answer：Okay, so I have this problem where there's a series of numbers written in the base-60 system, and I need to show that these numbers form an arithmetic progression and then write them in the decimal system. The numbers given are: 1.36; 1.52; 2.8; 2.24; 2.40; 2.56; 3·12; 3·28; 3·44; 4.First, I need to understand what it means for numbers to be in base-60. I know that in base-10, each digit represents a power of 10, like units, tens, hundreds, etc. Similarly, in base-60, each position represents a power of 60. So, for example, a number like 1.36 in base-60 would be 1 times 60 plus 36 in base-10. That makes sense because the first digit is the "60s" place and the second digit is the units place.So, I think the first step is to convert each of these base-60 numbers into decimal (base-10) numbers. Once I have them in decimal, I can check if they form an arithmetic progression by seeing if the difference between consecutive terms is constant.Let me start by converting each number one by one.1. **1.36 in base-60:** - The first digit is 1, which is in the 60s place, so that's 1 * 60. - The second digit is 36, which is in the units place, so that's 36 * 1. - Adding them together: 1 * 60 + 36 = 60 + 36 = 96.2. **1.52 in base-60:** - First digit: 1 * 60 = 60. - Second digit: 52 * 1 = 52. - Total: 60 + 52 = 112.3. **2.8 in base-60:** - First digit: 2 * 60 = 120. - Second digit: 8 * 1 = 8. - Total: 120 + 8 = 128.4. **2.24 in base-60:** - First digit: 2 * 60 = 120. - Second digit: 24 * 1 = 24. - Total: 120 + 24 = 144.5. **2.40 in base-60:** - First digit: 2 * 60 = 120. - Second digit: 40 * 1 = 40. - Total: 120 + 40 = 160.6. **2.56 in base-60:** - First digit: 2 * 60 = 120. - Second digit: 56 * 1 = 56. - Total: 120 + 56 = 176.7. **3·12 in base-60:** - Here, the notation is a bit different. It says 3·12, which I think means 3 multiplied by 12 in base-60. - But wait, in base-60, 3·12 would actually be 3 in the 60s place and 12 in the units place. - So, first digit: 3 * 60 = 180. - Second digit: 12 * 1 = 12. - Total: 180 + 12 = 192.8. **3·28 in base-60:** - Similar to the previous one. - First digit: 3 * 60 = 180. - Second digit: 28 * 1 = 28. - Total: 180 + 28 = 208.9. **3·44 in base-60:** - Again, same approach. - First digit: 3 * 60 = 180. - Second digit: 44 * 1 = 44. - Total: 180 + 44 = 224.10. **4 in base-60:** - This is a single digit, so it's just 4 in base-10.So, after converting all these numbers, I have the following decimal numbers:96, 112, 128, 144, 160, 176, 192, 208, 224, 4.Now, I need to check if these numbers form an arithmetic progression. An arithmetic progression is a sequence where the difference between consecutive terms is constant.Let me list the numbers in order:96, 112, 128, 144, 160, 176, 192, 208, 224, 4.Wait a minute, the last number is 4, which is much smaller than the previous numbers. That seems out of place. Maybe I made a mistake in converting the last number.Looking back at the original numbers, the last one is just "4" in base-60. Since it's a single digit, it should indeed be 4 in base-10. But in the context of the sequence, it seems like an outlier because all the other numbers are much larger.Perhaps the sequence is supposed to be cyclic or something? Or maybe the last number is a typo or misplacement. Alternatively, maybe I'm supposed to ignore the last number or consider it separately.But the problem says to show that these numbers represent an arithmetic progression, so I think the last number should fit into the sequence. Maybe I need to reconsider how I'm interpreting the numbers.Wait, let me double-check the conversions:1.36: 1*60 + 36 = 961.52: 1*60 + 52 = 1122.8: 2*60 + 8 = 1282.24: 2*60 + 24 = 1442.40: 2*60 + 40 = 1602.56: 2*60 + 56 = 1763·12: 3*60 + 12 = 1923·28: 3*60 + 28 = 2083·44: 3*60 + 44 = 2244: 4So, the conversions seem correct. The last number is indeed 4, which is much smaller. Maybe the sequence wraps around? Like, after 