) ) goes from 0 up to 1 and back to 0. So, the product ( (5 - x) sin(pi x) ) will start at 0, rise to a maximum, and then fall back to 0. Since the maximum value of ( sin(pi x) ) is 1, the maximum of the product is ( (5 - x) ), which is 5 at ( x = 0 ) and decreases to 4 at ( x = 1 ). So, the product can reach up to 5, which is much larger than ( frac{1}{2} ). Therefore, there should be two points in this interval where the product equals ( frac{1}{2} ).Similarly, between ( x = 1 ) and ( x = 2 ), ( (5 - x) ) is still positive, decreasing from 4 to 3, and ( sin(pi x) ) goes from 0 down to -1 and back to 0. So, the product will go from 0 down to a minimum and back to 0. The minimum value will be negative, so the product will cross ( frac{1}{2} ) twice: once on the way down and once on the way up. Wait, but ( frac{1}{2} ) is positive, so actually, when the product is negative, it won't cross ( frac{1}{2} ). Hmm, maybe only once? Or maybe not at all?Wait, let me think again. Between ( x = 1 ) and ( x = 2 ), ( sin(pi x) ) is negative, so the product ( (5 - x) sin(pi x) ) is negative. Since ( frac{1}{2} ) is positive, the product won't equal ( frac{1}{2} ) in this interval. So, actually, there are no solutions between ( x = 1 ) and ( x = 2 ).Similarly, between ( x = 2 ) and ( x = 3 ), ( sin(pi x) ) is positive again, increasing from 0 to 1 and back to 0. So, the product ( (5 - x) sin(pi x) ) will go from 0 up to a maximum and back to 0. The maximum value here is ( (5 - x) ) at ( x = 2.5 ), which is 2.5. So, the product can reach up to 2.5, which is greater than ( frac{1}{2} ). Therefore, there should be two points in this interval where the product equals ( frac{1}{2} ).Wait, but ( (5 - x) ) is decreasing from 3 to 2 as ( x ) goes from 2 to 3. So, the maximum of the product is actually ( (5 - x) ) times 1, which is decreasing. So, the maximum value is 3 at ( x = 2 ) and 2 at ( x = 3 ). So, the product will go from 0 up to 3 and back to 0. So, it will cross ( frac{1}{2} ) twice in this interval.Similarly, between ( x = 3 ) and ( x = 4 ), ( sin(pi x) ) is negative, so the product will be negative, and since ( frac{1}{2} ) is positive, there are no solutions here.Continuing this pattern, I can see that in each interval ( [2k, 2k+1] ) where ( k ) is an integer, the product ( (5 - x) sin(pi x) ) will be positive, and in each interval ( [2k+1, 2k+2] ), it will be negative. Therefore, solutions will only occur in the intervals where the product is positive, i.e., ( [0,1] ), ( [2,3] ), ( [4,5] ), ( [6,7] ), ( [8,9] ), and ( [10,11] ). But since our interval is up to 10, we'll consider up to ( [8,9] ) and ( [9,10] ).Wait, actually, at ( x = 10 ), ( (5 - 10) sin(10pi) = (-5)(0) = 0 ). So, the last interval is ( [9,10] ). Let me check the behavior there.Between ( x = 9 ) and ( x = 10 ), ( (5 - x) ) is negative, decreasing from -4 to -5, and ( sin(pi x) ) goes from 0 up to 1 and back to 0. So, the product ( (5 - x) sin(pi x) ) will be negative, going from 0 down to a minimum and back to 0. Since ( frac{1}{2} ) is positive, there are no solutions in this interval.So, in total, the intervals where solutions can occur are ( [0,1] ), ( [2,3] ), ( [4,5] ), ( [6,7] ), ( [8,9] ), and ( [10,11] ). But since we're only considering up to ( x = 10 ), the last interval is ( [8,9] ).Wait, but ( [10,11] ) is outside our interval, so we can ignore that. So, we have five intervals where solutions can occur: ( [0,1] ), ( [2,3] ), ( [4,5] ), ( [6,7] ), ( [8,9] ).In each of these intervals, we expect two solutions because the product ( (5 - x) sin(pi x) ) starts at 0, reaches a maximum, and then returns to 0, crossing ( frac{1}{2} ) twice. So, that would be 2 solutions per interval, times 5 intervals, giving 10 solutions. But wait, let me check.Wait, actually, between ( x = 0 ) and ( x = 1 ), the product goes from 0 up to 5 and back to 0. So, it will cross ( frac{1}{2} ) twice. Similarly, between ( x = 2 ) and ( x = 3 ), the product goes from 0 up to 3 and back to 0, crossing ( frac{1}{2} ) twice. Between ( x = 4 ) and ( x = 5 ), the product goes from 0 up to 1 and back to 0, so it will cross ( frac{1}{2} ) twice. Between ( x = 6 ) and ( x = 7 ), the product goes from 0 up to -1 and back to 0, but wait, that's negative. Wait, no, ( (5 - x) ) is negative here, so ( (5 - x) sin(pi x) ) is negative times positive, which is negative. So, the product is negative, so it won't cross ( frac{1}{2} ). Wait, that contradicts my earlier thought.Wait, let me clarify. Between ( x = 4 ) and ( x = 5 ), ( (5 - x) ) is positive, decreasing from 1 to 0, and ( sin(pi x) ) is positive, increasing from 0 to 1 and back to 0. So, the product is positive, going from 0 up to 1 and back to 0. So, it will cross ( frac{1}{2} ) twice.Between ( x = 5 ) and ( x = 6 ), ( (5 - x) ) is negative, and ( sin(pi x) ) is positive, so the product is negative. Therefore, no solutions here.Between ( x = 6 ) and ( x = 7 ), ( (5 - x) ) is negative, and ( sin(pi x) ) is negative, so the product is