🧠:[<think>Okay, so I need to find the point where the line intersects the plane. Let me think. The line is given in that symmetric equation form, which I remember is like (x - x0)/a = (y - y0)/b = (z - z0)/c. So, the direction vector of the line is <4, -3, -2>, and it passes through the point (2, 1, -3). The plane is given by 3x - y + 4z = 0. Hmm, to find the intersection, I need a point that lies on both the line and the plane. Since the line is parametric, maybe I should write it in parametric form and plug those into the plane equation. Let's try that.Starting with the symmetric equations: (x - 2)/4 = (y - 1)/(-3) = (z + 3)/(-2) = t, where t is a parameter. So, each part equals t. That means I can express x, y, z in terms of t.So, x = 2 + 4ty = 1 - 3tz = -3 - 2tRight? Because for x, subtracting 2 and dividing by 4 gives t, so multiplying back gives x = 2 + 4t. Similarly for y and z. Now, these parametric equations should satisfy the plane equation 3x - y + 4z = 0. So I can substitute x, y, z into the plane equation and solve for t.Let's do that step by step. Substitute:3x = 3*(2 + 4t) = 6 + 12t-y = -(1 - 3t) = -1 + 3t4z = 4*(-3 - 2t) = -12 - 8tAdding them all together: 6 + 12t -1 + 3t -12 -8t = 0Combine like terms:(6 -1 -12) + (12t + 3t -8t) = 0Calculating the constants: 6 -1 is 5, 5 -12 is -7Variables: 12t +3t is 15t, 15t -8t is 7tSo overall: -7 + 7t = 0Solving for t: 7t = 7 → t = 1Okay, so t is 1. Now plug t = 1 back into the parametric equations for x, y, z.x = 2 + 4(1) = 6y = 1 - 3(1) = 1 -3 = -2z = -3 -2(1) = -5So the point of intersection is (6, -2, -5). Let me check if this point satisfies both the line and the plane equations.First, check the line equations. Let's see:(x -2)/4 = (6 -2)/4 = 4/4 = 1(y -1)/(-3) = (-2 -1)/(-3) = (-3)/(-3) = 1(z +3)/(-2) = (-5 +3)/(-2) = (-2)/(-2) = 1All equal to 1, which is the t value we found. So that's good.Now check the plane equation: 3x - y +4z3*6 = 18-(-2) = +24*(-5) = -20Adding them: 18 +2 -20 = 0. Perfect, that equals 0. So the point (6, -2, -5) is indeed on both the line and the plane. I think that's the answer. But let me just recap to make sure I didn't skip any steps or make a mistake in arithmetic. The parametric equations were set up correctly from the symmetric form. Then substitution into the plane equation was done term by term, combined, solved for t. Then plugged back in. Checked both equations. Seems solid. Wait, another way to approach this might be to use the direction vector and see if the line is parallel to the plane or not. The direction vector of the line is <4, -3, -2>. The normal vector of the plane is <3, -1, 4>. The line is not parallel to the plane if the direction vector is not perpendicular to the normal vector. To check that, take the dot product of direction vector and normal vector:4*3 + (-3)*(-1) + (-2)*4 = 12 +3 -8 = 7. Since it's not zero, the line is not parallel, so they must intersect at exactly one point. Which is consistent with our result. So that's another confirmation that the intersection exists and we found it correctly.So, yeah, confident that (6, -2, -5) is the correct point of intersection.Final AnswerThe point of intersection is boxed{(6, -2, -5)}.</think>To find the point of intersection of the line (frac{x-2}{4} = frac{y-1}{-3} = frac{z+3}{-2}) and the plane (3x - y + 4z = 0), we start by expressing the line in parametric form. The line can be parameterized as:[x = 2 + 4t, quad y = 1 - 3t, quad z = -3 - 2t]Next, substitute these parametric equations into the plane equation (3x - y + 4z = 0):[3(2 + 4t) - (1 - 3t) + 4(-3 - 2t) = 0]Simplifying each term:[3(2 + 4t) = 6 + 12t][-(1 - 3t) = -1 + 3t][4(-3 - 2t) = -12 - 8t]Combine all terms:[6 + 12t - 1 + 3t - 12 - 8t = 0]Combine like terms:[(6 - 1 - 12) + (12t + 3t - 8t) = 0][-7 + 7t = 0]Solving for (t):[7t = 7 implies t = 1]Substitute (t = 1) back into the parametric equations:[x = 2 + 4(1) = 6][y = 1 - 3(1) = -2][z = -3 - 2(1) = -5]The point of intersection is ((6, -2, -5)). Verification confirms this point satisfies both the line and the plane equations.The point of intersection is (boxed{(6, -2, -5)}).,<think>Alright, so I need to find the point where the given line intersects the plane. Let me start by recalling what I know. The line is given in the symmetric form, which I think is like (x - x₀)/a = (y - y₀)/b = (z - z₀)/c. Here, the point (x₀, y₀, z₀) is (2, 1, -3), and the direction ratios are a=4, b=-3, c=-2. That means the direction vector of the line is <4, -3, -2>. The plane equation is 3x - y + 4z = 0. To find the intersection, I probably need to parameterize the line and then substitute those parametric equations into the plane equation to solve for the parameter. Let me check if that makes sense.So, if I set each part of the symmetric equation equal to a parameter t, I can write x, y, z in terms of t. Let me try that:(x - 2)/4 = t (y - 1)/(-3) = t (z + 3)/(-2) = tTherefore, solving for each coordinate:x = 2 + 4t y = 1 - 3t z = -3 - 2tOkay, so those are the parametric equations of the line. Now, substitute x, y, z into the plane equation 3x - y + 4z = 0.Let me plug them in:3*(2 + 4t) - (1 - 3t) + 4*(-3 - 2t) = 0Now, expand each term:3*2 + 3*4t - 1 + 3t + 4*(-3) + 4*(-2t) Which is 6 + 12t -1 + 3t -12 -8tCombine like terms:First, constants: 6 -1 -12 = 6 -13 = -7 Variables: 12t + 3t -8t = 7tSo the equation becomes: -7 + 7t = 0Solving for t: 7t = 7 => t = 1Hmm, so t = 1. Then substitute back into the parametric equations to get the point.x = 2 + 4*(1) = 6 y = 1 - 3*(1) = -2 z = -3 -2*(1) = -5Wait, so the point would be (6, -2, -5). Let me verify if this point satisfies both the line equation and the plane equation.First, check the plane equation: 3x - y + 4z = 0Plug in (6, -2, -5):3*6 - (-2) + 4*(-5) = 18 + 2 -20 = 0. Yes, that works.Now check if the point lies on the line. The symmetric equations for the line:(x - 2)/4 = (6 - 2)/4 = 4/4 = 1 (y - 1)/(-3) = (-2 -1)/(-3) = (-3)/(-3) = 1 (z +3)/(-2) = (-5 +3)/(-2) = (-2)/(-2) = 1All equal to 1, which is the parameter t we found. So that checks out.Hmm, seems like everything is consistent. Let me just recap the steps to make sure I didn't skip anything.1. Converted the symmetric equations to parametric form by introducing parameter t.2. Expressed x, y, z in terms of t.3. Substituted those expressions into the plane equation.4. Solved for t.5. Plugged t back into parametric equations to get the point.6. Verified the point lies on both the line and the plane.I don't see any errors here. The calculations seem straightforward. Let me just redo the substitution step to double-check:3*(2 + 4t) = 6 +12t - (1 -3t) = -1 +3t 4*(-3 -2t) = -12 -8t Adding them up: 6 +12t -1 +3t -12 -8t = (6 -1 -12) + (12t +3t -8t) = (-7) +7t =0 Yep, same result. t=1.Therefore, the point of intersection is (6, -2, -5). Seems solid. I think that's the answer.Final AnswerThe point of intersection is boxed{(6, -2, -5)}.</think>To find the point of intersection of the line (frac{x-2}{4} = frac{y-1}{-3} = frac{z+3}{-2}) and the plane (3x - y + 4z = 0), we start by converting the symmetric equations of the line into parametric form. Setting each part of the symmetric equation equal to a parameter (t), we get:[frac{x-2}{4} = t, quad frac{y-1}{-3} = t, quad frac{z+3}{-2} = t]Solving for (x), (y), and (z) in terms of (t):[x = 2 + 4t, quad y = 1 - 3t, quad z = -3 - 2t]Next, substitute these parametric equations into the plane equation (3x - y + 4z = 0):[3(2 + 4t) - (1 - 3t) + 4(-3 - 2t) = 0]Expanding and simplifying:[6 + 12t - 1 + 3t - 12 - 8t = 0][-7 + 7t = 0]Solving for (t):[7t = 7 implies t = 1]Substituting (t = 1) back into the parametric equations:[x = 2 + 4(1) = 6, quad y = 1 - 3(1) = -2, quad z = -3 - 2(1) = -5]Verification shows that the point ((6, -2, -5)) satisfies both the line and the plane equations. Therefore, the point of intersection is (boxed{(6, -2, -5)}).]

🔑:## SolutionLet's write the parametric equations of the line.begin{aligned}& frac{x-2}{4}=frac{y-1}{-3}=frac{z+3}{-2}=t Rightarrow & left{begin{array}{l}x=2+4 t y=1-3 t z=-3-2 tend{array}right.end{aligned}Substitute into the equation of the plane:3(2+4 t)-(1-3 t)+4(-3-2 t)=06+12 t-1+3 t-12-8 t=07 t-7=0t=1Find the coordinates of the intersection point of the line and the plane:left{begin{array}{l}x=2+4 cdot 1=6 y=1-3 cdot 1=-2 z=-3-2 cdot 1=-5end{array}right.We get:(6 ;-2 ;-5)## Problem Kuznetsov Analytic Geometry 14-27

🧠:[<think>Okay, let's see. The problem says that √2, √3, and √x form a geometric sequence. Then I need to find log base (3/√2) of x. Hmm, okay. Let me recall what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a constant called the common ratio. So, if √2, √3, √x are in a geometric sequence, then the ratio between the second and first term should be the same as the ratio between the third and second term. So, the common ratio (r) would be √3 divided by √2, right? Because the second term is √3 and the first term is √2. Then the ratio between the third and second term should also be the same r. So, √x divided by √3 should equal to r. Therefore, √x / √3 = √3 / √2. That seems like the key equation here.Let me write that down step by step. First term: √2Second term: √3Third term: √xCommon ratio (r) = second term / first term = √3 / √2Similarly, common ratio (r) = third term / second term = √x / √3Since they are equal, set them equal to each other:√x / √3 = √3 / √2Now, solve for x. Let's cross-multiply to get rid of the fractions. So:√x * √2 = √3 * √3Simplify the right side: √3 * √3 is (√3)^2 which is 3. So,√x * √2 = 3Combine the square roots on the left: √(2x) = 3Now, square both sides to eliminate the square root:(√(2x))^2 = 3^2Which simplifies to:2x = 9Divide both sides by 2:x = 9/2So x is 9/2. Then the problem asks for log base (3/√2) of x. So that's log base (3/√2) of 9/2. Let me write that as log_{3/√2} (9/2). Now, I need to compute this logarithm. Let me recall logarithm properties. The logarithm log_b a is the exponent you need to raise b to get a. So, if I let y = log_{3/√2} (9/2), then by definition, (3/√2)^y = 9/2. So, I need to solve for y such that (3/√2)^y = 9/2.Let me write that equation:(3 / √2)^y = 9 / 2I can express both sides in terms of prime factors to see if that helps. Let's see. 3 is prime, √2 is 2^(1/2), 9 is 3^2, and 2 is prime. So let's rewrite everything with exponents:Left side: (3 / 2^(1/2))^yRight side: 3^2 / 2So, left side becomes 3^y / 2^(y/2). Right side is 3^2 / 2^1. Therefore, set the exponents equal. For the bases 3 and 2:For base 3: exponent on left is y, exponent on right is 2. So y must equal 2.For base 2: exponent on left is y/2, exponent on right is 1. So y/2 = 1 => y = 2.Wait, both equations give y = 2. That's consistent. So y = 2. Therefore, the value of the logarithm is 2.Wait, so that's straightforward. Let me double-check.Alternatively, maybe using logarithm change of base formula. log_{3/√2} (9/2) = ln(9/2) / ln(3/√2). Let's compute that.First, ln(9/2) = ln(9) - ln(2) = 2 ln(3) - ln(2)Then, ln(3/√2) = ln(3) - ln(√2) = ln(3) - (1/2) ln(2)So the ratio is [2 ln(3) - ln(2)] / [ln(3) - (1/2) ln(2)]Let me factor numerator and denominator.Let me factor out ln(3) from numerator and denominator:Numerator: 2 ln(3) - ln(2) = ln(3)*(2) - ln(2)Denominator: ln(3) - (1/2) ln(2) = ln(3)*(1) - (1/2) ln(2)Hmm, not obvious. Maybe substitute variables. Let me let a = ln(3) and b = ln(2). Then:Numerator: 2a - bDenominator: a - (1/2)bSo the ratio is (2a - b)/(a - 0.5b). Let's see if this simplifies. Let me try to divide these. Let's perform polynomial division or see if there's a scalar multiple.Let me see if denominator * 2 = 2a - b. Let's check:Denominator * 2: 2*(a - 0.5b) = 2a - b. Exactly the numerator. Therefore, (2a - b)/(a - 0.5b) = 2. So the ratio is 2. Therefore, the logarithm is 2. So the answer is 2.Therefore, regardless of the method, the result is 2. So the answer is boxed{2}.But just to recap, in case I made a miscalculation:We found that x is 9/2. Then we needed to compute log base 3/√2 of 9/2. Let's check with exponents:(3/√2)^2 = 9 / ( (√2)^2 ) = 9 / 2. Which is exactly 9/2. Therefore, (3/√2)^2 = 9/2. Therefore, log_{3/√2} (9/2) = 2. Correct.So everything checks out. So yes, the answer is 2.Final Answerboxed{2}</think>Given that (sqrt{2}), (sqrt{3}), and (sqrt{x}) form a geometric sequence, we need to find (log_{frac{3}{sqrt{2}}} x).In a geometric sequence, the common ratio (r) between consecutive terms is constant. Therefore, we have:[frac{sqrt{3}}{sqrt{2}} = frac{sqrt{x}}{sqrt{3}}]Solving for (x), we cross-multiply:[sqrt{x} cdot sqrt{2} = (sqrt{3})^2][sqrt{2x} = 3]Squaring both sides:[2x = 9][x = frac{9}{2}]Next, we need to compute (log_{frac{3}{sqrt{2}}} frac{9}{2}). Let (y = log_{frac{3}{sqrt{2}}} frac{9}{2}). By definition of logarithms:[left(frac{3}{sqrt{2}}right)^y = frac{9}{2}]Expressing both sides in terms of prime factors:[left(frac{3}{2^{1/2}}right)^y = frac{3^2}{2}][frac{3^y}{2^{y/2}} = frac{3^2}{2^1}]Equating the exponents of the corresponding bases:- For base 3: (y = 2)- For base 2: (frac{y}{2} = 1 implies y = 2)Thus, (y = 2). Therefore, the value of (log_{frac{3}{sqrt{2}}} x) is (boxed{2}).,<think>Okay, so the problem says that √2, √3, and √x form a geometric sequence. I need to find log base (3/√2) of x. Hmm, let's start by recalling what a geometric sequence is. In a geometric sequence, each term after the first is found by multiplying the previous term by a common ratio, right? So, if these three terms form a geometric sequence, the ratio between the second and the first should be the same as the ratio between the third and the second.Let me write that out. The common ratio r would be √3 divided by √2, because that's the second term divided by the first. Then, the ratio between the third term and the second term should be the same. So, √x divided by √3 should equal the same ratio. Therefore, √3 / √2 = √x / √3. That equation should help me solve for x.Let me set up the equation:√3 / √2 = √x / √3To solve for x, I can cross-multiply. Multiply both sides by √3 and by √2 to get rid of the denominators. So, √3 * √3 = √x * √2. That simplifies to (√3)^2 = √(2x). Since (√3)^2 is 3, that becomes 3 = √(2x). Then, to solve for x, I can square both sides of the equation. Squaring the left side gives 9, and squaring the right side gives 2x. So, 9 = 2x, which means x = 9/2. So x is 4.5 or 9/2. Wait, let me check that again. If √3 divided by √2 is the common ratio, then multiplying √3 by that ratio should give me √x. Let's see: √3 * (√3 / √2) = (√3 * √3) / √2 = 3 / √2. But √x should equal 3/√2. Then, squaring both sides gives x = (3/√2)^2. That would be 9 / 2, which is indeed 4.5. So, that checks out. So x is 9/2. Now, the question is asking for log base (3/√2) of x. So, log_(3/√2) (9/2). Let me recall how logarithms work. Log base a of b is equal to c if a^c = b. So, we need to find the exponent c such that (3/√2)^c = 9/2.Let me write that equation: (3 / √2)^c = 9 / 2. Let's express both sides with exponents that might make it easier to compare. Let's break down the numbers into prime factors. First, 3 is already a prime. √2 is 2^(1/2). So, 3 / √2 is 3 * 2^(-1/2). Then, raising that to the power of c gives (3^c) * (2^(-c/2)).On the right side, 9 is 3^2, and 2 is 2^1. So, 9/2 is 3^2 * 2^(-1).Therefore, setting the exponents equal, since the bases 3 and 2 are different and prime, their exponents must match. So, for the base 3: c = 2. For the base 2: -c/2 = -1. Let's check that.If c = 2, then for the base 2 part: -2/2 = -1, which matches. So, both exponents are satisfied when c = 2. Therefore, log base (3/√2) of (9/2) is 2. So, the answer is 2. Let me verify once again to make sure I didn't make any mistakes.Starting with the geometric sequence: √2, √3, √x. The ratio between √3 and √2 is √3/√2. Then, multiplying √3 by this ratio gives √3 * √3 / √2 = 3/√2, which is the third term. Therefore, √x = 3/√2, so squaring gives x = 9/2. Correct.Then, log base 3/√2 of 9/2. Let's compute (3/√2)^2. That is 9 / ( (√2)^2 ) = 9 / 2. Which is exactly the argument of the logarithm. So, yes, the exponent is 2. Therefore, the logarithm is indeed 2.I think that's correct. I don't see any mistakes here. So the answer should be 2.Final Answerboxed{2}</think>Given that (sqrt{2}), (sqrt{3}), and (sqrt{x}) form a geometric sequence, we need to find (log_{frac{3}{sqrt{2}}} x).In a geometric sequence, the common ratio (r) between consecutive terms is constant. Therefore, the ratio of the second term to the first term must equal the ratio of the third term to the second term:[frac{sqrt{3}}{sqrt{2}} = frac{sqrt{x}}{sqrt{3}}]Cross-multiplying gives:[sqrt{3} cdot sqrt{3} = sqrt{x} cdot sqrt{2}]Simplifying both sides, we get:[3 = sqrt{2x}]Squaring both sides to solve for (x):[9 = 2x implies x = frac{9}{2}]Next, we need to find (log_{frac{3}{sqrt{2}}} frac{9}{2}). Let (c = log_{frac{3}{sqrt{2}}} frac{9}{2}). By the definition of logarithms, this means:[left( frac{3}{sqrt{2}} right)^c = frac{9}{2}]Expressing both sides with prime factors, we have:[left( 3 cdot 2^{-1/2} right)^c = 3^2 cdot 2^{-1}]Equating the exponents of the prime factors, we get:[3^c = 3^2 quad text{and} quad (2^{-1/2})^c = 2^{-1}]From the first equation, (c = 2). From the second equation, (-c/2 = -1) which also gives (c = 2). Both conditions are satisfied, confirming that (c = 2).Thus, the final answer is:[boxed{2}]]

🔑:This problem primarily tests the concept of geometric sequences and the operations of logarithms, which is a basic question.First, from the given information, we have (sqrt{3})^2 = sqrt{2} cdot sqrt{x}. This simplifies to x = frac{9}{2}.Next, we want to find log_{frac{3}{sqrt{2}}} x. Substituting x = frac{9}{2}, we get log_{frac{3}{sqrt{2}}} frac{9}{2}.Notice that frac{9}{2} = left(frac{sqrt{3}}{2}right)^2 and frac{3}{sqrt{2}} = frac{sqrt{3}}{2} cdot sqrt{2} = frac{sqrt{3}}{sqrt{1}} = frac{sqrt{3}}{2}.Using the change of base formula, we can rewrite the logarithm as log_{frac{sqrt{3}}{2}} left(frac{sqrt{3}}{2}right)^2.Since log_b b^n = n, we have log_{frac{sqrt{3}}{2}} left(frac{sqrt{3}}{2}right)^2 = 2.Therefore, the answer is boxed{2}.

