answer:Since sin 2α= frac {2}{3}, we can use the double angle formula for sine to express sin ^{2}(α+ frac {π}{4}) as follows: sin ^{2}(α+ frac {π}{4}) = frac{1 - cos(2α + frac{π}{2})}{2}. Now, we can apply the cofunction identity for cosine, which states that cos(θ) = sin(90° - θ) or in radians cos(θ) = sin(π/2 - θ), to transform the cosine term: cos(2α + frac{π}{2}) = sin(2α). Substituting the given value for sin 2α, we have sin ^{2}(α+ frac {π}{4}) = frac{1 - sin(2α)}{2} = frac{1 + frac{2}{3}}{2} = frac{5}{6}. Therefore, the correct answer is boxed{D: frac {5}{6}}. This problem primarily assesses the application of double angle formulas and cofunction identities, making it a fundamental trigonometry question.

answer:Solution: (Ⅰ) By taking the derivative, we get f'(x) = frac{(x-a)(x-1)}{x} (x > 0) (1) When 0 < a < 1, let f'(x) < 0, we get a < x < 1, Since x > 0, we have a < x < 1; let f'(x) > 0, we get x < a or x > 1, Since x > 0, we have 0 < x < a or x > 1 Therefore, the function f(x) is increasing on (0, a) and (1, +infty), and decreasing on (a, 1); (2) When a = 1, f'(x) geq 0, the function is increasing on (0, +infty); (3) When a > 1, let f'(x) < 0, we get 1 < x < a, Since x > 0, we have 1 < x < a; let f'(x) > 0, we get x > a or x < 1, Since x > 0, we have 0 < x < 1 or x > a Therefore, the function f(x) is increasing on (0, 1) and (a, +infty), and decreasing on (1, a); (Ⅱ) When a = -1, f(x) = -ln{x} + frac{x^2}{2}, (x > 0), f'(x) = frac{(x+1)(x-1)}{x}, Let f'(x) > 0, we solve to get: x > 1, let f'(x) < 0, we solve to get: 0 < x < 1, Therefore, f(x) is decreasing on (0, 1) and increasing on (1, +infty), Therefore, when x = 1, f(x) reaches its minimum value which is f(1) = frac{1}{2}, Hence, f(x) geq frac{1}{2} is valid. Thus, the final answers are: - For (Ⅰ), the function f(x) is increasing on (0, a) and (1, +infty), and decreasing on (a, 1) when 0 < a < 1; increasing on (0, +infty) when a = 1; increasing on (0, 1) and (a, +infty), and decreasing on (1, a) when a > 1. - For (Ⅱ), we have boxed{f(x) geq frac{1}{2}}.

answer:To determine the values of parameters (a) and (b) such that the inequality [ left| x^2 + ax + b right| leq frac{1}{8} ] holds for all (x in [0, 1]), follow the steps below: 1. **Critical Points and Boundary Values:** - Evaluate the function ( f(x) = left| x^2 + a x + b right| ) at critical points and at the boundaries ( x = 0 ) and ( x = 1 ). - Since the polynomial ( x^2 + ax + b ) is quadratic, the critical points inside the interval ([0, 1]) are the roots of its derivative ( f'(x) ). - ( f'(x) ) is given by: [ f'(x) = frac{d}{dx} left| x^2 + ax + b right| cdot text{sign}(x^2 + ax + b) ] - Setting ( x^2 + a x + b = 0 ), we find the critical points as the roots of the quadratic equation: [ x = frac{-a pm sqrt{a^2 - 4b}}{2} ] 2. **Evaluating at the Boundaries:** - Evaluate ( f(x) ) at the boundaries ( x = 0 ) and ( x = 1 ): - At ( x = 0 ): [ f(0) = |b| leq frac{1}{8} ] Hence: [ -frac{1}{8} leq b leq frac{1}{8} ] - At ( x = 1 ): [ f(1) = |1 + a + b| leq frac{1}{8} ] 3. **Combining Inequalities:** - Combine the inequalities obtained: [ -frac{1}{8} leq b leq frac{1}{8} ] [ |1 + a + b| leq frac{1}{8} ] 4. **Critical Point Condition:** - To ensure the inequality at the critical points derived from ( x = frac{-a}{2} ): [ left| -frac{a^2}{4} + b right| leq frac{1}{8} ] 5. **System of Inequalities:** - Set up the system of inequalities: [ begin{cases} -frac{1}{8} leq b leq frac{1}{8} |1 + a + b| leq frac{1}{8} left| -frac{a^2}{4} + b right| leq frac{1}{8} end{cases} ] 6. **Solving the System:** - From ( -frac{1}{8} leq b leq frac{1}{8} ) and ( |1 + a + b| leq frac{1}{8} ): [ -frac{9}{8} leq a + b leq -frac{7}{8} ] - And from ( -frac{1}{8} leq left( b - frac{a^2}{4} right) leq frac{1}{8} ): [ -frac{1}{8} leq -frac{a^2}{4} + b leq frac{1}{8} ] - Combining these inequalities and solving, we find that the only value satisfying all conditions is ( a = -1 ) and ( b = frac{1}{8} ). # Conclusion: [ boxed{a = -1, , b = frac{1}{8}} ]

answer:To find the intersection A cap B, we need to understand the domain of x for each set A and B based on the given conditions. For set A, we have y = sqrt{3-2x}. The expression under the square root, 3-2x, must be non-negative for the square root to be real. Therefore, we set up the inequality: [3-2x geqslant 0] Solving for x, we get: [3 geqslant 2x] [x leqslant dfrac{3}{2}] This means the domain of x for set A is x leqslant dfrac{3}{2}. For set B, we have y = lg(2^x - 1). The argument of the logarithm, 2^x - 1, must be positive for the logarithm to be defined. Therefore, we set up the inequality: [2^x - 1 > 0] Solving for x, we get: [2^x > 1] Since 2^x is always greater than 1 for x > 0, the domain of x for set B is x > 0. The intersection A cap B would then be the set of x values that satisfy both conditions: x leqslant dfrac{3}{2} and x > 0. This leads us to: [0 < x leqslant dfrac{3}{2}] Therefore, the correct answer is boxed{text{C: }left{xmid 0 < xleqslant dfrac{3}{2}right}}.