answer：Okay, so I have this problem here about a lyceum where 76% of the students didn't do their homework at least once, and 5/37 sometimes forget their second pair of shoes. I need to find the number of students in the lyceum, given that it's more than 1000 but fewer than 2000. Hmm, let me try to figure this out step by step.First, let's parse the information given. We have two percentages or fractions related to the students:1. 76% of the students did not do their homework at least once.2. 5/37 of the students sometimes forget their second pair of shoes.Wait, actually, the second one is a fraction, not a percentage. So, 5/37 is approximately 0.135, which is about 13.5%. So, both of these are fractions of the total student population. But the question is asking for the number of students, which is between 1000 and 2000. So, we need to find a number N such that N is between 1000 and 2000, and both 76% of N and 5/37 of N are whole numbers because you can't have a fraction of a student.So, essentially, N must be a number where:1. 76% of N is an integer. That means N must be divisible by 100/ gcd(76,100). Let me think about that. 76% is the same as 76/100, which simplifies to 19/25. So, 76% of N is (19/25)*N, which must be an integer. Therefore, N must be a multiple of 25 because 25 is the denominator after simplifying 76%.2. Similarly, 5/37 of N must be an integer. So, N must be a multiple of 37 because 37 is the denominator here.Therefore, N must be a common multiple of 25 and 37. Since 25 and 37 are coprime (they have no common factors other than 1), their least common multiple (LCM) is 25*37, which is 925. But 925 is less than 1000, so we need the next multiple. Let me calculate 25*37=925, then 2*925=1850, which is still less than 2000. The next one would be 3*925=2775, which is way above 2000. So, the possible Ns are 925, 1850, 2775, etc. But since N must be between 1000 and 2000, the only possible value is 1850.Wait, hold on. Let me verify that. If N is 1850, then 76% of 1850 is 0.76*1850. Let me compute that. 0.76*1850. 1850*0.7 is 1295, and 1850*0.06 is 111. So, 1295+111=1406. So, 1406 students didn't do their homework at least once. That's an integer, which is good.Now, 5/37 of 1850 is (5/37)*1850. Let me compute that. 1850 divided by 37. Let me see, 37*50 is 1850. So, 1850/37=50. Then, 5*50=250. So, 250 students sometimes forget their second pair of shoes. That's also an integer. Perfect.So, N=1850 satisfies both conditions. Is there any other number between 1000 and 2000 that is a multiple of both 25 and 37? Well, since 25 and 37 are coprime, their LCM is 925, so the next multiple is 1850, and then 2775, which is beyond 2000. So, 1850 is the only number in that range.But just to make sure I didn't make a mistake, let me think again. So, 76% as a fraction is 19/25, so N must be a multiple of 25. 5/37 is already a fraction, so N must be a multiple of 37. Therefore, N must be a multiple of LCM(25,37)=925. So, 925 is too low, next is 1850, which is within the range. So, 1850 is the answer.Wait, but let me check if 1850 is indeed between 1000 and 2000. Yes, 1850 is more than 1000 and less than 2000. So, that's correct.Alternatively, maybe I can approach this problem by setting up equations. Let me denote N as the total number of students. Then, 0.76*N must be an integer, and (5/37)*N must also be an integer.So, 0.76*N = (76/100)*N = (19/25)*N must be integer. Therefore, 25 divides N. Similarly, (5/37)*N must be integer, so 37 divides N. Therefore, N must be a multiple of both 25 and 37. So, N must be a multiple of LCM(25,37)=925. So, N=925*k, where k is a positive integer.Given that N is between 1000 and 2000, let's find k such that 1000 < 925*k < 2000.Dividing all parts by 925: 1000/925 ≈1.0816 < k < 2000/925≈2.162. So, k must be 2 because k must be an integer. Therefore, N=925*2=1850.Yes, that's consistent with my earlier conclusion. So, N=1850 is the only number in that range that satisfies both conditions.Just to double-check, let me compute 76% of 1850 again. 1850*0.76. Let me compute 1850*0.7=1295, 1850*0.06=111, so 1295+111=1406. That's an integer.And 5/37 of 1850: 1850 divided by 37 is 50, because 37*50=1850. Then, 5*50=250. That's also an integer. So, both conditions are satisfied.Therefore, the number of students in the lyceum is 1850.**Final Answer**The number of students in the lyceum is boxed{1850}.

