answer:To determine if the squirrel can catch the nut, we need to consider the motion of the nut under the action of gravity and the conditions of the squirrel's jumping capabilities. 1. **Initial Positions and Equations of Motion:** - The initial position of the squirrel is (x = a, y = 0). - The nut is thrown horizontally with an initial velocity ( V_0 = 5 , text{m/s} ). - The equations for the coordinates of the nut as functions of time ( t ) are: [ x(t) = V_0 t, quad y(t) = frac{g t^2}{2} ] where ( g = 10 , text{m/s}^2 ). 2. **Distance Function:** - The square of the distance ( r(t) ) between the position of the squirrel and the flying nut at time ( t ) is given by: [ r^2 = (x(t) - a)^2 + (y(t))^2 ] Substituting the motion equations, we get: [ r^2 = (V_0 t - a)^2 + left(frac{g t^2}{2}right)^2 ] Simplifying further: [ r^2 = (5t - 3.75)^2 + left(frac{10 t^2}{2}right)^2 ] [ r^2 = (5t - 3.75)^2 + (5t^2)^2 ] [ r^2 = (5t - 3.75)^2 + 25t^4 ] Expanding and combining like terms: [ r^2 = 25t^2 - 37.5t + 14.0625 + 25t^4 ] 3. **Minimization of Distance:** - To minimize the distance ( r(t) ), we need to minimize ( r^2(t) ), which is the function: [ f(t) = 25 t^4 + 25 t^2 - 37.5 t + 14.0625 ] 4. **Finding the Critical Points:** - Take the derivative of ( f(t) ) with respect to ( t ): [ f'(t) = 100 t^3 + 50 t - 37.5 ] - Set the derivative to zero to find critical points: [ 100 t^3 + 50 t - 37.5 = 0 ] Simplifying: [ 8t^3 + 4t - 3 = 0 ] 5. **Solving the Cubic Equation:** - Factoring the cubic equation: [ 8t^3 + 4t - 3 = (2t - 1)(4t^2 + 2t + 3) ] - The quadratic term (4t^2 + 2t + 3) has no real roots as its discriminant (D = 2^2 - 4 cdot 4 cdot 3 = 4 - 48 = -44) is negative. - Therefore, we solve (2t - 1 = 0) for (t): [ 2t - 1 = 0 implies t = frac{1}{2} ] 6. **Verifying Minimum Distance:** - Substitute ( t = frac{1}{2} ) back into the distance function to find ( r_{min} ): [ r^2 left( frac{1}{2} right) = 25 left(frac{1}{2}right)^4 + 25 left(frac{1}{2}right)^2 -37.5 left(frac{1}{2}right) + 14.0625 ] [ = 25 cdot frac{1}{16} + 25 cdot frac{1}{4} - 18.75 + 14.0625 ] [ = frac{25}{16} + frac{25}{4} - 18.75 + 14.0625 ] [ = frac{25}{16} + frac{100}{16} - frac{300}{16} + frac{225}{16} ] [ = frac{25 + 100 - 300 + 225}{16} ] [ = frac{50}{16} = frac{25}{8} ] Hence, [ r_{min} = sqrt{frac{25}{8}} = frac{5}{2 sqrt{2}} = frac{5sqrt{2}}{4} ] 7. **Comparison to the Squirrel's Range:** - The distance the squirrel can jump is (1.7 , text{m} = frac{17}{10} , text{m}). - Comparing: [ frac{5sqrt{2}}{4} quad text{vs} quad frac{17}{10} ] [ frac{5sqrt{2}}{4} approx 1.77 , text{m} quad > quad frac{17}{10} = 1.7 , text{m} ] Therefore, the nut will be out of reach for the squirrel. **Conclusion:** (boxed{text{No}})