positive. So, the product goes from 0 up to a maximum and back to 0. The maximum value is ( (5 - x) ) times ( sin(pi x) ). At ( x = 6.5 ), ( (5 - 6.5) = -1.5 ), and ( sin(6.5pi) = sin(pi/2) = 1 ). So, the product is -1.5 * 1 = -1.5, but wait, that's negative. Wait, no, ( sin(6.5pi) = sin(pi/2) = 1 ), but ( (5 - x) ) is negative, so the product is negative. Wait, but in the interval ( [6,7] ), ( sin(pi x) ) goes from 0 to -1 and back to 0. So, the product ( (5 - x) sin(pi x) ) is negative times negative, which is positive. So, the product goes from 0 up to a maximum and back to 0. The maximum occurs at ( x = 6.5 ), where ( (5 - 6.5) = -1.5 ) and ( sin(6.5pi) = 1 ), so the product is -1.5 * 1 = -1.5, but wait, that's negative. Hmm, I'm confused.Wait, no, ( sin(6.5pi) = sin(pi/2 + 6pi) = sin(pi/2) = 1 ). But ( (5 - x) ) at ( x = 6.5 ) is ( -1.5 ). So, the product is ( -1.5 * 1 = -1.5 ). But since ( sin(pi x) ) is negative in ( [6,7] ), because ( x ) is between 6 and 7, ( pi x ) is between 6π and 7π, which is between 180π and 210π, so ( sin(pi x) ) is negative in this interval. Wait, no, ( sin(pi x) ) at ( x = 6.5 ) is ( sin(6.5pi) = sin(pi/2) = 1 ), but that's positive. Wait, no, ( 6.5pi ) is equivalent to ( pi/2 ) plus 6π, which is the same as ( pi/2 ), so it's positive. But ( x = 6.5 ) is in the interval ( [6,7] ), where ( sin(pi x) ) is positive? Wait, no, ( sin(pi x) ) is positive in ( [6,7] ) because ( pi x ) is between 6π and 7π, which is equivalent to 0 to π, so ( sin(pi x) ) is positive in ( [6,7] ). Wait, but ( x = 6.5 ) is in the middle of ( [6,7] ), so ( sin(6.5pi) = sin(pi/2) = 1 ), which is positive. So, the product ( (5 - x) sin(pi x) ) is negative times positive, which is negative. So, the product is negative in ( [6,7] ), meaning it goes from 0 down to a minimum and back to 0. Therefore, it won't cross ( frac{1}{2} ), which is positive. So, no solutions in ( [6,7] ).Similarly, in ( [7,8] ), ( (5 - x) ) is negative, and ( sin(pi x) ) is negative, so the product is positive. So, the product goes from 0 up to a maximum and back to 0. The maximum occurs at ( x = 7.5 ), where ( (5 - 7.5) = -2.5 ) and ( sin(7.5pi) = sin(3pi/2) = -1 ). So, the product is ( -2.5 * -1 = 2.5 ). So, the product goes from 0 up to 2.5 and back to 0. Therefore, it will cross ( frac{1}{2} ) twice in this interval.Wait, but ( [7,8] ) is outside our initial consideration because we were only looking at intervals where the product is positive. But actually, ( [7,8] ) is within our interval ( [0,10] ), so we need to consider it.Wait, let me clarify. The intervals where the product is positive are ( [0,1] ), ( [2,3] ), ( [4,5] ), ( [6,7] ), ( [8,9] ), and ( [10,11] ). But since we're only considering up to ( x = 10 ), the last interval is ( [8,9] ).Wait, but in ( [6,7] ), the product is negative, so no solutions. In ( [7,8] ), the product is positive, so solutions exist. So, I need to adjust my earlier thought.So, actually, the intervals where the product is positive are ( [0,1] ), ( [2,3] ), ( [4,5] ), ( [7,8] ), ( [9,10] ). Wait, but in ( [9,10] ), ( (5 - x) ) is negative, and ( sin(pi x) ) is positive, so the product is negative. So, no solutions there.Wait, this is getting confusing. Maybe a better approach is to consider the function ( f(x) = 2(5 - x) sin(pi x) - 1 ) and analyze its behavior.Let me plot the function mentally. At ( x = 0 ), ( f(0) = 2*5*0 - 1 = -1 ). At ( x = 1 ), ( f(1) = 2*4*0 - 1 = -1 ). At ( x = 2 ), ( f(2) = 2*3*0 - 1 = -1 ). Similarly, at ( x = 3 ), ( f(3) = 2*2*0 - 1 = -1 ), and so on up to ( x = 10 ), where ( f(10) = 2*(-5)*0 - 1 = -1 ).But between these integer points, the function oscillates. For example, between ( x = 0 ) and ( x = 1 ), ( sin(pi x) ) goes from 0 to 1 and back to 0, so ( f(x) ) goes from -1 up to a maximum and back to -1. The maximum occurs at ( x = 0.5 ), where ( sin(pi x) = 1 ), so ( f(0.5) = 2*(5 - 0.5)*1 - 1 = 2*4.5 - 1 = 9 - 1 = 8 ). So, the function reaches 8 at ( x = 0.5 ), which is way above zero. Therefore, it must cross zero twice in this interval.Similarly, between ( x = 1 ) and ( x = 2 ), ( sin(pi x) ) is negative, so ( f(x) ) goes from -1 down to a minimum and back to -1. The minimum occurs at ( x = 1.5 ), where ( sin(pi x) = -1 ), so ( f(1.5) = 2*(5 - 1.5)*(-1) - 1 = 2*3.5*(-1) - 1 = -7 - 1 = -8 ). So, the function reaches -8 at ( x = 1.5 ), which is below zero. Therefore, it crosses zero twice in this interval as well.Wait, but earlier I thought that in intervals where the product is negative, there are no solutions. But here, the function ( f(x) ) crosses zero twice in both positive and negative intervals. So, maybe my earlier reasoning was flawed.Wait, let me clarify. The equation ( f(x) = 0 ) is ( 2(5 - x) sin(pi x) - 1 = 0 ), which is equivalent to ( 2(5 - x) sin(pi x) = 1 ). So, the product ( (5 - x) sin(pi x) ) must equal ( frac{1}{2} ). Therefore, regardless of the sign, as long as the product equals ( frac{1}{2} ), it's a solution.But in the interval ( [1,2] ), ( (5 - x) ) is positive, and ( sin(pi x) ) is negative, so the product is negative. Therefore, ( (5 - x) sin(pi x) = frac{1}{2} ) would require a negative product to equal a positive ( frac{1}{2} ), which is impossible. Therefore, there are no solutions in ( [1,2] ).Wait, but earlier I thought that ( f(x) ) crosses zero twice in ( [1,2] ). But if ( f(x) = 2(5 - x) sin(pi x) - 1 ), and in ( [1,2] ), ( (5 - x) sin(pi x) ) is negative, so ( f(x) ) is negative minus 1, which is more negative. So, actually, ( f(x) ) goes from -1 down to -8 and back to -1, so it doesn't cross zero at all in ( [1,2] ).Wait, that contradicts my earlier thought. Let me recast this.At ( x = 0.5 ), ( f(x) = 8 ), which is positive. At ( x = 1 ), ( f(x) = -1 ). So, between ( x = 0.5 ) and ( x = 1 ), the function goes from 8 to -1, crossing zero once. Similarly, between ( x = 1 ) and ( x = 1.5 ), the function goes from -1 to -8, so it doesn't cross zero. Then, between ( x = 1.5 ) and ( x = 2 ), it goes from -8 back to -1, so it doesn't cross zero again. Therefore, in ( [0,1] ), there is one zero crossing, and in ( [1,2] ), none.Wait, but earlier I thought there were two crossings in ( [0,1] ). Maybe I need to think more carefully.Let me consider the function ( f(x) = 2(5 - x) sin(pi x) - 1 ).At ( x = 0 ), ( f(0) = -1 ).At ( x = 0.5 ), ( f(0.5) = 8 ).At ( x = 1 ), ( f(1) = -1 ).So, the function goes from -1 at 0, up to 8 at 0.5, and back to -1 at 1. Therefore, it must cross zero twice in this interval: once between 0 and 0.5, and once between 0.5 and 1.Similarly, in ( [1,2] ), the function goes from -1 at 1, down to -8 at 1.5, and back to -1 at 2. So, it doesn't cross zero in this interval.In ( [2,3] ), the function goes from -1 at 2, up to some maximum, and back to -1 at 3. Let's check the maximum at ( x = 2.5 ):( f(2.5) = 2*(5 - 2.5)*sin(2.5pi) - 1 = 2*2.5*(-1) - 1 = -5 - 1 = -6 ). Wait, that's negative. So, the function goes from -1 at 2, down to -6 at 2.5, and back to -1 at 3. So, it doesn't cross zero in this interval.Wait, but that contradicts my earlier thought. Let me recast.Wait, ( sin(2.5pi) = sin(pi/2 + 2pi) = sin(pi/2) = 1 ). Wait, no, ( 2.5pi ) is equivalent to ( pi/2 ) plus ( 2pi ), which is ( pi/2 ), so ( sin(2.5pi) = 1 ). But ( (5 - 2.5) = 2.5 ), so ( f(2.5) = 2*2.5*1 - 1 = 5 - 1 = 4 ). Wait, that's positive. So, the function goes from -1 at 2, up to 4 at 2.5, and back to -1 at 3. Therefore, it crosses zero twice in this interval: once between 2 and 2.5, and once between 2.5 and 3.Similarly, in ( [3,4] ), the function goes from -1 at 3, down to some minimum, and back to -1 at 4. Let's check the minimum at ( x = 3.5 ):( f(3.5) = 2*(5 - 3.5)*sin(3.5pi) - 1 = 2*1.5*(-1) - 1 = -3 - 1 = -4 ). So, the function goes from -1 at 3, down to -4 at 3.5, and back to -1 at 4. Therefore, it doesn't cross zero in this interval.In ( [4,5] ), the function goes from -1 at 4, up to some maximum, and back to -1 at 5. Let's check the maximum at ( x = 4.5 ):( f(4.5) = 2*(5 - 4.5)*sin(4.5pi) - 1 = 2*0.5*1 - 1 = 1 - 1 = 0 ). So, the function reaches zero at ( x = 4.5 ). Therefore, it crosses zero once in this interval.Wait, but let's see: at ( x = 4 ), ( f(4) = -1 ); at ( x = 4.5 ), ( f(4.5) = 0 ); and at ( x = 5 ), ( f(5) = -1 ). So, it goes from -1 to 0 and back to -1, crossing zero once.Similarly, in ( [5,6] ), the function goes from -1 at 5, down to some minimum, and back to -1 at 6. Let's check the minimum at ( x = 5.5 ):( f(5.5) = 2*(5 - 5.5)*sin(5.5pi) - 1 = 2*(-0.5)*(-1) - 1 = 1 - 1 = 0 ). So, the function reaches zero at ( x = 5.5 ). Therefore, it crosses zero once in this interval.Wait, but ( (5 - 5.5) = -0.5 ), and ( sin(5.5pi) = sin(pi/2 + 5pi) = sin(pi/2) = 1 ). So, ( f(5.5) = 2*(-0.5)*1 - 1 = -1 - 1 = -2 ). Wait, that contradicts my earlier calculation. Let me recast.Wait, ( sin(5.5pi) = sin(pi/2 + 5pi) = sin(pi/2) = 1 ). But ( (5 - 5.5) = -0.5 ), so ( f(5.5) = 2*(-0.5)*1 - 1 = -1 - 1 = -2 ). So, the function goes from -1 at 5, down to -2 at 5.5, and back to -1 at 6. Therefore, it doesn't cross zero in this interval.Wait, but earlier I thought it reached zero at 5.5. That was a mistake. So, in ( [5,6] ), the function doesn't cross zero.In ( [6,7] ), the function goes from -1 at 6, up to some maximum, and back to -1 at 7. Let's check the maximum at ( x = 6.5 ):( f(6.5) = 2*(5 - 6.5)*sin(6.5pi) - 1 = 2*(-1.5)*1 - 1 = -3 - 1 = -4 ). Wait, that's negative. So, the function goes from -1 at 6, down to -4 at 6.5, and back to -1 at 7. Therefore, it doesn't cross zero in this interval.In ( [7,8] ), the function goes from -1 at 7, up to some maximum, and back to -1 at 8. Let's check the maximum at ( x = 7.5 ):( f(7.5) = 2*(5 - 7.5)*sin(7.5pi) - 1 = 2*(-2.5)*(-1) - 1 = 5 - 1 = 4 ). So, the function goes from -1 at 7, up to 4 at 7.5, and back to -1 at 8. Therefore, it crosses zero twice in this interval: once between 7 and 7.5, and once between 7.5 and 8.In ( [8,9] ), the function goes from -1 at 8, down to some minimum, and back to -1 at 9. Let's check the minimum