🧠:[<think>Okay, let's tackle this problem step by step. There are two parts: first, finding the general formulas for the geometric sequence {a_n} and the arithmetic sequence {b_n}, and second, finding T_n which is the sum of the products a_k b_k from k=1 to n. Let me start with part (1).Starting with the geometric sequence {a_n}. We know that a_1 = 1, which is the first term, and a_6 = 243. Since it's a geometric sequence, each term is multiplied by a common ratio, r. The general formula for a geometric sequence is a_n = a_1 * r^(n-1). Given that a_1 is 1, the formula simplifies to a_n = r^(n-1). Now, we need to find r. We know that a_6 = 243. Plugging into the formula: a_6 = r^(6-1) = r^5 = 243. To find r, we can take the fifth root of 243. Let me calculate that. 3^5 is 243 because 3*3=9, 9*3=27, 27*3=81, 81*3=243. So r = 3. Therefore, the general formula for {a_n} is a_n = 3^(n-1). Got that part.Now moving on to the arithmetic sequence {b_n}. The problem states that b_1 = 3 and S_5 = 35, where S_n is the sum of the first n terms. For an arithmetic sequence, the general formula for the nth term is b_n = b_1 + (n-1)d, where d is the common difference. The sum of the first n terms, S_n, is given by the formula S_n = n/2 * [2b_1 + (n-1)d] or equivalently S_n = n*(b_1 + b_n)/2. We need to find d. We know S_5 = 35. Let's use the sum formula. Plugging in n=5:S_5 = 5/2 * [2b_1 + (5-1)d] = 5/2 * [2*3 + 4d] = 5/2 * [6 + 4d]. We know S_5 = 35, so:5/2 * (6 + 4d) = 35.Multiply both sides by 2/5 to solve for (6 + 4d):(6 + 4d) = 35 * (2/5) = 14.So, 6 + 4d = 14. Subtract 6 from both sides: 4d = 8. Therefore, d = 8/4 = 2.So the common difference d is 2. Therefore, the general formula for {b_n} is:b_n = b_1 + (n-1)d = 3 + (n-1)*2 = 3 + 2n - 2 = 2n + 1.Wait, let me check that arithmetic again. (n-1)*2 is 2n - 2, so 3 + 2n - 2 is 2n + 1. Yes, that's correct. So b_n = 2n + 1.Let me verify this with S_5. If each term is 2n + 1, then the first five terms are:b_1 = 2*1 +1 = 3,b_2 = 2*2 +1 =5,b_3=2*3 +1=7,b_4=9,b_5=11.Sum these: 3+5=8, 8+7=15, 15+9=24, 24+11=35. Perfect, that matches S_5=35. So the formula for {b_n} is indeed b_n=2n+1.So part (1) is done: a_n=3^(n-1) and b_n=2n+1.Now part (2): Find T_n = a_1b_1 + a_2b_2 + ... + a_n b_n.So T_n is the sum from k=1 to n of a_k b_k. Let's write that out:T_n = Σ_{k=1}^n [3^{k-1} * (2k + 1)].Hmm, this looks like a sum involving terms of the form 3^{k-1} multiplied by a linear term in k. So, perhaps we can split the sum into two parts: Σ 2k*3^{k-1} + Σ 3^{k-1}. Let's write that:T_n = 2 Σ_{k=1}^n k*3^{k-1} + Σ_{k=1}^n 3^{k-1}.We can compute each sum separately. Let's start with the second sum, which is simpler: Σ_{k=1}^n 3^{k-1}. That's a geometric series with first term 3^{0}=1 and ratio 3. The sum of the first n terms is (3^n - 1)/(3 - 1) = (3^n -1)/2.Now the first sum: 2 Σ_{k=1}^n k*3^{k-1}. Let's denote S = Σ_{k=1}^n k*3^{k-1}. Then the first part is 2S. So we need to find S.To compute S = Σ_{k=1}^n k*3^{k-1}, there is a standard formula for sums of the form Σ k r^{k}. Let me recall. The sum Σ_{k=0}^n k r^{k} = r/(1 - r)^2 - (n + 1) r^{n + 1}/(1 - r) + r^{n + 2}/(1 - r)^2. But our sum is Σ_{k=1}^n k*3^{k -1}. Let's adjust the indices.First, note that 3^{k -1} = 3^{k}/3. So S = (1/3) Σ_{k=1}^n k*3^{k}. Let’s set r = 3. Then Σ_{k=1}^n k r^{k} = r(1 - (n + 1) r^n + n r^{n +1}) / (1 - r)^2. This is a known formula for the sum of k r^k from k=1 to n.Let me verify that formula. Let's consider the sum S = Σ_{k=1}^n k r^k.We can derive this using generating functions or by manipulating the sum. Let me recall the method.Let S = r + 2 r^2 + 3 r^3 + ... + n r^n.Multiply both sides by r:rS = r^2 + 2 r^3 + 3 r^4 + ... + n r^{n + 1}.Subtract the second equation from the first:S - rS = (r + r^2 + r^3 + ... + r^n) - n r^{n +1}.Left side: S(1 - r).Right side: Σ_{k=1}^n r^k - n r^{n +1} = [r(1 - r^n)/(1 - r)] - n r^{n +1}.Therefore,S = [r(1 - r^n)/(1 - r)^2 - n r^{n +1}/(1 - r)].Thus,Σ_{k=1}^n k r^k = r/(1 - r)^2 - (n +1) r^{n +1}/(1 - r)^2 + r^{n +2}/(1 - r)^2.Wait, actually, the standard formula is:Σ_{k=1}^n k r^k = r(1 - (n +1) r^n + n r^{n +1}) / (1 - r)^2.Yes, that's correct.So plugging r = 3 into this formula:Σ_{k=1}^n k*3^k = 3(1 - (n +1)3^n + n*3^{n +1}) / (1 - 3)^2.Compute denominator: (1 - 3)^2 = (-2)^2 =4.Numerator: 3[1 - (n +1)3^n + n*3^{n +1}].Let me expand the numerator:3[1 - (n +1)3^n + n*3^{n +1}] = 3[1 + 3^n(- (n +1) + n*3)].Simplify the terms inside:- (n +1) + 3n = -n -1 +3n = 2n -1.Therefore, numerator becomes 3[1 + 3^n(2n -1)].So Σ_{k=1}^n k*3^k = 3[1 + 3^n(2n -1)] /4.But remember, our S is (1/3) Σ_{k=1}^n k*3^k. So:S = (1/3) * [3(1 + 3^n(2n -1))/4] = [1 + 3^n(2n -1)] /4.Therefore, the first sum 2S = [1 + 3^n(2n -1)] /2.Then, T_n = 2S + Σ 3^{k-1} = [1 + 3^n(2n -1)] /2 + (3^n -1)/2.Combine the two fractions:[1 + 3^n(2n -1) + 3^n -1]/2 = [1 -1 +3^n(2n -1 +1)] /2 = [3^n(2n)] /2 = (2n *3^n)/2 = n*3^n.Wait, that's interesting. Let me check that again.So, first, 2S = [1 + 3^n(2n -1)] /2Wait, hold on. Wait, no:Wait, S = [1 + 3^n(2n -1)] /4Therefore, 2S = [1 + 3^n(2n -1)] /2Then, Σ_{k=1}^n 3^{k-1} = (3^n -1)/2Therefore, T_n = [1 + 3^n(2n -1)] /2 + (3^n -1)/2Adding the numerators:1 + 3^n(2n -1) + 3^n -1 = 1 -1 + 3^n(2n -1 +1) = 0 + 3^n(2n) = 2n*3^nThen divide by 2:(2n*3^n)/2 = n*3^nWow, so T_n simplifies to n*3^n. That's a surprisingly neat answer. Let me verify this with small n to make sure.Let's test n=1. Then T_1 = a1*b1 =1*3=3. According to the formula, n=1: 1*3^1=3. Correct.n=2: T_2 = a1b1 +a2b2 =1*3 + 3*5=3 +15=18. Formula:2*3^2=2*9=18. Wait, wait, no. Wait, according to the formula, n*3^n. For n=2, 2*3^2=2*9=18. Correct. T_2=18. Correct.n=3: T_3 = 3 +15 + a3*b3. a3=3^{2}=9, b3=2*3 +1=7, so 9*7=63. So T_3=3+15+63=81. Formula: 3*3^3=3*27=81. Correct.n=4: T_4=81 + a4*b4. a4=3^3=27, b4=2*4+1=9, so 27*9=243. T_4=81+243=324. Formula:4*3^4=4*81=324. Correct.So the formula works for these test cases. Therefore, T_n =n*3^n.So part (2) is T_n =n*3^n.Therefore, the answers are:1) a_n=3^{n-1}, b_n=2n+1.2) T_n= n*3^n.Final Answer(1) The general formulas are boxed{a_n = 3^{n-1}} and boxed{b_n = 2n + 1}. (2) The expression for T_n is boxed{n cdot 3^n}.</think>Given that {a_n} is a geometric sequence with a_1=1 and a_6=243, and S_n is the sum of the first n terms of the arithmetic sequence {b_n}, where b_1=3 and S_5=35:(1) To find the general formula for {a_n}:- The common ratio r is found using a_6 = 243 = r^5. Solving this gives r = 3.- Therefore, the general formula for the geometric sequence is a_n = 3^{n-1}.To find the general formula for {b_n}:- The sum of the first 5 terms S_5 = 35 is used. The formula for the sum of an arithmetic sequence gives S_5 = frac{5}{2} [2 cdot 3 + 4d] = 35. Solving this gives d = 2.- Therefore, the general formula for the arithmetic sequence is b_n = 3 + (n-1) cdot 2 = 2n + 1.(2) To find T_n = a_1b_1 + a_2b_2 + ldots + a_nb_n:- Express T_n as sum_{k=1}^n 3^{k-1} (2k + 1).- Split the sum into two parts: 2 sum_{k=1}^n k cdot 3^{k-1} + sum_{k=1}^n 3^{k-1}.- The second sum is a geometric series: sum_{k=1}^n 3^{k-1} = frac{3^n - 1}{2}.- The first sum is calculated using the formula for the sum of k cdot r^k: sum_{k=1}^n k cdot 3^{k-1} = frac{1 + 3^n (2n - 1)}{4}. Multiplying by 2 gives frac{1 + 3^n (2n - 1)}{2}.- Combining both parts results in T_n = frac{1 + 3^n (2n - 1)}{2} + frac{3^n - 1}{2} = n cdot 3^n.Final answers:(1) The general formulas are boxed{a_n = 3^{n-1}} and boxed{b_n = 2n + 1}.(2) The expression for T_n is boxed{n cdot 3^n}.,<think>Okay, let's tackle this problem step by step. There are two parts: first, finding the general formulas for the geometric sequence {a_n} and the arithmetic sequence {b_n}, and second, finding T_n which is the sum of the products a_k b_k from k=1 to n. Let me start with part (1).Starting with the geometric sequence {a_n}. We know that a_1 = 1 and a_6 = 243. In a geometric sequence, each term is the previous term multiplied by a common ratio r. So, the general formula for a geometric sequence is a_n = a_1 * r^(n-1). Given a_1 is 1, so a_n = r^(n-1). Then, a_6 would be r^(5) because n=6, so 6-1=5. They tell us a_6 = 243, so r^5 = 243. Let me solve for r. What number raised to the 5th power is 243? Let's see. 3^5 is 243 because 3*3=9, 9*3=27, 27*3=81, 81*3=243. So, r=3. Therefore, the general formula for {a_n} is a_n = 1 * 3^(n-1) = 3^(n-1). That should be the formula for the geometric sequence.Now, moving on to the arithmetic sequence {b_n}. We know that b_1 = 3 and S_5 = 35, where S_n is the sum of the first n terms. For an arithmetic sequence, the general formula is b_n = b_1 + (n-1)d, where d is the common difference. The sum of the first n terms of an arithmetic sequence is given by S_n = n/2 * [2b_1 + (n-1)d]. So, S_5 = 35. Let's plug in n=5, b_1=3 into the sum formula.S_5 = 5/2 * [2*3 + (5-1)d] = 35. Let's compute that step by step.First, 2*3 is 6. Then, (5-1)d is 4d. So inside the brackets, it's 6 + 4d. Then, multiplying by 5/2: (5/2)*(6 + 4d) = 35. Let's solve for d.Multiply both sides by 2: 5*(6 + 4d) = 70. Then, 30 + 20d = 70. Subtract 30: 20d = 40. Divide by 20: d = 2. So the common difference d is 2. Therefore, the general term for the arithmetic sequence is b_n = 3 + (n-1)*2. Simplifying that: 3 + 2(n-1) = 3 + 2n - 2 = 2n + 1. Wait, 3 - 2 is 1, so it's 2n + 1? Let me check.Wait, 3 + (n-1)*2 = 3 + 2n - 2 = 2n +1. Yes, that's correct. So b_n = 2n + 1. Let me verify with the sum S_5. If d=2, then the terms are b_1=3, b_2=5, b_3=7, b_4=9, b_5=11. Summing them up: 3 +5=8, +7=15, +9=24, +11=35. Perfect, that matches S_5=35. So that's confirmed.So part (1) is done. The general formulas are:a_n = 3^(n-1)b_n = 2n + 1Now, part (2): Let T_n = a_1b_1 + a_2b_2 + ... + a_nb_n. Find T_n.So T_n is the sum from k=1 to n of a_k * b_k. Let's express a_k and b_k in terms of k.We already have a_k = 3^(k-1) and b_k = 2k +1. Therefore, the term a_k b_k = 3^(k-1)*(2k +1). So T_n is the sum from k=1 to n of [3^(k-1)*(2k +1)].Hmm. This seems like a sum that might have a closed-form formula. Let's try to compute it. Let me write out the terms for a few n to see if I can spot a pattern or think of a way to sum it.For k=1: 3^(0)*(2*1 +1)=1*3=3k=2: 3^1*(2*2 +1)=3*5=15k=3: 3^2*(2*3 +1)=9*7=63k=4: 3^3*(2*4 +1)=27*9=243k=5: 3^4*(11)=81*11=891So each term is 3^(k-1)*(2k +1). The sum T_n is 3 + 15 + 63 + 243 + 891 + ... up to n terms. This seems like a geometric series but with each term multiplied by (2k +1). Hmm. To find the sum, maybe we can find a recurrence relation or use generating functions, or perhaps split the sum into two parts.Let me consider splitting the term 2k +1 into two parts: 2k and 1. So T_n = sum_{k=1}^n [3^{k-1}*(2k +1)] = sum_{k=1}^n [2k*3^{k-1} + 3^{k-1}]. Therefore, T_n = 2*sum_{k=1}^n k*3^{k-1} + sum_{k=1}^n 3^{k-1}.So if I can compute these two sums separately, I can combine them. The second sum is a geometric series. Let's compute that first.Sum_{k=1}^n 3^{k-1} = (3^n -1)/(3 -1) = (3^n -1)/2. Because the sum of a geometric series from k=0 to n-1 of 3^k is (3^n -1)/2, but here the sum starts at k=1, which is from 3^0 up to 3^{n-1}, so it's the same as the sum from k=0 to n-1, so indeed (3^n -1)/2.Now, the first sum is 2*sum_{k=1}^n k*3^{k-1}. Let's compute sum_{k=1}^n k*3^{k-1}.I remember that there is a formula for the sum of k*r^{k-1}. The standard formula is sum_{k=1}^n k*r^{k-1} = (1 - (n+1)r^n + n*r^{n+1}) ) / (1 - r)^2. Let me verify that. For r ≠ 1.In this case, r=3, so substituting r=3:sum_{k=1}^n k*3^{k-1} = [1 - (n+1)*3^n + n*3^{n+1}]/(1 -3)^2 = [1 - (n+1)*3^n + n*3^{n+1}]/4.Let me simplify numerator:1 - (n+1)3^n + n*3^{n+1} = 1 - (n+1)3^n + n*3*3^n = 1 - (n+1)3^n + 3n*3^n = 1 + [ - (n+1) + 3n ] 3^n = 1 + (2n -1)3^n.Therefore, the sum becomes [1 + (2n -1)3^n]/4.Therefore, sum_{k=1}^n k*3^{k-1} = [1 + (2n -1)3^n]/4.Therefore, multiplying by 2, the first part of T_n is 2*[1 + (2n -1)3^n]/4 = [1 + (2n -1)3^n]/2.Then, the second part of T_n is (3^n -1)/2.Therefore, T_n = [1 + (2n -1)3^n]/2 + (3^n -1)/2.Let's combine these terms. Both have denominator 2, so combine numerators:[1 + (2n -1)3^n + 3^n -1]/2 = [ (1 -1) + (2n -1)3^n + 3^n ] / 2 = [ (2n -1 +1)3^n ] /2 = [2n*3^n]/2 = n*3^n.Wait, that simplifies to n*3^n. Let me verify step by step.Original numerator:1 + (2n -1)3^n + 3^n -1= (1 -1) + (2n -1)3^n + 3^n= 0 + (2n -1)3^n + 3^nFactor out 3^n:= [ (2n -1) +1 ]3^n= 2n*3^nTherefore, numerator is 2n*3^n, divided by 2 gives n*3^n.So T_n = n*3^n.Wait, that's surprisingly simple. Let me check with the terms I calculated earlier.For n=1: T_1 = 3. According to the formula, 1*3^1=3. Correct.For n=2: T_2 = 3 +15=18. Formula: 2*3^2=2*9=18. Correct.For n=3: T_3=3+15+63=81. Formula: 3*3^3=3*27=81. Correct.n=4: T_4=3+15+63+243=324. Formula:4*3^4=4*81=324. Correct.n=5: T_5=3+15+63+243+891=1215. Formula:5*3^5=5*243=1215. Correct. So the formula works.Therefore, T_n = n*3^n.Hmm. That's elegant. So even though the individual terms look complicated, the sum telescopes or combines nicely into n*3^n.So, summarizing part (2), T_n is equal to n times 3 to the nth power.Therefore, the answers are:(1) a_n = 3^{n-1}, b_n = 2n +1(2) T_n = n*3^nFinal Answer(1) The general formulas are (boxed{a_n = 3^{n-1}}) and (boxed{b_n = 2n + 1}).(2) The sum (T_n) is (boxed{n cdot 3^n}).</think>Given that ({a_n}) is a geometric sequence with (a_1 = 1) and (a_6 = 243), and (S_n) is the sum of the first (n) terms of the arithmetic sequence ({b_n}), where (b_1 = 3) and (S_5 = 35):(1) Finding the general formula for ({a_n}) and ({b_n}):- For the geometric sequence ({a_n}): - The common ratio (r) is found by solving (a_6 = 243). Since (a_n = a_1 cdot r^{n-1}), we have (a_6 = 1 cdot r^5 = 243). Solving (r^5 = 243) gives (r = 3). - Therefore, the general formula is (a_n = 3^{n-1}).- For the arithmetic sequence ({b_n}): - The sum of the first 5 terms (S_5 = 35). Using the sum formula for an arithmetic sequence (S_n = frac{n}{2} [2b_1 + (n-1)d]), we solve for the common difference (d): [ 35 = frac{5}{2} [2 cdot 3 + 4d] implies 35 = frac{5}{2} [6 + 4d] implies 35 = 15 + 10d implies 20 = 10d implies d = 2 ] - The general term is (b_n = 3 + (n-1) cdot 2 = 2n + 1).(2) Finding (T_n = a_1b_1 + a_2b_2 + ldots + a_nb_n):- Express (T_n) as the sum (sum_{k=1}^n a_k b_k). Substituting (a_k = 3^{k-1}) and (b_k = 2k + 1), we get: [ T_n = sum_{k=1}^n 3^{k-1} (2k + 1) ]- Split the sum into two parts: [ T_n = 2 sum_{k=1}^n k cdot 3^{k-1} + sum_{k=1}^n 3^{k-1} ]- The second sum is a geometric series: [ sum_{k=1}^n 3^{k-1} = frac{3^n - 1}{3 - 1} = frac{3^n - 1}{2} ]- The first sum uses the formula for (sum_{k=1}^n k cdot r^{k-1}) with (r = 3): [ sum_{k=1}^n k cdot 3^{k-1} = frac{1 - (n+1)3^n + n cdot 3^{n+1}}{(1 - 3)^2} = frac{1 + (2n - 1)3^n}{4} ]- Combining these results: [ T_n = 2 cdot frac{1 + (2n - 1)3^n}{4} + frac{3^n - 1}{2} = frac{1 + (2n - 1)3^n}{2} + frac{3^n - 1}{2} = frac{2n cdot 3^n}{2} = n cdot 3^n ]Final answers:(1) The general formulas are (boxed{a_n = 3^{n-1}}) and (boxed{b_n = 2n + 1}).(2) The sum (T_n) is (boxed{n cdot 3^n}).]

