🧠:[<think>Okay, so I need to find the minimum value of the expression (frac{3a^2 - a}{1 + a^2} + frac{3b^2 - b}{1 + b^2} + frac{3c^2 - c}{1 + c^2}) where (a), (b), and (c) are positive real numbers that add up to 1. Hmm, let me start by understanding the problem better.First, since (a), (b), and (c) are positive and sum to 1, each of them is between 0 and 1. That might be useful because the denominators (1 + a^2), etc., are going to be between 1 and 2. Maybe that helps in approximating or simplifying?Looking at each term of the expression, like (frac{3a^2 - a}{1 + a^2}), I wonder if I can simplify this or find some property. Let me consider a function (f(x) = frac{3x^2 - x}{1 + x^2}). Then the problem is asking for the minimum of (f(a) + f(b) + f(c)) under the constraint (a + b + c = 1).Maybe I can analyze the function (f(x)) first. Let's compute its derivative to see if it's convex or concave. If the function is convex or concave, I can apply Jensen's inequality, which is often useful for such optimization problems with constraints.So, compute (f'(x)):(f(x) = frac{3x^2 - x}{1 + x^2}).Using the quotient rule: (f'(x) = frac{(6x - 1)(1 + x^2) - (3x^2 - x)(2x)}{(1 + x^2)^2}).Let me compute the numerator:First term: ((6x - 1)(1 + x^2) = 6x(1) + 6x(x^2) - 1(1) - 1(x^2) = 6x + 6x^3 - 1 - x^2)Second term: (-(3x^2 - x)(2x) = -2x(3x^2 - x) = -6x^3 + 2x^2)Adding both terms together:6x + 6x^3 -1 -x^2 -6x^3 + 2x^2 = 6x -1 + ( -x^2 + 2x^2 ) + (6x^3 -6x^3)Simplify:6x -1 + x^2 + 0 = x^2 + 6x -1So the derivative is (frac{x^2 + 6x -1}{(1 + x^2)^2}).Hmm, interesting. Now, to check the critical points, set numerator equal to zero:(x^2 + 6x -1 =0)Solving for x:(x = [-6 pm sqrt{36 + 4}]/2 = [-6 pm sqrt{40}]/2 = [-6 pm 2sqrt{10}]/2 = -3 pm sqrt{10})But since (x) is positive (given (a, b, c in mathbf{R}_+)), the only critical point in positive reals is (x = -3 + sqrt{10}). Let's approximate that:(sqrt{10} approx 3.1623), so ( -3 + 3.1623 approx 0.1623). So critical point at around 0.1623.Now, let's check the behavior of the derivative around that point. For (x < 0.1623), the numerator (x^2 +6x -1) would be negative (since at x=0, it's -1, and increasing as x increases). At x=0.1623, it's zero. Then for x >0.1623, it becomes positive. So the function f(x) is decreasing until x≈0.1623 and increasing after that. Therefore, the function f(x) has a minimum at x≈0.1623.Wait, but in our problem, variables a, b, c are positive and sum to 1. So each variable can be as small as approaching 0 or up to 1. So depending on the variable's value, the function f(x) might be increasing or decreasing. Hmm, this complicates things.Since each term has a minimum at around 0.1623, but variables have to sum to 1. So maybe if we set all variables equal to 1/3, which is approximately 0.333, which is greater than 0.1623, so in the increasing region. But if some variables are smaller, like approaching 0, then their f(x) would be in the decreasing region.But since variables are positive and sum to 1, if one variable is very small, the others have to be larger. So perhaps making one variable small (so f(x) decreases for that variable) but others are larger (so f(x) increases for them). The balance between these effects will determine the minimum.Alternatively, maybe the minimum occurs when two variables are equal and the third is different. Or maybe all variables are equal? Let's check.If we set a = b = c = 1/3, then each term is f(1/3):Compute f(1/3) = [3*(1/3)^2 - (1/3)] / [1 + (1/3)^2] = [3*(1/9) - 1/3] / [1 + 1/9] = [ (1/3 - 1/3) ] / (10/9) = 0 / (10/9) = 0.So each term is zero, so total sum is zero. Wait, but if that's the case, then the sum is zero. But is this the minimum?Wait, but can the sum be negative? Let me check with a different value. Let's take a = 0.1, b = 0.1, c = 0.8. Let's compute f(a), f(b), f(c).First, compute f(0.1):Numerator: 3*(0.1)^2 - 0.1 = 3*0.01 - 0.1 = 0.03 - 0.1 = -0.07Denominator: 1 + 0.01 = 1.01So f(0.1) = -0.07 / 1.01 ≈ -0.0693Similarly, f(0.1) again ≈ -0.0693Now compute f(0.8):Numerator: 3*(0.8)^2 - 0.8 = 3*0.64 - 0.8 = 1.92 - 0.8 = 1.12Denominator: 1 + 0.64 = 1.64So f(0.8) ≈ 1.12 / 1.64 ≈ 0.6829Total sum: 2*(-0.0693) + 0.6829 ≈ -0.1386 + 0.6829 ≈ 0.5443, which is higher than zero.Hmm, so even with two variables at 0.1, which is below the critical point, the third variable at 0.8 makes the overall sum positive. So maybe when variables are split into small and large, the large ones contribute more positive, so the total is higher.Wait, but maybe if we have one variable very close to 0, and the other two variables at around 0.5 each. Let's test.Let a approach 0, say a = 0.0001, then b + c ≈ 1. Let’s set b = c = 0.49995.Compute f(a):Numerator: 3*(0.0001)^2 - 0.0001 ≈ 0 - 0.0001 = -0.0001Denominator: 1 + (0.0001)^2 ≈ 1So f(a) ≈ -0.0001f(b) = f(0.49995):Numerator: 3*(0.49995)^2 - 0.49995 ≈ 3*(0.24995) - 0.49995 ≈ 0.74985 - 0.49995 ≈ 0.2499Denominator: 1 + (0.49995)^2 ≈ 1 + 0.24995 ≈ 1.24995So f(b) ≈ 0.2499 / 1.24995 ≈ 0.1999Similarly, f(c) is the same as f(b), so total sum ≈ -0.0001 + 0.1999 + 0.1999 ≈ 0.3997, which is still positive.Wait, but this is less than the previous case where two variables were 0.1. Hmm, so maybe distributing more towards the middle gives a lower total. Wait, but 0.3997 is still higher than zero. But in the case when all variables are 1/3, the total is zero. So maybe the minimum is zero? But wait, when variables are 1/3, each term is zero. Let me verify again.Wait, if a = 1/3, then:Numerator: 3*(1/3)^2 - (1/3) = 3*(1/9) - 1/3 = 1/3 - 1/3 = 0. So each term is zero, so total sum is zero.But when I tried other values, like a = 0.1, b = 0.1, c = 0.8, the total was positive. When I tried a approaching 0, b and c approaching 0.5 each, the total was still positive. So maybe the minimum is indeed zero, achieved when all variables are 1/3?But let me test another case where variables are different but sum to 1, maybe with one variable at the critical point x ≈0.1623.Let me compute x = -3 + sqrt(10) ≈0.1623. Let's set a = 0.1623, then b + c = 1 - 0.1623 ≈0.8377. Let me set b = c = 0.41885 each.Compute f(a):Numerator: 3*(0.1623)^2 - 0.1623 ≈ 3*(0.02635) - 0.1623 ≈ 0.07905 - 0.1623 ≈ -0.08325Denominator: 1 + (0.1623)^2 ≈1 + 0.02635 ≈1.02635So f(a) ≈ -0.08325 / 1.02635 ≈-0.0811f(b) for b=0.41885:Numerator: 3*(0.41885)^2 - 0.41885 ≈3*(0.1754) -0.41885≈0.5262 -0.41885≈0.1073Denominator:1 + (0.41885)^2≈1 +0.1754≈1.1754f(b)≈0.1073 /1.1754≈0.0913So total sum: -0.0811 +2*0.0913≈-0.0811 +0.1826≈0.1015, which is still positive. Hmm, so even when one variable is at the critical point where f(x) is minimized, the other variables are in the increasing region, and the total sum is still positive.Wait, but the total sum here is 0.1015, which is higher than the zero we get when all variables are 1/3. So that seems to confirm that when all variables are 1/3, the total is zero, which might be the minimum.But wait, let's check if there's a case where the sum is negative. Maybe if two variables are at the critical point? Let me see.Suppose two variables are at x ≈0.1623 each. Then the third variable is 1 - 2*0.1623 ≈1 -0.3246≈0.6754.Compute f(x) for each:First two variables:f(0.1623) ≈-0.0811 each, so total for two variables:≈-0.1622Third variable, x=0.6754:Numerator:3*(0.6754)^2 -0.6754≈3*(0.4562) -0.6754≈1.3686 -0.6754≈0.6932Denominator:1 + (0.6754)^2≈1 +0.4562≈1.4562f(x)≈0.6932 /1.4562≈0.476Total sum≈-0.1622 +0.476≈0.3138, still positive.Alternatively, maybe all three variables are at the critical point? But 3*0.1623≈0.4869, which is less than 1. So that's not possible. So we can't have all variables at the critical point.Alternatively, maybe having one variable at the critical point and others arranged in some way? But previous attempts didn't give a negative total. Hmm.Wait, let's try another case: two variables very small, like approaching zero, and the third variable approaching 1. Let's compute the limit as two variables go to 0.Suppose a approaches 0, b approaches 0, c approaches 1.Compute the sum: f(a) + f(b) + f(c)As a approaches 0:f(a) ≈ [0 - a] / [1 + 0] = -a → 0Similarly, f(b) ≈ -b → 0f(c) as c approaches 1:Numerator:3*(1)^2 -1=3 -1=2Denominator:1 +1=2So f(1)=2/2=1Thus, the total sum approaches 0 +0 +1=1. So the sum approaches 1 as two variables approach 0 and the third approaches 1. But 1 is larger than zero. So that's worse.Alternatively, what if we have one variable at the critical point and the other two variables arranged such that their f(x) is as low as possible? But they can't be lower than their own minimal values. Wait, but each term's minimal value is at x≈0.1623. So if we set two variables at that point, but the third variable is 1 - 2*0.1623≈0.6754, as before. Which gives a sum of≈0.3138, which is still higher than zero.Alternatively, maybe setting variables unequally but not at the critical points. Let's think. Suppose a = 0.2, b=0.2, c=0.6. Compute:f(0.2)= [3*(0.04) -0.2]/[1 +0.04]= [0.12 -0.2]/1.04= (-0.08)/1.04≈-0.0769f(0.2) same as above:≈-0.0769f(0.6)= [3*(0.36) -0.6]/[1 +0.36]= [1.08 -0.6]/1.36=0.48/1.36≈0.3529Total sum≈2*(-0.0769) +0.3529≈-0.1538 +0.3529≈0.1991, still positive.Hmm. So in all these cases, the sum is positive. But when all variables are 1/3, the sum is zero. So is zero the minimum?Wait, but wait, let me check another case where variables are different. Let me try a=0.25, b=0.25, c=0.5.Compute f(0.25):3*(0.0625) -0.25=0.1875 -0.25= -0.0625Denominator:1 +0.0625=1.0625f(0.25)= -0.0625 /1.0625≈-0.0588Similarly, another f(0.25)≈-0.0588f(0.5)= [3*(0.25) -0.5]/[1 +0.25]= [0.75 -0.5]/1.25=0.25/1.25=0.2Total sum≈2*(-0.0588) +0.2≈-0.1176 +0.2≈0.0824, still positive.Hmm. So it's still positive. What if I set two variables equal to 1/3, and the third variable 1/3 as well. Then all terms are zero. Wait, but that's the case when all variables are 1/3. So, that gives sum zero. If all variables are equal, the sum is zero. But can we get lower than zero?Wait, let me see if there's a case where individual terms are negative. For example, if a variable is less than 1/3, then numerator 3a² - a can be negative. For example, take a=0.1, then 3*(0.01) -0.1=0.03 -0.1=-0.07. So f(a) is negative. So if we have a term negative, but others might be positive. But in previous examples, when one term is negative, the others are positive enough to make the total positive. But is there a case where the negative terms outweigh the positive ones?Wait, suppose we have two variables at a value where f(x) is negative and the third variable such that the total sum is minimized. Let's try a=0.2, b=0.2, c=0.6. We did that earlier, total≈0.1991. Still positive.How about a=0.15, b=0.15, c=0.7.Compute f(0.15):3*(0.0225) -0.15=0.0675 -0.15=-0.0825Denominator:1 +0.0225=1.0225f(0.15)= -0.0825 /1.0225≈-0.0807Another term same:≈-0.0807f(0.7):3*(0.49) -0.7=1.47 -0.7=0.77Denominator:1 +0.49=1.49f(0.7)=0.77 /1.49≈0.5168Total sum≈2*(-0.0807) +0.5168≈-0.1614 +0.5168≈0.3554, still positive.Hmm. So even with more negative terms, the third term compensates.Alternatively, maybe have three variables all below 1/3? But since they sum to 1, if all are below 1/3, their total would be less than 1, which contradicts the constraint. Wait, no. If all are below 1/3, then a + b + c <1, but the problem states a + b + c =1. So we can't have all variables below 1/3. So at least one variable must be ≥1/3.Wait, that's a good point. So since a + b + c =1, and all are positive, by the pigeonhole principle, at least one variable must be ≥1/3. So that variable would be in the region where f(x) is increasing (since the critical point is ≈0.1623, so 1/3≈0.333>0.1623). Therefore, the term for that variable would be contributing a positive value. The other two variables can be less than 1/3, but their terms might be negative or positive.But in all the cases I tried, even when two variables are below the critical point (thus f(x) decreasing, so f(x) < f(0.1623)), but since their sum is less than 1, the third variable is forced to be greater than 1 - 2*0.1623≈0.6754, which is way above the critical point, leading to a large positive contribution. Hence, the total sum remains positive.Alternatively, suppose two variables are exactly at the critical point, x≈0.1623, then the third variable is 1 - 2*0.1623≈0.6754. Then, as calculated before, the sum is≈-0.0811*2 +0.476≈0.3138, which is still positive.Alternatively, maybe even if two variables are slightly above the critical point, but I think this might not lead to a lower sum.Wait, perhaps there's symmetry when all variables are equal, and that's the only case where the sum is zero. Since in all other cases, even with some negative terms, the total is positive, zero seems to be the minimum. But wait, let me check one more case where variables are not equal but close to 1/3.Let’s take a=0.3, b=0.3, c=0.4.Compute f(0.3):3*(0.09) -0.3=0.27 -0.3=-0.03Denominator:1 +0.09=1.09f(0.3)= -0.03 /1.09≈-0.0275Another term same:≈-0.0275f(0.4):3*(0.16) -0.4=0.48 -0.4=0.08Denominator:1 +0.16=1.16f(0.4)=0.08 /1.16≈0.0689Total sum≈2*(-0.0275) +0.0689≈-0.055 +0.0689≈0.0139, which is still positive, but closer to zero.So, as variables approach 1/3, the sum approaches zero from above. So if all variables are exactly 1/3, the sum is zero, which might be the minimum. But how can I be sure?Alternatively, perhaps using calculus to check if zero is indeed the minimum. Let's consider the case where all variables are equal, so a = b = c =1/3. Then, the sum is zero. To check if this is a minimum, maybe use Lagrange multipliers.Let’s set up the Lagrangian. Let’s denote the variables as (a, b, c), and the constraint (a + b + c =1). The function to minimize is (f(a) + f(b) + f(c)).The Lagrangian is ( mathcal{L} = frac{3a^2 -a}{1 + a^2} + frac{3b^2 -b}{1 + b^2} + frac{3c^2 -c}{1 + c^2} - lambda(a + b + c -1) ).Taking partial derivatives with respect to a, b, c, and set them to zero.For variable a:(frac{dmathcal{L}}{da} = f'(a) - lambda =0)Similarly for b and c:(f'(b) - lambda =0)(f'(c) - lambda =0)So at the minimum, we have (f'(a) = f'(b) = f'(c) = lambda).This implies that either all variables are equal (since the derivative is the same), or they are at different points but with the same derivative value.Earlier, we found that the derivative of f(x) is (frac{x^2 +6x -1}{(1 + x^2)^2}). Let's see if this derivative can take the same value at different x's.Suppose there are two different x and y such that f'(x) = f'(y). Let's check if that's possible.For example, take x =1/3, compute f'(1/3):Numerator: (1/3)^2 +6*(1/3) -1 =1/9 +2 -1=1/9 +1=10/9≈1.1111Denominator: (1 + (1/3)^2)^2=(10/9)^2≈1.2345So f'(1/3)= (10/9)/(100/81)= (10/9)*(81/100)= (810/900)= 9/10=0.9Wait, let me recompute:Wait, f'(x) = (x² +6x -1)/(1 +x²)^2.At x=1/3:Numerator: (1/9) + 6*(1/3) -1 = (1/9) + 2 -1 = (1/9) +1 =10/9≈1.1111Denominator: (1 +1/9)^2=(10/9)^2=100/81≈1.2345So f'(1/3)= (10/9)/(100/81)= (10/9)*(81/100)= (810/900)= 9/10=0.9So f'(1/3)=0.9If we set another x such that f'(x)=0.9. Let's suppose x is not 1/3.So solve (x² +6x -1)/(1 +x²)^2=0.9Multiply both sides by (1 +x²)^2:x² +6x -1 =0.9*(1 +x²)^2Bring all terms to left:x² +6x -1 -0.9*(1 +2x² +x^4)=0Expand the right term:x² +6x -1 -0.9 -1.8x² -0.9x^4=0Combine like terms:x² -1.8x² +6x -1 -0.9 -0.9x^4=0-0.8x² +6x -1.9 -0.9x^4=0Multiply both sides by -10 to eliminate decimals:8x² -60x +19 +9x^4=0Rearrange:9x^4 +8x² -60x +19=0This is a quartic equation. Solving quartic equations analytically is complicated. Maybe there's a double root at x=1/3? Let's check.Plug x=1/3 into the equation:9*(1/3)^4 +8*(1/3)^2 -60*(1/3) +19=9*(1/81) +8*(1/9) -20 +19= (1/9) + (8/9) -20 +19= (9/9) -1 =1 -1=0Yes, x=1/3 is a root. Therefore, (x -1/3) is a factor.Let's perform polynomial division.Divide 9x^4 +8x² -60x +19 by (x -1/3). Alternatively, since x=1/3 is a root, let's factor it.Alternatively, use synthetic division.But quartic divided by linear term:Let me write the polynomial as 9x^4 +0x^3 +8x^2 -60x +19.Divide by (x -1/3):Using synthetic division:Root: 1/3Coefficients: 9 | 0 |8 | -60 |19Bring down 9Multiply by 1/3: 3Add to next coefficient:0 +3=3Multiply by1/3:1Add to next coefficient:8 +1=9Multiply by1/3:3Add to next coefficient:-60 +3= -57Multiply by1/3: -19Add to next coefficient:19 + (-19)=0So the quotient polynomial is 9x^3 +3x^2 +9x -57So 9x^4 +8x² -60x +19= (x -1/3)(9x^3 +3x^2 +9x -57)Now, let's factor the cubic polynomial 9x^3 +3x^2 +9x -57.Perhaps factor out a 3:3(3x^3 +x^2 +3x -19)Trying rational roots for 3x^3 +x^2 +3x -19. Possible roots are ±1, ±19, ±1/3, etc.Test x=1: 3 +1 +3 -19= -12 ≠0x=19: too big.x=1/3: 3*(1/27) +1/9 +3*(1/3) -19= 1/9 +1/9 +1 -19≈ -17.888≠0Not promising. So likely, the cubic has one real root and two complex roots. Therefore, the quartic equation has two real roots: x=1/3 and another real root from the cubic, and two complex roots.Therefore, there exists another x≠1/3 where f'(x)=0.9. Let's approximate that root.Let’s denote g(x)=9x^4 +8x² -60x +19. We know x=1/3 is a root. The cubic after division is 9x^3 +3x^2 +9x -57.Let me define h(x)=3x^3 +x^2 +3x -19.Looking for real roots of h(x). Let's compute h(2):3*(8)+4 +6 -19=24 +4 +6 -19=15>0h(1):3 +1 +3 -19= -12<0So there's a root between 1 and 2.h(1.5):3*(3.375) +2.25 +4.5 -19=10.125 +2.25 +4.5 -19=16.875 -19= -2.125<0h(1.75):3*(5.359375) +3.0625 +5.25 -19≈16.0781 +3.0625 +5.25 -19≈24.3906 -19≈5.3906>0So root between 1.5 and1.75.h(1.6):3*(4.096)+2.56 +4.8 -19≈12.288 +2.56 +4.8≈19.648 -19=0.648>0h(1.55):3*(3.723)+2.4025 +4.65 -19≈11.169 +2.4025 +4.65≈18.2215 -19≈-0.7785<0So between 1.55 and1.6.h(1.575):3*(1.575)^3 + (1.575)^2 +3*(1.575) -19Compute 1.575^3≈1.575*1.575=2.4806*1.575≈3.9093*3.909≈11.727(1.575)^2≈2.48063*1.575≈4.725So total≈11.727 +2.4806 +4.725 -19≈19.9326 -19≈0.9326>0Wait, but h(1.575)=3*(1.575)^3 + (1.575)^2 +3*(1.575) -19Wait, actually, h(x)=3x^3 +x^2 +3x -19, so:At x=1.575:3*(1.575)^3 + (1.575)^2 +3*(1.575) -19Compute 1.575^3≈1.575*1.575=2.480625*1.575≈3.9093753*3.909375≈11.728125(1.575)^2≈2.4806253*1.575≈4.725Total:11.728125 +2.480625 +4.725 -19≈19.93375 -19≈0.93375>0Still positive. Let me try x=1.56:1.56^3≈1.56*1.56=2.4336*1.56≈3.8073*3.807≈11.4211.56^2≈2.43363*1.56≈4.68Total:11.421 +2.4336 +4.68 -19≈18.5346 -19≈-0.4654<0So between 1.56 and1.575. Let's use linear approximation.Between x=1.56 (h=-0.4654) and x=1.575 (h=0.93375). The difference in x is0.015, difference in h is0.93375 -(-0.4654)=1.39915We need to find delta such that h=0 at x=1.56 +delta*(0.015)delta≈0.4654 /1.39915≈0.3327Thus, x≈1.56 +0.015*0.3327≈1.56 +0.00499≈1.56499Approximately x≈1.565. So the cubic has a real root at≈1.565. Thus, the quartic equation has another real root at x≈1.565. So f'(x)=0.9 at x=1/3 and x≈1.565.But in our problem, variables a, b, c are positive and sum to1. So variables can't be as large as1.565. Therefore, the only feasible critical point is x=1/3. Therefore, when using Lagrange multipliers, the only critical point within the domain is when all variables are equal to1/3. Hence, the minimum occurs at a=b=c=1/3, and the sum is zero. Therefore, the minimum value is zero.But wait, to confirm, we need to ensure that this is indeed a minimum. Since in all other cases we checked, the sum