🧠:[<think>Okay, let's see. I need to find the value of a + b given that the function f(x) = ax² + (b - 3)x + 3 is an even function on the interval [a² - 2, a]. The options are A through D, with numerical answers from 1 to 4. Alright, let me start by recalling what an even function is.An even function satisfies the condition f(x) = f(-x) for all x in its domain. So, for every x in the interval [a² - 2, a], the equation ax² + (b - 3)x + 3 = a(-x)² + (b - 3)(-x) + 3 should hold true. Let me write that out:ax² + (b - 3)x + 3 = a(x²) + (b - 3)(-x) + 3Simplify both sides:Left side: ax² + (b - 3)x + 3Right side: ax² - (b - 3)x + 3Subtract the right side from the left side to see what the difference is:[ax² + (b - 3)x + 3] - [ax² - (b - 3)x + 3] = 2(b - 3)xFor the function to be even, this difference must be zero for all x in the domain. So 2(b - 3)x = 0 for all x in [a² - 2, a]. The only way this can be true for all x is if the coefficient of x is zero. Therefore:2(b - 3) = 0 ⇒ b - 3 = 0 ⇒ b = 3Okay, so that gives me b = 3. Then, a + b would be a + 3. So if I can find the value of a, I can compute a + 3 and select the correct option.But there's another condition here: the function is defined on the interval [a² - 2, a]. For the function to be even, the domain must also be symmetric about the origin. That is, the interval [a² - 2, a] must satisfy that for every x in the interval, -x is also in the interval. So, the domain itself has to be symmetric around zero.For the interval [a² - 2, a] to be symmetric about zero, the lower bound must be the negative of the upper bound. In other words:a² - 2 = -aSo, solving this equation:a² + a - 2 = 0Factor this quadratic equation. Let's see, looking for two numbers that multiply to -2 and add to +1. Hmm, 2 and -1.Therefore:(a + 2)(a - 1) = 0So the solutions are a = -2 or a = 1.Wait, but we need to check if these solutions make sense in the context of the interval [a² - 2, a]. Let's check both possibilities.First, a = -2:If a = -2, then the interval becomes [(-2)² - 2, -2] = [4 - 2, -2] = [2, -2]. Wait, but this interval is from 2 to -2, which is invalid because the lower bound should be less than the upper bound. Since 2 > -2, this interval is empty or invalid. So a = -2 is not acceptable.Now check a = 1:If a = 1, then the interval is [1² - 2, 1] = [1 - 2, 1] = [-1, 1]. This is a valid interval, and it is symmetric about zero, as required for an even function. So a = 1 is valid.Therefore, a = 1 and b = 3. Hence, a + b = 1 + 3 = 4, which is option D. Wait, but let me double-check because sometimes there might be an oversight.Wait, the problem says x ∈ [a² - 2, a]. So when a = 1, the interval is [-1, 1], which is symmetric around 0, which is necessary for the function to be even. If the domain wasn't symmetric, even if the algebraic condition f(x) = f(-x) is satisfied, the function wouldn't be even because the domain itself isn't symmetric. So in this case, since a = 1 gives a symmetric interval, and we already found b = 3, that's correct.Therefore, a + b = 1 + 3 = 4. So the answer should be D.But wait, let me check once more. Let me confirm by plugging the values back into the original function.Given a = 1, b = 3, then f(x) = 1*x² + (3 - 3)x + 3 = x² + 0x + 3 = x² + 3. This is indeed an even function because x² is even, and 3 is a constant term, which is also even. The domain is [-1, 1], which is symmetric about 0, so it's valid.If I had chosen a = -2, even though the interval [2, -2] is invalid, if for some reason we considered it as [-2, 2], but that's not the case here because a is -2, so a² - 2 is 4 - 2 = 2, and the upper limit is a = -2, so the interval is [2, -2], which is not valid. Therefore, a = -2 is rejected.So yes, a = 1 is the only valid solution, leading to a + b = 4. Therefore, the correct answer is D:4.Final Answerboxed{D}</think>Given the function ( f(x) = ax^2 + (b - 3)x + 3 ) defined on the interval ([a^2 - 2, a]) is an even function, we need to find the value of (a + b).An even function satisfies ( f(x) = f(-x) ) for all ( x ) in its domain. Therefore, we equate the function to its reflection:[ax^2 + (b - 3)x + 3 = ax^2 - (b - 3)x + 3]Subtracting the right side from the left side, we get:[2(b - 3)x = 0]For this to hold for all ( x ) in the domain, the coefficient must be zero:[2(b - 3) = 0 implies b - 3 = 0 implies b = 3]Next, the interval ([a^2 - 2, a]) must be symmetric about the origin. Thus, the lower bound must be the negative of the upper bound:[a^2 - 2 = -a]Solving this quadratic equation:[a^2 + a - 2 = 0]Factoring, we get:[(a + 2)(a - 1) = 0 implies a = -2 text{ or } a = 1]Checking the validity of these solutions:- For ( a = -2 ), the interval is ([2, -2]), which is invalid.- For ( a = 1 ), the interval is ([-1, 1]), which is valid and symmetric.Thus, ( a = 1 ) is the valid solution. Therefore, ( a + b = 1 + 3 = 4 ).The final answer is (boxed{D}).,<think>Okay, let's see. I need to find the value of a + b given that the function f(x) = ax² + (b - 3)x + 3 is an even function on the interval [a² - 2, a]. The options are A through D with numbers 1 to 4. Hmm, alright. Let me recall what an even function is. An even function satisfies the condition f(x) = f(-x) for all x in its domain. So, for every x in [a² - 2, a], the equation ax² + (b - 3)x + 3 = a(-x)² + (b - 3)(-x) + 3 should hold true. Let me write that down.First, let's substitute -x into the function:f(-x) = a(-x)² + (b - 3)(-x) + 3 = a x² - (b - 3)x + 3Since the function is even, f(x) must equal f(-x) for all x in the domain. Therefore:ax² + (b - 3)x + 3 = a x² - (b - 3)x + 3Subtracting ax² and 3 from both sides gives:(b - 3)x = - (b - 3)xBring all terms to one side:(b - 3)x + (b - 3)x = 02(b - 3)x = 0For this equation to hold true for all x in the interval [a² - 2, a], the coefficient of x must be zero. So:2(b - 3) = 0=> b - 3 = 0=> b = 3Alright, so we found that b must be 3. Now, moving on to the domain. The function is defined on [a² - 2, a]. For the function to be even, the domain must be symmetric about the origin. That is, for every x in the domain, -x must also be in the domain. So, the left endpoint of the interval should be the negative of the right endpoint. Let me check that.Given the interval [a² - 2, a], symmetry about the origin requires that:a² - 2 = -aSo solving this equation:a² + a - 2 = 0This is a quadratic equation. Let's factor it or use the quadratic formula. Let's see if it factors:Looking for two numbers that multiply to -2 and add to 1. Hmm, 2 and -1. Yes.(a + 2)(a - 1) = 0Therefore, a = -2 or a = 1.Wait, but we need to check if these solutions make sense in terms of the interval. The interval is [a² - 2, a]. The left endpoint must be less than or equal to the right endpoint. So, for a = -2:Left endpoint: (-2)^2 - 2 = 4 - 2 = 2Right endpoint: a = -2So, the interval would be [2, -2], which doesn't make sense because 2 > -2. Intervals are written with the smaller number first. So this is invalid. Therefore, a = -2 is not a valid solution.Now check a = 1:Left endpoint: (1)^2 - 2 = 1 - 2 = -1Right endpoint: a = 1So the interval is [-1, 1]. That's valid because -1 ≤ 1, and it is symmetric around 0. Because if you take any x in [-1, 1], -x is also in [-1, 1]. So, this works. Therefore, a = 1 is the valid solution.Therefore, a = 1 and b = 3. So a + b = 1 + 3 = 4. Wait, but the options are A:1, B:2, C:3, D:4. So the answer should be D:4.Wait, but let me double-check. Because sometimes there might be some gotchas here. Let me verify the function with a=1 and b=3. Then f(x) = 1*x² + (3 -3)x +3 = x² + 0x +3 = x² +3. This is indeed an even function because x² is even, and 3 is a constant term, which is even. The domain is [-1,1], which is symmetric about 0. So everything checks out. If a was -2, the interval would be [2, -2], which is invalid, so a must be 1. Therefore, a + b = 1 +3=4, so D.But wait, let me check if there are any other constraints. The problem states that x is in [a² -2, a]. So when a=1, the interval is [-1,1], which is okay. When a=-2, the interval is [2, -2], which is not a valid interval, so we can't have a=-2. So only a=1 is valid, so the answer is D:4. Hmm. But the options given include D:4, so that's the answer. But let me just think again. The problem states that the function is even on that interval, which requires both that the domain is symmetric and that the function is even. We already ensured that. So with a=1, the domain is symmetric, and the function becomes f(x)=x² +3, which is even. Therefore, yes, everything checks out. So a + b =1 +3=4. So the answer is D. I think that's correct.Final Answerboxed{D}</think>Given the function ( f(x) = ax^2 + (b - 3)x + 3 ) defined on the interval ([a^2 - 2, a]), we need to determine the value of (a + b) given that the function is even.An even function satisfies ( f(x) = f(-x) ) for all (x) in its domain. Substituting (-x) into the function, we get:[f(-x) = a(-x)^2 + (b - 3)(-x) + 3 = ax^2 - (b - 3)x + 3]For (f(x)) to be even, we must have:[ax^2 + (b - 3)x + 3 = ax^2 - (b - 3)x + 3]Subtracting (ax^2) and 3 from both sides, we get:[(b - 3)x = - (b - 3)x]This simplifies to:[2(b - 3)x = 0]For this to hold for all (x) in the domain, the coefficient must be zero:[2(b - 3) = 0 implies b - 3 = 0 implies b = 3]Next, the domain ([a^2 - 2, a]) must be symmetric about the origin, meaning (a^2 - 2 = -a). Solving this quadratic equation:[a^2 + a - 2 = 0]Factoring, we get:[(a + 2)(a - 1) = 0 implies a = -2 text{ or } a = 1]Checking the validity of these solutions:- For (a = -2), the interval is ([2, -2]), which is invalid.- For (a = 1), the interval is ([-1, 1]), which is valid and symmetric.Thus, (a = 1) is the valid solution. With (a = 1) and (b = 3), we find (a + b = 1 + 3 = 4).The final answer is (boxed{D}).]

🔑:Since the domain should be symmetric about the origin, we have a^2 - 2 = -a, which gives a = 1 or a = -2 (discard -2). Furthermore, since f(-x) = f(x) always holds, we have: ax^2 - (b-3)x + 3 = ax^2 + (b-3)x + 3, thus, b = 3. Therefore, a + b = 4. Hence, the correct choice is: boxed{D}. First, from "the domain should be symmetric about the origin", and since f(-x) = f(x) always holds, we can determine b using the method of undetermined coefficients. This problem mainly examines the definition of the even and odd functions, starting with the domain being symmetric about the origin, and secondly discussing the relationship between f(x) and f(-x), which is a medium-level question.

🧠:[<think>Okay, let's see. I need to determine which of the given functions can be considered a generalized "distance" based on the three properties listed: non-negativity, symmetry, and the triangle inequality. Let me go through each function one by one.Starting with function (1): f(x, y) = |x - y|.First, non-negativity. The absolute value is always non-negative, and |x - y| equals zero only when x = y. So that checks out.Second, symmetry. |x - y| is the same as |y - x| because subtraction inside the absolute value just changes the sign, but the absolute value doesn't care about the sign. So yes, symmetric.Third, the triangle inequality. The standard triangle inequality for absolute values says that |x - y| ≤ |x - z| + |z - y|. Which is exactly what's required here. So function (1) satisfies all three properties. So (1) is good.Moving on to function (2): f(x, y) = (x - y)^2.First, non-negativity. Squaring any real number gives a non-negative result, and (x - y)^2 is zero only when x = y. So that's satisfied.Second, symmetry. (x - y)^2 is the same as (y - x)^2 because squaring negates the sign. So that's symmetric. Good so far.Third, the triangle inequality. Hmm, this is trickier. We need to check if (x - y)^2 ≤ (x - z)^2 + (z - y)^2 for any real z.Let me test with some numbers. Let's take x = 0, y = 2, z = 1. Then:Left side: (0 - 2)^2 = 4.Right side: (0 - 1)^2 + (1 - 2)^2 = 1 + 1 = 2.But 4 ≤ 2? No, 4 is not less than or equal to 2. So the triangle inequality fails here. Therefore, function (2) does not satisfy the triangle inequality. So (2) is out.Wait, but maybe I should check another example to be sure. Let's take x = 1, y = 3, z = 2.Left side: (1 - 3)^2 = 4.Right side: (1 - 2)^2 + (2 - 3)^2 = 1 + 1 = 2. Again, 4 ≤ 2 is false. So definitely, the triangle inequality doesn't hold for squared differences. Therefore, function (2) is not a generalized distance.Now function (3): f(x, y) = sqrt(x - y).Wait, hold on. The square root function here. First, the expression inside the square root must be non-negative, right? Because the square root of a negative number isn't a real number. So x - y ≥ 0, which implies x ≥ y. But the function is supposed to be defined for all real numbers x and y, right? Because A and B are subsets of R, and it's a binary function for any x in A and y in B. So if x < y, then sqrt(x - y) would be taking the square root of a negative number, which is not a real number. But the problem statement says "for real numbers x and y", so the function must output a real number. Therefore, f(x, y) is only defined when x ≥ y. But the problem requires that for any x and y in R, f(x, y) is determined. However, if x < y, sqrt(x - y) is not a real number. Therefore, this function isn't properly defined for all real numbers x and y. So already, function (3) fails because it's not defined for all x and y. Therefore, it can't be a generalized distance.But even if we ignore that and consider where it is defined, let's check the properties.First, non-negativity. The square root is always non-negative when defined. So if x ≥ y, sqrt(x - y) ≥ 0. However, equality holds only when x - y = 0, which is x = y. So that part is okay when x ≥ y, but since it's not defined for x < y, the non-negativity is only partial. But the problem requires the function to be defined for any x and y. So this is a problem.Second, symmetry. If we consider x and y where x ≥ y, then f(x, y) = sqrt(x - y). But if we switch x and y, then if y < x, we get f(y, x) = sqrt(y - x), which is not a real number. So f(y, x) is not defined. Therefore, the function isn't symmetric. Even if we restrict to x ≥ y, the symmetry is not there because f(y, x) would not be the same as f(x, y). For example, x=2, y=1: f(2,1)=sqrt(1)=1, but f(1,2)=sqrt(-1), which is undefined. So symmetry fails.Third, the triangle inequality. Even if we ignore the previous issues, suppose we try to check. For example, take x=3, y=1, z=2. Then f(3,1)=sqrt(3-1)=sqrt(2). Then f(3,2)=sqrt(1), f(2,1)=sqrt(1). So sqrt(2) ≤ 1 + 1? sqrt(2) ≈1.414 ≤ 2. That's true. But this is just one case. Let's try another. Take x=4, y=1, z=3. Then f(4,1)=sqrt(3)≈1.732. f(4,3)=sqrt(1)=1, f(3,1)=sqrt(2)≈1.414. So 1.732 ≤ 1 + 1.414 ≈2.414. That's true. But maybe another case where it might fail. Let's take x=5, y=0, z=3. Then f(5,0)=sqrt(5)≈2.236. f(5,3)=sqrt(2)≈1.414, f(3,0)=sqrt(3)≈1.732. Then 2.236 ≤ 1.414 + 1.732 ≈3.146. That holds. Hmm, maybe in cases where the function is defined, the triangle inequality holds? Wait, but if I pick another z. Let's say x=5, y=0, z=4. Then f(5,0)=sqrt(5), f(5,4)=sqrt(1)=1, f(4,0)=sqrt(4)=2. Then 1 + 2 =3 ≥ sqrt(5)≈2.236. Still holds.But wait, maybe if z is not between x and y. Let me think. Suppose x=5, y=0, z=2. Then same as before. Wait, regardless of where z is, as long as x >= z >= y, but z can be anything. Wait, but z can be any real number. Suppose z=6. Then f(5,6)=sqrt(-1), which is undefined. So in this case, the triangle inequality can't be checked because f(5,6) is not defined. Therefore, since the function isn't defined for all z, the triangle inequality can't hold for all real z. Thus, function (3) fails the triangle inequality because it's not defined for all z. Therefore, even if we tried to fix the domain, the function isn't properly a binary function over all real numbers.Therefore, function (3) doesn't satisfy all three properties, primarily because it's not defined for all real numbers x and y, and it's not symmetric. So (3) is invalid.So the only one that works is (1).Final AnswerThe correct sequence numbers are boxed{1}.