at ( x = 8.5 ):( f(8.5) = 2*(5 - 8.5)*sin(8.5pi) - 1 = 2*(-3.5)*(-1) - 1 = 7 - 1 = 6 ). Wait, that's positive. So, the function goes from -1 at 8, up to 6 at 8.5, and back to -1 at 9. Therefore, it crosses zero twice in this interval: once between 8 and 8.5, and once between 8.5 and 9.In ( [9,10] ), the function goes from -1 at 9, down to some minimum, and back to -1 at 10. Let's check the minimum at ( x = 9.5 ):( f(9.5) = 2*(5 - 9.5)*sin(9.5pi) - 1 = 2*(-4.5)*(-1) - 1 = 9 - 1 = 8 ). Wait, that's positive. So, the function goes from -1 at 9, up to 8 at 9.5, and back to -1 at 10. Therefore, it crosses zero twice in this interval: once between 9 and 9.5, and once between 9.5 and 10.Wait, but earlier I thought that in ( [9,10] ), the product ( (5 - x) sin(pi x) ) is negative, so the function ( f(x) ) would be negative. But according to this calculation, ( f(9.5) = 8 ), which is positive. So, there must be a mistake in my earlier reasoning.Wait, let's recast ( sin(9.5pi) ). ( 9.5pi = 10pi - 0.5pi = 0 - 0.5pi = -pi/2 ), but sine is periodic with period ( 2pi ), so ( sin(9.5pi) = sin(-pi/2) = -1 ). Wait, no, ( 9.5pi = 9pi + 0.5pi = (4*2pi + pi) + 0.5pi = pi + 0.5pi = 1.5pi ), so ( sin(1.5pi) = -1 ). Therefore, ( f(9.5) = 2*(5 - 9.5)*(-1) - 1 = 2*(-4.5)*(-1) - 1 = 9 - 1 = 8 ). So, it's correct. Therefore, the function goes from -1 at 9, up to 8 at 9.5, and back to -1 at 10. So, it crosses zero twice in this interval.Wait, but this contradicts the earlier thought that in ( [9,10] ), the product ( (5 - x) sin(pi x) ) is negative. Let me check:At ( x = 9.5 ), ( (5 - x) = -4.5 ), and ( sin(pi x) = sin(9.5pi) = -1 ). So, the product ( (5 - x) sin(pi x) = (-4.5)*(-1) = 4.5 ), which is positive. Therefore, ( f(x) = 2*4.5 - 1 = 9 - 1 = 8 ). So, the function is positive at 9.5, meaning it crosses zero twice in this interval.Therefore, summarizing:- ( [0,1] ): 2 zeros- ( [1,2] ): 0 zeros- ( [2,3] ): 2 zeros- ( [3,4] ): 0 zeros- ( [4,5] ): 1 zero- ( [5,6] ): 0 zeros- ( [6,7] ): 0 zeros- ( [7,8] ): 2 zeros- ( [8,9] ): 2 zeros- ( [9,10] ): 2 zerosWait, that adds up to 2 + 0 + 2 + 0 + 1 + 0 + 0 + 2 + 2 + 2 = 11 zeros. But earlier, I thought there were 12 zeros. Hmm, maybe I missed one.Wait, in ( [4,5] ), the function reaches zero at ( x = 4.5 ), which is one zero. In ( [5,6] ), the function reaches zero at ( x = 5.5 ), but earlier I thought it didn't. Wait, let me recast.At ( x = 5.5 ), ( f(5.5) = 2*(5 - 5.5)*sin(5.5pi) - 1 = 2*(-0.5)*(-1) - 1 = 1 - 1 = 0 ). So, it does cross zero at ( x = 5.5 ). Therefore, in ( [5,6] ), there is one zero.So, correcting the earlier count:- ( [0,1] ): 2- ( [1,2] ): 0- ( [2,3] ): 2- ( [3,4] ): 0- ( [4,5] ): 1- ( [5,6] ): 1- ( [6,7] ): 0- ( [7,8] ): 2- ( [8,9] ): 2- ( [9,10] ): 2Total zeros: 2 + 0 + 2 + 0 + 1 + 1 + 0 + 2 + 2 + 2 = 12 zeros.Okay, so there are 12 zeros in total in the interval ( [0,10] ).Now, the question is to find the sum of all these zeros. That is, if the zeros are ( x_1, x_2, ldots, x_{12} ), then we need to find ( x_1 + x_2 + ldots + x_{12} ).This seems challenging because finding the exact values of these zeros would require solving the equation ( 2(5 - x) sin(pi x) = 1 ), which is transcendental and likely doesn't have a closed-form solution. Therefore, we need another approach.I recall that for functions with symmetry, the sum of zeros can sometimes be found using properties of the function. Let me check if this function has any symmetry.The function is ( f(x) = 2(5 - x) sin(pi x) - 1 ). Let's consider the transformation ( x to 10 - x ). Let's see what happens to ( f(x) ):( f(10 - x) = 2(5 - (10 - x)) sin(pi (10 - x)) - 1 = 2(x - 5) sin(10pi - pi x) - 1 ).Now, ( sin(10pi - pi x) = sin(pi x) ) because ( sin(2pi n - theta) = sin(theta) ) for integer ( n ). Wait, no, ( sin(10pi - pi x) = sin(pi x) ) because ( sin(2pi n - theta) = sin(theta) ) for integer ( n ). Since 10 is even, ( 10pi = 5*2pi ), so ( sin(10pi - pi x) = sin(pi x) ).Therefore, ( f(10 - x) = 2(x - 5) sin(pi x) - 1 ).Compare this to ( f(x) = 2(5 - x) sin(pi x) - 1 ). Notice that ( f(10 - x) = -2(5 - x) sin(pi x) - 1 = -f(x) - 2 ).Wait, let's compute:( f(10 - x) = 2(x - 5) sin(pi x) - 1 = -2(5 - x) sin(pi x) - 1 = -f(x) - 2 ).Therefore, ( f(10 - x) = -f(x) - 2 ).This is an interesting relationship. Let me see if this can help us find the sum of the zeros.Suppose ( x ) is a zero of ( f(x) ), so ( f(x) = 0 ). Then, substituting into the equation above:( f(10 - x) = -0 - 2 = -2 neq 0 ).So, ( 10 - x ) is not a zero of ( f(x) ). Therefore, the zeros are not symmetric around 5 in the sense that if ( x ) is a zero, then ( 10 - x ) is not necessarily a zero. However, maybe there's another relationship.Alternatively, perhaps the function has some other symmetry or property that can be exploited.Wait, let's consider the function ( g(x) = 2(5 - x) sin(pi x) ). Then, ( f(x) = g(x) - 1 ). The zeros of ( f(x) ) are the points where ( g(x) = 1 ).Now, ( g(x) = 2(5 - x) sin(pi x) ). Let's analyze ( g(x) ).We can note that ( g(x) ) is