was positive, and the second derivative test could help, but it might be complicated.Alternatively, we can use the method of convexity. However, since the function f(x) is not convex over the entire interval (0,1), as we saw the derivative first decreases then increases, making the function first concave then convex? Wait, let's check the second derivative.Wait, computing the second derivative of f(x) might be cumbersome, but given that the first derivative has a minimum at x≈0.1623, maybe the function f(x) is neither convex nor concave over the entire interval. Therefore, Jensen's inequality might not apply directly unless we can segment the interval.Alternatively, another approach. Let's consider that for each x in (0,1), can we show that f(x) ≥0? If that's true, then the sum would be ≥0, with equality when each term is zero, i.e., when a=b=c=1/3.Wait, let's check for x in (0,1):Is ( frac{3x^2 -x}{1 +x^2} geq 0 )?Multiply both sides by (1 +x^2) which is positive:3x^2 -x ≥0 ?Which is equivalent to 3x^2 -x ≥0 →x(3x -1)≥0.Since x>0, this inequality holds when 3x -1 ≥0 →x≥1/3.When x<1/3, 3x -1 <0, so the expression is negative.Therefore, f(x) ≥0 when x ≥1/3, and f(x) <0 when x <1/3.But in our problem, since a + b + c =1, at least one variable must be ≥1/3 (as noted earlier). That variable will have f(x) ≥0, while the others could be <1/3 and have f(x) <0. However, the question is whether the sum is minimized when all variables are equal.But in the previous examples, even with two variables contributing negative terms, the third variable's positive contribution outweighs them, leading to a positive total.Alternatively, perhaps the sum is minimized when all variables are equal, and this is the only point where the negative and positive contributions balance out to zero.Alternatively, let's consider the function for each variable:For x ≥1/3, f(x) ≥0For x <1/3, f(x) <0But if we have a variable x <1/3, its contribution is negative, but to compensate, another variable must be greater than1/3, making its contribution positive. The question is whether the positive contribution from the larger variable can be less than the negative contributions from the smaller variables.But according to the function's behavior, for x <1/3, f(x) is decreasing, so making x smaller makes f(x) more negative. However, the other variable must be larger, but since f(x) is increasing for x >0.1623, making that variable larger would make its f(x) larger.So the trade-off is between getting more negative from the small x and more positive from the large x. The key is to see if the increase in the positive term can be outweighed by the decrease in the negative term.But given that f(x) is convex or concave in certain intervals, maybe we can apply some inequality.Alternatively, let's consider the function f(x) + f(y) where x + y = t. Then, find the minimum of f(x) + f(t -x). Maybe this is a simpler case.Wait, but the problem has three variables. Alternatively, maybe use the method of smoothing: if two variables are unequal, adjusting them to be closer while maintaining their sum can decrease the total. If this is true, then the minimum occurs when all variables are equal.Let’s consider two variables a and b such that a ≠ b. Replace them with (a + b)/2 each. Let's see if the sum f(a) + f(b) decreases.Define g(a, b) = f(a) + f(b). Let’s set a = c - d and b = c + d, where c = (a + b)/2 and d is the deviation. Then see if g(c - d, c + d) ≤ g(a, b) for d >0.Compute the difference g(c - d, c + d) - g(a, b) = [f(c - d) + f(c + d)] - [f(a) + f(b)].If this difference is ≤0, then replacing a and b with their average reduces the sum, implying convexity. If the difference is ≥0, then it implies concavity.Compute the second derivative of f(x) to check convexity.But computing the second derivative might be tedious. Alternatively, compute the difference for specific values.Take c =1/3 and d =0.1, so a=1/3 -0.1≈0.2333, b=1/3 +0.1≈0.4333.Compute f(a) + f(b):f(0.2333)= [3*(0.0544) -0.2333]/[1 +0.0544]≈[0.1632 -0.2333]/1.0544≈(-0.0701)/1.0544≈-0.0665f(0.4333)= [3*(0.1877) -0.4333]/[1 +0.1877]≈[0.5631 -0.4333]/1.1877≈0.1298/1.1877≈0.1093Sum≈-0.0665 +0.1093≈0.0428Now, if we replace a and b with c=1/3≈0.3333:f(0.3333)=0, as before.So the sum becomes 0 +0=0, which is less than0.0428. Therefore, replacing unequal a and b with their average reduces the sum. This suggests that the function is convex around this region? Or that the sum is convex?Alternatively, this suggests that the minimal occurs when variables are equal.Similarly, try another example: take a=0.1, b=0.3, sum=0.4. Replace with c=0.2 each.Compute f(0.1)=≈-0.07/1.01≈-0.0693f(0.3)=≈-0.03/1.09≈-0.0275Sum≈-0.0693 -0.0275≈-0.0968After replacement:f(0.2) + f(0.2)=2*(-0.0769)≈-0.1538Which is lower than -0.0968. Wait, but this contradicts the previous result. Wait, here, replacing a=0.1 and b=0.3 with two variables of 0.2 each leads to a lower sum. So the sum decreased.But in the previous case, where variables were around 1/3, replacing them with equal variables increased the sum. Wait, no, in the first example, the sum was positive, and replacing with equal variables made the sum zero, which is lower. Wait, in the first example, replacing a≈0.2333 and b≈0.4333 (sum≈0.6666) with two variables of0.3333 each led to the sum going from≈0.0428 to0, which is a decrease.In the second example, replacing a=0.1 and b=0.3 (sum0.4) with two variables of0.2 each led to the sum going from≈-0.0968 to≈-0.1538, which is also a decrease.Hmm, so in both cases, replacing two variables with their average (keeping the sum the same) decreases the total sum. This suggests that the function f(x) is concave in the region x <1/3, because the sum is decreased by moving towards the center. Wait, but how does this align with the derivative?Alternatively, maybe the function f(x) is concave for x <c and convex for x >c, leading to the optimal at the boundaries or at equal points. But I might need a different approach.Alternatively, using the tangent line method. Suppose we can find a linear function that bounds f(x) from below, and then apply the condition a + b + c=1 to find the minimum.Alternatively, consider the following identity or manipulation:Note that ( frac{3x^2 -x}{1 +x^2} = 3 - frac{3 +x}{1 +x^2} )Wait, let's verify:(3 - frac{3 +x}{1 +x^2} = frac{3(1 +x^2) - (3 +x)}{1 +x^2} = frac{3 +3x^2 -3 -x}{1 +x^2} = frac{3x^2 -x}{1 +x^2}). Yes, that's correct.Therefore, each term can be rewritten as (3 - frac{3 +x}{1 +x^2}). Therefore, the sum becomes:(sum left(3 - frac{3 +x}{1 +x^2}right) = 9 - sum frac{3 +x}{1 +x^2})Therefore, the original expression is equal to (9 - left( frac{3 +a}{1 +a^2} + frac{3 +b}{1 +b^2} + frac{3 +c}{1 +c^2} right)).Hence, minimizing the original expression is equivalent to maximizing the sum (S = frac{3 +a}{1 +a^2} + frac{3 +b}{1 +b^2} + frac{3 +c}{1 +c^2}).Therefore, our problem reduces to maximizing S under the constraint a + b + c =1, a, b, c >0.So, if we can find the maximum of S, then subtract it from9 to get the minimum of the original expression.Now, let's focus on maximizing S.Perhaps now it's easier since the function (g(x) = frac{3 +x}{1 +x^2}) may have nicer properties. Let's analyze g(x).Compute its derivative:(g(x) = frac{3 +x}{1 +x^2})Using quotient rule:(g'(x) = frac{(1)(1 +x^2) - (3 +x)(2x)}{(1 +x^2)^2} = frac{1 +x^2 -6x -2x^2}{(1 +x^2)^2} = frac{1 -6x -x^2}{(1 +x^2)^2})Set numerator equal to zero:( -x^2 -6x +1 =0 rightarrow x^2 +6x -1 =0 )Which is the same equation as before! Solutions are x = [-6 ±√40]/2 = -3 ±√10. The positive root is x≈0.1623.Therefore, critical point at x≈0.1623, which we saw earlier.Compute the second derivative to check concavity, but maybe it's easier to note the behavior.For x <0.1623, g'(x) = [1 -6x -x²]/denominator. The numerator is positive when 1 -6x -x² >0. Let's see:At x=0, numerator=1>0.At x approaching0.1623, the numerator approaches zero.For x>0.1623, numerator becomes negative. Therefore, g(x) is increasing on (0,0.1623) and decreasing on (0.1623, ∞). Therefore, g(x) attains its maximum at x≈0.1623.Thus, the function g(x) has a single peak at x≈0.1623, and decreases on either side.Therefore, to maximize the sum S = g(a) +g(b)+g(c), we need to set as many variables as possible to the maximum point x≈0.1623, given the constraint a +b +c=1.However, since the variables must be positive and sum to1, we can try to set two variables to0.1623 and the third to1 -2*0.1623≈0.6754.But let's compute S in this case:g(0.1623)= (3 +0.1623)/(1 +0.1623²)≈3.1623 /1.0263≈3.081Two variables give≈3.081*2≈6.162Third variable:g(0.6754)= (3 +0.6754)/(1 +0.6754²)≈3.6754 /1.4562≈2.524Total S≈6.162 +2.524≈8.686Alternatively, if we set all three variables to1/3≈0.3333:g(1/3)= (3 +1/3)/(1 +1/9)= (10/3)/(10/9)= (10/3)*(9/10)=3Sum S=3*3=9Wait, that's higher. But earlier, when we set two variables to0.1623 and one to0.6754, the sum was≈8.686, which is less than9.But according to our previous transformation, the original expression is9 -S. So if S is maximized when variables are1/3, leading to S=9, then the original expression is9 -9=0, which is the minimum.Wait, but this contradicts the previous calculation where setting variables to0.1623 and0.6754 gives S≈8.686, which is less than9. But how?Wait, if S is maximized when variables are1/3, then the maximum S is9, leading to the original expression being0. But if when variables are set to0.1623 and0.6754, S is≈8.686, which is less than9, then the original expression would be9 -8.686≈0.314, which is positive. However, when variables are1/3, S=9, so original expression=0.But this suggests that the maximum of S is9, achieved when all variables are1/3, hence the original expression attains its minimum at0. Therefore, this confirms that the minimum value is0.But why when we tried setting two variables to0.1623 and third to0.6754, the sum S was less than9? Because in that case, variables are not at the maximum point of g(x). The maximum of g(x) is at≈0.1623, but when variables are1/3, which is past the maximum, but summing three terms at g(1/3)=3 each gives a higher total.Wait, but how? If the function g(x) peaks at0.1623, then g(0.1623)≈3.081, which is higher thang(1/3)=3. So if we can set more variables to0.1623, we could get a higher sum. However, due to the constraint a +b +c=1, we can't set three variables to0.1623, since 3*0.1623≈0.4869<1. If we set two variables to0.1623, the third must be0.6754, which has g(x)=≈2.524, so total≈3.081*2 +2.524≈6.162 +2.524≈8.686.If we set one variable to0.1623, and two variables to(1 -0.1623)/2≈0.4189 each:Compute g(0.1623)≈3.081g(0.4189)= (3 +0.4189)/(1 +0.4189²)≈3.4189 /1.175≈2.909Sum≈3.081 +2*2.909≈3.081 +5.818≈8.899, which is still less than9.Alternatively, set all three variables to1/3≈0.3333:g(1/3)=3. Therefore, sum=9.Even though g(1/3)=3 <g(0.1623)=≈3.081, the sum of three terms at3 each gives9, which is greater than distributing variables to mix of0.1623 and higher values.Therefore, despite individual terms being lower than the peak value, the sum is maximized when all variables are equal. This suggests that the function S has its maximum when variables are equal due to the symmetry and the constraint a +b +c=1.Therefore, the maximum of S is9, achieved at a=b=c=1/3, leading to the original expression being9 -9=0, which is the minimum value.Therefore, the minimum value is0, achieved when a=b=c=1/3.Final AnswerThe minimum value is boxed{0}.</think>Given (a, b, c in mathbf{R}_{+}) such that (a + b + c = 1), we need to find the minimum value of the expression:frac{3a^2 - a}{1 + a^2} + frac{3b^2 - b}{1 + b^2} + frac{3c^2 - c}{1 + c^2}First, we analyze the function (f(x) = frac{3x^2 - x}{1 + x^2}). We compute its derivative:f'(x) = frac{(6x - 1)(1 + x^2) - (3x^2 - x)(2x)}{(1 + x^2)^2} = frac{x^2 + 6x - 1}{(1 + x^2)^2}The critical points are found by solving (x^2 + 6x - 1 = 0), yielding (x = -3 pm sqrt{10}). The positive critical point is (x approx 0.1623).Next, we consider the behavior of (f(x)). For (x < 0.1623), (f(x)) is decreasing, and for (x > 0.1623), (f(x)) is increasing. However, since (a, b, c) are positive and sum to 1, we need to check if the minimum value occurs when all variables are equal.We test the case (a = b = c = frac{1}{3}):fleft(frac{1}{3}right) = frac{3left(frac{1}{3}right)^2 - frac{1}{3}}{1 + left(frac{1}{3}right)^2} = frac{0}{1 + frac{1}{9}} = 0Thus, the sum is (0 + 0 + 0 = 0).We also consider the transformation of the original expression:frac{3a^2 - a}{1 + a^2} + frac{3b^2 - b}{1 + b^2} + frac{3c^2 - c}{1 + c^2} = 9 - left( frac{3 + a}{1 + a^2} + frac{3 + b}{1 + b^2} + frac{3 + c}{1 + c^2} right)To minimize the original expression, we need to maximize the sum ( frac{3 + a}{1 + a^2} + frac{3 + b}{1 + b^2} + frac{3 + c}{1 + c^2} ).The function (g(x) = frac{3 + x}{1 + x^2}) has a maximum at (x approx 0.1623), but due to the constraint (a + b + c = 1), the maximum sum is achieved when (a = b = c = frac{1}{3}), giving each term as 3. Thus, the sum is (3 + 3 + 3 = 9).Therefore, the minimum value of the original expression is:9 - 9 = 0The minimum value is boxed{0}.,<think>Okay, so I have this problem here: given positive real numbers a, b, c such that a + b + c = 1, I need to find the minimum value of the expression S = (3a² - a)/(1 + a²) + (3b² - b)/(1 + b²) + (3c² - c)/(1 + c²). Hmm, let me try to figure out how to approach this.First, since the variables are positive and sum to 1, each of a, b, c must be between 0 and 1. The expression S is a sum of three similar terms, each involving a quadratic function over 1 plus the square of the variable. So maybe symmetry can be used here? But since it's asking for the minimum, maybe the extremum occurs when all variables are equal? Let me check that.If a = b = c = 1/3, then each term becomes (3*(1/3)² - 1/3)/(1 + (1/3)²) = (3*(1/9) - 1/3)/(1 + 1/9) = (1/3 - 1/3)/(10/9) = 0/(10/9) = 0. So each term is zero, so S = 0 + 0 + 0 = 0. Hmm, so when all variables are equal, the expression is zero. But is this the minimum?Wait, maybe we can get negative values? Let's see. Let's consider a single term: (3x² - x)/(1 + x²). Let me analyze this function for x in (0,1). Let me define f(x) = (3x² - x)/(1 + x²). What's the range of this function?Compute f(0): (0 - 0)/(1 + 0) = 0.Compute f(1): (3*1 -1)/(1 +1) = (3 -1)/2 = 1.But maybe there's a minimum in between? Let's take the derivative of f(x) to find critical points.f(x) = (3x² - x)/(1 + x²). Let's compute f'(x):Using the quotient rule: [(6x -1)(1 + x²) - (3x² -x)(2x)] / (1 + x²)²Let's compute numerator:First term: (6x -1)(1 + x²) = 6x*(1) + 6x*x² -1*(1) -1*x² = 6x + 6x³ -1 -x²Second term: -(3x² -x)(2x) = - [6x³ - 2x²] = -6x³ + 2x²So adding both terms:6x + 6x³ -1 -x² -6x³ + 2x² = (6x) + (6x³ -6x³) + (-x² + 2x²) -1 = 6x + x² -1Therefore, f'(x) = (x² + 6x -1)/(1 + x²)²Set numerator equal to zero: x² + 6x -1 = 0.Solve for x: x = [-6 ± sqrt(36 +4)]/2 = [-6 ± sqrt(40)]/2 = [-6 ± 2*sqrt(10)]/2 = -3 ± sqrt(10)Since x is positive, only x = -3 + sqrt(10) ≈ -3 + 3.16 ≈ 0.16 is valid.So critical point at x ≈ 0.16. Let's compute f at this point.Compute x ≈ 0.16, then f(x) = (3*(0.16)^2 -0.16)/(1 + (0.16)^2). Let's calculate:3*(0.0256) = 0.0768, 0.0768 -0.16 = -0.0832Denominator: 1 + 0.0256 = 1.0256So f(x) ≈ -0.0832 /1.0256 ≈ -0.0811. So the function f(x) has a minimum of approximately -0.081 at x ≈ 0.16.Therefore, each term can be as low as approximately -0.081. So if all three variables are at this critical point, then the total sum would be around -0.243. But wait, the problem states that a + b + c =1. If each variable is approximately 0.16, then the sum would be 0.48, which is less than 1. Therefore, we can't have all three variables at the critical point. So maybe we need to set two variables at some value and the third variable adjusted so that the sum is 1.Alternatively, perhaps the minimum occurs when one variable is as large as possible, and the others are as small as possible. Let me test some cases.Case 1: Let one variable approach 1, and the other two approach 0. Let's say a ≈1, b≈0, c≈0.Compute S ≈ (3*1 -1)/(1 +1) + 2*(0 -0)/(1 +0) = (2)/2 + 0 +0 =1. So S≈1. So in this case, S is 1.Case 2: Let two variables be equal and the third adjusted. Let me try a = b = t, so c =1 -2t. Then compute S = 2*(3t² -t)/(1 + t²) + (3(1 -2t)^2 - (1 -2t))/(1 + (1 -2t)^2). Let's try t = 0.1, then c = 0.8.Compute first term: 2*(3*0.01 -0.1)/ (1 +0.01) = 2*(0.03 -0.1)/1.01 ≈2*(-0.07)/1.01≈-0.1386Second term: (3*0.64 -0.8)/(1 +0.64) = (1.92 -0.8)/1.64 ≈1.12/1.64≈0.682Total S≈-0.1386 +0.682≈0.543. Which is better than 1, but still positive.Wait, but if we take t closer to the critical point 0.16, maybe?Let me set t =0.16, then c =1 - 2*0.16 =0.68.First term: 2*(3*(0.16)^2 -0.16)/(1 + (0.16)^2) ≈2*(0.0768 -0.16)/1.0256≈2*(-0.0832)/1.0256≈-0.162Second term: (3*(0.68)^2 -0.68)/(1 + (0.68)^2) ≈(3*0.4624 -0.68)/1.4624≈(1.3872 -0.68)/1.4624≈0.7072/1.4624≈0.483Total S≈-0.162 +0.483≈0.321. Still positive. So even when two variables are at the critical point, the third variable is 0.68, which gives a positive term. So total S is still positive.Wait, but earlier when all variables are equal, S is zero, so maybe zero is the minimum? But when I tried a single variable at critical point, I saw that the function can be negative. But since the variables are constrained to sum to 1, perhaps the minimal value is when two variables are at the critical point, but that seems not possible as the sum would be too low. Maybe only one variable is at the critical point, and the other two variables sum to 1 - x. But then, how do we distribute between the other two variables? If we set one variable at the critical point x ≈0.16, and the other two variables sum to 0.84. If we set the other two variables as 0.42 each, then compute S.So, let x =0.16, then a=0.16, b=0.42, c=0.42.Compute S = f(0.16) + 2*f(0.42). From earlier, f(0.16)≈-0.081, and f(0.42) is:(3*(0.42)^2 -0.42)/(1 + (0.42)^2)= (3*0.1764 -0.42)/1.1764≈(0.5292 -0.42)/1.1764≈0.1092/1.1764≈0.0928.Thus, S≈-0.081 + 2*0.0928≈-0.081 +0.1856≈0.1046. Still positive.Alternatively, maybe set one variable at the critical point and the other two variables at different values. Maybe one variable at critical point and another variable also at critical point, but then the third variable would be 1 - 2x. But if x≈0.16, 1 -2x≈0.68. As tried before, which still gives a positive sum.Alternatively, maybe set one variable to be higher than the critical point and another lower? Let me think.Wait, the function f(x) = (3x² -x)/(1 +x²). Let me analyze its behavior.From x=0 to x≈0.16, f(x) is decreasing, reaching a minimum at x≈0.16, then increasing afterwards. So for x <0.16, the function is decreasing, and for x>0.16, it's increasing. At x=0, f(x)=0. At x=1, f(x)=1. The minimum is around x≈0.16, f(x)≈-0.081.So if we can set two variables at x≈0.16 and the third variable at 1 -2x≈0.68, but as we saw, the sum S is positive. Alternatively, maybe set two variables less than 0.16? But then their f(x) would be higher (since decreasing to x=0.16), so their terms would be more negative? Wait, no. Wait, if x is less than 0.16, since f is decreasing from 0 to 0.16, so lower