answer:1. **Introduction:** - We need to determine the maximum number of distinct n-tuples with integer entries such that any two form an exquisite pair. - An exquisite pair is defined by the inequality: [ left| a_1 b_1 + a_2 b_2 + cdots + a_n b_n right| leq 1 ] 2. **Construction of Example with Maximum Number:** - We demonstrate that it is possible to construct ( n^2 + n + 1 ) such n-tuples, where each pair satisfies the exquisite pair condition. - Consider a construction where tuples are formed such that at most two products a_i b_i are non-zero and check if their sum magnitude remains within the bound leq 1. - Example: For n = 2, consider the tuples: [ (0, 0), (0, 1), (1, 0), (0, -1), (-1, 0), (1, 1), (1, -1) ] - We can generalize this construction to higher n using * to represent any number of zeroes, ensuring each tuple formed fits the condition. - Count of tuples: (1 + n + n + binom{n}{2} + binom{n}{2} = n^2 + n + 1). 3. **Lemma to Establish an Upper Bound:** - **Lemma:** Any set of 2n + 1 distinct non-zero n-tuples of real numbers includes two tuples (a_1, ldots, a_n) and (b_1, ldots, b_n) such that: [ a_1 b_1 + ldots + a_n b_n > 0 ] 4. **Proof of Lemma:** - **Base Case (n = 1):** - For three non-zero numbers, two must share the same sign, thus their product is positive. - **Inductive Step:** - Assume it holds for n-1. - Consider 2n + 1 tuples with n entries. - Claim: We may assume one tuple is (0, 0,ldots, 0, -1). - Separate out tuples based on the sign of their last entry, ensuring non-negative last entries for remaining tuples. - Reduce problem to (n-1)-dimensional case using induction hypothesis. - Prove that resulting products and manipulations do not violate exquisite pair conditions, preserving sign criterion. 5. **Conclusion from Lemma:** - We construct disjoint subsets (A_1, A_2, ldots, A_n) each limited to (2i) such that: [ |A_i| leq 2i quad implies quad 2+4+ldots+2n = n^2 + n ] 6. **Proof of Maximum:** - Contradiction: If |A_i| geq 2i + 1, subsets produce sums geq 2, violating exquisite pair definition. - Therefore, no more than n^2 + n + 1 tuples may exist such that all pairs satisfy the condition. # Conclusion: [ boxed{n^2 + n + 1} ]

answer:# Solution Details Part (Ⅰ) Given the function f(x) = sqrt{3}sin omega x - cos omega x, let's first rewrite it in a form that will make it easier to work with: [f(x) = sqrt{3}sin omega x - cos omega x = 2sinleft(omega x - frac{pi}{6}right).] Given that the distance between two adjacent symmetric axes of its graph is frac{pi}{2}, we understand that the graph's symmetry implies a periodic nature. Thus, the minimum positive period of f(x) can be deduced to be pi. Consequently, the relationship between omega and the period T of the sine function, which is T = frac{2pi}{omega}, leads us to: [frac{2pi}{omega} = pi.] Solving this equation for omega gives us: [omega = 2.] Therefore, we rewrite f(x) with omega = 2 as: [f(x) = 2sinleft(2x - frac{pi}{6}right).] When we set f(x) = 1, we get: [2sinleft(2x - frac{pi}{6}right) = 1.] This equation can be solved for x in terms of k (where k in mathbb{Z}), giving us: [2x - frac{pi}{6} = 2kpi + frac{pi}{6}] or [2x - frac{pi}{6} = 2kpi + frac{5pi}{6}.] Solving these equations for x yields: [x = kpi + frac{pi}{6}] or [x = kpi + frac{pi}{2},] where k in mathbb{Z}. Part (Ⅱ) When the graph of f(x) is shifted to the left by m units, the new function can be expressed as: [y = 2sinleft(2x + 2m - frac{pi}{6}right).] Given that this new graph coincides with the graph of y = 2cos 2x, we set the phase shift equal to the argument of the cosine function: [2m - frac{pi}{6} = 2kpi + frac{pi}{2},] where k in mathbb{Z}. Solving this equation for m gives: [m = kpi + frac{pi}{3},] where k in mathbb{Z}. Since m > 0, the smallest positive value of m occurs when k = 0, which yields: [m = frac{pi}{3}.] Therefore, the minimum value of m is boxed{frac{pi}{3}}.

answer:Let's denote the number of balls selected as ( n ). Given that the probability of picking an odd-numbered ball first is ( frac{2}{3} ) (or 0.6666666666666666), we can set up the following equation: [ P(text{First ball is odd}) = frac{text{Number of odd balls}}{text{Total number of balls}} ] Since there are 100 balls in total, with 50 odd and 50 even, the probability of picking an odd ball first is: [ P(text{First ball is odd}) = frac{50}{100} = frac{1}{2} ] However, the given probability is ( frac{2}{3} ), which means that the selection of balls must be influencing this probability. The only way the probability of picking an odd ball first can be ( frac{2}{3} ) is if the selection of balls is such that there are two odd balls and one even ball. Let's denote the number of odd balls selected as ( O ) and the number of even balls selected as ( E ). We know that ( O = 2 ) and ( E = 1 ). The total number of balls selected is ( n = O + E = 2 + 1 = 3 ). Therefore, boxed{3} balls are selected.

answer:**Analysis** This question tests the method of finding the number of elements in the intersection, which is a basic problem. When solving it, one should carefully read the question and correctly apply the definition of intersection. **Answer** Given A={1,2,3,4} and B={2,4,6,8}, Therefore, A cap B = {2,4}, Thus, the number of elements in A cap B is 2. Hence, the correct choice is boxed{B}.