answer：Alright, so I have this problem where I need to split the natural numbers from 1 to 12 into six pairs such that the sum of each pair is a different prime number. Hmm, okay, primes are numbers greater than 1 that have no divisors other than 1 and themselves. So, I need to find six different prime numbers that can be formed by adding two numbers from 1 to 12.First, maybe I should list out all the possible prime numbers that can be formed by adding two numbers between 1 and 12. Let's see:- The smallest sum is 1 + 2 = 3, which is prime.- Then, 1 + 4 = 5, which is also prime.- 2 + 3 = 5, but we already have 5 from 1 + 4.- 1 + 6 = 7, which is prime.- 2 + 5 = 7, again, already have 7.- 3 + 4 = 7, same as above.- 1 + 12 = 13, prime.- 2 + 11 = 13, same as above.- 3 + 10 = 13, same.- 4 + 9 = 13, same.- 5 + 8 = 13, same.- 6 + 7 = 13, same.- 5 + 6 = 11, prime.- 7 + 8 = 15, not prime.- 9 + 10 = 19, prime.- 11 + 12 = 23, prime.So, the possible prime sums are: 3, 5, 7, 11, 13, 17, 19, 23. Wait, I missed 17. How can I get 17? Let's see:- 8 + 9 = 17, yes, that's prime.Okay, so the primes we can use are: 3, 5, 7, 11, 13, 17, 19, 23.But we need only six different primes for six pairs. So, I need to choose six primes from this list and make sure that each pair sums to one of these primes, and all numbers from 1 to 12 are used exactly once.Let me try to find such pairs.Starting with the smallest number, 1. What can I pair it with to get a prime?- 1 + 2 = 3- 1 + 4 = 5- 1 + 6 = 7- 1 + 12 = 13So, options are 2, 4, 6, or 12.Let's pick 1 + 2 = 3. Now, 3 is used.Next, number 3. What can I pair it with?- 3 + 4 = 7- 3 + 8 = 11- 3 + 10 = 13So, options are 4, 8, or 10.Let's pick 3 + 8 = 11. Now, 11 is used.Next, number 4. It was paired with 1 earlier, but wait, I paired 1 with 2, so 4 is still available. What can I pair 4 with?- 4 + 9 = 13- 4 + 3 = 7 (but 3 is already paired with 8)- 4 + 5 = 9, not prime- 4 + 7 = 11 (but 11 is already used)- 4 + 11 = 15, not prime- 4 + 12 = 16, not primeHmm, only 4 + 9 = 13 is available. So, pair 4 with 9, sum is 13.Now, number 5. What can I pair it with?- 5 + 2 = 7 (but 2 is paired with 1)- 5 + 6 = 11 (but 11 is used)- 5 + 8 = 13 (but 13 is used)- 5 + 12 = 17So, pair 5 with 12, sum is 17.Next, number 6. What can I pair it with?- 6 + 7 = 13 (used)- 6 + 11 = 17 (used)- 6 + 1 = 7 (1 is paired)- 6 + 5 = 11 (used)- 6 + 10 = 16, not prime- 6 + 9 = 15, not primeWait, 6 is left. Maybe I made a wrong pairing earlier. Let me backtrack.Instead of pairing 1 with 2, let's try pairing 1 with 4.So, 1 + 4 = 5.Then, number 2. What can I pair it with?- 2 + 3 = 5 (used)- 2 + 5 = 7- 2 + 11 = 13Let's pick 2 + 5 = 7.Now, number 3. What can I pair it with?- 3 + 8 = 11- 3 + 10 = 13Let's pick 3 + 8 = 11.Number 6. What can I pair it with?- 6 + 7 = 13- 6 + 11 = 17Let's pick 6 + 7 = 13.Number 9. What can I pair it with?- 9 + 4 = 13 (used)- 9 + 8 = 17 (used)- 9 + 10 = 19So, pair 9 with 10, sum is 19.Number 12. What can I pair it with?- 12 + 1 = 13 (used)- 12 + 5 = 17 (used)- 12 + 11 = 23So, pair 12 with 11, sum is 23.Let me check all pairs:- (1,4) = 5- (2,5) = 7- (3,8) = 11- (6,7) = 13- (9,10) = 19- (11,12) = 23All sums are different primes: 5,7,11,13,19,23. Perfect!Now, for part b, splitting numbers 1 to 22 into eleven pairs with distinct prime sums.First, the total sum from 1 to 22 is (22*23)/2 = 253.If we have eleven distinct primes, their sum must be 253.Let me list primes up to 43 (since 22+21=43):Primes: 3,5,7,11,13,17,19,23,29,31,37,41,43.We need eleven distinct primes from this list that sum to 253.Let me try to select the largest primes first to see if their sum is 253.43 + 41 + 37 + 31 + 29 + 23 + 19 + 17 + 13 + 11 + 7 = let's calculate:43+41=8484+37=121121+31=152152+29=181181+23=204204+19=223223+17=240240+13=253253+11=264264+7=271Wait, that's