x gives higher f(x). For example, at x=0.1, f(x)= (0.03 -0.1)/1.01≈-0.07. At x=0.16, f(x)= -0.081. So indeed, the minimum is at x≈0.16.Therefore, to minimize the sum S, we need to have as many variables as possible at the critical point x≈0.16. But the problem is that the sum of the variables must be 1. So if we set two variables at x≈0.16, the third variable is 0.68, but then f(0.68) is positive. If we set one variable at x≈0.16 and the other two variables sum to 0.84. If we set both of them higher than 0.16, their f(x) would be positive. Alternatively, maybe setting one variable at x≈0.16 and another variable higher, and the third variable lower? Wait, but variables have to be positive. If we set one variable at 0.16, another variable as high as possible, but the third variable would be 1 -0.16 -high variable. Hmm, maybe not helpful.Alternatively, maybe set two variables at 0.16, third at 0.68. But as we saw, sum S is positive. Alternatively, set three variables at different points. Hmm. Maybe using Lagrange multipliers?Let me try that. Let's denote S = f(a) + f(b) + f(c), where f(x) = (3x² -x)/(1 +x²). We need to minimize S subject to a + b + c =1, with a,b,c >0.Set up Lagrangian: L = f(a) + f(b) + f(c) - λ(a + b + c -1)Take partial derivatives:∂L/∂a = f’(a) - λ =0Similarly for ∂L/∂b and ∂L/∂c.Therefore, at the minimum, we have f’(a) = f’(b) = f’(c) = λ.So either all variables are equal (a = b = c) or some variables are equal and others take different values with the same derivative.We saw that when a = b = c =1/3, the derivative f’(1/3) = ?Compute f’(x) = (x² +6x -1)/(1 +x²)^2.So at x=1/3, f’(1/3) = ( (1/9) + 6*(1/3) -1 ) / (1 +1/9)^2 = (1/9 + 2 -1)/ (10/9)^2 = (1/9 +1)/ (100/81) = (10/9)/(100/81) = (10/9)*(81/100)= (9/10) = 0.9.So f’(1/3) = 0.9. If variables are equal, their derivatives are 0.9 each.Alternatively, if we have two variables equal and the third different, say a = b ≠ c. Then f’(a) = f’(b) = λ, and f’(c)=λ. So either a = b = c, or a = b ≠ c with f’(a) = f’(c).But the equation f’(x) = λ can have two solutions for x, given λ. Because f’(x) = (x² +6x -1)/(1 +x²)^2. The derivative of f(x) first increases, reaches a maximum, then decreases? Wait, need to analyze f’(x).Wait, f’(x) = (x² +6x -1)/(1 +x²)^2. Let me check the behavior of f’(x):Let me compute f’(x) for x from 0 to 1.At x=0: f’(0) = (0 +0 -1)/(1 +0)^2 = -1/1 = -1.At x=0.16 (critical point of f(x)): f’(x)=0.At x=1: f’(1) = (1 +6 -1)/(1 +1)^2=6/4=1.5.So f’(x) starts at -1 when x=0, increases to 0 at x≈0.16, then continues increasing to 1.5 at x=1.Wait, so f’(x) is increasing from x=0 to x=1? Because the numerator is x² +6x -1, which is a quadratic function opening upwards. The critical point of f’(x) would be where its derivative is zero. Let's compute f''(x) to check concavity, but this might get complicated. Alternatively, notice that the numerator x² +6x -1 is increasing for x > -3 (since derivative 2x +6 >0 for x > -3). Since x is positive here, x² +6x -1 is increasing. Therefore, f’(x) numerator is increasing, denominator (1 +x²)^2 is also increasing. So overall, f’(x) is increasing from x=0 to x=1. Therefore, f’(x) is strictly increasing on [0,1]. Therefore, for each λ between -1 and 1.5, there is exactly one x in [0,1] such that f’(x)=λ. Therefore, the only way to have f’(a) = f’(b) = f’(c) is if a = b = c. Hence, the only critical point under the constraint is when all variables are equal. Therefore, the minimal value is when a = b = c =1/3, giving S=0.Wait, but this contradicts the earlier thought that individual terms can be negative. So according to Lagrange multipliers, the minimal value is 0, achieved at a = b = c =1/3. But when I considered variables at the critical point x≈0.16, which gives a negative term, but then the other terms become positive, leading to a total S that is still positive. Wait, but if all variables are equal, each term is zero, so S=0, but maybe there's a way to have some negative terms and some positive terms such that the total sum is less than zero?Wait, let's test another case. Suppose a is at the critical point x≈0.16, then f(a)≈-0.081. Then suppose the other two variables are b and c, with b + c =1 -0.16=0.84. Let me set b=0.5 and c=0.34. Then compute f(b)= (3*(0.5)^2 -0.5)/(1 +0.25)= (0.75 -0.5)/1.25=0.25/1.25=0.2. f(c)= (3*(0.34)^2 -0.34)/(1 +0.1156)= (3*0.1156 -0.34)/1.1156≈(0.3468 -0.34)/1.1156≈0.0068/1.1156≈0.0061. So total S≈-0.081 +0.2 +0.0061≈0.1251. Still positive.Alternatively, set a=0.16, then set b=0.16, c=0.68. Then f(a)=f(b)=≈-0.081 each, and f(c)= (3*(0.68)^2 -0.68)/(1 +0.4624)≈(1.3872 -0.68)/1.4624≈0.7072/1.4624≈0.483. So total S≈-0.162 +0.483≈0.321. Still positive.Alternatively, set a=0.16, b=0.7, c=0.14. Then f(a)=≈-0.081, f(b)= (3*0.49 -0.7)/1.49≈(1.47 -0.7)/1.49≈0.77/1.49≈0.517, f(c)= (3*0.0196 -0.14)/1.0196≈(0.0588 -0.14)/1.0196≈-0.0812/1.0196≈-0.0796. So total S≈-0.081 +0.517 -0.0796≈0.356. Still positive.Wait, but in this case, two terms are negative: a≈0.16 and c≈0.14. But their sum is 0.3, so b=0.7. But f(b)=0.517, which is positive. So even with two negative terms and one positive term, the total is still positive.Alternatively, can we get three negative terms? If all variables are less than the critical point 0.16. But then a + b + c =1, which would require at least one variable being greater than 1/3≈0.333, which is way above the critical point 0.16. So such a variable would have a positive term. So even if two variables are below 0.16 (with negative terms), the third variable would be above 0.68, giving a large positive term, dominating the sum. Therefore, it's impossible to have all three terms negative. The maximum number of negative terms is two, but even then the third term is positive enough to make the total sum positive. Therefore, maybe the minimal value is indeed 0, achieved when all variables are equal.But wait, when all variables are equal, each term is zero. So if we deviate from equality, some terms become negative but others become positive. However, the total sum seems to stay positive. Therefore, maybe the minimal value is 0.But let me check another case. Suppose a=0.5, b=0.5, c=0. But c must be positive. So c approaching 0. Then S = f(0.5) + f(0.5) + f(0). Compute f(0.5)= (3*(0.25) -0.5)/1.25=(0.75 -0.5)/1.25=0.25/1.25=0.2. f(0)=0. So S≈0.2 +0.2 +0=0.4. Still positive.Alternatively, a=0.4, b=0.4, c=0.2. Then S=2*f(0.4) +f(0.2). f(0.4)= (3*0.16 -0.4)/1.16≈(0.48 -0.4)/1.16≈0.08/1.16≈0.0689. f(0.2)= (0.12 -0.2)/1.04≈-0.08/1.04≈-0.0769. So total S≈2*0.0689 -0.0769≈0.1378 -0.0769≈0.0609. Still positive. Even when one term is negative, the other two terms compensate.Alternatively, let me try a=0.25, b=0.25, c=0.5. Then S=2*f(0.25) +f(0.5). f(0.25)= (3*0.0625 -0.25)/1.0625=(0.1875 -0.25)/1.0625≈(-0.0625)/1.0625≈-0.0588. f(0.5)=0.2. So S≈2*(-0.0588) +0.2≈-0.1176 +0.2≈0.0824. Still positive.Wait, even when two terms are negative and one is positive, the total is still positive. Hmm. What if we have one term at the critical point (≈-0.081) and another term slightly negative, but the third term is large positive. Let's see.Suppose a=0.16 (f≈-0.081), b=0.2 (f≈-0.0769), c=0.64 (f≈(3*0.4096 -0.64)/1.4096≈(1.2288 -0.64)/1.4096≈0.5888/1.4096≈0.4175). Then S≈-0.081 -0.0769 +0.4175≈0.2596. Still positive.Alternatively, a=0.16, b=0.16, c=0.68. Then S≈-0.081*2 +0.483≈-0.162 +0.483≈0.321. Still positive.Alternatively, if we set two variables at 0.16, and the third at 0.68, but then even with two negative terms, the positive term dominates. So it seems that regardless of how we set the variables (as long as they sum to 1), the total S remains non-negative, and the minimal value is 0 when all variables are equal.But wait, another approach. Let me consider the function f(x) = (3x² -x)/(1 +x²). Can we show that for x in (0,1), f(x) ≥ -1/4? Let me check the minimal value.Earlier, we found that f(x) has a minimum at x= (-3 + sqrt(10))≈0.162. Let's compute the exact minimal value.Let x0 = -3 + sqrt(10). Then f(x0)= (3x0² -x0)/(1 +x0²). Let's compute this.First, compute x0²:x0 = sqrt(10) -3, so x0² = (sqrt(10) -3)^2 =10 -6*sqrt(10) +9=19 -6*sqrt(10).Then 3x0² -x0 =3*(19 -6*sqrt(10)) - (sqrt(10)-3)=57 -18*sqrt(10) -sqrt(10) +3=60 -19*sqrt(10).Denominator:1 +x0²=1 +19 -6*sqrt(10)=20 -6*sqrt(10).Thus, f(x0)= (60 -19*sqrt(10))/(20 -6*sqrt(10)).Multiply numerator and denominator by (20 +6*sqrt(10)):Numerator: (60 -19√10)(20 +6√10) =60*20 +60*6√10 -19√10*20 -19√10*6√10=1200 +360√10 -380√10 -19*6*10=1200 -20√10 -1140=60 -20√10Denominator: (20)^2 - (6√10)^2=400 -360=40Therefore, f(x0)= (60 -20√10)/40= (60/40) - (20√10)/40= 1.5 -0.5√10≈1.5 -0.5*3.16≈1.5 -1.58≈-0.08.So the minimal value of f(x) is approximately -0.08. So each term can be as low as ≈-0.08, but when we have three variables, but the sum is constrained to 1, we can't have all three variables at the minimal point. So the maximum negative contribution would be when two variables are at the minimal point. But even then, the third variable would be 1 -2x0≈1 -2*(0.162)=0.676. Then f(0.676)= (3*(0.676)^2 -0.676)/(1 + (0.676)^2). Compute that:3*(0.457)≈1.371, 1.371 -0.676≈0.695. Denominator:1 +0.457≈1.457. So f(0.676)≈0.695/1.457≈0.477.So total S≈2*(-0.08) +0.477≈-0.16 +0.477≈0.317.Still positive. Therefore, even if two variables are at their minimal points, the third variable's term is positive enough to make the total sum positive.Therefore, perhaps the minimal value is indeed 0, achieved when all variables are equal. Therefore, the answer should be 0.But let me verify with another example. Suppose a=1/2, b=1/2, c=0. But c must be positive. Let's take c approaching 0. Then S=2*(3*(1/2)^2 -1/2)/(1 + (1/2)^2) +0≈2*(0.75 -0.5)/1.25≈2*(0.25)/1.25≈0.4. Positive.Alternatively, take a=0.9, b=0.05, c=0.05. Then compute S:f(0.9)= (3*0.81 -0.9)/ (1 +0.81)= (2.43 -0.9)/1.81≈1.53/1.81≈0.845f(0.05)= (3*0.0025 -0.05)/1.0025≈(0.0075 -0.05)/1.0025≈-0.0425/1.0025≈-0.0424So S≈0.845 +2*(-0.0424)≈0.845 -0.0848≈0.7602. Still positive.Alternatively, set a=0.8, b=0.1, c=0.1.f(0.8)= (3*0.64 -0.8)/1.64≈(1.92 -0.8)/1.64≈1.12/1.64≈0.6829f(0.1)= (0.03 -0.1)/1.01≈-0.07/1.01≈-0.0693So S≈0.6829 +2*(-0.0693)≈0.6829 -0.1386≈0.5443. Still positive.Alternatively, set a=0.7, b=0.15, c=0.15.f(0.7)= (3*0.49 -0.7)/1.49≈(1.47 -0.7)/1.49≈0.77/1.49≈0.517f(0.15)= (3*0.0225 -0.15)/1.0225≈(0.0675 -0.15)/1.0225≈-0.0825/1.0225≈-0.0807So S≈0.517 +2*(-0.0807)≈0.517 -0.1614≈0.3556. Still positive.So in all these cases, even when two variables are somewhat small (but not at the minimal point), the third variable is large enough to make the total sum positive.Moreover, according to the Lagrange multipliers method, the critical point is when all variables are equal, which gives S=0. Since all other configurations give S>0, then 0 is the minimal value.But to confirm, let me check if there's any possible configuration where S is negative. Suppose we set two variables at x=0.162 (critical point) and third at 1 -2*0.162≈0.676. Then compute S=2*(-0.08) +f(0.676). As before, f(0.676)≈0.477. So S≈-0.16 +0.477≈0.317>0.Alternatively, set one variable at x=0.162, another variable at x=0.162, and the third at 0.676. Then same as above.Alternatively, set one variable at x=0.162, another variable at x=0.1, and the third at 0.738. Then compute S= f(0.162) + f(0.1) + f(0.738).f(0.162)=≈-0.081, f(0.1)=≈-0.07, f(0.738)= (3*0.738² -0.738)/(1 +0.738²)= (3*0.544 -0.738)/1.544≈(1.632 -0.738)/1.544≈0.894/1.544≈0.579. So total S≈-0.081 -0.07 +0.579≈0.428>0.Still positive. Hmm. So regardless of how I arrange the variables, the total sum S remains non-negative. Therefore, the minimal value is indeed 0, achieved when a = b = c =1/3.Therefore, the answer should be 0.Final AnswerThe minimum value is boxed{0}.</think>Given positive real numbers (a, b, c) such that (a + b + c = 1), we need to find the minimum value of the expression:[frac{3a^2 - a}{1 + a^2} + frac{3b^2 - b}{1 + b^2} + frac{3c^2 - c}{1 + c^2}]First, we consider the function (f(x) = frac{3x^2 - x}{1 + x^2}) for (x in (0, 1)). By analyzing the critical points of this function, we find that it has a minimum value at (x = -3 + sqrt{10} approx 0.16), where the function value is approximately (-0.081). However, due to the constraint (a + b + c = 1), we cannot have all three variables at this critical point simultaneously.Testing various configurations, including cases where one variable approaches 1 and the others approach 0, or different distributions of variables, we find that the expression remains non-negative. For example, when (a = b = c = frac{1}{3}), each term in the expression evaluates to 0, resulting in a total sum of 0.Using Lagrange multipliers, we confirm that the critical point occurs when all variables are equal, leading to the minimum value of the expression. Further analysis shows that any deviation from equal distribution results in positive contributions from larger variables, which dominate any negative contributions from smaller variables.Thus, the minimum value of the given expression is achieved when (a = b = c = frac{1}{3}), resulting in:[boxed{0}]]

🔑:( Hint: First prove frac{3 a^{2}-a}{1+a^{2}} geqslant frac{9}{10}left(a-frac{1}{3}right). Similarly, obtain the other two inequalities. Then add the three inequalities, to get the minimum value of 0. )

🧠:[<think>Okay, let's try to work through these three parts step by step. I'm a bit nervous because calculus problems can be tricky, especially involving logarithms and tangents. Let's take it one at a time.Part I: If a=2, find the equation of the tangent line to the curve y=f(x) at x=1Alright, so first, when a=2, the function becomes f(x) = 2x + ln x. I need to find the tangent line at x=1. To do that, I remember that the equation of a tangent line at a point x=a is given by y = f(a) + f’(a)(x - a). Here, a is given as 1, so x=1.First, let's compute f(1). Plugging in x=1: f(1) = 2*1 + ln 1. Since ln 1 is 0, that simplifies to f(1) = 2. So the point is (1, 2).Next, find the derivative f’(x) to get the slope at x=1. The derivative of 2x is 2, and the derivative of ln x is 1/x. So f’(x) = 2 + 1/x. Then, f’(1) = 2 + 1/1 = 3. So the slope is 3.Now, using the point-slope form. Wait, actually, the formula I mentioned earlier is already the point-slope form. So plugging in, the equation is y = 2 + 3(x - 1). Let's expand that: y = 2 + 3x - 3 = 3x - 1. So the tangent line should be y = 3x - 1.Let me double-check. At x=1, y=3*1 -1=2, which matches the point. The slope is 3, which we found correctly. So I think that's right.Part II: Find the monotonic intervals of f(x)Now, this is for general a, since the problem says "the function f(x)=ax + ln x (a∈R)". Monotonic intervals depend on the derivative. So first, let's compute f’(x). The derivative of ax is a, and the derivative of ln x is 1/x. So f’(x) = a + 1/x.To find where the function is increasing or decreasing, we need to analyze the sign of f’(x). The function is increasing when f’(x) > 0 and decreasing when f’(x) < 0.So let's set f’(x) = a + 1/x = 0. Solving for x: 1/x = -a => x = -1/a. But x must be in the domain of f(x), which is (0, +∞) because of the ln x term. Therefore, x has to be positive. So, x = -1/a must be positive, which implies that -1/a > 0 => a < 0.So, if a is negative, there is a critical point at x = -1/a. If a is non-negative (a ≥ 0), then f’(x) = a + 1/x is always positive because both a and 1/x are positive for x > 0. So in that case, the function is always increasing.But if a is negative, then there is a critical point at x = -1/a. Let's see:- For x < -1/a: Since x is in (0, ∞), and x < -1/a. Since a is negative, -1/a is positive, so x is between 0 and -1/a.In that interval, let's pick a test point. Let's say x is very small, approaching 0. Then 1/x becomes very large positive, so f’(x) = a + 1/x. Since a is negative, but 1/x is large positive, so f’(x) > 0 here. Wait, that's conflicting with what I thought earlier.Wait, perhaps I need to re-examine.If a is negative, then f’(x) = a + 1/x.At x approaching 0 from the right, 1/x approaches +infty, so f’(x) approaches +infty. At x approaching +infty, 1/x approaches 0, so f’(x) approaches a (which is negative). Therefore, the derivative starts positive, decreases, crosses zero at x = -1/a, and becomes negative beyond that. Wait, but if x = -1/a is the critical point, then:- For x < -1/a: Since x is in (0, -1/a) when a is negative (since -1/a is positive), so f’(x) = a + 1/x. Let's take a test point in (0, -1/a). Let's say x = -1/(2a). Wait, but a is negative, so -1/(2a) is positive. Then, plugging into f’(x): a + 1/(-1/(2a)) = a + (-2a) = -a. But a is negative, so -a is positive. Therefore, f’(x) is positive in (0, -1/a). Similarly, for x > -1/a: take x = -2/a (positive since a is negative). Then f’(x) = a + 1/(-2/a) = a + (-a/2) = a/2, which is negative because a is negative. Therefore, the function is increasing on (0, -1/a) and decreasing on (-1/a, +infty) when a is negative.So putting it all together:- If a ≥ 0: f’(x) > 0 for all x > 0, so f(x) is increasing on (0, +∞).- If a < 0: f(x) is increasing on (0, -1/a) and decreasing on (-1/a, +∞).Hence, the monotonic intervals are as above.Wait, the problem just says "find the monotonic intervals", so we need to write the answer depending on the value of a.So for part II, the answer would be:- When a ≥ 0, f(x) is increasing on (0, +∞).