224, it goes back to 4? But that would mean the difference is negative, which isn't typical for an arithmetic progression.Alternatively, maybe the last number is supposed to be 4 in base-60, which is 4*60^0 = 4, but perhaps it's part of a larger sequence where after 3·44, it goes to 4·something, but in this case, it's just 4.Wait, maybe the last number is supposed to be 4 in base-60, which is 4 in decimal, but perhaps it's the next term after 3·44. Let's see:The previous term is 3·44 = 224. If we add the common difference, which we'll figure out, to 224, we should get the next term. But the next term is 4, which is way smaller. That doesn't make sense.Alternatively, maybe the sequence is cyclic in base-60, meaning that after 3·44, it goes back to 4, which is like 4·00 in base-60, making the sequence wrap around. But in decimal, that would mean the sequence goes from 224 back to 4, which is a decrease of 220, which is not consistent with the previous differences.So, perhaps the last number is a mistake or an outlier. Maybe it's supposed to be 4·something, but it's just 4. Alternatively, maybe the sequence is supposed to be 96, 112, 128, 144, 160, 176, 192, 208, 224, and then the next term would be 240, but instead, it's given as 4, which is 240 - 236 = 4. That doesn't make sense either.Wait, maybe I'm overcomplicating it. Let's just look at the first nine numbers:96, 112, 128, 144, 160, 176, 192, 208, 224.These numbers have a common difference of 16:112 - 96 = 16128 - 112 = 16144 - 128 = 16160 - 144 = 16176 - 160 = 16192 - 176 = 16208 - 192 = 16224 - 208 = 16So, the first nine numbers form an arithmetic progression with a common difference of 16. The last number, 4, seems to be out of place. Maybe it's a mistake or perhaps it's part of a different sequence.Given that the problem asks to show that these numbers represent an arithmetic progression, it's likely that the last number is either a typo or should be ignored. Therefore, focusing on the first nine numbers, they form an arithmetic progression in decimal.So, to summarize, after converting each base-60 number to decimal, we get:96, 112, 128, 144, 160, 176, 192, 208, 224, 4.Ignoring the last number, the sequence 96, 112, 128, 144, 160, 176, 192, 208, 224 is an arithmetic progression with a common difference of 16.I think that's the solution. The key steps were converting each base-60 number to decimal and then checking the differences between consecutive terms to verify the arithmetic progression.

answer：Okay, so I need to figure out the total cost for 2 sandwiches and 4 sodas. Each sandwich costs 2.45, and each soda is 0.87. Hmm, let me break this down step by step.First, I'll calculate the cost of the sandwiches. There are 2 sandwiches, and each one is 2.45. So, I need to multiply the number of sandwiches by the cost per sandwich. That would be 2 times 2.45. Let me do that multiplication: 2 times 2 is 4, and 2 times 0.45 is 0.90. So, adding those together, that's 4.90 for the sandwiches.Next, I'll calculate the cost of the sodas. There are 4 sodas, and each one is 0.87. So, I need to multiply the number of sodas by the cost per soda. That would be 4 times 0.87. Let me do that multiplication: 4 times 0.80 is 3.20, and 4 times 0.07 is 0.28. Adding those together, that's 3.48 for the sodas.Now, to find the total cost, I need to add the cost of the sandwiches and the cost of the sodas together. That would be 4.90 plus 3.48. Let me add those: 4 plus 3 is 7, and 0.90 plus 0.48 is 1.38. So, adding those together, that's 8.38.Wait a minute, let me double-check my calculations to make sure I didn't make any mistakes. For the sandwiches: 2 times 2.45 is indeed 4.90. For the sodas: 4 times 0.87 is 3.48. Adding those together: 4.90 plus 3.48 equals 8.38. Yep, that seems correct.I think that's the total cost for 2 sandwiches and 4 sodas. It looks like everything adds up correctly, so I'm confident that 8.38 is the right answer.

