answer:1. First, find the largest power of 12 that is less than 1025. We start by calculating powers of 12: - 12^1 = 12 - 12^2 = 144 - 12^3 = 1728 As 1728 exceeds 1025, we use 12^2 = 144. 2. Next, calculate how many times 144 fits into 1025: - 1025 div 144 approx 7.11806 Thus, the largest whole number is 7. 3. So, 7 times 144 = 1008 is the largest multiple of 144 that is less than 1025. Therefore, the first digit of 1025_{10} when converted to base 12 is boxed{7}.

answer:Let the point of tangency be P(x_0, y_0), Since y=ax^2, we have y'=2ax, Thus, we have x_0 - y_0 - 1 = 0 (the tangent point lies on the tangent line)①; y_0 = ax_0^2 (the tangent point lies on the curve)② 2ax_0 = 1 (the derivative of the x-coordinate of the tangent point equals the slope of the tangent line)③; Solving ①, ②, and ③, we get a= boxed{frac{1}{4}}. First, assume the coordinates of the tangent point, then differentiate the equation of the parabola, and substitute the tangent point into both the line equation and the parabola equation to solve for a. This problem mainly examines the application of parabolas and tests the students' ability to comprehensively use their knowledge.

answer:(I) Since f(x) = (x-k)e^x, we have f'(x) = (x-k+1)e^x. Setting f'(x) = 0, we solve to find x = k-1. From f'(x) > 0, we get x < k-1; from f'(x) < 0, we get x > k-1. Therefore, the interval of increase for f(x) is (k-1, +infty), and the interval of decrease is (-infty, k-1). (II) When k-1 leq 0, i.e., k leq 1, the function f(x) is increasing on the interval [0, 1], thus, min f(x) = f(0) = -k; When 0 < k-1 leq 1, i.e., 1 < k leq 2, from (I), we know that the function f(x) is decreasing on the interval [0, k-1] and increasing on (k-1, 1], thus, min f(x) = f(k-1) = -e^{k-1}. When k-1 > 1, i.e., k > 2, the function f(x) is decreasing on the interval [0, 1], thus, min f(x) = f(1) = (1-k)e. (III) From (II), when k leq 1, for f(x) > k^2 - 2 to always hold on the interval [0, 1], it is only necessary that when k leq 1, the minimum value of f(x) on the interval [0, 1], -k > k^2 - 2, thus, -2 < k < 1. Therefore, the range of values for k is boxed{-2 < k < 1}.

answer:Solution: (1) |M|=-3, from the formula of the inverse matrix, we know M^{-1}= begin{bmatrix} -frac{1}{3} & -frac{2}{3} -frac{2}{3} & -frac{1}{3} end{bmatrix}. (2) The characteristic polynomial of matrix M is f(lambda) = begin{vmatrix} lambda-1 & 2 2 & lambda-1 end{vmatrix} = (lambda-1)^2 - 4. Let f(lambda) = 0, we get lambda_1 = 3, lambda_2 = -1. When lambda_1 = 3, begin{cases} 2x+2y=0 2x+2y=0 end{cases}, take x=1, then y=-1, the corresponding eigenvector is begin{bmatrix} 1 -1 end{bmatrix}; When lambda_2 = -1, begin{cases} -2x+2y=0 2x-2y=0 end{cases}, take x=1, then y=1, the corresponding eigenvector is begin{bmatrix} 1 1 end{bmatrix}. Therefore, the eigenvalues of matrix M are lambda_1 = 3, lambda_2 = -1, and the corresponding eigenvectors are begin{bmatrix} 1 -1 end{bmatrix} and begin{bmatrix} 1 1 end{bmatrix} respectively. (3) alpha= begin{bmatrix} 3 1 end{bmatrix} = begin{bmatrix} 1 -1 end{bmatrix} + 2 begin{bmatrix} 1 1 end{bmatrix}, Therefore, M^{20}alpha= 3^{20} begin{bmatrix} 1 -1 end{bmatrix} + 2(-1)^{20} begin{bmatrix} 1 1 end{bmatrix} = begin{bmatrix} 3^{20}+2 -3^{20}+2 end{bmatrix}. The final answers are: (1) M^{-1} = boxed{begin{bmatrix} -frac{1}{3} & -frac{2}{3} -frac{2}{3} & -frac{1}{3} end{bmatrix}} (2) Eigenvalues: lambda_1 = boxed{3}, lambda_2 = boxed{-1}; Eigenvectors: boxed{begin{bmatrix} 1 -1 end{bmatrix}} and boxed{begin{bmatrix} 1 1 end{bmatrix}} (3) M^{20}a = boxed{begin{bmatrix} 3^{20}+2 -3^{20}+2 end{bmatrix}}