- When a < 0, f(x) is increasing on (0, -1/a) and decreasing on (-1/a, +∞).But perhaps we need to write it more formally. For example, using intervals:If a ≥ 0, the function is strictly increasing on (0, ∞).If a < 0, the function is strictly increasing on (0, -1/a) and strictly decreasing on (-1/a, ∞).Yes, that sounds right.Part III: If f(x) < 2 holds for any x ∈ (0,+∞), find the range of values for the real number a.So, we need to find all real numbers a such that ax + ln x < 2 for all x > 0.This is a bit more complex. Let's think about it. For f(x) = ax + ln x to be always less than 2, we need to ensure that the maximum value of f(x) is less than 2. Since if there is a maximum, it must be below 2. If the function tends to negative infinity or something else, but the maximum is the key.But first, we need to check the behavior of f(x) as x approaches 0 and x approaches infinity.As x approaches 0 from the right:ln x approaches -infty, and ax approaches 0 (if a is finite). So f(x) approaches -infty. So near zero, the function is very negative, so f(x) < 2 holds there.As x approaches infinity:If a > 0: Then ax dominates, so f(x) approaches +infty. Therefore, for a > 0, f(x) will exceed 2 for large x. Hence, such a cannot be in the range.If a = 0: Then f(x) = ln x. As x approaches infinity, ln x approaches +infty, which again exceeds 2. So a=0 is invalid.If a < 0: Then ax approaches -infty as x approaches infinity. But ln x approaches +infty, so the behavior is a competition between -infty and +infty. Which one dominates? Let's see:We can write f(x) = ax + ln x. For a < 0, let's see the limit as x approaches infinity:lim_{x→infty} (ax + ln x). Since a is negative, ax is like -|a|x. Compare the growth rates: linear term vs. logarithmic. The linear term dominates, so the entire expression tends to -infty. Therefore, for a < 0, as x approaches infinity, f(x) approaches -infty, so f(x) will eventually be less than 2. However, in between, the function might have a maximum. So the maximum value of f(x) occurs at x = -1/a (from part II), since when a < 0, the critical point is at x = -1/a.Therefore, to ensure that f(x) < 2 for all x > 0, the maximum value of f(x) must be less than 2. So compute f(-1/a) and set it less than 2.So let's compute f(-1/a). Since x = -1/a:f(-1/a) = a*(-1/a) + ln(-1/a) = -1 + ln(-1/a). Wait, but a is negative, so -1/a is positive, so ln(-1/a) is valid.So f(-1/a) = -1 + ln(-1/a). We need this to be less than 2:-1 + ln(-1/a) < 2.Solving for a:ln(-1/a) < 3.Exponentiate both sides:-1/a < e^3.But since a is negative, multiplying both sides by a (negative) reverses the inequality:-1 > a e^3.Wait, let's do it step by step.Starting from ln(-1/a) < 3.Exponentiate both sides: e^{ln(-1/a)} < e^3 => -1/a < e^3.Multiply both sides by a (which is negative, so inequality flips):-1 > a e^3.Divide both sides by e^3 (positive, so inequality remains):-1/e^3 > a.Which is equivalent to a < -1/e^3.But we already have a < 0 from earlier. So combining these, the range is a < -1/e^3.Wait, but let's check this again.Wait, starting from:ln(-1/a) < 3.Let me solve for a.First, ln(-1/a) < 3.Exponentiate both sides:-1/a < e^3.Multiply both sides by -1 (remember to reverse inequality):1/a > -e^3.But 1/a > -e^3. Since a is negative, let's write a = -b, where b > 0.Then 1/(-b) > -e^3 => -1/b > -e^3 => Multiply both sides by -1 (reverse inequality):1/b < e^3 => b > 1/e^3 => Since b = -a, then -a > 1/e^3 => a < -1/e^3.Yes, that's correct. Therefore, the range of a is a < -1/e^3.But let's verify this.Suppose a = -1/e^3. Then, the critical point is at x = -1/a = e^3. Compute f(e^3):f(e^3) = a*e^3 + ln(e^3) = (-1/e^3)*e^3 + 3 = -1 + 3 = 2. So f(e^3) = 2. But the problem states f(x) < 2 for all x > 0. So if a = -1/e^3, the maximum value is exactly 2, which doesn't satisfy the strict inequality. Therefore, a must be less than -1/e^3.So the range is a ≤ -1/e^3? Wait, no. Wait, when a = -1/e^3, f(x) reaches 2 at x = e^3. So to have f(x) < 2 for all x > 0, a must be less than -1/e^3. Because at a = -1/e^3, the maximum is exactly 2, which violates the strict inequality. Therefore, the range is a < -1/e^3.But let's check with a value slightly less than -1/e^3, say a = -2/e^3. Then, critical point at x = -1/a = e^3 / 2. Compute f(e^3/2):f(e^3/2) = a*(e^3/2) + ln(e^3/2) = (-2/e^3)*(e^3/2) + ln(e^3) - ln 2 = -1 + 3 - ln 2 = 2 - ln 2 ≈ 2 - 0.693 ≈ 1.307 < 2. So that works.If a is less than -1/e^3, then the maximum at x = -1/a is f(-1/a) = -1 + ln(-1/a). Let's see:Let a = -k where k > 0. Then the maximum is at x = 1/k. Then f(1/k) = -k*(1/k) + ln(1/k) = -1 + (-ln k). So we need -1 - ln k < 2 => -ln k < 3 => ln k > -3 => k > e^{-3} => since k = -a, then -a > e^{-3} => a < -e^{-3} = -1/e^3. Exactly. So yes, that's the condition.Therefore, the range is a < -1/e^3. So part III's answer is a ∈ (-∞, -1/e³).Summary of Answers:I. The tangent line at x=1 when a=2 is y = 3x - 1.II. Monotonic intervals:- If a ≥ 0, f(x) is strictly increasing on (0, +∞).- If a < 0, f(x) is strictly increasing on (0, -1/a) and strictly decreasing on (-1/a, +∞).III. The range of a for which f(x) < 2 for all x > 0 is a < -1/e³.Final Answer(I) The equation of the tangent line is boxed{y = 3x - 1}.(II) The monotonic intervals are:- When (a geq 0), (f(x)) is strictly increasing on ((0, +infty));- When (a < 0), (f(x)) is strictly increasing on (left(0, -frac{1}{a}right)) and strictly decreasing on (left(-frac{1}{a}, +inftyright)).(III) The range of values for (a) is boxed{(-infty, -frac{1}{e^3})}).</think>Given the function ( f(x) = ax + ln x ) where ( a in mathbb{R} ):(I) If ( a = 2 ), find the equation of the tangent line to the curve ( y = f(x) ) at ( x = 1 ):1. Compute ( f(1) ): [ f(1) = 2 cdot 1 + ln 1 = 2 + 0 = 2 ] The point is ( (1, 2) ).2. Compute the derivative ( f'(x) ): [ f'(x) = 2 + frac{1}{x} ]3. Evaluate the derivative at ( x = 1 ): [ f'(1) = 2 + frac{1}{1} = 3 ] The slope of the tangent line is 3.4. Form the equation of the tangent line using point-slope form: [ y - 2 = 3(x - 1) implies y = 3x - 1 ]The equation of the tangent line is (boxed{y = 3x - 1}).(II) Find the monotonic intervals of ( f(x) ):1. Compute the derivative ( f'(x) ): [ f'(x) = a + frac{1}{x} ]2. Analyze the sign of ( f'(x) ): - If ( a geq 0 ), ( f'(x) > 0 ) for all ( x > 0 ), so ( f(x) ) is strictly increasing on ( (0, +infty) ). - If ( a < 0 ), solve ( f'(x) = 0 ): [ a + frac{1}{x} = 0 implies x = -frac{1}{a} ] The critical point is ( x = -frac{1}{a} ). For ( x in (0, -frac{1}{a}) ), ( f'(x) > 0 ) and for ( x in (-frac{1}{a}, +infty) ), ( f'(x) < 0 ). Thus, ( f(x) ) is strictly increasing on ( (0, -frac{1}{a}) ) and strictly decreasing on ( (-frac{1}{a}, +infty) ).The monotonic intervals are:- When ( a geq 0 ), ( f(x) ) is strictly increasing on ( (0, +infty) );- When ( a < 0 ), ( f(x) ) is strictly increasing on ( left(0, -frac{1}{a}right) ) and strictly decreasing on ( left(-frac{1}{a}, +inftyright) ).(III) If ( f(x) < 2 ) holds for any ( x in (0, +infty) ), find the range of values for ( a ):1. Analyze the behavior of ( f(x) ): - As ( x to 0^+ ), ( ln x to -infty ), so ( f(x) to -infty ). - As ( x to +infty ): - If ( a geq 0 ), ( f(x) to +infty ), which violates ( f(x) < 2 ). - If ( a < 0 ), ( f(x) to -infty ).2. Find the maximum value of ( f(x) ) when ( a < 0 ): - The critical point is ( x = -frac{1}{a} ). - Compute ( fleft(-frac{1}{a}right) ): [ fleft(-frac{1}{a}right) = a left(-frac{1}{a}right) + lnleft(-frac{1}{a}right) = -1 + lnleft(-frac{1}{a}right) ] - Set ( -1 + lnleft(-frac{1}{a}right) < 2 ): [ lnleft(-frac{1}{a}right) < 3 implies -frac{1}{a} < e^3 implies a < -frac{1}{e^3} ]The range of values for ( a ) is (boxed{(-infty, -frac{1}{e^3})}).,<think>Okay, let me try to tackle these three parts one by one. Starting with part (I). The problem says that when a is 2, find the equation of the tangent line to the curve y = f(x) at x = 1. Alright, so first, when a = 2, the function becomes f(x) = 2x + ln x. To find the tangent line at x = 1, I remember that the equation of the tangent line at a point x = c is given by y = f(c) + f'(c)(x - c). So here, c is 1. First, let's compute f(1). Plugging x = 1 into the function: 2*1 + ln 1. Since ln 1 is 0, this simplifies to 2. So the point is (1, 2). Next, we need the derivative f'(x) to find the slope of the tangent at that point. The derivative of 2x is 2, and the derivative of ln x is 1/x. So f'(x) = 2 + 1/x. Then, evaluating the derivative at x = 1: f'(1) = 2 + 1/1 = 3. Therefore, the slope of the tangent line is 3. Now, putting it all together into the tangent line equation: y = 3(x - 1) + 2. Let's simplify that. Distribute the 3: y = 3x - 3 + 2, which simplifies to y = 3x - 1. So I think that's the equation of the tangent line. Let me double-check: when x = 1, y should be 2. Plugging in x = 1: 3(1) -1 = 2, which matches. And the slope is indeed 3. Okay, part (I) seems done.Moving on to part (II): Find the monotonic intervals of f(x). The function is f(x) = a x + ln x, where a is a real number. Monotonic intervals depend on the derivative. So first, let's find the derivative f'(x). The derivative of a x is a, and the derivative of ln x is 1/x. So f'(x) = a + 1/x.To find monotonic intervals, we need to determine where the derivative is positive or negative. So set f'(x) > 0 and f'(x) < 0. Let's solve for x.First, f'(x) = a + 1/x. Let's analyze this expression. The domain of f(x) is x > 0 because of the ln x term. So x is in (0, +∞). Case 1: If a + 1/x > 0. Then, 1/x > -a, which implies x < -1/a. But wait, this only makes sense if -a is positive, so a must be negative. Wait, let me think again. Let's solve the inequality step by step.Starting with a + 1/x > 0. Then 1/x > -a. Multiply both sides by x (which is positive, so inequality direction doesn't change): 1 > -a x. Then, 1 > -a x → a x > -1. Hmm, maybe this approach complicates. Alternatively, let's solve 1/x > -a.If a is positive, then -a is negative. Since 1/x is always positive for x > 0, so if -a is negative, then 1/x > -a is always true. Therefore, if a is positive, then f'(x) = a + 1/x is always positive because both terms are positive. Therefore, the function is increasing on the entire domain (0, +∞).If a is zero, then f'(x) = 1/x, which is always positive, so again the function is increasing on (0, +∞).If a is negative, then -a is positive. So 1/x > -a → x < 1/(-a). So in this case, when x is less than 1/(-a), the derivative is positive, and when x is greater than 1/(-a), the derivative is negative. Therefore, the function is increasing on (0, 1/(-a)) and decreasing on (1/(-a), +∞). However, note that 1/(-a) is positive only when a is negative, which is the case here.So summarizing:- If a ≥ 0, then f'(x) > 0 for all x > 0, so f(x) is increasing on (0, +∞).- If a < 0, then f(x) is increasing on (0, -1/a) and decreasing on (-1/a, +∞).Therefore, the monotonic intervals are as above. Let me check if this makes sense. For example, if a is negative, say a = -1, then the critical point is at x = 1/(-(-1)) = 1. So at x =1, the derivative is -1 +1/1=0. Then, for x <1, derivative is -1 +1/x. If x is 0.5, then 1/0.5=2, so derivative is -1 +2=1>0. For x=2, derivative is -1 +0.5=-0.5<0. So indeed increasing up to x=1, decreasing after. So the logic holds. Therefore, part (II)'s answer should be:If a ≥ 0, f(x) is increasing on (0, +∞). If a <0, f(x) is increasing on (0, -1/a) and decreasing on (-1/a, +∞).Moving on to part (III): If f(x) < 2 holds for any x ∈ (0, +∞), find the range of values for the real number a. So we need to ensure that a x + ln x < 2 for all x >0. We need to find all a such that this inequality is satisfied for all positive x.Let me think. Since the inequality must hold for all x >0, especially as x approaches 0+ and x approaches +∞. Let's analyze the behavior of f(x) as x approaches 0 and +∞.First, as x approaches 0+:ln x approaches -infty. So f(x) = a x + ln x. The term a x approaches 0, so the entire expression approaches -infty. Therefore, near x=0, f(x) is very negative, so the inequality f(x) <2 is certainly satisfied here. So the problem is not near x=0, but probably at some maximum point where f(x) could reach 2. Similarly, as x approaches +infty:If a >0: Then a x term dominates, so f(x) approaches +infty. So if a >0, then for large x, f(x) will exceed 2, violating the inequality. Therefore, a must be ≤0.Wait, but let's check. If a ≤0, then for a =0, f(x) = ln x. But ln x approaches +infty as x approaches +infty, so a=0 also causes f(x) to exceed 2 for large x, which is not allowed. So a=0 is invalid. So maybe a needs to be negative. Let's see.If a <0: Then as x approaches +infty, a x approaches -infty, and ln x approaches +infty, but which term dominates? Let's compare the growth rates. The term a x is linear, negative coefficient, so it goes to -infty, while ln x grows to +infty, but much slower. So the linear term dominates. Therefore, overall, f(x) approaches -infty as x approaches +infty. Therefore, for a <0, as x approaches +infty, f(x) tends to -infty, which is less than 2. So for a <0, the potential problem is not at infinity, but somewhere in the middle.But wait, if a <0, then as x approaches 0+, f(x) approaches -infty, which is less than 2. As x approaches +infty, f(x) approaches -infty, so also less than 2. So the maximum of f(x) must be somewhere in the middle. So we need to ensure that the maximum value of f(x) is less than 2.From part (II), we saw that if a <0, the function f(x) first increases on (0, -1/a) and then decreases on (-1/a, +∞). Therefore, the maximum of f(x) occurs at x = -1/a. Therefore, to ensure that f(x) <2 everywhere, we need the maximum value f(-1/a) <2.Therefore, compute f(-1/a) and set it less than 2.But wait, x = -1/a. Let's compute f at that point:f(-1/a) = a*(-1/a) + ln(-1/a) = -1 + ln(-1/a). Wait, since a is negative, -1/a is positive, so ln(-1/a) is defined.So we have -1 + ln(-1/a) < 2.Solving for a:ln(-1/a) < 3Exponentiate both sides:-1/a < e^3Multiply both sides by a. But a is negative, so inequality reverses:-1 > a e^3Divide both sides by e^3 (positive, so inequality direction remains):-1/e^3 > aTherefore, a < -1/e^3.But since a is negative, the upper bound is -1/e^3. So the range of a is a ≤ -1/e^3? Wait, let's check.Wait, we had:ln(-1/a) <3 → -1/a < e^3 → -1/a < e^3 → multiplying both sides by a (negative) reverses inequality: -1 > a e^3 → a < -1/e^3.Therefore, the maximum value occurs at x = -1/a, and the maximum value is -1 + ln(-1/a). To make this less than 2, we get a < -1/e^3.But we need to verify if this is indeed the case.Let me check with a = -1/e^3. Then, x = -1/a = e^3. Then f(e^3) = (-1/e^3)*e^3 + ln(e^3) = -1 + 3 = 2. So when a = -1/e^3, the maximum value is exactly 2. But the problem states that f(x) <2 for all x>0. Therefore, a must be less than -1/e^3 so that the maximum is less than 2. Therefore, the range is a ≤ -1/e^3? Wait, but if a = -1/e^3, then f(x) reaches 2 at x = e^3, which violates the condition f(x) <2 for all x>0. Therefore, a must be strictly less than -1/e^3.But the problem says "find the range of values for the real number a such that f(x) <2 holds for any x∈(0,+∞)". Therefore, a must be less than -1/e^3. So the range is (-infty, -1/e^3).Wait, let me confirm with an example. Suppose a = -1/e^3 - ε, where ε >0. Then, the maximum at x = -1/a = 1/(1/e^3 + ε). Then, compute f(-1/a) = a*(-1/a) + ln(-1/a) = -1 + ln(-1/a). Let's compute ln(-1/a):a = -1/e^3 - ε, so -1/a = 1/(1/e^3 + ε). So ln(-1/a) = ln[1/(1/e^3 + ε)] = -ln(1/e^3 + ε). Then, f(-1/a) = -1 - ln(1/e^3 + ε). If ε approaches 0, then 1/e^3 + ε approaches 1/e^3, so ln(1/e^3) = -3, so f(-1/a) approaches -1 - (-3) = 2. But if ε is positive, then 1/e^3 + ε >1/e^3, so ln(1/e^3 + ε) > ln(1/e^3) = -3. Therefore, -ln(1/e^3 + ε) <3. Thus, f(-1/a) = -1 + [something less than 3]. Wait, wait.Wait, no. Let me re-examine. If a = -1/e^3 - ε, then -1/a =1/(1/e^3 + ε). Let's call that value k =1/(1/e^3 + ε). Then, ln(k) = ln[1/(1/e^3 + ε)] = -ln(1/e^3 + ε). Therefore, f(-1/a) = a*(-1/a) + ln(-1/a) = -1 + ln(k) = -1 - ln(1/e^3 + ε). Now, since 1/e^3 + ε >1/e^3, ln(1/e^3 + ε) > ln(1/e^3) = -3. Therefore, -ln(1/e^3 + ε) <3. Therefore, f(-1/a) = -1 - ln(1/e^3 + ε) < -1 - (-3) =2. So if a is less than -1/e^3, then f(-1/a) <2. Hence, the maximum value is less than 2, so the entire function is below 2.If a = -1/e^3, then f(-1/a) =2, which is not allowed because we need f(x) <2 for all x. Therefore, a must be less than -1/e^3. Therefore, the range is a ∈ (-infty, -1/e^3).But let me check for another a. Suppose a is very negative, say a = -100. Then x = -1/a =1/100. Then f(1/100) = -100*(1/100) + ln(1/100) = -1 + (-4.605) ≈ -5.605 <2. And as x approaches 0+, f(x) approaches -infty, and as x approaches +infty, f(x) approaches -infty. The maximum is at x=1/100, which is -1 + ln(1/100) ≈-5.605, which is less than 2. So even though a is very negative, the maximum is still less than 2. Wait, but wait. Wait, when a is more negative than -1/e^3, for example, a = -2/e^3. Then, x = -1/a = e^3/2. Then, f(x) = (-2/e^3)*(e^3/2) + ln(e^3/2) = -1 + 3 - ln2 ≈2 -0.693≈1.307 <2. So that's okay.But wait, if a is less than -1/e^3, then x = -1/a is less than e^3. Let me see. If a < -1/e^3, then -1/a < e^3. Wait, for example, a = -2/e^3, then -1/a = e^3/2. Which is less than e^3. Therefore, the maximum occurs at x = e^3/2. So as a becomes more negative, the position of the maximum x = -1/a becomes smaller. Wait, but the maximum value is still -1 + ln(-1/a). So when a is more negative, ln(-1/a) becomes ln(something smaller), so it's more negative. Therefore, f(-1/a) = -1 + ln(-1/a) becomes less than 2. Wait, but if a is approaching -infty, ln(-1/a) approaches ln(0+) which is -infty. So f(-1/a) approaches -1 -infty = -infty. So even as a becomes more negative, the maximum of f(x) becomes more negative. Therefore, the most restrictive case is when a approaches -1/e^3 from below. When a approaches -1/e^3 from the left, the maximum approaches 2. Therefore, the critical value is a = -1/e^3. So to ensure that f(x) <2 for all x>0, a must be less than -1/e^3.Therefore, the answer for part (III) is a < -1/e³, or in interval notation, (-∞, -1/e³).But let me check again with a = -1/e³. Then, f(x) = (-1/e³)x + lnx. Then, the maximum at x = e³, f(e³) = (-1/e³)(e³) + ln(e³) = -1 +3 =2. So at x = e³, f(x)=2. But the problem requires f(x) <2 for all x in (0, +infty). Therefore, a must be strictly less than -1/e³. Hence, the range is a < -1/e³. So the answer is a ∈ (-∞, -1/e³).Therefore, summarizing all three parts:(I) The tangent line is y = 3x -1.(II) Monotonic intervals: If a ≥0, increasing on (0, ∞); if a <0, increasing on (0, -1/a) and decreasing on (-1/a, ∞).(III) The range of a is a < -1/e³.Final Answer(I) The equation of the tangent line is boxed{y = 3x - 1}.(II) The monotonic intervals of ( f(x) ) are:- When ( a geq 0 ), ( f(x) ) is increasing on ( (0, +infty) ).- When ( a < 0 ), ( f(x) ) is increasing on ( left(0, -frac{1}{a}right) ) and decreasing on ( left(-frac{1}{a}, +inftyright) ).(III) The range of values for ( a ) is boxed{(-infty, -frac{1}{e^3})}.</think>Given the function ( f(x) = ax + ln x ) where ( a in mathbb{R} ):(I) When ( a = 2 ), the function becomes ( f(x) = 2x + ln x ). To find the equation of the tangent line at ( x = 1 ):1. Calculate ( f(1) ): [ f(1) = 2 cdot 1 + ln 1 = 2 + 0 = 2 ]2. Compute the derivative ( f'(x) ): [ f'(x) = 2 + frac{1}{x} ]3. Evaluate the derivative at ( x = 1 ): [ f'(1) = 2 + frac{1}{1} = 3 ]4. Form the tangent line equation using the point-slope form: [ y - 2 = 3(x - 1) implies y = 3x - 1 ](II) To determine the monotonic intervals of ( f(x) ):1. Compute the derivative ( f'(x) ): [ f'(x) = a + frac{1}{x} ]2. Analyze the sign of ( f'(x) ): - If ( a geq 0 ), ( f'(x) > 0 ) for all ( x > 0 ), so ( f(x) ) is increasing on ( (0, +infty) ). - If ( a < 0 ), solve ( a + frac{1}{x} = 0 ) to find the critical point ( x = -frac{1}{a} ). The function is increasing on ( left(0, -frac{1}{a}right) ) and decreasing on ( left(-frac{1}{a}, +inftyright) ).(III) To ensure ( f(x) < 2 ) for all ( x in (0, +infty) ):1. Analyze the behavior of ( f(x) ) as ( x to 0^+ ) and ( x to +infty ): - As ( x to 0^+ ), ( ln x to -infty ), so ( f(x) to -infty ). - As ( x to +infty ), if ( a geq 0 ), ( f(x) to +infty ), which violates the condition. Thus, ( a ) must be negative.2. For ( a < 0 ), the maximum value occurs at ( x = -frac{1}{a} ): [ fleft(-frac{1}{a}right) = -1 + lnleft(-frac{1}{a}right) ]3. Set the maximum value less than 2: [ -1 + lnleft(-frac{1}{a}right) < 2 implies lnleft(-frac{1}{a}right) < 3 implies -frac{1}{a} < e^3 implies a < -frac{1}{e^3} ]Final Answers:(I) The equation of the tangent line is boxed{y = 3x - 1}.(II) The monotonic intervals of ( f(x) ) are:- When ( a geq 0 ), ( f(x) ) is increasing on ( (0, +infty) ).- When ( a < 0 ), ( f(x) ) is increasing on ( left(0, -frac{1}{a}right) ) and decreasing on ( left(-frac{1}{a}, +inftyright) ).(III) The range of values for ( a ) is boxed{(-infty, -frac{1}{e^3})}.]

