answer：Okay, so I have this problem about a triangle. The area is 30, one side has length 10, and the median to that side is 9. They want me to find sin(theta), where theta is the acute angle between that side and the median. The answer choices are given, so that might help if I get stuck.First, let me recall what a median is. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. So, in this case, the median is drawn to the side of length 10, which means it splits that side into two equal parts of 5 each.Now, the triangle has an area of 30. I remember that the area of a triangle can be calculated in a few ways. One common formula is (1/2)*base*height. Another formula that might be useful here is using two sides and the sine of the included angle: (1/2)*a*b*sin(theta). Maybe that can help since we have a side, a median, and the angle between them.Let me visualize the triangle. Let's call the triangle ABC, where side BC is the side of length 10. The median from vertex A to side BC will meet BC at its midpoint, which we can call point D. So, BD = DC = 5. The length of the median AD is 9. We need to find sin(theta), where theta is the acute angle between side BC and median AD.So, in triangle ABD, we have sides AB, BD, and AD. Wait, but I don't know AB. Hmm. Maybe I can use the area formula with sides AB and AD and the angle between them? But I don't know AB or the angle. Hmm.Wait, maybe I can use the area of the whole triangle ABC. Since D is the midpoint, the area of triangle ABD is half of the area of ABC, so 15. So, area of ABD is 15.In triangle ABD, sides BD = 5, AD = 9, and angle between them is theta. So, area = (1/2)*BD*AD*sin(theta). Plugging in the numbers: 15 = (1/2)*5*9*sin(theta). Let me compute that.15 = (1/2)*5*9*sin(theta)15 = (45/2)*sin(theta)So, sin(theta) = 15 * (2/45) = 30/45 = 2/3.Wait, that seems straightforward. So, sin(theta) is 2/3. Let me check if that makes sense.Wait, hold on. Is theta the angle between BD and AD? Yes, because BD is part of BC, so the angle between BC and AD is the same as the angle between BD and AD. So, yes, that should be correct.But let me make sure I didn't make a mistake in the area. The area of triangle ABC is 30, so the area of ABD is half of that, which is 15. Then, using the formula (1/2)*a*b*sin(theta) with a=5, b=9, and angle theta between them, we get 15 = (1/2)*5*9*sin(theta). Solving for sin(theta) gives 15 = (45/2)*sin(theta), so sin(theta) = (15*2)/45 = 30/45 = 2/3. So, that's 2/3.Looking at the answer choices, option D is 2/3, so that should be the answer.Wait, just to make sure, is there another way to approach this? Maybe using coordinates or vectors?Let me try that. Let me place point B at (0,0), point C at (10,0), so midpoint D is at (5,0). Then, point A is somewhere in the plane, and the median AD has length 9. So, coordinates of A satisfy the distance from D(5,0) to A(x,y) is 9, so sqrt((x-5)^2 + y^2) = 9. Also, the area of triangle ABC is 30, which can be calculated using the base BC (length 10) and the height from A to BC.The area is (1/2)*base*height = 30, so (1/2)*10*height = 30 => 5*height = 30 => height = 6. So, the y-coordinate of point A must be 6 or -6, since the height is the vertical distance from A to BC, which is along the x-axis.So, point A is at (x,6) or (x,-6). But since theta is the acute angle between BC and AD, I think we can assume it's above the x-axis, so y=6.So, coordinates of A are (x,6). Then, the distance from D(5,0) to A(x,6) is 9, so sqrt((x-5)^2 + (6)^2) = 9. Squaring both sides: (x-5)^2 + 36 = 81 => (x-5)^2 = 45 => x-5 = sqrt(45) or x-5 = -sqrt(45). So, x = 5 + 3*sqrt(5) or x = 5 - 3*sqrt(5).So, coordinates of A are either (5 + 3√5, 6) or (5 - 3√5, 6). Now, let's find the angle theta between BC and AD.Vector BC is from B(0,0) to C(10,0), so it's (10,0). Vector AD is from A(x,6) to D(5,0), so it's (5 - x, -6).Wait, actually, theta is the angle between BC and