answer：Okay, so I have this problem here: Given a complex number ( z ) that satisfies ( sqrt{2}i cdot z = 1 + i ), I need to find the magnitude ( |z| ). Hmm, complex numbers can be a bit tricky, but let me see if I can work through this step by step.First, I remember that when dealing with equations involving complex numbers, one common approach is to solve for ( z ) directly. So, if ( sqrt{2}i cdot z = 1 + i ), then I can isolate ( z ) by dividing both sides of the equation by ( sqrt{2}i ). That should give me ( z ) in terms of known quantities.Let me write that down:[z = frac{1 + i}{sqrt{2}i}]Okay, so now I have ( z ) expressed as a fraction. The next step is to simplify this expression. I recall that when you have a complex number in the denominator, it's often helpful to multiply the numerator and the denominator by the complex conjugate of the denominator to make the denominator real. However, in this case, the denominator is purely imaginary (( sqrt{2}i )), so maybe there's a simpler way to handle it.Wait, actually, I remember that multiplying by ( i ) has a specific effect on complex numbers. Specifically, multiplying by ( i ) rotates the complex number by 90 degrees in the complex plane. But I'm not sure if that helps here. Maybe I should just go ahead and multiply the numerator and denominator by ( i ) to eliminate the imaginary unit from the denominator.Let me try that:[z = frac{(1 + i) cdot i}{sqrt{2}i cdot i}]Multiplying out the numerator:[(1 + i) cdot i = 1 cdot i + i cdot i = i + i^2]I know that ( i^2 = -1 ), so this simplifies to:[i + (-1) = -1 + i]Now, the denominator:[sqrt{2}i cdot i = sqrt{2} cdot i^2 = sqrt{2} cdot (-1) = -sqrt{2}]So putting it all together, we have:[z = frac{-1 + i}{-sqrt{2}} = frac{-1}{-sqrt{2}} + frac{i}{-sqrt{2}} = frac{1}{sqrt{2}} - frac{i}{sqrt{2}}]Hmm, that looks a bit neater. I can also rationalize the denominators if needed, but since we're dealing with magnitudes, maybe I don't need to go that far. Let me just write ( z ) as:[z = frac{1}{sqrt{2}} - frac{1}{sqrt{2}}i]Now, to find the magnitude ( |z| ), I remember that for any complex number ( a + bi ), the magnitude is given by:[|a + bi| = sqrt{a^2 + b^2}]So applying this formula to our ( z ):[|z| = sqrt{left(frac{1}{sqrt{2}}right)^2 + left(-frac{1}{sqrt{2}}right)^2}]Calculating each term:[left(frac{1}{sqrt{2}}right)^2 = frac{1}{2}][left(-frac{1}{sqrt{2}}right)^2 = frac{1}{2}]Adding them together:[frac{1}{2} + frac{1}{2} = 1]Taking the square root of 1:[|z| = sqrt{1} = 1]So, the magnitude of ( z ) is 1. Wait, that seems straightforward. Let me double-check my steps to make sure I didn't make any mistakes.1. I started with ( sqrt{2}i cdot z = 1 + i ).2. Divided both sides by ( sqrt{2}i ) to get ( z = frac{1 + i}{sqrt{2}i} ).3. Multiplied numerator and denominator by ( i ) to simplify, resulting in ( z = frac{-1 + i}{-sqrt{2}} ).4. Simplified further to ( z = frac{1}{sqrt{2}} - frac{1}{sqrt{2}}i ).5. Calculated the magnitude using ( sqrt{a^2 + b^2} ), which gave me 1.Everything seems to check out. Another way to think about this is using the property of magnitudes in complex numbers: ( |ab| = |a||b| ). So, if ( sqrt{2}i cdot z = 1 + i ), then taking magnitudes on both sides:[|sqrt{2}i cdot z| = |1 + i|]Which simplifies to:[|sqrt{2}i| cdot |z| = |1 + i|]Calculating each magnitude:[|sqrt{2}i| = sqrt{2}][|1 + i| = sqrt{1^2 + 1^2} = sqrt{2}]So, plugging back in:[sqrt{2} cdot |z| = sqrt{2}]Dividing both sides by ( sqrt{2} ):[|z| = 1]Yep, that confirms my earlier result. So, regardless of the method, the magnitude of ( z ) is indeed 1. I think I'm confident with this answer now.

