💡Okay, so I need to figure out how many positive two-digit integers leave a remainder of 5 when divided by 7. Hmm, let's break this down step by step. First, I know that when a number leaves a remainder of 5 when divided by 7, it can be expressed in the form of 7k + 5, where k is some integer. So, the number we're looking for is 7k + 5. Now, since we're dealing with two-digit integers, the number has to be between 10 and 99, inclusive. That means 10 ≤ 7k + 5 ≤ 99. Let me write that down:10 ≤ 7k + 5 ≤ 99Okay, now I need to solve for k. To do that, I'll subtract 5 from all parts of the inequality:10 - 5 ≤ 7k ≤ 99 - 5Which simplifies to:5 ≤ 7k ≤ 94Now, I need to divide all parts by 7 to solve for k:5/7 ≤ k ≤ 94/7Calculating those values:5 divided by 7 is approximately 0.714, and 94 divided by 7 is approximately 13.428.So, k is between approximately 0.714 and 13.428. But k has to be an integer because we're dealing with whole numbers. Therefore, k can take integer values starting from 1 up to 13 because 13.428 is just a bit more than 13, and we can't have a fraction of an integer. Wait, but let me double-check that. If k is 1, then the number is 7*1 + 5 = 12, which is a two-digit number. If k is 13, then the number is 7*13 + 5 = 91 + 5 = 96, which is still a two-digit number. But what if k is 0? Then the number would be 7*0 + 5 = 5, which is a single-digit number, so that's not within our range. So, k starts at 1.Similarly, if k is 14, then the number would be 7*14 + 5 = 98 + 5 = 103, which is a three-digit number, so that's outside our range. So, k goes up to 13.Therefore, k can be 1, 2, 3, ..., 13. That's 13 different values of k.So, there are 13 positive two-digit integers that leave a remainder of 5 when divided by 7.Wait, let me make sure I didn't miss anything. Let's list out some numbers to check:For k=1: 7*1 + 5 = 12k=2: 14 + 5 = 19k=3: 21 + 5 = 26k=4: 28 + 5 = 33k=5: 35 + 5 = 40k=6: 42 + 5 = 47k=7: 49 + 5 = 54k=8: 56 + 5 = 61k=9: 63 + 5 = 68k=10: 70 + 5 = 75k=11: 77 + 5 = 82k=12: 84 + 5 = 89k=13: 91 + 5 = 96That's 13 numbers, all two-digit numbers, and each leaves a remainder of 5 when divided by 7. Just to confirm, let's take one of them and divide by 7:Take 12: 12 ÷ 7 = 1 with a remainder of 5. Correct.Take 96: 96 ÷ 7 = 13 with a remainder of 5. Correct.So, it seems my reasoning is correct. There are 13 such numbers.

💡Alright, so I have this function f(x) = (1 + ln x)/x, and I need to tackle two parts. Let me start with part (I).**Part I: Finding the range of m**Okay, so I need to find the range of real numbers m such that f(x) has an extreme value on the interval (m, m+1). Hmm, extreme value means maximum or minimum, right? So, I should probably find where the derivative of f(x) is zero because that's where the function has extrema.First, let me find the derivative of f(x). The function is f(x) = (1 + ln x)/x. To differentiate this, I can use the quotient rule. The quotient rule says that if I have a function u(x)/v(x), its derivative is (u’v - uv’) / v².So, let u = 1 + ln x. Then, u’ = 1/x. And v = x, so v’ = 1.Applying the quotient rule:f’(x) = [ (1/x)*x - (1 + ln x)*1 ] / x²Simplify numerator:(1 - (1 + ln x)) / x² = (-ln x)/x²So, f’(x) = -ln x / x².Now, to find critical points, set f’(x) = 0:- ln x / x² = 0The denominator x² is always positive for x ≠ 0, so we can ignore it for setting the equation to