AD. So, vectors BC is (10,0) and AD is from A to D, which is (5 - x, -6). Alternatively, if we consider the angle at D, but actually, the angle between BC and AD is at point D, right? Wait, no, theta is the angle formed by side BC and median AD, so it's at point B or point C? Wait, no, the side BC is a side, and the median is from A to BC, so the angle is at point D? Wait, no, the angle between side BC and median AD is at point B or point C?Wait, maybe I'm overcomplicating. Let me think. The side BC is along the x-axis from (0,0) to (10,0). The median AD goes from A(x,6) to D(5,0). So, the angle between BC and AD is the angle between vector BC (10,0) and vector AD (5 - x, -6). So, we can compute the angle between these two vectors.But actually, theta is the acute angle between BC and AD, so regardless of the direction, we can compute the angle between the vectors BC and AD.Wait, vector BC is (10,0), and vector AD is (5 - x, -6). So, the angle between them can be found using the dot product formula:cos(theta) = (BC · AD) / (|BC| |AD|)Compute BC · AD: (10,0) · (5 - x, -6) = 10*(5 - x) + 0*(-6) = 50 - 10x|BC| is 10, |AD| is 9.So, cos(theta) = (50 - 10x)/(10*9) = (50 - 10x)/90 = (5 - x)/9But we know from earlier that x = 5 ± 3√5. So, let's plug that in.Case 1: x = 5 + 3√5cos(theta) = (5 - (5 + 3√5))/9 = (-3√5)/9 = (-√5)/3Case 2: x = 5 - 3√5cos(theta) = (5 - (5 - 3√5))/9 = (3√5)/9 = (√5)/3Since theta is the acute angle, we take the positive value, so cos(theta) = √5 / 3. Therefore, sin(theta) can be found using sin^2(theta) + cos^2(theta) = 1.So, sin(theta) = sqrt(1 - (√5 / 3)^2) = sqrt(1 - 5/9) = sqrt(4/9) = 2/3.So, sin(theta) is 2/3, which is option D. That matches my earlier result, so I think that's correct.Alternatively, maybe I can use the formula for the area in terms of the median. Wait, is there a formula that relates the area, side, median, and angle?I think the formula I used earlier is the most straightforward: area of triangle ABD is 15, which is (1/2)*5*9*sin(theta). So, 15 = (45/2)*sin(theta), so sin(theta) = 2/3. Yep, same answer.I think that's solid. So, I'm confident the answer is D, 2/3.**Final Answer**boxed{D}

answer：Alright, so I have this problem here: I need to find the number of integers ( x ) for which the value of ( frac{-6}{x+1} ) is an integer. The options are (A) 8, (B) 9, (C) 2, (D) 6, (E) 7. Hmm, okay, let me try to figure this out step by step.First, let me understand what the problem is asking. It says that ( frac{-6}{x+1} ) should be an integer. So, that means when I divide -6 by ( x+1 ), the result is a whole number, not a fraction or a decimal. So, ( x+1 ) must be a divisor of -6. Got it.So, if ( x+1 ) divides -6, then ( x+1 ) must be one of the integers that can multiply by another integer to get -6. In other words, ( x+1 ) is a divisor of -6. Therefore, I need to find all the integer divisors of -6 and then subtract 1 from each to get the corresponding ( x ) values.Wait, but hold on, divisors can be positive or negative, right? So, both positive and negative divisors of -6 will work here. Let me list all the divisors of -6.The divisors of 6 are 1, 2, 3, 6, and their negatives: -1, -2, -3, -6. So, for -6, the divisors are the same because divisors can be positive or negative. So, the divisors of -6 are: -6, -3, -2, -1, 1, 2, 3, 6. That's 8 divisors in total.So, each of these divisors is a possible value for ( x+1 ). Therefore, ( x+1 ) can be -6, -3, -2, -1, 1, 2, 3, or 6. So, to find ( x ), I need to subtract 1 from each of these.Let me compute each ( x ):1. If ( x+1 = -6 ), then ( x = -6 - 1 = -7 )2. If ( x+1 = -3 ), then ( x = -3 - 1 = -4 )3. If ( x+1 = -2 ), then ( x = -2 - 1 = -3 )4. If ( x+1 = -1 ), then ( x = -1 - 1 = -2 )5. If ( x+1 = 1 ), then ( x = 1 - 1 = 0 )6. If ( x+1 = 2 ), then ( x = 2 - 1 = 1 )7. If ( x+1 = 3 ), then ( x = 3 - 1 = 2 )8. If ( x+1 = 6 ), then ( x = 6 - 1 = 5 )So, the possible integer values for ( x ) are: -7, -4, -3, -2, 0, 1, 2, 5. Let me count them: that's 8 values. Hmm, so that would make the answer (A) 8, right?Wait, but let me double-check. Maybe I missed something. The problem is asking for integers ( x ), so I need to make sure that ( x+1 ) is not zero because division by zero is undefined. So, ( x+1 ) cannot be zero, which would mean ( x ) cannot be -1. But in my list above, I didn't include -1 because ( x+1 ) can't be zero. So, that's correct.But let me think again: are there only 8 divisors? Let me list them again:Divisors of -6 are: -6, -3, -2, -1, 1, 2, 3, 6. So, 8 divisors. Each gives a unique ( x ) when subtracting 1. So, 8 values of ( x ). So, 8 is the answer, which is option (A).Wait, but hold on, let me check if each of these ( x ) values actually gives an integer when plugged back into ( frac{-6}{x+1} ).Let me test each one:1. ( x = -7 ): ( frac{-6}{-7 + 1} = frac{-6}{-6} = 1 ) which is an integer.2. ( x = -4 ): ( frac{-6}{-4 + 1} = frac{-6}{-3} = 2 ) which is an integer.3. ( x = -3 ): ( frac{-6}{-3 + 1} = frac{-6}{-2} = 3 ) which is an integer.4. ( x = -2 ): ( frac{-6}{-2 + 1} = frac{-6}{-1} = 6 ) which is an integer.5. ( x = 0 ): ( frac{-6}{0 + 1} = frac{-6}{1} = -6 ) which is an integer.6. ( x = 1 ): ( frac{-6}{1 + 1} = frac{-6}{2} = -3 ) which is an integer.7. ( x = 2 ): ( frac{-6}{2 + 1} = frac{-6}{3} = -2 ) which is an integer.8. ( x = 5 ): ( frac{-6}{5 + 1} = frac{-6}{6} = -1 ) which is an integer.Okay, so all of them work. So, that's 8 integers. So, why is option (E) 7? Maybe I made a mistake somewhere.Wait, let me check if I considered all possible divisors. The divisors of -6 are indeed -6, -3, -2, -1, 1, 2, 3, 6. So, 8 divisors. So, 8 x's. So, 8 is the answer, which is option (A). So, why is (E) 7? Maybe I miscounted.Wait, let me recount the x's:-7, -4, -3, -2, 0, 1, 2, 5. That's 8 numbers. So, 8 is correct.Alternatively, maybe the problem is considering only positive divisors? But no, because -6 is negative, so negative divisors are also valid.Wait, another thought: maybe when ( x+1 = 1 ), ( x = 0 ), but is 0 considered? Yes, because ( x ) is an integer, so 0 is allowed.Wait, but let me think about the definition of divisors. When we talk about divisors, negative numbers are included, so both positive and negative divisors are considered. So, that's 8.Alternatively, maybe the question is considering only positive integers ( x )? But the problem says "the number of integers ( x )", so it should include both positive and negative.Wait, let me check the options again: (A) 8, (B)9, (C)2, (D)6, (E)7. So, 8 is an option, so I think 8 is correct.Wait, but wait, hold on, another thought: is 0 considered a divisor? No, because division by zero is undefined, but in this case, ( x+1 ) can't be zero, so ( x ) can't be -1. But in our list, we didn't include -1, so that's fine.So, yeah, I think 8 is correct. So, the answer is (A) 8.Wait, but hold on, let me think again. Maybe I made a mistake in counting the divisors. So, the number of divisors of a number is calculated by its prime factorization. So, for 6, the prime factors are 2 and 3, so the number of positive divisors is (1+1)(1+1) = 4. So, positive divisors are 1,2,3,6. Negative divisors would be -1,-2,-3,-6. So, total 8 divisors. So, that's correct.Therefore, 8 is the correct number of integers ( x ). So, the answer is (A) 8.Wait, but in the initial problem, the options are (A)8, (B)9, (C)2, (D)6, (E)7. So, 8 is an option, so I think that's correct.Wait, but hold on, another thought: is ( x+1 ) allowed to be 1? Because ( x = 0 ) is allowed, right? So, yeah, 0 is an integer, so that's fine.Wait, let me think about whether I considered all possible cases. So, ( x+1 ) must divide -6, so ( x+1 ) can be any integer that is a divisor of -6. So, the divisors are as I listed: -6, -3, -2, -1, 1, 2, 3, 6. So, 8 divisors, so 8 x's. So, 8 is correct.Therefore, I think the answer is (A) 8.But wait, hold on, the options are (A)8, (B)9, (C)2, (D)6, (E)7. So, 8 is an option, so I think that's correct.Wait, but let me think again: is ( x+1 ) allowed to be 0? No, because division by zero is undefined, so ( x+1 ) cannot be 0, so ( x ) cannot be -1. So, in our list, we didn't include ( x = -1 ), so that's correct.So, all right, I think I've considered all the possibilities, and the answer is 8, which is option (A).**Final Answer**boxed{A}