</think>To determine which functions can serve as a generalized "distance" for real numbers (x) and (y), we need to check if they satisfy the three properties: non-negativity, symmetry, and the triangle inequality.1. Function (f(x, y) = |x - y|): - Non-negativity: (|x - y| geq 0) and equality holds if and only if (x = y). This is satisfied. - Symmetry: (|x - y| = |y - x|). This is satisfied. - Triangle inequality: (|x - y| leq |x - z| + |z - y|) for any real (z). This is satisfied by the standard triangle inequality for absolute values.2. Function (f(x, y) = (x - y)^2): - Non-negativity: ((x - y)^2 geq 0) and equality holds if and only if (x = y). This is satisfied. - Symmetry: ((x - y)^2 = (y - x)^2). This is satisfied. - Triangle inequality: Testing with specific values (e.g., (x = 0), (y = 2), (z = 1)) shows that ((x - y)^2 leq (x - z)^2 + (z - y)^2) does not hold. This is not satisfied.3. Function (f(x, y) = sqrt{x - y}): - Non-negativity: (sqrt{x - y}) is non-negative only when (x geq y). However, it is not defined for (x < y), so it fails to be a real number for all (x, y). - Symmetry: (sqrt{x - y}) is not symmetric because (sqrt{y - x}) is not defined for (x < y). This is not satisfied. - Triangle inequality: Since the function is not defined for all (x, y), the triangle inequality cannot be verified for all real (z). This is not satisfied.Thus, the only function that satisfies all three properties is (f(x, y) = |x - y|).The correct sequence numbers are boxed{1}.,<think>Okay, let's see. I need to figure out which of the given functions qualify as generalized "distance" based on the three properties: non-negativity, symmetry, and triangle inequality. The options are (1) absolute difference, (2) squared difference, and (3) square root of (x - y). Hmm.Starting with the first function, f(x, y) = |x - y|. Let's check each property.1. Non-negativity: Absolute value is always non-negative, and |x - y| = 0 if and only if x = y. That checks out.2. Symmetry: |x - y| is the same as |y - x|, so symmetry holds.3. Triangle inequality: The regular triangle inequality for absolute values states that |x - y| ≤ |x - z| + |z - y|, which is exactly what's required. So yes, this satisfies all three properties. So (1) is a generalized distance.Moving on to the second function, f(x, y) = (x - y)^2.1. Non-negativity: A square is always non-negative, and (x - y)^2 = 0 only when x = y. So that's good.2. Symmetry: (x - y)^2 is the same as (y - x)^2, so symmetry holds here too.3. Triangle inequality: Hmm, does (x - y)^2 ≤ (x - z)^2 + (z - y)^2 hold for all real z? Let me test with some numbers. Let's take x = 0, y = 2, z = 1. Then left side is (0 - 2)^2 = 4. Right side is (0 - 1)^2 + (1 - 2)^2 = 1 + 1 = 2. But 4 ≤ 2? No, that's false. So the triangle inequality doesn't hold here. Therefore, (2) is not a generalized distance.Third function, f(x, y) = sqrt(x - y). Wait, hold on. The square root is only defined when x - y ≥ 0. So this function isn't even defined for all real numbers x and y, right? If x < y, then we have a square root of a negative number, which isn't real. So already, this function isn't a valid binary function for all real x and y. But the problem statement says "for real numbers x and y". Therefore, if the function isn't defined for all real x and y, it can't satisfy the conditions. So even before checking the properties, this is invalid.But let's check the properties for cases where it is defined (x ≥ y). However, even then, the problem requires the function to be defined for any x ∈ A, y ∈ B where A and B are subsets of R. If the function isn't defined for all x and y, then it can't satisfy the first condition, because the problem states "for any x ∈ A, y ∈ B". So unless A and B are restricted to specific sets where x ≥ y always, but the question doesn't specify that. So I think (3) is invalid because it's not defined for all real numbers.Additionally, even if we consider x ≥ y, let's check symmetry. sqrt(x - y) vs sqrt(y - x). If x ≠ y, then one of them would be undefined. If x = y, both are zero. But since in cases where x > y, sqrt(x - y) is defined, but sqrt(y - x) isn't. So symmetry fails. For example, take x = 3, y = 1. Then f(3,1) = sqrt(2), but f(1,3) = sqrt(-2), which isn't real. So symmetry isn't satisfied. Therefore, (3) fails both because it's not always defined and because symmetry doesn't hold.So summarizing:Function (1) passes all three properties.Function (2) fails the triangle inequality.Function (3) isn't properly defined for all real numbers and fails symmetry.Therefore, the correct answer is only (1).Final AnswerThe correct sequence numbers are boxed{1}.</think>To determine which of the given functions can serve as a generalized "distance" for real numbers (x) and (y), we need to check if they satisfy the three properties: non-negativity, symmetry, and the triangle inequality.1. Function (f(x, y) = |x - y|): - Non-negativity: (|x - y| geq 0) with equality if and only if (x = y). This holds. - Symmetry: (|x - y| = |y - x|). This holds. - Triangle inequality: (|x - y| leq |x - z| + |z - y|) for any real (z). This holds by the standard triangle inequality for absolute values. - Conclusion: This function satisfies all properties.2. Function (f(x, y) = (x - y)^2): - Non-negativity: ((x - y)^2 geq 0) with equality if and only if (x = y). This holds. - Symmetry: ((x - y)^2 = (y - x)^2). This holds. - Triangle inequality: Check if ((x - y)^2 leq (x - z)^2 + (z - y)^2) for all real (z). Testing with (x = 0), (y = 2), (z = 1) gives (4 leq 1 + 1 = 2), which is false. This does not hold. - Conclusion: This function fails the triangle inequality.3. Function (f(x, y) = sqrt{x - y}): - Non-negativity: The square root is only defined when (x geq y), and (sqrt{x - y} geq 0). However, it is not defined for all real (x) and (y) (e.g., when (x < y)). - Symmetry: (sqrt{x - y}) is not equal to (sqrt{y - x}) when (x neq y), and the latter is not defined if (x > y). This fails. - Conclusion: This function is not defined for all real numbers and fails symmetry.Thus, the correct sequence number is (boxed{1}).]

🔑:Solution: Only (1) satisfies all three properties; where (2) does not satisfy property (3), (3) does not satisfy property (2), thus the correct answer is (1).

🧠:[<think>Okay, let's see. I need to find the modulus of the complex number z, which is given as z = -1 + sqrt(3)i. Hmm, modulus of a complex number... right, modulus is like the distance from the origin in the complex plane. So if z is a + bi, then the modulus |z| is sqrt(a² + b²). Yeah, that formula rings a bell. Let me confirm that. So, in this case, the real part a is -1, and the imaginary part b is sqrt(3). Therefore, substituting into the formula, |z| should be sqrt((-1)² + (sqrt(3))²). Let me calculate each part step by step. First, (-1) squared. Well, negative one times negative one is positive one. So that's 1. Then, sqrt(3) squared. The square of a square root cancels out, right? So sqrt(3) squared is just 3. Then, adding those two results together: 1 + 3 = 4. So then the modulus is the square root of 4. The square root of 4 is 2. Wait, that seems straightforward. Let me just double-check. Maybe I mixed up the real and imaginary parts? The complex number is written as -1 + sqrt(3)i, so the real part is indeed -1 and the imaginary coefficient is sqrt(3). So plugging into sqrt(a² + b²) is correct. Squaring them gives 1 and 3, sum is 4, square root is 2. Yeah, that's right. Alternatively, maybe there's another way to think about it. Sometimes with complex numbers, converting to polar form helps. The modulus is the same as the magnitude in polar coordinates, so maybe if I tried converting z to polar form, I would get the modulus directly. Let's try that. To convert a complex number to polar form, you need the modulus and the argument. The modulus is what we're trying to find, so that's circular. The argument is the angle, which isn't necessary here, but maybe going through the process would reinforce the modulus. The modulus is sqrt(a² + b²), which again is sqrt(1 + 3) = sqrt(4) = 2. So regardless of the method, the modulus is 2. Wait, another way to visualize this. If I plot the complex number z = -1 + sqrt(3)i on the complex plane, the real part is -1 on the real axis, and the imaginary part is sqrt(3) on the imaginary axis. So the point is in the second quadrant. The distance from the origin to this point would be the hypotenuse of a right triangle with legs of length 1 and sqrt(3). Wait, hypotenuse squared is 1² + (sqrt(3))² = 1 + 3 = 4, so hypotenuse is 2. So that's the modulus. Alternatively, maybe recognizing that this complex number is related to some standard angles. For instance, 1 + sqrt(3)i is a complex number with modulus 2 and argument 60 degrees (pi/3 radians). But here it's -1 + sqrt(3)i, which is like reflecting the 1 + sqrt(3)i across the imaginary axis, moving it to the second quadrant. The modulus would still be the same because reflecting doesn't change the distance from the origin. Wait, no, actually, reflecting changes the real part's sign, but the modulus depends on the squares, so (-1)^2 is the same as 1^2. So modulus is the same. So modulus is indeed 2. Alternatively, if I remember that modulus is multiplicative, but since we aren't multiplying complex numbers here, maybe that's not helpful. Alternatively, maybe using vector magnitude. The complex number can be seen as a vector in 2D space with components (-1, sqrt(3)). The magnitude of this vector is sqrt((-1)^2 + (sqrt(3))^2) which again is sqrt(1 + 3) = sqrt(4) = 2. So, no matter how I approach this, the modulus seems to be 2. Wait, let me check once more to be absolutely sure. Maybe I made a mistake in squaring the terms. The real part is -1, so square is (-1)^2 = 1. The imaginary part is sqrt(3), so square is (sqrt(3))^2 = 3. Sum is 1 + 3 = 4. Square root of 4 is 2. Correct. Alternatively, is there a possibility that the question is asking for something else? For example, the conjugate or the argument? But the question specifically says |z|, which is the modulus. So modulus is the right thing to compute here. Alternatively, maybe the problem is a trick question? For example, if z was given in a different form, like polar form, but here it's given in standard rectangular form, so straightforward. Alternatively, maybe there's a typo in the problem? But as given, z = -1 + sqrt(3)i. So no, that's correct. Therefore, after verifying through multiple methods: using the modulus formula, converting to polar form, visualizing on the complex plane, and checking the vector magnitude, all lead to the modulus being 2. Final AnswerThe modulus of the complex number ( z ) is boxed{2}.</think>To find the modulus of the complex number ( z = -1 + sqrt{3}i ), we use the formula for the modulus of a complex number ( a + bi ), which is ( sqrt{a^2 + b^2} ).1. Identify the real part ( a = -1 ) and the imaginary part ( b = sqrt{3} ).2. Calculate the square of the real part: ( (-1)^2 = 1 ).3. Calculate the square of the imaginary part: ( (sqrt{3})^2 = 3 ).4. Sum these squares: ( 1 + 3 = 4 ).5. Take the square root of the sum: ( sqrt{4} = 2 ).Thus, the modulus of the complex number ( z ) is (boxed{2}).,<think>Okay, so I need to find the modulus of the complex number z, which is given as z = -1 + √3 i. Hmm, modulus... right, the modulus of a complex number is like its distance from the origin in the complex plane. I remember there's a formula for that. Let me think. If a complex number is written as a + bi, then the modulus is the square root of (a squared plus b squared). Yeah, that sounds right. So in this case, a is -1 and b is √3. Wait, so does that mean I just plug those into the formula? Let's try that. First, square the real part, which is -1. Squaring -1 gives 1, right? Because negative times negative is positive. Then the imaginary part is √3, so squaring that would be (√3)^2. That should be 3, since squaring a square root cancels them out. So adding those two results together: 1 + 3 equals 4. Then take the square root of 4, which is 2. So the modulus |z| should be 2? That seems straightforward.But let me double-check to make sure I didn't mix up the real and imaginary parts. The complex number is -1 + √3 i, so the real part is indeed -1 and the imaginary coefficient is √3. Yeah, the formula is sqrt(a² + b²), so plugging in a = -1 and b = √3. Yep, (-1)^2 is 1, (√3)^2 is 3, sum is 4, square root of 4 is 2. That seems correct.Wait, another way to think about modulus is the distance from the point (-1, √3) to the origin (0,0) on the coordinate plane. So using the distance formula, which is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Here, the points are (0,0) and (-1, √3). So plugging into the distance formula: sqrt[(-1 - 0)^2 + (√3 - 0)^2] which is sqrt[1 + 3] = sqrt[4] = 2. So that confirms it again. The modulus is 2.Is there any chance I could have made a mistake here? Let me see. If I confused the real and imaginary parts... but no, the real part is clearly -1, and the imaginary part is √3. Squaring them both gives positive numbers, so their sum is positive. Square root of a positive number is positive, so modulus is a positive real number. So 2 makes sense. Another check: sometimes when dealing with complex numbers, especially when converting between rectangular and polar forms, modulus is important. If I recall, the modulus is also the hypotenuse of the right triangle formed by the real and imaginary parts. So in this case, the legs of the triangle would be 1 and √3. Wait, 1 and √3... Hmm, that triangle is actually a 30-60-90 triangle. In a 30-60-90 triangle, the sides are in the ratio 1 : √3 : 2. So if the legs are 1 and √3, the hypotenuse should be 2. Which matches our previous result. So that's another confirmation.Wait, but in this case, the real part is -1, so the triangle is in the second quadrant. But does that affect the modulus? No, because modulus is just the distance, regardless of direction. So even though the real part is negative, squaring it removes the sign. So modulus is still positive. Therefore, even if the complex number is in a different quadrant, the modulus calculation remains the same. So that's not an issue here.I think I've checked it multiple ways: using the formula, using the distance formula, relating it to a known triangle ratio. All lead to the modulus being 2. I don't see any errors in my reasoning. So I think the answer is 2.Final AnswerThe modulus of ( z ) is boxed{2}.</think>To find the modulus of the complex number ( z = -1 + sqrt{3}i ), we use the formula for the modulus of a complex number ( a + bi ), which is ( sqrt{a^2 + b^2} ).1. Identify the real part ( a ) and the imaginary part ( b ): - Real part ( a = -1 ) - Imaginary part ( b = sqrt{3} )2. Square both parts: - ( (-1)^2 = 1 ) - ( (sqrt{3})^2 = 3 )3. Sum the squares: - ( 1 + 3 = 4 )4. Take the square root of the sum: - ( sqrt{4} = 2 )Thus, the modulus of ( z ) is (boxed{2}).]

🔑:Given z = -1 + sqrt{3}i,we want to find the modulus of z, denoted as |z|.Recall that the modulus of a complex number z = a + bi is given by sqrt{a^2 + b^2}. Applying this formula to our given complex number, we have:|z| = sqrt{(-1)^2 + (sqrt{3})^2}|z| = sqrt{1 + 3}|z| = sqrt{4}|z| = 2Thus, the modulus of the complex number is boxed{2}.