too much. I need to sum only eleven primes to get 253.Wait, maybe I need to adjust.Alternatively, maybe it's not possible because the sum of the smallest eleven primes is 3+5+7+11+13+17+19+23+29+31+37= let's calculate:3+5=88+7=1515+11=2626+13=3939+17=5656+19=7575+23=9898+29=127127+31=158158+37=195195 is less than 253. So, we need larger primes.But the sum of the largest eleven primes:43+41+37+31+29+23+19+17+13+11+7= let's calculate:43+41=8484+37=121121+31=152152+29=181181+23=204204+19=223223+17=240240+13=253253+11=264264+7=271Wait, that's twelve primes. I need eleven.So, if I take the largest eleven primes: 43,41,37,31,29,23,19,17,13,11,7.Sum is 271, which is more than 253.So, maybe I need to replace some larger primes with smaller ones to reduce the total sum to 253.Let me see: 271 - 253 = 18. So, I need to reduce the sum by 18.Perhaps replace the largest prime, 43, with a smaller prime that is 43 - x = ?Wait, maybe replace 43 with a smaller prime such that the difference is 18.But 43 - 18 = 25, which is not prime.Alternatively, replace two primes: say, replace 43 and 41 with smaller primes.But this might complicate.Alternatively, maybe it's not possible because the sum of eleven distinct primes cannot be exactly 253.Moreover, considering that in part a, we had to carefully choose pairs, and for part b, with more numbers, it's likely not possible.So, probably, the answer is no.

answer：Okay, so I have this problem here: Let ( S_n ) be the sum of the first ( n ) terms of an arithmetic sequence ( {a_n} ). It says that ( 2S_3 = 3S_2 + 6 ), and I need to find the common difference ( d ). Hmm, alright, let's break this down step by step.First, I remember that in an arithmetic sequence, each term increases by a constant difference ( d ). So, the first term is ( a_1 ), the second term is ( a_1 + d ), the third term is ( a_1 + 2d ), and so on. That makes sense.Now, ( S_n ) is the sum of the first ( n ) terms. I think the formula for the sum of the first ( n ) terms of an arithmetic sequence is ( S_n = frac{n}{2}(2a_1 + (n-1)d) ). Let me verify that. Yeah, that seems right because it's the average of the first and last term multiplied by the number of terms. So, for ( S_3 ), it would be ( frac{3}{2}(2a_1 + 2d) ), which simplifies to ( frac{3}{2} times 2(a_1 + d) = 3(a_1 + d) ). Similarly, ( S_2 ) would be ( frac{2}{2}(2a_1 + d) = (a_1 + frac{d}{2}) times 2 = 2a_1 + d ). Wait, no, actually, ( S_2 ) is just the sum of the first two terms, so ( a_1 + (a_1 + d) = 2a_1 + d ). Yeah, that's correct.So, plugging these into the given equation: ( 2S_3 = 3S_2 + 6 ). Let's substitute the expressions for ( S_3 ) and ( S_2 ).( 2 times 3(a_1 + d) = 3 times (2a_1 + d) + 6 )Simplifying both sides:Left side: ( 6(a_1 + d) )Right side: ( 6a_1 + 3d + 6 )So, expanding the left side:( 6a_1 + 6d = 6a_1 + 3d + 6 )Hmm, okay, now I can subtract ( 6a_1 ) from both sides to simplify:( 6d = 3d + 6 )Subtract ( 3d ) from both sides:( 3d = 6 )Divide both sides by 3:( d = 2 )Wait, so the common difference ( d ) is 2? Let me double-check my steps to make sure I didn't make a mistake.1. I used the formula for the sum of an arithmetic sequence: ( S_n = frac{n}{2}(2a_1 + (n-1)d) ). That seems correct.2. Calculated ( S_3 = 3(a_1 + d) ) and ( S_2 = 2a_1 + d ). That also seems right.3. Substituted into the equation: ( 2 times 3(a_1 + d) = 3 times (2a_1 + d) + 6 ). Yep.4. Expanded both sides: ( 6a_1 + 6d = 6a_1 + 3d + 6 ). Correct.5. Subtracted ( 6a_1 ): ( 6d = 3d + 6 ). Then subtracted ( 3d ): ( 3d = 6 ). Divided by 3: ( d = 2 ). All steps look good.I think I did everything correctly. So, the common difference ( d ) is indeed 2.**Final Answer**The common difference ( d ) is boxed{2}.