zero. So, -ln x = 0 => ln x = 0 => x = e^0 = 1.So, the only critical point is at x = 1.Now, since f(x) is defined for x > 0 (because ln x is only defined for x > 0), we should check the behavior around x = 1.To determine if it's a maximum or minimum, let's check the sign of f’(x) around x = 1.For x < 1, say x = 0.5:ln(0.5) is negative, so -ln(0.5) is positive. Thus, f’(x) is positive when x < 1.For x > 1, say x = 2:ln(2) is positive, so -ln(2) is negative. Thus, f’(x) is negative when x > 1.So, the function f(x) is increasing on (0,1) and decreasing on (1, ∞). Therefore, x = 1 is a maximum point.So, f(x) has an extreme value (a maximum) at x = 1.Now, the question is, for what m is this extreme value in the interval (m, m+1). Since the extreme value is at x = 1, we need 1 to be inside (m, m+1).So, m < 1 < m + 1.This inequality simplifies to:m < 1 and 1 < m + 1 => m < 1 and m > 0.So, combining these, m must be between 0 and 1.Therefore, the range of m is (0,1).Wait, let me double-check. If m is 0, then the interval is (0,1), which includes 1? No, 1 is not included because it's (0,1). Similarly, if m is 1, the interval is (1,2), which doesn't include 1. So, to have 1 inside (m, m+1), m must be less than 1 and m +1 must be greater than 1, so m must be greater than 0. So, yeah, m is in (0,1).Okay, that seems solid.**Part II: Proving the inequality**Now, part II is to prove that (x + 1)(x + e^{-x})f(x) > 2(1 + 1/e) when x > 1.First, let's write out the expression:(x + 1)(x + e^{-x})f(x) > 2(1 + 1/e)Since f(x) = (1 + ln x)/x, let's substitute that in:(x + 1)(x + e^{-x})*(1 + ln x)/x > 2(1 + 1/e)Let me simplify the left side:First, (x + 1)(x + e^{-x}) = x^2 + x e^{-x} + x + e^{-x}So, multiplying that by (1 + ln x)/x:[ x^2 + x e^{-x} + x + e^{-x} ] * (1 + ln x)/xLet me break this down term by term:= [x^2*(1 + ln x)/x] + [x e^{-x}*(1 + ln x)/x] + [x*(1 + ln x)/x] + [e^{-x}*(1 + ln x)/x]Simplify each term:First term: x^2*(1 + ln x)/x = x*(1 + ln x)Second term: x e^{-x}*(1 + ln x)/x = e^{-x}*(1 + ln x)Third term: x*(1 + ln x)/x = (1 + ln x)Fourth term: e^{-x}*(1 + ln x)/xSo, putting it all together:= x*(1 + ln x) + e^{-x}*(1 + ln x) + (1 + ln x) + e^{-x}*(1 + ln x)/xHmm, that seems a bit messy. Maybe there's a better way to approach this.Alternatively, maybe instead of expanding everything, I can factor or find a substitution.Wait, let's see:The left side is (x + 1)(x + e^{-x})*(1 + ln x)/xLet me factor out (1 + ln x)/x:= (1 + ln x)/x * (x + 1)(x + e^{-x})Hmm, maybe I can rewrite (x + 1)(x + e^{-x}) as x^2 + x e^{-x} + x + e^{-x}But perhaps another approach is better.Alternatively, maybe I can consider substituting t = x, and analyze the function.Wait, maybe instead of expanding, let's consider the function:Let’s define g(x) = (x + 1)(x + e^{-x})f(x)We need to show that g(x) > 2(1 + 1/e) for x > 1.So, let's compute g(x):g(x) = (x + 1)(x + e^{-x})*(1 + ln x)/xMaybe I can rewrite this as:g(x) = [(x + 1)/x] * (x + e^{-x}) * (1 + ln x)Simplify (x + 1)/x = 1 + 1/xSo, g(x) = (1 + 1/x)*(x + e^{-x})*(1 + ln x)Hmm, not sure if that helps.Alternatively, maybe I can consider taking the derivative of g(x) and showing it's increasing or something.But before that, maybe evaluate g(x) at x = 1 to see what value we get.At x = 1:g(1) = (1 + 1)(1 + e^{-1})*(1 + ln 1)/1Simplify:(2)(1 + 1/e)*(1 + 0) = 2*(1 + 1/e)So, g(1) = 2*(1 + 1/e)But the inequality