🧠:[<think>Okay, let's see. I need to figure out what happens when the graph of a function y = f(x), which has an inverse, is rotated 90 degrees counterclockwise around the origin. The options are given as different transformations involving the inverse function. Hmm, let's start by recalling some basics.First, rotating a graph 90 degrees counterclockwise around the origin. I remember that in coordinate geometry, rotating a point (a, b) 90 degrees counterclockwise around the origin gives the point (-b, a). Is that right? Let me verify. If you have a point (1, 0), rotating it 90 degrees CCW should take it to (0, 1), then another 90 to (-1, 0), etc. So yes, the formula for rotation is (x, y) becomes (-y, x). So if a point (a, b) is on the original graph, after rotation, it becomes (-b, a). Now, the original graph is of y = f(x), so any point on it is (x, f(x)). After rotation, this point becomes (-f(x), x). So the new graph has points (-f(x), x). Let's denote the new function as y = g(x). So, for each point (X, Y) on the new graph, we have X = -f(x) and Y = x. Wait, but we need to express Y in terms of X. Let me write that again. The original point is (x, f(x)). After rotation, it's (-f(x), x). So the new coordinates are X = -f(x) and Y = x. So, to express Y in terms of X, we need to solve for x in terms of X. Since X = -f(x), we can write f(x) = -X. Then, applying the inverse function f^{-1} to both sides, x = f^{-1}(-X). But Y = x, so Y = f^{-1}(-X). So the equation of the new graph is Y = f^{-1}(-X). In terms of X and Y, but we can replace X with x and Y with y for the function notation. So the new function is y = f^{-1}(-x). Hmm, that would correspond to option D: y = f^{-1}(-x). But wait, let me check again.Wait, but maybe I made a mistake in the direction of rotation. Let me confirm the rotation again. If I rotate the graph 90 degrees counterclockwise, then each point (x, y) is transformed to (-y, x). So if the original graph is y = f(x), then the transformed graph is obtained by replacing x with -y and y with x. So, substituting into the original equation: x = f(-y). Wait, that might be another approach. Let me think.Alternatively, suppose we consider the rotation as a transformation of coordinates. If we rotate the coordinate system 90 degrees clockwise, which is equivalent to rotating the graph 90 degrees counterclockwise. The transformation matrix for a 90 degree CCW rotation is [0, -1; 1, 0]. So, if (X, Y) is the new coordinate after rotation, then:X = x*0 - y*1 = -yY = x*1 + y*0 = xTherefore, to express the original coordinates in terms of the new ones:From X = -y ⇒ y = -XFrom Y = x ⇒ x = YSo substituting into the original equation y = f(x):- X = f(Y)Therefore, Y = f^{-1}(-X)Which is the same as before. So, replacing X with x and Y with y, the equation is y = f^{-1}(-x). So that's option D. But wait, hold on. Let me check with an example to verify.Suppose f(x) = x. Then f^{-1}(x) = x. Then, rotating the graph y = x 90 degrees CCW. The original graph is a line through the origin with slope 1. Rotating 90 CCW should give a line through the origin with slope -1, but since rotation might flip axes. Wait, actually, rotating y = x 90 degrees CCW would result in the line y = -x. Let's see: the original points (a, a) become (-a, a). So when X = -a, Y = a. Then Y = -X. So the equation is y = -x. Which is f^{-1}(-x) since f^{-1}(x) = x, so f^{-1}(-x) = -x. So in this case, the rotated graph is y = -x, which matches y = f^{-1}(-x). So that seems correct.Another example: suppose f(x) = 2x. Then f^{-1}(x) = x/2. Rotating the graph y = 2x 90 degrees CCW. The original points are (a, 2a). After rotation, they become (-2a, a). So X = -2a, Y = a. So Y = (-X)/2. Therefore, the equation is y = (-x)/2. On the other hand, f^{-1}(-x) = (-x)/2. So again, the rotated graph is y = f^{-1}(-x). So that seems to work.Wait, but let me check with a function that's not linear. Let's take f(x) = x³. Then f^{-1}(x) = x^{1/3}. The original graph is y = x³. Rotating it 90 CCW. So points (a, a³) become (-a³, a). So in terms of the new coordinates, X = -a³, Y = a. Therefore, Y = a, and X = -a³ ⇒ X = -Y³ ⇒ Y = (-X)^{1/3} = - (X)^{1/3}. But f^{-1}(-x) = (-x)^{1/3} = -x^{1/3}. Wait, but according to our previous conclusion, the rotated graph should be y = f^{-1}(-x) = (-x)^{1/3} = -x^{1/3}. But in this example, we have Y = (-X)^{1/3} which is the same as y = (-x)^{1/3} = -x^{1/3}. So that's exactly f^{-1}(-x). So that also matches. Therefore, seems like option D is correct.But wait, let me check another way. Suppose we start with the inverse function. If the original function is f, and it has an inverse f^{-1}, then the graph of f^{-1} is the reflection of the graph of f over the line y = x. So if we perform a rotation on the graph of f, how does it relate to the inverse function?Alternatively, let's think about transformations step by step. Rotating the graph 90 CCW is equivalent to a combination of reflection and scaling? Wait, rotation matrix is [0, -1; 1, 0], which is a linear transformation. So in general, rotating the graph of y = f(x) 90 CCW would result in a new graph. But since functions must pass the vertical line test, rotating a graph might not result in a function unless the original function is invertible. Since the problem states that f has an inverse, so the rotated graph is also a function.Alternatively, let's consider parameterizing the original graph as (t, f(t)). After rotation, the coordinates become (-f(t), t). So we can consider this as (X, Y) = (-f(t), t). To express Y as a function of X, we need to express t in terms of X. Since X = -f(t), we have f(t) = -X. Therefore, t = f^{-1}(-X). Then, Y = t = f^{-1}(-X). Therefore, Y = f^{-1}(-X), so the equation is y = f^{-1}(-x). That's the same as before.Alternatively, maybe the answer is A: y = -f^{-1}(-x). Wait, but in our examples, like with f(x) = x, rotating gives y = -x, which is f^{-1}(-x) since f^{-1}(x) = x. So f^{-1}(-x) = -x. If it were option A, it would be -f^{-1}(-x) which would be -(-x) = x, which is not correct. So that's not. Similarly, with f(x) = 2x, f^{-1}(-x) = -x/2, which matches the rotated graph. If it were option A, -f^{-1}(-x) = -(-x/2) = x/2, which is not the case. So A is incorrect.Option B is y = -f^{-1}(x). For the first example, f(x)=x, this would be -x, which is the same as D. Wait, but in the first example, both D and B give -x. Wait, but in our first example, f(x) = x, so f^{-1}(x) = x. Then, D is f^{-1}(-x) = -x, which is correct. Option B is -f^{-1}(x) = -x, which is same as D here. But in the second example, f(x) = 2x. Then D is f^{-1}(-x) = -x/2, whereas B is -f^{-1}(x) = -x/2. Wait, same result. Wait, but in this case, they are the same. Wait, but in the first example, they are same too. Wait, but when is there a difference?Wait, if the function is even, for example. Wait, but the function has an inverse, so it can't be even unless it's f(x) = x, which is its own inverse. Wait, let's take another function where f and its inverse are different. For example, f(x) = e^x, then f^{-1}(x) = ln(x). So let's check with this function.Original function: y = e^x. Rotate 90 CCW. Points (a, e^a) become (-e^a, a). So to express this as a function, we need to write y in terms of x. So x = -e^a, y = a. Therefore, a = y, so x = -e^y. Therefore, solving for y: y = ln(-x). But the domain of ln is positive numbers, so x must be negative. So the function is y = ln(-x), which is defined for x < 0. Now, f^{-1}(-x) is ln(-x). So the rotated graph is y = f^{-1}(-x). So that's D. If we do option B, which is y = -f^{-1}(x), that would be -ln(x). Which is different. So in this case, D is correct. So in the case of f(x) = e^x, the rotated graph is y = ln(-x) which is f^{-1}(-x), so D is correct. If we had chosen B, it would be y = -ln(x), which is not the same.Therefore, in this case, D is correct, not B. So the answer should be D. Let me confirm again with this example. Original function y = e^x. Rotated 90 CCW, the points (a, e^a) become (-e^a, a). So the new graph is x = -e^y. Solving for y gives y = ln(-x), which is f^{-1}(-x) since f^{-1}(x) = ln(x), so f^{-1}(-x) = ln(-x). Therefore, D is correct.Therefore, after going through different examples and methods, it seems the correct answer is D: y = f^{-1}(-x).Final Answerboxed{D}</think>To determine the effect of rotating the graph of ( y = f(x) ) 90 degrees counterclockwise around the origin, we start by noting that each point ((x, y)) on the original graph is transformed to ((-y, x)) after rotation. 1. Original Point Transformation: For a point ((x, f(x))) on the graph of ( y = f(x) ), after rotation, the new coordinates become ((-f(x), x)).2. New Coordinates: Let the new coordinates be ((X, Y)). Then, we have: [ X = -f(x) quad text{and} quad Y = x ]3. Expressing ( Y ) in terms of ( X ): We need to solve for ( x ) in terms of ( X ). From ( X = -f(x) ), we get: [ f(x) = -X ] Applying the inverse function ( f^{-1} ) to both sides, we obtain: [ x = f^{-1}(-X) ] Since ( Y = x ), we substitute to get: [ Y = f^{-1}(-X) ]4. Function Notation: Replacing ( X ) with ( x ) and ( Y ) with ( y ) in the function notation, the equation of the new graph is: [ y = f^{-1}(-x) ]This confirms that the new graph represents the function ( y = f^{-1}(-x) ), which corresponds to option D.[boxed{D}],<think>Okay, so I need to figure out which of the given options represents the function obtained when the graph of y = f(x) is rotated 90 degrees counterclockwise around the origin. The function f has an inverse, which is important because we have options involving f inverse. Let me think through this step by step.First, I remember that rotating a graph involves transforming all the points (x, y) on the original graph to new points (x', y') according to the rotation. For a 90-degree counterclockwise rotation around the origin, the standard transformation is (x, y) becomes (-y, x). Let me verify that. If I take a point (1, 0) and rotate it 90 degrees counterclockwise, it should go to (0, 1), right? Applying the transformation (-y, x) to (1, 0) gives (0, 1), which is correct. Similarly, a point (0, 1) would rotate to (-1, 0). So yes, the rotation formula seems correct: (x, y) → (-y, x).Now, the original graph is y = f(x). So any point (a, b) on this graph satisfies b = f(a). After rotation, the point (a, b) moves to (-b, a). So the new coordinates after rotation are (x', y') = (-b, a). Therefore, for the rotated graph, each point (x', y') satisfies x' = -b and y' = a. But since b = f(a), we can substitute: x' = -f(a) and y' = a. Let's express a and b in terms of x' and y'. From y' = a, we have a = y'. Then x' = -f(a) = -f(y'). Therefore, x' = -f(y'). To write this as a function y' in terms of x', we can rearrange: f(y') = -x', which implies y' = f^{-1}(-x'). So, substituting back into variables x and y (since the axes remain the same, just the graph is rotated), the equation of the rotated graph is y = f^{-1}(-x). But let me check the options given.Wait, the options are:(A) y = -f^{-1}(-x)(B) y = -f^{-1}(x)(C) y = f^{-1}(x)(D) y = f^{-1}(-x)Hmm, so according to my derivation, the answer would be (D) y = f^{-1}(-x). But let me verify again to make sure I didn't make any mistakes.Original point: (a, f(a)) on y = f(x). After rotation, it becomes (-f(a), a). So the new x-coordinate is -f(a) and the new y-coordinate is a. So in the rotated graph, the x is -f(a) and y is a. To express y as a function of x, we need to solve for y in terms of x. So x = -f(y) => f(y) = -x => y = f^{-1}(-x). Yes, that seems correct. So the equation is y = f^{-1}(-x), which is option D.But wait, let me consider another approach. Sometimes when rotating graphs, especially non-injective functions, things can get tricky, but since f has an inverse, it's bijective, so that should be okay.Alternatively, we can think about the inverse function. The graph of the inverse function is the reflection over the line y = x. So if we have y = f(x), then y = f^{-1}(x) is its reflection over y = x. But rotation is different from reflection. A 90-degree rotation is a more drastic transformation.Alternatively, perhaps using transformation matrices. A 90-degree counterclockwise rotation matrix is [0, -1; 1, 0]. So applying this to a vector [x; y] gives [ -y; x ]. So any point (x, y) on the original graph is transformed to (-y, x). Therefore, for the equation y = f(x), replacing x with -y' and y with x', where (x', y') are the new coordinates after rotation. So x' = -y and y' = x. Hence, substituting into the original equation: y = f(x) => x' = -y => y = -x' and y' = x => x = y'. Therefore, substituting into y = f(x), we get -x' = f(y'). Therefore, solving for y', we have y' = f^{-1}(-x'). So again, the equation is y' = f^{-1}(-x'), which in terms of the original coordinate system (since x' and y' are just the new coordinates, which we can rename to x and y) would be y = f^{-1}(-x). So that's still option D.But let me check with a simple example. Suppose f(x) = x. Then f is its own inverse. The graph is the line y = x. If we rotate this 90 degrees counterclockwise, we should get the line y = -x. Let's see: applying the rotation, each point (a, a) becomes (-a, a). So for example, (1,1) becomes (-1,1), (2,2) becomes (-2,2), etc. The set of all points (-a, a) is the line y = -x. If we use the formula from option D: f^{-1}(-x) = -x, so y = f^{-1}(-x) = -x, which is indeed the line y = -x. So that works.Another example: let f(x) = 2x. Then the inverse function is f^{-1}(x) = x/2. Rotating the graph y = 2x 90 degrees counterclockwise. Let's take a point (1,2) on the original graph. After rotation, it becomes (-2, 1). Similarly, (2,4) becomes (-4,2). So the rotated graph consists of points (-2,1), (-4,2), etc., which would satisfy the equation y = (-x)/2, because if x = -2, y = 1; x = -4, y = 2. So y = (-x)/2. According to option D: f^{-1}(-x) = (-x)/2, which is exactly y = (-x)/2. So that works as well.What about option A? For the first example where f(x) = x, option A would give y = -f^{-1}(-x) = -(-x) = x, which is not the line y = -x. So A is incorrect. Option B: y = -f^{-1}(x) = -x, which is the same as D in this case, but wait, in the first example, f(x) = x, so f^{-1}(x) = x. So option D gives y = f^{-1}(-x) = -x, which is correct. Option B gives y = -f^{-1}(x) = -x, which is also correct? Wait, that seems conflicting.Wait, hold on, in the first example, f(x) = x, then rotating the graph y = x 90 degrees counterclockwise gives y = -x, which is achieved by both options B and D? But that's not possible because the answer should be unique. Wait, let's check again.If f(x) = x, then f^{-1}(x) = x. Then:Option A: y = -f^{-1}(-x) = -(-x) = x.Option B: y = -f^{-1}(x) = -x.Option C: y = f^{-1}(x) = x.Option D: y = f^{-1}(-x) = -x.So both B and D give y = -x. But in reality, rotating y = x 90 degrees counterclockwise gives y = -x, so both B and D would work here? But that can't be. Wait, but according to our previous derivation, it should be D. But in this case, both B and D are the same. So maybe in some cases they coincide, but in general, which one is correct?Wait, let's test with another function where f and its inverse are different. Let me take f(x) = 2x + 1. Then f is invertible. Let's compute f inverse. Let y = 2x + 1. Solving for x: x = (y - 1)/2. So f^{-1}(x) = (x - 1)/2.Now, the original graph is y = 2x + 1. Let's take a point (0,1) on this graph. Rotating 90 degrees counterclockwise around the origin gives (-1, 0). Another point (1,3) would rotate to (-3,1). Let's see what equation these points satisfy.Original function: y = 2x + 1.After rotation, points (x, y) become (-y, x). So the original point (0,1) becomes (-1,0). The original point (1,3) becomes (-3,1). Let's find the equation connecting (-1,0) and (-3,1). The slope is (1 - 0)/(-3 - (-1)) = 1 / (-2) = -1/2. So the line would be y - 0 = -1/2 (x + 1), so y = -1/2 x - 1/2.Alternatively, using the formula from option D: y = f^{-1}(-x) = (-x - 1)/2 = (-x)/2 - 1/2, which is y = -1/2 x - 1/2. That's exactly the equation we derived. So that works.Option B: y = -f^{-1}(x) = -[(x - 1)/2] = (-x + 1)/2 = -1/2 x + 1/2. So the equation would be y = -1/2 x + 1/2, which is different. If we check with the rotated points:For (-1,0): plug into option B: y = -1/2*(-1) + 1/2 = 1/2 + 1/2 = 1 ≠ 0. So option B does not satisfy this point. Whereas option D: y = -1/2*(-1) - 1/2 = 1/2 - 1/2 = 0, which works. For (-3,1): option D: y = -1/2*(-3) - 1/2 = 3/2 - 1/2 = 1, which is correct. So in this case, option D is correct while option B is not. So even though in the first example both B and D gave the same result, in general, option D is the correct one.Therefore, the answer should be D.Final Answerboxed{D}</think>To determine the function represented by the new graph after rotating the graph of ( y = f(x) ) 90 degrees counterclockwise around the origin, we need to consider the effect of this rotation on the coordinates of the points on the graph. A 90-degree counterclockwise rotation transforms any point ((x, y)) to ((-y, x)). For a point ((a, b)) on the graph ( y = f(x) ), this rotation results in the new coordinates ((-b, a)). Given ( b = f(a) ), the new coordinates become ((-f(a), a)). To express this as a function, we let ( x' = -f(a) ) and ( y' = a ). Solving for ( a ) in terms of ( x' ), we get ( a = y' ). Substituting this into ( x' = -f(a) ) gives ( x' = -f(y') ), which can be rearranged to ( f(y') = -x' ). Therefore, ( y' = f^{-1}(-x') ).Renaming ( x' ) and ( y' ) back to ( x ) and ( y ) for the new coordinates, the equation of the rotated graph is ( y = f^{-1}(-x) ).Thus, the correct answer is (boxed{D}).]

🔑:[Solution] Note that the graph of the function y=f(x) is symmetric to the graph of x=f(y) with respect to the line y=x.The graph of the function -x=f(y) is symmetric to the graph of x=f(y) with respect to the y-axis.Therefore, the graph of the function -x=f(y) can be regarded as the result of rotating the graph of the function y=f(x) counterclockwise by 90^{circ} around the origin O in the x O y plane.Also, -x=f(y) is usually written as quad y=f^{-1}(-x).Hence, the answer is (D).