🔑:(1) Let the common ratio of the geometric sequence be q. Since {a_n} is a geometric sequence with a_1=1 and a_6=243, the common ratio q=3. Therefore, a_n=3^{n-1}. Let the common difference of the arithmetic sequence {b_n} be d. Since S_n is the sum of the first n terms of the arithmetic sequence {b_n}, with b_1=3 and S_5=35, we have 15+10d=35, thus d=2. Therefore, b_n=2n+1.(2) T_n=a_1b_1+a_2b_2+ldots+a_nb_n=3times1+5times3+ldots+(2n-1)times3^{n-2}+(2n+1)times3^{n-1} 3T_n=3times3+5times3^2+ldots+(2n-1)times3^{n-1}+(2n+1)times3^n Subtracting the two equations, we get: -2T_n=3+2times(3+3^2+ldots+3^{n-1})-(2n+1)times3^n Therefore, T_n=boxed{ncdot3^n}.

🧠:[<think>Okay, so I need to solve the equation (frac{1}{log_{2}a} + frac{1}{log_{3}a} = 2). Hmm, let's think about how to approach this. I remember that logarithms can be converted from one base to another using the change of base formula. Maybe that could help here?First, let me recall the change of base formula: (log_{b}a = frac{ln a}{ln b}) or (log_{b}a = frac{log a}{log b}) if I prefer using common logarithms. Either way, the idea is that I can express the logarithms with different bases in terms of natural logs or common logs. Since both terms in the equation have logs with different bases (2 and 3), converting them to a common base might make it easier to combine the terms.Let me try applying the change of base formula to each term. Starting with (frac{1}{log_{2}a}), using the natural logarithm for the conversion:[frac{1}{log_{2}a} = frac{1}{frac{ln a}{ln 2}} = frac{ln 2}{ln a}]Similarly, for the second term:[frac{1}{log_{3}a} = frac{1}{frac{ln a}{ln 3}} = frac{ln 3}{ln a}]So substituting these back into the original equation, we get:[frac{ln 2}{ln a} + frac{ln 3}{ln a} = 2]Since both terms on the left have the same denominator, I can combine them:[frac{ln 2 + ln 3}{ln a} = 2]Simplifying the numerator, (ln 2 + ln 3 = ln (2 times 3) = ln 6). So the equation becomes:[frac{ln 6}{ln a} = 2]Now, solving for (ln a). Multiply both sides by (ln a):[ln 6 = 2 ln a]Then, divide both sides by 2:[frac{ln 6}{2} = ln a]This means that (ln a = frac{ln 6}{2}). To solve for (a), I can exponentiate both sides with base (e) to eliminate the natural log:[a = e^{frac{ln 6}{2}}]Simplifying the exponent, I know that (e^{ln x} = x), so:[a = 6^{frac{1}{2}} = sqrt{6}]Wait, let me check that again. If (ln a = frac{ln 6}{2}), then (a = e^{frac{ln 6}{2}}). And since (e^{ln 6^{1/2}} = 6^{1/2}), that's correct. So (a = sqrt{6}).But let me verify this solution to make sure I didn't make any mistakes. Let's substitute (a = sqrt{6}) back into the original equation.First, compute (log_{2} sqrt{6}). Since (sqrt{6} = 6^{1/2}), then:[log_{2} 6^{1/2} = frac{1}{2} log_{2} 6]Therefore, (frac{1}{log_{2} sqrt{6}} = frac{2}{log_{2} 6}).Similarly, (log_{3} sqrt{6} = frac{1}{2} log_{3} 6), so:[frac{1}{log_{3} sqrt{6}} = frac{2}{log_{3} 6}]Adding these together:[frac{2}{log_{2} 6} + frac{2}{log_{3} 6}]But we can simplify each term. Let's use the change of base formula again. For the first term, (log_{2}6 = frac{ln 6}{ln 2}), so:[frac{2}{frac{ln 6}{ln 2}} = 2 times frac{ln 2}{ln 6}]Similarly, the second term:[frac{2}{frac{ln 6}{ln 3}} = 2 times frac{ln 3}{ln 6}]Adding them together:[2 left( frac{ln 2 + ln 3}{ln 6} right)]But (ln 2 + ln 3 = ln 6), so this simplifies to:[2 times frac{ln 6}{ln 6} = 2 times 1 = 2]Which matches the original equation. So yes, (a = sqrt{6}) is indeed a solution.Wait a second, but when dealing with logarithms, we have to make sure that (a) is positive and not equal to 1. Since (sqrt{6}) is approximately 2.45, which is positive and certainly not 1, so that's fine. Also, the logarithms (log_{2}a) and (log_{3}a) are defined because (a > 0) and (a ne 1). So everything checks out.Is there another possible solution? Let me think. When I solved the equation step by step, each step was an equivalence, so unless there was a step where I squared both sides or something that could introduce extraneous solutions, there shouldn't be another solution. Let me check each step again.Original equation:[frac{1}{log_{2}a} + frac{1}{log_{3}a} = 2]Converted each term using change of base:[frac{ln 2}{ln a} + frac{ln 3}{ln a} = 2]Combined terms:[frac{ln 6}{ln a} = 2]Solved for (ln a):[ln a = frac{ln 6}{2}]Exponentiated to find (a):[a = e^{frac{ln 6}{2}} = sqrt{6}]All steps seem reversible and no operations that would introduce extraneous solutions. So yes, (sqrt{6}) is the only solution.Alternatively, I can think about using another approach. Let's let (x = log_{6}a). Then, using the change of base formula again, (log_{6}a = frac{ln a}{ln 6}), so (ln a = x ln 6). But I'm not sure if this substitution helps here. Let me see.Alternatively, perhaps writing both terms with the same base. Let's suppose I use base 2 for both logarithms.Wait, (frac{1}{log_{2}a}) is straightforward, and (frac{1}{log_{3}a}) can be expressed in terms of base 2. Using change of base formula:[log_{3}a = frac{log_{2}a}{log_{2}3}]Therefore, (frac{1}{log_{3}a} = frac{log_{2}3}{log_{2}a})So substituting back into the original equation:[frac{1}{log_{2}a} + frac{log_{2}3}{log_{2}a} = 2]Factor out (frac{1}{log_{2}a}):[frac{1 + log_{2}3}{log_{2}a} = 2]Then, multiply both sides by (log_{2}a):[1 + log_{2}3 = 2 log_{2}a]So, (2 log_{2}a = log_{2}3 + 1)Wait, 1 can be written as (log_{2}2), so:[2 log_{2}a = log_{2}3 + log_{2}2 = log_{2}(3 times 2) = log_{2}6]Therefore, (2 log_{2}a = log_{2}6), which implies:[log_{2}a^2 = log_{2}6]Since the logarithms are equal, their arguments must be equal:[a^2 = 6 implies a = sqrt{6}]Since (a) must be positive, we discard the negative root.So this method also leads to (a = sqrt{6}), confirming the previous result. Good, so regardless of the approach, substitution or changing the base, we end up with the same answer.Alternatively, another way to think about the original equation: (frac{1}{log_{2}a} + frac{1}{log_{3}a} = 2). Recognizing that (frac{1}{log_{b}a} = log_{a}b), because of the reciprocal of the logarithm. Let me verify that:Yes, (log_{a}b = frac{ln b}{ln a}), and (frac{1}{log_{b}a} = frac{1}{frac{ln a}{ln b}} = frac{ln b}{ln a} = log_{a}b). So that's correct.Therefore, the original equation can be rewritten as:[log_{a}2 + log_{a}3 = 2]Using the logarithm addition rule, which states that (log_{a}b + log_{a}c = log_{a}(b times c)), we can combine these terms:[log_{a}(2 times 3) = log_{a}6 = 2]This equation means that (a^2 = 6), so (a = sqrt{6}), same as before. This is perhaps the most straightforward method once you recognize the reciprocal logarithm relationship.So in all three approaches I tried, changing the base to natural logs, converting both terms to the same base, and using the reciprocal logarithm identity, I arrived at the same answer. Therefore, I can be confident that (a = sqrt{6}) is the correct solution.Just to recap, the key steps were recognizing that the reciprocals of logarithms can be expressed as logarithms with the base and argument swapped, then using properties of logarithms to combine the terms and solve for (a). It's important to check the solution to ensure it's valid in the original equation, especially since logarithms have specific domains, but in this case, everything works out.Another point to note is that if the equation had different constants, the same approach would apply. For example, if we had (frac{1}{log_{b}a} + frac{1}{log_{c}a} = k), we could convert each term to (log_{a}b) and (log_{a}c), combine them to (log_{a}(bc)), set that equal to (k), then solve for (a) by exponentiating. This seems like a generalizable method.So, in conclusion, by leveraging logarithmic identities and properties, the problem simplifies nicely, leading us to the solution (a = sqrt{6}).Final AnswerThe value of (a) is boxed{sqrt{6}}.</think>Given the equation (frac{1}{log_{2}a} + frac{1}{log_{3}a} = 2), we need to find the value of (a).First, we use the change of base formula to convert the logarithms. The change of base formula states that (log_{b}a = frac{ln a}{ln b}). Applying this to each term, we get:[frac{1}{log_{2}a} = frac{ln 2}{ln a} quad text{and} quad frac{1}{log_{3}a} = frac{ln 3}{ln a}]Substituting these into the original equation, we have:[frac{ln 2}{ln a} + frac{ln 3}{ln a} = 2]Combining the terms on the left-hand side:[frac{ln 2 + ln 3}{ln a} = 2]Simplifying the numerator using the property (ln 2 + ln 3 = ln 6):[frac{ln 6}{ln a} = 2]Solving for (ln a):[ln 6 = 2 ln a]Dividing both sides by 2:[frac{ln 6}{2} = ln a]Exponentiating both sides to solve for (a):[a = e^{frac{ln 6}{2}} = sqrt{6}]We verify this solution by substituting (a = sqrt{6}) back into the original equation. Using the reciprocal logarithm identity (frac{1}{log_{b}a} = log_{a}b), we rewrite the equation as:[log_{a}2 + log_{a}3 = 2]Combining the logarithms:[log_{a}(2 times 3) = log_{a}6 = 2]This implies (a^2 = 6), leading to (a = sqrt{6}).Thus, the value of (a) is (boxed{sqrt{6}}).,<think>Okay, so I need to solve this equation: 1 divided by log base 2 of a plus 1 divided by log base 3 of a equals 2. Hmm, let me write that down to make sure I have it right. The equation is:1/(log₂a) + 1/(log₃a) = 2Alright, first step, maybe I can recall some logarithm properties. I remember that 1 divided by log_b a is equal to log_a b. Wait, is that right? Let me check. Yes, the change of base formula says log_b a = ln a / ln b, so 1/(log_b a) would be ln b / ln a, which is log_a b. So that's correct. So applying that here, I can rewrite each term.So 1/(log₂a) is log_a 2, and 1/(log₃a) is log_a 3. Therefore, the equation becomes:log_a 2 + log_a 3 = 2Hmm, combining those two logarithms. I remember that log_a b + log_a c = log_a (b*c). So applying that property here:log_a (2*3) = 2Which simplifies to:log_a 6 = 2Okay, so log base a of 6 equals 2. That means that a squared equals 6, right? Because log_a 6 = 2 implies a^2 = 6. So solving for a, take the square root of both sides.a = sqrt(6)But wait, logarithms have some restrictions on their bases. The base a must be positive and not equal to 1. Since sqrt(6) is positive and definitely not 1, that should be valid. Let me double-check by plugging it back into the original equation.First, compute log base 2 of sqrt(6). Let's recall that log base 2 of 6^(1/2) is (1/2) log₂6. Similarly, log base 3 of sqrt(6) is (1/2) log₃6.So the first term is 1 / [(1/2) log₂6] = 2 / log₂6The second term is 1 / [(1/2) log₃6] = 2 / log₃6Adding those together: 2/log₂6 + 2/log₃6Factor out the 2: 2[1/log₂6 + 1/log₃6]Wait a second, but according to the original equation, this sum should equal 2, so:2[1/log₂6 + 1/log₃6] = 2Divide both sides by 2: [1/log₂6 + 1/log₃6] = 1But let me compute 1/log₂6 and 1/log₃6. Using the same property as before, 1/log_b a = log_a b. Therefore:1/log₂6 = log_6 21/log₃6 = log_6 3So then log_6 2 + log_6 3 = log_6 (2*3) = log_6 6 = 1Which is correct! So indeed, the equation holds true. Therefore, a = sqrt(6) is the valid solution.But hold on, when we solve equations involving logarithms, sometimes there can be extraneous solutions. But in this case, since we squared both sides but a must be positive, and we only took the principal root, there's no negative solution here. So sqrt(6) is the only solution.Let me just recap the steps:1. Recognize that 1/log_b a = log_a b2. Convert the original equation into log_a 2 + log_a 3 = 23. Combine the logs to get log_a 6 = 24. Solve for a: a^2 = 6 => a = sqrt(6)5. Verify by substitution that this works and no extraneous solutions were introduced.Therefore, the value of a is the square root of 6.Final AnswerThe value of ( a ) is boxed{sqrt{6}}.</think>Given the equation:[frac{1}{log_{2}a} + frac{1}{log_{3}a} = 2]We use the property of logarithms that (frac{1}{log_{b}a} = log_{a}b). Applying this property, we convert each term:[log_{a}2 + log_{a}3 = 2]Using the logarithm addition rule (log_{a}b + log_{a}c = log_{a}(bc)), we combine the terms:[log_{a}(2 cdot 3) = 2 implies log_{a}6 = 2]This implies that:[a^2 = 6]Solving for (a), we take the positive square root (since (a) must be positive and not equal to 1):[a = sqrt{6}]Verification by substituting (a = sqrt{6}) back into the original equation confirms the solution is correct. Therefore, the value of (a) is:[boxed{sqrt{6}}]]

🔑:We start with the given equation,frac{1}{log_{2}a} + frac{1}{log_{3}a} = 2Using the change of base formula, we can rewrite the equation aslog_{a}2 + log_{a}3 = 2By applying the product rule of logarithms, we combine the terms on the left side:log_{a}6 = 2Now we exponentiate both sides to remove the logarithm:a^2 = 6Since a must be positive (as it is the base of a logarithm), we take the positive square root:a = sqrt{6}Therefore, the solution is boxed{a = sqrt{6}}.

🧠:[<think>Okay, let's try to tackle this problem step by step. So, we have two parts here. The first one is to find the analytical expression of f(x) and its extreme values, given that f(x) is defined in terms of its derivative at x=1 and its value at x=0. The second part is about finding the maximum value of [b(a+1)]/2 given that f(x) is greater than or equal to g(x), which is a quadratic function. Let's start with the first part.Starting with the given function:f(x) = f’(1) e^{x−1} − f(0) x + (1/2)x^2.We need to find f(x) explicitly. The problem here is that f(x) is defined in terms of its derivative at 1 and its value at 0. So, we probably need to find f’(x) first and then solve for the constants f’(1) and f(0).Let me try differentiating f(x). Let's compute f’(x):f’(x) = d/dx [f’(1) e^{x−1} - f(0)x + (1/2)x^2]The derivative of f’(1) e^{x−1} is f’(1) e^{x−1} (since the derivative of e^{x−1} is e^{x−1}).The derivative of -f(0)x is -f(0).The derivative of (1/2)x^2 is x.So, putting it all together:f’(x) = f’(1) e^{x−1} - f(0) + x.But wait, the problem mentions that f’(x) is the derivative of f(x), which we just computed. However, we have f’(x) expressed in terms of f’(1) and f(0). So, perhaps we can substitute x=1 into this derivative to find an equation involving f’(1) and f(0).Let's compute f’(1):f’(1) = f’(1) e^{1−1} - f(0) + 1.Simplify e^{0} is 1, so:f’(1) = f’(1) * 1 - f(0) + 1.Which simplifies to:f’(1) = f’(1) - f(0) + 1.Hmm, subtract f’(1) from both sides:0 = -f(0) + 1.Therefore:f(0) = 1.Okay, so we found that f(0) is 1. Now, let's use the original function f(x) to compute f(0):From f(x) = f’(1) e^{x−1} - f(0)x + (1/2)x^2,Set x=0:f(0) = f’(1) e^{−1} - f(0)*0 + (1/2)(0)^2.Therefore:f(0) = f’(1)/e.But we already found that f(0) = 1. So:1 = f’(1)/e.Therefore:f’(1) = e.So now we have both f(0) = 1 and f’(1) = e. Now we can substitute these back into the original f(x) expression to get the analytical form of f(x).Original f(x):f(x) = f’(1) e^{x−1} - f(0)x + (1/2)x^2.Substituting f’(1) = e and f(0) = 1:f(x) = e * e^{x−1} - 1 * x + (1/2)x^2.Simplify e * e^{x−1}:e^{1} * e^{x−1} = e^{1 + x -1} = e^{x}.So:f(x) = e^{x} - x + (1/2)x^2.Alright, so that's the analytical expression for f(x). Let me just verify if this makes sense. Let's check f(0):f(0) = e^0 - 0 + 0 = 1, which matches. Now check f’(x):f’(x) = derivative of e^x is e^x, derivative of -x is -1, derivative of (1/2)x^2 is x. So f’(x) = e^x -1 + x. Then f’(1) = e^1 -1 + 1 = e, which matches. So that's correct.Now, part 1 also asks for the extreme values of f(x). To find extreme values, we need to find critical points by setting f’(x) = 0 and solve for x. Then determine if those points are minima or maxima.So, f’(x) = e^x + x -1.Set that equal to zero:e^x + x -1 = 0.Hmm, solving e^x + x -1 = 0. This is a transcendental equation and might not have an analytical solution. Let's check.Let me consider the function h(x) = e^x + x -1.We can analyze its behavior. Let's compute h(x) at some points.At x=0: h(0) = 1 + 0 -1 = 0. So x=0 is a root.Wait, so h(0) = 0. Therefore, x=0 is a critical point.But let's check if there are others. Let's compute the derivative of h(x):h’(x) = e^x +1. Since e^x is always positive, so h’(x) = e^x +1 >0 for all x. Therefore, h(x) is strictly increasing. Since h(x) is strictly increasing and h(0)=0, then x=0 is the only root. Therefore, the only critical point is at x=0.Therefore, f(x) has only one critical point at x=0. Now, to determine if it's a minimum or maximum, we can check the second derivative or analyze the sign change of the first derivative.Compute f''(x):f''(x) = derivative of f’(x) = e^x +1.At x=0, f''(0) = 1 +1 = 2 >0, so the function is concave upward at x=0, which means it's a local minimum.Therefore, the only extreme value is a local minimum at x=0 with value f(0)=1.Wait, but let me check. Wait, the critical point is at x=0, which is a local minimum, but let me confirm if there are any other critical points. Since h(x) is strictly increasing and crosses zero only at x=0, there are no other critical points. Therefore, the only extreme value is the local minimum at x=0, which is 1. However, since the function f(x) tends to infinity as x approaches infinity (since e^x dominates) and as x approaches negative infinity, the (1/2)x^2 term dominates, so the function goes to positive infinity as x approaches both infinities. Therefore, the function has a global minimum at x=0, and no maxima.So, the extreme value is a global minimum at (0,1).Therefore, part 1 answer is f(x) = e^x - x + (1/2)x^2, and its only extreme value is a global minimum at x=0 with value 1.Moving on to part 2: If f(x) ≥ g(x), find the maximum value of [b(a+1)]/2.Given that g(x) = (1/2)x^2 + a x + b.We need f(x) ≥ g(x) for all x, which implies that f(x) - g(x) ≥0 for all x. Let's compute f(x) - g(x):f(x) - g(x) = [e^x - x + (1/2)x^2] - [(1/2)x^2 + a x + b] = e^x -x + (1/2)x^2 - (1/2)x^2 -a x -b = e^x -x -a x -b = e^x - (1 + a)x - b.So, the inequality is e^x - (1 + a)x - b ≥0 for all x ∈ ℝ.We need this expression to be non-negative for all real numbers x. So, the problem reduces to finding the maximum value of [b(a+1)]/2, given that e^x - (1 + a)x - b ≥0 for all x.Therefore, first, we need to find the conditions on a and b such that e^x - (1 + a)x - b ≥0 for all x. Then, given those conditions, maximize [b(a +1)]/2.So, to ensure that e^x - (1 + a)x - b is always non-negative, the function h(x) = e^x - (1 + a)x - b must lie above or touch the x-axis everywhere. This happens when the minimum of h(x) is zero. Therefore, we need to find a and b such that h(x) has its minimum value equal to zero. Because if the minimum is greater than or equal to zero, but to maximize [b(a +1)]/2, we probably need the minimal case where the minimum is exactly zero (touching the x-axis), as otherwise, if the minimum is higher, then b could be smaller, but since we want to maximize [b(a+1)]/2, which involves b and a, it's likely the maximum occurs when h(x) touches the x-axis at its minimum point.So, first, let's find the minimum of h(x) = e^x - (1 + a)x - b.Compute h’(x) = derivative of h(x) = e^x - (1 + a).Set h’(x) = 0:e^x - (1 + a) =0 → e^x = 1 + a → x = ln(1 + a), provided that 1 + a >0, so a > -1.Therefore, the critical point is at x = ln(1 + a). Then, the minimum of h(x) occurs at x = ln(1 + a) when a > -1. If a ≤ -1, then h’(x) = e^x - (1 + a). Since 1 + a ≤0, so e^x - (1 + a) is always positive (since e^x >0 and -(1 + a) ≥0), so h’(x) >0 for all x, meaning h(x) is increasing. Therefore, when a ≤ -1, h(x) is increasing, so its minimum would be at x approaching -infty. But as x approaches -infty, e^x approaches 0, and -(1 + a)x approaches infinity if a < -1 (since 1 + a is negative, so -(1 + a) is positive, multiplied by x approaching -infty, so it approaches -infty). Wait, actually:Wait, h(x) = e^x - (1 + a)x - b.If a ≤ -1, then 1 + a ≤0. Let's suppose a < -1, so 1 + a <0.Then, -(1 + a) is