answer:Let t= sin x, t in (0,1], then the range of the function frac {sin x+a}{sin x}(0 < x < π) is the same as the range of the function y=1+ frac {a}{t}, t in (0,1]. Since a > 0, y=1+ frac {a}{t}, t in (0,1], is a decreasing function. When t=1, the function has a minimum value 1+a, and the function has no maximum value. Therefore, the correct answer is B. By substitution, let t= sin x, t in (0,1], the function is transformed into y=1+ frac {a}{t}, t in (0,1]. By studying the monotonicity and range of the function y=1+ frac {a}{t} with respect to t, it can be concluded that the original function is a decreasing function, thus obtaining the correct answer. This problem tests the understanding of the maximum and minimum values of a function and their geometric meaning. It is a moderate difficulty problem. When solving the problem, pay attention to the application of the substitution method and the technique of using the monotonicity of the inverse proportional function. Solution Enhancement: Let's first substitute t = sin x, where t in (0, 1], since sin x is positive and less than or equal to 1 for x in (0, π). This substitution transforms our original function f(x)= frac {sin x+a}{sin x}(0 < x < π) into y = 1 + frac{a}{t}, where t in (0, 1]. Now, let's analyze the transformed function y = 1 + frac{a}{t}. Since a > 0, as t approaches 0, y will increase indefinitely, and as t approaches 1, y will approach its minimum value 1 + a. Thus, the function has a minimum value of boxed{1 + a}, but it does not have a maximum value since it keeps increasing as t approaches 0. The final answer is B: It has a minimum value but no maximum value.

answer:We calculate the first few terms as follows: [ a_1 = 2023, a_2 = 2024, ] [ a_3 = -4044 + (1+2) = -4041 text{ (since } 2023 + 2024 + a_3 = 1+2) ] [ a_4 = 2024 text{ (since } 2024 - 4041 + a_4 = 2+2) ] [ a_5 = -4040 + (3+2) = -4037 text{ (since } -4041 + 2024 + a_5 = 3+2) ] It appears that each term is 1 greater than the number three terms previous, i.e., a_{n+3} = a_n + 1. This pattern results from the adjusted recurrence relation, where: [ a_n + a_{n+1} + a_{n+2} = n+2 ] [ a_{n+1} + a_{n+2} + a_{n+3} = n+3 ] Subtracting the first from the second gives: [ a_{n+3} - a_n = 1 ] Thus, we find: [ a_1 = 2023, a_4 = 2024, a_7 = 2025, ldots ] [ a_{1000} = 2023 + left(frac{1000-1}{3}right) = 2023 + 333 = boxed{2356} ]

answer:Solution: (I) f(x)=2sin^2left(frac{pi}{4}+omega xright)-sqrt{3}cos 2omega x-1 =-cosleft(frac{pi}{2}+2omega xright)-sqrt{3}cos 2omega x =sin 2omega x-sqrt{3}cos 2omega x=2sinleft(2omega x-frac{pi}{3}right) (omega > 0) Since the smallest positive period of f(x) is frac{2pi}{3}, therefore frac{2pi}{2omega}=frac{2pi}{3}, therefore omega=frac{3}{2}. So, boxed{omega=frac{3}{2}}. (II) From (I), we know f(x)=2sinleft(3x-frac{pi}{3}right). When xinleft[frac{pi}{6}, frac{pi}{2}right], we have 3x-frac{pi}{3}inleft[frac{pi}{6}, frac{7pi}{6}right], then f(x)in[-1,2]. therefore If the inequality |f(x)-m| < 2 always holds for xinleft[frac{pi}{6}, frac{pi}{2}right], then -2 < f(x)-m < 2, i.e., f(x)-2 < m < f(x)+2 always holds for xinleft[frac{pi}{6}, frac{pi}{2}right], therefore (f(x)-2)_{max} < m < (f(x)+2)_{min}, f(x)_{max}-2 < m < f(x)_{min}+2 therefore boxed{0 < m < 1}.

answer:We need to find the smallest palindrome greater than 134,782 and calculate the positive difference between this number and 134,782. 1. The number 134,782 is a six-digit number. The next possible palindrome should have the same number of digits or more. We start by considering six-digit palindromes. 2. The palindrome must be greater than 134,782. We look for the smallest six-digit palindrome starting with '134' or '135'. Since no six-digit palindrome starts with '134' that is greater than 134,782, we consider '135'. 3. The smallest six-digit palindrome starting with '135' is 135531. 4. Calculate the difference between 134,782 and 135531: 135531 - 134782 = 749. Thus, the least natural number that can be added to 134,782 to create a palindrome is boxed{749}.