🧠:[<think>Okay, so I have this problem where there are three real numbers a, b, c. They satisfy two conditions: their sum is 2 (a + b + c = 2), and their product is 4 (abc = 4). The problem has two parts. The first part asks for the minimum value of the maximum of a, b, c. So, basically, among all possible triples (a, b, c) that satisfy those two conditions, I need to find the smallest possible value that the largest of the three numbers can take. The second part is to find the minimum value of the sum of the absolute values of a, b, c. That is, minimize |a| + |b| + |c|. Alright, let's start with the first part.Starting with part (1): Find the minimum value of the maximum of a, b, c. Let me denote the maximum of a, b, c as M. We need to find the minimal possible M such that there exist real numbers a, b, c with a + b + c = 2, abc = 4, and all of them are less than or equal to M. Since we need the minimum such M, we need to arrange the numbers such that the largest among them is as small as possible, while still satisfying the given conditions.First, let's think about the constraints. If all three numbers were equal, we would have a = b = c. Then, a + b + c = 3a = 2, so a = 2/3. Then the product would be (2/3)^3 = 8/27 ≈ 0.296, which is way less than 4. So, clearly, the numbers can't all be equal. Since the product needs to be much larger, at least one of the numbers has to be significantly larger. But since their sum is fixed at 2, if one number is larger, the others might have to be smaller, possibly negative. However, if some numbers are negative, that complicates the product, because multiplying two negatives gives a positive, but if only one is negative, the product would be negative. But in our case, the product is positive (4), so either all three numbers are positive, or exactly two of them are negative. But if two are negative and one is positive, then their sum would be (negative) + (negative) + (positive). To get a sum of 2, the positive number would have to be quite large to offset the two negatives. Let's check both possibilities.Case 1: All three numbers are positive. Then, since all are positive, we can apply the AM ≥ GM inequality. Wait, but AM of a, b, c is 2/3, and GM is (4)^(1/3) ≈ 1.587. Since 1.587 > 2/3 ≈ 0.666, which contradicts AM ≥ GM because AM should be greater than or equal to GM. But here, AM is less than GM, which is impossible. Therefore, it's impossible for all three numbers to be positive. Therefore, exactly two of them must be negative, and one positive. Let's confirm that.Suppose two numbers are negative, and one is positive. Let's denote the positive number as c, and a and b as negative. Then, a + b + c = 2, with a, b < 0, c > 0. Then, c = 2 - a - b. Since a and b are negative, -a and -b are positive, so c = 2 + (-a) + (-b). Therefore, c must be greater than 2. Also, the product abc = 4. Since a and b are negative, their product is positive, so abc = (positive)*(positive) = positive, which matches the given product of 4.So, in this scenario, the positive number c has to be greater than 2, and the two negative numbers a and b. So, in terms of the maximum of a, b, c, since c is positive and a, b are negative, the maximum is c. Therefore, if we can minimize c, then we minimize the maximum of a, b, c. Therefore, part (1) reduces to finding the minimal possible c >= 2 such that there exist real numbers a, b < 0 with a + b = 2 - c and ab = 4/c.Wait, let's see. Since a + b + c = 2, then a + b = 2 - c. And abc = 4, so ab = 4/c. So, given c, the numbers a and b are the roots of the quadratic equation x^2 - (2 - c)x + (4/c) = 0. For real roots, the discriminant must be non-negative. So discriminant D = (2 - c)^2 - 4*(4/c) >= 0. Therefore, (2 - c)^2 - 16/c >= 0.So, the problem reduces to finding the minimal c >= 2 such that (2 - c)^2 - 16/c >= 0. Let's compute this.First, let's denote f(c) = (c - 2)^2 - 16/c. We need f(c) >= 0. So, we need to solve (c - 2)^2 >= 16/c. Multiply both sides by c (since c > 0), we get c(c - 2)^2 >= 16.So, we need to find the minimal c >= 2 such that c(c - 2)^2 >= 16.Let me expand the left-hand side:c(c - 2)^2 = c*(c^2 - 4c + 4) = c^3 - 4c^2 + 4c.So, the inequality becomes c^3 - 4c^2 + 4c - 16 >= 0.Let me compute c^3 - 4c^2 + 4c - 16. Let's try to factor this polynomial. Maybe c = 4 is a root? Let's check:4^3 - 4*(4)^2 + 4*4 - 16 = 64 - 64 + 16 -16 = 0. Yes, c = 4 is a root. So, we can factor out (c - 4):Using polynomial division or synthetic division:Divide c^3 - 4c^2 + 4c - 16 by (c - 4). Coefficients: 1 | -4 | 4 | -16Bring down 1. Multiply by 4: 4. Add to next coefficient: -4 + 4 = 0. Multiply by 4: 0. Add to next coefficient: 4 + 0 = 4. Multiply by 4: 16. Add to last coefficient: -16 +16=0. So, the quotient is c^2 + 0c + 4 = c^2 + 4. Therefore, c^3 -4c^2 +4c -16 = (c -4)(c^2 + 4). So, the inequality becomes (c -4)(c^2 +4) >=0. Since c^2 +4 is always positive for real c, the inequality is equivalent to c -4 >=0, i.e., c >=4.Therefore, the minimal c satisfying this inequality is c=4. Therefore, the minimal maximum value M is 4. Wait, but let's check this.Wait, when c=4, then a + b = 2 -4 = -2, and ab = 4 /4 =1. So the quadratic equation is x^2 +2x +1=0, which factors as (x +1)^2=0. Therefore, a = b = -1. So the triplet is (-1, -1, 4). Indeed, sum is -1 -1 +4=2, product is (-1)(-1)(4)=4. So this works. So the maximum here is 4.But the question is, is 4 the minimal possible maximum? Because maybe there's a c between 2 and 4 where the discriminant is non-negative? Wait, according to the inequality, c must be >=4. Because (c -4)(c^2 +4) >=0 implies c >=4. So for c <4, the left-hand side is negative. Therefore, for c <4, the discriminant is negative, meaning that there are no real roots a and b. Therefore, there are no real numbers a, b when c <4. Therefore, the minimal possible c is 4. Therefore, the minimal value of the maximum of a, b, c is 4. So part (1) answer is 4.But let's see, is there a way to have a lower maximum? Suppose, for example, two numbers are equal. Suppose two of them are equal. Let's assume a = b. Then, 2a + c =2, so c = 2 -2a. Then, the product is a^2 c =4. Substitute c: a^2 (2 -2a)=4. So 2a^2 -2a^3 =4. Divide both sides by 2: a^2 -a^3=2. Rearranged: a^3 -a^2 +2=0. Wait, let's check possible real roots. Maybe a= -1: (-1)^3 - (-1)^2 +2= -1 -1 +2=0. So a=-1 is a root. Therefore, factor as (a +1)(a^2 -2a +2)=0. So roots are a=-1, and a=(2 ±sqrt(4 -8))/2, which are complex. Therefore, the only real root is a=-1. Then c=2 -2*(-1)=4. So again, we get a=-1, b=-1, c=4. So same solution. Therefore, even if we assume two variables are equal, we still get the same result.Alternatively, suppose all three numbers are distinct. Let's see. If we take c slightly larger than 4, then a and b would be real numbers. For example, c=5. Then a + b =2 -5= -3, ab=4/5=0.8. Then quadratic equation x^2 +3x +0.8=0. The roots are (-3 ±sqrt(9 -3.2))/2= (-3 ±sqrt(5.8))/2. So real roots. So yes, c can be larger than 4, but we are to find the minimal c. So minimal c is 4.Alternatively, maybe if we take two numbers as positive? Wait, but earlier analysis suggested that it's impossible for all three to be positive. Wait, but maybe two positive and one negative? Let's see. Suppose two numbers are positive, one is negative. Then, sum is 2. Let's say a and b are positive, c is negative. Then, a + b + c =2. Since c is negative, a + b =2 -c >2. The product abc=4. Since c is negative, then ab must be negative. But a and b are positive, so ab is positive. Therefore, abc is negative, which contradicts abc=4>0. Therefore, this case is impossible. Therefore, only possible case is two negatives and one positive, as we considered.Therefore, conclusion is that the minimal maximum is 4. So part (1) answer is 4.Now, moving to part (2): Find the minimum value of |a| + |b| + |c|.Given that a + b + c =2, abc=4. Again, as before, we deduced that two of them must be negative and one positive. Let's denote again c as the positive number, and a, b as negative. Then, |a| + |b| + |c| = (-a) + (-b) + c = c - (a + b). But since a + b =2 - c, this becomes c - (2 - c) = 2c -2. So, |a| + |b| + |c| = 2c -2.But in this case, we need to minimize 2c -2. Therefore, to minimize this expression, we need to minimize c. But from part (1), the minimal possible c is 4. Therefore, plugging c=4, we get |a| + |b| + |c| =2*4 -2=8 -2=6. So is the answer 6?Wait, but hold on. Wait, is this the only scenario? Because |a| + |b| + |c| could also be considered if, for example, all three numbers are positive. But we saw that all three can't be positive. If two are positive and one is negative, as before, but that case is impossible because product would be negative. Alternatively, if all three are negative, sum would be negative, but the sum is 2, so that's impossible. So only possible case is two negatives and one positive. Therefore, in that case, the expression |a| + |b| + |c| is indeed 2c -2, as we derived.But wait, is this the only way? Let's confirm. Suppose a and b are negative, c is positive. Then, |a| = -a, |b| = -b, |c| = c. Therefore, |a| + |b| + |c| = (-a) + (-b) + c = c - (a + b). Since a + b =2 - c, then this is c - (2 - c)=2c -2. Therefore, yes, the expression is linear in c with positive coefficient. Therefore, to minimize it, we need to minimize c. Since minimal c is 4, then the minimal sum is 6. However, wait, is there a way to get a smaller sum?Wait, hold on. Maybe if we take c less than 4? But from part (1), we saw that if c <4, there are no real solutions. So for c >=4, we have real numbers a, b. Therefore, the minimal sum would be when c is minimal, which is 4, leading to 2*4 -2=6. Therefore, the answer is 6. But let's verify with the example we have.When c=4, a=b=-1. Then, |a| + |b| + |c|=1 +1 +4=6. Correct. But what if there is another triplet where the sum is smaller?Suppose we take c=5. Then, a + b=2 -5=-3, ab=4/5=0.8. Then, the quadratic equation is x^2 +3x +0.8=0. The roots are x=(-3 ±sqrt(9 -3.2))/2=(-3 ±sqrt(5.8))/2. Approximately sqrt(5.8)=2.408, so roots are (-3 +2.408)/2≈(-0.592)/2≈-0.296 and (-3 -2.408)/2≈-5.408/2≈-2.704. Then, |a| + |b| + |c|≈0.296 +2.704 +5=8. So that's larger than 6. Similarly, if we take c=6, sum |a| + |b| + |c|=2c -2=12 -2=10, which is even larger. Therefore, increasing c increases the sum. Therefore, the minimal sum is at c=4, which gives 6. Therefore, answer is 6.But wait, is there another scenario where |a| + |b| + |c| could be smaller? For example, if one of the numbers is zero? But abc=4, so none of them can be zero. If two numbers are zero, product is zero. So all three must be non-zero. So all have absolute value at least something. But since two are negative, one is positive.Alternatively, is there a case where one of the negative numbers has a smaller absolute value, thus making the total sum smaller? But if we decrease c, which would require increasing the absolute values of a and b. Wait, but c cannot be decreased below 4. Because if c is less than 4, then there are no real solutions. Therefore, the minimal c is 4, leading to a and b being -1 each. So their absolute values are 1 each, which are already as small as possible. Therefore, in that case, |a| + |b| =2, and |c|=4, giving total 6. If we try to make a or b less negative (i.e., closer to zero), then c would have to increase to compensate in the product. For example, suppose a is -0.5, then let's see what happens.Suppose a=-0.5. Then, b + c =2 -(-0.5)=2.5. The product abc= (-0.5)*b*c=4. Therefore, bc= -8. Also, b +c=2.5. Therefore, bc=-8. So quadratic equation x^2 -2.5x -8=0. Discriminant=6.25 +32=38.25. sqrt(38.25)=6.186. So roots=(2.5 ±6.186)/2. That gives (2.5 +6.186)/2≈4.343 and (2.5 -6.186)/2≈-1.843. So b≈4.343, c≈-1.843. Wait, but then in this case, the numbers would be a=-0.5, b≈4.343, c≈-1.843. But then, c is supposed to be the positive number. But here, c is negative. So this contradicts our initial assumption. Wait, but hold on, in this case, a is -0.5, b is 4.343, c is -1.843. So the positive number is b=4.343, and a and c are negative. Then, |a| + |b| + |c|=0.5 +4.343 +1.843≈6.686, which is more than 6. Alternatively, if we set a to be less negative, but then the other negative number becomes more negative. Wait, maybe if we set a closer to zero, but then the other negative number has to be more negative to compensate. Let me check.Suppose a approaches zero from the negative side. Let's say a= -ε, where ε is very small positive number. Then, b + c =2 + ε. Product abc= (-ε)bc=4 => bc= -4/ε. Then, since b + c=2 + ε and bc=-4/ε. The quadratic equation becomes x^2 - (2 + ε)x -4/ε=0. The discriminant is (2 + ε)^2 +16/ε. For very small ε, the discriminant is dominated by 16/ε, which is very large. So roots are approximately x≈ [ (2 + ε) ±sqrt(16/ε) ] /2 ≈ ±sqrt(4/ε). Therefore, one root is approximately sqrt(4/ε) and the other is approximately -sqrt(4/ε). But since b + c=2 + ε, and one root is positive and one is negative. Let's denote b≈sqrt(4/ε), c≈-sqrt(4/ε) + (2 + ε). Wait, this is getting complicated. But essentially, as ε approaches zero, sqrt(4/ε) approaches infinity. Therefore, the positive root b approaches infinity, and the negative root c approaches negative infinity. But their sum is 2 + ε, so c=2 + ε -b. If b approaches infinity, c approaches negative infinity. Therefore, |a| + |b| + |c|= ε + sqrt(4/ε) + |2 + ε - sqrt(4/ε)|. As ε approaches zero, this expression behaves like sqrt(4/ε) + | - sqrt(4/ε)| = sqrt(4/ε) + sqrt(4/ε)= 2*sqrt(4/ε)=4/sqrt(ε), which approaches infinity. Therefore, making a closer to zero actually increases the sum |a| + |b| + |c|. Therefore, the minimal sum occurs at a= -1, b=-1, c=4, giving sum 6.Alternatively, let's see if there's another case where two numbers are negative and one positive, but not equal. For example, suppose a=-2, b=-0.5, then c=2 - (-2) - (-0.5)=2 +2 +0.5=4.5. Then product abc= (-2)(-0.5)(4.5)=1*4.5=4.5≠4. Not equal to 4. So that's not valid. Let's try a=-3, b=..., Wait, this might not be straightforward.Alternatively, suppose we set a=-d, b=-e, c=f, where d, e >0, and f=2 + d + e. Then, abc= (-d)(-e)f=def=4. So def=4. We need to minimize |a| + |b| + |c|=d + e + f= d + e + (2 + d + e)=2 + 2d + 2e. So we need to minimize 2 + 2d + 2e subject to def=4, where d, e >0. Let's set variables as d and e, and f=2 + d + e. Then, def=4. So we need to minimize 2 + 2(d + e) given that d e (2 + d + e)=4. Let me denote S = d + e. Then, we have d e (2 + S)=4. We need to minimize 2 + 2S. So, subject to d e (2 + S)=4. Since d and e are positive numbers, by AM ≥ GM, S =d + e ≥2√(de). Let's denote de =k. Then, S >=2√k. Then, the constraint is k(2 + S)=4. Since S >=2√k, then 2 + S >=2 + 2√k. Therefore, k(2 + S)>=k(2 + 2√k). But k(2 + 2√k)=2k + 2k^(3/2). This must be equal to 4. So, 2k + 2k^(3/2)=4. Let's divide both sides by 2: k + k^(3/2)=2. Let’s set t=√k, so k = t^2. Then, equation becomes t^2 + t^3 =2. So, t^3 + t^2 -2=0. Let's try to factor this. Possible rational roots are t=1: 1 +1 -2=0. So t=1 is a root. Therefore, factor as (t -1)(t^2 + 2t +2)=0. The other roots are complex, so the only real root is t=1. Therefore, t=1, so √k=1 => k=1. Therefore, equality occurs when d=e=1, since de=1 and d + e=2. So S=2, and f=2 +2=4. Therefore, in this case, the minimal value of 2 + 2S=2 +4=6. Therefore, this is the minimal value. Therefore, this confirms that the minimal sum is 6, achieved when d=e=1, f=4, i.e., a=-1, b=-1, c=4.Therefore, answer to part (2) is 6.Wait, but just to check, is there any other way where the sum |a| + |b| + |c| can be smaller? For example, if the two negative numbers are not equal. Let's suppose a=-1.5, b=-0.5, then c=2 - (-1.5) - (-0.5)=2 +1.5 +0.5=4. Then, product abc=(-1.5)(-0.5)(4)=3.0. Not equal to 4. So that's not valid. If a=-2, b=-0.5, then c=2 +2 +0.5=4.5, product abc=(-2)(-0.5)(4.5)=4.5≠4. Not valid. If a=-1.2, b=-0.8, then c=2 +1.2 +0.8=4.0, product abc=(-1.2)(-0.8)(4)=3.84≠4. Close, but not exactly 4. So, it's seems that only when a=b=-1, c=4, the product is 4, and the sum is 6. Therefore, yes, 6 is the minimal value.Alternatively, if we set a and b to different negative numbers. Let's set a=-1 -t, b=-1 +t, for some t. Then, c=2 -a -b=2 -(-1 -t) - (-1 +t)=2 +1 +t +1 -t=4. So c=4 regardless of t. Then, product abc=(-1 -t)(-1 +t)(4)=[(1 +t)(1 -t)]*4=(1 -t^2)*4=4(1 -t^2). Set equal to 4: 4(1 -t^2)=4 => 1 -t^2=1 => t^2=0 => t=0. Therefore, only possible when t=0, i.e., a=-1, b=-1, c=4. Therefore, no other solutions in this neighborhood. Therefore, uniqueness.Therefore, both parts have answers 4 and 6.Final Answer(1) The minimum value of the maximum of (a, b, c) is (boxed{4}).(2) The minimum value of (|a| + |b| + |c|) is (boxed{6}).