answer:(1) Since f'(x)=2ax+frac{b}{x}. Given that f(x) has an extreme value of frac{1}{2} at x=1, we have begin{cases}f(1)=frac{1}{2} f'(1)=0end{cases}, which leads to begin{cases}a=frac{1}{2} 2a+b=0end{cases}. Solving these equations, we obtain a=frac{1}{2} and b=-1. (2) From (1), we know f(x)=frac{1}{2} x^{2}-ln x, whose domain is (0,+infty), and f'(x)=x-frac{1}{x} =frac{left(x+1right)left(x-1right)}{x}. When f'(x) < 0, we have 0 < x < 1; when f'(x) > 0, we have x > 1. Therefore, the function y=f(x) is decreasing on (0,1) and increasing on (1,+infty). (3) According to (2), the function is decreasing on [frac{1}{e},1] and increasing on [1,e]. The minimum value is f(1)= frac{1}{2}. It is easy to obtain f(e)=frac{1}{2}{e}^{2}-1 and f(frac{1}{e} )=frac{1}{2{e}^{2}}+1. Since f(e)-f(frac{1}{e})= frac{1}{2}{e}^{2}- frac{1}{2{e}^{2}}-2 > frac{1}{2}×6- frac{1}{2}-2 > 0, the maximum value is boxed{f(e)= frac{1}{2}{e}^{2}-1}.

answer:To find the speed of the slower train, we first need to determine the relative speed at which the two trains cross each other. Since they are moving in opposite directions, their relative speeds add up. The total distance covered when the two trains cross each other is the sum of their lengths: Total distance = Length of Train 1 + Length of Train 2 Total distance = 140 m + 210 m Total distance = 350 m The time taken to cross each other is given as 12.59899208063355 seconds. We can calculate the relative speed (V_rel) using the formula: V_rel = Total distance / Time taken to cross each other V_rel = 350 m / 12.59899208063355 s Now, let's calculate the relative speed: V_rel ≈ 350 m / 12.59899208063355 s ≈ 27.783 m/s We need to convert this relative speed from meters per second to kilometers per hour to match the units of the given speed of the faster train: V_rel (in km/hr) = V_rel (in m/s) × (3600 s/hr) / (1000 m/km) V_rel (in km/hr) ≈ 27.783 m/s × 3.6 V_rel (in km/hr) ≈ 99.9788 km/hr Now we know the relative speed of the two trains is approximately 99.9788 km/hr. The faster train runs at 60 km/hr, so we can find the speed of the slower train (V_slower) by subtracting the speed of the faster train from the relative speed: V_rel = V_faster + V_slower V_slower = V_rel - V_faster V_slower = 99.9788 km/hr - 60 km/hr V_slower ≈ 39.9788 km/hr Therefore, the speed of the slower train is approximately boxed{39.9788} km/hr.

answer:We know g(n) = m if and only if [m - frac{1}{2} < sqrt[5]{n} < m + frac{1}{2},] leading to [left(m - frac{1}{2}right)^5 < n < left(m + frac{1}{2}right)^5.] The essential range for n is: [m^5 - frac{5}{2}m^4 + frac{5}{2}m^3 - frac{5}{8}m^2 + frac{1}{8}m - frac{1}{32} < n < m^5+ frac{5}{2}m^4 + frac{5}{2}m^3 + frac{5}{8}m^2 + frac{1}{8}m + frac{1}{32}.] Simplifying, the difference between upper and lower bounds is [5m^4 + m]. There are precisely 5m^4 + m values of n that g(n) = m. Each m value contributes (5m^4 + m) cdot frac{1}{m} = 5m^3 + 1 to the sum. Expanding until m = 7 (since 6.5^5 = 8913 frac{1}{32}): [ sum_{m=1}^6 (5m^3 + 1) = 5(1+8+27+64+125+216) + 6 = 2226. ] The remaining terms for m=7 stretch from n = 8914 to 4095: [ 4095 - 8914 + 1 = -4818. ] Negative count implies miscalculation since m = 7 is too large to fit our range. Thus, consider up to m = 6 only. Therefore, the solution is boxed{2226}.

answer:Let's calculate how much money Jason spent on books first. He spent 1/4 of his initial money, which is: 1/4 * 320 = 80 He also spent an additional 10 on books, so in total, he spent: 80 + 10 = 90 on books. After buying books, he had: 320 - 90 = 230 left. Next, let's calculate how much money he spent on DVDs. He spent 2/5 of the remaining money, which is: 2/5 * 230 = 92 He also spent an additional 8 on DVDs, so in total, he spent: 92 + 8 = 100 on DVDs. After buying DVDs, he had: 230 - 100 = 130 left. So, Jason was left with boxed{130} after buying books and DVDs.