answer:Solution. By definitionint_{0}^{+infty} frac{d x}{1+x^{2}}=lim _{b rightarrow+infty} int_{0}^{b} frac{d x}{1+x^{2}}Sincebegin{aligned}& lim _{b rightarrow+infty} int_{0}^{b} frac{d x}{1+x^{2}}=left.lim _{b rightarrow+infty} operatorname{arctg} xright|_{0} ^{b}= = & lim _{b rightarrow+infty}(operatorname{arctg} b-operatorname{arctg} 0)=frac{pi}{2}-0=frac{pi}{2}end{aligned}then the integral converges.

answer:[Solution 1] Consider two cases:(1) If x geqslant y geqslant 0, x^{3}, y^{3} are equivalent to the volumes of cubes with edge lengths x, y respectively. For a cube with a total volume of 2, consider two extreme scenarios:(1) x=sqrt[3]{2}, y=0, quad in this case x+y=sqrt[3]{2};(2) x=y=1, in this case x+y=2.Thus, x+y varies between sqrt[3]{2} and 2.(2) If x>0, y>0.In summary, x+y in(0,2].Therefore, the answer is (C).[Solution 2] Given x^{3}+y^{3}=2,which is (x+y)left(x^{2}-x y+y^{2}right)=2,so x+y=frac{2}{x^{2}-x y+y^{2}}.If x>0, y>0, then when x=y=1, (x+y)_{text {max }}=2.If x>0, y>0, then t=x+y>0 always holds (t in R),therefore t(t-2) leqslant 0, i.e., 0 leqslant t leqslant 2.When t=0, x=-y, then x^{3}+y^{3}=0 which contradicts the given condition.Thus, 0 < t leqslant 2.Since x>0, y>0, quad therefore quad x+y>0, thus options (A),(B) can be ruled out.If x=0, y=sqrt[3]{2}, then x+y=sqrt[3]{2} in(0,2] which rules out (D). Therefore, the answer is (C).

answer:6. A.Obviously, p neq 2.When the prime p geqslant 3, obviously,p>y>x,and p(p-1)=2(y+x)(y-x).From p mid 2(y+x)(y-x), we know p mid(y+x).Then p leqslant x+y<2 p Rightarrow p=x+ybegin{array}{l}Rightarrow p-1=2(y-x) Rightarrow p=4 x-1 Rightarrow 4 x-1-1=2 x^{2} Rightarrow x=1 Rightarrow p=3 .end{array}Upon verification, p=3 meets the conditions.

answer:Since G is the centroid of triangle ABC, we have overrightarrow {GA} + overrightarrow {GB} + overrightarrow {GC} = overrightarrow {0}, which implies overrightarrow {CG} = overrightarrow {GA} + overrightarrow {GB}. Given that a overrightarrow {GA} + b overrightarrow {GB} + frac { sqrt {3}}{3}c overrightarrow {GC} = overrightarrow {0}, by rearranging, we get overrightarrow {CG} = frac { sqrt {3}a}{c} overrightarrow {GA} + frac { sqrt {3}b}{c} overrightarrow {GB}. According to the fundamental theorem of plane vectors, we have frac { sqrt {3}a}{c} = frac { sqrt {3}b}{c} = 1, which leads to a=b= frac { sqrt {3}}{3}c. Let c= sqrt {3}, then we get a=b=1. By the law of cosines, we have cos A = frac {b^{2}+c^{2}-a^{2}}{2bc} = frac {1+3-1}{2times 1times sqrt {3}} = frac { sqrt {3}}{2}, Since A is an internal angle of the triangle, we have 0°<A<180°, thus A=30°. Therefore, the correct choice is: boxed{D} By utilizing the properties of the centroid of a triangle, we derive overrightarrow {GA} + overrightarrow {GB} + overrightarrow {GC} = overrightarrow {0}, which implies overrightarrow {CG} = overrightarrow {GA} + overrightarrow {GB}. From the given vector equation, by rearranging, we obtain overrightarrow {CG} = frac { sqrt {3}a}{c} overrightarrow {GA} + frac { sqrt {3}b}{c} overrightarrow {GB}. According to the fundamental theorem of plane vectors, we find frac { sqrt {3}a}{c} = frac { sqrt {3}b}{c} = 1, leading to a=b= frac { sqrt {3}}{3}c. Finally, by applying the law of cosines, we calculate the size of angle A. This problem involves the properties of the centroid of a triangle, the fundamental theorem of plane vectors, and the application of the law of cosines to solve for an angle in a triangle, making it a medium-level question.

