answer:To solve, we find M cap N = {1,2,3} cap {2,3,4} = {2,3} Therefore, the correct choice is boxed{C}. **Analysis:** By using the direct method, we can solve by finding the intersection and union of the two sets and observing the inclusion relationship between the two sets.

answer:Since the function f(x) is monotonically increasing in the interval [2,4], it follows that f′(x)geqslant 0 holds true in the interval [2,4], which means (x-1)e^{x}+ageqslant 0 holds true in the interval [2,4], Let g(x)=(x-1)e^{x}+a, then g(x)_{min}geqslant 0, g′(x)=xe^{x}, since xin[2,4], it follows that g′(x) > 0, thus g(x) is increasing in [2,4], hence g(x)_{min}=g(2)=e^{2}+ageqslant 0, Solving this gives: ageqslant -e^{2}, Therefore, the range of the real number a is: ageqslant -e^{2}. Thus, the answer is: boxed{[-e^{2},+infty)}. The problem is transformed into (x-1)e^{x}+ageqslant 0 holding true in the interval [2,4], and by determining the range of a based on the monotonicity of the function. This question examines the monotonicity and extremum of functions, the application of derivatives, and the idea of classification discussion and transformation, making it a medium-level problem.

answer:Given: f(x)=x^{2}-2x+3 with the axis of symmetry at x=1. We have three cases to consider: 1. When 1 leq a-2, or equivalently a geq 3, the minimum value occurs at the left endpoint a-2. Thus, f(a-2)=6 Rightarrow a=5. 2. When a+2 leq 1, or equivalently a leq -1, the minimum value occurs at the right endpoint a+2. Thus, f(a+2)=6 Rightarrow a=-3. 3. When a-2 < 1 < a+2, the minimum value of f(x) is 2, which contradicts the given minimum value of 6. So, we disregard this case. Therefore, the set of possible values for a is boxed{{-3, 5}}. This problem primarily tests your understanding of the properties of quadratic functions, the relationship between the axis of symmetry and the interval, and set expressions.

answer:Firstly, we rearrange and complete the square to bring the equation to its standard form: [ begin{aligned} 3(x^2+2x) - (y^2+4y) + 8 &= 0 3(x^2 + 2x + 1) - (y^2 + 4y + 4) + 8 &= 3 - 4 3(x+1)^2 - (y+2)^2 &= -1 frac{(x+1)^2}{1/3} - frac{(y+2)^2}{1} &= 1. end{aligned} ] This fits the standard hyperbola form: [ frac{(x-h)^2}{a^2} - frac{(y-k)^2}{b^2} = 1, ] where a^2 = frac{1}{3}, b^2 = 1, h = -1, k = -2. Thus, the hyperbola is centered at (-1, -2). Given that a = frac{1}{sqrt{3}}, b = 1, compute c (the distance from the center to the foci): [ c = sqrt{a^2 + b^2} = sqrt{frac{1}{3} + 1} = sqrt{frac{4}{3}} = frac{2}{sqrt{3}} = frac{2sqrt{3}}{3}. ] The foci are vertically aligned. Therefore: [ (-1, -2 pm frac{2sqrt{3}}{3}). ] Thus, choosing one of the foci we have: [ boxed{(-1, -2+frac{2sqrt{3}}{3})}. ]