answer:To analyze the function f(x)=ln x+ax^{2}-2 and determine the range of a for which f(x) is monotonically decreasing in the interval (1,2), we proceed as follows: 1. **Find the derivative of f(x):** [ f'(x) = frac{d}{dx}(ln x + ax^2 - 2) = frac{1}{x} + 2ax. ] Simplifying, we get: [ f'(x) = frac{2ax^2 + 1}{x}. ] 2. **Analyze the sign of f'(x) for different values of a:** - **When a geq 0:** Since 2ax^2 + 1 > 0 for all x > 0, it follows that f'(x) > 0 for all x > 0. Thus, f(x) is monotonically increasing on (0, +infty), which contradicts the requirement for f(x) to be monotonically decreasing in the interval (1,2). - **When a < 0:** To find the critical points, we set f'(x) = 0: [ frac{2ax^2 + 1}{x} = 0 implies 2ax^2 + 1 = 0. ] Solving for x, we get: [ x = pmsqrt{-frac{1}{2a}}. ] Since x > 0, we discard the negative root and consider x = sqrt{-frac{1}{2a}}. - For x in (0, sqrt{-frac{1}{2a}}), we have f'(x) > 0, indicating that f(x) is monotonically increasing. - For x in (sqrt{-frac{1}{2a}}, +infty), we have f'(x) < 0, indicating that f(x) is monotonically decreasing. 3. **Ensure f(x) is monotonically decreasing in (1,2):** For f(x) to be monotonically decreasing in (1,2), we must have sqrt{-frac{1}{2a}} < 2. Solving this inequality for a, we get: [ a < -frac{1}{8}. ] Therefore, the range of real number a for which f(x) is monotonically decreasing in the interval (1,2) is ({-infty, -frac{1}{8}}). Hence, the correct answer is boxed{D}.

answer:Let's start by calculating the amount of sugar in the original 440-liter solution. Since the solution is 88% water and 8% concentrated kola, the remaining percentage for sugar is 100% - 88% - 8% = 4%. The amount of sugar in the original solution is therefore 4% of 440 liters, which is: 0.04 * 440 = 17.6 liters When 3.2 liters of sugar are added, the new amount of sugar becomes: 17.6 liters + 3.2 liters = 20.8 liters The new total volume of the solution after adding 3.2 liters of sugar, 10 liters of water, and a certain amount of concentrated kola (let's call this x liters) is: 440 liters + 3.2 liters + 10 liters + x liters = 453.2 liters + x liters According to the problem, after these additions, sugar makes up 4.521739130434784% of the solution. So, we can set up the equation: 20.8 liters = 0.04521739130434784 * (453.2 liters + x liters) Now, let's solve for x: 20.8 = 0.04521739130434784 * (453.2 + x) 20.8 / 0.04521739130434784 = 453.2 + x 460.00000000000006 ≈ 453.2 + x 460.00000000000006 - 453.2 = x x ≈ 6.8 liters Therefore, approximately boxed{6.8} liters of concentrated kola were added to the solution.

