answer:First, we calculate g(-x): [ g(-x) = log left( -x + 2 + sqrt{1 + (-x + 2)^2} right) ] [ = log left( -x + 2 + sqrt{1 + x^2 - 4x + 4} right) ] [ = log left( -x + 2 + sqrt{(x-2)^2 + 1} right) ] To see if g(x) is odd or even, we evaluate: If g(x) is odd, g(-x) should equal -g(x): [ g(-x) = log left( -x + 2 + sqrt{(x-2)^2 + 1} right) ] [ -g(x) = -log left( x + 2 + sqrt{(x+2)^2 + 1} right) ] We find that these are not simply negatives of each other, indicating that g(x) is not odd. To see if g(x) is even, g(-x) should equal g(x): [ g(-x) = log left( -x + 2 + sqrt{(x-2)^2 + 1} right) ] [ g(x) = log left( x + 2 + sqrt{(x+2)^2 + 1} right) ] These expressions are not the same, indicating that g(x) is not even. Therefore, g(x) is boxed{text{neither}}.

answer:**Analysis** This question tests the application of coordinate operations of spatial vectors and the formula for the magnitude of a vector. It is a basic problem. According to the coordinate operations of vectors, we calculate the minimum value of ((vec{b} - vec{a})^2), from which the result can be obtained. **Solution** Given: (vec{b} - vec{a} = (1-t, 2t-1, 0) - (2, t, t) = (1+t, 1-t, t)), Therefore, ((vec{b} - vec{a})^2 = (1+t)^2 + (1-t)^2 + t^2 = 3t^2 + 2 geqslant 2), The minimum value of ((vec{b} - vec{a})^2) is (2), which occurs when and only when (t=0), Thus, the minimum value of (|vec{b} - vec{a}|) is (sqrt{2}), Therefore, the correct answer is boxed{text{C}}.

answer:1. Since S_n = n^2 + kn, S_n - 5kn = n^2 - 4kn = (n - 2k)^2 - 4k^2, where k in mathbb{N}^*. The minimum value of S_n - 5kn occurs when n = 2k, which is -4k^2 = -4. Solving for k, we get k = 1. Therefore, S_n = n^2 + n. When n = 1, a_1 = S_1 = 2. For n geq 2, a_n = S_n - S_{n-1} = n^2 + n - [(n-1)^2 + (n-1)] = 2n. Thus, the general formula for the n-th term of the sequence {a_n} is a_n = 2n. 2. b_n = frac{a_n}{4^n} = frac{2n}{4^n}, Let T_n denote the sum of the first n terms of the sequence {b_n}. T_n = frac{2}{4} + frac{2 cdot 2}{4^2} + frac{2 cdot 3}{4^3} + dots + frac{2n}{4^n}, frac{1}{4}T_n = frac{2}{4^2} + frac{2 cdot 2}{4^3} + dots + frac{2(n-1)}{4^n} + frac{2n}{4^{n+1}}, Subtracting the two equations, we get frac{3}{4}T_n = frac{2}{4} + 2(frac{1}{4^2} + frac{1}{4^3} + dots + frac{1}{4^n}) - frac{2n}{4^{n+1}}. Simplifying, we get T_n = frac{8}{9} - frac{8+6n}{9} cdot frac{1}{4^n}. The final answers are: 1. boxed{a_n = 2n} 2. boxed{T_n = frac{8}{9} - frac{8+6n}{9} cdot frac{1}{4^n}}

answer:Let W(t) represent the daily sales revenue (in yuan) of the product as a function of time t. Then we have: W(t) = f(t)g(t) = begin{cases} (frac{1}{3}t + 8)(30 - t), & 0 leqslant t < 15, t in mathbb{N}^{+} (-frac{1}{3}t + 18)(30 - t), & 15 leqslant t leqslant 30, t in mathbb{N}^{+} end{cases} = begin{cases} -frac{1}{3}t^2 + 2t + 240, & 0 leqslant t < 15, t in mathbb{N}^{+} frac{1}{3}t^2 - 28t + 540, & 15 leqslant t leqslant 30, t in mathbb{N}^{+} end{cases} = begin{cases} -frac{1}{3}(t - 3)^2 + 243, & 0 leqslant t < 15, t in mathbb{N}^{+} frac{1}{3}(t - 42)^2 - 48, & 15 leqslant t leqslant 30, t in mathbb{N}^{+} end{cases} When 0 leqslant t < 15, t in mathbb{N}^{+}, it is clear that W(t) reaches its maximum at t = 3, and the maximum value is W(3) = 243. When 15 leqslant t leqslant 30, t in mathbb{N}^{+}, the function is decreasing, so W(t) reaches its maximum at t = 15, and the maximum value is W(15) = 195. Therefore, the maximum daily sales revenue for this product is boxed{243} yuan when t = 3.