answer：First, I need to define the variables for the problem. Let’s let ( w ) represent the width of the garden in feet. Since the length is "Si" feet longer than the width, and "Si" means four in Chinese, the length can be expressed as ( w + 4 ) feet.Next, I'll set up the equation for the area of the garden. The area of a rectangle is calculated by multiplying its length by its width. Given that the area is 60 square feet, the equation becomes:[w(w + 4) = 60]Expanding this equation, I get:[w^2 + 4w = 60]To form a standard quadratic equation, I'll subtract 60 from both sides:[w^2 + 4w - 60 = 0]Now, I'll solve this quadratic equation using the quadratic formula:[w = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Here, ( a = 1 ), ( b = 4 ), and ( c = -60 ). Plugging these values into the formula:[w = frac{-4 pm sqrt{16 + 240}}{2} = frac{-4 pm sqrt{256}}{2} = frac{-4 pm 16}{2}]This gives two potential solutions:[w = frac{12}{2} = 6 quad text{and} quad w = frac{-20}{2} = -10]Since a negative width doesn't make sense in this context, the width of the garden is 6 feet.Finally, I'll determine the length by adding 4 feet to the width:[text{Length} = 6 + 4 = 10 text{ feet}]To verify, I'll calculate the area:[6 times 10 = 60 text{ square feet}]This confirms that the dimensions are correct.

answer：Okay, so I have this problem about a rectangular sign that's 9 meters by 16 meters. They want to paint a square advertisement on it, but there has to be a border around the square that's at least 1.5 meters wide. I need to find the area of the largest square that can fit on the sign with that border. The options are given, so maybe I can figure it out step by step.First, let me visualize the sign. It's a rectangle, longer in one side—16 meters—and shorter on the other—9 meters. So, it's like a landscape-oriented rectangle. Now, they want to put a square advertisement on it with a border all around. The border has to be at least 1.5 meters wide. That means on each side of the square, there's 1.5 meters of space left on the sign.Hmm, so the square can't be too big because it needs that border. I need to figure out the maximum size of the square that can fit within the rectangle while leaving at least 1.5 meters on all sides.Let me think about the dimensions. The sign is 9 meters tall and 16 meters wide. If I place a square on it, the square will have the same height and width. But since the sign is longer in width, maybe the square is limited by the shorter side? Wait, but the border is on all sides, so both the height and width of the sign will affect the size of the square.Let me denote the side length of the square as 's'. Since the border is at least 1.5 meters wide on all sides, the square can't extend beyond that border. So, on the height side, the square plus the border on top and bottom can't exceed 9 meters. Similarly, on the width side, the square plus the border on the left and right can't exceed 16 meters.So, for the height: the square's height plus two times the border (top and bottom) must be less than or equal to 9 meters. That gives me the equation:s + 2*(1.5) ≤ 9Similarly, for the width: the square's width plus two times the border (left and right) must be less than or equal to 16 meters. So,s + 2*(1.5) ≤ 16Wait, but since the square has the same height and width, both these conditions must be satisfied. So, I can write both inequalities:1. s + 3 ≤ 92. s + 3 ≤ 16Let me solve the first inequality:s + 3 ≤ 9Subtract 3 from both sides:s ≤ 6And the second inequality:s + 3 ≤ 16Subtract 3:s ≤ 13So, from the first inequality, s must be less than or equal to 6 meters, and from the second, it's less than or equal to 13 meters. But since both conditions must be satisfied, the stricter condition is s ≤ 6 meters.Therefore, the largest possible square that can fit on the sign with a 1.5-meter border on all sides has a side length of 6 meters. Then, the area would be 6 meters times 6 meters, which is 36 square meters.Wait, but hold on. Let me double-check because 6 meters seems a bit small, and one of the options is 144, which is 12 squared. Maybe I made a mistake.Let me think again. If the border is at least 1.5 meters, does that mean the border can be more? Or is it exactly 1.5 meters? The problem says "at least," so the border can be more, but the square can be smaller. But we want the largest possible square, so we need the border to be exactly 1.5 meters on all sides. Otherwise, if the border is more, the square would be smaller.So, if the border is exactly 1.5 meters, then the square's side length is 9 - 2*(1.5) = 9 - 3 = 6 meters on the height side, and 16 - 2*(1.5) = 16 - 3 = 13 meters on the width side. Wait, that doesn't make sense because the square can't have different side lengths.Ah, right! The square has to have the same side length on both height and width. So, even though the width allows for a larger square, the height restricts it. So, the maximum square we can have is limited by the shorter side after subtracting the borders.So, on the height side, after subtracting 3 meters (1.5 on top and bottom), we have 6 meters left. On the width side, after subtracting 3 meters (1.5 on each side), we have 13 meters left. But since it's a square, both sides have to be equal. So, the square can't be larger than 6 meters on either side because otherwise, it would exceed the height.Therefore, the maximum square is 6 meters by 6 meters, which gives an area of 36 square meters. That's option C.Wait, but let me think again. Maybe I can place the square differently? Like, rotated or something? But the problem says it's a square advertisement, so I think it has to be axis-aligned with the sign. Otherwise, if it's rotated, the required border might be less, but I don't think that's the case here.Alternatively, maybe the border doesn't have to be 1.5 meters on all sides? But the problem says "the border around the square must be at least 1.5 meters wide." So, that should mean on all sides, the border is at least 1.5 meters. So, I think my initial reasoning is correct.So, if the square