answer：To determine the upper bound ( U ) of the range of the expression ( 4n + 7 ) where there are 25 integer values of ( n ) such that ( 4n + 7 ) is greater than 1 and less than ( U ), follow these steps:1. **Identify the smallest ( n ) such that ( 4n + 7 > 1 ):** [ 4n + 7 > 1 implies 4n > -6 implies n > -frac{3}{2} ] Since ( n ) must be an integer, the smallest value of ( n ) is ( n = -1 ).2. **Determine the 25th value of ( n ):** Starting from ( n = -1 ), the 25th integer value of ( n ) is ( n = 23 ).3. **Calculate the corresponding value of ( 4n + 7 ) for ( n = 23 ):** [ 4(23) + 7 = 92 + 7 = 99 ]4. **Find the upper bound ( U ):** The upper bound ( U ) is the next integer after the 25th value of ( 4n + 7 ), which is ( 99 + 1 = 100 ).Therefore, the upper bound of the range of the expression is (boxed{100}).

answer：Alright, so I have this function f(x) = ln x + (1/2)x² - a x + a, and I need to solve two parts. Let me start with part (1).**Problem (1):** If f(x) is monotonically increasing on (0, +∞), find the range of the real number a.Okay, so for a function to be monotonically increasing, its derivative should be non-negative for all x in the interval. So, first, I need to find the derivative of f(x).f'(x) = d/dx [ln x + (1/2)x² - a x + a]Let me compute that term by term:- The derivative of ln x is 1/x.- The derivative of (1/2)x² is x.- The derivative of -a x is -a.- The derivative of a constant a is 0.So, putting it all together:f'(x) = 1/x + x - aSimplify that:f'(x) = (1 + x²)/x - aWait, actually, that's not the most straightforward way. Let me just write it as:f'(x) = (1/x) + x - aSo, for f(x) to be monotonically increasing on (0, +∞), we need f'(x) ≥ 0 for all x > 0.So, (1/x) + x - a ≥ 0 for all x > 0.Which can be rewritten as:a ≤ (1/x) + x for all x > 0.So, a has to be less than or equal to the minimum value of the function g(x) = x + 1/x on (0, +∞).Hmm, I remember that the function g(x) = x + 1/x has a minimum at x = 1, and the minimum value is 2. Let me verify that.Compute the derivative of g(x):g'(x) = 1 - 1/x²Set g'(x) = 0:1 - 1/x² = 0 => 1 = 1/x² => x² = 1 => x = 1 (since x > 0)So, at x = 1, g(x) = 1 + 1/1 = 2.And since the second derivative g''(x) = 2/x³, which is positive for x > 0, so it's a minimum.Therefore, the minimum value of g(x) is 2, so a must be ≤ 2.So, the range of a is (-∞, 2].Wait, but let me double-check. If a is less than or equal to 2, then (1/x) + x - a is always non-negative because (1/x) + x is always at least 2, so subtracting a number less than or equal to 2 will keep it non-negative.Yes, that makes sense.So, for part (1), the range of a is all real numbers less than or equal to 2.**Problem (2):** If f(x) has extreme values at x = x₁ and x = x₂, where x₂ ≥ e x₁ (e is the base of the natural logarithm), find the maximum value of f(x₂) - f(x₁).Alright, so f(x) has two critical points, x₁ and x₂, which are the solutions to f'(x) = 0.From part (1), we know that f'(x) = (1/x) + x - a.So, setting f'(x) = 0:(1/x) + x - a = 0Multiply both sides by x to eliminate the denominator:1 + x² - a x = 0So, x² - a x + 1 = 0Thus, the critical points are the roots of the quadratic equation x² - a x + 1 = 0.Let me denote the roots as x₁ and x₂. So, from quadratic equation theory:x₁ + x₂ = ax₁ x₂ = 1So, that's two equations.Given that x₂ ≥ e x₁, and since x₁ and x₂ are positive (because the domain is x > 0), we can consider t = x₂ / x₁ ≥ e.Let me set t = x₂ / x₁, so t ≥ e.Given that x₁ x₂ = 1, and x₂ = t x₁, so substituting:x₁ * (t x₁) = 1 => t x₁² = 1 => x₁² = 1/t => x₁ = 1/√tSimilarly, x₂ = t x₁ = t / √t = √tSo, x₁ = 1/√t and x₂ = √t.Also, from x₁ + x₂ = a:a = x₁ + x₂ = 1/√t + √tSo, a = √t + 1/√tNow, we need to compute f(x₂) - f(x₁).Let me write f(x) again:f(x) = ln x + (1/2)x² - a x + aSo, f(x₂) - f(x₁) = [ln x₂ + (1/2)x₂² - a x₂ + a] - [ln x₁ + (1/2)x₁² - a x₁ + a]Simplify this:= ln(x₂ / x₁) + (1/2)(x₂² - x₁²) - a(x₂ - x₁) + (a - a)= ln(x₂ / x₁) + (1/2)(x₂² - x₁²) - a(x₂ - x₁)Now, let's express this in terms of t.We have x₂ = √t and x₁ = 1/√t.So, x₂ / x₁ = √t / (1/√t) = tThus, ln(x₂ / x₁) = ln tNext, compute