answer：First, I need to find the smallest positive angle that shares the same terminal side as -2012 degrees. To do this, I'll determine how many full 360-degree rotations fit into -2012 degrees.Dividing -2012 by 360 gives approximately -5.588. Since we're looking for the smallest positive angle, I'll take the ceiling of this value, which is -5. This means there are 5 full rotations in the negative direction.Next, I'll calculate the remaining angle after these 5 rotations. Multiplying 5 by 360 degrees gives 1800 degrees. Subtracting this from -2012 degrees results in -212 degrees. However, this is still a negative angle, so I'll add another full rotation of 360 degrees to make it positive.Adding 360 degrees to -212 degrees gives 148 degrees. This is the smallest positive angle that has the same terminal side as -2012 degrees.

answer：Alright, so I have this problem with two complex numbers, z₁ and z₂. They're given as z₁ = sin(2x) + λi and z₂ = m + (m - √3 cos(2x))i. The problem says that z₁ equals z₂, which means their real parts and imaginary parts must be equal. Starting with part 1, where λ is 0 and x is between 0 and π. Since z₁ equals z₂, their real parts must be equal, so sin(2x) should equal m. Similarly, the imaginary parts must be equal, so λ equals m minus √3 cos(2x). But since λ is 0, that equation becomes 0 = m - √3 cos(2x). Wait, but from the real part, m is equal to sin(2x). So substituting m into the imaginary part equation, we get 0 = sin(2x) - √3 cos(2x). That simplifies to sin(2x) = √3 cos(2x). Dividing both sides by cos(2x), we get tan(2x) = √3. Hmm, tan(2x) equals √3. I know that tan(π/3) is √3, so 2x must be π/3 plus some multiple of π. But since x is between 0 and π, 2x is between 0 and 2π. So the solutions for 2x would be π/3 and 4π/3. Therefore, x would be π/6 and 2π/3. Let me check if these values are within the given range. π/6 is approximately 0.523 radians, which is between 0 and π. 2π/3 is approximately 2.094 radians, also between 0 and π. So both solutions are valid. Moving on to part 2, λ is now a function of x, so λ = f(x). It's given that when x is α, λ is 1/2. From part 1, we have λ = sin(2x) - √3 cos(2x). So substituting x with α, we get 1/2 = sin(2α) - √3 cos(2α). I think I can express this as a single sine function using the amplitude-phase form. The expression sin(2x) - √3 cos(2x) can be written as 2 sin(2x - π/3). Let me verify that: Using the identity A sinθ + B cosθ = C sin(θ + φ), where C = √(A² + B²) and tanφ = B/A. Here, A is 1 and B is -√3, so C is √(1 + 3) = 2. Then tanφ = -√3/1, so φ is -π/3. Therefore, sin(2x) - √3 cos(2x) = 2 sin(2x - π/3). So, 2 sin(2α - π/3) = 1/2, which means sin(2α - π/3) = 1/4. Therefore, 2α - π/3 = arcsin(1/4) or π - arcsin(1/4). But the problem asks for cos(4α + π/3). Let me express 4α + π/3 in terms of 2α - π/3. 4α + π/3 is equal to 2*(2α) + π/3. Since 2α - π/3 = arcsin(1/4), then 2α = arcsin(1/4) + π/3. Therefore, 4α = 2*arcsin(1/4) + 2π/3. So, 4α + π/3 = 2*arcsin(1/4) + 2π/3 + π/3 = 2*arcsin(1/4) + π. Hmm, cos(4α + π/3) = cos(2*arcsin(1/4) + π). Using the identity cos(A + π) = -cos(A), so this becomes -cos(2*arcsin(1/4)). Now, cos(2θ) = 1 - 2 sin²θ. Let θ = arcsin(1/4), so cos(2θ) = 1 - 2*(1/4)² = 1 - 2*(1/16) = 1 - 1/8 = 7/8. Therefore, cos(4α + π/3) = -7/8. Wait, let me double-check that. If sin(2α - π/3) = 1/4, then 2α - π/3 = arcsin(1/4). So, 2α = arcsin(1/4) + π/3. Then, 4α = 2*arcsin(1/4) + 2π/3. Adding π/3 gives 4α + π/3 = 2*arcsin(1/4) + π. Yes, so cos(4α + π/3) = cos(2*arcsin(1/4) + π) = -cos(2*arcsin(1/4)). And cos(2θ) where θ = arcsin(1/4) is 1 - 2*(1/4)^2 = 7/8. So, -7/8 is the result. I think that makes sense. So, the final answer for part 2 is -7/8.