is 6 meters on each side, the border on the top and bottom would be 1.5 meters each, and on the sides, it would also be 1.5 meters each. So, that fits within the 9 by 16 meter sign.Wait, let me confirm the math:Height of the sign: 9 meters.Border on top and bottom: 1.5 + 1.5 = 3 meters.So, the square's height: 9 - 3 = 6 meters.Width of the sign: 16 meters.Border on left and right: 1.5 + 1.5 = 3 meters.Square's width: 16 - 3 = 13 meters.But since it's a square, the side length has to be the smaller of the two, which is 6 meters.Therefore, the area is 6*6=36 m², which is option C.But wait, another thought: is the border required to be at least 1.5 meters, meaning it could be more? So, if we make the square smaller, the border would be more than 1.5 meters. But since we want the largest square, we need the border to be exactly 1.5 meters on all sides. So, that's why we subtract 3 meters from both height and width.Alternatively, if we didn't make the border exactly 1.5 meters, but more, the square could be smaller, but that's not what we want. We want the largest square, so we need the minimal border, which is 1.5 meters.So, I think 6 meters is correct, so 36 m² is the answer.But hold on, the options include 144 m², which is 12x12. Maybe I miscalculated something.Wait, 144 is 12 squared. Let me see: if the square is 12 meters on each side, then the border on the height side would be (9 - 12)/2, which is negative. That doesn't make sense. So, that can't be.Alternatively, if the square is 12 meters on the width, but then the height would require a border of (9 - 12)/2, which is negative. So, that's impossible.Wait, maybe I'm confusing the sides. Let me think again.If the square is placed on the sign, the height of the sign is 9 meters, so the square can't exceed 9 meters in height. Similarly, the width is 16 meters, so the square can't exceed 16 meters in width. But the square has equal sides, so the maximum possible square is 9 meters on each side, but wait, that would leave no border on the height side, which is not allowed because the border must be at least 1.5 meters.So, if the square is 9 meters, the border on the height side would be zero, which is less than 1.5 meters. So, that's not acceptable.Therefore, the square must be smaller. So, subtracting 3 meters from the height gives 6 meters, as I did before.Wait, but if I subtract 3 meters from the width, I get 13 meters, which is more than 6 meters. So, the square is limited by the height.So, 6 meters is correct.But let me think about another approach. Maybe the square can be placed such that the border is 1.5 meters on the top and bottom, but more on the sides? But no, the problem says the border must be at least 1.5 meters wide. So, it can be more, but not less. So, if I make the border wider on the sides, the square can be smaller, but we want the largest square, so we need to minimize the border.Therefore, the minimal border is 1.5 meters on all sides, so the square is 6x6.Wait, but maybe the square can be placed in such a way that the border is 1.5 meters on the top and bottom, but on the sides, it's more. But then, the square can be larger in width? Wait, no, because the square has equal sides. So, if the square is 6 meters in height, it's also 6 meters in width, leaving 16 - 6 = 10 meters on the width side, which is split equally on both sides as 5 meters each. But 5 meters is more than 1.5 meters, which is acceptable because the border just needs to be at least 1.5 meters.Wait, so in that case, the square is 6 meters in height, which requires a 1.5-meter border on top and bottom, but on the sides, the border is 5 meters each, which is more than 1.5 meters. So, that's acceptable.Alternatively, if I try to make the square larger, say 7 meters on each side, then the border on the height side would be (9 - 7)/2 = 1 meter on top and bottom, which is less than 1.5 meters. That's not acceptable because the border must be at least 1.5 meters. So, 7 meters is too big.Similarly, 6 meters is the maximum where the border on the height side is exactly 1.5 meters, and on the width side, it's 5 meters, which is more than enough.Therefore, 6 meters is correct, and the area is 36 m², which is option C.But wait, another thought: is the border required to be 1.5 meters on all four sides, or is it 1.5 meters in total? The problem says "a border around the square must be at least 1.5 meters wide." So, I think it means that the border is at least 1.5 meters on each side. So, that's what I considered earlier.Alternatively, if it meant that the total border width is 1.5 meters, that would be different, but that's not how it's worded. It says "a border around the square must be at least 1.5 meters wide," which I interpret as each side of the border must be at least 1.5 meters.So, given that, my calculation of 6 meters is correct.Wait, but let me check the options again. The options are:(A) 78 m²(B) 144 m²(C) 36 m²(D) 9 m²(E) 56.25 m²So, 36 is an option, but 56.25 is also there. Maybe I made a mistake in interpreting the border.Wait, 56.25 is 7.5 squared. Maybe if the border is 1.5 meters on the top and bottom, but not on the sides? But the problem says "around the square," which implies all sides.Alternatively, maybe the border is 1.5 meters in total around the square, meaning the sum of all borders is 1.5 meters. But that doesn't make much sense because the border is around the square, so it's on all four sides.Alternatively, maybe the border is 1.5 meters on the top and bottom, but not on the sides? But the problem says "around the square," which should include all sides.Wait, perhaps the border is 1.5 meters on the top and bottom, and the sides can have a different border? But the problem says "a border around the square must be at least 1.5 meters wide," which is a bit ambiguous. It could mean that the border is uniformly 1.5 meters on all sides, or that the border is at least 1.5 meters on each side.But in any case, the minimal border is 1.5 meters on each side, so the square is limited by the shorter side.Wait, another approach: