🧠:[<think>Alright, let's see. The question is asking which of the options cannot be the sum of the interior angles of a polygon. The options are A: 720°, B: 2160°, C: 1080°, and D: 800°. Hmm, okay. I remember that the formula for the sum of interior angles of a polygon is (n-2)*180°, where n is the number of sides. So, each of these angles should be a multiple of 180°, but adjusted by adding 360° or something? Wait, no, the formula is (n-2)*180. So, the sum must be a multiple of 180°, right? Because 180° multiplied by an integer (n-2). Let me check each option.First, option A: 720°. Let's divide 720 by 180. 720 ÷ 180 = 4. Then, according to the formula, (n-2) = 4, so n would be 6. A hexagon. That makes sense. Hexagons have 6 sides, so the sum is 720°. So A is possible.Option B: 2160°. Let's do the same. 2160 ÷ 180. Let me compute that. 180*12 is 2160, right? 180*10=1800, 180*2=360, so 1800+360=2160. So (n-2)=12, so n=14. A 14-sided polygon. That's possible. So B is okay.Option C: 1080°. Divide by 180. 1080 ÷ 180 = 6. So (n-2)=6, which gives n=8. An octagon. Yeah, octagons have a sum of 1080°. So C is valid.Option D: 800°. Let's divide 800 by 180. 180*4=720, 180*5=900. So 800 is between 4 and 5 times 180. So 800 ÷ 180 = 4.444... So (n-2) would be approximately 4.444. But n has to be an integer, right? Because the number of sides can't be a fraction. So 4.444 isn't an integer. Therefore, there's no integer n such that (n-2)*180 = 800°. So D is not possible. Therefore, the answer should be D.Wait, but let me double-check. Maybe I made a mistake in calculation. Let me verify each one again. For 800°, 800 divided by 180. Let's do 800 ÷ 180. 180*4=720, subtract that from 800: 800-720=80. 80/180=4/9. So total is 4 and 4/9. So 4.444..., which is not an integer. So yeah, n-2 would have to be 4.444, which is not possible. Therefore, D can't be the sum. The other options all gave integer values for n-2, so they are okay. Therefore, the answer is D:800°.Final Answerboxed{D}</think>To determine which of the given options cannot be the sum of the interior angles of a polygon, we use the formula for the sum of the interior angles: ((n-2) times 180^circ), where (n) is the number of sides. The sum must be a multiple of 180°, and (n-2) must be an integer.- Option A: 720° (720 div 180 = 4) ((n-2) = 4 implies n = 6) A hexagon has 6 sides, so this is possible.- Option B: 2160° (2160 div 180 = 12) ((n-2) = 12 implies n = 14) A 14-sided polygon is possible.- Option C: 1080° (1080 div 180 = 6) ((n-2) = 6 implies n = 8) An octagon has 8 sides, so this is possible.- Option D: 800° (800 div 180 = 4.444ldots) ((n-2) = 4.444ldots) is not an integer, so this is not possible.Thus, the answer is (boxed{D}).,<think>Okay, let's see. The question is asking which of the options cannot be the sum of the interior angles of a polygon. The options are A: 720°, B: 2160°, C: 1080°, and D: 800°. Hmm, I remember that the formula for the sum of the interior angles of a polygon is (n-2) multiplied by 180°, where n is the number of sides. So, maybe I need to check each of these options and see if they fit into that formula. First, let's recall the formula: Sum = (n - 2) × 180°, right? So, if we rearrange the formula to solve for n, we can check if each given sum corresponds to an integer number of sides. Because n has to be an integer (a polygon can't have a fractional number of sides), if the result isn't an integer, then that sum isn't possible.Let's start with option A: 720°. Let me plug that into the formula. So, 720 = (n - 2) × 180. If I divide both sides by 180, that gives me 720 ÷ 180 = n - 2. Calculating that, 720 divided by 180 is 4. So, 4 = n - 2. Then, adding 2 to both sides, n = 6. 6 is a whole number, so a 6-sided polygon (hexagon) would have a sum of interior angles equal to 720°. So that works. Therefore, option A is possible.Moving on to option B: 2160°. Again, using the formula: 2160 = (n - 2) × 180. Divide both sides by 180: 2160 ÷ 180 = n - 2. Let me compute that. 2160 divided by 180. Hmm, 180 times 12 is 2160, right? Because 180 x 10 is 1800, plus 180 x2 is 360, so 1800 + 360 = 2160. So, 12 = n - 2. Then n = 14. 14 is an integer, so a 14-sided polygon would have an interior angle sum of 2160°. So B is possible.Option C: 1080°. Applying the formula again: 1080 = (n - 2) × 180. Dividing both sides by 180: 1080 ÷ 180 = n - 2. 1080 divided by 180 is 6, because 180 x6 = 1080. So, 6 = n - 2, meaning n = 8. An octagon has 8 sides, which is a whole number. So, 1080° is possible. Thus, C is okay.Now option D: 800°. Let's plug this into the formula. 800 = (n - 2) × 180. Divide both sides by 180: 800 ÷ 180 = n - 2. Let me calculate that. 800 divided by 180. Hmm, 180 x4 = 720, so 800 - 720 = 80. So that's 4 and 80/180, which simplifies to 4 and 4/9. So, 800 divided by 180 is 4.444... So, n - 2 = 4.444..., which means n = 6.444... But n has to be a whole number, right? You can't have a polygon with 6.444 sides. That doesn't make sense. So 800° isn't a possible sum for the interior angles of a polygon. Therefore, D is the answer.Let me double-check my calculations to make sure I didn't make any mistakes. For D: 800 divided by 180. Let's see: 180 x4 = 720, subtract that from 800, you get 80. 80/180 is 4/9, which is approximately 0.444. So yes, 4.444... So n would need to be 6.444..., which isn't possible. So, D can't be the sum. The other options all gave whole numbers for n, so they are valid. Therefore, the answer must be D.</think>To determine which option cannot be the sum of the interior angles of a polygon, we use the formula for the sum of interior angles: ((n - 2) times 180^circ), where (n) is the number of sides. Each sum must correspond to an integer (n).- Option A: 720° [ 720 = (n - 2) times 180 implies n - 2 = frac{720}{180} = 4 implies n = 6 ] (n = 6) (hexagon) is valid.- Option B: 2160° [ 2160 = (n - 2) times 180 implies n - 2 = frac{2160}{180} = 12 implies n = 14 ] (n = 14) is valid.- Option C: 1080° [ 1080 = (n - 2) times 180 implies n - 2 = frac{1080}{180} = 6 implies n = 8 ] (n = 8) (octagon) is valid.- Option D: 800° [ 800 = (n - 2) times 180 implies n - 2 = frac{800}{180} approx 4.444 implies n approx 6.444 ] (n) is not an integer, so this sum is impossible.Answer: boxed{D}]

🔑:Analysis: The sum of the interior angles of an n-sided polygon (n≥3) is (n-2) times 180°. Therefore, the sum of the interior angles of a polygon must be an integer multiple of 180°.Since 800° is not an integer multiple of 180°, the correct answer is boxed{text{D}}.

🧠:[<think>Okay, so I have this problem about the function f(x) = (a x² + x + a)/(2 e^x). There are two parts: part I asks for the value of a (which is non-negative) such that the maximum value of f(x) is 5/(2e). Part II is about finding the range of b when a is non-positive, such that f(x) ≤ (b ln(x+1))/2 for all x in [0, infinity). Let me tackle them one by one.Starting with part I. Since a is non-negative, and we need to find a such that the maximum of f(x) is 5/(2e). To find the maximum of f(x), I'll need to take its derivative, set it equal to zero, and solve for x. Then, substitute that x back into f(x) and set it equal to 5/(2e) to solve for a.First, compute f'(x). Let's recall that the derivative of a function u(x)/v(x) is (u’v - uv’)/v². Here, u = a x² + x + a, so u’ = 2a x + 1. The denominator v = 2 e^x, so v’ = 2 e^x. Therefore, f’(x) = [(2a x + 1)(2 e^x) - (a x² + x + a)(2 e^x)] / (2 e^x)^2. Wait, hold on. Let me write that more carefully:f(x) = (a x² + x + a)/(2 e^x). So, the numerator is (a x² + x + a) and the denominator is 2 e^x. Therefore, the derivative f’(x) is [d/dx (a x² + x + a) * 2 e^x - (a x² + x + a) * d/dx (2 e^x)] / (2 e^x)^2. Wait, actually, applying the quotient rule correctly: (u’v - uv’) / v². Here, u = a x² + x + a, so u’ = 2a x + 1. v = 2 e^x, so v’ = 2 e^x. Then, f’(x) = [ (2a x + 1)(2 e^x) - (a x² + x + a)(2 e^x) ] / (2 e^x)^2.Simplify numerator: Factor out 2 e^x:Numerator = 2 e^x [ (2a x + 1) - (a x² + x + a) ].Let me compute the expression inside the brackets:(2a x + 1) - (a x² + x + a) = 2a x + 1 - a x² - x - a = -a x² + (2a - 1) x + (1 - a).Therefore, f’(x) = [2 e^x (-a x² + (2a - 1) x + (1 - a))] / (4 e^(2x)) ) = [ -a x² + (2a - 1)x + (1 - a) ] / (2 e^x).So, f’(x) = [ -a x² + (2a - 1)x + (1 - a) ] / (2 e^x).To find critical points, set numerator equal to zero: -a x² + (2a - 1)x + (1 - a) = 0.Multiply both sides by -1: a x² - (2a - 1)x - (1 - a) = 0.Which simplifies to a x² - (2a - 1)x + (a - 1) = 0.So quadratic equation: a x² - (2a - 1)x + (a - 1) = 0.Let me write that again: a x² - (2a - 1)x + (a - 1) = 0.Let me compute discriminant D:D = [-(2a - 1)]² - 4 * a * (a - 1) = (2a - 1)^2 - 4a(a - 1).Expand (2a - 1)^2: 4a² -4a +1.4a² -4a +1 -4a(a -1) = 4a² -4a +1 -4a² +4a = 1.So discriminant D = 1. So roots are x = [ (2a -1) ± sqrt(1) ] / (2a).Thus, x = [2a -1 ±1]/(2a).Therefore, two critical points:x1 = [2a -1 +1]/(2a) = (2a)/(2a) = 1.x2 = [2a -1 -1]/(2a) = (2a -2)/(2a) = (a -1)/a.But wait, since a is a real number and a >=0 for part I. But a could be zero? Let's check. If a =0, then f(x) = (0 +x +0)/(2 e^x) = x/(2 e^x). Then, the maximum of x/(2 e^x) occurs at x=1, since derivative of x/e^x is (1 -x)/e^x, set to zero when x=1. Then f(1) = 1/(2 e). But in the problem, the maximum is 5/(2 e). So a cannot be zero. Therefore, a>0.Therefore, a is positive. Then, critical points are x=1 and x=(a -1)/a.But (a -1)/a. Let's see, since a>0, (a -1)/a = 1 - 1/a. Depending on whether a>1 or a<1, this can be positive or negative. Since x is in [0, infinity), we need to check if (a -1)/a is in [0, infinity). If a >=1, then (a -1)/a >=0. If a <1, then (a -1)/a <0. Therefore, for a >=1, two critical points: x=1 and x=(a -1)/a. For 0 <a <1, only x=1 is in the domain [0, infinity), because x=(a -1)/a would be negative.Therefore, depending on the value of a, the critical points inside the domain [0, infinity) are:- For a >=1: x=1 and x=(a -1)/a.- For 0 <a <1: only x=1.Therefore, to find the maximum of f(x), we need to evaluate f at these critical points and check which one gives the maximum value. Also, we should check the behavior as x approaches infinity. Since f(x) is (a x² + x +a)/(2 e^x). As x approaches infinity, the denominator grows exponentially, while the numerator is quadratic, so f(x) tends to zero. At x=0, f(0) = (0 +0 +a)/(2 e^0) = a/2.So, the maximum is either at x=1, or at x=(a -1)/a (if a >=1), or at x=0. Wait, but when a >=1, x=(a -1)/a is in [0, infinity), so we need to compare f(1) and f((a -1)/a) and f(0). But f(0) = a/2, and as a >=1, a/2 is at least 1/2. But the maximum is supposed to be 5/(2e) ≈5/(2*2.718)≈0.92, which is larger than 1/2=0.5. So possible.But the problem states that the maximum is 5/(2e). So, let's compute f(1):f(1) = (a*1 +1 +a)/(2 e^1) = (2a +1)/(2 e). If this is the maximum, then we set (2a +1)/(2e) =5/(2e). Therefore, 2a +1=5, so 2a=4, a=2. So that's possible.But also, if a >=1, then there's another critical point at x=(a -1)/a. Let's compute f at that point.Let me denote x0=(a -1)/a. Then,f(x0)= [a x0² +x0 +a]/[2 e^{x0}]First, compute x0: x0=(a -1)/a=1 -1/a.Let me compute numerator:a x0² +x0 +a = a*(1 -2/a +1/a²) + (1 -1/a) +aLet's expand:a*(1 -2/a +1/a²) = a -2 +1/aThen adding (1 -1/a) +a:Total numerator: (a -2 +1/a) +1 -1/a +a = a -2 +1/a +1 -1/a +a = 2a -1.Therefore, numerator is 2a -1.Therefore, f(x0) = (2a -1)/(2 e^{x0}) = (2a -1)/(2 e^{1 -1/a}) = (2a -1)/(2 e * e^{-1/a}) ) = (2a -1) e^{1/a}/(2 e).Therefore, f(x0) = (2a -1) e^{1/a}/(2 e).So, for a >=1, we have two critical points, x=1 and x0=1 -1/a, with f(1)=(2a +1)/(2e) and f(x0)=(2a -1) e^{1/a}/(2e). We need to compare these two to see which one is larger.If a=2, for example, then f(1)=(4 +1)/(2e)=5/(2e), which is given as the maximum. Let's check f(x0) when a=2:x0=1 -1/2=1/2f(x0)= (4 -1) e^{1/2}/(2e)=3 e^{1/2}/(2e)=3/(2 e^{1/2})≈3/(2*1.648)≈3/3.296≈0.91. But 5/(2e)≈5/5.436≈0.92. So they are close, but f(x0) is actually a bit less. Wait, maybe exactly?Wait, compute 3/(2 sqrt(e)) versus 5/(2 e). Let's see:sqrt(e)≈1.6487, e≈2.7183.3/(2*1.6487)=3/3.2974≈0.915/(2*2.7183)=5/5.436≈0.919.So, 0.91 vs 0.919. So f(1) is slightly larger. So when a=2, the maximum is at x=1. But maybe for other a>1, f(x0) could be larger than f(1). Let's check.Suppose a=3. Then f(1)=(6 +1)/(2e)=7/(2e)≈7/5.436≈1.287. But f(x0)= (6 -1) e^{1/3}/(2e)=5 e^{1/3}/(2e). e^{1/3}≈1.3956, so 5*1.3956≈6.978, divided by 2e≈5.436, so 6.978/5.436≈1.283. So f(1)≈1.287 vs f(x0)≈1.283. So f(1) is still slightly larger. Hmm. Maybe as a increases, f(1) increases while f(x0) also increases but less?Wait, when a approaches infinity, f(1)=(2a +1)/(2e) ≈a/e, which goes to infinity. But f(x0)= (2a -1) e^{1/a}/(2e). As a approaches infinity, 1/a approaches 0, e^{1/a}≈1 +1/a. So (2a -1)(1 +1/a)/2e≈ (2a -1 +2 -1/a)/2e≈(2a +1)/2e≈same as f(1). So, as a becomes large, f(x0) approaches f(1). But for finite a>1, which one is larger?Take a=1.5. Then f(1)= (3 +1)/2e=4/(2e)=2/e≈0.7358. f(x0)=(3 -1)e^{2/3}/(2e)=2 e^{2/3}/(2e)=e^{2/3}/e≈e^{-1/3}≈0.7165. So here, f(1) is larger.Another test: a=1. Let's check. When a=1, x0=(1-1)/1=0. So f(x0)=f(0)=1/2. f(1)=(2 +1)/2e=3/(2e)≈0.551. So f(0)=0.5 and f(1)≈0.551. So maximum is at x=1. But when a=1, the critical points are x=1 and x=0. So f(0)=a/2=0.5, f(1)=3/(2e). So in this case, maximum is at x=1.Wait, but for a=1, the other critical point is x=0. So x=0 is a critical point. So maybe for a=1, we need to check x=0 and x=1. But since a=1 is included in the case a >=1. So for a=1, x0=(1-1)/1=0, so both critical points are at x=1 and x=0.But when a=1, f(x0)=f(0)=1/2=0.5, and f(1)=3/(2e)≈0.551. So maximum is at x=1. So in all cases where a >=1, the maximum is at x=1. Wait, but when a=2, we saw that f(1)=5/(2e)≈0.919 and f(x0)=3/(2 e^{1/2})≈0.91. So very close, but f(1) is slightly larger. So maybe for a >=1, the maximum is at x=1. If that's the case, then the maximum value is (2a +1)/(2e). Therefore, set (2a +1)/(2e)=5/(2e), so 2a +1=5 => a=2. That's straightforward.But wait, to confirm that for a >=1, the maximum is at x=1, we can check the derivative's behavior. Let me check the sign of f’(x) around x=1 and x0.Alternatively, since for a >=1, x0=(a -1)/a is in [0,1). Let's check the values of f(x) at x0 and x=1. For example, with a=2, as above, x0=1 -1/2=1/2. f(1/2)= [a*(1/4) +1/2 +a]/(2 e^{1/2})= [ (2*(1/4) +1/2 +2) ]/(2 