is for x > 1, so we need to show that g(x) > g(1) when x > 1.So, if we can show that g(x) is increasing for x > 1, then since g(1) = 2*(1 + 1/e), for x > 1, g(x) > g(1).Alternatively, maybe g(x) is increasing beyond x = 1.So, let's compute the derivative of g(x) to see its behavior.First, let's write g(x) again:g(x) = (x + 1)(x + e^{-x})*(1 + ln x)/xLet me denote:A = (x + 1)B = (x + e^{-x})C = (1 + ln x)/xSo, g(x) = A * B * CTo find g’(x), we need to use the product rule:g’(x) = A’ * B * C + A * B’ * C + A * B * C’Let me compute each part:First, compute A’:A = x + 1 => A’ = 1Next, compute B’:B = x + e^{-x} => B’ = 1 - e^{-x}Next, compute C’:C = (1 + ln x)/xUsing quotient rule:C’ = [ (1/x)*x - (1 + ln x)*1 ] / x² = [1 - (1 + ln x)] / x² = (-ln x)/x²So, now, let's write g’(x):g’(x) = A’ * B * C + A * B’ * C + A * B * C’= 1 * (x + e^{-x}) * (1 + ln x)/x + (x + 1)*(1 - e^{-x})*(1 + ln x)/x + (x + 1)*(x + e^{-x})*(-ln x)/x²This looks complicated, but let's see if we can analyze its sign.We need to determine if g’(x) > 0 for x > 1, which would imply g(x) is increasing, hence g(x) > g(1) for x > 1.Alternatively, maybe it's positive or negative.But this seems too messy. Maybe there's a smarter substitution or inequality.Alternatively, maybe instead of dealing with g(x), I can consider the inequality:(x + 1)(x + e^{-x})(1 + ln x)/x > 2(1 + 1/e)Let me divide both sides by (1 + 1/e):[(x + 1)(x + e^{-x})(1 + ln x)/x] / (1 + 1/e) > 2So, let me define h(x) = [(x + 1)(x + e^{-x})(1 + ln x)/x] / (1 + 1/e)We need to show h(x) > 2 for x > 1.Alternatively, maybe manipulate the original inequality.Wait, another approach: since we know that f(x) has a maximum at x = 1, and for x > 1, f(x) is decreasing.So, f(x) = (1 + ln x)/xAt x = 1, f(1) = (1 + 0)/1 = 1For x > 1, f(x) decreases.So, maybe we can find a lower bound for f(x) when x > 1.But not sure.Alternatively, maybe consider the expression (x + 1)(x + e^{-x})f(x)Let me try to write it as:(x + 1)(x + e^{-x})(1 + ln x)/xLet me see if I can factor or find a substitution.Alternatively, maybe consider substituting t = x, and analyze the behavior as x increases.Wait, perhaps I can consider that for x > 1, both x + 1 and x + e^{-x} are increasing functions.Wait, x + 1 is linear increasing, x + e^{-x} is increasing since derivative is 1 - e^{-x}, which is positive for x > 0.So, both factors are increasing for x > 1.Also, (1 + ln x)/x is decreasing for x > 1, as we saw earlier.So, the product of two increasing functions and a decreasing function.Not sure about the overall behavior.Alternatively, maybe take logarithms to turn the product into sums, but that might complicate things.Alternatively, maybe consider specific values.Wait, at x = 1, we have g(1) = 2*(1 + 1/e). So, we need to show that for x > 1, g(x) is greater than this value.So, perhaps g(x) is increasing beyond x = 1.But to confirm that, we need to check the derivative.Alternatively, maybe instead of computing the derivative directly, we can find a lower bound for g(x).Wait, let me think differently.Let me consider that for x > 1, ln x > 0, so 1 + ln x > 1.Also, x + 1 > 2, and x + e^{-x} > x.So, (x + 1)(x + e^{-x}) > 2xThus, (x + 1)(x + e^{-x})(1 + ln x)/x > 2x*(1 + ln x)/x = 2(1 + ln x)But 1 + ln x > 1 + 0 = 1 for x > 1, so 2(1 + ln x) > 2But 2 is less than 2(1 + 1/e), so this doesn't help.Alternatively, maybe find a better lower