maybe the border is 1.5 meters on the top and bottom, but on the sides, it can be more, but not less. So, if we make the square as large as possible on the width side, but still have at least 1.5 meters on the top and bottom.Wait, let me think about this.If I ignore the side borders for a moment, and just ensure that the top and bottom borders are at least 1.5 meters, then the square can be as wide as possible, limited by the height.So, the height of the sign is 9 meters. If the top and bottom borders are each 1.5 meters, then the square's height is 9 - 2*1.5 = 6 meters. So, the square can be 6 meters tall, and as wide as possible. But since it's a square, the width must also be 6 meters. So, that's the same as before.Alternatively, if I don't care about the side borders, but the problem says "a border around the square," which should include all sides. So, the side borders must also be at least 1.5 meters.Therefore, the square is limited by both the height and the width.Wait, but let me think about the width. If the square is 6 meters wide, then the side borders are (16 - 6)/2 = 5 meters, which is more than 1.5 meters, so that's acceptable.But if I try to make the square larger, say 7 meters, then the side borders would be (16 - 7)/2 = 4.5 meters, which is still more than 1.5 meters, but the top and bottom borders would be (9 - 7)/2 = 1 meter, which is less than 1.5 meters. So, that's not acceptable.Similarly, if I make the square 8 meters, the top and bottom borders would be (9 - 8)/2 = 0.5 meters, which is too small.So, the maximum square that allows the top and bottom borders to be at least 1.5 meters is 6 meters.Therefore, the area is 6*6=36 m², which is option C.Wait, but hold on, another thought. Maybe the border doesn't have to be uniform on all sides? Like, maybe the top border is 1.5 meters, the bottom border is 1.5 meters, but the side borders can be more. But the problem says "a border around the square must be at least 1.5 meters wide." So, I think that means all sides must have at least 1.5 meters. So, the minimal border on each side is 1.5 meters.Therefore, the square must be placed such that on all four sides, there's at least 1.5 meters of border. So, the square can't extend beyond 1.5 meters from any edge.Therefore, the maximum square is 9 - 2*1.5 = 6 meters in height, and 16 - 2*1.5 = 13 meters in width. But since it's a square, the side length is the minimum of these two, which is 6 meters.Therefore, the area is 36 m², which is option C.Wait, but in that case, why is 56.25 an option? 56.25 is 7.5 squared. Maybe if the border is only on the top and bottom, but not on the sides? Let me check.If the border is only on the top and bottom, each 1.5 meters, then the square's height is 9 - 3 = 6 meters, and the width can be up to 16 meters. But since it's a square, the width would have to be 6 meters as well, leaving 10 meters on the sides, which is more than 1.5 meters. So, that still gives 6 meters.Alternatively, if the border is only on the sides, each 1.5 meters, then the square's width is 16 - 3 = 13 meters, and the height would have to be 13 meters, but the sign is only 9 meters tall, so that's impossible.Alternatively, maybe the border is 1.5 meters on the top, bottom, left, and right, but not necessarily the same on all sides. Wait, no, the problem says "a border around the square must be at least 1.5 meters wide." So, each side must have at least 1.5 meters.Therefore, the square must be placed such that on all four sides, there's at least 1.5 meters. So, the maximum square is 6 meters on each side.Therefore, 36 m² is the answer, which is option C.But wait, let me think again because 56.25 is 7.5 squared, which is larger than 36. Maybe if the border is not uniform, but the problem says "a border around the square must be at least 1.5 meters wide." So, maybe the border can be more on some sides, but at least 1.5 on all.Wait, but if the square is 7.5 meters on each side, then the top and bottom borders would be (9 - 7.5)/2 = 0.75 meters, which is less than 1.5 meters. So, that's not acceptable.Similarly, if the square is 12 meters on each side, the top and bottom borders would be (9 - 12)/2 = negative, which is impossible.Therefore, 7.5 meters is too big because the top and bottom borders would be less than 1.5 meters.So, 6 meters is the maximum.Therefore, the area is 36 m², which is option C.Wait, but let me check the math again. 9 meters height, subtract 3 meters for the borders, gives 6 meters for the square. 16 meters width, subtract 3 meters for the borders, gives 13 meters for the square. But since it's a square, the side length is the minimum of 6 and 13, which is 6. So, 6x6=36.Yes, that seems correct.Alternatively, maybe the square can be placed such that the border is 1.5 meters on the top and bottom, but on the sides, it's more, but the square is as wide as possible. But since the square is limited by the height, it can't be wider than 6 meters.Wait, but if the square is 6 meters tall, it can be 6 meters wide, leaving 10 meters on the sides, which is more than 1.5 meters. So, that's acceptable.Therefore, 6 meters is correct.So, the answer is C, 36 m².**Final Answer**boxed{C}

answer：To solve the first part, I need to find the number of projects ( p ) where the annual costs of both software options are equal. The cost equations are ( C_1 = 500 + 100p ) for VFX Pro and ( C_2 = 150p ) for CinematicFX. Setting these equal to each other gives:[500 + 100p = 150p]Subtracting ( 100p ) from both sides:[500 = 50p]Dividing both sides by 50:[p = 10]So, the breakeven point is at 10 projects.For the second part, I need to determine the maximum number of projects each software can handle within a 2,000 budget.Starting with VFX Pro:[500 + 100p leq 2000]Subtracting 500:[100p leq 1500]Dividing by 100:[p leq 15]So, with VFX Pro, the company can handle up to 15 projects.For CinematicFX:[150p leq 2000]Dividing by 150:[p leq frac{2000}{150} approx 13.33]Since the number of projects must be a whole number, the company can handle up to 13 projects with CinematicFX.