answer：Alright, so I have this problem where I need to find the values of ( p ) and ( q ) for the parabola ( y = x^2 + p x + q ). The parabola is tangent to the line ( y = 2x - 3 ) at the point ( M(2, 1) ). Hmm, okay, let me try to figure this out step by step.First, I know that if the parabola is tangent to the line at point ( M(2, 1) ), that means two things: 1. The point ( M(2, 1) ) lies on both the parabola and the line.2. The slope of the parabola at point ( M ) is equal to the slope of the line.So, let me start by verifying that ( M(2, 1) ) is indeed on both the parabola and the line.For the line ( y = 2x - 3 ):If I plug in ( x = 2 ), then ( y = 2*2 - 3 = 4 - 3 = 1 ). So, yes, ( M(2, 1) ) is on the line.For the parabola ( y = x^2 + p x + q ):Plugging in ( x = 2 ) and ( y = 1 ), we get:( 1 = (2)^2 + p*(2) + q )Simplify that:( 1 = 4 + 2p + q )So, ( 2p + q = 1 - 4 = -3 )That's our first equation: ( 2p + q = -3 ) (Equation 1)Now, moving on to the second condition: the slopes must be equal at point ( M ).The slope of the line ( y = 2x - 3 ) is 2, as it's in the form ( y = mx + b ) where ( m ) is the slope.For the parabola ( y = x^2 + p x + q ), the derivative ( dy/dx ) gives the slope at any point ( x ). Let's compute that.( dy/dx = 2x + p )At point ( M(2, 1) ), ( x = 2 ), so the slope is:( 2*2 + p = 4 + p )Since the parabola is tangent to the line at this point, their slopes must be equal. Therefore:( 4 + p = 2 )Solving for ( p ):( p = 2 - 4 = -2 )Alright, so ( p = -2 ). Now, let's plug this back into Equation 1 to find ( q ).From Equation 1:( 2p + q = -3 )Substitute ( p = -2 ):( 2*(-2) + q = -3 )( -4 + q = -3 )Adding 4 to both sides:( q = -3 + 4 = 1 )So, ( q = 1 ).Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, plugging ( p = -2 ) and ( q = 1 ) into the parabola equation:( y = x^2 - 2x + 1 )At ( x = 2 ):( y = 4 - 4 + 1 = 1 ), which matches point ( M(2, 1) ). Good.Now, checking the derivative:( dy/dx = 2x - 2 )At ( x = 2 ):( dy/dx = 4 - 2 = 2 ), which matches the slope of the line. Perfect.So, both conditions are satisfied. Therefore, ( p = -2 ) and ( q = 1 ).But just to be thorough, let me also think about another approach. Since the line is tangent to the parabola, the system of equations should have exactly one solution at ( x = 2 ). So, setting the equations equal to each other:( x^2 + p x + q = 2x - 3 )Bring all terms to one side:( x^2 + (p - 2)x + (q + 3) = 0 )Since the line is tangent to the parabola, this quadratic equation should have exactly one solution, meaning the discriminant is zero.The discriminant ( D ) of ( ax^2 + bx + c = 0 ) is ( D = b^2 - 4ac ).Here, ( a = 1 ), ( b = (p - 2) ), and ( c = (q + 3) ).So, discriminant:( D = (p - 2)^2 - 4*1*(q + 3) = 0 )We already found ( p = -2 ) and ( q = 1 ), let's plug them in:( D = (-2 - 2)^2 - 4*(1 + 3) = (-4)^2 - 4*4 = 16 - 16 = 0 )Which confirms that the discriminant is zero, so the line is indeed tangent to the parabola. That's a good check.Alternatively, if I didn't know ( p ) and ( q ), I could set up the discriminant equation:( (p - 2)^2 - 4*(q + 3) = 0 ) (Equation 2)And from earlier, we had Equation 1: ( 2p + q = -3 )So, now we have a system of two equations:1. ( 2p + q = -3 )2. ( (p - 2)^2 - 4(q + 3) = 0 )Let me solve this system to double-check.From Equation 1: ( q = -3 - 2p )Substitute into Equation 2:( (p - 2)^2 - 4*(-3 - 2p + 3) = 0 )Simplify inside the brackets:( -3 - 2p + 3 = -2p )So, Equation 2 becomes:( (p - 2)^2 - 4*(-2p) = 0 )Simplify:( (p - 2)^2 + 8p = 0 )Expand ( (p - 2)^2 ):( p^2 - 4p + 4 + 8p = 0 )Combine like terms:( p^2 + 4p + 4 = 0 )Factor:( (p + 2)^2 = 0 )Thus, ( p = -2 )Then, substitute back into Equation 1:( 2*(-2) + q = -3 )( -4 + q = -3 )( q = 1 )So, same result. That's reassuring.Alternatively, another way to think about this is that since the parabola is tangent to the line at ( x = 2 ), ( x = 2 ) is a repeated root of the equation ( x^2 + p x + q = 2x - 3 ). So, the quadratic equation ( x^2 + (p - 2)x + (q + 3) = 0 ) has a double root at ( x = 2 ). Therefore, it can be written as ( (x - 2)^2 = 0 ).Expanding ( (x - 2)^2 ):( x^2 - 4x + 4 = 0 )Comparing this with ( x^2 + (p - 2)x + (q + 3) = 0 ), we can equate coefficients:1. Coefficient of ( x^2 ): 1 = 1 (already matches)2. Coefficient of ( x ): ( p - 2 = -4 ) => ( p = -2 )3. Constant term: ( q + 3 = 4 ) => ( q = 1 )Same result again. So, regardless of the method, I end up with ( p = -2 ) and ( q = 1 ).Just to visualize, the parabola ( y = x^2 - 2x + 1 ) can be rewritten as ( y = (x - 1)^2 ). So, it's a parabola opening upwards with vertex at ( (1, 0) ). The line ( y = 2x - 3 ) is a straight line with slope 2. They touch at ( (2, 1) ), which makes sense because plugging ( x = 2 ) into the parabola gives ( y = (2 - 1)^2 = 1 ), and the line at ( x = 2 ) is also 1. The slope of the parabola at that point is 2, matching the line's slope, so they are indeed tangent there.I think I've covered all the bases here. I used multiple methods: plugging in the point, equating slopes, using the discriminant, and factoring the quadratic. All approaches led me to the same values for ( p ) and ( q ). I don't see any mistakes in my reasoning, so I'm confident that ( p = -2 ) and ( q = 1 ) are the correct answers.**Final Answer**The values of ( p ) and ( q ) are (boxed{-2}) and (boxed{1}) respectively.

answer：Alright, let me try to figure out this problem step by step. So, we have a bag with red and yellow balls. When 60 balls are taken out, exactly 56 are red. Then, every time 18 balls are taken out, 14 of them are red, until the last batch of 18 balls is taken out. Also, the total number of red balls is four-fifths of the total number of balls. We need to find how many red balls are in the bag.First, let me parse the problem. There are two stages of taking balls out: the first time, 60 balls are taken, with 56 red. Then, repeatedly, 18 balls are taken out, each time with 14 red, until the last batch of 18 is taken. So, the total number of balls taken out is 60 plus some multiple of 18. Also, the total red balls are four-fifths of the total balls in the bag. So, let me denote the total number of balls as N, and the total number of red balls as R. Then, R = (4/5)N.Now, let's think about the process of taking out the balls. The first time, 60 balls are taken, 56 red and 4 yellow. Then, each subsequent time, 18 balls are taken, 14 red and 4 yellow. So, each time after the first, the ratio of red to yellow taken out is 14:4, which simplifies to 7:2. So, the proportion of red balls being taken out is 14/18 = 7/9, and yellow is 4/18 = 2/9.Wait, but the problem says that the total number of red balls is four-fifths of the total. So, maybe the ratio of red to yellow in the bag is 4:1? Because four-fifths red and one-fifth yellow.Hmm, but when we take out the first 60 balls, we have 56 red and 4 yellow, which is a ratio of 14:1. Then, each subsequent 18 balls have 14 red and 4 yellow, which is 7:2. So, the ratio of red to yellow is changing as we take out balls.Wait, so maybe the composition of the bag changes as we take out balls? Or is the bag being taken out without replacement? I think it's without replacement because we're taking balls out of the bag, so the composition changes each time.So, perhaps we can model this as a hypergeometric process, but it might get complicated. Alternatively, maybe we can think in terms of proportions.Let me try