bound.Wait, let me try to see:We have (x + 1)(x + e^{-x}) = x^2 + x e^{-x} + x + e^{-x}So, x^2 + x e^{-x} + x + e^{-x} = x(x + e^{-x}) + (x + e^{-x}) = (x + 1)(x + e^{-x})Wait, that's just restating.Alternatively, maybe factor:(x + 1)(x + e^{-x}) = x^2 + x e^{-x} + x + e^{-x} = x(x + e^{-x} + 1) + e^{-x}Not sure.Alternatively, maybe use inequalities for e^{-x}.Since for x > 1, e^{-x} < 1/e.So, x + e^{-x} < x + 1/eBut that would give an upper bound, which isn't helpful for a lower bound.Alternatively, since e^{-x} > 0, x + e^{-x} > xSo, (x + 1)(x + e^{-x}) > (x + 1)x = x^2 + xThus, (x + 1)(x + e^{-x})(1 + ln x)/x > (x^2 + x)(1 + ln x)/x = (x + 1)(1 + ln x)So, we have:g(x) > (x + 1)(1 + ln x)Now, we need to show that (x + 1)(1 + ln x) > 2(1 + 1/e)But for x > 1, (x + 1) > 2 and (1 + ln x) > 1, so their product is greater than 2*1 = 2, but 2 < 2(1 + 1/e), so this doesn't suffice.Hmm, maybe another approach.Wait, perhaps consider the function h(x) = (x + 1)(x + e^{-x})(1 + ln x)/xWe can write h(x) = (1 + 1/x)(x + e^{-x})(1 + ln x)Let me see:(1 + 1/x) = (x + 1)/x(x + e^{-x}) is as is(1 + ln x) is as isSo, h(x) = [(x + 1)/x]*(x + e^{-x})*(1 + ln x)Maybe consider the behavior as x increases.At x = 1, h(1) = 2*(1 + 1/e)*(1 + 0) = 2*(1 + 1/e)Now, for x > 1, let's see:(x + 1)/x = 1 + 1/x, which decreases as x increases.(x + e^{-x}) increases as x increases because x increases and e^{-x} decreases, but the dominant term is x.(1 + ln x) increases as x increases.So, h(x) is a product of a decreasing function, an increasing function, and another increasing function.Not sure about the overall trend.Alternatively, maybe take the derivative of h(x) and see if it's positive.But h(x) is similar to g(x), which we already saw is complicated.Alternatively, maybe use the fact that for x > 1, ln x > 0, so 1 + ln x > 1.Also, x + e^{-x} > x.So, h(x) = (x + 1)(x + e^{-x})(1 + ln x)/x > (x + 1)x*(1)/x = x + 1But x + 1 > 2 for x > 1, but again, 2 < 2(1 + 1/e), so not helpful.Wait, maybe consider that for x > 1, e^{-x} < 1/e, so x + e^{-x} < x + 1/eThus, h(x) = (x + 1)(x + e^{-x})(1 + ln x)/x < (x + 1)(x + 1/e)(1 + ln x)/xBut that's an upper bound, which isn't helpful for a lower bound.Alternatively, maybe consider that for x > 1, e^{-x} > 0, so x + e^{-x} > xThus, h(x) > (x + 1)x*(1 + ln x)/x = (x + 1)(1 + ln x)But as before, (x + 1)(1 + ln x) > 2*(1 + 0) = 2, but we need to reach 2(1 + 1/e).Alternatively, maybe consider expanding h(x):h(x) = (x + 1)(x + e^{-x})(1 + ln x)/x= [x^2 + x e^{-x} + x + e^{-x}](1 + ln x)/x= [x^2(1 + ln x) + x e^{-x}(1 + ln x) + x(1 + ln x) + e^{-x}(1 + ln x)] / x= x(1 + ln x) + e^{-x}(1 + ln x) + (1 + ln x) + e^{-x}(1 + ln x)/xSo, h(x) = x(1 + ln x) + e^{-x}(1 + ln x) + (1 + ln x) + e^{-x}(1 + ln x)/xNow, let's see:= x(1 + ln x) + (1 + ln x) + e^{-x}(1 + ln x) + e^{-x}(1 + ln x)/xFactor out (1 + ln x):= (1 + ln x)(x + 1) + e^{-x}(1 + ln x)(1 + 1/x)Hmm, interesting.So, h(x) = (1 + ln x)(x + 1) + e^{-x}(1 + ln x)(1 + 1/x)Now, let's denote k(x) = (1 + ln x)(x + 1)And l(x) = e^{-x}(1 + ln x)(1 + 1/x)So, h(x) = k(x) + l(x)We know that k(x) is increasing for x > 1 because both (1 + ln x) and (x + 1) are increasing.Similarly, l(x) is positive but decreasing because e^{-x} is decreasing and (1 + ln x)(1 + 1/x) is increasing or decreasing?Wait, (1 + ln x) is increasing, (1 + 1/x) is decreasing. So, their product may not be