answer：Okay, so I have this function f(x) = ln(e^x + e^{-x}) + x², and I need to find the range of x where f(2x) > f(x + 3). Hmm, let me try to break this down step by step.First, I remember that ln(e^x + e^{-x}) looks familiar. Isn't that related to hyperbolic functions? Oh yeah, ln(e^x + e^{-x}) is actually equal to ln(2*cosh(x)), since cosh(x) = (e^x + e^{-x}) / 2. So, f(x) can be rewritten as ln(2*cosh(x)) + x². That might make it easier to analyze.Now, I need to compare f(2x) and f(x + 3). Maybe I should first find the derivative of f(x) to understand its behavior. Let's compute f'(x):f'(x) = d/dx [ln(e^x + e^{-x})] + d/dx [x²]The derivative of ln(e^x + e^{-x}) is (e^x - e^{-x}) / (e^x + e^{-x}), which simplifies to tanh(x). And the derivative of x² is 2x. So, f'(x) = tanh(x) + 2x.Hmm, tanh(x) is an odd function, meaning tanh(-x) = -tanh(x), and it's symmetric around the origin. Also, tanh(x) approaches 1 as x approaches infinity and -1 as x approaches negative infinity. So, for positive x, tanh(x) is positive but less than 1, and for negative x, it's negative but greater than -1.Now, let's think about f'(x). For x > 0, tanh(x) is positive, so f'(x) = tanh(x) + 2x is definitely positive because both terms are positive. For x < 0, tanh(x) is negative, but 2x is also negative. So, f'(x) is negative for x < 0. At x = 0, f'(0) = tanh(0) + 0 = 0.This tells me that f(x) has a minimum at x = 0. So, f(x) is decreasing for x < 0 and increasing for x > 0. Therefore, f(x) is an even function because it's symmetric around x = 0. Wait, is that true? Let me check.An even function satisfies f(-x) = f(x). Let's see:f(-x) = ln(e^{-x} + e^{x}) + (-x)^2 = ln(e^x + e^{-x}) + x² = f(x). Yes, so f(x) is indeed even. That's useful because it means the function behaves the same on both sides of the y-axis.So, since f(x) is even and has a minimum at x = 0, it's symmetric and increasing for x > 0, decreasing for x < 0. Therefore, f(a) > f(b) if |a| > |b| because the function increases as you move away from zero in either direction.Wait, is that always true? Let me think. If a and b are both positive, then yes, if a > b, f(a) > f(b). Similarly, if a and b are both negative, since f is even, f(a) = f(|a|) and f(b) = f(|b|), so again, if |a| > |b|, f(a) > f(b). So, in general, f(a) > f(b) if |a| > |b|.Therefore, f(2x) > f(x + 3) is equivalent to |2x| > |x + 3|. That's a key insight. So, instead of dealing with the logarithmic and quadratic terms, I can just compare the absolute values.So, let's write that inequality:|2x| > |x + 3|To solve this inequality, I can square both sides because both sides are non-negative. Squaring both sides:(2x)^2 > (x + 3)^2Which simplifies to:4x² > x² + 6x + 9Subtract x² + 6x + 9 from both sides:4x² - x² - 6x - 9 > 0Simplify:3x² - 6x - 9 > 0Divide both sides by 3:x² - 2x - 3 > 0Now, let's factor this quadratic:x² - 2x - 3 = (x - 3)(x + 1)So, the inequality becomes:(x - 3)(x + 1) > 0To solve this, we can find the critical points where the expression equals zero: x = 3 and x = -1. These points divide the number line into three intervals:1. x < -12. -1 < x < 33. x > 3We can test each interval to see where the product is positive.1. For x < -1: Let's pick x = -2. (-2 - 3)(-2 + 1) = (-5)(-1) = 5 > 0. So, positive.2. For -1 < x < 3: Let's pick x = 0. (0 - 3)(0 + 1) = (-3)(1) = -3 < 0. So, negative.3. For x > 3: Let's pick x = 4. (4 - 3)(4 + 1) = (1)(5) = 5 > 0. So, positive.Therefore, the inequality (x - 3)(x + 1) > 0 holds when x < -1 or x > 3. So, the solution is x ∈ (-∞, -1) ∪ (3, ∞).Wait, let me double-check if squaring both sides introduced any extraneous solutions. Since both |2x| and |x + 3| are non-negative, squaring shouldn't introduce any issues. So, the solution should be correct.Therefore, the range of x for which f(2x) > f(x + 3) is (-∞, -1) ∪ (3, ∞), which corresponds to option D.**Final Answer**boxed{D}

answer：Okay, so I have this problem here where I need to find the sum of all possible values of ( k ) such that the cubic polynomial ( x^3 - 3x^2 + 2x ) and the quadratic polynomial ( x^2 + 3x + k ) have a common root. Hmm, let me try to figure this out step by step.First, I remember that if two polynomials have a common root, then that root must satisfy both equations. So, if there's a common root ( r ), then plugging ( r ) into both polynomials should give me zero. That makes sense.So, let me write down both polynomials:1. Cubic polynomial: ( x^3 - 3x^2 + 2x )2. Quadratic polynomial: ( x^2 + 3x + k )Since ( r ) is a common root, substituting ( r ) into both should equal zero. So:1. ( r^3 - 3r^2 + 2r = 0 )2. ( r^2 + 3r + k = 0 )I think the strategy here is to find the roots of the cubic polynomial first because if I know the possible roots, I can then plug them into the quadratic equation to find the corresponding ( k ) values. Then, I can sum those ( k ) values to get the answer.Let me factor the cubic polynomial. Maybe that will make it easier to find the roots. So, starting with ( x^3 - 3x^2 + 2x ), I can factor out an ( x ) first:( x(x^2 - 3x + 2) )Now, I need to factor the quadratic part ( x^2 - 3x + 2 ). Let me think... I need two numbers that multiply to 2 and add up to -3. Hmm, -1 and -2. Yes, that works.So, factoring ( x^2 - 3x + 2 ) gives me ( (x - 1)(x - 2) ). Therefore, the cubic polynomial factors completely as:( x(x - 1)(x - 2) )So, the roots of the cubic polynomial are ( x = 0 ), ( x = 1 ), and ( x = 2 ). Got that.Now, since these are the possible common roots with the quadratic polynomial, I can substitute each of these roots into the quadratic equation ( x^2 + 3x + k = 0 ) and solve for ( k ) each time.Let me do that one by one.1. **First root: ( x = 0 )** Plugging into the quadratic: ( 0^2 + 3(0) + k = 0 ) Simplifying: ( 0 + 0 + k = 0 ) So, ( k = 0 ).2. **Second root: ( x = 1 )** Plugging into the quadratic: ( 1^2 + 3(1) + k = 0 ) Simplifying: ( 1 + 3 + k = 0 ) So, ( 4 + k = 0 ) which means ( k = -4 ).3. **Third root: ( x = 2 )** Plugging into the quadratic: ( 2^2 + 3(2) + k = 0 ) Simplifying: ( 4 + 6 + k = 0 ) So, ( 10 + k = 0 ) which means ( k = -10 ).Okay, so the possible values of ( k ) are 0, -4, and -10. Now, the question asks for the sum of all possible values of ( k ). So, I need to add these together.Calculating the sum:( 0 + (-4) + (-10) = 0 - 4 - 10 = -14 )So, the sum of all possible values of ( k ) is -14.Wait, let me double-check to make sure I didn't make any mistakes. I factored the cubic correctly, right? Yes, ( x^3 - 3x^2 + 2x ) factors into ( x(x - 1)(x - 2) ), so the roots are indeed 0, 1, and 2.Then, substituting each root into the quadratic equation:- For ( x = 0 ): ( 0 + 0 + k = 0 ) gives ( k = 0 ). That seems correct.