answer：First, I recognize that there are 30 students in total.Out of these, 12 students have cool dads, and 15 have cool moms. However, 9 students have both cool dads and cool moms.To find out how many students have either a cool dad or a cool mom, I use the principle of inclusion-exclusion. This means I add the number of students with cool dads and cool moms and then subtract those who have both to avoid double-counting.So, 12 plus 15 equals 27, and subtracting the 9 students who have both gives me 18 students who have either a cool dad or a cool mom.Finally, to find the number of students whose both parents are uncool, I subtract this number from the total number of students. Therefore, 30 minus 18 equals 12 students.

answer：Okay, so I have this problem where I need to find a polynomial f(x) and a constant C such that the equation∫₀ˣ f(y) dy + ∫₀¹ (x + y)² f(y) dy = x² + Cholds true. Hmm, let me try to break this down step by step.First, I know that f(x) is a polynomial, so let's assume it's of degree n. Since the integral of f(y) from 0 to x will be a polynomial of degree n+1, and the other integral involves (x + y)² multiplied by f(y), which when expanded will have terms up to x² times f(y). So, when we integrate that from 0 to 1, it will result in a quadratic polynomial in x, because the highest power of x in (x + y)² is x².Looking at the right-hand side, it's x² + C, which is a quadratic polynomial. So, the left-hand side must also be a quadratic polynomial. That suggests that f(y) must be such that when we take the integrals, the highest degree term is x².Let me think about the degrees. If f(y) is a quadratic polynomial, say f(y) = ay² + by + c, then the integral ∫₀ˣ f(y) dy will be (a/3)x³ + (b/2)x² + cx. But on the right-hand side, the highest degree is x², so the x³ term must cancel out. That means a must be zero. So, f(y) is at most a linear polynomial.Wait, but if f(y) is linear, say f(y) = by + c, then ∫₀ˣ f(y) dy = (b/2)x² + cx. The other integral, ∫₀¹ (x + y)² f(y) dy, let's expand that:(x + y)² = x² + 2xy + y², so multiplying by f(y) = by + c gives:x²(by + c) + 2xy(by + c) + y²(by + c)Which is:b x² y + c x² + 2b x y² + 2c x y + b y³ + c y²Now, integrating term by term from 0 to 1:∫₀¹ b x² y dy = b x² ∫₀¹ y dy = b x² (1/2) = (b/2) x²∫₀¹ c x² dy = c x² ∫₀¹ dy = c x² (1) = c x²∫₀¹ 2b x y² dy = 2b x ∫₀¹ y² dy = 2b x (1/3) = (2b/3) x∫₀¹ 2c x y dy = 2c x ∫₀¹ y dy = 2c x (1/2) = c x∫₀¹ b y³ dy = b ∫₀¹ y³ dy = b (1/4) = b/4∫₀¹ c y² dy = c ∫₀¹ y² dy = c (1/3) = c/3So, putting all these together, the second integral becomes:(b/2) x² + c x² + (2b/3) x + c x + b/4 + c/3Simplify:[(b/2 + c)] x² + [(2b/3 + c)] x + (b/4 + c/3)So, now, let's write the entire left-hand side:First integral: (b/2) x² + c xSecond integral: (b/2 + c) x² + (2b/3 + c) x + (b/4 + c/3)Adding them together:[(b/2) + (b/2 + c)] x² + [c + (2b/3 + c)] x + (b/4 + c/3)Simplify each coefficient:For x²: (b/2 + b/2 + c) = b + cFor x: (c + 2b/3 + c) = 2c + 2b/3Constants: b/4 + c/3So, the left-hand side is:(b + c) x² + (2c + 2b/3) x + (b/4 + c/3)And this is supposed to equal x² + C.Therefore, we can set up equations by equating coefficients:1. Coefficient of x²: b + c = 12. Coefficient of x: 2c + (2b)/3 = 03. Constant term: b/4 + c/3 = CSo, now we have a system of three equations:1. b + c = 12. (2b)/3 + 2c = 03. (b)/4 + (c)/3 = CLet me write these equations more clearly:Equation 1: b + c = 1Equation 2: (2/3) b + 2c = 0Equation 3: (1/4) b + (1/3) c = CSo, let's solve Equations 1 and 2 for b and c.From Equation 1: c = 1 - bPlug into Equation 2:(2/3) b + 2(1 - b) = 0Simplify:(2/3) b + 2 - 2b = 0Combine like terms:(2/3 b - 2b) + 2 = 0Convert 2b to thirds: 2b = 6/3 bSo, (2/3 - 6/3) b + 2 = 0 => (-4/3) b + 2 = 0Solve for b:(-4/3) b = -2 => b = (-2) * (-3/4) = 6/4 = 3/2So, b = 3/2Then, from Equation 1: c = 1 - b = 1 - 3/2 = -1/2So, c = -1/2Now, let's find C from Equation 3:C = (1/4) b + (1/3) c = (1/4)(3/2) + (1/3)(-1/2) = (3/8) - (1/6)To subtract these, find a common denominator, which is 24:3/8 = 9/241/6 = 4/24So, 9/24 - 4/24 = 5/24Therefore, C = 5/24So, f(y) is a linear polynomial: f(y) = by + c = (3/2)y - 1/2Let me write that as f(y) = (3/2)y - 1/2Let me check if this works.Compute the first integral: ∫₀ˣ f(y) dy = ∫₀ˣ (3/2 y - 1/2) dy= (3/4 y² - 1/2 y) evaluated from 0 to x= (3/4 x² - 1/2 x) - (0 - 0) = 3/4 x² - 1/2 xSecond integral: ∫₀¹ (x + y)² f(y) dyWe already expanded this earlier, but let me compute it with b = 3/2 and c = -1/2.From earlier, the second integral is:(b/2 + c) x² + (2b/3 + c) x + (b/4 + c/3)Plugging in b = 3/2 and c = -1/2:First term: ( (3/2)/2 + (-1/2) ) x² = (3/4 - 1/2) x² = (3/4 - 2/4) x² = 1/4 x²Second term: (2*(3/2)/3 + (-1/2)) x = ( (3)/3 - 1/2 ) x = (1 - 1/2) x = 1/2 xThird term: ( (3/2)/4 + (-1/2)/3 ) = (3/8 - 1/6) = (9/24 - 4/24) = 5/24So, the second integral is (1/4)x² + (1/2)x + 5/24Adding the first integral and the second integral:First integral: 3/4 x² - 1/2 xSecond integral: 1/4 x² + 1/2 x + 5/24Adding together:(3/4 + 1/4) x² + (-1/2 + 1/2) x + 5/24Simplify:1 x² + 0 x + 5/24Which is x² + 5/24So, that's equal to x² + C, so C = 5/24. Perfect, that matches.Therefore, f(y) = (3/2)y - 1/2 and C = 5/24.So, writing f(x) as a polynomial, it's (3/2)x - 1/2, which can also be written as (3x - 1)/2.So, f(x) = (3x - 1)/2 and C = 5/24.I think that's the solution. Let me just recap to make sure I didn't make any mistakes.We assumed f(x) is linear because the right-hand side is quadratic, and if f(x) were quadratic, the first integral would give a cubic term, which isn't present on the right. So, f(x) must be linear.Then, we set f(y) = by + c, expanded both integrals, and equated coefficients to solve for b, c, and C. The calculations led us to b = 3/2, c = -1/2, and C = 5/24. Then, plugging back into the integrals confirmed that it works.Therefore, I'm confident that this is the correct solution.**Final Answer**The polynomial is ( f(x) = frac{3x - 1}{2} ) and the constant is ( C = frac{5}{24} ). Thus, the solution is ( boxed{f(x) = frac{3x - 1}{2}} ) and ( boxed{C = frac{5}{24}} ).However, since the problem asks to find both f(x) and C, and the instructions specify to put the final answer within boxes, I think it's appropriate to present both in separate boxes as above.**Final Answer**The polynomial is ( boxed{frac{3x - 1}{2}} ) and the constant is ( boxed{frac{5}{24}} ).

