answer:To solve this problem, we start by considering an integer root r of the first polynomial: [p(x) = a_{10} x^{10} + a_9 x^9 + dots + a_2 x^2 + a_1 x + a_0 = 0.] Given that r is an integer root, we have: [a_{10} r^{10} + a_9 r^9 + dots + a_1 r + a_0 = 0.] Since a_0 neq 0, it implies r neq 0. Therefore, we can divide both sides of the equation by r^{10}, yielding: [a_{10} + a_9 cdot frac{1}{r} + dots + a_1 cdot frac{1}{r^9} + a_0 cdot frac{1}{r^{10}} = 0.] This shows that frac{1}{r} is a root of the second polynomial: [q(x) = a_0 x^{10} + a_1 x^9 + a_2 x^8 + dots + a_8 x^2 + a_9 x + a_{10} = 0.] For frac{1}{r} to also be an integer, r must be either 1 or -1 because these are the only integers for which their reciprocal is also an integer. Given that r = frac{1}{r} for r = 1 and r = -1, the roots of p(x) and q(x) can only be 1 and -1. The possible multisets of roots are those that contain k instances of 1 and 10 - k instances of -1, where 0 leq k leq 10. This gives us a total of 11 possible values for k, corresponding to 11 possible multisets. Therefore, the number of possible multisets S = {r_1, r_2, dots, r_{10}} is boxed{11}.

answer:Given the sequence {a_{n}} where a_{n+1}=a_{n}+2 and a_{3}=5, we need to find a_{2}+a_{4} and the sum of the first n terms, S_{n}. First, let's establish the nature of the sequence. Since a_{n+1}=a_{n}+2, we have: [a_{n+1}-a_{n}=2] This indicates that {a_{n}} is an arithmetic sequence with a common difference d=2. To find a_{1}, we use the given a_{3}=5 and the formula for an arithmetic sequence: [a_{1}=a_{3}-2d=5-4=1] Now, we can express any term a_{n} in terms of n: [a_{n}=a_{1}+(n-1)d=1+2(n-1)=2n-1] To find a_{2}+a_{4}, we use the expressions for a_{2} and a_{4}: begin{align*} a_{2}+a_{4} &= (a_{3}-d) + (a_{3}+d) &= 5-2 + 5+2 &= 10 end{align*} For the sum of the first n terms, S_{n}, we use the formula for the sum of an arithmetic sequence: [S_{n}=frac{n(a_{1}+a_{n})}{2}=frac{n(1+2n-1)}{2}=frac{ncdot 2n}{2}=n^{2}] Therefore, the answers are: [a_{2}+a_{4}=boxed{10}] [S_{n}=boxed{n^{2}}]

answer:Given that f(x) = f(2-x), the graph of f(x) is symmetric about the line x=1. Since f'(x) > 0 in the interval [1, 2], the function is increasing in the interval [1, 2]. Therefore, the function is decreasing in the interval [0, 1]. Since f(x) is an even function defined on mathbb{R}, it is decreasing in the interval [-2, -1] and the function's increasing/decreasing behavior is reversed in the interval [-1, 0]. Given f(x) = f(2-x) = f(2-(2-x)) = f(x+4), f(x) is a periodic function with a period of 4. Therefore, f(x) is increasing in the interval [3, 4]. Hence, the correct option is boxed{text{C}}.

answer:1. Given that ( c ) is a prime number, and (11c + 1) is the square of a positive integer, we start by letting this square be ( m^2 ): [ 11c + 1 = m^2 ] 2. Rearrange the equation to isolate the term involving ( c ): [ m^2 - 1 = 11c ] 3. Recognize that the left-hand side of the equation can be factored using the difference of squares formula: [ m^2 - 1 = (m + 1)(m - 1) ] 4. Substitute this factorization into the equation: [ (m + 1)(m - 1) = 11c ] 5. Notice that since ( 11 ) and ( c ) are prime, ( 11c ) must have exactly these two factors, which are primes. We need to determine how ( (m + 1) ) and ( (m - 1) ) can be factored into ( 11 ) and ( c ). We'll consider the possible assignments: - If ( m - 1 = 1 ) and ( m + 1 = 11c ), then ( m = 2 ) and ( 2 = 11c - 1), which is not possible since ( 2 neq 11c -1 ). - If ( m - 1 = 11 ) and ( m + 1 = c ), then ( m = 12 ) and ( c = 13 ). 6. Given ( m = 12 ), check if these values satisfy the initial equation: [ (12 + 1)(12 - 1) = 11 times 13 ] [ 13 times 11 = 143 ] 7. Substitute back into the initial equation to verify: [ 11c + 1 = m^2 ] [ 11 cdot 13 + 1 = 143 + 1 = 144 ] [ 12^2 = 144 ] 8. Since all conditions fit, the prime number ( c ) satisfies the equation when ( c = 13 ). (boxed{13})