x₂² - x₁²:x₂² = (√t)² = tx₁² = (1/√t)² = 1/tSo, x₂² - x₁² = t - 1/tThus, (1/2)(x₂² - x₁²) = (1/2)(t - 1/t)Now, compute a(x₂ - x₁):We have a = √t + 1/√tx₂ - x₁ = √t - 1/√tSo, a(x₂ - x₁) = (√t + 1/√t)(√t - 1/√t) = (√t)^2 - (1/√t)^2 = t - 1/tTherefore, putting it all together:f(x₂) - f(x₁) = ln t + (1/2)(t - 1/t) - (t - 1/t)Simplify:= ln t + (1/2)(t - 1/t) - t + 1/t= ln t - (1/2)(t - 1/t)Wait, let me compute step by step:First, (1/2)(t - 1/t) - (t - 1/t) = (1/2)(t - 1/t) - (t - 1/t) = (-1/2)(t - 1/t)So, f(x₂) - f(x₁) = ln t - (1/2)(t - 1/t)Alternatively, that's ln t - (1/2)t + (1)/(2t)So, f(x₂) - f(x₁) = ln t - (t)/2 + 1/(2t)Let me denote this as h(t) = ln t - (t)/2 + 1/(2t)We need to find the maximum value of h(t) for t ≥ e.So, to find the maximum of h(t) on [e, ∞), we can take the derivative of h(t) and analyze its behavior.Compute h'(t):h'(t) = d/dt [ln t - (t)/2 + 1/(2t)]= (1/t) - (1)/2 - (1)/(2t²)Simplify:= (1/t) - 1/2 - 1/(2t²)Let me write it as:h'(t) = (2t - t² - 1) / (2t²)Wait, let me compute step by step:First term: derivative of ln t is 1/t.Second term: derivative of -(t)/2 is -1/2.Third term: derivative of 1/(2t) is -1/(2t²).So, altogether:h'(t) = 1/t - 1/2 - 1/(2t²)To combine these terms, let me get a common denominator, which would be 2t².So:= (2t)/ (2t²) - (t²)/(2t²) - 1/(2t²)= [2t - t² - 1] / (2t²)So, h'(t) = (-t² + 2t - 1) / (2t²)Factor the numerator:-t² + 2t - 1 = -(t² - 2t + 1) = -(t - 1)^2So, h'(t) = -(t - 1)^2 / (2t²)Since t > 0, the denominator is always positive. The numerator is -(t - 1)^2, which is always non-positive because (t - 1)^2 is non-negative, and we have a negative sign.Therefore, h'(t) ≤ 0 for all t > 0.Moreover, h'(t) = 0 only when (t - 1)^2 = 0, i.e., t = 1.But in our case, t ≥ e, which is approximately 2.718, so t is always greater than 1.Therefore, h'(t) is negative for all t ≥ e, meaning that h(t) is strictly decreasing on [e, ∞).Thus, the maximum value of h(t) on [e, ∞) occurs at t = e.Therefore, the maximum value of f(x₂) - f(x₁) is h(e).Compute h(e):h(e) = ln e - (e)/2 + 1/(2e)We know that ln e = 1, so:h(e) = 1 - e/2 + 1/(2e)Simplify:= 1 - (e)/2 + 1/(2e)We can write this as:= 1 - (e)/2 + (1)/(2e)Alternatively, factor out 1/2:= 1 - (1/2)(e - 1/e)But perhaps it's better to leave it as is.So, the maximum value is 1 - e/2 + 1/(2e).Let me compute this numerically to check:e ≈ 2.71828So, e/2 ≈ 1.359141/(2e) ≈ 0.18394Thus, 1 - 1.35914 + 0.18394 ≈ 1 - 1.35914 + 0.18394 ≈ 1 - 1.1752 ≈ -0.1752Wait, that's negative. But f(x₂) - f(x₁) is the difference between two function values. Since x₂ is a local maximum or minimum?Wait, hold on. Let me think.Wait, f'(x) = 0 at x₁ and x₂. Since f'(x) is a quadratic in x, and the quadratic opens upwards (coefficient of x² is positive), so the function f(x) has a local minimum at x = x₁ and a local maximum at x = x₂? Or vice versa?Wait, let me think about the derivative.f'(x) = (1/x) + x - aAs x approaches 0 from the right, f'(x) approaches +∞ because 1/x dominates.As x approaches +∞, f'(x) behaves like x, which goes to +∞.So, the derivative starts at +∞, decreases to a minimum, then increases back to +∞.Therefore, the critical points x₁ and x₂ are such that x₁ is a local minimum and x₂ is a local maximum.Wait, no, actually, if f'(x) goes from +∞ to some minimum and back to +∞, then the critical points are a local minimum and a local maximum.Wait, actually, when f'(x) decreases, crosses zero from above, reaches a minimum, then increases back, crossing zero from below. So, the first critical point is a local maximum, and the second is a local minimum.Wait, let me test with a specific a.Suppose a = 3, which is greater than 2. Then f'(x) = (1/x) + x - 3.Set to zero: x + 1/x = 3.Multiply by x: x² - 3x + 1 = 0.Solutions: x = [3 ± sqrt(9 - 4)] / 2 = [3 ± sqrt(5)] / 2 ≈ (3 ± 2.236)/2 ≈ 2.618 and 0.382.So, x₁ ≈ 0.382, x₂ ≈ 2.618.Compute f'(x) around x₁: For x < x₁, say x = 0.1, f'(0.1) = 10 + 0.1 - 3 = 7.1 > 0.At x = x₁, f'(x₁) = 0.For x between x₁ and x₂, say x = 1, f'(1) = 1 + 1 - 3 = -1 < 0.For x > x₂, say x = 3, f'(3) = 1/3 + 3 - 3 = 1/3 > 0.So, f'(x) is positive before