positive. So, the term -(1 + a)x is equal to positive * x. As x approaches -infty, this term tends to -infty (since x is negative, multiplied by positive gives negative infinity). Adding e^x (which tends to 0) and -b, so overall h(x) tends to -infty as x approaches -infty. Therefore, if a < -1, h(x) tends to -infty as x approaches -infty, so h(x) cannot be non-negative for all x. Similarly, if a = -1, then 1 + a =0, so h(x) = e^x -0 -b = e^x - b. For h(x) = e^x - b to be non-negative everywhere, we need b ≤0, since the minimum of e^x is 0 (as x approaches -infty). But e^x approaches 0, so if b ≤0, then e^x -b ≥ -b ≥0. However, if a = -1, then the expression to maximize is [b(a +1)]/2 = [b(0)]/2 =0. So, maximum value would be 0. But we need to check if a can be -1. If a =-1, then h(x) = e^x - b must be ≥0 for all x, which requires that b ≤0. Then [b(a +1)]/2 =0. So the maximum in this case is 0, but perhaps we can get a higher value when a > -1.Therefore, the only viable case is when a > -1, where the function h(x) has a critical point at x = ln(1 + a). For h(x) to be non-negative everywhere, the minimum at x = ln(1 + a) must be zero.So, set h(ln(1 + a)) =0.Compute h(ln(1 + a)):h(ln(1 + a)) = e^{ln(1 + a)} - (1 + a)ln(1 + a) - b.Simplify:e^{ln(1 + a)} = 1 + a.Therefore:h(ln(1 + a)) = (1 + a) - (1 + a)ln(1 + a) - b =0.Thus:b = (1 + a) - (1 + a)ln(1 + a).So, we have b in terms of a:b = (1 + a)(1 - ln(1 + a)).Therefore, given that a > -1, the relationship between a and b is b = (1 + a)(1 - ln(1 + a)).Now, we need to maximize the expression [b(a +1)]/2.Substitute b:[b(a +1)]/2 = [ (1 + a)(1 - ln(1 + a)) * (a +1) ] /2 = [ (1 + a)^2 (1 - ln(1 + a)) ] /2.Therefore, we need to maximize the function:F(a) = [ (1 + a)^2 (1 - ln(1 + a)) ] /2, where a > -1.So, our goal is to find the maximum of F(a) over a > -1.To find the maximum, we can take the derivative of F(a) with respect to a, set it equal to zero, and solve for a.Let’s denote t = 1 + a, where t >0.Then F(a) can be rewritten as F(t) = [ t^2 (1 - ln t) ] /2, with t >0.So, we can instead find the maximum of F(t) = (t^2 (1 - ln t))/2.Take the derivative of F(t) with respect to t:F’(t) = [ d/dt (t^2 (1 - ln t)) ] /2.First compute d/dt [t^2 (1 - ln t)]:Use product rule: derivative of t^2 * (1 - ln t) + t^2 * derivative of (1 - ln t).Derivative of t^2 is 2t.Derivative of (1 - ln t) is - (1/t).Therefore:d/dt [t^2 (1 - ln t)] = 2t(1 - ln t) + t^2 (-1/t) = 2t(1 - ln t) - t.Simplify:= 2t - 2t ln t - t = t - 2t ln t.Therefore, F’(t) = (t - 2t ln t)/2 = t(1 - 2 ln t)/2.Set F’(t) =0:t(1 - 2 ln t)/2 =0.Since t >0, the solutions come from 1 - 2 ln t =0:1 - 2 ln t =0 → 2 ln t =1 → ln t =1/2 → t= e^{1/2} = sqrt(e).Therefore, the critical point is at t= sqrt(e).Now, we need to check if this is a maximum. Let's compute the second derivative or check the sign change of F’(t).Alternatively, note that when t approaches 0+, F(t) approaches 0 (since t^2 approaches 0 and (1 - ln t) approaches infinity, but t^2 goes to zero faster). As t approaches infinity, t^2 grows but (1 - ln t) becomes negative and dominates, so F(t) tends to negative infinity. Hence, there must be a maximum at t = sqrt(e).Thus, the maximum of F(t) occurs at t = sqrt(e).Therefore, the maximum value is F(sqrt(e)) = [ (sqrt(e))^2 (1 - ln sqrt(e)) ] /2.Compute:(sqrt(e))^2 = e.ln sqrt(e) = ln e^{1/2} = (1/2) ln e =1/2.Thus:1 - ln sqrt(e) =1 -1/2 =1/2.Therefore:F(sqrt(e)) = [ e * (1/2) ] /2 = (e/2)/2 = e/4.So, the maximum value of [b(a +1)]/2 is e/4.Therefore, the answer is e/4.But let me check this calculation again.Wait, F(t) = [ t^2 (1 - ln t) ] /2.At t= sqrt(e):t^2 = e.1 - ln t =1 - ln(e^{1/2}) =1 - (1/2) =1/2.So, F(t)= (e * 1/2)/2= e/4. Yes, that's correct.Therefore, the maximum value is e/4.So, part 2 answer is e/4.Therefore, summarizing:1. f(x) = e^x -x + (1/2)x^2, with a global minimum at x=0, f(0)=1.2. The maximum value of [b(a +1)]/2 is e/4.Final Answer1. The analytical expression of ( f(x) ) is (boxed{e^x - x + frac{1}{2}x^2}) with a minimum value of (boxed{1}).2. The maximum value of (frac{b(a+1)}{2}) is (boxed{dfrac{e}{4}}).</think>Given the function ( f(x) = f'(1)e^{x-1} - f(0)x + frac{1}{2}x^2 ), we need to find its analytical expression and extreme values. We also need to find the maximum value of ( frac{b(a+1)}{2} ) given that ( f(x) geq g(x) ) where ( g(x) = frac{1}{2}x^2 + ax + b ).1. Finding the analytical expression of ( f(x) ) and its extreme values: - First, we compute the derivative of ( f(x) ): [ f'(x) = f'(1)e^{x-1} - f(0) + x ] - Evaluating ( f'(1) ): [ f'(1) = f'(1)e^{0} - f(0) + 1 implies 0 = -f(0) + 1 implies f(0) = 1 ] - Evaluating ( f(0) ) from the original function: [ f(0) = f'(1)e^{-1} implies 1 = frac{f'(1)}{e} implies f'(1) = e ] - Substituting ( f'(1) = e ) and ( f(0) = 1 ) into ( f(x) ): [ f(x) = e cdot e^{x-1} - x + frac{1}{2}x^2 = e^x - x + frac{1}{2}x^2 ] - Finding the critical points by setting ( f'(x) = 0 ): [ f'(x) = e^x + x - 1 ] Solving ( e^x + x - 1 = 0 ), we find ( x = 0 ) is the only critical point. The second derivative ( f''(x) = e^x + 1 ) is positive at ( x = 0 ), indicating a local minimum. Since ( f(x) ) tends to infinity as ( x to pminfty ), the minimum at ( x = 0 ) is a global minimum. - Answer: The analytical expression of ( f(x) ) is (boxed{e^x - x + frac{1}{2}x^2}) with a minimum value of (boxed{1}).2. Finding the maximum value of ( frac{b(a+1)}{2} ): - Given ( f(x) geq g(x) ), we have: [ e^x - (1 + a)x - b geq 0 quad forall x in mathbb{R} ] - The function ( h(x) = e^x - (1 + a)x - b ) must be non-negative for all ( x ). The minimum occurs at ( x = ln(1 + a) ) for ( a > -1 ). Setting ( h(ln(1 + a)) = 0 ): [ (1 + a) - (1 + a)ln(1 + a) - b = 0 implies b = (1 + a)(1 - ln(1 + a)) ] - Maximizing ( frac{b(a+1)}{2} ): [ frac{(1 + a)^2 (1 - ln(1 + a))}{2} ] - Let ( t = 1 + a ), then we maximize ( frac{t^2 (1 - ln t)}{2} ). The critical point is ( t = sqrt{e} ), leading to: [ frac{(sqrt{e})^2 (1 - frac{1}{2})}{2} = frac{e cdot frac{1}{2}}{2} = frac{e}{4} ] - Answer: The maximum value of ( frac{b(a+1)}{2} ) is (boxed{dfrac{e}{4}}).,<think>Okay, let's tackle this problem step by step. First, I need to find the analytical expression of the function f(x) and its extreme values. Then, I have to find the maximum value of (b(a+1))/2 given that f(x) is always greater than or equal to g(x). Alright, let's start with the first part.So, the function f(x) is given as f(x) = f’(1)e^{x−1} − f(0)x + (1/2)x². Hmm, interesting. It seems that f(x) is defined in terms of its derivative at x=1 and its value at x=0. So, to find f(x), I probably need to solve for f’(1) and f(0). That might involve setting up some equations by differentiating f(x) and plugging in specific values of x.Let me write down the given function again:f(x) = f’(1)e^{x−1} − f(0)x + (1/2)x².First, let's find f(0). Plugging x=0 into the equation:f(0) = f’(1)e^{-1} − f(0)*0 + (1/2)(0)^2f(0) = f’(1)e^{-1}So, f(0) is equal to f’(1)/e. Got that. Let's keep that in mind: f(0) = f’(1)/e.Next, let's find the derivative f’(x). Differentiating f(x):f’(x) = d/dx [f’(1)e^{x−1}] − d/dx [f(0)x] + d/dx [(1/2)x²]f’(x) = f’(1)e^{x−1}*(1) − f(0) + xSo, f’(x) = f’(1)e^{x−1} − f(0) + x.Now, the problem mentions f’(1), so let's compute f’(1):f’(1) = f’(1)e^{1−1} − f(0) + 1f’(1) = f’(1)e^{0} − f(0) + 1Since e^0 is 1, this simplifies to:f’(1) = f’(1) - f(0) + 1Hmm, let's rearrange this equation:f’(1) - f’(1) = -f(0) + 10 = -f(0) + 1So, f(0) = 1.Wait, that's nice. So earlier we had f(0) = f’(1)/e, and now we found f(0) = 1. Therefore, f’(1)/e = 1, which implies f’(1) = e.Great, so now we can substitute back f’(1) = e and f(0) = 1 into the original function f(x):f(x) = e * e^{x−1} - 1 * x + (1/2)x²Simplify e * e^{x−1}: that's e^{1} * e^{x−1} = e^{x}. So:f(x) = e^{x} - x + (1/2)x².Alright, so that's the analytical expression for f(x). Let me double-check that.Original expression:f(x) = f’(1)e^{x−1} − f(0)x + (1/2)x²We found f’(1) = e, f(0)=1, so substituting:f(x) = e * e^{x−1} - 1 * x + (1/2)x² = e^{x} - x + (1/2)x². Yep, that seems right.Now, let's check the derivative f’(x) again with these values. Earlier, we had:f’(x) = f’(1)e^{x−1} - f(0) + x = e * e^{x−1} - 1 + x = e^{x} -1 + x.Wait, but if f(x) = e^x - x + (1/2)x², then the derivative should be f’(x) = e^x -1 + x, which matches. So that's consistent. Also, at x=1, f’(1) = e^1 -1 +1 = e, which matches the given f’(1)=e. And f(0) = e^0 -0 +0 =1, which matches f(0)=1. Perfect, so the function is correctly derived.Now, part 1 also asks for its extreme values. To find extreme values, we need to find critical points by setting the derivative equal to zero and solve for x.So, set f’(x) = e^x -1 + x = 0.Wait, solve e^x + x -1 =0. Hmm, that's a transcendental equation. Let's see if we can find real solutions.Let me denote the equation as e^x + x -1 =0. Let's check x=0: e^0 +0 -1=1-1=0. So x=0 is a solution. Wait, that's interesting. So x=0 is a critical point.Is there another critical point? Let's check.Let me analyze the function f’(x) = e^x + x -1.We can check its behavior. The derivative of f’(x) with respect to x is f''(x) = e^x +1, which is always positive since e^x >0 for all x. Therefore, f’(x) is strictly increasing. Since f’(x) is strictly increasing and we found that at x=0, f’(0)=0. Therefore, x=0 is the only critical point. Because if f’(x) is strictly increasing, it can cross zero at most once. Since it does cross at x=0, that's the only critical point.Therefore, the function f(x) has an extreme value at x=0. Since the second derivative f''(x) = e^x +1 is always positive, this critical point is a local minimum. Moreover, since it's the only critical point, it's the global minimum.Therefore, the extreme value is a minimum at x=0, and the value is f(0)=1.Wait, but let me confirm. If we have a critical point at x=0, and since f''(x) is always positive, then yes, it's a local minimum. But let me compute f''(x):Yes, f''(x) is the derivative of f’(x):f’(x) = e^x + x -1f''(x) = e^x +1, which is always greater than zero. So indeed, x=0 is the only critical point and it's a local minimum. Since the function is convex (second derivative always positive), this is the global minimum.Therefore, the function f(x) has a global minimum at x=0 with value 1. Are there any other extreme values? Since there's only one critical point, and the function tends to infinity as x approaches infinity (since e^x dominates) and as x approaches negative infinity, the x^2 term dominates, so the function goes to infinity in both directions. Therefore, the only extreme value is the minimum at x=0.Wait, but hold on. Wait, f’(x) = e^x + x -1. At x=0, derivative is zero, so that's the critical point. Since the function is convex, that's the only minimum. So, I think that's correct.So, for part 1, the analytical expression of f(x) is e^x - x + (1/2)x², and it has a global minimum at x=0 with value 1. That's the only extreme value.Moving on to part 2: If f(x) ≥ g(x), find the maximum value of (b(a+1))/2.Given that g(x) = (1/2)x² + a x + b, and f(x) is e^x -x + (1/2)x².So, the condition is that for all real numbers x, e^x -x + (1/2)x² ≥ (1/2)x² + a x + b.Simplifying this inequality: subtract (1/2)x² from both sides:e^x -x ≥ a x + bWhich simplifies to:e^x -x - a x - b ≥ 0Combine like terms:e^x - (1 + a)x - b ≥ 0 for all x ∈ ℝ.So, we need the function h(x) = e^x - (1 + a)x - b to be non-negative for all x. The problem is to find the maximum value of (b(a + 1))/2 given this condition.Our goal is to maximize (b(a + 1))/2, given that e^x - (1 + a)x - b ≥ 0 for all x.So, first, we need to find the conditions on a and b such that h(x) = e^x - (1 + a)x - b is always non-negative. Then, given those conditions, find the maximum value of (b(a + 1))/2.This seems like an optimization problem with constraints. The function h(x) must be tangent to the x-axis at its minimum point; otherwise, if it dips below, the inequality would not hold. So, to ensure h(x) ≥ 0 for all x, the minimal value of h(x) must be zero. That occurs when h(x) touches the x-axis at its minimum. Therefore, we need to find a and b such that h(x) has a minimum value of zero.So, let's proceed step by step.First, find the minimum of h(x) = e^x - (1 + a)x - b.To find the critical points, compute h’(x):h’(x) = e^x - (1 + a).Set h’(x) = 0:e^x - (1 + a) = 0 → e^x = 1 + a → x = ln(1 + a).Therefore, the function h(x) has a critical point at x = ln(1 + a), provided that 1 + a > 0, since the argument of the natural logarithm must be positive. So, 1 + a > 0 → a > -1.If 1 + a ≤ 0, then the equation e^x = 1 + a has no real solution, meaning h’(x) = e^x - (1 + a) is always positive (since e^x > 0 and 1 + a ≤ 0). Therefore, h(x) is strictly increasing if a ≤ -1. But if h(x) is strictly increasing, then to have h(x) ≥ 0 for all x, we need the limit as x approaches -infty to be ≥ 0. However, as x approaches -infty, e^x approaches 0, and -(1 + a)x approaches positive infinity if a < -1 (since 1 + a < 0, so -(1 + a)x is positive when x approaches -infty). Wait, but let's check:If a < -1, then 1 + a < 0. So, -(1 + a)x = |1 + a| * x, but x approaches -infty, so |1 + a| * (-infty) approaches -infty. Therefore, h(x) = e^x - (1 + a)x - b ≈ - (1 + a)x - b. Since x approaches -infty, and 1 + a is negative, -(1 + a) is positive, so -(1 + a)x ≈ positive infinity * x, which is negative infinity. Therefore, h(x) approaches -infty as x approaches -infty if a < -1. Therefore, h(x) cannot be non-negative for all x if a < -1. Similarly, if a = -1, then h(x) = e^x - 0 * x - b = e^x - b. For this to be non-negative for all x, we need b ≤ 0, because the minimum of e^x is 0 (as x approaches -infty). But e^x approaches 0 as x approaches -infty, so h(x) approaches -b. Therefore, to have h(x) ≥ 0, we need -b ≥ 0 → b ≤ 0. Then, the maximum value of (b(a + 1))/2 when a = -1 would be (b(0))/2 = 0. But we are to maximize this expression, so maybe this case gives zero, which is less than other possible cases.Therefore, the only possible case where h(x) ≥ 0 for all x is when a > -1, so that the critical point x = ln(1 + a) exists, and at that critical point, the minimum value of h(x) is zero.Therefore, we must have:h(x) ≥ 0 for all x, which requires that the minimum of h(x) is zero. Therefore, the minimum occurs at x = ln(1 + a), and at that point, h(x) = 0.So, let's compute h(ln(1 + a)):h(ln(1 + a)) = e^{ln(1 + a)} - (1 + a)ln(1 + a) - b = (1 + a) - (1 + a)ln(1 + a) - b.Set this equal to zero:(1 + a) - (1 + a)ln(1 + a) - b = 0Solve for b:b = (1 + a) - (1 + a)ln(1 + a) = (1 + a)(1 - ln(1 + a))So, given that, we have b = (1 + a)(1 - ln(1 + a)). Therefore, our goal is to maximize the expression:(b(a + 1))/2 = [ (1 + a)(1 - ln(1 + a)) * (a + 1) ] / 2 = (1 + a)^2(1 - ln(1 + a)) / 2.So, let’s denote t = 1 + a, where t > 0 (since a > -1). Then, the expression becomes:(t)^2(1 - ln t) / 2.So, we need to maximize the function F(t) = (t²(1 - ln t)) / 2 for t > 0.Therefore, the problem reduces to finding the maximum of F(t) = (t²(1 - ln t))/2 over t > 0.Let’s compute the derivative of F(t) with respect to t to find critical points.First, write F(t):F(t) = (t²(1 - ln t)) / 2.Compute F’(t):Use product rule:d/dt [t²(1 - ln t)] = 2t(1 - ln t) + t²(-1/t) = 2t(1 - ln t) - t.Therefore, F’(t) = [2t(1 - ln t) - t] / 2 = [2t(1 - ln t) - t] / 2 = [2t - 2t ln t - t] / 2 = [t - 2t ln t] / 2 = t(1 - 2 ln t)/2.Set F’(t) = 0:t(1 - 2 ln t)/2 = 0.Since t > 0, the solutions are when 1 - 2 ln t = 0:1 - 2 ln t = 0 → 2 ln t = 1 → ln t = 1/2 → t = e^{1/2} = √e.Therefore, the critical point is at t = √e. We need to check whether this is a maximum.Second derivative test or sign changes.Let’s check the sign of F’(t) around t = √e.For t < √e, say t = 1 (since √e ≈ 1.648), ln t = 0, so 1 - 2 ln t = 1 > 0. Wait, no. For t < √e, ln t < 1/2, so 1 - 2 ln t > 0. Therefore, F’(t) = t(1 - 2 ln t)/2 > 0 when t < √e.For t > √e, ln t > 1/2, so 1 - 2 ln t < 0, hence F’(t) < 0.Therefore, F(t) increases from t=0 to t=√e and decreases from t=√e to infinity. Therefore, t=√e is the point of maximum.Therefore, the maximum of F(t) occurs at t=√e. Let's compute F(√e):F(√e) = ( (√e)^2 (1 - ln √e ) ) / 2.Compute each part:(√e)^2 = e.ln √e = ln e^{1/2} = 1/2.So, 1 - ln √e = 1 - 1/2 = 1/2.Therefore, F(√e) = (e * 1/2) / 2 = e / 4.Therefore, the maximum value of (b(a + 1))/2 is e/4.Wait, let me verify the calculation again:F(√e) = ( (√e)^2 * (1 - ln √e ) ) / 2(√e)^2 = e.ln √e = 1/2.Thus, 1 - 1/2 = 1/2.Therefore, F(√e) = e * (1/2) / 2 = e / 4.Yes, that's correct.Therefore, the maximum value of (b(a +1))/2 is e/4.But let's confirm this result by backtracking the variables.Given t = 1 + a = √e, so a = √e -1.Then, b = (1 + a)(1 - ln(1 + a)) = √e (1 - ln √e) = √e (1 - 1/2) = √e * 1/2 = √e / 2.Therefore, (b(a +1))/2 = ( (√e / 2) * √e ) / 2 = ( (e / 2 ) ) / 2 = e /4. Yep, same result.So, this checks out.Therefore, the maximum value is e/4.Hence, the answer to part 2 is e/4.But let me just recap the entire process to ensure I didn't miss any steps.We started with the condition f(x) ≥ g(x) for all x, which led us to e^x - (1 + a)x - b ≥ 0 for all x. We realized that for this inequality to hold, the function h(x) must touch the x-axis at its minimum point, leading us to find a and b in terms of t = 1 + a, then express b in terms of t. Substituting back, we converted the problem into maximizing a function of t, found the critical point via differentiation, verified it's a maximum, and computed the value.Everything seems to check out. So, the maximum value of (b(a +1))/2 is e/4.Final Answer1. The analytical expression of ( f(x) ) is (boxed{e^x - x + frac{1}{2}x^2}) with a global minimum at ( x = 0 ) of value (boxed{1}).2. The maximum value of ( frac{b(a+1)}{2} ) is (boxed{dfrac{e}{4}}).</think>Given the function ( f(x) = f'(1)e^{x-1} - f(0)x + frac{1}{2}x^2 ), we need to find its analytical expression and extreme values. Then, given ( f(x) geq g(x) ) where ( g(x) = frac{1}{2}x^2 + ax + b ), we need to find the maximum value of ( frac{b(a+1)}{2} ).1. Finding the analytical expression of ( f(x) ) and its extreme values: - First, we find ( f(0) ) by substituting ( x = 0 ): [ f(0) = f'(1)e^{-1} ] - Next, we compute the derivative ( f'(x) ): [ f'(x) = f'(1)e^{x-1} - f(0) + x ] - Evaluating ( f'(1) ): [ f'(1) = f'(1)e^{0} - f(0) + 1 implies f'(1) = f'(1) - f(0) + 1 implies f(0) = 1 ] - Since ( f(0) = frac{f'(1)}{e} ), we get ( f'(1) = e ). Substituting back into ( f(x) ): [ f(x) = e cdot e^{x-1} - x + frac{1}{2}x^2 = e^x - x + frac{1}{2}x^2 ] - To find the extreme values, we solve ( f'(x) = e^x - 1 + x = 0 ). The solution ( x = 0 ) is a critical point. The second derivative ( f''(x) = e^x + 1 ) is always positive, indicating a local minimum at ( x = 0 ). This is the global minimum since the function is convex. The value at ( x = 0 ) is ( f(0) = 1 ).2. Finding the maximum value of ( frac{b(a+1)}{2} ): - The condition ( f(x) geq g(x) ) simplifies to ( e^x - (1 + a)x - b geq 0 ). The function ( h(x) = e^x - (1 + a)x - b ) must be non-negative for all ( x ). - The minimum of ( h(x) ) occurs at ( x = ln(1 + a) ) (for ( a > -1 )). Setting ( h(ln(1 + a)) = 0 ): [ (1 + a) - (1 + a)ln(1 + a) - b = 0 implies b = (1 + a)(1 - ln(1 + a)) ] - We maximize ( frac{b(a + 1)}{2} = frac{(1 + a)^2(1 - ln(1 + a))}{2} ). Let ( t = 1 + a ), then maximize ( frac{t^2(1 - ln t)}{2} ). - The critical point ( t = sqrt{e} ) gives the maximum value. Substituting back: [ frac{(sqrt{e})^2(1 - ln sqrt{e})}{2} = frac{e cdot frac{1}{2}}{2} = frac{e}{4} ] Final Answer1. The analytical expression of ( f(x) ) is (boxed{e^x - x + frac{1}{2}x^2}) with a global minimum at ( x = 0 ) of value (boxed{1}).2. The maximum value of ( frac{b(a+1)}{2} ) is (boxed{dfrac{e}{4}}).]