</think>Given real numbers (a, b, c) that satisfy (a + b + c = 2) and (abc = 4):1. Finding the minimum value of the maximum of (a, b, c): - Since (abc = 4) is positive, either all three numbers are positive or exactly two are negative and one is positive. However, if all three were positive, the arithmetic mean would be less than the geometric mean, which is impossible. Thus, two numbers must be negative and one positive. - Let (c) be the positive number and (a, b) be the negative numbers. We need to minimize (c). - Given (a + b + c = 2), we have (a + b = 2 - c). The product (abc = 4) implies (ab = frac{4}{c}). - The quadratic equation for (a) and (b) is (x^2 - (2 - c)x + frac{4}{c} = 0). The discriminant must be non-negative for real roots: [ (2 - c)^2 - frac{16}{c} geq 0 ] - Simplifying, we get (c(c - 2)^2 geq 16). Solving this, we find (c geq 4). The minimum value of (c) is 4, achieved when (a = b = -1) and (c = 4). Therefore, the minimum value of the maximum of (a, b, c) is (boxed{4}).2. Finding the minimum value of (|a| + |b| + |c|): - Given (a) and (b) are negative and (c) is positive, (|a| + |b| + |c| = -a - b + c). - Substituting (a + b = 2 - c), we get (|a| + |b| + |c| = c - (2 - c) = 2c - 2). - To minimize this expression, we need the smallest (c). From part (1), the minimum (c) is 4. Substituting (c = 4), we get: [ 2 cdot 4 - 2 = 6 ] Therefore, the minimum value of (|a| + |b| + |c|) is (boxed{6}).,<think>Alright, let's tackle this problem. We have three real numbers a, b, c such that their sum is 2 and their product is 4. There are two parts to the problem: 1. Find the minimum value of the maximum of a, b, c. 2. Find the minimum value of |a| + |b| + |c|.Hmm, okay. Let's start with the first part. So, we need to minimize the maximum value among a, b, c. In other words, we want the largest of the three numbers to be as small as possible, given the constraints a + b + c = 2 and abc = 4. First, maybe it's helpful to think about the relationships between these variables. Since their product is 4, which is positive, there must be either all three positive numbers or one positive and two negative numbers. But since their sum is 2, which is positive, it's more likely that there's one positive and two negative numbers, because if all three are positive, their product is 4, which is possible, but maybe the sum would be too big? Wait, but their sum is 2. Hmm, maybe we need to check both possibilities.Wait, let's see. If all three numbers are positive, then their sum is 2, and product is 4. Let's check if that's possible. For positive numbers, by AM ≥ GM, the arithmetic mean is (a + b + c)/3 = 2/3, and the geometric mean is (abc)^(1/3) = 4^(1/3) ≈ 1.587. Since AM < GM here (2/3 ≈ 0.666 < 1.587), this violates the AM-GM inequality. Therefore, it's impossible for all three numbers to be positive. Therefore, there must be exactly one positive number and two negative numbers.Okay, that's a key insight. So, one of a, b, c is positive, and the other two are negative. Let's denote the positive one as, say, a, and the other two as b and c, which are negative. So, a > 0, b < 0, c < 0. Then, their sum is a + b + c = 2, which can be rewritten as a = 2 - (b + c). Since b and c are negative, (b + c) is negative, so a = 2 - (negative) = 2 + positive, so a must be greater than 2. That's interesting. So, the positive number is greater than 2. Then, the product is a * b * c = 4. Since a > 2, and b and c are negative, their product bc is positive. Therefore, abc = a * (bc) = 4. Since a > 2, then bc = 4/a < 2. Hmm.But maybe we can model this with equations. Let's let a be the maximum of the three numbers. Since we want to minimize the maximum, we can assume a is the maximum and find the minimal possible a. Then, the problem reduces to finding the minimal a such that there exist real numbers b and c with b + c = 2 - a, bc = 4/a, and b, c ≤ a (since a is the maximum). But since a is positive and greater than 2, and b and c are negative, actually, b and c being negative would automatically be less than a, which is positive. So, the condition that b and c ≤ a is automatically satisfied. Therefore, we just need to find the minimal a ≥ 2 such that there exist real numbers b and c with b + c = 2 - a and bc = 4/a.So, given that, let's consider the quadratic equation whose roots are b and c. The quadratic would be x^2 - (b + c)x + bc = 0. Substituting the known values, that's x^2 - (2 - a)x + 4/a = 0. For this quadratic to have real roots, the discriminant must be non-negative. So, discriminant D = (2 - a)^2 - 4*(1)*(4/a) ≥ 0.Therefore, we have the inequality:(2 - a)^2 - 16/a ≥ 0Let's expand (2 - a)^2:= (4 - 4a + a^2) - 16/a ≥ 0So, the inequality becomes:a^2 - 4a + 4 - 16/a ≥ 0Hmm, this is a bit complex. Let's multiply both sides by a (since a > 0, this is allowed without changing the inequality direction):a^3 - 4a^2 + 4a - 16 ≥ 0So, we need to solve the inequality:a^3 - 4a^2 + 4a - 16 ≥ 0Let me try to factor this cubic equation. Let's check rational roots using Rational Root Theorem. The possible rational roots are factors of 16 over factors of 1, so ±1, ±2, ±4, ±8, ±16.Testing a = 4:4^3 - 4*4^2 + 4*4 -16 = 64 - 64 + 16 -16 = 0. So, a = 4 is a root. Therefore, we can factor (a - 4) from the cubic:Using polynomial division or synthetic division:Divide a^3 -4a^2 +4a -16 by (a -4):Coefficients: 1 | -4 | 4 | -16Bring down 1. Multiply by 4: 4. Add to -4: 0. Multiply by 4: 0. Add to 4: 4. Multiply by 4: 16. Add to -16: 0. Therefore, the cubic factors as (a -4)(a^2 + 0a +4) = (a -4)(a^2 +4). So, the cubic factors into (a -4)(a^2 +4). Therefore, the inequality becomes:(a -4)(a^2 +4) ≥ 0Since a^2 +4 is always positive (as a^2 is non-negative and we add 4), the sign of the entire expression depends on (a -4). Therefore, (a -4) ≥ 0. Hence, a ≥ 4.But wait, so the discriminant is non-negative only when a ≥4? That seems to indicate that the minimal value of a is 4. Therefore, the minimal maximum of a, b, c is 4. Let's check that.If a =4, then b + c = 2 -4 = -2, and bc =4/4=1. So, the quadratic equation is x^2 +2x +1 =0, which factors as (x +1)^2 =0. So, b = c = -1. Then, the numbers are 4, -1, -1. Sum: 4 -1 -1=2. Product: 4*(-1)*(-1)=4. So, this works. So, when a=4, we have two variables equal to -1, and the maximum is 4. If we tried a value less than 4, say a=3, then we need to check if the discriminant is non-negative. Let's compute D for a=3:(2 -3)^2 -16/3 = ( -1)^2 -16/3 =1 -16/3= -13/3 <0. So, no real roots. Therefore, a=3 is not possible. Similarly, for a=4, discriminant is zero, so that's the minimal a where real roots exist. Therefore, the minimal maximum is 4.Therefore, part (1) answer is 4.Now, part (2): Find the minimum value of |a| + |b| + |c|.Hmm. So, we need to minimize the sum of absolute values of a, b, c, given that a + b + c=2 and abc=4.Again, since we know from part (1) that one of the numbers must be positive and the other two negative, let's denote a >0, b <0, c <0. Then, |a| + |b| + |c| = a - b - c (since b and c are negative). But since a + b + c =2, then b + c =2 - a. Therefore, |a| + |b| + |c| = a - (b + c) = a - (2 -a) = 2a -2.So, we need to minimize 2a -2, where a is the positive number, and b and c are negative numbers such that a + b + c =2 and abc=4.But from part (1), we found that the minimal possible a is 4, leading to |a| + |b| + |c| =2*4 -2=6. But maybe there's a lower value? Wait, but how?Wait, maybe the minimal value of |a| + |b| + |c| is not necessarily achieved when a is minimized. Because 2a -2 is a linear function in a, so to minimize it, we need to minimize a. Since in part (1), the minimal a is 4, so maybe the minimal |a| + |b| + |c| is 6. But let's check if there are other possibilities.Wait, but maybe there are other configurations where not all variables are real? Wait, no, the problem states a, b, c are real numbers. So, given that, in the case where a is minimized, which is 4, then the sum is 6. However, perhaps there's another case where the variables are arranged differently? Wait, but we already concluded that there must be exactly one positive and two negatives. Therefore, the expression |a| + |b| + |c| is equal to a - b - c =2a -2.Therefore, if we can have a smaller a, then the sum would be smaller. However, from part (1), we found that a cannot be smaller than 4. Therefore, the minimal value would be 2*4 -2=6.But wait, is this correct? Let me verify. Suppose there exists a case where a is not the maximum. Wait, but in the problem statement, we need to compute |a| + |b| + |c| regardless of which variable is the maximum. But in our previous analysis, we assumed a is the maximum (the positive one). But perhaps, if another variable is the maximum, even though it's negative? Wait, but the maximum of three numbers is the largest among them. If two are negative and one is positive, then the positive one is the maximum. So, in all cases, the maximum is the positive number. Therefore, the expression |a| + |b| + |c| is always equal to a - b - c =2a -2, given a is positive and b, c are negative. Therefore, the sum is linear in a, and since a must be at least 4, the minimal sum is 2*4 -2=6.But wait, let's check if there's another configuration where maybe two numbers are positive? Wait, earlier we saw that if all three numbers are positive, their product is 4, but by AM-GM, their arithmetic mean is 2/3, and geometric mean is cube root of 4, which is higher. Therefore, it's impossible. So, there must be exactly one positive and two negative numbers.Therefore, the sum |a| + |b| + |c| is indeed a - b - c = 2a -2, so to minimize it, we need to minimize a. And from part (1), the minimal a is 4. Therefore, the minimal sum is 6.But let me test with a=4, b=c=-1. Then |4| + | -1| + | -1| =4 +1 +1=6. Correct.But perhaps there is a different set of numbers where a is not the maximum but the sum of absolute values is less? But since in the valid configurations, the maximum is a, which is positive, and the others are negative, so the sum is fixed as 2a -2. Therefore, minimal when a is minimal, which is 4, leading to 6.Alternatively, maybe there are cases where the variables are complex? But no, the problem states real numbers.Wait, another thought: maybe if two of the numbers are zero? But their product is 4, so none can be zero. So, all three are non-zero. So, we can't have zeros.Alternatively, maybe one variable is very large positive, and the other two variables are negative but close to zero. But their product has to be 4. Let's see. Suppose a approaches infinity, then b + c approaches -infinity, but bc =4/a approaches zero. But if a approaches infinity, then bc approaches zero, but b + c approaches -infinity. However, if b and c are both negative and their sum is approaching -infinity, but their product is approaching zero, that requires one of them approaches zero and the other approaches -infinity. For example, let’s say b approaches 0 from the negative side, then c = (2 -a) - b. If a is large, 2 -a is large negative. If b approaches 0, then c approaches (2 -a) -0 = - (a -2), which approaches -infinity as a approaches infinity. Then, bc = (approaching 0)(approaching -infinity) approaches 0. But we need bc =4/a, which approaches 0 as a approaches infinity, which is consistent. But then |a| + |b| + |c| =a + |b| + |c|. If a approaches infinity, |c| approaches infinity, so the sum |a| + |b| + |c| would approach infinity. Therefore, the sum can be made arbitrarily large, but we are looking for the minimum. Alternatively, if we decrease a below 4, but as we saw in part (1), when a is less than 4, the discriminant is negative, so no real solutions. So, there's no such real numbers. Therefore, the minimal possible a is indeed 4, leading to sum 6. Therefore, the answer to part (2) is 6.But wait, let me check another example. Suppose we have a=5, then b + c=2 -5=-3, bc=4/5=0.8. Then, solving the quadratic x^2 +3x +0.8=0. The roots are [-3 ± sqrt(9 - 3.2)]/2 = [-3 ± sqrt(5.8)]/2. sqrt(5.8)≈2.408, so roots≈(-3 ±2.408)/2. So, approx (-3 +2.408)/2≈-0.296, and (-3 -2.408)/2≈-2.704. Then, |a| + |b| + |c|=5 +0.296 +2.704=8, which is more than 6. So, that's worse. If we take a=4, sum is 6. If a is higher, sum is higher. If a is lower but impossible, sum would be lower. So, 6 is indeed the minimum.Alternatively, let's suppose that two variables are equal. Let’s assume that b = c. Then, since a + 2b =2 => a=2 -2b. Also, abc=4 => (2 -2b)*b^2=4. So, (2 -2b)*b^2=4 => 2b^2 -2b^3=4 => -2b^3 +2b^2 -4=0 => 2b^3 -2b^2 +4=0 => b^3 -b^2 +2=0. Let's try to solve this equation. Possible rational roots are ±1, ±2. Test b=1: 1 -1 +2=2≠0. b=-1: -1 -1 +2=0. So, b=-1 is a root. Therefore, factor (b +1) from the cubic:Using synthetic division:Coefficients: 1 | -1 | 0 | 2Wait, original cubic is b^3 - b^2 + 0b +2. Testing root b=-1:Bring down 1. Multiply by -1: -1. Add to -1: -2. Multiply by -1: 2. Add to 0:2. Multiply by -1: -2. Add to 2:0. Therefore, the cubic factors as (b +1)(b^2 -2b +2). So, b=-1 or roots of b^2 -2b +2=0. The quadratic equation has discriminant 4 -8= -4 <0, so no real roots. Therefore, the only real solution is b=-1. Then, a=2 -2*(-1)=2 +2=4. Then, c=b=-1. So, we get a=4, b=c=-1, which is the same solution as before. Therefore, the case where two variables are equal gives us the minimal sum.Therefore, the minimal value of |a| + |b| + |c| is indeed 6.But let me check another case where variables are not symmetric. Suppose a=5, then as before, sum |a| + |b| + |c|=8. If we take a=6, then b +c= -4, bc=4/6=2/3. Then, solving x^2 +4x +2/3=0. The roots are [-4 ± sqrt(16 - 8/3)]/2 = [-4 ± sqrt(40/3)]/2 = [-4 ± (2*sqrt(30)/3)]/2 ≈ [-4 ± 3.415]/2. So, roots≈ (-4 +3.415)/2 ≈-0.292 and (-4 -3.415)/2≈-3.707. Then, the sum |a| + |b| + |c|=6 +0.292 +3.707≈10. So, again larger.Alternatively, take a=3. But as we saw before, when a=3, discriminant is negative, so no real solutions. Therefore, the minimal possible a is indeed 4, leading to sum 6. Therefore, answer to part (2) is 6.But wait, let me think again. Suppose that instead of two negative numbers and one positive, there is a different sign configuration. But we already ruled out all positives, as it's impossible by AM-GM. What about two positives and one negative? Let's check if that's possible.Suppose two positives and one negative. Let’s say a, b >0, c <0. Then, sum a + b + c=2, product abc=4. Since a and b are positive, c=2 -a -b. Then, abc=ab(2 -a -b)=4. Let’s see if this is possible.But here, we have two positive numbers and one negative. Let's check the product. Since c is negative, 2 -a -b must be negative. Therefore, a + b >2. So, the sum of two positive numbers is greater than 2, and their product with a negative number (c) is 4. Let’s see if such numbers exist.Suppose a and b are positive, c=2 -a -b negative. Then, let's set variables x=a, y=b, so x + y >2. Then, we have x y (2 -x -y)=4. Let's check if this is possible.For example, take x=3, y=1. Then, x + y=4>2, c=2 -4=-2. Then, product=3*1*(-2)=-6≠4. Not good.Take x=4, y=1. Then, x + y=5>2, c=2 -5=-3. Product=4*1*(-3)=-12≠4.Take x=2, y=2. Then, x + y=4>2, c=2 -4=-2. Product=2*2*(-2)=-8≠4.Take x=1, y=3. Then, product=1*3*(-2)=-6≠4.Wait, seems difficult. Let's try x=5, y=0.5. Then, x + y=5.5>2, c=2 -5.5=-3.5. Product=5*0.5*(-3.5)=5*0.5=2.5; 2.5*(-3.5)= -8.75≠4.Alternatively, maybe non-integers. Let’s suppose x=3, then y + c= -1. Wait, but maybe set up equations. Let’s denote s =x + y, which is >2, and p=xy. Then, c=2 -s. Then, the product xyc= p*(2 -s)=4. Also, we have s>2, and x, y>0. So, p*(2 -s)=4. Also, since x and y are positive, by AM-GM, p ≤ (s/2)^2. So, (s/2)^2*(2 -s) ≥4. Let’s see if this is possible.Let’s define f(s) = (s^2/4)*(2 -s). We need f(s)=4. So, (s^2/4)*(2 -s)=4 => s^2(2 -s)=16.Let’s check for s>2. Let’s try s=4: 16*(2 -4)=16*(-2)=-32≠16. s=3: 9*(2 -3)=9*(-1)=-9≠16. s=2.5: 6.25*(-0.5)=-3.125≠16. So, no solution here. Therefore, there is no solution in this case. Therefore, two positives and one negative is impossible. Therefore, the only possible configuration is one positive and two negatives, so our previous analysis holds. Therefore, the minimal sum is 6.Therefore, both answers are 4 and 6.Final Answer(1) boxed{4}(2) boxed{6}</think>Given real numbers (a, b, c) such that (a + b + c = 2) and (abc = 4):1. Finding the minimum value of the maximum of (a, b, c): - Since the product (abc = 4) is positive and the sum (a + b + c = 2) is positive, there must be one positive number and two negative numbers. - Let (a) be the positive number and (b, c) be the negative numbers. We need to minimize (a) such that (a + b + c = 2) and (abc = 4). - The quadratic equation for (b) and (c) is (x^2 - (2 - a)x + frac{4}{a} = 0). For real roots, the discriminant must be non-negative: [ (2 - a)^2 - frac{16}{a} geq 0 ] - Simplifying this inequality leads to: [ a^3 - 4a^2 + 4a - 16 geq 0 ] - Factoring the cubic equation, we find (a = 4) is a root. Thus, the minimal value of (a) is 4. Verifying, when (a = 4), (b = c = -1) satisfy the conditions. Therefore, the minimum value of the maximum of (a, b, c) is (boxed{4}).2. Finding the minimum value of (|a| + |b| + |c|): - Given one positive number (a) and two negative numbers (b, c), we have (|a| + |b| + |c| = a - b - c). - Since (a + b + c = 2), we get (a - (2 - a) = 2a - 2). - To minimize (2a - 2), we need the minimum (a). From part (1), the minimum (a) is 4. Thus, the minimum value is (2 times 4 - 2 = 6). Therefore, the minimum value of (|a| + |b| + |c|) is (boxed{6}).]

