answer:Let's denote the number of people born every two seconds as B. Given that the death rate is 1 person every two seconds, the net increase in population every two seconds is B - 1. We want to find the net increase in population over one day (24 hours). There are 60 minutes in an hour and 60 seconds in a minute, so there are 60 * 60 = 3600 seconds in an hour. Therefore, in 24 hours, there are 24 * 3600 = 86400 seconds. Since the net increase occurs every two seconds, we need to divide the total number of seconds in a day by 2 to find out how many times this net increase happens in one day: 86400 seconds / 2 seconds = 43200 times. The estimated size of the population net increase in one day is 259200, and this happens over the course of 43200 times (every two seconds). Therefore, the net increase per two-second interval is: 259200 net increase / 43200 times = 6 people net increase every two seconds. Since this is the net increase, and we know that 1 person dies every two seconds, we can add that 1 person back to the net increase to find out the total number of births every two seconds: 6 people net increase + 1 person death = 7 people born every two seconds. Therefore, boxed{7} people are born every two seconds in the city.

answer:From 2x + frac{π}{4} = frac{π}{2}, we get x = frac{π}{8}, ∴ x₁ + x₂ = 2 × frac{π}{8} = frac{π}{4}, From 2x + frac{π}{4} = frac{3π}{2}, we get x = frac{5π}{8}, ∴ x₂ + x₃ = 2 × frac{5π}{8} = frac{5π}{4}, ∴ 2x₁ + 3x₂ + x₃ = 2(x₁ + x₂) + x₂ + x₃ = 2 × frac{π}{4} + frac{5π}{4} = frac{7π}{4}, Hence, the answer is: boxed{text{D}}. First, find the first and second symmetry axes on the right side of the y-axis, which are x = frac{π}{8} and x = frac{5π}{4}, respectively. Then, calculate 2x₁ + 3x₂ + x₃ = 2(x₁ + x₂) + x₂ + x₃ using these symmetry axes. This question tests your understanding of the sine function's graph. It is of medium difficulty.

answer:Differentiate f(x) = x^3 + ax^2 + (a-2)x to obtain f'(x) = 3x^2 + 2ax + (a-2) Since f'(x) is an even function, we have f'(x) = f'(-x) Substituting into the equation, we get 3x^2 + 2ax + (a-2) = 3x^2 - 2ax + (a-2) Simplifying, we find a = 0 Therefore, the answer is boxed{a=0}.

answer:1. **Define Set (D) and (D')**: Let (D) be the set of all prime divisors of (n). Notice that (D') is the set of prime divisors of (n) that are greater than some prime (q). 2. **Restate the Problem**: We are given that some numbers in the set ({1, 2, 3, ldots, n}) are colored red. The condition specifies that if a number (a) from the set is red and (a neq 1), then any multiple of (a) up to (n) must also be red. We aim to demonstrate that the number of red numbers does not exceed Euler's totient function (varphi(n)). 3. **Restate the Lemma**: (D) is a set of distinct primes dividing (n). The number of integers not exceeding (n) and not divisible by any prime in (D) equals [ n prod_{p in D} left(1 - frac{1}{p}right) ] According to Euler's totient function: [ varphi(n) = n prod_{p in D} left(1 - frac{1}{p}right) ] 4. **Assume For Contradiction**: Assume the number of red numbers is greater than (varphi(n)). Therefore, among the red numbers, there will exist at least one number that shares a common prime factor with (n) due to the pigeonhole principle. 5. **Selection of Prime (q)**: Let (q) be the largest prime factor of (n). A red number (a) is considered such that it is divisible by (q). 6. **Consider Modular Congruence**: By the assumption, we can find distinct red numbers (b) and (c) such that: [ b equiv c pmod{frac{n}{q}} ] This shows that these red numbers share the same remainder when divided by (frac{n}{q}). 7. **Application of the Lemma**: The lemma implies that the count of natural numbers less than or equal to (n) that are not divisible by any element in (D') (a subset of (D)) is: [ n prod_{p in D', p neq q} left(1 - frac{1}{p}right) < varphi(n) ] These numbers represent the possible values modulo (frac{n}{q}) which are fewer than needed for more than (varphi(n)) possibilities. 8. **Contradiction**: Since (varphi(n)) cannot be exceeded, by our assumption of having more red numbers, we create a contradiction. Every value modulo (frac{n}{q}) must have at least one corresponding red number, indicating a discrepancy in our calculation due to the unique factorization theorem if (varphi(n)) constraint is exceeded. # Conclusion: Hence, the number of red numbers does not exceed (varphi(n)). [ boxed{text{Therefore, the number of red numbers is at most } varphi(n).} ]