answer:Since sinalpha = -frac{4}{5}, alpha in left(-frac{pi}{2}, frac{pi}{2}right), it follows that alpha in left(-frac{pi}{2}, 0right) and cosalpha = frac{3}{5}, therefore, sin 2alpha = 2sinalphacosalpha = -frac{24}{25}． Hence, the correct choice is boxed{A}． First, find cosalpha from sinalpha, then use the double angle formula for sine. This question tests the relationship between sine and cosine of the same angle and the double angle formula for sine.

answer:Solution: 1) When a= frac {1}{2}, the inequality becomes f(x) = x^{2} - frac {5}{2}x + 10, Therefore, (x - frac {1}{2})(x - 2) leq 0, From the inequality f(x) = (x - frac {1}{a})(x - a) leq 0, The solution set is: x in left[frac {1}{2}, 2right] 2) When a > 1, since frac {1}{a} < a, the solution set of the inequality is left{x mid frac {1}{a} leq x leq aright}; When a = 1, the solution is x = 1. Therefore, the final answers are: - For a = frac {1}{2}, the solution set is boxed{x in left[frac {1}{2}, 2right]} - For a > 1, the solution set is boxed{left{x mid frac {1}{a} leq x leq aright}} - For a = 1, the solution is boxed{x = 1}.

answer:Given that the two distinct square roots of a positive number x are m+3 and 2m-15, we can set up the equation based on the property that the sum of the roots equals zero when they are considered as roots of a quadratic equation. Thus, we have: [ m + 3 + 2m - 15 = 0 ] Simplifying this equation step by step: [ 3m - 12 = 0 ] [ 3m = 12 ] [ m = 4 ] Having found m = 4, we can substitute it back into either of the expressions for the square roots to find the value of x. Using m + 3: [ m + 3 = 4 + 3 = 7 ] Since this is a square root of x, squaring both sides gives us: [ x = 7^2 = 49 ] Therefore, the positive number x is boxed{49}.

answer:Solution: (1) Substituting the line y=x+m into the ellipse equation yields: x^{2}+4(x+m)^{2}=4, which simplifies to: 5x^{2}+8mx+4m^{2}-4=0, and the discriminant triangle =(8m)^{2}-4×5×(4m^{2}-4)=-16m^{2}+80=0, solving this gives: m=± sqrt {5}; (2) Let the line intersect the ellipse at two points A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}, x_{2} are the roots of the equation 5x^{2}+8mx+4m^{2}-4=0, by Vieta's formulas, we have: x_{1}+x_{2}=- dfrac {8m}{5}, x_{1}⋅x_{2}= dfrac {4m^{2}-4}{5}, thus, |AB|= sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}= sqrt {1+k^{2}} sqrt {(x_{1}+x_{2})^{2}-4x_{1}x_{2}} = sqrt {2} sqrt {(- dfrac {8m}{5})^{2}-4× dfrac {4m^{2}-4}{5}}=2; therefore, m=± dfrac { sqrt {30}}{4}. So, the answers are m=± sqrt {5} for part (1) and m=± dfrac { sqrt {30}}{4} for part (2). Thus, the final answers are boxed{m=± sqrt {5}} for the first part and boxed{m=± dfrac { sqrt {30}}{4}} for the second part.