answer:**Analysis** This problem examines the application of piecewise functions and the judgment of the number of zeros of a function. If g(x) = ln x, then the range of the function is mathbb{R}, and the function is monotonic. Therefore, the equation g(x) = t has exactly one root, so f(t) = 1. If g(x) = begin{cases} x, & xleqslant 0, -x^{2}+2ax+a, & x > 0, end{cases} (ain mathbb{R}), and there exists t such that f(t+2) > f(t) holds, then when x > 0, the maximum value of the function is greater than 2, and the axis of symmetry is to the right of the y-axis, from which the answer can be derived. **Solution** If g(x) = ln x, then the range of the function is mathbb{R}, and the function is monotonic. Therefore, the equation g(x) = t has exactly one root, So, f(t) = 1. For g(x) = begin{cases} x, & xleqslant 0, -x^{2}+2ax+a, & x > 0, end{cases}, When tleqslant 0, f(t) = 1 always holds. If there exists t such that f(t+2) > f(t) holds, Then when x > 0, the maximum value of the function is greater than 2, and the axis of symmetry is to the right of the y-axis, That is, begin{cases} a > 0, dfrac{-4a-4a^{2}}{-4} > 2, end{cases} Solving this gives: a > 1, Therefore, the answer is a > 1. So, for the first part, f(t) = boxed{1}; for the second part, the range of values for a is boxed{a > 1}.

answer:Given the interior angle ratio of a triangle is (3:3:4), we need to find the measure of the largest interior angle. 1. Let's denote the three angles by (3x), (3x), and (4x) where (x) is a common variable. 2. The sum of the interior angles of any triangle is always (180^circ). Thus, we can set up the following equation: [ 3x + 3x + 4x = 180^circ ] 3. Combine the terms: [ 10x = 180^circ ] 4. Solve for (x): [ x = frac{180^circ}{10} = 18^circ ] 5. Now, the measures of the angles are: [ 3x = 3 times 18^circ = 54^circ ] [ 3x = 3 times 18^circ = 54^circ ] [ 4x = 4 times 18^circ = 72^circ ] 6. The largest interior angle is (72^circ). # Conclusion: [boxed{text{D}}]

answer:Solution: (1) By solving the system of equations begin{cases} 3x+y=0 x+y=2 end{cases}, we get begin{cases} x=-1 y=3 end{cases}, which means the intersection point is C(-1,3). The equation of line l is frac{x}{4} + frac{y}{3} = 1, Let the equation of line l_1 be frac{x}{4} + frac{y}{3} = m. Since line l_1 passes through the intersection point C(-1,3), we have m= frac{-1}{4} + frac{3}{3} = frac{3}{4}. Therefore, the equation of line l_1 is frac{x}{4} + frac{y}{3} = frac{3}{4}, which simplifies to 3x+4y-9=0. (2) Let the equation of line l_1 be frac{x}{3} - frac{y}{4} = n. When x=0, y=-4n; when y=0, x=3n. The area of the triangle formed by line l_1 and the coordinate axes is S= frac{1}{2}|3n| cdot |-4n| = 6, which leads to n^2=1. Solving this gives n=pm1. Hence, the equation of line l_1 is frac{x}{3} - frac{y}{4} = pm1, which simplifies to 4x-3y-12=0 or 4x-3y+12=0. Therefore, the answers are: - For part (1), the equation of line l_1 is boxed{3x+4y-9=0}. - For part (2), the equations of line l_1 are boxed{4x-3y-12=0} or boxed{4x-3y+12=0}.

answer:Given U={-1, 0, 1, 2} and A={-1, 1}, Therefore, complement_U A={0, 2}. Hence, the answer is boxed{{0, 2}}.