to denote the total number of red balls as R and yellow balls as Y. So, R = (4/5)N and Y = (1/5)N, where N = R + Y.Now, when we take out 60 balls, 56 are red and 4 are yellow. So, the number of red balls remaining is R - 56, and yellow balls remaining is Y - 4.Then, each subsequent time, we take out 18 balls, with 14 red and 4 yellow. So, each time, we remove 14 red and 4 yellow. But this process continues until the last batch of 18 balls is taken out.Wait, so how many times do we take out 18 balls after the first 60? Let me denote the number of times we take out 18 balls as k. So, total balls taken out are 60 + 18k. But the total number of balls in the bag is N, so 60 + 18k = N. So, N must be equal to 60 + 18k, where k is an integer.But also, the total number of red balls is R = 56 + 14k, because each time we take out 14 red balls, and we do this k times. Similarly, the total number of yellow balls is Y = 4 + 4k.But we also know that R = (4/5)N and Y = (1/5)N.So, substituting R and Y:R = 56 + 14k = (4/5)NY = 4 + 4k = (1/5)NBut N = R + Y = (56 + 14k) + (4 + 4k) = 60 + 18kWhich matches the earlier equation N = 60 + 18k.So, now, let's write the equations:From R: 56 + 14k = (4/5)(60 + 18k)From Y: 4 + 4k = (1/5)(60 + 18k)Let me check both equations to see if they give the same k.First, let's solve the Y equation:4 + 4k = (1/5)(60 + 18k)Multiply both sides by 5:20 + 20k = 60 + 18kSubtract 18k and subtract 20:2k = 40So, k = 20.Now, let's check the R equation:56 + 14k = (4/5)(60 + 18k)Substitute k = 20:Left side: 56 + 14*20 = 56 + 280 = 336Right side: (4/5)(60 + 18*20) = (4/5)(60 + 360) = (4/5)(420) = 336Yes, both sides equal 336, so k = 20 is correct.Therefore, total number of red balls R = 56 + 14*20 = 56 + 280 = 336.Total number of balls N = 60 + 18*20 = 60 + 360 = 420.Let me verify that R is indeed four-fifths of N:(4/5)*420 = 336, which matches R.Also, Y = 4 + 4*20 = 4 + 80 = 84, and (1/5)*420 = 84, which also matches.So, the total number of red balls is 336.Wait, but let me think again. Is there another way this could be interpreted? For example, does the last batch of 18 balls have exactly 14 red and 4 yellow, or could it be different?The problem says "until the last batch of 18 balls is taken out." So, it implies that the last batch is also 18 balls, so the total number of balls must be a multiple of 18 after the initial 60. So, N = 60 + 18k, which we have already considered.Therefore, I think the solution is correct.**Final Answer**The number of red balls in the bag is boxed{336}.

answer：Okay, so I have this function f(x) = 2√3 sin x cos x + 2 cos²x - 1, and I need to find its smallest positive period and then calculate f(π/3). Hmm, let me start by understanding what the function is composed of.First, looking at the function, I see it's made up of sine and cosine terms. The first term is 2√3 sin x cos x, which reminds me of the double-angle identity for sine. I remember that sin(2x) = 2 sin x cos x, so maybe I can rewrite that term using that identity.Let me try that. If I factor out the 2√3, I can write 2√3 sin x cos x as √3 * 2 sin x cos x, which is √3 sin(2x). So that term becomes √3 sin(2x).Next, the second term is 2 cos²x. I recall that there's a double-angle identity for cosine squared as well. The identity is cos²x = (1 + cos(2x))/2. So if I substitute that in, 2 cos²x becomes 2*(1 + cos(2x))/2, which simplifies to 1 + cos(2x).So putting it all together, the function f(x) can be rewritten as:f(x) = √3 sin(2x) + 1 + cos(2x) - 1.Wait, the +1 and -1 cancel each other out, so we're left with:f(x) = √3 sin(2x) + cos(2x).Okay, that's simpler. Now, this looks like a combination of sine and cosine functions with the same argument, 2x. I remember that such expressions can be combined into a single sine or cosine function using the amplitude-phase form. The general formula is A sin(θ) + B cos(θ) = C sin(θ + φ), where C = √(A² + B²) and tan φ = B/A.Let me