answer：Alright, so I've got this problem about a company called Cymru Industries, and I need to figure out two things. First, I need to find the maximum stock price and when it happens. Second, I need to determine when the stock price will be at least 30 so I can decide when to invest. Let me take this step by step.Starting with the first part: finding the maximum stock price and the time it takes to reach that maximum. The stock price is modeled by the quadratic equation ( P(t) = -2t^2 + 12t + 20 ). Hmm, okay, quadratic equations have either a maximum or a minimum value depending on the coefficient of the ( t^2 ) term. Since the coefficient here is -2, which is negative, the parabola opens downward, meaning it has a maximum point. That makes sense because the stock price can't just keep increasing forever; it has to peak and then come back down.To find the maximum, I remember that for a quadratic equation in the form ( at^2 + bt + c ), the vertex occurs at ( t = -frac{b}{2a} ). So in this case, ( a = -2 ) and ( b = 12 ). Plugging those into the formula: ( t = -frac{12}{2*(-2)} ). Let me compute that. The denominator is 2 times -2, which is -4. So, ( t = -frac{12}{-4} ), which simplifies to ( t = 3 ). So, the maximum stock price occurs at 3 years.Now, to find the maximum stock price itself, I need to plug ( t = 3 ) back into the equation ( P(t) ). Let's do that:( P(3) = -2*(3)^2 + 12*(3) + 20 )Calculating each term step by step:First, ( (3)^2 = 9 ), so ( -2*9 = -18 ).Next, ( 12*3 = 36 ).So, putting it all together: ( -18 + 36 + 20 ).Adding those up: ( -18 + 36 = 18 ), then ( 18 + 20 = 38 ).So, the maximum stock price is 38, and it happens at 3 years. That seems straightforward.Moving on to the second part: I want to invest when the stock price is at least 30. So, I need to solve the inequality ( -2t^2 + 12t + 20 geq 30 ). Let me write that down:( -2t^2 + 12t + 20 geq 30 )To solve this, I should bring all terms to one side to form a standard quadratic inequality. Subtract 30 from both sides:( -2t^2 + 12t + 20 - 30 geq 0 )Simplify that:( -2t^2 + 12t - 10 geq 0 )Hmm, okay, so now I have ( -2t^2 + 12t - 10 geq 0 ). It might be easier to work with positive coefficients, so let me multiply both sides by -1. But wait, I remember that multiplying both sides of an inequality by a negative number reverses the inequality sign. So, doing that:( 2t^2 - 12t + 10 leq 0 )Alright, now I have a quadratic inequality ( 2t^2 - 12t + 10 leq 0 ). To find the values of t that satisfy this, I need to find the roots of the quadratic equation ( 2t^2 - 12t + 10 = 0 ) and then determine the intervals where the quadratic is less than or equal to zero.Let me try to factor this quadratic, but I don't think it factors nicely. Let me check the discriminant to see if it can be factored or if I need to use the quadratic formula. The discriminant ( D = b^2 - 4ac ). Here, ( a = 2 ), ( b = -12 ), ( c = 10 ).Calculating the discriminant:( D = (-12)^2 - 4*2*10 = 144 - 80 = 64 )Oh, 64 is a perfect square, so that means the quadratic can be factored or solved easily with the quadratic formula. Let's use the quadratic formula:( t = frac{-b pm sqrt{D}}{2a} )Plugging in the values:( t = frac{-(-12) pm sqrt{64}}{2*2} = frac{12 pm 8}{4} )So, two solutions:1. ( t = frac{12 + 8}{4} = frac{20}{4} = 5 )2. ( t = frac{12 - 8}{4} = frac{4}{4} = 1 )So, the roots are at t = 1 and t = 5. Since the quadratic ( 2t^2 - 12t + 10 ) opens upwards (because the coefficient of ( t^2 ) is positive), the quadratic will be below or equal to zero between its roots. So, the solution to the inequality ( 2t^2 - 12t + 10 leq 0 ) is ( 1 leq t leq 5 ).But wait, let me double-check that. If the quadratic opens upwards, it will be below zero between the two roots. So, yes, t between 1 and 5. That makes sense because the original quadratic equation ( P(t) = -2t^2 + 12t + 20 ) is a downward opening parabola, so it will be above 30 between t = 1 and t = 5.So, the stock price is at least 30 from year 1 to year 5. Therefore, I should consider investing between these years.Let me just visualize this to make sure. The original stock price equation is a downward opening parabola with vertex at t = 3, P = 38. It crosses the P = 30 line at t = 1 and t = 5. So, between t = 1 and t = 5, the stock price is above 30, which is when I want to invest. That seems correct.So, summarizing:1. The maximum stock price is 38, occurring at t = 3 years.2. The stock price is at least 30 from t = 1 to t = 5 years.I think that's all. Let me just recap to ensure I didn't make any calculation errors.For the first part, vertex at t = -b/(2a) = -12/(2*(-2)) = 3. Plugging back in, P(3) = -2*(9) + 36 + 20 = -18 + 36 + 20 = 38. Correct.For the second part, solving -2t² + 12t + 20 ≥ 30 leads to -2t² + 12t -10 ≥ 0, which becomes 2t² -12t +10 ≤ 0. Solving 2t² -12t +10 = 0 gives t = [12 ± sqrt(144 - 80)] / 4 = [12 ± 8]/4, so t = 5 and t = 1. Since the quadratic opens upwards, it's ≤ 0 between 1 and 5. So, the investment range is between 1 and 5 years. That all checks out.**Final Answer**1. The maximum stock price is boxed{38} dollars, reached after boxed{3} years.2. The stock price is at least 30 from year boxed{1} to year boxed{5}.

