answer:1. When m=1, we have f(x)=frac{1}{x}+xln x. Differentiating f(x) gives f'(x)=frac{1}{x^{2}}+ln x+1. Since f'(x) is monotonically increasing in (0, +infty) and f'(1)=0, for x > 1, we have f'(x) > 0; for 0 < x < 1, f'(x) < 0. Thus, the function f(x) is increasing on the interval (1, +infty). 2. We compute h(x)=frac{m}{x}+2x-sqrt{2}, and its derivative h'(x)=frac{2x^2-m}{x^2}. Setting h'(x)=0 yields x=sqrt{frac{m}{2}}. When 0 < x < sqrt{frac{m}{2}}, we have h'(x) < 0, so h(x) is decreasing on (0, sqrt{frac{m}{2}}). When x > sqrt{frac{m}{2}}, h'(x) > 0 and h(x) is increasing on (sqrt{frac{m}{2}}, +infty). The minimum value of h(x) is boxed{h(sqrt{frac{m}{2}})=2sqrt{2}m-sqrt{2}}. For case (i) when sqrt{2}(2m-1) geq sqrt{frac{m}{2}}, i.e., m geq frac{4}{9}, we need h(2sqrt{2}m-sqrt{2})=sqrt{2}left[frac{m}{2(2sqrt{m}-1)} + 2(2sqrt{m}-1)-1right]=frac{3}{2}sqrt{2}. Solving gives sqrt{m}=1 or sqrt{m}=frac{9}{17} (discard), thus m=1. For case (ii) when 0 < sqrt{2}(2sqrt{m}-1) < sqrt{frac{m}{2}}, that is frac{1}{4} < m < frac{4}{9}, the function y=h(h(x)) has the minimum value h(sqrt{frac{m}{2}})=sqrt{2}(2sqrt{m}-1)=frac{3}{2}sqrt{2}. This yields sqrt{m}=frac{5}{4} (discard), hence m=1 is the only value. 3. By the given condition, K_{OA}=frac{m}{x^{2}}+ln x and K_{OB}=frac{ln x-2}{x}. Considering the function y=frac{ln x-2}{x}, its derivative y'=frac{3-ln x}{x^2} is always positive in [1, e]. Therefore, y=frac{ln x-2}{x} is increasing on [1, e], so K_{OB}in [-2, -frac{1}{e}], and K_{OA}in [frac{1}{2}, e]. This implies frac{1}{2}leqfrac{m}{x^{2}}+ln xleq e should always hold in [1, e], yielding frac{x^{2}}{2}-x^{2}ln xleq m leq x^{2}(e-ln x) for all x in [1, e]. Let p(x)=frac{x^2}{2}-x^2ln x. Its derivative, p'(x)=-2ln xleq 0 on [1, e], implies that p(x) is decreasing, so m geq p(1)=frac{1}{2}. Let q(x)=x^2(e-ln x) with q'(x)=x(2e-1-2ln x) > 0 on [1, e], indicating that q(x) is increasing. Hence, m leq q(1)=e. Combining the two results, we find that the range of m is boxed{[frac{1}{2}, e]}.

answer:Since overrightarrow{a} and overrightarrow{b} are both unit vectors and the angle between them is 120^{circ}, we have (overrightarrow{a} - overrightarrow{b}) cdot overrightarrow{b} = overrightarrow{a} cdot overrightarrow{b} - |overrightarrow{b}|^{2} = 1 times 1 times (-frac{1}{2}) - 1 = -frac{3}{2} And, |overrightarrow{a} - overrightarrow{b}|^{2} = |overrightarrow{a}|^{2} - 2overrightarrow{a} cdot overrightarrow{b} + |overrightarrow{b}|^{2} = 1 - 2 times 1 times 1 times (-frac{1}{2}) + 1 = 3 Thus, |overrightarrow{a} - overrightarrow{b}| = sqrt{3} Let theta be the angle between overrightarrow{a} - overrightarrow{b} and overrightarrow{b}, then cos theta = frac{(overrightarrow{a} - overrightarrow{b}) cdot overrightarrow{b}}{|overrightarrow{b}| cdot |overrightarrow{a} - overrightarrow{b}|} = frac{-frac{3}{2}}{1 times sqrt{3}} = -frac{sqrt{3}}{2} Given that 0^{circ} leqslant theta leqslant 180^{circ}, we have theta = 150^{circ} Therefore, the angle between overrightarrow{a} - overrightarrow{b} and overrightarrow{b} is boxed{150^{circ}}. This problem can be solved by applying the formulas for the dot product of vectors and the magnitude of a vector. The key to solving such problems is being proficient in the properties and formulas of dot product operations and the application of vector magnitude formulas.

answer:Given n has the same remainder when divided by 7 or 9, we can write n = 7a + r = 9b + r where 0 leq r leq 6. This implies 7a = 9b. Therefore, a must be a multiple of 9, so we let a = 9k for some integer k. Substituting back, we get n = 63k + r. Given 150 < n < 250, this translates to 150 < 63k + r < 250. Subtracting r yields 150 - r < 63k < 250 - r. Since 0 leq r leq 6, the bounds on k vary slightly depending on r. To find integer k values, consider the extreme cases for r: - If r = 0, then 150 < 63k < 250, leading to frac{150}{63} < k < frac{250}{63}, approximately 2.38 < k < 3.97. Thus, k = 3. - If r = 6, then 144 < 63k < 244, leading to frac{144}{63} < k < frac{244}{63}, approximately 2.29 < k < 3.87. Thus, k = 3. For k = 3, n = 63 times 3 + r = 189 + r. The possible values for n are 189, 190, 191, 192, 193, 194, 195, all within the range 150 < n < 250. Hence, there are boxed{7} possible values for n.

answer:1. First, we need to identify the prime factorization of ( m = 30030 ). We have: [ 30030 = 2 times 3 times 5 times 7 times 11 times 13 ] This means ( m ) has six distinct prime factors. 2. We need to find the set ( M ) of positive divisors of ( m ) that have exactly 2 prime factors. These divisors are the products of any two distinct primes from the prime factorization of ( m ). Therefore, the elements of ( M ) are: [ M = {2 times 3, 2 times 5, 2 times 7, 2 times 11, 2 times 13, 3 times 5, 3 times 7, 3 times 11, 3 times 13, 5 times 7, 5 times 11, 5 times 13, 7 times 11, 7 times 13, 11 times 13} ] This gives us 15 elements in ( M ). 3. We need to determine the smallest positive integer ( n ) such that for any choice of ( n ) numbers from ( M ), there exist 3 numbers ( a, b, c ) among them satisfying ( abc = m ). 4. We can represent the problem using graph theory. Consider a graph where each vertex represents a prime factor of ( m ), and each edge represents a product of two primes (an element of ( M )). We need to find the minimum number of edges (elements from ( M )) such that any subset of ( n ) edges contains a triangle (a set of three edges forming a cycle). 5. The problem can be reduced to finding the minimum number of edges in a graph with 6 vertices (the primes) such that any subset of ( n ) edges contains a triangle. This is a known problem in extremal graph theory. 6. According to Turán's theorem, the minimum number of edges required to ensure a triangle in a graph with 6 vertices is given by: [ T(6, 3) = leftlceil frac{5 times 4}{2} rightrceil + 1 = 11 ] This means that if we have 11 edges, any subset of 11 edges will contain a triangle. 7. Therefore, the smallest positive integer ( n ) such that any choice of ( n ) numbers from ( M ) contains 3 numbers ( a, b, c ) satisfying ( abc = m ) is ( n = 11 ). The final answer is ( boxed{11} ).