answer：First, I recognize that the average of 10 numbers is calculated by dividing the total sum of the numbers by 10.Given that the correct average is 51, I can determine the correct total sum of the numbers by multiplying 51 by 10, which equals 510.It's discovered that one of the numbers, 75, was incorrectly read as 25. This means the incorrect total sum used in the initial calculation was 510 minus 75 plus 25, resulting in 460.Finally, to find the incorrect average, I divide the incorrect total sum of 460 by 10, which gives an incorrect average of 46.

answer：First, I identify the number of tomatoes each plant bore based on the given information.The first plant bore 12 tomatoes.The second plant bore 6 tomatoes less than twice the number of tomatoes that the first plant bore. So, I calculate this as (2 * 12) - 6 = 18 tomatoes.Next, I find the combined total of the first two plants: 12 + 18 = 30 tomatoes.The third plant bore half the number of tomatoes that the first two plants bore combined. Therefore, it bore 30 / 2 = 15 tomatoes.The fourth and fifth plants together bore three times the number of tomatoes as the first two plants combined. This means they bore 3 * 30 = 90 tomatoes each. Since there are two such plants, the total for both is 90 * 2 = 180 tomatoes.The sixth and seventh plants each bore 1.5 times the number of tomatoes as the first two plants had combined. Thus, each of these plants bore 1.5 * 30 = 45 tomatoes. With two such plants, the total is 45 * 2 = 90 tomatoes.Finally, I add up all the tomatoes from all seven plants to find the total number of tomatoes Mabel has.

answer：Okay, so I have this geometry problem here about triangle ABC. Let me try to visualize it first. We know that angle C is 120 degrees, which is pretty big, more than a right angle. The area of the triangle is given as 36 square units. Now, there are a few points mentioned: M is the midpoint of AB, D is a point on BC such that BD:DC is 3:2, and E is the midpoint of AC. The question is asking for the area of triangle BME. Hmm, okay. First, let me recall some properties of triangles and midpoints. Since M is the midpoint of AB and E is the midpoint of AC, that might mean that ME is a midline of the triangle. I remember that the midline theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. So, ME should be parallel to BC and half its length. But wait, we're dealing with triangle BME, not just ME. So, maybe I need to find some relationship between the areas of these smaller triangles. Given that the area of ABC is 36, perhaps I can use some area ratio properties. Since M and E are midpoints, maybe the area of BME is a fraction of the area of ABC. Let me think about the coordinates. Maybe assigning coordinates to the triangle could help. Let me place point C at the origin (0,0) for simplicity. Since angle C is 120 degrees, I can use the Law of Cosines if needed, but maybe coordinates will make it easier. Let me denote point C as (0,0). Let me let point B be somewhere on the x-axis, say at (b,0), and point A somewhere in the plane. Since angle C is 120 degrees, the coordinates of A can be determined if I know the lengths of sides AC and BC. But I don't have the lengths yet. Wait, maybe I can use the area formula. The area of triangle ABC is 36, and it's given by (1/2)*ab*sin(theta), where a and b are the lengths of sides AC and BC, and theta is the angle between them, which is 120 degrees. So, 36 = (1/2)*AC*BC*sin(120). Sin(120) is sqrt(3)/2, so plugging that in: 36 = (1/2)*AC*BC*(sqrt(3)/2). Simplifying, 36 = (sqrt(3)/4)*AC*BC. Therefore, AC*BC = (36*4)/sqrt(3) = 144/sqrt(3) = 48*sqrt(3). Hmm, okay, so the product of AC and BC is 48*sqrt(3). Not sure if that helps directly, but maybe. Since M is the midpoint