related to ( g(10 - x) ):( g(10 - x) = 2(5 - (10 - x)) sin(pi (10 - x)) = 2(x - 5) sin(10pi - pi x) = 2(x - 5) sin(pi x) = -2(5 - x) sin(pi x) = -g(x) ).So, ( g(10 - x) = -g(x) ).Therefore, if ( g(x) = 1 ), then ( g(10 - x) = -1 ). So, if ( x ) is a solution to ( g(x) = 1 ), then ( 10 - x ) is a solution to ( g(x) = -1 ).But our equation is ( g(x) = 1 ), so the zeros of ( f(x) ) are the solutions to ( g(x) = 1 ). Therefore, if ( x ) is a solution, then ( 10 - x ) is a solution to ( g(x) = -1 ), which is not our equation. Therefore, ( 10 - x ) is not a solution to ( g(x) = 1 ).However, perhaps we can consider that for each solution ( x ) of ( g(x) = 1 ), there is a corresponding solution ( 10 - x ) of ( g(x) = -1 ). But since we're only interested in solutions where ( g(x) = 1 ), we can't directly pair the zeros in a symmetric way.But wait, perhaps if we consider that the function ( g(x) ) has a certain symmetry, the sum of the zeros can be related to the number of zeros times 5, due to the symmetry around 5.Wait, let's think about it differently. Suppose we have a function that is symmetric around ( x = 5 ), meaning ( g(5 + a) = g(5 - a) ). But in our case, ( g(10 - x) = -g(x) ), which is not the same as symmetry around 5. However, perhaps there's a way to pair the zeros such that their average is 5.Wait, if ( x ) is a zero, then ( 10 - x ) is not a zero, but perhaps there's another zero ( y ) such that ( x + y = 10 ). But I don't see a direct relationship.Alternatively, perhaps the function ( f(x) ) has an odd number of zeros, but in our case, we have 12 zeros, which is even. So, maybe they can be paired such that each pair sums to 10.Wait, let's consider that for each zero ( x ), there is another zero ( 10 - x ). But earlier, we saw that if ( x ) is a zero, then ( 10 - x ) is not a zero. However, perhaps the function ( f(x) ) has a property that for each zero ( x ), there is another zero ( y ) such that ( x + y = 10 ).Wait, let's test this with the zeros we have. For example, in ( [0,1] ), there are two zeros, say ( x_1 ) and ( x_2 ). If the function is symmetric around 5, then there should be corresponding zeros in ( [9,10] ) such that ( x_1 + x_3 = 10 ) and ( x_2 + x_4 = 10 ). Similarly, in ( [2,3] ), the zeros would correspond to zeros in ( [7,8] ), and in ( [4,5] ), the zero would correspond to a zero in ( [5,6] ).Wait, in ( [4,5] ), we have a zero at ( x = 4.5 ), and in ( [5,6] ), we have a zero at ( x = 5.5 ). So, ( 4.5 + 5.5 = 10 ). Similarly, in ( [0,1] ), suppose the zeros are ( x_1 ) and ( x_2 ), then in ( [9,10] ), the zeros would be ( 10 - x_1 ) and ( 10 - x_2 ). Similarly, in ( [2,3] ), the zeros would correspond to zeros in ( [7,8] ).Therefore, if this pairing holds, then each pair of zeros sums to 10. Since there are 12 zeros, which is 6 pairs, the total sum would be ( 6 * 10 = 60 ).But wait, let's verify this with the zeros we have. For example, in ( [4,5] ), the zero is at ( x = 4.5 ), and in ( [5,6] ), the zero is at ( x = 5.5 ). ( 4.5 + 5.5 = 10 ). Similarly, in ( [0,1] ), suppose the zeros are at ( x = a ) and ( x = b ), then in ( [9,10] ), the zeros would be at ( x = 10 - a ) and ( x = 10 - b ). Therefore, ( a + (10 - a) = 10 ) and ( b + (10 - b) = 10 ). Similarly, in ( [2,3] ), zeros at ( x = c ) and ( x = d ), corresponding to zeros at ( x = 10 - c ) and ( x = 10 - d ) in ( [7,8] ).Therefore, each pair of zeros sums to 10, and since there are 6 such pairs, the total sum is 60.Therefore, the sum of all the zeros of the function ( y = 2(5 - x) sin(pi x) - 1 ) in the interval ( 0 leq x leq 10 ) is 60.</think>

🧠:<think>Okay, so I have this problem about two overlapping circular discs, each with a radius of 2 units. Their combined covered area is 6π square units. I need to find the distance between the centers of these circles and express it as a fraction with a denominator of 100, where the numerators are integers that differ by 1. Hmm, okay, let me try to break this down step by step.First, I know that each circle has a radius of 2 units, so the area of one circle is πr², which is π*(2)² = 4π. Since there are two circles, the total area without any overlap would be 2*4π = 8π. But the problem says the combined covered area is 6π, which is less than 8π. That means the overlapping area is 8π - 6π = 2π. So, the overlapping area is 2π.Now, I need to find the distance between the centers of the two circles. Let's denote this distance as d. I remember that the area of overlap between two circles can be calculated using the formula involving the radius and the distance between the centers. The formula for the area of overlap is a bit complicated, but I think it involves the radii and the angle between the points of intersection.Let me recall the formula for the area of overlap between two circles. If two circles of radius r are separated by a distance d, the area of their intersection is given by:2r² cos⁻¹(d/(2r)) - (d/2)√(4r² - d²)In this case, both circles have radius 2, so r = 2. Plugging that into the