e^{0.5})= [0.5 +0.5 +2]/ (2*1.6487)=3/3.297≈0.91. Whereas f(1)=5/(2e)≈0.919. So indeed, f(1) is larger. Similarly, for a=3, x0=2/3, f(x0)=5 e^{1/3}/(2e)≈5*1.3956/5.436≈6.978/5.436≈1.283, while f(1)=7/(2e)=7/5.436≈1.287. So f(1) is still slightly larger. So maybe in general, for a >=1, the maximum is at x=1. Therefore, setting f(1)=5/(2e) gives a=2. Therefore, the answer is a=2.But wait, just to be thorough, let's check if there's any a>1 where f(x0) is larger than f(1). Suppose a=1.1. Then, x0=(1.1 -1)/1.1≈0.0909. Compute f(x0):Numerator: a x0² +x0 +a =1.1*(0.0909)^2 +0.0909 +1.1≈1.1*0.00826 +0.0909 +1.1≈0.00909 +0.0909 +1.1≈1.1999. Denominator:2 e^{0.0909}≈2*1.095≈2.19. So f(x0)=1.1999/2.19≈0.548. f(1)= (2*1.1 +1)/(2e)=3.2/(2*2.718)≈1.6/2.718≈0.589. So f(1) is still larger. Therefore, for a=1.1, maximum is at x=1. So seems like for all a >=1, maximum is at x=1.Therefore, if the maximum of f(x) is 5/(2e), then set f(1)=5/(2e). As above, (2a +1)/(2e)=5/(2e) => 2a +1=5 => a=2. Therefore, answer for part I is a=2.Now part II: For any a <=0, f(x) <= (b ln(x+1))/2 holds true for x in [0, infinity). Find the range of real number b.So, for all a <=0 and for all x >=0, (a x² +x +a)/(2 e^x) <= (b ln(x+1))/2. Multiply both sides by 2 e^x (since e^x >0):a x² +x +a <= b e^x ln(x+1).But since a <=0, we have that a x² +x +a <=x +a, because a x² <=0. Also, but x +a <=x, since a <=0. But x is in [0, infinity). So the left-hand side a x² +x +a is <=x +a. But we need to find the minimal b such that for all a <=0, the inequality holds. Wait, no. The problem says "for any a <=0, f(x) <= (b ln(x+1))/2". So for each x in [0, infinity), and for all a <=0, (a x² +x +a)/(2 e^x) <= (b ln(x+1))/2. Therefore, we can rearrange this as:For all x >=0, and for all a <=0,(a x² +x +a) <= b e^x ln(x+1).But since this must hold for all a <=0, we need the maximum of (a x² +x +a) over a <=0 to be <= b e^x ln(x+1).So for each x, find sup_{a <=0} [a x² +x +a] = sup_{a <=0} [a(x² +1) +x]. Since a is a variable <=0, and x is fixed. The expression is linear in a: a(x² +1) +x. Since x² +1 >=1 >0, the coefficient of a is positive. Therefore, the supremum over a <=0 is when a is as large as possible, which is a=0. Therefore, the supremum is x. Therefore, sup_{a <=0} [a x² +x +a] =x. Therefore, the inequality becomes x <= b e^x ln(x+1). Therefore, for all x >=0, x <= b e^x ln(x+1). Therefore, b >= x/(e^x ln(x+1)) for all x >=0. Therefore, the minimal such b is the supremum of x/(e^x ln(x+1)) over x >=0.Therefore, the range of b is [b0, infinity), where b0 is the supremum of the function g(x)=x/(e^x ln(x+1)) for x >=0.Thus, we need to find the maximum value of g(x)=x/(e^x ln(x+1)) on x >=0. Then, b must be at least this maximum value.Therefore, compute sup_{x >=0} [x / (e^x ln(x+1))].To find the maximum of g(x), compute its derivative and find critical points.Let me denote g(x)=x / [e^x ln(x+1)]. Let's write it as g(x)=x e^{-x} / ln(x+1).Compute derivative g’(x):Using quotient rule: if u =x e^{-x}, v=ln(x+1), then g= u/v.Thus, g’=(u’ v - u v’)/v².Compute u’:u =x e^{-x}, so u’=e^{-x} -x e^{-x}=e^{-x}(1 -x).v=ln(x+1), so v’=1/(x+1).Therefore,g’(x)= [e^{-x}(1 -x) * ln(x+1) - x e^{-x} * 1/(x+1)] / [ln(x+1)]²Factor out e^{-x} from numerator:g’(x)= e^{-x [ (1 -x) ln(x+1) - x/(x+1) ] / [ln(x+1)]²Set numerator equal to zero:(1 -x) ln(x+1) - x/(x+1) =0.Thus, (1 -x) ln(x+1)=x/(x+1).We need to solve this equation for x >=0.This seems challenging analytically. Let's try to find critical points numerically.First, check x=0:Left side: (1 -0) ln(1) -0/(0+1)=0 -0=0. So equality holds. x=0 is a critical point.Check x=1:Left side: (1 -1) ln(2) -1/2=0 -0.5= -0.5 <0.Right side is zero? Wait, equation is (1 -x) ln(x+1) -x/(x+1)=0. At x=1, left side is 0*ln(2) -1/2= -1/2≠0. So not zero. So g’(1)= negative.Check x approaching infinity: As x→infty, ln(x+1)~ln x, (1 -x) ln(x+1) ~ -x ln x, and x/(x+1) ~1. So left side ~ -x ln x -1 → -infty. So equation not satisfied.At x=0, derivative is zero. Let's check behavior near x=0. For x approaching 0 from the right:Compute limit as x→0+ of g’(x). Since numerator is e^{-x}[...] ≈1*( (1 -x)(x -x²/2 +...) -x/(1 +x) ) ≈ (1 -x)(x -x²/2) -x(1 -x +x² -...) ≈ (x -x²/2 -x² +x³/2) -x +x² -x³ ≈ x - (3/2)x² + (1/2)x³ -x +x² -x³ ≈ (-1/2)x² - (1/2)x³. So near x=0, numerator is approx. - (1/2)x², denominator is [ln(x+1)]² ≈x². So g’(x) ≈ (-1/2 x²)/x²= -1/2. Therefore, near x=0, derivative is approaching -1/2. So x=0 is a critical point but the derivative goes from 0 (at x=0) to negative immediately. So x=0 is a local maximum? Wait, but wait. At x=0, g(x)=0 / [1 * ln1] but ln1=0. So g(x) is 0/0 undefined. Wait, need to compute the limit as x approaches 0+.Original function g(x)=x/(e^x ln(x+1)). As x approaches 0, e^x≈1 +x, ln(x+1)≈x -x²/2. So g(x)=x / [ (1 +x)(x -x²/2) ] ≈x / [x (1 +x)(1 -x/2) ] ≈1 / [ (1 +x)(1 -x/2) ]≈1 / [1 +x -x/2 +...]≈1 / (1 +x/2 +...)→1 as x→0. Therefore, the limit as x approaches 0 of g(x) is 1. But at x=0, g(0)=0/(1*0) undefined, but the limit is 1.Therefore, the function g(x) approaches 1 as x→0+, has a critical point at x=0 (but the function is extended continuously to 1 at x=0). Then, the derivative near x=0 is approaching -1/2, so decreasing. Let's check at x=0.5:Compute g(0.5)=0.5/(e^{0.5} ln1.5)≈0.5/(1.6487*0.4055)≈0.5/(0.668)≈0.748.Compute derivative at x=0.5:Numerator: (1 -0.5) ln(1.5) -0.5/(1.5)=0.5*0.4055 -0.333≈0.2027 -0.333≈-0.1303. Denominator: [ln(1.5)]²≈0.1644. So g’(0.5)= e^{-0.5}*(-0.1303)/0.1644≈0.6065*(-0.1303)/0.1644≈-0.6065*0.792≈-0.48. So negative derivative. So decreasing.At x=0.5, g(x)≈0.748. At x=1, g(1)=1/(e ln2)≈1/(2.718*0.693)≈1/1.885≈0.530. So decreasing from x=0.5 to x=1.Check at x=0.2:g(0.2)=0.2/(e^{0.2} ln1.2)≈0.2/(1.2214*0.1823)≈0.2/(0.2227)≈0.9.Compute derivative at x=0.2:Numerator: (1 -0.2) ln(1.2) -0.2/(1.2)=0.8*0.1823 -0.1667≈0.1458 -0.1667≈-0.0209.Denominator: [ln(1.2)]²≈0.0332.So g’(0.2)= e^{-0.2}*(-0.0209)/0.0332≈0.8187*(-0.0209)/0.0332≈-0.8187*0.629≈-0.515. Still negative. So decreasing.Wait, but as x approaches 0, g(x) approaches 1, then decreases. So maximum at x=0? But the function is defined as limit 1 at x=0, then decreasing. So the supremum is 1? But wait, when x approaches 0, the limit is 1. But let's check for x>0, the function is less than 1. Because near x=0, g(x)≈1, but is the function ever greater than 1? Let's compute the limit as x approaches 0:Using Taylor series:ln(x+1)=x -x²/2 +x³/3 -...e^{-x}=1 -x +x²/2 -x³/6 +...So g(x)=x/[ (1 -x +x²/2 -...)(x -x²/2 +x³/3 -...) ]=x/[ (x -x²/2 +x³/3 -...)(1 -x +x²/2 -...) ].Multiply the denominators:(x)(1 -x +x²/2) -x²/2(1 -x) +x³/3(1) +...≈x -x² +x³/2 -x²/2 +x³/2 +...=x - (3/2)x² +x³ +...Therefore, denominator≈x - (3/2)x² +x³.Thus, g(x)=x / [x - (3/2)x² +x³]≈1 / [1 - (3/2)x +x²]. So as x→0+, this approaches 1 from above? Wait, denominator is 1 - (3/2)x +x², so 1/[1 - (3/2)x +x²] ≈1 + (3/2)x +... So g(x)≈1 + (3/2)x. Therefore, near x=0, g(x)≈1 + (3/2)x, which is slightly above 1. So the function g(x) actually exceeds 1 near x=0. But since the limit as x→0+ is 1, but for x>0, approaching 0, it might go above 1. Wait, this contradicts? Wait, let me check again.Wait, let me compute g(x) for very small x. Let x=0.0001:g(0.0001)=0.0001/(e^{0.0001} ln(1.0001))≈0.0001/(1.0001 * 0.0001). Since ln(1.0001)≈0.0001 -0.000000005 +...≈0.000099995. So denominator≈1.0001*0.000099995≈0.000100005. Therefore, g(x)≈0.0001/0.000100005≈0.99995. So slightly less than 1.Wait, maybe my previous approximation was wrong. Let's recast:g(x)=x/(e^x ln(x+1)).Let x approach 0. Let me write ln(x+1)=x -x²/2 +x³/3 -x^4/4 +...e^x=1 +x +x²/2 +x³/6 +...Therefore, e^x ln(x+1)= (1 +x +x²/2 +...)(x -x²/2 +x³/3 -...)=x(1 +x +x²/2)(1 -x/2 +x²/3 -x³/4 +...)Multiply term by term:First, multiply 1*(1 -x/2 +x²/3 -x³/4)=1 -x/2 +x²/3 -x³/4Then, x*(1 -x/2 +x²/3)=x -x²/2 +x³/3Then, x²/2*(1 -x/2)=x²/2 -x³/4Adding up:1 -x/2 +x²/3 -x³/4 +x -x²/2 +x³/3 +x²/2 -x³/4Combine like terms:Constant term:1x terms: (-x/2 +x)=x/2x² terms: (x²/3 -x²/2 +x²/2)=x²/3x³ terms: (-x³/4 +x³/3 -x³/4)= (-1/4 -1/4 +1/3)x³= (-1/2 +1/3)x³= -1/6 x³Higher terms: negligible.So e^x ln(x+1)≈x/2 +x²/3 - (1/6)x³ +...Therefore, g(x)=x/[x/2 +x²/3 - (1/6)x³]≈1/[1/2 +x/3 -x²/6]≈2/[1 + (2x)/3 - (x²)/3]≈2[1 - (2x)/3 + (x²)/3 + (4x²)/9 +...] by expanding 1/(1 +ε)≈1 -ε +ε².Therefore,≈2[1 - (2x)/3 + (x²)/3 + (4x²)/9]≈2 - (4x)/3 + (2x²)/3 + (8x²)/9≈2 - (4x)/3 + (14x²)/9.Therefore, g(x)≈2 - (4x)/3 + (14x²)/9 as x approaches 0.Wait, this contradicts previous numeric. Wait, when x=0.0001,g(x)≈2 -4/3*0.0001 +14/9*(0.0001)^2≈2 -0.0001333≈1.9998667. But numerically, when x=0.0001, g(x)=0.0001/(e^{0.0001} ln(1.0001))≈0.0001/(1.0001*0.000099995)≈0.0001/0.000100005≈0.99995. So there's a discrepancy. Therefore, my series expansion must be wrong.Wait, perhaps I made a mistake in expanding e^x ln(x+1). Let me recompute:ln(x+1)=x -x²/2 +x³/3 -x^4/4 +...e^x=1 +x +x²/2 +x³/6 +x^4/24 +...Multiply them together:e^x * ln(x+1)= [1 +x +x²/2 +x³/6 +...][x -x²/2 +x³/3 -x^4/4 +...]Multiply term by term:1*x =x1*(-x²/2)= -x²/21*x³/3= x³/31*(-x^4/4)= -x^4/4x*x= x²x*(-x²/2)= -x³/2x*x³/3= x^4/3x*(-x^4/4)= -x^5/4x²/2*x= x³/2x²/2*(-x²/2)= -x^4/4x²/2*x³/3= x^5/6x³/6*x= x^4/6...Collecting terms up to x³:x -x²/2 +x³/3 +x² -x³/2 +x³/2 +...Wait:x term: xx² terms: -x²/2 +x²= (1 -1/2)x²= x²/2x³ terms: x³/3 -x³/2 +x³/2= x³/3So up to x³: e^x ln(x+1)=x +x²/2 +x³/3 +...Therefore, e^x ln(x+1)=x +x²/2 +x³/3 + higher terms.Thus, g(x)=x / [x +x²/2 +x³/3 +...] =1 / [1 +x/2 +x²/3 +...] ≈1 -x/2 -x²/3 + (x/2)^2 +... by expanding 1/(1 +ε)≈1 -ε +ε².Thus, g(x)≈1 -x/2 -x²/3 +x²/4=1 -x/2 -x²/12.So as x approaches 0 from the right, g(x)≈1 -x/2 -x²/12. Therefore, slightly less than 1. So the function approaches 1 from below. Which matches the numeric result of g(0.0001)≈0.99995.Therefore, the maximum value of g(x)=x/(e^x ln(x+1)) occurs at x=0 with limit 1, and decreases afterwards. However, wait, but if for x>0, g(x)<1, then the supremum is 1. But according to the previous calculation, near x=0, g(x) is approaching 1 from below. But when x approaches 0, from the right, the function approaches 1. So the supremum is 1. However, the function never actually reaches 1, but gets arbitrarily close. So the supremum is 1. Therefore, to satisfy x <=b e^x ln(x+1) for all x>=0, b must be >=1. Because if b=1, then x <=e^x ln(x+1). But does this hold?Wait, check for x=0: 0<=1*1*ln1=0, equality holds. For x>0, need x <=e^x ln(x+1). Let's check at x=1: 1 <=e*ln2≈2.718*0.693≈1.885. True. At x=0.5:0.5 <=e^{0.5} ln1.5≈1.648*0.405≈0.668. 0.5<=0.668, true. At x approaching infinity: x <=e^x ln(x+1). As x→infty, e^x grows exponentially, ln(x+1) grows logarithmically, so RHS ~e^x ln x, which tends to infinity. So holds. Therefore, for all x>=0, x<=e^x ln(x+1). Is this true?Wait, actually, let's consider function h(x)=e^x ln(x+1) -x. We need to check if h(x)>=0 for all x>=0.Compute h(0)=1*0 -0=0.Compute derivative h’(x)=e^x ln(x+1) +e^x/(x+1) -1.At x=0: h’(0)=0 +1 -1=0.Compute second derivative h’’(x)=e^x ln(x+1) +e^x/(x+1) +e^x/(x+1) -e^x/(x+1)^2.But this might be complicated. Let's check h’(x):h’(x)=e^x ln(x+1) +e^x/(x+1) -1.At x=0: h’(0)=0 +1 -1=0.Compute h’(0.1):e^0.1 ln1.1 + e^0.1 /1.1 -1≈1.105*0.0953 +1.105/1.1 -1≈0.105 +1.0045 -1≈0.1095>0.So h’(x) becomes positive after x=0. Therefore, h(x) is increasing after x=0. Since h(0)=0 and h’(x) becomes positive, h(x)>=0 for x>=0.Therefore, e^x ln(x+1) >=x for all x>=0. Thus, if we take b=1, the inequality x <=b e^x ln(x+1) holds for all x>=0. Moreover, since the supremum of g(x)=x/(e^x ln(x+1)) is 1, then the minimal such b is 1. Therefore, the range of b is [1, infinity).Wait, but earlier, the supremum of g(x) was approaching 1 as x approaches 0. So in order to have x/(e^x ln(x+1)) <=b for all x>=0, we need b>=1, because for any b<1, there exists x near 0 such that g(x)=x/(e^x ln(x+1))≈1, which would exceed b. Therefore, the minimal b is 1, hence the range is [1, ∞).Therefore, answer for part II is b >=1, so the range is [1, infinity).Final Answer(I) boxed{2}(II) boxed{[1,+infty)}</think>Given the function ( f(x) = dfrac{ax^2 + x + a}{2e^x} ):(I) To find the value of ( a ) such that the maximum value of ( f(x) ) is ( dfrac{5}{2e} ) when ( a geq 0 ):1. Compute the derivative ( f'(x) ) using the quotient rule: [ f'(x) = frac{(2ax + 1)2e^x - (ax^2 + x + a)2e^x}{(2e^x)^2} = frac{-ax^2 + (2a - 1)x + (1 - a)}{2e^x} ]2. Set the numerator equal to zero to find critical points: [ -ax^2 + (2a - 1)x + (1 - a) = 0 ]3. Solve the quadratic equation. The discriminant is 1, leading to roots: [ x = 1 quad text{and} quad x = frac{a - 1}{a} ]4. For ( a geq 1 ), both critical points are valid. For ( 0 < a < 1 ), only ( x = 1 ) is valid.5. Evaluate ( f(x) ) at critical points. For ( x = 1 ): [ f(1) = frac{2a + 1}{2e} ]6. Set ( frac{2a + 1}{2e} = frac{5}{2e} ) to find ( a = 2 ).Thus, the value of ( a ) is (boxed{2}).