💡Okay, so I have this problem with three parts. Let me go through each part step by step. First, part (1) asks for the value of f’(1), where f(x) is given as (e^x - 2)/x. Hmm, I need to find the derivative of f(x) and then evaluate it at x=1. Alright, to find f’(x), I can use the quotient rule. The quotient rule is (low d high minus high d low) over low squared. So, f(x) is numerator e^x - 2 over denominator x. So, f’(x) would be [x*(d/dx)(e^x - 2) - (e^x - 2)*(d/dx)(x)] / x^2. Calculating the derivatives: d/dx(e^x - 2) is e^x, and d/dx(x) is 1. So plugging those in, f’(x) becomes [x*e^x - (e^x - 2)*1] / x^2. Simplify the numerator: x*e^x - e^x + 2. So, f’(x) = (x*e^x - e^x + 2) / x^2. Now, evaluate this at x=1. Plugging in 1 for x, we get (1*e^1 - e^1 + 2) / 1^2. Simplify the numerator: e - e + 2 = 2. So, f’(1) is 2/1 = 2. Okay, that seems straightforward. So, part (1) is done, and the answer is 2.Moving on to part (2): Find the maximum value of g(x) on the interval [1,2], where g(x) = a ln x - 1/x, and a is a real number. Hmm, to find the maximum value of a function on a closed interval, I remember that we need to check the critical points inside the interval and also evaluate the function at the endpoints. First, let's find the derivative of g(x). g(x) = a ln x - 1/x. The derivative, g’(x), is a*(1/x) - (-1/x^2) = a/x + 1/x^2. So, g’(x) = (a x + 1)/x^2. To find critical points, set g’(x) = 0. So, (a x + 1)/x^2 = 0. Since x^2 is always positive for x ≠ 0, the numerator must be zero: a x + 1 = 0. So, x = -1/a. But x is in [1,2], so we need to check if -1/a is in [1,2]. Wait, but a is a real number, so depending on the value of a, -1/a could be positive or negative. If a is positive, then -1/a is negative, which is not in [1,2]. So, in that case, there are no critical points in [1,2], and the maximum must occur at the endpoints. If a is negative, then -1/a is positive. So, we need to check if -1/a is in [1,2]. So, if a is negative, and -1/a is between 1 and 2, that is, if 1 ≤ -1/a ≤ 2. Let me solve for a: 1 ≤ -1/a ≤ 2. Since a is negative, multiplying all parts by a (which is negative) reverses the inequalities: a ≤ -1 and a ≥ -1/2. Wait, that seems a bit confusing. Let me think again. If 1 ≤ -1/a ≤ 2, and a is negative, then multiplying all parts by a (negative) reverses the inequalities. So, 1 ≤ -1/a ≤ 2. First inequality: 1 ≤ -1/a. Since a is negative, multiply both sides by a (negative), inequality reverses: a ≤ -1. Second inequality: -1/a ≤ 2. Multiply both sides by a (negative), inequality reverses: -1 ≥ 2a, so a ≤ -1/2. Wait, that doesn't seem right. Let me try another approach. Let me solve 1 ≤ -1/a ≤ 2 for a. First, 1 ≤ -1/a. Since a is negative, let me write a = -b where b > 0. Then, 1 ≤ 1/b ≤ 2. So, 1 ≤ 1/b ≤ 2. Which implies 1/2 ≤ b ≤ 1. Since b = -a, so 1/2 ≤ -a ≤ 1, which implies -1 ≤ a ≤ -1/2. So, when a is between -1 and -1/2, then -1/a is between 1 and 2, so the critical point x = -1/a is in [1,2]. Therefore, for a in (-1, -1/2), the critical point is inside [1,2], so we need to evaluate g(x) at x=1, x=2, and x=-1/a. For a ≤ -1, -1/a ≤ 1, so the critical point is at x=1 or less, which is outside the interval [1,2]. So, in that case, the maximum is at x=1 or x=2. Similarly, for a ≥ -1/2, -1/a ≥ 