🔑:14. (1) Without loss of generality, let a be the maximum of a, b, c, i.e., a geqslant b, a geqslant c. From the problem, we know a>0, and b+c=2-a, bc=frac{4}{a}. Therefore, b, c are the two real roots of the quadratic equation x^{2}-(2-a)x+frac{4}{a}=0, thenbegin{array}{l}Delta=(2-a)^{2}-4 times frac{4}{a} geqslant 0, a^{3}-4 a^{2}+4 a-16 geqslant 0, left(a^{2}+4right)(a-4) geqslant 0 .end{array}Thus, a geqslant 4.When a=4, b=c=-1, the conditions are satisfied.Therefore, the minimum value of the maximum among a, b, c is 4.(2) Since abc>0, a, b, c are either all positive or one positive and two negative.(i) If a, b, c are all positive, then from (1), the maximum of a, b, c is not less than 4, which contradicts a+b+c=2.(ii) If a, b, c are one positive and two negative, let a>0, b<0, c<0, thenbegin{array}{l}|a|+|b|+|c|=a-b-c =a-(2-a)=2a-2 .end{array}From (1), we know a geqslant 4, hence 2a-2 geqslant 6.When a=4, b=c=-1, the conditions are satisfied and the equality holds. Therefore, the minimum value of |a|+|b|+|c| is 6.

🧠:[<think>Okay, let's tackle this problem step by step. So, the function is f(x) = a e^{a x} - ln x, and we need to find the range of values for a such that f(x) is always greater than or equal to zero for any x > 1. Hmm, interesting. Let me think about how to approach this.First, since the problem is about ensuring that f(x) is non-negative for all x > 1, I need to analyze the behavior of the function f(x) in that interval. Maybe I should start by understanding the components of the function. The function has two parts: a e^{a x} and -ln x. The first term is exponential, and the second term is logarithmic. Depending on the value of a, the exponential term could be increasing or decreasing. Let's not jump to conclusions yet.Let me note that a e^{a x} is an exponential function. If a is positive, then as x increases, e^{a x} grows very rapidly, so a e^{a x} would dominate over the logarithmic term, which grows much slower. However, if a is negative, then e^{a x} decays as x increases, so the term a e^{a x} becomes negative (since a is multiplied by a decaying exponential). In that case, the function f(x) would be a negative term minus ln x, which is also negative if x > 1 (since ln x is positive for x > 1). Wait, but ln x is positive for x > 1, so -ln x is negative. So if a is negative, then a e^{a x} is negative (since a is negative and e^{a x} is positive), so f(x) would be negative (from a e^{a x}) minus another negative (from -ln x). Hmm, so f(x) would be negative plus a positive? Wait, let me clarify:If a is negative, then a e^{a x} is negative because a is negative and e^{a x} is positive. Then, -ln x is also negative because ln x is positive for x > 1. So f(x) = negative + negative, which is more negative. That can't be non-negative. Therefore, if a is negative, f(x) would be negative for x > 1, right? Because both terms are negative. So maybe a has to be non-negative? Let's test this intuition.Suppose a = 0. Then f(x) = 0 - ln x = -ln x. But for x > 1, ln x is positive, so f(x) would be negative. Therefore, a = 0 is not allowed. So a must be positive? Because if a is positive, then a e^{a x} is positive and growing, while -ln x is negative. So the positive term might dominate the negative term. But we need to ensure that for all x > 1, the positive term is at least as large as the negative term.So the problem reduces to finding all positive a such that a e^{a x} ≥ ln x for all x > 1. Alternatively, a e^{a x} ≥ ln x for all x > 1. To find the range of a where this inequality holds.Now, the standard approach for such problems is to find the minimum value of f(x) on the interval x > 1 and ensure that this minimum is non-negative. So, we can compute the derivative of f(x) with respect to x, find its critical points, evaluate f(x) at those points, and set the minimum to be ≥ 0.Let's try that.First, compute the derivative f’(x):f(x) = a e^{a x} - ln xSo, f’(x) = d/dx [a e^{a x}] - d/dx [ln x]The derivative of a e^{a x} with respect to x is a * a e^{a x} = a² e^{a x}The derivative of -ln x is -1/xTherefore, f’(x) = a² e^{a x} - 1/xWe need to find critical points by setting f’(x) = 0:a² e^{a x} - 1/x = 0So, a² e^{a x} = 1/xBut solving this equation for x in terms of a might be difficult. Let's see. Maybe we can denote t = a x, so then x = t/a. Then the equation becomes:a² e^{t} = a / tWait, substituting x = t/a into the equation:a² e^{a*(t/a)} = 1/(t/a)Simplify:a² e^{t} = a / tDivide both sides by a:a e^{t} = 1 / tHmm, so we have a e^{t} = 1 / t, where t = a x. But this seems like a transcendental equation and might not have an analytical solution. So perhaps we need to approach this differently.Alternatively, perhaps we can analyze the behavior of f’(x). Let's consider the derivative f’(x) = a² e^{a x} - 1/x.Since x > 1, 1/x is between 0 and 1. For a > 0, a² e^{a x} is always positive and increasing because as x increases, e^{a x} increases exponentially. Therefore, the term a² e^{a x} is increasing, while 1/x is decreasing. Therefore, the derivative f’(x) is increasing for x > 1 because the positive term is increasing and the negative term is decreasing. Therefore, f’(x) is increasing on x > 1.If the derivative is increasing, then there can be at most one critical point where f’(x) = 0. So, either f’(x) is always positive (if even the minimal value of f’(x) is positive) or it starts negative and becomes positive after some critical point. If there's a critical point, then the function f(x) would have a minimum there. So, to find whether there's a critical point, we can check the limit of f’(x) as x approaches 1 from the right and as x approaches infinity.As x approaches 1 from the right:f’(1+) = a² e^{a * 1} - 1/1 = a² e^{a} - 1As x approaches infinity:f’(inf) = lim_{x→∞} [a² e^{a x} - 1/x] = infinity - 0 = infinitySo, since f’(x) tends to infinity as x approaches infinity, and f’(x) is increasing, the behavior at x approaching 1 is important. If f’(1) ≥ 0, then since f’(x) is increasing, the derivative is always non-negative, so f(x) is non-decreasing on x > 1. Therefore, the minimal value of f(x) would occur at x = 1. But the problem states x > 1, so actually, x approaching 1 from the right. But wait, the function is defined for x > 1, but the interval is x > 1. So the minimal value would be at the infimum of the interval, which is x approaching 1 from the right, and then increasing thereafter.But in this case, if f’(x) is always non-negative for x > 1, then f(x) is increasing on x > 1. Therefore, the minimal value of f(x) on x > 1 would be lim_{x→1+} f(x). Let's compute that.lim_{x→1+} f(x) = a e^{a * 1} - ln 1 = a e^{a} - 0 = a e^{a}So, if f’(x) ≥ 0 for all x > 1, then f(x) is increasing, and the minimal value is a e^{a}. Therefore, to ensure f(x) ≥ 0 for all x > 1, we need a e^{a} ≥ 0. But since a > 0, a e^{a} is always positive. Wait, that can't be right. Wait, but the problem states that for x > 1, f(x) ≥ 0. If f(x) is increasing on x > 1, then the minimal value is at x approaching 1 from the right, which is a e^{a}, and if we ensure that a e^{a} ≥ 0, which it is since a > 0, then f(x) would be ≥ 0 for all x > 1. But this contradicts the earlier case when a = 0, but we already saw that a = 0 is invalid. Wait, maybe there's a mistake here.Wait, actually, if f’(x) is always positive for x > 1, then f(x) is increasing on x > 1. Therefore, the minimal value is at x approaching 1 from the right. So, if lim_{x→1+} f(x) = a e^{a}, and since we need f(x) ≥ 0 for all x > 1, then we just need a e^{a} ≥ 0. But since a > 0, a e^{a} is always positive. Therefore, in this case, as long as f’(x) is always positive, which requires that f’(1) ≥ 0, i.e., a² e^{a} - 1 ≥ 0. Because if f’(1) ≥ 0, then since f’(x) is increasing, f’(x) ≥ 0 for all x > 1. Therefore, the condition reduces to a² e^{a} - 1 ≥ 0, which is a² e^{a} ≥ 1. So solving this inequality would give the required range of a.But wait, let's verify. If a² e^{a} ≥ 1, then f’(1) ≥ 0, so f’(x) is increasing and stays non-negative, hence f(x) is increasing on x > 1. Therefore, the minimal value is a e^{a}, which is already positive. Therefore, f(x) is always ≥ a e^{a} ≥ 0. Wait, but a e^{a} is positive, so f(x) would be greater than or equal to a positive number. But the problem requires f(x) ≥ 0 for all x > 1, which would automatically hold since f(x) is increasing and its minimal value is positive. Therefore, if we can ensure that a² e^{a} ≥ 1, then the function f(x) is always increasing on x > 1 with minimal value a e^{a} > 0, hence f(x) ≥ 0.But what if a² e^{a} < 1? Then, f’(1) = a² e^{a} - 1 < 0. Since f’(x) is increasing (as we established earlier), but starts negative at x = 1 and goes to infinity as x approaches infinity. Therefore, there must be some critical point x = c > 1 where f’(c) = 0. At that critical point, f(x) attains a local minimum. Therefore, to ensure that f(x) ≥ 0 for all x > 1, we need to ensure that this local minimum is also ≥ 0.Therefore, the problem splits into two cases:1. If a² e^{a} ≥ 1, then f’(x) ≥ 0 for all x > 1, so f(x) is increasing, and the minimal value is a e^{a} ≥ 0, which is already satisfied. Therefore, in this case, f(x) ≥ 0 for all x > 1.2. If a² e^{a} < 1, then there exists a critical point x = c > 1 where f’(c) = 0, and at that point, f(c) must be ≥ 0. Therefore, we need to find the minimal value of f(x) at this critical point and set it ≥ 0.Therefore, the range of a is the union of the solutions from both cases. However, in case 1, when a² e^{a} ≥ 1, the minimal value is a e^{a} ≥ 0, which is automatically true since a > 0. Wait, but a e^{a} is positive. But the problem requires f(x) ≥ 0 for all x > 1. If the function is increasing, starting at a e^{a} (when x approaches 1) and goes to infinity as x approaches infinity, then yes, it's always non-negative. Therefore, in case 1, all a such that a² e^{a} ≥ 1 will satisfy f(x) ≥ 0.But in case 2, where a² e^{a} < 1, we need to ensure that at the critical point x = c, f(c) ≥ 0. So, the problem requires that even if there's a critical point (a local minimum), that minimum is still non-negative.Therefore, the full solution requires solving both cases. However, solving case 2 seems challenging because it involves solving for c where f’(c) = 0 and then setting f(c) ≥ 0. Let's formalize this.Let’s denote that at the critical point x = c, we have:f’(c) = a² e^{a c} - 1/c = 0 => a² e^{a c} = 1/c.Also, we need f(c) = a e^{a c} - ln c ≥ 0.So, from the critical point equation: a² e^{a c} = 1/c => a e^{a c} = 1/(a c).Substituting this into f(c):f(c) = a e^{a c} - ln c = 1/(a c) - ln c ≥ 0.Therefore, we have the inequality:1/(a c) - ln c ≥ 0 => 1/(a c) ≥ ln c => a ≤ 1/(c ln c).But we also have from the critical point equation:a² e^{a c} = 1/c.But this seems a bit tangled. Let me see if we can relate a and c here.From the critical point equation:a² e^{a c} = 1/c.Let’s take natural logarithm on both sides:ln(a²) + a c = -ln c.So,2 ln a + a c = -ln c.This equation relates a and c. It's a transcendental equation and might not have a closed-form solution, so we might need to find a relationship or parameterize it.Alternatively, let's consider substituting a c = t. Then c = t/a. Let's substitute into the equation:2 ln a + a*(t/a) = -ln(t/a)Simplify:2 ln a + t = -ln t + ln aSo,2 ln a + t = ln a - ln tSubtract ln a from both sides:ln a + t = -ln tTherefore,ln a + t + ln t = 0ln(a t) + t = 0Hmm, this seems complicated. Alternatively, let's try to express a in terms of c.From the critical point equation:a² e^{a c} = 1/cWe can write this as:a² = e^{-a c}/cTake square roots:a = e^{-a c/2}/sqrt(c)But not sure if this helps.Alternatively, from the critical point equation and the expression for f(c):We have f(c) = 1/(a c) - ln c ≥ 0.But from the critical point equation, a² e^{a c} = 1/c, so we can write e^{a c} = 1/(a² c). Take natural logarithm:a c = -ln(a² c) = - (2 ln a + ln c)So,a c + 2 ln a + ln c = 0But we have f(c) = 1/(a c) - ln c ≥ 0. Let's denote t = a c. Then, since c = t/a, we have:f(c) = 1/t - ln(t/a) ≥ 0= 1/t - ln t + ln a ≥ 0But from the equation above: a c + 2 ln a + ln c = 0With t = a c, and c = t/a, ln c = ln t - ln a. Substitute into the equation:t + 2 ln a + ln t - ln a = 0Simplify:t + ln a + ln t = 0So,ln(a t) + t = 0Therefore, ln(a t) = -tExponentiating both sides:a t = e^{-t}Therefore,a = e^{-t}/tSo, a is expressed in terms of t. Also, from f(c) = 1/t - ln c ≥ 0. But c = t/a, so:ln c = ln(t/a) = ln t - ln aFrom the equation above, ln(a t) = -t, so ln a + ln t = -t. Therefore, ln a = -t - ln tThus, ln c = ln t - ln a = ln t - (-t - ln t) = ln t + t + ln t = 2 ln t + tWait, that seems complicated. Wait, let me redo that:From ln(a t) = -t, so ln a + ln t = -t, so ln a = -t - ln tTherefore, ln c = ln(t/a) = ln t - ln a = ln t - (-t - ln t) = ln t + t + ln t = 2 ln t + tWait, that's not possible. Wait, perhaps I made a miscalculation.Wait, ln c = ln(t/a) = ln t - ln a. But ln a = -t - ln t. Therefore,ln c = ln t - (-t - ln t) = ln t + t + ln t = t + 2 ln tYes, that's correct.Therefore, f(c) = 1/t - ln c = 1/t - (t + 2 ln t)But we need f(c) ≥ 0:1/t - t - 2 ln t ≥ 0So, 1/t - t - 2 ln t ≥ 0Let’s denote this as:g(t) = 1/t - t - 2 ln t ≥ 0We need to find t > 0 such that g(t) ≥ 0, and a = e^{-t}/t.But also, since c = t/a and c > 1 (because x > 1), we have c = t/a > 1 => t/a > 1 => t > a. But a = e^{-t}/t, so t > e^{-t}/t => t² > e^{-t}But since t > 0, this inequality might hold for certain t. However, this might complicate things. Maybe instead of focusing on t, let's analyze the inequality g(t) = 1/t - t - 2 ln t ≥ 0.So, we need to find the values of t where 1/t - t - 2 ln t ≥ 0. Then, for each such t, a is given by a = e^{-t}/t. The corresponding a will be part of the solution.This seems like a difficult equation to solve analytically. Maybe we can analyze the function g(t) = 1/t - t - 2 ln t.Compute its derivative:g’(t) = -1/t² - 1 - 2/tSet derivative to zero to find critical points:-1/t² -1 -2/t = 0Multiply both sides by -t²:1 + t² + 2 t = 0Which is t² + 2 t + 1 = 0 => (t + 1)^2 = 0 => t = -1But t > 0, so there are no critical points in t > 0. Therefore, g(t) is decreasing for all t > 0, since g’(t) = -1/t² -1 -2/t < 0 for all t > 0. Therefore, g(t) is strictly decreasing on t > 0.Now, let's find the value of t where g(t) = 0. Since g(t) is strictly decreasing, there can be at most one solution. Let's check the limits.As t approaches 0+:1/t tends to infinity, -t tends to 0, -2 ln t tends to infinity. So overall, g(t) tends to infinity.As t approaches infinity:1/t tends to 0, -t tends to -infinity, -2 ln t tends to -infinity. So overall, g(t) tends to -infinity.Therefore, since g(t) is continuous on t > 0 and strictly decreasing from infinity to -infinity, there exists a unique t0 > 0 where g(t0) = 0. For t ≤ t0, g(t) ≥ 0; for t > t0, g(t) < 0.Therefore, the inequality g(t) ≥ 0 holds only when t ≤ t0, where t0 is the unique solution to 1/t - t - 2 ln t = 0.Therefore, in the case where a² e^{a} < 1 (i.e., case 2), we need t ≤ t0, which translates to a = e^{-t}/t ≤ e^{-t0}/t0. Wait, since a = e^{-t}/t and t ≤ t0, but since a = e^{-t}/t, how does a vary with t? Let's analyze the function a(t) = e^{-t}/t.Compute derivative of a(t) with respect to t:a’(t) = (-e^{-t} * t - e^{-t} * 1)/t² = -e^{-t}(t + 1)/t² < 0 for all t > 0.Therefore, a(t) is strictly decreasing in t > 0. Therefore, as t increases, a decreases. So, if t ≤ t0, then a = e^{-t}/t ≥ e^{-t0}/t0. Since a(t) is decreasing, larger t gives smaller a. Therefore, the minimal a in this case is a_min = e^{-t0}/t0, and the maximal a is... Well, when t approaches 0, a(t) approaches infinity, but since t ≤ t0, but in our case, t is related to the critical point. However, actually, when considering case 2, which is when a² e^{a} < 1, but combined with the requirement that t ≤ t0, which gives a ≥ e^{-t0}/t0. Therefore, in case 2, a must satisfy e^{-t0}/t0 ≤ a < ... Well, the upper bound here is the a where a² e^{a} = 1. Let's call that a1. So, a1 is the solution to a² e^{a} = 1.So, to summarize:The full range of a is:- For a ≥ a1, where a1 is the solution to a² e^{a} = 1, we are in case 1, and f(x) is increasing, so f(x) ≥ 0.