answer：Okay, so I have this problem where I'm given a function f(x) = sin x cos x - sqrt(3) sin²x. Then, the graph of y = f(x) is translated π/12 units to the right and sqrt(3)/2 units upward to get the graph of y = g(x). I need to find the value of g(π/4). The options are A: 0, B: 1, C: sqrt(3)/2, D: sqrt(3). First, I think I should try to simplify the function f(x) as much as possible. Maybe I can use some trigonometric identities to rewrite it in a more manageable form. Let me recall some identities. I know that sin x cos x can be rewritten using the double-angle identity: sin(2x) = 2 sin x cos x, so sin x cos x = (1/2) sin(2x). That might help simplify the first term.For the second term, sqrt(3) sin²x, I remember that sin²x can be expressed using the identity sin²x = (1 - cos(2x))/2. So, substituting that in, the second term becomes sqrt(3) * (1 - cos(2x))/2, which simplifies to (sqrt(3)/2) - (sqrt(3)/2) cos(2x).Putting it all together, f(x) becomes (1/2) sin(2x) - (sqrt(3)/2) + (sqrt(3)/2) cos(2x). Wait, no, actually, the second term is subtracted, so it's (1/2) sin(2x) - sqrt(3) sin²x, which after substitution becomes (1/2) sin(2x) - (sqrt(3)/2)(1 - cos(2x)). So that's (1/2) sin(2x) - sqrt(3)/2 + (sqrt(3)/2) cos(2x).Now, combining the terms, I have (1/2) sin(2x) + (sqrt(3)/2) cos(2x) - sqrt(3)/2. Hmm, this looks like it could be expressed as a single sine function with a phase shift. The general form is A sin(2x + φ) + C, where A is the amplitude, φ is the phase shift, and C is a constant.To combine the sine and cosine terms into a single sine function, I can use the identity A sin θ + B cos θ = C sin(θ + φ), where C = sqrt(A² + B²) and tan φ = B/A. In this case, A is 1/2 and B is sqrt(3)/2. So, C would be sqrt((1/2)² + (sqrt(3)/2)²) = sqrt(1/4 + 3/4) = sqrt(1) = 1. Then, tan φ = (sqrt(3)/2) / (1/2) = sqrt(3). So φ = arctan(sqrt(3)) = π/3. Therefore, the expression (1/2) sin(2x) + (sqrt(3)/2) cos(2x) can be written as sin(2x + π/3). So, f(x) simplifies to sin(2x + π/3) - sqrt(3)/2. Now, the function g(x) is obtained by translating f(x) π/12 units to the right and sqrt(3)/2 units upward. Translating a function to the right by π/12 means replacing x with x - π/12 in the function. Translating it upward by sqrt(3)/2 means adding sqrt(3)/2 to the entire function.So, g(x) = f(x - π/12) + sqrt(3)/2. Substituting f(x) into this, we get:g(x) = [sin(2(x - π/12) + π/3) - sqrt(3)/2] + sqrt(3)/2.Simplifying inside the sine function: 2(x - π/12) = 2x - π/6. Adding π/3 gives 2x - π/6 + π/3 = 2x + π/6. So, g(x) = sin(2x + π/6) - sqrt(3)/2 + sqrt(3)/2. The -sqrt(3)/2 and +sqrt(3)/2 cancel each other out, so g(x) = sin(2x + π/6).Now, I need to find g(π/4). Plugging π/4 into g(x):g(π/4) = sin(2*(π/4) + π/6) = sin(π/2 + π/6).Simplifying the angle: π/2 + π/6 = (3π/6 + π/6) = 4π/6 = 2π/3.So, g(π/4) = sin(2π/3). I know that sin(2π/3) is equal to sin(π - π/3) = sin(π/3) = sqrt(3)/2.Therefore, the value of g(π/4) is sqrt(3)/2, which corresponds to option C.