🔑:(I) If a=2, then f(x)=2x+ln x. The derivative of the function is f′(x)=2+ frac {1}{x}. The slope of the tangent line to the curve y=f(x) at x=1 is 3. The point of tangency is (1,2). Therefore, the equation of the tangent line is y-2=3(x-1), which simplifies to boxed{3x-y-1=0}.(II) The derivative of the function f(x)=ax+ln x is f′(x)=a+ frac {1}{x}= frac {ax+1}{x}, where x > 0.- When ageqslant 0, f′(x) > 0, and f(x) is increasing on (0,+∞).- When a - frac {1}{a}, and f′(x) > 0 which gives 0 e^{3}, g′(x) > 0, and g(x) is increasing;- When 0 < x < e^{3}, g′(x) < 0, and g(x) is decreasing.The function g(x) attains its minimum value at x=e^{3}, and the minimum value is - frac {1}{e^{3}}. Therefore, we have a < - frac {1}{e^{3}}. Thus, the range of values for the real number a is boxed{(-infty,- frac {1}{e^{3}})}.

🧠:[<think>Okay, let's try to solve this problem: Determine all positive integers ( n ) such that the cube root of ( n ) is obtained by removing the last three decimal digits of ( n ). Hmm, interesting. So, if I understand correctly, if we take the number ( n ), remove its last three digits, we get the cube root of ( n ). Let me rephrase that to make sure. Suppose ( sqrt[3]{n} = m ), where ( m ) is the number obtained by removing the last three digits of ( n ). That means if ( n ) is a four-digit number or more, then chopping off the last three digits gives ( m ), which is the cube root of ( n ). So, mathematically, this relationship can be written as:( m = sqrt[3]{n} )But also, ( n ) can be expressed in terms of ( m ) by appending three digits to ( m ). Wait, but appending three digits isn't just multiplying by 1000, because appending three digits could be any digits. But in this case, since we're removing the last three digits to get ( m ), that means ( n = m times 1000 + k ), where ( k ) is the number formed by the last three digits. However, the problem states that removing the last three digits gives the cube root. Therefore, the cube root is exactly ( m ), so ( n = m^3 ). Wait, so putting those together:( n = m^3 )and( n = 1000m + k )where ( k ) is an integer between 0 and 999 (inclusive). So, combining these two equations:( m^3 = 1000m + k )Since ( k ) is between 0 and 999, we have:( 0 leq m^3 - 1000m leq 999 )So, ( m^3 - 1000m geq 0 ) and ( m^3 - 1000m leq 999 ). Let's analyze these inequalities.First inequality: ( m^3 - 1000m geq 0 )Factor out ( m ):( m(m^2 - 1000) geq 0 )Since ( m ) is a positive integer, ( m geq 1 ). Therefore, the inequality simplifies to:( m^2 - 1000 geq 0 )So,( m^2 geq 1000 )Taking square roots:( m geq sqrt{1000} approx 31.62 )Therefore, ( m geq 32 )Second inequality: ( m^3 - 1000m leq 999 )So,( m^3 - 1000m leq 999 )Rearranged:( m^3 - 1000m - 999 leq 0 )Hmm, this is a cubic equation. Let's see. Let's denote ( f(m) = m^3 - 1000m - 999 ). We need to find ( m ) such that ( f(m) leq 0 ).Since ( m geq 32 ), let's compute ( f(32) ):( f(32) = 32^3 - 1000*32 - 999 )Compute 32^3: 32*32 = 1024, 1024*32 = 327681000*32 = 32000So, 32768 - 32000 - 999 = 32768 - 32000 = 768; 768 - 999 = -231So, ( f(32) = -231 leq 0 )Good. Now check ( f(33) ):33^3 = 359371000*33 = 33000So, 35937 - 33000 - 999 = 35937 - 33000 = 2937; 2937 - 999 = 1938 > 0So, ( f(33) = 1938 > 0 )Similarly, check ( f(34) ):34^3 = 3930439304 - 34000 - 999 = 39304 - 34000 = 5304; 5304 - 999 = 4305 > 0So, ( f(m) ) is negative at m=32, positive at m=33, and increasing for m > some point. Since the cubic term dominates, as m increases, ( f(m) ) will keep increasing. So, the only possible value for m is 32, since at m=32, the inequality holds, but for m=33 and above, it doesn't. Therefore, the only possible m is 32.Therefore, n = m^3 = 32^3 = 32768Wait, but let me check. If n = 32768, then removing the last three digits would be 32.768, but since n is an integer, removing the last three digits would give 32 (since 32768 is 32 followed by 768). So, yes, removing the last three digits (768) gives 32, which is the cube root of 32768. Let's verify:32^3 = 32*32*32 = 1024*32 = 32768. Correct.So, that works. Are there any other possible m? From the above analysis, it seems m must be 32. Let's check m=31. Wait, earlier we found m must be at least 32. Because when m=31:Check first inequality: m=31, m^2 = 961 < 1000, so m^3 - 1000m = 29791 - 31000 = -1209 < 0, which would mean k is negative, but k can't be negative. So, m cannot be 31. So, m starts from 32.Similarly, m=32 gives k= -231, but wait, in our previous calculation, k = m^3 - 1000m = 32768 - 32000 = 768. Wait, hold on, maybe I confused k. Wait, from n = m^3 and n = 1000m + k. So, k = m^3 - 1000m. So, when m=32, k= 32768 - 32000 = 768. So, k is 768, which is a valid three-digit number. So, n = 1000*32 + 768 = 32768, which is correct.Wait, but earlier when I thought of m=31, k would be 29791 - 31000 = -1209, which is negative, so invalid. So, m must be at least 32. And for m=32, k=768, which is valid. For m=33, k=35937 - 33000= 2937. But k must be less than 1000. But 2937 is more than 999, so invalid. Similarly, for m=34, k=39304 - 34000= 5304, which is way too big. So, indeed, only m=32 gives k in the range 0-999.Therefore, the only solution is n=32768.Wait, but let me think again. Is there a possibility that m is larger? For example, m=100. Then n=100^3=1,000,000. Removing the last three digits would give 1000, not 100. So, that doesn't work. Similarly, if m=999, n=999^3=997002999. Removing the last three digits gives 997002, which is not 999. So, the relationship only holds when n is just m followed by three digits. But in general, the cube of a number is going to be much larger than the number times 1000 plus some three digits. So, the only possible case is when m^3 is equal to 1000m + k, where k is three digits. So, m^3 must be slightly larger than 1000m, but not by more than 999. That seems possible only around when m^3 ≈ 1000m, which is when m^2 ≈ 1000, so m≈32, as we had before.Therefore, m=32 is the only solution, leading to n=32^3=32768.But let me check m=10. m=10, n=1000. Removing last three digits gives 1, but cube root of 1000 is 10. So, 1 ≠ 10. Doesn't work.Similarly, m=99. n=970299. Remove last three digits: 970. Cube root of 970299 is 99. So, 970 ≠ 99. Not equal.Another example: m=5. n=125. Remove last three digits: 0 (since it's a three-digit number). But cube root of 125 is 5. 0≠5. Doesn't work.Wait, the problem says "removing the last three decimal digits". So, if n is a three-digit number, removing the last three digits would result in 0. But cube root of n is 0? That's not possible unless n=0, but n must be positive. So, n must have at least four digits. So, n must be a four-digit number or more. Therefore, m must be such that n = m^3 is at least a four-digit number, so m is at least 10 (since 10^3=1000). But as we saw, m=10 gives n=1000, which when removing the last three digits gives 1, not 10. So, doesn't work.So, only m=32 gives n=32768, which is a five-digit number. Removing the last three digits (768) gives 32, which is indeed the cube root. So, that works.Is there another m where m^3 is equal to 1000m + k, with 0 ≤ k ≤ 999? Let's check m=32, which works. What about m=0? But n must be positive. So, m must be at least 1, but m=1 would give n=1, removing three digits would give 0. Not equal. So, no. Therefore, the only possible solution is n=32768.Wait, but let me check m=00. Wait, m can't be zero. So, the only possible value is m=32. Therefore, the answer is boxed{32768}Final Answerboxed{32768}</think>To determine all positive integers ( n ) such that the cube root of ( n ) is obtained by removing the last three decimal digits of ( n ), we start by noting that if ( sqrt[3]{n} = m ), then ( n ) can be expressed as ( m^3 ). Additionally, removing the last three digits of ( n ) gives ( m ), which means ( n = 1000m + k ) where ( 0 leq k leq 999 ).Combining these two expressions for ( n ), we get:[m^3 = 1000m + k]with ( 0 leq k leq 999 ). This implies:[0 leq m^3 - 1000m leq 999]First, solving the inequality ( m^3 - 1000m geq 0 ):[m(m^2 - 1000) geq 0]Since ( m ) is a positive integer, this simplifies to ( m^2 geq 1000 ), giving ( m geq sqrt{1000} approx 31.62 ). Thus, ( m geq 32 ).Next, solving the inequality ( m^3 - 1000m leq 999 ):[m^3 - 1000m - 999 leq 0]Checking values of ( m ) starting from 32:- For ( m = 32 ): [ 32^3 - 1000 cdot 32 = 32768 - 32000 = 768 ] Here, ( k = 768 ), which is valid since ( 0 leq 768 leq 999 ). Thus, ( n = 32^3 = 32768 ).Checking ( m = 33 ) and higher values results in ( k ) exceeding 999, making them invalid. Therefore, the only valid solution is ( m = 32 ).Verification shows that removing the last three digits of ( 32768 ) gives ( 32 ), which is indeed the cube root of ( 32768 ).Thus, the only positive integer ( n ) satisfying the condition is (boxed{32768}).,<think>Okay, so I need to find all positive integers n such that the cube root of n is obtained by removing the last three digits of n. Hmm, let's break this down. First, let me make sure I understand the problem correctly. If I have a number n, and if I remove its last three digits, I get the cube root of n. So, for example, if n was some number like 123456, then removing the last three digits would give me 123, and that should be equal to the cube root of 123456. But of course, 123 cubed is 123*123*123, which is probably not 123456. So, the task is to find all such numbers where this condition holds.Let me formalize this. Let’s denote the cube root of n as k. So, k = ∛n, which implies that n = k³. Now, the problem states that if we remove the last three digits of n, we get k. Removing the last three digits is equivalent to dividing the number by 1000 and taking the integer part, right? So, mathematically, we can write this as:k = floor(n / 1000)But since n = k³, substituting that in:k = floor(k³ / 1000)But since k is an integer, floor(k³ / 1000) is just the integer division of k³ by 1000. However, since k must be equal to floor(k³ / 1000), this implies that k³ / 1000 must be at least k but less than k+1. So:k ≤ k³ / 1000 < k + 1Multiplying all parts by 1000:1000k ≤ k³ < 1000(k + 1)Let’s divide these inequalities by k (assuming k > 0, which it is because n is positive):1000 ≤ k² < 1000(k + 1)/kSimplify the right side:1000 ≤ k² < 1000(1 + 1/k)So, the left inequality tells us that k² ≥ 1000, so k ≥ √1000 ≈ 31.62. Since k is an integer, k must be at least 32.The right inequality is k² < 1000 + 1000/k. Since k ≥ 32, 1000/k is at most 1000/32 ≈ 31.25. So, k² < 1000 + 31.25 = 1031.25. Therefore, k² < 1031.25, which implies k < √1031.25 ≈ 32.11. Since k must be an integer, this means k ≤ 32.So combining the two inequalities, k must be between 32 and 32, inclusive. So k = 32.Wait, so k must be exactly 32. Then n = 32³ = 32768. Let's check. If we remove the last three digits of 32768, we get 32. Which is exactly the cube root. So, 32 is the cube root of 32768, and removing the last three digits gives 32. So that works.But the problem says "determine all positive integers n". So is 32768 the only one? Let me confirm.Wait, let's check if maybe there are more possibilities. Maybe my reasoning is missing something. Let me go back.The inequalities I had were:1000k ≤ k³ < 1000(k + 1)Which simplifies to 1000 ≤ k² < 1000 + 1000/kSo, solving for k.Given that k is integer ≥1, but as we saw, k must be ≥32 and ≤32. Hence, only k=32. Therefore, n=32³=32768. So that's the only solution.Wait, but let me check for k=0. Wait, no, n is positive integer, so k must be at least 1. But k=32 is the only solution here.But let me think again. Suppose that when we remove the last three digits, it's exactly k, not just the floor. So, is there a case where n is of the form k followed by three zeros? For example, k=1000, then n=1000³=1000000000, removing the last three digits gives 1000000, which is not 1000. Wait, so that's not correct.Wait, but in this problem, the cube root is k, and n is k³. So, when you remove the last three digits of k³, you get k. So, k³ must be a number that when you remove the last three digits, you get k. So, for example, if k=32, then k³=32768, removing last three digits (768) gives 32. Which matches.But if k=100, then k³=1,000,000. Removing the last three digits would give 1000, but k=100. So that's not equal.Alternatively, maybe k³ has at least three digits, so that you can remove three digits. So, k must be such that k³ has at least three digits. So, since k is at least 32, which gives a 5-digit number, 32³=32768. So that's okay.But let's check if k=31. Then k³=29791. Removing the last three digits would give 29. But k=31, which is not 29. So that doesn't work. Similarly, k=33: 33³=35937. Remove the last three digits: 35. But k=33. Not equal.But according to our inequality, k must be exactly 32 because the lower bound was 32 and upper bound was ~32.11. So only 32. So, that's the only solution.Wait, but perhaps there are numbers where after removing three digits, you get the cube root. But what if the number has more than five digits? For example, suppose n is a 6-digit number, like abcdef. Removing the last three digits would give abc. Then, abc should be the cube root of abcdef. Is that possible?But according to our initial equation, n = k³, and k is equal to floor(n / 1000). So, for example, if n is a 6-digit number, k would be a 3-digit number. But let's see. Let's denote n as 1000k + m, where m is the last three digits, so 0 ≤ m < 1000. Then the condition is that k = ∛(1000k + m). So, k³ = 1000k + m. Since m is between 0 and 999, we have k³ - 1000k must be between 0 and 999. So, 0 ≤ k³ - 1000k ≤ 999.So, we can write:k³ - 1000k ≥ 0 => k(k² - 1000) ≥ 0. Since k is positive, this implies k² - 1000 ≥ 0 => k ≥ √1000 ≈31.62, so k ≥32.Also, k³ - 1000k ≤ 999 => k³ - 1000k - 999 ≤ 0. Let's solve this inequality.So, for k=32: 32³ -1000*32 = 32768 - 32000 = 768 ≤ 999. So, that's okay.For k=33: 33³ -1000*33 = 35937 - 33000 = 2937. Which is greater than 999. So, the inequality k³ -1000k ≤999 is only true for k=32.So, only k=32 satisfies both inequalities, so m =768, so n=32768.Hence, the only solution is 32768. So, I think that's the answer.But let me check if there's another possible k where k³ -1000k is between 0 and 999.Check k=10: 1000 -10000= -9000, which is negative. So no.k=20: 8000 -20000= -12000. Negative.k=31: 29791 -31000= -1209. Negative.k=32: 32768 -32000=768. Positive and less than 999.k=33: 35937 -33000=2937>999. So, no.Thus, only k=32.Therefore, the answer is 32768.But wait, let me check again. Suppose n is a 7-digit number. Let's say k is a 4-digit number. For example, k=1000, then n=1000³=1000000000, which is a 10-digit number. Removing the last three digits would give 1000000, which is not equal to k=1000.Alternatively, maybe k is a 4-digit number, such that when you cube it, you get a 7-digit number. Wait, 1000³ is 1,000,000,000, which is 10 digits. 100³ is 1,000,000, which is 7 digits. So, if k is 100, then n=1,000,000. Remove the last three digits: 1000. But k=100, so 1000≠100. Not equal.Wait, but if k is a three-digit number, like 123. Then n=123³=1860867. Remove last three digits: 1860. Which is not 123. So that's not equal.Alternatively, maybe there's another k where the cube is such that when you remove the last three digits, you get k. Let's check k=999. 999³=997002999. Remove last three digits: 997002. Which is not 999. Doesn't work.Alternatively, think of the equation k³ = 1000k + m, where 0 ≤ m ≤999. So, k³ -1000k = m. So, for m to be an integer between 0 and 999 inclusive, k must satisfy that equation.So, solving k³ -1000k = m, with m in 0 to 999. So, k³ -1000k must be between 0 and 999.As above, for k=32, 32³ -32*1000=32768-32000=768. So m=768. So, n=1000*32 +768=32768.For k=31: 31³ -31*1000=29791 -31000= -1209. Which is negative, so invalid.For k=33:33³ -33*1000=35937 -33000=2937>999. So invalid.Thus, only k=32 is valid. Therefore, the only n is 32768. So, that's the only solution.Therefore, the answer is boxed{32768}.Final Answerboxed{32768}</think>To determine all positive integers ( n ) such that the cube root of ( n ) is obtained by removing the last three decimal digits of ( n ), we start by denoting the cube root of ( n ) as ( k ). This implies ( n = k^3 ). The problem states that removing the last three digits of ( n ) gives ( k ), which is equivalent to:[ k = leftlfloor frac{n}{1000} rightrfloor ]Since ( n = k^3 ), substituting this into the equation gives:[ k = leftlfloor frac{k^3}{1000} rightrfloor ]This implies:[ k leq frac{k^3}{1000} < k + 1 ]Multiplying through by 1000, we get:[ 1000k leq k^3 < 1000(k + 1) ]Dividing the left inequality by ( k ):[ 1000 leq k^2 ]Thus, ( k geq sqrt{1000} approx 31.62 ), so ( k geq 32 ).For the right inequality, we have:[ k^3 < 1000(k + 1) ]Dividing by ( k ):[ k^2 < 1000 + frac{1000}{k} ]Since ( k geq 32 ), ( frac{1000}{k} leq frac{1000}{32} approx 31.25 ), so:[ k^2 < 1000 + 31.25 = 1031.25 ]Thus, ( k < sqrt{1031.25} approx 32.11 ), so ( k leq 32 ).Combining these results, ( k ) must be 32. Checking this, we find:[ n = 32^3 = 32768 ]Removing the last three digits of 32768 gives 32, which is indeed the cube root of 32768. Therefore, the only solution is:[boxed{32768}]]

🔑:Answer: 32768 is the only such integer.If n=m^{3} is a solution, then m satisfies 1000 m leqslant m^{3}<1000(m+1). From the first inequality, we get m^{2} geqslant 1000, or m geqslant 32. By the second inequality, we then havem^{2}<1000 cdot frac{m+1}{m} leqslant 1000 cdot frac{33}{32}=1000+frac{1000}{32} leqslant 1032,or m leqslant 32. Hence, m=32 and n=m^{3}=32768 is the only solution.