- For e^{-t0}/t0 ≤ a < a1, where t0 is the solution to 1/t - t - 2 ln t = 0, we are in case 2, and the minimum of f(x) at x = c is non-negative.Therefore, the total range is a ≥ e^{-t0}/t0. But we need to find the value of e^{-t0}/t0 and a1.But this seems complicated because t0 is the solution to 1/t - t - 2 ln t = 0, and a1 is the solution to a² e^{a} = 1. However, maybe these two values are related? Let's check.Wait, if we look back at the equation for t0: 1/t0 - t0 - 2 ln t0 = 0. Let me see if t0 = 1. Let's test t = 1:g(1) = 1/1 - 1 - 2 ln 1 = 1 - 1 - 0 = 0. Oh! So t0 = 1.Wait, really? Let's confirm:g(1) = 1/1 - 1 - 2 ln 1 = 1 - 1 - 0 = 0. Yes, t0 = 1.Therefore, the critical t0 is 1. So, the inequality g(t) ≥ 0 holds for t ≤ 1. Therefore, in case 2, we have t ≤ 1. But t is related to a via a = e^{-t}/t. Since a(t) is decreasing, t ≤ 1 corresponds to a ≥ e^{-1}/1 = 1/e ≈ 0.3679.Therefore, in case 2, a must satisfy a ≥ 1/e. But also, in case 2, we have a < a1, where a1 is the solution to a² e^{a} = 1.Therefore, the total range of a is a ≥ 1/e, since for a ≥ 1/e, either a ≥ a1 (case 1) or 1/e ≤ a < a1 (case 2). But we need to check if a1 is equal to 1/e or not.Wait, let's compute a1, which is the solution to a² e^{a} = 1.Is a1 equal to 1? Let's check a = 1:1² e^1 = e ≈ 2.718 > 1. So, a1 must be less than 1.Wait, let's solve a² e^{a} = 1 for a.Let’s consider the function h(a) = a² e^{a}. We need to find a such that h(a) = 1.h(a) is increasing for a > 0 because its derivative h’(a) = 2a e^{a} + a² e^{a} = a e^{a} (2 + a) > 0 for a > 0.Therefore, h(a) is strictly increasing from h(0) = 0 to h(∞) = ∞. Therefore, there exists a unique a1 > 0 such that h(a1) = 1.We can approximate a1 numerically.Let’s try a = 0.5:h(0.5) = (0.25) e^{0.5} ≈ 0.25 * 1.6487 ≈ 0.4122 < 1a = 0.7:h(0.7) = 0.49 * e^{0.7} ≈ 0.49 * 2.0138 ≈ 0.986 ≈ 0.986 < 1Close to 1. Let's try a = 0.71:h(0.71) = 0.71² * e^{0.71} ≈ 0.5041 * 2.0335 ≈ 1.025 ≈ 1.025 > 1Therefore, a1 is between 0.7 and 0.71.Use linear approximation:At a = 0.7, h(a) ≈ 0.986At a = 0.71, h(a) ≈ 1.025The difference needed is 1 - 0.986 = 0.014 over an interval of 0.01 in a.So, approximately, a1 ≈ 0.7 + (0.014 / (1.025 - 0.986)) * 0.01 ≈ 0.7 + (0.014 / 0.039) * 0.01 ≈ 0.7 + 0.0036 ≈ 0.7036. So, a1 ≈ 0.7036.But for the purposes of this problem, we might not need the exact value. However, the key point is that a1 is approximately 0.7036, and we found that in case 2, a must be ≥ 1/e ≈ 0.3679. But since a1 ≈ 0.7036 > 1/e, the total range of a is a ≥ 1/e. Because:- For a ≥ a1 ≈ 0.7036, we are in case 1 (f’(x) ≥ 0), so f(x) is increasing and ≥ 0.- For 1/e ≈ 0.3679 ≤ a < a1 ≈ 0.7036, we are in case 2, and the minimal value of f(x) at the critical point is ≥ 0.But wait, how do we know that when a ≥ 1/e, the minimal value in case 2 is non-negative?Wait, earlier we found that in case 2, when a² e^{a} < 1, which corresponds to a < a1, but a must be ≥ 1/e to ensure that f(c) ≥ 0. Therefore, the total range is a ≥ 1/e. Because even though a1 is approximately 0.7036, the lower bound for a in case 2 is 1/e, so combining both cases, the minimal a is 1/e, and for a ≥ 1/e, either in case 1 or case 2, f(x) ≥ 0.But let's verify this. Suppose a = 1/e ≈ 0.3679. Then, what happens?First, check if a² e^{a} = (1/e)^2 e^{1/e} ≈ (1/e²) * e^{0.3679} ≈ (0.1353) * 1.444 ≈ 0.195 < 1. Therefore, it's in case 2. Then, t0 = 1, so a = e^{-1}/1 = 1/e. Therefore, at a = 1/e, the critical point occurs at t = 1, so c = t/a = 1/(1/e) = e ≈ 2.718. Then, f(c) = 1/(a c) - ln c = 1/( (1/e)*e ) - ln e = 1/1 - 1 = 0. Therefore, at a = 1/e, the minimal value of f(x) is zero, occurring at x = e.Therefore, for a = 1/e, f(x) ≥ 0 for all x > 1 because the minimal value is zero. For a > 1/e, in case 2, since a increases beyond 1/e, the minimal value f(c) becomes positive. Wait, no. Wait, when a increases, how does f(c) behave?Wait, when a increases above 1/e, then in case 2 (a < a1), but since a(t) is decreasing, a higher a corresponds to a lower t. Wait, this is getting a bit confusing. Let's recap:From the earlier analysis, when a = 1/e, the minimal value f(c) = 0. If a increases beyond 1/e, then since we are still in case 2 (until a reaches a1), we need to check if f(c) remains non-negative.Wait, when a increases above 1/e, then since a(t) = e^{-t}/t is decreasing, to get a higher a, we need a lower t. For example, t < 1. But when t < 1, from the equation g(t) = 1/t - t - 2 ln t ≥ 0, since t < 1, ln t is negative, so -2 ln t is positive. 1/t is greater than 1, -t is greater than -1. So, overall, g(t) is positive for t < 1. Wait, but we established that g(t) is strictly decreasing from infinity to -infinity, crossing zero at t0 = 1. Therefore, for t < 1, g(t) > 0. Therefore, when a > 1/e, which corresponds to t < 1 (since a(t) = e^{-t}/t, which is decreasing), then g(t) = 1/t - t - 2 ln t > 0. Therefore, f(c) = g(t) > 0. Hence, f(c) > 0 for a > 1/e.Wait, that makes sense. So, when a = 1/e, f(c) = 0, and when a > 1/e (but still in case 2, i.e., a < a1), f(c) > 0. Therefore, the minimal value is non-negative for all a ≥ 1/e, regardless of whether they are in case 1 or case 2.Therefore, the critical value is a = 1/e. Therefore, the range of a is [1/e, ∞).But wait, but earlier we had a1 ≈ 0.7036, which is greater than 1/e ≈ 0.3679. So, when a increases from 1/e to a1, we are in case 2 where there is a critical point, but the minimal value is positive. When a ≥ a1, we are in case 1 where the function is increasing, so the minimal value is a e^{a} which is also positive. Therefore, combining both cases, the total range is a ≥ 1/e.But to confirm, let's test a = 1/2 ≈ 0.5, which is greater than 1/e ≈ 0.3679 but less than a1 ≈ 0.7036. So, in case 2. Let's see if f(x) ≥ 0 for all x > 1.Compute the critical point x = c where f’(c) = 0. From a = 0.5:a² e^{a c} = 1/c => (0.25) e^{0.5 c} = 1/c.This equation is transcendental, so let's approximate c.Let's guess c = 2:Left side: 0.25 e^{1} ≈ 0.25 * 2.718 ≈ 0.6795Right side: 1/2 = 0.50.6795 > 0.5, so need higher c.c = 3:Left: 0.25 e^{1.5} ≈ 0.25 * 4.4817 ≈ 1.1204Right: 1/3 ≈ 0.3331.1204 > 0.333Too big. Wait, but as x increases, left side (a² e^{a x}) increases exponentially, while right side (1/x) decreases. Wait, but we are solving 0.25 e^{0.5 c} = 1/c.Let me rearrange:e^{0.5 c} = 4 / cTake natural logarithm:0.5 c = ln(4) - ln cSo,0.5 c + ln c = ln 4This is still transcendental. Let's attempt to approximate.Let’s try c = 2:0.5*2 + ln 2 ≈ 1 + 0.693 ≈ 1.693 < ln 4 ≈ 1.386. Wait, ln 4 ≈ 1.386. Wait, wait, 0.5 c + ln c = 1 + 0.693 = 1.693, but right side is ln 4 ≈ 1.386. So left > right. Need to decrease c.Wait, that seems contradictory. Wait, if c increases, 0.5 c increases, ln c increases, so 0.5 c + ln c increases. Therefore, to get 0.5 c + ln c = ln 4 ≈ 1.386, we need to find c such that this equality holds.Wait, let's try c = 1:0.5*1 + ln 1 = 0.5 + 0 = 0.5 < 1.386c = 1.5:0.5*1.5 + ln 1.5 ≈ 0.75 + 0.405 ≈ 1.155 < 1.386c = 1.6:0.8 + ln 1.6 ≈ 0.8 + 0.470 ≈ 1.270 < 1.386c = 1.7:0.85 + ln 1.7 ≈ 0.85 + 0.5306 ≈ 1.3806 ≈ 1.3806 < 1.386Very close. c = 1.71:0.855 + ln 1.71 ≈ 0.855 + 0.536 ≈ 1.391 > 1.386Therefore, c ≈ 1.705Therefore, the critical point is at x ≈ 1.705.Now compute f(c):f(c) = a e^{a c} - ln c = 0.5 e^{0.5 * 1.705} - ln 1.705Compute e^{0.8525} ≈ e^{0.85} ≈ 2.34, so 0.5 * 2.34 ≈ 1.17ln 1.705 ≈ 0.534Therefore, f(c) ≈ 1.17 - 0.534 ≈ 0.636 > 0.Therefore, f(c) > 0 when a = 0.5, which is greater than 1/e. Hence, it satisfies the condition.Therefore, this example supports the conclusion that a ≥ 1/e is the valid range.Similarly, take a = 1/e ≈ 0.3679. Then, the critical point is at x = e ≈ 2.718. Then, f(e) = (1/e) e^{(1/e)*e} - ln e = (1/e) e^{1} - 1 = (e/e) - 1 = 1 - 1 = 0. Therefore, at a = 1/e, the minimal value is zero, which is acceptable.If we take a < 1/e, say a = 0.3, which is less than 1/e ≈ 0.3679. Then, check if there's a critical point and whether f(c) ≥ 0.First, check if a² e^{a} = 0.09 e^{0.3} ≈ 0.09 * 1.349 ≈ 0.1214 < 1. So, in case 2.Compute t such that a = e^{-t}/t => 0.3 = e^{-t}/t.Solve for t:t e^{t} = 1/0.3 ≈ 3.333.This is a classic equation, solution is t = W(3.333), where W is the Lambert W function.Compute W(3.333). Since 3.333 is greater than e (~2.718), the principal branch W0(3.333) is approx let's see:We know that W(3) ≈ 1.0495, W(4) ≈ 1.202. So, interpolating, W(3.333) ≈ 1.13.Therefore, t ≈ 1.13.Thus, c = t/a ≈ 1.13 / 0.3 ≈ 3.767.Now compute f(c) = 1/(a c) - ln c = 1/(0.3 * 3.767) - ln 3.767 ≈ 1/1.13 ≈ 0.885 - 1.326 ≈ -0.441 < 0.Therefore, f(c) < 0, which violates the condition. Therefore, a = 0.3 is invalid.Therefore, the minimal a is indeed 1/e, since for a < 1/e, the minimal value of f(x) becomes negative, whereas for a ≥ 1/e, the minimal value is non-negative.Therefore, the range of a is [1/e, ∞).Final AnswerThe range of values for ( a ) is (boxed{[dfrac{1}{e}, +infty)}).</think>Given the function ( f(x) = ae^{ax} - ln x ), we need to find the range of values for ( a ) such that ( f(x) geq 0 ) for all ( x > 1 ).1. Analyzing the function components: - The function ( f(x) ) consists of an exponential term ( ae^{ax} ) and a logarithmic term ( -ln x ). - For ( x > 1 ), ( ln x ) is positive, making ( -ln x ) negative. Therefore, ( ae^{ax} ) must be sufficiently large to ensure ( f(x) geq 0 ).2. Derivative analysis: - Compute the derivative ( f'(x) = a^2 e^{ax} - frac{1}{x} ). - The critical points occur where ( a^2 e^{ax} = frac{1}{x} ).3. Behavior of the derivative: - For ( x > 1 ), ( a^2 e^{ax} ) is increasing and ( frac{1}{x} ) is decreasing. - If ( f'(x) geq 0 ) for all ( x > 1 ), ( f(x) ) is non-decreasing, ensuring ( f(x) geq 0 ).4. Critical points and minimum value: - If ( a^2 e^a geq 1 ), then ( f'(x) geq 0 ) for all ( x > 1 ), making ( f(x) ) increasing with the minimum value at ( x to 1^+ ), which is ( ae^a geq 0 ). - If ( a^2 e^a < 1 ), there exists a critical point ( x = c ) where ( f(c) geq 0 ).5. Solving for critical points: - The critical point equation ( a^2 e^{ac} = frac{1}{c} ) leads to ( f(c) = frac{1}{ac} - ln c geq 0 ). - Solving ( frac{1}{ac} geq ln c ) and using the Lambert W function, we find ( a geq frac{1}{e} ).6. Verification: - For ( a = frac{1}{e} ), the critical point ( x = e ) results in ( f(e) = 0 ). - For ( a > frac{1}{e} ), the minimum value of ( f(x) ) is positive.Thus, the range of values for ( a ) is (boxed{[dfrac{1}{e}, +infty)}).,<think>Alright, let's tackle this problem. So, we have the function f(x) = a e^{a x} - ln x, and we need to find the range of values for a such that f(x) ≥ 0 for all x > 1. Hmm, okay. Let me start by understanding what the question is asking. For any x greater than 1, this function must be non-negative. So, whatever a is, when we plug in any x bigger than 1, the value of a e^{a x} has to be at least as big as ln x. Interesting.First, maybe I should analyze the function f(x) itself. Let's see. The function is composed of two parts: a e^{a x} and -ln x. The first term is exponential, and the second is logarithmic. Depending on the value of a, the exponential term could be growing or decaying. Since x > 1, ln x is positive but grows slowly. If a is positive, then e^{a x} will grow exponentially, which might dominate over ln x. If a is negative, then e^{a x} will decay, possibly making the first term small or even negative, which could be problematic because we need the whole expression to be non-negative.Wait, but if a is negative, then a e^{a x} would be negative because a is negative and e^{a x} is positive. Then we have negative minus ln x, which would be even more negative. So that can't be. So maybe a has to be positive? Because if a is positive, then a e^{a x} is positive, and ln x is positive, but we need the positive term to be bigger than ln x. Hmm.So perhaps a must be positive? Let me check that. If a is zero, then the first term is zero, and f(x) = -ln x, which is negative for x > 1, so a can't be zero. If a is negative, as I thought, the first term is negative and