answer：Okay, so I have this problem where I need to show that ( f(1987) = 1 ) given some conditions on the function ( f ). Let me try to break this down step by step.First, let's understand the given functional equation:[ fleft(left(frac{x_1^m + x_2^m + cdots + x_n^m}{n}right)right) = left(frac{|f(x_1)|^m + cdots + |f(x_n)|^m}{n}right) ]Here, ( m ) is a fixed even positive integer, and ( n > 1 ) is also fixed. The function ( f ) is defined on the non-negative real numbers. So, the function ( f ) takes an average of the ( m )-th powers of some non-negative real numbers ( x_1, x_2, ldots, x_n ), and then applies ( f ) to that average. On the right side, it's taking the average of the ( m )-th powers of the absolute values of ( f ) evaluated at each ( x_i ).Given that ( m ) is even, ( x_i^m ) is always non-negative, and so is ( |f(x_i)|^m ). That makes sense because the function is defined on non-negative reals.We are also given that ( f(1988) ) is non-zero and ( f(1986) - 1986 ) is non-zero. So, ( f(1988) neq 0 ) and ( f(1986) neq 1986 ).Our goal is to show that ( f(1987) = 1 ).Hmm, okay. Let's think about possible forms of ( f ). Since the functional equation involves averages of ( m )-th powers, maybe ( f ) is a linear function or something involving powers.Let me consider if ( f ) could be a constant function. Suppose ( f(x) = c ) for all ( x ). Then, substituting into the equation:Left side: ( fleft(frac{x_1^m + cdots + x_n^m}{n}right) = c )Right side: ( frac{|c|^m + cdots + |c|^m}{n} = frac{n|c|^m}{n} = |c|^m )So, we have ( c = |c|^m ). Since ( m ) is even, ( |c|^m = c^m ). So, ( c = c^m ). The solutions to this are ( c = 0 ) or ( c^{m-1} = 1 ). Since ( m ) is even, ( m-1 ) is odd, so the only real solution is ( c = 1 ).But wait, we are given that ( f(1988) ) is non-zero. So, if ( f ) were constant, it would have to be 1. But then ( f(1986) = 1 ), which would make ( f(1986) - 1986 = 1 - 1986 = -1985 neq 0 ). So, that's consistent with the given condition. However, if ( f ) is constant 1, then ( f(1987) = 1 ), which is what we need to show. But is ( f ) necessarily constant?Wait, maybe not. Let's see if there are other possible functions.Suppose ( f(x) = kx ) for some constant ( k ). Let's test this.Left side: ( fleft(frac{x_1^m + cdots + x_n^m}{n}right) = k cdot frac{x_1^m + cdots + x_n^m}{n} )Right side: ( frac{|k x_1|^m + cdots + |k x_n|^m}{n} = frac{k^m (x_1^m + cdots + x_n^m)}{n} )So, equating left and right sides:( k cdot frac{x_1^m + cdots + x_n^m}{n} = frac{k^m (x_1^m + cdots + x_n^m)}{n} )Assuming ( x_1^m + cdots + x_n^m neq 0 ), we can divide both sides by that term:( k = k^m )So, ( k^m - k = 0 ) => ( k(k^{m-1} - 1) = 0 ). So, ( k = 0 ) or ( k^{m-1} = 1 ). Since ( m ) is even, ( m-1 ) is odd, so the real solutions are ( k = 0 ) or ( k = 1 ).But ( f(1988) neq 0 ), so ( k = 1 ). So, ( f(x) = x ). But then ( f(1986) = 1986 ), which contradicts the given condition that ( f(1986) - 1986 neq 0 ). Therefore, ( f ) cannot be linear of the form ( f(x) = x ).Hmm, so maybe ( f ) is a constant function 1? But earlier, I thought that might be possible, but let me check again.If ( f(x) = 1 ) for all ( x ), then:Left side: ( fleft(frac{x_1^m + cdots + x_n^m}{n}right) = 1 )Right side: ( frac{|1|^m + cdots + |1|^m}{n} = frac{n}{n} = 1 )So, that works. Also, ( f(1988) = 1 neq 0 ), which is okay, and ( f(1986) = 1 neq 1986 ), which is also okay. So, ( f(x) = 1 ) is a solution.But is that the only solution? Or are there other functions that satisfy the equation?Wait, let's think about functions of the form ( f(x) = |x|^k ). Let's test this.Let ( f(x) = |x|^k ). Then:Left side: ( fleft(frac{x_1^m + cdots + x_n^m}{n}right) = left|frac{x_1^m + cdots + x_n^m}{n}right|^k )Right side: ( frac{|f(x_1)|^m + cdots + |f(x_n)|^m}{n} = frac{(|x_1|^k)^m + cdots + (|x_n|^k)^m}{n} = frac{x_1^{km} + cdots + x_n^{km}}{n} )So, we