answer:(I) From the problem, we know that x_1 and x_2 are two distinct positive real roots of the equation ln x = ax. Let g(x)=ln x and h(x)=ax (x > 0). 1. When a leqslant 0, g(x) and h(x) have at most one intersection point, so a leqslant 0 does not meet the requirements. 2. When a > 0, suppose y=kx (k > 0) is the tangent line of g(x)=ln x, and the tangent point is (x_0, y_0). Then, k=frac{1}{x_0}. So begin{cases} y_0 = kx_0 = 1 y_0 = ln x_0 end{cases}, which means x_0 = e and k = frac{1}{x_0} = frac{1}{e}. Thus, 0 < a < frac{1}{e}. In conclusion, the range of a is (0, frac{1}{e}). (II) The conclusion is frac{2}{x_1+x_2} < a, and the proof is as follows: From the problem, we have begin{cases} ln x_1 = ax_1 ln x_2 = ax_2 end{cases}. Then, a = frac{ln x_2 - ln x_1}{x_2 - x_1}. We only need to prove that frac{ln x_2 - ln x_1}{x_2 - x_1} > frac{2}{x_1 + x_2}, which is equivalent to proving ln frac{x_2}{x_1} > frac{2(x_2 - x_1)}{x_2 + x_1} = frac{2(frac{x_2}{x_1} - 1)}{frac{x_2}{x_1} + 1}. Let t = frac{x_2}{x_1}, where t > 1. We need to prove that ln t > frac{2(t-1)}{t+1}, which is equivalent to proving frac{1}{2} ln t - 1 + frac{2}{t+1} > 0. Let k(t) = frac{1}{2} ln t - 1 + frac{2}{t+1}, where t > 1. Then, k'(t) = frac{1}{2t} - frac{2}{(t+1)^2} = frac{(t-1)^2}{2t(t+1)^2} > 0. So, k(t) is strictly increasing on (1, +infty). Thus, k(t) > k(1) = 0. In conclusion, boxed{frac{2}{x_1+x_2} < a}.

answer:First, understand that the number of rectangles that can be formed between any two rows and any two columns in a 5x5 grid of dots can be determined by choosing 2 rows and 2 columns from the 5 available. For rows, binom{5}{2} choices are possible, and similarly for columns, binom{5}{2} choices are possible. The total number of rectangles is the product of these two values since the choice of rows and columns is independent: [ text{Number of rectangles} = binom{5}{2} times binom{5}{2} = left(frac{5 times 4}{2 times 1}right) times left(frac{5 times 4}{2 times 1}right) = 10 times 10 = 100 ] Therefore, the total number of rectangles that can be formed is boxed{100}.

answer:From the given conditions, we have the discriminant Delta = 4m^{2}-8m > 0, which yields m < 0 (rejected) or m > 2. Let A=(a,b). Since set A contains exactly two integers, we have |b-a|leqslant 3, which implies: left|frac{m+sqrt{m^{2}-2m}}{2}-frac{m-sqrt{m^{2}-2m}}{2}right|=sqrt{m^{2}-2m}leqslant 3 Solving this inequality, we get 2 < mleqslant 1+sqrt{10}. Thus, the axis of symmetry lies in the range 1 < frac{m}{2}leqslant frac{1+sqrt{10}}{2} < frac{5}{2}. Let f(x)=2x^{2}-2mx+m. We have f(4)=32-7m > 0, f(0)=m > 0, and f(1)=2-m < 0. Therefore, the two integers in set A are 1 and 2, which implies f(2) < 0 and f(3)geqslant 0. Solving these inequalities, we get begin{cases} 8-3m < 0 18-5mgeqslant 0end{cases}, which yields frac{8}{3} < mleqslant frac{18}{5}. Hence, the range of the real number m is (frac{8}{3}, frac{18}{5}]. So, the answer is boxed{D}. From the discriminant, we get m > 2. Using the fact that set A contains exactly two integers, we derive an inequality for m. By considering the symmetry and properties of the quadratic function, we obtain the range of m. This problem tests the method of solving one-variable quadratic inequalities and requires understanding the properties of quadratic functions. It reflects mathematical transformational thinking and is of medium difficulty.