answer：Okay, so I have this trigonometry problem here, and I need to find the value of this expression: cos(α - 3π/10) divided by sin(α - π/5). They also tell me that tan(α) is equal to 2 times tan(π/5). Hmm, okay. Let me try to unpack this step by step.First, I know that tan(α) = 2 tan(π/5). Maybe I can use this information to express tan(α) in terms of tan(π/5), which might help simplify the expression later on. So, let's note that down: tan(α) = 2 tan(π/5).Now, looking at the expression I need to evaluate: cos(α - 3π/10) / sin(α - π/5). I remember that there are angle subtraction formulas for sine and cosine. Maybe I can apply those here. Let me recall them:- cos(A - B) = cos A cos B + sin A sin B- sin(A - B) = sin A cos B - cos A sin BSo, if I apply these formulas to both the numerator and the denominator, I can expand them and see if something cancels out or simplifies.Starting with the numerator: cos(α - 3π/10). Let me set A = α and B = 3π/10. So, expanding that:cos(α - 3π/10) = cos α cos(3π/10) + sin α sin(3π/10)Similarly, for the denominator: sin(α - π/5). Let me set A = α and B = π/5. So, expanding that:sin(α - π/5) = sin α cos(π/5) - cos α sin(π/5)So now, my expression becomes:[cos α cos(3π/10) + sin α sin(3π/10)] / [sin α cos(π/5) - cos α sin(π/5)]Hmm, this looks a bit complicated, but maybe I can relate some of these angles or find a relationship between them. Let me think about the angles involved.I notice that 3π/10 is equal to π/2 - π/5 because π/2 is 5π/10, so 5π/10 - 2π/10 = 3π/10. So, 3π/10 = π/2 - π/5. That might be useful because I know that cos(π/2 - x) = sin x and sin(π/2 - x) = cos x. Maybe I can use that identity to rewrite cos(3π/10) and sin(3π/10).Let me try that:cos(3π/10) = cos(π/2 - π/5) = sin(π/5)sin(3π/10) = sin(π/2 - π/5) = cos(π/5)So, substituting these back into the numerator:cos(α - 3π/10) = cos α sin(π/5) + sin α cos(π/5)Wait a second, that looks familiar. That's actually the sine addition formula: sin(A + B) = sin A cos B + cos A sin B. So, cos(α - 3π/10) is equal to sin(α + π/5). Let me confirm:sin(α + π/5) = sin α cos(π/5) + cos α sin(π/5)Yes, that's exactly what we have in the numerator. So, cos(α - 3π/10) = sin(α + π/5). That's a nice simplification.So, now, my expression becomes:sin(α + π/5) / sin(α - π/5)Okay, so now I have sin(α + π/5) divided by sin(α - π/5). Maybe I can express this ratio in terms of tan(α) since I know tan(α) is given.Let me recall that sin(A + B) = sin A cos B + cos A sin B and sin(A - B) = sin A cos B - cos A sin B. So, if I write both numerator and denominator using these formulas, I can perhaps factor out sin A and cos A.Let me write them out:Numerator: sin(α + π/5) = sin α cos(π/5) + cos α sin(π/5)Denominator: sin(α - π/5) = sin α cos(π/5) - cos α sin(π/5)So, the expression becomes:[sin α cos(π/5) + cos α sin(π/5)] / [sin α cos(π/5) - cos α sin(π/5)]Hmm, this looks like a fraction where both numerator and denominator have similar terms. Maybe I can factor out sin α cos(π/5) from both, but I'm not sure if that helps. Alternatively, I can divide both numerator and denominator by cos α cos(π/5) to express everything in terms of tan α.Let me try that. Dividing numerator and denominator by cos α cos(π/5):Numerator: [sin α cos(π/5) + cos