of AB and E is the midpoint of AC, perhaps I can use vectors or coordinate geometry to find the coordinates of M and E, then compute the area of triangle BME. Let me assign coordinates more specifically. Let me let point C be at (0,0). Let me let point B be at (c,0), and point A at (d,e). Then, the area of triangle ABC can be calculated using the determinant formula: (1/2)*| (c*e - 0*d) | = (1/2)*|c*e| = 36. So, |c*e| = 72. But I also know that angle C is 120 degrees. Using the Law of Cosines, AB² = AC² + BC² - 2*AC*BC*cos(120). Wait, cos(120) is -0.5, so AB² = AC² + BC² + AC*BC. But without knowing the actual lengths, this might not help directly. Maybe I can express AB in terms of AC and BC. Alternatively, maybe using vectors. Let me consider vectors from point C. Let me denote vector CB as (c,0) and vector CA as (d,e). Then, the angle between vectors CB and CA is 120 degrees. The dot product formula says that CB · CA = |CB||CA|cos(theta). So, (c,0) · (d,e) = c*d + 0*e = c*d = |CB||CA|cos(120). |CB| is sqrt(c² + 0²) = c, and |CA| is sqrt(d² + e²). So, c*d = c*sqrt(d² + e²)*(-0.5). Simplifying, d = -0.5*sqrt(d² + e²). Hmm, that's an equation relating d and e. Also, from the area, we have |c*e| = 72, so c*e = ±72. This is getting a bit complicated. Maybe there's a simpler way. Since M is the midpoint of AB and E is the midpoint of AC, perhaps I can use the concept of similar triangles or area ratios. Let me recall that the area of a triangle formed by midpoints is 1/4 of the original triangle. But here, we're dealing with triangle BME, not the medial triangle. Wait, maybe I can decompose the triangle ABC into smaller triangles and find the area ratios. Let me consider the midpoints. Since E is the midpoint of AC, the line BE divides ABC into two smaller triangles of equal area. So, area of ABE is 18, and area of CBE is 18. Similarly, since M is the midpoint of AB, the line CM divides ABC into two triangles of equal area, each 18. But we need the area of BME. Let me see. Point M is the midpoint of AB, and point E is the midpoint of AC. So, the segment ME connects midpoints of AB and AC, so it's a midline and is parallel to BC and half its length. So, triangle BME is similar to triangle BAC, but scaled down by a factor of 1/2. Wait, no, because ME is parallel to BC, but BM is half of AB, so maybe the scaling factor is different. Wait, let me think. Since ME is parallel to BC and half its length, and BM is half of AB, then triangle BME is similar to triangle BAC with a scaling factor of 1/2. Therefore, the area would be (1/2)^2 = 1/4 of the original area. So, 36*(1/4) = 9. But wait, is that correct? Because triangle BME is not the medial triangle, which would be connecting all midpoints. Instead, it's just connecting M and E to B. Alternatively, maybe I can use coordinate geometry. Let me assign coordinates again. Let me place point C at (0,0), point B at (2,0), and point A somewhere. Since angle C is 120 degrees, I can use the Law of Cosines to find coordinates of A. Wait, maybe it's better to assign specific lengths. Let me assume BC = 5 units, so that BD:DC = 3:2, meaning BD = 3 and DC = 2. Then, using the area formula, I can find the lengths of other sides. Wait, the area is 36, so using the formula (1/2)*ab*sin(theta) = 36. If I let BC = 5, then (1/2)*AC*5*sin(120) = 36. So, AC = (36*2)/(5*(sqrt(3)/2)) = (72)/(5*(sqrt(3)/2)) = (72*2)/(5*sqrt(3)) = 144/(5*sqrt(3)) = (144*sqrt(3))/(15) = (48*sqrt(3))/5. Hmm, okay, so AC is (48*sqrt(3))/5. Now, coordinates: Let me place C at (0,0), B at (5,0). Point A is somewhere. Let me find coordinates of A. Since AC = (48*sqrt(3))/5, and angle at C is 120 degrees, I can use polar coordinates. From point C, point A is at a distance of (48*sqrt(3))/5 at an