answer:Since the range of the function f(x) is mathbb{R}, Let the radicand t = x^2 + 2ax + 1, which is a quadratic function regarding x, and let its range be M. It must be that (0, +infty) subseteq M. Since the graph of the quadratic function t = x^2 + 2ax + 1 is an upward-opening parabola, We have Delta = 4a^2 - 4 geq 0 Rightarrow a^2 geq 1. Furthermore, since the base of the logarithm is a, and a > 0 and a neq 1, We conclude that a > 1. Therefore, the answer is: boxed{(1, +infty)}.

answer:Given: There are 63 fruits on a miraculous tree. On the first day, 1 fruit falls off the tree, and starting from the second day, the number of fallen fruits each day increases by 1 compared to the previous day. If there are fewer fruits on the tree than the number of fruits expected to fall on any given day, the process restarts with 1 fruit falling that day, and the pattern continues. To determine: The day when all fruits have fallen off the tree. 1. **Calculate the total number of fallen fruits in the first cycle:** [ text{Sum of the first 10 days} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 ] Using the sum of an arithmetic series formula ( S_n = frac{n(n+1)}{2} ): [ S_{10} = frac{10(10+1)}{2} = frac{10 cdot 11}{2} = 55 text{ fruits} ] 2. **Calculate the remaining fruits after the first cycle:** [ text{Remaining fruits} = 63 - 55 = 8 text{ fruits} ] 3. **Restart the dropping sequence for the remaining 8 fruits:** The sequence restarts with 1 fruit on the first day: [ 1 + 2 + 3 = 6 text{ fruits} ] 4. **Calculate the remaining fruits after the secondary cycle:** [ text{Remaining fruits} = 8 - 6 = 2 text{ fruits} ] 5. **Continue with the reset procedure:** There are 2 remaining fruits. We start again with 1 falling fruit per day: - Day 1: 1 fruit falls. - Remaining fruits = 2 - 1 = 1. - Restart again with 1 fruit falling: - Day 1: 1 fruit falls. 6. **Count the total number of days:** [ text{Total days} = 10 text{ (first cycle)} + 3 text{ (second cycle)} + 2 text{ (final resets)} = 15 text{ days} ] # Conclusion: [ boxed{15} ]

answer:First, simplify the given expression as follows: 3^{2005} times 7^{2007} times 2 = (3 times 7)^{2005} times 7^2 times 2 = 21^{2005} times 49 times 2. Now, compute 49 times 2 = 98. Therefore, the expression becomes: 21^{2005} times 98. This is further simplified by expressing 21^{2005} as 21^{2005}, recognizing that 21^{2005} is an odd number and ends with its least significant digit as the least significant digit of 21. Since 21 ends in 1, 21^{2005} also ends in 1. Multiplying by 98, we determine that 21^{2005} times 98 maintains 98 as the final two digits: text{The final digits of } 21^{2005} times 98 text{ are } 98. Therefore, the decimal representation of 21^{2005} times 98 is similar to that of a very large number ending in 98, and the sum of the digits is: 9 + 8 = boxed{17}. Conclusion: The sum of the digits of the decimal representation of 3^{2005} times 7^{2007} times 2 is boxed{17}.

answer:According to the problem, we can obtain: S_{22} = 1-5+9-13+17-21+…+81-85 = (1-5)+(9-13)+(17-21)+…+(81-85) = (-4) times 11 = -44 S_{17} = 1-5+9-13+17-21+…+57-61+65 = (1-5)+(9-13)+(17-21)+…+(57-61)+65 = (-4) times 8 + 65 = 33 Thus, S_{17} - S_{22} = 33 - (-44) = boxed{77}. Therefore, the answer is (C). From the given conditions, we can deduce that S_{22} = (-4) times 11 = -44, S_{17} = (-4) times 8 + 65 = 33, and hence we can find the value of S_{17} - S_{22}. This problem tests the ability to find the sum of a sequence. Pay attention to choosing the appropriate method according to the characteristics of the sequence. In this problem, the sum of every two adjacent terms is -4, and we can use the grouping method, but note the difference between an odd number and an even number of terms when solving.