α sin(π/5)] / [cos α cos(π/5)] = [sin α / cos α + sin(π/5)/cos(π/5)] = tan α + tan(π/5)Denominator: [sin α cos(π/5) - cos α sin(π/5)] / [cos α cos(π/5)] = [sin α / cos α - sin(π/5)/cos(π/5)] = tan α - tan(π/5)So, now, the entire expression simplifies to:(tan α + tan(π/5)) / (tan α - tan(π/5))That's much simpler! Now, I know that tan α is given as 2 tan(π/5). So, let's substitute that in:tan α = 2 tan(π/5)So, replacing tan α with 2 tan(π/5):[2 tan(π/5) + tan(π/5)] / [2 tan(π/5) - tan(π/5)]Simplify numerator and denominator:Numerator: 2 tan(π/5) + tan(π/5) = 3 tan(π/5)Denominator: 2 tan(π/5) - tan(π/5) = tan(π/5)So, the expression becomes:3 tan(π/5) / tan(π/5) = 3So, the value of the expression is 3.Wait, let me double-check to make sure I didn't make any mistakes. Starting from the beginning, I used the angle subtraction formulas correctly, recognized the complementary angle, converted cos(3π/10) and sin(3π/10) into sin(π/5) and cos(π/5), respectively. Then, I noticed that the numerator became sin(α + π/5) and the denominator remained sin(α - π/5). Then, I expanded both using sine addition and subtraction formulas, divided numerator and denominator by cos α cos(π/5) to get tan α + tan(π/5) over tan α - tan(π/5). Substituted tan α = 2 tan(π/5), which gave me 3 tan(π/5) over tan(π/5), simplifying to 3. That seems correct.I think I did everything right. So, the answer should be 3, which is option C.**Final Answer**boxed{C}

answer：Okay, so I have this problem where I need to find the mass of a plate D with a given surface density. The plate is bounded by some curves, and the density function is μ = x²/(x² + y²). Hmm, okay. Let me try to break this down step by step.First, I remember that the mass of a plate can be found by integrating the density function over the area of the plate. So, the mass m is given by the double integral of μ over the region D. That makes sense. So, m = ∬_D (x²/(x² + y²)) dx dy.Now, looking at the region D, it's bounded by the curves y² - 4y + x² = 0, y² - 8y + x² = 0, y = x/√3, and x = 0. I need to figure out what this region looks like. Maybe I can rewrite those circle equations to understand them better.Let's take the first equation: y² - 4y + x² = 0. If I complete the square for the y terms, I get y² - 4y + 4 + x² = 4, which simplifies to (y - 2)² + x² = 2². So, that's a circle centered at (0, 2) with radius 2. Similarly, the second equation y² - 8y + x² = 0 can be rewritten as (y - 4)² + x² = 4². So, that's a circle centered at (0, 4) with radius 4.Okay, so the region D is between these two circles. The other boundaries are y = x/√3, which is a straight line with a slope of 1/√3, and x = 0, which is the y-axis. So, the region is kind of a lens-shaped area between the two circles, bounded on the right by the line y = x/√3 and on the left by the y-axis.Since the density function μ = x²/(x² + y²) involves both x and y, and the region is bounded by circles and lines, it might be easier to switch to polar coordinates. In polar coordinates, x = ρ cos φ and y = ρ sin φ, so x² + y² = ρ². That should simplify the density function to μ = (ρ² cos² φ)/ρ² = cos² φ. That seems manageable.Also, the circles in polar coordinates can be expressed as ρ = 4 sin φ and ρ = 8 sin φ. Let