answer:Solution: For the function y=sin x, the intervals where it is increasing correspond to: [2kpi- frac {pi}{2},2kpi+ frac {pi}{2}], where kin mathbb{Z}. Letting k=0, we get - frac {pi}{2} leqslant x leqslant frac {pi}{2}, thus, the correct option is: boxed{B}. This problem can be solved by understanding the properties of the sine function and how to determine intervals of increase. It examines the graph of the sine function and the method of finding intervals of increase, which is quite basic.

answer:Let N = a_n a_{n-1} ldots a_1 a_0 be the decimal representation of a number. We express this number mathematically as: [ N = sum_{i=0}^n a_i cdot 10^i ] 1. **Transformation Definition**: We are given a transformation rule that replaces the number N = 10a + b with N' = a + 2b. Our goal is to apply this transformation iteratively until the resulting number is less than or equal to 19 and prove the condition involving divisibility by 19. 2. **Process Insight**: For the transformation rule N = 10a + b converted to N' = a + 2b, observe that: [ N - N' = 10a + b - (a + 2b) = 9a - b ] 3. **Decreasing Function**: We need to show that this transformation decreases the value of the number if N > 19. Suppose N > 19. If N = 10a + b, here a and b are its respective digit components. We assume the inequality: [ 10a + b > a + 2b ] This simplifies to: [ 9a > b ] Since 9a - b must be non-negative, it establishes that N' = a + 2b will be lesser than or equal to 19 after sufficient transformations, as each transformation will decrease the number. This progressive decrease moves us towards a number that is less than or equal to 19 eventually. 4. **Divisibility Condition**: To prove that the original number N is divisible by 19 if and only if the resulting number after several transformations is 19, we proceed as follows: - **Forward Direction**: Assume that N is divisible by 19. Suppose: [ N = 10a + b quad text{is divisible by 19} ] Applying the transformation, the new number becomes: [ N' = a + 2b ] We want to show that if N is divisible by 19, then so is N'. Rewrite the transformation in a congruence form: [ 20a + 2b equiv 0 pmod{19} ] Since 20 equiv 1 pmod{19}, this reduces to: [ a + 2b equiv 0 pmod{19} ] Hence, a + 2b is indeed divisible by 19. - **Backward Direction**: Conversely, suppose the iterative transformation of N = 10a + b results in the final number 19. Since the transformations have maintained the property that: [ a + 2b equiv 0 pmod{19} quad text{(before hitting 19)} ] Considering N' = 19 equiv 0 pmod{19}, this implies the initial number N should also adhere to the property of divisibility by 19. # Conclusion: By mathematical induction via transformation, this proves that the original number is divisible by 19 if and only if, after applying the transformation multiple times, the final resulting number is exactly 19. [ boxed{N text{ is divisible by 19 if and only if the final resulting number is 19}} ]

answer:To find f'(0), we first need to differentiate f(x) = sin(x + e^x). Using the chain rule, we get: f'(x) = cos(x + e^x) cdot (1 + e^x) Substituting x = 0 into the derivative, we have: f'(0) = cos(0 + e^0) cdot (1 + e^0) = cos(1) cdot 2 Since cos(1) is a non-zero value and the question is a multiple-choice one, we focus on the coefficient of cos(1), which is 2. Therefore, the correct answer is: boxed{B}

answer:# Problem 1: Given real numbers x > 0, y > 0, and frac{2}{x} + y = 1, find the minimum value of 2x + frac{1}{3y}. **Solution:** 1. Given that x > 0, y > 0, and frac{2}{x} + y = 1, we start by multiplying 2x + frac{1}{3y} with the given equation: 2x + frac{1}{3y} = left(2x + frac{1}{3y}right)left(frac{2}{x} + yright) 2. Expanding the right-hand side, we get: = 2xy + frac{2}{3xy} + frac{13}{3} 3. By applying the AM-GM inequality, where AM ge GM, for the terms 2xy and frac{2}{3xy}: 2xy + frac{2}{3xy} + frac{13}{3} ge 2sqrt{2xy cdot frac{2}{3xy}} + frac{13}{3} = frac{13 + 4sqrt{3}}{3} 4. The equality holds if and only if 2xy = frac{2}{3xy}, which simplifies to xy = frac{sqrt{3}}{3}. 5. Given x > 0, y > 0, and frac{2}{x} + y = 1, and solving these with xy = frac{sqrt{3}}{3}, we find that x = 2 + frac{sqrt{3}}{3} and y = frac{2sqrt{3} - 1}{11} satisfy the conditions for equality. 6. Therefore, the minimum value of 2x + frac{1}{3y} is boxed{frac{13 + 4sqrt{3}}{3}}. # Problem 2: Given real numbers x and y satisfying x + y = 1, x > 0, y > 0, find the minimum value of frac{1}{2x} + frac{x}{y + 1}. **Solution:** 1. With x + y = 1, x > 0, and y > 0, we express the given function in terms of x and y: frac{1}{2x} + frac{x}{y + 1} = frac{x + y}{2x} + frac{x}{x + 2y} = frac{1}{2} + frac{y}{2x} + frac{1}{1 + frac{2y}{x}} 2. Letting frac{y}{x} = t > 0, we can rewrite the expression as: = frac{1}{2} + frac{t}{2} + frac{1}{1 + 2t} = frac{2t^2 + 3t + 3}{2(1 + 2t)} 3. Simplifying further: = frac{1}{4} cdot frac{(1 + 2t)^2 + 2t + 1 + 4}{1 + 2t} = frac{1}{4} left[(1 + 2t) + frac{4}{1 + 2t} + 1right] 4. Applying the AM-GM inequality: ge frac{1}{4} times 2sqrt{(1 + 2t) cdot frac{4}{1 + 2t}} + frac{1}{4} = boxed{frac{5}{4}} 5. The equality holds if and only if 1 + 2t = frac{4}{1 + 2t}, which simplifies to t = frac{y}{x} = frac{1}{2}. 6. Given x + y = 1, solving with t = frac{1}{2}, we find that x = frac{2}{3} and y = frac{1}{3} satisfy the equality conditions. 7. Hence, the minimum value of frac{1}{2x} + frac{x}{y + 1} is boxed{frac{5}{4}}.