- For ( x = 1 ): ( 1 + 3 + k = 0 ) gives ( k = -4 ). That also looks right.- For ( x = 2 ): ( 4 + 6 + k = 0 ) gives ( k = -10 ). Yep, that's correct too.Adding them up: 0 + (-4) + (-10) = -14. Hmm, that seems straightforward.I don't think I missed anything here. The logic is solid: find the roots of the cubic, plug them into the quadratic to find corresponding ( k ) values, then sum those ( k ) values. So, I feel confident that the answer is -14.**Final Answer**boxed{A}

answer：Okay, so I have this problem about an ellipse. Let me try to understand it step by step. First, part (1) asks to find the equation of the ellipse M. The ellipse has the right focus at F(1,0) and passes through the point D(1, 3/2). The standard equation of an ellipse is given as x²/a² + y²/b² = 1, where a > b > 0. I remember that for an ellipse, the distance from the center to each focus is c, and c² = a² - b². Since the right focus is at (1,0), the center of the ellipse must be at (h, k). But wait, in the standard form, the center is at (0,0) if it's written as x²/a² + y²/b² = 1. So, does that mean the center is at (0,0)? But if the focus is at (1,0), then c = 1. So, c² = 1 = a² - b². So, a² - b² = 1. Also, the ellipse passes through the point D(1, 3/2). So, substituting x = 1 and y = 3/2 into the ellipse equation should satisfy it. So, plugging in, we get (1)²/a² + (3/2)²/b² = 1. That simplifies to 1/a² + 9/(4b²) = 1. So, now we have two equations: 1. a² - b² = 12. 1/a² + 9/(4b²) = 1I need to solve these two equations to find a² and b². Let me denote a² as A and b² as B for simplicity. Then, the equations become:1. A - B = 12. 1/A + 9/(4B) = 1From the first equation, A = B + 1. Substitute this into the second equation:1/(B + 1) + 9/(4B) = 1Let me find a common denominator for the left side. The common denominator would be 4B(B + 1). So, rewriting each term:- 1/(B + 1) = 4B / [4B(B + 1)]- 9/(4B) = 9(B + 1) / [4B(B + 1)]Adding them together:[4B + 9(B + 1)] / [4B(B + 1)] = 1Simplify the numerator:4B + 9B + 9 = 13B + 9So, (13B + 9) / [4B(B + 1)] = 1Multiply both sides by 4B(B + 1):13B + 9 = 4B(B + 1)Expand the right side:13B + 9 = 4B² + 4BBring all terms to one side:4B² + 4B - 13B - 9 = 04B² - 9B - 9 = 0Now, solve this quadratic equation for B. Using the quadratic formula:B = [9 ± sqrt(81 + 144)] / 8Because discriminant D = (-9)² - 4*4*(-9) = 81 + 144 = 225So, sqrt(225) = 15Thus, B = [9 ± 15]/8We have two solutions:1. B = (9 + 15)/8 = 24/8 = 32. B = (9 - 15)/8 = (-6)/8 = -3/4But since B is b², it must be positive. So, B = 3. Then, from A = B + 1, A = 3 + 1 = 4. So, a² = 4 and b² = 3.Therefore, the equation of the ellipse is x²/4 + y²/3 = 1.Okay, that seems straightforward. So, part (1) is done.Now, part (2) is more complicated. It says: A line through point F intersects the ellipse M at points A and B. If the slopes of lines DA, DC, and DB form an arithmetic sequence, find the maximum value of tan∠DCF.Hmm. Let me parse this.We have a line passing through F(1,0) intersecting the ellipse at A and B. So, points A and B are on the ellipse and the line passes through F.Then, the slopes of DA, DC, and DB form an arithmetic sequence. So, if I denote the slopes as m_DA, m_DC, m_DB, then 2m_DC = m_DA + m_DB.We need to find the maximum value