🔑:1. From the given information, we have f'(x) = f'(1)e^{x-1} - f(0) + x. Setting x = 1, we get f'(0) = f'(0) - f(0) + 1, which implies f(0) = 1. Also, f(0) = frac{f'(1)}{e}, so f'(1) = e. Therefore, f(x) = e^x - x + frac{1}{2}x^2. Differentiating f(x), we get f'(x) = e^x + x - 1. Since f'(x) is increasing on mathbb{R} and f'(0) = 0, we have f'(x) 0 when x > 0. Thus, x = 0 is the minimum point of f(x), and the minimum value of f(x) is f(0) = 1.2. From f(x) geq g(x), we get e^x - (a+1)x - b geq 0. Let h(x) = e^x - (a+1)x - b. Then, h'(x) = e^x - (a+1). When a + 1 leq 0, h'(x) > 0, so h(x) is increasing on mathbb{R}. But lim_{x to -infty} h(x) = -infty, which contradicts h(x) geq 0. When a + 1 > 0, from h'(x) > 0, we get x > ln(a+1), and from h'(x) 0. Let F(x) = x^2 - x^2ln x, for x > 0. Then, F'(x) = x(1 - 2ln x). From F'(x) > 0, we get 0 sqrt{e}. Therefore, the maximum value of F(x) is achieved at x = sqrt{e}, and F_{max}(x) = F(sqrt{e}) = frac{e}{2}. Thus, when a = sqrt{e} - 1 and b = frac{sqrt{e}}{2}, the maximum value of (a+1)b is frac{e}{2}, and the maximum value of frac{(a+1)b}{2} is boxed{frac{e}{4}}.

🧠:[<think>Okay, so I need to solve this problem about a hyperbola. Let me start by understanding the given information and what's being asked. The hyperbola is given by the equation (x^2 - frac{y^2}{3} = 1). The left focus is F, and we draw a line AB through F that intersects the hyperbola at points A and B. Then, we construct the tangents (l_1) and (l_2) at points A and B, respectively, and they intersect at point M. The two parts are: (1) Find the angle between the vector MF and AB; (2) Find the minimum value of the area of triangle MAB.First, let me recall some properties of hyperbolas. The standard form is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1). Comparing this with the given equation (x^2 - frac{y^2}{3} = 1), we can see that (a^2 = 1) so (a = 1), and (b^2 = 3) so (b = sqrt{3}). For hyperbolas, the distance to the foci is given by (c^2 = a^2 + b^2). So here, (c^2 = 1 + 3 = 4), so (c = 2). The foci are at ((pm c, 0)), which are ((pm 2, 0)). Since we need the left focus, F is at (-2, 0).Now, the line AB passes through F(-2, 0) and intersects the hyperbola at points A and B. The first part is to find the angle between vector MF and AB. The second part is the minimum area of triangle MAB.Let me tackle part (1) first. To find the angle between vector MF and line AB, we need to find the direction of MF and the direction of AB. The angle between them can be found using the dot product formula if we can express these vectors in coordinate form.But since we don't know the specific line AB (it can be any line through F), this angle might be a constant, independent of the choice of AB. Maybe there's a geometric property here. Alternatively, we might need to parametrize the line AB and find expressions for M, then compute the vectors.Alternatively, maybe using the reflection property of hyperbolas? Wait, hyperbolas have reflection properties similar to ellipses but different. For hyperbolas, the tangent at a point makes equal angles with the lines from that point to each focus. But here, we are dealing with tangents at A and B, intersecting at M. Maybe M has some relation to the other focus? Let me think.Alternatively, perhaps using poles and polars. The intersection of the tangents at A and B is the pole of the chord AB with respect to the hyperbola. So if AB is a chord through F, then the pole of AB is M. The polar of F with respect to the hyperbola might be a specific line, and maybe there's a relation here.Alternatively, let's parametrize the line AB. Let's assume the line passes through F(-2, 0). Let me parametrize AB with a slope. Let’s suppose the line AB has a slope m, so its equation is (y = m(x + 2)). This line intersects the hyperbola (x^2 - frac{y^2}{3} = 1). Let's substitute y from the line into the hyperbola equation.Substituting (y = m(x + 2)) into the hyperbola equation:(x^2 - frac{[m(x + 2)]^2}{3} = 1)Expanding that:(x^2 - frac{m^2(x^2 + 4x + 4)}{3} = 1)Multiply through by 3 to eliminate the denominator:(3x^2 - m^2(x^2 + 4x + 4) = 3)Expand:(3x^2 - m^2x^2 - 4m^2x - 4m^2 = 3)Combine like terms:((3 - m^2)x^2 - 4m^2x - (4m^2 + 3) = 0)This is a quadratic equation in x. Let me denote coefficients:A = (3 - m^2)B = -4m^2C = -(4m^2 + 3)So quadratic equation is (Ax^2 + Bx + C = 0)The solutions are x = [-B ± sqrt(B^2 - 4AC)]/(2A)But since the line passes through F(-2, 0), which is a focus, perhaps this quadratic equation has two roots, one of which is x = -2? Wait, but if the line passes through F(-2, 0), then substituting x = -2 into the line equation gives y = 0, so the point (-2, 0) is on the line. But does this point lie on the hyperbola? Let's check:Plugging into hyperbola equation: (-2)^2 - (0)^2 /3 = 4 - 0 = 4 ≠ 1. So F is not on the hyperbola. Therefore, the line AB passes through F(-2, 0) and intersects the hyperbola at two distinct points A and B. Therefore, the quadratic equation above will have two roots corresponding to x-coordinates of A and B.So the quadratic equation (3 - m^2)x^2 -4m^2x - (4m^2 +3) =0. Let me call the roots x1 and x2. Then the sum x1 + x2 = -B/A = (4m^2)/(3 - m^2), and the product x1x2 = C/A = [-(4m^2 +3)]/(3 - m^2)But maybe instead of working with x-coordinates, it's easier to parametrize the line AB with parameter t. Alternatively, use parametric equations for the hyperbola. Wait, hyperbola parametric equations are usually (a secθ, b tanθ) or (a cosh t, b sinh t). Maybe using the secθ and tanθ parametrization might be useful here.But perhaps proceed step by step. Let's find coordinates of points A and B. Since the line intersects the hyperbola at A and B, and passes through F(-2,0). Let's denote the parametric equations for line AB. Let’s set parameter t such that when t=0, we are at F(-2, 0). Then, for some direction vector (dx, dy), the parametric equations are x = -2 + t*dx, y = 0 + t*dy. But maybe better to parametrize using slope.Alternatively, let's let the line AB have a direction vector (1, m), so parametric equations starting from F(-2, 0):x = -2 + ty = 0 + mtThen substitute into hyperbola equation:(-2 + t)^2 - (mt)^2 /3 =1Expand:(4 -4t + t^2) - (m^2 t^2)/3 =1Combine terms:t^2(1 - m^2/3) -4t + (4 -1) =0So:t^2(1 - m^2/3) -4t +3=0Multiply through by 3 to eliminate denominator:t^2(3 - m^2) -12t +9=0So quadratic equation in t: (3 - m^2)t^2 -12t +9=0Solutions:t = [12 ± sqrt(144 - 4*(3 - m^2)*9)]/(2*(3 - m^2))Calculate discriminant:144 - 36*(3 - m^2) =144 -108 +36m^2=36 +36m^2=36(1 +m^2)Thus,t = [12 ±6sqrt(1 + m^2)]/(2*(3 - m^2)) = [6 ±3sqrt(1 + m^2)]/(3 - m^2)Therefore, t = [6 +3sqrt(1 + m^2)]/(3 - m^2) and t = [6 -3sqrt(1 + m^2)]/(3 - m^2)Therefore, the points A and B are at:x = -2 + ty = mtSo coordinates of A and B depend on t. But this seems a bit messy. Maybe there's a better way.Alternatively, since the problem involves tangents at A and B intersecting at M, maybe we can use the equation of the tangent to the hyperbola at a point (x0, y0). The tangent line equation is (xx_0 - frac{yy_0}{3} =1).Given that, if A is (x1, y1), then tangent l1 is (x x1 - frac{y y1}{3} =1), and similarly for l2 at B(x2, y2): (x x2 - frac{y y2}{3} =1). The intersection point M of l1 and l2 can be found by solving these two equations.To find M, we need to solve:(x x1 - frac{y y1}{3} =1)(x x2 - frac{y y2}{3} =1)Let me write these as:1) x x1 - (y y1)/3 =12) x x2 - (y y2)/3 =1We can solve this system for x and y. Let's denote variables as x and y. Let me arrange the equations:From equation 1:x x1 =1 + (y y1)/3Similarly, equation 2:x x2 =1 + (y y2)/3Let me solve for x from both equations and set them equal:From equation 1: x = [1 + (y y1)/3]/x1From equation 2: x = [1 + (y y2)/3]/x2Therefore:[1 + (y y1)/3]/x1 = [1 + (y y2)/3]/x2Cross-multiplying:x2 [1 + (y y1)/3] = x1 [1 + (y y2)/3]Multiply out:x2 + (x2 y y1)/3 = x1 + (x1 y y2)/3Bring terms with y to one side:(x2 y y1 - x1 y y2)/3 = x1 - x2Factor y:y [ (x2 y1 - x1 y2)/3 ] = x1 - x2Therefore,y = [3(x1 - x2)] / [x2 y1 - x1 y2]Similarly, once we have y, we can substitute back into one of the equations to find x.But this seems complicated. Maybe there's a better way. Alternatively, perhaps using the fact that points A and B lie on the hyperbola and line AB passes through F(-2, 0). Also, since M is the intersection of tangents at A and B, perhaps we can use some properties of pole and polar.In hyperbola theory, the pole of the line AB with respect to the hyperbola is the point M, since the polar of M is the line AB. The equation of the polar line of a point M(u, v) with respect to the hyperbola (x^2 - y^2/3 =1) is given by (u x - (v y)/3 =1). But if line AB is the polar of M, then AB has the equation (u x - (v y)/3 =1). But we know that line AB passes through F(-2, 0). So substituting F into the polar equation:u*(-2) - (v*0)/3 =1 => -2u =1 => u= -1/2Therefore, the x-coordinate of M is -1/2. Wait, this seems important. So if M is the pole of AB, then the polar of M is AB, which passes through F(-2,0), so substituting F into the polar equation gives u*(-2) =1 => u = -1/2. Therefore, M must lie on x = -1/2. Wait, that is, the x-coordinate of M is fixed at -1/2, regardless of the line AB? That's interesting.Therefore, regardless of the line AB through F, the intersection point M of the two tangents at A and B lies on the vertical line x = -1/2. So M is (-1/2, y_M) for some y_M. That seems like a crucial insight.So for part (1), we need to find the angle between vector MF and AB. Since F is at (-2,0) and M is at (-1/2, y_M), the vector MF is from M to F: (-2 - (-1/2), 0 - y_M) = (-3/2, -y_M). The line AB has some slope, which depends on the choice of AB, but since all lines AB pass through F, and M lies on x = -1/2, maybe there's a geometric relation here.Alternatively, perhaps the angle is always 90 degrees? Let me check with a specific case. Let's take AB as the horizontal line passing through F(-2,0). But the hyperbola is (x^2 - y^2/3 =1). A horizontal line through F is y=0. Substituting y=0 into the hyperbola equation gives x^2 =1, so x=±1. So points A and B would be (-1,0) and (1,0), but the line passes through F(-2,0). Wait, but if the line is y=0, it passes through F(-2,0) and intersects the hyperbola at (-1,0) and (1,0). Wait, but that line passes through (-2,0), (-1,0), (1,0). But the hyperbola is only at x=±1 when y=0. So points A and B are (-1,0) and (1,0). Then, the tangents at A and B. The tangent at A(-1,0) is given by substituting into the tangent equation: x*(-1) - y*0/3 =1 => -x =1 => x = -1. Similarly, tangent at B(1,0) is x*1 - y*0/3 =1 => x=1. So the tangents are x=-1 and x=1, which intersect at... Wait, x=-1 and x=1 are parallel? No, they are vertical lines at x=-1 and x=1, which are parallel? No, vertical lines are parallel? Wait, vertical lines are all parallel to each other. Therefore, they don't intersect unless they are the same line. But x=-1 and x=1 are distinct and parallel, so they do not intersect. Therefore, in this case, there is no point M. That's a problem. So my assumption that taking AB as the x-axis leads to tangents that don't intersect is contradictory. Therefore, this case is invalid. Therefore, the line AB cannot be the x-axis, because then the tangents at A and B are vertical lines which do not intersect. Hence, the line AB cannot be horizontal. Therefore, the line AB must have a slope such that the tangents at A and B intersect.Therefore, in the previous parametrization, we must have m ≠0. Hmm. So perhaps the line AB cannot be horizontal. Similarly, maybe vertical lines? Let me check a vertical line through F(-2,0). A vertical line is x = -2. Substitute into hyperbola equation: (-2)^2 - y^2/3 =1 =>4 - y^2/3 =1 => y^2=9 => y=±3. Therefore, points A and B are (-2,3) and (-2,-3). Then, the tangents at A and B. For point A(-2,3), the tangent is x*(-2) - y*3/3 =1 => -2x - y =1. Similarly, tangent at B(-2,-3) is x*(-2) - y*(-3)/3 =1 => -2x + y =1. So equations: -2x - y =1 and -2x + y =1. Solving these: subtract the two equations: (-2x - y) - (-2x + y) =1 -1 => -2y=0 => y=0. Then substituting back: -2x -0=1 => x= -1/2. Therefore, intersection point M is (-1/2, 0). So M is (-1/2,0). Then vector MF is from M(-1/2,0) to F(-2,0): (-2 - (-1/2), 0 -0)= (-3/2, 0). The line AB is vertical (x=-2). So AB is vertical, vector MF is horizontal. Therefore, the angle between them is 90 degrees. So the angle is 90 degrees in this case.Similarly, take another case. Let's choose a different line AB. Let me choose a line with slope 1 through F(-2,0). So equation y = x + 2. Substitute into hyperbola:x^2 - (x + 2)^2 /3 =1Multiply through by 3:3x^2 - (x^2 +4x +4) =3So 3x^2 -x^2 -4x -4 =32x^2 -4x -7=0Solutions x = [4 ± sqrt(16 +56)]/4 = [4 ± sqrt(72)]/4 = [4 ±6sqrt(2)]/4 = [2 ±3sqrt(2)]/2So x=1 + (3sqrt(2))/2 ≈1 +2.121≈3.121 and x=1 - (3sqrt(2))/2≈1 -2.121≈-1.121Thus, points A and B have x-coordinates approximately 3.121 and -1.121. The exact coordinates are (1 + (3√2)/2, (1 + (3√2)/2) +2) and (1 - (3√2)/2, (1 - (3√2)/2) +2). Simplify y-coordinates: For x=1 + (3√2)/2, y= x +2= 3 + (3√2)/2; for x=1 - (3√2)/2, y=3 - (3√2)/2.Therefore, points A and B are (1 + (3√2)/2, 3 + (3√2)/2) and (1 - (3√2)/2, 3 - (3√2)/2).Now, find the equations of the tangents at A and B.Tangent at A: x*x1 - (y*y1)/3 =1Substituting x1=1 + (3√2)/2, y1=3 + (3√2)/2:x*(1 + (3√2)/2) - [y*(3 + (3√2)/2)]/3 =1Similarly for tangent at B: x*(1 - (3√2)/2) - [y*(3 - (3√2)/2)]/3 =1These equations look complicated, but maybe we can simplify them.Let me compute the tangent at A:First term: x*(1 + (3√2)/2)Second term: [y*(3 + (3√2)/2)]/3 = y*(1 + (√2)/2)Therefore, tangent equation:x*(1 + (3√2)/2) - y*(1 + (√2)/2) =1Similarly for tangent at B:x*(1 - (3√2)/2) - y*(1 - (√2)/2) =1Now, to find the intersection point M of these two tangents, we need to solve:1) x*(1 + (3√2)/2) - y*(1 + (√2)/2) =12) x*(1 - (3√2)/2) - y*(1 - (√2)/2) =1Let me denote these as equations (1) and (2). Let's subtract equation (2) from equation (1):[1 + (3√2)/2 - (1 - (3√2)/2)]x - [1 + (√2)/2 - (1 - (√2)/2)]y =0Compute coefficients:For x: [1 -1 + (3√2)/2 + (3√2)/2] =3√2For y: -[1 -1 + (√2)/2 + (√2)/2] = -√2Thus, 3√2 x - √2 y =0Simplify by dividing both sides by √2:3x - y =0 => y=3xNow substitute y=3x into one of the equations, say equation (1):x*(1 + (3√2)/2) -3x*(1 + (√2)/2) =1Factor x:x[1 + (3√2)/2 -3 - (3√2)/2] =1Simplify inside the brackets:1 -3 + (3√2)/2 - (3√2)/2 = -2Thus:-2x =1 => x= -1/2Therefore, x= -1/2, y=3*(-1/2)= -3/2So point M is (-1/2, -3/2). Wait, but earlier when we considered the vertical line AB, M was at (-1/2,0). So here, M is at (-1/2, -3/2). So indeed, the x-coordinate of M is always -1/2, confirming the previous conclusion from the polar line.Therefore, regardless of the line AB, M lies on x= -1/2. Therefore, the locus of M is the vertical line x= -1/2.Now, back to part (1): the angle between vector MF and AB. Let's compute vector MF and the direction of AB.Given point M(-1/2, y_M) and F(-2,0). Vector MF is F - M: (-2 - (-1/2), 0 - y_M) = (-3/2, -y_M). The line AB has direction determined by points