(II) To find the range of ( b ) such that ( f(x) leq dfrac{b ln(x+1)}{2} ) for all ( x in [0, +infty) ) and ( a leq 0 ):1. Rearrange the inequality: [ a x^2 + x + a leq b e^x ln(x+1) ]2. For ( a leq 0 ), the left-hand side is maximized at ( a = 0 ), giving ( x ).3. The inequality reduces to ( x leq b e^x ln(x+1) ) for all ( x geq 0 ).4. Find the supremum of ( frac{x}{e^x ln(x+1)} ). Analysis shows the supremum is 1 as ( x to 0 ).5. Therefore, ( b geq 1 ).The range of ( b ) is (boxed{[1, +infty)}).,<think>Okay, let me try to solve these two parts. Starting with part (I). So we have the function f(x) = (a x² + x + a)/(2 e^x), and we need to find the value of a (where a ≥ 0) such that the maximum value of f(x) is 5/(2e). Hmm.First, since we need to find the maximum value of f(x), I should probably find its derivative and set it equal to zero to find critical points. Then check those points to find the maximum. Let me compute f'(x).The function f(x) is (a x² + x + a) divided by 2 e^x. So, using the quotient rule for derivatives: if f(x) = u/v, then f'(x) = (u'v - uv')/v². Alternatively, maybe it's easier to write it as (a x² + x + a) * (2 e^x)^{-1} and use the product rule. Let me see.But actually, 2 is a constant, so f(x) = (1/2) * (a x² + x + a)/e^x. So maybe compute the derivative of (a x² + x + a)/e^x first, then multiply by 1/2. Let's do that.Let u = a x² + x + a, v = e^x. Then the derivative of u/v is (u'v - uv')/v². So u' is 2a x + 1, and v' is e^x. So:f'(x) = (1/2) * [ (2a x + 1)e^x - (a x² + x + a)e^x ] / e^(2x)Simplify numerator: e^x [ (2a x + 1) - (a x² + x + a) ] = e^x [2a x + 1 - a x² - x - a]Simplify inside the brackets: -a x² + (2a x - x) + (1 - a) = -a x² + x(2a - 1) + (1 - a)Therefore, f'(x) = (1/2) * [ -a x² + (2a - 1)x + (1 - a) ] / e^xSet this derivative equal to zero to find critical points. Since e^x is never zero, the numerator must be zero:-a x² + (2a - 1)x + (1 - a) = 0Multiply both sides by -1 to make it easier:a x² - (2a - 1)x - (1 - a) = 0So quadratic equation: a x² - (2a -1)x + (a -1) = 0Wait, let me check the signs. Original equation after multiplying by -1:a x² - (2a -1)x - (1 - a) = 0But - (1 - a) is a -1 + a, right? Wait, no: - (1 - a) = -1 + a. So the constant term is a -1.Therefore, the quadratic equation is a x² - (2a -1)x + (a -1) = 0We can write this as a x² - (2a -1)x + (a -1) = 0Let me see if this quadratic can be factored. Let's check discriminant first.Discriminant D = [ - (2a -1) ]² - 4 * a * (a -1 )Compute D:= (2a -1)^2 -4a(a -1)= 4a² -4a +1 -4a² +4aSimplify: 4a² -4a +1 -4a² +4a = 1So discriminant D =1. Therefore, roots are:x = [ (2a -1) ± sqrt(1) ] / (2a)So x = [2a -1 ±1]/(2a)Thus, two roots:x1 = [2a -1 +1]/(2a) = (2a)/(2a) = 1x2 = [2a -1 -1]/(2a) = (2a -2)/(2a) = (a -1)/aSo critical points at x =1 and x=(a -1)/aBut note that (a -1)/a is equal to 1 - 1/a. Since a ≥0, but if a=0, this would be undefined. Wait, when a=0, the original function becomes f(x)=(0 +x +0)/(2e^x)=x/(2e^x). Let's check if a=0 is allowed here. The problem says a ≥0, so a=0 is allowed. But when a=0, the critical points would be different, since the original quadratic becomes 0 x² + x +0, which is just x. So f(x)=x/(2e^x). Let me check its derivative.f'(x) = (1*e^x - x e^x)/(2e^{2x}) ) = (1 - x)/(2 e^x)Setting derivative to zero: 1 -x=0 => x=1. So for a=0, critical point is x=1. But in our general case, when a ≠0, we have two critical points: x=1 and x=(a -1)/a. However, when a approaches 0 from the positive side, (a -1)/a approaches -infty. So for a>0, we have two critical points, but one of them is at x=(a -1)/a. But since x is in [0, +infty), because in part (I) we need to find the maximum over all real x? Wait, the problem says "the maximum value of the function f(x)", but it doesn't specify the domain. Maybe it's over all real numbers? But given that f(x) is defined for all real x, but e^x grows without bound as x approaches infinity, so for x->infty, the denominator 2e^x dominates, so f(x) tends to zero. Similarly, as x approaches -infty, the numerator is a quadratic (if a>0) which tends to +infty if a>0, and denominator tends to zero (since e^x tends to zero), so f(x) tends to +infty if a>0 as x->-infty. But the problem states a ≥0, so if a>0, then as x approaches -infty, f(x) goes to infinity. But the problem is asking for the maximum value. However, if a>0, then f(x) can go to infinity as x approaches -infty, so the maximum would be infinity. But the problem says the maximum is 5/(2e). Therefore, there must be a restriction on the domain. Wait, maybe I misread.Wait, the problem says "the maximum value of the function f(x) is 5/(2e)". So if a>0, then as x approaches -infty, f(x) approaches infinity, which contradicts the maximum being finite. Therefore, the problem must implicitly assume that the domain is x ≥0? Or maybe part (I) is for x≥0? Wait, but the problem statement is:"(I) If a ≥0, the maximum value of the function f(x) is 5/(2e), find the value of the real number a;"It doesn't specify the domain, but considering part (II) mentions x ∈ [0, +infty), maybe part (I) is also over x ∈ [0, +infty). That would make sense. Otherwise, for a>0, as x approaches -infty, f(x) would go to infinity, making the maximum infinity. So perhaps the problem assumes x ≥0 for part (I). Let me check.The problem statement for part (I) doesn't specify the domain, but part (II) does. So maybe the original function is defined for x in [0, +infty) for both parts? Wait, the problem defines f(x) = (a x² +x +a)/(2 e^x). If there's no restriction, the domain is all real numbers. However, in part (II), they specify x ∈ [0, +infty). But part (I) doesn't. Hmm. This is confusing.Wait, but since part (I) says "the maximum value of the function f(x)", which, if a>0, would be infinity unless restricted to x ≥0. Therefore, probably part (I) is also over x ≥0. Otherwise, the problem is inconsistent. Let me assume that both parts are over x ∈ [0, +infty). Otherwise, part (I) would be impossible for a>0. So proceeding under that assumption.So critical points in [0, +infty). So x=1 is a critical point. The other critical point is x=(a -1)/a. So we need to check whether (a -1)/a is in [0, +infty).Compute (a -1)/a. If a >0, then (a -1)/a = 1 - 1/a. So if a ≥1, then 1 -1/a ≥0, so x=(a -1)/a is in [0, +infty). If 0 <a <1, then (a -1)/a is negative, so not in the domain. Therefore, for a ≥1, there are two critical points: x=1 and x=(a -1)/a. For 0 ≤a <1, only x=1 is in the domain.Therefore, for a ≥1, critical points at x=1 and x=(a -1)/a. For 0 ≤a <1, critical point only at x=1. We need to evaluate f(x) at these critical points and endpoints to find the maximum.But the endpoints are x=0 and x approaching infinity. As x approaches infinity, f(x) tends to zero. At x=0, f(0) = (0 +0 +a)/(2 e^0) = a/2. So f(0)=a/2.Therefore, the possible maximum is either at x=0, x=1, or x=(a -1)/a (if a ≥1). So we need to compare these values.First, let's compute f(1). f(1) = (a*1 +1 +a)/(2 e^1) = (2a +1)/(2e). So f(1)=(2a +1)/(2e)For a ≥1, we also have f((a -1)/a). Let me compute that. Let x=(a -1)/a. Then substitute into f(x):f(x) = [a x² +x +a]/(2 e^x)Compute x = (a -1)/a. So x =1 -1/a.So compute numerator: a*(1 -1/a)^2 + (1 -1/a) +aFirst term: a*(1 - 2/a +1/a²) = a -2 +1/aSecond term: 1 -1/aThird term: aSo adding them up:(a -2 +1/a) + (1 -1/a) +a = a -2 +1/a +1 -1/a +a = 2a -1Therefore, numerator is 2a -1. Denominator is 2 e^{x} = 2 e^{1 -1/a}Therefore, f((a -1)/a) = (2a -1)/(2 e^{1 -1/a})So for a ≥1, the critical points are x=1 and x=1 -1/a, and we need to compare f(1), f(1 -1/a), and f(0)=a/2.We need to find the maximum value of f(x) over x ≥0. So let's analyze for different ranges of a.Case 1: a ≥1. Then critical points at x=1 and x=1 -1/a. Also check x=0.Compute f(1) = (2a +1)/(2e), f(1 -1/a) = (2a -1)/(2 e^{1 -1/a}), and f(0)=a/2.We need to compare these three values. The maximum of these is given to be 5/(2e). So need to set the maximum among them equal to 5/(2e) and solve for a.Case 2: 0 ≤a <1. Then only critical point at x=1. So compare f(0)=a/2, f(1)=(2a +1)/(2e), and as x approaches infinity, f(x) approaches zero. So maximum is max(a/2, (2a +1)/(2e)). We need to check when this maximum equals 5/(2e). So set max(a/2, (2a +1)/(2e))=5/(2e). Therefore, either a/2=5/(2e) which gives a=5/e ≈1.839, but since a <1 in this case, this is impossible. Or (2a +1)/(2e)=5/(2e), which implies 2a +1=5 => a=2. But a=2 is in the previous case (a ≥1). Therefore, for 0 ≤a <1, the maximum cannot be 5/(2e). Hence, the solution must be in the case a ≥1.Therefore, focus on a ≥1. So we need to compare f(1), f(1 -1/a), and f(0)=a/2.First, let's check f(0)=a/2. Since a ≥1, f(0) ≥1/2. But 5/(2e) ≈5/(5.436)≈0.919, so 1/2=0.5 <0.919. Therefore, f(0) is less than 5/(2e) when a=5/(2e)*2=5/e≈1.839. Wait, maybe not. Wait, f(0)=a/2. So if a/2=5/(2e), then a=5/e≈1.839. So if a=5/e, then f(0)=5/(2e). But in that case, we need to check whether the maximum is at x=0 or somewhere else. Wait, but when a=5/e, which is approximately1.839, which is greater than1, so in the case a ≥1.But we need to see whether f(0)=5/(2e) is actually the maximum. However, if a=5/e, then f(0)=5/(2e), and f(1)=(2*(5/e)+1)/(2e)= (10/e +1)/(2e). Let's compute 10/e≈3.679. So 10/e +1≈4.679. Divided by 2e≈5.436, so 4.679/5.436≈0.861, which is less than 5/(2e)≈0.919. Similarly, f(1 -1/a) when a=5/e:First, compute x=1 -1/a=1 - e/5≈1 -0.551≈0.449Compute f(x)= (2a -1)/(2 e^{1 -1/a})= (2*(5/e) -1)/(2 e^{1 -e/5})Calculate numerator: 10/e -1≈3.679 -1=2.679Denominator: 2 e^{1 -e/5}≈2 e^{1 -0.551}=2 e^{0.449}≈2*1.567≈3.134Therefore, f(x)≈2.679/3.134≈0.855, which is still less than 5/(2e). So when a=5/e, the maximum is at x=0, which is 5/(2e). But is this the case? Wait, but when a increases, f(0)=a/2 increases. However, when a increases, the critical points might have higher values. Wait, perhaps for some a, the maximum is at x=1 -1/a. Hmm. Let's check.Alternatively, maybe the maximum is achieved at x=1. Let's see. Suppose a is such that f(1)=5/(2e). Then:(2a +1)/(2e)=5/(2e) => 2a +1=5 => 2a=4 =>a=2.So if a=2, then f(1)=5/(2e). Then, check other critical points. For a=2, x=1 -1/2=0.5. Compute f(0.5):Numerator=2*(0.5)^2 +0.5 +2=2*(0.25)+0.5+2=0.5+0.5+2=3. Denominator=2 e^{0.5}≈2*1.6487≈3.297. So f(0.5)=3/3.297≈0.91, which is less than 5/(2e)≈0.919. Also, f(0)=2/2=1. So f(0)=1≈1.0, which is greater than 5/(2e). But wait, this contradicts. If a=2, then f(0)=2/2=1, which is greater than 5/(2e). But the problem states that the maximum value is 5/(2e). Therefore, a=2 cannot be the answer because f(0)=1 which is greater. So this suggests inconsistency. Wait, perhaps the problem is only considering local maxima? Or maybe I made a wrong assumption.Wait, hold on. If a=2, then the function f(x) has critical points at x=1 and x=0.5. At x=0, f(0)=1, which is higher than the critical value at x=1, which is 5/(2e)≈0.919. Therefore, the maximum would actually be 1, not 5/(2e). Therefore, there's a problem here. So the initial thought that setting f(1)=5/(2e) gives a=2, but then f(0) is higher. Therefore, that can't be the correct value of a.Therefore, perhaps the maximum is achieved at x=1 -1/a. Let's suppose that for some a, f(1 -1/a)=5/(2e). Then:(2a -1)/(2 e^{1 -1/a})=5/(2e)Multiply both sides by 2 e^{1 -1/a}:2a -1=5 e^{-1/a}So equation: 2a -1=5 e^{-1/a}This is a transcendental equation and might not have an algebraic solution. We might need to solve this numerically. Let me check possible values.Let me test a=2:Left side: 2*2 -1=3. Right side:5 e^{-1/2}≈5*0.6065≈3.032. Close. So 3 vs 3.032. So left≈3, right≈3.032. So 2a -1≈5 e^{-1/a} when a≈2. So a≈2 is close.Let me compute more accurately. Let a=2:Left:4 -1=3. Right:5 e^{-0.5}≈5*0.60653066≈3.0326533. So difference:3 -3.0326533≈-0.0326533. So left is less than right.So need a slightly larger than 2. Let's try a=2.01:Left:2*2.01 -1=4.02 -1=3.02Right:5 e^{-1/2.01}≈5 e^{-0.497512}≈5*0.608≈5*0.608=3.04So left=3.02, right≈3.04. Still left < right.a=2.02:Left=2*2.02 -1=4.04 -1=3.04Right=5 e^{-1/2.02}≈5 e^{-0.4950}≈5*0.610≈3.05Still left < right.a=2.05:Left=2*2.05 -1=4.1 -1=3.1Right=5 e^{-1/2.05}≈5 e^{-0.4878}≈5*0.613≈3.065Left=3.1 > right≈3.065. So cross-over between a=2.02 and a=2.05.Use linear approximation:At a=2.02, left=3.04, right≈3.05At a=2.03:Left=2*2.03 -1=4.06 -1=3.06Right=5 e^{-1/2.03}≈5 e^{-0.4926}≈5*0.610≈3.05So left=3.06, right≈3.05. Therefore, left > right.Therefore, the solution is between a=2.02 and 2.03.Wait, but this seems messy. Alternatively, maybe there is an exact solution. Let me check.Equation: 2a -1 =5 e^{-1/a}Let me let t=1/a, so a=1/t. Then equation becomes:2*(1/t) -1=5 e^{-t}Multiply both sides by t:2 - t =5 t e^{-t}So 2 - t =5 t e^{-t}This is still transcendental. Not sure.Alternatively, maybe assume that a=2 is the answer, even though when a=2, f(0)=1 which is larger. But the problem says the maximum value is 5/(2e). So if a=2, then maximum value is 1, which is not 5/(2e). So that can't be.Wait, maybe my assumption about the domain is wrong. If the domain is all real numbers, then for a>0, the maximum is at x approaching -infty, which is infinity. But the problem states the maximum is 5/(2e). Therefore, the domain must be restricted. Perhaps part (I) is over x ≥0? Then, even if a=2, the maximum at x=0 is 1, which is more than 5/(2e). Hence, a=2 is invalid. Therefore, contradiction arises.Wait, maybe the problem is considering local maxima? But the problem says "the maximum value of the function f(x)", which would be the global maximum. Therefore, there must be an error in my approach.Wait, let's re-examine. If a=5/(2e), then f(0)=a/2=5/(4e)≈0.46, which is less than 5/(2e). Wait, but earlier we thought a=5/e≈1.839 gives f(0)=5/(2e). Let me check:If a=5/(2e), then f(0)= (5/(2e))/2=5/(4e). Which is less than 5/(2e). So to have f(0)=5/(2e), need a=5/e. Then f(0)=5/(2e). So a=5/e≈1.839.But then for a=5/e, compute f(1)= (2a +1)/(2e)= (2*(5/e)+1)/(2e)= (10/e +1)/(2e). Compute 10/e≈3.678, so 10/e +1≈4.678. Divided by 2e≈5.436, gives≈0.861, which is less than 5/(2e)≈0.919. Then compute f(1 -1/a). For a=5/e, x=1 - e/5≈1 -0.551≈0.449.Compute f(0.449)= [a x² +x +a]/(2 e^x). Let's compute:a=5/e≈1.839, x≈0.449a x²≈1.839*(0.449)^2≈1.839*0.201≈0.369x≈0.449a≈1.839Sum:0.369 +0.449 +1.839≈2.657Denominator:2 e^{0.449}≈2*1.567≈3.134So f(x)≈2.657/3.134≈0.848, which is less than 5/(2e). Therefore, when a=5/e, the maximum is at x=0, which is 5/(2e). But when a=5/e, which is approximately1.839, which is greater than1, so in the case a ≥1. Therefore, even though there is another critical point at x=1 -1/a≈0.449, f(x) at that point is less than 5/(2e). So f(0)=5/(2e), and since for x>0, f(x) decreases from x=0 to some point and then increases? Wait, no. Wait, if a=5/e, then f(x) has critical points at x=1 and x=1 -1/a≈0.449. So between x=0 and x≈0.449, the function might be increasing or decreasing? Wait, let's compute the derivative sign.For a=5/e≈1.839, critical points at x=1 and x≈0.449. Let's check the derivative.f'(x) = (1/2) * [ -a x² + (2a -1)x + (1 -a) ] / e^xAt a=5/e≈1.839, the quadratic in the numerator is -1.839 x² + (2*1.839 -1)x + (1 -1.839)≈-1.839x² +2.678x -0.839The roots of this quadratic are x=1 and x=(a -1)/a≈0.449.Therefore, the derivative changes sign as follows: For x <0.449, derivative is negative? Let's pick x=0:Plug x=0 into the numerator quadratic: -1.839*0 +2.678*0 -0.839= -0.839 <0. Therefore, derivative is negative at x=0. So function is decreasing at x=0.Between x=0.449 and x=1, the quadratic is positive, since the quadratic opens downward (coefficient of x² is negative). So between the roots, quadratic is positive, so derivative is positive. Therefore, function is increasing from x≈0.449 to x=1.Then, for x>1, quadratic becomes negative again, so derivative negative, function decreasing.Therefore, for a=5/e, the function decreases from x=0 to x≈0.449, then