answer:# Step-by-Step Solution: Part 1: Finding the Cost Price per Piece Let's denote the cost price of type A souvenir as x yuan, and the cost price of type B souvenir as y yuan. From the given information, we can set up the following system of equations: 1. For 1 piece of type A and 5 pieces of type B: x + 5y = 52 2. For 3 pieces of type A and 4 pieces of type B: 3x + 4y = 68 Solving this system, we start by multiplying the first equation by 3 to eliminate x: - 3(x + 5y) = 3(52) which simplifies to 3x + 15y = 156 Subtracting the second original equation from this result: - (3x + 15y) - (3x + 4y) = 156 - 68 - 11y = 88 - y = 8 Substituting y = 8 into the first original equation: - x + 5(8) = 52 - x + 40 = 52 - x = 12 Therefore, the cost price per piece of type A and type B souvenirs are x = 12 yuan and y = 8 yuan, respectively. boxed{text{The cost price of type A souvenir is 12 yuan, and type B souvenir is 8 yuan.}} Part 2: Finding the Number of Purchasing Plans Let t be the number of type A souvenirs purchased, then (100 - t) is the number of type B souvenirs purchased. Given the total cost is between 992 yuan and 1002 yuan, we can write the inequality as: 992 leqslant 12t + 8(100 - t) leqslant 1002 Simplifying the inequality: 992 leqslant 12t + 800 - 8t leqslant 1002 Combining like terms: 192 leqslant 4t leqslant 202 Dividing through by 4: 48 leqslant t leqslant 50.5 Since t must be an integer (you can't purchase a fraction of a souvenir), t can be 48, 49, or 50. Thus, there are three purchasing plans: 1. 48 pieces of type A and 52 pieces of type B. 2. 49 pieces of type A and 51 pieces of type B. 3. 50 pieces of type A and 50 pieces of type B. boxed{text{There are three purchasing plans.}}

answer:Given that f(1-x)=f(x), we have a symmetry around x=frac{1}{2}. This implies that f'(frac{1}{2})=0. Since (x-2)f''(x)>0, there are two cases to consider: 1. x<2 and f''(x)>0, which means f(x) is concave up on (-infty, 2). 2. x>2 and f''(x)<0, which means f(x) is concave down on (2, infty). Now, let's examine the possible values of x_1 and x_2. Since x_1 < x_2 and x_1+x_2 > 1, either both x_1 and x_2 are greater than frac{1}{2}, or one of them is less than frac{1}{2} and the other is greater than frac{1}{2}. Case 1: Both x_1 and x_2 are greater than frac{1}{2}. In this case, f(x) is concave up on the interval between x_1 and x_2. As x_1 < x_2, we have f(x_1) < f(x_2). Case 2: One of x_1 and x_2 is less than frac{1}{2} and the other is greater than frac{1}{2}. Without loss of generality, let x_1 < frac{1}{2} < x_2. Since f(x) is symmetric around x=frac{1}{2}, and f(x) is increasing on (frac{1}{2}, 2) (due to being concave up), we have f(x_1) = f(1-x_1) < f(x_2). Therefore, the correct answer is: boxed{f(x_1) < f(x_2)}.

answer:Given that (a + b + c + d = 200), we seek to maximize the expression (2ab + 3bc + 4cd). First, express (2ab + 3bc + 4cd) in a similar form as the original problem, (ab + bc + cd), but with adjusted coefficients: [2ab + 3bc + 4cd.] We can estimate a bound by adding terms to form groups: [2ab + 3bc + 4cd le 2ab + 3bc + 4cd + ad = 2(ab + ad + bc + cd) + bc ] However, this doesn't simplify nicely. We rearrange using AM-GM Inequality: [ab + ad + bc + cd le left(frac{a + b + c + d}{2}right)^2 = left(frac{200}{2}right)^2 = 100^2 = 10000.] For (2ab + 3bc + 4cd + ad le 2 cdot 10000 + bc), we adjust the simplest case where one term dominates; Set (a = d = 0, b = 100, c = 100), then (2ab + 3bc + 4cd = 3(100 times 100) = 30000.) Thus, we hypothesize the maximum value occurs when (b = c = 100) and (a = d = 0). Here, (2ab + 3bc + 4cd = 30000.) Conclusion: The largest possible value of (2ab + 3bc + 4cd) given (a + b + c + d = 200) is (boxed{30000}).

answer:The probability of selecting any of the three teams is frac{1}{3}. Let n be the number of students in the selected team. For a team with n students and 3 co-captains, the number of ways to choose three students is binom{n}{3}, and the number of ways to specifically choose the 3 co-captains is binom{3}{3} = 1. Hence, the probability of choosing the 3 co-captains, given that a team has been chosen, is: P(text{selecting 3 co-captains} | text{team chosen}) = frac{1}{binom{n}{3}} = frac{6}{n(n-1)(n-2)} Calculating for each team: - For the team with 6 students: frac{6}{6 cdot 5 cdot 4} = frac{6}{120} = frac{1}{20} - For the team with 9 students: frac{6}{9 cdot 8 cdot 7} = frac{6}{504} = frac{1}{84} - For the team with 10 students: frac{6}{10 cdot 9 cdot 8} = frac{6}{720} = frac{1}{120} Summing these probabilities, weighted by the 1/3 chance of selecting each team: frac{1}{3} left(frac{1}{20} + frac{1}{84} + frac{1}{120}right) = frac{1}{3} left(frac{25.2 + 6 + 4.2}{840}right) = frac{1}{3} left(frac{35.4}{840}right) = frac{35.4}{2520} = frac{177}{12600} Thus, the total probability is boxed{frac{177}{12600}}.