me check that. For the first circle, (y - 2)² + x² = 4. Substituting x = ρ cos φ and y = ρ sin φ, we get (ρ sin φ - 2)² + (ρ cos φ)² = 4. Expanding that, ρ² sin² φ - 4ρ sin φ + 4 + ρ² cos² φ = 4. Simplifying, ρ² (sin² φ + cos² φ) - 4ρ sin φ + 4 = 4. Since sin² + cos² = 1, this becomes ρ² - 4ρ sin φ + 4 = 4. Subtracting 4 from both sides, ρ² - 4ρ sin φ = 0, so ρ(ρ - 4 sin φ) = 0. So, ρ = 4 sin φ. Similarly, the second circle will give ρ = 8 sin φ. That makes sense.The line y = x/√3 in polar coordinates is ρ sin φ = (ρ cos φ)/√3, which simplifies to tan φ = 1/√3, so φ = π/6. And x = 0 is the y-axis, which is φ = π/2 in polar coordinates. So, the region D in polar coordinates is bounded by ρ from 4 sin φ to 8 sin φ, and φ from π/6 to π/2.So, now I can set up the double integral in polar coordinates. The mass m becomes the integral over φ from π/6 to π/2, and for each φ, ρ goes from 4 sin φ to 8 sin φ. The integrand is μ times the Jacobian determinant, which is ρ. So, m = ∫_{π/6}^{π/2} ∫_{4 sin φ}^{8 sin φ} cos² φ * ρ dρ dφ.Let me write that out:m = ∫_{π/6}^{π/2} cos² φ [ ∫_{4 sin φ}^{8 sin φ} ρ dρ ] dφFirst, I can compute the inner integral with respect to ρ. The integral of ρ dρ is (1/2)ρ². Evaluating from 4 sin φ to 8 sin φ gives:(1/2)(8 sin φ)^2 - (1/2)(4 sin φ)^2 = (1/2)(64 sin² φ - 16 sin² φ) = (1/2)(48 sin² φ) = 24 sin² φ.So, the mass integral simplifies to:m = ∫_{π/6}^{π/2} cos² φ * 24 sin² φ dφ = 24 ∫_{π/6}^{π/2} sin² φ cos² φ dφ.Hmm, integrating sin² φ cos² φ. I remember there are some trigonometric identities that can help simplify this. Let me recall that sin² φ cos² φ can be written as (1/4) sin² 2φ, because sin 2φ = 2 sin φ cos φ, so sin² 2φ = 4 sin² φ cos² φ. Therefore, sin² φ cos² φ = (1/4) sin² 2φ.So, substituting that in, the integral becomes:24 ∫_{π/6}^{π/2} (1/4) sin² 2φ dφ = 6 ∫_{π/6}^{π/2} sin² 2φ dφ.Now, I can use another identity for sin² 2φ. The identity is sin² x = (1 - cos 2x)/2. So, sin² 2φ = (1 - cos 4φ)/2.Substituting that in, the integral becomes:6 ∫_{π/6}^{π/2} (1 - cos 4φ)/2 dφ = 3 ∫_{π/6}^{π/2} (1 - cos 4φ) dφ.Now, let's split the integral:3 [ ∫_{π/6}^{π/2} 1 dφ - ∫_{π/6}^{π/2} cos 4φ dφ ].Compute each integral separately.First integral: ∫ 1 dφ from π/6 to π/2 is just the difference in φ, so π/2 - π/6 = π/3.Second integral: ∫ cos 4φ dφ. The integral of cos(ax) dx is (1/a) sin(ax). So, ∫ cos 4φ dφ = (1/4) sin 4φ. Evaluated from π/6 to π/2.So, let's compute that:At φ = π/2: (1/4) sin(4*(π/2)) = (1/4) sin(2π) = 0.At φ = π/6: (1/4) sin(4*(π/6)) = (1/4) sin(2π/3) = (1/4)*(√3/2) = √3/8.So, the second integral is 0 - √3/8 = -√3/8.Putting it all together:3 [ (π/3) - (-√3/8) ] = 3 [ π/3 + √3/8 ].Simplify this:3*(π/3) = π, and 3*(√3/8) = 3√3/8.So, the total mass m is π + 3√3/8.Wait, let me double-check that. The integral of cos 4φ from π/6 to π/2 is [ (1/4) sin 4φ ] evaluated from π/6 to π/2. At π/2, sin(4*(π/2)) = sin(2π) = 0. At π/6, sin(4*(π/6)) = sin(2π/3) = √3/2. So, it's (1/4)*(√3/2) = √3/8. So, the integral is 0 - √3/8 = -√3/8. So, when subtracting, it becomes -(-√3/8) = +√3/8. So, yes, that part is correct.Then, 3*(π/3 + √3/8) = π + 3√3/8. That seems right.So, the mass of the plate D is π + 3√3/8.**Final Answer**The mass of the plate is boxed{pi + dfrac{3sqrt{3}}{8}}.

