answer:Since {a_n} is an arithmetic sequence with a common difference d, we have: a_n = a_1 + (n - 1)d For n = 3, 4, 5, and 6, we have: a_3 = a_1 + 2d a_4 = a_1 + 3d a_5 = a_1 + 4d a_6 = a_1 + 5d The equation provided in the problem can be written as: 3a_6 = a_3 + a_4 + a_5 + 6 Substituting the expressions for a_3, a_4, a_5, and a_6 into the equation yields: 3(a_1 + 5d) = (a_1 + 2d) + (a_1 + 3d) + (a_1 + 4d) + 6 Expanding both sides gives: 3a_1 + 15d = 3a_1 + 9d + 6 By simplifying and moving terms with d to one side and constants to the other, we get: 15d - 9d = 6 Reducing the equation by combining like terms, we find: 6d = 6 Dividing both sides by 6, we find the common difference d: d = boxed{1}

answer:Since the range of the function f(x)= begin{cases} (1-2a)x+3a, & x < 1 ln x, & xgeqslant 1end{cases} is mathbb{R}: 1. For x < 1, the linear function (1-2a)x+3a needs to satisfy the following: - It should not have a horizontal line, which implies 1-2a neq 0, otherwise its range would not be mathbb{R}. - Thus, we must have 1-2a > 0, from which we get a < frac{1}{2}. 2. For xgeqslant 1, the function ln x has range (0, infty). To connect smoothly with the linear part for x < 1, the linear function has to approach a value greater than or equal to 0 as x approaches 1 from the left. So we have: - (1-2a)cdot 1 + 3a geqslant 0 - Simplifying this yields a geqslant -1. Combining both conditions, the fact that the range of the function is mathbb{R} leads to -1 leqslant a < frac{1}{2}. Therefore, the correct option is boxed{A}.

answer:1. Let ( x, y in mathbb{R}^{-} ), the set of negative real numbers. 2. We want to prove the following inequality: [ frac{x^4}{y^4} + frac{y^4}{x^4} - frac{x^2}{y^2} - frac{y^2}{x^2} + frac{x}{y} + frac{y}{x} geq 2 ] 3. Let's consider rewriting and simplifying the given expression: [ frac{x^4}{y^4} + frac{y^4}{x^4} - frac{x^2}{y^2} - frac{y^2}{x^2} + frac{x}{y} + frac{y}{x} - 2 ] 4. Group the terms as follows: [ = left[ left( frac{x^2}{y^2} right)^2 - 2 frac{x^2}{y^2} + 1 right] + left[ left( frac{y^2}{x^2} right)^2 - 2 frac{y^2}{x^2} + 1 right] ] [ + left[ left( frac{x}{y} right)^2 - 2 frac{x}{y} cdot frac{y}{x} + left( frac{y}{x} right)^2 right] + left[ left( sqrt{frac{x}{y}} right)^2 - 2 sqrt{frac{x}{y}} cdot sqrt{frac{y}{x}} + left( sqrt{frac{y}{x}} right)^2 right] ] 5. Recognize the patterns inside each pair of square brackets. Each of these is a perfect square of a binomial: [ = left( frac{x^2}{y^2} - 1 right)^2 + left( frac{y^2}{x^2} - 1 right)^2 + left( frac{x}{y} - frac{y}{x} right)^2 + left( sqrt{frac{x}{y}} - sqrt{frac{y}{x}} right)^2 ] 6. Since each term is a square of a real number, each term is non-negative. Thus, their sum is non-negative: [ left( frac{x^2}{y^2} - 1 right)^2 + left( frac{y^2}{x^2} - 1 right)^2 + left( frac{x}{y} - frac{y}{x} right)^2 + left( sqrt{frac{x}{y}} - sqrt{frac{y}{x}} right)^2 geq 0 ] 7. Therefore, the expression simplifies to: [ frac{x^4}{y^4} + frac{y^4}{x^4} - frac{x^2}{y^2} - frac{y^2}{x^2} + frac{x}{y} + frac{y}{x} - 2 geq 0 ] 8. Adding 2 to both sides gives us the desired inequality: [ frac{x^4}{y^4} + frac{y^4}{x^4} - frac{x^2}{y^2} - frac{y^2}{x^2} + frac{x}{y} + frac{y}{x} geq 2 ] Conclusion: [ boxed{frac{x^4}{y^4} + frac{y^4}{x^4} - frac{x^2}{y^2} - frac{y^2}{x^2} + frac{x}{y} + frac{y}{x} geq 2} ]

answer:To solve for l^2, where l is the length of the longest line segment that can be drawn in one of the three equal-sized sector-shaped pieces of a circular pie with a diameter of 12text{ cm}, we follow these steps: 1. **Identify the Longest Segment**: The longest segment in a sector is the arc's chord, which, in this case, is the line segment AB. 2. **Form Right Triangles**: By drawing the perpendicular bisector from O (the center of the circle) to AB, we split the chord AB into two equal parts and form two right triangles, triangle MOB and triangle MOA. 3. **Calculate angle MOB**: Since angle AOB represents a third of a full circle (360^circ), angle AOB = 120^circ. The perpendicular bisector creates two 60^circ angles, so angle MOB = 60^circ. This makes triangle MOB a 30-60-90 triangle. 4. **Determine Side Lengths of triangle MOB**: With OB as the hypotenuse of triangle MOB and knowing the diameter of the pie is 12text{ cm}, we have OB = 6text{ cm}. In a 30-60-90 triangle, the sides are in the ratio 1:sqrt{3}:2. Therefore, MO = OB/2 = 6/2 = 3text{ cm}, and MB = MOsqrt{3} = 3sqrt{3}text{ cm}. 5. **Find AB**: Since AB is twice the length of MB (because we bisected AB), we have AB = 2 cdot MB = 2 cdot 3sqrt{3} = 6sqrt{3}text{ cm}. Thus, l = 6sqrt{3}text{ cm}. 6. **Calculate l^2**: Finally, l^2 = (6sqrt{3})^2 = 36 cdot 3 = 108. Therefore, the length of the longest line segment squared, l^2, is boxed{108}.