🧠:[<think>Okay, let's see. I have this problem about a quadratic function f(x) = ax² + bx + c. The conditions are that when x is between -1 and 1, the absolute value of f(x) is less than or equal to 1. There are two parts: first, I need to prove that the absolute value of b is less than or equal to 1. Second, if f(0) = -1 and f(1) = 1, find the value of a. Hmm. Let me start with part (1).First, part (1): Prove |b| ≤ 1. So, given that |f(x)| ≤ 1 for all x in [-1,1], need to show that the coefficient b of the linear term is bounded in absolute value by 1. Since f(x) is a quadratic, maybe I can use some properties of quadratics or inequalities. Alternatively, perhaps use specific values of x to get information about the coefficients.Since the function is bounded between -1 and 1 on the interval [-1,1], maybe evaluating the function at certain points will give me inequalities involving a, b, and c. For example, plugging in x = 0: f(0) = c. So |c| ≤ 1. That's straightforward. Then maybe other points like x = 1 and x = -1. Let's check that.At x = 1: f(1) = a + b + c. Since |f(1)| ≤ 1, we have |a + b + c| ≤ 1.At x = -1: f(-1) = a - b + c. Similarly, |a - b + c| ≤ 1.Also, since we know |c| ≤ 1 from x=0. But how do we get b from these? Maybe subtract or add these equations? Let me see.If I take f(1) - f(-1) = (a + b + c) - (a - b + c) = 2b. So 2b = f(1) - f(-1). Taking absolute values: |2b| = |f(1) - f(-1)| ≤ |f(1)| + |f(-1)| ≤ 1 + 1 = 2. Therefore, |2b| ≤ 2, so |b| ≤ 1. That seems straightforward! Wait, is that valid?Yes, because by the triangle inequality, |f(1) - f(-1)| ≤ |f(1)| + |f(-1)|, and since each of those is ≤1, the sum is ≤2. Then dividing by 2 gives |b| ≤1. So that proves part (1). Cool, that wasn't too bad. So maybe that's the approach.Alternatively, another way might involve derivatives? Because the maximum or minimum of the quadratic could be related to the coefficients. The derivative of f(x) is 2a x + b. The extremum is at x = -b/(2a). But maybe that's more complicated. The first approach using specific points seems simpler and works directly.So part (1) is done. Now part (2): If f(0) = -1 and f(1) = 1, find the value of a. Given these specific conditions, and the general condition that |f(x)| ≤1 on [-1,1].First, let's note that f(0) = c = -1. So c = -1. Then f(1) = a + b + c = a + b -1 = 1. Therefore, a + b = 2. So equation (1): a + b = 2.We need to find the value of a. But we have to use the condition that |f(x)| ≤1 for all x in [-1,1]. So even though we have c and a relation between a and b, we need to ensure that the quadratic doesn't exceed 1 or go below -1 in that interval. So perhaps we can write the quadratic in terms of a and b, substitute c = -1, then use some extremum conditions.So f(x) = a x² + b x -1. We need |a x² + b x -1| ≤1 for all x in [-1,1]. Which is equivalent to -1 ≤ a x² + b x -1 ≤1. Adding 1 to all parts: 0 ≤ a x² + b x ≤2. So for all x in [-1,1], a x² + b x is between 0 and 2.Alternatively, the original inequality is |a x² + b x -1| ≤1. So, that means that the maximum value of |a x² + b x -1| over x in [-1,1] is at most 1. Therefore, the maximum of a x² + b x -1 is ≤1, and the minimum is ≥-1.So we can write two inequalities:1. a x² + b x -1 ≤1 => a x² + b x ≤2 for all x in [-1,1]2. a x² + b x -1 ≥-1 => a x² + b x ≥0 for all x in [-1,1]So combining these, 0 ≤ a x² + b x ≤2 for all x in [-1,1].Given that a + b =2, from f(1)=1. So we can express b =2 - a.So substitute b =2 -a into the inequalities:0 ≤ a x² + (2 -a) x ≤2 for all x in [-1,1].So we need to ensure that for all x in [-1,1], the expression a x² + (2 -a)x is between 0 and 2.So perhaps find the maximum and minimum of the function g(x) = a x² + (2 -a)x over the interval [-1,1], and set those to be within [0,2].To find the extrema, we can take the derivative of g(x):g'(x) = 2a x + (2 -a).Setting derivative to zero:2a x + (2 -a) =0 => x = (a -2)/(2a).But this critical point may or may not lie within [-1,1]. So we need to check if the critical point is inside the interval. Also, we need to check the endpoints x=-1 and x=1.So let's compute g(-1) and g(1):g(-1) = a*(1) + (2 -a)*(-1) = a - (2 -a) = a -2 +a = 2a -2.g(1) = a*(1) + (2 -a)*(1) = a +2 -a = 2.So at x=1, g(1)=2, which is exactly the upper bound, which is allowed. At x=-1, g(-1)=2a -2. Since we need g(-1) ≥0, because 0 ≤g(x) ≤2. Therefore:2a -2 ≥0 => 2a ≥2 => a ≥1.Also, since g(-1) must also be ≤2:2a -2 ≤2 => 2a ≤4 => a ≤2.So from x=-1, we have 1 ≤a ≤2.Additionally, we need to check if the critical point x=(a-2)/(2a) is within [-1,1].Let's compute x_c = (a -2)/(2a). Let's see for a in [1,2]:When a=1: x_c=(1-2)/(2*1)= (-1)/2= -0.5, which is in [-1,1].When a=2: x_c=(2-2)/(2*2)=0/4=0, which is in [-1,1].For a between 1 and 2, let's see:For a in (1,2), x_c = (a -2)/(2a). Since a>0, denominator is positive. Numerator: a-2. When a <2, numerator is negative; when a=2, numerator=0. So x_c is between (-1)/2 and 0. So x_c ∈ [-0.5,0], which is within [-1,1]. Therefore, the critical point is always in [-1,1] for a ∈ [1,2].Thus, we need to compute the value of g(x) at the critical point x_c.Compute g(x_c):g(x_c) = a x_c² + (2 -a) x_c.First, let's compute x_c:x_c = (a -2)/(2a)Then x_c² = [(a -2)/(2a)]² = (a² -4a +4)/(4a²)Therefore:g(x_c) = a*(a² -4a +4)/(4a²) + (2 -a)*( (a -2)/(2a) )Simplify term by term:First term: a*(a² -4a +4)/(4a²) = (a² -4a +4)/(4a)Second term: (2 -a)*(a -2)/(2a) = -(a -2)^2/(2a)Therefore, adding them together:(a² -4a +4)/(4a) - (a -2)^2/(2a)Note that (a -2)^2 = a² -4a +4, so:= (a² -4a +4)/(4a) - (a² -4a +4)/(2a)= [1/(4a) - 2/(4a)]*(a² -4a +4)= (-1/(4a))*(a² -4a +4)= -(a² -4a +4)/(4a)So g(x_c) = - (a² -4a +4)/(4a)But since we need g(x_c) ≥0, because the lower bound is 0.Therefore:- (a² -4a +4)/(4a) ≥0Multiply both sides by -1 (inequality reverses):(a² -4a +4)/(4a) ≤0Note that denominator 4a is positive since a ≥1, so the sign is determined by the numerator:a² -4a +4 ≤0But a² -4a +4 = (a -2)^2. So (a -2)^2 ≤0But a square is always non-negative, so equality holds only when a=2.Thus, unless a=2, the inequality (a -2)^2 ≤0 is not satisfied. Therefore, the only case when g(x_c) ≥0 is when a=2, which gives (a -2)^2=0, so g(x_c)=0. If a≠2, then g(x_c) <0, which would violate the condition that g(x) ≥0 for all x in [-1,1]. Therefore, the only possible value of a is 2.Wait, that seems to be the case. Let me verify.If a=2, then b=2 -a=0. So the function is f(x)=2x² -1. Let's check if this satisfies |f(x)| ≤1 on [-1,1].At x=0: f(0)=-1. Okay.At x=1: f(1)=2*1 -1=1. Good.At x=-1: f(-1)=2*1 -1=1. So f(-1)=1. So between x=-1 and 1, the function is 2x² -1. Since x² is between 0 and1, so 2x² is between 0 and2, so 2x² -1 is between -1 and1. Hence, |f(x)| ≤1. Perfect. So a=2 is valid.But let's check another a, say a=1. Then b=2 -1=1. So f(x)=x² +x -1. Let's check if this satisfies |f(x)| ≤1 on [-1,1].At x=1: 1 +1 -1=1. Okay.At x=-1: 1 -1 -1=-1. Okay.At x=0: -1. Okay.What about x=0.5: f(0.5)=0.25 +0.5 -1= -0.25. That's within [-1,1].What about x=0.75: f(0.75)=0.5625 +0.75 -1=0.3125. Still within.Wait, but we need to check the maximum and minimum. Let's compute the critical point. The derivative is 2a x +b=2*1*x +1=2x +1. Setting to zero: 2x +1=0 => x=-0.5. So critical point at x=-0.5. Compute f(-0.5)= (-0.5)^2 + (-0.5) -1=0.25 -0.5 -1= -1.25. Which is less than -1. That's a problem! So |f(-0.5)|=1.25 >1, which violates the condition. Therefore, a=1 is invalid.So even though a=1 satisfies the endpoint conditions and the derivative critical point at x=-0.5 gives a value outside the allowed range, so a=1 is invalid. Therefore, the only possible value is a=2.Hence, the answer for part (2) is a=2.But let's see if there are other possible a's. Wait, according to the earlier analysis, only a=2 works because otherwise, the critical point gives a negative value, which violates the lower bound. Therefore, a must be 2. Therefore, the answer is 2.Final Answer(1) boxed{1}(2) boxed{2}</think>Given the function ( f(x) = ax^2 + bx + c ) where ( a, b, c in mathbf{R} ) and ( |f(x)| leqslant 1 ) for ( x in [-1, 1] ):1. Prove ( |b| leqslant 1 ): - Evaluate ( f(1) ) and ( f(-1) ): [ |f(1)| = |a + b + c| leqslant 1 ] [ |f(-1)| = |a - b + c| leqslant 1 ] - Subtract these two inequalities: [ |f(1) - f(-1)| = |2b| leqslant |f(1)| + |f(-1)| leqslant 2 ] - Dividing by 2 gives: [ |b| leqslant 1 ] - Hence, ( |b| leqslant 1 ) is proven.2. Find the value of ( a ) given ( f(0) = -1 ) and ( f(1) = 1 ): - From ( f(0) = -1 ), we get ( c = -1 ). - From ( f(1) = 1 ), we get ( a + b + c = 1 ). Substituting ( c = -1 ): [ a + b - 1 = 1 implies a + b = 2 ] - The function becomes ( f(x) = ax^2 + (2 - a)x - 1 ). - We need ( |ax^2 + (2 - a)x - 1| leqslant 1 ) for ( x in [-1, 1] ). - Check the critical point ( x = frac{a - 2}{2a} ): - For ( a in [1, 2] ), the critical point lies within ([-1, 1]). - Evaluate ( g(x) = ax^2 + (2 - a)x ) at critical point: [ gleft(frac{a - 2}{2a}right) = -frac{(a - 2)^2}{4a} ] - For ( g(x) geq 0 ), we need ( -frac{(a - 2)^2}{4a} geq 0 ), which is only true if ( a = 2 ). - Verify ( a = 2 ): - Function becomes ( f(x) = 2x^2 - 1 ). - Check ( |2x^2 - 1| leqslant 1 ) for ( x in [-1, 1] ), which holds true.Thus, the value of ( a ) is (boxed{2}). Final Answer(1) (boxed{1})(2) (boxed{2}),<think>Okay, so I have this problem about a quadratic function f(x) = ax² + bx + c. The conditions are that when x is between -1 and 1, the absolute value of f(x) is less than or equal to 1. There are two parts: first, I need to prove that the absolute value of b is less than or equal to 1. Second, given that f(0) = -1 and f(1) = 1, I have to find the value of a. Let me start with part (1).First, part (1) says to prove |b| ≤ 1. Since f(x) is a quadratic function, and we know that |f(x)| ≤ 1 on the interval [-1, 1], maybe I can use some specific values of x to find bounds on the coefficients a, b, c. Since we need to find |b|, perhaps evaluating the function at certain points where the coefficient b is prominent would help. Let's think: if I plug in x = 0, then f(0) = c. So |c| ≤ 1. But that's not directly helpful for b. What if I consider the derivative? The derivative of f(x) is f’(x) = 2ax + b. At x = 0, the derivative is b. But does the derivative have any relation to the maximum or minimum values? Maybe, but since the function is bounded, maybe the derivative can't be too large? Hmm, not sure. Maybe another approach.Alternatively, maybe using the fact that for x in [-1,1], |f(x)| ≤1, so evaluating at x = 1 and x = -1. Let's see:f(1) = a + b + c, and |f(1)| ≤1.f(-1) = a - b + c, and |f(-1)| ≤1.So we have two inequalities:|a + b + c| ≤1,|a - b + c| ≤1.But how can I get b from these? If I subtract these two equations, maybe. Let's compute f(1) - f(-1) = (a + b + c) - (a - b + c) = 2b. So 2b = f(1) - f(-1). Therefore, |2b| = |f(1) - f(-1)| ≤ |f(1)| + |f(-1)| ≤ 1 + 1 = 2. Therefore, |2b| ≤2 implies |b| ≤1. That works! So part (1) is proved. Nice.Now part (2): If f(0) = -1, f(1) = 1, find the value of a. Let's note that f(0) = c = -1. So c is known: c = -1. Then f(1) = a + b + c = a + b -1 =1. Therefore, a + b = 2. So equation (1): a + b = 2.But we also know that |f(x)| ≤1 on [-1,1]. So we have to use this condition to find a. Let's recall that f(x) = ax² + bx -1, and for all x in [-1,1], |ax² + bx -1| ≤1. So the maximum of |ax² + bx -1| over x in [-1,1] is at most 1. So we need to find a such that this holds, given that a + b =2.But how to relate a and b here? Maybe we can express b in terms of a: since a + b =2, so b=2 -a. Therefore, f(x) = ax² + (2 -a)x -1.Now, we need to ensure that |ax² + (2 -a)x -1| ≤1 for all x in [-1,1]. So let's write f(x) as a(x² -x) +2x -1. Hmm, maybe that's not helpful. Alternatively, since f(x) is quadratic, perhaps the maximum occurs at one of the endpoints or at the vertex.The vertex of the parabola is at x = -B/(2A), where A = a and B = (2 -a). So the vertex is at x = -(2 -a)/(2a) = (a -2)/(2a). Let me compute where this vertex is. Depending on the value of a, the vertex could be inside or outside the interval [-1,1]. If the vertex is inside the interval, then the maximum or minimum of f(x) would be at the vertex. So maybe I need to check the maximum of |f(x)| at x = -1, x =1, and at the vertex if it's inside [-1,1].Alternatively, maybe use specific points. Let's check x =0: f(0) = -1, which is given. So that's already at the lower bound. Let's check x =1: f(1)=1, which is at the upper bound. So maybe the maximums occur at other points. Let's check x = -1: f(-1) = a*(-1)^2 + (2 -a)*(-1) -1 = a + (-2 +a) -1 = 2a -3. So f(-1) = 2a -3. But |f(-1)| ≤1. Therefore, |2a -3| ≤1. Therefore, -1 ≤2a -3 ≤1. Let's solve this inequality.First, 2a -3 ≥ -1: 2a ≥2 => a ≥1.Second, 2a -3 ≤1: 2a ≤4 => a ≤2.So from x=-1, we get that a must be between 1 and 2. So 1 ≤a ≤2.Also, since we have the vertex at x=(a -2)/(2a). Let's see if this is in [-1,1]. Let's compute (a -2)/(2a). Let's plug in a=1: (1 -2)/(2*1)= -1/2, which is in [-1,1]. For a=2: (2 -2)/(2*2)=0, which is also in the interval. So for a between 1 and 2, (a -2)/(2a) ranges from -1/2 to 0, so it is inside [-1,1]. Therefore, the vertex is inside the interval, so we need to check the value at the vertex as well.So the maximum of |f(x)| could be at x= -1, x=1, or at the vertex. We already know that f(1)=1 and f(-1)=2a -3. Also, f(0)=-1. Let's check the vertex.The value at the vertex x=(a-2)/(2a):First, compute f((a-2)/(2a)).Let me compute f(x) at x = (a-2)/(2a):f(x) = a x² + (2 -a)x -1.Let’s compute x² and x:x = (a -2)/(2a)x² = (a -2)^2/(4a²)So f(x) = a*(a -2)^2/(4a²) + (2 -a)*(a -2)/(2a) -1Simplify term by term:First term: a*(a² -4a +4)/(4a²) = (a² -4a +4)/(4a)Second term: (2 -a)*(a -2)/(2a) = (- (a -2)^2)/(2a)Third term: -1So f(x) = (a² -4a +4)/(4a) - (a -2)^2/(2a) -1Let’s factor (a -2)^2:= ( (a -2)^2 )/(4a) - 2*(a -2)^2/(4a) -1= (- (a -2)^2)/(4a) -1Therefore, f at the vertex is - (a -2)^2/(4a) -1So that's the value at the vertex. Let me write that as:f_vertex = - [(a - 2)^2 / (4a) + 1]But since |f(x)| ≤1 for all x in [-1,1], then |f_vertex| ≤1.Therefore,| - [(a -2)^2/(4a) +1 ] | ≤1Which implies:| (a -2)^2/(4a) +1 | ≤1But (a -2)^2/(4a) +1 is always positive? Let's check.Since a is between 1 and 2, as established earlier, so a is positive. (a -2)^2 is positive, and 4a is positive. So (a -2)^2/(4a) is non-negative. Then adding 1, so the expression inside the absolute value is positive. Therefore, the inequality simplifies to:(a -2)^2/(4a) +1 ≤1But (a -2)^2/(4a) +1 ≤1 implies (a -2)^2/(4a) ≤0. But since (a -2)^2 is non-negative and 4a is positive, this fraction is non-negative. The only way it can be ≤0 is if it is zero. Therefore:(a -2)^2/(4a) =0Which implies (a -2)^2=0 => a=2.Therefore, the only possible value of a is 2. Let me check this.If a=2, then b=2 -a=0. So the function is f(x)=2x² +0x -1=2x² -1.Check if |2x² -1| ≤1 for x in [-1,1].At x=0: 2*0 -1=-1, | -1 | =1.At x=1: 2*1 -1=1, |1|=1.At x=-1: 2*1 -1=1, same.The vertex is at x=0 (since b=0), so f(0)=-1. The maximum of 2x² -1 occurs at x=±1, which is 1. So indeed, |f(x)| ≤1 on [-1,1]. So a=2 is valid.But wait, the earlier step when we considered the vertex leading to a=2 seems like the only solution. However, let's check the other constraints. We had from x=-1: |2a -3| ≤1 leading to a between 1 and 2.But if a=2 is the only solution, then that's the answer. Let me see. Wait, but when a=2, we already satisfy the other conditions. If a=1, what happens?If a=1, then b=2 -1=1. So f(x)=x² +x -1.Check |f(x)| on [-1,1].At x=0: -1, okay.At x=1: 1 +1 -1=1, okay.At x=-1: 1 -1 -1=-1, okay.But the vertex is at x=(1 -2)/(2*1)= -1/2. Let's compute f(-1/2): (-1/2)^2 + (-1/2) -1 = 1/4 -1/2 -1 = -5/4. So |-5/4|=5/4>1. Which violates the condition. Therefore, a=1 is invalid. So even though a=1 satisfies the endpoint conditions, the function exceeds the bound at the vertex. Therefore, a must be 2. Therefore, the answer is a=2.Alternatively, maybe another approach. Since f(1)=1, f(0)=-1, f(-1)=2a -3. Also, since the vertex is inside [-1,1], we need to ensure that the value at the vertex doesn't exceed the bound. So we derived that the vertex value is - [(a -2)^2/(4a) +1]. The absolute value of this is [(a -2)^2/(4a) +1], and this must be ≤1. So [(a -2)^2/(4a) +1] ≤1 => (a -2)^2/(4a) ≤0. Which as before, only possible when a=2.Hence, the value of a is 2.Final Answer(1) boxed{|b| leq 1}(2) boxed{2}</think>Given the quadratic function ( f(x) = ax^2 + bx + c ) with the condition that ( |f(x)| leq 1 ) for ( x in [-1, 1] ):1. Prove ( |b| leq 1 ): - Evaluate ( f(1) ) and ( f(-1) ): [ |f(1)| = |a + b + c| leq 1 quad text{and} quad |f(-1)| = |a - b + c| leq 1 ] - Subtract these two inequalities: [ |f(1) - f(-1)| = |2b| leq |f(1)| + |f(-1)| leq 2 ] - Therefore, ( |2b| leq 2 ) implies ( |b| leq 1 ).2. Find the value of ( a ) given ( f(0) = -1 ) and ( f(1) = 1 ): - From ( f(0) = -1 ), we get ( c = -1 ). - From ( f(1) = 1 ), we get ( a + b - 1 = 1 ) leading to ( a + b = 2 ). Thus, ( b = 2 - a ). - The function becomes ( f(x) = ax^2 + (2 - a)x - 1 ). - Check the value at ( x = -1 ): [ f(-1) = a + (-2 + a) - 1 = 2a - 3 ] This must satisfy ( |2a - 3| leq 1 ), leading to ( 1 leq a leq 2 ). - Check the vertex at ( x = frac{a - 2}{2a} ): - The value at the vertex is: [ fleft( frac{a - 2}{2a} right) = -frac{(a - 2)^2}{4a} - 1 ] - This must satisfy ( left| -frac{(a - 2)^2}{4a} - 1 right| leq 1 ), leading to ( frac{(a - 2)^2}{4a} + 1 leq 1 ), which simplifies to ( frac{(a - 2)^2}{4a} leq 0 ). The only solution is ( a = 2 ).Thus, the answers are:(1) (boxed{|b| leq 1})(2) (boxed{2})]

🔑:text { Three, 13. (1) Since } f(1)=a+b+c, f(-1)=a-b+c, then b=frac{1}{2}(f(1)-f(-1)).By the problem, |f(1)| leqslant 1,|f(-1)| leqslant 1.Thus, |b|=frac{1}{2}|f(1)-f(-1)|leqslant frac{1}{2}(|f(1)|+|f(-1)|) leqslant 1 text {. }(2) From f(0)=-1, f(1)=1, we get c=-1, b=2 boldsymbol{a}. Thenbegin{array}{l}f(x)=a x^{2}+(2-a) x-1 =aleft(x-frac{a-2}{2 a}right)^{2}-frac{(a-2)^{2}}{4 a}-1 .end{array}From |f(-1)| leqslant 1, we get |2 a-3| leqslant 1, i.e., 1 leqslant a leqslant 2.Therefore, frac{a-2}{2 a}=frac{1}{2}-frac{1}{a} in[-1,1].Thus, left|fleft(frac{a-2}{2 a}right)right| leqslant 1.begin{array}{l}text { Hence }left|frac{(a-2)^{2}}{4 a}+1right| leqslant 1 . text { Also, } frac{(a-2)^{2}}{4 a} geqslant 0, frac{(a-2)^{2}}{4 a}+1 geqslant 1, text { then } frac{(a-2)^{2}}{4 a}+1=1, text { i.e., } a=2 .end{array}

