answer:Let y = f(x) cdot g(x), then y' = f'(x) cdot g(x) + f(x) cdot g'(x), Since f'(x)g(x) + f(x)g'(x) < 0, it follows that y is monotonically decreasing on mathbb{R}, and since x < b, we have f(x)g(x) > f(b)g(b). Therefore, the correct choice is boxed{text{A}}.

answer:Since f(x+3) = left lfloor frac{x+4}{3} - left lfloor frac{x+3}{3} right rfloor right rfloor = left lfloor frac{x+1}{3} + 1 - left lfloor frac{x}{3} + 1 right rfloor right rfloor = f(x), f(x) is a periodic function, 3 is one of its periods, so statement 1 is correct. f(x) = left lfloor frac{x+1}{3} - left lfloor frac{x}{3} right rfloor right rfloor = begin{cases} 0, & x in [0,2) 1, & x in [2,3) end{cases}. According to the periodicity of the function, the range of the function is {0, 1}, so the function is not an even function, statement 2 is incorrect, and statement 3 is correct. f(x) = left lfloor frac{x+1}{3} - left lfloor frac{x}{3} right rfloor right rfloor = begin{cases} 0, & x in [0,2) cup [3, pi) 1, & x in [2,3) end{cases}. Therefore, g(x) = f(x) - cos pi x has three different zero points frac{1}{2}, frac{3}{2}, 2 in the interval (0, pi), so statement 4 is incorrect. So the correct statements are 1 and 3. Answer: boxed{1, 3}

answer:To find the width of the metallic sheet, we need to use the information given about the volume of the box that is formed after cutting off the squares from each corner. Let's denote the width of the metallic sheet as ( w ) meters. When the squares with a side length of 8 m are cut from each corner, the new length and width of the box will be reduced by 16 m (since 8 m is cut from both sides of the length and width). So, the new dimensions of the box will be: - Length: ( 50 - 16 = 34 ) m - Width: ( w - 16 ) m - Height: ( 8 ) m (equal to the side of the square that is cut off) The volume of the box is given by the formula: [ text{Volume} = text{Length} times text{Width} times text{Height} ] Given that the volume of the box is 5440 m³, we can set up the equation: [ 5440 = 34 times (w - 16) times 8 ] Now, let's solve for ( w ): [ 5440 = 272 times (w - 16) ] [ frac{5440}{272} = w - 16 ] [ 20 = w - 16 ] [ w = 20 + 16 ] [ w = 36 ] Therefore, the width of the metallic sheet is boxed{36} meters.

answer:1. **Geometric Locus of Lines Forming Equal Angles with Given Planes**: Consider two non-parallel planes (mathbf{S}_1) and (mathbf{S}_2). The locus of all lines that form equal angles with these planes are exactly the lines that are parallel to the angle bisector planes of (mathbf{S}_1) and (mathbf{S}_2). Let's denote these two bisector planes as (mathbf{F}_1) and (mathbf{F}_2). 2. **Geometric Locus of Lines Equidistant from Given Parallel Lines**: Let (a) and (b) be two parallel lines ((a parallel b)). There are two primary sets of lines at equal distances from (a) and (b). One set comprises the lines that intersect the perpendicular bisector of the segment connecting any point on (a) to the corresponding point on (b). The other set includes lines that are parallel to both (a) and (b) and lie within a plane ( mathbf{F} ), which is perpendicular to (a) and (b) and equidistant from them. 3. **Identification of Suitable Planes Through Point (P)**: Let (P) be a given point in space. Construct planes through (P) that are parallel to the angle bisector planes (mathbf{F}_1) and (mathbf{F}_2). Name these planes as (mathbf{U}_1) and (mathbf{U}_2). 4. **Finding the Intersection Lines**: The planes (mathbf{F}_1), (mathbf{F}_2), (mathbf{U}_1), and (mathbf{U}_2) intersect, forming lines. As we look for lines that intersect point (P) and satisfy the given conditions: begin{aligned} &left| mathbf{F}_1 cap mathbf{U}_1 right|, &left| mathbf{F}_1 cap mathbf{U}_2 right|, &left| mathbf{F}_2 cap mathbf{U}_1 right|, &left| mathbf{F}_2 cap mathbf{U}_2 right| end{aligned} These intersections represent the lines that form equal angles with the planes (mathbf{S}_1) and (mathbf{S}_2) and are equidistant from the parallel lines (a) and (b) without being parallel to (a) and (b). 5. **Special Cases**: - If the intersection line of (mathbf{S}_1) and (mathbf{S}_2) is parallel to (a) and (b), all four lines coincide into a single line that is parallel to (a) and (b). Thus, there is only one solution if (P) lies on the plane (mathbf{F}). - If (mathbf{F}_1) coincides with any of the planes (mathbf{U}_1) or (mathbf{U}_2), or similarly (mathbf{F}_2) with (mathbf{U}_1) or (mathbf{U}_2), infinite solutions are derived from the intersection, providing a set of lines (pencil of lines) that meet the conditions. This does not affect other distinct solutions. # Conclusion: Generally, there are four distinct lines that satisfy the given conditions. However, these may reduce to one or create an infinite set under specific geometric configurations. boxed{}