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 greater than \(k\). If \(k\) is the least number with this property,…

The value of $k$ is $9$. Setting $y=\frac{3x^2-6x+12}{x^2+2x+4}$ gives $(y-3)x^2+(2y+6)x+(4y-12)=0$, which has a real root exactly when $(y+3)^2\ge4(y-3)^2$, that is $1\le y\le9$. The greatest value is attained at $x=-2$, so the least admissible $k$ is $9$. The value $1$ is the lower bound, and $7$ is the value of the expression at $x=-1$.