answer：Okay, so I need to figure out where the function ( f(x) = x^3 - 15x^2 - 33x + 6 ) is decreasing. I remember from my calculus class that to find where a function is increasing or decreasing, I need to look at its first derivative. If the derivative is positive, the function is increasing, and if it's negative, the function is decreasing.Alright, let's start by finding the first derivative of ( f(x) ). The function is a polynomial, so I can differentiate term by term. The derivative of ( x^3 ) is ( 3x^2 ), the derivative of ( -15x^2 ) is ( -30x ), the derivative of ( -33x ) is ( -33 ), and the derivative of the constant term ( 6 ) is 0. So, putting it all together, the first derivative ( f'(x) ) is:[ f'(x) = 3x^2 - 30x - 33 ]Now, I need to determine where this derivative is negative because that's where the function is decreasing. So, I need to solve the inequality:[ 3x^2 - 30x - 33 < 0 ]Hmm, this is a quadratic inequality. I think the best way to approach this is to first find the roots of the quadratic equation ( 3x^2 - 30x - 33 = 0 ). Once I have the roots, I can determine the intervals where the quadratic expression is negative.To find the roots, I'll use the quadratic formula:[ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ]In this equation, ( a = 3 ), ( b = -30 ), and ( c = -33 ). Plugging these values into the formula:[ x = frac{-(-30) pm sqrt{(-30)^2 - 4 cdot 3 cdot (-33)}}{2 cdot 3} ][ x = frac{30 pm sqrt{900 + 396}}{6} ][ x = frac{30 pm sqrt{1296}}{6} ][ x = frac{30 pm 36}{6} ]So, the two roots are:1. ( x = frac{30 + 36}{6} = frac{66}{6} = 11 )2. ( x = frac{30 - 36}{6} = frac{-6}{6} = -1 )Alright, so the roots are ( x = -1 ) and ( x = 11 ). These roots divide the real number line into three intervals:1. ( (-infty, -1) )2. ( (-1, 11) )3. ( (11, infty) )Now, I need to test each interval to see where the quadratic expression ( 3x^2 - 30x - 33 ) is negative. I'll pick a test point from each interval and plug it into the expression.- For the interval ( (-infty, -1) ), let's choose ( x = -2 ): [ 3(-2)^2 - 30(-2) - 33 = 3(4) + 60 - 33 = 12 + 60 - 33 = 39 ] Since 39 is positive, the expression is positive in this interval.- For the interval ( (-1, 11) ), let's choose ( x = 0 ): [ 3(0)^2 - 30(0) - 33 = 0 - 0 - 33 = -33 ] Since -33 is negative, the expression is negative in this interval.- For the interval ( (11, infty) ), let's choose ( x = 12 ): [ 3(12)^2 - 30(12) - 33 = 3(144) - 360 - 33 = 432 - 360 - 33 = 39 ] Since 39 is positive, the expression is positive in this interval.So, the quadratic expression ( 3x^2 - 30x - 33 ) is negative only in the interval ( (-1, 11) ). Therefore, the function ( f(x) ) is decreasing on this interval.Just to make sure I didn't make any mistakes, let me double-check my calculations. The derivative was calculated correctly, and the quadratic formula was applied properly. The test points also seem to give the correct signs for the expression. So, I think my conclusion is correct.**Final Answer**The function is decreasing on the interval boxed{(-1, 11)}.