answer：Okay, so I have this problem where I need to find the real number ( a ) such that the curves ( y = frac{1}{2e}x^2 ) and ( y = a ln x ) have a common tangent at their common point ( P(s, t) ). Hmm, let me try to figure this out step by step.First, I know that if two curves have a common tangent at a point, that means they not only intersect at that point but also have the same slope there. So, two conditions must be satisfied:1. The curves intersect at point ( P(s, t) ). So, substituting ( x = s ) into both equations should give the same ( y )-value ( t ).2. The derivatives of both curves at ( x = s ) must be equal. This ensures that the slopes of the tangents are the same at that point.Alright, let's write down these conditions.**Condition 1: Intersection at ( P(s, t) )**For the first curve: ( t = frac{1}{2e}s^2 )For the second curve: ( t = a ln s )So, setting them equal:[ frac{1}{2e}s^2 = a ln s ]Let me note this as equation (1):[ frac{s^2}{2e} = a ln s ]**Condition 2: Equal derivatives at ( x = s )**First, find the derivatives of both curves.For the first curve ( y = frac{1}{2e}x^2 ):The derivative ( y' ) is ( frac{d}{dx} left( frac{1}{2e}x^2 right) = frac{1}{e}x ). So, at ( x = s ), the slope is ( frac{s}{e} ).For the second curve ( y = a ln x ):The derivative ( y' ) is ( frac{d}{dx} (a ln x) = frac{a}{x} ). So, at ( x = s ), the slope is ( frac{a}{s} ).Setting these equal:[ frac{s}{e} = frac{a}{s} ]Let me note this as equation (2):[ frac{s}{e} = frac{a}{s} ]Now, I have two equations:1. ( frac{s^2}{2e} = a ln s ) (Equation 1)2. ( frac{s}{e} = frac{a}{s} ) (Equation 2)I need to solve these two equations for ( a ) and ( s ). Since I need to find ( a ), I can express ( a ) from equation (2) and substitute into equation (1).From equation (2):Multiply both sides by ( s ):[ frac{s^2}{e} = a ]So, ( a = frac{s^2}{e} )Now, substitute this expression for ( a ) into equation (1):[ frac{s^2}{2e} = left( frac{s^2}{e} right) ln s ]Simplify the right-hand side:[ frac{s^2}{2e} = frac{s^2}{e} ln s ]Hmm, let's see. Let me write it as:[ frac{s^2}{2e} = frac{s^2}{e} ln s ]I can divide both sides by ( frac{s^2}{e} ) to simplify, assuming ( s neq 0 ) (since ( ln s ) is only defined for ( s > 0 ), so ( s ) can't be zero anyway).Dividing both sides by ( frac{s^2}{e} ):[ frac{1}{2} = ln s ]So, ( ln s = frac{1}{2} )To solve for ( s ), exponentiate both sides:[ s = e^{frac{1}{2}} = sqrt{e} ]Great, so ( s = sqrt{e} ). Now, let's find ( a ).From equation (2), ( a = frac{s^2}{e} ). Plugging ( s = sqrt{e} ):[ a = frac{(sqrt{e})^2}{e} = frac{e}{e} = 1 ]So, ( a = 1 ).Wait, let me double-check my steps to make sure I didn't make a mistake.1. Found the intersection condition: ( frac{s^2}{2e} = a ln s ).2. Found the derivative condition: ( frac{s}{e} = frac{a}{s} ), which gives ( a = frac{s^2}{e} ).3. Substituted ( a ) into the first equation: ( frac{s^2}{2e} = frac{s^2}{e} ln s ).4. Simplified to ( frac{1}{2} = ln s ), so ( s = sqrt{e} ).5. Plugged back to find ( a = 1 ).Seems solid. Let me just verify by plugging ( s = sqrt{e} ) and ( a = 1 ) into both original equations.First curve at ( s = sqrt{e} ):( y = frac{1}{2e} (sqrt{e})^2 = frac{1}{2e} times e = frac{1}{2} ).Second curve at ( s = sqrt{e} ):( y = 1 times ln sqrt{e} = ln e^{1/2} = frac{1}{2} ).So, both curves have the point ( (sqrt{e}, frac{1}{2}) ).Now, check the derivatives:First curve's derivative at ( s = sqrt{e} ):( frac{s}{e} = frac{sqrt{e}}{e} = frac{1}{sqrt{e}} ).Second curve's derivative at ( s = sqrt{e} ):( frac{a}{s} = frac{1}{sqrt{e}} ).Yes, both derivatives are equal. So, the slopes are the same, which confirms that they have a common tangent at that point.Therefore, the value of ( a ) is indeed 1.**Final Answer**The real number ( a ) is boxed{1}.

