answer:1. **Understanding the Problem:** We need to place the integers (1, 2, ldots, n^2) on an (n times n) chessboard such that any two fields that share a common edge or vertex have numbers differing by at most (n + 1). 2. **Path Analysis:** We define a "path" as a sequence of squares such that each adjacent pair of squares in the sequence are also adjacent on the board. For any path of length (k), the difference between the value of the first and last square is at most ((k - 1)(n + 1)). 3. **Connecting Any Two Squares:** Any two squares on the board are connected by at least one path of length (n). The only pairs of squares connected by a unique path of length (n) are two opposite corners. 4. **Placement of 1 and (n^2):** Consider the locations of 1 and (n^2). Any path connecting them must have length (n) and must consist of the squares numbered (1, n + 2, 2n + 3, ldots, n^2 - n - 1, n^2). This path must be unique, so we have four choices of a corner in which to place the number 1. These numbers lie along the main diagonal. Thus, our board looks like: [ begin{bmatrix} 1 & & & & & & & & & & & & & & & & & & & n^2 end{bmatrix} ] 5. **Placement of 2:** The number 2 cannot be placed such that there is a path of length (n - 1) joining 2 to (n^2). Therefore, 2 must lie along the top or left edge. There are (n - 1) possible paths joining 2 to (n^2), and it turns out that 2 must be adjacent to 1. This gives us two choices for the location of 2. The board now looks like: [ begin{bmatrix} 1 & 2 & & & & & & & & & & & & & & & & & & n^2 end{bmatrix} ] 6. **Continuing the Pattern:** By continuing this pattern, we see that 3 must be adjacent to 2, and so on. Similarly, the entry (n^2 - 1) must be adjacent to (n^2), forcing a lower diagonal. 7. **Final Configuration:** In the end, nearly everything is forced, and only the last two entries in the first and the last row are free to be switched. This gives us the final configuration: [ begin{bmatrix} 1 & 2 & 3 & * & * & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & * & * & 23 & 24 & 25 end{bmatrix} ] 8. **Counting the Placements:** The total number of such placements is given by: [ 4 cdot 2 cdot 2^2 = 32 ] The final answer is (boxed{32})

answer:This problem involves the monotonicity of a function and the properties of a quadratic function. First, we need to find the derivative of the given function: y'=x^2+2bx+(b+6) For the function to have three monotonic intervals, its derivative must have two distinct real roots. This means the discriminant of the quadratic function must be greater than 0. Delta=4b^2-4(b+6) > 0 Solving the inequality, we get b < -2 text{ or } b > 3 Therefore, the answer is option D: boxed{b < -2 text{ or } b > 3}.

answer:(1) We first compute each term separately: - f(1) = frac{1}{1+1^2} = frac{1}{2} - f(2) = frac{1}{1+2^2} = frac{1}{5} - f(3) = frac{1}{1+3^2} = frac{1}{10} - f(frac{1}{2}) = frac{1}{1+(frac{1}{2})^2} = frac{4}{5} - f(frac{1}{3}) = frac{1}{1+(frac{1}{3})^2} = frac{9}{10} Now, we add these values together: frac{1}{2} + frac{1}{5} + frac{1}{10} + frac{4}{5} + frac{9}{10} = boxed{frac{5}{2}} (2) For the range of f(x), we note that since x^2 geq 0 for all real numbers x, we have 1 + x^2 geq 1, and thus 0 < frac{1}{1+x^2} leq 1. Therefore, the range of f(x) is (0, 1].

answer:According to the problem, when x < 1, f'(x) > 0, which means f(x) is an increasing function; Since the function is symmetric about x=1, we have f(x)=f(2-x), Therefore, f(3)=f(-1), Since -3 < -1 < 0, we have f(-3) < f(-1) < f(0), which means f(-3) < f(3) < f(0), Thus, b < c < a. Therefore, the correct answer is: boxed{text{D}}. This problem examines the relationship between the monotonicity of a function and its derivative, testing computational skills.