angle of 120 degrees. So, coordinates of A would be ( (48*sqrt(3)/5)*cos(120), (48*sqrt(3)/5)*sin(120) ). Cos(120) is -0.5, sin(120) is sqrt(3)/2. So, coordinates of A are ( (48*sqrt(3)/5)*(-0.5), (48*sqrt(3)/5)*(sqrt(3)/2) ) = ( -24*sqrt(3)/5, (48*3)/10 ) = ( -24*sqrt(3)/5, 144/10 ) = ( -24*sqrt(3)/5, 14.4 ). Okay, so A is at (-24√3/5, 14.4). Now, point M is the midpoint of AB. Coordinates of B are (5,0), coordinates of A are (-24√3/5, 14.4). So, midpoint M has coordinates: x-coordinate: (5 + (-24√3/5))/2 = (25/5 -24√3/5)/2 = (25 -24√3)/10 y-coordinate: (0 + 14.4)/2 = 7.2 So, M is at ( (25 -24√3)/10 , 7.2 ). Point E is the midpoint of AC. Coordinates of A are (-24√3/5, 14.4), coordinates of C are (0,0). So, midpoint E has coordinates: x-coordinate: (-24√3/5 + 0)/2 = -12√3/5 y-coordinate: (14.4 + 0)/2 = 7.2 So, E is at (-12√3/5, 7.2). Now, we need the area of triangle BME. Points B, M, E. Coordinates: B: (5,0) M: ( (25 -24√3)/10 , 7.2 ) E: (-12√3/5, 7.2 ) Let me write down the coordinates more clearly: B: (5, 0) M: ( (25 -24√3)/10 , 7.2 ) E: (-12√3/5, 7.2 ) To find the area of triangle BME, I can use the shoelace formula. First, let me list the coordinates in order: B, M, E, B. So, x1 = 5, y1 = 0 x2 = (25 -24√3)/10, y2 = 7.2 x3 = -12√3/5, y3 = 7.2 Compute the area as (1/2)| (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) | Plugging in the values: = (1/2)| 5*(7.2 - 7.2) + (25 -24√3)/10*(7.2 - 0) + (-12√3/5)*(0 - 7.2) | Simplify term by term: First term: 5*(0) = 0 Second term: (25 -24√3)/10 *7.2 Third term: (-12√3/5)*(-7.2) Compute second term: (25 -24√3)/10 *7.2 = (25*7.2 -24√3*7.2)/10 25*7.2 = 180 24*7.2 = 172.8 So, 180 -172.8√3 Divide by 10: 18 -17.28√3 Third term: (-12√3/5)*(-7.2) = (12√3/5)*7.2 12*7.2 = 86.4 So, 86.4√3 /5 = 17.28√3 Now, add all terms: 0 + (18 -17.28√3) +17.28√3 = 18 So, the area is (1/2)*|18| = 9 Therefore, the area of triangle BME is 9 square units. Wait, that seems straightforward in the end. So, the answer is 9, which is option D. But let me double-check. Maybe I made a mistake in calculations. Wait, the shoelace formula gave me 18 inside the absolute value, so half of that is 9. That seems correct. Alternatively, since ME is parallel to BC and half its length, and BM is half of AB, the area ratio would be 1/4, but since we're only considering triangle BME, which is a smaller triangle within ABC, the area is 1/4 of the original, which is 9. Yes, that matches. So, I think 9 is the correct answer.

answer：First, I need to understand the problem where 2340 is divided into three parts such that the fifth part of the first, the seventh part of the second, and the eleventh part of the third are equal. My goal is to find the smallest part among them.I'll start by denoting the three parts as ( A ), ( B ), and ( C ). According to the problem, we have:[frac{A}{5} = frac{B}{7} = frac{C}{11} = x]From these equations, I can express ( A ), ( B ), and ( C ) in terms of ( x ):[A = 5x, quad B = 7x, quad C = 11x]Next, I know that the sum of ( A ), ( B ), and ( C ) is 2340:[A + B + C = 2340][5x + 7x + 11x = 2340][23x = 2340][x = frac{2340}{23} approx 101.7391]Now, I'll calculate the values of ( A ), ( B ), and ( C ):[A = 5x approx 5 times 101.7391 approx 508.6957][B = 7x approx 7 times 101.7391 approx 712.1739][C = 11x approx 11 times 101.7391 approx 1119.1304]Since we're dealing with whole numbers, I'll round ( x ) to the nearest whole number:[x = 102]Now, recalculating the parts:[A = 5 times 102 = 510][B = 7 times 102 = 714][C = 11 times 102 = 1122]However, the sum of these parts exceeds 2340:[510 + 714 + 1122 = 2346]To correct this, I'll subtract the excess from the largest part:[C = 1122 - 6 = 1116]Now, the sum is:[510 + 714 + 1116 = 2340]Finally, the smallest part is ( A = 510 ).