increases to x=1, then decreases to zero as x approaches infinity.Therefore, the maximum value is either at x=0 or at x=1.But f(0)=5/(2e)≈0.919, and f(1)≈0.861. Therefore, the maximum is at x=0.Therefore, if we set a=5/e≈1.839, then the maximum of f(x) is indeed 5/(2e). Therefore, the answer is a=5/e.Wait, but why earlier when I thought a=5/e gives f(0)=5/(2e) and that this is the maximum. Then, that would be the value. So perhaps the answer is a=5/e.But let's verify if there's another a where the maximum at x=1 -1/a equals 5/(2e). Suppose that the maximum is at x=1 -1/a, which for a ≥1. So set f(1 -1/a)=5/(2e). Then (2a -1)/(2 e^{1 -1/a})=5/(2e). Multiply both sides by 2e^{1 -1/a}:2a -1=5 e^{-1/a}So equation 2a -1=5 e^{-1/a}This is a transcendental equation. Let me see if a=2 is a solution. Then left=3, right=5 e^{-0.5}≈3.032. Close. But not exact. So need to solve numerically.Let me try a=2.02:Left=2*2.02 -1=4.04 -1=3.04Right=5 e^{-1/2.02}≈5 e^{-0.495}≈5*0.610≈3.05Close.a=2.02 gives left=3.04, right≈3.05a=2.03:Left=2*2.03 -1=4.06 -1=3.06Right=5 e^{-1/2.03}≈5 e^{-0.4926}≈5*0.611≈3.055Still left < right.a=2.04:Left=3.08Right≈5 e^{-0.4902}≈5*0.612≈3.06Now left=3.08, right≈3.06. Therefore, left > right.So solution is between a=2.03 and 2.04.Using linear approximation:At a=2.03: left=3.06, right≈3.055At a=2.03, left - right≈0.005At a=2.04: left=3.08, right≈3.06, left - right≈0.02Wait, but actually, perhaps more precise calculation:Compute right at a=2.03:1/2.03≈0.4926e^{-0.4926}≈e^{-0.5}*e^{0.0074}≈0.6065*1.0075≈0.611Therefore, 5*0.611≈3.055Left at a=2.03 is 3.06. So 3.06 vs 3.055. So f(1 -1/a)≈3.06/(2 e^{1 -1/a})≈3.06/(2 e^{0.5074})≈3.06/(2*1.66)≈3.06/3.32≈0.921. Which is close to 5/(2e)≈0.919. So even closer.Similarly, at a=2.02:Right=5 e^{-1/2.02}=5 e^{-0.495}≈5*0.610≈3.05Left=3.04. So f(1 -1/a)=3.04/(2 e^{0.505})≈3.04/(2*1.657)≈3.04/3.314≈0.917, which is≈5/(2e). So maybe a≈2.02 gives f≈0.917, close to 0.919. So a≈2.02. But 5/(2e)≈0.919, so the exact value would require a≈2.02.But this seems complicated. However, the problem states that the maximum is 5/(2e). But when a=5/e≈1.839, the maximum is at x=0 with value 5/(2e). Therefore, maybe this is the intended answer. However, when a=5/e, even though there is another critical point, the maximum is still at x=0. Therefore, the answer is a=5/e.But let me check if for a=5/e, whether there is any x where f(x) is greater than 5/(2e). For example, take x slightly negative. Wait, but the domain. If the problem is over x ∈ [0, +infty), then x=0 is the left endpoint, and the function decreases from x=0 to x≈0.449, then increases to x=1, then decreases. Therefore, maximum at x=0.But if the problem is over all real x, then as x approaches -infty, f(x) tends to infinity for a>0, making the maximum infinity. But since the problem states the maximum is 5/(2e), the domain must be x ≥0.Therefore, the answer is a=5/e.But let's check units. 5/(2e) is a constant. If a=5/e, then yes, f(0)=a/2=5/(2e), and since for x>0, the function is first decreasing then increasing then decreasing, but the maximum at x=0. Therefore, the answer is a=5/e.But 5/e is approximately1.839, which is greater than1, so in the case a≥1. However, for this a, the maximum is at x=0. Therefore, the answer is a=5/e.But let me confirm this by checking the derivative. If a=5/e, then f'(x)=0 at x=1 and x=1 -e/5≈0.449. As we saw, at x=0, the derivative is negative, so the function is decreasing at x=0. Therefore, the maximum at x=0 is a local maximum? Wait, but if the function is decreasing at x=0, how can x=0 be a maximum? Wait, no. If the derivative at x=0 is negative, that means the function is decreasing as we move from left to right at x=0. But since x=0 is the left endpoint, the only way to have a maximum at x=0 is if the function decreases to the right of x=0. Therefore, even though the derivative at x=0 is negative, since we can't go left of x=0, the maximum is at x=0. That makes sense.Therefore, if a=5/e, then f(0)=5/(2e), and for all x>0, f(x) <5/(2e). Therefore, this satisfies the problem's condition.Therefore, the answer is a=5/e. Therefore, boxed answer: boxed{dfrac{5}{e}}.Wait, but let's check once more. When a=5/e, f(0)=a/2=5/(2e), which is the given maximum. For x>0, as established, f(x) first decreases then increases then decreases, but since f(1)= (2a +1)/(2e)= (10/e +1)/(2e). Since 10/e≈3.679, so 10/e +1≈4.679, divided by 2e≈5.436, gives≈0.861, which is less than 5/(2e). Similarly, the other critical point at x≈0.449 gives f(x)≈0.848, which is also less. Therefore, the maximum is indeed at x=0, and the value is 5/(2e). Therefore, the correct value of a is 5/e.Now part (II): If for any a ≤0, f(x) ≤ (b ln(x+1))/2 holds for x ∈ [0, +infty), find the range of real number b.So for all a ≤0 and all x ≥0, (a x² +x +a)/(2 e^x) ≤ (b ln(x+1))/2. Multiply both sides by 2:(a x² +x +a)/e^x ≤ b ln(x+1)So we need to find the minimal b such that for all a ≤0 and all x ≥0, (a x² +x +a)/e^x ≤ b ln(x+1)Wait, but since a ≤0, the left-hand side is (a x² +x +a)/e^x. Let me analyze this expression.Given that a ≤0, so a x² ≤0, a ≤0. Therefore, the numerator is a x² +x +a = a(x² +1) +x. Since a ≤0, the term a(x² +1) is non-positive. So numerator ≤x.Therefore, (a x² +x +a)/e^x ≤x/e^x. Therefore, the maximum possible value of the left-hand side is x/e^x. But x/e^x has a maximum at x=1, where it's 1/e≈0.3679. Therefore, if we can ensure that b ln(x+1) is always ≥x/e^x, then the inequality holds. But the problem states for any a ≤0, so the left-hand side can be less than x/e^x, but we need to find b such that even the maximum of (a x² +x +a)/e^x over a ≤0 is ≤b ln(x+1). Therefore, for each x, find the maximum of (a x² +x +a)/e^x over a ≤0, and set that ≤b ln(x+1). Then find the minimal b such that this holds for all x ≥0.Therefore, for each x ≥0, the maximum of f(a) = (a x² +x +a)/e^x over a ≤0 is the maximum of a(x² +1) +x over a ≤0. Since this is linear in a, the maximum occurs at the endpoint. Since the coefficient of a is (x² +1). If x² +1 >0, which it always is, then if we are maximizing over a ≤0, the maximum occurs at a=0, because if the coefficient of a is positive, then increasing a increases the expression. But since a ≤0, the maximum is at a=0. Therefore, for each x ≥0, the maximum of (a x² +x +a)/e^x over a ≤0 is (0 +x +0)/e^x =x/e^x. Therefore, the inequality x/e^x ≤b ln(x+1) must hold for all x ≥0. Therefore, we need to find the minimal b such that x/(e^x) ≤b ln(x+1) for all x ≥0. Hence, b ≥ sup_{x ≥0} [x/(e^x ln(x+1))].Therefore, the range of b is [b0, +infty), where b0 is the supremum of x/(e^x ln(x+1)) over x ≥0.Therefore, we need to compute the maximum value of the function g(x)=x/(e^x ln(x+1)) for x ≥0, and then b must be at least this maximum.Therefore, the problem reduces to finding the maximum of g(x)=x/(e^x ln(x+1)) on x ≥0.To find the maximum of g(x), compute its derivative and find critical points.Compute g'(x):g(x)=x / [e^x ln(x+1)]Use quotient rule: g'(x)= [ (1) * e^x ln(x+1) - x * (e^x ln(x+1) + e^x/(x+1)) ] / [e^{2x} ln²(x+1)]Wait, wait, better compute step by step.Let me write g(x)=x e^{-x} / ln(x+1)Therefore, g'(x)= [ (e^{-x} -x e^{-x}) * ln(x+1) - x e^{-x} * (1/(x+1)) ] / [ln²(x+1)]Simplify numerator:e^{-x}(1 -x) ln(x+1) - x e^{-x}/(x +1)Factor out e^{-x}:e^{-x} [ (1 -x) ln(x+1) -x/(x +1) ]Set numerator equal to zero:(1 -x) ln(x+1) -x/(x +1)=0So equation: (1 -x) ln(x+1)=x/(x +1)This equation is likely transcendental and needs to be solved numerically.Let me analyze the function h(x)= (1 -x) ln(x+1) -x/(x +1). We need to find x ≥0 where h(x)=0.First, check x=0:h(0)=1*ln(1) -0=0 -0=0. So x=0 is a root.Check x=1:h(1)= (1 -1) ln(2) -1/2=0 -0.5= -0.5 <0Check x approaching infinity:As x→∞, (1 -x) ln(x+1) ~ -x ln x, and x/(x +1) ~1. So h(x)~ -x ln x -1→-infty. So h(x)→-infty.At x=0.5:h(0.5)=0.5 ln(1.5) -0.5/(1.5)=0.5*(0.4055) -0.333≈0.2027 -0.333≈-0.1303 <0At x=0.1:h(0.1)=0.9 ln(1.1) -0.1/1.1≈0.9*0.0953 -0.0909≈0.0858 -0.0909≈-0.0051 <0At x=0.05:h(0.05)=0.95 ln(1.05) -0.05/1.05≈0.95*0.04879 -0.0476≈0.0463 -0.0476≈-0.0013 <0At x=0.01:h(0.01)=0.99 ln(1.01) -0.01/1.01≈0.99*0.00995 -0.0099≈0.00985 -0.0099≈-0.00005 <0So as x approaches 0 from the right, h(x) approaches 0 from below. At x=0, h(x)=0. So perhaps x=0 is the only root. Wait, but at x=0, g(x)=0 / [something]=0. Wait, but original function g(x)=x/(e^x ln(x+1)). At x=0, it's 0/(1*0) which is 0/0. So need to compute the limit as x approaches0+.Compute lim_{x→0+} x/(e^x ln(x+1)). Using L’Hospital's Rule:lim_{x→0} [x]/[e^x ln(x+1)] = lim_{x→0} [1]/[e^x (1/(x+1)) + e^x ln(x+1) ].But denominator derivative:d/dx [e^x ln(x+1)] = e^x ln(x+1) + e^x/(x+1)At x=0, this is e^0 ln(1) + e^0 /1=0 +1=1. Therefore, limit is 1/1=1.So g(x) approaches1 as x→0. At x=0, indeterminate, but limit is1.Therefore, function g(x) starts at1 (limit as x→0), decreases to some value, and we need to see if it has a maximum at x=0 or somewhere else.Wait, since h(x)=0 only at x=0 (from the previous analysis), but h(x) is negative for x>0. Therefore, the derivative g'(x) is negative for x>0. Therefore, function g(x) is decreasing on x≥0. Therefore, its maximum is at x→0+, which is1. However, we need to check this.Wait, but when we computed h(x) at x=0.01, h(x)≈-0.00005 <0, so g'(x) negative. Therefore, g(x) decreasing for all x>0. Therefore, the maximum of g(x) is1 (limit as x→0), and the function decreases to0 as x→∞.But wait, at x=1, g(1)=1/(e * ln2)≈1/(2.718*0.693)≈1/1.885≈0.530At x=0.5, g(0.5)=0.5/(sqrt(e) * ln1.5)≈0.5/(1.6487 *0.4055)≈0.5/0.668≈0.748At x=0.1, g(x)=0.1/(e^{0.1} ln1.1)≈0.1/(1.105 *0.0953)≈0.1/0.105≈0.952So approaching1 as x→0.But according to derivative, since g'(x) is negative for all x>0, the function is decreasing on x>0. Hence, the maximum is1 (attained as x approaches0), and the supremum is1. Therefore, to satisfy x/e^x ≤b ln(x+1) for all x ≥0, we need b ≥1.Wait, but at x=0, the inequality is0 ≤b ln1=0, which holds for any b. The problem is for x>0. But since the maximum of g(x)=x/(e^x ln(x+1)) is1, then b must be ≥1. Wait, but when x approaches0, g(x) approaches1, so b must be at least1. For example, at x approaching0, the inequality x/e^x ≤b ln(x+1) ≈x -x²/2 +... So x/e^x ≈x(1 -x +x²/2 -...), and ln(x+1)≈x -x²/2 +x³/3 -... So b must satisfy x(1 -x +...) ≤b(x -x²/2 +...). Dividing both sides by x (x≠0), get (1 -x +...) ≤b(1 -x/2 +...). Taking x→0, we get1 ≤b*1 => b≥1.Similarly, for other x, since g(x) ≤1, then b=1 satisfies the inequality. Therefore, the minimal b is1, so the range is b ≥1, i.e., [1, +infty).But let me verify with x=1, which requires1/e ≤b ln2. Since1/e≈0.3679, and ln2≈0.6931. So0.3679 ≤b*0.6931 => b≥0.3679/0.6931≈0.530. But since we need b≥1 to satisfy near x=0, then b=1 is sufficient. For example, for x=1,1/e≈0.3679 ≤1*ln2≈0.6931, which holds.Therefore, the minimal b is1.Therefore, the range of real number b is [1, +infty). Therefore, boxed answer: boxed{[1,+infty)}.Final Answer(I) boxed{dfrac{5}{e}};(II) boxed{[1,+infty)}</think>Given the function ( f(x) = dfrac{ax^2 + x + a}{2e^x} ):(I) To find the value of ( a ) such that the maximum value of ( f(x) ) is ( dfrac{5}{2e} ) when ( a geq 0 ):1. Compute the derivative ( f'(x) ) using the quotient rule.2. Set the derivative equal to zero to find critical points.3. Analyze the critical points and endpoints to determine the maximum value.4. For ( a geq 1 ), the critical points are ( x = 1 ) and ( x = 1 - frac{1}{a} ). For ( 0 leq a < 1 ), the only critical point is ( x = 1 ).5. Evaluate ( f(x) ) at these points and check the maximum value.6. Determine that the maximum value occurs at ( x = 0 ) when ( a = dfrac{5}{e} ), giving ( f(0) = dfrac{5}{2e} ).Thus, the value of ( a ) is (boxed{dfrac{5}{e}}).(II) To find the range of ( b ) such that ( f(x) leq dfrac{b ln(x+1)}{2} ) for any ( a leq 0 ) and ( x in [0, +infty) ):1. Recognize that for ( a leq 0 ), the numerator ( a x^2 + x + a leq x ).2. The inequality simplifies to ( dfrac{x}{e^x} leq b ln(x+1) ).3. Find the maximum value of ( dfrac{x}{e^x ln(x+1)} ) for ( x geq 0 ).4. Analyze the function and determine that the maximum value is 1 as ( x to 0 ).5. Conclude that ( b geq 1 ) ensures the inequality holds for all ( x geq 0 ).Thus, the range of ( b ) is (boxed{[1, +infty)}).]

🔑:Solution:(I) According to the problem, f'(x)=dfrac{1}{2}left[ (2ax+1)e^{-x}-(ax^{2}+x+a)e^{-x} right]=-dfrac{1}{2}e^{-x}left[ ax^{2}+(1-2a)x+a-1 right]=-dfrac{1}{2}e^{-x}(x-1)(ax+1-a).(1) When a=0, f'(x)=-dfrac{1}{2}e^{-x}(x-1), let f'(x) > 0, we get x 1,Thus, f(x) is increasing in (-infty, 1) and decreasing in (1, +infty).Therefore, the maximum value of f(x) is f(1)=dfrac{1}{2e}neq dfrac{5}{2e}, which contradicts the problem statement.(2) When a > 0, 1-dfrac{1}{a} 0, we get 1-dfrac{1}{a} 1,Thus, f(x) is increasing in (1-dfrac{1}{a}, 1), and decreasing in (-infty, 1-dfrac{1}{a}) and (1, +infty).Therefore, the maximum value of f(x) is f(1)=dfrac{2a+1}{2e}=dfrac{5}{2e}, we get a=2.In summary, boxed{a=2}.(II) Let g(a)=dfrac{(x^{2}+1)a}{2e^{x}}+dfrac{x}{2e^{x}}, ain (-infty, 0], when xin [0,+infty), dfrac{x^{2}+1}{2e^{x}}geqslant 0,Then g(a)leqslant dfrac{bln(x+1)}{2},For forall ain (-infty, 0] always holds equivalent to g(a)leqslant g(0)leqslant dfrac{bln(x+1)}{2},That is, dfrac{x}{e^{x}}leqslant bln(x+1),For xin [0,+infty) always holds.(1) When bleqslant 0, for forall xin (0,+infty), bln(x+1) 0,At this time, dfrac{x}{e^{x}} > bln(x+1), which contradicts the problem statement.(2) When b > 0, let h(x)=bln(x+1)-dfrac{x}{e^{x}}, xin [0,+infty),Then h'(x)=dfrac{b}{x+1}-left( e^{-x}-xe^{-x} right)=dfrac{be^{x}+x^{2}-1}{(x+1)e^{x}}, where (x+1)e^{x} > 0, forall xin [0,+infty),Let p(x)=be^{x}+x^{2}-1,xin [0,+infty), then p(x) is increasing in the interval [0,+infty),① When bgeqslant 1, p(x)geqslant p(0)=b-1geqslant 0,So for forall xin [0,+infty), h'(x)geqslant 0,Thus, h(x) is increasing in [0,+infty),So for any xin [0,+infty), h(x)geqslant h(0)=0,That is, the inequality bln(x+1)geqslant x{e^{-x}} always holds in [0,+infty).② When 0 0 and p(x) is increasing in the interval [0,+infty),There exists a unique x_{0}in (0,1) such that p(x_{0})=0, and for xin (0,x_{0}), p(x_{0}) < 0.Thus, for xin (0,x_{0}), h'(x) < 0,So, h(x) is decreasing in the interval (0,x_{0}),Then for xin (0,x_{0}), h(x) < h(0)=0,That is, bln(x+1) < x{e^{-x}},Which contradicts the problem statement. In summary, boxed{bgeqslant 1}.

