answer:Solution: (I) The domain of the function f(x) is (0, +infty), f′(x) = frac{-ax^2 + ax + 1}{x}. Let varphi(x) = -ax^2 + ax + 1, Since a < 0, the quadratic function varphi(x) opens upwards, its axis of symmetry is x = frac{1}{2} > 0, and it always passes through the point (0,1). ① When Delta = a^2 + 4a leqslant 0, i.e., -4 leqslant a < 0, varphi(x) geqslant 0 always holds in (0, +infty). At this time, the function f(x) is monotonically increasing in (0, +infty), with no extreme points. ② When Delta = a^2 + 4a > 0, i.e., a < -4, varphi(x) = 0 has two real roots x_1 and x_2 (assuming x_1 < x_2). For x in (0, x_1) and (x_2, +infty), f′(x) > 0; for x in (x_1, x_2), f′(x) < 0. Therefore, f(x) is increasing in (0, x_1) and (x_2, +infty), and decreasing in (x_1, x_2). At this time, the function has two extreme points x_1 and x_2, where x_1 is a local maximum and x_2 is a local minimum. In summary: When -4 leqslant a < 0, there are no extreme points; when a < -4, the function has two extreme points. (II) Let g(x) = f(x) - 2ax + x + 1 = ln x - frac{1}{2}ax^2 + (1-a)x + 1 (for x > 0) g′(x) = frac{-ax^2 + (1-a)x + 1}{x} ① When a > 0, g′(x) = frac{-(x - frac{1}{a})(x + 1)}{x}, for x in (0, frac{1}{a}), g′(x) > 0, g(x) is increasing in (0, frac{1}{a}); for x in (frac{1}{a}, +infty), g′(x) < 0, g(x) is decreasing in (frac{1}{a}, +infty), Thus, the maximum value of the function g(x) is g(frac{1}{a}) = ln frac{1}{a} - frac{1}{2}a(frac{1}{a})^2 + (1-a)frac{1}{a} + 1 = frac{1}{2a} - ln a Let h(a) = frac{1}{2a} - ln a, then h′(a) = -frac{1}{2a^2} - frac{1}{a} < 0 (for a > 0) always holds, thus h(a) is monotonically decreasing in (0, +infty), and since h(1) = frac{1}{2} > 0, h(2) = frac{1}{4} - ln 2 < 0, Therefore, when a geqslant 2, h(a) leqslant h(2) < 0, thus, when the inequality f(x) leqslant 2ax - x - 1 always holds, the minimum value of the integer a is 2; ② When a leqslant 0, g′(x) > 0, g(x) is monotonically increasing in (0, +infty), and g(1) = 2 - frac{3}{2}a > 0 Therefore, the inequality f(x) leqslant 2ax - x - 1 does not always hold, In summary: When the inequality f(x) leqslant 2ax - x - 1 always holds, the minimum value of the integer a is boxed{2}; (III) frac{f(x_2) - f(x_1)}{x_2 - x_1} = frac{ln x_2 - ln x_1}{x_2 - x_1} - frac{1}{2}a(x_1 + x_2) + a Since f′(x) = frac{1}{x} - ax + a, thus f′left( frac{x_1 + x_2}{2}right) = frac{2}{x_1 + x_2} - afrac{x_1 + x_2}{2} + a. Therefore, frac{f(x_1) - f(x_2)}{x_2 - x_1} - f′left( frac{x_1 + x_2}{2}right) = frac{ln x_2 - ln x_1}{x_2 - x_1} - frac{2}{x_1 + x_2} = frac{1}{x_2 - x_1}left[ln frac{x_2}{x_1} - frac{2(x_2 - x_1)}{x_2 + x_1}right] = frac{1}{x_2 - x_1}left(ln frac{x_2}{x_1} - 2frac{frac{x_2}{x_1} - 1}{frac{x_2}{x_1} + 1}right) Assuming x_1 < x_2, let t = frac{x_2}{x_1} (t > 1), then ln frac{x_2}{x_1} - 2frac{frac{x_2}{x_1} - 1}{frac{x_2}{x_1} + 1} = ln t - 2frac{t - 1}{t + 1} (t > 1) Let G(t) = ln t - 2frac{t - 1}{t + 1} (t > 1), then G′(t) = frac{(t - 1)^2}{t(t + 1)} > 0, thus G(t) is monotonically increasing in (1, +infty). Therefore, G(t) > G(1) = 0, i.e., ln frac{x_2}{x_1} - 2frac{frac{x_2}{x_1} - 1}{frac{x_2}{x_1} + 1} > 0, Since x_2 - x_1 > 0, therefore frac{f(x_1) - f(x_2)}{x_2 - x_1} > f′left( frac{x_1 + x_2}{2}right).

