answer:1. **Graph Representation**: Consider the graph ( T ) with ( n ) vertices and ( n-2 ) edges. There exists an edge ((i,j)) if and only if the transposition ((i,j)) is one of the ( n-2 ) transpositions from the hypothesis. 2. **Graph Properties**: Since ( T ) has ( n ) vertices and ( n-2 ) edges, it is not a connected graph. The minimal connected graph (a tree) with ( n ) vertices has ( n-1 ) edges. Therefore, ( T ) must have at least one disconnected component. 3. **Set ( S(k) )**: The set ( S(k) ) is defined as ( {sigma(k) : sigma in G} ). We claim that ( S(k) ) is the set of vertices ( x ) such that there is a path from ( k ) to ( x ) in the graph ( T ). 4. **Path and Transpositions**: Any permutation ( sigma in G ) can be written as a finite composition of transpositions. If ( sigma(k) = x ), then there are transpositions that will map successively ( k rightarrow k_1 rightarrow k_2 ldots rightarrow x ), which corresponds to a path in the graph ( T ). 5. **Disconnected Components**: Since ( T ) is not connected, there are multiple components. Each component is a subgraph of ( T ) and contains fewer than ( n ) vertices. Specifically, each component can have at most ( n-1 ) vertices because ( T ) has ( n-2 ) edges and is not connected. 6. **Conclusion**: Therefore, for any ( k ), the set ( S(k) ) is contained within one of these components. Since each component has at most ( n-1 ) vertices, it follows that ( |S(k)| leq n-1 ). (blacksquare) The final answer is ( boxed{ |S(k)| leq n-1 } )

answer:To solve this problem, let's break it down step by step, following the given solution closely. 1. **Given Equation Analysis**: - We start with the given equation |overrightarrow{a}+overrightarrow{b}|=|overrightarrow{a}-2overrightarrow{b}|. - Squaring both sides to remove the absolute value, we get: [ (overrightarrow{a}+overrightarrow{b})^2 = (overrightarrow{a}-2overrightarrow{b})^2 ] - Expanding both sides, we have: [ overrightarrow{a}^2 + overrightarrow{b}^2 + 2overrightarrow{a}cdotoverrightarrow{b} = overrightarrow{a}^2 + 4overrightarrow{b}^2 - 4overrightarrow{a}cdotoverrightarrow{b} ] - Using the dot product to express overrightarrow{a}cdotoverrightarrow{b} in terms of |overrightarrow{a}|, |overrightarrow{b}|, and costheta, we get: [ overrightarrow{a}^2 + overrightarrow{b}^2 + 2|overrightarrow{a}| cdot |overrightarrow{b}| cos theta = overrightarrow{a}^2 + 4overrightarrow{b}^2 - 4|overrightarrow{a}| cdot |overrightarrow{b}| cos theta ] 2. **Projection Vector Analysis**: - The projection of overrightarrow{b} onto overrightarrow{a} is given as frac{2}{3}overrightarrow{a}. - This means: [ |overrightarrow{b}| cos theta = frac{2}{3}|overrightarrow{a}| ] - Substituting cos theta = frac{|overrightarrow{b}|}{2|overrightarrow{a}|} from the given solution, we have: [ frac{|overrightarrow{b}|^2}{2|overrightarrow{a}|^2} = frac{2}{3} ] - Solving for frac{|overrightarrow{a}|}{|overrightarrow{b}|}, we get: [ left(frac{|overrightarrow{a}|}{|overrightarrow{b}|}right)^2 = frac{3}{4} ] - Taking the square root of both sides, we find: [ frac{|overrightarrow{a}|}{|overrightarrow{b}|} = frac{sqrt{3}}{2} ] Therefore, the correct answer is boxed{B: frac{sqrt{3}}{2}}.

answer:We need to find the range of ( x ) such that its square falls between 121 and 225. We start by finding the square roots of the boundary values: - ( 121 = 11^2 ), so the smallest value of ( x ) is 11. - ( 225 = 15^2 ), so the largest value of ( x ) is 15. Calculating the squares for integers between 11 and 15: [ 11^2 = 121, quad 12^2 = 144, quad 13^2 = 169, quad 14^2 = 196, quad 15^2 = 225 ] All these squares fall within the range [121, 225]. Thus, the integers ( x ) that satisfy ( 121 leq x^2 leq 225 ) are ( x = 11, 12, 13, 14, 15 ). Counting these integers gives us 5 values. Conclusion: There are (boxed{5}) positive integers ( x ) such that ( 121 leq x^2 leq 225 ).

answer:Since f'(x)=3^{x}ln 3+4 > 0, f(x) is increasing on R, and f(1)=-1 < 0, f(2)=9 > 0, therefore, f(x) has a zero point in [1,2], hence, k=1, hence, g(x)=x-e^{x}, g'(x)=1-e^{x}, let g'(x) > 0, solve for x < 0, let g'(x) < 0, solve for x > 0, therefore, g(x) is increasing on (-infty,0) and decreasing on (0,+infty), hence, the maximum value of g(x) is g(0)=-1, thus, the answer is: boxed{C}. To solve this problem, we first determine the value of k by finding the range of the zero point of the function f(x). Then, we derive the expression for g(x) and determine its maximum value by examining its monotonicity. This problem tests the understanding of function monotonicity, extreme value problems, the application of derivatives, and the zero point problem of functions. It is a moderately difficult problem.