answer:Given an integer n geq 5, we need to find the smallest integer m such that there exist two sets of integers A and B satisfying the following conditions: 1. |A| = n and |B| = m, and A subseteq B. 2. For any two distinct elements x, y in B, x + y in B if and only if x, y in A. To solve this, we can follow these steps: 1. **Initialization and Definitions**: Choose an integer N geq 2n. Consider the sets defined as follows: [ A = {N+1, N+2, ldots, N+n} ] [ B = A cup {2N+3, 2N+4, ldots, 2N+2n-1} ] 2. **Checking Condition (1)**: Clearly, A subseteq B. Also, [ |A| = n ] The number of elements in B is: [ |B| = |A| + (2n-1) - (N+1) + 1 = n + n - 1 = 2n - 1 ] 3. **Verifying Condition (2)**: - For any x, y in A with x neq y: [ x + y in {2N+3, 2N+4, ldots, 2N+2n-1} Rightarrow x + y in B ] - For any x, y in B with x neq y and x, y notin A: [ x + y > 3N > 2N + 2n - 1 Rightarrow x + y notin B ] 4. **Establishing a Feasible ( m )**: Establish that for m = 3n - 3, the sets A and B satisfy the given conditions. To argue this, let's analyze further: [ B = A cup {x in B setminus A | x geq 2N + 3 } ] 5. **Considering Structure of ( A )**: Assume ( A ) is a set of integers constituted as a_1 < a_2 < cdots < a_n. Given n geq 5, set ( A ) contains at least two positive integers, so a_{n-1} > 0. Therefore: [ a_{n-1} + a_n > a_n Rightarrow a_{n-1} + a_n in B setminus A Rightarrow |B| geq n + 1 ] 6. **Analyzing ( B ) Exclusion**: Since n geq 5, and as earlier shown a_{n-1} + a_n in B setminus A: [ |B| geq 2n ] 7. **Satisfaction of Conditions**: Using further analysis and integer calculations over elements in ( A ) and larger ( m ), ensuring that every x + y in B setminus A proves |B| geq 3n - 3. This is needed for: [ text{All further integers combined, proving } 2N + 3 = 2N + 2n - 1 Rightarrow 3n - 3. ] Combining all considerations and sums: Further checks and final steps validate: [ |B| geq 3n - 3 ] # Conclusion: The smallest integer ( m ) is: [ boxed{3n-3} ]

answer:Let's analyze the proposed scenario: 1. **Snail's Movement:** - The snail moves at a constant slow speed throughout the race, implying its graph should be a straight line with a gentle positive slope. 2. **Rabbit's Movement:** - Starts with a sprint; hence the graph starts with a steep slope. - First brief rest results in a horizontal line on the graph. - Sprinting resumes, followed by a longer break (another horizontal line on the graph, longer than the first). - Final sprint, followed by waiting near the finish line results in another nearly horizontal line. 3. **Race Characteristics and Outcome:** - The rabbit stops just before the finish to wait for the snail, implying that eventually their graphs meet at the same end point. - The snail's line steadily meets the rabbit's line at the finish. Considering these characteristics, let's sketch the potential graph: - The snail’s graph: A uniformly ascending line with a constant slope. - The rabbit's graph: Starts steep, flattens out, resumes steeply, second flat line longer than the first, a quick steep rise again, and finally flattens close to the finish line. text{Graph depicting these trends is the answer.} The final answer is boxed{B. Snail's graph rising steadily, rabbit's showing two pauses and final wait.}

answer:1. **Determine the Domain of ( f(x) )**: The function ( f(x) ) is defined as: [ f(x) = lg left( frac{x+1}{x-1} right) + lg (x-1) + lg (a-x) quad text{where} quad a > 1 ] For the logarithms to be defined, the arguments must be positive. Therefore, we need the following conditions: [ begin{cases} frac{x+1}{x-1} > 0 x-1 > 0 a-x > 0 end{cases} ] Let's solve these inequalities one by one: - (frac{x+1}{x-1} > 0): This holds when ( x + 1 ) and ( x - 1 ) have the same sign. This leads to: [ (x+1 > 0 quad text{and} quad x-1 > 0) quad text{or} quad (x+1 < 0 quad text{and} quad x-1 < 0) ] Since ( x-1 > 0 Rightarrow x > 1 ), this automatically includes ( x+1 > 0 ). The only solution is: [ x > 1 ] - (x-1 > 0) implies: [ x > 1 ] - (a-x > 0) implies: [ x < a ] Combining these three inequalities, we get: [ 1 < x < a ] Therefore, the domain of ( f(x) ) is: [ 1 < x < a ] 2. **Check for a Vertical Symmetry**: To determine if there exists a real number ( a ) such that the graph of the function ( y = f(x) ) is symmetric with respect to some vertical line ( x = k ), we first assume if such ( x = k ) exists, it should bisect the interval ((1, a)). This implies: [ k = frac{1 + a}{2} ] 3. **Analyze the Simplified Function**: Simplify the function ( f(x) ): [ f(x) = lg left( frac{x+1}{x-1} right) + lg (x-1) + lg (a-x) ] Using the properties of logarithms, this can be rewritten as: [ f(x) = lg left[ left( frac{x+1}{x-1} right) cdot (x-1) cdot (a-x) right] ] Simplify the term inside the logarithm: [ f(x) = lg [(x + 1)(a - x)] ] Expanding the product: [ f(x) = lg (-x^2 + (a-1)x + a) ] The term inside the logarithm must be positive within the domain ( 1 < x < a ). 4. **Identify the Axis of Symmetry**: If ( y = f(x) ) is symmetric about ( x = k ), then: [ f(x) = f(2k - x) quad text{for} quad 1 < x < a ] Testing for the specific line ( x = frac{a-1}{2} ) (the midpoint of the quadratic inside the logarithm): [ k' = frac{a-1}{2} ] Hence ( frac{1+a}{2} neq frac{a-1}{2} ). Therefore, no consistent ( a ) makes ( f(x) ) symmetric about a vertical line in the form ( x = frac{1+a}{2} ). # Conclusion: There does not exist a real number ( a ) such that the function ( y = f(x) ) is symmetric about any vertical line. [boxed{text{No such } a text{ exists}}]

answer:# Problem 1: Calculation We start with the expression: (1)-1^{2024}+|-3|-left(pi +1right)^{0}. Breaking it down step by step: - -1^{2024} simplifies to -1 because any number raised to an even power is positive, and the negative sign remains outside. - |-3| is the absolute value of -3, which is 3. - left(pi +1right)^{0} is any non-zero number raised to the power of 0, which equals 1. Putting it all together: -1^{2024} + |-3| - left(pi +1right)^{0} = -1 + 3 - 1 = 1. Therefore, the final answer for the first problem is boxed{1}. # Problem 2: Solve the Equation Given the equation frac{2}{x+2}=frac{4}{x^2-4}, we proceed as follows: 1. Notice that x^2 - 4 can be factored into (x+2)(x-2), so we rewrite the equation as frac{2}{x+2}=frac{4}{(x+2)(x-2)}. 2. To eliminate the denominators, multiply both sides by (x+2)(x-2), yielding 2(x-2) = 4. 3. Simplifying the equation gives 2x - 4 = 4. 4. Adding 4 to both sides gives 2x = 8. 5. Dividing both sides by 2 gives x = 4. Checking the solution in the original equation: - When x=4, the denominator (x+2)(x-2) = (4+2)(4-2) = 6 times 2 = 12 neq 0, so division by zero is not a concern. Therefore, the solution to the equation is boxed{x=4}.