formula, the area of overlap becomes:2*(2)² cos⁻¹(d/(2*2)) - (d/2)*√(4*(2)² - d²)= 8 cos⁻¹(d/4) - (d/2)*√(16 - d²)We know the area of overlap is 2π, so:8 cos⁻¹(d/4) - (d/2)*√(16 - d²) = 2πHmm, that's a bit complicated. Maybe I can simplify this equation or find another way to express it. Alternatively, I remember that the area of overlap can also be expressed in terms of the angle θ (in radians) between the radii at the points of intersection. The formula for the area of overlap is:2r² (θ - sinθ)/2 = r² (θ - sinθ)So, in this case, it would be:(2)² (θ - sinθ) = 4(θ - sinθ)And this should equal 2π, so:4(θ - sinθ) = 2π=> θ - sinθ = π/2Okay, so now I have the equation θ - sinθ = π/2. I need to solve for θ. This seems like a transcendental equation, which means it can't be solved algebraically and I'll need to use numerical methods or approximation.Let me think about how to approach this. Maybe I can use the Newton-Raphson method to approximate θ. First, I need an initial guess. Let's consider the range of θ. Since the circles overlap, θ must be between 0 and π. Let me try θ = π/2 as an initial guess.Plugging θ = π/2 into the equation:π/2 - sin(π/2) = π/2 - 1 ≈ 1.5708 - 1 = 0.5708But we need θ - sinθ = π/2 ≈ 1.5708. So, 0.5708 is much less than 1.5708. That means θ needs to be larger. Let me try θ = π.Plugging θ = π:π - sinπ = π - 0 ≈ 3.1416That's larger than π/2. So, the solution lies between π/2 and π. Let me try θ = 2.θ = 2 radians is approximately 114.59 degrees.Plugging θ = 2:2 - sin(2) ≈ 2 - 0.9093 ≈ 1.0907Still less than π/2 ≈ 1.5708. Let's try θ = 3.θ = 3 radians is approximately 171.89 degrees.Plugging θ = 3:3 - sin(3) ≈ 3 - 0.1411 ≈ 2.8589That's way larger than π/2. So, the solution is between 2 and 3 radians. Wait, but 2 radians is about 114 degrees, and 3 radians is about 171 degrees. Hmm, but the equation is θ - sinθ = π/2 ≈ 1.5708.Wait, when θ = 2, θ - sinθ ≈ 1.0907, which is less than 1.5708. When θ = 3, it's 2.8589, which is more. So, the solution is between 2 and 3 radians. Let me try θ = 2.5.θ = 2.5 radians is about 143.24 degrees.Plugging θ = 2.5:2.5 - sin(2.5) ≈ 2.5 - 0.5985 ≈ 1.9015Still less than 1.5708? Wait, no, 1.9015 is more than 1.5708. So, the solution is between 2 and 2.5 radians.Wait, hold on, when θ = 2, θ - sinθ ≈ 1.0907, which is less than 1.5708. When θ = 2.5, it's 1.9015, which is more. So, the solution is between 2 and 2.5.Wait, but 1.0907 is less than 1.5708, and 1.9015 is more. So, the solution is between 2 and 2.5. Let me try θ = 2.2.θ = 2.2 radians is about 126 degrees.Plugging θ = 2.2:2.2 - sin(2.2) ≈ 2.2 - 0.8085 ≈ 1.3915Still less than 1.5708. Let's try θ = 2.3.θ = 2.3 radians is about 131.78 degrees.Plugging θ = 2.3:2.3 - sin(2.3) ≈ 2.3 - 0.7457 ≈ 1.5543Closer to 1.5708. Let's try θ = 2.35.θ = 2.35 radians is about 134.7 degrees.Plugging θ = 2.35:2.35 - sin(2.35) ≈ 2.35 - 0.7024 ≈ 1.6476That's more than 1.5708. So, the solution is between 2.3 and 2.35.Let me try θ = 2.32.θ = 2.32 radians is about 133 degrees.Plugging θ = 2.32:2.32 - sin(2.32) ≈ 2.32 - 0.7272 ≈ 1.5928Still more than 1.5708. Let's try θ = 2.31.θ = 2.31 radians is about 132.4 degrees.Plugging θ = 2.31:2.31 - sin(2.31) ≈ 2.31 - 0.7334 ≈ 1.5766Still a bit more. Let's try θ = 2.305.θ = 2.305 radians is about 132 degrees.Plugging θ = 2.305:2.305 - sin(2.305) ≈ 2.305 - 0.7353 ≈ 1.5697That's very close to 1.5708. Let me try θ = 2.306.θ = 2.306 radians.Plugging θ = 2.306:2.306 - sin(2.306) ≈ 2.306 - sin(2.306)I need to calculate sin(2.306). Let me use a calculator.sin(2.306) ≈ sin(2.306) ≈ 0.7356So, 2.306 - 0.7356 ≈ 1.5704That's very close to 1.5708. So, θ ≈ 2.306 radians.So, θ ≈ 2.306 radians.Now, I need to relate this angle back to the distance d between the centers.I remember that in the formula for the area of overlap, θ is the angle at the center of one of the circles, subtended by the chord connecting the points of intersection. So, in the triangle formed by the centers of the circles and one of the points of intersection, we can relate θ and d.In the triangle, the sides are both radii (2 units) and the base is d. The angle opposite the base is θ. So, using the law of cosines:d² = 2² + 2² - 2*2*2*cosθ= 4 + 4 - 8cosθ= 8(1 - cosθ)So, d = sqrt(8(1 - cosθ)) = 2*sqrt(2(1 - cosθ))Alternatively, I can use the half-angle identity:1 - cosθ = 2sin²(θ/2)So, d = 2*sqrt(2*2sin²(θ/2)) = 2*sqrt(4sin²(θ/2)) = 2*2|sin(θ/2)| = 4sin(θ/2)Since θ is between 0 and π, sin(θ/2) is positive, so d = 4sin(θ/2)We found θ ≈ 2.306 radians, so θ/2 ≈ 1.153 radians.Now, let's compute sin(1.153). Using a calculator:sin(1.153) ≈ sin(1.153) ≈ 0.913So, d ≈ 4*0.913 ≈ 3.652Wait, that can't be right because the maximum distance between centers for overlapping circles is less than 4 (since each radius is 2). Wait, no, actually, if the circles are just touching, the distance is 4. If they are overlapping, the distance is less than 4. But 3.652 is less than 4, so that's okay.But wait, let me double-check. If θ ≈ 2.306 radians, then θ/2 ≈ 1.153 radians, and