answer：Okay, so I have this function f(x) = √3 sin 2x + 2 sin²x. I need to find its smallest positive period and then determine its maximum and minimum values on the interval [0, π/2]. Hmm, let me start with part (I), the period.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. So, for functions like sin and cos, the standard period is 2π, but when they're multiplied by coefficients inside the argument, the period changes. Specifically, for sin(kx) or cos(kx), the period is 2π / |k|. Looking at f(x), it's composed of two terms: √3 sin 2x and 2 sin²x. I need to find the periods of each term separately and then find the least common multiple (LCM) of those periods to get the overall period of f(x).Starting with the first term, √3 sin 2x. Here, k is 2, so the period is 2π / 2 = π. So, the period of the first term is π.Now, the second term is 2 sin²x. Hmm, sin²x can be rewritten using a double-angle identity to make it easier to find the period. I recall that sin²x = (1 - cos 2x)/2. So, substituting that in, 2 sin²x becomes 2*(1 - cos 2x)/2, which simplifies to 1 - cos 2x. So, now the second term is 1 - cos 2x. The period of cos 2x is 2π / 2 = π, just like the first term. So, the second term also has a period of π.Wait a second, both terms have the same period, which is π. So, does that mean the overall period of f(x) is π? Because if both components have the same period, their sum should also have that period. Let me check that.Let me test f(x + π). f(x + π) = √3 sin[2(x + π)] + 2 sin²(x + π). Simplify each term:sin[2(x + π)] = sin(2x + 2π) = sin 2x, since sine has a period of 2π, but here it's 2π inside, so it's just sin 2x.Similarly, sin(x + π) = -sin x, so sin²(x + π) = (-sin x)² = sin²x. Therefore, 2 sin²(x + π) = 2 sin²x.So, f(x + π) = √3 sin 2x + 2 sin²x = f(x). So, yes, the period is π. But wait, is there a smaller period?Let me see. Suppose T is a period of f(x). Then, T must be a period for both terms. Since both terms have period π, the fundamental period can't be smaller than π. Because if T were smaller than π, say T = π/2, then let's check if f(x + π/2) = f(x).Compute f(x + π/2):First term: √3 sin[2(x + π/2)] = √3 sin(2x + π) = √3*(-sin 2x) = -√3 sin 2x.Second term: 2 sin²(x + π/2) = 2 cos²x, since sin(x + π/2) = cos x. So, 2 cos²x.Therefore, f(x + π/2) = -√3 sin 2x + 2 cos²x. Is this equal to f(x)? Let's see:Original f(x) = √3 sin 2x + 2 sin²x.So, unless √3 sin 2x = -√3 sin 2x and 2 sin²x = 2 cos²x, which would require sin 2x = 0 and sin²x = cos²x, which isn't true for all x. So, f(x + π/2) ≠ f(x). Therefore, π/2 is not a period.Similarly, let's check T = π/3. Probably not, but just to be thorough.f(x + π/3) = √3 sin[2(x + π/3)] + 2 sin²(x + π/3).Compute sin[2(x + π/3)] = sin(2x + 2π/3). That's not equal to sin 2x in general.Similarly, sin²(x + π/3) is not equal to sin²x. So, f(x + π/3) ≠ f(x). Therefore, the period can't be smaller than π. So, the smallest positive period is π.Wait, but hold on. The second term, after rewriting, is 1 - cos 2x. So, f(x) can be rewritten as √3 sin 2x + 1 - cos 2x. So, f(x) = √3 sin 2x - cos 2x + 1.Hmm, maybe I can combine the first two terms into a single sinusoidal function. That might help in analyzing the period, but since both sin 2x and cos 2x have the