answer:Consider the given regular triangular prism ABC - A_{1}B_{1}C_{1} where the height, h, is 2 and the base edge length is 1. The centroid of the top face A_{1}B_{1}C_{1} is P. 1. **Construct the Problem**: - Assume a spatial coordinate system. - Place points at their respective coordinates based on the given conditions: - A (0, 0, 0), - B (1, 0, 0), - C left( frac{1}{2}, -frac{sqrt{3}}{2}, 0 right), - A_{1} (0, 0, 2), - B_{1} (1, 0, 2), - C_{1} left( frac{1}{2}, -frac{sqrt{3}}{2}, 2 right). - The centroid of the top face P: [ P left( frac{1}{2}, -frac{sqrt{3}}{6}, 2 right) ] 2. **Find the coordinates of D**: - Given the perpendicular plane BCD perp AP and let AD = h: [ D (0, 0, h) ] 3. **Use the Perpendicular Condition**: - Vector overrightarrow{AP} = left( frac{1}{2}, -frac{sqrt{3}}{6}, 2 right) and overrightarrow{BD} = (-1, 0, h). - As AP perp BCD: [ overrightarrow{AP} cdot overrightarrow{BD} = 0 ] - Compute the dot product: [ left( frac{1}{2} right)(-1) + left( -frac{sqrt{3}}{6} right)(0) + (2)(h) = 0 ] [ - frac{1}{2} + 2h = 0 implies h = frac{1}{4} ] Hence, D left( 0, 0, frac{1}{4} right). 4. **Calculate distances**: - Calculate CD: [ CD = sqrt{left( frac{1}{2} - 0 right)^2 + left( -frac{sqrt{3}}{2} - 0 right)^2 + left( 0 - frac{1}{4} right)^2} ] [ = sqrt{left( frac{1}{2} right)^2 + left( -frac{sqrt{3}}{2} right)^2 + left( -frac{1}{4} right)^2} ] [ = sqrt{ frac{1}{4} + frac{3}{4} + frac{1}{16} } ] [ = sqrt{ 1 + frac{1}{16} } ] [ = sqrt{ frac{16}{16} + frac{1}{16} } ] [ = sqrt{ frac{17}{16} } ] [ = frac{sqrt{17}}{4} ] 5. **Calculate BD**: - Calculate BD: [ BD = sqrt{ (1 - 0)^2 + (0 - 0)^2 + left( 0 - frac{1}{4} right)^2 } ] [ = sqrt{ 1 + left( frac{1}{4} right)^2 } ] [ = sqrt{ 1 + frac{1}{16} } ] [ = sqrt{ frac{16}{16} + frac{1}{16} } ] [ = sqrt{ frac{17}{16} } ] [ = frac{sqrt{17}}{4} ] 6. **Calculate the semi-perimeter**: - Using Heron's formula: - Side lengths: BC = 1, CD = frac{sqrt{17}}{4}, and BD = frac{sqrt{17}}{4}. - Semi-perimeter: [ p = frac{1 + frac{sqrt{17}}{4} + frac{sqrt{17}}{4}}{2} ] [ = frac{1 + frac{2sqrt{17}}{4}}{2} ] [ = frac{1 + frac{sqrt{17}}{2}}{2} ] [ = frac{2 + sqrt{17}}{4} ] 7. **Using Heron's Formula for the Area**: - Compute the area: [ S = sqrt{ p(p-a)(p-b)(p-c) } ] - Where: [ p = frac{2 + sqrt{17}}{4}, quad a = 1, quad b = frac{sqrt{17}}{4}, quad c = frac{sqrt{17}}{4} ] - Compute p-a: [ p - a = frac{2 + sqrt{17}}{4} - 1 = frac{2 + sqrt{17} - 4}{4} = frac{sqrt{17} - 2}{4} ] - Compute p-b: [ p - b = frac{2 + sqrt{17}}{4} - frac{sqrt{17}}{4} = frac{2}{4} = frac{1}{2} ] - Compute p-c: [ p - c = frac{2 + sqrt{17}}{4} - frac{sqrt{17}}{4} = frac{2}{4} = frac{1}{2} ] 8. **Calculate S**: - Substitute: [ S = sqrt{ frac{2 + sqrt{17}}{4} cdot frac{sqrt{17} - 2}{4} cdot frac{1}{2} cdot frac{1}{2} } ] Simplifying the product inside the square root: [ = sqrt{ left( frac{2 + sqrt{17}}{4} right) left( frac{sqrt{17} - 2}{4} right) cdot frac{1}{4} } ] [ = sqrt{ frac{(2 + sqrt{17})(sqrt{17} - 2)}{64} } ] [ = sqrt{ frac{2sqrt{17} - 4 + 17 - 2sqrt{17}}{64} } ] [ = sqrt{ frac{13}{64} } ] [ = sqrt{ frac{13}{8^2} } ] [ = frac{sqrt{13}}{8} ] **Conclusion**: [ boxed{ frac{sqrt{13}}{8} } ]

