If $f(x) = \left| \begin{array}{ccc} \sin x & \cos x & \tan x \\ x^3 & x^2 & x \\ 2x & 1 & x \end{array} \right|$,then $\lim_{x \to 0} \frac{f(x)}{x^2}$ is equal to

Let $A=\begin{bmatrix} -1 & -2 & -3 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}$,$B=\begin{bmatrix} 1 & -2 \\ -1 & 2 \end{bmatrix}$ and $C=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$. If $a, b$ and $c$ respectively denote the ranks of $A, B$ and $C$,then the correct order of these numbers is:

If $D(x) = \begin{vmatrix} x - 1 & (x - 1)^2 & x^3 \\ x - 1 & x^2 & (x + 1)^3 \\ x & (x + 1)^2 & (x + 1)^3 \end{vmatrix}$,then the coefficient of $x$ in $D(x)$ is

The rank of the matrix $\begin{bmatrix} 2 & -3 & 4 & 0 \\ 5 & -4 & 2 & 1 \\ 1 & -3 & 5 & -4 \end{bmatrix}$ is

The rank of the matrix $A=\begin{bmatrix} 1 & -1 & 0 & -2 \\ -4 & 4 & 0 & 8 \\ -2 & 1 & 2 & 4 \end{bmatrix}$ is

The determinant $\left| {\begin{array}{ccc} 4 + {x^2} & -6 & -2 \\ -6 & 9 + {x^2} & 3 \\ -2 & 3 & 1 + {x^2} \end{array}} \right|$ is not divisible by