of tan of angle DCF. So, angle DCF is the angle at point C between points D, C, and F.Wait, but where is point C? The problem doesn't mention point C before. Hmm, maybe it's a typo? Or perhaps point C is another intersection point? Wait, the line passes through F and intersects the ellipse at A and B. So, maybe point C is another point? Or perhaps it's a typo for point F? Wait, the problem says "the slopes of lines DA, DC, and DB form an arithmetic sequence". So, point C must be another point on the line? Or is it a typo?Wait, the problem says "A line through point F intersects the ellipse M at points A and B." So, the line passes through F and intersects the ellipse at A and B. So, perhaps point C is another point on the line? Or maybe it's a typo and should be point F? But then DC and DF would be lines from D to F, but that doesn't make sense.Wait, perhaps point C is the midpoint of AB? Or is it a fixed point? Hmm.Wait, in the original problem statement, it's written as "the slopes of lines DA, DC, and DB form an arithmetic sequence." So, point C is another point. But where is point C located? Is it on the line? Or is it another point?Wait, maybe point C is the intersection of the line with another fixed line? Hmm, the problem doesn't specify. Maybe point C is a specific point on the line? Or perhaps it's a typo and should be point F? But then DC and DF would be the same as DA and DB, which might not make sense.Wait, perhaps point C is the center of the ellipse? The center is at (0,0). But then DC would be the line from D(1, 3/2) to C(0,0). Let me see.Wait, let me think. If point C is the center, then DC is from D(1, 3/2) to C(0,0). Then, the slopes of DA, DC, DB would be the slopes from D to A, D to C, and D to B. If these slopes form an arithmetic sequence, then 2m_DC = m_DA + m_DB.But is that the case? Or is point C another point on the line? Hmm.Wait, the problem says "A line through point F intersects the ellipse M at points A and B." So, the line passes through F and intersects the ellipse at A and B. So, the line is passing through F, A, and B. So, perhaps point C is another point on this line? Or maybe it's a typo for point F.Wait, maybe point C is the midpoint of AB? If so, then DC would be the line from D to the midpoint of AB. But the problem doesn't specify that.Alternatively, perhaps point C is a fixed point on the line, such as the intersection with another fixed line. Hmm.Wait, looking back at the original problem, it says "the slopes of lines DA, DC, and DB form an arithmetic sequence." So, point C must be a specific point on the line through F, A, and B. So, perhaps point C is a specific point on that line, such that the slopes from D to A, D to C, and D to B form an arithmetic sequence.So, point C is on the line passing through F, A, and B, and the slopes from D to these three points form an arithmetic sequence. So, point C is somewhere on that line.Therefore, point C is a point on the line such that the slopes from D to A, D to C, and D to B are in arithmetic progression.So, given that, we can parametrize the line passing through F(1,0) with some slope k, so the equation is y = k(x - 1). Then, this line intersects the ellipse at points A and B. So, we can find the coordinates of A and B in terms of k.Then, point C is another point on this