🧠:[<think>Okay, so I need to compare the sizes of a, b, and c where a is 1.5 raised to the 0.3 power, b is the logarithm of 6 with base 7, and c is the tangent of 300 degrees. Hmm, let me start by understanding each of these individually.First, let's tackle a = 1.5^0.3. I know that exponents between 0 and 1 will result in roots. So 0.3 is the same as 3/10, which means 1.5^(3/10) is the 10th root of 1.5 cubed. But maybe it's easier to think of this in terms of natural exponents or logarithms. Alternatively, maybe I can approximate the value. Let's see, 1.5 is the same as 3/2. So 1.5^0.3. Maybe taking the natural logarithm: ln(1.5^0.3) = 0.3 * ln(1.5). Calculating ln(1.5) is approximately 0.4055, so 0.3 * 0.4055 ≈ 0.1216. Then exponentiating, e^0.1216 ≈ 1.129. So a ≈ 1.129. Let me check that again. Wait, if I use a calculator, 1.5^0.3. Let me see: 1.5^0.3. Let me use logarithms. Alternatively, maybe use common logarithms. Wait, but maybe I can use the fact that 1.5^0.3 is the same as e^{0.3 ln 1.5}. So, ln 1.5 is about 0.4055, so 0.3*0.4055 = 0.12165. Then e^0.12165 is approximately 1.129. So, a ≈ 1.129. That seems reasonable.Next, b = log_7 6. This is the logarithm of 6 with base 7. To compare this with a number, I can convert it to natural log or common log. Remember that log_b a = ln a / ln b. So log_7 6 = ln6 / ln7. Let's calculate ln6 ≈ 1.7918 and ln7 ≈ 1.9459. So 1.7918 / 1.9459 ≈ 0.920. So b ≈ 0.920. Wait, is that right? Let me check with another method. Alternatively, since 7^1 = 7 and 6 is less than 7, so log_7 6 must be less than 1. Also, 7^0.9 is about e^{0.9 ln7} ≈ e^{0.9*1.9459} ≈ e^{1.751} ≈ 5.77. So 7^0.9 ≈ 5.77, which is less than 6. So log_7 6 is a bit higher than 0.9. Maybe 0.92? Let's compute 7^0.92. Take ln7 ≈ 1.9459, so 0.92 * 1.9459 ≈ 1.790. Then e^1.790 ≈ 5.99. So 7^0.92 ≈ 5.99, which is very close to 6. Therefore, log_7 6 ≈ 0.92. So b ≈ 0.92. So that seems correct.Now, c = tan 300 degrees. Let's recall that 300 degrees is in the fourth quadrant. Tangent in the fourth quadrant is negative because sine is negative and cosine is positive there. So tan 300° is negative. To compute its exact value, we can find the reference angle. 300 degrees is 360 - 60, so the reference angle is 60 degrees. Therefore, tan 300° = -tan 60°. Tan 60° is √3 ≈ 1.732, so tan 300° ≈ -1.732. So c ≈ -1.732.So now, putting it all together:a ≈ 1.129,b ≈ 0.92,c ≈ -1.732.Therefore, comparing them: c is negative, so the smallest, then b is approximately 0.92, and a is approximately 1.129. So from smallest to largest: c < b < a.Wait, but let me make sure I didn't make any calculation errors. Let me verify each step again.First, a = 1.5^0.3. So 1.5 is 3/2. Let's compute 1.5^0.3. Let me use logarithms again. Log base 10: log(1.5) ≈ 0.1761. So 0.3 * log(1.5) ≈ 0.3 * 0.1761 ≈ 0.0528. Then 10^0.0528 ≈ 1.129. That's the same as before. So a ≈ 1.129. Correct.For b = log_7 6. Let's use change of base formula: log_7 6 = (ln6)/(ln7). Let me compute this using calculator-like precision. ln6 ≈ 1.791759, ln7 ≈ 1.945910. So 1.791759 / 1.945910 ≈ 0.920782. So approximately 0.9208. So that's more precise. So b ≈ 0.9208.And c = tan 300°, which is -tan60° = -√3 ≈ -1.73205.So now, the values are:a ≈ 1.129,b ≈ 0.9208,c ≈ -1.732.Therefore, ordering them: c is clearly the smallest because it's negative. Then between a and b, a is about 1.129 and b is about 0.9208, so a is larger than b. So the order is c < b < a.Wait, but just to make sure about a and b. Maybe my approximations are off. Let me check with another method. For example, for a = 1.5^0.3, can I approximate this without logarithms?Let me think. 1.5^0.3. Let's write 0.3 as 3/10. So it's 1.5^(3/10). Alternatively, 1.5^(1/3.333). Hmm, maybe not helpful. Alternatively, since 1.5 is 3/2, perhaps express in terms of e: 1.5^x = e^{x ln(1.5)}. So with x = 0.3, as before.Alternatively, maybe compare 1.5^0.3 with, say, 1.2. Let's see: 1.5^0.3. If x=0.3, perhaps approximate using a Taylor series expansion around x=0?Wait, maybe not. Alternatively, compare with known exponents. For example, 1.5^0.3. Let's note that 1.5^1 = 1.5, so 1.5^0.3 should be less than 1.5, which it is (1.129). Also, 1.5^0.5 would be sqrt(1.5) ≈ 1.2247. So 1.5^0.3 is between 1.1 and 1.2. Which is consistent with the previous result of 1.129.Alternatively, use logarithms with base 10. log10(1.5) ≈ 0.1761. Then 0.3*0.1761 ≈ 0.0528, so 10^0.0528. 10^0.05 is approximately 1.122, and 10^0.0028 is approximately 1 + 0.0028*ln10 ≈ 1 + 0.0028*2.3026 ≈ 1 + 0.0064 ≈ 1.0064. So 1.122 * 1.0064 ≈ 1.129. So same result.Therefore, a ≈ 1.129 is correct.For b, log_7 6. Let's confirm again. Since 7^1 = 7, and 7^0.9 is approximately 5.77 as calculated before, and 7^0.92 ≈ 5.99, which is very close to 6, so log_7 6 is approximately 0.92. That's accurate.Therefore, a ≈ 1.129, which is more than 1, and b ≈ 0.92, less than 1. So a > b. And c is negative, so c is the smallest.Thus, the order is c < b < a.I think that's solid. Let me just check if there's any possible mistake here. For example, could I have messed up the reference angle for 300 degrees? 300 degrees is in the fourth quadrant, 360 - 60, so reference angle is 60 degrees. So tan300° = -tan60° = -√3. Correct. So c is indeed about -1.732. Negative. So that's right.Alternatively, could a be less than 1? If the exponent were negative, but 0.3 is positive, and 1.5 >1, so any positive exponent would result in a number greater than 1. Therefore, a is greater than 1, which is more than b (~0.92). So yes, a > b. And c is negative, so c < b < a.Therefore, the answer should be c < b < a.Final AnswerThe size relationship is boxed{c < b < a}.</think>To compare the sizes of (a = 1.5^{0.3}), (b = log_{7} 6), and (c = tan 300^circ):1. Calculating (a): - (a = 1.5^{0.3}) - Using natural logarithms: (ln(1.5^{0.3}) = 0.3 ln(1.5)) - (ln(1.5) approx 0.4055), so (0.3 times 0.4055 approx 0.12165) - Exponentiating: (e^{0.12165} approx 1.129) - Therefore, (a approx 1.129)2. Calculating (b): - (b = log_{7} 6) - Using the change of base formula: (log_{7} 6 = frac{ln 6}{ln 7}) - (ln 6 approx 1.7918) and (ln 7 approx 1.9459) - (frac{1.7918}{1.9459} approx 0.9208) - Therefore, (b approx 0.9208)3. Calculating (c): - (c = tan 300^circ) - (300^circ) is in the fourth quadrant with a reference angle of (60^circ) - (tan 300^circ = -tan 60^circ = -sqrt{3} approx -1.732) - Therefore, (c approx -1.732)Comparing the values:- (a approx 1.129)- (b approx 0.9208)- (c approx -1.732)Thus, the order from smallest to largest is (c < b < a).[boxed{c < b < a}],<think>Okay, so I need to compare the sizes of a, b, and c where a is 1.5 raised to the 0.3 power, b is the logarithm of 6 with base 7, and c is the tangent of 300 degrees. Hmm, let me start by understanding each of these one by one.First, let's tackle a = 1.5^{0.3}. I know that exponentials can sometimes be tricky. Since the base is 1.5, which is greater than 1, and the exponent is 0.3, which is a positive number less than 1, the result should be a number greater than 1 but not too large. Maybe I can approximate this value. Let me recall that 1.5 is the same as 3/2. So, (3/2)^{0.3} is the same as e^{0.3 * ln(3/2)}. Let me calculate ln(3/2). ln(3) is approximately 1.0986, ln(2) is about 0.6931, so ln(3/2) = ln(3) - ln(2) ≈ 1.0986 - 0.6931 = 0.4055. Then, 0.3 * 0.4055 ≈ 0.1216. So, e^{0.1216}. I know that e^{0.1} ≈ 1.1052, e^{0.12} ≈ 1.1275, so maybe e^{0.1216} is roughly 1.129. Therefore, a ≈ 1.129. Let me check with a calculator if possible. Wait, I don't have a calculator here, but these approximations should be close enough for comparison purposes.Next, b = log_7 6. Logarithms can be converted using change of base formula. So, log_7 6 = ln(6)/ln(7). Let me compute ln(6) and ln(7). ln(6) is ln(2*3) = ln(2) + ln(3) ≈ 0.6931 + 1.0986 ≈ 1.7917. ln(7) is approximately 1.9459. Therefore, log_7 6 ≈ 1.7917 / 1.9459 ≈ 0.920. So, b ≈ 0.920. That seems less than a, which we approximated as 1.129. Wait, but let me confirm. If a is approximately 1.129 and b is approximately 0.920, then a is larger than b. Hmm, okay.Now, c = tan 300 degrees. Let me recall the unit circle. 300 degrees is in the fourth quadrant, where tangent is negative because tangent is sine over cosine, and in the fourth quadrant, sine is negative while cosine is positive. The reference angle for 300 degrees is 60 degrees, since 360 - 300 = 60. So tan(300 degrees) = -tan(60 degrees). Tan(60 degrees) is √3 ≈ 1.732, so tan(300 degrees) is -√3 ≈ -1.732. Therefore, c ≈ -1.732. So, c is negative, which is definitely less than both a and b, which are positive. So, c is the smallest. Then, comparing a and b, since a is approximately 1.129 and b is approximately 0.920, then a > b > c. But let me double-check my calculations to make sure I didn't make a mistake.Starting with a = 1.5^{0.3}. Let me verify using another method. Alternatively, 0.3 is 3/10, so 1.5^{3/10} is the 10th root of 1.5^3. Let's compute 1.5^3 first. 1.5^2 = 2.25, so 1.5^3 = 2.25 * 1.5 = 3.375. Then, taking the 10th root of 3.375. Hmm, the 10th root of 2 is approximately 1.0718, and the 10th root of 3.375 should be larger. Let's recall that 3.375 is 2^1.5 (since 2^1.5 ≈ 2.828, which is actually less than 3.375). Wait, maybe another way. Let's take natural logs again. ln(3.375) ≈ ln(3) + ln(1.125) ≈ 1.0986 + 0.1178 ≈ 1.2164. Then, ln(a) = (1/10)*ln(3.375) ≈ 0.1216, so a ≈ e^{0.1216} ≈ 1.129. Same result as before. So, that seems consistent.For b, log_7 6. Alternatively, since 7^1 = 7, which is greater than 6, so log_7 6 must be less than 1. Therefore, b is between 0 and 1, which matches our earlier calculation of approximately 0.920. Let's see if we can approximate it better. Let me think. 7^0.9 is e^{0.9 ln 7} ≈ e^{0.9*1.9459} ≈ e^{1.7513} ≈ 5.77. But 7^0.9 ≈ 5.77, which is less than 6. Then 7^0.92: Let's compute 0.92*ln7 ≈ 0.92*1.9459 ≈ 1.7902, so e^1.7902 ≈ 5.99. So 7^0.92 ≈ 5.99, which is very close to 6. Therefore, log_7 6 ≈ 0.92. That aligns with our previous result. Therefore, b ≈ 0.92, which is indeed less than a ≈ 1.129.Then, c is tan 300°, which we determined is -√3 ≈ -1.732. So, negative. So, c is definitely the smallest. So, ordering from smallest to largest: c < b < a. Wait, but b is approximately 0.92 and a is approximately 1.129, so that would be c < b < a? Wait, but wait, c is negative, so it's definitely the smallest. Then, between a and b, which are both positive. Since a ≈1.129 and b≈0.92, then a > b. So, the order is c < b < a. Wait, but hold on, let me check again. Wait, 0.92 is less than 1.129, so yes, b is less than a. So, c is less than b, which is less than a. Therefore, c < b < a.But wait, let me check again for a. 1.5^0.3. Let's see, maybe another approach. Let's compare 1.5^0.3 and 1.5^0.333..., which would be 1.5^(1/3). Since 0.3 is 3/10, which is less than 1/3 ≈ 0.333... So, since the exponent is smaller, and the base is greater than 1, 1.5^0.3 < 1.5^(1/3). What is 1.5^(1/3)? Let's approximate. 1.5^1 = 1.5, 1.5^(1/3) is the cube root of 1.5. The cube root of 1 is 1, cube root of 1.728 is 1.2, since 1.2^3 = 1.728. So, 1.5 is between 1.728 and 1, so cube root of 1.5 is between 1.14 and 1.2. Maybe approximately 1.1447. So, 1.5^(1/3) ≈1.1447. Then, 1.5^0.3 is less than that. Wait, but earlier calculation gave a ≈1.129. That seems reasonable.Alternatively, 1.5^0.3 can be thought as e^{0.3 ln 1.5} ≈ e^{0.3*0.4055} ≈ e^{0.12165} ≈1.129, as before. So, that seems consistent.Alternatively, maybe use logarithms to compare a and b. Let's see. Take log base 10 of a: log10(a) = 0.3 * log10(1.5). log10(1.5) ≈0.1761. So, 0.3 * 0.1761 ≈0.0528. Therefore, a ≈10^{0.0528}≈1.129. Same as before. For b, log7 6 = log10(6)/log10(7) ≈0.7782/0.8451≈0.920. So, same result.Therefore, a ≈1.129, b≈0.920, c≈-1.732. Therefore, c < b < a. So, the answer should be c < b < a. Let me just verify once again to make sure there's no mistake.Wait, maybe check the value of a with another method. Let's use logarithm tables. Wait, not really necessary. Alternatively, approximate 1.5^0.3. Let me think in terms of exponents. 1.5 is 3/2, so (3/2)^0.3. Take natural log: 0.3*(ln3 - ln2) ≈0.3*(1.0986 -0.6931)=0.3*0.4055≈0.12165, so exponentiate that: e^{0.12165}≈1.1295. So, a≈1.1295.Similarly, for b, log7 6. Let me check if 7^0.92 is approximately 6. Let's compute 7^0.92. Take ln(7^0.92)=0.92*ln7≈0.92*1.9459≈1.7902. Then, e^{1.7902}≈5.987≈6. So, 7^0.92≈6, so log7 6≈0.92. Therefore, b≈0.92.Yes, seems right. So, c is -1.732, which is clearly less than both a and b. So, the order is c < b < a.Wait, but hold on, the problem says "compare the size relationship among a, b, and c". So, maybe write the order from smallest to largest. So, c is the smallest, then b, then a. So, c < b < a.But let me check once more if I didn't mix up any values. For example, maybe I miscalculated the logarithm? Let me verify log7 6 again. log7 6 = ln6 / ln7 ≈1.7918 /1.9459≈0.9207. Yes, approximately 0.92. So, correct. And a is about 1.129. So, yes, 0.92 <1.129. So, order is c < b < a.Alternatively, perhaps I made a mistake with the tangent? Let me confirm tan 300 degrees. 300 degrees is in the fourth quadrant. The reference angle is 60 degrees. Tan is negative there. Tan 60 is sqrt(3)≈1.732. Therefore, tan 300 is -sqrt(3)≈-1.732. So, yes, correct.Therefore, the conclusion is that c is the smallest, then b, then a. So, c < b < a. Therefore, the answer should be c < b < a.Final AnswerThe size relationship is boxed{c < b < a}.</think>Given (a = 1.5^{0.3}), (b = log_{7} 6), and (c = tan 300^circ), we need to compare their sizes.1. Calculating (a = 1.5^{0.3}): - Convert to natural logarithm: (a = e^{0.3 ln(1.5)}). - (ln(1.5) approx 0.4055), so (0.3 times 0.4055 approx 0.1216). - Exponentiating: (e^{0.1216} approx 1.129). - Therefore, (a approx 1.129).2. Calculating (b = log_{7} 6): - Use the change of base formula: (log_{7} 6 = frac{ln 6}{ln 7}). - (ln 6 approx 1.7917) and (ln 7 approx 1.9459). - Divide: (frac{1.7917}{1.9459} approx 0.920). - Therefore, (b approx 0.920).3. Calculating (c = tan 300^circ): - (300^circ) is in the fourth quadrant, reference angle (60^circ). - (tan 300^circ = -tan 60^circ = -sqrt{3} approx -1.732). - Therefore, (c approx -1.732).Comparing the values:- (c approx -1.732) (negative) is the smallest.- (b approx 0.920) is less than (a approx 1.129).Thus, the size relationship is (boxed{c < b < a}).]

🔑:First, let's analyze each part separately:1. For a = 1.5^{0.3}: Since 1.5>1 and the exponent 0.3>0, we have a>1.2. For b = log_{7}6: By definition, log_{7}6 represents the exponent to which 7 must be raised to produce 6. Since 6 1Therefore, boxed{c < b < a}.This problem assesses one's understanding of properties related to logarithmic, exponential, and trigonometric functions.

🧠:[<think>Okay, so I need to find the value of f(2) given that f(x) is an increasing function on all real numbers, and for any real x, f(f(x) - 3^x) = 4. Hmm, let's start by understanding what this equation means. Since the function f is applied to (f(x) - 3^x) and the result is always 4, regardless of x, that suggests that the input to f, which is (f(x) - 3^x), must be a constant value. Because if f is a function that's increasing, it can't take multiple different inputs to the same output, right? Because an increasing function is one-to-one, meaning each input has a unique output. So if the output is always 4, the input must be the same every time. So, that means f(x) - 3^x is a constant. Let me check that logic again. If f is increasing, it's injective (one-to-one), so f(a) = f(b) implies a = b. Therefore, since f(something) = 4 for all x, that something must be the same for all x. Therefore, f(x) - 3^x = c, where c is a constant. Then, f(x) = 3^x + c. Then, substituting back into the original equation: f(f(x) - 3^x) = f(c) = 4. So f(c) = 4. But since f(x) = 3^x + c, then f(c) = 3^c + c = 4. So, we have 3^c + c = 4. Now, we need to solve for c. Let's think. Let's try plugging in some integer values for c. If c = 1, then 3^1 + 1 = 3 + 1 = 4. That works! So c = 1. Let me confirm if there are any other solutions. Let's check c=0: 3^0 +0=1+0=1≠4. c=2: 3^2 +2=9+2=11≠4. c= -1: 3^(-1)+(-1)=1/3 -1= -2/3≠4. How about c=1? As above, that gives 4. Since f is increasing, and 3^x is an exponential function which is also increasing. So f(x) =3^x +1. Therefore, f(x) is the sum of two increasing functions, so it's increasing. That makes sense. Therefore, f(x) =3^x +1. Therefore, f(2)=3^2 +1=9 +1=10. Therefore, the answer is 10. Wait, but before I conclude, let me check everything again to make sure I didn't skip any steps or make any wrong assumptions. First, the key step was recognizing that since f is increasing, it's injective, so f(f(x)-3^x)=4 for all x implies f(x)-3^x is constant. That seems right. Then, setting f(x)=3^x +c. Then f(c)=4 gives 3^c +c=4. Solving that gives c=1. Then f(2)=3^2 +1=10. Seems solid. Let me check another possible c. Suppose c=1.2: 3^1.2≈3*3^0.2≈3*1.2457≈3.737, plus 1.2≈4.937>4. If c=0.5: 3^0.5≈1.732 +0.5≈2.232<4. So the function 3^c +c is increasing in c, since both 3^c and c are increasing. Therefore, there's only one solution to 3^c +c=4, which is c=1. Therefore, this is the only solution. So f(x)=3^x +1, hence f(2)=10. Seems correct. So the answer is 10.Final AnswerThe value of ( f(2) ) is boxed{10}.</think>Given that ( f(x) ) is an increasing function on (mathbf{R}), and for any ( x in mathbf{R} ), we have[fleft(f(x) - 3^xright) = 4.]Since ( f ) is an increasing function, it is injective (one-to-one). This implies that the input to ( f ) must be a constant value for the output to always be 4. Let this constant be ( c ). Therefore, we have:[f(x) - 3^x = c quad text{for all } x in mathbf{R}.]This implies:[f(x) = 3^x + c.]Substituting this back into the given equation, we get:[f(c) = 4.]Since ( f(c) = 3^c + c ), we have:[3^c + c = 4.]Solving this equation, we test ( c = 1 ):[3^1 + 1 = 3 + 1 = 4,]which satisfies the equation. Checking other values confirms that ( c = 1 ) is the only solution because ( 3^c + c ) is an increasing function in ( c ).Thus, the function ( f(x) ) is:[f(x) = 3^x + 1.]Therefore, evaluating ( f(2) ):[f(2) = 3^2 + 1 = 9 + 1 = 10.]The value of ( f(2) ) is (boxed{10}).,<think>Okay, let's see. I need to find f(2) given that f(x) is an increasing function on all real numbers, and for any real x, f(f(x) - 3^x) = 4. Hmm, interesting. So, since f is increasing, that means if a < b, then f(a) < f(b). That's good to know because it might help me figure out something about the function's behavior.First, let's parse the equation f(f(x) - 3^x) = 4. For every real number x, when I take f(x), subtract 3^x from it, and then apply f again, I get 4. So, the output is always 4, regardless of the input x. That seems like a constant function, but it's inside f. Maybe there's a specific value of y such that f(y) = 4 for all y? Wait, no, because y here would be f(x) - 3^x, which probably varies with x. But since f is applied to y and gives 4 every time, that suggests that y must be a constant value, right? Because if f is injective (which it is, since it's strictly increasing), then f(a) = f(b) implies a = b. So, if f(f(x) - 3^x) is always 4, then f(x) - 3^x must be the same for all x. Is that right?Let me check that. Suppose that f is injective. Then f(a) = f(b) implies a = b. So, if f(something) = 4 for all x, then that something must be the same for all x. Therefore, f(x) - 3^x is a constant. Let's call that constant c. So, f(x) - 3^x = c for all x. Then, f(x) = 3^x + c. Then, substitute back into the original equation: f(f(x) - 3^x) = f(c) = 4. So, f(c) = 4. But since f(x) = 3^x + c, then f(c) = 3^c + c. Therefore, 3^c + c = 4. So, we have to solve for c in the equation 3^c + c = 4. Hmm, that seems tricky. Let's see if we can find such a c.Let me try plugging in some numbers. If c = 1: 3^1 + 1 = 3 + 1 = 4. Oh, that works! So c = 1. Therefore, f(x) = 3^x + 1. Then, f(2) = 3^2 + 1 = 9 + 1 = 10. So, is the answer 10?Wait, but let's verify this. Let me check if this function satisfies all the given conditions. First, f(x) = 3^x + 1. Is this function increasing? Well, 3^x is an exponential function which is always increasing, and adding 1 doesn't change the increasing nature. So yes, f(x) is increasing on R. Then, check if f(f(x) - 3^x) = 4. Let's compute f(f(x) - 3^x). Since f(x) = 3^x + 1, then f(x) - 3^x = (3^x + 1) - 3^x = 1. Therefore, f(1) = 3^1 + 1 = 4, which matches the given equation. So, everything checks out. Therefore, f(2) = 10. So, the answer is 10. That seems straightforward once I realized that f(x) - 3^x must be a constant. But let me make sure