2, so the critical point is at x=2 or more, outside [1,2]. So, maximum is at x=1 or x=2. So, let's summarize: - If a ≤ -1: Critical point x=-1/a ≤1, so evaluate at x=1 and x=2. - If -1 < a < -1/2: Critical point x=-1/a is in (1,2), so evaluate at x=1, x=2, and x=-1/a. - If a ≥ -1/2: Critical point x=-1/a ≥2, so evaluate at x=1 and x=2. Now, let's compute g(x) at these points. First, g(1) = a ln 1 - 1/1 = 0 - 1 = -1. g(2) = a ln 2 - 1/2. And for the critical point x=-1/a, g(-1/a) = a ln(-1/a) - 1/(-1/a) = a ln(-1/a) + a. So, now we need to compare these values depending on the value of a. Case 1: a ≤ -1. In this case, the critical point is at x=-1/a ≤1, so we only evaluate at x=1 and x=2. Compute g(1) = -1, g(2) = a ln2 - 1/2. Since a ≤ -1, a ln2 is ≤ -ln2 ≈ -0.693. So, g(2) = a ln2 - 1/2 ≤ -0.693 - 0.5 = -1.193. Compare with g(1) = -1. So, g(2) is less than g(1). Therefore, the maximum is at x=1, which is -1. Case 2: -1 < a < -1/2. Here, the critical point x=-1/a is in (1,2). So, we need to evaluate g at x=1, x=2, and x=-1/a. Compute g(1) = -1, g(2) = a ln2 -1/2, and g(-1/a) = a ln(-1/a) + a. We need to see which of these is the largest. Let me consider the behavior of g(x). Since g’(x) changes sign from positive to negative at x=-1/a, because for a <0, the derivative g’(x) = (a x +1)/x^2. When x < -1/a, a x +1 >0, so g’(x) >0. When x > -1/a, a x +1 <0, so g’(x) <0. Therefore, x=-1/a is a maximum point. Therefore, in this case, the maximum is at x=-1/a, which is g(-1/a) = a ln(-1/a) + a. Case 3: a ≥ -1/2. Here, the critical point x=-1/a ≥2, so outside the interval. Therefore, evaluate at x=1 and x=2. Compute g(1) = -1, g(2) = a ln2 -1/2. Since a ≥ -1/2, a ln2 is ≥ (-1/2) ln2 ≈ -0.3466. So, g(2) = a ln2 -1/2 ≥ -0.3466 -0.5 = -0.8466. Compare with g(1) = -1. So, g(2) is greater than g(1). Therefore, the maximum is at x=2, which is a ln2 -1/2. So, summarizing: - If a ≤ -1, maximum is -1 at x=1. - If -1 < a < -1/2, maximum is a ln(-1/a) + a at x=-1/a. - If a ≥ -1/2, maximum is a ln2 -1/2 at x=2. Okay, that seems to cover all cases. Now, part (3): When a=1, prove that for any x in (0, +∞), the inequality f(x) > g(x) - cosx /x holds. So, given a=1, g(x) = ln x -1/x. So, the inequality is f(x) > g(x) - cosx /x. Substitute f(x) and g(x): (e^x -2)/x > (ln x -1/x) - cosx /x. Simplify the right-hand side: ln x -1/x - cosx /x = ln x - (1 + cosx)/x. So, the inequality becomes: (e^x -2)/x > ln x - (1 + cosx)/x. Multiply both sides by x (since x >0, inequality direction remains the same): e^x -2 > x ln x - (1 + cosx). Bring all terms to one side: e^x -2 -x ln x +1 + cosx >0. Simplify: e^x -x ln x -1 + cosx >0. So, we need to show that e^x -x ln x -1 + cosx >0 for all x >0. Hmm, let's define h(x) = e^x -x ln x -1 + cosx. We need to show h(x) >0 for all x >0. Let me analyze h(x). First, let's check the behavior as x approaches 0+ and as x approaches infinity. As x approaches 0+: e^x approaches 1. x ln x approaches 0 (since x approaches 0 and ln x approaches -infty, but x ln x approaches 0). cosx approaches 1. So, h(x) approaches 1 -0 -1 +1 =1. So, h(x) approaches 1 as x approaches 0+. As x approaches infinity: e^x grows exponentially. x ln x grows slower than e^x. cosx oscillates between -1 and 1. So, e^x dominates, so