same frequency, their combination will also have the same period, which is π. So, that confirms again that the period is π.So, part (I) is done. The smallest positive period is π.Now, moving on to part (II): finding the maximum and minimum values of f(x) on the interval [0, π/2].First, let's write f(x) again:f(x) = √3 sin 2x + 2 sin²x.Alternatively, as I did earlier, f(x) can be rewritten as √3 sin 2x - cos 2x + 1. Maybe that form is easier to handle because it's a combination of sine and cosine with the same argument, which can be expressed as a single sine or cosine function with a phase shift.Let me try that. So, f(x) = √3 sin 2x - cos 2x + 1.Let me denote A = √3, B = -1, so f(x) = A sin 2x + B cos 2x + 1.I can write A sin θ + B cos θ as C sin(θ + φ), where C = √(A² + B²) and tan φ = B / A.Wait, actually, the formula is A sin θ + B cos θ = C sin(θ + φ), where C = √(A² + B²) and φ = arctan(B / A). Wait, no, actually, it's φ = arctan(B / A) if we express it as sin(θ + φ), but sometimes it's expressed as cos φ = A / C and sin φ = B / C.Let me recall the exact identity. The identity is:A sin θ + B cos θ = C sin(θ + φ), where C = √(A² + B²), and φ is such that cos φ = A / C and sin φ = B / C.Alternatively, it can also be written as C cos(θ - φ'), where φ' is another phase shift.But regardless, let's compute C first.Given A = √3, B = -1.So, C = √[(√3)² + (-1)²] = √[3 + 1] = √4 = 2.So, C = 2.Now, let's find φ such that:cos φ = A / C = √3 / 2,sin φ = B / C = (-1) / 2.So, cos φ = √3 / 2 and sin φ = -1/2.Looking at the unit circle, the angle φ that satisfies this is in the fourth quadrant. Specifically, φ = -π/6 or 11π/6.Therefore, we can write:√3 sin 2x - cos 2x = 2 sin(2x - π/6).Therefore, f(x) = 2 sin(2x - π/6) + 1.That's a much simpler expression. So, f(x) = 2 sin(2x - π/6) + 1.Now, since this is a sine function with amplitude 2, shifted vertically by 1, its maximum value is 2 + 1 = 3, and its minimum value is -2 + 1 = -1. However, this is over the entire real line. But we are restricted to the interval [0, π/2].So, we need to find the maximum and minimum of f(x) on [0, π/2].To do this, we can analyze the function f(x) = 2 sin(2x - π/6) + 1 on [0, π/2].First, let's find the critical points by taking the derivative and setting it equal to zero.f'(x) = derivative of 2 sin(2x - π/6) + 1.The derivative is 2 * cos(2x - π/6) * 2 = 4 cos(2x - π/6).Set f'(x) = 0:4 cos(2x - π/6) = 0 => cos(2x - π/6) = 0.So, 2x - π/6 = π/2 + kπ, where k is an integer.Solving for x:2x = π/2 + π/6 + kπ = (3π/6 + π/6) + kπ = (4π/6) + kπ = (2π/3) + kπ.Therefore, x = (2π/3)/2 + (kπ)/2 = π/3 + (kπ)/2.Now, we need to find all x in [0, π/2] that satisfy this.Let's consider k = 0: x = π/3 ≈ 1.047, which is less than π/2 ≈ 1.5708. So, x = π/3 is in the interval.k = 1: x = π/3 + π/2 = (2π/6 + 3π/6) = 5π/6 ≈ 2.617, which is greater than π/2, so it's outside the interval.k = -1: x = π/3 - π/2 = -π/6, which is negative, so also outside the interval.Therefore, the only critical point in [0, π/2] is x = π/3.So, now, to find the maximum and minimum, we need to evaluate f(x) at the critical point x = π/3 and at the endpoints x = 0 and x = π/2.Let's compute f(0):f(0) = 2 sin(2*0 - π/6) + 1 = 2 sin(-π/6) + 1 = 2*(-1/2) + 1 = -1 + 1 = 0.f(π/3):First, compute 