have:( left(frac{x_1^m + cdots + x_n^m}{n}right)^k = frac{x_1^{km} + cdots + x_n^{km}}{n} )This needs to hold for all non-negative ( x_i ). Let's consider the case where all ( x_i ) are equal, say ( x_i = a ) for all ( i ).Then, left side: ( left(frac{n a^m}{n}right)^k = (a^m)^k = a^{mk} )Right side: ( frac{n a^{km}}{n} = a^{km} )So, equality holds in this case. But does it hold in general?Suppose ( x_1 = a ) and ( x_2 = b ), and the rest ( x_3, ldots, x_n ) are zero. Let's see:Left side: ( left(frac{a^m + b^m + 0 + cdots + 0}{n}right)^k = left(frac{a^m + b^m}{n}right)^k )Right side: ( frac{a^{km} + b^{km} + 0 + cdots + 0}{n} = frac{a^{km} + b^{km}}{n} )So, we need:( left(frac{a^m + b^m}{n}right)^k = frac{a^{km} + b^{km}}{n} )Is this true for all ( a, b geq 0 )?Let me test with specific values. Let ( n = 2 ), ( m = 2 ), and ( k = 1 ):Left side: ( left(frac{a^2 + b^2}{2}right)^1 = frac{a^2 + b^2}{2} )Right side: ( frac{a^{2} + b^{2}}{2} )So, equality holds.What if ( k = 2 ), ( m = 2 ), ( n = 2 ):Left side: ( left(frac{a^2 + b^2}{2}right)^2 = frac{(a^2 + b^2)^2}{4} )Right side: ( frac{a^{4} + b^{4}}{2} )These are equal only if ( (a^2 + b^2)^2 = 2(a^4 + b^4) ), which simplifies to ( a^4 + 2a^2b^2 + b^4 = 2a^4 + 2b^4 ), leading to ( 2a^2b^2 = a^4 + b^4 ), which is not true in general. For example, take ( a = 1 ), ( b = 1 ): ( 2*1*1 = 1 + 1 ) => 2 = 2, which holds. But ( a = 1 ), ( b = 0 ): Left side: ( (1 + 0)^2 = 1 ), Right side: ( (1 + 0)/2 = 0.5 ). So, 1 ≠ 0.5. So, equality doesn't hold.Therefore, ( f(x) = |x|^k ) only works if ( k = 1 ) or ( k = 0 ), but ( k = 0 ) would make ( f(x) = 1 ) for all ( x ), which is a constant function, as we saw earlier.Wait, but earlier when we considered ( f(x) = x ), it didn't satisfy the condition because ( f(1986) = 1986 ), which is not allowed. So, maybe the only possible function is the constant function 1.But let's check if there are other possibilities. Suppose ( f ) is not constant but satisfies the given equation.Let me consider the case where all ( x_i ) except one are zero. Let ( x_1 = a ), and ( x_2 = x_3 = cdots = x_n = 0 ).Then, left side: ( fleft(frac{a^m + 0 + cdots + 0}{n}right) = fleft(frac{a^m}{n}right) )Right side: ( frac{|f(a)|^m + |f(0)|^m + cdots + |f(0)|^m}{n} = frac{|f(a)|^m + (n-1)|f(0)|^m}{n} )So, we have:( fleft(frac{a^m}{n}right) = frac{|f(a)|^m + (n-1)|f(0)|^m}{n} )This must hold for all ( a geq 0 ).Let me denote ( c = f(0) ). Then, the equation becomes:( fleft(frac{a^m}{n}right) = frac{|f(a)|^m + (n-1)c^m}{n} )Let me consider ( a = 0 ):Left side: ( f(0) = c )Right side: ( frac{|f(0)|^m + (n-1)c^m}{n} = frac{c^m + (n-1)c^m}{n} = frac{n c^m}{n} = c^m )So, ( c = c^m ). Since ( m ) is even, ( c^m = |c|^m ). So, ( c = |c|^m ). The solutions to this are ( c = 0 ) or ( c = 1 ) because ( m geq 2 ).Case 1: ( c = 0 )Then, from the earlier equation:( fleft(frac{a^m}{n}right) = frac{|f(a)|^m + 0}{n} = frac{|f(a)|^m}{n} )But let's also consider the case where all ( x_i ) are equal to ( a ). Then, as before, we have:Left side: ( f(a^m) )Right side: ( frac{n |f(a)|^m}{n} = |f(a)|^m )So, ( f(a^m) = |f(a)|^m )But from the case where one ( x_i = a ) and others zero, we have:( fleft(frac{a^m}{n}right) = frac{|f(a)|^m}{n} )Let me denote ( b = frac{a^m}{n} ). Then, ( a = left(n bright)^{1/m} ). So,( f(b) = frac{|fleft( (n b)^{1/m} right)|^m}{n} )But from the all-equal case, ( f(a^m) = |f(a)|^m ). Let me set ( a = (n b)^{1/m} ), so ( a^m = n b ). Then,( f(n b) = |fleft( (n b)^{1/m} right)|^m )But from the previous equation, ( f(b) = frac{|fleft( (n b)^{1/m} right)|^m}{n} ), so ( |fleft( (n b)^{1/m} right)|^m = n f(b) ). Substituting into the equation above:( f(n b) = n f(b) )So, we have ( f(n b) = n f(b) ). Let me denote ( d = b ), so ( f(n d) = n f(d) ). This suggests that ( f ) is linear on the scale of ( n ). But we also have ( f(a^m) = |f(a)|^m ).Let