answer:Solution: (I) Since f(x)=xsin x+cos x, therefore f'(x)=sin x+xcos x-sin x=xcos x, xin(0,2pi), f'(x)=0, therefore x= dfrac {pi}{2} or dfrac {3pi}{2}, therefore f(x) is increasing in (0, dfrac {pi}{2}) and ( dfrac {3pi}{2},2pi), and decreasing in ( dfrac {pi}{2}, dfrac {3pi}{2}), therefore f(x)_{text{min}}=- dfrac {3pi}{2}, f(x)_{text{max}}= dfrac {pi}{2}; (II) Proof: From f'(x)=0 and x > 0, we get x_{i}= dfrac {(2i-1)pi}{2}, (ninmathbb{N}^*), Since dfrac {1}{x_{i}^{2}}= dfrac {4}{(2i-1)^{2}pi ^{2}} < dfrac {2}{pi ^{2}}cdot dfrac {2}{(2i-1)^{2}-1}= dfrac {2}{pi ^{2}}cdot( dfrac {1}{2i-2}- dfrac {1}{2i}), (igeqslant 2,iinmathbb{N}^*), therefore dfrac {1}{x_{2}^{2}}+ dfrac {1}{x_{3}^{2}}+ldots+ dfrac {1}{x_{n}^{2}} < dfrac {2}{pi ^{2}}left[ ( dfrac {1}{2}- dfrac {1}{4})+( dfrac {1}{4}- dfrac {1}{6})+( dfrac {1}{6}- dfrac {1}{8})+ldots+( dfrac {1}{2n-2}- dfrac {1}{2n})right] = dfrac {2}{pi ^{2}}( dfrac {1}{2}- dfrac {1}{2n}) < dfrac {2}{pi ^{2}}cdot dfrac {1}{2}= dfrac {1}{pi ^{2}} < dfrac {1}{9}. Thus, the final answer for (I) is f(x)_{text{min}}=- dfrac {3pi}{2}, f(x)_{text{max}}= dfrac {pi}{2}, and for (II) the proof shows that dfrac {1}{ x_{ 2 }^{ 2 }}+ dfrac {1}{ x_{ 3 }^{ 2 }}+ldots+ dfrac {1}{ x_{ n }^{ 2 }} < dfrac {1}{9}, which can be summarized as boxed{dfrac {1}{ x_{ 2 }^{ 2 }}+ dfrac {1}{ x_{ 3 }^{ 2 }}+ldots+ dfrac {1}{ x_{ n }^{ 2 }} < dfrac {1}{9}}.

answer:To find out how many ear muffs customers bought before December, we can subtract the number of ear muffs bought during December from the total number of ear muffs bought. Total ear muffs bought = Ear muffs bought before December + Ear muffs bought during December Given that the total ear muffs bought is 7,790 and the ear muffs bought during December is 6,444, we can set up the equation: 7,790 = Ear muffs bought before December + 6,444 Now, we can solve for the ear muffs bought before December: Ear muffs bought before December = 7,790 - 6,444 Ear muffs bought before December = 1,346 So, customers bought boxed{1,346} ear muffs before December.

answer:To find the difference between B's and C's shares, we first need to determine the total number of parts in the given proportion and then calculate the individual shares of B and C. The proportion is given as 2 : 3 : 5 : 4. Let's add up the parts: A's share = 2 parts B's share = 3 parts C's share = 5 parts D's share = 4 parts Total parts = 2 + 3 + 5 + 4 = 14 parts Now, we know the total profit is Rs. 20000, and it is to be divided into 14 parts. Value of each part = Total profit / Total parts Value of each part = Rs. 20000 / 14 Value of each part = Rs. 1428.57 (rounded to two decimal places) Now, let's find B's and C's shares: B's share = 3 parts C's share = 5 parts B's share in rupees = 3 parts * Rs. 1428.57/part B's share in rupees = Rs. 4285.71 C's share in rupees = 5 parts * Rs. 1428.57/part C's share in rupees = Rs. 7142.85 Now, let's find the difference between B's and C's shares: Difference = C's share - B's share Difference = Rs. 7142.85 - Rs. 4285.71 Difference = Rs. 2857.14 So, the difference between B's and C's shares is Rs. boxed{2857.14} .

answer:The contrapositive of a proposition is obtained by negating both the hypothesis and the conclusion of the original proposition and then swapping them. ∴ The contrapositive of the proposition "If ab=0, then a=0 or b=0" is If a neq 0 and b neq 0, then ab neq 0 Therefore, the correct option is boxed{C}.