sin(1.153) ≈ 0.913. So, 4*0.913 ≈ 3.652. Hmm, that seems a bit high because if the distance is 3.652, the overlapping area might be smaller. Let me verify.Alternatively, maybe I made a mistake in the calculation. Let me recalculate sin(1.153).Using a calculator, sin(1.153) is approximately sin(1.153) ≈ 0.913. So, 4*0.913 ≈ 3.652. Hmm, okay, maybe that's correct.But let's think about it. If the distance is 3.652, which is quite close to 4, the overlapping area would be small. But in our case, the overlapping area is 2π, which is about 6.283. Wait, no, the overlapping area is 2π, which is about 6.283, but the area of each circle is 4π, so 6.283 is about half the area of one circle. So, maybe the distance is indeed around 3.652.Wait, but let me check with the area formula. If d ≈ 3.652, then the area of overlap should be 2π. Let me plug d back into the area formula to verify.The area of overlap is 4(θ - sinθ), which we set equal to 2π. We found θ ≈ 2.306, so 4(2.306 - sin(2.306)) ≈ 4(2.306 - 0.7356) ≈ 4*(1.5704) ≈ 6.2816, which is approximately 2π (since 2π ≈ 6.2832). So, that checks out.Therefore, the distance d ≈ 3.652 units.But the problem asks for the distance as a fraction with denominator 100, where the numerators are integers differing by 1. So, I need to express 3.652 as a fraction over 100. Let's see:3.652 is approximately 365.2/100. Since we need integer numerators, we can consider 365/100 and 366/100. But 365/100 = 3.65 and 366/100 = 3.66. The actual value is approximately 3.652, so it's between 3.65 and 3.66.Wait, but the problem says "an interval between two fractions, with denominators of 100, integer numerators, and a difference of 1 between their numerators, that includes the distance between the centers of the circles."So, the distance is approximately 3.652, which is between 3.65 and 3.66. Therefore, the interval is (365/100, 366/100), or in other words, the distance is between 365/100 and 366/100.But let me make sure. Is there a more precise way to express this? Maybe I should check if 3.652 is indeed between 3.65 and 3.66. Yes, because 3.65 is 3.650 and 3.66 is 3.660, so 3.652 is in between.Alternatively, if I want to be more precise, I can check if 3.652 is closer to 3.65 or 3.66. 3.652 - 3.65 = 0.002, and 3.66 - 3.652 = 0.008. So, it's closer to 3.65. But the problem just asks for an interval that includes the distance, so 365/100 to 366/100 is sufficient.Wait, but let me think again. The distance d is 4sin(θ/2), and θ ≈ 2.306 radians. So, θ/2 ≈ 1.153 radians, and sin(1.153) ≈ 0.913, so d ≈ 4*0.913 ≈ 3.652. So, that's correct.Alternatively, maybe I can use a better approximation for sin(1.153). Let me use a Taylor series expansion around π/2 (1.5708 radians) to approximate sin(1.153). Wait, 1.153 is less than π/2, so maybe expanding around 0.Wait, no, 1.153 is about 66 degrees, so it's in the first quadrant. The Taylor series for sin(x) around 0 is x - x³/6 + x⁵/120 - ... Let's compute sin(1.153) using this.But 1.153 is a bit large for a good approximation with just a few terms. Alternatively, maybe use a calculator for a more accurate value.Using a calculator, sin(1.153) ≈ sin(1.153) ≈ 0.913. So, that's consistent with my earlier calculation.Therefore, d ≈ 3.652, which is between 3.65 and 3.66. So, the interval is 365/100 to 366/100.Wait, but 365/100 is 3.65, and 366/100 is 3.66. So, the distance is between these two fractions.Alternatively, if I want to express it as an interval, it's (365/100, 366/100). But the problem says "an interval between two fractions," so I think that's acceptable.But let me double-check if there's a more precise way to narrow it down. Maybe I can calculate sin(1.153) more accurately.Using a calculator, sin(1.153) ≈ 0.91305. So, d ≈ 4*0.91305 ≈ 3.6522. So, 3.6522 is approximately 3.6522, which is 365.22/100. So, it's between 365.22/100 and 365.22/100, but since we need integer numerators, 365 and 366 are the closest integers. Therefore, the interval is between 365/100 and 366/100.Wait, but 365.22 is closer to 365 than to 366, but the problem just asks for an interval that includes the distance, not necessarily the tightest interval. So, 365/100 to 366/100 is correct.Alternatively, if I want to be more precise, maybe I can find a better approximation for θ and then for d. Let me try to refine θ.Earlier, I found θ ≈ 2.306 radians, which gave θ - sinθ ≈ 1.5704, which is very close to π/2 ≈ 1.5708. So, θ ≈ 2.306 is accurate enough.Therefore, d ≈ 4sin(θ/2) ≈ 4sin(1.153) ≈ 4*0.913 ≈ 3.652.So, the distance is approximately 3.652, which is between 3.65 and 3.66. Therefore, the interval is 365/100 to 366/100.Wait, but let me think again. The problem says "an interval between two fractions, with denominators of 100, integer numerators, and a difference of 1 between their numerators, that includes the distance between the centers of the circles."So, the two fractions must have denominators of 100, integer numerators, and the numerators must differ by 1. So, the fractions are like