A and B. However, since AB passes through F(-2,0), the direction vector can be parametrized as (x_A +2, y_A) for point A, or similarly for point B.Alternatively, the direction vector of AB can be found from the slope. For example, in the previous case where AB was the vertical line x=-2, direction vector is (0,1). When AB had slope 1, direction vector was (1,1). But in general, the direction of AB is variable, so the angle between MF and AB might not be fixed. However, in the two specific cases we checked:1. When AB was vertical (x=-2), M was (-1/2,0). So vector MF was (-3/2,0), direction of AB was vertical. The angle between horizontal vector (-3/2,0) and vertical line is 90 degrees.2. When AB was the line y =x +2, M was (-1/2, -3/2). Vector MF is (-3/2, 3/2). The direction of AB is (1,1). Let's compute the angle between (-3/2, 3/2) and (1,1). The vectors are (-3/2, 3/2) and (1,1). The dot product is (-3/2)(1) + (3/2)(1) =0. Therefore, the vectors are perpendicular. Hence, angle is 90 degrees. Wait, that's interesting. In this case, even though the direction of AB was (1,1), the vector MF was (-3/2, 3/2), which is (-1,1) scaled by 3/2. The direction vector of AB is (1,1). The dot product is (-1)(1) + (1)(1) =0. Therefore, they are perpendicular. So angle is 90 degrees.So in both cases, the angle is 90 degrees. Therefore, is the angle always 90 degrees? Let's test another case.Let me take another line AB through F(-2,0). Let's choose a line with slope m=√3. Then the equation is y=√3(x +2). Substitute into hyperbola:x² - [3(x +2)²]/3 =1 =>x² - (x² +4x +4)=1 => -4x -4=1 => -4x=5 =>x= -5/4Wait, but this is a linear equation, which suggests the line is tangent to the hyperbola. But the line intersects the hyperbola at only one point. However, we need the line AB to intersect the hyperbola at two points A and B. If m=√3, the line is tangent, so this is not allowed. Therefore, slopes for which the line is tangent are excluded. Let's try m=1/√3.So line AB: y=(1/√3)(x +2). Substitute into hyperbola:x² - [(1/√3)(x +2)]² /3 =1 =>x² - [(x +2)^2 /9] =1Multiply through by 9:9x² - (x² +4x +4) =9 =>8x² -4x -13=0Solutions x=(4 ±sqrt(16 +416))/16=(4 ±sqrt(432))/16=(4 ±12sqrt(3))/16=(1 ±3sqrt(3))/4Therefore, points A and B have x-coordinates (1 +3sqrt(3))/4 and (1 -3sqrt(3))/4. Then, y-coordinates are (1/√3)(x +2). Let's compute for x=(1 +3√3)/4:y=(1/√3)[(1 +3√3)/4 +2] = (1/√3)[(1 +3√3 +8)/4] = (1/√3)(9 +3√3)/4 = [9 +3√3]/(4√3) = [3(3 +√3)]/(4√3) = [3/(4√3)](3 +√3) = [√3/4](3 +√3) = (3√3 +3)/4Similarly for the other point x=(1 -3√3)/4:y=(1/√3)[(1 -3√3)/4 +2] = (1/√3)(9 -3√3)/4 = [9 -3√3]/(4√3) = [3(3 -√3)]/(4√3) = [3/(4√3)](3 -√3) = [√3/4](3 -√3) = (3√3 -3)/4Thus, points A and B are:A: [(1 +3√3)/4, (3√3 +3)/4]B: [(1 -3√3)/4, (3√3 -3)/4]Now, find the equations of the tangents at A and B.Tangent at A: x*x1 - (y*y1)/3 =1Substituting x1=(1 +3√3)/4, y1=(3√3 +3)/4:x*(1 +3√3)/4 - y*(3√3 +3)/12 =1Multiply through by 12 to eliminate denominators:3x*(1 +3√3) - y*(3√3 +3) =12Similarly, tangent at B:x*(1 -3√3)/4 - y*(3√3 -3)/12 =1Multiply through by 12:3x*(1 -3√3) - y*(3√3 -3) =12Now, solve these two equations:1) 3x(1 +3√3) - y(3√3 +3) =122) 3x(1 -3√3) - y(3√3 -3) =12Let me expand both equations:1) 3x +9√3 x -3√3 y -3y =122) 3x -9√3 x -3√3 y +3y =12Let me write these as:1) (3x -3y) + (9√3 x -3√3 y) =122) (3x +3y) + (-9√3 x -3√3 y) =12Hmm, this seems messy. Maybe subtract the two equations:Equation1 - Equation2:[3x -3y -3x -3y] + [9√3 x -3√3 y +9√3 x +3√3 y] =12 -12Simplify:(-6y) + (18√3 x) =0 =>18√3 x -6y=0 =>3√3 x - y=0 =>y=3√3 xNow substitute y=3√3 x into one of the equations, say Equation1:3x +9√3 x -3√3*(3√3 x) -3*(3√3 x) =12Compute term by term:3x +9√3 x -3√3*3√3 x -9√3 x =12Simplify:3x +9√3 x -27x -9√3 x =12Combine like terms:(3x -27x) + (9√3 x -9√3 x) =12-24x +0=12 =>x= -12/24= -1/2Therefore, x= -1/2, then y=3√3*(-1/2)= -3√3/2Therefore, point M is (-1/2, -3√3/2). Then vector MF is F - M = (-2 - (-1/2), 0 - (-3√3/2))= (-3/2, 3√3/2)The direction vector of AB: Let's take points A and B. For point A: [(1 +3√3)/4, (3√3 +3)/4]. The vector from F(-2,0) to A is [(1 +3√3)/4 +2, (3√3 +3)/4 -0] = [(9 +3√3)/4, (3√3 +3)/4]. Similarly, direction vector AB can be taken as ( (1 +3√3)/4 - (1 -3√3)/4 , (3√3 +3)/4 - (3√3 -3)/4 ) = (6√3/4, 6/4 ) = ( (3√3)/2, 3/2 )Therefore, direction vector of AB is (3√3/2, 3/2). Vector MF is (-3/2, 3√3/2). Let's compute the dot product:(-3/2)*(3√3/2) + (3√3/2)*(3/2) = (-9√3)/4 + (9√3)/4=0Therefore, the vectors are perpendicular. Hence, angle is 90 degrees.So in all the cases I checked, the angle between vector MF and AB is 90 degrees. Therefore, the answer to part (1) is 90 degrees or π/2 radians.For part (2), we need to find the minimum area of triangle MAB. Let's recall that the area of triangle MAB can be calculated as (1/2)*|AB|*d, where d is the distance from M to line AB. Alternatively, since we have coordinates for M, A, and B, we can use the formula:Area = (1/2)| (x_A(y_B - y_M) + x_B(y_M - y_A) + x_M(y_A - y_B)) |But perhaps it's more straightforward to compute vectors MA and MB and take half the magnitude of their cross product.But given that we found M lies on x=-1/2, and F is at (-2,0), and we have some parametrization, maybe there's a smarter way.Alternatively, since AB is a chord passing through F(-2,0), and M is the pole of AB, we can exploit properties of poles and polars. However, maybe parametrizing AB with a parameter and expressing the area in terms of that parameter, then minimizing.Alternatively, note that in all cases, vector MF is perpendicular to AB. Therefore, the distance from M to line AB is |MF|, since MF is perpendicular to AB. Wait, if MF is perpendicular to AB, then the distance from M to AB is indeed the length of MF. Therefore, the area of triangle MAB is (1/2)*|AB|*|MF|.Since angle between MF and AB is 90 degrees, yes, then the area is (1/2)*|AB|*|MF|.Therefore, to minimize the area, we can express |AB| and |MF| in terms of a parameter (like slope m) and find the minimum.Alternatively, since |MF| is the distance between M and F, which are fixed points? Wait, no. Wait, M is varying depending on AB. Wait, M is on x=-1/2, so coordinates of M are (-1/2, y_M). Therefore, |MF| is the distance between (-1/2, y_M) and (-2,0):|MF| = sqrt[ (-2 +1/2)^2 + (0 - y_M)^2 ] = sqrt[ (-3/2)^2 + y_M^2 ]=sqrt(9/4 + y_M^2 )So |MF|=sqrt(9/4 + y_M^2 )Also, the length |AB| can be expressed in terms of coordinates of A and B. Alternatively, since AB is a chord passing through F, we can parametrize AB with a parameter and find |AB| in terms of that parameter.Alternatively, let's use the parametrization of the line AB as we did earlier. Let me recall that when we parametrized AB as y=m(x +2), and substituted into the hyperbola, we had quadratic equation in x:(3 -m²)x² -4m²x - (4m² +3)=0Let me denote the roots x1 and x2. Then, the coordinates of A and B are (x1, m(x1 +2)) and (x2, m(x2 +2)). The distance between A and B is sqrt[(x1 -x2)^2 + (m(x1 +2) -m(x2 +2))^2] = sqrt[(x1 -x2)^2 + m²(x1 -x2)^2] = |x1 -x2|sqrt(1 +m²)Also, from the quadratic equation, the difference of roots is sqrt[(sum)^2 -4*product]. Sum x1 +x2 =4m²/(3 -m²), product x1x2= -(4m² +3)/(3 -m²). Therefore, (x1 -x2)^2=(x1 +x2)^2 -4x1x2= [16m^4/(3 -m²)^2] -4*(-(4m² +3)/(3 -m²))=16m^4/(3 -m²)^2 +4(4m² +3)/(3 -m²)Factor out 4/(3 -m²)^2:[16m^4 +4(4m² +3)(3 -m²)]/(3 -m²)^2Compute numerator:16m^4 +4[12m² -4m^4 +9 -3m²]=16m^4 +4[9 +9m² -4m^4]=16m^4 +36 +36m² -16m^4=36 +36m²Therefore, (x1 -x2)^2=(36(1 +m²))/(3 -m²)^2Therefore, |AB|=|x1 -x2|sqrt(1 +m²)= [6sqrt(1 +m²)/|3 -m²| ] * sqrt(1 +m²)=6(1 +m²)/|3 -m²|Therefore, |AB|=6(1 +m²)/|3 -m²|Also, |MF|=sqrt(9/4 + y_M^2 )But y_M is the y-coordinate of M, which we need to find in terms of m. Recall that M is the intersection of the two tangents at A and B. From previous examples, when we solved for M, we found that M is (-1/2, y_M). Let me find y_M in terms of m.From the earlier parametrization, when we solved the system of tangent equations for a general line AB with slope m, we found that y=3x for one case and y=3√3 x for another, but those were specific cases. Wait, actually in the general case, when we had the line AB with slope m, we found that M has coordinates (-1/2, y_M). Let me see if we can find y_M in terms of m.Alternatively, recall that in the general case when solving for M, we had y = [3(x1 - x2)] / [x2 y1 -x1 y2]But given that points A(x1, y1) and B(x2, y2) lie on the hyperbola and on the line y=m(x +2). Therefore, y1=m(x1 +2), y2=m(x2 +2). Therefore:Denominator: x2 y1 -x1 y2 =x2*m(x1 +2) -x1*m(x2 +2)=m[x2 x1 +2x2 -x1 x2 -2x1]=m[2x2 -2x1]=2m(x2 -x1)Numerator:3(x1 -x2)Thus, y=3(x1 -x2)/[2m(x2 -x1)]= -3(x2 -x1)/[2m(x2 -x1)]= -3/(2m)Therefore, y_M= -3/(2m)Therefore, coordinates of M are (-1/2, -3/(2m))Therefore, |MF|=sqrt[ (-2 +1/2)^2 + (0 - (-3/(2m)) )^2 ]=sqrt[ (-3/2)^2 + (3/(2m))^2 ]=sqrt[9/4 +9/(4m²)]=(3/2)sqrt(1 +1/m²)Thus, |MF|=(3/2)sqrt(1 +1/m²)Therefore, area of triangle MAB is (1/2)*|AB|*|MF|=(1/2)*(6(1 +m²)/|3 -m²|)*(3/2)sqrt(1 +1/m²)Simplify:(1/2)*(6)*(3/2)*(1 +m²)/|3 -m²|*sqrt(1 +1/m²)= (9/2)*(1 +m²)/|3 -m²|*sqrt(1 +1/m²)Let me simplify sqrt(1 +1/m²)=sqrt((m² +1)/m²)=sqrt(m² +1)/|m|Therefore, Area=(9/2)*(1 +m²)/|3 -m²|*(sqrt(m² +1)/|m| )Note that |3 -m²|=|m² -3|, and since m² can be greater or less than 3, but the area is positive, so we can write |3 -m²|=|m² -3|Also, |m|=sqrt(m²)Assuming m ≠0 (since if m=0, the line is horizontal, but earlier we saw that m=0 leads to parallel tangents which don't intersect). Therefore, m≠0.Therefore, Area=(9/2)*(1 +m²)*sqrt(1 +m²)/( |m² -3|*|m| )But note that (1 +m²)*sqrt(1 +m²)= (1 +m²)^(3/2)Thus, Area=(9/2)*(1 +m²)^(3/2)/( |m|*|m² -3| )This is a function of m. To find its minimum, we can consider m>0 and m<0 separately due to the absolute values. Since the expression is even in m (since replacing m by -m doesn't change |m| or |m² -3| or (1 +m²)^(3/2)), we can assume m>0 without loss of generality.Therefore, let m>0. Then |m|=m, |m² -3|=|m² -3|. Depending on whether m²>3 or m²<3, |m² -3|=m² -3 or 3 -m².Let me define t=m². Then t>0, and Area expression becomes:Area=(9/2)*(1 +t)^(3/2)/( sqrt(t)*|t -3| )We need to minimize this function for t>0, t≠3.Let me write this as:A(t) = (9/2) * (1 + t)^(3/2) / [ sqrt(t) |t -3| ]We can consider two cases: t<3 and t>3.Case 1: t<3. Then |t -3|=3 -t.A(t) = (9/2) * (1 +t)^(3/2)/( sqrt(t)(3 -t) )Case 2: t>3. Then |t -3|=t -3.A(t) = (9/2) * (1 +t)^(3/2)/( sqrt(t)(t -3) )We need to find the minimum of A(t) over t>0, t≠3.Let me analyze both cases.First, consider Case 1: t ∈ (0,3)Let me set f(t) = (1 + t)^(3/2)/( sqrt(t)(3 -t) )We need to minimize f(t).Take natural logarithm to make differentiation easier:ln f(t) = (3/2) ln(1 + t) - (1/2) ln t - ln(3 -t)Differentiate:f’/f = (3/2)*(1/(1 +t)) - (1/2)*(1/t) - (-1/(3 -t))Thus,f’/f = 3/(2(1 +t)) -1/(2t) +1/(3 -t)Set derivative to zero:3/(2(1 +t)) -1/(2t) +1/(3 -t) =0Multiply through by 2t(1 +t)(3 -t) to eliminate denominators:3* t(3 -t) - (1 +t)(3 -t) +2t(1 +t) =0Expand each term:First term:3t(3 -t)=9t -3t²Second term: - (1 +t)(3 -t)= - [3 -t +3t -t²] = -3 -2t +t²Third term:2t(1 +t)=2t +2t²Combine all terms:(9t -3t²) -3 -2t +t² +2t +2t² =0Simplify:9t -3t² -3 -2t +t² +2t +2t²=0Combine like terms:(-3t² +t² +2t²) + (9t -2t +2t) + (-3)=(0t²) + (9t) + (-3)=9t -3=0Thus, 9t -3=0 =>t=1/3Thus, critical point at t=1/3 in case 1.Check second derivative or endpoints. Also, need to check behavior as t→0+ and t→3-.As t→0+, f(t)= (1)^(3/2)/(0*3)= which tends to infinity.As t→3-, denominator approaches 0, so f(t)→infty.Thus, critical point at t=1/3 is the only minimum in case 1.Now, check case 2: t>3.Define f(t)= (1 +t)^(3/2)/( sqrt(t)(t -3) )Similarly, take ln f(t)= (3/2)ln(1 +t) - (1/2)ln t -ln(t -3)Differentiate:f’/f=(3/(2(1 +t))) -1/(2t) -1/(t -3)Set to zero:3/(2(1 +t)) -1/(2t) -1/(t -3)=0Multiply through by 2t(t -3)(1 +t):3*t(t -3) - (t -3)(1 +t) -2t(1 +t)=0Expand each term:First term:3t(t -3)=3t² -9tSecond term: - (t -3)(t +1)= - [t² +t -3t -3]= -t² +2t +3Third term:-2t(t +1)= -2t² -2tCombine all terms:3t² -9t -t² +2t +3 -2t² -2t=0Simplify:(3t² -t² -2t²)+ (-9t +2t -2t) +3=0t² -9t +3= -9t +3=0 =>t=1/3But t>3 in this case, so no critical points in case 2.Therefore, only critical point at t=1/3 in case 1.Thus, the minimum area occurs at t=1/3, which is m²=1/3, so m=1/√3 or m=-1/√3 (but since we considered m>0, m=1/√3).Compute the minimal area:A(t)= (9/2)*(1 +1/3)^(3/2)/( sqrt(1/3)*(3 -1/3)) )Compute step by step:1 +1/3=4/3(4/3)^(3/2)= (4^(3/2))/(3^(3/2))= (8)/(3*√3)=8/(3√3)sqrt(1/3)=1/√3Denominator: sqrt(1/3)*(3 -1/3)= (1/√3)*(8/3)=8/(3√3)Therefore,A(t)=(9/2)*(8/(3√3)) / (8/(3√3)) )= (9/2)*1=9/2Wait, that can't be right. Wait, numerator is (1 +t)^(3/2)= (4/3)^(3/2)=8/(3√3)Denominator is sqrt(t)*(3 -t)=sqrt(1/3)*(8/3)= (1/√3)*(8/3)=8/(3√3)Thus, the fraction is [8/(3√3)] / [8/(3√3)]=1Therefore, A(t)=(9/2)*1=9/2But wait, when we computed the area for the case where m=1/√3 (which corresponds to t=1/3), we found M at (-1/2, -3/(2m))= (-1/2, -3/(2*(1/√3)) )= (-1/2, - (3√3)/2 )Then |MF|=sqrt[ ( -2 +1/2 )^2 + (0 +3√3/2 )^2 ]=sqrt[ ( -3/2 )^2 + (3√3/2 )^2 ]=sqrt[9/4 +27/4]=sqrt[36/4]=sqrt[9]=3Length |AB|=6(1 +m²)/|3 -m²|=6*(1 +1/3)/|3 -1/3|=6*(4/3)/(8/3)=6*(4/3)*(3/8)=6*(4/8)=6*(1/2)=3Therefore, area=(1/2)*3*3=9/2=4.5Therefore, the minimal area is 9/2.But let's check another case. For example, when m=1 (previous case), we found area as:In that case, points A and B were (1 + (3√2)/2, 3 + (3√2)/2) and (1 - (3√2)/2, 3 - (3√2)/2). Let's compute |AB|.Compute the distance between A and B:Δx=(1 +3√2/2)-(1 -3√2/2)=3√2Δy=(3 +3√2/2)-(3 -3√2/2)=3√2Therefore, |AB|=sqrt( (3√2)^2 + (3√2)^2 )=sqrt(18 +18)=sqrt(36)=6|MF|=sqrt( (-3/2)^2 + ( -3/2 )^2 )=sqrt(9/4 +9/4 )=sqrt(18/4)=3√2/2Area=(1/2)*6*(3√2/2 )=(1/2)*6*(3√2/2 )=(18√2)/4=(9√2)/2≈6.364, which is larger than 9/2=4.5, so indeed the minimum occurs at m=1/√3.Therefore, the minimal area is 9/2.Hence, the answers are:1) The angle is 90 degrees, so (boxed{dfrac{pi}{2}})2) The minimal area is (boxed{dfrac{9}{2}})Final Answer(1) The angle between the vector (MF) and (AB) is (boxed{dfrac{pi}{2}}).(2) The minimum value of the area of (triangle MAB) is (boxed{dfrac{9}{2}}).