answer:Part (1): Given that ((1 + x^2)(2 - x) < M) holds for any (0 < x < 1), let's find the smallest value of (M). 1. Start by simplifying the given inequality: [ (1 + x^2)(2 - x) < M ] Expand the left-hand side: [ 1 cdot 2 + 1 cdot (-x) + x^2 cdot 2 + x^2 cdot (-x) = 2 - x + 2x^2 - x^3 ] Therefore, the inequality becomes: [ 2 + 2x^2 - x - x^3 < M ] 2. We aim to find the maximum of (f(x) = 2 + 2x^2 - x - x^3) for (0 < x < 1). 3. To do so, calculate the derivative (f'(x)) and find its critical points: [ f(x) = 2 + 2x^2 - x - x^3 ] [ f'(x) = 4x - 1 - 3x^2 ] 4. Setting (f'(x) = 0) to find the critical points: [ 4x - 1 - 3x^2 = 0 implies 3x^2 - 4x + 1 = 0 ] Solve the quadratic equation using the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}) where (a = 3), (b = -4), and (c = 1): [ x = frac{4 pm sqrt{16 - 12}}{6} = frac{4 pm 2}{6} ] [ x = 1 text{ or } x = frac{1}{3} ] 5. Evaluate (f(x)) at the critical points and endpoints (0) and (1): [ fleft(frac{1}{3}right) = 2 + 2left(frac{1}{3}right)^2 - left(frac{1}{3}right) - left(frac{1}{3}right)^3 = 2 + 2left(frac{1}{9}right) - frac{1}{3} - frac{1}{27} ] [ = 2 + frac{2}{9} - frac{3}{9} - frac{1}{27} = 2 - frac{1}{9} - frac{1}{27} = 2 - frac{3}{27} - frac{1}{27} = 2 - frac{4}{27} = frac{54 - 4}{27} = frac{50}{27} ] At (x = 0): [ f(0) = 2 ] At (x = 1): [ f(1) = 2 + 2(1) - 1 - 1 = 2 + 2 - 1 - 1 = 2 ] 6. Therefore, the maximum value of (f(x)) in the interval (0 < x < 1) is 2. Hence, the smallest value for (M) is: [ M = 2 ] **Conclusion:** [ boxed{2} ] Part (2): Given (x_1, x_2, ldots, x_n in mathbf{R}^+) and (x_1 + x_2 + cdots + x_n = 1), we need to prove: [ frac{1}{1 + x_1^2} + frac{1}{1 + x_2^2} + cdots + frac{1}{1 + x_n^2} > frac{2n - 1}{2} ] From part (1), we have the inequality (0 < x < 1) which implies: [ -frac{1}{2}x + 1 < frac{1}{1 + x^2} ] 1. Applying the above inequality to each (x_i): [ -frac{1}{2} x_1 + 1 < frac{1}{1 + x_1^2} ] [ -frac{1}{2} x_2 + 1 < frac{1}{1 + x_2^2} ] [ vdots ] [ -frac{1}{2} x_n + 1 < frac{1}{1 + x_n^2} ] 2. Summing up these inequalities: [ sum_{i = 1}^{n} left( -frac{1}{2} x_i + 1 right) < sum_{i = 1}^{n} frac{1}{1 + x_i^2} ] [ sum_{i = 1}^{n} (-frac{1}{2} x_i) + sum_{i = 1}^{n} 1 < sum_{i=1}^{n} frac{1}{1 + x_i^2} ] [ -frac{1}{2} (x_1 + x_2 + cdots + x_n) + n < frac{1}{1 + x_1^2} + frac{1}{1 + x_2^2} + cdots + frac{1}{1 + x_n^2} ] 3. Using the given (x_1 + x_2 + cdots + x_n = 1): [ -frac{1}{2} cdot 1 + n < frac{1}{1 + x_1^2} + frac{1}{1 + x_2^2} + cdots + frac{1}{1 + x_n^2} ] [ n - frac{1}{2} < frac{1}{1 + x_1^2} + frac{1}{1 + x_2^2} + cdots + frac{1}{1 + x_n^2} ] 4. Rearranging the inequality gives: [ frac{1}{1 + x_1^2} + frac{1}{1 + x_2^2} + cdots + frac{1}{1 + x_n^2} > n - frac{1}{2} ] [ frac{1}{1 + x_1^2} + frac{1}{1 + x_2^2} + cdots + frac{1}{1 + x_n^2} > frac{2n - 1}{2} ] **Conclusion:** [ blacksquare ]

answer:1. **Prime Numbers**: For any prime p, f(p) = p + 1. We need p+1 to be prime. Checking within the range: - n = 2: f(2) = 3 (prime) - n = 3: f(3) = 4 (not prime) - n = 5: f(5) = 6 (not prime) - n = 7: f(7) = 8 (not prime) - n = 11: f(11) = 12 (not prime) - n = 13: f(13) = 14 (not prime) - n = 17: f(17) = 18 (not prime) - n = 19: f(19) = 20 (not prime) - n = 23: f(23) = 24 (not prime) - n = 29: f(29) = 30 (not prime) - n = 31: f(31) = 32 (not prime) - n = 37: f(37) = 38 (not prime) - n = 41: f(41) = 42 (not prime) - n = 43: f(43) = 44 (not prime) - n = 47: f(47) = 48 (not prime) Only n = 2 works. 2. **Prime Powers**: Check powers of smaller primes up to n = 50. - f(4) = 7 (prime) - f(8) = 15 (not prime) - f(16) = 31 (prime) - f(9) = 13 (prime) - f(25) = 31 (prime) - f(27) = 40 (not prime) - f(32) = 63 (not prime) n = 4, 16, 9, 25 work. 3. **Composite numbers (products of distinct primes)**: Since f(n) for a product of two distinct primes is the product of two integers greater than 1, f(n) is always composite. So, n values that work are 2, 4, 9, 16, 25 for a total of boxed{5} values.