x₁, negative between x₁ and x₂, and positive after x₂.Therefore, f(x) is increasing on (0, x₁), decreasing on (x₁, x₂), and increasing on (x₂, ∞).Therefore, x₁ is a local maximum, and x₂ is a local minimum.Wait, that contradicts my earlier thought. So, x₁ is a local maximum, x₂ is a local minimum.But in the problem statement, it says "extreme values at x = x₁ and x = x₂". So, both are extreme points, one maximum, one minimum.Therefore, f(x₂) - f(x₁) is the difference between a local minimum and a local maximum, so it's negative.But in our computation, h(e) ≈ -0.1752, which is negative.But the problem asks for the maximum value of f(x₂) - f(x₁). So, the maximum value would be the least negative value, which occurs at the smallest t, which is t = e.Wait, since h(t) is decreasing, its maximum on [e, ∞) is at t = e, which is the highest point.So, even though it's negative, it's the maximum (least negative) value.So, the maximum value is 1 - e/2 + 1/(2e).Let me just check if this is correct.Alternatively, perhaps I made a mistake in the sign when computing f(x₂) - f(x₁).Wait, let me re-examine the computation.f(x₂) - f(x₁) = [ln x₂ + (1/2)x₂² - a x₂ + a] - [ln x₁ + (1/2)x₁² - a x₁ + a]= ln(x₂ / x₁) + (1/2)(x₂² - x₁²) - a(x₂ - x₁)But since x₂ is a local minimum and x₁ is a local maximum, f(x₂) < f(x₁), so f(x₂) - f(x₁) is negative.But the problem asks for the maximum value of f(x₂) - f(x₁), so the maximum (i.e., the least negative) occurs when t is smallest, which is t = e.So, yes, h(t) is decreasing, so h(e) is the maximum.Therefore, the maximum value is 1 - e/2 + 1/(2e).Let me write that as 1 - (e)/2 + 1/(2e).Alternatively, factor out 1/2:= 1 - (e - 1/e)/2But perhaps it's better to leave it as is.So, the maximum value is 1 - e/2 + 1/(2e).I think that's the answer.**Final Answer**(1) boxed{(-infty, 2]}(2) boxed{1 - frac{e}{2} + frac{1}{2e}}

answer：Okay, so I have this problem about two villages, Upper Vasyuki and Lower Vasyuki, which are on a riverbank. There's a steamboat and a boat that travel between these two villages. The steamboat takes one hour to go from Upper to Lower, and the boat takes 45 minutes. It's also given that the speed of the boat in still water is twice the speed of the steamboat in still water. I need to find out how long it would take a raft to float downstream from Upper Vasyuki to Lower Vasyuki.Alright, first, I need to understand what's given and what I need to find. So, we have two vessels: a steamboat and a boat. The steamboat takes 1 hour, and the boat takes 45 minutes to cover the same distance. The boat's speed in still water is twice that of the steamboat's speed in still water. I need to find the time it takes for a raft to float downstream. Since a raft just floats with the current, its speed would be equal to the speed of the river current.Let me denote some variables to make this clearer. Let's say:- Let ( D ) be the distance between Upper Vasyuki and Lower Vasyuki.- Let ( v_s ) be the speed of the steamboat in still water.- Let ( v_b ) be the speed of the boat in still water.- Let ( v_c ) be the speed of the river current.Given that the boat's speed in still water is twice that of the steamboat, so ( v_b = 2v_s ).Now, when the steamboat is going downstream, its effective speed would be ( v_s + v_c ) because the current aids its movement. Similarly, when the boat is going downstream, its effective speed would be ( v_b + v_c ).We know the time taken by the steamboat is 1 hour, so:[ D = (v_s + v_c) times 1 ][ D = v_s + v_c ]Similarly, for the boat, the time taken is 45 minutes, which is ( frac{3}{4} ) of an hour. So:[ D = (v_b + v_c) times frac{3}{4} ][ D = (2v_s + v_c) times frac{3}{4} ]Since both expressions equal ( D ), I can set them equal to each other:[ v_s + v_c = (2v_s + v_c) times frac{3}{4} ]Now, I need to solve for ( v_s ) and ( v_c ). Let me write that equation again:[ v_s + v_c = frac{3}{4}(2v_s + v_c) ]Let me expand the right side:[ v_s + v_c = frac{3}{4} times 2v_s + frac{3}{4} times v_c ][ v_s + v_c = frac{3}{2}v_s + frac{3}{4}v_c ]Now, let's bring all terms to one side to solve for ( v_s ) and ( v_c ):[ v_s + v_c - frac{3}{2}v_s - frac{3}{4}v_c = 0 ]Combine like terms:[ (v_s - frac{3}{2}v_s) + (v_c - frac{3}{4}v_c) = 0 ][ (-frac{1}{2}v_s) + (frac{1}{4}v_c) = 0 ]So,[ -frac{1}{2}v_s + frac{1}{4}v_c = 0 ]Let me solve for ( v_c ):[ frac{1}{4}v_c = frac{1}{2}v_s ][ v_c = 2 times frac{1}{2}v_s ][ v_c = v_s ]Wait, that's interesting. So the speed of the current is equal to the speed of the steamboat in still water. Let me check my calculations to make sure I didn't make a mistake.Starting from:[ v_s + v_c = frac{3}{4}(2v_s + v_c) ]Expanding the right side:[ v_s + v_c = frac{3}{4} times 2v_s + frac{3}{4} times v_c ][ v_s + v_c = frac{3}{2}v_s + frac{3}{4}v_c ]Subtracting ( frac{3}{2}v_s + frac{3}{4}v_c ) from both sides:[ v_s + v_c - frac{3}{2}v_s - frac{3}{4}v_c = 0 ][ (1 - frac{3}{2})v_s + (1 - frac{3}{4})v_c = 0 ][ (-frac{1}{2})v_s + (frac{1}{4})v_c = 0 ]So,[ -frac{1}{2}v_s + frac{1}{4}v_c = 0 ]Multiplying both sides by 4 to eliminate denominators:[ -2v_s + v_c = 0 ][ v_c = 2v_s ]Ah, I see. I made a mistake in my earlier calculation when solving for ( v_c ). It should be ( v_c = 2v_s ), not ( v_c = v_s ). That makes more sense because the current is aiding the boats, so it's reasonable that the current's speed is significant.So, ( v_c = 2v_s ).Now, going back to the first equation:[ D = v_s + v_c ][ D = v_s + 2v_s ][ D = 3v_s ]So, the distance ( D ) is three times the speed of the steamboat in still water.Now, I need to find the time it takes for a raft to float downstream. A raft moves at the speed of the current, which is ( v_c = 2v_s ).So, the time ( t ) taken by the raft is:[ t = frac{D}{v_c} ][ t = frac{3v_s}{2v_s} ][ t = frac{3}{2} text{ hours} ][ t = 1.5 text{ hours} ][ t = 90 text{ minutes} ]So, it would take the raft 90 minutes to float downstream from Upper Vasyuki to Lower Vasyuki.Let me just verify this result to make sure it makes sense. If the raft is moving at the speed of the current, and the current is twice the speed of the steamboat, then the raft should be moving faster than the steamboat. The steamboat takes 1 hour, so the raft should take less time, which aligns with 90 minutes being less than an hour? Wait, no, 90 minutes is more than an hour. That doesn't make sense because the raft should be moving faster, so it should take less time, not more.Wait, I think I have a confusion here. If the raft is moving at the speed of the current, and the current is aiding the movement, then the raft should be moving faster than the steamboat, which is also moving downstream. But according to my calculation, the raft is moving at ( v_c = 2v_s ), and the steamboat is moving at ( v_s + v_c = 3v_s ). So, the steamboat is actually moving faster than the raft, which contradicts the initial assumption that the raft should be moving faster.Wait, no, the steamboat is a powered vessel, so its speed in still water is ( v_s ), but when going downstream, it's ( v_s + v_c ). The raft is just moving at ( v_c ). So, if ( v_c = 2v_s ), then the steamboat's downstream speed is ( v_s + 2v_s = 3v_s ), and the raft's speed is ( 2v_s ). So, the steamboat is indeed faster than the raft, which makes sense because it's a powered vessel. The raft is slower than the steamboat, so it takes longer to cover the same distance.But in the problem, the boat (which is the powered vessel) takes 45 minutes, which is less than the steamboat's 1 hour. Wait, that doesn't make sense either because the boat is supposed to be faster. Maybe I have misread the problem.Wait, the problem says the steamboat takes 1 hour, and the boat takes 45 minutes. So, the boat is faster than the steamboat. But according to my variables, the boat's speed in still water is twice that of the steamboat's. So, maybe my earlier calculations are correct, and the raft takes 90 minutes, which is longer than both the steamboat and the boat, which makes sense because the raft is slower than both.Yes, that makes sense. The boat is the fastest, taking 45 minutes, then the steamboat takes 1 hour, and the raft takes the longest time, 90 minutes. So, my final answer should be 90 minutes.