🧠:[<think>Okay, let's see. I need to find the value of tan A given that equation involving cos A and sin A. The equation is (sqrt(3) cos A + sin A) divided by (sqrt(3) sin A - cos A) equals tan(-7/12 pi). Hmm. Alright, let's start by understanding each part step by step.First, the left-hand side (LHS) of the equation is a fraction where both the numerator and denominator are linear combinations of cos A and sin A. The right-hand side (RHS) is a tangent of some angle, specifically -7/12 pi. Since tangent has a period of pi, tan(-7pi/12) is the same as tan(pi - 7pi/12) = tan(5pi/12) but negative. Wait, actually, tan is periodic with period pi, so tan(theta + pi) = tan theta. But here it's negative, so let's compute tan(-7pi/12). Since tangent is an odd function, tan(-x) = -tan x, so this is equal to -tan(7pi/12). So maybe I should compute tan(7pi/12) first.But 7pi/12 is equal to pi/3 + pi/4. Because pi/3 is 4pi/12, pi/4 is 3pi/12, so together 7pi/12. Therefore, tan(7pi/12) = tan(pi/3 + pi/4). Using the tangent addition formula: tan(a + b) = (tan a + tan b)/(1 - tan a tan b). So, tan(pi/3 + pi/4) = (tan(pi/3) + tan(pi/4))/(1 - tan(pi/3)tan(pi/4)). Let me compute that.tan(pi/3) is sqrt(3), tan(pi/4) is 1. So, numerator: sqrt(3) + 1, denominator: 1 - sqrt(3)*1 = 1 - sqrt(3). Therefore, tan(7pi/12) = (sqrt(3) + 1)/(1 - sqrt(3)). To rationalize the denominator, multiply numerator and denominator by (1 + sqrt(3)):[(sqrt(3) + 1)(1 + sqrt(3))]/[(1 - sqrt(3))(1 + sqrt(3))]Calculate numerator: sqrt(3)*1 + sqrt(3)*sqrt(3) + 1*1 + 1*sqrt(3) = sqrt(3) + 3 + 1 + sqrt(3) = 4 + 2 sqrt(3)Denominator: 1 - (sqrt(3))^2 = 1 - 3 = -2Therefore, tan(7pi/12) = (4 + 2 sqrt(3))/(-2) = - (4 + 2 sqrt(3))/2 = - (2 + sqrt(3))Therefore, tan(-7pi/12) = - tan(7pi/12) = -(- (2 + sqrt(3))) = 2 + sqrt(3). So the RHS is 2 + sqrt(3). Therefore, the equation simplifies to:[sqrt(3) cos A + sin A]/[sqrt(3) sin A - cos A] = 2 + sqrt(3)Now, I need to solve for tan A. Let's denote t = tan A. Then, since tan A = sin A / cos A, perhaps I can express the numerator and denominator in terms of t.Let me write the numerator and denominator:Numerator: sqrt(3) cos A + sin ADenominator: sqrt(3) sin A - cos ADivide numerator and denominator by cos A (assuming cos A ≠ 0, which I think is safe here because if cos A were zero, then sin A would be ±1, but plugging into the original equation would probably not satisfy the RHS):Numerator becomes: sqrt(3) + tan ADenominator becomes: sqrt(3) tan A - 1Therefore, the entire left-hand side becomes [sqrt(3) + t]/[sqrt(3) t - 1], where t = tan A.So we have the equation:[sqrt(3) + t]/[sqrt(3) t - 1] = 2 + sqrt(3)Now, solve for t. Let's cross-multiply:sqrt(3) + t = (2 + sqrt(3))(sqrt(3) t - 1)Expand the right-hand side:First, multiply 2 by (sqrt(3) t - 1): 2 sqrt(3) t - 2Then multiply sqrt(3) by (sqrt(3) t - 1): sqrt(3)*sqrt(3) t - sqrt(3)*1 = 3 t - sqrt(3)So altogether, RHS is (2 sqrt(3) t - 2) + (3 t - sqrt(3)) = 2 sqrt(3) t + 3 t - 2 - sqrt(3)Therefore, equation becomes:sqrt(3) + t = (2 sqrt(3) + 3) t - (2 + sqrt(3))Bring all terms to the left side:sqrt(3) + t - (2 sqrt(3) + 3) t + (2 + sqrt(3)) = 0Combine like terms:sqrt(3) + 2 + sqrt(3) + t - (2 sqrt(3) + 3) t = 0Simplify constants: sqrt(3) + sqrt(3) + 2 = 2 sqrt(3) + 2Simplify t terms: t - (2 sqrt(3) + 3) t = t(1 - 2 sqrt(3) - 3) = t(-2 sqrt(3) - 2)So equation becomes:2 sqrt(3) + 2 + t(-2 sqrt(3) - 2) = 0Factor out 2 from the constants and the t terms:2(sqrt(3) + 1) - 2 t(sqrt(3) + 1) = 0Factor out 2(sqrt(3) + 1):2(sqrt(3) + 1)(1 - t) = 0Since sqrt(3) + 1 is not zero (sqrt(3) is about 1.732, so sqrt(3)+1 is about 2.732), we can divide both sides by 2(sqrt(3) + 1):1 - t = 0Thus, t = 1Wait, so tan A = 1? Let me check if that makes sense. If tan A is 1, then A is pi/4 or 45 degrees. Let me substitute t = 1 back into the original equation to verify.Original equation: [sqrt(3) cos A + sin A]/[sqrt(3) sin A - cos A] = 2 + sqrt(3)If tan A = 1, then sin A = cos A = sqrt(2)/2 (assuming A is in the first quadrant, but even if not, the ratio might still hold).Compute numerator: sqrt(3)*sqrt(2)/2 + sqrt(2)/2 = sqrt(2)/2 (sqrt(3) + 1)Denominator: sqrt(3)*sqrt(2)/2 - sqrt(2)/2 = sqrt(2)/2 (sqrt(3) - 1)Therefore, the ratio is [sqrt(3) + 1]/[sqrt(3) - 1]. Multiply numerator and denominator by [sqrt(3) + 1]:Numerator: (sqrt(3) + 1)^2 = 3 + 2 sqrt(3) + 1 = 4 + 2 sqrt(3)Denominator: (sqrt(3))^2 - 1^2 = 3 - 1 = 2So ratio is (4 + 2 sqrt(3))/2 = 2 + sqrt(3), which matches the RHS. Therefore, t = 1 is indeed a solution. So tan A = 1.But wait, is that the only solution? Let's check if there are any restrictions. When I divided by cos A earlier, I assumed that cos A ≠ 0, which is valid unless A is pi/2 or 3pi/2, but in those cases, the original equation would involve division by zero in the denominator sqrt(3) sin A - cos A. For A = pi/2, sin A = 1, cos A = 0, denominator becomes sqrt(3)*1 - 0 = sqrt(3), numerator is sqrt(3)*0 + 1 = 1, so ratio is 1/sqrt(3) ≈ 0.577, which is not equal to 2 + sqrt(3) ≈ 3.732. Similarly, for A = 3pi/2, sin A = -1, cos A = 0, numerator is sqrt(3)*0 + (-1) = -1, denominator is sqrt(3)*(-1) - 0 = -sqrt(3), ratio is (-1)/(-sqrt(3)) = 1/sqrt(3) ≈ 0.577, still not equal to RHS. So no, A can't be pi/2 or 3pi/2, so our solution tan A = 1 is valid. Therefore, the answer is 1.Wait, but let me make sure that there isn't another solution. When we solved the equation step by step, we arrived at t = 1. So unless there was a mistake in the algebra steps, that should be the only solution. Let's retrace the steps.Starting from [sqrt(3) + t]/[sqrt(3) t - 1] = 2 + sqrt(3)Cross-multiplying gives sqrt(3) + t = (2 + sqrt(3))(sqrt(3) t - 1)Expanding the RHS: 2*sqrt(3) t - 2*1 + sqrt(3)*sqrt(3) t - sqrt(3)*1 = 2 sqrt(3) t - 2 + 3 t - sqrt(3)So RHS is (2 sqrt(3) t + 3 t) - (2 + sqrt(3)) = t(2 sqrt(3) + 3) - (2 + sqrt(3))Then moving all terms to the left: sqrt(3) + t - t(2 sqrt(3) + 3) + 2 + sqrt(3) = 0Which simplifies to sqrt(3) + 2 + sqrt(3) + t(1 - 2 sqrt(3) - 3) = 0Which is 2 sqrt(3) + 2 + t(-2 - 2 sqrt(3)) = 0Factor out 2(sqrt(3) + 1) - 2 t(sqrt(3) + 1) = 0Then 2(sqrt(3) + 1)(1 - t) = 0, leading to t = 1.Yes, seems correct. So tan A = 1. So the answer is 1. Therefore, the final answer is boxed{1}Final Answerboxed{1}</think>Given the equation in triangle (ABC):[frac{sqrt{3}cos A + sin A}{sqrt{3}sin A - cos A} = tan left(-frac{7}{12}piright)]First, we simplify the right-hand side (RHS). Using the properties of the tangent function, we know that (tan(-theta) = -tan(theta)). Therefore, (tanleft(-frac{7}{12}piright) = -tanleft(frac{7}{12}piright)). Next, we use the angle addition formula for tangent. Since (frac{7}{12}pi = frac{pi}{3} + frac{pi}{4}), we have:[tanleft(frac{7}{12}piright) = tanleft(frac{pi}{3} + frac{pi}{4}right) = frac{tanfrac{pi}{3} + tanfrac{pi}{4}}{1 - tanfrac{pi}{3}tanfrac{pi}{4}} = frac{sqrt{3} + 1}{1 - sqrt{3}}]Rationalizing the denominator:[frac{sqrt{3} + 1}{1 - sqrt{3}} cdot frac{1 + sqrt{3}}{1 + sqrt{3}} = frac{(sqrt{3} + 1)(1 + sqrt{3})}{1 - 3} = frac{4 + 2sqrt{3}}{-2} = - (2 + sqrt{3})]Thus, (tanleft(-frac{7}{12}piright) = 2 + sqrt{3}).Now, we equate the left-hand side (LHS) to (2 + sqrt{3}):[frac{sqrt{3}cos A + sin A}{sqrt{3}sin A - cos A} = 2 + sqrt{3}]Let (t = tan A). Dividing the numerator and denominator by (cos A):[frac{sqrt{3} + t}{sqrt{3}t - 1} = 2 + sqrt{3}]Cross-multiplying gives:[sqrt{3} + t = (2 + sqrt{3})(sqrt{3}t - 1)]Expanding the right-hand side:[sqrt{3} + t = 2sqrt{3}t - 2 + 3t - sqrt{3}]Combining like terms:[sqrt{3} + t = (2sqrt{3}t + 3t) - (2 + sqrt{3})]Rearranging all terms to the left-hand side:[sqrt{3} + t - 2sqrt{3}t - 3t + 2 + sqrt{3} = 0]Simplifying:[2sqrt{3} + 2 + t(1 - 2sqrt{3} - 3) = 0][2(sqrt{3} + 1) - 2t(sqrt{3} + 1) = 0]Factoring out (2(sqrt{3} + 1)):[2(sqrt{3} + 1)(1 - t) = 0]Since (sqrt{3} + 1 neq 0), we have (1 - t = 0), leading to (t = 1). Therefore, (tan A = 1).The final answer is (boxed{1}).,<think>Okay, so I need to find the value of tan A given this equation involving trigonometric functions. Let me start by writing down the problem again to make sure I have it right:In triangle ABC, the three interior angles are A, B, and C. The equation given is:(√3 cos A + sin A) / (√3 sin A - cos A) = tan(-7/12 π)We need to find tan A.First, let me recall that tan(-θ) is equal to -tan θ, so tan(-7π/12) is the same as -tan(7π/12). That might be useful later.Now, looking at the left side of the equation: (√3 cos A + sin A) divided by (√3 sin A - cos A). Hmm, this looks a bit complicated, but maybe I can simplify it by expressing the numerator and denominator in terms of sine and cosine of some angle. Let's see.Let me consider the numerator: √3 cos A + sin A. If I factor out a 2, perhaps? Let's check:√3 cos A + sin A = 2*( (√3/2 cos A) + (1/2 sin A) )Hmm, √3/2 and 1/2 are familiar. Those are the cosine and sine of π/6, right? Because cos(π/6) = √3/2 and sin(π/6) = 1/2. So maybe I can write this as:2*(cos(π/6) cos A + sin(π/6) sin A) = 2*cos(A - π/6)Using the cosine addition formula: cos(A - B) = cos A cos B + sin A sin B. Wait, but here it's cos(π/6) cos A + sin(π/6) sin A, which is cos(A - π/6). So the numerator simplifies to 2 cos(A - π/6).Similarly, the denominator: √3 sin A - cos A. Let me try the same approach. Factor out a 2:√3 sin A - cos A = 2*( (√3/2 sin A) - (1/2 cos A) )Again, √3/2 is cos(π/6) and 1/2 is sin(π/6). But here we have sin A multiplied by cos(π/6) and cos A multiplied by sin(π/6), but with a negative sign. Let's see:2*( sin A cos(π/6) - cos A sin(π/6) ) = 2*sin(A - π/6)Using the sine subtraction formula: sin(A - B) = sin A cos B - cos A sin B. So yes, the denominator simplifies to 2 sin(A - π/6).Therefore, the left side of the equation becomes:[2 cos(A - π/6)] / [2 sin(A - π/6)] = cos(A - π/6)/sin(A - π/6) = cot(A - π/6)But cotangent is the reciprocal of tangent, so cot(A - π/6) = 1/tan(A - π/6). Therefore, the left side is equal to cot(A - π/6), and the equation becomes:cot(A - π/6) = tan(-7π/12)But tan(-7π/12) is equal to -tan(7π/12). Let me write that:cot(A - π/6) = -tan(7π/12)Now, cot θ is equal to tan(π/2 - θ). So:tan(π/2 - (A - π/6)) = -tan(7π/12)Simplify the left side:tan(π/2 - A + π/6) = tan(2π/3 - A) = -tan(7π/12)So:tan(2π/3 - A) = -tan(7π/12)But tan is periodic with period π, so tan(θ) = tan(θ + kπ) for any integer k. Also, tan(-θ) = -tan θ. So maybe I can write:tan(2π/3 - A) = tan(-7π/12 + kπ)But since tangent is periodic with period π, the general solution would be:2π/3 - A = -7π/12 + kπSolving for A:A = 2π/3 + 7π/12 - kπConvert 2π/3 to twelfths: 2π/3 = 8π/12, so:A = 8π/12 + 7π/12 - kπ = 15π/12 - kπ = 5π/4 - kπBut since A is an angle in a triangle, it must be between 0 and π. Let's check possible integer values for k.If k = 0, then A = 5π/4 ≈ 225 degrees, which is more than 180 degrees. Not possible in a triangle.If k = 1, then A = 5π/4 - π = π/4 ≈ 45 degrees. That's valid.If k = 2, then A = 5π/4 - 2π = -3π/4, which is negative. Not valid.So the only valid solution is A = π/4. Therefore, tan A = tan(π/4) = 1. But wait, hold on. Let me check this.Wait, if A = π/4, then let's verify the original equation:Left side: (√3 cos(π/4) + sin(π/4)) / (√3 sin(π/4) - cos(π/4))Compute cos(π/4) and sin(π/4): both are √2/2.So numerator: √3*(√2/2) + √2/2 = √2/2 (√3 + 1)Denominator: √3*(√2/2) - √2/2 = √2/2 (√3 - 1)So left side: [√2/2 (√3 + 1)] / [√2/2 (√3 - 1)] = (√3 + 1)/(√3 - 1)Multiply numerator and denominator by (√3 + 1):[(√3 + 1)^2] / [(√3)^2 - 1^2] = (3 + 2√3 + 1)/(3 - 1) = (4 + 2√3)/2 = 2 + √3Right side: tan(-7π/12) = -tan(7π/12). Let's compute tan(7π/12). 7π/12 is 105 degrees. tan(105 degrees) = tan(60 + 45) degrees.Using tan(A + B) = (tan A + tan B)/(1 - tan A tan B)tan(60° + 45°) = (tan 60° + tan 45°)/(1 - tan 60° tan 45°) = (√3 + 1)/(1 - √3*1) = (√3 + 1)/(1 - √3)Multiply numerator and denominator by (1 + √3):[(√3 + 1)(1 + √3)] / [(1)^2 - (√3)^2] = [ (√3*1 + √3*√3 + 1*1 + 1*√3) ] / (1 - 3) = [ (√3 + 3 + 1 + √3) ] / (-2) = (4 + 2√3)/(-2) = -2 - √3Therefore, tan(7π/12) = -2 - √3, so tan(-7π/12) = 2 + √3.Wait, but the left side was (√3 + 1)/(√3 - 1) = 2 + √3, and the right side is tan(-7π/12) = 2 + √3. So they are equal. So A = π/4 is correct. Therefore, tan A = 1.Wait, but this contradicts my previous thought where I thought A = π/4. But according to the equation, yes, if A is π/4, then tan A = 1. But let me check the steps again.Wait, when I solved for A, I found A = 5π/4 - kπ. Then for k=1, A= π/4. That seems correct.But let me think again: If A is π/4, then the other angles B and C must sum to 3π/4. But there's no restriction given in the problem about the other angles, so it's possible. The equation only relates to angle A, so perhaps tan A is 1.But let me verify once more. Let's compute tan A if A=π/4. tan(π/4)=1, which would be the answer. However, let me check another approach to see if I get the same result.Alternative approach: Let me set t = tan A. Then, express the left side in terms of t. Let's see.Given (√3 cos A + sin A)/(√3 sin A - cos A). Let me divide numerator and denominator by cos A:[√3 + tan A] / [√3 tan A - 1]So that expression becomes (√3 + t)/(√3 t - 1). According to the problem, this equals tan(-7π/12) which we found earlier is 2 + √3.So set (√3 + t)/(√3 t - 1) = 2 + √3Solve for t:Multiply both sides by (√3 t - 1):√3 + t = (2 + √3)(√3 t - 1)Expand the right side:(2 + √3)(√3 t) - (2 + √3)(1)= 2√3 t + 3 t - 2 - √3So the equation becomes:√3 + t = 2√3 t + 3 t - 2 - √3Bring all terms to the left side:√3 + t - 2√3 t - 3 t + 2 + √3 = 0Combine like terms:(√3 + √3) + (t - 2√3 t - 3 t) + 2 = 02√3 + (1 - 2√3 - 3)t + 2 = 0Simplify coefficients:2√3 + 2 + (-2 - 2√3)t = 0Wait, let's check the coefficients step by step:Original left side after expansion:√3 + t - 2√3 t - 3 t + 2 + √3Combine constants: √3 + √3 + 2 = 2√3 + 2Combine t terms: t - 2√3 t - 3 t = t(1 - 2√3 - 3) = t(-2 - 2√3)Therefore, equation becomes:2√3 + 2 + (-2 - 2√3)t = 0Bring constants to the right side:(-2 - 2√3)t = - (2√3 + 2)Factor out -2 on both sides:-2(1 + √3)t = -2(√3 + 1)Divide both sides by -2(1 + √3):t = (-2(√3 + 1)) / (-2(1 + √3)) = (√3 + 1)/(1 + √3) = 1Because (√3 + 1)/(1 + √3) is 1, since numerator and denominator are the same.Therefore, t = 1. So tan A = 1. Therefore, the answer is 1.Wait, so both methods confirm that tan A = 1. But let me just make sure there isn't a mistake in my alternative approach. Because when I solved for A earlier, I got A = π/4, which indeed gives tan A = 1. And when I substituted into the original equation, both sides equal 2 + √3, which matches. So that's consistent.Therefore, the answer should be 1, which is tan(π/4). So I think that is correct. But let me just check if there's any other possible solution for A.In the equation tan(2π/3 - A) = -tan(7π/12), we considered the principal solution where 2π/3 - A = -7π/12 + kπ. But let me check if there are other angles in the range of a triangle (0 to π) that could satisfy this.We found A = 5π/4 - kπ. For k=1, A=π/4. For k=0, A=5π/4 (invalid). For k=2, A=5π/4 - 2π= -3π/4 (invalid). So only A=π/4 is valid. Therefore, tan A=1.Hence, the answer is 1.Final AnswerThe value of tan A is boxed{1}.</think>Given the equation in triangle (ABC):[frac{sqrt{3}cos A + sin A}{sqrt{3}sin A - cos A} = tan left(-frac{7}{12}piright)]First, we simplify the left-hand side. We recognize that the numerator (sqrt{3}cos A + sin A) and the denominator (sqrt{3}sin A - cos A) can be expressed using trigonometric identities. Factoring out 2 from both the numerator and the denominator, we get:[sqrt{3}cos A + sin A = 2 left( cosleft(A - frac{pi}{6}right) right)][sqrt{3}sin A - cos A = 2 left( sinleft(A - frac{pi}{6}right) right)]Thus, the left-hand side simplifies to:[frac{2 cosleft(A - frac{pi}{6}right)}{2 sinleft(A - frac{pi}{6}right)} = cotleft(A - frac{pi}{6}right)]We know that (cot theta = tanleft(frac{pi}{2} - thetaright)), so:[cotleft(A - frac{pi}{6}right) = tanleft(frac{pi}{2} - left(A - frac{pi}{6}right)right) = tanleft(frac{2pi}{3} - Aright)]Given the right-hand side is (tanleft(-frac{7pi}{12}right)), we use the identity (tan(-theta) = -tan(theta)):[tanleft(-frac{7pi}{12}right) = -tanleft(frac{7pi}{12}right)]Equating both sides, we have:[tanleft(frac{2pi}{3} - Aright) = -tanleft(frac{7pi}{12}right)]Using the periodicity of tangent, we get:[frac{2pi}{3} - A = -frac{7pi}{12} + kpi]Solving for (A):[A = frac{2pi}{3} + frac{7pi}{12} - kpi = frac{15pi}{12} - kpi = frac{5pi}{4} - kpi]Since (A) must be an angle in a triangle (between 0 and (pi)), we check valid values for (k):- For (k = 1), (A = frac{5pi}{4} - pi = frac{pi}{4}), which is valid.- Other values of (k) result in angles outside the valid range.Thus, (A = frac{pi}{4}), and (tan A = tanleft(frac{pi}{4}right) = 1).The value of (tan A) is (boxed{1}).]

🔑:AnalysisThis question mainly tests the application of basic trigonometric relationships for the same angle, the formula for the tangent of the sum of two angles, and the properties of the tangent function graph. It is a basic question.By applying the basic trigonometric relationships for the same angle, we can obtain frac{sqrt{3}+tan A}{1-sqrt{3}tan A}=tan frac{7}{12}pi. Using the formula for the tangent of the sum of two angles, we can obtain tan left(A+ frac{pi}{3} right)=tan frac{7pi}{12}. Considering the range of angle A, we can solve for A and thus find the value of tan A.SolutionGiven: frac{sqrt{3}cos A+sin A}{sqrt{3}sin A-cos A}=tan left(-frac{7}{12}piright),Rightarrow frac{sqrt{3}+tan A}{1-sqrt{3}tan A}=tan frac{7}{12}pi,Rightarrow tan left(A+ frac{pi}{3} right)=tan frac{7pi}{12},Since frac{pi}{3} < A+ frac{pi}{3} < frac{4pi}{3},We can deduce: A+ frac{pi}{3} = frac{7pi}{12}, solving for A gives A= frac{pi}{4},Therefore, tan A=1.Hence, the answer is boxed{1}.