there are no other solutions.Suppose there is another c such that 3^c + c = 4. Let's see, for c=0: 3^0 + 0 = 1 + 0 = 1 ≠ 4. c=1: 4. c=2: 9 + 2 = 11. c=3: 27 + 3 = 30. So, as c increases, 3^c + c increases. For c <1, say c=0.5: 3^0.5 ≈1.732 +0.5 ≈2.232 <4. So, between c=0.5 and c=1, it goes from ~2.232 to 4. So, is there a solution between 0.5 and 1? Wait, but c=1 already gives 4. Since 3^c is increasing and c is also increasing, the sum is strictly increasing. Therefore, the function 3^c + c is strictly increasing, so there's only one solution at c=1. Therefore, c=1 is the only solution. Therefore, f(x) =3^x +1 is the only function that satisfies the conditions, and thus f(2)=10.Therefore, the answer should be boxed{10}Final Answerboxed{10}</think>Given that ( f(x) ) is an increasing function on (mathbf{R}), and for any ( x in mathbf{R} ), we have[fleft(f(x) - 3^xright) = 4.]Since ( f ) is injective (strictly increasing), the equation ( fleft(f(x) - 3^xright) = 4 ) implies that ( f(x) - 3^x ) must be a constant. Let this constant be ( c ). Therefore, we have:[f(x) = 3^x + c.]Substituting this back into the given equation, we get:[f(c) = 4.]Using the expression for ( f(x) ), we find:[f(c) = 3^c + c = 4.]Solving for ( c ), we test ( c = 1 ):[3^1 + 1 = 3 + 1 = 4,]which satisfies the equation. Therefore, ( c = 1 ).Thus, the function ( f(x) ) is:[f(x) = 3^x + 1.]To find ( f(2) ):[f(2) = 3^2 + 1 = 9 + 1 = 10.]The final answer is (boxed{10}).,<think>Okay, so I need to find the value of f(2) given that f(x) is an increasing function on all real numbers, and for any real number x, f(f(x) - 3^x) = 4. Hmm, let's try to break this down step by step.First, since f is increasing, that means if a < b, then f(a) < f(b). This property might be useful later. The equation given is f(f(x) - 3^x) = 4 for any x. So no matter what x we choose, when we plug f(x) - 3^x into f, we always get 4. That seems like a constant result, so maybe f(x) - 3^x is a constant value? Wait, let me check.If f(f(x) - 3^x) = 4 for all x, then maybe f is a constant function? But no, the problem states that f is increasing, so it can't be constant. Therefore, the argument inside f must be a constant value that maps to 4. Since f is injective (because it's strictly increasing), the input to f must be the same for all x. That is, there exists some constant c such that f(x) - 3^x = c for all x. Then, f(c) = 4. Wait, is that right?Let me write that down. Suppose that for all x, f(x) - 3^x = c. Then, substituting back into the original equation, f(c) = 4. So if we can find c such that f(c) = 4, and also f(x) = 3^x + c for all x. But wait, does that hold?Wait, if f(x) = 3^x + c, then f is certainly increasing because 3^x is increasing and adding a constant doesn't change the monotonicity. So that satisfies the condition that f is increasing. Then, substituting back into the original equation, f(f(x) - 3^x) = f(c) = 4. Therefore, f(c) = 4. But f(c) = 3^c + c, right? Because f(c) = 3^c + c. Therefore, 3^c + c = 4. So we need to solve 3^c + c = 4 for c. Hmm, this seems tricky. Let's see.Let me try plugging in some integer values for c. If c = 1, 3^1 +1 = 3 +1 = 4. Oh! That works. So c=1. Then, if c=1, then f(x) = 3^x +1. Therefore, f(2) = 3^2 +1 = 9 +1 =10. So is the answer 10? Wait, but let me verify.Wait, let's check if c=1 is the only solution. Let me see. Suppose we define the function g(c) = 3^c + c. Then we need to find c such that g(c) =4. We found c=1 gives 3 +1=4. What about c=0? 1 +0=1. c=2: 9 +2=11. Negative c? Let's try c= -1: 3^{-1} + (-1) = 1/3 -1 = -2/3. So as c decreases, 3^c approaches 0, and c becomes more negative. For c approaching negative infinity, 3^c approaches 0, and c approaches negative infinity, so g(c) approaches negative infinity. For c approaching positive infinity, 3^c dominates and g(c) approaches infinity. Since g(c) is continuous and strictly increasing (because derivative g’(c)=ln3 *3^c +1 >0 for all c), so there is exactly one real solution. We found c=1, so that's the only solution.Therefore, c=1, so f(x)=3^x +1. Then f(2)=3^2 +1=9+1=10. Let's double-check if this works with the original functional equation. Let’s take any x, compute f(x) -3^x= (3^x +1)-3^x=1, so f(1)=3^1 +1=4, which matches the equation f(f(x)-3^x)=4. So yes, for any x, f(x)-3^x=1, so f(1)=4. So this works. Therefore, the answer is 10. So f(2)=10.But just to make sure there's no mistake here. Let's recap. The key idea was to recognize that since f is injective (due to being strictly increasing), the equation f(f(x)-3^x)=4 for all x implies that f(x)-3^x is constant for all x. Because if f(a) = f(b) implies a = b for injective functions. Therefore, if f(k) =4 for all x, but since f is injective, k must be the same for all x, which gives us f(x)-3^x =k for some constant k. Then solving for k gives us the value, and then building f(x) accordingly. Then evaluating at x=2. All steps seem logical. The key was using the injectivity of f to conclude that the argument inside f must be constant. So yes, I think this is correct. Therefore, f(2)=10.Final Answerboxed{10}</think>Given that ( f(x) ) is an increasing function on (mathbf{R}), and for any ( x in mathbf{R} ), we have[fleft(f(x) - 3^xright) = 4.]Since ( f ) is increasing, it is injective. This implies that the argument inside the function ( f ) must be a constant for all ( x ). Let this constant be ( c ). Therefore, we have:[f(x) - 3^x = c quad text{for all} quad x in mathbf{R}.]Substituting this back into the given equation, we get:[f(c) = 4.]From the equation ( f(x) = 3^x + c ), we substitute ( c ) into the function:[f(c) = 3^c + c.]We need to solve ( 3^c + c = 4 ). Trying integer values, we find that ( c = 1 ) satisfies this equation:[3^1 + 1 = 3 + 1 = 4.]Thus, the function ( f(x) ) is:[f(x) = 3^x + 1.]To find ( f(2) ):[f(2) = 3^2 + 1 = 9 + 1 = 10.]Therefore, the final answer is:[boxed{10}],<think>Okay, let's try to figure out how to solve this problem. We have a function f(x) that's increasing on all real numbers. For any real number x, the equation f(f(x) - 3^x) = 4 holds. We need to find f(2). Hmm, interesting. Let me break this down step by step.First, since f is increasing, that means if a < b, then f(a) < f(b). So it's a strictly increasing function, right? Which also implies that it's one-to-one, so it has an inverse function. Maybe that's useful here. But I'm not sure yet. Let's see.The given equation is f(f(x) - 3^x) = 4 for any real x. Since this is true for any x, maybe we can choose a specific value of x that would make things simpler. The question is asking for f(2), so perhaps we can find an x such that f(x) - 3^x equals 2, because then f(2) would be 4. Wait, that might not be straightforward. Let's think.Let me denote y = f(x) - 3^x. Then, according to the equation, f(y) = 4. So for any x, if I compute y as f(x) - 3^x, then f(y) is 4. Since f is increasing, it can take the value 4 at most once. Wait, but the equation says that for any x, y = f(x) - 3^x is mapped to 4. So this suggests that y is the same for all x? Because otherwise, if different y's are mapped to 4, which is the same value, that would contradict f being increasing (since it has to be injective). Therefore, y must be a constant. Let me check that.Suppose y is not constant. Then there exist x1 and x2 such that y1 = f(x1) - 3^{x1} and y2 = f(x2) - 3^{x2}, with y1 ≠ y2. But f(y1) = f(y2) = 4. Since f is increasing, if y1 ≠ y2, then f(y1) ≠ f(y2). But here f(y1) and f(y2) are both 4. Therefore, this would only be possible if y1 = y2. Therefore, for all x, f(x) - 3^x must be equal to some constant c. So f(x) = 3^x + c, where c is a constant. Then, f(c) = 4, because substituting into the original equation: f(f(x) - 3^x) = f(c) = 4. So c is the value such that f(c) = 4. But since f(x) = 3^x + c, then f(c) = 3^c + c. So 3^c + c = 4. Now we need to solve 3^c + c = 4. Hmm, this is an equation in c. Let me solve this.Let me try plugging in some values. Let's test c=1: 3^1 +1 =3 +1=4. That works! So c=1. Therefore, f(x)=3^x +1. Let me check if this works. If f(x)=3^x +1, then f(f(x) -3^x)=f( (3^x +1) -3^x )=f(1)=3^1 +1=4, which matches the given equation. Great! So f(x)=3^x +1. Therefore, f(2)=3^2 +1=9 +1=10. So the answer should be 10. Wait, but let me make sure there are no other solutions for c. Suppose there's another c where 3^c +c=4. Let's check c=0: 1 +0=1≠4. c=1:4, which works. c=2:9 +2=11≠4. c= -1: 1/3 + (-1)= -2/3≠4. So it seems c=1 is the only solution. Hence, f(x)=3^x +1 and f(2)=10. That seems correct.But let me verify again. Since f is supposed to be increasing. Is 3^x +1 increasing? Yes, because the derivative of 3^x is ln3 *3^x, which is always positive, and the constant 1 doesn't affect the increasing nature. So yes, f(x) is indeed increasing. So all conditions are satisfied.Therefore, the answer is 10.Final Answerboxed{10}</think>Given that ( f(x) ) is an increasing function on (mathbf{R}), and for any ( x in mathbf{R} ), we have[fleft(f(x) - 3^xright) = 4.]We need to find ( f(2) ).Since ( f ) is an increasing function, it is injective. The equation ( fleft(f(x) - 3^xright) = 4 ) implies that ( f(x) - 3^x ) must be a constant value for all ( x ). Let this constant be ( c ). Therefore, we have:[f(x) = 3^x + c.]Substituting this into the given equation, we get:[fleft(f(x) - 3^xright) = f(c) = 4.]Since ( f(c) = 3^c + c ), we need to solve:[3^c + c = 4.]Testing ( c = 1 ):[3^1 + 1 = 3 + 1 = 4,]which satisfies the equation. Therefore, ( c = 1 ).Thus, the function is ( f(x) = 3^x + 1 ). To find ( f(2) ):[f(2) = 3^2 + 1 = 9 + 1 = 10.]The final answer is:[boxed{10}]]

🔑:8. 10 .From the problem, we know that f(x)-3^{x} is a constant. Let's assume f(x)-3^{x}=m.Then f(m)=4, f(x)=3^{x}+m.Therefore, 3^{m}+m=4 Rightarrow 3^{m}+m-4=0.It is easy to see that the equation 3^{m}+m-4=0 has a unique solution m=1.Thus, f(x)=3^{x}+1,and hence, f(2)=3^{2}+1=10.

🧠:[<think>Okay, so I need to figure out the values of the binomial coefficients for n choose 0, n choose 1, and n choose 2. Hmm, let's start by recalling what a binomial coefficient actually is. The notation is (binom{n}{k}), which represents the number of ways to choose k elements from a set of n elements without considering the order. Right, so combinations, not permutations.First, let's tackle (binom{n}{0}). That's the number of ways to choose 0 elements from n. Hmm, how does that work? Well, if you think about it, there's exactly one way to choose nothing from a set, right? No matter how many elements are in the set, not picking any is just one scenario. So I think (binom{n}{0} = 1) for any n. Let me verify that. The formula for binomial coefficients is (binom{n}{k} = frac{n!}{k!(n - k)!}). If we plug in k = 0, we get (frac{n!}{0!n!}). Since 0! is defined as 1, this simplifies to 1. Yep, that checks out. So (binom{n}{0} = 1).Next up is (binom{n}{1}). This is the number of ways to choose 1 element from n. Well, if there are n elements, and you want to pick just one, there should be exactly n ways to do that. Each element individually can be chosen. Let's confirm with the formula. (binom{n}{1} = frac{n!}{1!(n - 1)!}). The n! in the numerator divided by (n - 1)! leaves us with n, since n! = n × (n - 1)! So, (frac{n × (n - 1)!}{1 × (n - 1)!} = n). That's correct. So (binom{n}{1} = n).Now for (binom{n}{2}). This is the number of ways to choose 2 elements from n. How do we calculate that? The formula would be (frac{n!}{2!(n - 2)!}). Let's compute that. The numerator is n × (n - 1) × (n - 2)! and the denominator is 2 × 1 × (n - 2)! So the (n - 2)! cancels out from numerator and denominator, leaving (frac{n(n - 1)}{2}). Let me think of another way to see this. When you choose 2 elements from n, the first element can be any of the n elements, and the second element can be any of the remaining (n - 1) elements. But since order doesn't matter here (combinations), each pair is counted twice. For example, choosing element A then element B is the same as choosing B then A. Therefore, we divide by 2 to account for the duplicates. So indeed, it's (frac{n(n - 1)}{2}). That makes sense. For example, if n = 4, then (binom{4}{2} = 6), which is (4×3)/2 = 6. Yep, that's correct.Let me just check all three results again with small n to ensure I haven't made a mistake. Let's take n = 3. Then:- (binom{3}{0}) should be 1. Correct, there's one way to choose nothing.- (binom{3}{1}) should be 3. Indeed, each of the three elements can be chosen individually.- (binom{3}{2}) should be 3. Since choosing two elements from three is the same as leaving out one element, so three ways. The formula gives (3×2)/2 = 3. Correct.Another example, n = 2:- (binom{2}{0}) = 1- (binom{2}{1}) = 2- (binom{2}{2}) = 1 (Wait, using the formula for (binom{2}{2}) would be (2×1)/2 = 1. Correct.)So the formulas hold here as well. So I think the reasoning is solid. Just to recap:- (binom{n}{0}) is 1 because there's only one way to choose nothing.- (binom{n}{1}) is n because there are n single elements you can pick.- (binom{n}{2}) is (frac{n(n - 1)}{2}) because you pick two elements, considering order doesn't matter, so you divide by 2 to eliminate duplicates.I don't see any mistakes here, so I think these are the correct expressions for the binomial coefficients.Final Answer(boxed{1}), (boxed{n}), and (boxed{frac{n(n-1)}{2}})</think>To calculate the binomial coefficients (binom{n}{0}), (binom{n}{1}), and (binom{n}{2}):1. For (binom{n}{0}): - The number of ways to choose 0 elements from (n) is 1, as there is exactly one way to choose nothing. - Using the formula: (binom{n}{0} = frac{n!}{0!n!} = frac{1}{1 cdot 1} = 1). - Therefore, (binom{n}{0} = 1).2. For (binom{n}{1}): - The number of ways to choose 1 element from (n) is (n), as there are (n) individual elements to choose from. - Using the formula: (binom{n}{1} = frac{n!}{1!(n-1)!} = frac{n cdot (n-1)!}{1 cdot (n-1)!} = n). - Therefore, (binom{n}{1} = n).3. For (binom{n}{2}): - The number of ways to choose 2 elements from (n) is (frac{n(n-1)}{2}), because each pair is counted twice when considering order, so we divide by 2. - Using the formula: (binom{n}{2} = frac{n!}{2!(n-2)!} = frac{n(n-1)(n-2)!}{2 cdot (n-2)!} = frac{n(n-1)}{2}). - Therefore, (binom{n}{2} = frac{n(n-1)}{2}). Final Answer(boxed{1}), (boxed{n}), and (boxed{frac{n(n-1)}{2}}),<think>Okay, so I need to calculate the binomial coefficients for n choose 0, n choose 1, and n choose 2. Hmm, let me start by recalling what a binomial coefficient actually is. The binomial coefficient, written as (binom{n}{k}), represents the number of ways to choose k elements from a set of n elements without considering the order. The formula for it is (frac{n!}{k!(n - k)!}), where "!" denotes factorial, which is the product of all positive integers up to that number. First, let's tackle (binom{n}{0}). Plugging into the formula, that would be (frac{n!}{0!(n - 0)!}). Now, I remember that 0! is defined as 1. So simplifying the denominator, we have 1 multiplied by n!, which is just n!. So the expression becomes (frac{n!}{1 times n!}). The n! in the numerator and denominator cancel each other out, leaving us with 1. Therefore, (binom{n}{0}) should be 1. That makes sense because there's exactly one way to choose nothing from a set, right? Just by doing nothing.Moving on to (binom{n}{1}). Using the same formula, it would be (frac{n!}{1!(n - 1)!}). Since 1! is also 1, the denominator simplifies to 1 multiplied by (n - 1)!, which is (n - 1)!. So now we have (frac{n!}{(n - 1)!}). Let's compute that. n! is n multiplied by (n - 1)!, so when you divide n! by (n - 1)!, the (n - 1)! terms cancel out, leaving just n. So (binom{n}{1}) equals n. That also makes intuitive sense because if you have n elements, there are n different ways to choose one element from them.Now the last one, (binom{n}{2}). Applying the formula again, this is (frac{n!}{2!(n - 2)!}). Let's simplify the denominator first. 2! is 2, so the denominator becomes 2 multiplied by (n - 2)!. So the expression is (frac{n!}{2(n - 2)!}). Let's work out the numerator. n! is n times (n - 1) times (n - 2)!, so substituting that in, we have (frac{n times (n - 1) times (n - 2)!}{2(n - 2)!}). The (n - 2)! in the numerator and denominator cancels out, leaving (frac{n(n - 1)}{2}). Therefore, (binom{n}{2}) is (frac{n(n - 1)}{2}). This result also makes sense because when choosing 2 elements from n, the first element can be chosen in n ways, and the second in (n - 1) ways, but since order doesn't matter here (i.e., choosing element A then B is the same as B then A), we divide by 2 to account for the duplicates, hence the division by 2.Let me just verify these results with some small numbers to make sure. For example, take n = 5. - (binom{5}{0}) should be 1. According to the formula, yes, it's 1. Makes sense.- (binom{5}{1}) is 5. That's correct, as there are 5 different single elements to choose.- (binom{5}{2}) should be (frac{5 times 4}{2} = 10). If I list them out: there are 10 pairs possible from 5 elements. That checks out.Another test case: n = 3.- (binom{3}{0}) is 1. Correct.- (binom{3}{1}) is 3. Yep, three elements.- (binom{3}{2}) is (frac{3 times 2}{2} = 3). Indeed, there are 3 ways to choose 2 elements from 3.So, all test cases worked. Therefore, the general formulas I derived seem correct.Wait, just to be thorough, what if n is 0 or 1? Let me check edge cases.For n = 0:- (binom{0}{0}) is 1. The formula gives (frac{0!}{0!0!} = frac{1}{1 times 1} = 1). Correct.- (binom{0}{1}) would be 0, since you can't choose 1 element from an empty set. Using the formula (frac{0!}{1!(-1)!}), but wait, negative factorial is undefined. Hmm, but in our general formula, (binom{n}{1} = n), so if n = 0, then it's 0. Which is correct. But the factorial formula would require (n - 1)! which is (-1)! when n=0, which is undefined. So maybe the general formula (binom{n}{1} = n) still holds here even if the factorial expression runs into undefined terms? Interesting. So perhaps in cases where k > n, the binomial coefficient is 0. But for n=0 and k=1, that is indeed 0.Similarly, for n=1:- (binom{1}{0}) is 1. Correct.- (binom{1}{1}) is 1. The formula gives (frac{1!}{1!0!} = 1). Correct.- (binom{1}{2}) would be 0, since you can't choose 2 elements from 1. But according to the formula (frac{1 times 0}{2} = 0). So 0. Correct.Therefore, even in edge cases, the formulas hold. Therefore, the results are consistent.So, summarizing:- (binom{n}{0} = 1) for any n ≥ 0.- (binom{n}{1} = n) for any n ≥ 1, and 0 when n < 1 (but in the general case, since n here is the set size, if we consider n to be a non-negative integer, then (binom{n}{1} = n) which is 0 when n=0 as per formula).- (binom{n}{2} = frac{n(n - 1)}{2}) for any n ≥ 2, 0 otherwise. But if n is a non-negative integer, then even for n=0 or 1, the formula gives (frac{0 times (-1)}{2} = 0) and (frac{1 times 0}{2} = 0), which is correct.Therefore, the general formulas derived are valid for any non-negative integer n, giving the correct binomial coefficients even when n is 0 or 1. So, all cases are covered.Final Answer(binom{n}{0} = boxed{1}), (binom{n}{1} = boxed{n}), and (binom{n}{2} = boxed{frac{n(n-1)}{2}})</think>To calculate the binomial coefficients (binom{n}{0}), (binom{n}{1}), and (binom{n}{2}):1. For (binom{n}{0}): [ binom{n}{0} = frac{n!}{0!(n-0)!} = frac{n!}{1 cdot n!} = 1 ] Therefore, (binom{n}{0} = boxed{1}).2. For (binom{n}{1}): [ binom{n}{1} = frac{n!}{1!(n-1)!} = frac{n cdot (n-1)!}{1 cdot (n-1)!} = n ] Therefore, (binom{n}{1} = boxed{n}).3. For (binom{n}{2}): [ binom{n}{2} = frac{n!}{2!(n-2)!} = frac{n cdot (n-1) cdot (n-2)!}{2 cdot (n-2)!} = frac{n(n-1)}{2} ] Therefore, (binom{n}{2} = boxed{frac{n(n-1)}{2}}).These results are verified for various values of (n) and edge cases, confirming their correctness. Final Answer(binom{n}{0} = boxed{1}), (binom{n}{1} = boxed{n}), and (binom{n}{2} = boxed{frac{n(n-1)}{2}})]

🔑:There is only one subset with 0 elements (the empty set), so binom{n}{0}=1.There are as many subsets with 1 element (called singletons) as there are elements, so binom{n}{1}=n.To obtain a subset with 2 elements, we start by choosing an element a (n possibilities) and then an element b neq a (n-1 possibilities). However, we have counted each set {a, b} and {b, a} twice. Consequently, binom{n}{2}=frac{n(n-1)}{2}.## Classical IdentitiesIn this section, we derive some classical identities for binomial coefficients using the principle of "double-counting": we consider a set and count its number of elements in two different ways. It will then suffice to equate the two values obtained to get (without further calculation) a new identity.

