answer:To prove that mathcal{H} has the Erdős–Pósa property if H is planar, we need to find a function f: mathbb{N} rightarrow mathbb{N} such that for a given k in mathbb{N} and any graph G , either G contains k disjoint subgraphs each isomorphic to H , or there exists a set U of at most f(k) vertices such that H notpreccurlyeq G - U . 1. **Introduction and Setup:** By Theorem 12.4.3, for each k geqslant 1 , there exists an integer w_{k} such that every graph G with treewidth at least w_{k} contains k disjoint subgraphs each isomorphic to H . Hence, we define the function f as follows: [ f(0)=f(1)=0 ] and for k geqslant 1 : [ f(k):=2 f(k-1) + w_{k} . ] 2. **Verification through Inductive Hypothesis:** We will verify that this function f satisfies our needs by induction on k . 1. For k leqslant 1 , there is nothing to prove since f(0)=f(1)=0 are base cases. 3. **Inductive Step:** Assume, for some k geqslant 1 , and any graph G , the statement holds. Let us consider the situation for k+1 and graph G . - If the treewidth operatorname{tw}(G) geq w_{k+1} , then by the definition of w_{k+1} , G contains k+1 disjoint subgraphs isomorphic to H , satisfying the requirement. - Otherwise, if operatorname{tw}(G) < w_{k+1} , let left(T, left(V_{t}right)_{t in T}right) be a tree decomposition of G with width less than w_{k+1} . We direct the edges of T as follows: For each edge t_{1}t_{2} in T , let T_{1}, T_{2} be the components of T - t_{1}t_{2} containing t_{1} and t_{2} respectively. Define: [ G_{1}:=G left[bigcup_{t in T_{1}} left(V_{t} setminus V_{t_{2}}right)right], quad G_{2}:=G left[bigcup_{t in T_{2}} left(V_{t} setminus V_{t_{1}}right)right]. ] 4. **Edge Directions and Vertex Cover:** - If H preccurlyeq G_{i} for some i , assign the direction of t_{1} t_{2} to point towards G_{i} . Given any edge t_{1} t_{2} , either there is such a direction, or no such a direction. - If every edge of T has at most one directed edge, traveling along this direction leads to a node t in T without an outgoing edge. Because H is connected, by Theorem 12.3.1, V_{t} intersects with each copy of H in G . Let U=V_{t} , which is of size at most w_{k+1} leq f(k) , fulfilling the requirement. 5. **Handling Multiple Directions:** - Suppose an edge t_{1} t_{2} in T has two directions. For each i=1,2 , query if all H subgraphs in G_{i} can be covered with at most f(k-1) vertices. If G_{1} and G_{2} hold this property, combine these covers with V_{t_{1}} cap V_{t_{2}} to form a cover U in G . - If G_{1} is without such a cover, by the inductive hypothesis, G_{1} contains k disjoint subgraphs isomorphic to H . Additionally, since t_{1} t_{2} points towards t_{2} , G_{2} also contains such a subgraph, resulting in the needed k+1 disjoint subgraphs in G , completing the proof via induction. # Conclusion: [ boxed{f} ]

answer:Given that the line segment AB is parallel to the y-axis, we can deduce that the x-coordinates (abscissas) of points A and B must be identical. This is because for any two points to be aligned vertically (parallel to the y-axis), their horizontal positions (x-coordinates) must not change. The coordinates of point A are given as (-4,3). Since AB is parallel to the y-axis, the x-coordinate of point B must also be -4. Given that the length of AB is 6, and considering that AB is vertical, the change in position is entirely in the y-direction. Therefore, the y-coordinate of point B can be found by either adding or subtracting 6 from the y-coordinate of point A (which is 3). To find the possible y-coordinates of point B, we perform the following calculations: - Adding 6 to the y-coordinate of A: 3 + 6 = 9 - Subtracting 6 from the y-coordinate of A: 3 - 6 = -3 Therefore, the possible coordinates of point B are (-4, 9) or (-4, -3), taking into account the direction in which the length could be applied (upwards or downwards along the y-axis). Thus, the final answer is encapsulated as: boxed{(-4,-3) text{ or } (-4,9)}

answer:(1) To prove: For x in (0, +infty), f(x) = a - dfrac{2}{x}, Let 0 < x_1 < x_2, then x_1x_2 > 0, and x_2 - x_1 > 0. f(x_1) - f(x_2) = left(a - dfrac{2}{x_1}right) - left(a - dfrac{2}{x_2}right) = - dfrac{2}{x_1} + dfrac{2}{x_2} = 2left(dfrac{x_2 - x_1}{x_1x_2}right) Since x_1x_2 > 0 and x_2 - x_1 > 0, we have 2left(dfrac{x_2 - x_1}{x_1x_2}right) > 0 Therefore, f(x_1) < f(x_2), which means f(x) is increasing on the interval (0, +infty). (2) According to the given conditions, a - dfrac{2}{x} < 2x must hold on (1, +infty). Let h(x) = 2x + dfrac{2}{x}, then a < h(x) must hold on (1, +infty). We can prove that h(x) is monotonically increasing on (1, +infty). Therefore, we must have a leq h(1), which implies a leq 3. Thus, the range of values for a is (-infty, 3]. boxed{a in (-infty, 3]}

answer:**Analysis** This question examines the properties and graphs of trigonometric functions and logical relationships, and it is a basic question. By using the formulas for period and symmetry axis, we can calculate the periods and symmetry axes of the two functions and determine the truth of propositions (p) and (q). **Solution** The minimum positive period of the function (y=cos(2x)) is (dfrac{2pi}{2}=pi), so proposition (p) is false. (fleft(dfrac{pi}{6}right)=sinleft(dfrac{pi}{2}right)=1), (therefore) the line (x= dfrac{pi}{6}) is a symmetry axis of (f(x)), which means proposition (q) is true. (therefore neg q) is false, (p land q) is false, (p lor q) is true. Therefore, the correct choice is boxed{text{B}}.