</think>Given the hyperbola (x^2 - frac{y^2}{3} = 1) with the left focus (F(-2, 0)), we need to find the angle between the vector (MF) and the line (AB), and the minimum value of the area of triangle (MAB).1. Angle between vector (MF) and line (AB): - The line (AB) passes through the focus (F(-2, 0)) and intersects the hyperbola at points (A) and (B). - The tangents at (A) and (B) intersect at point (M), which lies on the line (x = -frac{1}{2}) (the polar of the focus (F)). - The vector (MF) is from (M(-frac{1}{2}, y_M)) to (F(-2, 0)), which is ((- frac{3}{2}, -y_M)). - The direction vector of line (AB) is found to be perpendicular to the vector (MF) in all cases checked, leading to the conclusion that the angle between them is (90^circ) or (frac{pi}{2}) radians.2. Minimum value of the area of (triangle MAB): - The area of (triangle MAB) is given by (frac{1}{2} times |AB| times |MF|). - Using parametrization and optimization, the minimum area is found when the slope (m) of line (AB) is (frac{1}{sqrt{3}}) or (-frac{1}{sqrt{3}}). - The minimum area is calculated to be (frac{9}{2}). Final Answers1. The angle between the vector (MF) and (AB) is (boxed{dfrac{pi}{2}}).2. The minimum value of the area of (triangle MAB) is (boxed{dfrac{9}{2}}).,<think>Alright, so I have this problem about a hyperbola. Let me first try to understand the setup and what's being asked. The hyperbola is given by the equation ( x^2 - frac{y^2}{3} = 1 ). The left focus is F, and we have to draw a line AB through F that intersects the hyperbola at points A and B. Then, construct the tangents ( l_1 ) and ( l_2 ) at A and B, which intersect at point M. The questions are: (1) Find the angle between the vector MF and AB; (2) Find the minimum value of the area of triangle MAB.First, I need to recall some properties of hyperbolas. The standard form is ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ). Comparing with the given equation, ( a^2 = 1 ), so ( a = 1 ), and ( b^2 = 3 ), so ( b = sqrt{3} ). The foci of a hyperbola are located at ( (pm c, 0) ), where ( c^2 = a^2 + b^2 ). Calculating that, ( c^2 = 1 + 3 = 4 ), so ( c = 2 ). Therefore, the left focus F is at (-2, 0).Next, the line AB passes through F (-2, 0) and intersects the hyperbola at A and B. Since the line passes through the focus, maybe there's some special property here? Hmm. Then, we have to find the tangents at A and B. The tangents to a hyperbola at a point (x₀, y₀) can be given by ( frac{xx₀}{a^2} - frac{yy₀}{b^2} = 1 ). So, substituting a² and b², the tangent at A (x₁, y₁) is ( x x₁ - frac{y y₁}{3} = 1 ), and similarly for B (x₂, y₂).These two tangents intersect at point M. Then, we need to find the angle between vector MF and AB, and the minimal area of triangle MAB.For part (1), the angle between MF and AB. Since AB is a line, the direction of AB is given by the slope of the line AB. The vector MF would be from M to F, so we need the direction of MF and the direction of AB to find the angle between them.For part (2), the area of triangle MAB. The area can be calculated using the determinant formula: ( frac{1}{2} | vec{AB} times vec{AM} | ), but maybe there's a better way. Alternatively, if we can find the coordinates of M, A, B, then we can compute the area.To approach both problems, perhaps parametrize the line AB. Let me think.Let’s first parametrize line AB. Since it passes through F(-2, 0), let's assume the line has a slope m. Then, the equation of AB is ( y = m(x + 2) ). This line intersects the hyperbola ( x^2 - frac{y^2}{3} = 1 ). Let's substitute y from the line equation into the hyperbola.So substituting ( y = m(x + 2) ) into the hyperbola:( x^2 - frac{[m(x + 2)]^2}{3} = 1 )Expand that:( x^2 - frac{m^2(x^2 + 4x + 4)}{3} = 1 )Multiply through by 3 to eliminate denominator:( 3x^2 - m^2(x^2 + 4x + 4) = 3 )Group like terms:( (3 - m^2)x^2 - 4m^2 x - (4m^2 + 3) = 0 )This is a quadratic equation in x. Let me denote this as:( [3 - m^2]x^2 - [4m^2]x - [4m^2 + 3] = 0 )Let’s denote coefficients:A = 3 - m²B = -4m²C = -4m² - 3The solutions are given by quadratic formula:x = [4m² ± sqrt(16m^4 + 4(3 - m²)(4m² + 3))]/[2(3 - m²)]Let’s compute discriminant D:D = 16m^4 + 4(3 - m²)(4m² + 3)Expand the second term:4*(3*(4m² + 3) - m²*(4m² + 3)) = 4*(12m² + 9 -4m^4 -3m²) = 4*(9 +9m² -4m^4) = 36 + 36m² -16m^4Therefore, D = 16m^4 +36 +36m² -16m^4 = 36m² +36 = 36(m² +1)Thus, sqrt(D) = 6sqrt(m² +1)So, x = [4m² ±6sqrt(m² +1)]/[2(3 - m²)] = [2m² ±3sqrt(m² +1)]/(3 - m²)Therefore, the x-coordinates of A and B are [2m² +3sqrt(m² +1)]/(3 - m²) and [2m² -3sqrt(m² +1)]/(3 - m²)But since the line passes through F(-2,0), and intersects the hyperbola at A and B, perhaps both points are on the hyperbola. However, since hyperbola has two branches, but given that the line passes through the left focus (-2,0), which is on the left side, the line could intersect both branches? Wait, the hyperbola ( x² - y²/3 =1 ) opens left and right. The left focus is at (-2,0). So a line passing through (-2,0) might intersect the left branch once and the right branch once? Or maybe both intersections are on the left? Wait, if the slope is steep enough, maybe it only intersects once? But since the problem states that line AB intersects the hyperbola at points A and B, so two distinct points. So, the line must intersect both branches? Wait, but a hyperbola has two separate branches. If the line passes through the left focus, which is on the left side, maybe the line intersects the left branch at F and another point? But the problem says "intersecting the hyperbola at points A and B", so two distinct points, so maybe the line passes through the left focus and intersects the left branch at A and the right branch at B? Or both on left?Wait, let me check with the equation. Let's pick a horizontal line through F: y=0. Then, substituting into hyperbola: x² =1, so x=1 and x=-1. But the line passes through F(-2,0), so when y=0, x=-2. But the hyperbola at y=0 has x=±1, so the line y=0 intersects hyperbola at (1,0) and (-1,0), but it also passes through (-2,0). Wait, but a line can only intersect a hyperbola at two points. So maybe in this case, if the line passes through F, which is outside the hyperbola, then the line would intersect the hyperbola at two points A and B. Wait, but the hyperbola is x² - y²/3 =1. The foci are at (±2,0). So the foci are outside the hyperbola's vertices at (±1,0). So a line passing through F(-2,0) would intersect the hyperbola at two points: one close to the left, maybe, and another on the right? Let's check with vertical line x=-2. But x=-2 is a vertical line. Substituting x=-2 into the hyperbola: 4 - y²/3 =1 ⇒ y²=9 ⇒ y=±3. So points (-2,3) and (-2,-3). So in this case, line x=-2 intersects hyperbola at two points. But in the case of horizontal line y=0, the line passes through (-2,0) but intersects the hyperbola at (1,0) and (-1,0). Wait, but (-2,0) is not on the hyperbola. So the line y=0 passes through F(-2,0) and intersects the hyperbola at (1,0) and (-1,0). So two points. Similarly, a line with slope m passing through (-2,0) would intersect the hyperbola at two points A and B. So the parametrization I did earlier gives those two points.But let's think about this. If the line is not tangent, then it will intersect the hyperbola at two distinct points. Since the problem states that AB intersects the hyperbola at points A and B, we can assume it's a secant line. So, parametrizing the line as y = m(x + 2), then the two intersection points A and B have coordinates as found above.But perhaps instead of parametrizing by slope m, it's better to use parametric equations for the line. Alternatively, use a parameter t for the line AB. Let's consider parametric equations. Let’s let AB be a line passing through F(-2,0) with direction vector (cosθ, sinθ). Then, parametric equations for AB are x = -2 + t cosθ, y = 0 + t sinθ. Substitute into hyperbola equation:(-2 + t cosθ)^2 - (t sinθ)^2 /3 =1Expanding:4 -4 t cosθ + t² cos²θ - (t² sin²θ)/3 =1Simplify:4 -4 t cosθ + t² (cos²θ - (sin²θ)/3) -1 =0So:3 -4 t cosθ + t² [ (3 cos²θ - sin²θ)/3 ] =0Multiply through by 3:9 -12 t cosθ + t² (3 cos²θ - sin²θ) =0This is a quadratic in t:[3 cos²θ - sin²θ] t² -12 cosθ t +9=0Let me denote A = 3 cos²θ - sin²θ, B = -12 cosθ, C=9.Then discriminant D = B² -4AC = 144 cos²θ -4*(3 cos²θ - sin²θ)*9Compute:144 cos²θ -36*(3 cos²θ - sin²θ) =144 cos²θ -108 cos²θ +36 sin²θ=36 cos²θ +36 sin²θ=36(cos²θ + sin²θ)=36*1=36Thus sqrt(D)=6Thus solutions:t = [12 cosθ ±6]/[2*(3 cos²θ - sin²θ)] = [6 cosθ ±3]/[3 cos²θ - sin²θ]Therefore, t = [6 cosθ +3]/[3 cos²θ - sin²θ] and t = [6 cosθ -3]/[3 cos²θ - sin²θ]Therefore, the two points A and B correspond to parameters t1 and t2:t1 = [6 cosθ +3]/D, t2 = [6 cosθ -3]/D, where D=3 cos²θ - sin²θHence, the coordinates of A and B are:For t1: x = -2 + t1 cosθ, y = t1 sinθSimilarly for t2: x = -2 + t2 cosθ, y = t2 sinθThis seems a bit complicated, but maybe manageable.Alternatively, since we need to construct the tangents at A and B, which intersect at M. Then, perhaps there is a property that could relate point M to focus F, which could help in determining the angle.Alternatively, perhaps use pole-polar concepts. The pole of line AB with respect to the hyperbola is the point M where the tangents from M touch the hyperbola at A and B. Wait, yes! In projective geometry, the pole of the line AB with respect to the hyperbola is the point M, since the polar line of M is AB, as MA and MB are the tangents from M.Given that, perhaps there is a relationship between point M and focus F.Alternatively, maybe there's a property specific to hyperbolas regarding the foci and the tangents. For example, in an ellipse, the tangents make equal angles with the lines to the foci, but I'm not sure about hyperbola.Alternatively, maybe use coordinates. Let's suppose that line AB has slope m, parametrized as before. Then, find coordinates of A and B, then find equations of tangents at A and B, find their intersection M, then compute vector MF and angle with AB. Then compute area of triangle MAB.This seems computational but feasible.Let me try this approach.First, let's parametrize line AB as y = m(x + 2). Then, the intersection points A and B are found by solving the equation as before. The x-coordinates are [2m² ±3√(m² +1)]/(3 - m²). Let me denote sqrt(m² +1) as s for simplicity. Then, x-coordinates are [2m² ±3s]/(3 - m²). Then, the coordinates of A and B are:For x1 = [2m² +3s]/(3 - m²), y1 = m(x1 + 2) = m([2m² +3s]/(3 - m²) + 2) = m([2m² +3s + 2(3 - m²)]/(3 - m²)) = m([2m² +3s +6 -2m²]/(3 - m²)) = m(3s +6)/(3 - m²) = 3m(s +2)/(3 - m²)Similarly, for x2 = [2m² -3s]/(3 - m²), y2 = m(x2 + 2) = m([2m² -3s + 2(3 - m²)]/(3 - m²)) = m([2m² -3s +6 -2m²]/(3 - m²)) = m(6 -3s)/(3 - m²) = 3m(2 - s)/(3 - m²)Therefore, points A and B are:A: ( [2m² +3s]/(3 - m²), 3m(s +2)/(3 - m²) )B: ( [2m² -3s]/(3 - m²), 3m(2 - s)/(3 - m²) )Where s = sqrt(m² +1)Now, we need to find the equations of the tangents at A and B.The tangent at point (x₁, y₁) on hyperbola x² - y²/3 =1 is given by:x x₁ - (y y₁)/3 =1So, for point A (x₁, y₁):Tangent l₁: x x₁ - (y y₁)/3 =1Similarly, for point B (x₂, y₂):Tangent l₂: x x₂ - (y y₂)/3 =1We need to find the intersection M of l₁ and l₂.To find M, solve the two equations:x x₁ - (y y₁)/3 =1x x₂ - (y y₂)/3 =1This is a system of linear equations in x and y. Let's write them as:x x₁ - (y y₁)/3 =1 ...(1)x x₂ - (y y₂)/3 =1 ...(2)Subtracting equation (2) from equation (1):x(x₁ - x₂) - y(y₁ - y₂)/3 =0So,x = [ y(y₁ - y₂)/3 ] / (x₁ - x₂)Let me denote:Let’s compute x₁ - x₂:From coordinates of A and B:x₁ - x₂ = [2m² +3s - (2m² -3s)]/(3 - m²) = (6s)/(3 - m²)Similarly, y₁ - y₂ = 3m(s +2 - (2 -s))/(3 - m²) = 3m(2s)/(3 - m²) = 6m s/(3 - m²)Therefore,x = [ y * (6m s/(3 - m²)) /3 ] / (6s/(3 - m²)) )Simplify:Numerator: y*(6m s /3(3 - m²)) = y*(2m s)/(3 - m²)Denominator: 6s/(3 - m²)Thus,x = [ y*(2m s)/(3 - m²) ] / [6s/(3 - m²) ] = y*(2m s)/(3 - m²) * (3 - m²)/(6s) ) = y*(2m)/6 = y*(m)/3Therefore, x = (m/3) yNow, substitute x = (m/3)y into one of the tangent equations, say equation (1):x x₁ - (y y₁)/3 =1Substitute x:(m/3)y x₁ - (y y₁)/3 =1Factor y/3:y/3 [m x₁ - y₁] =1Thus,y = 3 / [m x₁ - y₁]Then, x = (m/3)y = (m/3)(3 / [m x₁ - y₁]) ) = m / [m x₁ - y₁]Therefore, coordinates of M are ( m / [m x₁ - y₁], 3 / [m x₁ - y₁] )Compute denominator m x₁ - y₁:From coordinates of A:x₁ = [2m² +3s]/(3 - m²)y₁ = 3m(s +2)/(3 - m²)Thus,m x₁ - y₁ = m*[2m² +3s]/(3 - m²) - 3m(s +2)/(3 - m²) = [2m³ +3m s -3m s -6m]/(3 - m²) = [2m³ -6m]/(3 - m²) = 2m(m² -3)/(3 - m²) = -2m(m² -3)/(m² -3) = -2mWait, that's interesting. Let me check:Numerator:2m³ +3m s -3m s -6m = 2m³ -6m = 2m(m² -3)Denominator: (3 - m²)Thus,m x₁ - y₁ = [2m(m² -3)] / (3 - m²) = -2m(m² -3)/(m² -3) = -2mBut wait, (m² -3) cancels with (3 - m²) which is -(m² -3). So yes, simplifies to -2m.Therefore, m x₁ - y₁ = -2mTherefore, the coordinates of M are:x = m / (-2m) = -1/2y = 3 / (-2m) = -3/(2m)Therefore, point M is ( -1/2, -3/(2m) )Wow, that's a significant simplification. So regardless of the slope m (as long as the line AB is not vertical, which would correspond to m approaching infinity, but vertical case can be handled separately), point M is fixed at (-1/2, -3/(2m)).Wait, but m is the slope of AB. So M is dependent on m, but its x-coordinate is fixed at -1/2? Wait, no, wait:Wait, in the computation above, after simplifying, we found that m x₁ - y₁ = -2m, so denominator is -2m. Thus, coordinates of M:x = m / (-2m) = -1/2y = 3 / (-2m) = -3/(2m)So indeed, regardless of m (except when m=0, which would be a horizontal line). So the x-coordinate of M is always -1/2, and the y-coordinate is -3/(2m). So as m varies, M moves along the vertical line x = -1/2, with y-coordinate inversely proportional to m.This is a crucial result. So point M is always on the line x = -1/2, which is a vertical line passing through the midpoint between the center (0,0) and the left vertex (-1,0). Interesting.Now, the focus F is at (-2,0). So vector MF is from M(-1/2, -3/(2m)) to F(-2,0). The components of vector MF are:Δx = -2 - (-1/2) = -2 +1/2 = -3/2Δy = 0 - (-3/(2m)) = 3/(2m)Therefore, vector MF is (-3/2, 3/(2m))The direction of AB is given by the slope m. The vector along AB can be taken as (1, m), since the line AB is y = m(x +2). Therefore, the direction vector of AB is (1, m). To find the angle between vector MF and AB, we can use the dot product formula.The angle θ between vector MF and direction vector AB is given by:cosθ = (MF • AB) / (|MF| |AB|)Compute MF • AB:(-3/2)(1) + (3/(2m))(m) = -3/2 + 3/2 = 0Wait, that's zero. Therefore, the angle between MF and AB is 90 degrees, i.e., π/2 radians. Therefore, the angle is 90 degrees, which is constant regardless of the slope m. So answer to part (1) is 90 degrees or π/2 radians.But wait, let me verify this calculation.MF vector is (-3/2, 3/(2m))AB direction vector is (1, m)Dot product: (-3/2)(1) + (3/(2m))(m) = -3/2 + 3/2 = 0Yes, so the dot product is zero, meaning they are perpendicular. Therefore, the angle is 90 degrees.Wow, so that's part (1). The angle is always 90 degrees, regardless of the line AB. So the answer is 90 degrees, or π/2 radians.Now, part (2): Find the minimum value of the area of triangle MAB.First, let's find the coordinates of points A, B, M.We have M at (-1/2, -3/(2m))Points A and B are given as:A: ( [2m² +3s]/(3 - m²), 3m(s +2)/(3 - m²) )B: ( [2m² -3s]/(3 - m²), 3m(2 - s)/(3 - m²) )Where s = sqrt(m² +1)Alternatively, perhaps we can express the area in terms of m and then minimize it.Alternatively, since we have coordinates for A, B, M, we can compute vectors MA and MB, then compute the area using the cross product.But maybe a better approach is to use coordinates.The area of triangle MAB can be calculated using the formula:Area = (1/2)| (x_A - x_M)(y_B - y_M) - (x_B - x_M)(y_A - y_M) |Alternatively, use determinant formula:Area = (1/2)| x_A(y_B - y_M) + x_B(y_M - y_A) + x_M(y_A - y_B) | But this might get messy. Alternatively, compute vectors AB and AM and take half the magnitude of their cross product.First, let's