answer：Alright, so I've got this problem with three parts, all involving these functions f(x) and g(x). Let me try to unpack each part step by step. I'll start with part 1 since it seems foundational.**Problem 1: Find the extreme values of g(x).**Okay, g(x) is given as (e x)/e^x. Hmm, that simplifies to (e x)/e^x, which is the same as x e^{1 - x}, right? Because e divided by e^x is e^{1 - x}. So, g(x) = x e^{1 - x}.To find the extreme values, I need to find where the derivative g'(x) is zero or undefined. Since g(x) is defined for all real numbers, I just need to compute the derivative.Let me compute g'(x). Using the product rule: if g(x) = u(x) v(x), then g'(x) = u'(x) v(x) + u(x) v'(x). Here, u(x) = x and v(x) = e^{1 - x}.So, u'(x) = 1, and v'(x) = derivative of e^{1 - x} is -e^{1 - x} because the derivative of (1 - x) is -1.Putting it together:g'(x) = 1 * e^{1 - x} + x * (-e^{1 - x}) = e^{1 - x} - x e^{1 - x} = e^{1 - x}(1 - x).So, g'(x) = e^{1 - x}(1 - x). To find critical points, set g'(x) = 0.e^{1 - x} is never zero, so 1 - x = 0 => x = 1.So, the only critical point is at x = 1. Now, to determine if it's a maximum or minimum, I can look at the sign of g'(x) around x = 1.For x < 1, say x = 0.5: 1 - 0.5 = 0.5 > 0, so g'(x) > 0. So, the function is increasing before x = 1.For x > 1, say x = 2: 1 - 2 = -1 < 0, so g'(x) < 0. So, the function is decreasing after x = 1.Therefore, x = 1 is a maximum point. So, the maximum value of g(x) is g(1) = (e * 1)/e^1 = e/e = 1.As for minimum values, since g(x) approaches 0 as x approaches infinity (because e^{-x} dominates) and as x approaches negative infinity, g(x) also approaches 0 (since x e^{1 - x} tends to 0 as x becomes very negative). So, there's no global minimum, just approaching 0 asymptotically.So, for part 1, the extreme value is a maximum at x = 1 with g(1) = 1.**Problem 2: If m = 1, α < 0, and for any x₁, x₂ ∈ [3,4] (x₁ ≠ x₂), the inequality |f(x₂) - f(x₁)| < |1/g(x₂) - 1/g(x₁)| always holds, find the minimum value of a.**Wait, hold on. The problem mentions α < 0, but in the function f(x), it's written as f(x) = m x - α ln x - m. So, α is a parameter. But the question is asking for the minimum value of "a". Hmm, maybe that's a typo? It should probably be α, right? Because "a" isn't defined anywhere else. So, I think it's a typo, and it should be α.So, assuming that, we need to find the minimum value of α (which is negative) such that for any x₁, x₂ in [3,4], the inequality |f(x₂) - f(x₁)| < |1/g(x₂) - 1/g(x₁)| holds.Given that m = 1, so f(x) = x - α ln x - 1.First, let's analyze f(x) and 1/g(x).Compute 1/g(x): since g(x) = x e^{1 - x}, so 1/g(x) = e^{x - 1}/x.So, 1/g(x) = e^{x - 1}/x.Let me denote h(x) = 1/g(x) = e^{x - 1}/x.So, the inequality becomes |f(x₂) - f(x₁)| < |h(x₂) - h(x₁)| for any x₁, x₂ in [3,4].This looks like a condition on the derivatives of f and h. Specifically, for the inequality to hold for all x₁, x₂, the function f must be "less steep" than h. In other words, the derivative of f must be bounded by the derivative of h in absolute value.So, perhaps we can use the Mean Value Theorem. For any x₁, x₂, there exists a c between x₁ and x₂ such that |f'(c)| < |h'(c)|. So, if we can ensure that |f'(x)| < |h'(x)| for all x