answer：Okay, so I have this problem here about quadratic inequalities. Let me try to figure it out step by step. The problem says that the solution set of the inequality ( ax^2 + bx + c > 0 ) is ( {x | -1 < x < 2} ). I need to determine the solution set for another inequality: ( a(x^2 + 1) + b(x - 1) + c > 2ax ).First, let me understand what the first inequality tells me. The solution set ( -1 < x < 2 ) means that the quadratic equation ( ax^2 + bx + c = 0 ) has roots at ( x = -1 ) and ( x = 2 ). Also, since the inequality is greater than zero and the solution is between the roots, the parabola must open downwards. That tells me that the coefficient ( a ) is negative.So, from the roots, I can write the quadratic in its factored form: ( a(x + 1)(x - 2) ). Let me expand this to find expressions for ( a ), ( b ), and ( c ).Expanding ( a(x + 1)(x - 2) ):[a(x^2 - 2x + x - 2) = a(x^2 - x - 2) = ax^2 - ax - 2a]So, comparing this to ( ax^2 + bx + c ), I can see that:- ( b = -a )- ( c = -2a )Alright, so I have expressions for ( b ) and ( c ) in terms of ( a ). Now, moving on to the second inequality: ( a(x^2 + 1) + b(x - 1) + c > 2ax ).Let me simplify this inequality step by step. First, distribute the terms:[a x^2 + a + b x - b + c > 2a x]Now, substitute ( b = -a ) and ( c = -2a ) into the inequality:[a x^2 + a + (-a) x - (-a) + (-2a) > 2a x]Simplify each term:- ( a x^2 ) remains as is.- ( a ) remains as is.- ( (-a) x ) is ( -a x ).- ( -(-a) ) is ( +a ).- ( -2a ) remains as is.So putting it all together:[a x^2 + a - a x + a - 2a > 2a x]Combine like terms:- The constant terms: ( a + a - 2a = 0 )- The ( x ) terms: ( -a x )So, the left side simplifies to:[a x^2 - a x]So now, the inequality is:[a x^2 - a x > 2a x]Let me bring all terms to one side to make it easier to handle:[a x^2 - a x - 2a x > 0]Combine the ( x ) terms:[a x^2 - 3a x > 0]Factor out ( a x ):[a x (x - 3) > 0]Hmm, so the inequality simplifies to ( a x (x - 3) > 0 ). Now, I know that ( a ) is negative because the parabola opens downward in the original inequality. So, ( a < 0 ).Let me write this as:[a x (x - 3) > 0]Since ( a ) is negative, I can factor that into the inequality. Let me consider the product ( x (x - 3) ) and the sign of ( a ).First, let's analyze the expression ( x (x - 3) ). This is a quadratic that opens upwards (since the coefficient of ( x^2 ) is positive). Its roots are at ( x = 0 ) and ( x = 3 ). So, the expression ( x (x - 3) ) is positive when ( x < 0 ) or ( x > 3 ), and negative when ( 0 < x < 3 ).But since we have ( a ) multiplied by this expression, and ( a ) is negative, the inequality ( a x (x - 3) > 0 ) will reverse the sign of the intervals where the expression is positive or negative.So, let's break it down:1. When ( x < 0 ): ( x (x - 3) ) is positive. Multiplying by ( a ) (negative) makes it negative. So, ( a x (x - 3) < 0 ). This doesn't satisfy the inequality.2. When ( 0 < x < 3 ): ( x (x - 3) ) is negative. Multiplying by ( a ) (negative) makes it positive. So, ( a x (x - 3) > 0 ). This satisfies the inequality.3. When ( x > 3 ): ( x (x - 3) ) is positive. Multiplying by ( a ) (negative) makes it negative. So, ( a x (x - 3) < 0 ). This doesn't satisfy the inequality.Therefore, the solution set is ( 0 < x < 3 ).Wait, let me double-check my steps to make sure I didn't make a mistake.Starting from the original inequality:( a(x^2 + 1) + b(x - 1) + c > 2ax )Substituted ( b = -a ) and ( c = -2a ), which seems correct because from the original quadratic, ( b = -a ) and ( c = -2a ).Then, expanding and simplifying led me to ( a x^2 - 3a x > 0 ), which factors to ( a x (x - 3) > 0 ). Since ( a < 0 ), the inequality flips signs in the intervals.Yes, that seems correct. So, the solution is ( 0 < x < 3 ).I think that's it. I don't see any mistakes in my reasoning.**Final Answer**The solution set is boxed{{x | 0 < x < 3}}.

