answer:# Problem: 求实数 a 的取值范围, 使得对任意实数 x 和任意 theta inleft[0, frac{pi}{2}right] 恒有: [ (x+3+2 sin theta cos theta)^{2}+(x+a sin theta+a cos theta)^{2} geqslant frac{1}{8}. ] 为了求解此不等式恒成立时 a 的取值范围，我们可以分析其对 x 的二次不等式。将给出的条件展开得： [ (x+3+2 sin theta cos theta)^{2}+(x+a sin theta+a cos theta)^{2} geqslant frac{1}{8}. ] 首先展开并整理后，我们得到以下的二次不等式： [ x^{2} + (3+2 sin theta cos theta + a sin theta + a cos theta)x + frac{1}{2}(3+2 sin theta cos theta)^{2} + frac{1}{2}(a sin theta + a cos theta)^{2} - frac{1}{16} geqslant 0. ] 为了使得对任意 theta in left[0, frac{pi}{2}right]，此不等式恒成立，我们要求判别式 Delta 满足 Delta leqslant 0. 判别式 Delta 的计算如下： 我们从二次项系数 A=1，B = 3 + 2sinthetacostheta + asintheta + acostheta，C = frac{1}{2}(3 + 2sinthetacostheta)^2 + frac{1}{2}(asintheta + acostheta)^2-frac{1}{16}。 所以： [ Delta = B^2 - 4AC leqslant 0 ] 具体展开 Delta 得: [ (3 + 2sinthetacostheta + asintheta + acostheta)^2 geqslant 2left( frac{1}{8} - frac{1}{2}(3+2sinthetacostheta)^2 - frac{1}{2}(asintheta + acostheta)^2 right) ] 可以分解整理得到: [ (3 + 2sinthetacostheta - asintheta - acostheta)^2 leqslant frac{1}{4} ] 接下来，我们将不等式对 theta in left[0, frac{pi}{2}right] 进行分析。 由这些计算我们可以得到： [ a geq frac{3 + 2sinthetacostheta + frac{1}{2}}{sintheta + costheta}, quad text{或者} quad a leq frac{3 + 2sinthetacostheta - frac{1}{2}}{sintheta + costheta} ] 为了找到 a 的取值范围，这里需要考虑 sintheta + costheta 的最大值和最小值。 **分析最优值：** 对于 sintheta + costheta 在 theta in left[0, frac{pi}{2}right] 时的范围，可以得知： [ 1 leq sintheta + costheta leq sqrt{2} ] 利用这个范围来获得最终解得不等式： 1. 从 sintheta + costheta geq 1 可得： [ a geq 1 + frac{5}{2} = frac{7}{2} ] 2. 从 sintheta + costheta leq sqrt{2} 可得： [ a leq sqrt{2} cdot frac{3}{2} + frac{3}{2} = frac{6sqrt{2}}{2} = sqrt{6} ] 综上所述，a 的取值范围为： [ a geq frac{7}{2} quad text{或者} quad a leq sqrt{6} ] boxed{frac{7}{2} leq a text{ 或 } a leq sqrt{6}}

answer:1. Calculate the side length of the square using the distance formula between the two given points, ((-2, -1)) and ( (3, 6) ). 2. The distance (or side length (s)) between these points is: [ s = sqrt{((-2) - 3)^2 + ((-1) - 6)^2} = sqrt{(-2 - 3)^2 + (-1 - 6)^2} = sqrt{(-5)^2 + (-7)^2} = sqrt{25 + 49} = sqrt{74} ] 3. Consequently, the area (A) of the square is (s^2), or: [ A = ( sqrt{74} )^2 = 74 ] Thus, the area of the square is boxed{74}.

answer:Let's denote the cost of a large monkey doll as ( L ) dollars. Since the small monkey dolls are 2 cheaper, the cost of a small monkey doll would be ( L - 2 ) dollars. If Dhoni buys only large monkey dolls, the number of large dolls he can buy is ( frac{300}{L} ). If he buys only small monkey dolls, the number of small dolls he can buy is ( frac{300}{L - 2} ). According to the problem, the number of small dolls he can buy is 25 more than the number of large dolls he can buy. So we can write the equation: [ frac{300}{L - 2} = frac{300}{L} + 25 ] To solve for ( L ), we first get rid of the denominators by multiplying both sides of the equation by ( L(L - 2) ): [ 300L = 300(L - 2) + 25L(L - 2) ] Expanding the right side of the equation: [ 300L = 300L - 600 + 25L^2 - 50L ] Now, we can move all terms to one side to set the equation to zero: [ 25L^2 - 50L - 600 = 0 ] Divide the entire equation by 25 to simplify: [ L^2 - 2L - 24 = 0 ] This is a quadratic equation that can be factored: [ (L - 6)(L + 4) = 0 ] The solutions to this equation are ( L = 6 ) and ( L = -4 ). Since the cost of a doll cannot be negative, we discard ( L = -4 ). Therefore, the cost of a large monkey doll is ( L = boxed{6} ).

answer:**Analysis** This problem examines the comprehensive application of derivatives and the judgment and application of the monotonicity of functions. It also involves the application of the idea of case discussion, making it a medium-level question. It is easy to see that when x=0, |frac{1}{2}x^{3}-ax|=0 < 1; when 0 < x leqslant 1, |frac{1}{2}x^{3}-ax|=x|frac{1}{2}x^{2}-a|, thus it can be transformed into |frac{1}{2}x^{2}-a| leqslant frac{1}{x}. Then, by discussing based on a, the monotonicity and value of the function can be determined, from which the solution can be derived. **Solution** When x=0, |frac{1}{2}x^{3}-ax|=0 < 1, When 0 < x leqslant 1, |frac{1}{2}x^{3}-ax|=x|frac{1}{2}x^{2}-a|, Therefore, |frac{1}{2}x^{3}-ax| leqslant 1 can be transformed into |frac{1}{2}x^{2}-a| leqslant frac{1}{x}; ① When a leqslant 0, |frac{1}{2}x^{2}-a| leqslant frac{1}{x} can be transformed into frac{1}{2}x^{2}-a leqslant frac{1}{x}, That is, a geqslant frac{1}{2}x^{2}-frac{1}{x}, It is easy to know that y=frac{1}{2}x^{2}-frac{1}{x} is an increasing function on (0,1], Therefore, it only needs to make a geqslant frac{1}{2}(-1)=-frac{1}{2}; ② When 0 < a leqslant 1, |frac{1}{2}x^{2}-a| leqslant 1 leqslant frac{1}{x}, so it holds; ③ When a > 1, |frac{1}{2}x^{2}-a| leqslant frac{1}{x} can be transformed into -frac{1}{2}x^{2}+a leqslant frac{1}{x}, That is, a leqslant frac{1}{2}x^{2}+frac{1}{x}, (frac{1}{2}x^{2}+frac{1}{x})' = x-frac{1}{x^{2}} leqslant 0, Therefore, frac{1}{2}x^{2}+frac{1}{x} is a decreasing function on (0,1], Therefore, a leqslant frac{1}{2}+1=frac{3}{2}; In summary, -frac{1}{2} leqslant a leqslant frac{3}{2}; Thus, the answer is boxed{[-frac{1}{2}, frac{3}{2}]}.