answer:To find the range of the function ( f(x, y) = frac{x + y}{[x][y] + [x] + [y] + 1} ) under the constraint ( x y = 1 ) and ([x]) denotes the floor function, we proceed as follows: 1. **Assumption and Initial Case:** - Assume, without loss of generality, ( x geqslant y ). 2. **Case when ( x = 1, y = 1 ):** - Calculate: [ f(1, 1) = frac{1 + 1}{[1][1] + [1] + [1] + 1} = frac{2}{1 cdot 1 + 1 + 1 + 1} = frac{2}{4} = frac{1}{2} ] 3. **Case when ( x > 1 ):** - Let ( x = n + alpha ) where ( n = [x] ) (the greatest integer less than or equal to ( x )). - Given ( xy = 1 ) and thus ( y = frac{1}{x} = frac{1}{n + alpha} ). - We analyze the function: [ f(x, y) = frac{n + alpha + frac{1}{n + alpha}}{n + 1} ] 4. **Analysis of the Function ( g(x) = x + frac{1}{x} ):** - Note ( g(x) ) is increasing for ( x geqslant 1 ). - We assess the min and max values over the interval ([n, n+1)). 5. **Bounds Calculation:** - Minimum value: [ a_n = frac{n + frac{1}{n}}{n + 1} = frac{n^2 + 1}{n^2 + n} ] - Maximum value: [ b_n = frac{n + 1 + frac{1}{n + 1}}{n + 1} = 1 + frac{1}{(n + 1)^2} ] 6. **Behavior for different ( n ):** - We observe: - ( a_1 > a_2 = a_3, a_3 < a_4 < cdots ) - ( b_1 > b_2 > b_3 > cdots ) - Specifically: [ a_1 = frac{1 + 1}{1 + 1} = 1, quad a_2 = frac{2 + frac{1}{2}}{2 + 1} = frac{5}{6}, quad b_1 = 1 + frac{1}{4} = frac{5}{4} ] 7. **Conclusion for ( x > 1 ):** - The range of ( f(x, y) ) for ( x > 1 ) is: [ left[ a_2, b_1 right) = left[ frac{5}{6}, frac{5}{4} right) ] 8. **Overall Conclusion:** - Combining outcomes, the range of ( f(x, y) ) is: [ f(x, y) in left{ frac{1}{2} right} cup left[ frac{5}{6}, frac{5}{4} right) ] (boxed{left{ frac{1}{2} right} cup left[ frac{5}{6}, frac{5}{4} right)})

answer:Let's denote the original rate of interest as R% and the sum of money as P. According to the given information, if the sum had been put at a (R + 7.5)% rate, it would have fetched 630 more in interest over 12 years. The interest for the original rate R% for 12 years is: I = P * R * 12 / 100 The interest for the increased rate (R + 7.5)% for 12 years is: I' = P * (R + 7.5) * 12 / 100 According to the problem, the difference in interest (I' - I) is 630: I' - I = 630 Substituting the values of I and I' we get: P * (R + 7.5) * 12 / 100 - P * R * 12 / 100 = 630 Now, let's simplify the equation: P * 12 * (R + 7.5) / 100 - P * 12 * R / 100 = 630 P * 12 * R / 100 + P * 12 * 7.5 / 100 - P * 12 * R / 100 = 630 P * 12 * 7.5 / 100 = 630 Now, we can solve for P: P = 630 * 100 / (12 * 7.5) P = 63000 / 90 P = 700 Therefore, the sum of money (P) is boxed{700} .

answer:We will analyze each proposition to determine its validity. (1) The negation of the proposition "For all x in (0,2), 3^x > x^3" is indeed "There exists x_0 in (0,2) such that 3^{x_0} leq x_0^3". This statement is correct. (2) If f(x) = 2^x - 2^{-x}, then f(-x) = 2^{-x} - 2^x. Since 2^{-x} = frac{1}{2^x}, we have f(-x) = frac{1}{2^x} - 2^x = -(2^x - frac{1}{2^x}) = -f(x). Thus, the proposition is correct. (3) If f(x) = x + frac{1}{x+1}, then f(x) = 1 implies x + frac{1}{x+1} = 1, which leads to x = 0. However, x = 0 is not in the domain of f(x). Therefore, the proposition is false. (4) In triangle ABC, if angle A > angle B, then side a > side b (opposite to the respective angles). Using the law of sines, we have frac{a}{sin A} = frac{b}{sin B} = 2R, where R is the circumradius. Therefore, a > b Rightarrow 2R sin A > 2R sin B Rightarrow sin A > sin B. Thus, the proposition is correct. The true propositions are boxed{1, 2, 4}.

answer:1. Since sin A + sin B = (cos A + cos B) sin C, by applying the cosine rule and sine rule, we have a + b = (frac{b^2 + c^2 - a^2}{2bc} + frac{c^2 + a^2 - b^2}{2ca})c quad ...(2) Simplifying and rearranging the equation, we get (a + b)(a^2 + b^2) = (a + b)c^2. Since a + b > 0, it implies that a^2 + b^2 = c^2 quad ...(4). Therefore, triangle ABC is a right triangle with angle C = 90^circ quad ...(6). 2. Given that a + b + c = 1 + sqrt{2} and a^2 + b^2 = c^2, we have 1 + sqrt{2} = a + b + sqrt{a^2 + b^2} geq 2sqrt{ab} + sqrt{2ab} = (2 + sqrt{2})sqrt{ab}. The equality holds if and only if a = b. Thus, we get sqrt{ab} leq frac{sqrt{2}}{2} quad ...(8). Therefore, the area of triangle ABC is S_{triangle ABC} = frac{1}{2}ab leq frac{1}{2} times (frac{sqrt{2}}{2})^2 = frac{1}{4}. Hence, the maximum area of triangle ABC is boxed{frac{1}{4}} quad ...(12).