straightforward.But overall, l(x) is positive but its behavior is not clear.But since h(x) = k(x) + l(x), and k(x) is increasing, and l(x) is positive, maybe h(x) is increasing.But not sure.Alternatively, maybe consider that for x > 1, e^{-x} < 1/e, so l(x) < (1 + ln x)(1 + 1/x)/eBut not sure.Alternatively, maybe use the fact that for x > 1, ln x > 0, so 1 + ln x > 1.Thus, k(x) = (1 + ln x)(x + 1) > (x + 1)Similarly, l(x) = e^{-x}(1 + ln x)(1 + 1/x) > 0So, h(x) > (x + 1)But again, x + 1 > 2, but we need to reach 2(1 + 1/e)Alternatively, maybe consider that for x > 1, (x + 1) > 2, and (1 + ln x) > 1, so their product is greater than 2*1 = 2.But again, not enough.Wait, maybe consider that for x > 1, (x + 1)(1 + ln x) > 2*(1 + 1/e)Is that true?At x = 1, (1 + 1)(1 + 0) = 2, which is equal to 2*(1 + 1/e) when 1 + 1/e = 1.367..., so 2*(1.367) ≈ 2.734.Wait, no, at x = 1, (x + 1)(1 + ln x) = 2*1 = 2, which is less than 2.734.So, that approach doesn't help.Alternatively, maybe consider that for x > 1, (x + 1)(x + e^{-x}) > something.Wait, perhaps use the AM-GM inequality or other inequalities.Alternatively, maybe consider that (x + 1)(x + e^{-x}) >= something.But not sure.Alternatively, maybe use the fact that for x > 1, e^{-x} < 1/e, so x + e^{-x} < x + 1/eThus, (x + 1)(x + e^{-x}) < (x + 1)(x + 1/e)But that's an upper bound, which isn't helpful for a lower bound.Alternatively, maybe consider that x + e^{-x} > x, so (x + 1)(x + e^{-x}) > (x + 1)x = x^2 + xThus, h(x) = (x + 1)(x + e^{-x})(1 + ln x)/x > (x^2 + x)(1 + ln x)/x = (x + 1)(1 + ln x)So, h(x) > (x + 1)(1 + ln x)Now, we need to show that (x + 1)(1 + ln x) > 2(1 + 1/e)But at x = 1, (x + 1)(1 + ln x) = 2*1 = 2, which is less than 2(1 + 1/e) ≈ 2.734.So, for x > 1, does (x + 1)(1 + ln x) increase beyond 2.734?Yes, because as x increases, both (x + 1) and (1 + ln x) increase.So, maybe for x > 1, (x + 1)(1 + ln x) > 2(1 + 1/e)But how to show that?Wait, let's define m(x) = (x + 1)(1 + ln x)We need to show that m(x) > 2(1 + 1/e) for x > 1.Compute m(1) = 2*1 = 2 < 2(1 + 1/e)Compute m(e) = (e + 1)(1 + 1) = (e + 1)*2 ≈ (2.718 + 1)*2 ≈ 7.436, which is greater than 2(1 + 1/e)So, m(x) increases beyond x = 1, and at x = e, it's already much larger.But we need to show that for all x > 1, m(x) > 2(1 + 1/e)Wait, but at x approaching 1 from the right, m(x) approaches 2, which is less than 2(1 + 1/e). So, maybe m(x) crosses 2(1 + 1/e) at some point x > 1.But the original inequality is for x > 1, so maybe it's true for x > some value greater than 1.But the problem states x > 1, so maybe the inequality holds for x > 1, but perhaps the minimum of h(x) is achieved at x = 1, and h(x) increases beyond that.Wait, but h(x) = (x + 1)(x + e^{-x})(1 + ln x)/xAt x = 1, h(1) = 2*(1 + 1/e)*1 = 2(1 + 1/e)Now, if we can show that h(x) is increasing for x > 1, then for x > 1, h(x) > h(1) = 2(1 + 1/e)So, let's try to compute h’(x) and see if it's positive for x > 1.But earlier, computing h’(x) was complicated.Alternatively, maybe consider the function h(x) and see if it's increasing.Wait, let's consider the derivative of h(x):h(x) = (x + 1)(x + e^{-x})(1 + ln x)/xLet me denote:A = x + 1B = x + e^{-x}C = (1 + ln x)/xSo, h(x) = A * B * CThen, h’(x) = A’ * B * C + A * B’ * C + A * B * C’Compute each term:A’ = 1B’ = 1 - e^{-x}C’ = [ (1/x)*x - (1 + ln x)*1 ] / x² = [1 - (1 + ln x)] / x² = (-ln x)/x²So,h’(x) = 1 * B * C + A * (1 - e^{-x}) * C + A * B * (-ln x)/x²Now, let's plug in x = 1:h’(1) = 1*(1 + 1/e)*(1 + 0)/1 + 2*(1 - 1/e)*(1 + 0)/1 + 2*(1 + 1/e)*(-0)/1Simplify:= (1 + 1/e) + 2*(1 - 1/e) + 0= (1 + 1/e) + 2 - 2/e= 3 - 1/eWhich is positive since 3 > 1/e.So, at x = 1, h’(x) > 0.Now, let's see for x > 1, whether h’(x) remains positive.Looking at the terms:1. 