compute coordinates for A and B.Recall:A = ( [2m² +3s]/(3 - m²), 3m(s +2)/(3 - m²) )B = ( [2m² -3s]/(3 - m²), 3m(2 - s)/(3 - m²) )M = (-1/2, -3/(2m))Let’s compute coordinates difference for vectors.First, compute vectors AB and AM.Vector AB: from A to B is (x_B - x_A, y_B - y_A)x_B - x_A = [ (2m² -3s) - (2m² +3s) ] / (3 - m²) = (-6s)/(3 - m²)y_B - y_A = [3m(2 -s) -3m(s +2) ] / (3 - m²) = 3m[2 -s -s -2]/(3 - m²) = 3m(-2s)/(3 - m²) = -6m s/(3 - m²)Thus, vector AB is ( -6s/(3 - m²), -6m s/(3 - m²) )Factor out -6s/(3 - m²):AB = (-6s/(3 - m²)) (1, m)Similarly, vector AM: from A to M is ( -1/2 - x_A, -3/(2m) - y_A )Compute x_A:x_A = [2m² +3s]/(3 - m²)Similarly, y_A = 3m(s +2)/(3 - m²)Thus,x-component of AM: -1/2 - [2m² +3s]/(3 - m²) = [ - (3 - m²)/2 -2m² -3s ] / (3 - m² )Wait, perhaps compute numerator:= (-1/2)(3 - m²) -2m² -3s= (-3/2 + m²/2) -2m² -3s= -3/2 - (3m²)/2 -3sSimilarly, y-component of AM:-3/(2m) - [3m(s +2)]/(3 - m²) = [ -3(3 - m²) -6m²(s +2) ] / [2m(3 - m²) ]Wait, this seems complicated. Maybe there's a smarter way.Alternatively, since AB is a line passing through F, and M is a point such that MF is perpendicular to AB, as we found in part (1). Therefore, the area of triangle MAB is (1/2)*|AB| * |MF|. Because the area is 1/2 * base * height, where base is AB and height is the distance from M to line AB. But since MF is perpendicular to AB, then the distance from M to AB is |MF|. Therefore, area is (1/2)*|AB|*|MF|.Therefore, Area = (1/2)*|AB|*|MF|So maybe compute |AB| and |MF|, multiply, and then minimize.Let’s compute |AB|:From vector AB above, which is (-6s/(3 - m²), -6m s/(3 - m²)), so its magnitude is:|AB| = sqrt[ ( (-6s/(3 - m²))^2 + (-6m s/(3 - m²))^2 ) ] = (6s/(3 - m²)) sqrt(1 + m²)But sqrt(1 + m²) = s, so |AB| = 6s² / |3 - m²|But s = sqrt(m² +1), so s² = m² +1. Therefore,|AB| = 6(m² +1)/|3 - m²|Similarly, compute |MF|:Vector MF is (-3/2, 3/(2m)), so its magnitude is sqrt[ (9/4) + (9)/(4m²) ) ] = (3/2)sqrt(1 + 1/m² ) = (3/2)sqrt( (m² +1)/m² ) = (3/2)(sqrt(m² +1)/|m| )But since m can be positive or negative, but area is positive, so we can take m >0 for simplicity (since hyperbola is symmetric with respect to x-axis).Therefore, |MF| = (3/2)(s/|m| ) where s = sqrt(m² +1)Therefore, Area = (1/2)*(6(m² +1)/|3 - m²| )*(3/2)(s/|m| )Simplify:= (1/2)*(6(m² +1)/|3 - m²| )*(3/2)(sqrt(m² +1)/|m| )= (1/2)*(6*3/2)*( (m² +1)sqrt(m² +1) ) / ( |3 - m²||m| )= (9/2)*( (m² +1)^(3/2) ) / ( |3 - m²||m| )Therefore, Area = (9/2) * (m² +1)^(3/2) / ( |3 - m²||m| )But this expression is quite complex. Let's note that since the hyperbola is x² - y²/3 =1, the line AB intersects the hyperbola, so the quadratic equation in x had real roots, so discriminant was positive. As we found before, discriminant D was 36(m² +1), which is always positive. So for any real m, except when the denominator 3 - m² =0, which would be m=±sqrt(3). At m=±sqrt(3), the denominator becomes zero, so lines with slope sqrt(3) or -sqrt(3) would be asymptotes? Wait, the asymptotes of the hyperbola are y=±sqrt(3)x. So slope sqrt(3) and -sqrt(3) are the asymptotes. Therefore, lines with slope m=±sqrt(3) are asymptotes, which do not intersect the hyperbola, but in our case, the line AB is passing through focus F(-2,0), so even if m=sqrt(3), the line would intersect the hyperbola at two points, one being F? Wait, no. Wait, the asymptotes pass through the center (0,0), not the focus. So a line with slope sqrt(3) passing through F(-2,0) is not an asymptote. The asymptotes are y=±sqrt(3)x, which pass through the center. Therefore, the lines with slope sqrt(3) through F(-2,0) would intersect the hyperbola at two points. Therefore, m can be any real number except where 3 -m²=0, which would make the denominator zero in |AB|. Wait, but in the expression for |AB|, denominator is |3 - m²|. If m²=3, denominator is zero. But when m²=3, the original quadratic equation in x would have leading coefficient A=3 -m²=0, making it a linear equation. Therefore, for m²=3, the line AB is tangent to the hyperbola? Wait, let's check when m²=3, i.e., m=±sqrt(3).The line y=±sqrt(3)(x +2). Let's substitute into hyperbola:x² - [3(x +2)^2]/3 =x² - (x +2)^2 =x² - (x² +4x +4)= -4x -4=1 ⇒ -4x -4=1 ⇒-4x=5⇒x= -5/4.Thus, it intersects the hyperbola at x=-5/4, then y=±sqrt(3)(-5/4 +2)=±sqrt(3)(3/4)=±(3 sqrt(3))/4. So only one point? Wait, but quadratic equation reduces to linear, so only one intersection point? That would imply the line is tangent. But substitution shows that at m=±sqrt(3), the line intersects the hyperbola at one point (-5/4, ±3sqrt(3)/4). Hence, the line is tangent at that point. Therefore, for m=±sqrt(3), the line AB is tangent, but the problem states that AB is a secant line intersecting at two points A and B. Hence, m cannot be ±sqrt(3). Therefore, m² ≠3, so denominator 3 -m² ≠0, hence valid.Therefore, returning to Area expression:Area = (9/2) * (m² +1)^(3/2) / ( |3 - m²||m| )To find the minimum area, we need to minimize this expression with respect to m.Let’s note that m can be positive or negative, but since the expression involves |m| and even powers of m, the function is even in m. Therefore, we can assume m >0, and find the minimum for m >0, which will correspond to the overall minimum.Let’s set m >0, so |m| =m, and |3 -m²|=|m² -3|. Therefore, since m >0, depending on whether m² >3 or m² <3, |3 -m²|=3 -m² if m² <3, and m² -3 if m² >3.Hence, split into two cases:Case 1: 0 < m² <3Then |3 -m²|=3 -m²Area = (9/2)*(m² +1)^(3/2)/( (3 -m²)m )Case 2: m² >3Then |3 -m²|=m² -3Area = (9/2)*(m² +1)^(3/2)/( (m² -3)m )We need to analyze both cases for minima.Alternatively, perform substitution to make it easier. Let’s let t = m², where t >0, t ≠3.Thus, Area can be expressed as:For t <3:Area = (9/2)*(t +1)^(3/2)/( (3 -t)sqrt(t) )For t >3:Area = (9/2)*(t +1)^(3/2)/( (t -3)sqrt(t) )Therefore, we can write the Area as a function of t:A(t) = (9/2) * (t +1)^(3/2)/( |t -3| sqrt(t) )To find the minimum, we can consider t ∈ (0,3) and t ∈ (3, ∞) separately.Let’s first consider t ∈ (0,3):A(t) = (9/2)*(t +1)^(3/2)/( (3 -t)sqrt(t) )Let’s set f(t) = (t +1)^(3/2)/( (3 -t)sqrt(t) )We need to minimize f(t).Similarly, for t ∈ (3, ∞):A(t) = (9/2)*(t +1)^(3/2)/( (t -3)sqrt(t) )Let’s set g(t) = (t +1)^(3/2)/( (t -3)sqrt(t) )Need to minimize g(t).We can compute derivatives for both functions and find critical points.First, consider t ∈ (0,3):Let’s compute f(t):f(t) = (t +1)^{3/2} / [ (3 -t) t^{1/2} ]Take natural logarithm to simplify differentiation:ln f(t) = (3/2) ln(t +1) - ln(3 -t) - (1/2) ln tDifferentiate with respect to t:f’/f = (3/2)(1/(t +1)) - (-1)/(3 -t) - (1/2)(1/t)Simplify:= (3/(2(t +1))) + (1/(3 -t)) - (1/(2t))Set derivative to zero:(3/(2(t +1))) + (1/(3 -t)) - (1/(2t)) =0Multiply through by 2t(t +1)(3 -t) to eliminate denominators:3t(3 -t) + 2t(t +1) - (t +1)(3 -t) =0Expand each term:First term: 3t(3 -t) =9t -3t²Second term:2t(t +1)=2t² +2tThird term: -(t +1)(3 -t) = -[3t -t² +3 -t] = -[ -t² +2t +3 ]= t² -2t -3Combine all terms:9t -3t² +2t² +2t + t² -2t -3 =0Simplify:( -3t² +2t² +t² ) + (9t +2t -2t ) + (-3) =00t² +9t -3=0 →9t -3=0 →t=1/3Therefore, critical point at t=1/3 in (0,3). Check if this is a minimum.Second derivative test or check sign changes.For t approaching 0+:f(t) ≈ (1)^{3/2}/(3*sqrt(t)) =1/(3 sqrt(t)) →∞At t=1/3:Compute f(1/3):(1/3 +1)^{3/2}/( (3 -1/3)sqrt(1/3) )=(4/3)^{3/2}/( (8/3)(1/sqrt(3)) ) )= ( (4/3)^{3/2} * sqrt(3) ) / (8/3 )= ( (8/(3*sqrt(3))) * sqrt(3) ) / (8/3 )Wait:(4/3)^{3/2} = (4/3)*sqrt(4/3) = (4/3)*(2/sqrt(3)) )=8/(3 sqrt(3))Then, denominator: (8/3)(1/sqrt(3)) =8/(3 sqrt(3))Thus, f(1/3)= [8/(3 sqrt(3)) ] / [8/(3 sqrt(3)) ]=1Wait, that can't be right. Wait:Wait, f(t) = (t +1)^{3/2} / [ (3 -t)sqrt(t) ]At t=1/3:Numerator: (1/3 +1)^{3/2}=(4/3)^{3/2}= (4/3)*sqrt(4/3)= (4/3)*(2/sqrt(3))=8/(3 sqrt(3))Denominator: (3 -1/3)sqrt(1/3)= (8/3)*(1/sqrt(3))=8/(3 sqrt(3))Therefore, f(1/3)= [8/(3 sqrt(3))]/[8/(3 sqrt(3))]=1Therefore, at t=1/3, f(t)=1.But how does this relate to the original Area expression?Original Area is (9/2)*f(t). So at t=1/3, Area=(9/2)*1=9/2.But need to check behavior around t=1/3 to confirm if it's a minimum.Take t=1/3 ±ε.At t=1/3 +ε, compute derivative:If t=1/3 +ε, derivative is negative? Or positive?Wait, the derivative was zero at t=1/3. Let's check the sign of f’(t) for t <1/3 and t >1/3.Take t=0.2 (t <1/3≈0.333):Compute (3/(2(t +1))) + (1/(3 -t)) - (1/(2t))3/(2*1.2) +1/(2.8) -1/(0.4)= (3/2.4) + (1/2.8) -2.5≈1.25 +0.357 -2.5≈-0.893<0At t=0.5 (t >1/3):Compute:3/(2*1.5) +1/(2.5) -1/(1)= (3/3) +0.4 -1=1 +0.4 -1=0.4>0Therefore, derivative changes from negative to positive at t=1/3, indicating a minimum.Therefore, in the interval (0,3), the minimum occurs at t=1/3, with Area=9/2.Now, consider t ∈ (3, ∞). Let’s compute g(t):g(t) = (t +1)^{3/2}/( (t -3)sqrt(t) )Similarly, take ln g(t)= (3/2) ln(t +1) - ln(t -3) - (1/2) ln tDifferentiate:g’/g= (3/(2(t +1))) - (1/(t -3)) - (1/(2t))Set to zero:3/(2(t +1)) -1/(t -3) -1/(2t)=0Multiply through by 2t(t +1)(t -3):3t(t -3) -2t(t +1) - (t +1)(t -3)=0Expand:First term:3t² -9tSecond term:-2t² -2tThird term: -[t² -3t +t -3]= -t² +2t +3Combine all terms:3t² -9t -2t² -2t -t² +2t +3= (3t² -2t² -t²) + (-9t -2t +2t) +3=0t² -9t +3= -9t +3=0 ⇒-9t +3=0⇒t=1/3But t=1/3 is not in (3, ∞). Therefore, no critical points in t ∈ (3, ∞). Therefore, function g(t) has no critical points in this interval. Therefore, minima occurs at the boundary.But as t approaches 3 from above, g(t) tends to infinity (since denominator approaches 0). As t approaches infinity, let's see behavior:g(t)= (t +1)^{3/2}/( (t -3)sqrt(t) )≈ t^{3/2}/( t * sqrt(t) )=t^{3/2}/( t^{3/2} )=1. So as t→∞, g(t)→1. Therefore, the function decreases from infinity to 1 as t increases from 3 to ∞. Therefore, the minimum in this interval is approaching 1, but never actually reaching it. Therefore, there is no minimum in this interval; the infimum is 1, but it’s not achieved.Therefore, comparing the two cases:- For t ∈ (0,3), the minimal Area is 9/2 at t=1/3.- For t ∈ (3, ∞), Area approaches 9/2 *1=9/2 as t→∞, but since it never actually reaches it, the minimum remains 9/2 achieved at t=1/3.Therefore, the minimum area is 9/2.But let me verify this calculation, because when I computed f(1/3)=1, and the Area is (9/2)*f(t)=9/2.But let's verify by plugging m²=1/3, so m=1/√3.Compute coordinates of M, A, B.M is at (-1/2, -3/(2m))=(-1/2, -3/(2*(1/√3)))=(-1/2, - (3√3)/2 )Now, points A and B:s=sqrt(m² +1)=sqrt(1/3 +1)=sqrt(4/3)=2/sqrt(3)Compute x_A = [2*(1/3) +3*(2/sqrt(3)) ]/(3 -1/3)= [2/3 +6/sqrt(3) ]/(8/3)= [2/3 +2√3 ]/(8/3)= [2 +6√3]/8= [1 +3√3]/4Similarly, y_A =3m(s +2)/(3 -m²)=3*(1/√3)*(2/sqrt(3) +2)/(8/3)= (3/√3)*(2/sqrt(3) +2)*(3/8)= (sqrt(3))*(2/sqrt(3) +2)*(3/8)= [2 +2sqrt(3)]*(3/8)= (6 +6sqrt(3))/8= (3 +3sqrt(3))/4Similarly, x_B = [2*(1/3) -3*(2/sqrt(3)) ]/(8/3)= [2/3 -6/sqrt(3) ]/(8/3)= [2/3 -2√3 ]/(8/3)= [2 -6√3]/8= [1 -3√3]/4y_B=3m(2 -s)/(8/3)=3*(1/√3)*(2 -2/sqrt(3))*(3/8)= (sqrt(3))*(2 -2/sqrt(3))*(3/8)= [2sqrt(3) -2]*(3/8)= (6sqrt(3) -6)/8= (3sqrt(3) -3)/4Therefore, points:A: ( (1 +3√3)/4, (3 +3√3)/4 )B: ( (1 -3√3)/4, (3sqrt(3) -3)/4 )M: (-1/2, -3√3/2 )Compute vectors MA and MB or use coordinates to compute area.Using coordinates, the area can be calculated via the shoelace formula.Coordinates:M: (-1/2, -3√3/2 )A: ( (1 +3√3)/4, (3 +3√3)/4 )B: ( (1 -3√3)/4, (3√3 -3)/4 )Compute area:Area = (1/2)| (x_A -x_M)(y_B -y_M) - (x_B -x_M)(y_A -y_M) |Compute x_A -x_M = [ (1 +3√3)/4 +1/2 ]= [ (1 +3√3) +2 ]/4 = (3 +3√3)/4Similarly, y_B -y_M = [ (3√3 -3)/4 +3√3/2 ]= [ (3√3 -3) +6√3 ]/4 = (9√3 -3)/4Similarly, x_B -x_M = [ (1 -3√3)/4 +1/2 ]= [ (1 -3√3) +2 ]/4 = (3 -3√3)/4y_A -y_M = [ (3 +3√3)/4 +3√3/2 ]= [ (3 +3√3) +6√3 ]/4 = (3 +9√3)/4Then, compute (x_A -x_M)(y_B -y_M) = (3 +3√3)/4 * (9√3 -3)/4= [3(1 +√3)/4] * [3(3√3 -1)/4]= 9/16 (1 +√3)(3√3 -1)Similarly, compute (x_B -x_M)(y_A -y_M)= (3 -3√3)/4 * (3 +9√3)/4= [3(1 -√3)/4] * [3(1 +3√3)/4]=9/16 (1 -√3)(1 +3√3)Now, compute the difference:9/16 [ (1 +√3)(3√3 -1) - (1 -√3)(1 +3√3) ]Expand both products:First product: (1)(3√3) +1(-1) +√3(3√3) +√3(-1) =3√3 -1 +9 -√3= (3√3 -√3) + ( -1 +9 )=2√3 +8Second product: (1)(1) +1(3√3) -√3(1) -√3(3√3)=1 +3√3 -√3 -9= (3√3 -√3) + (1 -9)=2√3 -8Therefore, difference:[2√3 +8 - (2√3 -8)] =16Thus, Area= (1/2)|9/16 *16|= (1/2)*9=9/2Therefore, the area is indeed 9/2 when m²=1/3, confirming our previous result.Therefore, the minimal area is 9/2.Hence, answers:1. The angle is 90 degrees, so π/2.2. The minimum area is 9/2.Final Answer(1) The angle between the vector ( MF ) and ( AB ) is (boxed{dfrac{pi}{2}}).(2) The minimum value of the area of (triangle MAB) is (boxed{dfrac{9}{2}}).</think>Given the hyperbola ( x^2 - frac{y^2}{3} = 1 ), we know the left focus ( F ) is at ((-2, 0)). A line ( AB ) through ( F ) intersects the hyperbola at points ( A ) and ( B ). Tangents at ( A ) and ( B ) intersect at point ( M ).1. Angle between vector ( MF ) and ( AB ): - The line ( AB ) is parametrized as ( y = m(x + 2) ). - Solving the intersection of this line with the hyperbola, we find the coordinates of points ( A ) and ( B ). - The tangents at ( A ) and ( B ) intersect at point ( M ), which is found to be ((-1/2, -3/(2m))). - Vector ( MF ) is calculated as ((-3/2, 3/(2m))). - The direction vector of ( AB ) is ((1, m)). - The dot product of ( MF ) and ( AB ) is zero, indicating they are perpendicular. Thus, the angle is ( frac{pi}{2} ).2. Minimum value of the area of ( triangle MAB ): - Using the coordinates of ( A ), ( B ), and ( M ), the area of ( triangle MAB ) is calculated using the determinant formula. - The area is found to be ( frac{9}{2} times frac{(m^2 + 1)^{3/2}}{|3 - m^2||m|} ). - By analyzing the function and finding critical points, the minimum area is determined to be ( frac{9}{2} ) when ( m^2 = frac{1}{3} ). Final Answers(1) The angle between the vector ( MF ) and ( AB ) is (boxed{dfrac{pi}{2}}).(2) The minimum value of the area of (triangle MAB) is (boxed{dfrac{9}{2}}).]

🔑:11. Solution: (1) It is easy to know that F(-2,0), let the equation of the line be x=m y-2, Aleft(x_{A}, y_{A}right), Bleft(x_{B}, y_{B}right), Mleft(x_{M}, y_{M}right),Substituting into the hyperbola equation yields: left(m^{2}-frac{1}{3}right) y^{2}-4 m y+3=0, hence y_{A}+y_{B}=frac{12 m}{3 m^{2}-1}, y_{A} y_{B}=frac{9}{3 m^{2}-1}; the equation of line l_{1} is: x x_{A}-frac{y y_{A}}{3}=1, the equation of line l_{2} is: x x_{B}-frac{y y_{B}}{3}=1;Since M F bullet A B=0, the angle between them is 90^{circ};(2) S_{triangle M A B}=frac{1}{2} A B bullet M F=frac{3}{4}left(m^{2}+1right)left|y_{A}-y_{B}right|=frac{9}{2}left(m^{2}+1right) frac{sqrt{m^{2}+1}}{left|3 m^{2}-1right|}Let t=frac{1}{sqrt{m^{2}+1}}, then t in(0,1], S_{triangle M A B}=frac{9}{2} frac{1}{left|3 t-4 t^{3}right|}, it is easy to know that f(t)=3 t-4 t^{3} is monotonically increasing on left(0, frac{1}{2}right) and monotonically decreasing on left(frac{1}{2}, 1right), hence when t in(0,1], f(t) in[-1,1], i.e., S_{triangle M A B} inleft[frac{9}{2},+inftyright), when m= pm sqrt{3}, 0, i.e., the equation of line A B is x= pm sqrt{3} y-2 or x=-2, S_{triangle M A B} reaches its minimum value frac{9}{2}.