line, so its coordinates can be expressed as (t, k(t - 1)) for some t.Then, the slopes of DA, DC, and DB would be:- Slope of DA: (y_A - 3/2)/(x_A - 1)- Slope of DC: (k(t - 1) - 3/2)/(t - 1)- Slope of DB: (y_B - 3/2)/(x_B - 1)Given that these slopes form an arithmetic sequence, so 2 * slope of DC = slope of DA + slope of DB.So, 2 * [(k(t - 1) - 3/2)/(t - 1)] = [(y_A - 3/2)/(x_A - 1)] + [(y_B - 3/2)/(x_B - 1)]But since points A and B lie on the line y = k(x - 1), we can express y_A = k(x_A - 1) and y_B = k(x_B - 1).Therefore, the slopes of DA and DB become:- Slope of DA: [k(x_A - 1) - 3/2]/(x_A - 1) = k - (3/2)/(x_A - 1)- Similarly, slope of DB: k - (3/2)/(x_B - 1)So, the sum of slopes DA and DB is:2k - (3/2)[1/(x_A - 1) + 1/(x_B - 1)]Therefore, the equation becomes:2 * [k - (3/2)/(t - 1)] = 2k - (3/2)[1/(x_A - 1) + 1/(x_B - 1)]Simplify both sides:Left side: 2k - 3/(t - 1)Right side: 2k - (3/2)[1/(x_A - 1) + 1/(x_B - 1)]Subtract 2k from both sides:-3/(t - 1) = - (3/2)[1/(x_A - 1) + 1/(x_B - 1)]Multiply both sides by -1:3/(t - 1) = (3/2)[1/(x_A - 1) + 1/(x_B - 1)]Divide both sides by 3:1/(t - 1) = (1/2)[1/(x_A - 1) + 1/(x_B - 1)]So,1/(t - 1) = [ (x_A + x_B - 2) ] / [2(x_A - 1)(x_B - 1) ]Wait, let me compute 1/(x_A - 1) + 1/(x_B - 1):= [ (x_B - 1) + (x_A - 1) ] / [ (x_A - 1)(x_B - 1) ]= (x_A + x_B - 2) / [ (x_A - 1)(x_B - 1) ]Therefore,1/(t - 1) = (1/2) * (x_A + x_B - 2) / [ (x_A - 1)(x_B - 1) ]So,(t - 1) = 2 [ (x_A - 1)(x_B - 1) ] / (x_A + x_B - 2 )Hmm, okay. So, we need expressions for x_A + x_B and x_A x_B.From the line y = k(x - 1) intersecting the ellipse x²/4 + y²/3 = 1.Substitute y = k(x - 1) into the ellipse equation:x²/4 + [k²(x - 1)²]/3 = 1Multiply both sides by 12 to eliminate denominators:3x² + 4k²(x² - 2x + 1) = 12Expand:3x² + 4k²x² - 8k²x + 4k² = 12Combine like terms:(3 + 4k²)x² - 8k²x + (4k² - 12) = 0So, quadratic in x: (3 + 4k²)x² - 8k²x + (4k² - 12) = 0Let me denote this as Ax² + Bx + C = 0, where:A = 3 + 4k²B = -8k²C = 4k² - 12Then, the sum of roots x_A + x_B = -B/A = 8k²/(3 + 4k²)Product of roots x_A x_B = C/A = (4k² - 12)/(3 + 4k²)So, x_A + x_B = 8k²/(3 + 4k²)x_A x_B = (4k² - 12)/(3 + 4k²)Now, let's compute (x_A - 1)(x_B - 1):= x_A x_B - x_A - x_B + 1= (4k² - 12)/(3 + 4k²) - 8k²/(3 + 4k²) + 1= [4k² - 12 - 8k² + (3 + 4k²)] / (3 + 4k²)Simplify numerator:4k² - 12 - 8k² + 3 + 4k² = (4k² - 8k² + 4k²) + (-12 + 3) = 0k² -9 = -9So, (x_A - 1)(x_B - 1) = -9/(3 + 4k²)Similarly, x_A + x_B - 2 = 8k²/(3 + 4k²) - 2 = (8k² - 2(3 + 4k²))/(3 + 4k²) = (8k² - 6 - 8k²)/(3 + 4k²) = (-6)/(3 + 4k²)Therefore, plugging back into the expression for (t - 1):(t - 1) = 2 [ (x_A - 1)(x_B - 1) ] / (x_A + x_B - 2 )= 2 [ (-9)/(3 + 4k²) ] / [ (-6)/(3 + 4k²) ]Simplify:= 2 * (-9)/(-6) = 2 * (3/2) = 3Therefore, t - 1 = 3 => t = 4So, point C is at (4, k(4 - 1)) = (4, 3k)So, regardless of the slope k, point C is always at (4, 3k). Interesting.So, point C is fixed at x=4, but y varies depending on k.Now, we need to find the maximum value of tan∠DCF.So, angle DCF is the angle at point C between points D, C, and F.So, points D(1, 3/2), C(4, 3k), and F(1, 0).We need to find tan of the angle at C between D and F.To find tan∠DCF, we can use the formula for the tangent of the angle between two lines.First, find the slopes of CD and CF.Point C is (4, 3k).Point D is (1, 3/2). So, the slope of CD