answer：Okay, so I have this equation here: 3x plus y equals 1. The problem is asking me to express y in terms of x. Hmm, I think that means I need to solve for y, right? Let me see. Alright, starting with the equation: 3x + y = 1. I need to get y by itself on one side of the equation. So, how do I do that? Well, if I subtract 3x from both sides, that should help me isolate y. Let me try that.So, subtracting 3x from both sides gives me: 3x + y - 3x = 1 - 3x. Simplifying the left side, 3x minus 3x cancels out, leaving just y. So now I have y equals 1 minus 3x. Wait, is that right? Let me double-check. If I start with 3x + y = 1 and subtract 3x from both sides, yes, that should give me y = 1 - 3x. But sometimes, people prefer writing the equation with the x term first. So maybe I should write it as y = -3x + 1. Is that the same thing? Yeah, because 1 - 3x is the same as -3x + 1. Let me think if there's another way to approach this. Maybe I could also add something to both sides instead of subtracting. But no, since I want to get rid of the 3x term on the left side, subtracting is the right move. Is there a way to check if this is correct? Maybe by plugging values back into the original equation. Let's say x is 0. Then y should be 1 - 3(0) = 1. Plugging into the original equation: 3(0) + 1 = 1, which works. What if x is 1? Then y would be 1 - 3(1) = -2. Plugging back in: 3(1) + (-2) = 3 - 2 = 1, which also works. Okay, so it seems like my solution is correct. Expressing y in terms of x gives me y = -3x + 1. That makes sense because for every increase in x by 1, y decreases by 3, which matches the slope-intercept form of a linear equation. I think I've got it. So, the final answer should be y equals negative three x plus one. Yeah, that sounds right.

answer：Okay, so I have this equation: x squared plus y squared plus 6x equals 2y plus 10. Hmm, I need to find the area of the region defined by this equation. I remember that equations involving x squared and y squared usually define circles, ellipses, hyperbolas, or parabolas. Since both x and y are squared and have the same sign, it might be a circle or an ellipse.Let me try to rearrange the equation to see if I can get it into a more familiar form. So, starting with the given equation:x² + y² + 6x = 2y + 10I think I should move all the terms to one side to see if I can complete the square for both x and y. Let me subtract 2y and 10 from both sides to get:x² + 6x + y² - 2y = 10Okay, now I have x² + 6x and y² - 2y on the left side. I remember that completing the square involves creating perfect square trinomials. For the x terms, I have x² + 6x. To complete the square, I take half of the coefficient of x, which is 6, so half of that is 3, and then square it, which gives 9. So, I can write x² + 6x as (x + 3)² minus 9 because (x + 3)² = x² + 6x + 9, so to balance it, I need to subtract 9.Similarly, for the y terms, I have y² - 2y. Half of -2 is -1, and squaring that gives 1. So, I can write y² - 2y as (y - 1)² minus 1 because (y - 1)² = y² - 2y + 1, so I need to subtract 1 to keep the equation balanced.Let me substitute these back into the equation:(x + 3)² - 9 + (y - 1)² - 1 = 10Now, combine the constants on the left side:(x + 3)² + (y - 1)² - 10 = 10Wait, that doesn't seem right. Let me check my steps again. I had:x² + 6x + y² - 2y = 10After completing the square:(x + 3)² - 9 + (y - 1)² - 1 = 10So, combining the constants:-9 -1 = -10So, the equation becomes:(x + 3)² + (y - 1)² - 10 = 10Now, add 10 to both sides to isolate the squared terms:(x + 3)² + (y - 1)² = 20Ah, okay, that looks better. So, now the equation is in the standard form of a circle:(x - h)² + (y - k)² = r²Where (h, k) is the center and r is the radius. Comparing, I have:h = -3, k = 1, and r² = 20So, the radius r is the square root of 20. To find the area of the circle, I use the formula A = πr². Since r² is 20, the area is 20π.Wait, let me make sure I didn't make any mistakes in completing the square. For the x terms: x² + 6x, half of 6 is 3, squared is 9, so (x + 3)² - 9. For the y terms: y² - 2y, half of -2 is -1, squared is 1, so (y - 1)² - 1. Adding those together:(x + 3)² - 9 + (y - 1)² - 1 = 10Which simplifies to:(x + 3)² + (y - 1)² - 10 = 10Then adding 10 to both sides:(x + 3)² + (y - 1)² = 20Yes, that seems correct. So, the radius squared is 20, so the area is 20π.I think that's it. The region defined by the equation is a circle with radius squared equal to 20, so the area is 20π.