h(x) approaches infinity. Therefore, h(x) tends to infinity as x approaches infinity. Now, let's check h(1): h(1) = e^1 -1*ln1 -1 + cos1 = e -0 -1 + cos1 ≈ 2.718 -1 + 0.540 ≈ 2.258 >0. So, h(1) is positive. Now, let's check if h(x) ever becomes zero or negative. To ensure h(x) >0 for all x >0, we can check its derivative. Compute h’(x): h(x) = e^x -x ln x -1 + cosx. h’(x) = e^x - [ln x + x*(1/x)] - sinx = e^x - ln x -1 - sinx. So, h’(x) = e^x - ln x -1 - sinx. We need to analyze h’(x). If h’(x) is always positive, then h(x) is increasing, and since h(x) approaches 1 as x approaches 0+, and h(x) is increasing, it would stay above 1. But let's check h’(x). At x=1: h’(1) = e -0 -1 -0 = e -1 ≈ 1.718 >0. As x approaches 0+: e^x approaches 1. ln x approaches -infty, so -ln x approaches +infty. -sinx approaches 0. So, h’(x) approaches 1 - (-infty) -1 -0 = +infty. Wait, that can't be right. Wait, h’(x) = e^x - ln x -1 - sinx. As x approaches 0+, e^x approaches 1, ln x approaches -infty, so -ln x approaches +infty, -1 is -1, sinx approaches 0. So, h’(x) approaches 1 +infty -1 +0 = +infty. So, h’(x) approaches +infty as x approaches 0+. As x approaches infinity: e^x dominates, so h’(x) approaches infinity. Now, let's check if h’(x) ever becomes zero or negative. Suppose h’(x) =0: e^x - ln x -1 - sinx =0. But e^x grows exponentially, while ln x grows very slowly. So, for large x, e^x dominates, so h’(x) is positive. For x near 0, h’(x) is positive. At x=1, h’(x) is positive. Is there any x where h’(x) could be negative? Let me check x=π/2 ≈1.5708, where sinx=1. h’(π/2) = e^{π/2} - ln(π/2) -1 -1 ≈ e^{1.5708} ≈4.810, ln(π/2)≈0.4515, so 4.810 -0.4515 -1 -1 ≈2.358 >0. What about x=π ≈3.1416, sinx=0. h’(π) = e^{π} - lnπ -1 -0 ≈23.1407 -1.144 -1 ≈21.0 >0. What about x=2: h’(2)= e^2 - ln2 -1 - sin2 ≈7.389 -0.693 -1 -0.909 ≈7.389 -2.602≈4.787>0. Hmm, seems positive. Wait, maybe check x=0.5: h’(0.5)= e^{0.5} - ln0.5 -1 - sin0.5 ≈1.6487 - (-0.6931) -1 -0.4794 ≈1.6487 +0.6931 -1 -0.4794≈1.6487+0.6931=2.3418 -1.4794≈0.8624>0. So, h’(x) is positive at x=0.5. What about x=0.1: h’(0.1)= e^{0.1}≈1.1052, ln0.1≈-2.3026, sin0.1≈0.0998. So, h’(0.1)=1.1052 - (-2.3026) -1 -0.0998≈1.1052 +2.3026 -1 -0.0998≈3.4078 -1.0998≈2.308>0. So, h’(x) is positive everywhere. Therefore, h(x) is strictly increasing on (0, ∞). Since h(x) approaches 1 as x approaches 0+, and h(x) is increasing, h(x) >1 for all x>0. Wait, but earlier, when I simplified, I had h(x)=e^x -x ln x -1 + cosx. Wait, but when x approaches 0+, h(x) approaches 1, but h(x) is increasing, so h(x) >1 for x>0. But wait, the original inequality was f(x) > g(x) - cosx /x, which simplifies to h(x) >0. But if h(x) >1 for all x>0, then certainly h(x) >0. Wait, but let me double-check. Wait, when x approaches 0+, h(x) approaches 1, and h(x) is increasing, so h(x) ≥1 for all x>0. Therefore, h(x) >0 for all x>0. Hence, the inequality holds. So, that completes the proof.**Final Answer**(1) boxed{2}(2) The maximum value of ( g(x) ) on ([1, 2]) is:- (boxed{-1}) when ( a leq -1 ),- (boxed{a lnleft(-frac{1}{a}right) + a}) when ( -1 < a < -frac{1}{2} ),- (boxed{a ln 2 - frac{1}{2}}) when ( a geq -frac{1}{2} ).(3) The inequality ( f(x) > g(x) - frac{cos x}{x} ) holds for all ( x in (0, +infty) ) when ( a = 1 ).