me see if ( f ) can be linear. Suppose ( f(x) = k x ). Then,( f(a^m) = k a^m )And ( |f(a)|^m = |k a|^m = k^m a^m )So, ( k a^m = k^m a^m ) => ( k = k^m ). As before, ( k = 0 ) or ( k = 1 ). But ( f(1988) neq 0 ), so ( k = 1 ). But then ( f(1986) = 1986 ), which is not allowed. So, ( f ) cannot be linear.Alternatively, suppose ( f(x) = 1 ) for all ( x ). Then,( f(a^m) = 1 )And ( |f(a)|^m = 1^m = 1 )So, ( 1 = 1 ), which holds. Also, ( f(n b) = 1 = n cdot 1 ) only if ( n = 1 ), but ( n > 1 ). Wait, that's a problem.Wait, if ( f(x) = 1 ), then ( f(n b) = 1 ), but ( n f(b) = n cdot 1 = n ). So, ( 1 = n ), which is not true since ( n > 1 ). Therefore, ( f(x) = 1 ) cannot satisfy ( f(n b) = n f(b) ) unless ( n = 1 ), which it's not.Hmm, this is a contradiction. So, if ( c = 0 ), we end up with ( f(n b) = n f(b) ), but also ( f(a^m) = |f(a)|^m ). If ( f ) is linear, it leads to a contradiction with the given conditions. If ( f ) is constant 1, it also leads to a contradiction because ( f(n b) = 1 ) but ( n f(b) = n ).Therefore, maybe ( c neq 0 ). Let's consider Case 2: ( c = 1 ).So, ( f(0) = 1 ).Then, from the earlier equation:( fleft(frac{a^m}{n}right) = frac{|f(a)|^m + (n-1) cdot 1^m}{n} = frac{|f(a)|^m + (n-1)}{n} )Also, from the all-equal case, ( f(a^m) = |f(a)|^m ).Let me denote ( b = frac{a^m}{n} ). Then, ( a = (n b)^{1/m} ). So,( f(b) = frac{|fleft( (n b)^{1/m} right)|^m + (n-1)}{n} )But from the all-equal case, ( f(a^m) = |f(a)|^m ). Let ( a = (n b)^{1/m} ), so ( a^m = n b ). Then,( f(n b) = |fleft( (n b)^{1/m} right)|^m )From the equation above, ( f(b) = frac{|fleft( (n b)^{1/m} right)|^m + (n-1)}{n} ), which can be rewritten as:( |fleft( (n b)^{1/m} right)|^m = n f(b) - (n - 1) )Substituting into the all-equal case equation:( f(n b) = n f(b) - (n - 1) )So, ( f(n b) = n f(b) - (n - 1) )This is a functional equation for ( f ). Let me see if I can solve it.Let me denote ( d = b ), so the equation becomes:( f(n d) = n f(d) - (n - 1) )This is a linear functional equation. Let me see if I can find a solution.Suppose ( f(d) = k d + c ). Let's test this.Then,( f(n d) = k (n d) + c = n k d + c )On the other hand,( n f(d) - (n - 1) = n(k d + c) - (n - 1) = n k d + n c - n + 1 )So, equating:( n k d + c = n k d + n c - n + 1 )Simplify:( c = n c - n + 1 )( 0 = (n - 1)c - n + 1 )( (n - 1)c = n - 1 )If ( n neq 1 ), which it isn't, then ( c = 1 ).So, ( f(d) = k d + 1 )Now, let's substitute back into the functional equation ( f(n d) = n f(d) - (n - 1) ):Left side: ( f(n d) = k (n d) + 1 = n k d + 1 )Right side: ( n f(d) - (n - 1) = n(k d + 1) - (n - 1) = n k d + n - n + 1 = n k d + 1 )So, both sides are equal. Therefore, any function of the form ( f(d) = k d + 1 ) satisfies the functional equation ( f(n d) = n f(d) - (n - 1) ).But we also have another condition from the all-equal case: ( f(a^m) = |f(a)|^m ).Let me substitute ( f(a) = k a + 1 ) into this:( f(a^m) = k a^m + 1 )And ( |f(a)|^m = |k a + 1|^m )So, we have:( k a^m + 1 = |k a + 1|^m )This must hold for all ( a geq 0 ).Let me analyze this equation.First, since ( m ) is even, ( |k a + 1|^m = (k a + 1)^m ) because ( k a + 1 ) is non-negative for ( a geq 0 ) if ( k geq 0 ). But ( k ) could be negative, but let's see.Wait, ( f ) is defined on non-negative reals, but its output can be any real number, since it's real-valued. However, in the functional equation, we have ( |f(x_i)|^m ), so the sign of ( f(x_i) ) doesn't matter because it's raised to an even power. But in the equation ( f(a^m) = |f(a)|^m ), the left side is ( f(a^m) ), which could be negative, but the right side is non-negative because it's an even power.Wait, actually, ( f(a^m) ) must equal ( |f(a)|^m ), which is non-negative. Therefore, ( f(a^m) geq 0 ) for all ( a geq 0 ). So, ( f ) must map non-negative reals to non-negative reals, because ( a^m ) is non-negative, and ( f(a^m) ) is