answer：Okay, so I have this function f(x) = a ln x + (1/2)x² + (a + 1)x + 1, and there are three parts to the problem. Let me tackle each part step by step.**Part (1): When a = -1, find the interval of monotonic increase for f(x).**Alright, so first, I need to substitute a = -1 into the function. Let me write that out:f(x) = (-1) ln x + (1/2)x² + (-1 + 1)x + 1.Simplifying that, the term with x becomes 0 because (-1 + 1) is 0. So,f(x) = -ln x + (1/2)x² + 1.Now, to find the intervals where the function is increasing, I need to compute its derivative and find where it's positive.f'(x) = derivative of -ln x is -1/x, derivative of (1/2)x² is x, and the derivative of 1 is 0. So,f'(x) = -1/x + x.I need to find where f'(x) > 0. Let's set up the inequality:-1/x + x > 0.To combine these terms, I'll get a common denominator:(x² - 1)/x > 0.So, the inequality becomes (x² - 1)/x > 0.Now, let's solve this inequality. The numerator is x² - 1, which factors into (x - 1)(x + 1). The denominator is x.So, the critical points are x = -1, x = 0, and x = 1. But since the original function f(x) involves ln x, the domain is x > 0. So, we only consider x > 0.Thus, the critical points in the domain are x = 1 and x = 0. But x = 0 isn't in the domain, so we focus on x = 1.Now, let's test intervals around x = 1.- For x in (0, 1): Let's pick x = 0.5. Plugging into (x² - 1)/x: (0.25 - 1)/0.5 = (-0.75)/0.5 = -1.5 < 0.- For x in (1, ∞): Let's pick x = 2. (4 - 1)/2 = 3/2 = 1.5 > 0.So, f'(x) is positive when x > 1. Therefore, the function is increasing on (1, ∞).**Part (2): If f(x) is increasing on (0, ∞), find the range of a.**Okay, so f(x) is increasing everywhere on its domain. That means its derivative f'(x) must be non-negative for all x > 0.First, compute f'(x):f'(x) = derivative of a ln x is a/x, derivative of (1/2)x² is x, derivative of (a + 1)x is (a + 1), and derivative of 1 is 0. So,f'(x) = a/x + x + (a + 1).Simplify that:f'(x) = (a/x) + x + a + 1.Let me combine terms:f'(x) = (a/x) + x + a + 1.Hmm, maybe I can write this as a single fraction:f'(x) = (a + x² + a x + x)/x.Wait, let me check that:Wait, a/x + x + a + 1 = (a + x² + a x + x)/x.Yes, that's correct. So,f'(x) = (x² + (a + 1)x + a)/x.Now, since x > 0, the denominator x is positive. So, for f'(x) ≥ 0, the numerator must be ≥ 0 for all x > 0.So, the numerator is x² + (a + 1)x + a. Let me denote this as N(x) = x² + (a + 1)x + a.We need N(x) ≥ 0 for all x > 0.Let me analyze N(x). It's a quadratic in x.Quadratic equation: x² + (a + 1)x + a.We can factor this quadratic:Looking for two numbers that multiply to a and add to (a + 1). Hmm, let's see:x² + (a + 1)x + a = (x + 1)(x + a).Yes, because (x + 1)(x + a) = x² + (a + 1)x + a.So, N(x) = (x + 1)(x + a).Since x > 0, x + 1 is always positive (since x > 0, x + 1 > 1 > 0).Therefore, the sign of N(x) depends on (x + a).To have N(x) ≥ 0 for all x > 0, we need (x + a) ≥ 0 for all x > 0.But x > 0, so x + a ≥ 0 for all x > 0.The minimal value of x + a occurs as x approaches 0 from the right. So, as x approaches 0+, x + a approaches a.Therefore, to have x + a ≥ 0 for all x > 0, we need a ≥ 0.So, the range of a is [0, ∞).Wait, let me double-check. If a is negative, say a = -1, then near x = 0, x + a would be negative, making N(x) negative, which would make f'(x) negative, contradicting the requirement that f'(x) ≥ 0 for all x > 0. So, yes, a must be ≥ 0.**Part (3): If a > 0, and for any x₁, x₂ ∈ (0, ∞), x₁ ≠ x₂, it always holds that |f(x₁) - f(x₂)| > 2|x₁ - x₂|. Find the minimum value of a.