in [3,4], then the inequality will hold.So, let's compute f'(x) and h'(x).First, f(x) = x - α ln x - 1. So, f'(x) = 1 - α / x.Similarly, h(x) = e^{x - 1}/x. So, h'(x) = [e^{x - 1} * x - e^{x - 1} * 1] / x² = e^{x - 1}(x - 1)/x².So, h'(x) = e^{x - 1}(x - 1)/x².Since x is in [3,4], x - 1 is positive, so h'(x) is positive. Similarly, since α < 0, f'(x) = 1 - α / x. Since α is negative, -α / x is positive, so f'(x) is 1 + |α| / x, which is greater than 1.Wait, but the inequality is |f(x₂) - f(x₁)| < |h(x₂) - h(x₁)|. Since both f and h are increasing functions on [3,4], because f'(x) > 0 and h'(x) > 0, we can drop the absolute values:f(x₂) - f(x₁) < h(x₂) - h(x₁).Which implies that f(x₂) - h(x₂) < f(x₁) - h(x₁).So, the function u(x) = f(x) - h(x) must be decreasing on [3,4]. Therefore, u'(x) must be ≤ 0 for all x in [3,4].Compute u(x) = f(x) - h(x) = x - α ln x - 1 - e^{x - 1}/x.Then, u'(x) = f'(x) - h'(x) = [1 - α / x] - [e^{x - 1}(x - 1)/x²].So, u'(x) = 1 - α / x - e^{x - 1}(x - 1)/x².We need u'(x) ≤ 0 for all x in [3,4].So, 1 - α / x - e^{x - 1}(x - 1)/x² ≤ 0.Let me rearrange this:- α / x ≤ e^{x - 1}(x - 1)/x² - 1.Multiply both sides by x (which is positive, so inequality direction remains):- α ≤ e^{x - 1}(x - 1)/x - x.Multiply both sides by -1 (which reverses the inequality):α ≥ x - e^{x - 1}(x - 1)/x.So, α must be greater than or equal to x - e^{x - 1}(x - 1)/x for all x in [3,4].Therefore, the minimum value of α is the maximum of the function v(x) = x - e^{x - 1}(x - 1)/x over [3,4].So, we need to find the maximum of v(x) on [3,4].Compute v(x) = x - e^{x - 1}(x - 1)/x.Simplify:v(x) = x - e^{x - 1}(x - 1)/x.Let me compute v(x) at x = 3 and x = 4.First, at x = 3:v(3) = 3 - e^{2}(2)/3 = 3 - (2/3)e².Similarly, at x = 4:v(4) = 4 - e^{3}(3)/4 = 4 - (3/4)e³.Now, since e² ≈ 7.389 and e³ ≈ 20.085.Compute v(3):3 - (2/3)(7.389) ≈ 3 - (2/3)(7.389) ≈ 3 - 4.926 ≈ -1.926.Compute v(4):4 - (3/4)(20.085) ≈ 4 - (15.064) ≈ -11.064.So, v(3) ≈ -1.926 and v(4) ≈ -11.064.But we need to check if v(x) has a maximum somewhere inside [3,4]. So, let's compute the derivative of v(x) to see if it's increasing or decreasing.Compute v'(x):v(x) = x - e^{x - 1}(x - 1)/x.Let me denote the second term as w(x) = e^{x - 1}(x - 1)/x.So, v(x) = x - w(x).Compute w'(x):w(x) = e^{x - 1}(x - 1)/x.Use the product rule:w'(x) = derivative of e^{x - 1} times (x - 1)/x + e^{x - 1} times derivative of (x - 1)/x.Compute derivative of e^{x - 1}: that's e^{x - 1}.Derivative of (x - 1)/x: Let me compute that.Let me write (x - 1)/x = 1 - 1/x. So, derivative is 0 - (-1/x²) = 1/x².So, putting it together:w'(x) = e^{x - 1} * (x - 1)/x + e^{x - 1} * (1/x²).Factor out e^{x - 1}:w'(x) = e^{x - 1} [ (x - 1)/x + 1/x² ].Simplify inside the brackets:(x - 1)/x + 1/x² = (x(x - 1) + 1)/x² = (x² - x + 1)/x².So, w'(x) = e^{x - 1} (x² - x + 1)/x².Therefore, v'(x) = derivative of x - w'(x) = 1 - w'(x).So,v'(x) = 1 - e^{x - 1} (x² - x + 1)/x².We need to see if v'(x) is positive or negative in [3,4].Compute v'(x):v'(x) = 1 - e^{x - 1} (x² - x + 1)/x².Let me compute this at x = 3:v'(3) = 1 - e^{2} (9 - 3 + 1)/9 = 1 - e² (7)/9 ≈ 1 - (7/9)(7.389) ≈ 1 - 5.825 ≈ -4.825.Similarly, at x = 4:v'(4) = 1 - e^{3} (16 - 4 + 1)/16 = 1 - e³ (13)/16 ≈ 1 - (13/16)(20.085) ≈ 1 - 16.655 ≈ -15.655.So, v'(x) is negative at both ends, and since v'(x) is decreasing (because e^{x - 1} is increasing and the other terms are positive), v'(x) is negative throughout [3,4]. Therefore, v(x) is decreasing on [3,4].Therefore, the maximum of v(x) on [3,4] is at x = 3, which is approximately -1.926.But since α must be greater than or equal to v(x) for all x in [3,4], the minimum value of α is the maximum of v(x), which is at x = 3: 3 - (2/3)e².So, α must be ≥ 3 - (2/3)e². Since α is negative, this is the minimum value of α.Therefore, the minimum value of α is 3 - (2/3)e².**Problem 3: If α = 2, and for any given x₀ ∈ (0,e], there always exist t₁, t₂ (t₁ ≠ t₂) in the interval (0,e] such that f(t₁) = f(t₂) = g(x₀) holds, find the range of m.**Okay, so α = 2, so f(x) = m x - 2 ln x - m.We need that for any x₀ in (0,e], the equation f(t) = g(x₀) has at least two distinct solutions t₁, t₂ in (0,e].So, essentially, the function f(t) must take every value in the range of g(x₀) at least twice in (0,e].First, let's understand the range of g(x₀). From part 1, we know that g(x) = x e^{1 - x}, which has a maximum at x = 1, g(1) = 1, and it approaches 0 as x approaches 0 or infinity. But since x₀ ∈ (0,e], the range of g(x₀) is (0,1].So, for any y ∈ (0,1], the equation f(t) = y must have at least two distinct solutions t₁, t₂ in (0,e].Therefore, f(t) must be such that it is not monotonic on (0,e], and it must attain every value in (0,1] at least twice.So, first, let's analyze f(t) = m t - 2 ln t - m.Compute the derivative f'(t) = m - 2 / t.Set f'(t) = 0: m - 2 / t = 0 => t = 2 / m.So, f(t) has a critical point at t = 2/m.For f(t) to have a critical point in (0,e], we need 2/m ∈ (0,e], which implies m > 2/e.If m ≤ 2/e, then f'(t) = m - 2/t is always negative on (0,e], because m ≤ 2/e and t ≤ e, so 2/t ≥ 2/e ≥ m, so f'(t) ≤ 0. Therefore, f(t) is decreasing on (0,e], which would mean it can't take any value more than once. So, m must be > 2/e.But the problem states that for any x₀ ∈ (0,e], there exist t₁, t₂ such that f(t₁) = f(t₂) = g(x₀). So, f(t) must be surjective onto (0,1] and must have at least two pre-images for each y ∈ (0,1].Therefore, f(t) must have a minimum value less than or equal to 0 and a maximum value greater than or equal to 1, but wait, actually, since g(x₀) ∈ (0,1], f(t) needs to cover (0,1] with at least two t's for each y.Wait, but f(t) is defined as m t - 2 ln t - m.Let me compute the limits of f(t) as t approaches 0+ and t approaches e.As t approaches 0+, ln t approaches -infty, so -2 ln t approaches +infty. Therefore, f(t) approaches +infty.As t approaches e, f(e) = m e - 2 ln e - m = m e - 2 - m = m(e - 1) - 2.So, f(e) = m(e - 1) - 2.We need f(t) to attain every value in (0,1] at least twice. So, the minimum value of f(t) must be less than or equal to 0, and the maximum value must be at least 1.Wait, but as t approaches 0+, f(t) approaches +infty, and as t approaches e, f(e) = m(e - 1) - 2.We need f(t) to have a minimum somewhere in (0,e]. The critical point is at t = 2/m. So, f(t) has a minimum at t = 2/m.Compute f(2/m):f(2/m) = m*(2/m) - 2 ln(2/m) - m = 2 - 2 ln(2/m) - m.Simplify:f(2/m) = 2 - m - 2 ln(2/m).We need this minimum value to be ≤ 0 because we need f(t) to