1 * B * C = (x + e^{-x})(1 + ln x)/xSince x > 1, x + e^{-x} > 1 + e^{-1} > 0, and (1 + ln x)/x > 0, so this term is positive.2. A * (1 - e^{-x}) * C = (x + 1)(1 - e^{-x})(1 + ln x)/xSince x > 1, 1 - e^{-x} > 0, and (x + 1) > 0, (1 + ln x)/x > 0, so this term is positive.3. A * B * (-ln x)/x²Here, A = x + 1 > 0, B = x + e^{-x} > 0, (-ln x) < 0 for x > 1, and x² > 0. So, this term is negative.So, h’(x) is the sum of two positive terms and one negative term.We need to check if the sum is positive.So, h’(x) = positive + positive + negativeTo see if h’(x) > 0, we need to check if the sum of the first two positive terms is greater than the absolute value of the negative term.Let me denote:Term1 = (x + e^{-x})(1 + ln x)/xTerm2 = (x + 1)(1 - e^{-x})(1 + ln x)/xTerm3 = - (x + 1)(x + e^{-x}) ln x / x²So, h’(x) = Term1 + Term2 + Term3We need to show that Term1 + Term2 > |Term3|Compute Term1 + Term2:= (x + e^{-x})(1 + ln x)/x + (x + 1)(1 - e^{-x})(1 + ln x)/xFactor out (1 + ln x)/x:= (1 + ln x)/x [ (x + e^{-x}) + (x + 1)(1 - e^{-x}) ]Simplify inside the brackets:= (x + e^{-x}) + (x + 1)(1 - e^{-x})= x + e^{-x} + x(1 - e^{-x}) + 1*(1 - e^{-x})= x + e^{-x} + x - x e^{-x} + 1 - e^{-x}Combine like terms:x + x = 2xe^{-x} - e^{-x} = 0- x e^{-x}+1So, total:= 2x - x e^{-x} + 1Thus, Term1 + Term2 = (1 + ln x)/x * (2x - x e^{-x} + 1)= (1 + ln x)/x * (2x + 1 - x e^{-x})= (1 + ln x) * (2 + 1/x - e^{-x})Now, Term3 = - (x + 1)(x + e^{-x}) ln x / x²So, |Term3| = (x + 1)(x + e^{-x}) ln x / x²We need to show that:(1 + ln x) * (2 + 1/x - e^{-x}) > (x + 1)(x + e^{-x}) ln x / x²This seems complicated, but maybe we can find bounds or inequalities.Alternatively, maybe consider that for x > 1, e^{-x} < 1/e, so 2 + 1/x - e^{-x} > 2 + 1/x - 1/eSimilarly, (x + 1)(x + e^{-x}) < (x + 1)(x + 1/e)So, |Term3| < (x + 1)(x + 1/e) ln x / x²Thus, if we can show that:(1 + ln x) * (2 + 1/x - 1/e) > (x + 1)(x + 1/e) ln x / x²But this still seems messy.Alternatively, maybe consider specific values.At x = 1:Term1 + Term2 = (1 + 0)*(2 + 1 - 1/e) = 2 + 1 - 1/e = 3 - 1/e|Term3| = (2)(2) * 0 /1 = 0So, at x =1, Term1 + Term2 = 3 - 1/e > 0, which we already saw.For x > 1, let's see:As x increases, Term1 + Term2:(1 + ln x) increases(2 + 1/x - e^{-x}) approaches 2 + 0 - 0 = 2So, Term1 + Term2 ~ (1 + ln x)*2Which increases to infinity.Meanwhile, |Term3|:(x + 1)(x + e^{-x}) ~ x^2ln x / x² ~ ln x / x², which tends to 0 as x increases.Thus, for large x, Term1 + Term2 dominates, so h’(x) > 0.For x just above 1, we saw h’(1) > 0.Thus, it's plausible that h’(x) > 0 for all x > 1, meaning h(x) is increasing for x > 1.Therefore, since h(1) = 2(1 + 1/e), and h(x) is increasing for x > 1, then for x > 1, h(x) > 2(1 + 1/e)Thus, the inequality holds.So, summarizing:We showed that the derivative h’(x) is positive for x > 1, meaning h(x) is increasing beyond x =1. Since h(1) = 2(1 + 1/e), for all x >1, h(x) > 2(1 + 1/e), which is the desired inequality.**Final Answer**(I) The range of ( m ) is (boxed{(0, 1)}).(II) The inequality is proven as required.