non-negative.Therefore, ( f(x) geq 0 ) for all ( x geq 0 ). So, ( f ) is non-negative.Given that, ( f(a) = k a + 1 geq 0 ) for all ( a geq 0 ). So, ( k a + 1 geq 0 ). Since ( a geq 0 ), this implies that ( k geq 0 ), because if ( k ) were negative, for large enough ( a ), ( k a + 1 ) would become negative, which is not allowed.Therefore, ( k geq 0 ).So, we have ( f(a) = k a + 1 ), with ( k geq 0 ), and:( k a^m + 1 = (k a + 1)^m )Let me expand the right side using the binomial theorem:( (k a + 1)^m = sum_{i=0}^m binom{m}{i} (k a)^i cdot 1^{m-i} = sum_{i=0}^m binom{m}{i} k^i a^i )So, we have:( k a^m + 1 = sum_{i=0}^m binom{m}{i} k^i a^i )Let me write this as:( sum_{i=0}^m binom{m}{i} k^i a^i - k a^m - 1 = 0 )Simplify:For ( i = 0 ): ( binom{m}{0} k^0 a^0 = 1 )For ( i = 1 ): ( binom{m}{1} k a )...For ( i = m-1 ): ( binom{m}{m-1} k^{m-1} a^{m-1} )For ( i = m ): ( binom{m}{m} k^m a^m = k^m a^m )So, subtracting ( k a^m ) and 1:( sum_{i=1}^{m-1} binom{m}{i} k^i a^i + (k^m a^m - k a^m) = 0 )Factor ( a^m ):( sum_{i=1}^{m-1} binom{m}{i} k^i a^i + (k^m - k) a^m = 0 )This must hold for all ( a geq 0 ). The only way this polynomial can be zero for all ( a ) is if all coefficients are zero.So, for each ( i ) from 1 to ( m-1 ):( binom{m}{i} k^i = 0 )And for ( i = m ):( k^m - k = 0 )But ( binom{m}{i} ) is positive for ( i = 1, 2, ldots, m-1 ), and ( k geq 0 ). So, ( binom{m}{i} k^i = 0 ) implies ( k = 0 ) for each ( i geq 1 ).Therefore, ( k = 0 ).Substituting ( k = 0 ) into the equation for ( i = m ):( 0^m - 0 = 0 ), which holds.So, the only solution is ( k = 0 ), which gives ( f(a) = 1 ) for all ( a geq 0 ).Wait, but earlier I thought that ( f(x) = 1 ) leads to a contradiction because ( f(n b) = 1 ) but ( n f(b) - (n - 1) = n cdot 1 - (n - 1) = 1 ). So, actually, it does satisfy ( f(n b) = 1 = n cdot 1 - (n - 1) = 1 ). So, no contradiction here.Wait, earlier I thought that ( f(n b) = n f(b) ) when ( c = 0 ), but in this case, with ( c = 1 ), we have ( f(n b) = n f(b) - (n - 1) ). So, when ( f(b) = 1 ), ( f(n b) = n cdot 1 - (n - 1) = 1 ), which is consistent.So, ( f(x) = 1 ) is indeed a solution.But earlier, when I considered ( f(x) = 1 ), I thought that ( f(n b) = 1 ) but ( n f(b) = n ), which would imply ( 1 = n ), but that was under the assumption that ( c = 0 ). However, in this case, with ( c = 1 ), the equation is ( f(n b) = n f(b) - (n - 1) ), which is satisfied by ( f(b) = 1 ).Therefore, the only solution is ( f(x) = 1 ) for all ( x geq 0 ).But wait, let's check the given conditions:1. ( f(1988) ) is non-zero: ( f(1988) = 1 neq 0 ). Okay.2. ( f(1986) - 1986 ) is non-zero: ( f(1986) = 1 ), so ( 1 - 1986 = -1985 neq 0 ). Okay.Therefore, ( f(x) = 1 ) satisfies all the given conditions, and in particular, ( f(1987) = 1 ).Is there any other possible function? From the above analysis, it seems that ( f(x) = 1 ) is the only solution. Because any other form either leads to contradictions or doesn't satisfy the functional equation for all ( x_i ).Therefore, the conclusion is ( f(1987) = 1 ).

answer：First, I need to calculate the number of episodes in the 5th season using the given polynomial equation ( E(n) = 2n^2 + 3n + 1 ). I'll substitute ( n = 5 ) into the equation.Next, I have to determine the range of days ( x ) for which the number of episodes watched in a week, given by ( W(x) = x^2 + 4x + 3 ), does not exceed 30 episodes. This involves solving the inequality ( x^2 + 4x + 3 leq 30 ).To solve the inequality, I'll first bring all terms to one side to form a quadratic inequality: ( x^2 + 4x - 27 leq 0 ). Then, I'll find the roots of the corresponding quadratic equation ( x^2 + 4x - 27 = 0 ) using the quadratic formula. Once I have the roots, I'll determine the intervals where the quadratic expression is less than or equal to zero. Finally, I'll consider the practical context to determine the valid range of ( x ) values.