answer：Alright, so I have this problem about curvature between two points on the curve ( f(x) = x^3 + 2 ). The curvature is defined as ( varphi(M,N) = frac{|k_M - k_N|}{|MN|} ), where ( k_M ) and ( k_N ) are the slopes of the tangents at points ( M ) and ( N ) respectively, and ( |MN| ) is the distance between these two points. Additionally, it's given that ( x_1 x_2 = 1 ), where ( x_1 ) and ( x_2 ) are the x-coordinates of points ( M ) and ( N ).First, I need to understand what exactly is being asked. The problem is asking for the range of ( varphi(M,N) ) given the condition ( x_1 x_2 = 1 ). So, I need to express ( varphi(M,N) ) in terms of ( x_1 ) and ( x_2 ), and then find its possible values given that ( x_1 x_2 = 1 ).Let me start by recalling that the slope of the tangent to the curve at any point ( x ) is given by the derivative of ( f(x) ). So, let's compute the derivative:( f'(x) = 3x^2 ).Therefore, the slopes at points ( M ) and ( N ) are ( k_M = 3x_1^2 ) and ( k_N = 3x_2^2 ) respectively.Next, I need to compute the absolute difference between these slopes:( |k_M - k_N| = |3x_1^2 - 3x_2^2| = 3|x_1^2 - x_2^2| ).Now, I need to find the distance between points ( M ) and ( N ). The coordinates of ( M ) are ( (x_1, f(x_1)) = (x_1, x_1^3 + 2) ) and similarly for ( N ): ( (x_2, x_2^3 + 2) ).The distance ( |MN| ) can be calculated using the distance formula:( |MN| = sqrt{(x_2 - x_1)^2 + (f(x_2) - f(x_1))^2} ).Let's compute ( f(x_2) - f(x_1) ):( f(x_2) - f(x_1) = x_2^3 + 2 - (x_1^3 + 2) = x_2^3 - x_1^3 ).So, the distance becomes:( |MN| = sqrt{(x_2 - x_1)^2 + (x_2^3 - x_1^3)^2} ).Now, let's factor ( x_2^3 - x_1^3 ). I remember that ( a^3 - b^3 = (a - b)(a^2 + ab + b^2) ), so:( x_2^3 - x_1^3 = (x_2 - x_1)(x_2^2 + x_1 x_2 + x_1^2) ).Therefore, the distance can be rewritten as:( |MN| = sqrt{(x_2 - x_1)^2 + [(x_2 - x_1)(x_2^2 + x_1 x_2 + x_1^2)]^2} ).Factor out ( (x_2 - x_1)^2 ) from both terms inside the square root:( |MN| = sqrt{(x_2 - x_1)^2 [1 + (x_2^2 + x_1 x_2 + x_1^2)^2]} ).Take ( |x_2 - x_1| ) out of the square root:( |MN| = |x_2 - x_1| sqrt{1 + (x_2^2 + x_1 x_2 + x_1^2)^2} ).Since ( |x_2 - x_1| ) is positive (as ( x_1 ) and ( x_2 ) are different points), we can write:( |MN| = |x_2 - x_1| sqrt{1 + (x_2^2 + x_1 x_2 + x_1^2)^2} ).Now, going back to the curvature ( varphi(M,N) ):( varphi(M,N) = frac{3|x_1^2 - x_2^2|}{|x_2 - x_1| sqrt{1 + (x_2^2 + x_1 x_2 + x_1^2)^2}} ).Simplify the numerator:( |x_1^2 - x_2^2| = |(x_1 - x_2)(x_1 + x_2)| = |x_1 - x_2||x_1 + x_2| ).So, substituting back:( varphi(M,N) = frac{3|x_1 - x_2||x_1 + x_2|}{|x_2 - x_1| sqrt{1 + (x_2^2 + x_1 x_2 + x_1^2)^2}} ).Notice that ( |x_1 - x_2| = |x_2 - x_1| ), so they cancel out:( varphi(M,N) = frac{3|x_1 + x_2|}{sqrt{1 + (x_2^2 + x_1 x_2 + x_1^2)^2}} ).Now, we have the expression for ( varphi(M,N) ) in terms of ( x_1 ) and ( x_2 ). But we know that ( x_1 x_2 = 