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question:The function f(x) satisfies f(x+5)=f(x) for all real numbers x, and when x in [0,5), f(x)=x^2x. Find the value of f(2008).
answer:Since f(x) satisfies f(x+5)=f(x) for all real numbers x, we have: f(2008)=f(5times401+3) =f(3), because when x in [0,5), f(x)=x^2-x. Therefore, f(2008)=f(3)=9-3=6. Hence, the answer is boxed{6}.
question:Let alpha and beta be the roots of x^2 + px 1 = 0, and let gamma and delta be the roots of x^2 + qx + 1 = 0. Express [(alpha gamma)(beta gamma)(alpha + delta)(beta + delta)] in terms of p and q.
answer:Since alpha and beta are the roots of x^2 + px - 1 = 0, we have [(x - alpha)(x - beta) = x^2 + px - 1.] Setting x = gamma, we get [(gamma - alpha)(gamma - beta) = gamma^2 + pgamma - 1.] Thus, (alpha - gamma)(beta - gamma) = gamma^2 + pgamma - 1. Setting x = -delta, we find [(-delta - alpha)(-delta - beta) = delta^2 - p delta - 1,] which simplifies to (alpha + delta)(beta + delta) = delta^2 - p delta - 1. As gamma and delta are the roots of x^2 + qx + 1 = 0, gamma^2 + qgamma + 1 = 0 and delta^2 + qdelta + 1 = 0. Then [gamma^2 + pgamma - 1 = (p-q) gamma - (2)] and [delta^2 - p delta - 1 = -(p+q) delta - (2).] Multiplying these results, we have: [(p-q)gamma cdot (-(p+q)delta - 2) = -(p-q)(p+q)gamma delta - 2(p-q)gamma.] Using Vieta's formulas, gamma delta = -1, so [-(p-q)(p+q)(gamma delta) - 2(p-q)gamma = (p^2 - q^2) + 2(p-q)gamma.] Since gamma is not specified, the final expression simplifies to: [boxed{p^2 - q^2}.]
question:Let S<sub>n</sub> be the sum of the first n terms of the sequence {a<sub>n</sub>}, satisfying a<sub>1</sub> = frac {3}{2}, 2a<sub>n+1</sub> + 3S<sub>n</sub> = 3 (n ∈ N*), and S<sub>n</sub> + frac {2}{S_{n}} ≤ M holds for any n ∈ N*. Find the minimum value of the real number M. A: 2sqrt {2} B: frac {17}{6} C: frac {41}{12} D: 4
answer:From a<sub>1</sub> = frac {3}{2} and 2a<sub>n+1</sub> + 3S<sub>n</sub> = 3 (n ∈ N*), we have: 2a<sub>n</sub> + 3S<sub>n-1</sub> = 3. Subtracting the two equations, we get: 2a<sub>n+1</sub> - 2a<sub>n</sub> + 3a<sub>n</sub> = 0, which simplifies to: frac {a_{n+1}}{a_{n}}=- frac {1}{2}=q. Given a<sub>1</sub> = frac {3}{2}, we have: a<sub>n</sub> = frac {3}{2}(- frac {1}{2})^{n-1} = 3•2<sup>-n</sup>. Thus, the sum S<sub>n</sub> is: S<sub>n</sub> = frac { frac {3}{2}(1-(- frac {1}{2})^{n})}{1+ frac {1}{2}} = 1-(- frac {1}{2})^{n}. So, frac {3}{4} ≤ S<sub>n</sub> ≤ frac {3}{2}. To ensure that S<sub>n</sub> + frac {2}{S_{n}} ≤ M holds for any n ∈ N*, we utilize the property of the function f(x) = x + frac{2}{x}. When S<sub>n</sub> = frac {3}{4}, the function attains its maximum value of frac {41}{12}. Therefore, the minimum value of the real number M is frac {41}{12}. Answer: boxed{C: frac {41}{12}}
question:Calculate the base six product of the numbers 132_{6} and 14_{6}.
answer:First, convert the numbers from base six to base ten. - 132_6 = 1 times 6^2 + 3 times 6^1 + 2 times 6^0 = 1 times 36 + 3 times 6 + 2 times 1 = 36 + 18 + 2 = 56_{10} - 14_6 = 1 times 6^1 + 4 times 6^0 = 1 times 6 + 4 times 1 = 6 + 4 = 10_{10} Next, multiply these numbers in base ten: - 56_{10} times 10_{10} = 560_{10} Now, convert 560_{10} back to base six. We divide by 6 and consider remainders: - 560 div 6 = 93 remainder 2 (least significant digit) - 93 div 6 = 15 remainder 3 - 15 div 6 = 2 remainder 3 - 2 div 6 = 0 remainder 2 (most significant digit) Thus, 560_{10} = 2332_6. Conclusion: The base six product of 132_6 and 14_6 is boxed{2332}_6.