answer:Solution.Domain of definition: left{begin{array}{l}cos 3 x neq 0, cos x neq 0 .end{array}right.Using the formula operatorname{tg} 3 alpha=frac{3 operatorname{tg} alpha-operatorname{tg}^{3} alpha}{1-3 operatorname{tg}^{2} alpha}, rewrite the equation as37 cdot frac{3 operatorname{tg} x-operatorname{tg}^{3} x}{1-3 operatorname{tg}^{2} x}-11 operatorname{tg} x=0 Leftrightarrow operatorname{tg} x cdotleft(frac{111-37 operatorname{tg}^{2} x-11+33 operatorname{tg}^{2} x}{1-3 operatorname{tg}^{2} x}right)=0 LeftrightarrowLeftrightarrow operatorname{tg} x cdotleft(frac{100-4 operatorname{tg}^{2} x}{1-3 operatorname{tg}^{2} x}right)=0 Leftrightarrowleft{begin{array} { l } { [ begin{array} { l } { operatorname { tg } x = 0 } { operatorname { tg } ^ { 2 } x = 2 5 }end{array} } { operatorname { tg } ^ { 2 } x neq frac { 1 } { 3 } }end{array} Leftrightarrow left{begin{array}{l}{left[begin{array}{l}operatorname{tg} x=0 operatorname{tg} x= pm 5end{array}right.} operatorname{tg} x neq pm frac{1}{sqrt{3}}end{array}right.right.The last solution does not satisfy the original equation. Therefore, x_{1}=pi k, k in Z ; x_{2}=operatorname{arctg}( pm 5)+pi n= pm operatorname{arctg} 5+pi n, n in Z.Answer: x_{1}=pi k, k in Z ; x_{2}= pm operatorname{arctg} 5+pi n, n in Z.

answer:## Solution.a) With each operation, the sum of the numbers increases by 9. This means that after k operations, the sum of the numbers on the board will be 9 cdot k. The sum of the smallest eight consecutive natural numbers is 0+1+2+ldots+7=28, which is not divisible by 9, while 1+2+3+ldots+8=36. Since 9 cdot 4=36, we deduce that the minimum number of operations required to obtain eight consecutive numbers is 4. Here is one such possibility:| Initial | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 || :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: || Operation 1 | 1 | 2 | | | | | 3 | 3 || Operation 2 | | | 2 | 1 | | | 3 | 3 || Operation 3 | | | | | 3 | 3 | 1 | 2 || Operation 4 | | | 1 | 3 | 2 | 3 | | || Total | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |....................................................................................................... 2 pb) As observed in part a), after k operations, the sum of the 8 numbers is 9 cdot k. Assuming that after a certain number of operations all 8 numbers are equal to 2015, then their sum is 8 cdot 2015. Since 8 cdot 2015 is not divisible by 9, we conclude that it is not possible for all the numbers on the board to be equal to 2015 after any number of operations .................................................... 2pc) We have 2145=3 cdot 5 cdot 11 cdot 13. If the numbers on the board are 3,5,11,13,1,1,1,1, then 3+5+11+13+1+1+1+1=36, which means that 4 operations are required to reach these eight numbers. Since we cannot obtain 13 from 4 operations, we deduce that this variant is not possible.We cannot have 6 or more numbers equal to 1 on the board, since in any selection, three of the numbers are greater than 1. If there are 5 numbers equal to 1 on the board, it means that 5 operations have been performed. This results in a+b+c+1+1+1+1+1=45, where a is the product of two of the numbers 3,5,11 or 13, and b and c are the remaining two numbers. In none of these cases is the sum a+b+c equal to 40. Therefore, this variant is also not possible. In conclusion, it is not possible for the product of the numbers on the board to be 2145 after any sequence of operations.