1 ). Let's denote ( x_1 x_2 = 1 ), so ( x_2 = frac{1}{x_1} ).Let me substitute ( x_2 = frac{1}{x_1} ) into the expression. Let's denote ( x = x_1 ), so ( x_2 = frac{1}{x} ).Thus, ( x_1 + x_2 = x + frac{1}{x} ), and ( x_1^2 + x_1 x_2 + x_2^2 = x^2 + x cdot frac{1}{x} + left(frac{1}{x}right)^2 = x^2 + 1 + frac{1}{x^2} ).So, substituting these into ( varphi(M,N) ):( varphi(M,N) = frac{3|x + frac{1}{x}|}{sqrt{1 + left(x^2 + 1 + frac{1}{x^2}right)^2}} ).Simplify the denominator:First, compute ( left(x^2 + 1 + frac{1}{x^2}right)^2 ):Let me denote ( A = x^2 + frac{1}{x^2} ), so ( A + 1 = x^2 + 1 + frac{1}{x^2} ).Then, ( (A + 1)^2 = A^2 + 2A + 1 ).Compute ( A^2 = left(x^2 + frac{1}{x^2}right)^2 = x^4 + 2 + frac{1}{x^4} ).So, ( (A + 1)^2 = x^4 + 2 + frac{1}{x^4} + 2(x^2 + frac{1}{x^2}) + 1 = x^4 + 2 + frac{1}{x^4} + 2x^2 + frac{2}{x^2} + 1 = x^4 + 2x^2 + 3 + frac{2}{x^2} + frac{1}{x^4} ).Therefore, the denominator becomes:( sqrt{1 + x^4 + 2x^2 + 3 + frac{2}{x^2} + frac{1}{x^4}} = sqrt{x^4 + 2x^2 + 4 + frac{2}{x^2} + frac{1}{x^4}} ).Hmm, this seems a bit complicated. Maybe there's a better way to simplify this.Alternatively, let's consider that ( x^2 + frac{1}{x^2} = left(x + frac{1}{x}right)^2 - 2 ). Let me denote ( t = x + frac{1}{x} ). Then, ( x^2 + frac{1}{x^2} = t^2 - 2 ).So, substituting back into the denominator:( left(x^2 + 1 + frac{1}{x^2}right)^2 = (t^2 - 2 + 1)^2 = (t^2 - 1)^2 ).Therefore, the denominator becomes:( sqrt{1 + (t^2 - 1)^2} = sqrt{1 + t^4 - 2t^2 + 1} = sqrt{t^4 - 2t^2 + 2} ).So, now, ( varphi(M,N) ) can be written as:( varphi(M,N) = frac{3|t|}{sqrt{t^4 - 2t^2 + 2}} ).Since ( t = x + frac{1}{x} ), and ( x ) is a real number (since we're dealing with points on the curve), ( t ) can take certain values. Let's analyze the possible values of ( t ).Note that ( x + frac{1}{x} ) is a standard expression. For real ( x neq 0 ), the expression ( x + frac{1}{x} ) satisfies ( |x + frac{1}{x}| geq 2 ) by the AM-GM inequality. Specifically, ( x + frac{1}{x} geq 2 ) if ( x > 0 ), and ( x + frac{1}{x} leq -2 ) if ( x < 0 ). So, ( |t| geq 2 ).Therefore, ( t ) can be any real number such that ( |t| geq 2 ).So, now, our problem reduces to finding the range of the function:( varphi(t) = frac{3|t|}{sqrt{t^4 - 2t^2 + 2}} ),where ( |t| geq 2 ).Since ( |t| geq 2 ), we can consider ( t geq 2 ) and ( t leq -2 ). However, since ( varphi(t) ) is an even function (because both numerator and denominator are even in ( t )), we can just consider ( t geq 2 ) and the range will be the same for ( t leq -2 ).So, let's consider ( t geq 2 ), and define:( varphi(t) = frac{3t}{sqrt{t^4 - 2t^2 + 2}} ).We need to find the range of ( varphi(t) ) as ( t ) varies over ( [2, infty) ).To find the range, let's analyze the behavior of ( varphi(t) ) as ( t ) increases from 2 to infinity.First, compute ( varphi(2) ):( varphi(2) = frac{3 times 2}{sqrt{16 - 8 + 