answer:(I) Let the semi-focal length of ellipse M be c, which implies c = 1. Given the eccentricity e = frac{1}{2}, we have frac{c}{a} = frac{1}{2}. This gives us a = 2 and b^{2} = a^{2} - c^{2} = 3. Therefore, the equation of the ellipse M is frac{x^{2}}{4} + frac{y^{2}}{3} = 1. (II) Let the equation of line l be x = my - 1, with C(x_{1}, y_{1}) and D(x_{2}, y_{2}). Solving the system of equations obtained by substituting x = my - 1 into the equation of ellipse M and eliminating x, we get (3m^{2} + 4)y^{2} - 6my - 9 = 0. From this, we have y_{1} + y_{2} = frac{6m}{3m^{2} + 4} and y_{1}y_{2} = - frac{9}{3m^{2} + 4} < 0. Since S_{1} = S_{triangle ABC} = frac{1}{2}|AB| cdot |y_{1}| and S_{2} = S_{triangle ABD} = frac{1}{2}|AB| cdot |y_{2}|, with y_{1} and y_{2} having opposite signs, we have |S_{1} - S_{2}| = frac{1}{2}|AB| cdot |y_{1} + y_{2}| = frac{12|m|}{3m^{2} + 4} = frac{12}{3m + frac{4}{|m|}}. As 3|m| + frac{4}{|m|} geq 4sqrt{3}, with equality holding if and only if 3|m| = frac{4}{|m|}, i.e., m = pm frac{2}{sqrt{3}}, the maximum value of |S_{1} - S_{2}| is boxed{frac{12}{4sqrt{3}} = sqrt{3}}. In this case, the equation of line l is boxed{sqrt{3}x pm 2y + sqrt{3} = 0}.

answer:1. Let S be the area of the third polygon, R the radius of the circle, and n the number of sides of the first polygon. Since each subsequent polygon has twice as many sides as the previous one, the number of sides of the second and third polygons are 2n and 4n respectively. 2. Denote the central angle of the first polygon as alpha = frac{pi}{n}, then the central angles of the second and third polygons are frac{pi}{2n} and frac{pi}{4n} respectively. 3. The area of a regular polygon with n sides and circumradius R can be expressed as: [ S = frac{1}{2} n R^2 sin left(frac{2pi}{n}right) ] 4. Hence, the area of the first polygon is: [ S_1 = frac{1}{2} n R^2 sin left(frac{2pi}{n}right) ] 5. The area of the second polygon (with 2n sides) is: [ S_2 = frac{1}{2} (2n) R^2 sin left(frac{pi}{n}right) = n R^2 sin left(frac{pi}{n}right) ] 6. For the third polygon (with 4n sides), the area is: [ S = frac{1}{2} (4n) R^2 sin left(frac{pi}{2n}right) = 2n R^2 sin left(frac{pi}{2n}right) ] 7. We need to express this in terms of S_1 and S_2. Note: [ sinleft(frac{pi}{2n}right) = sqrt{frac{1 - cosleft(frac{pi}{n}right)}{2}} ] from the double-angle formula for cosine. 8. We can also use: [ sinleft(frac{pi}{n}right) = 2 sin left(frac{pi}{2n}right) cos left(frac{pi}{2n}right) ] 9. Since S_1 = frac{1}{2} n R^2 sin left(frac{2pi}{n}right): [ sin left(frac{2pi}{n}right) = 2 sin left(frac{pi}{n}right) cos left(frac{pi}{n}right) ] [ S_1 = frac{1}{2} n R^2 cdot 2 sin left(frac{pi}{n}right) cos left(frac{pi}{n}right) = n R^2 sin left(frac{pi}{n}right) cos left(frac{pi}{n}right) ] 10. Using S_1 and S_2: [ S_1 = n R^2 sin left(frac{pi}{n}right) cos left(frac{pi}{n}right) ] [ S_2 = n R^2 sin left(frac{pi}{n}right) ] 11. Finally, we find the area S using the area ratio: [ cos left(frac{pi}{2n}right) = sqrt{frac{S_2 + S_1}{2 S_2}} ] [ S = 2n R^2 sin left(frac{pi}{2n}right) = 2n R^2 cos left(frac{pi}{2n}right) sin left(frac{pi}{2n}right) ] [ S = n R^2 frac{1 - cosleft(frac{pi}{n}right)}{2} ] Thus, we conclude: [ S = sqrt{frac{2 S_{2}^{3}}{S_{1}+S_{2}}} ] Conclusion: [ boxed{S = sqrt{frac{2 S_{2}^{3}}{S_{1}+S_{2}}}} ]