apply that here. Here, A is √3 and B is 1. So, C would be √[(√3)² + 1²] = √(3 + 1) = √4 = 2. That's nice, so the amplitude is 2.Now, to find the phase shift φ, we have tan φ = B/A = 1/√3. I know that tan(π/6) = 1/√3, so φ = π/6.Therefore, f(x) can be rewritten as:f(x) = 2 sin(2x + π/6).Wait, hold on. Let me make sure. The formula is A sin θ + B cos θ = C sin(θ + φ). So, plugging in, √3 sin(2x) + 1 cos(2x) = 2 sin(2x + π/6). Let me verify that.Using the sine addition formula: sin(a + b) = sin a cos b + cos a sin b. So, 2 sin(2x + π/6) = 2 [sin(2x) cos(π/6) + cos(2x) sin(π/6)].Compute cos(π/6) and sin(π/6): cos(π/6) is √3/2 and sin(π/6) is 1/2.So, substituting back, we get:2 [sin(2x)*(√3/2) + cos(2x)*(1/2)] = 2*(√3/2 sin(2x) + 1/2 cos(2x)) = √3 sin(2x) + cos(2x).Yes, that matches our original expression. So, f(x) simplifies to 2 sin(2x + π/6).Now, for the first part, finding the smallest positive period of f(x). Since f(x) is expressed as a sine function with argument 2x + π/6, the period of sin(kx + c) is 2π / |k|. Here, k is 2, so the period is 2π / 2 = π.Therefore, the smallest positive period of f(x) is π.Moving on to the second part, calculating f(π/3). Let me substitute x = π/3 into the original function or the simplified version. Maybe the simplified version is easier.Using f(x) = 2 sin(2x + π/6), plug in x = π/3:f(π/3) = 2 sin(2*(π/3) + π/6) = 2 sin(2π/3 + π/6).Let me compute the angle inside the sine function. 2π/3 is equal to 4π/6, so adding π/6 gives 5π/6.So, f(π/3) = 2 sin(5π/6).I know that sin(5π/6) is sin(π - π/6) = sin(π/6) = 1/2. But wait, sin(5π/6) is actually 1/2 because it's in the second quadrant where sine is positive.Therefore, f(π/3) = 2*(1/2) = 1.Alternatively, if I use the original function, let me verify:f(π/3) = 2√3 sin(π/3) cos(π/3) + 2 cos²(π/3) - 1.Compute each term:sin(π/3) = √3/2, cos(π/3) = 1/2.First term: 2√3*(√3/2)*(1/2) = 2√3*(√3/4) = 2*(3)/4 = 6/4 = 3/2.Second term: 2*(1/2)^2 = 2*(1/4) = 1/2.Third term: -1.So, adding them up: 3/2 + 1/2 - 1 = (3/2 + 1/2) - 1 = 2 - 1 = 1.Same result. So, f(π/3) is indeed 1.Wait, hold on, in the original function, when I computed the first term, let me double-check:2√3 sin(π/3) cos(π/3) = 2√3*(√3/2)*(1/2) = 2√3*(√3)/4 = (2*3)/4 = 6/4 = 3/2. That's correct.Second term: 2 cos²(π/3) = 2*(1/2)^2 = 2*(1/4) = 1/2.Third term: -1.So, 3/2 + 1/2 = 2, minus 1 is 1. Correct.So, both methods give the same result, which is reassuring.Therefore, the smallest positive period is π, and f(π/3) is 1.**Final Answer**1. The smallest positive period of the function is boxed{pi}.2. The value of ( fleft( frac{pi}{3} right) ) is boxed{1}.

answer：Okay, so I have this problem where I need to find the limit as x approaches 1 of p(x) over q(x). Both p(x) and q(x) are degree 4 polynomials with leading coefficient 1. The roots of p(x) are 1, 2, 3, and 4, while the roots of q(x) are 1, 1/2, 1/3, and 1/4. First, I should probably write down what p(x) and q(x) look like. Since they're both degree 4 polynomials with leading coefficient 1, they can be expressed in their factored forms. For p(x), since the roots are 1, 2, 3, and 4, it should be:p(x) = (x - 1)(x - 2)(x - 3)(x - 4)Similarly, for q(x), the roots are 1, 1/2, 1/3, and 1/4, so:q(x) = (x - 1)(x - 1/2)(x - 1/3)(x - 1/4)So, the limit we need to find is:lim(x→1) [p(x)/q(x)] = lim(x→1) [(x - 1)(x - 2)(x - 3)(x - 4) / (x - 1)(x - 1/2)(x - 1/3)(x - 1/4)]Hmm, I notice that both the numerator and the denominator have a factor of (x - 1). So, as x approaches 1, both the numerator and denominator approach 0, which means we might have an indeterminate form of 0/0. In such cases, L’Hôpital’s Rule can be applied, which involves taking derivatives of the numerator and denominator until the limit can be evaluated.But before jumping into taking derivatives, maybe I can simplify the expression first. Since both have a (x - 1) term, I can cancel them out:lim(x→1) [(x - 2)(x - 3)(x - 