If $\text{⨍ }\left(\mathrm{x}\right)\text{=}\left|\begin{array}{ccc}\mathrm{x}& {\mathrm{x}}^2& {\mathrm{x}}^3\\ 1& 2\mathrm{x}& 3{\mathrm{x}}^2\\ 0& 2& 6\mathrm{x}\end{array}\right|$ then $\mathrm{A} : {\text{⨍ }}^{\text{'}}\left(\frac{1}{2}\right)$ then the ascending order of A, B, C, D is :

$\text{⨍ }\left(\mathrm{x}\right)\text{=}\left|\begin{array}{ccc}\mathrm{x}& {\mathrm{x}}^2& {\mathrm{x}}^3\\ 1& 2\mathrm{x}& 3{\mathrm{x}}^2\\ 0& 2& 6\mathrm{x}\end{array}\right|$

${\mathrm{f}}^1\left(\mathrm{x}\right)=\left|\begin{array}{ccc}1& 2\mathrm{x}& 3{\mathrm{x}}^2\\ 1& 2\mathrm{x}& 3{\mathrm{x}}^2\\ 0& 2& 6\mathrm{x}\end{array}\right|+\left|\begin{array}{ccc}\mathrm{x}& {\mathrm{x}}^2& {\mathrm{x}}^3\\ 0& 2& 6\mathrm{x}\\ 0& 2& 6\mathrm{x}\end{array}\right|+\left|\begin{array}{ccc}\mathrm{x}& {\mathrm{x}}^2& {\mathrm{x}}^3\\ 1& 2\mathrm{x}& 3{\mathrm{x}}^2\\ 0& 0& 6\end{array}\right|$

$$\begin{gathered} {\mathrm{f}}^1\left(\mathrm{x}\right)=0+0+6(2{\mathrm{x}}^2-{\mathrm{x}}^2) \\ =6{\mathrm{x}}^2 \\ {\mathrm{f}}^1\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=6\left(\frac{1}{4}\right)=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ {\mathrm{f}}^1\left(0\right)=0 \\ {\mathrm{f}}^1\left(1\right)=6 \\ {\mathrm{f}}^1\left(2\right)=6\left(4\right)=24 \\ \mathrm{Ascending} \mathrm{order}=\mathrm{ADBC} \end{gathered}$$