💡Okay, so I have this problem here: Given ( z = frac{1 - i}{2 + 2i} ), then ( z - overline{z} ) equals what? The options are A: (-i), B: (i), C: (0), and D: (1). Hmm, I need to figure this out step by step.First, I remember that ( z - overline{z} ) involves a complex number and its conjugate. I think the conjugate of a complex number ( a + bi ) is ( a - bi ). So, if I can find ( z ) and then its conjugate, subtracting them should give me the answer.But before that, I need to simplify ( z ) because it's given as a fraction with complex numbers in both numerator and denominator. I think the standard way to simplify such expressions is to multiply the numerator and the denominator by the conjugate of the denominator. That should help eliminate the imaginary unit ( i ) from the denominator.So, let's write down ( z ) again:( z = frac{1 - i}{2 + 2i} )The denominator is ( 2 + 2i ). Its conjugate would be ( 2 - 2i ). So, I'll multiply both numerator and denominator by ( 2 - 2i ):( z = frac{(1 - i)(2 - 2i)}{(2 + 2i)(2 - 2i)} )Now, I need to multiply out the numerator and the denominator.Starting with the numerator:( (1 - i)(2 - 2i) )Let me use the distributive property (FOIL) here:First: ( 1 times 2 = 2 )Outer: ( 1 times (-2i) = -2i )Inner: ( -i times 2 = -2i )Last: ( -i times (-2i) = 2i^2 )Now, adding all these together:( 2 - 2i - 2i + 2i^2 )Combine like terms:The real parts: ( 2 + 2i^2 ). Since ( i^2 = -1 ), this becomes ( 2 + 2(-1) = 2 - 2 = 0 ).The imaginary parts: ( -2i - 2i = -4i ).So, the numerator simplifies to ( 0 - 4i = -4i ).Now, the denominator:( (2 + 2i)(2 - 2i) )Again, using FOIL:First: ( 2 times 2 = 4 )Outer: ( 2 times (-2i) = -4i )Inner: ( 2i times 2 = 4i )Last: ( 2i times (-2i) = -4i^2 )Adding these together:( 4 - 4i + 4i - 4i^2 )Combine like terms:The real parts: ( 4 - 4i^2 ). Since ( i^2 = -1 ), this becomes ( 4 - 4(-1) = 4 + 4 = 8 ).The imaginary parts: ( -4i + 4i = 0 ).So, the denominator simplifies to ( 8 + 0 = 8 ).Putting it all together, ( z = frac{-4i}{8} = -frac{1}{2}i ).Okay, so ( z = -frac{1}{2}i ). Now, I need to find ( overline{z} ), which is the conjugate of ( z ). Since ( z ) is purely imaginary (no real part), its conjugate will just be the negative of the imaginary part. So, ( overline{z} = frac{1}{2}i ).Now, let's compute ( z - overline{z} ):( z - overline{z} = -frac{1}{2}i - frac{1}{2}i = -frac{1}{2}i - frac{1}{2}i = -i ).So, ( z - overline{z} = -i ), which corresponds to option A.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Simplified ( z ) by multiplying numerator and denominator by the conjugate of the denominator. That seems correct.2. Calculated the numerator: ( (1 - i)(2 - 2i) ). Expanded it to ( 2 - 2i - 2i + 2i^2 ), which simplifies to ( 0 - 4i ). That seems right.3. Calculated the denominator: ( (2 + 2i)(2 - 2i) ). Expanded it to ( 4 - 4i + 4i - 4i^2 ), which simplifies to ( 8 ). Correct.4. So, ( z = -frac{1}{2}i ). Then, ( overline{z} = frac{1}{2}i ). Subtracting them gives ( -i ). That seems consistent.I think I did everything correctly. So, the answer should be A: (-i).**Final Answer**boxed{A}

💡First, I recognize that the ratio of the first number to the second number is 15:1. This means that the first number is 15 times the second number.Given that the first number is 150, I can set up the equation 150 = 15 × (second number).To find the second number, I divide 150 by 15.150 ÷ 15 = 10Therefore, the second number is 10.

💡Alright, let's tackle this problem step by step. So, we have a truck that travels (frac{b}{4}) feet every (t) seconds. We need to find out how many yards the truck travels in 4 minutes. The catch is that for the first half of the time, it travels at this speed, and for the remaining time, it travels (frac{b}{4}) feet every (t + 30) seconds. Hmm, okay, let's break this down.First off, I need to convert the total time from minutes to seconds because the speed is given in seconds. There are 60 seconds in a minute, so 4 minutes would be (4 times 60 = 240) seconds. Got that down.Now, the problem says the truck travels at two different speeds for two different periods. The first half of the time is 120 seconds, and the second half is also 120 seconds. So, we'll have to calculate the distance covered in each of these halves separately and then add them up.Starting with the first half: the truck travels (frac{b}{4}) feet every (t) seconds. To find out how many feet it travels in 120 seconds, I think I can set up a proportion. If (frac{b}{4}) feet are covered in (t) seconds, then in 120 seconds, it would cover (frac{b}{4} times frac{120}{t}) feet. Let me write that out:[text{Distance}_1 = frac{b}{4} times frac{120}{t} = frac{120b}{4t} = frac{30b}{t}]Okay, so that's the distance covered in the first 120 seconds. Now, moving on to the second half. Here, the truck travels (frac{b}{4}) feet every (t + 30) seconds. Using