attain values down to 0.So, 2 - m - 2 ln(2/m) ≤ 0.Let me write this as:2 - m ≤ 2 ln(2/m).Divide both sides by 2:1 - m/2 ≤ ln(2/m).Let me denote k = m/2, so m = 2k.Then, 1 - k ≤ ln(2/(2k)) = ln(1/k).So, 1 - k ≤ -ln k.Multiply both sides by -1 (inequality reverses):k - 1 ≥ ln k.So, k - 1 - ln k ≥ 0.Let me define a function h(k) = k - 1 - ln k.We need h(k) ≥ 0.Compute h(k):h(k) = k - 1 - ln k.Compute h(1): 1 - 1 - 0 = 0.Compute derivative h'(k) = 1 - 1/k.Set h'(k) = 0: 1 - 1/k = 0 => k = 1.So, h(k) has a critical point at k = 1.For k < 1, h'(k) = 1 - 1/k < 0, so h(k) is decreasing.For k > 1, h'(k) = 1 - 1/k > 0, so h(k) is increasing.At k = 1, h(k) = 0.So, h(k) ≥ 0 for all k > 0, because it's minimized at k = 1 with h(1) = 0.Wait, but that can't be right because h(k) = k - 1 - ln k.Wait, let me check h(k) at k approaching 0+:As k approaches 0+, ln k approaches -infty, so h(k) approaches 0 - 1 - (-infty) = +infty.At k = 1, h(k) = 0.For k > 1, h(k) increases from 0 to +infty.For k < 1, h(k) decreases from +infty to 0.So, h(k) ≥ 0 for all k > 0.Therefore, the inequality k - 1 - ln k ≥ 0 is always true, which means our earlier inequality 1 - m/2 ≤ ln(2/m) is always true for m > 0.Wait, but that can't be, because if m is too large, f(e) = m(e - 1) - 2 might be too large, but we need f(t) to attain values in (0,1].Wait, perhaps I need to ensure that the maximum value of f(t) is at least 1.Wait, f(t) approaches +infty as t approaches 0+, so it definitely exceeds 1. But we need f(t) to attain every value in (0,1] at least twice.So, the minimum value of f(t) is f(2/m) = 2 - m - 2 ln(2/m). We need this minimum to be ≤ 0, which we already have because h(k) ≥ 0.But also, we need that f(t) can attain up to 1. Wait, actually, since f(t) approaches +infty as t approaches 0, and f(t) has a minimum at t = 2/m, which is ≤ 0, and then increases to f(e) = m(e - 1) - 2.So, to ensure that f(t) can attain all values in (0,1], we need that the minimum value is ≤ 0 and that f(e) ≥ 1.Because if f(e) < 1, then f(t) would only attain up to f(e), which is less than 1, so it couldn't cover the entire (0,1].Therefore, we need f(e) ≥ 1.Compute f(e):f(e) = m(e - 1) - 2 ≥ 1.So,m(e - 1) - 2 ≥ 1 => m(e - 1) ≥ 3 => m ≥ 3 / (e - 1).So, m must be ≥ 3 / (e - 1).Additionally, we need that the function f(t) is not monotonic, which requires m > 2/e, as we saw earlier.But 3/(e - 1) is approximately 3/(1.718) ≈ 1.746, while 2/e ≈ 0.735. So, 3/(e - 1) is greater than 2/e, so the stricter condition is m ≥ 3/(e - 1).Therefore, the range of m is [3/(e - 1), ∞).Let me double-check:If m = 3/(e - 1), then f(e) = 3/(e - 1)*(e - 1) - 2 = 3 - 2 = 1.So, f(e) = 1.And f(2/m) = 2 - m - 2 ln(2/m).Plug m = 3/(e - 1):f(2/m) = 2 - 3/(e - 1) - 2 ln(2/(3/(e - 1))) = 2 - 3/(e - 1) - 2 ln(2(e - 1)/3).We need this to be ≤ 0.Compute numerically:e ≈ 2.718, so e - 1 ≈ 1.718.So, 3/(e - 1) ≈ 1.746.So, f(2/m) ≈ 2 - 1.746 - 2 ln(2*1.718/3) ≈ 0.254 - 2 ln(3.436/3) ≈ 0.254 - 2 ln(1.145) ≈ 0.254 - 2*(0.135) ≈ 0.254 - 0.27 ≈ -0.016.So, it's approximately -0.016, which is ≤ 0. So, it works.If m increases beyond 3/(e - 1), f(e) increases beyond 1, which is fine because we still need to cover (0,1], and the minimum remains ≤ 0.Therefore, the range of m is [3/(e - 1), ∞).