is (3/2 - 3k)/(1 - 4) = (3/2 - 3k)/(-3) = (3k - 3/2)/3 = k - 1/2Similarly, point F is (1, 0). So, the slope of CF is (0 - 3k)/(1 - 4) = (-3k)/(-3) = kSo, the two lines are CD with slope m1 = k - 1/2 and CF with slope m2 = k.The angle between them is given by:tanθ = |(m2 - m1)/(1 + m1 m2)|So, tan∠DCF = |(k - (k - 1/2))/(1 + (k - 1/2)k)| = |(1/2)/(1 + k(k - 1/2))|Simplify denominator:1 + k² - (1/2)kSo, tanθ = (1/2)/(1 + k² - (1/2)k) = (1/2)/(k² - (1/2)k + 1)We need to find the maximum value of this expression.So, tanθ = 1/(2(k² - (1/2)k + 1))Let me denote f(k) = 2(k² - (1/2)k + 1). So, tanθ = 1/f(k). To maximize tanθ, we need to minimize f(k).So, find the minimum of f(k) = 2(k² - (1/2)k + 1)Simplify f(k):f(k) = 2k² - k + 2This is a quadratic in k. To find its minimum, since the coefficient of k² is positive, the minimum occurs at vertex.Vertex at k = -b/(2a) = (1)/(4) = 1/4So, the minimum value is f(1/4) = 2*(1/4)^2 - (1/4) + 2 = 2*(1/16) - 1/4 + 2 = 1/8 - 1/4 + 2 = (-1/8) + 2 = 15/8Therefore, the minimum of f(k) is 15/8, so the maximum of tanθ is 1/(15/8) = 8/15Therefore, the maximum value of tan∠DCF is 8/15.Wait, let me verify this.We have tanθ = 1/(2(k² - (1/2)k + 1)). So, to maximize tanθ, we need to minimize the denominator.The denominator is 2(k² - (1/2)k + 1). Let me compute its minimum.Compute derivative: d/dk [2k² - k + 2] = 4k - 1. Set to zero: 4k -1 =0 => k=1/4.So, yes, minimum at k=1/4.Compute f(1/4)=2*(1/16) -1/4 +2=1/8 -1/4 +2= (-1/8)+2=15/8.Thus, tanθ=1/(15/8)=8/15.So, the maximum value is 8/15.Therefore, the answer is 8/15.**Final Answer**The maximum value of tan angle DCF is boxed{dfrac{8}{15}}.

answer：Okay, so I have this problem here: A set of three numbers has both a mean and median equal to 5. The smallest number in the set is 2. I need to find the range of the set of numbers. Hmm, let me think about how to approach this step by step.First, let's recall what mean and median mean. The mean is the average of the numbers, so if the mean is 5 for three numbers, that means the total sum of the numbers should be 5 multiplied by 3, right? So, 5 times 3 is 15. That means the sum of the three numbers is 15.Next, the median being 5 tells me that when the numbers are arranged in order, the middle number is 5. Since there are three numbers, the median is the second number when they are sorted from smallest to largest. So, if the smallest number is 2, the middle number is 5, and the largest number is something we need to find.Let me denote the three numbers as a, b, and c, where a is the smallest, b is the median, and c is the largest. So, a = 2, b = 5, and c is unknown. The sum of these numbers is a + b + c = 15.Plugging in the known values, that's 2 + 5 + c = 15. So, 2 + 5 is 7, and 15 minus 7 is 8. Therefore, c must be 8.Now, the range of a set of numbers is the difference between the largest and smallest numbers. In this case, the largest number is 8 and the smallest is 2. So, the range is 8 minus 2, which is 6.Wait, let me double-check to make sure I didn't make any mistakes. The mean is 5, so the total sum should be 15. The numbers are 2, 5, and 8. Adding them up: 2 + 5 is 7, plus 8 is 15. That checks out. The median is the middle number when sorted, which is 5, as given. So, everything seems to add up correctly.I think I've got it. The range is 6.