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question:Given that m,n are two different lines, and alpha ,beta are two different planes, which of the following statements is correct? A: If mparallelalpha and mparallelbeta, then alpha parallelbeta B: If mperpalpha and mparallelbeta, then alpha parallelbeta C: If mperpalpha and nparallelalpha, then mparallel n D: If mperpalpha and nperpalpha, then mparallel n
answer:**Analysis** This question examines the determination of parallel planes, the properties of a line perpendicular to a plane, and the spatial relationship between lines. Each option can be judged accordingly. **Answer** For option A, if mparallelalpha and mparallelbeta, then alpha and beta can intersect, so it is incorrect. For option B, if mperpalpha and mparallelbeta, then alpha and beta can intersect, so it is incorrect. For option C, if mperpalpha and nparallelalpha, then mperp n, so it is incorrect. For option D, since two lines that are perpendicular to the same plane are parallel, if mperpalpha and nperpalpha, then mparallel n, so it is correct. Therefore, the correct answer is boxed{D}.
question:Given the function f(x)=|3x1|+ax+3, where ainmathbb{R}. (1) If a=1, solve the inequality f(x)leqslant 4; (2) If the function f(x) has a minimum value, find the range of a.
answer:(1) When a=1, the function becomes f(x)=|3x-1|+x+3. (a) If xgeqslant frac{1}{3}, the inequality f(x)leqslant 4 simplifies to 3x-1+x+3leqslant 4, which solves to frac{1}{3}leqslant xleqslant frac{1}{2}; (b) If x < frac{1}{3}, the inequality f(x)leqslant 4 simplifies to -3x+1+x+3leqslant 4, which solves to 0leqslant x < frac{1}{3}; Therefore, the solution set of the original inequality is [0, frac{1}{2}]. (2) The function f(x)=|3x-1|+ax+3 can be written as: [ f(x) = begin{cases} (3+a)x+2, & xgeqslant frac{1}{3} (a-3)x+4, & x < frac{1}{3} end{cases} ] The necessary and sufficient condition for the function f(x) to have a minimum value is: [ begin{cases} a+3geqslant 0 a-3leqslant 0 end{cases} ] This simplifies to -3leqslant aleqslant 3. Hence, the range of a is boxed{[-3,3]}.
question:Let f(x) = x+1. What is the value of f(f(f(f(f(3)))))?
answer:Apply the function f(x) = x+1 repeatedly: 1. f(3) = 3 + 1 = 4. 2. f(f(3)) = f(4) = 4 + 1 = 5. 3. f(f(f(3))) = f(5) = 5 + 1 = 6. 4. f(f(f(f(3)))) = f(6) = 6 + 1 = 7. 5. f(f(f(f(f(3))))) = f(7) = 7 + 1 = 8. Therefore, f(f(f(f(f(3))))) = boxed{8}.
question:Given 10 natural numbers ( a_{1} < a_{2} < cdots < a_{10} ), prove that their least common multiple is not less than ( 10a_{1} ).
answer:1. Let a_1, a_2, ldots, a_{10} be 10 natural numbers in increasing order, such that a_1 < a_2 < ldots < a_{10}. 2. We need to prove that the least common multiple (LCM) of these numbers, denoted as A, satisfies A geq 10 a_1. 3. Express the least common multiple of a_1, a_2, ldots, a_{10} as: [ A = [a_1, a_2, ldots, a_{10}] ] 4. By the definition of LCM, there exist integers k_i such that: [ A = k_1 a_1 = k_2 a_2 = cdots = k_{10} a_{10} ] 5. Since a_1 < a_2 < cdots < a_{10}, it follows that k_1 > k_2 > cdots > k_{10} because k_i are positive integers. 6. As k_i are distinct positive integers and arranged in descending order, the smallest value for k_1 occurs when k_i is at least 10, so: [ k_1 geq 10 ] 7. Therefore, we get: [ A = k_1 a_1 geq 10 a_1 ] # Conclusion [ boxed{A geq 10 a_1} ]