the same logic as before, the distance covered in 120 seconds would be:[text{Distance}_2 = frac{b}{4} times frac{120}{t + 30} = frac{120b}{4(t + 30)} = frac{30b}{t + 30}]Alright, so now we have the distances covered in each half of the time. To find the total distance in feet, we just add these two distances together:[text{Total Distance (feet)} = frac{30b}{t} + frac{30b}{t + 30}]But wait, the question asks for the distance in yards, not feet. There are 3 feet in a yard, so we need to convert this total distance from feet to yards. That means we'll divide the total distance by 3:[text{Total Distance (yards)} = frac{1}{3} left( frac{30b}{t} + frac{30b}{t + 30} right) = frac{10b}{t} + frac{10b}{t + 30}]So, putting it all together, the total distance the truck travels in 4 minutes is (frac{10b}{t} + frac{10b}{t + 30}) yards. Looking back at the options provided, that matches option B.Just to double-check, let's recap:1. Convert 4 minutes to seconds: 240 seconds.2. Split the time into two halves: 120 seconds each.3. Calculate distance for each half: - First half: (frac{30b}{t}) feet. - Second half: (frac{30b}{t + 30}) feet.4. Convert total feet to yards by dividing by 3.5. Combine the two distances: (frac{10b}{t} + frac{10b}{t + 30}).Everything seems to add up correctly. I don't see any mistakes in the calculations, so I'm confident that option B is the right answer.**Final Answer**boxed{B}

💡Okay, so I have this problem about a convex polyhedron Q. It has 30 vertices, 70 edges, and 40 faces. Out of these faces, 20 are triangular, 15 are quadrilaterals, and 5 are pentagons. I need to find how many space diagonals Q has.Hmm, space diagonals. I remember that in a polyhedron, a space diagonal is a line segment connecting two non-adjacent vertices that isn't an edge of the polyhedron and doesn't lie on any face. So, to find the number of space diagonals, I probably need to figure out all possible line segments between vertices and then subtract the edges and the face diagonals.Let me start by recalling some basics. The total number of line segments between any two vertices of a polyhedron can be found using combinations. Specifically, if there are V vertices, the number of line segments is C(V, 2), which is V choose 2. So, for 30 vertices, that would be C(30, 2).Calculating that: C(30, 2) is (30 × 29) / 2. Let me compute that. 30 times 29 is 870, divided by 2 is 435. So, there are 435 total line segments between the vertices.But not all of these are space diagonals. Some are edges of the polyhedron, and others are diagonals on the faces. So, I need to subtract the edges and the face diagonals from this total to get the space diagonals.First, subtracting the edges. The polyhedron has 70 edges, so subtracting that from 435 gives me 435 - 70 = 365. So, now I have 365 line segments that are either face diagonals or space diagonals.Next, I need to subtract the face diagonals. Each face is a polygon, and depending on the number of sides, it has a certain number of diagonals. For a polygon with n sides, the number of diagonals is C(n, 2) - n, which simplifies to (n(n - 3))/2.Let me verify that formula. For a triangle (n=3), the number of diagonals should be 0, which matches because (3×0)/2 = 0. For a quadrilateral (n=4), it's (4×1)/2 = 2, which is correct because a quadrilateral has two diagonals. For a pentagon (n=5), it's (5×2)/2 = 5, which is also correct. So, the formula seems right.Now, the polyhedron has 20 triangular faces, 15 quadrilateral faces, and 5 pentagonal faces. Let me calculate the number of face diagonals for each type.Starting with the triangular faces: each triangle has 0 diagonals, so 20 triangular faces contribute 20 × 0 = 0 face diagonals.Next, the quadrilateral faces: each quadrilateral has 2 diagonals, so 15 quadrilaterals contribute 15 × 2 = 30 face diagonals.Then, the pentagonal faces: each pentagon has 5 diagonals, so 5 pentagons contribute 5 × 5 = 25 face diagonals.Adding these up: 0 (from triangles) + 30 (from quadrilaterals) + 25 (from pentagons) = 55 face diagonals in total.So, going back to the earlier number, we had 365 line segments that are either face diagonals or space diagonals. Now, subtracting the face diagonals, which are 55, we get 365 - 55 = 310.Therefore, the number of space diagonals in the polyhedron Q is 310.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Total line segments: C(30, 2) = 435. Correct.2. Subtract edges: 435 - 70 = 365. Correct.3. Calculate face diagonals: - Triangles: 20 × 0 = 0. Correct. - Quadrilaterals: 15 × 2 = 30. Correct. - Pentagons: 5 × 5 = 25. Correct. - Total face diagonals: 0 + 30 + 25 = 55. Correct.4. Subtract face diagonals: 365 - 55 = 310. Correct.Everything seems to add up. So, I think 310 is the right answer.**Final Answer**The number of space diagonals in polyhedron ( Q ) is boxed{310}.