answer:The conditions on the function f imply [ x + 1 = f(x) + f(2x)] Since x and 2x are in the domain of f, f needs to work for any real x unless explicitly limited by encountering undefined expressions. Considering f(0) + f(0) = 0 + 1 = 1 suggests f(0) = frac{1}{2}. For any non-zero x, plug in -frac{1}{2}: [ -frac{1}{2} + 1 = fleft(-frac{1}{2}right) + f(-1) ] [ frac{1}{2} = fleft(-frac{1}{2}right) + f(-1) ] The existence of f(x) and f(2x), even after equating two expressions for non-zero x and x neq -frac{1}{2}, holds: As no other contradictions in domain requirements are evident, hence we deduce that the condition forces us to exclude -frac{1}{2} and 0 from the domain. Thus, the answer is (boxed{{xmid x neq -frac{1}{2} text{and} x neq 0}}) (Option (d)).

answer:We have f(x)=xln x, Thus, the derivative of the function is f′(x)=1+ln x, Let the coordinates of the tangent point be (x_0,x_0ln x_0), Hence, the equation of the tangent line to f(x)=xln x at (x_0,x_0ln x_0) is y-x_0ln x_0=(ln x_0+1)(x-x_0), Given that line l passes through point (0,-1), Thus, -1-x_0ln x_0=(ln x_0+1)(-x_0), Solving for x_0, we get x_0=1, Therefore, the equation of line l is: y=x-1. So the line equation is boxed{x-y-1=0}. We set the coordinates of the tangent point, find the derivative of the function, use the geometric meaning of the derivative to find the slope of the tangent line, and then use the point-slope form to find the equation of the tangent line. Substitute point (0,-1) and solve the equation to get the conclusion. This problem primarily tests the geometric meaning of derivatives and the form of linear equations. Finding the derivative of the function is the key to solving this problem.

answer:Given a quadratic equation ( ax^2 + bx + c = 0 ), where the coefficients satisfy the condition ( 2a + 3b + 6c = 0 ). We are to prove that this equation has a root in the interval ( (0, 1) ). 1. **Evaluate the integral**: To begin with, let's evaluate the definite integral of the quadratic expression from 0 to 1. [ int_{0}^{1}(ax^2 + bx + c) , dx ] We compute this integral as follows: [ int_{0}^{1}(ax^2 + bx + c) , dx = left. left( frac{a x^3}{3} + frac{b x^2}{2} + cx right) right|_{0}^{1} ] 2. **Calculate the definite integral**: Substituting the limits 0 and 1, we get: [ left( frac{a (1)^3}{3} + frac{b (1)^2}{2} + c(1) right) - left( frac{a (0)^3}{3} + frac{b (0)^2}{2} + c(0) right) ] [ = frac{a}{3} + frac{b}{2} + c ] 3. **Substitute the given condition**: By substituting ( 2a + 3b + 6c = 0 ), we simplify: [ frac{a}{3} + frac{b}{2} + c = frac{2a + 3b + 6c}{6} = frac{0}{6} = 0 ] 4. **Interpret the result**: Since the integral of ( ax^2 + bx + c ) from 0 to 1 equals 0, (int_{0}^{1}(ax^2 + bx + c) dx = 0), it implies that the quadratic function ( ax^2 + bx + c ) changes signs in the interval ([0,1]). - A continuous function that has zero integral over an interval must change signs. Therefore, there exists at least one point in the interval ((0,1)) where the quadratic function equals zero. # Conclusion: There must be at least one root of the equation ( ax^2 + bx + c = 0 ) in the interval ( (0, 1) ). [ boxed{text{Proven}} ]

answer:1. **Given Information**: - Difference in areas = 32 sq ft. - Ratio of areas = ( frac{A_1}{A_2} = 9 ). - One side of the smaller triangle = 4 ft. 2. **Set up the relation between areas**: [ A_1 - A_2 = 32 ] [ frac{A_1}{A_2} = 9 implies A_1 = 9A_2 ] 3. **Solve for ( A_2 )**: [ 9A_2 - A_2 = 32 implies 8A_2 = 32 implies A_2 = 4 ] 4. **Find the corresponding side of the larger triangle using the square root of the area ratio**: [ text{Ratio of the sides} = sqrt{9} = 3 ] [ text{Corresponding side of the larger triangle} = 4 times 3 = 12 text{ feet} ] 5. **Conclusion**: The corresponding side of the larger triangle is 12 feet. The final answer is boxed{12 text{ feet (B)}}