🧠:[<think>Okay, let's try to tackle this problem. So, we have three prime numbers a, b, c, all greater than 3, that satisfy the equation 2a + 5b = c. The theorem says that a + b + c is a multiple of some integer n, and we need to find the maximum possible value of n. Hmm, interesting.First, let's recall that all primes greater than 3 are of the form 6k ± 1 because any integer can be written as 6k, 6k ± 1, 6k + 2, 6k + 3, or 6k + 4. But primes greater than 3 can't be divisible by 2 or 3, so they must be 6k ± 1. So maybe modulo 6 considerations could help here?Given that a, b, c are primes greater than 3, let's think about their possible residues modulo 6. Since primes greater than 3 are congruent to 1 or 5 modulo 6, right? Because 6k + 1 and 6k + 5 are the only forms that aren't divisible by 2 or 3. So, primes can be 1 or 5 mod 6. Let's note that.So, let's denote:a ≡ 1 or 5 mod 6b ≡ 1 or 5 mod 6c ≡ 1 or 5 mod 6But c = 2a + 5b. Let's compute 2a + 5b modulo 6. Let's see:First, 2a: If a ≡ 1 mod 6, then 2a ≡ 2 mod 6. If a ≡ 5 mod 6, then 2a ≡ 10 ≡ 4 mod 6.Similarly, 5b: If b ≡ 1 mod 6, then 5b ≡ 5 mod 6. If b ≡ 5 mod 6, then 5b ≡ 25 ≡ 1 mod 6.So, 2a can be 2 or 4 mod 6, and 5b can be 5 or 1 mod 6.So, c = 2a + 5b mod 6 can be:Case 1: 2 + 5 = 7 ≡ 1 mod 6Case 2: 2 + 1 = 3 mod 6Case 3: 4 + 5 = 9 ≡ 3 mod 6Case 4: 4 + 1 = 5 mod 6But c must be a prime greater than 3, so c must be either 1 or 5 mod 6. So cases 2 and 3 give c ≡ 3 mod 6, which would mean c is divisible by 3, so c can't be a prime greater than 3 unless c = 3, which is excluded here. Therefore, c can only be 1 or 5 mod 6, so the valid cases are cases 1 and 4.So, possible combinations:Either:a ≡ 1 mod 6 and b ≡ 1 mod 6, leading to c ≡ 1 mod 6Ora ≡ 5 mod 6 and b ≡ 5 mod 6, leading to c ≡ 5 mod 6Wait, hold on. Let's check:Case 1: a ≡1, b≡1: 2(1) +5(1)=7≡1 mod6. Correct.Case 4: a≡5, b≡5: 2*5 +5*5 =10 +25=35≡35 mod6. 35 divided by 6 is 5*6=30, 35-30=5. So 35≡5 mod6. So c≡5 mod6. Correct.But if a and b are in different residues, then c would be 3 mod6, which is invalid, so that's not allowed. Therefore, in order for c to be a prime greater than 3, a and b must be congruent modulo 6. Either both 1 mod6 or both 5 mod6.Therefore, two possibilities:Either a ≡1 mod6 and b≡1 mod6, leading to c≡1 mod6,Or a≡5 mod6 and b≡5 mod6, leading to c≡5 mod6.Therefore, either all three primes are 1 mod6 or two are 5 mod6 and c is 5 mod6.Wait, no. If a and b are both 1 mod6, then c is 1 mod6. If a and b are both 5 mod6, then c is 5 mod6.So, in either case, a, b, c are all congruent mod6. Either all 1 mod6 or all 5 mod6? Wait, no. If a and b are both 1 mod6, then c=2a +5b=2*1 +5*1=7≡1 mod6, so c≡1 mod6. So a, b, c≡1 mod6. If a and b are both 5 mod6, then 2a=10≡4 mod6, 5b=25≡1 mod6, so c=4+1=5 mod6. So in this case, c≡5 mod6. So a≡5, b≡5, c≡5 mod6. So actually, in the first case, all three primes are 1 mod6, and in the second case, all three primes are 5 mod6? Wait, no. Wait, when a and b are 1 mod6, c is 1 mod6, so all three are 1 mod6. When a and b are 5 mod6, then c is 5 mod6, so all three are 5 mod6. So actually, either all three primes are 1 mod6 or all three are 5 mod6. Wait, is that true?Wait, let's check again.If a≡1 mod6, then 2a≡2 mod6. If b≡1 mod6, then 5b≡5 mod6. So 2 +5=7≡1 mod6, so c≡1 mod6. So a≡1, b≡1, c≡1.If a≡5 mod6, then 2a≡10≡4 mod6. If b≡5 mod6, then 5b≡25≡1 mod6. So 4 +1=5 mod6, so c≡5 mod6. So a≡5, b≡5, c≡5.Therefore, in each case, all three primes are congruent mod6. Either all 1 mod6 or all 5 mod6. That seems correct.Therefore, a, b, c are all ≡1 mod6 or all ≡5 mod6.Therefore, the sum a + b + c would be either 1 +1 +1=3≡3 mod6, or 5 +5 +5=15≡3 mod6. Wait, 1+1+1=3, which is 3 mod6, and 5+5+5=15, which is 3 mod6. So in both cases, a + b + c ≡3 mod6. So that means a + b + c is divisible by 3? Wait, 3 mod6 is not divisible by 6, but 3 divides 3. But 3 mod6 is 3, so the sum is divisible by 3. But wait, 3 mod6 is 3, so if the sum is ≡3 mod6, then the sum is 3 more than a multiple of 6, so sum =6k +3=3(2k +1). So sum is divisible by 3 but not necessarily by 6. Therefore, 3 divides a + b + c.Is that the case? Let's verify with actual primes.Take some examples. Let's try to find primes a, b, c >3 such that 2a +5b =c.Let me try a=7 (1 mod6), b=7 (1 mod6). Then c=2*7 +5*7=14 +35=49. But 49 is not a prime. So that's invalid. Hmm.Another example. Let's try a=13 (1 mod6), b=7 (1 mod6). Then c=2*13 +5*7=26 +35=61. 61 is a prime. So a=13, b=7, c=61. Then a + b + c=13 +7 +61=81. 81 is 9*9, which is 3^4, so 81 is divisible by 3, 9, 27, 81. So in this case, n could be 81? But maybe not, because we need it to hold for all possible such triples. Wait, the theorem says "if three prime numbers a, b, c greater than 3 satisfy the equation 2a +5b =c, then a + b + c is a multiple of n." So n must divide all possible a + b + c. Therefore, n is the greatest common divisor of all possible a + b + c. So the maximum possible n is the GCD of all such sums. Therefore, we need to find the GCD of all possible a + b + c where 2a +5b =c and a, b, c primes >3. So to find the maximum n, we need to find the greatest common divisor of all such sums. Therefore, we need to check several examples and find the GCD.Let's take the first example I found: a=13, b=7, c=61. Sum=81.Another example: Let's see, take a=7, b=11 (which is 5 mod6). Wait, but earlier we concluded that a and b must be both 1 mod6 or both 5 mod6. So if a is 7 (1 mod6), then b must be 1 mod6. Let's try a=7, b=13. Then c=2*7 +5*13=14 +65=79. 79 is prime. Then sum=7 +13 +79=99. 99=9*11. So 99 is divisible by 3, 9, 11, etc.Another example: a=19 (1 mod6), b=7 (1 mod6). Then c=2*19 +5*7=38 +35=73. 73 is prime. Sum=19 +7 +73=99. Same as before.Another example: a=31 (1 mod6), b=7 (1 mod6). Then c=2*31 +5*7=62 +35=97. 97 is prime. Sum=31 +7 +97=135. 135 is divisible by 3, 5, 9, 15, etc. 135=5*27=5*3^3.Wait, so sums are 81, 99, 99, 135. Let's check their GCD. The GCD of 81, 99, 135. Prime factors:81: 3^499: 3^2*11135: 3^3*5So GCD is 3^2=9. So if all sums are divisible by 9, then n could be 9. But wait, let's check another example.Take a=5 mod6. Let's find primes a and b both ≡5 mod6. Let's pick a=5, but wait, a must be greater than 3. So a=5 is allowed? Wait, no, primes greater than 3, so a=5 is okay. Wait, no, the problem says primes greater than 3. So 5 is allowed. Wait, but in the problem statement, the theorem is about primes a, b, c greater than 3. So 5 is allowed. Let me check.Wait, but in the original problem statement, it's given as the theorem: "If three prime numbers a, b, c greater than 3 satisfy the equation 2a +5b =c, then a + b + c is a multiple of the integer n." So primes greater than 3. So 5 is allowed. So let's try a=5, b=5. Then c=2*5 +5*5=10 +25=35, which is not prime. So invalid.Another example: a=5, b=11 (both 5 mod6). Then c=2*5 +5*11=10 +55=65, not prime.Another example: a=11, b=5. Then c=22 +25=47, which is prime. So a=11, b=5, c=47. Sum=11 +5 +47=63. 63=7*9=3^2*7. So 63 is divisible by 9.Another example: a=11, b=11. c=2*11 +5*11=22 +55=77, which is not prime.Another example: a=11, b=17 (both 5 mod6). Then c=2*11 +5*17=22 +85=107, which is prime. Sum=11 +17 +107=135. 135 as before.Another example: a=17, b=5. Then c=34 +25=59. Prime. Sum=17 +5 +59=81. So 81 again.Another example: a=17, b=11. c=34 +55=89. Prime. Sum=17 +11 +89=117. 117=13*9=13*3^2.Another example: a=23 (5 mod6), b=7 (1 mod6). Wait, but according to earlier analysis, a and b must both be 1 mod6 or 5 mod6. If a is 5 mod6 and b is 1 mod6, then 2a +5b would be 2*5 +5*1=10 +5=15≡3 mod6, which is invalid as c=15 is not prime. So such a combination is invalid. So, only when a and b are both 1 or both 5 mod6.So let's stick to that. So another example: a=17, b=17. c=2*17 +5*17=34 +85=119. Not prime (119=7*17). So invalid.So in the case where a and b are both 5 mod6, when does c become prime? Let's see:a=11, b=5: c=47 (prime)a=11, b=17: c=107 (prime)a=17, b=5: c=59 (prime)a=5, b=11: c=65 (not prime)a=5, b=17: c=10 +85=95 (not prime)a=23, b=5: c=46 +25=71 (prime). Sum=23 +5 +71=99.a=23, b=11: c=46 +55=101 (prime). Sum=23 +11 +101=135.a=29, b=5: c=58 +25=83 (prime). Sum=29 +5 +83=117.So, sums here are 63, 135, 81, 99, 135, 117, etc. All of these sums are divisible by 9. Let's check:63: 9*781:9*999:9*11117:9*13135:9*15So all these are divisible by 9. Also, in the previous cases with a and b ≡1 mod6, sums like 81, 99, 135, which are all divisible by 9. So is 9 the GCD? Let's check if all sums are divisible by 9.Wait, but let's check another example where maybe the sum is different. Let's see:Take a=7, b=7 (both 1 mod6). Then c=14 +35=49, which is not prime. So invalid.a=7, b=13 (both 1 mod6). c=14 +65=79, prime. Sum=7 +13 +79=99.a=7, b=19 (both 1 mod6). c=14 +95=109, prime. Sum=7 +19 +109=135.a=13, b=7: sum=13 +7 +61=81.a=13, b=13: c=26 +65=91, not prime.a=13, b=19: c=26 +95=121, not prime.a=19, b=7: c=38 +35=73. Sum=19 +7 +73=99.a=19, b=13: c=38 +65=103. Sum=19 +13 +103=135.So, all sums here are 81, 99, 135. All multiples of 9. Let's check if there is a case where sum is not divisible by 9.Wait, take a=43 (1 mod6), b=7 (1 mod6). Then c=86 +35=121, which is not prime.a=43, b=13. c=86 +65=151, which is prime. Sum=43 +13 +151=207. 207=9*23. Divisible by 9.a=43, b=19. c=86 +95=181, prime. Sum=43 +19 +181=243. 243=9*27=9*3^3. Divisible by 9.a=7, b=31. c=14 +155=169, not prime.a=7, b=37. c=14 +185=199, prime. Sum=7 +37 +199=243, again divisible by 9.Another example with a=5, b=23 (both 5 mod6). c=10 +115=125, not prime.a=5, b=29. c=10 +145=155, not prime.a=11, b=23. c=22 +115=137, prime. Sum=11 +23 +137=171. 171=9*19.So, in all these cases, the sum is divisible by 9. Therefore, it seems that 9 is a common divisor. Is there a higher possible n?Suppose we check if the sums are divisible by 9. Let's check with the examples:81: 9*999:9*11135:9*1563:9*7117:9*13243:9*27171:9*19So, all these sums are multiples of 9, but they are not all multiples of, say, 18 or 27. For instance, 81 is divisible by 27, but 99 is not. 99 is divisible by 9 but not by 27. Similarly, 63 is divisible by 9 but not by 27. Therefore, the greatest common divisor is 9.Therefore, the maximum possible value of n is 9.But wait, let's check if there is a theoretical reason why the sum is always divisible by 9. Earlier, we saw that a, b, c are all ≡1 mod6 or all ≡5 mod6.So let's consider two cases:Case 1: All primes a, b, c ≡1 mod6.Then a =6k +1, b=6m +1, c=6n +1.Sum: a + b + c =6(k + m + n) +3=3*(2(k + m + n) +1). So the sum is divisible by 3.Similarly, in case 2: All primes a, b, c ≡5 mod6.Then a=6k +5, b=6m +5, c=6n +5.Sum: a + b + c=6(k + m + n) +15=6(k + m + n +2) +3=3*(2(k + m + n +2) +1). Again, divisible by 3.But our examples show that the sum is divisible by 9, which is 3^2. So why is that?Let's analyze modulo 9.Since a, b, c ≡1 or 5 mod6, let's see what they are modulo 9.But primes can be in various residues modulo 9. For example, primes greater than 3 can be 1, 2, 4, 5, 7, or 8 mod9.But given that a, b, c are either all 1 mod6 or all 5 mod6, let's see what that implies modulo 3.Wait, 1 mod6 is equivalent to 1 mod3, since 6=2*3. Similarly, 5 mod6 is equivalent to 2 mod3. So:If a, b, c are all 1 mod6, then they are all 1 mod3.If a, b, c are all 5 mod6, then they are all 2 mod3.Therefore, in the first case:a ≡1 mod3, b≡1 mod3, c≡1 mod3.So sum a + b + c ≡1 +1 +1=3≡0 mod3. Therefore, sum divisible by 3.In the second case:a ≡2 mod3, b≡2 mod3, c≡2 mod3.Sum a + b + c ≡2 +2 +2=6≡0 mod3. So sum divisible by 3.But in both cases, the sum is 0 mod3. But in our examples, the sum is divisible by 9. So maybe modulo 9?Let's check.First, consider a, b, c all ≡1 mod6.So, possible residues modulo 9: primes can be 1, 7 mod9 (since 6k +1 can be 1,7,13≡4,19≡1, etc. mod9). Similarly, 5 mod6 is 5,11≡2,17≡8,23≡5,29≡2, etc. mod9. So primes can be in various residues modulo 9, but since they are ≡1 or 5 mod6, let's see:For primes ≡1 mod6:The residues mod9 can be 1, 7, 4, etc. For example:7≡7 mod913≡4 mod919≡1 mod925≡7 mod9 (but 25 is not prime)31≡4 mod937≡1 mod9Similarly, primes ≡5 mod6:5≡5 mod911≡2 mod917≡8 mod923≡5 mod929≡2 mod935≡8 mod9 (not prime)41≡5 mod9So primes ≡1 mod6 can be 1,4,7 mod9, and primes ≡5 mod6 can be 2,5,8 mod9.Therefore, even though a, b, c are ≡1 or 5 mod6, their residues mod9 can vary.However, given the equation 2a +5b =c, perhaps this imposes some restrictions mod9.Let's analyze the equation 2a +5b ≡c mod9.If we can find the possible residues of c given a and b, maybe we can find constraints on a + b + c mod9.Let's first consider the case where a, b, c are all ≡1 mod6 (so in mod3, they are 1). Then mod9:Let's suppose a ≡1 mod9, b≡1 mod9. Then c=2*1 +5*1=7 mod9. Then sum=1 +1 +7=9≡0 mod9.If a≡1 mod9, b≡4 mod9 (since primes ≡1 mod6 can be 1,4,7 mod9). Then c=2*1 +5*4=2 +20=22≡4 mod9. Sum=1 +4 +4=9≡0 mod9.If a≡1 mod9, b≡7 mod9. Then c=2*1 +5*7=2 +35=37≡1 mod9. Sum=1 +7 +1=9≡0 mod9.If a≡4 mod9, b≡1 mod9. c=2*4 +5*1=8 +5=13≡4 mod9. Sum=4 +1 +4=9≡0 mod9.a≡4, b≡4: c=2*4 +5*4=8 +20=28≡1 mod9. Sum=4 +4 +1=9≡0.a≡4, b≡7: c=2*4 +5*7=8 +35=43≡7 mod9. Sum=4 +7 +7=18≡0 mod9.a≡7, b≡1: c=2*7 +5*1=14 +5=19≡1 mod9. Sum=7 +1 +1=9≡0.a≡7, b≡4: c=2*7 +5*4=14 +20=34≡7 mod9. Sum=7 +4 +7=18≡0.a≡7, b≡7: c=2*7 +5*7=14 +35=49≡4 mod9. Sum=7 +7 +4=18≡0.So in all cases where a and b are ≡1 mod6, the sum a + b + c ≡0 mod9.Similarly, let's check the other case where a, b, c ≡5 mod6 (so ≡2 mod3).But modulo9, these primes can be 2,5,8 mod9.Let's take a≡5 mod9, b≡5 mod9. Then c=2*5 +5*5=10 +25=35≡8 mod9. Sum=5 +5 +8=18≡0 mod9.a≡5, b≡2: c=2*5 +5*2=10 +10=20≡2 mod9. Sum=5 +2 +2=9≡0.a≡5, b≡8: c=2*5 +5*8=10 +40=50≡5 mod9. Sum=5 +8 +5=18≡0.a≡2, b≡5: c=2*2 +5*5=4 +25=29≡2 mod9. Sum=2 +5 +2=9≡0.a≡2, b≡2: c=2*2 +5*2=4 +10=14≡5 mod9. Sum=2 +2 +5=9≡0.a≡2, b≡8: c=2*2 +5*8=4 +40=44≡8 mod9. Sum=2 +8 +8=18≡0.a≡8, b≡5: c=2*8 +5*5=16 +25=41≡5 mod9. Sum=8 +5 +5=18≡0.a≡8, b≡2: c=2*8 +5*2=16 +10=26≡8 mod9. Sum=8 +2 +8=18≡0.a≡8, b≡8: c=2*8 +5*8=16 +40=56≡2 mod9. Sum=8 +8 +2=18≡0.Therefore, in all cases where a and b are ≡5 mod6 (and hence c≡5 mod6), the sum a + b + c is ≡0 mod9.Therefore, regardless of whether a, b, c are all 1 mod6 or all 5 mod6, the sum a + b + c is always divisible by 9. Therefore, n must be at least 9.Moreover, since we found examples where the sum is 9 (but wait, in our examples, the smallest sum was 63, which is 9*7). But maybe there's no triplet where the sum is 9, given that primes are greater than 3. The smallest possible primes are 5,7,11,...Wait, the smallest primes greater than 3 are 5,7,11,13,...Let's check if there exists a triplet with a smaller sum. For example, take a=5, b=7. Then c=2*5 +5*7=10 +35=45, which is not prime.a=5, b=11: c=10 +55=65, not prime.a=5, b=17: c=10 +85=95, not prime.a=7, b=5: c=14 +25=39, not prime.a=7, b=7: c=14 +35=49, not prime.a=7, b=11: c=14 +55=69, not prime.a=7, b=13: c=14 +65=79, prime. Sum=7 +13 +79=99.So the smallest sum is 81, which is 9*9. So all sums we found are multiples of 9, but no smaller multiple. Hence, since 9 divides all possible sums, and there is no larger integer that divides all sums (since GCD of the sums is 9), then the maximum possible n is 9.Therefore, the answer is 9.Final AnswerThe maximum possible value of the integer ( n ) is boxed{9}.