4) / (x - 1/2)(x - 1/3)(x - 1/4)]Now, the expression is simplified, but as x approaches 1, the numerator becomes (1 - 2)(1 - 3)(1 - 4) and the denominator becomes (1 - 1/2)(1 - 1/3)(1 - 1/4). Let me compute these values.First, compute the numerator:(1 - 2) = -1(1 - 3) = -2(1 - 4) = -3Multiplying these together: (-1)(-2)(-3) = (-1)*(-2) = 2; 2*(-3) = -6Denominator:(1 - 1/2) = 1/2(1 - 1/3) = 2/3(1 - 1/4) = 3/4Multiplying these together: (1/2)*(2/3)*(3/4)Let me compute that step by step:(1/2)*(2/3) = (1*2)/(2*3) = 2/6 = 1/3Then, (1/3)*(3/4) = (1*3)/(3*4) = 3/12 = 1/4So, the denominator is 1/4.So, putting it together, the limit is (-6)/(1/4) = -6 * 4 = -24.Wait, is that right? Let me double-check my calculations.Numerator:(1 - 2) = -1(1 - 3) = -2(1 - 4) = -3Multiply: (-1)*(-2) = 2; 2*(-3) = -6. That seems correct.Denominator:(1 - 1/2) = 1/2(1 - 1/3) = 2/3(1 - 1/4) = 3/4Multiply: (1/2)*(2/3) = (1/2)*(2/3) = (1*2)/(2*3) = 2/6 = 1/3Then, (1/3)*(3/4) = (1*3)/(3*4) = 3/12 = 1/4. That also seems correct.So, the limit is (-6)/(1/4) = -6 * 4 = -24.But wait, is that the correct approach? Because when I canceled out the (x - 1) terms, I assumed that the limit as x approaches 1 would just be the value of the remaining expression at x=1. But is that always valid?Yes, because after canceling the common factor, the resulting function is continuous at x=1, so the limit is just the value of the function at x=1.Alternatively, if I didn't cancel the terms, I could have applied L’Hôpital’s Rule since both numerator and denominator approach 0 as x approaches 1.Let me try that approach as a check.So, if I don't cancel (x - 1), then both p(x) and q(x) have a root at x=1, so p(1)=0 and q(1)=0, meaning the limit is 0/0, which is indeterminate.Therefore, applying L’Hôpital’s Rule, we take the derivatives of the numerator and denominator.So, first, compute p'(x) and q'(x).But computing the derivatives of these polynomials might be a bit tedious, but let's see.Alternatively, since both p(x) and q(x) have a factor of (x - 1), we can write p(x) = (x - 1)*p1(x) and q(x) = (x - 1)*q1(x), where p1(x) and q1(x) are the remaining cubic polynomials.Therefore, p(x)/q(x) = (x - 1)*p1(x) / [(x - 1)*q1(x)] = p1(x)/q1(x) for x ≠ 1. So, the limit as x approaches 1 is p1(1)/q1(1).Which is exactly what I did earlier by canceling (x - 1). So, that's consistent.Therefore, computing p1(1) and q1(1) is equivalent to computing the limit.So, p1(x) is (x - 2)(x - 3)(x - 4). So, p1(1) = (1 - 2)(1 - 3)(1 - 4) = (-1)(-2)(-3) = -6.Similarly, q1(x) is (x - 1/2)(x - 1/3)(x - 1/4). So, q1(1) = (1 - 1/2)(1 - 1/3)(1 - 1/4) = (1/2)(2/3)(3/4) = 1/4.Thus, the limit is p1(1)/q1(1) = (-6)/(1/4) = -24.So, that's consistent with my earlier result.Alternatively, if I were to apply L’Hôpital’s Rule, I would compute the derivatives p'(x) and q'(x), evaluate them at x=1, and then take the ratio.Let me try that as another check.Compute p'(x):p(x) = (x - 1)(x - 2)(x - 3)(x - 4)To find p'(x), we can use the product rule, but that might be a bit involved. Alternatively, since we know p(x) is a polynomial, we can compute its derivative by expanding it, but that might take some time.Alternatively, since we are only interested in p'(1), maybe we can use the fact that p(x) = (x - 1)*p1(x), so p'(x) = p1(x) + (x - 1)*p1'(x). Then, evaluating at x=1, p'(1) = p1(1) + (1 - 1)*p1'(1) = p1(1) = -6.Similarly, q'(x) = q1(x) + (x - 1)*q1'(x), so q'(1) = q1(1) = 1/4.Therefore, applying L’Hôpital’s Rule once, the limit becomes p'(1)/q'(1) = (-6)/(1/4) = -24.So, same result.Alternatively, if I didn't think of factoring, I could have expanded both p(x) and q(x), then divided them, and then taken the limit as x approaches 1. But that seems more complicated.Alternatively, maybe using the concept of multiplicity of roots or something else, but I think the approaches I've taken are sufficient.So, in conclusion, the limit is -24.**Final Answer**boxed{-24}