🧠:[<think>Okay, so I have this problem here where points A, B, and C are on a line l. The vectors OA, OB, and OC satisfy the equation OA = [f(x) + 2f'(1)x] OB - ln x OC. I need to find the expression for the function y = f(x). Hmm, let me think about how to approach this.First, since points A, B, and C are on the same line l, their position vectors OA, OB, and OC must be collinear. That means they can be expressed as scalar multiples of each other. So maybe I can express OA in terms of OB and OC with some coefficients, and then compare that to the given equation to find f(x).The given equation is OA = [f(x) + 2f'(1)x] OB - ln x OC. Since OA, OB, and OC are collinear, their coefficients must satisfy some relationship. Let me recall that if three vectors are collinear, then any one of them can be expressed as a linear combination of the other two with coefficients that sum to 1. Wait, is that always true? Let me think.Suppose points A, B, C are collinear. Then, for any point on the line, say A, the vector OA can be written as a linear combination of OB and OC such that the coefficients add up to 1. That is, OA = λ OB + (1 - λ) OC for some scalar λ. But in the given equation, OA is expressed as [f(x) + 2f'(1)x] OB - ln x OC. So if I follow the collinearity condition, the coefficients of OB and OC should add up to 1. Therefore:[f(x) + 2f'(1)x] + (-ln x) = 1Is that right? Let me verify. If OA = λ OB + μ OC and points A, B, C are collinear, then λ + μ = 1. Yes, that seems correct. Because in affine combinations, the coefficients must sum to 1 for the combination to lie on the line through B and C. So in this case, the coefficients of OB and OC in the expression for OA must sum to 1. Therefore:f(x) + 2f'(1)x - ln x = 1So that's an equation involving f(x) and f'(1). But f'(1) is the derivative of f at x=1. Wait, so we need another equation to solve for f(x). Since we have f'(1), maybe we can take the derivative of the equation with respect to x and then evaluate at x=1 to get an equation involving f'(1).Let's try that. First, write down the equation:f(x) + 2f'(1)x - ln x = 1Take the derivative of both sides with respect to x:f'(x) + 2f'(1) - (1/x) = 0So:f'(x) = (1/x) - 2f'(1)Now, let's evaluate this derivative at x = 1:f'(1) = (1/1) - 2f'(1) => f'(1) = 1 - 2f'(1)Bring the 2f'(1) to the left side:f'(1) + 2f'(1) = 1 => 3f'(1) = 1 => f'(1) = 1/3Okay, so now we know that f'(1) = 1/3. Plugging this back into the original equation:f(x) + 2*(1/3)*x - ln x = 1Simplify:f(x) + (2/3)x - ln x = 1Therefore, solving for f(x):f(x) = 1 - (2/3)x + ln xWait, is that all? Let me check.So starting from the collinearity condition, we set the sum of coefficients equal to 1, got an equation involving f(x) and f'(1). Then we took the derivative to involve f'(x), evaluated at x=1 to find f'(1), and substituted back into the original equation to solve for f(x). That seems logical.Let me verify by plugging f(x) back into the original vector equation to see if everything holds.Given OA = [f(x) + 2f'(1)x] OB - ln x OCIf f(x) = 1 - (2/3)x + ln x, then f'(x) is derivative of that, which is -2/3 + 1/x. Then f'(1) = -2/3 + 1/1 = 1 - 2/3 = 1/3, which matches our previous result.Now, substitute f(x) and f'(1) into the expression [f(x) + 2f'(1)x]:[1 - (2/3)x + ln x + 2*(1/3)*x] = 1 - (2/3)x + ln x + (2/3)x = 1 + ln xSo OA = (1 + ln x) OB - ln x OCThen, the coefficients of OB and OC are (1 + ln x) and (-ln x), which sum to (1 + ln x) + (-ln x) = 1. So that satisfies the collinearity condition, since the coefficients add up to 1. Therefore, OA is an affine combination of OB and OC, meaning that point A lies on the line through B and C. That checks out.Therefore, the function f(x) is indeed 1 - (2/3)x + ln x. Wait, but let me check again if there are any other constraints or possibilities.Wait, when we derived the equation f(x) + 2f'(1)x - ln x = 1, and then substituted f'(1) = 1/3, that gives f(x) = 1 - (2/3)x + ln x. So unless there's another condition, that should be the unique solution.Is there a reason to think that there's an integration constant missing? Let me see. When we took the derivative of f(x), we got f'(x) = 1/x - 2f'(1). Then we solved for f'(x) and found f'(1) = 1/3, which is a constant. Then integrating f'(x) gives:f(x) = ∫ [1/x - 2*(1/3)] dx + C = ln x - (2/3)x + CThen plugging back into the original equation f(x) + (2/3)x - ln x = 1 gives:[ln x - (2/3)x + C] + (2/3)x - ln x = 1Simplifies to C = 1. So actually, the constant of integration is 1. Therefore, f(x) = ln x - (2/3)x + 1. Which is the same as 1 - (2/3)x + ln x. So that's consistent.Therefore, the function f(x) is 1 - (2/3)x + ln x. That should be the answer. Let me just make sure there are no mistakes in the steps.1. Recognized that collinearity implies the coefficients sum to 1. Correct.2. Set up the equation f(x) + 2f'(1)x - ln x = 1. Correct.3. Took derivative to get f'(x) + 2f'(1) - 1/x = 0. Correct.4. Evaluated at x=1 to solve for f'(1) = 1/3. Correct.5. Substituted back into the equation and solved for f(x). Correct.6. Checked the solution by plugging back into the original vector equation and verified the coefficients sum to 1. Correct.7. Also verified via integration, considering the constant, which turned out to be 1. Correct.So all steps seem solid. Therefore, the answer should be f(x) = 1 - (2/3)x + ln x.Final AnswerThe function ( y = f(x) ) is (boxed{1 - dfrac{2}{3}x + ln x}).</think>Given that points (A), (B), and (C) are collinear on line (l), the vectors (overrightarrow{OA}), (overrightarrow{OB}), and (overrightarrow{OC}) must satisfy the collinearity condition. This means the coefficients of (overrightarrow{OB}) and (overrightarrow{OC}) in the expression for (overrightarrow{OA}) must sum to 1. The given equation is:[overrightarrow{OA} = [f(x) + 2f'(1)x] overrightarrow{OB} - ln x cdot overrightarrow{OC}]Since the coefficients must sum to 1, we have:[f(x) + 2f'(1)x - ln x = 1]Taking the derivative of both sides with respect to (x):[f'(x) + 2f'(1) - frac{1}{x} = 0]Evaluating this derivative at (x = 1):[f'(1) + 2f'(1) - 1 = 0 implies 3f'(1) = 1 implies f'(1) = frac{1}{3}]Substituting (f'(1) = frac{1}{3}) back into the original equation:[f(x) + 2 left(frac{1}{3}right)x - ln x = 1]Simplifying this equation:[f(x) + frac{2}{3}x - ln x = 1 implies f(x) = 1 - frac{2}{3}x + ln x]Verification shows that the coefficients of (overrightarrow{OB}) and (overrightarrow{OC}) sum to 1, confirming the solution is correct. Therefore, the function (y = f(x)) is:[boxed{1 - dfrac{2}{3}x + ln x}],<think>Okay, so I need to find the expression for the function y = f(x) given that points A, B, and C are on line l, and the vectors OA, OB, OC satisfy the equation OA = [f(x) + 2f'(1)x] OB - ln x * OC. Hmm, let me try to break this down step by step.First, since A, B, and C are points on the same line l, their position vectors OA, OB, and OC must be collinear. That means each vector can be expressed as a scalar multiple of the others. So, for instance, OA can be written as some scalar multiple of OB, and OC can also be written as a scalar multiple of OB. But the given equation already expresses OA in terms of OB and OC. Let me note that.The equation given is:OA = [f(x) + 2f'(1)x] OB - (ln x) OC.Since all three points are collinear, the coefficients of OB and OC in this linear combination must satisfy some relationship. Because if OA is on the same line as OB and OC, then the combination of OB and OC here must result in a vector that's also on that line. However, since OA, OB, and OC are all on line l, which is a straight line, any linear combination of OB and OC should also lie on line l only if the combination is affine, i.e., the coefficients add up to 1. But wait, in vector terms, if we have OA expressed as a linear combination of OB and OC, then for OA to lie on the line through B and C, the coefficients must satisfy certain conditions.Alternatively, perhaps since all three points are on the same line, the vectors OA, OB, OC can be related by scalar multiples. Let me think. Let's assume that the line l has a direction vector v. Then OA, OB, OC can be expressed as OA = a * v, OB = b * v, OC = c * v, where a, b, c are scalars. Then, substituting into the equation:a * v = [f(x) + 2f'(1)x] * b * v - (ln x) * c * v.Since v is non-zero (as it's a direction vector), we can divide both sides by v to get:a = [f(x) + 2f'(1)x] * b - (ln x) * c.But here, a, b, c are scalars that correspond to the positions of points A, B, C along the line l. However, the problem is that OA, OB, OC are position vectors from the origin to points on line l. So unless the origin is also on line l, these position vectors OA, OB, OC are not necessarily scalar multiples of each other. Wait, that complicates things. Because if the origin is not on line l, then OA, OB, OC are vectors from the origin to points on l, which is a line not passing through the origin. So in that case, OA, OB, OC are not collinear vectors. But the problem says that points A, B, C are on line l, but it doesn't say the origin is on line l. So maybe OA, OB, OC are not collinear vectors. Hmm.But then how can the equation OA = [f(x) + 2f'(1)x] OB - ln x * OC hold? Because OA is expressed as a linear combination of OB and OC. If OA, OB, OC are vectors in, say, 3D space (but the problem says they are on a line, so probably in 2D or 1D), but regardless, the equation is a vector equation. Since A, B, C are colinear, the vectors OA, OB, OC must lie on a line in space. But OA is a linear combination of OB and OC. For OA to lie on the same line as OB and OC, the coefficients in the linear combination must satisfy some condition. In general, for three points to be colinear, the vectors must satisfy OA = λ OB + (1 - λ) OC for some scalar λ, right? That's the affine combination. So perhaps in this case, the coefficients [f(x) + 2f'(1)x] and -ln x must add up to 1? Wait, let's check.If OA is an affine combination of OB and OC, then the coefficients would satisfy [f(x) + 2f'(1)x] + (-ln x) = 1. But I don't know if that's necessarily the case here. Alternatively, maybe since OA, OB, OC are colinear points, the vectors OA - OB, OA - OC, etc., are scalar multiples. Let me think. If points A, B, C are colinear, then vectors AB and AC are parallel. So AB = k AC for some scalar k. Expressing vectors AB and AC in terms of OA, OB, OC:AB = OB - OAAC = OC - OASo, (OB - OA) = k (OC - OA). Then:OB - OA = k OC - k OAOB = OA + k OC - k OAOB = (1 - k) OA + k OCBut in our original equation, OA is expressed in terms of OB and OC. So maybe this is similar. Alternatively, let's try expressing OA in terms of OB and OC. The given equation is:OA = [f(x) + 2f'(1)x] OB - (ln x) OC.If points A, B, C are colinear, then OA can be written as a linear combination of OB and OC such that the combination coefficients sum to 1? Wait, that would be if the origin is on the line l. If the origin is not on line l, then affine combinations in terms of OA, OB, OC would require coefficients summing to 1. But since OA is expressed as a combination of OB and OC, regardless of where the origin is, maybe the coefficients have to satisfy some relation.Alternatively, let's consider that since A, B, C are colinear, the vectors OA, OB, OC must satisfy the relation that OA - OB is parallel to OB - OC, or something like that. But maybe there's a better approach.Let me think in terms of coordinates. Suppose line l is a line in the plane or in 3D space, and O is the origin. Let's assume we are in 2D for simplicity. Let me parameterize the line l. Let me suppose that line l can be written as OP = t*d + p0, where d is the direction vector, p0 is a point on the line, and t is a scalar parameter. Then, points A, B, C correspond to different values of t. Let's say OA = t_A*d + p0, OB = t_B*d + p0, OC = t_C*d + p0.Given that OA is expressed as [f(x) + 2f'(1)x] OB - ln x * OC, substitute the parameterizations:t_A*d + p0 = [f(x) + 2f'(1)x] (t_B*d + p0) - ln x (t_C*d + p0).Let's expand the right-hand side:= [f(x) + 2f'(1)x] t_B*d + [f(x) + 2f'(1)x] p0 - ln x * t_C*d - ln x * p0Now, group the terms with d and the terms with p0:= [ [f(x) + 2f'(1)x] t_B - ln x * t_C ] d + [ [f(x) + 2f'(1)x] - ln x ] p0But the left-hand side is t_A*d + p0. Therefore, equating coefficients:For the d component:t_A = [f(x) + 2f'(1)x] t_B - ln x * t_CFor the p0 component:1 = [f(x) + 2f'(1)x] - ln xWait, because the left-hand side's p0 coefficient is 1 (since it's p0), and the right-hand side's p0 coefficient is [f(x) + 2f'(1)x] - ln x. Therefore:1 = [f(x) + 2f'(1)x] - ln xSo that gives us an equation:f(x) + 2f'(1)x - ln x = 1So, f(x) = 1 + ln x - 2f'(1)xHmm, interesting. So this gives an expression for f(x) in terms of f'(1), which is a constant. So we need to determine f'(1) as well.But first, let's make sure this is correct. Since we assumed that line l can be parameterized as OP = t*d + p0, then OA, OB, OC can be written in terms of t_A, t_B, t_C. Then, by equating the components, we got two equations. The p0 component gives us the equation involving f(x), and the d component gives another equation involving t_A, t_B, t_C. But since A, B, C are arbitrary points on line l (except that they are fixed once the vectors are given), the coefficients for d must also hold. However, unless t_A, t_B, t_C are known, we can't get more information from that equation. But perhaps t_A, t_B, t_C are constants, but in the problem statement, is x a parameter? Wait, the vectors OA, OB, OC are given in terms of x? Wait, the problem says "vectors OA, OB, OC satisfy OA = [f(x) + 2f'(1)x] OB - ln x * OC", so OA is expressed in terms of OB and OC with coefficients depending on x. So, x is a variable here, and f is a function of x. So, in the equation, OA is a vector that depends on x, and so f(x) is the function we need to find.But how can OA depend on x? Because OA is the position vector of point A on line l. If OA depends on x, that would mean that point A moves along line l as x changes. Similarly, but points B and C are fixed on line l, right? Because their position vectors OB and OC are fixed. Wait, but in the problem statement, it just says A, B, C are three points on line l. It doesn't specify if they move or are fixed. But the equation OA = [f(x) + 2f'(1)x] OB - ln x * OC suggests that OA is varying with x, so point A is moving along line l as x changes, while points B and C are fixed. Therefore, for each x, we have a different point A on line l, expressed in terms of the fixed points B and C.Therefore, the equation must hold for all x (in some domain). So, given that, the equation we derived from the p0 component is:1 = [f(x) + 2f'(1)x] - ln xWhich rearranged gives:f(x) = 1 + ln x - 2f'(1)xNow, this is a differential equation because f'(1) is the derivative of f at x=1. But f'(1) is a constant, not a function of x. Therefore, in this expression, f(x) is expressed in terms of a constant f'(1). So, we can consider this as an expression for f(x) with an unknown constant f'(1). To determine f'(1), we can take the derivative of f(x) and evaluate it at x=1.Let me compute f'(x):f(x) = 1 + ln x - 2f'(1)xTherefore,f'(x) = (1/x) - 2f'(1)Then, evaluating at x=1:f'(1) = (1/1) - 2f'(1)So,f'(1) = 1 - 2f'(1)Bring the 2f'(1) to the left:f'(1) + 2f'(1) = 13f'(1) = 1Therefore,f'(1) = 1/3Then, substituting back into the expression for f(x):f(x) = 1 + ln x - 2*(1/3)*xSimplify:f(x) = 1 + ln x - (2/3)xTherefore, the function f(x) is 1 + ln x - (2/3)x.Wait, but let me check if this is consistent.So, f(x) = 1 + ln x - (2/3)xThen, f'(x) = 1/x - 2/3Therefore, f'(1) = 1 - 2/3 = 1/3, which matches our earlier result.Therefore, substituting back into the original equation, we need to verify that OA = [f(x) + 2f'(1)x] OB - ln x * OC.But OA, OB, OC are vectors on line l, which we parameterized earlier. But since in our derivation, we already used the condition from the p0 component, which gave us the equation for f(x), and the other equation involved t_A, t_B, t_C. However, unless there is more information about points A, B, C, we can't determine those t parameters. However, since the problem asks for the expression for f(x), and through the steps we derived f(x) = 1 + ln x - (2/3)x, and verified that f'(1) = 1/3, which is consistent, this should be the solution.Alternatively, let's check if there's another way to think about this problem without parameterizing the line.Since points A, B, C are colinear, the vectors OA, OB, OC satisfy the equation OA = λ OB + μ OC, where λ + μ = 1 (if considering affine combination). However, in the given equation, OA is expressed as [f(x) + 2f'(1)x] OB - ln x OC. If this were an affine combination, we would require that [f(x) + 2f'(1)x] + (-ln x) = 1. But that is exactly the equation we derived earlier from the p0 component. So, this is another way to see it: for OA to lie on the line through B and C, the coefficients in the linear combination must add to 1. Therefore:[f(x) + 2f'(1)x] - ln x = 1Which gives the same equation:f(x) = 1 + ln x - 2f'(1)xThen, proceeding as before, taking derivative, finding f'(1) = 1/3, and substituting back.Therefore, regardless of the approach, the solution seems to be f(x) = 1 + ln x - (2/3)x.Let me double-check the steps to make sure I didn't make a mistake.1. Recognize that OA, OB, OC are position vectors of colinear points A, B, C.2. Express OA as a linear combination of OB and OC: OA = α OB + β OC.3. Since A, B, C are colinear, the combination must be affine, i.e., α + β = 1. Wait, but affine combinations are when the coefficients sum to 1. However, in general, if the origin is not on the line, then the coefficients for an affine combination would need to sum to 1. But if the origin is on the line, then it's a linear combination. Wait, actually, in vector terms, if three points are colinear, then for any two points, say B and C, any other point A on the line can be expressed as A = B + t(C - B) = (1 - t)B + tC, which is an affine combination with coefficients (1 - t) and t summing to 1.Therefore, in the problem's equation OA = [f(x) + 2f'(1)x] OB - ln x OC, the coefficients must sum to 1 for OA to lie on the line through B and C (assuming origin is not on the line). So:[f(x) + 2f'(1)x] + (-ln x) = 1Which is exactly the equation we obtained earlier. Therefore, solving that gives f(x) = 1 + ln x - 2f'(1)x.Then, differentiating f(x):f'(x) = 1/x - 2f'(1)Evaluating at x=1:f'(1) = 1 - 2f'(1)Solving:3f'(1) = 1 => f'(1) = 1/3Thus, f(x) = 1 + ln x - (2/3)x.This seems consistent.Therefore, the final answer is f(x) = 1 + ln x - (2/3)x.Final AnswerThe expression for the function is boxed{1 + ln x - dfrac{2}{3}x}.</think>Given that points A, B, and C are collinear on line ( l ), and the vectors (overrightarrow{OA}), (overrightarrow{OB}), and (overrightarrow{OC}) satisfy the equation (overrightarrow{OA} = [f(x) + 2f'(1)x] overrightarrow{OB} - ln x cdot overrightarrow{OC}), we need to find the expression for the function ( y = f(x) ).1. Collinearity Condition: Since points A, B, and C are collinear, the vectors (overrightarrow{OA}), (overrightarrow{OB}), and (overrightarrow{OC}) must lie on a line. For (overrightarrow{OA}) to be expressed as a linear combination of (overrightarrow{OB}) and (overrightarrow{OC}) and still lie on the same line, the coefficients of the linear combination must sum to 1 (affine combination).2. Setting Up the Equation: The given equation is: [ overrightarrow{OA} = [f(x) + 2f'(1)x] overrightarrow{OB} - ln x cdot overrightarrow{OC} ] For (overrightarrow{OA}) to lie on the line through B and C, the coefficients must satisfy: [ [f(x) + 2f'(1)x] - ln x = 1 ]3. Solving for ( f(x) ): [ f(x) + 2f'(1)x - ln x = 1 ] Rearranging gives: [ f(x) = 1 + ln x - 2f'(1)x ]4. Finding ( f'(1) ): Differentiate ( f(x) ): [ f'(x) = frac{1}{x} - 2f'(1) ] Evaluating at ( x = 1 ): [ f'(1) = 1 - 2f'(1) ] Solving for ( f'(1) ): [ 3f'(1) = 1 implies f'(1) = frac{1}{3} ]5. Substituting ( f'(1) ) Back: Substitute ( f'(1) = frac{1}{3} ) into the expression for ( f(x) ): [ f(x) = 1 + ln x - 2 left( frac{1}{3} right) x ] Simplifying gives: [ f(x) = 1 + ln x - frac{2}{3}x ]Thus, the expression for the function ( y = f(x) ) is (boxed{1 + ln x - dfrac{2}{3}x}).]

🔑:Since points A, B, and C are collinear, and overrightarrow{OA}=[f(x)+2f'(1)x] overrightarrow{OB}-ln xcdot overrightarrow{OC},it follows that f(x)+2f'(1)x-ln x=1. Differentiating both sides gives f'(x)+2f'(1)-frac{1}{x}=0.Substituting x=1 yields f'(1)+2f'(1)-1=0, solving this gives f'(1)=frac{1}{3}.Therefore, f(x)+frac{2}{3}x-ln x=1, which simplifies to f(x)=ln x- frac{2x}{3}+1.Hence, the answer is: boxed{f(x)=ln x- frac{2x}{3}+1}.By knowing that the three points are collinear, we can derive f(x)+2f'(1)x-ln x=1. Differentiating and substituting x=1 gives the value of f'(1), which leads to the expression of the function. This problem tests the solution of function expressions, involving knowledge of vectors and derivatives, and is considered a basic question.