💡First, I need to define the variables for the prices of the tickets. Let ( A ) represent the price of an adult ticket, ( C ) the price of a child ticket, and ( S ) the price of a senior ticket. According to the problem, adult tickets cost twice as much as child tickets, so ( C = frac{A}{2} ). Additionally, senior tickets cost 1.50 less than adult tickets, so ( S = A - 1.50 ).Next, I'll set up an equation based on the total cost provided. The total cost for 6 adult tickets, 5 child tickets, and 3 senior tickets is 40.50. This gives me the equation:[6A + 5left(frac{A}{2}right) + 3(A - 1.50) = 40.50]I'll simplify this equation step by step:[6A + 2.5A + 3A - 4.50 = 40.50][11.5A - 4.50 = 40.50][11.5A = 45][A = frac{45}{11.5} approx 3.91]So, the price of an adult ticket is approximately 3.91.Now, I'll calculate the new total cost for 10 adult tickets, 8 child tickets, and 4 senior tickets:[10A + 8left(frac{A}{2}right) + 4(A - 1.50) = 10(3.91) + 8left(frac{3.91}{2}right) + 4(3.91 - 1.50)][= 39.10 + 15.64 + 9.64][= 64.38]Therefore, the total cost for 10 adult tickets, 8 child tickets, and 4 senior tickets is 64.38.

💡First, I need to determine how much the group of 10 people paid in total. Since each person pays 12 for admission and 6 for the tour, the total cost per person is 18. Therefore, the group of 10 people pays 10 multiplied by 18, which equals 180.Next, I'll subtract the amount paid by the group of 10 people from the total earnings of 240 to find out how much was earned from the people who only went to the aquarium. This calculation is 240 minus 180, resulting in 60.Finally, to find out how many people only went to the aquarium, I'll divide the remaining amount of 60 by the admission fee of 12 per person. This gives me 5 people who only went to the aquarium.

💡Okay, so I have this problem here about a random variable X that follows a normal distribution with mean 1 and standard deviation 3. The question is asking for the value of c such that the probability that X is less than or equal to c is equal to the probability that X is greater than c. The options given are A: 0, B: 1, C: 2, D: 3.Alright, let me try to unpack this step by step. First, I know that in a normal distribution, the data is symmetric around the mean. That means the mean is also the median and the mode. So, if I recall correctly, the median is the value where half of the data is less than or equal to it, and the other half is greater than it. In this problem, they're saying that P(X ≤ c) = P(X > c). That sounds exactly like the definition of the median. So, if c is the median, then it should be equal to the mean because, in a normal distribution, the mean and median are the same. Wait, let me make sure I'm not confusing anything here. The mean is 1, right? So, does that mean c is 1? That would make sense because if c is the mean, then half the distribution is on either side of it. So, P(X ≤ 1) should be 0.5, and P(X > 1) should also be 0.5. But let me think again to be thorough. The normal distribution is symmetric about its mean. So, if I imagine the bell curve, the peak is at the mean, which is 1 in this case. So, any value to the left of 1 is mirrored on the right side. Therefore, the point where the area under the curve is equal on both sides should be at the mean. Is there a way to verify this without just relying on the properties of the normal distribution? Maybe by using the Z-score formula or something like that? Let's try that. The Z-score formula is Z = (X - μ)/σ, where μ is the mean and σ is the standard deviation. Here, μ is 1 and σ is 3. So, if I plug in c into this formula, I get Z = (c - 1)/3. Now, since P(X ≤ c) = P(X > c), that means P(X ≤ c) = 0.5 because the total probability is 1, and it's split equally on both sides. So, I need to find c such that the cumulative distribution function (CDF) at c is 0.5. In standard normal distribution tables, the Z-score corresponding to a CDF of 0.5 is 0. That is, Z = 0. So, setting Z = 0 in the formula, we have 