2}} = frac{6}{sqrt{10}} = frac{6sqrt{10}}{10} = frac{3sqrt{10}}{5} approx 1.897 ).Next, consider the limit as ( t to infty ):( lim_{t to infty} varphi(t) = lim_{t to infty} frac{3t}{sqrt{t^4 - 2t^2 + 2}} ).Divide numerator and denominator by ( t^2 ):( lim_{t to infty} frac{3/t}{sqrt{1 - 2/t^2 + 2/t^4}} = frac{0}{sqrt{1}} = 0 ).So, as ( t ) increases, ( varphi(t) ) approaches 0.Now, we need to check if ( varphi(t) ) is decreasing for ( t geq 2 ). If it is, then the maximum value is at ( t = 2 ) and the minimum approaches 0 as ( t to infty ).To check if ( varphi(t) ) is decreasing, let's compute its derivative with respect to ( t ) and see if it's negative for ( t geq 2 ).Compute ( varphi(t) = frac{3t}{sqrt{t^4 - 2t^2 + 2}} ).Let me denote ( f(t) = 3t ) and ( g(t) = sqrt{t^4 - 2t^2 + 2} ), so ( varphi(t) = frac{f(t)}{g(t)} ).Using the quotient rule, the derivative is:( varphi'(t) = frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2} ).Compute ( f'(t) = 3 ).Compute ( g(t) = (t^4 - 2t^2 + 2)^{1/2} ), so ( g'(t) = frac{1}{2}(t^4 - 2t^2 + 2)^{-1/2} times (4t^3 - 4t) = frac{4t^3 - 4t}{2sqrt{t^4 - 2t^2 + 2}} = frac{2t^3 - 2t}{sqrt{t^4 - 2t^2 + 2}} ).Therefore, the derivative becomes:( varphi'(t) = frac{3 cdot sqrt{t^4 - 2t^2 + 2} - 3t cdot frac{2t^3 - 2t}{sqrt{t^4 - 2t^2 + 2}}}{t^4 - 2t^2 + 2} ).Simplify numerator:Multiply numerator and denominator by ( sqrt{t^4 - 2t^2 + 2} ):Numerator becomes:( 3(t^4 - 2t^2 + 2) - 3t(2t^3 - 2t) ).Compute each term:First term: ( 3t^4 - 6t^2 + 6 ).Second term: ( -3t(2t^3 - 2t) = -6t^4 + 6t^2 ).Combine both terms:( 3t^4 - 6t^2 + 6 - 6t^4 + 6t^2 = -3t^4 + 0t^2 + 6 = -3t^4 + 6 ).Therefore, the numerator is ( -3t^4 + 6 ), and the denominator is ( (t^4 - 2t^2 + 2)^{3/2} ).So, the derivative is:( varphi'(t) = frac{-3t^4 + 6}{(t^4 - 2t^2 + 2)^{3/2}} ).Factor numerator:( -3(t^4 - 2) ).Therefore,( varphi'(t) = frac{-3(t^4 - 2)}{(t^4 - 2t^2 + 2)^{3/2}} ).Now, let's analyze the sign of ( varphi'(t) ):The denominator ( (t^4 - 2t^2 + 2)^{3/2} ) is always positive because ( t^4 - 2t^2 + 2 = (t^2 - 1)^2 + 1 geq 1 > 0 ).The numerator is ( -3(t^4 - 2) ).So, the sign of ( varphi'(t) ) depends on ( -(t^4 - 2) ).Thus,- When ( t^4 - 2 > 0 ), i.e., ( t > sqrt[4]{2} approx 1.189 ), the numerator is negative, so ( varphi'(t) < 0 ).- When ( t^4 - 2 < 0 ), i.e., ( t < sqrt[4]{2} ), the numerator is positive, so ( varphi'(t) > 0 ).But in our case, ( t geq 2 ), which is greater than ( sqrt[4]{2} approx 1.189 ). Therefore, for ( t geq 2 ), ( varphi'(t) < 0 ), meaning ( varphi(t) ) is decreasing on ( [2, infty) ).Therefore, the maximum value of ( varphi(t) ) occurs at ( t = 2 ), which is ( frac{3sqrt{10}}{5} ), and as ( t ) increases beyond 2, ( varphi(t) ) decreases towards 0.Hence, the range of ( varphi(M,N) ) is ( (0, frac{3sqrt{10}}{5}) ).**Final Answer**The range of ( varphi(M,N) ) is boxed{left(0, dfrac{3sqrt{10}}{5}right)}.