n/100 and (n+1)/100, and the distance d must lie between them.Given that d ≈ 3.652, which is 365.2/100, so it's between 365/100 and 366/100.Therefore, the interval is (365/100, 366/100), or in other words, the distance is between 365/100 and 366/100.But let me make sure that this is the correct interval. If d ≈ 3.652, then 365/100 = 3.65 and 366/100 = 3.66. Since 3.652 is between 3.65 and 3.66, the interval is correct.Alternatively, if I want to express it as a closed interval, it's [365/100, 366/100], but since the problem says "an interval between two fractions," it's probably acceptable to write it as 365/100 and 366/100.Wait, but let me think if there's a more precise way to narrow it down. Maybe I can calculate d more accurately.Given θ ≈ 2.306 radians, let's compute sin(θ/2) more accurately.θ/2 ≈ 1.153 radians.Using a calculator, sin(1.153) ≈ 0.91305.So, d ≈ 4*0.91305 ≈ 3.6522.So, d ≈ 3.6522, which is 365.22/100. Therefore, it's between 365/100 and 366/100.Alternatively, if I want to express it as a decimal, it's approximately 3.6522, which is 3.65 when rounded down and 3.66 when rounded up. So, the interval is between 3.65 and 3.66.Therefore, the interval is 365/100 to 366/100.Wait, but let me think again. Is there a way to express this without using a calculator? Maybe using some trigonometric identities or approximations.Alternatively, maybe I can use the small angle approximation, but since θ is about 2.3 radians, which is not that small, the approximation might not be accurate.Alternatively, I can use the Newton-Raphson method to solve θ - sinθ = π/2 more accurately.Let me set f(θ) = θ - sinθ - π/2.We need to find θ such that f(θ) = 0.We can use the Newton-Raphson iteration:θ_{n+1} = θ_n - f(θ_n)/f'(θ_n)Where f'(θ) = 1 - cosθ.Starting with θ_0 = 2.306 (from earlier approximation).Compute f(θ_0) = 2.306 - sin(2.306) - π/2 ≈ 2.306 - 0.7356 - 1.5708 ≈ 2.306 - 2.3064 ≈ -0.0004f'(θ_0) = 1 - cos(2.306) ≈ 1 - (-0.680) ≈ 1.680Wait, cos(2.306) is negative because 2.306 radians is in the second quadrant (π/2 < 2.306 < π). So, cos(2.306) ≈ -0.680.Therefore, f'(θ_0) ≈ 1 - (-0.680) = 1.680.So, θ_{1} = θ_0 - f(θ_0)/f'(θ_0) ≈ 2.306 - (-0.0004)/1.680 ≈ 2.306 + 0.00024 ≈ 2.30624Now, compute f(θ_1):f(2.30624) = 2.30624 - sin(2.30624) - π/2Compute sin(2.30624):Using a calculator, sin(2.30624) ≈ sin(2.306) ≈ 0.7356 (since 2.30624 is very close to 2.306).So, f(θ_1) ≈ 2.30624 - 0.7356 - 1.5708 ≈ 2.30624 - 2.3064 ≈ -0.00016f'(θ_1) = 1 - cos(2.30624) ≈ 1 - (-0.680) = 1.680So, θ_{2} = θ_1 - f(θ_1)/f'(θ_1) ≈ 2.30624 - (-0.00016)/1.680 ≈ 2.30624 + 0.000095 ≈ 2.306335Compute f(θ_2):f(2.306335) = 2.306335 - sin(2.306335) - π/2Again, sin(2.306335) ≈ 0.7356So, f(θ_2) ≈ 2.306335 - 0.7356 - 1.5708 ≈ 2.306335 - 2.3064 ≈ -0.000065f'(θ_2) = 1 - cos(2.306335) ≈ 1.680θ_{3} = θ_2 - f(θ_2)/f'(θ_2) ≈ 2.306335 - (-0.000065)/1.680 ≈ 2.306335 + 0.000039 ≈ 2.306374Compute f(θ_3):f(2.306374) ≈ 2.306374 - 0.7356 - 1.5708 ≈ 2.306374 - 2.3064 ≈ -0.000026f'(θ_3) ≈ 1.680θ_{4} = θ_3 - f(θ_3)/f'(θ_3) ≈ 2.306374 - (-0.000026)/1.680 ≈ 2.306374 + 0.000015 ≈ 2.306389Compute f(θ_4):f(2.306389) ≈ 2.306389 - 0.7356 - 1.5708 ≈ 2.306389 - 2.3064 ≈ -0.000011f'(θ_4) ≈ 1.680θ_{5} = θ_4 - f(θ_4)/f'(θ_4) ≈ 2.306389 - (-0.000011)/1.680 ≈ 2.306389 + 0.0000065 ≈ 2.3063955Compute f(θ_5):f(2.3063955) ≈ 2.3063955 - 0.7356 - 1.5708 ≈ 2.3063955 - 2.3064 ≈ -0.0000045f'(θ_5) ≈ 1.680θ_{6} = θ_5 - f(θ_5)/f'(θ_5) ≈ 2.3063955 - (-0.0000045)/1.680 ≈ 2.3063955 + 0.0000027 ≈ 2.3063982Compute f(θ_6):f(2.3063982) ≈ 2.3063982 - 0.7356 - 1.5708 ≈ 2.3063982 - 2.3064 ≈ -0.0000018f'(θ_6) ≈ 1.680θ_{7} = θ_6 - f(θ_6)/f'(θ_6) ≈ 2.3063982 - (-0.0000018)/1.680 ≈ 2.3063982 + 0.0000011 ≈ 2.3063993Compute f(θ_7):f(2.3063993) ≈ 2.3063993 - 0.7356 - 1.5708 ≈ 2.3063993 - 2.3064 ≈ -0.0000007f'(θ_7) ≈ 1.680θ_{8} = θ_7 - f(θ_7)/f'(θ_7) ≈ 2.3063993 - (-0.0000007)/1.680 ≈ 2.3063993 + 0.0000004 ≈ 2.3063997Compute f(θ_8):f(2.3063997) ≈ 2.3063997 - 0.7356 - 1.5708 ≈ 2.3063997 - 2.3064 ≈ -0.0000003f'(θ_8) ≈ 1.680θ_{9} = θ_8 - f(θ_8)/f'(θ_8) ≈ 2.3063997 - (-0.0000003)/1.680 ≈ 2.3063997 + 0.00000018 ≈ 2.3063999Compute f(θ_9):f(2.3063999) ≈ 2.3063999 - 0.7356 - 1.5708 ≈ 2.3063999 - 2.3064 ≈ -0.0000001f'(θ_9) ≈ 1.680θ_{10} = θ_9 - f(θ_9)/f'(θ_9) ≈ 2.3063999 - (-0.0000001)/1.680 ≈ 2.3063999 + 0.00000006 ≈ 2.3064000Now, f(θ_{10}) = 2.3064 - sin(2.3064) - π/2Compute sin(2.3064):Using a calculator, sin(2.3064) ≈ 0.7356 (since 2.3064 is very close to 2.306).So, f(θ_{10}) ≈ 2.3064 - 0.7356 - 1.5708 ≈ 2.3064 - 2.3064 ≈ 0Therefore, θ ≈ 2.3064 radians.So, θ ≈ 2.3064 radians.Now, θ/2 ≈ 1.1532 radians.Compute sin(1.1532):Using a calculator, sin(1.1532) ≈ 0.91305.So, d = 4*sin(θ/2) ≈ 4*0.91305 ≈ 3.6522.So, d ≈ 3.6522, which is 365.22/100.Therefore, the distance is approximately 3.6522, which is between 3.65 and 3.66.So, the interval is between 365/100 and 366/100.Therefore, the distance between the centers of the circles is between 365/100 and 366/100.So, the final answer is that the distance is between 365/100 and 366/100.</think>