answer：Okay, so I have this function f(x) = 2 cos x sin(x + π/3) - √3/2. I need to find the smallest positive period and the symmetry center for part (I), and then for part (II), I have to find the range of f(x) on the interval [π/3, π]. Hmm, let me start with part (I).First, I remember that the period of a function is the smallest positive number T such that f(x + T) = f(x) for all x. Since f(x) is built from sine and cosine functions, which are periodic, the period of f(x) should be related to the periods of these components.Looking at f(x) = 2 cos x sin(x + π/3) - √3/2, it seems like it's a product of cos x and sin(x + π/3). Maybe I can simplify this expression using a trigonometric identity to make it easier to analyze.I recall that there's a product-to-sum identity: sin A cos B = [sin(A + B) + sin(A - B)] / 2. Let me try applying that here. So, if I let A = x + π/3 and B = x, then:sin(x + π/3) cos x = [sin((x + π/3) + x) + sin((x + π/3) - x)] / 2= [sin(2x + π/3) + sin(π/3)] / 2So, 2 cos x sin(x + π/3) becomes 2 * [sin(2x + π/3) + sin(π/3)] / 2 = sin(2x + π/3) + sin(π/3).Therefore, f(x) = sin(2x + π/3) + sin(π/3) - √3/2.Wait, sin(π/3) is √3/2, so f(x) simplifies to sin(2x + π/3) + √3/2 - √3/2 = sin(2x + π/3). Oh, that's much simpler! So f(x) is just sin(2x + π/3).Now, the period of sin(2x + π/3) is the same as the period of sin(2x), because the phase shift doesn't affect the period. The period of sin(kx) is 2π / |k|, so here k = 2, so the period is 2π / 2 = π. So the smallest positive period is π.Next, the symmetry center. For a sine function, which is symmetric about its midline, the midline is y = 0 in this case because there's no vertical shift. So the function f(x) = sin(2x + π/3) is symmetric about points where it crosses the midline, which is the x-axis.To find the symmetry centers, we need to find the points where the function crosses the x-axis. That is, where sin(2x + π/3) = 0. So, 2x + π/3 = kπ for some integer k. Solving for x:2x = kπ - π/3x = (kπ - π/3) / 2x = (kπ)/2 - π/6So, the symmetry centers are at points ( (kπ)/2 - π/6, 0 ) for all integers k. That makes sense because the sine function crosses the x-axis at regular intervals, and these points are spaced by half the period, which is π/2.Alright, so for part (I), the smallest positive period is π, and the symmetry centers are at ( (kπ)/2 - π/6, 0 ) for integers k.Moving on to part (II), I need to find the range of f(x) on the interval [π/3, π]. Since f(x) simplifies to sin(2x + π/3), I can analyze this sine function over the given interval.First, let's find the corresponding interval for the argument of the sine function. When x is in [π/3, π], then 2x is in [2π/3, 2π], so 2x + π/3 is in [2π/3 + π/3, 2π + π/3] = [π, 7π/3].So, the argument of the sine function, let's call it θ = 2x + π/3, ranges from π to 7π/3 as x goes from π/3 to π.Now, let's analyze sin(θ) over θ ∈ [π, 7π/3]. The sine function has a period of 2π, so 7π/3 is equivalent to 7π/3 - 2π = π/3. So, θ goes from π to 7π/3, which is the same as θ going from π to 2π and then from 0 to π/3.But wait, 7π/3 is actually 2π + π/3, so θ goes from π to 2π + π/3. Hmm, maybe it's better to just consider θ from π to 7π/3 without subtracting 2π.Let me think about the behavior of sin(θ) over θ from π to 7π/3.From θ = π to θ = 3π/2, sin(θ) decreases from 0 to -1.From θ = 3π/2 to θ = 2π, sin(θ) increases from -1 to 0.From θ = 2π to θ = 7π/3, sin(θ) increases from 0 to sin(π/3) = √3/2.Wait, but θ = 7π/3 is the same as θ = π/3, right? Because 7π/3 - 2π = π/3. So, sin(7π/3) = sin(π/3) = √3/2.So, putting it all together, as θ goes from π to 7π/3:- From π to 3π/2: sin(θ) decreases from 0 to -1.