</think>Given three prime numbers (a, b, c) greater than 3 that satisfy the equation (2a + 5b = c), we need to determine the maximum possible value of (n) such that (a + b + c) is a multiple of (n).1. Modulo 6 Analysis: - All primes greater than 3 are congruent to 1 or 5 modulo 6. - For (c) to be a prime greater than 3, (2a + 5b) must also be congruent to 1 or 5 modulo 6. - If (a equiv 1 pmod{6}) and (b equiv 1 pmod{6}), then (2a + 5b equiv 1 pmod{6}). - If (a equiv 5 pmod{6}) and (b equiv 5 pmod{6}), then (2a + 5b equiv 5 pmod{6}). - Therefore, (a, b, c) must all be congruent modulo 6 (either all 1 or all 5).2. Sum Analysis: - If (a, b, c equiv 1 pmod{6}), then (a + b + c equiv 3 pmod{6}). - If (a, b, c equiv 5 pmod{6}), then (a + b + c equiv 15 equiv 3 pmod{6}). - Both cases result in (a + b + c equiv 3 pmod{6}), indicating divisibility by 3.3. Modulo 9 Analysis: - Considering the equation (2a + 5b = c) modulo 9, we find that regardless of the specific residues of (a, b, c) modulo 9, the sum (a + b + c) is always divisible by 9. - Examples of valid triples ((a, b, c)) were checked, and all sums were found to be multiples of 9.4. Conclusion: - The greatest common divisor (GCD) of all possible sums (a + b + c) is 9, as verified by multiple examples and theoretical analysis.Thus, the maximum possible value of the integer (n) is (boxed{9}).,<think>Okay, let's try to figure out this problem. So, the theorem says that if three prime numbers a, b, c (all greater than 3) satisfy 2a + 5b = c, then a + b + c is a multiple of some integer n. We need to find the maximum possible value of n. Hmm, interesting.First, I remember that all primes greater than 3 are of the form 6k ± 1 because any integer can be written as 6k, 6k+1, 6k+2, 6k+3, 6k+4, or 6k+5, and primes greater than 3 can't be divisible by 2 or 3, so they can't be 6k, 6k+2, 6k+3, or 6k+4. So they must be 6k+1 or 6k-1. That seems right.So maybe I should express a, b, c in terms of 6k ± 1. Let me denote:a = 6m ± 1b = 6n ± 1c = 6p ± 1where m, n, p are integers. Then, substitute these into the equation 2a + 5b = c and see what happens.Let's try substituting a and b:2*(6m ± 1) + 5*(6n ± 1) = 6p ± 1Compute left side:12m ± 2 + 30n ± 5 = 6p ± 1Combine like terms:12m + 30n + (±2 ±5) = 6p ± 1So, 12m + 30n + (some combination of ±2 and ±5) equals 6p ± 1.Let me compute the possible combinations for the constants:±2 ±5 can be:2 + 5 = 72 - 5 = -3-2 + 5 = 3-2 -5 = -7So, the left side can be 12m + 30n + 7, 12m + 30n - 3, 12m + 30n + 3, or 12m + 30n -7.And the right side is 6p ±1, so 6p +1 or 6p -1.Therefore, equating left and right sides, we have four cases for the left and two cases for the right, so total eight possibilities? That seems a bit messy. Maybe there's a better way.Alternatively, maybe considering modulo 6? Because all primes greater than 3 are congruent to 1 or 5 mod 6. Let's try that.So, let's compute 2a + 5b mod 6. Since a and b are primes >3, they are either 1 or 5 mod 6.Case 1: a ≡1 mod6, b≡1 mod6Then 2a ≡ 2*1 ≡2 mod65b ≡5*1 ≡5 mod6Total: 2 +5 ≡7 ≡1 mod6. So c ≡1 mod6.Case 2: a ≡1 mod6, b≡5 mod62a ≡2 mod65b ≡5*5 ≡25 ≡1 mod6Total: 2 +1 ≡3 mod6. But c has to be a prime >3, but 3 mod6 is 3, which is not possible since primes greater than 3 can't be 0, 2, 3, or 4 mod6. So c ≡3 mod6 would mean c is divisible by 3, hence not prime (unless c=3, but c>3). So this case is invalid. So if a ≡1 mod6 and b≡5 mod6, then c would be 3 mod6, which is not allowed. Therefore, this combination is impossible.Case3: a ≡5 mod6, b≡1 mod62a ≡2*5=10≡4 mod65b ≡5*1=5 mod6Total:4 +5=9≡3 mod6. Again, same problem as case2. c≡3 mod6, which is invalid. So this case is impossible.Case4: a≡5 mod6, b≡5 mod62a≡2*5=10≡4 mod65b≡5*5=25≡1 mod6Total:4 +1=5 mod6. So c≡5 mod6. Which is valid, as primes can be 5 mod6.So, the only possible cases where c is a prime greater than 3 are when:Either (a≡1, b≡1, c≡1 mod6) or (a≡5, b≡5, c≡5 mod6).So, now, let's note that. Therefore, in these cases, a, b, c are either all 1 mod6 or all 5 mod6.Therefore, if we look at a + b + c, in each case:Case1: a≡1, b≡1, c≡1 mod6. Then a + b + c ≡1+1+1=3≡3 mod6.Case4: a≡5, b≡5, c≡5 mod6. Then a + b + c≡5+5+5=15≡3 mod6.So in either case, a + b + c ≡3 mod6. Therefore, 3 divides (a + b + c) - 3, so 3 divides (a + b + c). Wait, but 15≡3 mod6, but 15 is 15=6*2 +3, so 15≡3 mod6. So a + b + c is congruent to 3 mod6, which means that (a + b + c) = 6k +3 = 3(2k +1). Therefore, it's divisible by 3, but not necessarily 6. So n must be a multiple of 3. But the question is asking for the maximum possible n such that a + b + c is a multiple of n for all such primes a, b, c. So 3 is a candidate. But maybe there's a higher n.Wait, but maybe modulo other numbers?Alternatively, perhaps considering modulo 2 or 4?But primes greater than 3 are all odd. So a, b, c are all odd. Then 2a is even, 5b is odd (since 5 is odd, b is odd, 5b is odd). Then 2a +5b is even + odd = odd. So c must be odd, which is consistent with primes >3 being odd. So nothing new here.But let's think about modulo 3. Since primes greater than 3 are not divisible by 3, so they are either 1 or 2 mod3.Let me check the equation 2a +5b ≡c mod3.Express each prime modulo3:a can be 1 or 2 mod3b can be 1 or 2 mod3Similarly for c.Let's check all possibilities:Case1: a≡1, b≡1Then 2*1 +5*1=2 +5=7≡1 mod3. So c≡1 mod3.Case2: a≡1, b≡22*1 +5*2=2 +10=12≡0 mod3. But c cannot be 0 mod3, since it's a prime>3. So this case is invalid.Case3: a≡2, b≡12*2 +5*1=4 +5=9≡0 mod3. Again, c≡0 mod3, invalid.Case4: a≡2, b≡22*2 +5*2=4 +10=14≡2 mod3. So c≡2 mod3.So valid cases are when (a≡1, b≡1, c≡1) or (a≡2, b≡2, c≡2) mod3.Therefore, a + b + c in each case:Case1: 1 +1 +1=3≡0 mod3Case4:2 +2 +2=6≡0 mod3Therefore, regardless of the case, a + b + c ≡0 mod3. So 3 divides a + b + c.Earlier, we saw that modulo6, a + b + c ≡3 mod6, so 3 divides it, but 6 doesn't. So 3 is a factor.But let's check modulo 9. Is 9 a divisor?Wait, maybe 3 is the maximum, but maybe there's a larger number. Let's see.Alternatively, perhaps n is 6? But since a + b + c ≡3 mod6, then 6 doesn't divide it, but 3 does. So 3 is a factor, but 6 is not. So n can't be 6.But maybe 12? Let's check.Alternatively, maybe 15? Wait, but the problem states "a multiple of the integer n". The maximum possible n such that for all such a,b,c, a + b + c is a multiple of n. So we need the greatest common divisor of all possible a + b + c. So we need to find the GCD of all possible a + b + c where 2a +5b =c, a,b,c primes >3.So perhaps n is 3, but maybe higher. To check, we can find specific examples.Let me try to find some primes a, b, c >3 satisfying 2a +5b =c, and compute a + b + c, then see their GCD.First, let's find such primes.Start with small primes. Let's take a=5, b=7.Then 2*5 +5*7=10 +35=45, which is not prime. So invalid.a=5, b=5: 2*5 +5*5=10+25=35, not prime.a=7, b=5: 2*7 +5*5=14 +25=39, not prime.a=7, b=7: 2*7 +5*7=14 +35=49=7^2, not prime.a=11, b=7:2*11 +5*7=22 +35=57, not prime.a=11, b=11:2*11 +5*11=22 +55=77, not prime.a=5, b=11:2*5 +5*11=10 +55=65, not prime.Hmm, this is tricky. Maybe trying different combinations.Wait, let's see. Let me check a=7, b=5: 14 +25=39. Not prime.a=7, b=11:2*7 +5*11=14 +55=69, not prime.a=11, b=13:2*11 +5*13=22 +65=87, not prime.a=13, b=7:2*13 +5*7=26 +35=61. 61 is a prime. Okay, so here we have a=13, b=7, c=61.Check if all primes: 13,7,61 are primes>3. Yes. So that's a valid triplet.Compute a + b + c=13 +7 +61=81. 81 is 9*9. So 81 is divisible by 9, 3, etc.Another example. Let me find another triplet.a=17, b=7:2*17 +5*7=34 +35=69, not prime.a=19, b=5:2*19 +5*5=38 +25=63, not prime.a=7, b=13:2*7 +5*13=14 +65=79, which is prime.So a=7, b=13, c=79. Sum is 7 +13 +79=99. 99 is 9*11.Another example. a=5, b=13:2*5 +5*13=10 +65=75, not prime.a=5, b=17:10 +85=95, not prime.a=11, b=7:22 +35=57, not prime.a=11, b=13:22 +65=87, not prime.a=13, b=11:26 +55=81, not prime.a=17, b=11:34 +55=89, which is prime. So a=17, b=11, c=89.Sum:17 +11 +89=117. 117=9*13.Another example: a=19, b=7:38 +35=73 (prime). So sum=19+7+73=99 again.a=23, b=5:46 +25=71 (prime). Sum=23+5+71=99.a=29, b=5:58 +25=83 (prime). Sum=29+5+83=117.Hmm, so the sums we have are 81, 99, 117, 99, 99, 117. Let's calculate GCD of these numbers.First, 81,99,117.Prime factors:81=3^499=9*11=3^2*11117=9*13=3^2*13So GCD(81,99)=9, GCD(9,117)=9. So GCD is 9. So maybe 9 is the maximum n? But we need to check if all possible sums are divisible by 9.Wait, let's check another example.Take a=7, b=19:14 +95=109 (prime). Sum=7+19+109=135. 135 is divisible by 9 (135=15*9). Also, 135 is divisible by 5? 135=27*5, yes. But 135 is divisible by 9 and 5, but in previous examples, 81 isn't divisible by 5. So GCD of all sums would be 9. Let's check another one.a=13, b=19:26 +95=121, which is 11², not prime.a=11, b=17:22 +85=107 (prime). Sum=11+17+107=135. Again, 135.a=23, b=7:46 +35=81, which is not prime.Wait, 46 +35=81=9*9, not prime.Another one: a=31, b=5:62 +25=87, not prime.a=37, b=5:74 +25=99, not prime.a=7, b=17:14 +85=99, not prime.a=7, b=23:14 +115=129, not prime.a=11, b=19:22 +95=117, which is not prime.Wait, seems hard to find more examples, but from the ones we have, the sums are 81, 99, 117, 135. All of which are divisible by 9, but not by 18 or 27. For example, 81 is 9*9, 99 is 9*11, 117 is 9*13, 135 is 9*15. The GCD of these is 9. So perhaps the maximum n is 9. Let me check.But first, I need to ensure that all possible a + b + c are divisible by 9. Let's check with modulo 9.So, if we can show that a + b + c ≡0 mod9, then 9 divides the sum. Let's check.Given 2a +5b =c. So c=2a +5b. Therefore, a + b + c =a + b +2a +5b=3a +6b=3(a + 2b). So a + b + c=3(a + 2b). Therefore, the sum is divisible by 3. But we saw in examples that it's divisible by 9. So perhaps a +2b is divisible by 3?Wait, let's check.Since 3(a + 2b) is the sum. If a +2b is divisible by 3, then the sum is divisible by 9.So let's check if a + 2b ≡0 mod3.From earlier, we saw that for the equation 2a +5b =c to hold, we had constraints on a and b modulo3. Specifically, either both a and b ≡1 mod3 or both ≡2 mod3.Case1: a≡1 mod3, b≡1 mod3.Then a +2b ≡1 +2*1=1+2=3≡0 mod3. So a +2b≡0 mod3. Therefore, 3 divides (a +2b), so sum=3*(a +2b) is divisible by 9.Case2: a≡2 mod3, b≡2 mod3.Then a +2b≡2 +2*2=2 +4=6≡0 mod3. So again, a +2b≡0 mod3. Hence, sum divisible by 9.Therefore, in both valid cases (since other combinations of a and b mod3 lead to invalid c), the sum a +b +c=3(a +2b) is divisible by 9. Hence, 9 divides the sum. Therefore, n must be at least 9.But the examples show sums like 81,99,117,135, which are all divisible by 9 but not necessarily by higher numbers. For example, 81 is 9*9, 99=9*11, 117=9*13, 135=9*15. The GCD of these is 9. Therefore, the maximum possible n is 9.Therefore, the answer should be 9.Final AnswerThe maximum possible value of the integer ( n ) is boxed{9}.</think>Given the theorem that if three prime numbers (a, b, c) greater than 3 satisfy the equation (2a + 5b = c), then (a + b + c) is a multiple of the integer (n), we need to find the maximum possible value of (n).First, we note that all primes greater than 3 are of the form (6k pm 1). We analyze the equation (2a + 5b = c) modulo 6 and modulo 3.1. Modulo 6 Analysis: - Primes (a, b, c) can be either (1 mod 6) or (5 mod 6). - We find that valid cases for (a, b, c) are either all (1 mod 6) or all (5 mod 6). - In both cases, (a + b + c equiv 3 mod 6), indicating (a + b + c) is divisible by 3 but not 6.2. Modulo 3 Analysis: - Primes (a, b) can be (1 mod 3) or (2 mod 3). - Valid cases are when (a) and (b) are both (1 mod 3) or both (2 mod 3). - In both cases, (2a + 5b equiv c mod 3) results in (c) being (1 mod 3) or (2 mod 3), leading to (a + b + c equiv 0 mod 3).3. Sum Analysis: - The sum (a + b + c) can be expressed as (3(a + 2b)). - Since (a + 2b) is divisible by 3 (from modulo 3 analysis), the sum (3(a + 2b)) is divisible by 9.4. Examples and Verification: - Examples of valid primes (a, b, c) yield sums (81, 99, 117, 135), all divisible by 9. - The greatest common divisor of these sums is 9.Thus, the maximum possible value of (n) such that (a + b + c) is a multiple of (n) for all valid primes (a, b, c) is (boxed{9}).]

🔑:Solution: Let a=3 k_{1}+r_{1}, b=3 k_{2}+r_{2}. Thenbegin{array}{l}a+b+c=3(a+2 b) =3left(3 k_{1}+r_{1}+6 k_{2}+2 r_{2}right) =9left(k_{1}+2 k_{2}right)+3left(r_{1}+2 r_{2}right) .end{array}Since a, b are both primes greater than 3, we haver_{1} r_{2} neq 0.If r_{1} neq r_{2}, then r_{1}=1, r_{2}=2, or r_{1}=2, r_{2}=1. In this case,begin{aligned}c & =2 a+5 b=6 k_{1}+2 r_{1}+15 k_{2}+5 r_{2} & =3left(2 k_{1}+5 k_{2}+r_{2}right)+2left(r_{1}+r_{2}right) & =3left(2 k_{1}+5 k_{2}+r_{2}right)+6,end{aligned}which contradicts the fact that c is a prime. Therefore, r_{1}=r_{2}.begin{array}{l}text { Hence } a+b+c=9left(k_{1}+2 k_{2}right)+3left(r_{1}+2 r_{2}right) =9left(k_{1}+2 k_{2}+r_{1}right) .end{array}Thus, a+b+c is a multiple of 9, i.e., n=9 when the conclusion holds.Next, we prove: n leqslant 9.Take a=11, b=5, then c=2 a+5 b=47. In this case, a+b+c=63, so n mid 63.Take a=13, b=7, then c=2 a+5 b=61. In this case, a+b+c=81, so n mid 81.Therefore, n is a common divisor of 63 and 81.Hence n leqslant(63,81)=9.In conclusion, the maximum value of n is 9.