**Hmm, okay. So, for any two distinct points in the domain, the absolute difference in function values is greater than twice the absolute difference in x-values.This sounds like a condition on the derivative. Specifically, the Mean Value Theorem tells us that for some c between x₁ and x₂, |f(x₁) - f(x₂)| = |f'(c)| |x₁ - x₂|. So, if |f(x₁) - f(x₂)| > 2|x₁ - x₂|, then |f'(c)| > 2 for all c in (0, ∞).Therefore, we need f'(x) > 2 for all x > 0.Wait, but f'(x) is equal to (x + 1)(x + a)/x, as we found earlier.So, f'(x) = (x + 1)(x + a)/x.We need f'(x) > 2 for all x > 0.So, let's write that inequality:(x + 1)(x + a)/x > 2.Simplify this:Multiply both sides by x (since x > 0, inequality sign remains the same):(x + 1)(x + a) > 2x.Expand the left side:x² + (a + 1)x + a > 2x.Bring all terms to the left:x² + (a + 1)x + a - 2x > 0.Simplify:x² + (a - 1)x + a > 0.So, we have x² + (a - 1)x + a > 0 for all x > 0.We need this quadratic to be positive for all x > 0.Let me denote Q(x) = x² + (a - 1)x + a.We need Q(x) > 0 for all x > 0.Now, since Q(x) is a quadratic, its graph is a parabola opening upwards (since the coefficient of x² is 1 > 0). So, the minimum occurs at the vertex.The vertex is at x = -b/(2a) for Q(x) = ax² + bx + c. Wait, in our case, Q(x) = x² + (a - 1)x + a, so a = 1, b = (a - 1).So, the x-coordinate of the vertex is x = -(a - 1)/(2*1) = (1 - a)/2.Now, since we are concerned with x > 0, we need to check if the vertex is in the domain x > 0.Case 1: If (1 - a)/2 > 0, which implies a < 1.Then, the minimum of Q(x) occurs at x = (1 - a)/2.We need Q((1 - a)/2) > 0.Let me compute Q at x = (1 - a)/2.Q((1 - a)/2) = [(1 - a)/2]^2 + (a - 1)*[(1 - a)/2] + a.Let me compute each term:First term: [(1 - a)/2]^2 = (1 - 2a + a²)/4.Second term: (a - 1)*(1 - a)/2 = (a - 1)(- (a - 1))/2 = - (a - 1)^2 / 2.Third term: a.So, putting it all together:Q = (1 - 2a + a²)/4 - (a - 1)^2 / 2 + a.Let me simplify step by step.First, expand (a - 1)^2: (a - 1)^2 = a² - 2a + 1.So, the second term becomes - (a² - 2a + 1)/2.Now, let's write all terms with a common denominator of 4:First term: (1 - 2a + a²)/4.Second term: -2(a² - 2a + 1)/4.Third term: 4a/4.So, combining:[1 - 2a + a² - 2a² + 4a - 2 + 4a]/4.Simplify numerator:1 - 2a + a² - 2a² + 4a - 2 + 4a.Combine like terms:a² - 2a² = -a².-2a + 4a + 4a = 6a.1 - 2 = -1.So, numerator is -a² + 6a - 1.Thus, Q((1 - a)/2) = (-a² + 6a - 1)/4.We need this to be > 0:(-a² + 6a - 1)/4 > 0.Multiply both sides by 4 (positive, so inequality remains):-a² + 6a - 1 > 0.Multiply both sides by -1 (inequality sign reverses):a² - 6a + 1 < 0.Now, solve a² - 6a + 1 < 0.Find roots of a² - 6a + 1 = 0.Using quadratic formula:a = [6 ± sqrt(36 - 4)]/2 = [6 ± sqrt(32)]/2 = [6 ± 4*sqrt(2)]/2 = 3 ± 2*sqrt(2).So, the quadratic a² - 6a + 1 is less than 0 between its roots: 3 - 2√2 < a < 3 + 2√2.But in this case, we have a < 1 (from Case 1). So, the intersection of a < 1 and 3 - 2√2 < a < 3 + 2√2.Compute 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172.So, 0.172 < a < 1.Therefore, in this case, a must be in (3 - 2√2, 1).Case 2: If (1 - a)/2 ≤ 0, which implies a ≥ 1.In this case, the vertex is at x ≤ 0, so the minimum of Q(x) on x > 0 occurs at x approaching 0 from the right.Compute limit as x approaches 0+ of Q(x):lim x→0+ x² + (a - 1)x + a = 0 + 0 + a = a.Since a > 0, Q(x) approaches a, which is positive. Also, as x increases, Q(x) tends to infinity because it's a quadratic opening upwards.Therefore, in this case, Q(x) is always positive on x > 0.So, for a ≥ 1, Q(x) > 0 for all x > 0.Combining both cases:From Case 1: 3 - 2√2 < a < 1.From Case 2: a ≥ 1.So, overall, a > 3 - 2√2.But since a > 0, the minimal value of a is 3 - 2√2.Wait, but 3 - 2√2 is approximately 0.172, which is positive, so it's acceptable.Therefore, the minimal value of a is 3 - 2√2.Let me just verify this.If a = 3 - 2√2, then Q(x) = x² + (a - 1)x + a.Compute Q(x):x² + (3 - 2√2 - 1)x + (3 - 2√2) = x² + (2 - 2√2)x + (3 - 2√2).We can check if this quadratic is always positive.Compute its discriminant:D = [2 - 2√2]^2 - 4*1*(3 - 2√2).Compute [2 - 2√2]^2 = 4 - 8√2 + 8 = 12 - 8√2.Compute 4*1*(3 - 2√2) = 12 - 8√2.So, D = (12 - 8√2) - (12 - 8√2) = 0.So, the quadratic has a double root, meaning it touches the x-axis at one point, but since the coefficient of x² is positive, it's always non-negative. However, we need Q(x) > 0 for all x > 0.But since the discriminant is zero, it touches the x-axis at x = [-b/(2a)] = [-(2 - 2√2)]/(2*1) = (2√2 - 2)/2 = √2 - 1.So, x = √2 - 1 ≈ 0.414, which is in (0, ∞). So, at x = √2 - 1, Q(x) = 0.But we need Q(x) > 0 for all x > 0. So, if a = 3 - 2√2, then Q(x) is zero at x = √2 - 1, which is in the domain. Therefore, f'(x) would be 2 at that point, but the condition is |f(x₁) - f(x₂)| > 2|x₁ - x₂|, which requires f'(x) > 2 everywhere. So, at x = √2 - 1, f'(x) = 2, which doesn't satisfy f'(x) > 2.Therefore, a must be greater than 3 - 2√2 to ensure that Q(x) > 0 for all x > 0, making f'(x) > 2 for all x > 0.Hence, the minimal value of a is 3 - 2√2, but since at a = 3 - 2√2, f'(x) is equal to 2 at x = √2 - 1, which doesn't satisfy the strict inequality. Therefore, a must be greater than 3 - 2√2. But the question asks for the minimum value of a such that the condition holds. Since a can approach 3 - 2√2 from above, the infimum is 3 - 2√2, but since a must be greater than that, the minimal value in the permissible range is 3 - 2√2.Wait, but in the problem statement, it says "find the minimum value of the real number a". Since a must be greater than 3 - 2√2, the minimal value is 3 - 2√2, but technically, a must be greater than that. However, in optimization problems like this, the minimal value is often taken as the infimum, which is 3 - 2√2.Alternatively, perhaps I made a mistake in the earlier step. Let me re-examine.When a = 3 - 2√2, Q(x) has a double root at x = √2 - 1, so f'(x) = 2 at that point. But the condition is |f(x₁) - f(x₂)| > 2|x₁ - x₂|, which requires f'(x) > 2 everywhere. Therefore, a must be strictly greater than 3 - 2√2. However, the question asks for the minimum value of a, which would be the smallest a such that the condition holds. Since a can be as close to 3 - 2√2 as desired from above, the minimum value is 3 - 2√2.But in terms of real numbers, the minimum is achieved when a approaches 3 - 2√2 from above. However, since a must be greater than 3 - 2√2, the minimal value is 3 - 2√2, but it's not achieved; it's the infimum.Wait, but in the problem statement, it says "find the minimum value of the real number a". So, perhaps the answer is 3 - 2√2, even though at that exact value, the condition is not strictly satisfied, but it's the boundary case.Alternatively, maybe I need to ensure that f'(x) ≥ 2 for all x > 0, but the problem says |f(x₁) - f(x₂)| > 2|x₁ - x₂|, which is a strict inequality. Therefore, f'(x) must be strictly greater than 2 everywhere. So, a must be greater than 3 - 2√2. But the question asks for the minimum value of a, so the minimal a is 3 - 2√2, but since a must be greater, perhaps the answer is 3 - 2√2, considering it's the threshold.I think in such problems, the minimal value is taken as the boundary, so 3 - 2√2 is the answer.**Final Answer**(1) boxed{(1, +infty)}(2) boxed{[0, +infty)}(3) boxed{3 - 2sqrt{2}}

answer：First, determine how many questions Sasha has completed by subtracting the remaining questions from the total number of questions: 60 - 30 = 30 questions.Next, calculate the number of hours she worked by dividing the number of questions she completed by her rate of completing questions: 30 ÷ 15 = 2 hours.