answer:1. **Understanding the Set (A_k)**: The elements of the set (A_k) are defined as: [ A_{k} = left{2^{k-1} + sum_{i=0}^{k-2} a_{i} 2^{a_{i}+i} mid a_{i} in{0,1}, i=0,1,dots,k-2right} ] Here, (a_i) can either be 0 or 1. 2. **Calculating Elements of (A_k)**: When (a_i = 0), the term (a_i 2^{a_i+i} = 0). When (a_i = 1), the term (a_i 2^{a_i+i} = 2^{i+1}). Therefore, each element of (A_k) can be written as: [ x = 2^{k-1} + sum_{i=0}^{k-2} b_i 2^{i+1}, quad text{where } b_i in {0, 1} ] 3. **Summation in (A_k)**: The sum of all elements in (A_k), denoted as (n_k), is given by summing up all possible values of (x) within the set (A_k). 4. **Sum Calculation**: When the sum is calculated over all possible (b_i): [ n_k = 2^{k-1} left(2^{k-1}right) + sum_{b_i in {0,1}} left(2^{k-2} left(2^{k-1} + 2^{k-2} + cdots + 2right)right) ] Consider the series inside the sum: [ sum_{i=0}^{k-2} 2^{i+1} = 2 + 2^2 + 2^3 + cdots + 2^{k-1} = 2 left(1 + 2 + 2^2 + cdots + 2^{k-2}right) ] This series sums to: [ 2 left(frac{2^{k-1} - 1}{2-1}right) = 2 left(2^{k-1} - 1right) = 2^k - 2 ] 5. **Substituting Back**: Substituting the series sum back: [ n_k = 2^{2(k-1)} + 2^{k-2} left(2^k - 2right) = 2^{2k-2} + 2^{2k-2} - 2^{k-2+1} = 2^{2k-1} - 2^{k-1} ] 6. **Summation (sum_{k=1}^{2015} n_k)**: Summing (n_k) from (k=1) to (k=2015): [ sum_{k=1}^{2015} n_k = sum_{k=1}^{2015} left(2^{2k-1} - 2^{k-1}right) ] 7. **Simplifying the Series**: The geometric series for the powers of 2 can be summarized as: [ sum_{k=1}^{2015} 2^{2k-1} = 2 left(1 + 2^2 + 2^4 + cdots + 2^{4030}right) = 2 left(frac{2^{4030} - 1}{2^2 - 1}right) = frac{2^{4031} - 2}{3} ] Similarly, [ sum_{k=1}^{2015} 2^{k-1} = 1 + 2 + 2^1 + cdots + 2^{2014} = 2^{2015} - 1 ] 8. **Final Summation**: Combining these results, we get: [ sum_{k=1}^{2015} left(2^{2k-1} - 2^{k-1}right) = frac{2^{4031} - 2}{3} - left(2^{2015} - 1right) ] Simplifying further: [ = frac{2^{4031} - 2 - 3 cdot 2^{2015} + 3}{3} ] [ = frac{2^{4031} - 2^{2015+1} + 1}{3} = frac{2^{4031} + 1}{3} - 2^{2015} ] # Conclusion: [ boxed{frac{2^{4031}+1}{3}-2^{2015}} ]

answer:To solve for p(5) given that p(n) = frac{1}{n^2} for n = 1, 2, 3, and 4, we define a new polynomial q(x) = x^2 p(x) - 1. This transformation is chosen because q(n) = 0 for n = 1, 2, 3, and 4, which means q(x) has roots at these values of x. Since p(x) is a cubic polynomial, q(x) is a polynomial of degree 5. Given q(n) = 0 for n = 1, 2, 3, and 4, we can express q(x) as: [q(x) = (ax + b)(x - 1)(x - 2)(x - 3)(x - 4).] Here, a and b are constants that we need to determine. First, we find b by evaluating q(0): [q(0) = 0^2 cdot p(0) - 1 = -1.] Substituting x = 0 into our expression for q(x) gives: [q(0) = 24b = -1,] which implies that b = -frac{1}{24}. Next, we determine a by considering the coefficient of x in q(x) = x^2 p(x) - 1 is 0. The coefficient of x in our expression for q(x) is calculated as follows: begin{align*} &a(-1)(-2)(-3)(-4) + b(-2)(-3)(-4) + b(-1)(-3)(-4) &+ b(-1)(-2)(-4) + b(-1)(-2)(-3) &= 24a - 50b. end{align*} Since the coefficient of x in q(x) is 0, we have: [24a - 50b = 0,] which leads to a = frac{50b}{24} = -frac{25}{288}. Therefore, the polynomial q(x) can be expressed as: [q(x) = left( -frac{25}{288} x - frac{1}{24} right) (x - 1)(x - 2)(x - 3)(x - 4) = -frac{(25x + 12)(x - 1)(x - 2)(x - 3)(x - 4)}{288}.] To find p(5), we calculate q(5): [q(5) = -frac{137}{12}.] Finally, since q(x) = x^2 p(x) - 1, we solve for p(5): [p(5) = frac{q(5) + 1}{25} = boxed{-frac{5}{12}}.]