0 = (c - 1)/3. Solving for c, we get c = 1. Hmm, so that confirms it. Using both the symmetry property and the Z-score method, I arrive at c = 1. Wait, but just to be extra cautious, let me consider if there's any other value that might satisfy this condition. For example, could c be something else if the distribution wasn't symmetric? But no, the problem specifically states that X follows a normal distribution, which is symmetric. So, c has to be the mean. Also, looking at the options given: A is 0, which is less than the mean; C is 2, which is greater than the mean; D is 3, which is even further away. So, none of these other options make sense because they're not the mean. Therefore, I feel confident that the correct answer is B: 1.

💡Alright, so I've got this problem here: Given that ( a, b in mathbb{R} ) and ( a - |b| > 0 ), which of the following inequalities is correct? The options are A, B, C, and D. Let me try to figure this out step by step.First, let's understand what the given condition ( a - |b| > 0 ) means. This can be rewritten as ( a > |b| ). So, ( a ) is greater than the absolute value of ( b ). That tells me a couple of things. Since ( |b| ) is always non-negative, ( a ) must be positive because it's greater than a non-negative number. Also, ( b ) can be either positive or negative, but its absolute value is less than ( a ).Now, let's look at each option one by one.**Option A: ( b - a > 0 )**Hmm, if ( b - a > 0 ), that would mean ( b > a ). But wait, from the given condition ( a > |b| ), we know that ( a ) is greater than both ( b ) and ( -b ). So, ( a ) is definitely greater than ( b ), which means ( b - a ) must be negative. Therefore, ( b - a > 0 ) is not true. So, I can eliminate option A.**Option B: ( a^{3} + b^{3} < 0 )**Okay, let's think about this. Since ( a > |b| ), and ( a ) is positive, ( a^3 ) will also be positive because any positive number raised to an odd power remains positive. Now, ( b^3 ) could be positive or negative depending on whether ( b ) is positive or negative. However, since ( |b| < a ), the magnitude of ( b^3 ) will be less than ( a^3 ). Let me test this with some numbers to be sure. Suppose ( a = 2 ) and ( b = 1 ). Then ( a^3 = 8 ) and ( b^3 = 1 ), so ( a^3 + b^3 = 9 ), which is positive. If ( b ) is negative, say ( b = -1 ), then ( b^3 = -1 ), and ( a^3 + b^3 = 8 - 1 = 7 ), still positive. So, in both cases, the sum is positive, meaning ( a^3 + b^3 < 0 ) is false. So, option B is out.**Option C: ( a^{2} - b^{2} < 0 )**This one is interesting. Let's see. ( a^2 - b^2 ) can be factored as ( (a - b)(a + b) ). Since ( a > |b| ), both ( a - b ) and ( a + b ) are positive because ( a ) is greater than ( |b| ), so ( a - b ) is positive whether ( b ) is positive or negative, and ( a + b ) is also positive because ( a ) is greater than ( |b| ). Therefore, the product ( (a - b)(a + b) ) is positive, meaning ( a^2 - b^2 > 0 ). So, ( a^2 - b^2 < 0 ) is false. Option C is incorrect.**Option D: ( b + a > 0 )**Alright, let's analyze this. Since ( a > |b| ), we know that ( a ) is positive and greater than the absolute value of ( b ). If ( b ) is positive, then ( a + b ) is definitely positive because both ( a ) and ( b ) are positive. If ( b ) is negative, ( a + b ) would be ( a - |b| ), which is still positive because ( a > |b| ). Let me test this with numbers again. If ( a = 3 ) and ( b = 2 ), then ( a + b = 5 > 0 ). If ( b = -2 ), then ( a + b = 3 - 2 = 1 > 0 ). In both cases, the sum is positive. So, ( b + a > 0 ) is always true given the condition ( a > |b| ). Therefore, option D is correct.Wait, just to make sure I haven't missed anything, let me consider edge cases. What if ( b = 0 )? Then ( a > 0 ), and ( a + 0 = a > 0 ). Still holds. What if ( a ) is just slightly bigger than ( |b| )? Say ( a = 1.1 ) and ( b = 1 ). Then ( a + b = 2.1 > 0 ). If ( b = -1 ), ( a + b = 0.1 > 0 ). Yep, still works. So, after checking all options and testing with numbers, I'm confident that option D is the correct one.