2x - π/6 when x = π/3: 2*(π/3) - π/6 = 2π/3 - π/6 = (4π/6 - π/6) = 3π/6 = π/2.So, f(π/3) = 2 sin(π/2) + 1 = 2*1 + 1 = 3.f(π/2):Compute 2x - π/6 when x = π/2: 2*(π/2) - π/6 = π - π/6 = (6π/6 - π/6) = 5π/6.So, f(π/2) = 2 sin(5π/6) + 1 = 2*(1/2) + 1 = 1 + 1 = 2.So, summarizing:f(0) = 0,f(π/3) = 3,f(π/2) = 2.Therefore, on the interval [0, π/2], the maximum value of f(x) is 3, achieved at x = π/3, and the minimum value is 0, achieved at x = 0.Wait, but hold on. Let me double-check the computation for f(π/2). So, 2x - π/6 when x = π/2 is π - π/6 = 5π/6. Sin(5π/6) is 1/2, so 2*(1/2) is 1, plus 1 is 2. That seems correct.Similarly, f(0): 2 sin(-π/6) is 2*(-1/2) = -1, plus 1 is 0. Correct.And f(π/3): 2 sin(π/2) is 2*1 = 2, plus 1 is 3. Correct.So, the maximum is indeed 3, and the minimum is 0.But just to be thorough, let me also consider the behavior of the function. Since f(x) is a sinusoidal function with amplitude 2, shifted up by 1, its maximum possible value is 3 and minimum is -1. However, on the interval [0, π/2], the function doesn't reach -1 because the lowest point it reaches is 0 at x = 0.Is there a possibility that somewhere else in the interval, the function could dip below 0? Let's see.Looking at f(x) = 2 sin(2x - π/6) + 1.The sine function varies between -1 and 1, so 2 sin(...) varies between -2 and 2, plus 1 gives between -1 and 3.But on [0, π/2], let's see the range of 2x - π/6.When x = 0: 2*0 - π/6 = -π/6.When x = π/2: 2*(π/2) - π/6 = π - π/6 = 5π/6.So, 2x - π/6 ranges from -π/6 to 5π/6 as x goes from 0 to π/2.So, the argument of the sine function goes from -π/6 to 5π/6.So, sin(theta) where theta is from -π/6 to 5π/6.The sine function reaches its minimum at theta = -π/6, which is sin(-π/6) = -1/2, and its maximum at theta = π/2, which is 1.Therefore, 2 sin(theta) ranges from 2*(-1/2) = -1 to 2*1 = 2.Adding 1, f(x) ranges from -1 + 1 = 0 to 2 + 1 = 3.So, on [0, π/2], f(x) ranges from 0 to 3, confirming our earlier results.Therefore, the maximum is 3 and the minimum is 0.Wait, but hold on. The sine function reaches -1/2 at theta = -π/6, which is x = 0, giving f(x) = 0. Then, as theta increases from -π/6 to 5π/6, the sine function increases to 1 at theta = π/2 (x = π/3), then decreases back to sin(5π/6) = 1/2, giving f(x) = 2*(1/2) + 1 = 2.So, the function starts at 0, goes up to 3 at x = π/3, then comes back down to 2 at x = π/2. So, the minimum is indeed 0, and the maximum is 3.Therefore, the final answers are:(I) The smallest positive period is π.(II) The maximum value is 3, and the minimum value is 0 on the interval [0, π/2].**Final Answer**(I) The smallest positive period is boxed{pi}.(II) The maximum value is boxed{3} and the minimum value is boxed{0}.

answer：Let ( c ) represent the number of correct answers and ( i ) represent the number of incorrect answers.Since Olivia answered all 10 problems, we have:[c + i = 10]For the scoring, she earns 5 points for each correct answer and loses 2 points for each incorrect answer. Her total score is 29 points:[5c - 2i = 29]From the first equation, we can express ( i ) in terms of ( c ):[i = 10 - c]Substituting ( i ) into the second equation:[5c - 2(10 - c) = 29][5c - 20 + 2c = 29][7c - 20 = 29][7c = 49][c = 7]Therefore, Olivia had 7 correct answers.