the second term is negative, so overall f(x) would be negative, which doesn't satisfy the condition. So a must be positive. So that's a start.Therefore, a > 0. Now, we need to ensure that for all x > 1, a e^{a x} ≥ ln x. So, the question reduces to finding all positive a such that a e^{a x} is always greater than or equal to ln x for x > 1.Maybe it's helpful to consider the minimum of f(x) on the interval (1, ∞). If we can find the minimum value of f(x) and ensure that it's non-negative, then that would guarantee f(x) ≥ 0 for all x > 1.To find the minimum, we can take the derivative of f(x) with respect to x and set it equal to zero. Let's compute f'(x):f(x) = a e^{a x} - ln xf'(x) = a * a e^{a x} - (1/x) = a² e^{a x} - 1/xSet this equal to zero to find critical points:a² e^{a x} - 1/x = 0So, a² e^{a x} = 1/xHmm, solving this equation for x might be tricky. Let's see. Let's denote t = a x. Then, e^{t} = 1/(a² x) = 1/(a² (t/a)) ) = a/(a² t) ) = 1/(a t). So, e^{t} = 1/(a t)Hmm, this seems complicated. Maybe another approach. Let's rearrange the equation:a² e^{a x} = 1/xTake natural logarithm on both sides:ln(a²) + a x = -ln xSo:a x + ln(a²) + ln x = 0Wait, that's:a x + ln(a² x) = 0But this seems still difficult to solve for x. Maybe there's a substitution. Let me try substituting u = a x. Then x = u/a, and ln(a² x) = ln(a² * u/a) = ln(a u). So the equation becomes:u + ln(a u) = 0So:u + ln a + ln u = 0Hmm, but u = a x, which is a substitution variable. This equation relates u and a. It's still transcendental and can't be solved analytically. Maybe we need another approach.Alternatively, maybe instead of trying to find the critical points explicitly, we can analyze the behavior of f(x). Let's check the limits as x approaches 1 and infinity.First, as x approaches 1 from the right:f(1) = a e^{a} - ln 1 = a e^{a} - 0 = a e^{a}. So at x = 1, f(x) = a e^{a}, which is positive since a > 0.As x approaches infinity, let's see the behavior of f(x). The term a e^{a x} grows exponentially if a > 0, and ln x grows logarithmically. So, if a > 0, then as x approaches infinity, a e^{a x} will dominate over ln x, so f(x) tends to infinity. Thus, the potential problem area is not at the endpoints, but somewhere in between. So, the minimum of f(x) must occur somewhere in x > 1, and we need to ensure that this minimum is non-negative.Therefore, we need to find the minimal value of f(x) on (1, ∞) and set that to be greater than or equal to zero.But to find the minimal value, we need to find the critical point where f'(x) = 0. Since we can't solve for x explicitly, maybe we can parametrize the problem.Let me denote t = a x. Then, x = t/a. Let's rewrite the equation a² e^{a x} = 1/x in terms of t:a² e^{t} = a / tSimplify:a e^{t} = 1 / tSo:a = e^{-t} / tTherefore, for each critical point x, there is a corresponding t such that a = e^{-t}/t.But this seems to relate a and t. Let's consider this. So, if a = e^{-t}/t, then for each t > 0 (since x > 1 and a > 0, so t = a x > a*1 > 0), a is determined by t. But how does this help us?Alternatively, since we can't solve for x in terms of a, maybe we can express the minimal value of f(x) in terms of t. Let's try that.Given that at the critical point, a x = t, and a = e^{-t}/t. Then, x = t/a = t / (e^{-t}/t) ) = t^2 e^{t}.So x = t^2 e^{t}.But x must be greater than 1. Let's check if x = t^2 e^{t} > 1. Since t > 0, and t^2 e^{t} is increasing for t > 0, starting from 0 when t=0, and increasing to infinity as t approaches infinity. Therefore, there is a unique t for each x > 1, but this might not be helpful.Wait, perhaps another angle. Let's recall that at the critical point, the function f(x) reaches its minimum. So, we can express f(x) at the critical point in terms of t.Given that at critical point:x = t^2 e^{t} (from above)But also, f(x) = a e^{a x} - ln x. Let's substitute a = e^{-t}/t and x = t^2 e^{t}.Then:f(x) = (e^{-t}/t) * e^{ (e^{-t}/t) * t^2 e^{t} } - ln(t^2 e^{t})Simplify the exponent: (e^{-t}/t) * t^2 e^{t} = t e^{-t} * e^{t} = t.So, f(x) = (e^{-t}/t) * e^{t} - ln(t^2 e^{t}) = (e^{-t} * e^{t}) / t - [2 ln t + t]Simplify: 1/t - 2 ln t - tTherefore, f(x) at the critical point is 1/t - 2 ln t - tAnd we need this to be greater than or equal to zero, because the minimum value of f(x) must be non-negative.So, the problem reduces to finding t > 0 such that 1/t - 2 ln t - t ≥ 0, given that a = e^{-t}/t. Because each t corresponds to a value of a, so we need to find all a such that there exists a t where a = e^{-t}/t and 1/t - 2 ln t - t ≥ 0.Wait, but actually, for the minimal value of f(x) to be non-negative, the minimal value occurs when f(x) = 1/t - 2 ln t - t ≥ 0. So, for each a, which is e^{-t}/t, we need 1/t - 2 ln t - t ≥ 0.Therefore, the range of a is all a = e^{-t}/t where t satisfies 1/t - 2 ln t - t ≥ 0.Therefore, we need to find t > 0 such that 1/t - t - 2 ln t ≥ 0, and then a is e^{-t}/t for those t.So, our job now is to find all t > 0 where 1/t - t - 2 ln t ≥ 0, and then the corresponding a is e^{-t}/t. Therefore, the range of a is { e^{-t}/t | t > 0 and 1/t - t - 2 ln t ≥ 0 }.Therefore, we need to solve the inequality 1/t - t - 2 ln t ≥ 0 for t > 0.Let me define a function g(t) = 1/t - t - 2 ln t. We need to find t > 0 such that g(t) ≥ 0.Let's analyze g(t):First, compute the derivative of g(t):g'(t) = -1/t² - 1 - 2*(1/t) = -1/t² -1 - 2/tThis derivative is always negative for t > 0:Each term: -1/t² is negative, -1 is negative, -2/t is negative. So, g'(t) < 0 for all t > 0. Therefore, g(t) is strictly decreasing on t > 0.Now, let's check the limits:As t approaches 0+:1/t approaches infinity, -t approaches 0, -2 ln t approaches infinity (since ln t approaches -infty). So overall, g(t) approaches infinity.As t approaches infinity:1/t approaches 0, -t approaches -infty, -2 ln t approaches -infty. So g(t) approaches -infty.Since g(t) is strictly decreasing from infinity to -infty as t goes from 0 to infinity, there must be exactly one t0 where g(t0) = 0. Therefore, the inequality g(t) ≥ 0 holds for t ∈ (0, t0], where t0 is the unique solution to 1/t - t - 2 ln t = 0.Therefore, the values of t that satisfy the inequality are t ≤ t0, and since a = e^{-t}/t, and t ∈ (0, t0], then a = e^{-t}/t. Since t decreases from t0 to 0, a increases? Wait, let's see:Wait, when t increases, e^{-t}/t decreases because e^{-t} decays exponentially and t increases. So, as t increases from 0 to t0, a = e^{-t}/t decreases from infinity to e^{-t0}/t0. Wait, but t ∈ (0, t0], so as t approaches 0 from the right, e^{-t}/t approaches infinity (since e^{-t} approaches 1 and 1/t approaches infinity). As t approaches t0, a approaches e^{-t0}/t0. Therefore, the corresponding a values are a ≥ e^{-t0}/t0.But wait, since t is in (0, t0], and a = e^{-t}/t, which is a decreasing function of t on (0, ∞), because derivative of a with respect to t is:da/dt = d/dt [e^{-t}/t] = (-e^{-t} * t - e^{-t}) / t² = -e^{-t}(t + 1)/t² < 0.Therefore, as t increases from 0 to t0, a decreases from infinity to e^{-t0}/t0. Therefore, the possible values of a are [e^{-t0}/t0, ∞).But we need to find a such that for all x > 1, f(x) ≥ 0. And we found that this happens when a ≥ e^{-t0}/t0, where t0 is the solution to 1/t0 - t0 - 2 ln t0 = 0.Therefore, the range of a is [e^{-t0}/t0, ∞). So, we need to find the value of t0.But how do we find t0? It's the solution to 1/t - t - 2 ln t = 0.This is a transcendental equation, which can't be solved analytically. So, we need to approximate it numerically.Let me try to approximate t0.Let's define h(t) = 1/t - t - 2 ln t.We need to find t0 where h(t0) = 0.Let's compute h(t) for some values of t.Start with t = 1:h(1) = 1 - 1 - 0 = 0. Wait, h(1) = 1/1 - 1 - 2 ln 1 = 1 - 1 - 0 = 0.Oh! So t0 = 1 is a solution. Wait, but let's check:Wait, if t =1, then h(1) = 0. But since h(t) is strictly decreasing, and h(t) approaches infinity as t approaches 0 and -infty as t approaches infinity, then the only solution is t0=1. Wait, but when I plug t=1 into h(t), I get 0. Therefore, t0=1.Wait, is that correct?Wait, let's check another value. Let's try t=0.5:h(0.5) = 1/0.5 - 0.5 - 2 ln 0.5 = 2 - 0.5 - 2*(-0.6931) ≈ 1.5 + 1.3862 ≈ 2.8862 > 0.t=2:h(2) = 1/2 - 2 - 2 ln 2 ≈ 0.5 - 2 - 1.3863 ≈ -2.8863 < 0.So h(t) at t=1 is 0. Since h(t) is strictly decreasing, then t0=1 is the only solution. So t0=1.Wait, but when t=1, h(t)=0. Therefore, the inequality 1/t - t - 2 ln t ≥ 0 holds for t ∈ (0,1]. Therefore, the values of a are a = e^{-t}/t for t ∈ (0,1], which would mean a ≥ e^{-1}/1 = 1/e. Because as t decreases from 1 to 0, a = e^{-t}/t increases from 1/e to infinity. But since t ∈ (0,1], and a is decreasing as t increases, the minimal a is when t=1, which is a=1/e. Therefore, the range of a is [1/e, ∞).Wait, let's confirm this. If t0=1, then the minimal a is e^{-1}/1 = 1/e ≈ 0.3679. Therefore, the range of a is [1/e, ∞). But let's check if this is correct.Let me verify for a=1/e. Let's see if f(x) = (1/e) e^{(1/e)x} - ln x is non-negative for all x >1.At x=1: f(1) = (1/e)e^{1/e} - 0 ≈ (1/e)e^{0.3679} ≈ (0.3679)(1.445) ≈ 0.531, which is positive.At the critical point, which occurs when t=1 (since t0=1), so x = t^2 e^{t} = 1^2 e^1 = e ≈ 2.718.So at x=e, let's compute f(e):f(e) = (1/e)e^{(1/e)e} - ln e = (1/e)e^{1} - 1 = (e)/e -1 = 1 -1 =0.So at x=e, f(e)=0, which is the minimum. Since the function is minimized at x=e with value 0, and tends to infinity as x approaches 1 or infinity, then for a=1/e, f(x) ≥0 holds for all x>1. For a >1/e, the minimum value would be higher, so f(x) would still be non-negative.Therefore, the range of a is [1/e, ∞). Therefore, the answer is a ≥ 1/e, so the range is [1/e, ∞).But let me check another value. Suppose a=1/2, which is greater than 1/e≈0.367. Let's check f(x)=0.5 e^{0.5 x} - ln x. At x=2:f(2)=0.5 e^{1} - ln 2 ≈0.5*2.718 -0.693≈1.359 -0.693≈0.666>0.At critical point, which would be when t=1 for a=1/e, but for a=1/2, the critical point occurs at t where a=e^{-t}/t, which would be different. Wait, but according to our previous analysis, when a is larger than 1/e, the critical t would be less than 1, leading to higher minimal value. Wait, but perhaps our previous conclusion is still valid because when a increases beyond 1/e, the minimal value of f(x) increases. Wait, maybe the minimal value occurs at x=e when a=1/e, but for a>1/e, the minimal value would be at some x> e? Wait, no, actually, the critical point x depends on a. Let me see.Alternatively, since the minimal value of f(x) when a=1/e is zero at x=e, for a>1/e, the minimal value would be positive. Therefore, the minimal a is 1/e. Hence, the answer is a ≥1/e.Therefore, the range of a is [1/e, ∞).I think that's the conclusion. Let me just verify with a=1/e and another a>1/e.Take a=1. Then f(x)=1*e^{x} - ln x. For x>1, e^{x} grows rapidly, so certainly e^{x} > ln x for x>1, so f(x) is positive. For a=1/e, as we saw, the minimal value is zero at x=e. For a=0.5, which is greater than 1/e, the function is positive everywhere. For a=0.3, which is less than 1/e, let's check. Suppose a=0.3, f(x)=0.3 e^{0.3 x} - ln x. At x=e, f(e)=0.3 e^{0.3 e} -1 ≈0.3 e^{0.3*2.718}≈0.3 e^{0.815}≈0.3*2.26≈0.678 -1≈-0.322<0. So, for a=0.3 <1/e, f(e) is negative, which violates the condition. Therefore, a must be at least 1/e.Thus, the answer is a ≥1/e, or [1/e, ∞).Final AnswerThe range of values for ( a ) is (boxed{[dfrac{1}{e}, +infty)}).</think>Given the function ( f(x) = ae^{ax} - ln x ), we need to find the range of values for ( a ) such that ( f(x) geq 0 ) for all ( x > 1 ).1. Initial Analysis: - If ( a leq 0 ), the term ( ae^{ax} ) would be non-positive, leading ( f(x) ) to be non-positive for ( x > 1 ). Thus, ( a ) must be positive. - For ( a > 0 ), we need to ensure ( ae^{ax} geq ln x ) for all ( x > 1 ).2. Critical Points Analysis: - Compute the derivative ( f'(x) = a^2 e^{ax} - frac{1}{x} ). - Set ( f'(x) = 0 ) to find critical points: ( a^2 e^{ax} = frac{1}{x} ). - Solving this equation leads to the substitution ( t = ax ), resulting in ( a = frac{e^{-t}}{t} ).3. Function Behavior at Critical Points: - Express ( f(x) ) at the critical point in terms of ( t ): ( f(x) = frac{1}{t} - t - 2 ln t ). - The function ( g(t) = frac{1}{t} - t - 2 ln t ) must be non-negative for the minimum value of ( f(x) ) to be non-negative.4. Solving the Inequality: - Analyze ( g(t) ): It is strictly decreasing from ( infty ) to ( -infty ) as ( t ) increases from 0 to ( infty ). - The unique solution to ( g(t) = 0 ) is ( t = 1 ).5. Conclusion: - The corresponding ( a ) value for ( t = 1 ) is ( a = frac{1}{e} ). - For ( a geq frac{1}{e} ), the function ( f(x) ) is non-negative for all ( x > 1 ).Thus, the range of values for ( a ) is (boxed{[dfrac{1}{e}, +infty)}).]

🔑:Given the function f(x) = ae^{ax} - ln x, we need to find the range of values for a such that f(x) geqslant 0 for all x > 1.Step 1: Determine the effect of the sign of a on f(x).- For a leqslant 0, we notice that f(x) 1. Therefore, a leqslant 0 does not meet the requirements.- For a > 0, the term ae^{ax} is positive and grows exponentially, while -ln x remains negative. Therefore, we focus on a > 0.Step 2: Establish the inequality ae^{ax} geqslant ln x, which simplifies to axe^{ax} geqslant xln x.Step 3: Define g(x) = xln x and find its derivative, g'(x) = ln x + 1. Since g'(x) > 0 for x > 1, g(x) is monotonically increasing for x > 1.Step 4: Given that x > 1 and a > 0, it leads to e^{ax} > 1. Since axe^{ax} geqslant xln x, it implies that g(e^{ax}) geqslant g(x), and hence e^{ax} geqslant x.Step 5: From e^{ax} geqslant x, we deduce that ax geqslant ln x, which implies a geqslant frac{ln x}{x}.Step 6: Define h(x) = frac{ln x}{x} and find its derivative, h'(x) = frac{1 - ln x}{x^2}. Analyzing h'(x), we find that h(x) is monotonically increasing for 0 e.Step 7: The maximum of h(x) occurs at x = e, so h_{max}(x) = h(e) = frac{1}{e}.Step 8: Therefore, the range of values for a is [frac{1}{e}, +infty), because a must be at least frac{1}{e} to ensure that f(x) geqslant 0 for all x > 1.Final Answer: The range of values for a is boxed{[frac{1}{e}, +infty)}.