💡Okay, so I have this probability problem here: A family has two children. Let A be the event that there's at least one boy, and B be the event that both are boys. I need to find the probability of the intersection of A and B, which is written as P(B ∩ A). Hmm, okay.First, I should probably recall what the intersection of two events means. If I remember correctly, the intersection of events A and B, written as B ∩ A, is the event that both A and B occur at the same time. So, in this case, it's the probability that both children are boys and that there's at least one boy. Wait, but if both are boys, doesn't that automatically mean there's at least one boy? That seems redundant. Maybe I'm overcomplicating it.Let me think about the sample space for two children. Each child can be a boy (B) or a girl (G). So, the possible outcomes are:1. Boy and Boy (BB)2. Boy and Girl (BG)3. Girl and Boy (GB)4. Girl and Girl (GG)So, there are four possible outcomes, each equally likely if we assume that the probability of having a boy or a girl is the same, which is 1/2 for each. So, each outcome has a probability of 1/4.Now, let's define the events again:- Event A: At least one boy. So, this includes all outcomes except GG. That means A = {BB, BG, GB}.- Event B: Both are boys. So, B = {BB}.Now, the intersection of A and B, which is B ∩ A, should be the set of outcomes that are in both A and B. Since B is just {BB} and A includes {BB, BG, GB}, the intersection is just {BB}. So, B ∩ A is the same as B.Therefore, P(B ∩ A) is just the probability of B, which is the probability of both children being boys. Since there's only one outcome out of four that satisfies this, the probability should be 1/4.Wait, but let me double-check that. If I consider the definition of intersection, it's the probability that both events happen. So, in this case, both A and B happening would be the same as B happening because B is a subset of A. So, yes, P(B ∩ A) is indeed P(B), which is 1/4.Alternatively, I can think about it using the formula for conditional probability. The probability of B given A is P(B|A) = P(B ∩ A) / P(A). But since we already know that B is a subset of A, P(B ∩ A) is just P(B). So, if I calculate P(B|A), it would be P(B)/P(A). Let me compute P(A) first.P(A) is the probability of at least one boy, which is 1 minus the probability of no boys, which is 1 - P(GG). P(GG) is 1/4, so P(A) = 1 - 1/4 = 3/4.So, if I use the conditional probability formula, P(B|A) = P(B)/P(A) = (1/4)/(3/4) = 1/3. But wait, that's different from P(B ∩ A). So, I need to be careful here. P(B|A) is 1/3, but P(B ∩ A) is still 1/4 because it's just the probability of B.I think I confused myself for a second there, but now it makes sense. The intersection is just the probability of both events happening, which in this case is the same as the probability of B because B is entirely contained within A.So, to summarize:- Sample space: {BB, BG, GB, GG}- Event A: {BB, BG, GB}- Event B: {BB}- Intersection B ∩ A: {BB}- Probability: 1/4Yeah, that seems right. I don't think I made any mistakes this time. The key was recognizing that B is a subset of A, so their intersection is just B itself. Therefore, the probability is simply the probability of both children being boys, which is 1/4.