answer:(1) Proof: Let g(x) = x - f(x), then g'(x) = 1 - f'(x) Since 0 < f'(x) < 1, it follows that g'(x) = 1 - f'(x) > 0 Therefore, the function g(x) = x - f(x) is increasing on **R** Therefore, when x > a, g(x) = x - f(x) > a - f(a) = 0 Therefore, when x > a, it always holds that x > f(x) boxed{6text{ points}} (2) Proof: Since |x_1 - a| < 1, |x_2 - a| < 1 It follows that a - 1 < x_1 < a + 1; a - 1 < x_2 < a + 1 and 0 < f'(x) < 1 Therefore, f(x) is increasing on **R** Therefore, f(a - 1) < f(x_1) < f(a + 1); f(a - 1) < f(x_2) < f(a + 1) Therefore, f(a - 1) - f(a + 1) < f(x_1) - f(x_2) < f(a + 1) - f(a - 1) Therefore, |f(x_1) - f(x_2)| < f(a + 1) - f(a - 1) From (1) we know: f(a + 1) < a + 1; -f(a - 1) < -(a - 1) Therefore, |f(x_1) - f(x_2)| < f(a + 1) - f(a - 1) < 2 Therefore, |f(x_1) - f(x_2)| < 2. boxed{13text{ points}}

answer:To determine the values of the parameter ( a ) for which the inequality ( (x+2) sqrt{a x+x-x^{2}-a} geqslant 0 ) has exactly two solutions that are 4 units apart, we will proceed as follows: 1. **Simplify the Expression under the Square Root**: - Start with the expression inside the square root: ( ax + x - x^2 - a ). - Factor it: [ ax + x - x^2 - a = a(x - 1) + 2(x - 1) = (x - 1)(a + 2 - x) = -(x - a)(x - 1). ] - We need the argument of the square root to be non-negative, hence: [ -(x-a)(x-1) geqslant 0. ] 2. **Determine the Interval from the Inequality**: - The inequality (-(x-a)(x-1) geqslant 0) implies: [ (x-a)(x-1) leqslant 0. ] - This condition breaks down into: [ x in [min(a, 1), max(a, 1)]. ] - If ( a = 1 ): [ (x-1)^2 leqslant 0, ] which has a single solution ( x = 1 ), and this does not meet the problem's requirement of having two solutions. - For ( a ne 1 ): [ x in [a, 1] quad text{or} quad x in [1, a] quad text{depending on whether} quad a < 1 quad text{or} quad a > 1. ] 3. **Conditions for Exactly Two Solutions 4 Units Apart**: - We require the interval length to be at least 4 for there to be two solutions that differ by 4 units. - Thus, ( |a - 1| geq 4 ): [ a - 1 geq 4 quad text{or} quad 1 - a geq 4 ] which simplifies to: [ a geq 5 quad text{or} quad a leq -3. ] 4. **Final Consideration of Solutions**: - For ( a geq 5 ), ( x ) ranges from 1 to ( a ), covering values 1 and 5. - For ( a leq -3 ), the ( x ) interval similarly must include values such that the difference is 4 (e.g., if ( a = -6 ), then ( x = -6 ) and ( x = -2 )). In conclusion, the values of the parameter ( a ) meeting the condition are: [ boxed{a in [-6, -3] cup [5, +infty) } ]

answer:1. **Understand the Problem**: A cone-shaped mountain with a height of 12000 feet has frac{1}{5} of its volume above water. We want to determine the depth of the ocean at the mountain's base. 2. **Volume Formula**: The formula for the volume of a cone is ( V = frac{1}{3} pi r^2 h ), where ( r ) and ( h ) are the radius and height of the cone, respectively. 3. **Volume Scaling by Height**: For any cone, if the height is scaled by a factor ( k ), the radius scales similarly, affecting the volume by a factor of ( k^3 ): [ V' = k^3 V ] 4. **Fractional Volume above Water**: The top frac{1}{5} of the mountain is above water, implying the submerged volume fraction is frac{4}{5}. Let ( h' ) be the height of the submerged part, then: [ left(frac{h'}{12000}right)^3 = frac{4}{5} ] Solving for ( h' ): [ frac{h'}{12000} = sqrt[3]{frac{4}{5}} ] [ h' = 12000 times sqrt[3]{frac{4}{5}} approx 12000 times 0.928 = 11136 text{ feet} ] 5. **Calculating the Depth of the Ocean**: The depth of the ocean at the base of the mountain is the total height minus the submerged height: [ text{Depth of the ocean} = 12000 - 11136 = 864 text{ feet} ] [ 864 text{ feet} ] Conclusion: The depth of the ocean at the base of the mountain is 864 feet. The final answer is boxed{textbf{(B)} 864}