answer:Since f(x) is a decreasing function defined on R, we have the following conditions: - For x < 1, the coefficient of x in (2a-1)x+4a should be negative. - For x geq 1, the coefficient of x in -x+1 is already negative. This implies that: begin{cases} 2a-1 < 0 (2a-1)cdot 1 + 4a geq -1 + 1 end{cases} Solving the system of inequalities: begin{aligned} 2a-1 &< 0 a &< frac{1}{2} 2a-1+4a &geq 0 6a &geq 1 a &geq frac{1}{6} end{aligned} Combining both inequalities, we obtain the range of values for a: boxed{a in left[frac{1}{6}, frac{1}{2}right)} Therefore, the answer is A.

answer:1. **Rewrite the expression using geometric series**: [ left(frac{1 - x^{21}}{1 - x}right) left(frac{1 - x^{11}}{1 - x}right)^2 left(1 - x^3right) = frac{(1 - x^{21})(1 - x^{11})^2 (1 - x^3)}{(1 - x)^3}. ] 2. **Expand using binomial theorem**: We expand (frac{1}{(1 - x)^3}) using the binomial theorem: [ frac{1}{(1 - x)^3} = sum_{k=0}^{infty} binom{k+2}{2} x^k. ] Thus, our expression becomes: [ (1 - x^{21})(1 - x^{11})^2 (1 - x^3) sum_{k=0}^{infty} binom{k+2}{2} x^k. ] 3. **Determine the coefficient of (x^{21})**: Consider how (x^{21}) can be formed: - **Case 1**: (x^{21}) from the series term (binom{23}{2}), multiplied by coefficients from the other terms: - (1) from ((1 - x^{21})) - (1) from ((1 - x^{11})^2) - (1) from ((1 - x^3)) - Coefficient of (x^{21}) in the series: (binom{23}{2} = 253). - **Case 2**: (x^{18}) from the series term (binom{20}{2}), with (x^3) from (-(1 - x^3)): - (x^{18}) from the series: (binom{20}{2} = 190) - Coefficient is (-190) due to the (-x^3) term. Summing contributions: [ 253 - 190 = 63. ] 4. **Conclusion**: The coefficient of (x^{21}) in the expansion is (63). The final answer is boxed{B}.

answer:To find out how many ounces of the mixture the scientist wants to obtain, we need to calculate the total amount of salt in the final mixture and then determine what total volume would result in an 80% salt concentration. First, let's calculate the amount of salt in each solution: Amount of salt in Solution A: 40% of 28 ounces = 0.40 * 28 = 11.2 ounces Amount of salt in Solution B: 90% of 112 ounces = 0.90 * 112 = 100.8 ounces Now, let's add the amounts of salt from both solutions to get the total amount of salt in the mixture: Total salt = 11.2 ounces (from A) + 100.8 ounces (from B) = 112 ounces Since the scientist wants the final mixture to be 80% salt, we can set up the following equation to find the total volume of the mixture (let's call it V): 80% of V = 112 ounces 0.80 * V = 112 ounces V = 112 ounces / 0.80 V = 140 ounces Therefore, the scientist wants to obtain 140 ounces of the mixture that is boxed{80%} salt.