For all real values of x, $$\dfrac{3x^{2} - 6x + 12}{x^{2} + 2x + 4}$$ lies between 1 and k, and does not take any value above k. Then k equals...........

Let $$\dfrac{3x^2-6x+12}{x^2+2x+4}=y$$

$$3x^2-6x+12=x^2y+2xy+4y$$

$$x^2\left(y-3\right)+2x\left(y+3\right)+4y-12=0$$

For the real value of x, $$D\ge0$$ 

$$b^2-4ac\ge\ 0$$

$$\left[2\left(y+3\right)\right]^2-4\left(y-3\right)\left(4y-12\right)\ge\ 0$$

$$4\left(y^2+6y+9\right)-4\left(4y^2-24y+36\right)\ge\ 0$$

$$-12y^2+120y-108\ge\ 0$$

$$12y^2-120y+108\le\ \ 0$$

$$3y^2-30y+27\le\ \ 0$$

$$y^2-10y+9\le\ \ 0$$

$$\left(y-1\right)\left(y-9\right)\le\ 0$$

$$y\in\ \left[1,9\right]$$

Therefore, k = 9