answer:To solve this, we know that a plane is determined by three points that are not on the same line, so option A is incorrect; Two parallel or intersecting lines determine a plane, but two skew lines do not lie in any single plane, so option B is incorrect; A planar quadrilateral determines a plane, so option C is incorrect; However, for four non-coplanar points, any three of these points, which are not collinear, can determine a plane, and in total, they can determine C_{4}^{3}=4 planes, so option D is correct; Therefore, the correct choice is boxed{D} Based on Axiom 2: Through three non-collinear points, there exists exactly one plane, and related corollaries, we can judge the truth of options A, B, and C; by considering that any three of the four non-coplanar points are not collinear, we can judge the truth of option D, thereby obtaining the answer. This question tests knowledge of the basic properties of planes and related corollaries, where mastering Axiom 2 and its corollaries is key to solving this question.

answer:# Solution: Part (1): Given f(x)=ax+ln x for xin [1,2], we have {f'}(x)=a+frac{1}{x}. The condition |{f'}(x)| lt 1 for all xin D translates to |a+frac{1}{x}|＜1 for xin [1,2]. This implies: -1＜a+frac{1}{x}＜1 Rearranging, we get: -1-frac{1}{x}＜a＜1-frac{1}{x} for xin [1,2]. Let {y_1}=1-frac{1}{x} and {y_2}=-1-frac{1}{x}. Both {y_1} and {y_2} are monotonically increasing functions for xin [1,2]. Thus, the range of a is determined by evaluating these expressions at the endpoints of the interval [1,2], leading to: -frac{3}{2}＜a＜0 Therefore, the range of real number a is boxed{(-frac{3}{2},0)}. Part (2): Let g(x)=f(x)-x, then {g'}(x)={f'}(x)-1. Given that |{f'}(x)| lt 1, it follows that {g'}(x) lt 0 for all xin D. This means g(x) is strictly decreasing. A strictly decreasing function can cross the x-axis at most once, implying that the equation f(x)-x=0 has at most one real root. Thus, the statement is boxed{text{proved}}. Part (3): Let M and m be the maximum and minimum values of f(x) within one period, respectively. Assume f(a)=M and f(b)=m. Given that f(x) is periodic with period 2 and |{f'}(x)| lt 1, we have: |{frac{{f({{x_1}})-f({{x_2}})}}{{{x_1}-{x_2}}}}|＜1 If |a-b|leqslant 1, then: |f(x_{1})-f(x_{2})|leqslant M-m lt 1 If |a-b| gt 1, assume without loss of generality that a gt b, then: a+1 gt b+2 quad text{and} quad a lt b+2 This implies 0 lt b+2-a lt 1. Therefore: |f(x_{1})-f(x_{2})|leqslant M-m=f(a)-f(b+2)leqslant |a-(b+2)| lt 1 In conclusion, for any real numbers x_{1} and x_{2}, we have: |f(x_{1})-f(x_{2})|lt 1 Thus, the original statement is boxed{text{proved}}.

answer:Given that a_{2}+a_{8}=6 in the arithmetic sequence {a_{n}}, it implies that a_{1}+a_{9}=6. The sum of the first 9 terms of an arithmetic sequence can be calculated using the formula S_n = frac{n(a_{1}+a_{n})}{2}. Hence, the sum of the first 9 terms of the sequence {a_{n}} is S_9 = frac{9(a_{1}+a_{9})}{2} = 27. Therefore, the answer is boxed{27}. This problem tests the understanding of the general term formula and the sum formula of an arithmetic sequence, requiring both reasoning and computational skills. It is a moderately difficult problem.

answer:Since ( p(-6) = 0 ) and ( p(3) = 0 ), the quadratic polynomial ( p(x) ) can be expressed in the form: [ p(x) = c(x + 6)(x - 3) ] for some constant ( c ). To determine ( c ), we set ( x = 1 ) and use the fact that ( p(1) = -24 ): [ p(1) = c(1 + 6)(1 - 3) = c(7)(-2) = -14c ] Thus, setting this equal to -24, we find: [ -14c = -24 ] [ c = frac{-24}{-14} = frac{12}{7} ] Substituting back into the polynomial gives: [ p(x) = frac{12}{7}(x + 6)(x - 3) = frac{12}{7}(x^2 + 3x - 18) ] [ p(x) = frac{12}{7}x^2 + frac{36}{7}x - frac{216}{7} ] Concluding with: [ boxed{p(x) = frac{12}{7}x^2 + frac{36}{7}x - frac{216}{7}} ]