- From 3π/2 to 2π: sin(θ) increases from -1 to 0.- From 2π to 7π/3: sin(θ) increases from 0 to √3/2.Therefore, the minimum value of sin(θ) in this interval is -1, and the maximum value is √3/2.But wait, let me check the exact points:- When θ = π, sin(π) = 0.- When θ = 3π/2, sin(3π/2) = -1.- When θ = 2π, sin(2π) = 0.- When θ = 7π/3, sin(7π/3) = sin(π/3) = √3/2.So, the function f(x) = sin(θ) reaches its minimum at θ = 3π/2, which is -1, and its maximum at θ = 7π/3, which is √3/2.But I need to make sure that these θ values correspond to x values within [π/3, π].Let's solve for x when θ = 3π/2:θ = 2x + π/3 = 3π/22x = 3π/2 - π/3 = (9π/6 - 2π/6) = 7π/6x = 7π/12Is 7π/12 within [π/3, π]? Let's see: π/3 ≈ 1.047, 7π/12 ≈ 1.832, π ≈ 3.142. Yes, 7π/12 is within the interval.Similarly, when θ = 7π/3:θ = 2x + π/3 = 7π/32x = 7π/3 - π/3 = 6π/3 = 2πx = πWhich is the endpoint of the interval.So, the function f(x) reaches its minimum at x = 7π/12, which is within [π/3, π], and its maximum at x = π.Wait, but when θ = π, which corresponds to x = (π - π/3)/2 = (2π/3)/2 = π/3, which is the left endpoint. So, at x = π/3, f(x) = sin(π) = 0.So, putting it all together, the function f(x) starts at 0 when x = π/3, decreases to -1 at x = 7π/12, then increases back to 0 at x = π, but wait, at x = π, f(x) = sin(2π + π/3) = sin(π/3) = √3/2. Wait, that contradicts my earlier thought.Wait, let me recast that. When x = π, θ = 2π + π/3 = 7π/3, which is equivalent to π/3, and sin(7π/3) = sin(π/3) = √3/2. So, actually, f(π) = √3/2.Wait, so let me correct that. So, as x goes from π/3 to π, θ goes from π to 7π/3.At x = π/3: θ = π, sin(π) = 0.At x = 7π/12: θ = 3π/2, sin(3π/2) = -1.At x = π: θ = 7π/3, sin(7π/3) = √3/2.So, the function f(x) starts at 0, decreases to -1 at x = 7π/12, then increases to √3/2 at x = π.Therefore, the range of f(x) on [π/3, π] is from -1 to √3/2.Wait, but let me confirm if the function is indeed increasing from x = 7π/12 to x = π.The derivative of f(x) = sin(2x + π/3) is f'(x) = 2 cos(2x + π/3).At x = 7π/12, θ = 3π/2, so cos(3π/2) = 0. So, the derivative is 0 there.For x > 7π/12, let's pick x = 2π/3, which is approximately 2.094, which is greater than 7π/12 ≈ 1.832.θ = 2*(2π/3) + π/3 = 4π/3 + π/3 = 5π/3.cos(5π/3) = cos(π/3) = 0.5, which is positive. So, f'(x) = 2*0.5 = 1, which is positive. So, the function is increasing after x = 7π/12.Similarly, for x < 7π/12, say x = π/2, θ = 2*(π/2) + π/3 = π + π/3 = 4π/3.cos(4π/3) = -0.5, so f'(x) = 2*(-0.5) = -1, which is negative. So, the function is decreasing before x = 7π/12.Therefore, the function decreases from x = π/3 to x = 7π/12, reaching a minimum of -1, then increases from x = 7π/12 to x = π, reaching a maximum of √3/2.Hence, the range of f(x) on [π/3, π] is [-1, √3/2].Wait, but let me double-check the values at the endpoints and the critical point.At x = π/3:f(π/3) = sin(2*(π/3) + π/3) = sin(2π/3 + π/3) = sin(π) = 0.At x = 7π/12:f(7π/12) = sin(2*(7π/12) + π/3) = sin(14π/12 + 4π/12) = sin(18π/12) = sin(3π/2) = -1.At x = π:f(π) = sin(2π + π/3) = sin(7π/3) = sin(π/3) = √3/2.Yes, that's correct. So, the function starts at 0, goes down to -1, then up to √3/2. So, the range is all values from -1 to √3/2.Therefore, the range is [-1, √3/2].I think that's it. So, summarizing:(I) The smallest positive period is π, and the symmetry centers are at ( (kπ)/2 - π/6, 0 ) for all integers k.(II) The range of f(x) on [π/3, π] is [-1, √3/2].